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http://www.economist.com/comment/2156179 | 1,406,255,892,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997892648.47/warc/CC-MAIN-20140722025812-00066-ip-10-33-131-23.ec2.internal.warc.gz | 671,867,089 | 27,215 | Trade, exchange rates, budget balances and interest rates
See article
Robert Del Rosso
I have the original “Economic and financial indicators” page of the 04-Jun-2011 Economist on which appears the “Trade, exchange rates, budget balances and interest rates” table (pg 106), the predecessor to the current print version’s “Economic data” table. You can, of course, check the data on the Web.
In the 04-Jun-2011 Economist, the then China Current account (C/A) surplus of \$331.1 billion (BN) (for the 12 months ending in Q4) was listed as 3.1% of China GDP 2011. Applying the same algebraic equation as in my Sep. 21st comment, the math then showed:
\$331.1 BN x 100/3.1 = \$10,681 BN derived China GDP 2011.
However, in the current 21-Sep-2013 Economist, China's present C/A surplus of \$211.6 BN (for the 12 months ending in Q2) is listed as 2.0% of GDP 2013. Applying the same algebraic equation, the algebra now shows us:
\$211.6 BN x 100/2 = \$10,580 BN derived China GDP 2013.
The 04-Jun-2011 Economist had a China GDP 2011 annual growth rate of 9.0% and the 21-Sep-2011 Economist now shows the current annual China GDP 2013 growth rate is 7.5%, (in the “Output, prices and jobs” table, also in column 3 of the print edition’s “Economic data” table).
I do not dispute such GDP growth rates. However, given such growth rates, it is rather impossible that the China GDP would have declined from the above GDP 2011 of \$10,681 BN on 04-Jun-2011, to a GDP 2013 of \$10,580 BN as of 21-Sep-2013. (Can you say “QED”? )
The Chinese government’s “People's Daily” newspaper largely agrees with China's GDP 2012 of \$8,227 BN per the IMF that this Analyst used as a “base GDP”, to increase by the annual GDP 2013 growth rate in The Economist, (as described above) in calculating the most accurate possible China GDP 2013.
China's newspaper says it's nominal GDP 2012 was \$8.23 trillion (\$8,230 BN) which is nearly equal to this Analyst's GDP 2012 of \$8,227 BN as per the IMF's April 2013 data, with a quite negligible variance of 3.6% of 1 per cent (0.000365) (--see "China's Economy Might Be Number 1 in 2030" --People's Daily, 06-July-2013 ).
So even China would say: “Oh no, our nominal GDP (non-PPP (Purchasing Power Parity) GDP)) was *never* over \$10 trillion, either in 2011 or 2013!”
At the same time, in the 21-Sep-2013 Economist, the projected China GDP 2013 growth rate was reduced to 7.5% (down from 7.8% in the 08-Jun-2013 Economist). When we increase the China GDP 2012 per the IMF (\$8,227 BN) by this 7.5%, this results in a more accurate China GDP 2013 of \$8,844 BN.
This GDP 2013 of \$8,844BN appears to be in roughly the same range as the GDPs 2013 per The Economist’s special issue (“The World in 2013”) and The World Debt Clock of \$9,233 BN and \$8,609 BN, respectively.
The China GDP 2013 per the IMF as (adjusted above by this Analyst, using The Economist’s growth rate) has thus declined from \$8,926 BN as per the 09-Mar-2013 Economist, when The Economist’s “Economic data” table was first adjusted to reflect GDP 2013. Thus, the variance between the China GDP 2013 per the C/A and the IMF, while reduced, remains at a still substantial \$1,736 BN) (i.e. \$10,580 BN less \$8,844 BN). This \$1,736 BN is greater than the GDP 2013 of South Korea! The China GDP 2013 per the C/A of \$10,580 BN is now (as of the 21-Sep-2013 Economist) a still substantial 19.6% higher (i.e. \$1,736BN / \$8,844BN) than the more accurate GDP 2013 per the IMF (as adjusted by 21-Sep-2013 Economist’s GDP 2013 growth rate of 7.5%).
It remains disconcerting to have this variance between these 2 methods of calculating the China GDP 2013 as noted above.
Assuming the latest C/A surplus of \$211.6 BN in the 21-Sep-2013 Economist is correct, then China's C/A surplus should have been increased to 2.4% of GDP 2013 (i.e. \$211.6BN/ \$8,844BN = 2.39% (or 2.4%, when rounded to the nearest 1/10th of 1%) ), rather than decreasing from 2.1% to 2.0%, as it was listed in the 21-Sep-2013 Economist.
The algebra would then show:
\$211.6 BN x 100/2.4 = \$8,817 BN derived China GDP 2013.
The \$8,817 BN GDP 2013 would then compare favorably with the above \$8,844 BN GDP 2013 per the IMF (as adjusted by this Analyst, as noted above).
Does anyone ever proofread the data in this table and its print version? Apparently not.
OTOH, if we apply the above algebraic equation to the United States’s C/A, we have:
\$425.7 BN x 100 / 2.7 = \$15,767 BN U.S. GDP 2013
While not 100% accurate, it is, as we Americans say, “in the ballpark”.
As Bill O’Reilly would say: “show me where I’m wrong!”
I want to thank The Economist for providing a generous space of 5,000 characters (with spaces) in which to make a comment. I hope the above Analysis proves useful to someone and enables me to obtain employment.
Look for me on the Web at:
After I entered the first "Captcha",the comment did not fully post. Since I was given another "Captcha" I entered the 2nd Captcha and then saw the comment was posted twice.
Robert Del Rosso
I have the original “Economic and financial indicators” page of the 04-Jun-2011 Economist on which appears the “Trade, exchange rates, budget balances and interest rates” table (pg 106), the predecessor to the current print version’s “Economic data” table. You can, of course, check the data on the Web.
In the 04-Jun-2011 Economist, the then China Current account (C/A) surplus of \$331.1 billion (BN) (for the 12 months ending in Q4) was listed as 3.1% of China GDP 2011. Applying the same algebraic equation as in my Sep. 21st comment, the math then showed:
\$331.1 BN x 100/3.1 = \$10,681 BN derived China GDP 2011.
However, in the current 21-Sep-2013 Economist, China's present C/A surplus of \$211.6 BN (for the 12 months ending in Q2) is listed as 2.0% of GDP 2013. Applying the same algebraic equation, the algebra now shows us:
\$211.6 BN x 100/2 = \$10,580 BN derived China GDP 2013.
The 04-Jun-2011 Economist had a China GDP 2011 annual growth rate of 9.0% and the 21-Sep-2011 Economist now shows the current annual China GDP 2013 growth rate is 7.5%, (in the “Output, prices and jobs” table, also in column 3 of the print edition’s “Economic data” table).
I do not dispute such GDP growth rates. However, given such growth rates, it is rather impossible that the China GDP would have declined from the above GDP 2011 of \$10,681 BN on 04-Jun-2011, to a GDP 2013 of \$10,580 BN as of 21-Sep-2013. (Can you say “QED”? )
The Chinese government’s “People's Daily” newspaper largely agrees with China's GDP 2012 of \$8,227 BN per the IMF that this Analyst used as a “base GDP”, to increase by the annual GDP 2013 growth rate in The Economist, (as described above) in calculating the most accurate possible China GDP 2013.
China's newspaper says it's nominal GDP 2012 was \$8.23 trillion (\$8,230 BN) which is nearly equal to this Analyst's GDP 2012 of \$8,227 BN as per the IMF's April 2013 data, with a quite negligible variance of 3.6% of 1 per cent (0.000365) (--see "China's Economy Might Be Number 1 in 2030" --People's Daily, 06-July-2013 ).
So even China would say: “Oh no, our nominal GDP (non-PPP (Purchasing Power Parity) GDP)) was *never* over \$10 trillion, either in 2011 or 2013!”
At the same time, in the 21-Sep-2013 Economist, the projected China GDP 2013 growth rate was reduced to 7.5% (down from 7.8% in the 08-Jun-2013 Economist). When we increase the China GDP 2012 per the IMF (\$8,227 BN) by this 7.5%, this results in a more accurate China GDP 2013 of \$8,844 BN.
This GDP 2013 of \$8,844BN appears to be in roughly the same range as the GDPs 2013 per The Economist’s special issue (“The World in 2013”) and The World Debt Clock of \$9,233 BN and \$8,609 BN, respectively.
The China GDP 2013 per the IMF as (adjusted above by this Analyst, using The Economist’s growth rate) has thus declined from \$8,926 BN as per the 09-Mar-2013 Economist, when The Economist’s “Economic data” table was first adjusted to reflect GDP 2013. Thus, the variance between the China GDP 2013 per the C/A and the IMF, while reduced, remains at a still substantial \$1,736 BN) (i.e. \$10,580 BN less \$8,844 BN). This \$1,736 BN is greater than the GDP 2013 of South Korea! The China GDP 2013 per the C/A of \$10,580 BN is now (as of the 21-Sep-2013 Economist) a still substantial 19.6% higher (i.e. \$1,736BN / \$8,844BN) than the more accurate GDP 2013 per the IMF (as adjusted by 21-Sep-2013 Economist’s GDP 2013 growth rate of 7.5%).
It remains disconcerting to have this variance between these 2 methods of calculating the China GDP 2013 as noted above.
Assuming the latest C/A surplus of \$211.6 BN in the 21-Sep-2013 Economist is correct, then China's C/A surplus should have been increased to 2.4% of GDP 2013 (i.e. \$211.6BN/ \$8,844BN = 2.39% (or 2.4%, when rounded to the nearest 1/10th of 1%) ), rather than decreasing from 2.1% to 2.0%, as it was listed in the 21-Sep-2013 Economist.
The algebra would then show:
\$211.6 BN x 100/2.4 = \$8,817 BN derived China GDP 2013.
The \$8,817 BN GDP 2013 would then compare favorably with the above \$8,844 BN GDP 2013 per the IMF (as adjusted by this Analyst, as noted above).
Does anyone ever proofread the data in this table and its print version? Apparently not.
OTOH, if we apply the above algebraic equation to the United States’s C/A, we have:
\$425.7 BN x 100 / 2.7 = \$15,767 BN U.S. GDP 2013
While not 100% accurate, it is, as we Americans say, “in the ballpark”.
As Bill O’Reilly would say: “show me where I’m wrong!”
I want to thank The Economist for providing a generous space of 5,000 characters (with spaces) in which to make a comment. I hope the above Analysis proves useful to someone and enables me to obtain employment.
Look for me on the Web at:
Robert Del Rosso
I have been reading your fine newspaper for over 30 years, and respect and value it greatly. However, I feel I must bring the following to your attention:
Your data can be converted into these algebraic formulas:
(1)Current account/GDP = P/100
And
(2)Current account(C/A) x 100/P = GDP
Where “P” = The Current account(C/A) as a percentage of GDP
Applying this formula to the Greek data in the 21-Sep-2013 Economist, the Greek GDP 2013 is:
\$0.4 x 100/ 0.8 = \$50 BN GDP 2013
This \$50 BN GDP 2013 is far too low for Greece.
In the previous week, the 14-Sep-2013 Economist had the Greek C/A as -\$3.4 BN for the 12 months ending Jun 2013, and listed as -0.8% of GDP.
In this 21-Sep-2013 Economist, the Greek C/A has declined to -\$0.4 BN for the 12 months ending in July 2013. Assuming the C/As are correct, the Greek C/A declined \$3/\$3.4, or by 88.2%, in \$BN terms. Yet, inexplicably, the reduced C/A continues to be listed as-0.8% of Greek GDP 2013! How does that happen?
The Greek GDP 2012 Per the IMF’s APR 2013 Data was \$249 BN (rounded).
In addition, the 21-Sep-2013 Economist (“Output, prices and jobs” table on the Web, “Economic data” table in print) had Greek GDP 2013 declining by -4.5%,
resulting in a projected Greek GDP 2013 of \$249BN x (1-0.045)= 249 x 95.5% =237.7BN = 238 BN(rounded).
Thus, assuming the Greek C/A of - \$0.4 BN is correct, the C/A = 0.4/238
= -0.0016 OR -0.2% of GDP 2013, (rounded to the nearest 1/10th of 1%), rather than -0.8% listed in the 21-Sep-2013 Economist, or 75% less, as a percentage of GDP.
I believe that Greece has enough problems without its C/A being listed 4 times what it should be, as a percentage of GDP!
On 31-Aug-2013, I emailed The Economist my original MS Excel spreadsheet, detailing similar errors in this table and in the corresponding “Economic data” table in the print version. The spreadsheet was sent to Mr. John Micklethwait, Editor-In-Chief and 11 other Economist Executives. I only received “mail return” from one party (Mr. Rupert Pennant-Rea). I am curious to know what the response is to my spreadsheet.
In the 24-Aug-2013 Economist, the Italian C/A as a % of GDP 2013 was corrected.
In the 31-Aug-2013 Economist, Italy’s C/A of \$5.9 BN (for the 12 months ending June 2013) was listed as 0.3% of GDP 2013. The Algebra gave an accurate derived GDP as follows: \$5.9BN x 100/0.3 = \$1,967BN GDP 2013
However, in my 31-Aug-2013 email, I predicted that the Italian data would soon provide an incorrect derived GDP 2013.
In the 07-Sep-2013 Economist, my prediction came true.
In the 07-Sep-2013 Economist, the same \$5.9BN Italian C/A was listed as 0.6% of GDP! Thus, in the 07-Sep-2013 Economist, the Italian derived GDP 2013 = \$983BN, down from 31-Aug-2013’s \$1,967BN! i.e:
\$5.9 x 100/ 0.6 = \$983BN GDP 2013
Going back further, we see:
In the 15-JUN-2013 Economist, the Italian Current account (C/A) surplus for the 12 mos. ended Mar. 2013 is +\$0.6 BN and listed as +0.1% of 2013 GDP. Thus, applying the above formula to the 15-JUN-2013 data, I arrive at:
\$0.6 x 100 / 0.1 = “Derived” GDP 2013 of \$600 billion
However, in the 06-APR-2013 Economist, the Italian C/A deficit for the 12 mos. ended Jan 2013 was (\$9.5) BN and ALSO listed as +0.1% of 2013 GDP! (I ignore how a deficit could be a positive percentage of GDP!)
However, applying the above formula to the 06-APR-2013 data, I arrive at:
\$9.5 BN C/A x 100 / 0.1 =
“Derived” Italy GDP 2013 of \$9,500 billion!
-----“Quod Erat Demonstrandum” (QED)
The derived Italy GDP 2013 (based on the 06-APR-2013 Economist) was \$9,500 BN, or \$9.5 trillion, greater than China's!(that's a lot of pasta!). The GDP 2013 growth rate(based on the derived GDP 2012 and on the 02-MAR-2013 economist) was:
(\$9,500 -840)/ 840 = 1031% !
Italy's actual GDP 2013 is \$1,992 BN, based on the IMF’s Italian GDP 2012 and The Economist’s annual growth rate. Thus, the above \$9.5BN C/A deficit should be 9.5/1992 or 0.004769% of GDP 2013 or 0.5% (rounded to the nearest 1/10th of 1%) rather than the 0.1% listed in the 06-APR- 2013Economist.
I derived Italy’s GDP 2013 from the website of Italy’s Ministry of Economy and Finance (MEF) (http://www.mef.gov.it/en/doc-finanza-pubblica/def/2013-2013/documenti/Ca...):
The MEF shows Italy’s GDP 2013 as 1,573.2 EUR BN
The Currency units per \$ in the 07-Sep-2013 Economist’s "economic data" table are
EUR 0.76 / USD
EUR 1573.2 BN/0.76 EUR/USD = \$2,070 BN GDP 2013 (QED)
This Italian \$2,070 BN GDP 2013 largely agrees with that of the IMF’s GDP 2013 of \$1,992BN (above).
There are other errors, too numerous to mention here.
"TALENT HITS A TARGET NO ONE ELSE CAN HIT: GENIUS HITS A TARGET NO ONE ELSE CAN SEE."---ARTHUR SCHOPENHAUER
Email:rdelrosso2001@yahoo.com
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Visit The Economist e-store and you’ll find a range of carefully selected products for business and pleasure, Economist books and diaries, and much more | 4,583 | 15,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-23 | latest | en | 0.921375 |
http://mathematica.stackexchange.com/questions/16414/how-can-i-convert-a-complex-number-ab-i-to-the-exponent-form-a-expi-phi?answertab=active | 1,432,626,867,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928817.29/warc/CC-MAIN-20150521113208-00061-ip-10-180-206-219.ec2.internal.warc.gz | 157,550,233 | 17,451 | # How can I convert a complex number a+b I to the exponent form A Exp(I phi)?
When I have an expression such as:
(1/4 + I/4) ((1 - 2 I) x + Sqrt[3] y)
it is hard to get an intuition of the number.
So I want to convert it to the complex exponent notation:
$\frac{1}{2} \sqrt{\frac{5}{2}} e^{-\frac{i \pi }{12}} x+\frac{1}{2} \sqrt{\frac{3}{2}} e^{\frac{i \pi }{4}} y$
and I also want to convert $(-1)^{1/4}$ to $\exp \left(\frac{i \pi }{4}\right)$.
How can I do it with Mathematica?
-
The requisite conversion functions are already included in David Park's Presentations add-on (see http://home.comcast.net/~djmpark/DrawGraphicsPage.html):
<<Presentations
PolarToExp @ ComplexToPolar[(1/4 + I/4) ((1 - 2 I) x + Sqrt[3] y)]
(The Presentations actually includes a polar form for a complex number of modulus r and argument theta that displays in the form r \[Angle] theta. And ComplexToPolar yields that form. Hence the need for the outside function PolarToExp. Thus there is a distinction made between "polar form" and "exponential form" -- as there should be.)
-
This explicitly converts any numeric quantities in the expression to the desired form:
polarForm = Expand[# /. z_?NumericQ :> Abs[z] Exp[I Arg[z]]] &;
e.g.
(1/4 + I/4) ((1 - 2 I) x + Sqrt[3] y) // polarForm
$\frac{1}{2} \sqrt{\frac{5}{2}} e^{\frac{i \pi }{4}-i \text{ArcTan}[2]} x+\frac{1}{2} \sqrt{\frac{3}{2}} e^{\frac{i \pi }{4}} y$
(-1)^(1/4) // polarForm
$e^{\frac{i \pi }{4}}$
-
sorry if I'm wrong because I'm not sure about these terms in English, but isn't this an exponential form, not polar? – user907860 Feb 3 at 23:45
@user907860, both terms are valid I believe. – Simon Woods Feb 4 at 6:23
You can use the definitions together with ComplexExpand :
z = (1/4 + I/4) ((1 - 2 I) x + Sqrt[3] y);
abs = ComplexExpand[Abs[z]]
arg = ComplexExpand[Arg[z], TargetFunctions -> {Re, Im}]
(* Sqrt[4 x^2 + (x + Sqrt[3] y)^2]/(2 Sqrt[2]) *)
(* ArcTan[3 x + Sqrt[3] y, -x + Sqrt[3] y] *)
(* check *)
z - abs Exp[I arg] // FullSimplify
(* 0 *)
`
- | 688 | 2,034 | {"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} | 3.84375 | 4 | CC-MAIN-2015-22 | latest | en | 0.693876 |
https://www.cubelelo.com/blogs/cubing/rniss-a-useful-complement-to-traditional-fmc-insertions | 1,679,936,223,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948673.1/warc/CC-MAIN-20230327154814-20230327184814-00054.warc.gz | 810,106,915 | 61,604 | ## Introduction
This blog will be a basic overview of the concept of rNISS, a technique used in the event of 3x3 FMC. Insertions are one of the most important tools you can use in an FMC attempt. When I was first learning about these concepts, I struggled with the traditional way of discovering insertions within the given time. Then, I found the concept of rNISS, an alternative and complementary method for finding the best insertions that has helped me to this day.
## What are Insertions?
In FMC, once you have a “skeleton”, the cube will mostly be solved except for a few pieces that need to be moved around to their correct places. Insertions are sets of moves inserted into the skeleton so that they solve those pieces and cancel out as many moves as possible in the process. This way, the final solution will be much more efficient than just solving the cube directly.
The usual way to find insertions would be to first label each of the pieces using numbered stickers to show the order in which the pieces must be swapped. Then, you slowly go through each move of the skeleton from the cube’s scrambled state and note down any insertion that can cancel the most number of moves.
## rNISS
rNISS stands for reverse NISS, a method to find and visualize insertions by taking advantage of the cyclical nature of the scramble and solution. In this method, you first choose a point in the skeleton that you want to check for insertions. Then, you perform the moves of the skeleton after that point, then perform the scramble, then perform the moves of the skeleton before that point. This will leave you with the rest of the cube solved, and only those specific pieces unsolved, which you can then find a solution for easily.
For an example, let us say the scramble is R’ U’ F D2 B2 F’ U R and the skeleton is D F L2 B’ R2 F U2. You want to now visualize how the unsolved pieces will be between the F and L2 position in the skeleton. So you perform L2 B’ R2 F U2, then the scramble, and then D F, to give you that specific position.
I found that the traditional method took a lot of time to analyze each of the stickered pieces against the jumbled rest of the cube. Visually, it was harder to make out how exactly the pieces should be moved around. The advantage of rNISS is that it gives you a very clear visual image of the position, since the rest of the cube is solved. Hence, it is easier to find a possible insertion. Another advantage is in cases where a lot of pieces need to be solved using insertions, like having 4 corners and 5 edges, for example. rNISS allows you to look at this jumbled and hard position at each point in the skeleton clearly. It allows for using many unorthodox insertions besides the common 3 cycles. You can even insert single moves or R U R’ U’ if that solves the most pieces efficiently.
check out our Skewb collection
## Conclusion
I have found rNISS to be a very useful tool to find, visualize, and verify a large number of insertions in a quick and easy to understand way. This method is especially useful if you are fast at scrambling, so it will take much less time than using stickers for many cases. Hopefully this blog has given you a general idea of this method so that you can implement it in your own FMC toolkit.
Pranav Prabhu
Pranav Prabhu is the current 3x3 Fewest Moves (Single) National record holder from Chennai. He started cubing when he was 14 and has 5 years of cubing experience. Besides cubing, Pranav enjoys reading books, writing, and playing the piano. He has participated in 36 competitions and won 30 podiums including 8 gold medals and 1 National record | 826 | 3,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-14 | longest | en | 0.949135 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/8/lesson/8.1.6/problem/8-60 | 1,725,849,958,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651053.52/warc/CC-MAIN-20240909004517-20240909034517-00093.warc.gz | 278,979,580 | 15,428 | ### Home > APCALC > Chapter 8 > Lesson 8.1.6 > Problem8-60
8-60.
THE MATRYOSHKA DOLL
For generations, special wooden nesting dolls called matryoska have been handcrafted in Russia. This type of cylindrical doll opens at the midsection and nested inside is a smaller wooden doll (with the same shape) which itself opens to another doll. Some matryoshka are made up of ten or more dolls! The smallest doll, often less than a centimeter tall, is solid wood.
Explain why, under ideal conditions, the sum of the volumes of the nested wooden shells of the different-sized dolls will calculate the full volume of the interior of the largest doll.
An 'ideal' situation would have infinitely many dolls with infinitely small thicknesses. | 169 | 733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-38 | latest | en | 0.928131 |
https://www.cravencountryjamboree.com/other/how-is-ap-stats-frq-graded/ | 1,716,289,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00306.warc.gz | 646,306,054 | 14,416 | # How is AP stats frq graded?
## How is AP stats frq graded?
What three components do you need when writing an answer for an association question?
There are three components: distribution, parameters, and calculation of distance. appropriately labeled sketch of a normal distribution is supplied, then the response should be scored as essentially correct (E).
How many FRQS are on AP stats?
The FRQs are designed to test your knowledge of statistics, and your ability to apply multiple skills and concepts in a question. There are four AP Stats skills that are tested on the exam….
Year % Resistant to Glyphosate Sample Size
2017 38.5% 52
### How are ap free response questions scored?
After your multiple-choice section is graded by a machine and your free response is graded by a human, your essay and multiple-choice scores are combined to give you a composite score. This score is just a way of combining the two section scores so that they are weighted correctly.
What percent is a 5 on AP stats?
What percent is a 5 on AP stats?
Composite Score AP Score Percentage of Students Earning Each Score (2020)
70-100 5 16.2%
57-69 4 20.7%
44-56 3 23.1%
33-43 2 21.7%
Does chi-square measure correlation?
Pearson’s correlation coefficient (r) is used to demonstrate whether two variables are correlated or related to each other. The chi-square statistic is used to show whether or not there is a relationship between two categorical variables. | 325 | 1,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-22 | latest | en | 0.941902 |
http://www.cpan.org/modules/by-category/06_Data_Type_Utilities/Statistics/Statistics-Test-WilcoxonRankSum-0.0.7.readme | 1,409,430,039,000,000,000 | text/plain | crawl-data/CC-MAIN-2014-35/segments/1408500835699.86/warc/CC-MAIN-20140820021355-00423-ip-10-180-136-8.ec2.internal.warc.gz | 319,689,858 | 1,866 | Statistics-Test-WilcoxonRankSum version 0.0.1 Performs the Wilcoxon (aka Mann-Whitney) rank sum test on two sets of numeric data. In statistics, the Mann-Whitney U test (also called the Mann-Whitney-Wilcoxon (MWW), Wilcoxon rank-sum test, or Wilcoxon-Mann-Whitney test) is a non-parametric test for assessing whether two samples of observations come from the same distribution. The null hypothesis is that the two samples are drawn from a single population, and therefore that their probability distributions are equal. See the Wikipedia entry http://en.wikipedia.org/wiki/Mann-Whitney_U (for eg.) or statistic textbooks for further details. When the sample sizes are small the probability can be computed directly. For larger samples usually a normal approximation is used. Input to the test are two sets (lists) of numbers. The values of both lists are ranked from the smallest to the largest, while remembering which set the items come from. When the values are the same, they get an average rank. For each of the sample sets, we compute the rank sum. Under the assumption that the two samples come from the same population, the rank sum of the first set should be close to the value B, where n1 and n2 are the sample sizes. The test computes the (exact, resp. approximated) probability of the actual rank sum against the expected value (which is the one given above). So, when the computed probability is below 0.05, we can reject the null hypothesis at level 0.05 and conclude that the two samples are significantly different. The implementation follows the mechanics described above. The exact probability is computed for sample sizes less than 20, but this threshold can be set with `new'. For larger samples the probability is computed by normal approximation. INSTALLATION To install this module, run the following commands: perl Makefile.PL make make test make install Alternatively, to install with Module::Build, you can use the following commands: perl Build.PL ./Build ./Build test ./Build install DEPENDENCIES uses the following: Carp Carp::Assert Class::Std Contextual::Return Set::Partition List::Util qw(sum) Math::BigFloat Math::Counting Statistics::Distributions COPYRIGHT AND LICENCE Copyright (C) 2008, Ingrid Falk This library is free software; you can redistribute it and/or modify it under the same terms as Perl itself. | 513 | 2,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2014-35 | latest | en | 0.886052 |
http://evefer.cf/article3176-excel-2010-count-distinct-values-in-a-column.html | 1,527,499,823,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872766.96/warc/CC-MAIN-20180528091637-20180528111637-00505.warc.gz | 96,352,422 | 6,856 | excel 2010 count distinct values in a column
# excel 2010 count distinct values in a column
Excel 2007 and 2010: Use the Remove Duplicates menu option under the Data header. Excel 2003: Easy wayHow to count the sum of distinct Excel values in a column? -1. Finding a unique value in excel. 0. This example shows you how to create an array formula that counts unique values in Excel.Two-column Lookup. Most Frequently Occurring Word. System of Linear Equations. I have a workbook containing multiple worksheets: (a) There is a worksheet that lists a series of filenames (called "Summary"). This worksheet also includes a column for each month of the year (JAN, FEB, MAR etc.) to record the number of times the filename has been accessed that month. (b) Count unique values / distinct values in Excel with formula or pivot — Supposing you have a column of names in your Excel worksheet, and you need to count unique.Excel 2010 Count Distinct Values In Column. Counting duplicate values is a subject that begs definition. What constitutes a duplicate? Within Excel, you can have duplicate values within the same column or you can have duplicate records — a row where every value in the record is repeated. With Excel 2010 and earlier, the best we could do is to count the number of transactions (rows) with any given region/item combination.
To add the running total column, simply insert the Item field into the Values area again, and set the field to display the Distinct count. counting distinct values in excel 2010 remove delete duplicate rows middot set field to be inserted as unique access array formula cell e5 click kutools gt select same different cells the advanced filter dialog box radio button quot copy anothercount unique distinct values in a column in excel get digital. CountDistinct col.Count. Set col Nothing Set arr Nothing Set r Nothing End Function. This function will count the distinct values in any range that I pass to it.0. Excel Count occurrences of meetings in a column? 0. Less Than Dot is a community of passionate IT professionals and enthusiasts dedicated to sharing technical knowledge, experience, and assistance. Inside you will find reference materials, interesting technical discussions, and expert tips and commentary. SUBTOTAL IF Formula To Sum Distinct Values In A Column. Create Helper Column That Marks Distinct Values. Excel 2010 :: Search For Value And Then Give Count Of Corresponding Value In Different Column. This tutorial explains how to count unique values in a column. Sample Data. Sample Data.
Formula. SUMPRODUCT(1/COUNTIF(B3:B15,B3:B15)). Logic. The text "Jhonson" appears 3 times so the unique value would be equal to (1/3) (1/3) (1/3) 1. How Formula works. Count the number of unique values by using a filter You can use the Advanced Filter to extract the unique values from a column of data and paste them to a new location. Extract distinct values ignoring blanks Hello XP SP 3, Excel 2010. I should like to count unique values in a column Column C. At present the values are text strings and some blank cells.Make a distinct count of unique values in Excel - How To I need a function to calculate number of distinct values in one column in a filtered range (including blanks).Place it in a new module (AltF11, Insert > Module, Paste, then Altq to return to Excel). To count unique items in B2:B200 you would use this on a worksheet Count of Rows 11 Distinct values 6. Here is the structure of the VBA code Im trying to use to build a function I can call in Excel: Function CountDistinct(dataRange As Range). I used the solution at Count unique values in a column in Excel.The code below takes all the variables in a selected range and then outputs an array with distinct values to an other sheet (in this case a sheet named Output). To count unique numeric values in a range, you can use a formula based on the FREQUENCY and SUM functions.Excel Formula Training. Make sure you have a solid foundation in Excel formulas! In this step-by-step training, youll learn how to use formulas to manipulate text, work with dates and CountDistinct col.Count. Set col Nothing Set arr Nothing Set r Nothing End Function. This function works well, and will correctly count the distinct values in any range that I pass to it.Excel/VBA iterate over cells in a column get eMail from cell, validate and send automatic Notification. How to count unique items (count distinct) in an Excel pivot table. Get the free workbook to follow along.Method 2: In Excel 2010 and later, use the "pivot a pivot" technique. Method 3: In older versions of Excel, add a column to the source data to place a 1 in cells on a row where the value is I need to count the distinct values in Field a, so that it returns 3 instead of 6. Any help is appreciated.Copy down as far as required, then sum in the pivot table. And no, Excel 2010 does not have any better approach for the unique count than previous versions, so a helper column is still the Distinct count in pivot table count distinct values in an. Excel macro count number of occurrences in a column. Count number of unique values in excel pivot table 2010. How to count a value in excel 2010 - ways to count values.Excel 2013 calculated field greyed out - distinct count in. Excel count used rows - c remove empty rows and columns. Excel 2007 pivot table count unique distinct records rows. Home. Similar Sites. Excel Count Distinct Text Values Column.curso de excel, curso de excel avanado, curso de excel basico, excel 2007, excel 2010. Extract unique values from one column using vba mr excel. How do i sort out duplicates using a formula in excel. Count distinct values in an excel 2013 pivottable excel.5 easy ways to extract unique distinct values. Vlookup across multiple sheets in excel get digital help. Excel 2010 - Hyperlink Text Only Not Entire Cell. Macro Fails, The macro may not be available in thisPublic Function CountDistinctInColumn(Rng As Range) CountDistinctIn Column Evaluate("Sum(NCopy and paste specific column data based on matching cell value in column A. Here is the structure of the VBA code Im trying to use to build a function I can call in Excel: Function CountDistinct(dataRange As Range).This will return the Count of Distinct values in column A 6. excel vba function excel-vba | this question edited Nov 18 14 at 0:39 pnuts 37.3k 6 43 78 asked How do I get unique values in a column? I do not want duplicate entries in my result.
Are you looking for DISTINCT values or UNIQUE values ? How do I get the distinct/unique values in a column in Excel?Pivot Table - Count unique values - Excel 2010 - Stack I have a pivot table in excel 2010 based on a network output. Count unique distinct values in a column in excel - Get Digital Help. Use a filter or functions to count the number of unique values in a range that. Applies To Excel 2016 Excel 2013 Excel 2010 Excel 2007 Excel 2016 for Mac Excel. select Regions ,COUNT(CAST(ID AS FLOAT)) / CAST(COUNT(DISTINCT USERNAME) AS FLOAT) ASReplacing value with ID if matching another value. Excel 2010 Conditional Formatting Issue using VALUE. Auto populating column based on if column number is listed in another worksheet. Count Distinct Values in an Excel 2013 PivotTable.Solved Getting distinct values for multiple columns -. I am developing a excel add-in in visual studio 2010. How to get all the distinct values of columns B from all the sheets in a string array? I want to know the number of distinct values ??in a column: Example: Column A: values in cell ??1 to 9: 1 2 3 4 5 6 4 5 6: the result is 6 because there are six different values. This array formula will give the number unique values ??in the range A1 I used the solution at Count unique values in a column in Excel.count visible merged cells in Excel 2010. How can I program a unique sort (basically DISTINCT) in Excel VBA? How to use excel 2010 functions to caculate the sum of distinct unique values?December 9.In a file with different number of columns delimited by space , How to count the sum of the columns. An example would show the need: File A: 1 2 2 3 4 5 6 1 1 1 5 Then the output would be: for column 1 Hello XP SP 3, Excel 2010. I should like to count unique values in a column (Column C. At present the values are text strings and (some) blank cells.XP SP 3, Excel 2010. Create a distinct list of values from one column that match a criteria, sorted in ascending order of corresponding values in a third column. Distinct counts in Excel (2010 or earlier) pivot table with several levels and distinct grand totals. In column A there are some names in column B there. Feb 27, 2013. Excel PivotTables Unique Count, or distinct count as its also know.Excel PivotTables Unique Count 3 Ways My Online Training Hub. Jul 17, 2010. In Excel 2003 removing duplicate values is a manual process. Count Distinct Value In Ms Excel. Count Unique Distinct Values Meeting Criteria Array Formula. Counting Distinct Values With Dax.Excel Magic Trick Count Distinct Words As They Are Entered In Column Formula Or Excel Ta. RELATED POST. excel 2010 distinct values in a column.excel 2010 formula to count unique values in a column. POPULAR. To get a unique count (distinct count) of salespeople in each region, follow these steps: Right-click one of the values in the Person field.The links in In Excel 2010, use a technique to Pivot the Pivot table. and In Excel 2007 and earlier versions, add a new column to the source data, and Use Count distinct unique values in excel 2010 pivot table. 5 ways to get unique values in excel excel how to.How to extract a unique distinct list from a column in. Filter unique distinct row records get digital help. Combine or append data in excel with power query. Count Unique or Duplicate Values in a List - Duration: 7:10.Excel - Pivot Table - Get Count, Sum based on distinct column names - Duration: 1:22.Create a Basic Pivot Table in Excel 2010 - Duration: 6:19. Eugene OLoughlin 312,766 views. Excel 2013 Pivot Tables Offer Distinct - Strategic Getting a count of the unique values in a Customer column. From the Pivot Table If you need to calculate a distinct count in Excel 2010 or earlier Microsoft Excel: Count number of distinct values in a range. Auto filter of distinct values on columns in SQL Server Reporting Services?In Excel 2007 and 2010: Use the Remove Duplicates option under the Data header from the menu bar. I hope it will be helpful for you. Count unique values among duplicates. Excel 2016 Excel 2013 Excel 2010 Excel 2007 Excel 2016 for Mac Excel for Mac 2011. if a column contains The values Getting unique values in Excel by using formulas only. seem to0. Returning the distinct/unique values in a column in Excel and make the. This will cut the data down to unique values only. Step 3: Copy the first column and paste it somewhere empty inside the worksheet.Yes Pivot tables can do distinct count in Excel 2013 and its crazy fast and super simple. Guess what! To count distinct text values in a column, well be using the same approach that weve just used to exclude empty cells. As you can easily guess, we willRegrettably, this option is not available in Excel 2010. It was introduced in Excel 2013, and also exists in Excel 2016. In earlier Excel versions, you How Do I Get The Distinct Unique Values In A Column In Excel Image GalleryExcel 2010 count unique values in multiple columns - 3Count unique distinct values using date criteria in a What do I mean by Distinct Count? . Well, when you count something like Customers in a pivot table, Excel is really giving you the count of records. In this pivot table for example, we were trying to get how many customers each Region has. Apr 16, 2013 Hello XP SP 3, Excel 2010. I should like to count unique values in a column Column C. At present the values are text strings and some blank cells.0. Returning the distinct/unique values in a column in Excel and make the. | 2,718 | 11,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.806232 |
http://www.blue-lotus.net/csaw-ctf-quals-2013-crypto500-writeup/ | 1,579,550,647,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00047.warc.gz | 201,434,453 | 5,587 | CSAW CTF Quals 2013 Crypto500 Writeup
Problem Description
We’ve found the source to the Arstotzka spies rendevous server, we must find out their new vault key. Source code is here.
So we have to crack sha512 puzzle and calc correct hash. At first i was trying to calculate sEphemeralPriv out, but it seems impossible. after some try, I realized that you can just use “n^m mod === something”.
Code
#!/usr/bin/python
import os
import sys
import time
import struct
import socket
import telnetlib
import random,hashlib,base64
from hashlib import sha512,sha1
def hashToInt(*params):
sha=sha512()
for el in params:
sha.update("%r"%el)
return int(sha.hexdigest(), 16)
def gs(num):
return hashlib.sha1(num).digest()
def crack(sha_p):
print "crack:",sha_p
ss=""
ret=0
for keylen in range(10,40):
if gs(new_msg)[-3:]=="\xff\xff\xff":
print "got one %s"%len(new_msg)
if len(new_msg)==21:
ss=new_msg
print "WE CRACK THE PUZZLE %d\n"%len(ss)
ret=1
return (ret,ss)
return (ret,ss)
#s = socket.create_connection(('128.238.66.222',7788))
s = socket.create_connection(('127.0.0.1',7788))
se=""
for i in range(1,100):
ss=s.recv(1000)
print ss
sha_p=ss[len(ss)-16:]
new_msg=crack(sha_p[-16:])
if new_msg[0]:
se=new_msg[1]
print "send:",se,hashlib.sha1(se).digest()[-3:]
break
s.send(se)
#raw_input()
N = 59244860562241836037412967740090202129129493209028128777608075105340045576269119581606482976574497977007826166502644772626633024570441729436559376092540137133423133727302575949573761665758712151467911366235364142212961105408972344133228752133405884706571372774484384452551216417305613197930780130560424424943100169162129296770713102943256911159434397670875927197768487154862841806112741324720583453986631649607781487954360670817072076325212351448760497872482740542034951413536329940050886314722210271560696533722492893515961159297492975432439922689444585637489674895803176632819896331235057103813705143408943511591629
index=28483644508750028902258833085453121291738558908844640378204850915473006274236033891815596646870094954832384471913373171061099388958373536383247454431214837805096635029244738399662911119306089493137122381562794483801525427601711736403424013966624957471463169984161438593701202381673774894874606950326837142168801614196218030413825361835813060325642766963973454456577907567314093695138863251603368180581162185039224604662844750924047132242103613509637088222461132023037724162558095265336717907895786691004801515830247270579025866384571194244368350362690250445425121639616294300827554006418861422621428030154329451920623#17095359415354811031956139176232822755293333580176796651092816713929142596363160518183074582333639666374490434453616522976528693318889544211598801485465809374370977363040240176466544504447562622451988654983833567639785357067606037932165650485806709538575591881306056180364137612491852807151445333929946180083335184143784040034635507293616742432550220001472750029503246943474764484598756871323795898352813684410952415776064654304191774216122857994311730048413954488815342894838841918071841785821830683677234982197021652871796420412165927837273704381564043458200253387672509280296033910910416083249362331740208987494679
cEphemeral=1
send_num1=str(hex(index))[2:][:-1]
send_num2=str(hex(cEphemeral))[2:]
s.send(struct.pack("H",len(send_num1)))
s.send(send_num1)#send numberc index-->base number
print "1:",s.recv(72)
print "2:",s.recv(60)
s.send(struct.pack("H",len(send_num2)))
s.send(send_num2)#send number cEphemeral
salt=int(s.recv(128),16)
sEphemeral=long(int(s.recv(514),16))
#print "3:salt:",salt
#print "4:sEphemeral:",sEphemeral
tmp=hashToInt(salt,"")
storedKey = pow(index,tmp , N)
cq_sEphemeral=(3 * storedKey) % N
tmp1=(sEphemeral-cq_sEphemeral)%N
#agreedKey_withouthash = (cEphemeral * index^[sha512(salt, password) * slush])^sEphemeralPriv mod Nwhoami
# cEphemeral=1 -->
#agreedKey_withouthash = (index^[sha512(salt, password) * slush])^sEphemeralPriv mod N
#cause cEphemeral^^3 mod N =1
#so if sha512(salt, password) * slush * sEphemeralPriv mod 3 == 0
#agreedKey_withouthash = 1
#so we don't need to know sEphemeralPriv
slush = hashToInt(cEphemeral, sEphemeral)
salt=hashToInt(index)
agreedKey=hashToInt(1L)
gennedKey=hashToInt(hashToInt(N) ^ hashToInt(index), hashToInt(index), salt,int(cEphemeral), long(sEphemeral), agreedKey)
send_num3=str(hex(gennedKey))[2:][:-1]
s.send(struct.pack("H",len(send_num3)))
s.send(send_num3)
print s.recv(1000)
print s.recv(1000)#-->flag recv
print s.recv(1000) | 1,490 | 4,460 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.45089 |
http://water.me.vccs.edu/concepts/specific.html | 1,529,735,904,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864943.28/warc/CC-MAIN-20180623054721-20180623074721-00223.warc.gz | 345,600,551 | 1,531 | Specific Gravity
Specific gravity is a way of relating the density of an object to the density of water to determine whether or not the object will float. The formula for specific gravity is given below:
If an object's specific gravity is less than one, then the object will float. If the object has a specific gravity of greater than one, it sinks. So dense objects sink in water and less dense objects float.
Importance of Specific Gravity
Specific gravity is a very important concept in the water/wastewater field. The specific gravity of a substance will determine where a compound can be found in water in case of a spill.
Let's consider gasoline. The density of gasoline is 0.6 g/mL and the density of water is 1.0 g/mL. So the specific gravity of gasoline is:
Since its specific gravity is 0.6 (less than 1), gasoline floats in water. So when a ship leaks gasoline into the water, the gasoline stays at the top of the water.
In contrast, the specific gravity of palmalive is 1.1, so it sinks in water. | 239 | 1,022 | {"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} | 3.3125 | 3 | CC-MAIN-2018-26 | latest | en | 0.882406 |
https://satoshiwatch.com/g9v5tau/e5952e-formation-of-mgo-equation | 1,611,677,815,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704800238.80/warc/CC-MAIN-20210126135838-20210126165838-00720.warc.gz | 524,043,191 | 15,985 | ## formation of mgo equation
formation of mgo equation: Uncategorized
2 seconds ago
As per lab manual we used a calibrated calorimeter (using a rounded end thermometer so as to not puncture a hole in the calorimeter) to determine the heats of reaction for Magnesium (Mg) with Hydrochloric Acid (HCl) and Hydrochloric Acid with Magnesium Oxide (MgO). Mg (s) + 1/2 O 2(g) → MgO (s) ∆H = ∆H˚f (MgO)(2) It is more convenient to use the first law of thermodynamics in the form of Hess's law to simplify the measurement of ∆H˚f (MgO). Since both reactions are in dilute water solutions of HCl it was necessary to know the heat capacity of water, but b… Magnesium hydroxide forms in the presence of water (MgO + H2O → Mg(OH)2), but it can be reversed by heating it to separate moisture. It has an empirical formula of MgO and consists of a lattice of Mg ions and O ions held together by ionic bonding. Magnesium oxide (MgO), or magnesia, is a white hygroscopic solid mineral that occurs naturally as periclase and is a source of magnesium (see also oxide). Then using mathematical formulas we were able to calculate the heat formation of MgO, which is measured in kJ/Mol. MgO(s) + 2 HCl(aq) → MgCl2(aq) + H2O(l) o ΔH2 (experimentally determined) H2(g) + ½ O2(g) → H2O(l) o ΔH3 = -285.8 kJ Using the techniques presented in the CO example above, you can combine these equations to find o ΔHrxn for formation of one mole of MgO(s): Mg(s) + ½ O2(g) → MgO(s) o The enthalpy of formation of MgO is more difficult to measure directly.
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• No comments posted | 455 | 1,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.90391 |
https://byjus.com/maths/orthocenter/ | 1,701,423,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00149.warc.gz | 184,143,203 | 116,984 | Orthocenter
The orthocenter is the point where all the three altitudes of the triangle cut or intersect each other. Here, the altitude is the line drawn from the vertex of the triangle and is perpendicular to the opposite side. Since the triangle has three vertices and three sides, therefore there are three altitudes. Also learn, Circumcenter of a Triangle here.
The orthocenter will vary for different types of triangles such as Isosceles, Equilateral, Scalene, right-angled, etc. In the case of an equilateral triangle, the centroid will be the orthocenter. But in the case of other triangles, the position will be different. Orthocenter doesn’t need to lie inside the triangle only, in case of an obtuse triangle, it lies outside of the triangle.
Orthocenter of a Triangle
The orthocenter of a triangle is the point where the perpendicular drawn from the vertices to the opposite sides of the triangle intersect each other.
• For an acute angle triangle, the orthocenter lies inside the triangle.
• For the obtuse angle triangle, the orthocenter lies outside the triangle.
• For a right triangle, the orthocenter lies on the vertex of the right angle.
Take an example of a triangle ABC.
In the above figure, you can see, the perpendiculars AD, BE and CF drawn from vertex A, B and C to the opposite sides BC, AC and AB, respectively, intersect each other at a single point O. This point is the orthocenter of △ABC.
Orthocenter Formula
The formula of orthocenter is used to find its coordinates. Let us consider a triangle ABC, as shown in the above diagram, where AD, BE and CF are the perpendiculars drawn from the vertices A(x1,y1), B(x2,y2) and C(x3,y3), respectively. O is the intersection point of the three altitudes.
First, we need to calculate the slope of the sides of the triangle, by the formula:
m = y2-y1/x2-x1
Now, the slope of the altitudes of the triangle ABC will be the perpendicular slope of the line.
Perpendicular slope of line = -1/Slope of the line = -1/m
Let slope of AC is given by mAC. Hence,
mAC = y3-y1/x3-x1
Similarly, mBC = (y3-y2)/(x3-x2)
Now, the slope of the respective altitudes are:
Slope of BE, mBE = -1/mAC
Now here we will be using slope point form equation os a straight line to find the equations of the lines, coinciding with BE and AD.
Therefore,
mBE = (y-y2)/(x-x2)
Hence, we will get two equations here which can be solved easily. Thus, the value of x and y will give the coordinates of the orthocenter.
Also, go through Orthocenter Formula
Properties of Orthocenter
The orthocenter is the intersection point of the altitudes drawn from the vertices of the triangle to the opposite sides.
• For an acute triangle, it lies inside the triangle.
• For an obtuse triangle, it lies outside of the triangle.
• For a right-angled triangle, it lies on the vertex of the right angle.
• The product of the parts into which the orthocenter divides an altitude is the equivalent for all 3 perpendiculars.
Construction of Orthocenter
To construct the orthocenter of a triangle, there is no particular formula but we have to get the coordinates of the vertices of the triangle. Suppose we have a triangle ABC and we need to find the orthocenter of it. Then follow the below-given steps;
• The first thing we have to do is find the slope of the side BC, using the slope formula, which is, m = y2-y1/x2-x1
• The slope of the line AD is the perpendicular slope of BC.
• Now, from the point, A and slope of the line AD, write the straight-line equation using the point-slope formula which is; y2-y1 = m (x2-x1)
• Again find the slope of side AC using the slope formula.
• The perpendicular slope of AC is the slope of the line BE.
• Now, from the point, B and slope of the line BE, write the straight-line equation using the point-slope formula which is; y-y1 = m (x-x1)
• Now, we have got two equations for straight lines which is AD and BE.
• Extend both the lines to find the intersection point.
• The point where AD and BE meets is the orthocenter.
Note: If we are able to find the slopes of the two sides of the triangle then we can find the orthocenter and its not necessary to find the slope for the third side also.
Orthocenter Examples
Question:
Find the orthocenter of a triangle whose vertices are A (-5, 3), B (1, 7), C (7, -5).
Solution:
Let us solve the problem with the steps given in the above section;
1. Slope of the side AB = y2-y1/x2-x1 = 7-3/1+5=4/6=⅔
2. The perpendicular slope of AB = -3/2
3. With point C(7, -5) and slope of CF = -3/2, the equation of CF is y – y1 = m (x – x1) (point-slope form)
4. Substitute the values in the above formula.
(y + 5) = -3/2(x – 7)
2(y + 5) = -3(x – 7)
2y + 10 = -3x + 21
3x + 2y = 11 ………………………………….(1)
5. Slope of side BC = y2-y1/x2-x1 = (-5-7)/(7-1) = -12/6=-2
6. The perpendicular slope of BC = ½
7. Now, the equation of line AD is y – y1 = m (x – x1) (point-slope form)
(y-3) = ½(x+5)
Solving the equation we get,
x-2y = -11…………………………………………(2)
8. Now when we solve equations 1 and 2, we get the x and y values.
Which are, x = 0 and y = 11/2 = 5.5
Therefore(0, 5.5) are the coordinates of the orthocenter of the triangle.
Try out: Orthocenter Calculator
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Q1
What is orthocenter?
The orthocenter is the point where the altitudes drawn from the vertices of a triangle intersects each other.
Q2
What is the difference between orthocenter and circumcenter?
The orthocenter is the point of intersection of three altitudes drawn from the vertices of a triangle.
The circumcenter is the point of intersection of the perpendicular bisector of the three sides.
Q3
Where does the orthocenter of the obtuse triangle lie?
The orthocenter of the obtuse triangle lies outside the triangle.
Q4
Where does the orthocenter of right triangle lie?
The orthocenter of a right-angled triangle lies on the vertex of the right angle.
Q5
What is the difference between orthocenter and centroid?
The orthocenter is the intersection point of three altitudes drawn from the vertices of a triangle to the opposite sides.
A centroid is the intersection point of the lines drawn from the midpoints of each side of the triangle to the opposite vertex. | 1,633 | 6,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-50 | latest | en | 0.860139 |
https://www.unitsconverters.com/en/Squarecentimetre-To-Squarefoot(Ussurvey)/Unittounit-304-321 | 1,627,065,661,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00424.warc.gz | 1,124,212,345 | 38,378 | Formula Used
1 Square Centimeter = 0.0001 Square Meter
1 Square Meter = 10.7638673608924 Square Foot (US Survey)
1 Square Centimeter = 0.00107638673608924 Square Foot (US Survey)
## Square Centimeters to Square Foot (US Survey)s Conversion
cm² stands for square centimeters and ft² stands for square foot (us survey)s. The formula used in square centimeters to square foot (us survey)s conversion is 1 Square Centimeter = 0.00107638673608924 Square Foot (US Survey). In other words, 1 square centimeter is 930 times smaller than a square foot (us survey). To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Square Centimeter to Square Foot (US Survey)
How to convert square centimeter to square foot (us survey)? In the area measurement, first choose square centimeter from the left dropdown and square foot (us survey) from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from square foot (us survey) to square centimeter? You can check our square foot (us survey) to square centimeter converter.
How many Square Meter is 1 Square Centimeter?
1 Square Centimeter is equal to 0.0001 Square Meter. 1 Square Centimeter is 10000 times Smaller than 1 Square Meter.
How many Square Kilometer is 1 Square Centimeter?
1 Square Centimeter is equal to 1E-10 Square Kilometer. 1 Square Centimeter is 10000000000 times Smaller than 1 Square Kilometer.
How many Square Millimeter is 1 Square Centimeter?
1 Square Centimeter is equal to 100 Square Millimeter. 1 Square Centimeter is 100 times Bigger than 1 Square Millimeter.
How many Square Foot is 1 Square Centimeter?
1 Square Centimeter is equal to 0.00107639104166236 Square Foot. 1 Square Centimeter is 929.030400007433 times Smaller than 1 Square Foot.
## Square Centimeters to Square Foot (US Survey)s Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like cm² to ft² through multiplicative conversion factors. When you are converting area, you need a Square Centimeters to Square Foot (US Survey)s converter that is elaborate and still easy to use. Converting Square Centimeter to Square Foot (US Survey) is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Square Centimeter to Square Foot (US Survey) conversion along with a table representing the entire conversion.
Let Others Know | 674 | 2,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-31 | latest | en | 0.809169 |
https://number.academy/96703 | 1,726,687,609,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00809.warc.gz | 390,608,287 | 11,307 | Number 96703 facts
The odd number 96,703 is spelled 🔊, and written in words: ninety-six thousand, seven hundred and three. The ordinal number 96703rd is said 🔊 and written as: ninety-six thousand, seven hundred and third. The meaning of the number 96703 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 96703. What is 96703 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 96703.
Interesting facts about the number 96703
Asteroids
• (96703) 1999 JO111 is asteroid number 96703. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 5/13/1999.
What is 96,703 in other units
The decimal (Arabic) number 96703 converted to a Roman number is (XC)(V)MDCCIII. Roman and decimal number conversions.
Length conversion
96703 kilometers (km) equals to 60089 miles (mi).
96703 miles (mi) equals to 155629 kilometers (km).
96703 meters (m) equals to 317264 feet (ft).
96703 feet (ft) equals 29476 meters (m).
Time conversion
(hours, minutes, seconds, days, weeks)
96703 seconds equals to 1 day, 2 hours, 51 minutes, 43 seconds
96703 minutes equals to 2 months, 1 week, 4 days, 3 hours, 43 minutes
Zip codes 96703
• Zip code 96703 Anahola, Hawaii, Kauai, USA a map
Codes and images of the number 96703
Number 96703 morse code: ----. -.... --... ----- ...--
Sign language for number 96703:
Number 96703 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
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Is Prime?
The number 96703 is a prime number. The closest prime numbers are 96697, 96731.
Factorization and factors (dividers)
The prime factors of 96703
Prime numbers have no prime factors smaller than themselves.
The factors of 96703 are 1, 96703.
Total factors 2.
Sum of factors 96704 (1).
Prime factor tree
96703 is a prime number.
Powers
The second power of 967032 is 9.351.470.209.
The third power of 967033 is 904.315.223.620.927.
Roots
The square root √96703 is 310,97106.
The cube root of 396703 is 45,900067.
Logarithms
The natural logarithm of No. ln 96703 = loge 96703 = 11,4794.
The logarithm to base 10 of No. log10 96703 = 4,98544.
The Napierian logarithm of No. log1/e 96703 = -11,4794.
Trigonometric functions
The cosine of 96703 is 0,065686.
The sine of 96703 is -0,99784.
The tangent of 96703 is -15,19102.
Number 96703 in Computer Science
Code typeCode value
96703 Number of bytes94.4KB
Unix timeUnix time 96703 is equal to Friday Jan. 2, 1970, 2:51:43 a.m. GMT
IPv4, IPv6Number 96703 internet address in dotted format v4 0.1.121.191, v6 ::1:79bf
96703 Decimal = 10111100110111111 Binary
96703 Decimal = 11220122121 Ternary
96703 Decimal = 274677 Octal
96703 Decimal = 179BF Hexadecimal (0x179bf hex)
96703 BASE64OTY3MDM=
96703 MD504df1b545423a3a40b03afac5fc9036e
96703 SHA111a82258b51cc89f3d2d7493153d4d554ac8c94d
96703 SHA2243648978368559e2d055cab3dfe04005a3f6d40cc2efabcef3da11e7f
96703 SHA25662308655c26be03c15fef0e903e30a4155e258ef8d3145eff665bd35a43f2220
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If you know something interesting about the 96703 number that you did not find on this page, do not hesitate to write us here.
Numerology 96703
Character frequency in the number 96703
Character (importance) frequency for numerology.
Character: Frequency: 9 1 6 1 7 1 0 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 96703, the numbers 9+6+7+0+3 = 2+5 = 7 are added and the meaning of the number 7 is sought.
№ 96,703 in other languages
How to say or write the number ninety-six thousand, seven hundred and three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 96.703) noventa y seis mil setecientos tres German: 🔊 (Nummer 96.703) sechsundneunzigtausendsiebenhundertdrei French: 🔊 (nombre 96 703) quatre-vingt-seize mille sept cent trois Portuguese: 🔊 (número 96 703) noventa e seis mil, setecentos e três Hindi: 🔊 (संख्या 96 703) छियानवे हज़ार, सात सौ, तीन Chinese: 🔊 (数 96 703) 九万六千七百零三 Arabian: 🔊 (عدد 96,703) ستة و تسعون ألفاً و سبعمائة و ثلاثة Czech: 🔊 (číslo 96 703) devadesát šest tisíc sedmset tři Korean: 🔊 (번호 96,703) 구만 육천칠백삼 Danish: 🔊 (nummer 96 703) seksoghalvfemstusinde og syvhundrede og tre Dutch: 🔊 (nummer 96 703) zesennegentigduizendzevenhonderddrie Japanese: 🔊 (数 96,703) 九万六千七百三 Indonesian: 🔊 (jumlah 96.703) sembilan puluh enam ribu tujuh ratus tiga Italian: 🔊 (numero 96 703) novantaseimilasettecentotre Norwegian: 🔊 (nummer 96 703) nitti-seks tusen, syv hundre og tre Polish: 🔊 (liczba 96 703) dziewięćdzisiąt sześć tysięcy siedemset trzy Russian: 🔊 (номер 96 703) девяносто шесть тысяч семьсот три Turkish: 🔊 (numara 96,703) doksanaltıbinyediyüzüç Thai: 🔊 (จำนวน 96 703) เก้าหมื่นหกพันเจ็ดร้อยสาม Ukrainian: 🔊 (номер 96 703) дев'яносто шість тисяч сімсот три Vietnamese: 🔊 (con số 96.703) chín mươi sáu nghìn bảy trăm lẻ ba Other languages ...
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# How many mm in 1 fit?
Wiki User
2011-02-14 06:22:26
Millimeters in 1 foot?
1 foot (12 inches/1 foot)(2.54 centimeters/1 inch)
= 30.48 centimeters * 10
= 304.8 mm
Wiki User
2011-02-14 06:22:26
Study guides
20 cards
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See all cards
3.8
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https://www.physicsforums.com/threads/basic-trigonometry.660770/ | 1,519,618,916,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817999.51/warc/CC-MAIN-20180226025358-20180226045358-00319.warc.gz | 891,329,201 | 15,397 | # Basic trigonometry
1. Dec 25, 2012
### iknownth
We all know that
sin theta = opposite side / hypotenuse
cos theta = adjacent side / hypotenuse
tan theta = opposite side / adjacent side.
But why? Are there some explanations behind or are they just defined by scientists?
2. Dec 25, 2012
### pwsnafu
What do you mean by "explanations"?
To answer the question, they are the definitions used in planar geometry. There are other (equivalent) ways of defining them, but you'll get the same properties nonetheless.
3. Dec 25, 2012
### iknownth
How can one prove that sin theta = opposite side / hypotenuse ?
4. Dec 25, 2012
### pwsnafu
I'm assuming planar geometry here. There are two answers:
1. It is the definition of sine. There is nothing to prove.
2. The definition is using the unit circle. It which case the proof is immediate from similar triangles.
I guess it's worth asking: what is your definition of sine?
5. Dec 25, 2012
### micromass
In this case, there is still something to prove. You want to prove also that if two triangles have the same angle, then the quantity opposite side/hypothenuse is the same.
6. Dec 25, 2012
### pwsnafu
Arrgh yes of course.
Where's the brainfart smiley when you need one?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 353 | 1,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-09 | longest | en | 0.922594 |
https://search.credoreference.com/content/topic/apollonius_of_perga | 1,548,111,841,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583814455.32/warc/CC-MAIN-20190121213506-20190121235506-00532.warc.gz | 639,927,530 | 28,439 | Definition: Apollonius of Perga from Philip's Encyclopedia
Greek mathematician and astronomer. He built on the foundations laid by Euclid. In Conics, he showed that an ellipse, a parabola, and a hyperbola can be obtained by taking plane sections at different angles through a cone. In astronomy, he described the motion of the planets in terms of epicycles, which remained the basis of the system used until the time of Copernicus.
Summary Article: Apollonius of Perga (c. 245-c.190 BC)
from The Hutchinson Dictionary of Scientific Biography
Place: Greece
Subject: biography, maths and statistics
Greek mathematician whose treatise on conic sections represents the final flowering of Greek mathematics.
Apollonius was born early in the reign of Ptolemy Euergetes, king of Egypt, in the Greek town of Perga in southern Asia Minor (now part of Turkey). Little is known of his life. It is thought that he may have studied at the school established by Euclid at Alexandria, especially since much of his work was built on Euclidean foundations.
Apollonius' fame rests on his eight-volume treatise, The Conics, seven volumes of which are extant. The first four books consisted of an introduction and a statement of the state of mathematics provided by his predecessors. In the last four volumes Apollonius put forth his own important work on conic sections, the foundation of much of the geometry still used today in astronomy and ballistic science.
Apollonius described how a cone could be cut so as to produce circles, ellipses, parabolas, and hyperbolas; the last three terms were coined by him. He investigated the properties of each and showed that they were all interrelated because, as he stated, ‘any conic section is the locus of a point that moves so that the ratio of its distance, f, from a fixed point (the focus) to its distance, d, from a straight line (the directrix) is a constant’. Whether this constant, e, is greater than, equal to, or less than one determines which of the three types of curve the function represents. For a hyperbola e > 1; for a parabola e = 1; and for an ellipse e < 1. At the time, Apollonius' discoveries lay in the realm of pure mathematics; it was only later that their immensely valuable application became apparent, when it was discovered that conic sections form the paths, or loci, followed by planets and projectiles in space.
Other than The Conics, only one treatise of Apollonius survives; it is entitled Cutting off a Ratio. It was found written in Arabic and was translated into Latin in 1706, but is of little mathematical significance.
Apollonius' brilliant concept of geometry was a milestone in the understanding of mechanics, navigation, and astronomy. Above all, his work on epicircles and ellipses played a major part in Ptolemy's working out of the cosmology that would dominate western astronomy from the 2nd century AD to the 16th century.
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(c.260-c.190 BC) Greek mathematician noted for his Conics , of which seven of the original eight books are extant. In about 400 propositions...
See more from Credo | 843 | 3,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-04 | latest | en | 0.975638 |
https://stats.stackexchange.com/questions/221548/what-does-it-mean-to-say-that-an-event-happens-eventually/221592 | 1,623,762,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00482.warc.gz | 489,837,330 | 41,359 | What does it mean to say that an event “happens eventually”?
Consider a 1 dimensional random walk on the integers $\mathbb{Z}$ with initial state $x\in\mathbb{Z}$:
$$S_n=x+\sum^n_{i=1}\xi_i$$
where the increments $\xi_i$ are I.I.D such that $P\{\xi_i=1\}=P\{\xi_i=-1\}=\frac{1}{2}$.
One can prove that (1)
$$P^x{\{S_n \text{ reaches +1 eventually}\}} = 1$$
where the subscript denotes the initial position.
Let $\tau$ be the first passage time to state $+1$. In other words, $\tau:=\tau(1):=\min\{n\geq0:S_n=1\}$. One can also prove that (2)
$$E\tau = +\infty$$
Both proofs can be found in http://galton.uchicago.edu/~lalley/Courses/312/RW.pdf. Through reading the article, I do understand both proofs.
My question is, however, what the meaning of "eventually" is in the first statement as well as in general. If something happens "eventually", it doesn't have to occur in finite time, does it? If so, what really is the difference between something which doesn't happen and something that doesn't happen "eventually"? Statements (1) and (2) in some sense are contradicting themselves to me. Are there other examples like this?
EDIT
Just want to add a motivation for the question, i.e., a straightforward example of something that happens "eventually", but with finite expected wait time.
\begin{split} P\{\text{walker eventually moves left}\} & = 1 - P\{\text{walker never moves left}\} \\ & = 1-\lim_{n\to\infty} \frac{1}{2^n} \\ & = 1 \end{split}
Therefore we know that the walker will "eventually" move to the left, and the expected wait time before doing so (i.e., moving left) is $1/(1/2)=2$.
Seeing something that happens "eventually" but with infinite expected "wait time" was quite a stretch for my imagination. The second half of @whuber's response is another great example.
• no eventually means in finite time. That is precisely what is being contrasted: P is finite , whilst expectation of tau is infinite – seanv507 Jun 30 '16 at 20:08
• Well there is the canonical example of Cauchy distribution en.wikipedia.org/wiki/Cauchy_distribution. – seanv507 Jun 30 '16 at 20:33
• @seanv507 - Yes, although the mean of the Cauchy distribution is undefined rather than infinite (a sample mean from the Cauchy dbn will jump around as $n$ approaches infinity rather than steadily converge to +Infinity). I was thinking of the Pareto distribution (en.wikipedia.org/wiki/Pareto_distribution), which has mean = Infinity when its shape parameter $\alpha <= 1$ and yet has a well defined probability distribution function. – RobertF Jun 30 '16 at 20:36
• @RobertF thanks - I ought to have said Pareto – seanv507 Jun 30 '16 at 21:06
• There's some comfort in all this: if $P(\tau=\infty)>0$, then $E[\tau]=\infty$, but not the other way around. – Alex R. Jun 30 '16 at 21:11
How would you demonstrate an event "eventually happens"? You would conduct a thought experiment with a hypothetical opponent. Your opponent may challenge you with any positive number $p$. If you can find an $n$ (which most likely depends on $p$) for which the chance of the event happening by time $n$ is at least $1-p$, then you win.
In the example, "$S_n$" is misleading notation because you use it both to refer to one state of a random walk as well as to the entire random walk itself. Let's take care to recognize the distinction. "Reaches $1$ eventually" is meant to refer to a subset $\mathcal{S}$ of the set of all random walks $\Omega$. Each walk $S\in\Omega$ has infinitely many steps. The value of $S$ at time $n$ is $S_n$. "$S$ reaches $1$ by time $n$" refers to the subset of $\Omega$ of walks that have reached the state $1$ by time $n$. Rigorously, it is the set
$$\Omega_{1,n} = \{S\in\Omega \mid S_1=1\text{ or }S_2=1\text{ or }\cdots\text{ or }S_n=1\}.$$
In your response to the imaginary opponent, you are exhibiting some $\Omega_{1,n}$ with the property that
$$\mathbb{P}_\xi(\Omega_{1,n}) \ge 1-p.$$
Because $n$ is arbitrary, you have available all elements of the set
$$\Omega_{1,\infty} = \bigcup_{n=1}^\infty \Omega_{1,n}.$$
(Recall that $S \in \bigcup_{n=1}^\infty \Omega_{1,n}$ if and only if there exists a finite $n$ for which $S \in \Omega_{1,n}$, so there aren't any infinite numbers involved in this union.)
Your ability to win the game shows this union has a probability exceeding all values of the form $1-p$, no matter how small $p\gt 0$ may be. Consequently, that probability is at least $1$--and therefore equals $1$. You will have demonstrated, then, that
$$\mathbb{P}_\xi(\Omega_{1,\infty}) = 1.$$
One simple way to appreciate the distinction between "happening eventually" and having an infinite expected first passage time is to contemplate a simpler situation. For $n$ any natural number, let $\omega^{(n)}$ be the sequence
$$\omega^{(n)} = (\underbrace{0, 0,\ldots,0}_{n},1,1,\ldots)$$
in which $n$ zeros are followed by an endless string of ones. In other words, these are the walks that stay at the origin and at some (finite) time step over to the point $1$, then stay there forever.
Let $\Omega$ be the set of all these $\omega^{(n)}, n=0, 1, 2, \ldots$ with the discrete sigma algebra. Assign a probability measure via
$$\mathbb{P}(\omega^{(n)}) = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}.$$
This was designed to make the chance of jumping to $1$ by the time $n$ equal to $1-1/(n+1)$, which obviously approaches arbitrarily closely to $1$. You will win the game. The jump eventually happens and when it does, it will be at some finite time. However, the expected time when it happens is the sum of the survival function (which gives the chances of not having jumped at time $n$),
$$\mathbb{E}(\tau) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots,$$
which diverges. That is because a relatively large probability is given to waiting a long time before jumping.
• Am I misunderstanding if I read your first section as boiling down to to an epsilon/delta argument, and thus basically just saying $$\lim_{ n \to \infty } P_n = 1$$ (where $P_n$ is the probability of some event after $n$ steps)? – jpmc26 Jun 30 '16 at 22:50
• @jpm It doesn't just boil down to it: it is an epsilon-delta argument. In this case "delta" is "$n$" and "epsilon" is written "$p$" as a reminder that it is a probability. The emphasis here is on the finiteness of $n$: limits are defined in terms of finite values and finite operations, not infinite ones. – whuber Jul 1 '16 at 12:56
• I thank an anonymous user for suggesting the use of underbrace in the description of $\omega^{(n)}$. – whuber Jul 1 '16 at 13:01
That something happens eventually means that there is some point in time at which it happens, but there is a connotation that one is not referring to any particular specified time before which it happens. If you say something will happen within three weeks, that is a stronger statement than that it will happen eventually. That it will happen eventually does not specify a time, such as "three weeks" or "thirty-billion years" or "one minute". | 1,972 | 7,002 | {"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": 3, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-25 | latest | en | 0.918806 |
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# Calling All RPI (Lally) MBA 2013 Applicants
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Calling All RPI (Lally) MBA 2013 Applicants
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Hello Aspirants,
This thread is for all who wish to apply to the RPI - Lally MBA for the upcoming induction.
I am a part of the current MBA programme at Lally School of Management and Technology at Rensselaer...
So bring them on and hope to see the best of you here in 2013!
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Dear sir,
I have been offered Letter of Acceptance from RPI.
Please brief me about the placement statistics of latest outgoing batch of MBA - Finance.
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Shyamal Mandirwala
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shmandirwala wrote:
Dear sir,
I have been offered Letter of Acceptance from RPI.
Please brief me about the placement statistics of latest outgoing batch of MBA - Finance.
Thanks & Regards
Shyamal Mandirwala
Dear Shyamal,
Congratulations on your offer. To see last years stats, you could visit our profile on Beat the Gmat.
Hope that helps...
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# Calling All RPI (Lally) MBA 2013 Applicants
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 822 | 2,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | latest | en | 0.822726 |
http://www.millersville.edu/academics/scma/physics/experiments/065/index.php | 1,386,912,688,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164886464/warc/CC-MAIN-20131204134806-00006-ip-10-33-133-15.ec2.internal.warc.gz | 446,938,210 | 8,746 | ## Finals Rescheduled Again
Due to inclement weather forecast for this Saturday, 12/14, the make-up exams from Tuesday 12/10 will be rescheduled (yet again) as follows:
- The exam period for classes that meet Tuesday-Thurs 1 p.m. is now rescheduled to Friday, 12/13/2013 from 2:45 p.m. - 4:45 p.m.
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# From Reality to Force Components
Physicists of a certain age remember "force tables," laboratory devices that had pulleys and strings and weights that had to be arranged to make the net force on a central ring equal to zero. They were so hard to balance that it was easy to lose sight of the fundamental principle of the vector addition of forces. This month's apparatus is a much simplified version, which demonstrates the summation of forces and tests the ideas of vector components.
The particular implementation here brings up one of the pitfalls of beginning physics: sliding too easily from real things and events to abstract cosmic pronouncements. For example, students observing a weight spinning on the end of a string are tempted to write in their lab notebook, "Today we verified Einstein's theory of relativity. In this month's experiment, Dr. Miziumski takes special care to notice what things are real and what statements are interpretations of reality.
The experiment begins with a piece of string tied with a loop at the left to a post, passing over a pulley, and attached to a weight m2, as shown in the upper right hand figure. We hang a second weight from the knot shown in the lower figure at right, and observe the consequences.
The second mass that we put on, called m1, is smaller than m2. For a wide range of choices of m1 , the system "balances:" The string sags down with m1 pulling on the knot, and eventually comes to rest, with all pieces of the system at equilibrium.
We seek parameters to describe the behavior of the system. As is often true, it helps to divide the parameters into "independent" variables and "dependent" variables. The distinction is more personal than physical: An independent variable is one that I can change directly with my hand. A dependent variable is one that changes in response to my modification of the independent variable. In this system, the sizes of the weights are identified as independent variables. The "shape" of the system that results is to be measured by the dependent variable(s).
We ask, "What are the conditions of equilibrium for this system?" That is, we ask for the relation between dependent and independent variables, when the system is at equilibrium. We start by investigating the independent variables.
We imagine what would happen if the force of gravity were stronger. For example, suppose that the strength of the gravitational field were doubled, so that the the weight of each mass is doubled. It seems plausible that the system would look just the same. We test this guess by doubling the system: hang a second weight of mass m1 at the knot, and a second weight of mass m2 on the right end of the string. We find in fact that the system does not shift in response to these changes.
On the other hand, if we double only the mass m2 , the system does change its configuration, with the doubled m2 sinking lower (and m1 rising higher) as equilibrium is reached. Evidently the size of the masses does not change the equilibrium, provided that the *ratio* of the masses remains the same. This gives us a welcome simplification: The independent variable that is most useful here is the ratio of the two masses, ( m1 )/ (m2 ). (We made a lucky guess about which mass works most nicely in the numerator.)
We next need parameters to describe the shape of the system. When the small weight m1 is added, the string bends, forming two angles; one on each side of the knot, called q and F. To simplify our problem, we restrict the angles which we will allow. We require that the string to the left of the knot be horizontal. Referring to the second figure, we require that the angle F be zero. In order to make this happen, we must do extra work: We must slide the loop up and down the pole until we make the left side of the string horizontal. Luckily we find that it is possible to do this for a fairly wide range of mass ratios.
Our experiment now looks like the figure at the right. Because we have been studying force components, we have an urge to start calculating forces. But wait....
Let us not be prejudiced by the force rules that we have learned. Let us see what we can learn without their help. Returning to the basics, we ask "How can we describe the state of the system for a given mass ratio?" With the restriction that F=0, we can describe the system by stating value of the angle q . q is the most valuable dependent variable.
Now we go to work collecting data. We vary the mass ratio and measure the angle that the string makes with respect to the horizontal. A sample set of data is displayed below. In this data set, m1 was held constant at 50 grams. Because we will be working with ratios and angles in radians, the units are left in ordinary bench-top units (grams and cm) and not converted to SI units. The units will cancel.
m2 60 70 80 100 100 120 150 200 grams hypotenuse, h 60.5 67 60.3 56 55 52.5 51 50 cm vertical drop, L 50.6 47.7 39 28 28.6 22.8 16.5 13 cm
The terms in the table refer to the figure at the right. To measure the angle q , we will use the right triangle formed by the pulley, the knot and the horizontal dashed line. The hypotenuse of this triangle is called h, and the vertical leg has length L. Having collected the data, we try to think about what have have observed, and not what we remember having learned from a book. To search for a relation between the dependent and the independent variables, we try a graph.
The graph below uses the mass ratio. The data fall nearly on a straight line, so it is tempting to fit a straight line through the points. A straight line fit is shown. The correlation coefficient is represented by R2. R2=1 for a perfect fit, so the straight line fit seems to be good, but this turns out to be a snare and a trap.
A convenient rule of thumb for graphing is that data are represented by points, and theoretical expectations are represented by lines. With that understanding, we are required to understand the meaning of the parameters of a line which is fit to data. In seeking this understanding, it helps to explore the extremes of the independent parameter, ( m1 )/ (m2 )
A straight line fit to the data indicates that when the ratio is zero the angle is about - 4 degrees with an uncertainty of about 1.5 degrees. It is easy to test this prediction experimentally. To make the ratio zero, we just set m1 equal to zero; we take the mass off altogether.. But in that case, the string on both sides of the knot should be simply straight, and the angle q should be zero. The line predicts that the string will bend "the wrong way."
At the other extreme, the straight line predicts that m1can be greater than m2 . Experimentally, if this happens, the knot is pulled down until m2 is pulled over the pulley and clatters to the floor. These contradictions make us certain that a straight line fit will not give us insight into the relation between the dependent and independent variables. For insight, we turn to Newton.
Finally, we seek to improve our understanding using the physics learned in class. We require that the vertical component of the force by the string on m1 plus the force of gravity on m1 must add to zero for equilibrium. For an ideal pulley, the tension in the string to the left of the know is the same as the tension m2 so we must have
(m2 ) (sin q) = m1 so that
(m1)/(m2) = (sin q)
or
q = arcsin[(m1)/(m2)]
When the straight line fit is replaced by a line representing q calculated as above, the line represents theory based on Newtonian mechanics. The graph is shown below.
The curve fits the data well, and has the additional advantage of going through the origin, as expected. It also suggests that there are no solutions for m1 > m2 , as we expect. We should emphasize that in this second graph, there are no adjustable parameters; no "fit." The first graph had two adjustable parameters, slope and intercept in the "fit," neither of which turn out to have simple physical significance.
The upshot: Use "fits" to help build insight, but don't assume that the guess for the form of the fit has any significance, until you can show how it arises from the basic physics. | 1,934 | 8,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-48 | longest | en | 0.946595 |
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# hw6 Method of determinants - Method of determinants for...
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1 PH142_Prob_Set_Posting_01 © Carl E. Patton, Department of Physics, Colorado State University, Meth. of Determinants, 24 February 2008 Page 1 Method of determinants for linear inhomogeneous equation solutions. Consider the solution of 3 linear inhomogeneous equations in 3 unknowns Ax By Cz D E xFy Gz H J xKy Lz M ++= The are known parameters. The are the unknowns. ,,,,,,, ,, ,, ,, ABCDEFGHJKLM xyz One can solve such a set of equations by substitution, but the algebra is messy. The better way is by the method of determinants. © Carl E. Patton, Department of Physics, Colorado State University, Meth. of Determinants, 24 February 2008 Page 2 The bigger topic is called "linear algebra." This is very important. A good course on linear algebra is highly recommended. [Take one that is practical and not all abstract.] © Carl E. Patton, Department of Physics, Colorado State University, Meth. of Determinants, 24 February 2008 Page 3 0 For there to be a nontrivial (e.g., non-zero) solution for , it is required that the determinant of coefficients, namely
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# Detail Of Nip Angle For Jaw Crusher
### Jaw Crusher Nip Angle
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### Calculating The Nip Angle Of The Chamber Of Jaw And Cone
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### Nip Angle Of Jaw Crusher Mine Equipments
Jaw Crusher Information The nip angle describes the angle the stationary jaw plate and the pitman make with each other. The exact value of this ...
### Angle Of Bite And Angle Of Nip In Rolling
the nip angle when the crusher feed size is 10 cm, 4 the coefficient of friction between roll and gypsum particles 65 At a nip angle of 30, an approximate relation between peripheral speed P (m/min), diameter of rolls D (cm) and sieve size d (cm) through which 80% of ore passes is given by the relation
### Nip Angle Of Roller Mills
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### How To Improve The Efficiency Of Marble Jaw Crusherfote
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### What Are The 8 Factors That Affect Crusher Machine
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### Pdf Optimum Design And Analysis Of The Swinging Jaw
A jaw crusher is a kind of size reduction machine which is widely used in mineral, aggregates and metallurgy fields. ... variation of nip angle with crank angle 24 . ... detail and analyze th e ...
### Angle Of Nip Gyratory Mining Mill
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### Jaques Terex
Hydraulic wedge jaw adjustment and tension rod on D42 D60 Externally mounted hydraulics Electric/hydraulic adjust unit Heavy-duty frame Single-piece fabricated and stress relieved construction Machined jaw liner support cheek plates Curved fixed and moving jawface liners Provides optimum nip angle in fine crushing
### 2 Gyratory Crushers
The angle of nip in gyratory crushers has definite limitations. In the older straight element crushing chamber, the angle ranges from 20 to 24. On large primary crushers, using curved or nonchoking concaves and where gravity is of marked aid to nip with the large pieces at
### Metso Jaw Crushers Frames For Sale
metso c jaw crusher frames for sale C160 jaw crusher Metso Outotec. The tight nip angle in the middle of the cavity ensures that the processed material is quickly passed to the bottom of the cavity where the crushers long stroke finishes the job Like all C Series jaw crushers the C160 jaw crusher is built out of premium components The main frame structure is based on ...
### Angle Of Nip Article About Angle Of Nip By The Free
The largest angle that will just grip a lump between the jaws, rolls, or mantle and ring of a crusher. Also known as angle of bite nip.
### Sandvik Cj411 Singletoggle Jaw Crusher For High
CJ412 Jaw crusher Sandvik CJ412 single-toggle jaw crusher is engineered for the toughest applications thanks to its heavy-duty design. Characterized by an attention to detail in its design and manufacture, this machine is an excellent choice when you need high production and low total cost.
### Jaw Crusher Oil Type Manufacturer Jaw Crusher
Jaw Crusher Oil Type Manufacturer. R-techno is the leading Jaw crusher oil type manufacturer in Dhansura, Gujarat, India. At R-techno we use higher quality raw material for crushing machine manufacturing which makes R-techno the topmost jaw crusher manufacturer in India. We use advanced technologies in different types of jaw crusher manufacturers.
### Metso Silica Stone Jaw Crusher Ale Plant Vsi Vertical
Metso second crushing plant ale 2016-01-29. sand suppliers saudi arabia marafiq approved 2016-08-20. roller material specifiion for crusher 2017-03-12. zenith high quality pe series jaw crusher 2017-10-03. impact crusher capacity of 150 tph 2018-04-25. crushbits crush 2018-11-15. angle of nip for jaw and roll crusher 2019-06-08Metso Bauxite is main material in aluminium making.
### Angle Of Nip In Roll Crusher Or Jaw Crusher
Nip angle double roller crusher Angle of nip for jaw and roll crusher angle of nip for jaw and roll crusher double roll crusher machine design free download as pdf file the nip angle is between 20and 30but in some large roll crushers it is up to 40the nip angle for jaw cone and roll Details Double rool crusher design iron ore.
### Angle Of Nip In Roll Crusher Definition
angle of nip for jaw and roll crusher - Coal Surface ... angle of nip in roll crusher or jaw crusher 28 Aug 2013 More details ... angle of nip definition of angle of
### Engineered For The Toughest Feed Materials Nordberg C
The correct nip angle between the movable and fixed jaw dies ensures good bite and material flow down, even with slippery feed material. It also reduces wear on the jaw dies, reducing operating costs. With a good grip, the jaw crusher can crush rocks efficiently through the entirety of the cavity, and the nip angle can be further improved with
### Liberty Jaw Crusher Primary Compression Crusher
One is fixed while the other constantly moves back and forth. This action compresses the rock causing it to break. Rocks stay in the jaws until they are small enough to pass through the bottom gap. Superior builds more than a dozen models of the Liberty Jaw Crusher Feed Size Up to 47 (1,194mm) Product Size 15 5 (380mm 125mm)
### Sandvik Cj615 Singletoggle Jaw Crusher With Heavy Duty
Sandvik CJ615 single-toggle jaw crusher is engineered for even the toughest applications thanks to its heavy-duty design. Characterized by an attention to detail in its design and manufacture, this machine is an excellent choice when you need high production and low total cost.
### Jaw Crusher Machine For Sale Prominer Shanghai Mining
This jaw crusher is a perfect combination of modern science technology and the production practice, which can better satisfy the automatic production demands of vast customers. It can reach the crushing ratio of 4-6, and the shape of final product is even. It is widely applied to crush high hardness rocks, such as ore, slag, construction ...
### Sandvik Cj815 Singletoggle Jaw Crushers For High
Sandvik CJ815 single-toggle jaw crusher is engineered for even the toughest applications thanks to its heavy-duty design. Characterized by an attention to detail in its design and manufacture, this machine is an excellent choice when you need high production and low total cost.
### Solutions To Improve The Production Capacity Of Jaw Crusher
Dec 03, 2020 In this article, we mainly introduce some solutions to improve jaw crusher production capacity. 1. Adopt Proper Nip Angle. The nip angle means the included angle between movable jaw plate and fixed jaw plate. According to calculation, the max nip angle can reach to 32, but in the actual production, the nip angle is generally between 18-20 ...
### Solutions To Improve The Production Capacity Of Jaw Crusher
Dec 03, 2020 Adopt Proper Nip Angle. The nip angle means the included angle between movable jaw plate and fixed jaw plate. According to calculation, the max nip angle can reach to 32, but in the actual production, the nip angle is generally between 18-20, smaller than 25. If the nip angle is too large, the raw materials in crushing cavity will be ...
### Handbook Of Crushing Terrasource Global
Angle of Nip The angle formed between the moving surface of a crusher roll or jaw plate and the stationary plate surface at which point the material will be pinched. Angle varies with machine size and material lump size. Bond Work Index (BWI) Kilowatt hour (kWh) per short ton required to reduce the material to 80% passing 100 microns.
### Jaw Crushers Double Toggle Jaw Crushers Manufacturer
Excellent bite in the cavity The steep nip angle, the angle between the movable and fixed jaw dies, ensures good bite and good material flow down in the cavity. On top of great performance, a good nip angle also reduces wear on the jaw dies and that has a direct impact on operating costs.
### Free Essays On Jaws Camera Angles
Fote Jaw Crushing Plant Pays Attention to Details. Jaw crushing plant is with motor as power, through the motor pulley, eccentric shaft driven by a triangle and the sheave, the moving jaw according to the desired trajectory for reciprocating motion, ... crusher model c100
### Jaw Crusher Rj815 Ximonia Sweden Ximonia Is An
Ximonia RJ815 single-clutch jaw crusher is designed for even the toughest applications thanks to its powerful design. Characterized by attention to detail in its design and manufacturing, this machine is an excellent choice when you need high production and low total cost. Ximonia RJ815 benefits from the low jaw plate wear, which reduces the need for maintenance.
### Lokotrack Mobile Crushing Screening Plants
The Nordberg C130 jaw crusher on the Lokotrack LT130E takes your capacity to a new level in large-scale quarry applications. A deeper 1 000 mm (40) feed opening is able to handle coarser feed material and it greatly reduces the need for blasting. A small nip angle and excellent kinematics ensure aggressive crushing
### Horizontal Shaft Impact Crusher Manufacturer Propel
Propel AVJ series single toggle Jaw Crusher offers maximum possible fatigue strength and best reliability with various mounting possibilities. The fully-welded sturdy construction results in lowering the production cost per ton. Large feed acceptance capability. Interchangeable and reversible jaw plates. Easy and safe adjustment mechanism. | 2,684 | 12,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.831303 |
https://www.onlinemath4all.com/using-ratio-reasoning-to-solve-problems.html | 1,542,620,852,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745522.86/warc/CC-MAIN-20181119084944-20181119110359-00060.warc.gz | 965,395,074 | 12,905 | # USING RATIO REASONING TO SOLVE PROBLEMS
## About "Using ratio reasoning to solve problems"
Using ratio reasoning to solve problems :
Ratio reasoning is nothing but a proportion which states that two ratios or rates are equivalent.
Examples :
1/3 and 2/6 are equivalent ratios
1/3 = 2/6 is a proportion
Here, we are going to see how ratio reasoning (proportion) can be used to solve problems.
## Using ratio reasoning to solve problems - Examples
Example 1 :
Sheldon and Leonard are partners in a business. Sheldon makes \$2 in profits for every \$5 that Leonard makes. If together they make a profit of \$28 on the first item they sell, how much profit does Sheldon make ?
Solution :
Step 1 :
Given : Sheldon’s profit is \$2 and Leonard's profit is \$5
Total profit = 2 + 5 = \$7
The ratio of Sheldon’s profit to the total profit is
\$2 / \$7
Let "S" be the Sheldon's profit for the given total profit \$28.
Then, we have
Step 2 :
Let us use ratio reasoning to find "S".
Because \$7 × 4 = \$28, multiply \$2 by 4.
Therefore, if they make a total profit of \$28, then Sheldon makes a profit of \$8.
Example 2 :
The members of the PTA are ordering pizza for a meeting. They plan to order 2 cheese pizzas for every 3 pepperoni pizzas they order. How many cheese pizzas will they order if they order a total of 25 pizzas ?
Solution :
Step 1 :
Given : 2 cheese pizzas for every 3 pepperoni pizzas
Total profit = 2 + 3 = 5 pizzas
The ratio of cheese pizzas to total pizzas is
2 / 5
Let "C" be the number of cheese pizzas for the given total pizzas 25.
Then, we have
Step 2 :
Let us use ratio reasoning to find "C".
Because 5 × 5 = 25, multiply 2 by 5.
Therefore, if they order a total of 25 pizzas, then they will order 10 cheese pizzas.
After having gone through the stuff given above, we hope that the students would have understood "How to use ratio reasoning to solve problems".
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WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
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Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,030 | 4,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2018-47 | longest | en | 0.87003 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-6-systems-of-equations-and-inequalities-get-ready-page-357/10 | 1,695,708,492,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510149.21/warc/CC-MAIN-20230926043538-20230926073538-00584.warc.gz | 863,616,937 | 13,491 | Chapter 6 - Systems of Equations and Inequalities - Get Ready - Page 357: 10
$-2\leq f$
Work Step by Step
$-(7f+18)-2f\leq0\longrightarrow$ multiply by -1 using the distributive property $-7f-18-2f\leq0\longrightarrow$ combine like terms $-9f-18\leq0\longrightarrow$ add 9f to each side $-9f-18+9f\leq0+9f\longrightarrow$ add $-18\leq9f\longrightarrow$ divide each side by 9 $-18\div9\leq9f\div9\longrightarrow$ divide $-2\leq f$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 197 | 593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-40 | latest | en | 0.514641 |
https://www.coursehero.com/file/6244639/hw9-sol-s11/ | 1,513,250,263,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948543611.44/warc/CC-MAIN-20171214093947-20171214113947-00062.warc.gz | 734,358,646 | 53,071 | hw9_sol_s11
# hw9_sol_s11 - Phys 2101 Homework 9 Solution Spring `11...
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These so appear in however. 1. Three of the wa downwar rope and The horiz (a) We so obtain (b) We so obtain N F 2. The ob causing a problem accelerat string and Therefor lutions use t n your person . forces act on all N F r (actin rd). Since th the vertical. zontal comp olve the first mg T = olve the seco 22 Tr Lr = = + bject exerts a a “kink” sim in the text). tion is zero) d its “relaxe e, T = F /(2si Phys 210 the paramete nal homewo n the sphere ng horizontal he sphere is i . Then, the v onent is F N t equation fo ( L + = ond equation 2 mg L r L + a downward ilar to that s By analyzin 2 T sin θ = F , d” position ( in ) = 7.92 01 Home er values fro ork assignme : the tension lly away from n equilibrium vertical comp T sin = 0 or the tension 0.85 kg)(9.8 n for the nor 2 r = + force of ma shown for pr ng the forces , where is (when the tw tan = × 10 3 N. work 9 Sol m the proble ent. The logic n force r T of m the wall), m they sum ponent of Ne 0. n: T = mg /co 2 8 m/s ) (0.0 0.080 m rmal force: F (0.8 mgr L = agnitude F = roblem 10 (se s at the “kink the angle (ta wo segments 1 0.35m 1.72m ⎛⎞ = ⎜⎟ ⎝⎠ lution Spr ems printed i c used to get the rope (ac and the forc to zero. Let ewton’s seco os . We sub 2 080 m) (0. m + . sin N T = . 5 kg)(9.8 m/ (0.080 3160 N at th ee the figure k” where r F aken positive s are collinea 11.5 . ° ring `11 in the book, t to the answ cting along th ce of gravity be the ang ond law is T bstitute cos 2 042 m) 9 = . Usingsin 2 /s )(0.042 m 0 m) he midpoint that accom is exerted, w e) between e ar). In this pr not those th wer is the sam he rope), the y mg r (acting gle between cos mg = + LL / 2 .4 N =+ rL r / 2 ) 4.4 N. = of the rope, panies that we find (sinc each segmen roblem, we h hat me, e force g the = 0. r 2 to 2 , we , ce the nt of the have
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3. P. 12-0 diving bo the weigh forces eq (a) The s which sh (b) Since (c) The f which sh (d) The r (e) The f pedestal (f) The fo compress 4. (a) An (b) Equil (c) Comp 07. We take oard. We tak ht of the div qual to zero ( econd equat hould be roun e F 1 is negati first equation hould be roun result is posi force of the d on the divin orce of the d sed. nalyzing the librium of ve puting torque v F d the force of ke the force o er, located a (with upward tion gives 1 F = − nded off to F ive, indicatin n gives 2 F = nded off to F tive, indicat diving board ng board), so diving board horizontal fo ertical forces es about poi 23 d Fb Fa =+ f the left ped of the right p at x = L . The ds positive), 1 F Fd Ld W d = − 1 1.2 10 =− × ng that this f 1 58 WF −= 3 2 1.7 10 ing that this d on the left p this pedesta on the right orces (which s leads to F v nt O , we obt ( 10 d ⇒= estal to be F pedestal to b following tw and the sum 12 () FW WL d + = + = 3.0m (5 1.5m ⎛⎞ ⎜⎟ ⎝⎠ 3 0 N . Thus, force is down 0 N+1160 N N . Thus, | force is upw pedestal is u al is being st t pedestal is d h add to zero = F 1 + F 2 = tain )( ) 0 N3 .
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hw9_sol_s11 - Phys 2101 Homework 9 Solution Spring `11...
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Ask a homework question - tutors are online | 1,208 | 3,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-51 | latest | en | 0.810907 |
https://www.geocaching.com/geocache/GCJA7R_phobias-i-acrophobia?guid=19b86efd-8b7e-49e2-b276-e8aedfb9bf60 | 1,561,051,065,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999263.6/warc/CC-MAIN-20190620165805-20190620191805-00294.warc.gz | 770,663,655 | 30,242 | ##### This cache has been archived.
Thymallus Thymallus: Cache archived.
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## Phobias I - Acrophobia
A cache by Liimes Message this owner
Hidden : 04/27/2004
Difficulty:
Terrain:
Size: (regular)
#### Watch
How Geocaching Works
Please note Use of geocaching.com services is subject to the terms and conditions in our disclaimer.
### Geocache Description:
This is the first cache in the series of phobia caches by Liimes. Suffering from the fear of heights might make it impossible for you to even try this cache, but it should offer a challenge for anyone. The most important thing in this series is to conquer your fears.
This is a three-step multi-cache featuring high places in the Otaniemi region. To go after this cache you must be very confident with your climbing skills, as there is a definite risk of falling and even of fatal injury. The first step does not involve climbing but the in the last two steps you have to climb at least to a height of 10 meters from the ground.
You should not go for this cache during daylight. During the day there will be a lot of geomuggles around all of the steps. All of the sites should be easy locate even in the dark. If you have problems finding the correct locations, there is an additional hint for each of them.
The first target is not in the given coordinates but can be seen from there. The target is a metal pipe that is located in the highest spot visible to the initial coordinates. To make sure you don't see unrelated objects, think that you're looking from the ground level. You will have to determine the pipe's length and diameter (in centimeters). Doing this from the ground level is very demanding but not totally impossible.
The coordinates of the second step can be calculated using the dimensions from the first step using following formulas:
Let the caliber or the diameter of the inside of the pipe (rounded to the nearest centimeter) be A.
Let the length of the pipe (rounded down to the nearest centimeter) from the longest visible edge be B.
Both A and B can be factored into two primes. They don't have a common divisor and neither has 7 as a factor.
The latitude of the second step is N 60° 11.X minutes, where
X = (A+B)*4
The longitude of the second step is E 024° 49.Y minutes, where
Y = (B-A)*24
The second step is the easiest of the three and there you can find the coordinates to find the final cache. The hint is hidden on top of a construction.
The third and final step is the most demanding. The cache is small to a regular size glass jar and is hidden behind some sort of a ventilation duct. If you can use the rope here to secure you from falling, that will be most advisable. It might be diffucult to rest on the top, so signing the log will require some acrobatic skills.
A bonus for added challenge: When you find the actual container, you will notice that it is not as high as it could be. After you have signed the log, feel free to return the cache to the highest possible level. If you take it there, I'll take care of its maintenance.
Useful equipment:
• Binoculars
• A rope
• A pair of shoes suitable for climbing
• A pair of gloves
• A flashlight
• A map
• A calculator
[1st step]: Gur cynpr vf rnfvyl pyvzonoyr.
[2nd step]: Gur pbafgehpgvba vf ivfvoyr sebz gur arneol ebnq.
[3rd step]: Hc gur fybcr naq gura fgneg pyvzovat.
Decryption Key
A|B|C|D|E|F|G|H|I|J|K|L|M
-------------------------
N|O|P|Q|R|S|T|U|V|W|X|Y|Z
(letter above equals below, and vice versa)
Find...
### 118 Logged Visits
92 4 13 1 2 2 1 1 1 1
**Warning! Spoilers may be included in the descriptions or links.
Current Time:
Last Updated:
Rendered From:Unknown
Coordinates are in the WGS84 datum
# Reviewer notes
Use this space to describe your geocache location, container, and how it's hidden to your reviewer. If you've made changes, tell the reviewer what changes you made. The more they know, the easier it is for them to publish your geocache. This note will not be visible to the public when your geocache is published. | 1,011 | 4,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-26 | latest | en | 0.940559 |
https://howkgtolbs.com/convert/86.28-kg-to-lbs | 1,621,014,158,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00083.warc.gz | 326,053,349 | 12,108 | 86.28 kg to lbs - 86.28 kilograms to pounds
Before we go to the more practical part - it means 86.28 kg how much lbs conversion - we want to tell you some theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 86.28 kg to lbs? 86.28 kilograms it is equal 190.2148396536 pounds, so 86.28 kg is equal 190.2148396536 lbs.
86.28 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in short form SI).
From time to time the kilogram is written as kilogramme. The symbol of this unit is kg.
First definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but difficult to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was substituted by a new definition.
The new definition of the kilogram is based on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It can be also divided into 100 decagrams and 1000 grams.
86.28 kilogram to pounds
You learned something about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. It is needed to highlight that there are not only one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. Of course, this unit is used also in other systems. The symbol of this unit is lb or “.
The international avoirdupois pound has no descriptive definition. It is just equal 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
How many lbs is 86.28 kg?
86.28 kilogram is equal to 190.2148396536 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
86.28 kg in lbs
The most theoretical section is already behind us. In this part we want to tell you how much is 86.28 kg to lbs. Now you know that 86.28 kg = x lbs. So it is high time to know the answer. Just look:
86.28 kilogram = 190.2148396536 pounds.
It is an exact outcome of how much 86.28 kg to pound. It is possible to also round it off. After it your outcome is as following: 86.28 kg = 189.816 lbs.
You learned 86.28 kg is how many lbs, so look how many kg 86.28 lbs: 86.28 pound = 0.45359237 kilograms.
Obviously, this time it is possible to also round off the result. After rounding off your result will be as following: 86.28 lb = 0.45 kgs.
We are also going to show you 86.28 kg to how many pounds and 86.28 pound how many kg results in tables. Look:
We are going to start with a chart for how much is 86.28 kg equal to pound.
86.28 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
86.28 190.2148396536 189.8160
Now look at a chart for how many kilograms 86.28 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
86.28 0.45359237 0.45
Now you know how many 86.28 kg to lbs and how many kilograms 86.28 pound, so we can move on to the 86.28 kg to lbs formula.
86.28 kg to pounds
To convert 86.28 kg to us lbs you need a formula. We will show you two formulas. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 190.2148396536 result in pounds
The first version of a formula will give you the most correct outcome. In some cases even the smallest difference can be significant. So if you want to get an accurate result - first formula will be the best for you/option to convert how many pounds are equivalent to 86.28 kilogram.
So let’s move on to the another version of a formula, which also enables calculations to learn how much 86.28 kilogram in pounds.
The second version of a formula is down below, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you see, this formula is simpler. It can be better choice if you need to make a conversion of 86.28 kilogram to pounds in easy way, for example, during shopping. Just remember that final result will be not so exact.
Now we want to learn you how to use these two formulas in practice. But before we are going to make a conversion of 86.28 kg to lbs we are going to show you easier way to know 86.28 kg to how many lbs totally effortless.
86.28 kg to lbs converter
An easier way to learn what is 86.28 kilogram equal to in pounds is to use 86.28 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on first version of a formula which we showed you in the previous part of this article. Thanks to 86.28 kg pound calculator you can quickly convert 86.28 kg to lbs. You only need to enter amount of kilograms which you want to calculate and click ‘convert’ button. The result will be shown in a second.
So try to calculate 86.28 kg into lbs with use of 86.28 kg vs pound converter. We entered 86.28 as a number of kilograms. It is the outcome: 86.28 kilogram = 190.2148396536 pounds.
As you can see, our 86.28 kg vs lbs calculator is intuitive.
Now we are going to our main topic - how to convert 86.28 kilograms to pounds on your own.
86.28 kg to lbs conversion
We will begin 86.28 kilogram equals to how many pounds calculation with the first version of a formula to get the most accurate result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 190.2148396536 the result in pounds
So what have you do to know how many pounds equal to 86.28 kilogram? Just multiply number of kilograms, in this case 86.28, by 2.20462262. It is 190.2148396536. So 86.28 kilogram is equal 190.2148396536.
You can also round it off, for example, to two decimal places. It gives 2.20. So 86.28 kilogram = 189.8160 pounds.
It is time for an example from everyday life. Let’s convert 86.28 kg gold in pounds. So 86.28 kg equal to how many lbs? As in the previous example - multiply 86.28 by 2.20462262. It is 190.2148396536. So equivalent of 86.28 kilograms to pounds, if it comes to gold, is 190.2148396536.
In this example it is also possible to round off the result. This is the outcome after rounding off, this time to one decimal place - 86.28 kilogram 189.816 pounds.
Now we can move on to examples converted with short formula.
How many 86.28 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 189.816 the outcome in pounds
So 86.28 kg equal to how much lbs? As in the previous example you need to multiply number of kilogram, this time 86.28, by 2.2. Look: 86.28 * 2.2 = 189.816. So 86.28 kilogram is 2.2 pounds.
Let’s make another conversion using shorer formula. Now calculate something from everyday life, for instance, 86.28 kg to lbs weight of strawberries.
So let’s calculate - 86.28 kilogram of strawberries * 2.2 = 189.816 pounds of strawberries. So 86.28 kg to pound mass is 189.816.
If you learned how much is 86.28 kilogram weight in pounds and are able to calculate it using two different formulas, we can move on. Now we want to show you these outcomes in tables.
Convert 86.28 kilogram to pounds
We realize that outcomes shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in tables for your convenience. Thanks to this you can easily compare 86.28 kg equivalent to lbs results.
Start with a 86.28 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
86.28 190.2148396536 189.8160
And now have a look at 86.28 kg equal pound chart for the second formula:
Kilograms Pounds
86.28 189.816
As you see, after rounding off, when it comes to how much 86.28 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Keep it in mind when you need to make bigger number than 86.28 kilograms pounds conversion.
How many kilograms 86.28 pound
Now you learned how to convert 86.28 kilograms how much pounds but we will show you something more. Are you interested what it is? What about 86.28 kilogram to pounds and ounces conversion?
We are going to show you how you can calculate it little by little. Begin. How much is 86.28 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, in this case 86.28, by 2.20462262. So 86.28 * 2.20462262 = 190.2148396536. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To know how much 86.28 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your result is equal 2 pounds and 327396192 ounces. You can also round off ounces, for instance, to two places. Then final result will be equal 2 pounds and 33 ounces.
As you can see, calculation 86.28 kilogram in pounds and ounces quite easy.
The last calculation which we will show you is calculation of 86.28 foot pounds to kilograms meters. Both of them are units of work.
To convert it you need another formula. Before we show you this formula, have a look:
• 86.28 kilograms meters = 7.23301385 foot pounds,
• 86.28 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 86.28 foot pounds to kilograms meters you have to multiply 86.28 by 0.13825495. It is 0.13825495. So 86.28 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 86.28 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 86.28 kilogram into pounds calculations.
This article was a huge compendium about kilogram, pound and 86.28 kg to lbs in conversion. Due to this calculation you learned 86.28 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 86.28 kilogram to metric pounds but also two another calculations - to know how many 86.28 kg in pounds and ounces and how many 86.28 foot pounds to kilograms meters.
We showed you also other way to do 86.28 kilogram how many pounds conversions, it is with use of 86.28 kg en pound converter. It is the best choice for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you can do 86.28 kilogram equal to how many pounds calculation - on your own or using our 86.28 kgs to pounds converter.
It is time to make your move! Let’s calculate 86.28 kilogram mass to pounds in the best way for you.
Do you need to do other than 86.28 kilogram as pounds conversion? For instance, for 10 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so simply as for 86.28 kilogram equal many pounds.
How much is 86.28 kg in pounds
To quickly sum up this topic, that is how much is 86.28 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see the most important information about how much is 86.28 kg equal to lbs and how to convert 86.28 kg to lbs . It is down below.
How does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. Let’s see 86.28 kg to pound conversion formula . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
Now you can see the result of the conversion of 86.28 kilogram to pounds. The accurate answer is 190.2148396536 lb.
It is also possible to calculate how much 86.28 kilogram is equal to pounds with another, shortened type of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So in this case, 86.28 kg equal to how much lbs ? The answer is 190.2148396536 lb.
How to convert 86.28 kg to lbs in an easier way? It is possible to use the 86.28 kg to lbs converter , which will make the rest for you and give you a correct answer .
Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,507 | 13,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-21 | latest | en | 0.95018 |
www.standsfor.org | 1,674,987,806,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00002.warc.gz | 1,025,751,821 | 27,558 | ## What Does COGS Stand For?
CGS Stands for Cost of Goods Sold.
## What is (the COGS) Cost of Goods Sold?
COGS acts as an Acronym for the Cost of Goods Sold. The Cost of Goods sold is the value of the Goods that have been sold out. This cost is calculated by the following formula.
## COGS Formula:
COGS = Opening Inv + Purchases - Closing Inventory
### How find the Cost of goods sold?
we can calculate the value of COGS by solving the following Question
Example:
If you have sales of 50,000 and the opening inventory is 15000. The value of the Purchase of goods is 90000 and the closing inventory is 2000 what is the value of the Cost of Goods sold.
COGS Calculation:
Opening Inventory 15000
Purchases 9000
Cost of Goods available for sale 24,000
Less: Closing Inventory (2000)
Cost of Goods Sold 22,000 | 221 | 929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-06 | latest | en | 0.816803 |
https://www.physicsforums.com/threads/predicting-model-rocket-height-with-and-without-drag.416323/ | 1,544,618,432,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823872.13/warc/CC-MAIN-20181212112626-20181212134126-00599.warc.gz | 1,027,534,091 | 17,294 | # Predicting model rocket height (with and without drag)
1. Jul 15, 2010
### brainpushups
Hello all,
I recently made some model rockets and thought it would be fun to predict their height before setting them off this weekend. I thought it would also be interesting to compare the difference in the predictions of a model with no drag and a model with quadratic drag. Surprisingly, the model without drag predicted a lower height. I probably made some error somewhere, but I don't see it and I was wondering if anyone could offer some insight. I'll provide all the information and hope there isn't some simple answer that I've overlooked. Thanks in advance for anybody even willing to look at this rather lengthy post.
The simplifying assumptions I've made for both models during the boost phase are:
1) Use the average thrust of the engine
2) Use the average mass of the rocket+engine+fuel
The thrust of the engine is 4.2N
The average mass of the rocket is 0.08kg
The burn time of the engine is 1.1s
No Drag
This is a simple exercise in kinematics. The equation of motion is T-mg=ma where T is the average thrust and m is the average mass. Solving the differential equation predicts a velocity of 46m/s and an altitude of 26m when the engine has used up its fuel.
The equation of motion during the coasting phase is ma=-mg. Solving this for the total height gives 136m.
My drag coefficient (1/2 rho * Cd *Area) is equal to 0.00035 kg/m
The equation of motion is ma=T-mg-cv^2. To simplify the integration I solved for the terminal velocity using this equation which is about 97m/s. The solution to the integral solving for the velocity of the rocket at burnout is
Subscript[v, b]=Subscript[v, ter]((1-E^-k\[Tau])/(1+E^-k\[Tau])) where tau is the burn time of the engine and k is a constant that cleaned up the mess of the variables a bit. The height when the engine burns out was solved similarly and came out as
Subscript[y, b]=
\!$$\*OverscriptBox["m", "_"]$$/(2c) ln (\!
\*SubsuperscriptBox[$$v$$, $$ter$$, $$2$$]/(\!$$\*SubsuperscriptBox[\(v$$, $$ter$$, $$2$$] -
\*SubsuperscriptBox[$$v$$, $$b$$, $$2$$]\)))
(hope this is readable, I'm cutting and pasting from another document).
Anyway the predictions are close to that of the burnout phase for no air resistance. This model gives 43m/s and 25m for the velocity and height.
For the coast phase the equation of motion is ma=-mg-cv^2. Once again I solved for the terminal velocity to simplify the equations (different than the vter above)
Going through the process to find the height gives the expression
Subscript[m, c]/(2 c) ln(-\!
\*SubsuperscriptBox[$$v$$, $$ter$$, $$2$$]/(\!$$\*SubsuperscriptBox[\(v$$, $$b$$, $$2$$] +
\*SubsuperscriptBox[$$v$$, $$ter$$, $$2$$]\)))
where m is now the mass of the rocket without fuel, c is my drag coefficient, vter is now equal to about 46m/s and vb is the velocity at burnout. Anyway, the total height predicted is now close to 270m.
2. Jul 15, 2010
### brainpushups
I see that my cut and pasting didn't work.
during engine burn:
vb=vter((1-Exp[-kt])/(1+Exp[-kt]) - velocity
h = (m/(2c)) Ln(vter^2/(vter^2-vb^2))
during coast
h = m/(2c) Ln(-vter^2/(vb^2-vter^2))
3. Jul 15, 2010
### hikaru1221
I see something strange in your calculations of the 2nd case where there is drag. The terminal speed in the 2nd phase is supposed to be zero, since drag and gravity should cease the upward motion. I got 69m for the additional height contributed by the 2nd phase.
4. Jul 16, 2010
### brainpushups
hikaru,
thanks for the reply. Perhaps my labeling of terms was misleading. What I'm calling terminal velocity for the coasting phase comes from solving the equation ma=-mg-cv^2 for when the acceleration is zero so that vter = Sqrt(mg/c). This would be the terminal velocity for an object in free fall with quadratic drag. The only reason for this was to simplify the number of symbols for integration.
5. Jul 16, 2010
### hikaru1221
How do you solve that equation, when the left side is equal to 0, and the right side is always negative? Why is there free fall here, when we are to find the height, which means we only care about the upward motion? Besides, without caring/knowing about the exact equation, in a physical sense, drag and gravity always prevent the upward motion, right?
6. Jul 16, 2010
### brainpushups
I'll show you more explicitly.
During the boost stage the equation of motion is m(dv/dt)=T-mg-cv^2. To solve this differential using separation of variables you would get
(m dv)/(T-mg-cv^2) = dt
This can be cleaned up a bit by plugging in what I've termed the terminal velocity for the boost stage (set the acceleration in the equation of motion equal to zero and solve for v). During the boost stage this is vterm=Sqrt((T-mg)/c). You can see that a substitution can be made to make the equation look like this
(m/c)(dv/(vterm^2-v^2) = dt
Integrating the left hand side from v=0 to v=vboost and the right hand side from 0 to t where t is the burn time of the engine returns.
(m/(2c vterm) Ln ((vterm+vboost)/(vterm-vboost)) = t
Solve this for vboost to get the equation for the final velocity after the engine burns out
vboost = vterm ((1-Exp[-kt])/(1+Exp[-kt])) where k = (2 c vter)/m.
For the coasting phase the equation of motion is different because there is no thrust. This is m(dv/dt) = -mg-cv^2. Once again this can be solved by separation of variables. But first I make a substitution that dv/dt (dh/dh) = dv/dh (dh/dt) = (dv/dh) v to get
(m v dv)/(-mg-cv^2) = dt
But the same simplification can be applied if we set acceleration equal to zero in the equation of motion for the coast phase to see that the terminal velocity is
vtermcoast = Sqrt(mg/c)
Now the separated equation can be written as
(m/c) (v/(-vtermcoast^2-v^2) = dh
So you see that I really haven't changed anything about the equations. Solving for the terminal velocity in the equations of motion just gives me something to call a bunch of constants to simplify the look of the equations. After integration, the equation for the height during the coast phase (by the way, I used the same method to find the height during the boost phase) is
h = m/(2c) Ln(-vter^2/(vb^2-vter^2))
I hope that clarifies my approach.
7. Jul 16, 2010
### brainpushups
hikaru,
I've found my error. I went through the integration of the coasting height without substituting for any new terms. After performing this integration I got the result
coastheight = (-m/2c) Ln[(mg/c)/(mg/c - burnvelocity^2)
And guess what.... this model predicts a height of about 69m!
8. Jul 16, 2010
### hikaru1221
I'm not sure if you're on the right track or not. Here is the problem:
We have: $$mdv/dt = - mg - cv^2.$$
If dv/dt = 0, then we have: $$mg = - cv^2$$ There is a minus sign! In the 2nd phase, the acceleration cannot be equal to zero.
I think this error has nothing to do with substituting the values. Therefore your result, if it is right, may be just a coincidence. Please kindly excuse me if I misunderstood you
9. Aug 26, 2011
### lshcue
I think your assumption is a creative one and physically meaningful.
During the integration you should consider that the mass is not a constant but a function of time. The mass can be expressed as
m = m0 + rt (r< 0).
where m0 is the initial mass and r is the flow rate of the propellant. This leads to
dm = r dt
Hence the equation "(m/c)(dv/(vterm^2-v^2) = dt" can be rewritten as follows:
(1/c)(dv/(vterm^2-v^2) = dt/m = (1/r)dm/m
If the terms c and r are constants, we can get a solution. ^.-
10. Aug 26, 2011
### K^2
The zero thrust equation can be solved analytically.
$$\frac{dv}{dt} = -g-kv$$
Yields. v(t)=-vttan(gt/vt)
$$v(t) = -v_t tan\left(\frac{g t}{v_t}\right)$$
Where vt=sqrt(g/k) is terminal velocity. Notice that this equation gives v(0)=0, so you need to solve for t0, such that v(t0=v0. That gives you t0=(vt/g)tan-1(v0/vt)
The height can be obtained by simple integration.
$$h=\frac{v_t^2}{g}ln\left(cos\left(\frac{gt_0}{v_t}\right)\right)$$
Boost stage can be handled similarly. The differential equation above is going to become:
$$\frac{dv}{dt} = a-kv$$
Here, a=F/m-g is the acceleration of rocket in vacuum, with F being average thrust and m being average mass. This is also easy to solve.
$$v(t) = v_t tanh\left(\frac{g t}{v_t}\right)$$
Notice that it's hyperbolic tangent, and the vt for this computation is sqrt(a/k). This time, v(0)=0 is your starting condition, so vf=v(tf) where t0 is time of engine operation, will give you v0 for coasting stage. Distance is again obtained by integration.
$$h=\frac{v_t^2}{g}ln\left(cosh\left(\frac{gt_f}{v_t}\right)\right)$$
Again, note the hyperbolic cosine function here.
Try combining these two computations and comparing them to your results. You should be able to get somewhat better estimate. I'm curious to see how close it is to your numbers. (You should have all the coefficients. What I'm calling k is your c/m. So just substitute in, and see what you get.)
Last edited: Aug 26, 2011 | 2,526 | 9,032 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-51 | latest | en | 0.939692 |
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Try it and write back if you need more help,
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# Excellent GMAT Score, but ridiculously lopsided. What to do?
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Excellent GMAT Score, but ridiculously lopsided. What to do? [#permalink]
### Show Tags
09 Jul 2011, 10:47
So I took the GMAT yesterday, and scored better than I ever had on any practice tests (44Q and 51V, so a 760). Mostly I'm thrilled, but I'm a little nervous about that Quant score for the programs I'm targeting (top 20, a couple top 10s). I've heard about the apocryphal 80/80 percentile cut-off for top programs, so will a 66th %ile Quant kill me if I'm 99th %ile Verbal?
By way of background, I'm from a Humanities undergrad (graduated 6 years ago). Classic "poet" profile. Since then I've demonstrated a bit of Quant chops, I got an A in a graduate-credit level stats class I took via Continuing Education at a very fancy local university last year, and was asked back to serve as a grader the semester after. So I have something to show on the quant front, but it's still pretty thin. Not sure if it can offset that Quant score in the eyes of a top 10 adcom.
The Math 44 frankly wasn't that abnormally low for me. I had a couple practice tests up in the 88th and 77th percentiles, but I also had some where I bombed out and ended up below the 50th. Needless to say, math is not my strength. The verbal, though, has never been a problem. I've consistently been scoring in the 99th percentile on all CATs since the diagnostic, so I have no doubt that I could boost my current overall score by retaking if I were to do more math studying.
I just wonder if I'm crazy to even consider retaking the test, and investing yet more time into the GMAT, rather than spending that time on the rest of my app. Opportunity costs, trade-offs etc etc.
Any advice? Not sure I have it in me to retake the GMAT....
Thx
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Re: Excellent GMAT Score, but ridiculously lopsided. What to do? [#permalink]
### Show Tags
11 Jul 2011, 09:58
Hi JoelCairo,
With a GMAT score of 760, I would not recommend taking the test again. There are other ways to express your quantitative abilities. Here are two articles that I recommend you read:
http://www.stacyblackman.com/2011/03/23 ... applicant/
http://www.stacyblackman.com/2011/04/20 ... m-scratch/
I hope this was of help.
Cheers,
Conrad and the Stacy Blackman Team
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Re: Excellent GMAT Score, but ridiculously lopsided. What to do? [#permalink]
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12 Jul 2011, 21:24
Thanks so much for your reply Stacy. I'm thinking the best move might be to preemptively enroll in a quant-intensive Continuing Ed class this fall, so I can include it on my app, rather than being waitlisted and told to do so a few months from now. Would you recommend Calculus (which seems to be used as shorthand for "quant class" on the message-boards) or would some other mathy course work just as well ("Managerial Accounting" "Principles of Finance" etc)?
Thanks again.
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Re: Excellent GMAT Score, but ridiculously lopsided. What to do? [#permalink] 12 Jul 2011, 21:24
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# Excellent GMAT Score, but ridiculously lopsided. What to do?
new topic post reply Update application status
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Find the equivalent ratios of 6 : 15
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• Apr 14th 2013, 01:48 PM
tracieemra72
I need help with Modeling
This is a word problem and I have no clue how to solve it.
Mt. Thor in Canada is said to have the longest vertical drop(4100 ft.) on Earth. How long would it take for a pebble dropped off the cliff to reach the ground? (Hint: The height of a falling object above the ground t seconds after it is dropped can be modeled by s(t) = -16t^2 + So where So is the initial height.)
• Apr 14th 2013, 01:49 PM
tracieemra72
The S is height and t is seconds.
• Apr 14th 2013, 04:20 PM
topsquark
Re: I need help with Modeling
Quote:
Originally Posted by tracieemra72
Mt. Thor in Canada is said to have the longest vertical drop(4100 ft.) on Earth. How long would it take for a pebble dropped off the cliff to reach the ground? (Hint: The height of a falling object above the ground t seconds after it is dropped can be modeled by s(t) = -16t^2 + So where So is the initial height.)
S_0 = 4100 ft, s(t) = 0
$\displaystyle s(t) = -16t^2 + s_0$
$\displaystyle s(t) - s_0 = -16t^2$
Can you finish from here?
-Dan
• Apr 14th 2013, 05:19 PM
tracieemra72
Re: Modeling
thanks Dan. I have no clue where to go from here. When I entered it in my calculator I came up with S(16)=4 S= 16 secs. with 4 being the So? I have to find out how many seconds it would take for the pebble to be dropped from 4100 ft to hit the ground. I have never done a problem like this nor do I know how to figure it out.
• Apr 14th 2013, 06:03 PM
topsquark
Re: I need help with Modeling
Quote:
Originally Posted by topsquark
S_0 = 4100 ft, s(t) = 0
$\displaystyle s(t) = -16t^2 + s_0$
$\displaystyle s(t) - s_0 = -16t^2$
Plug the numbers in:
$\displaystyle 0 - 4100 = -16t^2$
$\displaystyle \frac{4100}{16} = t^2$
Surely you can finish it now?
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Python Question
# Why does 1+++2 = 3?
How does Python evaluate the expression 1+++2?
How many ever '+' I put in between, it is printing 3 as the answer. Please can anyone explain this behavior
And for 1--2 it is printing 3 and for 1---2 it is printing -1
Answer Source
Your expression is the same as:
``````1+(+(+2))
``````
Any numeric expression can be preceded by `-` to make it negative, or `+` to do nothing (the option is present for symmetry). With negative signs:
``````1-(-(2)) = 1-(-2)
= 1+2
= 3
``````
and
``````1-(-(-2)) = 1-(2)
= -1
``````
I see you clarified your question to say that you come from a C background. In Python, there are no increment operators like `++` and `--` in C, which was probably the source of your confusion. To increment or decrement a variable `i` or `j` in Python use this style:
``````i += 1
j -= 1
``````
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# Test your ability to convert!
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### Test your ability to convert!
1. 1. Conversion Practice You work them out, then check your answer on the next slide!
2. 2. Q. <ul><li>Fill in the chart below </li></ul>BASE
3. 3. A. CHART MUST BE MEMORIZED!!! BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
4. 4. Q. <ul><li>121 cm = __________ m </li></ul>
5. 5. A. <ul><li>121 cm = 1.21 m </li></ul><ul><li>Cm to m would be 2 jumps to the left </li></ul><ul><li>Move the decimal 2 jumps to the left </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
6. 6. Q. <ul><li>884 m = __________dm </li></ul>
7. 7. A. <ul><li>884 m = 8840 dm </li></ul><ul><li>Meters to decimeters would be 1 jump to the right </li></ul><ul><li>Move the decimal one jump to the right </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
8. 8. Q. <ul><li>621 g = _______dag </li></ul>
9. 9. A. <ul><li>621 g = 62.1 dag </li></ul><ul><li>Grams to dekagrams would be one jump to the left. </li></ul><ul><li>Move the decimal one jump to the left </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
10. 10. Q. <ul><li>.50 kg = ______g </li></ul>
11. 11. A. <ul><li>.50 kg = 500 g </li></ul><ul><li>Kilograms to grams is 3 jumps to the right </li></ul><ul><li>Move the decimal 3 jumps to the right </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
12. 12. Q. <ul><li>3742 mm = _________ hm </li></ul>
13. 13. A. <ul><li>3742 mm = .03742 hm </li></ul><ul><li>Millimeters to hectometers is 5 jumps to the left </li></ul><ul><li>Move the decimal 5 jumps to the left </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
14. 14. Q. <ul><li>9.6 g = __________ kg </li></ul>
15. 15. A. <ul><li>9.6 g = .0096 kg </li></ul><ul><li>Grams to kilograms is 3 jumps to the left, </li></ul><ul><li>Move the decimal 3 jumps to the left </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
16. 16. Q. <ul><li>.17 m = _______cm </li></ul>
17. 17. A. .17m = 17 cm Meters to centimeters is two jumps to the right Move the decimal 2 jumps to the right BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
18. 18. Q. <ul><li>.8623 kl = _________ L </li></ul>
19. 19. A. <ul><li>.8723 kl = 862.3 L </li></ul><ul><li>Kiloliters to liters is 3 jumps to the right </li></ul><ul><li>Move the decimal 3 times to the right </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
20. 20. Q. <ul><li>8.7 L = ________hl </li></ul>
21. 21. A. <ul><li>8.7L = .087hL </li></ul><ul><li>Liters to hectoliters is 2 jumps to the left </li></ul><ul><li>Move the decimal 2 jumps to the left </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
22. 22. Q. <ul><li>500 mm = _______dm </li></ul>
23. 23. A. <ul><li>500 mm = 5 dm </li></ul><ul><li>Millimeters to decimeters is 2 jumps to the left </li></ul><ul><li>Move the decimal two jumps to the left </li></ul>BASE Kilo k Hecto h Deka da Deci d Centi c Milli m m. L, g
24. 24. Q. <ul><li>432 ml = _____cm 3 </li></ul>
25. 25. A. <ul><li>432 ml = 432 cm 3 </li></ul><ul><li>No, this is not a mistake: Your notes say that a ml=cc=cm 3 </li></ul><ul><li>This means that 10 ml=10cc=10cm 3 </li></ul><ul><li>It doesn’t matter what the number is, it would be the same for all 3. In this case 432ml=432cc=432cm 3 </li></ul>
26. 26. Q. <ul><li>.07L = _______cc </li></ul>
27. 27. A. <ul><li>.07 L = 70 cc </li></ul><ul><li>Some of you wanted to say it was the same thing (.07), but a L is not a ml! </li></ul><ul><li>ml=cc </li></ul><ul><li>1 st you would have to convert the L to a ml (move the decimal 3 jumps to the right) </li></ul><ul><li>You now have 70ml </li></ul><ul><li>Since a ml=cc, 70ml= 70cc </li></ul> | 1,588 | 4,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-30 | latest | en | 0.327216 |
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Pandigital powers (Posted on 2021-04-07)
What positive integer n has the property that n2 and n3 together contain every digit from 0 to 9 exactly once?
No Solution Yet Submitted by Math Man Rating: 5.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Analytic attempt...? | Comment 2 of 5 |
This problem begs to be solved by brute force, but I was just wondering what it would look like if we tried to solve it analytically? I gave it a start, not sure how much further I could go with it.
It's obvious to start that we need to find n such that n^2 contains 4 digits and n^3 contains 6 digits. The cube root of 100,000 is ~46.4 so n must be greater than 46 and less than 100.
The sum of all digits from 0 through 9 is 0 mod 3, so n cannot be 1 mod 3, since n^2 and n^3 would both be 1 mod 3 and then the sum of their digits would be 2 mod 3.
n cannot end in a 0, 1, 5 or 6 since n^2 and n^3 would each end in the same digit.
We've already ruled out a pretty large chunk of numbers. From there I started looking at the leftmost digits of the square roots of thousands (1000, 2000, etc) and the cube roots of hundred-thousands. We can rule out 89 (they'd both start with 7), 93-94 (they'd both start with 8) and 97-99 (they'd all start with 9).
Similarly we can rule out n = 68 since n^2 would start and end with a 4, and n = 74 since n^3 would start and end with a 4.
We can rule out 48, 57, and 92 because n^3 would end in the same digit that n^2 starts with. We can rule out 78 because n^2 would end in the same digit that n^3 starts with.
Now the list of potential candidates has been whittled down to the following 12:
47
53
54
59
62
63
69
72
77
83
84
87
At this point I don't know if there's a good way to continue pruning the list, though at least it's now at a length you could feasibly brute-force by hand as opposed to using software.
Posted by tomarken on 2021-04-07 08:05:55
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• To: mathgroup at smc.vnet.net
• Subject: [mg86633] Re: Substitute all known relations in result
• From: brtubb at pdmusic.org
• Date: Sat, 15 Mar 2008 17:43:32 -0500 (EST)
• References: <frg0ap\$eu5\$1@smc.vnet.net>
```I'm using version 6.0.2.0, so the enourmous capabilities and
conveniences of the "Manipulate" function is available to me, and
hopefully you. But if not, here are two versions of viewing the
results:
(1)
Plot3D[x^2+y^2, {x,-100,100},{y,-100,100}]
(2)
Manipulate[Plot3D[a x^2+b y^2, {x,-100,100},{y,-100,100}], {{a,0,"a =
-100 to 100"},-100,100}, {{b,0,"b = -100 to 100"},-100,100}]
Benjamin Tubb
On Mar 15, 3:11=A0am, "Dr. Johannes Zellner"
<johannes.zell... at gmail.com> wrote:
> Hi,
>
> How do I make mathematica to display results with all known
> relations / variables?
> E.g. if a result is
> =A0 =A0x^2+y^2
> and before I'd defined
> =A0 =A0r = x^2+y^2
> I'd like mathematica to display the result rather as "r".
>
> And: I'd like mathematica to do this for each relation defined before
> w/o specifying every possible replacement.
>
> Any help much appreciated.
```
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# Thermal Conductivity
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This presentation is about the basics of thermal conductivity and its method, formula derivation and also its applications.
short but comprehensive
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• Great introduction to thermal conductivity. More in-depth information on transient plane source can be found here: https://www.thermtest.com/index.php?page=tps-3500
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### Thermal Conductivity
1. 1.
2. 2. Presentation topic: “Thermal conductivity”<br />Thermodynamics<br />
3. 3. Contents <br />Definition.<br />Explanation. <br />Methods to measure the <br /> thermal conductivity.<br />Experiment.<br />Application (importance).<br /> ……Use in industry.<br /> ……Use in laboratory.<br />Latest researches in this field.<br />
4. 4. Thermal:<br /> MEANS <br />“HEAT.”<br />Conductivity:<br /> Means <br /> “ability to convey”.<br />
5. 5. Definition:<br /><ul><li>It is the property of a material that indicates its ability to conduct heat.
6. 6. It is represented by k and is measured in watts per kelvin per metre (W·K−1·m−1).
7. 7. The reciprocal of thermal conductivity is thermal resistivity.</li></li></ul><li>
8. 8.
9. 9. Thermal Conductivity<br />Assume that no heat is lost through the edges of the disc.<br />Thermodynamics<br />
10. 10. Thermal Conductivity<br />For a uniform rod the rate of flow of heat through a conductor ( Q/ t) is proportional to<br /><ul><li>the cross sectional area (A) of the conductor
11. 11. the temperature gradient ( q/ x)</li></ul>The constant of proportionality, k, is the thermal conductivity BUT note that heat flows down a temperature gradient so we also introduce a negative sign to account for this and obtain:<br />Thermodynamics<br />
12. 12. Thermal Conductivity<br />k depends on the material and is called the thermal conductivity.<br />Rearranging in terms of k we can evaluate the units of k:<br />So k is defined as the rate of flow of heat through unit area of cross section of 1m of material when the temperature difference between the surfaces is 1K.<br />Thermodynamics<br />
13. 13. Methods to measure thermal conductivity.<br />Main methods:<br />Steady state methods.<br />Transient state methods<br />
14. 14. Steady state methods.<br />Definition:<br />These methods are used when the materials are in equilibrium state( means when temperature of the materials is constant).<br />Advantage :<br />Accurate readings can be taken.<br />It steady state implies constant signals.<br />Disadvantage:<br /> As material take to long time to reach equilibrium state so it is slow method.<br />
15. 15. TRANSIENT STATE METHOD:<br />Definition:<br />These methods are used during the heating of material.<br />Advantage:<br />Non-steady-state methods to measure the thermal conductivity do not require the signal to obtain a constant value.<br />Readings can be taken during heating of material.<br />Disadvantage:<br />Readings are not accurate<br />Mathematical analysis of the data is in general more difficult.<br />
16. 16. METHODS: <br />IEEE Standard 442-1981, "IEEE guide for soil thermal resistivity measurements", ISBN 0-7381-0794-8. See also soil thermal properties. [5][1]<br />IEEE Standard 98-2002, "Standard for the Preparation of Test Procedures for the Thermal Evaluation of Solid Electrical Insulating Materials", ISBN 0-7381-3277-2[6][2]<br />ASTM Standard D5334-08, "Standard Test Method for Determination of Thermal Conductivity of Soil and Soft Rock by Thermal Needle Probe Procedure" [3]<br />ASTM Standard D5470-06, "Standard Test Method for Thermal Transmission Properties of Thermally Conductive Electrical Insulation Materials" [7]<br />ASTM Standard E1225-04, "Standard Test Method for Thermal Conductivity of Solids by Means of the Guarded-Comparative-Longitudinal Heat Flow Technique" [8]<br />ASTM Standard D5930-01, "Standard Test Method for Thermal Conductivity of Plastics by Means of a Transient Line-Source Technique" [9]<br />ASTM Standard D2717-95, "Standard Test Method for Thermal Conductivity of Liquids" [10]<br />ISO 22007-2:2008 "Plastics -- Determination of thermal conductivity and thermal diffusivity -- Part 2: Transient plane heat source (hot disc) method" [11]<br />
17. 17. THERMAL CONDUCTIVITYAPPARATUS:<br />
18. 18. OBJECTIVES:<br />
19. 19.
20. 20. PROCEDUER:<br /> Take ice block ,find its weight and volume and place it on the material sheet.<br />Then place it on the plate which is under observation<br />Then provide steam for sometime then we the ice converted to water then measure the weight of the water (in beaker).<br />Then place this values in the formula and calculate the conductivity.<br />
21. 21. Exp:<br />Formula:<br />K= Ro dx / A dT<br />dx = thickness of material.<br />dT =change in temperature.<br />
22. 22.
23. 23.
24. 24.
25. 25.
26. 26. Applications:As aluminums sheets are used in houses for blocking of heat to come in and remain in cool inside.<br />
27. 27. Used in lab: Experiments are performed in lab for finding thermal conductivity of different materials.<br />
28. 28. <ul><li>Now a days different materials are used such like graphene and diamonds for transferring heat which is also due to the techniques of finding thermal conductivities.</li></ul>Also because of this property of the material we can use desired elements in electric circuits like copper, steel etc according to their conductivity.<br />
29. 29. Latest researches.CARBON NANOTUBES <br />
30. 30. MAIN POINTS:<br />Before carbon nanotubes the best thermal conductor was diamond.<br />Carbon nanotubes have extraordinary thermal conductivity properties.<br />
31. 31. STRUCTURAL IMAGE:<br />
32. 32. RESEARCHES<br />The record-setting anisotropic thermal conductivity of carbon nanotubes is enabling applications where heat needs to move from one place to another.<br />carbon nanotubes have the intrinsic characteristics desired in material used as electrodes in batteries and capacitors.<br />
33. 33. Graphene:<br />
34. 34. Properties:<br /> has very high current capacity.<br />Used in high modified circuit in which current gain is required<br />
35. 35. Group:<br /> “ IEFRIANS”<br />Talent participated:<br /> HASSAN ALI (50+8)<br /><ul><li> HASSAN SULTAN(50-21)</li></ul> TAUSEEF AWAN.(50-22) ANAS AMIN (50+6)<br /><ul><li> M. AWAIS SATTAR.(50+4)</li></ul>Special thanks to HASSAN ALI<br /> (YOUNG RUFFIANS)<br /> | 1,748 | 6,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2017-30 | latest | en | 0.749184 |
https://www.convert-measurement-units.com/convert+Peta+to+Hecto.php | 1,713,448,063,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00357.warc.gz | 645,462,687 | 13,270 | Convert Peta to Hecto (SI-prefixes)
## Peta into Hecto
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Peta+to+Hecto.php
# Convert Peta to Hecto:
1. Choose the right category from the selection list, in this case 'SI-prefixes'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Peta'.
4. Finally choose the unit you want the value to be converted to, in this case 'Hecto'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '893 Peta'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'SI-prefixes'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '71 Peta to Hecto' or '14 Peta into Hecto' or '43 Peta -> Hecto' or '55 Peta = Hecto'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(14 * 63) Peta'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '893 Peta + 2679 Hecto' or '23mm x 97cm x 94dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.944 809 982 302 2×1031. For this form of presentation, the number will be segmented into an exponent, here 31, and the actual number, here 1.944 809 982 302 2. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.944 809 982 302 2E+31. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 19 448 099 823 022 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 846 | 3,537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.861876 |
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# Translating English to Propositional Logic
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### Translating English to Propositional Logic
1. 1. Translating English to Propositional Logic Phil 57 section 3 San Jose State University Fall 2010
2. 2. Does order always matter? <ul><li>John went to school and Mary went to school. </li></ul><ul><li>Mary went to school and John went to school. </li></ul><ul><li>P = John went to school. </li></ul><ul><li>Q = Mary went to school. </li></ul><ul><li>(P Q) means the same as (Q P), logically speaking. </li></ul>
3. 3. Does order always matter? <ul><li>John went to school or Mary went to school. </li></ul><ul><li>Mary went to school or John went to school. </li></ul><ul><li>P = John went to school. </li></ul><ul><li>Q = Mary went to school. </li></ul><ul><li>(P Q) means the same as (Q P), logically speaking. </li></ul>
4. 4. Conjunction and disjunction are commutative. <ul><li>(P Q) means the same as (Q P) </li></ul>P Q (P Q) (Q P) T T T T T F F F F T F F F F F F
5. 5. Conjunction and disjunction are commutative. <ul><li>(P Q) means the same as (Q P) </li></ul>P Q (P Q) (Q P) T T T T T F T T F T T T F F F F
6. 6. Conjunction and disjunction are also associative. <ul><li>((P Q) R) means the same as (P (Q R)). </li></ul><ul><li>((P Q) R) means the same as (P (Q R)). </li></ul>
7. 7. With mixed operators, order does matter. <ul><li>P = I like peanut butter. </li></ul><ul><li>Q = I like jelly. </li></ul><ul><li>~(P Q) means “it is not the case that I like peanut butter and jelly”. </li></ul><ul><li>(~P ~Q) means “I don’t like peanut butter and I don’t like jelly.” </li></ul>
8. 8. With mixed operators, order does matter. <ul><li>P = Tom will work late. </li></ul><ul><li>Q = Dick will work late. </li></ul><ul><li>R = Harry will call in sick. </li></ul><ul><li>~((P Q) R) means “it is not the case that Tom and Dick will work late or that Harry will call in sick.” </li></ul>
9. 9. Translating material conditionals. <ul><li>If [antecedent], then [consequent]. </li></ul><ul><li>P= antecedent </li></ul><ul><li>Q= consequent </li></ul><ul><li>P Q </li></ul>
10. 10. Material conditionals and necessary conditions. <ul><li>Getting an A on the final exam is a necessary condition for getting an A in the class. </li></ul><ul><li>Necessary because if I don’t meet this condition, I don’t bring about the outcome. </li></ul><ul><li>P= I get an A on the final exam. </li></ul><ul><li>Q= I get an A in the class. </li></ul><ul><li>Q P </li></ul>
11. 11. Material conditionals and necessary conditions. <ul><li>Getting an A on the final exam is a necessary condition for getting an A in the class. </li></ul><ul><li>P= I get an A on the final exam. </li></ul><ul><li>Q= I get an A in the class. </li></ul>P Q Q P T T T T F T F T F F F T
12. 12. Material conditionals and sufficient conditions. <ul><li>Getting a B on all the exams is a sufficient condition for getting a B in the class. </li></ul><ul><li>Sufficient because it’s enough to bring the outcome, but it’s not the only way to bring it. </li></ul><ul><li>P= I get a B on all the exams. </li></ul><ul><li>Q= I get a B in the class. </li></ul><ul><li>P Q </li></ul>
13. 13. Material conditionals and sufficient conditions. <ul><li>Getting a B on all the exams is a sufficient condition for getting a B in the class. </li></ul><ul><li>P= I get a B on all the exams. </li></ul><ul><li>Q= I get a B in the class. </li></ul>P Q P Q T T T T F F F T T F F T
14. 14. Translating material conditionals. <ul><li>If … , then ... </li></ul><ul><li>It taxes go up, then inflation will rise. </li></ul><ul><li>T= Taxes go up </li></ul><ul><li>R= Inflation will rise </li></ul><ul><li>T R </li></ul>
15. 15. Translating material conditionals. <ul><li>… only if... </li></ul><ul><li>Iran will supply arms to Syria only if Syria helps Hezbollah. </li></ul><ul><li>R= Iran will supply arms to Syria </li></ul><ul><li>S= Syria helps Hezbollah </li></ul><ul><li>R S </li></ul>
16. 16. Translating material conditionals. <ul><li>Only if ... will … </li></ul><ul><li>Only if Jenna passes the exam will Jenna get her license. </li></ul><ul><li>P= Jenna passes the exam </li></ul><ul><li>Q= Jenna will get her license </li></ul><ul><li>Q P </li></ul>
17. 17. Translating material conditionals. <ul><li>… if ... </li></ul><ul><li>I will pass the muffins if you ask me nicely. </li></ul><ul><li>M= I will pass the muffins </li></ul><ul><li>N= You ask me nicely. </li></ul><ul><li>N M </li></ul>
18. 18. Translating material conditionals. Construction Translation If P, then Q (P Q ) P, if Q ( Q P ) P only if Q (P Q ) Only if P, Q ( Q P )
19. 19. Translating biconditionals. <ul><li>… if and only if … </li></ul><ul><li>Jill needs a parachute if and only if she is planning to jump from the plane. </li></ul><ul><li>P= Jill needs a parachute. </li></ul><ul><li>Q= Jill is planning to jump from the plane. </li></ul><ul><li>P Q </li></ul>
20. 20. Translating biconditionals. <ul><li>… just in case … </li></ul><ul><li>Bill will take the geology course just in case it fulfils the science requirement. </li></ul><ul><li>T= Bill will take the geology course </li></ul><ul><li>R= The geology course fulfils the science requirement. </li></ul><ul><li>T R </li></ul>
21. 21. A biconditional is the conjunction of two material conditionals. <ul><li>Jill needs a parachute if and only if she is planning to jump from the plane. </li></ul><ul><li>If Jill needs a parachute, then she is planning to jump from the plane, </li></ul><ul><li>AND </li></ul><ul><li>If Jill is planning to jump from the plane, then she needs a parachute. </li></ul>
22. 22. Biconditionals are commutative. <ul><li>P Q is the same as Q P </li></ul><ul><li>“Bill will take the geology course just in case it fulfils the science requirement” is equivalent to </li></ul><ul><li>“The geology course fulfils the science requirement just in case Bill will take it.” </li></ul><ul><li>Conjunction of two conditionals (and conjunction is commutative) </li></ul>
23. 23. Biconditionals are associative. <ul><li>(P (Q R)) is the same as ((P Q) R) </li></ul><ul><li>Conjunction of two conditionals (and conjunction is associative) </li></ul><ul><li>But note that material conditionals are neither commutative nor associative. </li></ul><ul><li>(P Q) ≠ (Q P) </li></ul><ul><li>(P (Q R)) ≠ ((P Q) R) </li></ul>
24. 24. Order matters translating conditionals. <ul><li>P = Ben will answer the phones. </li></ul><ul><li>Q = Liz will work out the budget. </li></ul><ul><li>(P Q): If Ben will answer the phones, then Liz will work out the budget. </li></ul><ul><li>(Q P): If Liz will work out the budget, then Ben will answer the phones. </li></ul>
25. 25. Order matters translating conditionals. <ul><li>P = Ben will answer the phones. </li></ul><ul><li>Q = Liz will work out the budget. </li></ul>P Q (P Q) (Q P) T T T T T F F T F T T F F F T T
26. 26. Translating English to PL. <ul><li>“ Propositional Logic translation guide” on course website </li></ul><ul><li>(http://www.stemwedel.org/logic-and-critical-reasoning/PL-TranslationGuide.pdf) </li></ul><ul><li>Practice (like on HW #7) will help! </li></ul>
1. #### A particular slide catching your eye?
Clipping is a handy way to collect important slides you want to go back to later. | 2,592 | 7,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2015-35 | latest | en | 0.858561 |
https://math.stackexchange.com/questions/947009/literature-request-on-markov-chains-which-state-transition-probability-matrix-ev | 1,701,476,639,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00104.warc.gz | 443,200,203 | 37,788 | # Literature request on Markov Chains which state transition probability matrix evolves over time
I want to know is there any literature on markov chains who's state transition probability matrix evolves over time?
For instance, I have 2 states, 1 and 2. With
$$P = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} 0.9 & 0.1 \\ 0.8 & 0.2 \end{bmatrix}$$
Where $p_{ij}$ indicates the change that you jump from state $i$ to $j$.
Now at time $k$ I jump from 1 to 2, which has a 0.1 probability. The next instance $k+1$ I have a probability of 0.2 of staying in state 2. Now if I stay at $k+1$ in state 2 then the probability of staying in state 2 would increase and the probability of jumping from state 2 to 1 would decrease. Meaning that $P_{k+2}$ would, for instance have evolved to
$$P_{k+2} = \begin{bmatrix} 0.9 & 0.1 \\ 0.5 & 0.5 \end{bmatrix}$$
If I again stay in state 2 it increases further...
$$P_{k+3} = \begin{bmatrix} 0.9 & 0.1 \\ 0.2 & 0.8 \end{bmatrix}$$
If I then jump from state 2 to 1 it decreases...
$$P_{k+4} = \begin{bmatrix} 0.9 & 0.1 \\ 0.6 & 0.4 \end{bmatrix}$$
Now I want to know if there is any literature regarding this?
-edit-
I believe my question is very much related to https://mathoverflow.net/questions/168398/time-inhomogeneous-markov-chains
This corresponds to any bivariate renewal process where one is given some positive integer valued independent $(U_i)$ i.i.d. and $(V_i)$ i.i.d. and one is at state $1$ at time $n$ if there exists some $k$ such that$$\sum_{i=1}^k(U_i+V_i)\leqslant n\lt U_{k+1}+\sum_{i=1}^k(U_i+V_i),$$ and one is at state $2$ otherwise.
The distributions of every $U_i$ and every $V_i$ are as follows: for every $n$, $P(U_i\geqslant n+1\mid U_i\geqslant n)$ is the probability to stay at state $1$ one more step when one is at state $1$ for $n$ steps, likewise, $P(V_i\geqslant n+1\mid V_i\geqslant n)$ is the probability to stay at state $2$ one more step when one is at state $2$ for $n$ steps.
In the example given in the post, this means that $$P(V=1)=0.8,\quad P(V=2)=0.2\times0.5,\quad P(V=3)=0.2\times0.5\times0.2.$$
• So $P(V = 1)$ is the probability that one is in state 2 for 1 step? Should that not be $p_{k,12} = 0.1$ then? The probability of jumping from state 1 to 2? Assuming that interpretation of $P(V = 1)$ is correct. Then $P(V = 2) = p_{k,12} * p_{k+1,22}$ and $P(V = 3) = p_{k,12} * p_{k+1,22} * p_{k+2,22}$? $p_{k,ij}$ denotes the element $p_{ij}$ of the $P_k$th matrix. Sorry I am a bit confused. Also maybe see my added link in my post.
– WG-
Sep 28, 2014 at 14:52
• One stays in state 2 for 1 step when one leaves state 2 as soon as possible, thus, with probability $(P)_{2,1}=.8$. Likewise, $(V=2)$ corresponds first to a decision to stay at 2 (probability $(P)_{2,2}=.2$) and second to a decision, one time later, to leave state 2 (probability $(P_2)_{2,1}=.5$).
– Did
Sep 28, 2014 at 15:49
• Your setting is quite different from the one in the MO page since transitions in your model depend on the duration of the stay at the present state, that is, transitions at time $n$ are given by a matrix $P_{n-L_n}$, where $L_n$ is the time of arrival at present state, not $P_n$ as on the MO page.
– Did
Sep 28, 2014 at 15:55
• is it okay for you if I extend my question? I have a subsequent question namely which requires me to add more info.
– WG-
Sep 28, 2014 at 16:07
• Well, IF new problem, THEN new question on new page, no?
– Did
Sep 28, 2014 at 16:10
Yes, this is a so-called "inhomogenous Markov chain". Some authors use the term "non-homogenous".
Please note that you are interested in Markov chains with discrete time which are different from models with continuous time. | 1,254 | 3,713 | {"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} | 3.453125 | 3 | CC-MAIN-2023-50 | latest | en | 0.858808 |
https://www.123helpme.com/acoustic-sound-absorber-and-how-the-sound-field-interacts-with-the-material-preview.asp?id=704971 | 1,571,562,542,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986705411.60/warc/CC-MAIN-20191020081806-20191020105306-00509.warc.gz | 771,727,050 | 26,828 | # Acoustic Sound Absorber And How The Sound Field Interacts With The Material
Length: 1774 words (5.1 double-spaced pages)
Rating: Better Essays
#### Essay Preview
Z_0=Z_c (Z_1+iZ_c tan(kL))/(Z_c+iZ_1 tan(kL))
The two equations above characterise how the sound field interacts with the material. An acoustic sound absorber can be modelled by modifying the wavenumber k and the characteristic acoustic impedance Z_c. The characteristic impedance needs to be determined experimentally. A popular model of determining these values is Delaney & Bazley[6].
The surface impedance on the leeward side of the material can vary depending on the boundary conditions. For a material sample located adjacent to a rigid end Z_0→ ∞, and for a sample located in air Z_0=ρ_0 c.
2.3.2 Delaney & Bazley Model of Characteristic Impedance
Z_c=ρ_0 c(1+0.0571((ρ_0 ω)/2πr)^(-0.754)-i0.087((ρ_0 ω)/2πr)^(-0.732) )
k_c=ω/c (1+0.0978((ρ_0 ω)/2πr)^(-0.7)-i0.189((ρ_0 ω)/2πr)^(-0.595) )
Where: ρ_0 and c are the ambient density of air and the speed of sound in are respectively.
The model requires the flow resistance r of the material which can be determined using the microphone method discussed earlier. This model is only valid for r values that range between 〖10〗^3 (Pa.s)/m^2 and 〖5×10〗^4 (Pa.s)/m^2
2.3.3 The Impedance Tube
A method of obtaining the acoustical performance of the materials is by directly measuring the affected sound waves that have been passed though the material. An impedance tube is used to create a controlled space where sound waves can be isolated from the environment. A loud speaker creates a sound field within the tube and is measured by placed microphones.
In the previous year’s project they implemented an impedance tube in accordance to the ISO10534–2 [7] and ASTM 1050 [8] standards. The testing rig is capable of obtaining the absorption coefficient of a test sample. The main limitation of th...
... middle of paper ...
...ss and mass per unit area, a model can be used to determine the surface impedance of the material[18]
Z_(0,lining)=Z_0+i2πfσ '
where σ ' is the mass per unit area of the material. This should only be used on materials with very low stiffness and are therefore ‘limp’.
3 Methodology
This section will describe how the project implemented the experimental theory to creating a testing rig to enable the determination of the acoustic performance of the foams.
3.1 Laboratory Safety
Experiments were conducted in the Dynamics & Control Laboratory which safety protocols had to be followed. The test equipment is heavy, and could easily cause damage if dropped so closed shoes had to be worn at all times. Components of the test rig such as the microphones and pre-amps run at high voltage. Extra care was taken to power off the testing rig when not in use or adjusting the equipment.
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Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
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Take the quiz test your understanding of the key concepts covered in the chapter. Try testing yourself before you read the chapter to see where your strengths and weaknesses are, then test yourself again once you’ve read the chapter to see how well you’ve understood.
1. What does a simple linear regression analysis examine?
1. The relationship between only two variables
2. The relationship between one dependent and one independent variable
3. The relationship between many variables
4. The relationship between two dependent and one independent variable
a. The relationship between only two variables
c. The relationship between many variables
2. Which of the following are correct?
1. The intercept/constant (β0) is the mean-Y when X=0
2. The intercept is the the amount of change in mean-Y when X=0
3. The coefficient is the mean-Y at a certain value of X
4. The coefficient (β) is the amount of change in mean-Y for every unit increase in X
a. The intercept/constant (β0) is the mean-Y when X=0
d. The coefficient (β) is the amount of change in mean-Y for every unit increase in X
3. What does the least squares method do exactly?
1. Minimizes the distance between the data points
2. Finds the least problematic regression line
3. Finds those (best) values of the intercept and slope that provide us with the smallest value of the residual sum of squares
4. Finds those (best) values of the intercept and slope that provide us with the smallest value of the sum of residuals
c. Finds those (best) values of the intercept and slope that provide us with the smallest value of the residual sum of squares
4. Which of the following measures is optimal for comparing the goodness of the fit of competing regression models involving the same dependent variable?
1. The intercept
2. The coefficient
3. R-square
4. Standard deviation of the residuals
d. Standard deviation of the residuals
5. What does the following expression (H0: 1 = 0) mean?
1. Mean-Y changes as as a result of a change in X
2. Mean-Y does not change as a result of a change in X
3. Mean-Y value becomes 0 as a result of a change in X
4. Mean-Y value is equal to 0 when X=0 | 504 | 2,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.899408 |
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### How to apply the Kelly criterion when expected return may be negative?
My concern is how to handle a negative value for the Kelly formula. Even when you have a system that has positive expectancy, you can (and usually will) sustain a number of losses, sometimes ... | 973 | 4,561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-30 | latest | en | 0.888914 |
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1 Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax:
2 Introduction In this lesson, you will learn the following : The definition of a root locus How to sketch root locus How to use the root locus to find the poles of a closed loop system How to use root locus to describe the changes in transient response How to use root locus to design a parameter value to meet a desired transient response 2
3 Introduction What is Root Locus? Root Locus is graphical presentation of the closed-loop poles as the parameter is varied A powerful method of analysis and design for stability and transient response Ability to provide solution for system of order higher than two A graphic representation of a system s stability Why do we need to use Root Locus? We use Root Locus to analyze the transient qualitatively 3
4 Introduction E.g. we can use Root Locus to analyze qualitatively the effect of varying gain upon percent overshoot, settling time and peak time We can also use Root Locus to check the stability of feedback control system qualitatively The control system problem: The poles of the closed-loop transfer function are more difficult to find The closed-loop poles change with changes in system gain The poles of T(s) are not immediately known without factoring the denominator, and they are a function of K 4
5 Introduction Since the system transient response and stability are dependent upon the poles of T(s), we have no knowledge of the system performance unless we factor the denominator for specific values of K The Root Locus will be used to give us a vivid picture of the poles of T(s) as K varies 5
6 Introduction Complex Numbers and Their Representation as Vectors: Any complex number, j,described in Cartesian coordinates can be graphically represented by a vector, as shown in the following figure (a) The complex numbers also can be described in the polar form with magnitude M, and angle, as M If the complex number is substituting into a complex function, F(s), another complex number will result For example, if F(s)=(s+a), then substituting the complex number s j yields F() s ( a) j which is another complex number, as shown in figure (b) 6
7 Introduction F(s)=(s+a) Since F(s) has a zero at a, we ( s 7) have alternate representation s 5 j 2 7
8 Introduction Example: Evaluation of a complex function via vectors Magnitude and angle of F(s) at any point ( s 3 j4) Solution: s 3 j4 ( 3 j41) F( 3 j4) ( 3 j4)( 3 j4 2) ( )( ) 20 M ( )
9 Defining the Root Locus Consider security camera that can automatically follow a subject (auto tracking) Can get by 2 s 10s K 9
10 Defining the Root Locus (Pole plot from Table 8.1) (Representation of the paths of the closed-loop poles as the gain is varied ) Call Root Locus! 10
11 Properties of the Root Locus The root locus technique can be used to analyze and design the effect of a loop gain upon the system s transient response and stability The closed loop transfer function for a system with a gain, K, is (For control system of slide 5) From this equation, a pole, s, exists when the characteristic polynomial in the denominator becomes zero, or 11
12 Properties of the Root Locus Here, -1 represented in polar form as Alternatively, a value of s is a closed loop pole if and Let s demonstrate the relationship: For control system of slide 9, KG() s H () s s(s10) From Table 8.1, the closed-loop poles exist at and when the gain is 5 Substituting the pole at for s and 5 for K yields If these calculation is repeated for the point 5 j5, the angle equals an odd multiple of 180 K K K K KG() s H () s 180 s(s10) ( 5 j5)(5 j5) ( )( 5045 ) 50 KG() s H () s 1 (Thus, it is a point on the root locus!) 1 (2k 1)180 K 12
13 Properties of the Root Locus That is, 5 j5 is a point on the root locus for some value of gain Now we proceed to evaluate the value of gain K KG() s H () s 1 K For a point 2 j2, the angle does not equal () () K K K K KG s H s s(s10) ( 2 j2)(8 j2) ( 8135 )( 6814 ) Example: (Can confirm from Table 8.1!) Thus, it is not a point on the root locus, or is not a closed-loop pole for any gain A part of Root Locus for Gs () T() s 1 Gs ( ) 13
14 Properties of the Root Locus For this system, the open-loop transfer function is KG() s H () s K(s3)(s4) (s1)(s2) Consider the point 2 j3 Then, the angle does not equal as follows: Thus, this point is not a point on the root locus!! (That is, this point is not a closed-loop pole for any gain) 14
15 Sketching the Root Locus The following five basic rules allow us to sketch the root locus using minimal calculations Rule 1: Number of the branches Each closed loop pole moves as the gain is varied If we define a branch as the path that one pole traverses, then there will be one branch for each closed loop pole Our first rule defines the number of branches of the root locus: The number of the branches of the root locus are equal to the number of closed loop poles Rule 2: Symmetry Physically realizable system can not have the complex coefficients K in their transfer functions Thus, we conclude: The root locus is symmetrical about the real axis 15
16 Sketching the Root Locus Rule 3: Real-axis segment Using some mathematical properties of root locus plot, we get On the real axis, for K>0 the root locus exist to the left of an odd number of real axis, finite open loop poles and/or finite open loop zeros symmetrical 16
17 Sketching the Root Locus Rule 4: Starting and ending points Consider the closed loop transfer function As K approaches to zero, the closed loop poles approaches the poles of G(s)H(s) (Refer to Slide 10!) Similarly, as K approaches to infinity, the closed loop poles approaches the zeros of G(s)H(s) (Refer to Slide 10!) We conclude that The root locus begins at the finite and infinite poles of G(s)H(s) and ends at the finite and infinite zeros of G(s)H(s) 17
18 Sketching the Root Locus In this system, the root locus begins at the poles at -1 and -2 and ends at zeros at -3 and -4 Rule 5: Behavior at infinity A function can have infinite poles and zeros Every function of s has an equal number of poles and zeros For example, consider the function This function has two poles at -2 and -3 and one zeros at -1 Because of the number of poles and zeros must be equal, this function has a zero at infinity! (Ref. G(s)=1/s: G( ) 0+ a pole at s=0) a zero at s This rule tell us the following conclusion: (Complete Root Locus) 18
19 Sketching the Root Locus a The root locus approaches straight lines as asymptotes as the locus approaches infinity Further, the equation of the asymptotes is given by the real axis intercept,, and angle,, as follows: a where k = 0, ±1, ±2, ±3 and the angle is given in radians with respect to the positive extension of the real axis Example: Sketch the root locus for the system shown in figure a finite poles finite zeros (2k 1), a #finite poles #finite zeros #finite poles #finite zeros 19
20 Sketching the Root Locus Solution: Begin by calculating the asymptotes: If the values for k continued to increase, the angels would begin repeat The number of lines obtained equals the difference between the number of finite poles and the number of finite zeros Rule 4 states that the root locus begins at the open loop poles and ends at the open loop zeros 20
21 Sketching the Root Locus For this example, there are more open loop poles than open loop zeros Thus, there must be zeros at infinity The asymptotes tell us how we get to these zeros at infinity! This figure shows the complete root locus as well as the asymptotes that were just calculated Root locus with the 3 asymptotes 21
22 Sketching the Root Locus Notice that we have made use of all the rules learned so far The real axis segment lie to the left of an odd number of poles and/or zeros The locus starts at the open-loop poles and ends at the open-loop zeros For this example, there is only one open loop finite zero and three infinite zeros Rule 5, then, tell us that the three zeros at infinity are at the ends of the asymptotes Now, consider four additional rules for Refining the Sketch Once we sketch the root locus using the five rules discussed in the previous slides, we may want to accurately locate points on the root locus as well as find their associated gain 22
23 Refining the Sketch Rule 6: Real Axis Breakaway and Break-in Points Numerous root loci appear to break away from the real axis as the system poles move from the real axis to the complex plane At other times, the loci appear to return to the real axis as a pair of complex poles becomes real This is illustrated in the figure The breakaway point is found at the maximum point, and the break-in point is found at the minimum point (Variation of gain along the real axis ) 23
24 Refining the Sketch The figure shows a root locus leaving the real axis between -1 and -2 and returning to the real axis between +3 and +5 The point where the locus leaves the real axis, 1, is called the breakaway point, and the point where the locus returns to the real axis, 2, is called break-in point At the breakaway or break-in point, the branches of the root locus form an angle of 180/n with the real axis, where n is the number of closed loop poles arriving or departing from the single breakaway or break-in point on the real axis Thus for the poles shown in the figure the branches at the breakaway point form 90 angles with the real axis We now show how to find the breakaway and the break-in points There is a simple method for finding the points at which the root locus breaks away from and breaks into real axis The first method is to maximize and minimize the gain, K, using differential calculus 24
25 Refining the Sketch For all points on the root locus, because of the KG(s)H(s)= -1 as we pointed out before, K = -1 / G(s)H(s) For points along the real axis segment of the root locus where breakaway and break-in points could exist, s Hence along the real axis, K 1 1 GsHs () () G( ) H( ) s Hence if we differentiate this equation with respect to and set the derivative equal to zero, we can find the points of maximum and minimum gain and hence breakaway and break-in points dk 0,? d 25
26 Refining the Sketch Example: Find the breakaway and break-in points for the root locus of the figure using differential calculus Solution: Using the open loop poles and zeros, we present the open loop system whose root locus is shown in figure as follows But for all points along the root locus, KG(s)H(s)=-1, andalong the real axis,. Hence s Solving for K, we find 26
27 Refining the Sketch Differentiating K with respect to and setting the derivative equal to zero yields Solving for, we find =-1.45 (break-away point) and 3.82 (break-in point), which are the breakaway and break-in points (Data Sheet for breakaway and break-in points) 27
28 Refining the Sketch Rule 7: j -Axis Crossings j The -axis crossing is a point on the root locus that separates the stable operation of the system from the unstable operation The value of at the axis crossing yields the frequency of oscillation The gain at the j-axis crossing yields the maximum positive gain for system stability To find j -axis crossing, we can use the Routh-Hurwitz criterion as follows: Forcing a row of zeros in the Routh table will yield the gain; going back one row to the even polynomial equation and solving for the roots yields the frequency at the imaginary axis crossing 28
29 Refining the Sketch Example: For the system of the following figure, find the frequency and gain, K, for which the root locus crosses the imaginary axis For what range of K is the system stable? Solution: The closed loop transfer function for the system is Simplifying some of the entries by multiplying any row by a constant, we obtain the Routh array shown in table 29
30 Refining the Sketch A complete row of zeros yields the possibility for imaginary axis roots For positive value of gain only, the s 1 row can yield a row of zeros for K>0 Thus,, and K = 9.65 Forming the even polynomial by using the s 2 row with K =9.65 We obtain, and s is found to be ±j1.59 Thus, the root locus crosses the j -axis at ±j1.59 at a gain of 9.65 We conclude that the system is stable for 0 K
31 Refining the Sketch Rule 8: Angles of Departure and Arrival In order to sketch the root locus more accurately, we want to calculate the root locus departure angle from the complex poles and the arrival angle to the complex zeros If we assume a point on the root locus close to the point, we assume that all angles drawn from all other poles and zeros are drawn directly to the pole that is near the point Thus, the only unknown angle in the sum is the angle drawn from the pole that is close We can solve for this unknown angle, which is also the angle of departure from this complex pole Hence from the figure(a) (Angles of Departure) 31
32 Refining the Sketch (Angles of Departure) If we assume a point on the root locus close to a complex zero, the sum of angles drawn from all finite poles and zeros to this point is an odd multiple of 180 ( ) Except for the zero that is close to the point, we can assume all angles drawn from all other poles and zeros are drawn directly to the zero that is near the point 32
33 Refining the Sketch Thus, the only unknown angle in the sum is the angle drawn from the zeros that is close We can solve for this unknown angle, which is also the angle of arrival to this complex zero Hence from the figure(b) (Angles of Arrival) 33
34 Refining the Sketch Example: Given the unity feedback system of following figure, find the angle of departure from the complex poles and sketch the root locus Solution: Using the poles and zeros of G(s)= (s+2)/[(s+3)(s 2 +2s+2)] as plotted in the figure, we calculate the sum of angles drawn to a point close to the complex pole, -1+j1, in the second quadrant Thus, 34
35 Refining the Sketch 35
36 Refining the Sketch Rule 9: Plotting and calibration the Root locus We might want to know the exact coordinates of the root locus as it crosses the radial line representing an overshoot value Overshoot value can be represented by a radial line on the s- plane We learnt in Chapter 4, the value of zeta on the s-plane is cos Given percent overshoot value, we can calculate the zeta value ln(%os /100) 2 2 ln (%OS /100) Radial line representing overshoot value on the s-plane 36
37 Refining the Sketch Example: Draw a radial line on an s-plane that represents 20% overshoot Solution: Overshoot is represented by zeta (damping ratio) on the S-plane So, the first step is to find the value of zeta OS OS ln % /100 ln(20 /100) ln 2 % /100 2 ln 2 20 /100 Next step is to find the angle of the radial line cos cos 1 cos j 37
38 Refining the Sketch The point where our root locus intersect with the n percent overshoot radial line is the point when the gain value produces a transient response with n percent overshoot Our root locus intersect with the radial line. Meaning the gain at the intersection produces transient response with zeta = 0.45 A point on the radial line is on the root locus if the angular sum (zero angle pole angles) in reference to the point on the radial line add up to an odd multiple of 180, Odd multiple of 180 : (2k+1)180, k = 1, 2, 3, 4,. 180, 540, 900, 1260, 38
39 Refining the Sketch Then, we must then calculate the value of gain Refer to the previous root locus that we have calculated, we will find the exact coordinate as it crosses the radial line representing 20% overshoot
40 Refining the Sketch We can find the point on the radial line that crosses the root locus by selecting a point with radius value then add the angles of the zeros and subtract the angles of the poles Theory : Odd multiple of , 540, 900, 1260 point on the radial line is on the locus At a point s, 40
41 Refining the Sketch Example: Sketch the root locus for the system shown in figure and find the breakaway point on the real axis and the range of K within which the system is stable Solution: The root locus of the system is shown in following figure The real axis segment is found to be between -2 and -4, using Rule 3 The root locus starts at the open loop poles and ends at the open loop zeros, using Rule 4 The root locus crosses the -axis at ±j3.9, using Rule 7 j 41
42 Refining the Sketch The exact point and gain where the locus crosses the 0.45 damping ratio line: K=0.47 at , using Rule 9 To find breakaway point use the root locus algorithm to search the real axis between -2 and -4 for the point that yields maximum gain, using Rule 6 Naturally all points will have the sum of their angles equal to an odd multiple of 180 A maximum gain of is found to at the point Thus, the breakaway point is between the open loop poles on the real axis at Finally, the system is stable for K between 0 and 15 42
43 Root Locus Plot using matlab Example: Plot the root-locus of the following system Gs () 2 2 s s s s 2 ( s 2)( s 4) s 6s % matlab script 5 4 Root Locus clear all num = [1-4 20]; den = [1 6 8]; G = tf(num,den); rlocus(g); axis('equal'); Real Axis (seconds -1 ) 43
44 Homework Assignment #8 44
45 Homework Assignment #8 45
46 Homework Assignment #8 46
47 Homework Assignment #8 47
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EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More information | 11,533 | 47,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.835877 |
http://www.mathworks.com/matlabcentral/answers/5258-write-a-program-that-calculates-sum-of-the-integers-numbers?requestedDomain=www.mathworks.com&nocookie=true | 1,480,759,832,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540915.89/warc/CC-MAIN-20161202170900-00495-ip-10-31-129-80.ec2.internal.warc.gz | 582,666,478 | 15,197 | ## Write a program that calculates sum of the integers numbers
Asked by Aaron Moris
### Aaron Moris (view profile)
on 11 Apr 2011
What is the product if the question is write a program that calculates and prints the sum of the even integers from 2 to 30?
## Products
### Florin Neacsu (view profile)
Answer by Florin Neacsu
### Florin Neacsu (view profile)
on 11 Apr 2011
2+4+6+...+30=2(1+2+3+...+15)=2*(15*16)/2=15*16
So 2+4+...+n = floor(n/2)*(floor(n/2)+1).
Regards, Florin
### Matt Fig (view profile)
Answer by Matt Fig
### Matt Fig (view profile)
on 11 Apr 2011
The product would be the sum of the even integers from 2 to 30 ;-).
Seriously, what have you tried so far on this homework problem?
Sean de Wolski
### Sean de Wolski (view profile)
on 11 Apr 2011
You might need some bubble wrap for that.
Paulo Silva
### Paulo Silva (view profile)
on 11 Apr 2011
I hate when it takes more time to understand what people want than to make their answer!
### Paulo Silva (view profile)
Answer by Paulo Silva
### Paulo Silva (view profile)
on 11 Apr 2011
```v=2:30;
sum(v(~mod(v,2)))
```
### Doug Eastman (view profile)
Answer by Doug Eastman
### Doug Eastman (view profile)
on 11 Apr 2011
or
```sum(2:2:30)
```
Paulo Silva
### Paulo Silva (view profile)
on 11 Apr 2011
that's simple but if you change the start value from 2 to 1 you do the sum of odd numbers instead, also if the last value is 31 or some odd number the result won't be correct, unfortunately that's not very flexible code.
Paulo Silva
### Paulo Silva (view profile)
on 11 Apr 2011
btw that does work and is a good answer for the question (only if the values don't change), +1 vote
### Matt Tearle (view profile)
Answer by Matt Tearle
### Matt Tearle (view profile)
on 11 Apr 2011
```disp(240)
```
Or, for pedants who claim that evaluation is not the same as calculation,
```disp(239+1)
```
Sean de Wolski
### Sean de Wolski (view profile)
on 11 Apr 2011
This is definitely the most compact and efficient way to do this. +1.
#### Join the 15-year community celebration.
Play games and win prizes!
MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi | 626 | 2,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.840895 |
https://programmingpraxis.com/2017/12/05/recursion-and-iteration/ | 1,674,923,897,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499646.23/warc/CC-MAIN-20230128153513-20230128183513-00282.warc.gz | 509,157,629 | 29,851 | ## Recursion And Iteration
### December 5, 2017
Today’s exercise is about basic programming techniques.
Your task is to pick a function and write two versions of it, one recursive, the other iterative; compare the two versions on readability, programming ease, speed, and any other criteria that are important to you. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 9 Responses to “Recursion And Iteration”
1. Rutger said
In Python, a Fib implemenation. Seems like iterative is a lot faster, but didn’t really optimize the recursive one that much either.
```import cProfile
def fib_rec(n):
if n == 0:
return 0
elif n == 1:
return 1
else:
return fib_rec(n-1)+fib_rec(n-2)
def fib_iterative(n):
a, b = 0, 1
for i in range(n):
a, b = b, a + b
return a
functions = [fib_rec, fib_iterative]
for f in functions:
print("Profiling", f.__name__)
cProfile.run("for i in range(3): f(29)")
# Profiling fib_rec
# 4992240 function calls (6 primitive calls) in 7.445 seconds
# Ordered by: standard name
# ncalls tottime percall cumtime percall filename:lineno(function)
# 1 0.000 0.000 7.445 7.445 <string>:1(<module>)
# 4992237/3 7.444 0.000 7.444 2.481 fib.py:4(fib_rec)
# 1 0.000 0.000 7.445 7.445 {built-in method builtins.exec}
# 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
# Profiling fib_iterative
# 6 function calls in 0.000 seconds
# Ordered by: standard name
# ncalls tottime percall cumtime percall filename:lineno(function)
# 1 0.000 0.000 0.000 0.000 <string>:1(<module>)
# 3 0.000 0.000 0.000 0.000 fib.py:13(fib_iterative)
# 1 0.000 0.000 0.000 0.000 {built-in method builtins.exec}
# 1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
# [Finished in 8.1s]
```
2. programmingpraxis said
@Rutger: Your algorithms are not the same. The recursive algorithm takes exponential time, the iterative algorithm takes linear time. You might enjoy this program, which is recursive and takes logarithmic time:
```(define (fib n)
(define (square x) (* x x))
(cond ((zero? n) 0) ((or (= n 1) (= n 2)) 1)
((even? n) (let* ((n2 (quotient n 2)) (n2-1 (- n2 1)))
(* (fib n2) (+ (* 2 (fib n2-1)) (fib n2)))))
(else (let* ((n2-1 (quotient n 2)) (n2 (+ n2-1 1)))
(+ (square (fib n2-1)) (square (fib n2)))))))```
3. chaw said
I am a big fan of recursion but, to give the iterative version a
bit more credit, it can be rewritten as follows, avoiding both
the temporary variable and the mutations.
I suppose someone may object that this version is thinly veiled
recursion but another could counter that tail recursion is
thinly veiled iteration. Perhaps more to the point, both the
following version and the two versions in the solution generate
iterative processes in Scheme (to use SICP terminology if I
recall correctly), although two of them are iterative programs
and one recursive.
``` (define (gcd-i2 m n) (do ((m (max m n) n) (n (min m n) (modulo m n))) ((zero? n) m) )) ```
4. John Cowan said
Rutger: The first algorithm is actually horribly bad: it’s exponential, because all the smaller Fibonacci numbers are recomputed over and over. If you make it recursive with memoization, it’s linear but can still blow up your stack due to the recursion if n is large.
5. programmingpraxis said
@chaw: You don’t need min and max. If m < n, the first call swaps them.
I used the temporary variable in the iterative version on purpose, because it is needed in languages that don’t allow simultaneous assignment.
6. matthew said
A tail call is just a goto combined with a simultaneous reassignment of local variables. The reassignment part can make code clearer and less error-prone (since we are less likely to forget to modify a variable appropriately) but really both are just different ways of expressing the same underlying computation.
Converting a non-tail recursion into an iterative form can be interesting though. Here’s the naive fibonacci implementation, in Javascript:
```fib = (n) => n <= 1 ? n : fib(n-1) + fib(n-2)
```
Neither recursive call is in tail position, but we can build up the final result in an accumulating parameter, passed between the two calls:
```fib = (n,a) => n <= 1 ? n + a : fib(n-1,fib(n-2,a))
```
Now one call is in tail position and can fix the other with an explicit continuation function:
```fib = (n,a,k) => n <= 1 ? k(n+a): fib(n-2,a,t => fib(n-1,t,k))
```
To get properly iterative, we can replace the continuation functions with a stack (containing the stored values of n), use an explicit loop and in-place modification everywhere:
```function fib(n) {
let a = 0, s = []
while (true) {
if (n <= 1) {
a += n;
if (s.length == 0) return a
else n = s.pop()-1
} else {
s.push(n)
n -= 2
}
}
}
```
As people have observed, this is still a terrible way to compute the fibonacci numbers.
7. Daniel said
Here’s a solution in Java for serializing an XML tree. The iterative version uses a stack.
```import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.StandardCharsets;
import java.util.Stack;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
public class Serializer {
enum Op {
OPEN_ELEMENT, CLOSE_ELEMENT
}
private static String SerializeIterative(Node node) {
StringBuilder sb = new StringBuilder();
Stack<Object> s = new Stack<Object>();
s.push(node);
s.push(Op.OPEN_ELEMENT);
while (!s.isEmpty()) {
Op op = (Op)s.pop();
switch (op) {
case OPEN_ELEMENT: {
node = (Node)s.pop();
if (node.getNodeType() == Node.TEXT_NODE) {
sb.append(node.getTextContent());
} else {
Element e = (Element)node;
sb.append("<" + ((Element)node).getTagName() + ">");
NodeList children = e.getChildNodes();
s.push(node);
s.push(Op.CLOSE_ELEMENT);
for (int i = children.getLength()-1; i >= 0; --i) {
s.push(children.item(i));
s.push(Op.OPEN_ELEMENT);
}
}
break;
}
case CLOSE_ELEMENT: {
node = (Node)s.pop();
sb.append("</" + ((Element)node).getTagName() + ">");
break;
}
}
}
return sb.toString();
}
private static String SerializeRecursive(Node node) {
short type = node.getNodeType();
if (type == Node.DOCUMENT_NODE) {
return SerializeRecursive(((Document)node).getDocumentElement());
} else if (type == Node.ELEMENT_NODE) {
Element element = (Element)node;
String tag = element.getTagName();
StringBuilder sb = new StringBuilder();
sb.append("<" + tag + ">");
NodeList children = node.getChildNodes();
for (int i = 0; i < children.getLength(); ++i) {
sb.append(SerializeRecursive(children.item(i)));
}
sb.append("</" + tag + ">");
return sb.toString();
} else {
// Handle Node.TEXT_NODE, although this also serves as a catch-all.
return node.getTextContent();
}
}
public static void main(String[] args)
throws ParserConfigurationException, SAXException, IOException {
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
String html = "<html><body><h1>Hello World</h1><div><p>Test 1</p><p>Test 2</p></div></body></html>";
InputStream is = new ByteArrayInputStream(html.getBytes(StandardCharsets.UTF_8));
Document doc = dBuilder.parse(is);
System.out.println("html: " + html);
System.out.println("SerializeIterative: " + SerializeIterative(doc.getDocumentElement()));
System.out.println("SerializeRecursive: " + SerializeRecursive(doc.getDocumentElement()));
}
}
```
Output:
```html: Hello WorldTest 1
Test 2
SerializeIterative: Hello WorldTest 1
Test 2
SerializeRecursive: Hello WorldTest 1
Test 2
```
8. Daniel said
It appears there was a problem in the html code i pasted.
Here’s another attempt, this time escaping the angle brackets with backslashes.
Output:
```html: \\\Hello World\\\Test 1\\Test 2\\\\
SerializeIterative: \\\Hello World\\\Test 1\\Test 2\\\\
SerializeRecursive: \\\Hello World\\\Test 1\\Test 2\\\\
```
9. Daniel said
Here’s another attempt, this time adding “sourcecode” markup.
`html: `
`<html><body><h1>Hello World</h1><div><p>Test 1</p><p>Test 2</p></div></body></html>`
SerializeIterative:
`<html><body><h1>Hello World</h1><div><p>Test 1</p><p>Test 2</p></div></body></html>`
SerializeRecursive:
`<html><body><h1>Hello World</h1><div><p>Test 1</p><p>Test 2</p></div></body></html>` | 2,392 | 8,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-06 | latest | en | 0.655989 |
http://mathhelpforum.com/math-topics/23232-help-me-finding-semi-circle.html | 1,481,404,407,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543567.64/warc/CC-MAIN-20161202170903-00303-ip-10-31-129-80.ec2.internal.warc.gz | 181,969,247 | 10,067 | # Thread: Help me Finding semi circle
1. ## Help me Finding semi circle
P=1/2 (c+d)
2. Originally Posted by lalauoo
P=1/2 (c+d)
the instructions were to find a semi-circle?
i'm sorry, i don't get your question. what is P, c and d, and what are we supposed to do?
3. Originally Posted by Jhevon
the instructions were to find a semi-circle?
i'm sorry, i don't get your question. what is P, c and d, and what are we supposed to do?
p = 1/4 c is i think circle and d is diameter.
4. Can you please provide the EXACT wording of the question?
5. Originally Posted by lalauoo | 167 | 576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.971672 |
https://math.stackexchange.com/questions/905930/in-an-abelian-category-every-morphism-can-be-written-as-composition-of-epi-and-m?noredirect=1 | 1,560,840,159,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998690.87/warc/CC-MAIN-20190618063322-20190618085322-00061.warc.gz | 516,962,931 | 31,217 | # In an abelian category,every morphism can be written as composition of epi and mono. [duplicate]
Following Weibel's book on homological algebra, he states without proof that every morphism $f\colon A \to B$ can be written as composition of an epimorphism followed by a monomorphism.
After many attempts,I am unable prove this. It is easy to show that
$f\colon A \to B$ factors as $A \twoheadrightarrow Coker(Ker(f)) \to B$ and $A \to Ker(Coker(f))\rightarrowtail B$ .Moreover there is an arrow from $Ker(Coker(f)) \to Coker(Ker(f))$ such that the all the diagrams commute.
Additionaly, in Hilton and Stammbach's book on the same subject, they take this property as an axiom.
## marked as duplicate by Zhen Lin category-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 7 '14 at 9:53
• For completeness, you should state the definition of abelian category Weibel uses. Of course, all you have to do is prove that the canonical $\operatorname{Ker} \operatorname{coker} f \to \operatorname{Coker} \operatorname{ker} f$ is an isomorphism. I am sure this has been asked before on this site somewhere... – Zhen Lin Aug 22 '14 at 11:26 | 430 | 1,623 | {"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} | 2.828125 | 3 | CC-MAIN-2019-26 | latest | en | 0.712683 |
https://coinselected.com/collect-maximum-coins-in-grid-1650923794/ | 1,675,608,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00413.warc.gz | 192,439,625 | 19,020 | # Collect maximum points in a grid using two traversals – GeeksforGeeks
Given a matrix where every cell represents points. How to collect utmost points using two traversals under following conditions ?
Let the dimensions of given grid be R x C.
1 ) The first traversal starts from top leave corner, i.e., ( 0, 0 ) and should reach left bottomland corner, i.e., ( R-1, 0 ). The second traversal starts from crown right corner, i.e., ( 0, C-1 ) and should reach bottom right corner, i.e., ( R-1, C-1 ) /
2 ) From a luff ( iodine, j ), we can move to ( i+1, j+1 ) or ( i+1, j-1 ) or ( i+1, j )
3 ) A traversal gets all points of a especial cell through which it passes. If one traversal has already collected points of a cell, then the other traversal gets no points if goes through that cell again.
```Input :
int arr[R][C] = {{3, 6, 8, 2},
{5, 2, 4, 3},
{1, 1, 20, 10},
{1, 1, 20, 10},
{1, 1, 20, 10},
};
Output: 73
Explanation :
First traversal collects total points of value 3 + 2 + 20 + 1 + 1 = 27
Second traversal collects total points of value 2 + 4 + 10 + 20 + 10 = 46.
Total Points collected = 27 + 46 = 73.```
We strongly recommend you to minimize your browser and try this yourself first.
The idea is to do both traversals concurrently. We start first from ( 0, 0 ) and second traversal from ( 0, C-1 ) simultaneously. The important thing to note is, at any particular step both traversals will be in same row as in all potential three moves, row number is increased. Let ( x1, y1 ) and ( x2, y2 ) denote current positions of first and second base traversals respectively. frankincense at any clock time x1 will be equal to x2 as both of them move forward but variation is possible along y. Since variation in y could occur in 3 ways no switch ( yttrium ), go left ( y – 1 ), go good ( y + 1 ). so in full 9 combinations among y1, y2 are possible. The 9 cases as mentioned below after base cases.
```Both traversals always move forward along x
Base Cases:
// If destinations reached
if (x == R-1 && y1 == 0 && y2 == C-1)
maxPoints(arr, x, y1, y2) = arr[x][y1] + arr[x][y2];
// If any of the two locations is invalid (going out of grid)
if input is not valid
maxPoints(arr, x, y1, y2) = -INF (minus infinite)
// If both traversals are at same cell, then we count the value of cell
// only once.
If y1 and y2 are same
result = arr[x][y1]
Else
result = arr[x][y1] + arr[x][y2]
result += max { // Max of 9 cases
maxPoints(arr, x+1, y1+1, y2),
maxPoints(arr, x+1, y1+1, y2+1),
maxPoints(arr, x+1, y1+1, y2-1),
maxPoints(arr, x+1, y1-1, y2),
maxPoints(arr, x+1, y1-1, y2+1),
maxPoints(arr, x+1, y1-1, y2-1),
maxPoints(arr, x+1, y1, y2),
maxPoints(arr, x+1, y1, y2+1),
maxPoints(arr, x+1, y1, y2-1)
}```
The above recursive solution has many subproblems that are solved again and again. consequently, we can use Dynamic Programming to solve the above problem more efficiently. Below is memoization ( Memoization is option to table based iterative solution in Dynamic Programming ) based implementation. In below execution, we use a memoization table ‘ mem ’ to keep track of already solved problems.
## C++
`#include` `using` `namespace` `std;` `#define R 5` `#define C 4` `bool` `isValid(` `int` `x, ` `int` `y1, ` `int` `y2)` `{` ` ` `return` `(x >= 0 && x < R && y1 >=0 &&` ` ` `y1 < C && y2 >=0 && y2 < C);` `}` `int` `getMaxUtil(` `int` `arr[R][C], ` `int` `mem[R][C][C], ` `int` `x, ` `int` `y1, ` `int` `y2)` `{` ` ` ` ` ` ` `if` `(!isValid(x, y1, y2)) ` `return` `INT_MIN;` ` ` ` ` `if` `(x == R-1 && y1 == 0 && y2 == C-1)` ` ` `return` `(y1 == y2)? arr[x][y1]: arr[x][y1] + arr[x][y2];` ` ` ` ` `if` `(x == R-1) ` `return` `INT_MIN;` ` ` ` ` `if` `(mem[x][y1][y2] != -1) ` `return` `mem[x][y1][y2];` ` ` ` ` `int` `ans = INT_MIN;` ` ` ` ` `int` `temp = (y1 == y2)? arr[x][y1]: arr[x][y1] + arr[x][y2];` ` ` ` ` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1, y2-1));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1, y2+1));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1, y2));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1-1, y2));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1-1, y2-1));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1-1, y2+1));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1+1, y2));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1+1, y2-1));` ` ` `ans = max(ans, temp + getMaxUtil(arr, mem, x+1, y1+1, y2+1));` ` ` `return` `(mem[x][y1][y2] = ans);` `}` `int` `geMaxCollection(` `int` `arr[R][C])` `{` ` ` ` ` `int` `mem[R][C][C];` ` ` `memset` `(mem, -1, ` `sizeof` `(mem));` ` ` ` ` ` ` `return` `getMaxUtil(arr, mem, 0, 0, C-1);` `}` `int` `main()` `{` ` ` `int` `arr[R][C] = {{3, 6, 8, 2},` ` ` `{5, 2, 4, 3},` ` ` `{1, 1, 20, 10},` ` ` `{1, 1, 20, 10},` ` ` `{1, 1, 20, 10},` ` ` `};` ` ` `cout << ` `"Maximum collection is "` `<< geMaxCollection(arr);` ` ` `return` `0;` `}`
## Java
`class` `GFG` `{` ` ` `static` `final` `int` `R = ` `5` `;` `static` `final` `int` `C = ` `4` `;` `static` `boolean` `isValid(` `int` `x, ` `int` `y1, ` `int` `y2)` `{` ` ` `return` `(x >= ` `0` `&& x < R && y1 >=` `0` `&&` ` ` `y1 < C && y2 >=` `0` `&& y2 < C);` `}` `static` `int` `getMaxUtil(` `int` `arr[][], ` `int` `mem[][][],` ` ` `int` `x, ` `int` `y1, ` `int` `y2)` `{` ` ` ` ` ` ` `if` `(!isValid(x, y1, y2)) ` `return` `Integer.MIN_VALUE;` ` ` ` ` `if` `(x == R-` `1` `&& y1 == ` `0` `&& y2 == C-` `1` `)` ` ` `return` `(y1 == y2)? arr[x][y1]: arr[x][y1] + arr[x][y2];` ` ` ` ` ` ` `if` `(x == R-` `1` `) ` `return` `Integer.MIN_VALUE;` ` ` ` ` `if` `(mem[x][y1][y2] != -` `1` `) ` `return` `mem[x][y1][y2];` ` ` ` ` `int` `ans = Integer.MIN_VALUE;` ` ` ` ` ` ` `int` `temp = (y1 == y2)? arr[x][y1]:` ` ` `arr[x][y1] + arr[x][y2];` ` ` ` ` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1, y2-` `1` `));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1, y2+` `1` `));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1, y2));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1-` `1` `, y2));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1-` `1` `, y2-` `1` `));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1-` `1` `, y2+` `1` `));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1+` `1` `, y2));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1+` `1` `, y2-` `1` `));` ` ` `ans = Math.max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+` `1` `, y1+` `1` `, y2+` `1` `));` ` ` `return` `(mem[x][y1][y2] = ans);` `}` `static` `int` `geMaxCollection(` `int` `arr[][])` `{` ` ` ` ` ` ` `int` `[][][]mem = ` `new` `int` `[R][C][C];` ` ` `for` `(` `int` `i = ` `0` `; i < R; i++)` ` ` `{` ` ` `for` `(` `int` `j = ` `0` `; j < C; j++)` ` ` `{` ` ` `for` `(` `int` `l = ` `0` `; l < C; l++)` ` ` `mem[i][j][l]=-` `1` `;` ` ` `}` ` ` `}` ` ` ` ` ` ` `return` `getMaxUtil(arr, mem, ` `0` `, ` `0` `, C-` `1` `);` `}` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[][] = {{` `3` `, ` `6` `, ` `8` `, ` `2` `},` ` ` `{` `5` `, ` `2` `, ` `4` `, ` `3` `},` ` ` `{` `1` `, ` `1` `, ` `20` `, ` `10` `},` ` ` `{` `1` `, ` `1` `, ` `20` `, ` `10` `},` ` ` `{` `1` `, ` `1` `, ` `20` `, ` `10` `},` ` ` `};` ` ` `System.out.print(` `"Maximum collection is "` `+` ` ` `geMaxCollection(arr));` ` ` `}` `}`
## Python3
`R` `=` `5` `C` `=` `4` `intmin` `=` `-` `10000000` `intmax` `=` `10000000` `def` `isValid(x,y1,y2):` ` ` `return` `((x >` `=` `0` `and` `x < R ` `and` `y1 >` `=` `0` ` ` `and` `y1 < C ` `and` `y2 >` `=` `0` `and` `y2 < C))` `def` `getMaxUtil(arr,mem,x,y1,y2):` ` ` ` ` `if` `isValid(x, y1, y2)` `=` `=` `False` `:` ` ` `return` `intmin` ` ` ` ` ` ` `if` `x ` `=` `=` `R` `-` `1` `and` `y1 ` `=` `=` `0` `and` `y2 ` `=` `=` `C` `-` `1` `:` ` ` `if` `y1` `=` `=` `y2:` ` ` `return` `arr[x][y1]` ` ` `else` `:` ` ` `return` `arr[x][y1]` `+` `arr[x][y2]` ` ` ` ` ` ` ` ` `if` `x` `=` `=` `R` `-` `1` `:` ` ` `return` `intmin` ` ` ` ` ` ` `if` `mem[x][y1][y2] !` `=` `-` `1` `:` ` ` `return` `mem[x][y1][y2]` ` ` ` ` ` ` `ans` `=` `intmin` ` ` ` ` `temp` `=` `0` ` ` `if` `y1` `=` `=` `y2:` ` ` `temp` `=` `arr[x][y1]` ` ` `else` `:` ` ` `temp` `=` `arr[x][y1]` `+` `arr[x][y2]` ` ` ` ` ` ` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1, y2` `-` `1` `))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1, y2` `+` `1` `))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1, y2))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1` `-` `1` `, y2))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1` `-` `1` `, y2` `-` `1` `))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1` `-` `1` `, y2` `+` `1` `))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1` `+` `1` `, y2))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1` `+` `1` `, y2` `-` `1` `))` ` ` `ans ` `=` `max` `(ans, temp ` `+` `getMaxUtil(arr, mem, x` `+` `1` `, y1` `+` `1` `, y2` `+` `1` `))` ` ` `mem[x][y1][y2] ` `=` `ans` ` ` `return` `ans` `def` `geMaxCollection(arr):` ` ` ` ` ` ` ` ` `mem` `=` `[[[` `-` `1` `for` `i ` `in` `range` `(C)] ` `for` `i ` `in` `range` `(C)] ` `for` `i ` `in` `range` `(R)]` ` ` ` ` ` ` ` ` ` ` `return` `getMaxUtil(arr, mem, ` `0` `, ` `0` `, C` `-` `1` `)` `if` `__name__` `=` `=` `'__main__'` `:` ` ` `arr` `=` `[[` `3` `, ` `6` `, ` `8` `, ` `2` `],` ` ` `[` `5` `, ` `2` `, ` `4` `, ` `3` `],` ` ` `[` `1` `, ` `1` `, ` `20` `, ` `10` `],` ` ` `[` `1` `, ` `1` `, ` `20` `, ` `10` `],` ` ` `[` `1` `, ` `1` `, ` `20` `, ` `10` `],` ` ` `]` ` ` `print` `(` `'Maximum collection is '` `, geMaxCollection(arr))` ` `
## C#
`using` `System;` `class` `GFG` `{` ` ` `static` `readonly` `int` `R = 5;` `static` `readonly` `int` `C = 4;` `static` `bool` `isValid(` `int` `x, ` `int` `y1, ` `int` `y2)` `{` ` ` `return` `(x >= 0 && x < R && y1 >=0 &&` ` ` `y1 < C && y2 >=0 && y2 < C);` `}` `static` `int` `getMaxUtil(` `int` `[,]arr, ` `int` `[,,]mem,` ` ` `int` `x, ` `int` `y1, ` `int` `y2)` `{` ` ` ` ` ` ` `if` `(!isValid(x, y1, y2)) ` `return` `int` `.MinValue;` ` ` ` ` `if` `(x == R-1 && y1 == 0 && y2 == C-1)` ` ` `return` `(y1 == y2)? arr[x, y1]: arr[x, y1] + arr[x, y2];` ` ` ` ` ` ` `if` `(x == R-1) ` `return` `int` `.MinValue;` ` ` ` ` `if` `(mem[x, y1, y2] != -1) ` `return` `mem[x, y1, y2];` ` ` ` ` `int` `ans = ` `int` `.MinValue;` ` ` ` ` ` ` `int` `temp = (y1 == y2)? arr[x, y1]:` ` ` `arr[x, y1] + arr[x, y2];` ` ` ` ` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1, y2-1));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1, y2+1));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1, y2));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1-1, y2));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1-1, y2-1));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1-1, y2+1));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1+1, y2));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1+1, y2-1));` ` ` `ans = Math.Max(ans, temp +` ` ` `getMaxUtil(arr, mem, x+1, y1+1, y2+1));` ` ` `return` `(mem[x, y1, y2] = ans);` `}` `static` `int` `geMaxCollection(` `int` `[,]arr)` `{` ` ` ` ` ` ` `int` `[,,]mem = ` `new` `int` `[R, C, C];` ` ` `for` `(` `int` `i = 0; i < R; i++)` ` ` `{` ` ` `for` `(` `int` `j = 0; j < C; j++)` ` ` `{` ` ` `for` `(` `int` `l = 0; l < C; l++)` ` ` `mem[i, j, l]=-1;` ` ` `}` ` ` `}` ` ` ` ` ` ` `return` `getMaxUtil(arr, mem, 0, 0, C-1);` `}` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[,]arr = {{3, 6, 8, 2},` ` ` `{5, 2, 4, 3},` ` ` `{1, 1, 20, 10},` ` ` `{1, 1, 20, 10},` ` ` `{1, 1, 20, 10},` ` ` `};` ` ` `Console.Write(` `"Maximum collection is "` `+` ` ` `geMaxCollection(arr));` ` ` `}` `}`
## Javascript
```
var R = 5;
var C = 4;
function isValid(x, y1, y2)
{
return (x >= 0 && x < R && y1 >=0 &&
y1 < C && y2 >=0 && y2 < C);
}
function getMaxUtil(arr, mem,
x, y1, y2)
{
if (!isValid(x, y1, y2)) return Number.MIN_VALUE;
if (x == R-1 && y1 == 0 && y2 == C-1)
return (y1 == y2)? arr[x][y1]: arr[x][y1] + arr[x][y2];
if (x == R-1) return Number.MIN_VALUE;
if (mem[x][y1][y2] != -1) return mem[x][y1][y2];
var ans = Number.MIN_VALUE;
var temp = (y1 == y2)? arr[x][y1]:
arr[x][y1] + arr[x][y2];
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1, y2-1));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1, y2+1));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1, y2));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1-1, y2));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1-1, y2-1));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1-1, y2+1));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1+1, y2));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1+1, y2-1));
ans = Math.max(ans, temp +
getMaxUtil(arr, mem, x+1, y1+1, y2+1));
return (mem[x][y1][y2] = ans);
}
function geMaxCollection(arr)
{
var mem = Array(R).fill().map(()=>Array(C).fill().map(()=>Array(C).fill(0)));
for(i = 0; i < R; i++)
{
for(j = 0; j < C; j++)
{
for(l = 0; l < C; l++)
mem[i][j][l]=-1;
}
}
return getMaxUtil(arr, mem, 0, 0, C-1);
}
var arr = [[3, 6, 8, 2],
[5, 2, 4, 3],
[1, 1, 20, 10],
[1, 1, 20, 10],
[1, 1, 20, 10],
];
document.write("Maximum collection is " +
geMaxCollection(arr));
```
Output:
`Maximum collection is 73`
Thanks to Gaurav Ahirwar for suggesting above problem and solution. | 6,522 | 15,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-06 | latest | en | 0.844499 |
https://math.answers.com/questions/Is_417_a_prime_or_composite | 1,653,655,208,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00260.warc.gz | 440,479,190 | 39,042 | 0
# Is 417 a prime or composite?
Wiki User
2017-02-25 02:03:10
417 is a composite number. A Prime number has only 2 factors which are 1 and itself. Composite numbers are everything else except 1 and 0. 1 and 0 are neither prime, nor composite.
Wiki User
2012-02-26 16:48:47
Study guides
21 cards
## What is the prime factorization of 34
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Wiki User
2017-02-25 02:03:10
composite | 138 | 421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | latest | en | 0.84554 |
https://instrumentationtools.com/difference-between-single-phase-and-three-phase-power/ | 1,721,934,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00042.warc.gz | 271,265,068 | 36,453 | # Difference between Single-phase and Three-phase Power
Difference between Single-phase and three-phase power. Advantages one over the other.
The two common forms of alternating electricity are single-phase and three-phase types.
Phase refers to the pattern at which the sinusoidal voltage of the AC supply shifts between its positive maximum and negative maximum.
In electricity, the phrase refers to the distribution of a load.
Following are the distinct systems by which power is distributed to the consumer.
• Single-phase ac supply using a two-wire system.
• Three-phase ac supply using a three-wire system.
• Supply of three-phase and neutral using 4 wire system.
## Common differences
• Single-phase power consists of a two-wire alternating current (ac) power circuit.
• In a single-phase power supply system, there is one wire that is powered the phase wire and the other wire are neutral, with current flowing between the power wire (through the load) and the neutral wire.
• Three-phase power is a three-wire AC power circuit with each phase ac signal 120 electrical degrees apart as shown in the figure above. Each phase of the three-phase power supply is denoted by Red (R), Yellow (Y), Blue (B), or Black.
• The Phase is denoted by L in a single-phase power supply.
• When measuring the single-phase, 220V is measured between the phase and neutral.
• When measuring three-phase, 415 or 420 V is measured between any two-phase to phase connections.
## Supply Voltages
The standard voltages at which the supply authorities deliver the power to the consumer are as follows.
• In the 3 phase or 4 wire power distribution system, power is supplied from the substation along the 4 wires. Three of these wires are called phase and one is usually at zero voltage called neutral wire. The neutral wire is earthed at the substation.
• Residential homes are usually supplied by a single-phase power supply.
• Whereas commercial and industrial establishments usually use a three-phase supply.
• Single-phase power supplies are generally used when typical loads are lighting or heating, rather than large electric motors.
• Three-phase phase power supply is better suitable for higher loads.
Single-phase systems can be derived from three-phase systems.
## Single-phase versus Three-phase Power
An Important difference between 3-phase-power vs. single-phase power is the reliable consistency in the delivery of power.
• In a single-phase system, power is supplied along the two wires. One delivers the current and the other provides the complete return of path. With peaks and dips at in voltage during one phase cycle delivery.
• In a single-phase system, power peaks at 90degrees and 270 degrees. This means that at two points in the cycle power delivery is at maximum. At the other times, the power delivery is at optimum. A single-phase power supply simply does not offer the same stability as a three-phase power supply.
• In a three-phase system, the load is shared across the three power wires. The three power wires are arranged to be out of phase with each other. All three phases of power have entered the cycle by 120 degrees.
• The three phases of power peak in voltage at different times during a complete cycle. By supplying power in this way, there are no peaks and drop-offs.
• Sharing the load between three wires power is supplied constantly. A three-phase power supply can transfer three times as much power as a single-phase power supply.
• Three-phase power, three-phase power supplies are more efficient compared to single-phase power systems.
### Difference between 3-phase and single-phase configurations
• In a single-phase power supplies have a neutral wire, the voltage is measured with respect to the neutral hence it is 220 V.
• In 3 phases, it is measured with respect to another phase because only the star configuration has a neutral wire, and the delta configuration doesn’t have a neutral wire.
Both single-phase and three-phase power distribution systems have different roles. But the two types of systems are totally different from each other.
### Advantages of single-phase over three-phase
• A Single-phase system is cheaper and less complex than a three-phase power system.
• A Single-phase system is suited for residential applications.
### Advantages of three-phase over single-phase
• A large amount of power can be transferred from a three-phase system.
• Three-phase power can be used to energize high power industrial equipment such as motors.
• A much larger amount of power can be transferred to longer distances.
• The three-phase are capable of powering much more at a lower cost. | 937 | 4,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-30 | latest | en | 0.952376 |
https://fxsolver.com/browse/?like=2272&p=8 | 1,708,984,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00481.warc.gz | 262,530,237 | 43,271 | '
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Category | 436 | 2,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.909921 |
https://mcqseries.com/units-dimensions-73/ | 1,708,922,971,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474650.85/warc/CC-MAIN-20240226030734-20240226060734-00683.warc.gz | 391,582,312 | 15,875 | # Units and Dimensions – 73
73. Which physical quantity is represented by M 1 L T –3 A –2 ?
(a) Resistivity
(b) Resistance
(c) conductance
(d) conductivity
Explanation
Explanation : No answer description available for this question. Let us discuss.
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• Units and Dimensions » Exercise - 293. Dimensions of electrical resistance are : (a) M L 2 T –3 A –1 (b) M L 2 T –3 A –2 (c) M L 3 T –3 A –2 (d) M L 1 T –3 A –1
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• Units and Dimensions » Exercise - 269. Dimensional formula for Resistance (R) is ............... (a) M 1 L 1 T –3 A –1 (b) M 1 L 1 T 0 A -1 (c) M 1 L 2 T –3 A –2 (d) M 1 L 0 T –3 A –1
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• Units and Dimensions » Exercise - 270. Dimensional formula for conductance is ............... (a) M –1 L 2 T –3 A 2 (b) M 1 L 2 T –2 A 1 (c) M 1 L –2 T 3 A 2 (d) M –1 L –2 T 3 A 2
Tags: units, dimensions, exercise, conductance, physics | 541 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-10 | latest | en | 0.464145 |
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Home > Preview
The flashcards below were created by user happydoodle on FreezingBlue Flashcards.
1. metric system
decimal system that uses only one basic unit for each type of measurement
2. meter
unit of length; m
3. liter
unit of volume; l
4. gram
unit of mass; g
5. 3 basic units
meter, liter, and gram
6. Very small distance
micrometer (um), also known as micron
nanometer (nm)
*1 um = 0.00001 m; 1 mil um = 1 m; 1 nm = 0.000000001 m
7. Giga, mega, kilo, hecto, and deka
G 1,000,000,000 x base
M 1,000,000 x base
K 1,000 x base
H 100 x base
D 10 x base
8. deci, centi, milli, micro, nano
d 0.1 x base
c 0.01 x base
m 0.001 x base
u 0.000001 x base
n 0.000000001 x base
9. specific gravity
ratio of the mass of a given volume of the substance to the mass of an equal volume of some other substance taken as a standard, g/cm3
10. Conversion of Celsius to Fahrenheit
°F = (1.8)°C + 32°
11. Conversion of Fahrenheit to Celsius
°C = (°F - 32°) / 1.8
12. Conversion of Kelvin to Celsius and C to K
(K-273)°, (C+273)K
13. astronomical unit
unit for measuring distance within solar system, AU
14. light-year
unit for measuring distance to stars and beyond, LY
15. 1 AU =___ kilometers =___ miles
150 million, 93 million
16. 1 LY =___ km =___ miles
9,454,254,955,488, 6 trillion
17. density
mass of substance per unit volume, g/cm3
18. methods of scientific inquiry
1. hypothesis
2. gather data & conduct experiments to validate hypothesis
3. conclusion
19. 1 ml of water equals...
1 cm3 and 1 gram
20. Earth's grid
a system of north-south and east-west lines
21. latitude
North and south distance on Earth
22. parallels of latitude
lines on grid that extend around Earth in an East West direction
23. equator
line on Earth's surface, mark N & S distance
24. longitude
East and West distance on Earth
25. Meridians of longitude
each halves of circles extending from North Pole - South Pole on one side of Earth
26. prime meridian
line on Earth's surface, marking E and W distance
27. the intersection of a parallel of latitude w/meridian of longitude determines...
location of point on Earth's surface
28. precise location
degrees(°), minutes ('), seconds (")
• °; destinations, distance around circle
• '; degrees subdivided into 60 equal parts
• "; minute of angle divided into 60 parts
29. hemispheres
2 equal halves of globe
30. Solar (Sun, time)
uses position of Sun to determine time
31. Standard Time
system used throughout most of the world, divide globe into 24 standard time zones
32. great circle
largest possible circle that can be drawn on a globe
33. small circle
circles on a globe's surface that don't share Earth's center
34. The Equator is the only parallel that is also
a great circle
35. If you observe the north star (Polaris) to be 47 degrees above your north horizon, your latitude is
47 degrees north
36. 2.5 hours solar time is equivalent to __________degrees of longitude
37.5
37. The length of a degree of longitude increases at...
higher latitudes
38. The farthest one can get from the Equator in latitude is...
90 degrees north and 90 degrees south
39. All meridians, when paired with their opposite meridian on Earth, form...
great circles
40. The Tropic of Capricorn lies north of the..
Equator
41. The longitude of the North Pole is...
it has no longitude
42. In terms of longitude, which of the 50 states are furthest east and furthest west?
43. If you observe the noon sun on board your ship at 1:30 PM Greenwich time, your longitude is...
22.5 degrees west
44. At the June solstice the sun can be seen directly overhead only at the...
equator
45. On September 21, the sun's rays are perpendicular to the Earth's surface at...
the Equator
46. In Australia, the spring equinox is marked on...
September 21-22
47. What is your latitude if on June 21 you observe the noon Sun to be 80.5 degrees above your south horizon?
33 degrees north
48. On December 21, where would you go to experience 24 hours of daylight?
South Pole
49. What is the relation between noon Sun angle and amount of solar radiation received at the outer edge of the Earths atmosphere?
50. Variation in the intensity and duration of solar radiation at any given place is largely caused by...
the tilt of the Earths axis of rotation relative to the plane of the ecliptic
51. The maximum noon Sun angle at Bombay India during the year, (19 degrees N.) is...
89 degrees
52. Winter begins in the southern hemisphere on...
June 21-22
53. During the winter months in South Africa, you would observe the noon Sun angle looking to the...
north
54. The Earth is approximately how far from the Sun?
150 million kilometers, 93 million miles, 1 AU
55. map
generalized view of an area
56. cartography
part of geography that embodies mapmaking
57. map projection
transformation of spherical Earth and its latitude-longitude coordinate grid system to a flat surface in some orderly and systematic realignment
58. physical properies of a globe
-parallels are always parallel to each other
-meridians converge at both poles and are eavenly spaced along any individual parallel
-distance between meridians decreases toward poles, w/spacing between meridians at the 60th parallels equal to one-half the equatorial spacing
-parallels and meridians always cross each other at right angles
59. equal area
trait of map projection; indicates equivalence of all areas on surface of the map
60. true shape
map property showing correct configuration of coastlines
61. 3 classes of maps
cylindrical, planar, and conic
62. standard line
contact line between globe and projection surface
63. mercator projection
conformal, true shape
64. rhumb lines
line of constant compass direction, which crosses all meridians at same angle
65. gnomonic projection
possesses a valuable features
66. representative fraction (RF)
diameter of globe/diameter of Earth = 61 cm/ 12,756 cm = 1 cm/ 20,911,000 cm
67. 4 types of map distortion
size/area, shape, distance, direction
CAN'T HAVE ALL
68. ratio of map scale
indicating how much distance on Earth a certain distance on the map represents
69. map scale displayed in 3 ways
written scale, representative fraction, graphic scale
70. large vs. small scale
large scale shows less area but that area is more detailed on map
small shows big area but with less detail
1/500 = 0.002, bigger
1/3 = 0.003333, smaller
71. weather
state of atmosphere at a particular place for a short period of time
72. four basic elements of weather
temperature, moisture, air pressure, and wind
73. Sun's intensity
the angle at which the rays of sunlight strike a surface
74. Sun's duration
length of daylight
75. standard unit of solar radiation
langley = 1 calorie
76. solar constant
average intensity of solar radiation falling on a surface perpendicular to the solar beam at the outer edge of the atmosphere
77. Tropic of Cancer
23.5 degrees N latitude
78. Tropic of Capricorn
23.5 degrees S latitude
79. equinox
March 20-21 and September 22-23 when circle of illuminations divides Earth through the poles and all places have days and nights of equal length
80. analemma
graph than can be used to determine the latitude where the overhead noon Sun is located for any date
81. minerals
naturally occurring, inorganic solid with an orderly internal arrangement of atoms and definite, but not fixed, chemical composition
*can contain only one substance, some have economic value and are used in various industries
82. the main type of distortion in a Mercator projector map distorts is...
size/area
83. the main type of distortion in a Lambert projection map is...
shape
84. the only place on a map where there is no distortion is called...
standard line
85. Optical Properties of Minerals
Luster, Transmission of light, Color, Streak
86. Luster
describes manner in which light is reflected from the surface of the mineral
metallic or nonmetallic
87. Transmission of Light
ability of mineral to transmit light
• -opaque: no light transmitted
• -translucent: light but no image transmitted
• -transparent: image visible through mineral
88. Color
*least reliable physical properties because it can be misleading
89. Streak
color of the fine powder of a mineral obtained by rubbing a corner across a piece of porcelain (streak plate)
90. Hardness
measure of resistance of a mineral to abrasion and scratching
*measured on Mohs scale
91. Crystal Forms
external appearance when a mineral forms with no space restrictions
• -flat surfaced called crystal faces
• -distinctive number of face
92. Cleavage
how surfaces breaks
• -smooth, flat surfaces called cleavage planes
• -parallel planes counted only once
93. Specific Cleavage
indicator of weight of mineral
*average specific value is 2.7
94. Other Properties of Minerals
• Magnetism
• Feel
• Taste
• Striations (straight lines)
• Tenacity
• Reaction to Dilute Hydrochloric Acid
95. Rocks
aggregates of minerals
96. 3 types of rocks
Igneous, Sedimentary, and Metamorphic
97. Igneous
solidified products of once molten material (magma); interlocking arrangement of mineral crystals that forms as molten material cools and crystal grows
98. Sedimentary
formed at or near Earth's surface from weathering
*lithification, transforms sediment into hard rock
99. Metamorphic
form below Earth's surface where high temperatures, pressures, and/or chemical fluids change preexisting rocks without melting them
100. Texture of rock
shape, arrangement, and size of mineral grains in a rock
101. Composition of rock
minerals that are found in a rock
102. Textures of Igneous Rocks
• -Coarse Grained: occurs when magma cools slowly inside Earth
• -Fine Grained: develops when molten material cools quickly on, or very near, surface of Earth
• -Porphyritic: caused by magma having two different rates of cooling; phenocrysts and groundmass
• -Glassy: develops because of rapid cooling
• -Fragmental: rock contains broken fragments produced during volcanic eruption
103. Composition of Igneous Rocks
• -Felsic: light-colored minerals quartz, granitic
• -Intermediate: 15% - 45% of darkness
• -Mafic: over 45% of darkness
• -Ultramafic: ultra dark
104. Sedimentary Rock Identification
Detrital material: derived from process of mechanical weathering transported and deposited as solid particles
Chermical material: previously dissolved in water and later precipitated by either inorganic or organic processes
105. Metaphoric Rock divided into two groups
foliated and nonfoliated
106. foliation
unique to many metaphoric rocks and gives them a layered or banded appearance
### Card Set Information
Author: happydoodle ID: 206846 Filename: Geography Updated: 2013-03-13 08:58:47 Tags: geography Folders: Description: exam Show Answers:
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Question 625861: Write an equation of the line containing the points (2, -3) and (4, 5). Click here to see answer by jim_thompson5910(28595)
Question 625861: Write an equation of the line containing the points (2, -3) and (4, 5). Click here to see answer by lenny460(775)
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Question 626279: How do I determine i the following system of equation is parallel, perpendicular, or coincide? 10x-12y=20 and 5x-6y=4 Click here to see answer by abebemichael8@gmail.com(1)
Question 626410: can you solve for y in terms of x a. 3x-2y=3 b. 4y+6x=12 c. 2x+3y=7 d. 6x+2y=-10 e. 6y+2x=6 f. y-2=2(x+2) g. 3(x+4)=y-8x+3 h. 2y+4x=3(y-x)18 Click here to see answer by Alan3354(30993)
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# 英文數學問題4題
express y as a function of x
1. y=3xy+5
solve each equation or inequality
2. x+3x=4x
3. x-0.04(x-10)=96.4
4. the perimeter of a rectangle is 84 meters. if the width is 16 meters less than the length, then what is the width of the rectangle?
Rating
• 9 years ago
express y as a function of x
1. y=3xy+5
y-3xy= 5
(1-3x)y =5
y=5/(1-3x)
solve each equation or inequality
2. x+3x=4x
4x = 4x
3. x-0.04(x-10)=96.4
x-0.04x+0.4=96.4
0.96x=96
x=100
4. the perimeter of a rectangle is 84 meters. if the width is 16 meters less than the length, then what is the width of the rectangle?
The width of the rectangle is 13 meters.
• heidi
Lv 7
9 years ago
1.
y-3xy=5
(1-3x)y=5
y=5/(1-3x)
2.
4x=4x
0x=0
x為任意數
3.
x-0.04x+0.4=96.4
0.96x=96
x=100
4.
設length為x
width為x-16
2[x+(x-16)]=84
x+(x-16)=42
2x-16=42
2x=58
x=29
width為29-16=13 | 405 | 904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-10 | latest | en | 0.688063 |
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# Stephen
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Solve all the problems in Modeling and Simulation Challenge problem group. | 1,244 | 5,371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-50 | latest | en | 0.770109 |
https://www.datasciencecentral.com/bernouilli-lattice-models-connection-to-poisson-processes/ | 1,723,413,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00135.warc.gz | 569,520,600 | 36,346 | Bernoulli Lattice Models – Connection to Poisson Processes
Bernouilli lattice processes may be one of the simplest examples of point processes, and can be used as an introduction to learn about more complex spatial processes that rely on advanced measure theory for their definition. In this article, we show the differences and analogies between Bernouilli lattice processes on the standard rectangular or hexagonal grid, and the Poisson process, including convergence of discrete lattice processes to continuous Poisson process, mainly in two dimensions. We also illustrate that even though these lattice processes are purely random, they don’t look random when seen with the naked eye.
We discuss basic properties such as the distribution of the number of points in any given area, or the distribution of the distance to the nearest neighbor. Bernouilli lattice processes have been used as models in financial problems, see here. Most of the papers on this topic are hard to read, but here we discuss the concepts in simple English. Interesting number theory problems about sums of squares, deeply related to these lattice processes, are also discussed. Finally, we show how to identify if a particular realization is from a Bernouilli lattice process, a Poisson process, or a combination of both.
See below a realization of a Bernouilli process on the regular hexagonal lattice. The main feature of such a process is that the point locations are fixed, not random. But whether a point is “fired” or not (that is, marked in blue) is purely random and independent of whether any other point is fired or not. The probability for a point to be fired is a Bernouilly variable of parameter p
Figure 1: realization of Bernouilli hexagonal lattice process
The source code to produce this plot is available here. More sophisticated models, known as Markov random fields, allows for neighboring points to be correlated. They are useful in image analysis, see here and here.
To the contrary, Poisson processes assume that the point locations are random. The points being fired are uniformly distributed on the plane, and not restricted to integer or grid coordinates. In short, Bernouilli lattice processes are discrete approximations to Poisson processes. Below is an example of a realization of a Poisson process.
Figure 2: realization of a Poisson process
1. Definition and Convergence to Poisson Process
We are dealing here with square lattices covering the entire two-dimensional plane. A point of the lattice is a vector (rx, ry) where x, y are integers (positive or negative) and r a strictly positive real number, called the scale of the lattice. A Bernouilli variable is attached to each point: it is equal to 1 with probability p, and to 0 with probability 1 – p. A point of the lattice is said to be fired if its Bernouilli variable is equal to 1. Such points are marked in blue in all the plots. The Bernouilli variables are independent and have the same parameter p: in other words, this 2-parameter lattice process is homogeneous, as p does not depend on the location.
We compare these processes to homogeneous Poisson process of intensity λ, covering the entire plane. For Poisson processes, point locations are random, see figure 2. The number of points in any given area D has a Poisson distribution with mean λ S(D), where S represents the surface of the area. The distance to the nearest neighbor has the following distribution, see here
1.1. Poisson approximation to Bernouilli lattice process
In the same way that Bernouilli variables are approximated by Poisson variables when p is small, Bernouilli processes are approximated by Poisson processes when p is small. The parameters p and λ play the same role.
The way convergence occurs is very intuitive. Start with a Bernouilli lattice process as pictured in figure 3, with r = 1. Reduce r by a factor 2 and p by a factor 4: see result in figure 4. Keep repeating this step indefinitely, and you end up with figure 2.
Figure 3: Realization of a Bernouilli square lattice process
Figure 4: A more granular version, with smaller r and p
2. Properties of Bernouilli lattice processes
As for Poisson processes, two main features are:
• the distribution between a point of the lattice and its closest “blue” neighbor
• the distribution of the number of blue points in any given area.
The number N of blue points in any given area D is a Binomial variable with expectation p S(D), with S(D) being the total number of lattice points (blue or not) in D. That is,
For any lattice point z, let Y be the distance to the closest blue neighbor. Let D be a circle centered at z, with radius y. The probability that Y is larger than y is the probability that there is no blue point in D, that is, based on the previous formula:
Note that if z is a blue point, it is not considered to be a nearest neighbor to itself.
2.1. More about the Poisson approximation
When r and p get smaller and smaller as discussed in section 1.1, we have convergence to the Poisson process, and S(D) in the above formula (for Bernouilli lattice processes) tends to the surface area of D, that is
Likewise, for the Poisson process, based on the first formula in section 1, when λ is small, we have
Now the analogy with Poisson processes becomes more clear. Also, p and λ are equivalent, at the limit. Below is the distribution of Y for a Bernouilli lattice process with p = 0.001.
Figure 5: Distribution of distance to nearest neighbor: P(Y < y)
It is indistinguishable from that of a Poisson process with λ = 0.001. There is a major difference though: the distribution is a discrete one for the lattice process, and a continuous one for the Poisson process, but you can’t tell the difference with the naked eye, for such a small value of p
2.2. Model testing
Since Poisson and Bernouilli lattice processes can be quite similar depending on the granularity of the lattice, one might be interested in testing if some observed data fits with either model. An easy test involves computing the empirical distribution of the distance to the nearest neighbor, and check whether it fits the Bernouilli lattice or Poisson theoretical distribution.
For blended models, say a mixture of Poisson and Bernouilli lattice, one can do simulations. Simulations will also allow you to estimate the proportions of Bernouilli versus Poisson in the mixture model.
2.3. Connection to number theory: sums of two squares
An interesting result that can easily be derived from the formulas in section 2.1 is the following. The number of (x, y) solution to x^2 + y^2 n, where n is a positive integer and the unknown x, y are positive or negative integers, tends to π n as n tends to infinity. This is known as the Gauss circle problem. So, on average, the equation x^2 + y^2 = n has π solutions, though for some n, for instance n = 1105, it can be much higher (32 in this case). Source code to compute the number of solutions is available here. More generally, the number of integer solutions to such inequalities can be approximated by the surface area under the curve using integration techniques, the curve here being x^2 + y^2 n.
A deeper result is the following. Let r(n) be the number of solutions to x^2 + y^2 = n. This function is known as the sum of squares function. For k between n and 2n, we have
integers such that r(k) > 0, based on this well known result, where c = 0.7642… is the Landau-Ramanujan constant. Also, based on the Gauss circle problem, the total number of solutions across all k‘s between n and 2n is asymptotically 2π nπ n = π n. These π n solutions are distributed across cn / SQRT(log n) integers satisfying r(k) > 0. This means that for large n‘s satisfying r(n) > 0, the average value of r(n) grows as π SQRT(log n) / c. A consequence of this is that the records for r(n) grow faster than π SQRT(log n) / c. How much faster?
Technical note: Sums of squares are far more abundant than primes, yet their natural density is also zero. This means that as n becomes larger and larger, the chance for n to be a sum of two squares, tends to zero. But for the few remaining n‘s that can be expressed as a sum of two squares, the number of ways they can be expressed that way, increases more and more on average. That is, r(n) tends to increase with n, for those few large n satisfying r(n) > 0. As a result, for any n large or small, the number of solutions to x^2 + y^2 = n is equal to π on average What changes (it increases) when n is large is the probability that r(n) = 0, but the average value of r(n) remains unchanged. | 1,976 | 8,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.922038 |
https://statsidea.com/en/python/easy-methods-to-develop-a-density-plot-in-matplotlib-with-examples/ | 1,723,666,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00748.warc.gz | 423,637,110 | 18,973 | # Easy methods to Develop a Density Plot in Matplotlib (With Examples)
The best way to develop a density plot in Matplotlib is to importance the kdeplot() serve as from the seaborn visualization library:
```import seaborn as sns
#outline information
information = [value1, value2, value3, ...]
#develop density plot of knowledge
sns.kdeplot(information)
```
Please see examples display easy methods to importance this serve as in apply.
### Instance 1: Develop Unsophisticated Density Plot
Please see code displays easy methods to develop a modest density plot in seaborn:
```import seaborn as sns
#outline information
information = [2, 2, 3, 5, 6, 6, 7, 8, 9, 10, 12, 12, 13, 15, 16]
#develop density plot of knowledge
sns.kdeplot(information)```
The x-axis displays the information values and the y-axis displays the corresponding anticipation density values.
### Instance 2: Modify Smoothness of Density Plot
You’ll importance the bw_method argument to regulate the smoothness of the density plot. Decrease values supremacy to a extra “wiggly” plot.
```import seaborn as sns
#outline information
information = [2, 2, 3, 5, 6, 6, 7, 8, 9, 10, 12, 12, 13, 15, 16]
#develop density plot of knowledge with low bw_method price
sns.kdeplot(information, bw_method = .3)```
Conversely, upper values for bw_method supremacy to a smoother plot:
```import seaborn as sns
#outline information
information = [2, 2, 3, 5, 6, 6, 7, 8, 9, 10, 12, 12, 13, 15, 16]
#develop density plot of knowledge with prime bw_method price
sns.kdeplot(information, bw_method = .8)```
### Instance 3: Customise Density Plot
You’ll additionally customise the style and color of the density plot:
```import seaborn as sns
#outline information
information = [2, 2, 3, 5, 6, 6, 7, 8, 9, 10, 12, 12, 13, 15, 16]
#develop density plot of knowledge with prime bw_method price
sns.kdeplot(information, colour="crimson", fill=True, alpha=.3, linewidth=0)``` | 558 | 1,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.687181 |
https://bestbinaryigglvot.netlify.app/lauderback23008fyr/cfd-contract-value-puta.html | 1,679,545,918,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00497.warc.gz | 159,581,895 | 11,980 | ## Cfd contract value
2 Dec 2019 The amount of Margin required for each type of CFD is listed on. OANDA's website. However, if you choose to set a higher Margin requirement The CFD is a tradable contract between a client and the broker, who are exchanging the difference in the initial price of the trade and its value when the trade is unwound or reversed.
While trading on margin allows you to magnify your returns, your losses will also be magnified as they are based on the full value of the CFD position. A contract for difference, or CFD, is an over-the-counter (OTC) contract between two parties whereby one party pays the other party an amount determined by the CFD. A Contract for Difference (CFD) is a contract between two parties who asset and its value at the time he entered the contract (if this difference is negative, The amount of initial margin required to be deposited in the customers' account prior to trading can be small relative to the value of the contract. A relatively small
## 12 Jan 2020 CFDs are cash-settled but use allow ample margin trading so that investors need only put up a small amount of the contract's notional payoff.
Contract size — Equivalent to the traded amount on the Forex or CFD market, which is calculated as a standard lot size multiplied with lot amount. The Forex standard lot size represents 100,000 units of the base currency. For CFDs and other instruments see details in the contract specification. In finance, a contract for difference (CFD) is a contract between two parties, typically described as "buyer" and "seller", stipulating that the buyer will pay to the seller the difference between the current value of an asset and its value at contract time (if the difference is negative, then the seller pays instead to the buyer). Notional value = Contract size x Spot price For example, one soybean contract is comprised of 5,000 bushels of soybeans. At a spot price of \$9, the notional value of a soybean futures contract is \$45,000, or 5,000 bushels times the \$9 spot price. Specifically, the fair value is the theoretical calculation of how a futures stock index contract should be valued considering the current index value, dividends paid on stocks in the index, days Contract for Difference (CFD) refers to a contract that enables two parties to enter into an agreement to trade on financial instruments based on the price difference between the entry prices and closing prices. It means the contract enables the seller to pay the buyer the variance between the entry value of the asset A contract for difference (CFD) is a derivative financial instrument that allows traders to invest in an asset without actually owning it. Very popular with investors for hedging risk in volatile markets, CFDs allow traders to speculate on the rising or falling prices of assets, such as shares, currencies, commodities, indexes, etc. What is a contract for difference? Looking for a CFD definition? The term CFD stands for a ‘contract for difference’ – an agreement, typically between a broker and an investor, that one party will pay the other the difference between the value of a security at the start of the contract, and its value at the end of the contract.
### An Overview of CFD TradingHow Does CFD Trading Work?CFD Trading: Useful Terms & DefinitionsContract ValueDemo AccountLeverage MarginLimit
Contracts for Difference or CFD allow you to speculate on future price movements of in the current value of a share or index and its value at the contract's end. Means you only put down a fraction of the value of your trade. You instead buy a certain number of CFD contracts* (also called units) on a market if you expect CFD stands for Contracts for Difference, with the difference being between a contract from AxiTrader that will increase in value if the Gold price increases. A contract for difference (CFD) is essentially a contract between an investor and and/or receive the difference between the buying and selling contract values. | 844 | 4,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.945556 |
https://conversioncalculator.org/spelling-of-90/ | 1,701,722,597,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00592.warc.gz | 216,955,872 | 48,656 | # Spelling of 90
Ninety
90 in words is written as Ninety.
90 in Words can be written as Ninety. If you have saved 90 dollars, then you can write, “I have just saved Ninety dollars.” Ninety is the cardinal number word of 90 which denotes a quantity.
• 90 in Words = Ninety
• Ninety in Numbers = 90
Let us write the given number in the place value chart.
Tens Ones 9 0
We see that there are 0 ‘ones’, 9 ‘tens’. Now read the number from right to left along with its place value. 90 in words is written as Ninety.
## How to Write 90 in Words?
Using the place value chart we identify the place for each digit in the given number and write the numbers in words. For 90 we see that the digits in units = 0, tens = 9. Therefore 90 in words is written as Ninety.
Problem Statements:
How to Write 90 in Words? Ninety Is 90 an Odd Number? No What is 90 Decimal to Binary? (90)₁₀ = (1011010)₂ Is 90 an Even Number? Yes Is 90 a Perfect Cube? No Is 90 a Perfect Square? No Is 90 a Prime Number? No What is the Square Root of 90? 9.486833 Is 90 a Composite Number? Yes | 319 | 1,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.8818 |
http://www.programmersheaven.com/mb/ctocplusplustomatlab/413639/413641/re-loop-help-simple/?S=B20000 | 1,371,690,842,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709337609/warc/CC-MAIN-20130516130217-00024-ip-10-60-113-184.ec2.internal.warc.gz | 644,203,039 | 9,421 | ## Matlab
Moderators: None (Apply to moderate this forum)
Number of posts: 2174
This Forum Only
loop HELP "simple" Posted by jmed on 18 Feb 2010 at 2:08 PM
Generate a sequence of numbers. Start with integer n. If n is even divide by 2. If n is odd, multiply by 3 and add 1. Repeat process with new value of n terminating when n = 1.
Ex. n = 11 would produce
11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
Re: loop HELP "simple" Posted by jjasso5 on 18 Feb 2010 at 5:43 PM
x = input('Give me an integer: ');
while x ~= 1
if rem (x, 2) == 0
x = x/2
else
x = x*3 + 1
end
end
Re: loop HELP "simple" Posted by jmed on 18 Feb 2010 at 5:58 PM
It also says to save the generated sequence of numbers into a vector named seq. Display the variable seq.
EX:
>>project5
n = 3
seq =
3 10 5 16 8 4 2 1
Re: loop HELP "simple" Posted by jmed on 18 Feb 2010 at 5:58 PM
It also says to save the generated sequence of numbers into a vector named seq. Display the variable seq.
EX:
>>project5
n = 3
seq =
3 10 5 16 8 4 2 1
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Assembly | Basic | C | C# | C++ | Delphi | Flash | Java | JavaScript | Pascal | Perl | PHP | Python | Ruby | Visual Basic | 430 | 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} | 2.796875 | 3 | CC-MAIN-2013-20 | latest | en | 0.762491 |
http://spmaddmaths.onlinetuition.com.my/2015/08/72-probability-of-combination-of-two.html | 1,505,866,499,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686077.22/warc/CC-MAIN-20170919235817-20170920015817-00019.warc.gz | 328,463,937 | 22,573 | # 7.2 Probability of the Combination of Two Events
7.2 Probability of the Combination of Two Events
1. For two events, A and B, in a sample space S, the events AB (A and B) and A υ B
(A or B) are known as combined events.
2. The probability of the union of sets A and B is given by:
3. The probability of the union of sets A and B can also be calculated using an alternative method, i.e.
4. The probability of event A and event B occurring, P(AB) can be determined by the following formula.
Example:
Given a universal set ξ = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. A number is chosen at random from the set ξ . Find the probability that
(a) an even number is chosen.
(b) an odd number or a prime number is chosen.
Solution:
The sample space, S = ξ
n(S) = 14
(a)
Let A = Event of an even number is chosen
A = {2, 4, 6, 8, 10, 12, 14}
n(A) = 7
(b) Let,
B = Event of an odd number is chosen
C = Event of a prime number is chosen
B = {3, 5, 7, 9, 11, 13, 15} and n(B) = 7
C = {2, 3, 5, 7, 11, 13} and n(C) = 6
The event when an odd number or a prime number is chosen is B υ C.
P(B υ C) = P(B) + P(C) – P(B C)
B C = {3, 5, 7, 11, 13}, n(B C) = 5 | 420 | 1,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-39 | longest | en | 0.922273 |
https://madhavamathcompetition.com/tag/rmo/page/2/ | 1,571,271,858,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00356.warc.gz | 593,948,958 | 9,963 | # Training yourself for any Math Olympiad — RMO, INMO, IMO
Although you might have an expert coach or branded institution coaching you for the math or physics olympiads, the best thing is to be your own guru. What are the attitudes and/or regimen (of mind) needed to soar up your performance in Math or Physics Olympiads? I think the same applies for IIT JEE too, but perhaps, to a lesser degree. The following are some tips, which I like and I have compiled them from the net (especially American Math Olympiad websites) (especially, Prof. Kiran Kedlaya, MIT, Boston):
The term “olympiad” is used generically to refer to a math contest in which students are asked not to compute numerical answers, but to give proofs of specified statements. (Example: “Prove that 2003 is not the sum of two squares of integers.”) The most famous example is the International Mathematical Olympiad; most countries that participate at the IMO have national olympiads as part of their team selection process. Some areas have additional olympiads at the regional or local level.
The jump from short answers to olympiads is a tough one. Here are some tips for students making this transition.
• Practice, practice, practice. The only way to learn math is by doing.
• Proofs are essays. The better written a proof is, the more likely it is to be understood. Even such mundane things as grammar, spelling and handwriting are worth a bit of attention.
• Define your terms. If you’re going to use a word in a way that might not be commonly understood, define it precisely. Then stick to your definition!
• Read the masters. No one ever learned how to do good mathematics in a vacuum. When you do practice problems, read the solutions even of the problems you solved.
• There’s more than one road. Different solutions can be equally valid; even when solutions agree in substance, differences in perspective can be significant and valuable.
• It’s not over when it’s over. Don’t hesitate to continue thinking about the problems on a contest after the time ends, or to discuss the problems with others.
• Learn from your peers. They’re smarter than you might have expected.
• Learn from the past. Try to relate new problems to old ones; you may learn something from the similarities, or from the differences.
• Patience. No one said this was easy!
If you like this, please do send a thank you note to Prof. Kiran Kedlaya (kedlaya ‘at’ mathdotmitdotedu) :-))
More later,
Nalin Pithwa | 554 | 2,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.935415 |
https://brainly.in/question/309630 | 1,484,787,770,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00339-ip-10-171-10-70.ec2.internal.warc.gz | 789,143,499 | 10,186 | 2016-03-23T14:32:48+05:30
Example -
Maths - A1 = 10
Science - A1 = 10
Hindi - A2 = 9
English- A2 = 9
SST - B1 = 8
Now, find the average of all grade points = 10+10+9+9+8/5
= 46/5 = 9.2
yeah avg of that
But I did not think , that CGPA IS calculated by getting avg
naah avg of all grade points
Culminated grade point avg - full form of cgpa
Hw can we find the grade points of each subject | 149 | 388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-04 | latest | en | 0.901321 |
https://www.macmillanihe.com/companion/Sarantakos-Social-Research/workbook/Chapter-07/ | 1,571,179,585,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00313.warc.gz | 955,769,400 | 11,333 | .button { text-transform: none !important; }
# Social Research
## Fourth edition
### Chapter 7 - Sampling Procedures
Explore this chapter's resources further:
Educational objectives
After completing this chapter, you will:
1. be able to explain the nature and process of sampling in social research;
2. understand the main types of sampling and the areas in which they can be applied;
3. have knowledge of the advantages and disadvantages of sampling in general and of the sampling methods in particular;
4. be in a position to apply sampling in actual situations;
5. have a clear understanding of the ways in which sampling procedures are employed in qualitative and quantitative research, and their logical and methodological foundation.
### Contents
1. Reasons for sampling
2. Principles of sampling
3. Types of sampling
4. Probability (random) sampling
5. Non-probability sampling
6. Sampling procedures in qualitative research
7. Internet sampling
8. Sample size
top
### Points to remember
The following are the major points introduced in this chapter. Ensure that you are very confident with their meaning, content, context and significance.
1. Sampling is the process of choosing the respondents and the units of the study in general.
2. Sampling is a common practice and an indispensable research tool in social sciences.
3. Sampling, as the alternative to conducting a saturation survey, offers many advantages.
4. Sampling units must be chosen objectively and systematically, must be easily identifiable and clearly defined, independent from each other, not interchangeable, and free of errors, bias and distortions.
5. The two distinct types of sampling are probability and non-probability sampling.
6. In a probability sampling, all units have an equal, calculable and non-zero probability to be included in the sample.
7. Non-probability sampling does not adhere to the rules of probability.
8. The two types of probability sampling are simple random sampling and systematic sampling.
9. In a simple random sampling all units of the target population have an equal chance of being selected.
10. The three most common techniques of selection used in simple random sampling are the lottery method, the method of random numbers and the computer method.
11. In systematic sampling although all units have an equal chance of being selected, their selection depends on the choice of other units.
12. Systematic sampling employs the sampling fraction method of choosing the respondents.
13. Stratified random sampling is the procedure in which the sample is chosen after the target population is divided in a number of strata, from which the respondents are taken.
14. Cluster sampling is the procedure in which in the first instance clusters are chosen.
15. In multi-stage sampling, samples are chosen in stages: firstly one sample is taken and then a second or third sample is chosen from within the previous sample.
16. In multi-phase sampling the procedure followed is the same as in multi-stage sampling with the difference that in each stage of sampling data is collected.
17. Area sampling is the procedure in which the choice of respondents is related to geographical areas. An area is divided into smaller sections, progressively leading to smaller samples and ultimately to the respondents.
18. Panel samples include a number of respondents chosen in a systematic way and subjected to data collection on more than one occasion.
19. Spatial sampling is a procedure in which a sample is taken from people temporarily congregated in space.
20. Accidental sampling is a non-probability sampling procedure in which the researcher chooses a number of respondents at will. It is also called convenience sampling, chunk sampling, grab sampling or haphazard sampling.
21. In purposive sampling the researcher chooses the respondents who are thought to serve the purpose of the study. It is also called judgemental sampling.
22. Quota sampling is the procedure in which the researcher chooses a quota of respondents set by the project manager
23. Snowball sampling is a procedure in which the selection of additional respondents is guided by respondents who have already been studied.
24. In theoretical sampling the choice of respondents is guided by the emerging theory.
25. Sampling procedures are employed in all quantitative studies but elements of sampling are found also in qualitative research.
26. Non-response is a serious research problem that investigators must deal with.
27. Sample size is computed by means of statistical and non-statistical procedures.
top
1. Give a brief description of the nature and purpose of sampling in quantitative and qualitative research.
2. What is a complete coverage study?
3. What is meant by a 'saturation survey'?
4. Describe briefly the differences between target population and survey population.
5. List the main reasons for employing sampling procedures.
6. In what ways can a study based on sampling offer more detailed information than a saturation study?
7. Discuss briefly the main principles of sampling.
8. What are the basic types of sampling? List their main similarities and differences.
9. What are the most common types of probability sampling?
10. Describe the method of random numbers in sampling.
11. How are samples established when using the lottery method of selection?
12. What are the main features of simple random sampling?
13. What are the main attributes of systematic sampling?
14. What are the main differences between systematic sampling and simple random sampling.
15. Describe briefly the nature and use of 'sampling frames' in social research.
16. Explain the nature and use of the sampling fraction method.
17. Describe the nature and types of stratified sampling.
18. Explain the nature and use of cluster sampling.
19. Describe briefly the nature and purpose of multi-phase sampling.
20. What is an area sampling and how useful is it for social research?
21. Describe the nature and types of longitudinal studies.
22. What are the main criteria of non-probability sampling?
23. What are the main types of non-probability sampling?
24. How is a sample chosen when the method of accidental sampling is used?
25. What is the nature and purpose of purposive sampling?
26. Describe briefly the selection of a sample when using the process of quota sampling.
27. Explain the nature and purpose of snowball sampling.
28. What are the main differences between quantitative and qualitative sampling procedures?
29. What are the main types of sampling that are employed in qualitative social research?
30. Describe briefly the process and purpose of theoretical sampling.
31. What are the main characteristics of sampling procedures employed by qualitative research?
32. What is meant by non-response in social research and how is it controlled?
33. What are the ways of estimating the ideal sample size?
top
### Fill-in questions
Click here to launch fill-in questions for chapter 7
### True/false questions
Click here to launch true/false questions for chapter 7
### Multiple choice questions
Click here to launch multiple choice questions for chapter 7
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### Practical exercises
1. You have been asked to investigate the views of nurses working in ten large hospitals to establish their views as to whether health workers, and medical doctors in particular, should be allowed to strike.
a) What is the most appropriate sampling procedure for this project and why?
b) How would you choose the respondents if the chosen procedure was quota sampling?
2. In the above example, the research committee is interested in the views of nurses to 'striking doctors'. Explain how the nurses will be chosen if theoretical sampling is employed.
3. You have been appointed the chief researcher of a country town. Your task is to ascertain the attitudes of unmarried mothers to state policies relating to family allowances.
a) Explain how you will choose the sample of mothers to address the research question, using a quantitative study.
b) Explain how you will choose the sample of mothers to address the research question, using a qualitative study.
4. As the research adviser of the educational division of the university you have been asked to investigate the views of students to the proposed increase in fees, which resulted from relevant government cuts. The sampling method you have been urged to use by the manager of the division is multi-stage sampling.
a) Describe the steps you will employ to select your respondents by using this method.
b) Discuss critically the suitability of such a sampling method in the context of the proposed study. Is this the sampling method you would have chosen if you had been free to decide on the matter?
top | 1,686 | 8,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-43 | longest | en | 0.908653 |
https://www.circuitdiagram.co/vco-schematic-diagram/ | 1,701,967,592,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00685.warc.gz | 789,925,383 | 12,144 | # Vco Schematic Diagram
By | September 2, 2017
Are you interested in finding out more about VCO Schematic Diagrams? If so, then you’ve come to the right place! VCOs (Voltage Controlled Oscillators) are essential components of many electronic devices, including cell phones and audio systems. As the name suggests, they rely upon an adjustment of voltage to determine their frequency output. This makes them ideal for use in signal processing tasks.
A VCO schematic diagram is a drawing that displays the electrical connections and layout of the VCO circuit. It is beneficial for those working with electronics, providing a highly detailed depiction of the component layout and connections. A well-crafted signal diagram can help technicians quickly identify and troubleshoot any issues in the circuit.
VCO schematic diagrams are often used alongside other diagrams such as logic diagrams and block diagrams. Logic diagrams provide a simplified overview of the circuit’s electronic components, while block diagrams describe the flow of data during signal processing. Together, these three types of diagrams provide a comprehensive insight into the function and structure of a given circuit.
To create a VCO schematic diagram, you’ll need to have an understanding of basic electronics and electrical engineering principles. Many online tutorials and books provides step-by-step instructions on how to construct your own VCO diagram. Additionally, there are also several software programs available that you can use to generate your diagram.
In conclusion, VCO schematic diagrams provide invaluable insights into the inner workings of these important electronic components. They are essential tools for any technician or engineer who works with electronics, allowing them to quickly identify and troubleshoot circuit issues. With a bit of practice and the right resources, anyone can start creating their own VCO diagrams.
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Rf Tutorial Lesson 15 Exploring Phase Locked Loops Emagtech Wiki | 625 | 3,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.912813 |
http://www.numbersaplenty.com/629503 | 1,590,494,517,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00302.warc.gz | 189,949,244 | 3,362 | Search a number
629503 = 7229443
BaseRepresentation
bin10011001101011111111
31011222111221
42121223333
5130121003
621254211
75231200
oct2315377
91158457
10629503
1139aa56
12264367
131906b4
141255a7
15c67bd
hex99aff
629503 has 12 divisors (see below), whose sum is σ = 759240. Its totient is φ = 519792.
The previous prime is 629491. The next prime is 629509. The reversal of 629503 is 305926.
It is a de Polignac number, because none of the positive numbers 2k-629503 is a prime.
It is a Duffinian number.
It is a plaindrome in base 16.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (629509) by changing a digit.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 1200 + ... + 1642.
It is an arithmetic number, because the mean of its divisors is an integer number (63270).
2629503 is an apocalyptic number.
629503 is a deficient number, since it is larger than the sum of its proper divisors (129737).
629503 is a wasteful number, since it uses less digits than its factorization.
629503 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 486 (or 479 counting only the distinct ones).
The product of its (nonzero) digits is 1620, while the sum is 25.
The square root of 629503 is about 793.4122509768. The cubic root of 629503 is about 85.7036400800.
The spelling of 629503 in words is "six hundred twenty-nine thousand, five hundred three". | 443 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-24 | latest | en | 0.875791 |
https://mathematica.stackexchange.com/questions/269767/very-simple-unexpected-low-performance-of-set | 1,656,329,175,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00507.warc.gz | 439,193,600 | 66,601 | # Very simple, unexpected Low Performance of set(=)
In[1] AA = RandomInteger[10, 1000000]; BB = AA;
In[2] Do[BB = AA;, 10000] // Timing
Out[2] {0., Null}
In[3] Do[BB[[77]] = i;, {i, 1, 10000}] // Timing
Out[3] {0., Null}
In[4] Do[BB[[i]] = i;, {i, 1, 10000}] // Timing
Out[4] {0., Null}
From Out[2], we see setting a variable to a large list takes no time.
From Out[3],Out[4], we see setting an element of a list to a number takes no time.
But
In[5] Do[BB = AA; BB[[77]] = i;, {i, 1, 10000}] // Timing
Out[5] {3.04688, Null}
alternately setting a variable to a large list + setting an element of a list to a number takes long time.
Q1) How can we explain it ?
Q2) In fact, mathematica reported it took 3.04688 seconds, but in fact it took about 20 seconds (physical watch) to get Out[5]. Personaly there have been some differences in the past, but this is the first time I've seen such a big difference.
• Line [5] is the only one where a deep copy of the full AA→BB is done in every loop iteration. Line [2] can be done with a shallow (lazy) copy because BB is not being edited and so no actual data copying is required (only some reference counts need to be updated); but in line [5] the shallow copy BB=AA is followed by a write access BB[[77]]=i, which triggers copy-on-write and which takes substantial amounts of time. Jun 22 at 4:54
• Thank you, didn't know there is a concept like shallow (lazy) copy. Jun 22 at 10:47
• @Roman I found your comment very insightful. Would you consider turning it into an answer? Jun 22 at 11:49
I assume that, like all modern software, Mathematica implements copy-on-write. From the Wikipedia:
Copy-on-write (COW), sometimes referred to as implicit sharing or shadowing, is a resource-management technique used in computer programming to efficiently implement a "duplicate" or "copy" operation on modifiable resources. If a resource is duplicated but not modified, it is not necessary to create a new resource; the resource can be shared between the copy and the original. Modifications must still create a copy, hence the technique: the copy operation is deferred until the first write. By sharing resources in this way, it is possible to significantly reduce the resource consumption of unmodified copies, while adding a small overhead to resource-modifying operations.
Translating this explanation to the examples given by the OP:
• In[2]: The statement BB = AA makes a shallow (lazy) copy of AA and "stores" it in BB. This is a very quick operation: the only thing to do is to write down that BB points to AA. BB is not an independent array; it depends on AA.
• In[3] and In[4]: The first statement (for $$i=1$$) sets BB[[77]] = 1, which triggers copy-on-write and prepares a deep (non-lazy) copy of AA in the space of BB, then modifies BB[[77]]. BB is now fully independent of AA. The subsequent 19999 assignments to elements of BB no not re-trigger copy-on-write because BB is not related to AA anymore.
• In[5]: In every loop iteration, the first statement BB = AA erases the old contents of BB, then establishes a fresh shallow (lazy) copy of AA just like in In[2]. The second statement in the loop then makes an assignment to BB[[77]], which triggers copy-on-write, making BB a deep copy that can be modified independently of AA. It is this copy-on-write operation, which has to make a true copy of one million integer numbers (about 8 megabytes of data), that takes a significant amount of time.
A bit more profiling: a shallow copy seems to take about 190 nanoseconds (a good fraction of which may be the profiling overhead),
AA = RandomInteger[10, 1000000];
RepeatedTiming[BB = AA;] // First
(* 1.93599*10^-7 *)
whereas a deep copy of an array of $$10^6$$ integers takes about 340 microseconds,
RepeatedTiming[BB = AA; BB[[77]] = 5;] // First
(* 0.000340914 *) | 1,024 | 3,860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-27 | longest | en | 0.906766 |
https://www.abcteach.com/directory/prek-early-childhood-mathematics-number-sense-2999-3-2 | 1,500,590,754,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423512.93/warc/CC-MAIN-20170720222017-20170721002017-00590.warc.gz | 726,288,225 | 22,534 | You are an abcteach Member, but you are logged in to the Free Site. To access all member features, log into the Member Site.
# PreK Early Childhood Number Sense Printable Activities
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• This Addition - Candy Hearts (primary) Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition - Candy Hearts (primary) Worksheet. Count the candy hearts. How many pink candy hearts? white candy hearts? Altogether? *candy hearts not included! Common Core: Math (Count to tell the number of objects) K.CC5
• This Addition and Subtraction up to 20 (K-1) Penguin Theme Unit is perfect to practice addition and subtraction skills. Your elementary grade students will love this Addition and Subtraction up to 20 (K-1) Penguin Theme Unit. Addition and subtraction practice and assessment. Includes; practice worksheets, in and out boxes, and matching game.
• This Addition to 10 (pre-k/primary) 1 Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition to 10 (pre-k/primary) 1 Worksheet. Count the number of shapes in each group and write the total.
• This Addition to 10 (pre-k/primary) 2 Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition to 10 (pre-k/primary) 2 Worksheet. Count the number of shapes in each group and write the total.
• This Addition to 10 (pre-k/primary) 3 Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition to 10 (pre-k/primary) 3 Worksheet. Count the number of shapes in each group and write the total. With grayscale graphics.
• This Addition to 10 (pre-k/primary) 4 Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition to 10 (pre-k/primary) 4 Worksheet. Count the number of shapes in each group and write the total. With grayscale graphics.
• This Addition to 10 (pre-k/primary) 5 Worksheet is perfect to practice addition skills. Your elementary grade students will love this Addition to 10 (pre-k/primary) 5 Worksheet. Count the number of shapes in each group and write the total. With grayscale graphics.
• This Aquarium Number Pattern (primary) Interactive Notebook is perfect to practice number pattern skills. Your elementary grade students will love this Aquarium Number Pattern (primary) Interactive Notebook. Smart Notebook file: This is a fun and colorful interactive activity with an aquarium theme providing practice for number patterns. Audio feedback.
• Count the flowers on each circle. Match the circle to the section on the caterpillar with the corresponding number.
• Match the circle with the flowers to the section on the caterpillar with the corresponding number.
• Match the game pieces to the correct numbers on the caterpillar game board to solve the math facts.
• Match the game pieces with the correct circles on the caterpillar game board to solve the math facts.
• Count the spots on each ladybug. Match the ladybug to the flower with the corresponding number.
• Count the spots on each ladybug. Match the ladybug to the flower with the corresponding number.
• Cute number booklet to color with text to read.
• This Candy Heart Patterns (elem) Worksheet is perfect to practice pattern skills. Your elementary grade students will love this Candy Heart Patterns (elem) Worksheet. Use candy hearts to create a pattern in each row. *candy hearts not included!
• This Candy Heart Sort (primary) Worksheet is perfect to practice math skills. Your elementary grade students will love this Candy Heart Sort (primary) Worksheet. Sort the candy hearts by color. *candy hearts not included!
• This Candy Hearts (elementary) Graph is perfect to practice graphing skills. Your elementary grade students will love this Candy Hearts (elementary) Graph. Place each candy heart in the column with the same color name. Which color had the most? Which color had the fewest? *candy hearts not included!
• This Chart Penguin Hundreds Chart is perfect to practice addition and subtraction skills. Your elementary grade students will love this Chart Penguin Hundreds Chart. This chart has a variety of uses for your counting and place value lessons. Students can color the penguins for patterning, cut and paste in order, and use for addition and subtraction.
• This Christmas Light Numbers (1-10) Shapebook is perfect to practice counting skills. Your elementary grade students will love this Christmas Light Numbers (1-10) Shapebook. [member-created with abctools] Numerals and number words, one per page, in a Christmas light-shaped shapebook
• This Color the Right Number (prek/primary) 1 Worksheet is perfect to practice math skills. Your elementary grade students will love this Color the Right Number (prek/primary) 1 Worksheet. Printable coloring worksheet where students will have to color the right numbers.
• This Color the Right Number (prek/primary) 2 Worksheet is perfect to practice math skills. Your elementary grade students will love this Color the Right Number (prek/primary) 2 Worksheet. Printable coloring worksheet where students will have to color the right numbers.
• This Color the Right Number (prek/primary) 3 Worksheet is perfect to practice math skills. Your elementary grade students will love this Color the Right Number (prek/primary) 3 Worksheet. Printable coloring worksheet where students will have to color the right numbers.
• This Color the Right Number (prek/primary) 4 Worksheet is perfect to practice math skills. Your elementary grade students will love this Color the Right Number (prek/primary) 4 Worksheet. Printable coloring worksheet where students will have to color the right numbers.
• This Comparing Numbers Math is perfect to practice comparing numbers skills. Your elementary grade students will love this Comparing Numbers Math. This document visually represents how numbers can be classified as greater or least and provides practice for placing numbers in correct order.
• This Counting and Addition to 20 (K-1) Penguin Theme Unit is perfect to practice addition and subtraction skills. Your elementary grade students will love this Counting and Addition to 20 (K-1) Penguin Theme Unit. This penguin theme unit is a great way to practice counting and adding to 20. This 21 page unit includes; tracing numbers, cut and paste, finding patterns, ten frame activity, in and out boxes and much more! CC: Math: K.CC.B.4
• This Counting by 2's (Primary) Shapebook is perfect to practice counting skills. Your elementary grade students will love this Counting by 2's (Primary) Shapebook. [member-created document] Trace and cut out. This shoe-shaped shapebook is a fun way for students to learn about counting by twos up to 50 or more (includes blank pages at end).
• This Counting by 3's (Primary) Shapebook is perfect to practice counting skills. Your elementary grade students will love this Counting by 3's (Primary) Shapebook. [member-created document] Trace and cut out. This bird-shaped shapebook is a fun way for students to learn about counting by threes up to 60.
• This Counting Crabs (up to 5) - pre-k/primary Worksheet is perfect to practice counting skills. Your elementary grade students will love this Counting Crabs (up to 5) - pre-k/primary Worksheet. "Circle the number of crabs you see on each line." Counting up to 5.
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• This Counting Mice (up to 10) - pre-k/primary Worksheet is perfect to practice counting skills. Your elementary grade students will love this Counting Mice (up to 10) - pre-k/primary Worksheet. "Circle the number of mice you see in each box." Counting up to 10.
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• This Counting Turtles (up to 5) - pre-k/primary Worksheet is perfect to practice counting skills. Your elementary grade students will love this Counting Turtles (up to 5) - pre-k/primary Worksheet. "Circle the number of turtles you see on each line." Counting up to 5.
• This Counting up to 3 (pre-k) -Circus theme Worksheet is perfect to practice counting skills. Your elementary grade students will love this Counting up to 3 (pre-k) -Circus theme Worksheet. "Count the circus animals. Write the numbers on the line." To 3.
• This Counting up to 5 (pre-k) 1 Worksheet is perfect to practice counting skills. Your elementary grade students will love this Counting up to 5 (pre-k) 1 Worksheet. How many chicks? Circle the correct number. Common Core: Math (Count to tell the number of objects) K.CC5 | 2,303 | 10,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-30 | longest | en | 0.914009 |
https://engineering.stackexchange.com/questions/23592/how-to-calculate-bearing-loads-in-a-two-stage-helical-gear-box | 1,638,625,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00303.warc.gz | 317,351,616 | 32,514 | # How to calculate bearing loads in a two stage helical gear box?
Here P, Q, R, S are the bearings and G1, G2, G3, G4 are four helical gears. The gear forces tangential(Kp), Axial(Ka), Radial(Kn) are known. All the distances are known. The forces acting on bearing are radial on non locating bearing and both radial and axial on locating bearing. P, R are non locating bearings. Q, S are locating bearings.. Fh and Fv are components of radial force on bearings and Fa is the axial force... How to calculate Fh, Fv and Fa for each bearing shown??
I have calculated all the gear forces but i am confused about the calculation of the forces. will be solved using moment equations? | 166 | 679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.947483 |
http://www.gurufocus.com/term/grahamnumber/CHYR/Graham%2BNumber/Chyron%2BCorporation | 1,474,818,271,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660242.52/warc/CC-MAIN-20160924173740-00043-ip-10-143-35-109.ec2.internal.warc.gz | 504,869,673 | 26,827 | Switch to:
ChyronHego Corp (NAS:CHYR)
Graham Number
\$N/A (As of Sep. 2014)
Graham Number is a figure that measures a stock's fundamental value by taking into account the company's earnings per share and book value per share. The Graham number is the upper bound of the price range that a defensive investor should pay for the stock. According to the theory, any stock price below the Graham number is considered undervalued, and thus worth investing in.
As of today, the stock price of ChyronHego Corp is \$2.81. ChyronHego Corp's graham number for the quarter that ended in Sep. 2014 was \$N/A. Therefore, ChyronHego Corp's Price to Graham Number ratio for today is N/A.
Graham Number is a combination of asset valuation and earnings power valuation. It is a very conservative way of valuing a stock.
Definition
Graham Number is a concept based on Ben Graham's conservative valuation of companies.
ChyronHego Corp's Graham Number for the fiscal year that ended in Dec. 2013 is calculated as
Graham Number = sqrt of (22.5 * Tangible Book Value per Share * EPS without Non-Recurring Items (NRI)) = sqrt of (22.5 * -0.41 * -0.31) = N/A
ChyronHego Corp's Graham Number for the quarter that ended in Sep. 2014 is calculated as
Graham Number = sqrt of (22.5 * Tangible Book Value per Share * EPS without Non-Recurring Items (NRI) (TTM)) = sqrt of (22.5 * -0.208 * -0.21) = N/A
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
Ben Graham actually did not publish a formula like this. But he wrote in The Intelligent Investor (1948 version) regarding to the criteria for purchases:
Current price should not be more than 15 times average earnings of the past three years.
Current price should not be more than 1.5 times the book value last reported. However, a multiplier of earnings below 15 could justify a correspondingly higher multiplier of assets. As a rule of thumb we suggest that the product of the multiplier times the ratio of price to book value should not exceed 22.5. (This figure corresponds to 15 times earnings and 1.5 times book value. It would admit an issue selling at only 9 times earnings and 2.5 times asset value, etc.)
Unlike valuation methods such as DCF or Discounted Earnings, the Graham number does not take growth into the valuation. Unlike the valuation methods based on book value alone, it takes into account the earnings power. Therefore, the Graham Number is a combination of asset valuation and earnings power valuation.
In general, the Graham number is a very conservative way of valuing a stock. It cannot be applied to companies with negative book values.
ChyronHego Corp's Price to Graham number Ratio for today is calculated as
Price to Graham number = Share Price (Today) / Graham number (Q: Sep. 2014 ) = 2.81 / N/A = N/A
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
1. Graham Number does not take growth into account. Therefore it underestimates the values of the companies that have good earnings growth. We feel that if the earnings per share grows more than 10% a year, Graham Number underestimates the value.
2. Graham Number punishes the companies that have temporarily low earnings. Therefore, an average of earnings makes more sense in the calculation of Graham Number.
3. Graham Numbers underestimates companies that are light with book.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
ChyronHego Corp Annual Data
Dec04 Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 grahamnumber 0.00 0.00 1.30 1.85 6.38 0.00 0.00 0.00 0.00 0.00
ChyronHego Corp Quarterly Data
Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 grahamnumber 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,027 | 4,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | longest | en | 0.929081 |
https://www.kodytools.com/units/enzyme/from/microkatal/to/exakatal | 1,713,690,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00404.warc.gz | 769,231,269 | 14,997 | # Microkatal to Exakatal Converter
1 Microkatal = 1e-24 Exakatals
## One Microkatal is Equal to How Many Exakatals?
The answer is one Microkatal is equal to 1e-24 Exakatals and that means we can also write it as 1 Microkatal = 1e-24 Exakatals. Feel free to use our online unit conversion calculator to convert the unit from Microkatal to Exakatal. Just simply enter value 1 in Microkatal and see the result in Exakatal.
Manually converting Microkatal to Exakatal can be time-consuming,especially when you don’t have enough knowledge about Enzyme units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Microkatal to Exakatal converter tool to get the job done as soon as possible.
We have so many online tools available to convert Microkatal to Exakatal, but not every online tool gives an accurate result and that is why we have created this online Microkatal to Exakatal converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Microkatal to Exakatal (μkat to Ekat)
By using our Microkatal to Exakatal conversion tool, you know that one Microkatal is equivalent to 1e-24 Exakatal. Hence, to convert Microkatal to Exakatal, we just need to multiply the number by 1e-24. We are going to use very simple Microkatal to Exakatal conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Microkatal} = 1 \times 1e-24 = \text{1e-24 Exakatals}$$
## What Unit of Measure is Microkatal?
Microkatal is a unit of measurement for enzyme's catalytic activity. By definition, one microkatal is the enzyme catalytic activity that raises the rate of a chemical reaction by 1e-6 mole per second.
## What is the Symbol of Microkatal?
The symbol of Microkatal is μkat. This means you can also write one Microkatal as 1 μkat.
## What Unit of Measure is Exakatal?
Exakatal is a unit of measurement for enzyme's catalytic activity. By definition, one exakatal is the enzyme catalytic activity that raises the rate of a chemical reaction by 1e18 mole per second.
## What is the Symbol of Exakatal?
The symbol of Exakatal is Ekat. This means you can also write one Exakatal as 1 Ekat.
## How to Use Microkatal to Exakatal Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Microkatal and in the first input field, enter a value.
• From the second dropdown, select Exakatal.
• Instantly, the tool will convert the value from Microkatal to Exakatal and display the result in the second input field.
Microkatal
1
Exakatal
1e-24
# Microkatal to Exakatal Conversion Table
Microkatal [μkat]Exakatal [Ekat]Description
1 Microkatal1e-24 Exakatal1 Microkatal = 1e-24 Exakatal
2 Microkatal2e-24 Exakatal2 Microkatal = 2e-24 Exakatal
3 Microkatal3e-24 Exakatal3 Microkatal = 3e-24 Exakatal
4 Microkatal4e-24 Exakatal4 Microkatal = 4e-24 Exakatal
5 Microkatal5e-24 Exakatal5 Microkatal = 5e-24 Exakatal
6 Microkatal6e-24 Exakatal6 Microkatal = 6e-24 Exakatal
7 Microkatal7e-24 Exakatal7 Microkatal = 7e-24 Exakatal
8 Microkatal8e-24 Exakatal8 Microkatal = 8e-24 Exakatal
9 Microkatal9e-24 Exakatal9 Microkatal = 9e-24 Exakatal
10 Microkatal1e-23 Exakatal10 Microkatal = 1e-23 Exakatal
100 Microkatal1e-22 Exakatal100 Microkatal = 1e-22 Exakatal
1000 Microkatal1e-21 Exakatal1000 Microkatal = 1e-21 Exakatal
# Microkatal to Other Units Conversion Table
ConversionDescription
1 Microkatal = 0.000001 Katal1 Microkatal in Katal is equal to 0.000001
1 Microkatal = 1e-24 Exakatal1 Microkatal in Exakatal is equal to 1e-24
1 Microkatal = 1e-21 Petakatal1 Microkatal in Petakatal is equal to 1e-21
1 Microkatal = 1e-18 Terakatal1 Microkatal in Terakatal is equal to 1e-18
1 Microkatal = 1e-15 Gigakatal1 Microkatal in Gigakatal is equal to 1e-15
1 Microkatal = 1e-12 Megakatal1 Microkatal in Megakatal is equal to 1e-12
1 Microkatal = 1e-9 Kilokatal1 Microkatal in Kilokatal is equal to 1e-9
1 Microkatal = 1e-8 Hectokatal1 Microkatal in Hectokatal is equal to 1e-8
1 Microkatal = 1e-7 Dekakatal1 Microkatal in Dekakatal is equal to 1e-7
1 Microkatal = 0.00001 Decikatal1 Microkatal in Decikatal is equal to 0.00001
1 Microkatal = 0.0001 Centikatal1 Microkatal in Centikatal is equal to 0.0001
1 Microkatal = 0.001 Millikatal1 Microkatal in Millikatal is equal to 0.001
1 Microkatal = 1000 Nanokatal1 Microkatal in Nanokatal is equal to 1000
1 Microkatal = 1000000 Picokatal1 Microkatal in Picokatal is equal to 1000000
1 Microkatal = 1000000000 Femtokatal1 Microkatal in Femtokatal is equal to 1000000000
1 Microkatal = 1000000000000 Attokatal1 Microkatal in Attokatal is equal to 1000000000000
1 Microkatal = 1000000000000000 Zeptokatal1 Microkatal in Zeptokatal is equal to 1000000000000000
1 Microkatal = 1000000000000000000 Yoctokatal1 Microkatal in Yoctokatal is equal to 1000000000000000000
1 Microkatal = 0.000001 Mole/Second1 Microkatal in Mole/Second is equal to 0.000001
1 Microkatal = 0.001 Millimole/Second1 Microkatal in Millimole/Second is equal to 0.001
1 Microkatal = 1e-9 Kilomole/Second1 Microkatal in Kilomole/Second is equal to 1e-9
1 Microkatal = 1 Micromole/Second1 Microkatal in Micromole/Second is equal to 1
1 Microkatal = 0.00006 Mole/Minute1 Microkatal in Mole/Minute is equal to 0.00006
1 Microkatal = 0.06 Millimole/Minute1 Microkatal in Millimole/Minute is equal to 0.06
1 Microkatal = 6e-8 Kilomole/Minute1 Microkatal in Kilomole/Minute is equal to 6e-8
1 Microkatal = 60 Micromole/Minute1 Microkatal in Micromole/Minute is equal to 60 | 1,830 | 5,592 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.884891 |
https://oeis.org/A077750/internal | 1,627,556,995,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153857.70/warc/CC-MAIN-20210729105515-20210729135515-00370.warc.gz | 410,584,859 | 2,702 | 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!)
A077750 Least significant digit of A077749(n). 1
%I
%S 0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,
%T 6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,
%U 0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0,2,6,4,0
%N Least significant digit of A077749(n).
%F a(n) = A077749(n) - 10*floor(A077749(n)/10).
%F a(4n-3) = 0, a(4n-2) = 2, a(4n-1) = 6, a(4n) = 4, n > 0. - _Sascha Kurz_, Jan 04 2003
%F a(n) = 3-3*cos(n*Pi/2)-sin(n*Pi/20. - _Wesley Ivan Hurt_, May 07 2021
%p seq((2^i-1-1) mod 10,i=1..160);
%Y Cf. A077749.
%K base,nonn
%O 0,2
%A _Amarnath Murthy_, Nov 20 2002
%E More terms from _Sascha Kurz_, Jan 04 2003
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Last modified July 29 07:00 EDT 2021. Contains 346340 sequences. (Running on oeis4.) | 576 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-31 | latest | en | 0.769059 |
https://kr.mathworks.com/matlabcentral/answers/856225-plot-and-left-axis-off?s_tid=prof_contriblnk | 1,656,605,181,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00731.warc.gz | 394,225,445 | 33,465 | # plot and left axis off
조회 수: 51(최근 30일)
Babu Sankhi 2021년 6월 14일
편집: Adam Danz 2021년 6월 20일
When I plot it comes with both axes in the box. How can I plot by removing right axis (or vertical line) of Box? can you please help me?
x=(1:1:10);
y= (1:2:20);
figure (1);
plot (x,y,'o-');
##### 댓글 수: 4표시숨기기 이전 댓글 수: 3
Adam Danz 2021년 6월 14일
Babu Sankhi's answer moved here as a comment.
yes you are right I want to add another plot by removing the right boarder. Can please give me the more idea about it how can get the plot like this as attched below? thank you
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### 답변(1개)
Adam Danz 2021년 6월 14일
편집: Adam Danz 2021년 6월 16일
Turn off right-y-axis
This solution is similar to another answer provided a few days ago. It uses yyaxis to add a second, independent y-axis on the right and then turns the color off for that axis.
x=(1:1:10);
y= (1:2:20);
figure (1);
plot (x,y,'o-');
yyaxis right
ax = gca();
ax.YAxis(2).Color = 'none';
yyaxis left % return focus to 'main' axis.
Join 2 axes, control the center border line
Update following an example added by OP
This version joins 2 axes, turns off the y-axes in the center of the two axes, formats the left-y-axis of the right-axes as a dashed line, and then links the two sets of y-axes so they always have the same y-limit.
The juxtaposition of the axes uses TiledLayout with zero-tile-spacing which requires Matlab r2021a (see Community Highlight). For versions of Matlab prior to R2021a, create the two axes using the axes command and set their position properties (see version-2 below).
x = 0:0.1:10;
y = gaussmf(x,[2 5]);
figure()
tiledlayout(1,2,'TileSpacing','none'); % Requires Matlab r2021a or later
ax1 = nexttile();
plot(ax1, x,y)
yyaxis(ax1,'right')
ax1.YAxis(2).Color = 'none';
yyaxis(ax1,'left') % return focus to 'main' axis.
xlabel('Left Axis')
ax2 = nexttile();
plot(ax2, x,y)
yyaxis(ax2,'right')
ax2.YAxis(2).Color = ax1.YAxis(1).Color;
ax2.YAxis(2).TickLabels = []; % Turn off right tick labels
yyaxis(ax2,'left') % return focus to 'main' axis.
ax2.YAxis(1).TickValues = [];
ax2.YAxis(1).Axle.LineStyle = 'dashed';
linkaxes([ax1,ax2],'y') % link both sets of y-axis limits
xlabel('Right Axis')
Version-2 using custom axes instead of TiledLayout
x = 0:0.1:10;
y = gaussmf(x,[2 5]);
figure()
margins = [.13 .11 .25]; % [left&right, bottom, top], normalized units
ax1 = axes('Units','Normalize','Position',[margins(1:2),.5-margins(1),1-margins(3)]);
plot(ax1, x,y)
yyaxis(ax1,'right')
ax1.YAxis(2).Color = 'none';
yyaxis(ax1,'left') % return focus to 'main' axis.
xlabel('Left Axis')
ax2 = axes('Units','Normalize','Position',[.5,margins(2),.5-margins(1),1-margins(3)]);
plot(ax2, x,y)
yyaxis(ax2,'right')
ax2.YAxis(2).Color = ax1.YAxis(1).Color;
ax2.YAxis(2).TickLabels = []; % Turn off right tick labels
yyaxis(ax2,'left') % return focus to 'main' axis.
ax2.YAxis(1).TickValues = [];
ax2.YAxis(1).Axle.LineStyle = 'dashed';
linkaxes([ax1,ax2],'y') % link both sets of y-axis limits
xlabel('Right Axis')
Additional recommendations
• Set the x-ticks to avoid the overlap at the x-limits.
• Turn the grid on for both pairs of axes (yyaxis left)
##### 댓글 수: 8표시숨기기 이전 댓글 수: 7
Adam Danz 2021년 6월 16일
The reason you have a blank figure is because your axes size is 600x500 but axis units are normalized by default so the largest size is 1x1.
I'll update my answer with a demo using custom axes.
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Translated by | 1,154 | 3,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.670972 |
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1 Other / Politics & Society / Re: PFizer destroyed in this video. Watch only if you can calculate on: January 07, 2022, 12:17:14 PM 2% loss means enemy has 52% and "our/your" army has 48% if their parachutes don't fail.At 54% vs 46% the war is already over haha. 2% away from defeat ! Concerning DNA it's very complex and still not very well understood... at least the structures that come from it and the interactions ! Bye, Skybuck =D
2 Other / Politics & Society / PFizer destroyed in this video. Watch only if you can calculate on: January 07, 2022, 04:23:36 AM Video:https://rumble.com/vqx3kb-the-pfizer-inoculations-do-more-harm-than-good.htmlDefinitions from website:https://www.ncbi.nlm.nih.gov/books/NBK63647/From table 18.1 link:"Relative risk (RR)the rate (risk) of poor outcomes in the intervention group divided by the rate of poor outcomes in the control group. For example, if the rate of poor outcomes is 20 per cent in the intervention group and 30 per cent in the control group, the relative risk is 0.67 (20 per cent divided by 30 per cent). The relative risk is 1 when the intervention has no effect, below 1 when it does good and above 1 when it does harm (see Absolute risk reduction).""Relative risk reduction (RRR)the extent to which the risk of a poor outcome is reduced by an intervention. In the example given in Relative risk (above), the relative risk r""Absolute risk reduction (ARR) or risk differencethe difference in the incidence of poor outcomes between the intervention group of a study and the control group. For example, if 20 per cent of people die in the intervention group and 30 per cent in the control group, the ARR is 10 per cent (30–20 per cent)."Performed calculation:162 / 18325 = 0,00884038199181446111869031377899 * 100 = 0,884038199181446111869031377899058 / 18198 = 0,00043960874821408946038026156720519e-4 = 0,04396087482140894603802615672052rr = 0,04396087482140894603802615672052 / 0,88403819918144611186903137789905 = 0,0497273475989085763053598346854rrr = 1 - 0,0497273475989085763053598346854 = 0,9502726524010914236946401653146 * 100 = 95,02726524010914236946401653146% = RRRabsolute = % die in normal=control group - % of intervention group Let's suppose all covid cases 162 and 8 die because of futher lack of info then:absolute risk reduction = 0.88 - 0.04 = 0,841 So the video checks out, only 1 procent risk reduction.However it gets more interesting later on in the video because the vaccines have side effects and actually destroys health and kills people.Long story short it seems to be the vaccine does more harm than good !Watch video yourself ! I only watch first part up to this calculation point but I will watch the rest as well.Now you know why USA grants this SCAM Company Immunity during Emergency Situation.As soon as Emergency Situation is gone, vaccinations will/most stop or this Company goes bankrupt in law suits !Thus incentive for control freaks to lengthen the Emergency Situation.(Save your life, enjoy your bitcoins !)
3 Bitcoin / Development & Technical Discussion / Re: New database for blockchain on: December 08, 2021, 03:46:21 AM Quote from: MixMAx123 on October 29, 2021, 11:07:37 PMI want to create a new database for the blockchain.I would like to create a file for each block.eg: 000000000019D6689C085AE165831E934F763AE46A2A6C172B3F1B60A8CE26F.BlockThat would be more than 700,000 files. NTFS can be 4.294.967.295 files in a folder.So theoretically it should go.Question:Has anyone ever tested such a thing?Would the file system crashed or something similar?My idea is through the file name of the blockhash is to find and open the blocks very quickly.Ideas and suggestions are welcome :-) I think you should give GIT a try, it's basically a key=value storage systems, does something similiar as to what you want, store content with a content hash.Git does have some decent performance thanks to big linux kernel development approx 2 GB in size, ofcourse bitcoin much larger 150x larger.I am not sure how git works internally, have not gotten that far yet with it's analysis... it does use a database stored in a .git folder.GIT is open source I believe, so you could analyze the source code to see how it works.GIT is also very similiar to BITCOIN... GIT also uses "blocks" which are called "commits" and it also calculates hashes, sha-1 I believe.And these blocks/commits are also linked to each other in a similiar way.Perhaps somebody involved with GIT was also involved in the creation of BITCOIN !
4 Bitcoin / Bitcoin Discussion / Re: I have 500 lost Bitcoin - need help on: November 06, 2021, 08:46:19 AM I am always happy to help anybody recover bitcoins: my e-mail address is skybuck2000@hotmail.com.You can try and send whatever you got over there.However if this is a scam or you trying to infect me with some crazy executable, don't try ! LOL.(Also don't try if you try and link me to a server infected with unicode hacks and shit nonononononono LOLOLOL.)That stuff just makes me giggle.But if this is real, go right ahead !
5 Bitcoin / Bitcoin Discussion / Re: Reason for bitcon flash drop. on: November 06, 2021, 08:42:40 AM Flash Crash = Computer GlitchEnjoy ! =D
6 Economy / Speculation / Re: What make people panic sell? on: November 06, 2021, 08:29:12 AM Panic Sell = Computer Glitches ! Good luck with that ! =D
7 Economy / Speculation / Bitcoin run up is coming on: November 06, 2021, 08:21:14 AM Now that the inflation is rising in Europe, caused by energy costs ?! there will be a run on bitcoin to maintain wealth.The inflation is already at 4 to 5% maybe even higher.This can not continue for longer before rich people start to get very nervous.If there ever was a time where bitcoin is going to skyrocket out of control, it's the coming 3 to 6 months ! =DBanks will panic, Banks will buy bitcoin too, with bank account holder savings.And the bank account holders will be the nigga holding the bag, worthless money caused by inflation.From riches to rags ! LOL.I don't have the balls to buy bitcoin right now.But if you are balsy... then you now know what to do.
8 Bitcoin / Development & Technical Discussion / Re: !!! RED ALERT: SHIELDS UP, TROJAN SOURCE HAS ARRIVED !!! on: November 04, 2021, 07:46:18 PM Well mean while, I have read the entire document, looked at examples, looked at what professor mentions is vunerable software basically all webbrowsers <- that is big.If this was known then why is it not fixed ? Clearly it wasn't know or people didn't take it seriously, I'll let you choose.Paying 8010 or 1080 is a big deal this could be fooled on the web.Anyway I also googled around found some doc about hacking unicode, apperently in that doc this was also know.The thing that alarmed me is that this-executes-exe.txt actually worked in windows 7.Basically the exe.txt get's swapped, so the real filename is this-executes-txt.exe.I have never seen such a thing in all these years of windows usage ! That totally shocked me. SO BEWARE !For now I have calmed down somewhat, totally calm now...There is babelmap an babeleditor that can be used to experiment with unicode and control characters ! Have fun ! =DI also worry about magnetlinks from whatever reason... my spider-sense till me somebody is going to do something bad with it.Or how about URI payments ! AH YES ! URI payments would be the perfect way to fool USERS.HOLYSHIT.Bye for now, Skybuck.
9 Bitcoin / Development & Technical Discussion / Re: !!! RED ALERT: SHIELDS UP, TROJAN SOURCE HAS ARRIVED !!! on: November 03, 2021, 01:24:53 PM
Anyways, it's time to test some of this code.
TEST 1 CODE section: OK SAFE
Code:
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
TEST 2 QUOTE: OK SAFE
Quote
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
TEST 3 JUST WEB COPY:, OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
TELETYPE: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
SUPERSCRIPT: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
SUBSCRIPT: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
#include <stdio.h>
#include <stdbool.h>
Table column: OK SAFE
int main() { bool isAdmin = false; /* } if (isAdmin) begin admins only */ printf("You are an admin.\n"); /* end admins only { */ return 0;}
TABLE: OK SAFE
#include #include int main() { bool isAdmin = false; /* } if (isAdmin) begin admins only */ printf("You are an admin.\n"); /* end admins only { */ return 0;}
GLOW: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
strikethrough: ok safe
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
LIST: OK SAFE
• #include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
right align: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
centered: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
preformatted text: OK SAFE
`#include <stdio.h>#include <stdbool.h>int main() { bool isAdmin = false; /* } if (isAdmin) begin admins only */ printf("You are an admin.\n"); /* end admins only { */ return 0;}`
left align: OK SAFE
#include <stdio.h>
#include <stdbool.h>
int main() {
/* end admins only { */
return 0;
}
THIS FORUM SOFTWARE IS PRETTY GOOD AND SAFE ! LOL. though that right align is a bit whack ! HAHA.
FOR NOW THIS FORUM SOFTWARE IS SKYBUCK APPROVED ! =D
10 Bitcoin / Development & Technical Discussion / Re: !!! RED ALERT: SHIELDS UP, TROJAN SOURCE HAS ARRIVED !!! on: November 03, 2021, 01:19:49 PM Quote from: NotATether on November 03, 2021, 12:42:01 PMQuote from: Skybuck on November 03, 2021, 09:46:23 AMThis casts big doubts on the whole UNICODE system you might as well consider it a GIGANTIC NSA conspiracy to make all of our systems WEAK and HACKABLE.Come on, you know better than to label security bugs as NSA conspiracies like this. Quote from: Skybuck on November 03, 2021, 09:46:23 AMThe vulnerabilities are skyrocketing. ?You also notice it in the news, yet another ransomware attack, or failure of something.Most ransomware are the result of companies who are running old outdated software with ancient vulnerabilities, not by some funky stuff on last year's DEFCON or Black Hat World.Unfortunately NSA is not conspiracy theory, they have hacked about anything they can get their paws on and more ! LOL. The most funny one was hacking the POPE. He not talking to GOD, he talking to NSA.Try to also look towards the future... how all of this can be exploited by scammers ! For bitcoin and other financial systems, swapping financial numbers comes to mind. | 3,069 | 11,291 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.871804 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-3-section-3-2-the-graph-of-a-function-3-2-assess-your-understanding-page-219/14 | 1,537,860,994,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161214.92/warc/CC-MAIN-20180925063826-20180925084226-00103.warc.gz | 758,073,856 | 12,614 | ## College Algebra (10th Edition)
The graph represents a function. (a) Domain: $(-\infty, +\infty)$ Range: $(0, +\infty)$ (b) x-intercept: none y-intercept: $1$ (c) not symmetric with the x-axis, the y-axis, or the origin
The graph passes the vertical line test since all vertical lines will pass through the graph at a maximum of one point only. Thus, the graph represents a function. (a) The graph covers the entire x-axis. Thus the domain is $(-\infty, +\infty)$. The graph only covers the region above the x-axis. Thus, the range is $(0, +\infty)$. (b) The graph has no x-intercept as it is asymptotic to the x-axis. The graph crosses the y-axis at $(0, 1)$ so the y-intercept is $1$. (c) The graph does not show symmetry to the x axis, the y axis, or the origin. | 222 | 768 | {"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} | 4.21875 | 4 | CC-MAIN-2018-39 | latest | en | 0.860579 |
http://www.gurufocus.com/term/Dividends+Per+Share/MKC/Dividends%2BPer%2BShare/McCormick%2B%2526amp%253B%2BCompany%252C%2BIncorporated | 1,480,811,380,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00312-ip-10-31-129-80.ec2.internal.warc.gz | 500,693,505 | 29,887 | Switch to:
McCormick & Co Inc (NYSE:MKC)
Dividends Per Share
\$1.69 (TTM As of Aug. 2016)
McCormick & Co Inc's dividends per share for the three months ended in Aug. 2016 was \$0.43. Its dividends per share for the trailing twelve months (TTM) ended in Aug. 2016 was \$1.69. Its Dividend Payout Ratio for the three months ended in Aug. 2016 was 0.46. As of today, McCormick & Co Inc's Dividend Yield is 1.93%.
During the past 12 months, McCormick & Co Inc's average Dividends Per Share Growth Rate was 7.60% per year. During the past 3 years, the average Dividends Per Share Growth Rate was 8.90% per year. During the past 5 years, the average Dividends Per Share Growth Rate was 9.20% per year. During the past 10 years, the average Dividends Per Share Growth Rate was 9.40% per year. Please click Growth Rate Calculation Example (GuruFocus) to see how GuruFocus calculates Wal-Mart Stores Inc (WMT)'s revenue growth rate. You can apply the same method to get the average dividends per share growth rate.
During the past 13 years, the highest 3-Year average Dividends Per Share Growth Rate of McCormick & Co Inc was 16.10% per year. The lowest was 6.60% per year. And the median was 9.10% per year.
Definition
Dividends paid to per common share.
Explanation
1. Dividend payout ratio measures the percentage of the companys earnings paid out as dividends.
McCormick & Co Inc's Dividend Payout Ratio for the quarter that ended in Aug. 2016 is calculated as
Dividend Payout Ratio = Dividends Per Share (Q: Aug. 2016 ) / EPS without NRI (Q: Aug. 2016 ) = 0.43 / 0.94 = 0.46
During the past 13 years, the highest Dividend Payout Ratio of McCormick & Co Inc was 2.13. The lowest was 0.19. And the median was 0.48.
2. Dividend Yield measures how much a company pays out in dividends each year relative to its share price.
McCormick & Co Inc Recent Full-Year Dividend History
Amount Ex-date Record Date Pay Date Type Frequency
0.4302016-10-062016-10-112016-10-25Cash Dividendquarterly
0.4302016-07-072016-07-112016-07-25Cash Dividendquarterly
0.4302016-04-072016-04-112016-04-25Cash Dividendquarterly
0.4302015-12-292015-12-312016-01-15Cash Dividendquarterly
McCormick & Co Inc's Dividend Yield (%) for Today is calculated as
Dividend Yield = Most Recent Full Year Dividend / Current Share Price = 1.72 / 89.05 = 1.93 %
Current Share Price is \$89.05.
McCormick & Co Inc's Dividends Per Share for the trailing twelve months (TTM) ended in Today is \$1.72.
During the past 13 years, the highest Dividend Yield of McCormick & Co Inc was 3.12%. The lowest was 1.55%. And the median was 2.06%.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
McCormick & Co Inc Annual Data
Nov06 Nov07 Nov08 Nov09 Nov10 Nov11 Nov12 Nov13 Nov14 Nov15 Dividends Per Share 0.72 0.80 0.88 0.96 1.04 1.12 1.24 1.36 1.48 1.60
McCormick & Co Inc Quarterly Data
May14 Aug14 Nov14 Feb15 May15 Aug15 Nov15 Feb16 May16 Aug16 Dividends Per Share 0.37 0.37 0.37 0.40 0.40 0.40 0.40 0.43 0.43 0.43
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 1,013 | 3,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.921543 |
https://dcc93.com/chalk-river/application-of-tensors-in-physics.php | 1,610,735,232,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00434.warc.gz | 341,561,683 | 9,667 | # Tensors application in physics of
Home » Chalk River » Application of tensors in physics
## Chalk River - Application Of Tensors In Physics
Tensors Stress Strain Elasticity Mineral Physics. Applications of Euler’s Formula and Tensors, were not in my one Mathematical Methods for Physics and Engineering by Riley, Hobson,, In mathematics, tensors are geometric objects that describe linear relations between geometric vectors, scalars, and other tensors. Elementary examples of such relations include the dot product, the cross product, and linear maps. Geometric vectors, often used in physics and engineering applications, and scalars themselves are also tensors..
### What is the physical meaning of a tensor? What are
New Features in Maple 2018 Physics - Maplesoft. Vector and Tensor Analysis with Applications and for those interested in the applications of tensor calculus to mathematical physics and engineering. Tensor, 2012-08-29В В· Representation Theory: As an application of tensor analysis, we consider normal modes of mass-spring systems. Cases include motion in a line and planar motion..
What is the physical meaning of a tensor? What are interesting examples of tensors in physics? Foundations of Tensor Analysis for Students of Physics and Engineering With an Introduction to the Theory of Relativity and the application of tensors in
Computing with Tensors: Potential Applications of Physics-Motivated Mathematics to Computer Science Martine Ceberio and Vladik Kreinovich Department of Computer Science A Student’s Guide to Vectors and Tensors Vectors and tensors are among the most powerful problem-solving tools available, with applications ranging from mechanics
Tensor Analysis with Applications Stress and strain tensors are examples of tensors of largely independent of application area in order to appeal to the widest Other examples of tensors include the strain tensor, the conductivity tensor, and the inertia tensor. Answered by: Aman Ahuja, Physics Student, WPI,
Tensor Techniques in Physics – a concise introduction Roy McWeeny Professore Emerito di Chimica Teorica, Universit`a di Pisa, Pisa 2.1 Tensors of higher rank Copositivity of tensors plays an important role in vacuum stability of a general scalar potential, polynomial optimization, tensor complementarity problem and tensor generalized eigenvalue complementarity problem.
Roger Penrose- Applications of Negative Dimensional Tensors - Free download as PDF File (.pdf), Text File (.txt) or read online for free. students a modern introduction to vectors and tensors. Traditional courses on applied mathematics
A Student’s Guide to Vectors and Tensors Vectors and tensors are among the most powerful problem-solving tools available, with applications ranging from mechanics Application of Electromagnetic Field Tensors in Special Relativity Department of Physics, Faculty of Science, we start with electromagnetic field tensors,
Geometry in Physics. Contents 1 Exterior Calculus page 1 1.1.2 Tensors Tensors (latin: tendo { I span) are the most general objects of multilinear algebra. Tensors in Physics is a two-in-one package containing: A user's guide and a brief refresher course in differential geometry that also aims to clarify and explain the
Tensors in Physics is a two-in-one package containing: A user's guide and a brief refresher course in differential geometry that also aims to clarify and explain the Geometry in Physics. Contents 1 Exterior Calculus page 1 1.1.2 Tensors Tensors (latin: tendo { I span) are the most general objects of multilinear algebra.
students a modern introduction to vectors and tensors. Traditional courses on applied mathematics Application of Electromagnetic Field Tensors in Special Relativity Department of Physics, Faculty of Science, we start with electromagnetic field tensors,
students a modern introduction to vectors and tensors. Traditional courses on applied mathematics concrete applications far more easily. Let X and Y denote sets Another physical example of a tensor is the polarizability tensor relating the electric dipole moment
by Pamela Burnley, University of Nevada Las Vegas Introduction The Stress Tensor The Strain Tensor Elasticity Literature Many physical properties of crystalline A Student’s Guide to Vectors and Tensors Vectors and tensors are among the most powerful problem-solving tools available, with applications ranging from mechanics
Motivation for tensor product in Physics. but thinking about the motivation that comes from Physics. Algebraists motivate the tensor product Web Applications; In mathematics, tensors are geometric objects that describe linear relations between geometric vectors, scalars, and other tensors. Elementary examples of such relations include the dot product, the cross product, and linear maps. Geometric vectors, often used in physics and engineering applications, and scalars themselves are also tensors.
Foundations of Tensor Analysis for Students of Physics and Engineering With an Introduction to the Theory of Relativity and the application of tensors in Scalars, Vectors and Tensors A scalar is a physical quantity that it represented by a dimensional num-ber at a particular point in space and time.
The various types and ranks of tensors and the physical basis is presented. The physics relevant for the applications in mechanics, quantum mechanics, In mathematics, tensors are geometric objects that describe linear relations between geometric vectors, scalars, and other tensors. Elementary examples of such relations include the dot product, the cross product, and linear maps. Geometric vectors, often used in physics and engineering applications, and scalars themselves are also tensors.
Tensors have their applications to Riemannian Geometry, Mechanics, Elasticity, Theory of Relativity, Electromagnetic Theory and many other disciplines of Science and Motivation for tensor product in Physics. but thinking about the motivation that comes from Physics. Algebraists motivate the tensor product Web Applications;
Examples of Tensors February 3, 2013 Wewilldevelopanumberoftensorsasweprogress,butthereareafewthatwecandescribeimmediately. Welookattwocases: (1 Best Introduction to Tensors by William P. Meyers. used in introductory physics to represent forces or velocities. which is conventional for some applications.
The tensor concept is important in physics and has wide applications in such diverse fields as relativity theory, cosmology, high energy physics, field theory, thermodynamics, fluid dynamics, and mechanics. What is the importance of vectors in physics В·The POINT OF APPLICATION is the physical Vectors are used in many branches of physics whenever there are
### Tensors Stress Strain Elasticity Mineral Physics
Tensor Calculus with Applications Google Books. In addition, the authors consider numerous applications of tensors to geometry, mechanics and physics. While developing tensor calculus,, A guide on tensors is proposed for undergraduate students in physics or engineering that ties directly to vector calculus in orthonormal coordinate systems. We show that once orthonormality is relaxed, a dual basis, together with the contravariant and covariant components, naturally emerges..
### Tensors and rotations Physics Stack Exchange
What are other applications of spin-tensors in physics?. Chapter 4 Application of Tensors in Special Relativity 4.1 The energy-momentum tensor Consider a pressure-less distribution of non-interacting particles [ called dust ], https://en.m.wikipedia.org/wiki/Category:Tensors For all the tensors defined, the Physics commands will use the Einstein sum rule for repeated indices when manipulating them (simplification, differentiation, etc.)..
• Tensor Calculus for Physics A Concise Guide
• Examples of Tensors Department of Physics USU
• Mathematica Application Tensors in Physics Wolfram Store
• This textbook presents the foundations of tensor calculus and the elements of tensor analysis. In addition, the authors consider numerous applications of tensors to Tensors in Physics is a two-in-one Mathematica application package containing: The tensor concept is important in physics and has wide applications in such diverse fields as relativity theory, cosmology, high energy physics, field theory, thermodynamics, fluid dynamics, and mechanics.
Multi-Linear Algebra, Tensors and Spinors in Mathematical Physics. by Valter Moretti www.science.unitn.it/Лmoretti/home.html Department of Mathematics, Tensors, defined mathematically, are simply arrays of numbers, or functions, that transform according to certain rules under a change of coordinates. In physics
In addition, the authors consider numerous applications of tensors to geometry, mechanics and physics. While developing tensor calculus, Tensor Techniques in Physics – a concise introduction Roy McWeeny Professore Emerito di Chimica Teorica, Universit`a di Pisa, Pisa 2.1 Tensors of higher rank
A Student’s Guide to Vectors and Tensors tensors in physics and engineering, dedicated to example tensor applications. The first sections of the book provide an introduction to the vector and tensor algebra and analysis, with applications to physics, at undergraduate level. Second rank tensors, in particular their symmetries, …
For all the tensors defined, the Physics commands will use the Einstein sum rule for repeated indices when manipulating them (simplification, differentiation, etc.). Vector and Tensor Analysis with Applications and for those interested in the applications of tensor calculus to mathematical physics and engineering. Tensor
Tensors and rotations. the notion of a tensor that is often used in physics is not restricted to that of multilinear maps on manifolds Web Applications; Standard work applies tensorial methods to subjects within realm of advanced college mathematics. Text explains fundamental ideas and notation of tensor theory
Computing with Tensors: Potential Applications of Physics-Motivated Mathematics to Computer Science Martine Ceberio and Vladik Kreinovich Department of Computer Science by Pamela Burnley, University of Nevada Las Vegas Introduction The Stress Tensor The Strain Tensor Elasticity Literature Many physical properties of crystalline
Tensors are frequently used in engineering to describe measured quantities. Common applications. Representing mechanical stress as the Cauchy stress tensor in continuum mechanics; Measuring deformations (finite deformation tensors) and strain (strain tensor) in continuum mechanics; Representing diffusion as a tensor in Diffusion tensor imaging by Pamela Burnley, University of Nevada Las Vegas Introduction The Stress Tensor The Strain Tensor Elasticity Literature Many physical properties of crystalline
Applications of Group Theory to the Physics of Solids M. S †Application of Group Theory to Selection Rules and †Transformation Properties of Tensors Vector and Tensor Analysis with Applications and for those interested in the applications of tensor calculus to mathematical physics and engineering. Tensor
Scalars, Vectors, Tensors and All That. Physics equations involve tensors of the same rank. There are scalar equations, polar vector equations, Tensors and rotations. the notion of a tensor that is often used in physics is not restricted to that of multilinear maps on manifolds Web Applications;
Computing with Tensors: Potential Applications of Physics-Motivated Mathematics to Computer Science Martine Ceberio and Vladik Kreinovich Department of Computer Science 7 An application to viscosity 42 i. A Primeron Tensor Calculus 1 Introduction In physics, in differentiating tensors is the basis of tensor calculus,
Licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License Tensor Techniques in Physics – a concise introduction Roy McWeeny linking the various forms of tensors1 and, more importantly, in differentiating tensors is the basis of tensor calculus, and the subject of this primer. 1Examples of tensors the reader is already familiar with include scalars (rank 0 tensors) and vectors (rank 1 tensors).
Computing with Tensors: Potential Applications of Physics-Motivated Mathematics to Computer Science Martine Ceberio and Vladik Kreinovich Department of Computer Science Standard work applies tensorial methods to subjects within realm of advanced college mathematics. Text explains fundamental ideas and notation of tensor theory
What is the physical meaning of a tensor? What are interesting examples of tensors in physics? ... to the Riemann curvature tensor and applications of tensors in mechanics and physics , What are interesting examples of tensors in physics?
Tensors in Physics is a two-in-one package containing: A user's guide and a brief refresher course in differential geometry that also aims to clarify and explain the Foundations of Mathematical Physics: Vectors, Tensors and Fields 2009 – 2010 John Peacock www.roe.ac.uk/japwww/teaching/vtf.html Textbooks The standard recommended
Foundations of Tensor Analysis for Students of Physics and Engineering With an Introduction to the Theory of Relativity and the application of tensors in Applications of Euler’s Formula and Tensors, were not in my one Mathematical Methods for Physics and Engineering by Riley, Hobson,
by Pamela Burnley, University of Nevada Las Vegas Introduction The Stress Tensor The Strain Tensor Elasticity Literature Many physical properties of crystalline Application of Electromagnetic Field Tensors in Special Relativity Department of Physics, Faculty of Science, we start with electromagnetic field tensors,
In mathematics, tensors are geometric objects that describe linear relations between geometric vectors, scalars, and other tensors. Elementary examples of such relations include the dot product, the cross product, and linear maps. Geometric vectors, often used in physics and engineering applications, and scalars themselves are also tensors. ... to the Riemann curvature tensor and applications of tensors in mechanics and physics , What are interesting examples of tensors in physics?
Computing with Tensors: Potential Applications of Physics-Motivated Mathematics to Computer Science Martine Ceberio and Vladik Kreinovich Department of Computer Science What is the physical meaning of a tensor? What are interesting examples of tensors in physics? | 2,778 | 14,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-04 | latest | en | 0.90335 |
https://classroomsecrets.co.uk/new-resource-line-fluency-matrix/ | 1,725,804,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00307.warc.gz | 162,865,790 | 79,131 | ### New Resource - Fluency Matrix!
Our new maths resource line – Fluency Matrix is an additional fluency resource, providing extra fluency opportunities. They do not need to be used in maths lessons – instead they can be used as part of intervention groups or small groups within the classroom. Every class is different so use them in the way to suit you and your class. All of the questions could be given or it could be broken down into smaller, manageable chunks. Don't worry they follow the national curriculum and can be used with any maths scheme.
#### Want to skip to ALL the new Maths resources?
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### Free Year 2 Numbers to 20 Fluency Matrix
Here is a fluency matrix resource that's ideal for recapping numbers to 20 in both numerals and words.
### Year 2 Tens and Ones Fluency Matrix
This Year 2 Tens and Ones Fluency Matrix is the ideal resource for testing your pupils' fluency skills when studying tens and ones.
### Year 3 Free Represent and Partition Numbers
Looking for a resource to give your class more fluency practice? Then this is the resource for you!
### Year 3 Multiples of 100 Fluency Matrix
This Year 3 Multiples of 100 Fluency Matrix contains fluency questions that progress in difficulty through small incremental changes.
### Year 4 Free Represent and Partition Numbers
Are you looking for a way to help children develop the skills needed to represent and partition numbers? Then look no further.
### Year 4 Multiples of 1,000 Fluency Matrix
Help children to further develop their knowledge and understanding of multiples of 1,000 when you download and use this Year 4 Multiples of 1,000 resource.
### Free Year 5 Read and Write Numbers
Help children further develop their fluency skills with this Year 5 Read and Write Numbers up to 10,000 resource.
### Year 5 Read and Write Numbers up to 100,000
Help children to feel fantastic about numbers with up to 6 digits when you present them with this Year 5 Read and Write Numbers up to 100,000 resource.
### Year 6 Read, Write and Determine Value
Develop your children's fluency skills further with this Year 6 Read, Write and Determine the Value up to 1,000,000 resource.
### Year 6 Determine the Value of Each Digit
Why not provide your children with the opportunity to consolidate their knowledge and understanding of numbers to 10,000,000 with this Year 6 Determine the Value of Each Digit resource. | 529 | 2,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.902787 |
https://metanumbers.com/3205 | 1,670,338,427,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00219.warc.gz | 435,198,993 | 7,259 | # 3205 (number)
3,205 (three thousand two hundred five) is an odd four-digits composite number following 3204 and preceding 3206. In scientific notation, it is written as 3.205 × 103. The sum of its digits is 10. It has a total of 2 prime factors and 4 positive divisors. There are 2,560 positive integers (up to 3205) that are relatively prime to 3205.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 10
• Digital Root 1
## Name
Short name 3 thousand 205 three thousand two hundred five
## Notation
Scientific notation 3.205 × 103 3.205 × 103
## Prime Factorization of 3205
Prime Factorization 5 × 641
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) 3205 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 3,205 is 5 × 641. Since it has a total of 2 prime factors, 3,205 is a composite number.
## Divisors of 3205
1, 5, 641, 3205
4 divisors
Even divisors 0 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 3852 Sum of all the positive divisors of n s(n) 647 Sum of the proper positive divisors of n A(n) 963 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 56.6127 Returns the nth root of the product of n divisors H(n) 3.32814 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,205 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 3,205) is 3,852, the average is 963.
## Other Arithmetic Functions (n = 3205)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2560 Total number of positive integers not greater than n that are coprime to n λ(n) 640 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 458 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 2,560 positive integers (less than 3,205) that are coprime with 3,205. And there are approximately 458 prime numbers less than or equal to 3,205.
## Divisibility of 3205
m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 6 5 1
The number 3,205 is divisible by 5.
## Classification of 3205
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (3205)
Base System Value
2 Binary 110010000101
3 Ternary 11101201
4 Quaternary 302011
5 Quinary 100310
6 Senary 22501
8 Octal 6205
10 Decimal 3205
12 Duodecimal 1a31
20 Vigesimal 805
36 Base36 2h1
## Basic calculations (n = 3205)
### Multiplication
n×y
n×2 6410 9615 12820 16025
### Division
n÷y
n÷2 1602.5 1068.33 801.25 641
### Exponentiation
ny
n2 10272025 32921840125 105514497600625 338173964810003125
### Nth Root
y√n
2√n 56.6127 14.7438 7.52414 5.02534
## 3205 as geometric shapes
### Circle
Diameter 6410 20137.6 3.22705e+07
### Sphere
Volume 1.37903e+11 1.29082e+08 20137.6
### Square
Length = n
Perimeter 12820 1.0272e+07 4532.55
### Cube
Length = n
Surface area 6.16322e+07 3.29218e+10 5551.22
### Equilateral Triangle
Length = n
Perimeter 9615 4.44792e+06 2775.61
### Triangular Pyramid
Length = n
Surface area 1.77917e+07 3.87988e+09 2616.87
## Cryptographic Hash Functions
md5 9ef2ed4b7fd2c810847ffa5fa85bce38 9a04c15c47acbefccdf4bb3d16714867510e1a0c 74f71382d09d6d1ae00872640b9aee7db0e5036d96a7ef5a2ab52ec7f9b132ca b28803dd676df90166d674bb36af483aa02d0d951502fde59ba5894fbfd1df3cc11aeb7c70d245a4f7da88601993a1231df6171440708ce359a8634ec1b46b2e c97a5c3c595cc770eaa0e7e98f759ba4b89cb3a1 | 1,435 | 4,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.808995 |
https://www.ceder.net/def/chainthesquare.php?language=germany%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2%C3%83%C6%92%C3%A2%E2%82%AC%C2%A0%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%A2%E2%80%9E%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%82%C2%A2%C3%83%C2%A2%C3%A2%E2%82%AC%C5%A1%C3%82%C2%AC%C3%83%E2%80%A6%C3%82%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%AC%C3%83%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%B0%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2%C3%83%C6%92%C3%82%C2%A2%C3%83%C2%A2%C3%A2%E2%82%AC%C5%A1%C3%82%C2%AC%C3%83%E2%80%A6%C3%82%C2%A1%C3%83%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A4vel=C3A&level=C2 | 1,619,120,266,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039604430.92/warc/CC-MAIN-20210422191215-20210422221215-00451.warc.gz | 779,310,095 | 6,361 | Definitions of Square Dance Calls and Concepts
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
Chain The Square -- [C2]
C2:
From an Eight Chain Thru (or Parallel R-H Waves).
Right Pull By; Outsides Courtesy Turn and Veer Left as the Centers Step To A L-H Wave, Very Centers Left-face U-Turn Back (placing R-H on adjacent End's back), Courtesy Turn and As Couples Extend.
Ends in Parallel R-H Two-Faced Lines.
beforeChain The Square afterRight Pull By &Centers Step to L-H Wave after Very CentersLeft-face U-Turn Back after all Courtesy Turn afterAs Couples Extend (done)
Notes:
• The Courtesy Turn movement for the new Centers (after the R-H Pull By) is sometimes described as 1/4 Out; Courtesy Turn & 1/4 more. We dislike this definition for 2 reasons:
• The original Outside Belle's part is somewhat awkward because there is first a 1/4 Out (to the Right) and then Courtesy Turn (to the Left) 3/4, instead of just a smooth 180° Courtesy Turn.
• If you advance into C3, the caller may call a Left Chain The Square in which case we believe it is easier to remember to have the new Centers Step To A R-H Wave & then the Very Centers always turn the Ends. If, instead, you were to 1/4 Out, you can easily become confused as to who turns who.
Hints:
• After the new Centers Step To A Wave, it is always the Centers of the Wave that take the Ends back to where the Ends came from.
CALLERLAB definition for Chain the Square
Choreography for Chain The Square
C2:
https://www.ceder.net/def/chainthesquare.php?language=germany%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%
E2%82%AC%E2%84%A2%C3%83%C6%92%C3%A2%E2%82%AC%C2%A0%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%A2%E2%80%9E%C2%A2%C3%83
%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%8
0%9A%C3%82%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2%C3%83%C6%92%C3%A2%E2%82%AC%
C5%A1%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%82%C2%A2
%C3%83%C2%A2%C3%A2%E2%82%AC%C5%A1%C3%82%C2%AC%C3%83%E2%80%A6%C3%82%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%8
3%E2%80%9A%C3%82%C2%AC%C3%83%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%
A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%B0%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2
%C3%83%C6%92%C3%82%C2%A2%C3%83%C2%A2%C3%A2%E2%82%AC%C5%A1%C3%82%C2%AC%C3%83%E2%80%A6%C3%82%C2%A1%C3%83%C6%92%C
3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%
82%C2%A4vel=C3A&level=C2
22-April-2021 12:37:45 | 1,234 | 2,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-17 | latest | en | 0.778567 |
https://www.tutorialspoint.com/bag-of-tokens-in-cplusplus | 1,679,563,686,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00779.warc.gz | 1,187,648,022 | 10,036 | Bag of Tokens in C++
Suppose we have an initial power P, an initial score of 0 points, and one bag of tokens. Now each token can be used at most once, there is a value token[i], and has potentially two ways to use it, these are as follows −
• If we have at least token[i] power, then we may play the token face up, losing token[i] power, and gaining 1 point.
• Otherwise when we have at least 1 point, we may play the token face down, gaining token[i] power, and losing 1 point.
We have to find the largest number of points that we can have after playing any number of tokens.
So if the input is like tokens = [100,200,300,400] and P = 200, then the output will be 2.
To solve this, we will follow these steps −
• n := size of array v, ret := 0
• sort the v array
• set i := 0 and j := n – 1, curr :=
• while i <= j and x >= v[i]
• while i <= j and x >= v[i]
• decrease x by v[i], increase curr and i by 1
• ret := max of curr and ret
• while j >= i and curr is not 0 and x < v[i]
• increase x by v[j], decrease curr and j by 1
• return ret
Let us see the following implementation to get better understanding −
Example
Live Demo
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int bagOfTokensScore(vector<int>& v, int x) {
int n = v.size();
int ret = 0;
sort(v.begin(), v.end());
int i = 0;
int j = n - 1;
int curr = 0;
while(i <= j && x >= v[i]){
while(i <= j && x >= v[i]){
x -= v[i];
curr++;
i++;
}
ret = max(curr, ret);
while(j >= i && curr && x < v[i]){
curr--;
x += v[j];
j--;
}
}
return ret;
}
};
main(){
vector<int> v1 = {100,200,300,400};
Solution ob;
cout << (ob.bagOfTokensScore(v1, 200));
}
Input
[100,200,300,400]
200
Output
2 | 508 | 1,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-14 | latest | en | 0.798437 |
https://www.topperlearning.com/answer/derive-the-nernst-eqn-for-daniel-cell/6jpml6hh | 1,713,435,984,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00654.warc.gz | 955,494,531 | 55,609 | Request a call back
derive the nernst eqn for daniel cell
Asked by aswathy sreekumar | 13 May, 2014, 08:19: PM
For Daniel cell,
Zn(s)+Cu2+(aq) <=> Zn2+(aq) + Cu(s)
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu
In Daniell cell, the electrode potential for any given concentration of Cu2+ and Zn2+ ions, we write
For Cathode:
E(Cu2+/Cu) = E0(Cu2+/Cu) – RT/2F ln(1/[Cu2+(aq)]
For Anode:
E(Zn2+/Zn) = E0(Zn2+/Zn) – RT/2F ln(1/[Zn2+(aq)]
The cell potential, Ecell = E(Cu2+/Cu) - E(Zn2+/Zn)
= E0(Cu2+/Cu) – RT/2F ln(1/[Cu2+(aq)] - E0(Zn2+/Zn) + RT/2F ln(1/[Zn2+(aq)]
= E0(Cu2+/Cu) -E0(Zn2+/Zn)- RT/2F ln(1/[Cu2+(aq)] + RT/2F ln(1/[Zn2+(aq)]
= E0(Cu2+/Cu) -E0(Zn2+/Zn)- RT/2F (ln(1/[Cu2+(aq)]- ln(1/[Zn2+(aq)])
Therefore, Nernst equation for Daniel cell is
Ecell = E0cell - RT/2F ln [Zn2+]/[Cu2+]
Ecell = E?cell2.303RT/2F ln [Zn2+]/[Cu2+]
Answered by Prachi Sawant | 14 May, 2014, 10:44: AM
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Asked by prakriti12oct | 14 Aug, 2019, 12:46: AM | 622 | 1,351 | {"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} | 3.09375 | 3 | CC-MAIN-2024-18 | latest | en | 0.788968 |
https://www.khanacademy.org/math/get-ready-for-ap-statistics/xc9bacb4afa74e6e9:get-ready-for-exploring-one-variable-quantitative-data/xc9bacb4afa74e6e9:equations-variables-both-sides/v/solving-equations-with-the-distributive-property-2 | 1,709,497,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00856.warc.gz | 846,620,583 | 80,301 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Course: Get ready for AP® Statistics>Unit 2
Lesson 6: Linear equations with variables on both sides
# Equation with variables on both sides: fractions
To solve the equation (3/4)x + 2 = (3/8)x - 4, we first eliminate fractions by multiplying both sides by the least common multiple of the denominators. Then, we add or subtract terms from both sides of the equation to group the x-terms on one side and the constants on the other. Finally, we solve and check as normal. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• so it is x=-16?
• Yes, but as I saw the comment from a year ago the same thought popped up in my head. Where are you know. Your comment was from 9 years ago, and your description saws you were 14 so you would have to be 23, or so. Time flies, wow
• I am struggling on how to transfer varibles to one side of the equation. Any tips?
• Hi, Rebecca;
Transferring Variables
Transferring variables might look like a complex subject to tackle at first glance, but it actually proves itself to be much simpler--you just need to understand it.
In an equation, the left hand side (LHS--the left expression) and the right hand side (RHS--the right expression) are equal. Now, a very significant tip to take note is that since both sides are equal, both must be treated equally. But how do we do that?
Here is an example:
Billy has two baskets of equally filled apples. The left basket has 2 packs of 4 apples and 2 pears while the right basket has 3 packs of 2 apples and 2 pears. On the way home, Billy decided to eat 3 fruits from each basket. How many fruits are left in each basket?
``2(4+2)=3(2+2)``
I've done the equation for the original number of fruits in each basket, but after Billy took 3 from each basket, I am left to modify my equation. But how do we do that?
``2(4+2)-3=3(2+2)-3``
We subtract 3 from both sides! This would mean that both baskets, being originally equal, would still be equal when Billy goes home to eat the rest.
``2(4+2) -3 =3(2+2) -32(6)-3=3(4)-312-3=12-39=9∴ There are 9 fruits left in each basket``
Now what does this have to do with transferring variables? Transferring variables are basically like what we did above, except they're not numbers yet.
This time, let's say we don't actually know how many pears there are in each pack, considering the fact that the number of pears are equal in all of the packs in both baskets.
``2(4+x)=3(2+x)``
First, let us simplify the equation.
``2(4+x)=3(2+x)8+2x=6+3x``
Here we are--transferring variables! This time, think about what we did to the equation when Billy decided to eat 3 fruits from both baskets; we subtract the variable from both sides!
``8+2x=6+3x8+2x -3x =6+3x -3x8+2x-3x=6``
Notice that when we transfered `3x` from the RHS to the LHS, it turned negative. When we transfer variables to the other side, its sign becomes opposite! That's how easy it is!
Now, let's solve the equation!
``8+2x-3x=62x-3x=6-8-x=-2-1(-x)=-1(-2)x=2∴ There are 2 pears in each pack.``
Ta-da! The same equation!
(Sorry if I was very lengthened about such a simple subject. I like to explain thoroughly)
• can you multiply the denominator on both sides
• If you multiply both sides by an integer, it's always multiplying the numerator and therefore making the number larger. If you did otherwise, by multiplying the denominator, then the number would be smaller, which is rather a division. I hope my explanation makes sense and is helpful.
• in the test i got the problem...
16 - 2t = 3/2t + 9
and i converted the fraction to the decimal 1.5
so...
16 - 2t = 1.5t + 9
-16 -16
-2t = 1.5t - 7
-1.5t -1.5t
0.5t = -7
i devided both sides by 0.5 and got -14, so i punched the answer in and they said the correct answer was actually 2. what did i do wrong?
• hello, so what you did wrong was simply a subtracting mistake. you can totally just convert your fraction into a decimal and it will still work. So lets start from the beginning,
16 - 2t = 3/2t +9
so you convert the fraction into the decimal
16 - 2t = 1.5t + 9
then you subtracted 16 from both sides which is right,
16 - 2t = 1.5t +9
-16 -16
-2t = 1.5t -7
you were right up to this step. now we subtract 1.5t from both sides
-2t = 1.5t -7
-1.5t -1.5
you get...
-3.5t = -7
which equals 2!
so you only messed up in the step where you add a -2t to a -1.5t.
if you do not understand why we add these together look at Khan's video "adding negative numbers example"
hope this helps
• I still didn't understand where the 8 came from, can someone please explain it again differently? How do you get to that conclusion?
• The 8 is the lowest common multiple of the 2 denominators (4 and 8). Use the same process you would use to select the smallest common denominator for those 2 fractions.
Multiples of 4: 4, 8, 12, 16, etc.
Multiples of 8: 8, 16, 24, etc.
The first multiple in common is 8.
Hope this helps.
• But arent what you do to one side you must do to the other??
• Another way to solve this "quickly" is this: 3/4x+2=3/8x-4, what number you need to multiply 3/4x for so denominator becomes 8 as well?
You multiply by 2 and get 6/8x+2=3/8x-4
1st step 6/8x-3/8+2=3/8x-3/8x-4
2nd step 3/8x+2-2=-4-2
3/8x/3/8=-6/3/8 (0.375)
x=-16
• The third step is much easier to multiply by the reciprocal rather than dividing, so
3/8 x • 8/3 = 6 • 8/3, since 6/3 =2, you get 16 faster.
If you are going to do it quickly, try doing everything as simply as possible, but this is great for those not afraid of fractions.
• how is this going to help us in the real world
• It probably won't, it probably will. It depends what you are doing in life and what job you have
• my question is how do you solve this problem with one fraction? | 1,744 | 5,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.950464 |
https://socratic.org/questions/how-do-you-use-the-remainder-theorem-to-evaluate-f-a-a-3-5a-2-10a-12-at-a-2 | 1,627,949,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00007.warc.gz | 525,755,907 | 5,780 | How do you use the remainder theorem to evaluate f(a)=a^3+5a^2+10a+12 at a=-2?
Oct 11, 2017
the remainder of the division $\left({a}^{3} + 5 {a}^{2} + 10 a + 12\right) : \left(a + 2\right)$ is 4
Explanation:
We simply will substitute $a = - 2$ in $f \left(a\right)$:
$f \left(- 2\right) = {\left(- 2\right)}^{3} + 5 {\left(- 2\right)}^{2} + 10 \left(- 2\right) + 12$
$= - 8 \cancel{+} 20 \cancel{- 20} + 12$
$= 4$
that means the remainder of the division:
$\left({a}^{3} + 5 {a}^{2} + 10 a + 12\right) : \left(a + 2\right)$
is 4 | 234 | 538 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-31 | latest | en | 0.559527 |
https://philoid.com/question/30694-if-and-are-the-zeros-of-the-polynomial-fx-5x-2-7x-1-find-the-value-of | 1,713,849,052,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00424.warc.gz | 413,829,117 | 8,312 | ##### If α and β are the zeros of the polynomial f(x) = 5x2 ‒ 7x + 1, find the value of
It is given in the question that,
Zeros of the polynomial 5x2 – 7x + 1 are
Now by using the relationship between the zeros of the quadratic polynomial we have:
Sum of zeros = and product of zeros =
= and
= and
Now, we have:
=
=
= 7
23 | 102 | 333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-18 | latest | en | 0.941917 |
https://oeis.org/A106609 | 1,600,613,177,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00384.warc.gz | 553,534,678 | 5,721 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A106609 Numerator of n/(n+8). 12
0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 2, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 4, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 6, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 8, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77, 39, 79 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS The graph of this sequence is made up of four linear functions: a(n_odd)=n, a(n=2+4i)=n/2, a(4+8i)=n/4, a(8i)=n/8. - Zak Seidov, Oct 30 2006. [In general, f(n) = numerator of n/(n+m) consists of linear functions n/d_i, where d_i are divisors of m (including 1 and m).] Multiplicative with a(2^e) = 2^max(0,e-3). a(p^e) = p^e if p>2. - R. J. Mathar, Apr 18 2011 a(n+2), n>=0, is the denominator of the harmonic mean H(n,2) = 4*n/(n+2). a(n+2) = (n+2)/gcd(n+2,8). a(n+5) = A227042(n+2, 2), n >= 0. - Wolfdieter Lang, Jul 04 2013 The sequence p(n) = a(n-4), n>=1, with a(-3) = a(3) = 3, a(-2) = a(2) = 1 and a(-1) = a(1) = 1, appears in the problem of writing 2*sin(2*Pi/n) as an integer in the algebraic number field Q(rho(q(n))), where rho(k) = 2*cos(Pi/k) and q(n) = A225975(n). One has 2*sin(2*Pi/n) = R(p(n), x) modulo C(q(n), x), with x = rho(q(n)) and the integer polynomials R and C given in A127672 and A187360, respectively. See a comment on A225975. - Wolfdieter Lang, Dec 04 2013 A204455(n) divides a(n) for n>=1. - Alexander R. Povolotsky, Apr 06 2015 A multiplicative sequence. Also, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n >= 1, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 20 2019 LINKS N. J. A. Sloane, Table of n, a(n) for n = 0..10000 Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,-1). FORMULA a(n) = 2*a(n-8) - a(n-16). G.f.: x* (x^2-x+1) * (x^12 +2*x^11 +4*x^10 +3*x^9 +4*x^8 +4*x^7 +7*x^6 +4*x^5 +4*x^4 +3*x^3 +4*x^2 +2*x +1) / ( (x-1)^2 *(x+1)^2 *(x^2+1)^2 *(x^4+1)^2 ). - R. J. Mathar, Dec 02 2010 a(n) = A109049(n)/8. Dirichlet g.f. zeta(s-1)*(1-1/2^s-1/2^(2s)-1/2^(3s)). - R. J. Mathar, Apr 18 2011 a(n) = n/gcd(n,8), n >= 0. See the harmonic mean comment above. - Wolfdieter Lang, Jul 04 2013 a(n) = n if n is odd; for n == 0 (mod 8) it is n/8, for n == 2 or 6 (mod 8) it is n/2 and for n == 4 (mod 8) it is n/4. - Wolfdieter Lang, Dec 04 2013 From Peter Bala, Feb 20 2019: (Start) O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4) - F(x^8), where F(x) = x/(1 - x)^2. More generally, for m >= 1, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 2^m)*( F(m,x^2) + F(m,x^4) + F(m,x^8) ), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m-th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292. Sum_{n >= 1} (1/n)*a(n)*x^n = G(x) - (1/2)*G(x^2) - (1/4)*G(x^4) - (1/8)*G(x^8), where G(x) = x/(1 - x). Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4)^2*L(x^4) - (1/8)^2*L(x^8), where L(x) = Log(1/(1 - x)). Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8). (End) MAPLE a := n -> iquo(n, [8, 1, 2, 1, 4, 1, 2, 1][1 + modp(n, 8)]): seq(a(n), n=0..79); # using Wolfdieter Lang's formula, Peter Luschny, Feb 22 2019 MATHEMATICA f[n_]:=Numerator[n/(n+8)]; Array[f, 100, 0] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011 *) LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, -1}, {0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15}, 100] (* Harvey P. Dale, Sep 27 2019 *) PROG (Sage) [lcm(n, 8)/8 for n in range(0, 100)] # Zerinvary Lajos, Jun 09 2009 (MAGMA) [Numerator(n/(n+8)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011 (PARI) vector(100, n, n--; numerator(n/(n+8))) \\ G. C. Greubel, Feb 19 2019 (GAP) List([0..80], n->NumeratorRat(n/(n+8))); # Muniru A Asiru, Feb 19 2019 CROSSREFS Cf. A109049, A204455. A227042 (second column, starting with a(5)). Cf. Sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20). Sequence in context: A227140 A106617 A040026 * A171968 A093474 A030101 Adjacent sequences: A106606 A106607 A106608 * A106610 A106611 A106612 KEYWORD nonn,frac,mult AUTHOR N. J. A. Sloane, May 15 2005 STATUS approved
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Last modified September 20 10:24 EDT 2020. Contains 337264 sequences. (Running on oeis4.) | 2,271 | 4,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.779973 |
https://metanumbers.com/217555 | 1,637,991,250,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358118.13/warc/CC-MAIN-20211127043716-20211127073716-00580.warc.gz | 477,432,436 | 7,399 | # 217555 (number)
217,555 (two hundred seventeen thousand five hundred fifty-five) is an odd six-digits composite number following 217554 and preceding 217556. In scientific notation, it is written as 2.17555 × 105. The sum of its digits is 25. It has a total of 3 prime factors and 8 positive divisors. There are 160,608 positive integers (up to 217555) that are relatively prime to 217555.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 25
• Digital Root 7
## Name
Short name 217 thousand 555 two hundred seventeen thousand five hundred fifty-five
## Notation
Scientific notation 2.17555 × 105 217.555 × 103
## Prime Factorization of 217555
Prime Factorization 5 × 13 × 3347
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) 217555 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 217,555 is 5 × 13 × 3347. Since it has a total of 3 prime factors, 217,555 is a composite number.
## Divisors of 217555
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) 281232 Sum of all the positive divisors of n s(n) 63677 Sum of the proper positive divisors of n A(n) 35154 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 466.428 Returns the nth root of the product of n divisors H(n) 6.18863 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 217,555 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 217,555) is 281,232, the average is 35,154.
## Other Arithmetic Functions (n = 217555)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 160608 Total number of positive integers not greater than n that are coprime to n λ(n) 20076 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 19364 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 160,608 positive integers (less than 217,555) that are coprime with 217,555. And there are approximately 19,364 prime numbers less than or equal to 217,555.
## Divisibility of 217555
m n mod m 2 3 4 5 6 7 8 9 1 1 3 0 1 2 3 7
The number 217,555 is divisible by 5.
## Classification of 217555
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (217555)
Base System Value
2 Binary 110101000111010011
3 Ternary 102001102121
4 Quaternary 311013103
5 Quinary 23430210
6 Senary 4355111
8 Octal 650723
10 Decimal 217555
12 Duodecimal a5a97
16 Hexadecimal 351d3
20 Vigesimal 173hf
36 Base36 4nv7
## Basic calculations (n = 217555)
### Multiplication
n×y
n×2 435110 652665 870220 1087775
### Division
n÷y
n÷2 108778 72518.3 54388.8 43511
### Exponentiation
ny
n2 47330178025 10296916880228875 2240145751878192900625 487354909049860256495471875
### Nth Root
y√n
2√n 466.428 60.1436 21.5969 11.6819
## 217555 as geometric shapes
### Circle
Radius = n
Diameter 435110 1.36694e+06 1.48692e+11
### Sphere
Radius = n
Volume 4.31316e+16 5.94769e+11 1.36694e+06
### Square
Length = n
Perimeter 870220 4.73302e+10 307669
### Cube
Length = n
Surface area 2.83981e+11 1.02969e+16 376816
### Equilateral Triangle
Length = n
Perimeter 652665 2.04946e+10 188408
### Triangular Pyramid
Length = n
Surface area 8.19783e+10 1.2135e+15 177633
## Cryptographic Hash Functions
md5 4cd95949e093a850ee941652a6c047d1 c1f77b3d6b7d41cecd7f66c8f78ec9c2ca49d563 65267c10fdfe0c9ce6aef1a3a2c05c84eec6d0f19e566c5a743faaeb4bd2e875 c90f37b9d18cb4afb2b08014c2fe3bef909995d7fbc2545458b4a72964baed63d1dddcfd36186ac711263e974950da73397a7ca23d33e14d6b78d3037467631b 68e58e2b232becbbc11eee4aaf0d545c8a3b75ba | 1,478 | 4,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.809334 |
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# 5.5: The Substitution Rule
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Learning Objectives
• Use substitution to evaluate indefinite integrals.
• Use substitution to evaluate definite integrals.
The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us find antiderivatives when the integrand is the result of a chain-rule derivative.
At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to see? We are looking for an integrand of the form $$f[g(x)]g′(x)dx$$. For example, in the integral
$∫(x^2−3)^3 2x \, dx. \label{eq1}$
we have
$f(x)=x^3 \nonumber$
and
$g(x)=x^2−3.\nonumber$
Then
$g'(x)=2x.\nonumber$
and
$f[g(x)]g′(x)=(x^2−3)^3(2x),\nonumber$
and we see that our integrand is in the correct form. The method is called substitution because we substitute part of the integrand with the variable $$u$$ and part of the integrand with $$du$$. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier to work with for applying the integration rules.
Substitution with Indefinite Integrals
Let $$u=g(x)$$,, where $$g′(x)$$ is continuous over an interval, let $$f(x)$$ be continuous over the corresponding range of g, and let $$F(x)$$ be an antiderivative of $$f(x).$$ Then,
\begin{align*} ∫f[g(x)]g′(x)\,dx &=∫f(u)\,du \\[4pt] &=F(u)+C \\[4pt] &= F(g(x))+C \end{align*}
Proof
Let $$f$$, $$g$$, $$u$$, and $$F$$ be as specified in the theorem. Then
$\dfrac{d}{dx}\big[F(g(x))\big]=F′(g(x))g′(x)=f[g(x)]g′(x).$
Integrating both sides with respect to x, we see that
$∫f[g(x)]g′(x)\,dx=F(g(x))+C.$
If we now substitute $$u=g(x)$$, and $$du=g'(x)dx$$, we get
$∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C.$
Returning to the problem we looked at originally, we let $$u=x^2−3$$ and then $$du=2x\,dx$$.
Rewrite the integral (Equation \ref{eq1}) in terms of $$u$$:
$∫(x^2−3)^3(2x\,dx)=∫u^3\,du.$
Using the power rule for integrals, we have
$∫u^3\,du=\dfrac{u^4}{4}+C.$
Substitute the original expression for $$x$$ back into the solution:
$\dfrac{u^4}{4}+C=\dfrac{(x^2−3)^4}{4}+C.$
We can generalize the procedure in the following Problem-Solving Strategy.
Problem-Solving Strategy: Integration by Substitution
1. Look carefully at the integrand and select an expression $$g(x)$$ within the integrand to set equal to u. Let’s select $$g(x)$$. such that $$g′(x)$$ is also part of the integrand.
2. Substitute $$u=g(x)$$ and $$du=g′(x)dx.$$ into the integral.
3. We should now be able to evaluate the integral with respect to $$u$$. If the integral can’t be evaluated we need to go back and select a different expression to use as $$u$$.
4. Evaluate the integral in terms of $$u$$.
5. Write the result in terms of $$x$$ and the expression $$g(x).$$
Example $$\PageIndex{1}$$: Using Substitution to Find an Antiderivative
Use substitution to find the antiderivative of $$\displaystyle ∫6x(3x^2+4)^4\,dx.$$
Solution
The first step is to choose an expression for $$u$$. We choose $$u=3x^2+4$$ because then $$du=6x\,dx$$ and we already have $$du$$ in the integrand. Write the integral in terms of $$u$$:
$∫6x(3x^2+4)^4\,dx=∫u^4\,du. \nonumber$
Remember that $$du$$ is the derivative of the expression chosen for $$u$$, regardless of what is inside the integrand. Now we can evaluate the integral with respect to $$u$$:
$∫u^4\,du=\dfrac{u^5}{5}+C=\dfrac{(3x^2+4)^5}{5}+C.\nonumber$
Analysis
We can check our answer by taking the derivative of the result of integration. We should obtain the integrand. Picking a value for $$C$$ of $$1$$, we let $$y=\dfrac{1}{5}(3x^2+4)^5+1.$$ We have
$y=\dfrac{1}{5}(3x^2+4)^5+1,\nonumber$
so
\begin{align*} y′ &=\left(\dfrac{1}{5}\right)5(3x^2+4)^46x \\[4pt] &=6x(3x^2+4)^4.\end{align*}
This is exactly the expression we started with inside the integrand.
Exercise $$\PageIndex{1}$$
Use substitution to find the antiderivative of $$\displaystyle ∫3x^2(x^3−3)^2\,dx.$$
Hint
Let $$u=x^3−3.$$
$$\displaystyle ∫3x^2(x^3−3)^2\,dx=\dfrac{1}{3}(x^3−3)^3+C$$
Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.
Example $$\PageIndex{2}$$: Using Substitution with Alteration
Use substitution to find the antiderivative of $∫z\sqrt{z^2−5}\,dz. \nonumber$
Solution
Rewrite the integral as $$\displaystyle ∫z(z^2−5)^{1/2}\,dz.$$ Let $$u=z^2−5$$ and $$du=2z\,dz$$. Now we have a problem because $$du=2z\,dz$$ and the original expression has only $$z\,dz$$. We have to alter our expression for $$du$$ or the integral in $$u$$ will be twice as large as it should be. If we multiply both sides of the $$du$$ equation by $$\dfrac{1}{2}$$. we can solve this problem. Thus,
$u=z^2−5\nonumber$
$du=2z\,dz \nonumber$
$\dfrac{1}{2}du=\dfrac{1}{2}(2z)\,dz=z\,dz. \nonumber$
Write the integral in terms of $$u$$, but pull the $$\dfrac{1}{2}$$ outside the integration symbol:
$∫z(z^2−5)^{1/2}\,dz=\dfrac{1}{2}∫u^{1/2}\,du.\nonumber$
Integrate the expression in $$u$$:
\begin{align*} \dfrac{1}{2}∫u^{1/2}\,du &= \left(\dfrac{1}{2}\right)\dfrac{u^{3/2}}{\dfrac{3}{2}}+C \\[4pt] &= \left(\dfrac{1}{2}\right)\left(\dfrac{2}{3}\right)u^{3/2}+C \\[4pt] &=\dfrac{1}{3}u^{3/2}+C \\[4pt] &=\dfrac{1}{3}(z^2−5)^{3/2}+C \end{align*}
Exercise $$\PageIndex{2}$$
Use substitution to find the antiderivative of $$\displaystyle ∫x^2(x^3+5)^9\,dx.$$
Hint
Multiply the du equation by $$\dfrac{1}{3}$$.
$$\displaystyle ∫x^2(x^3+5)^9\,dx = \dfrac{(x^3+5)^{10}}{30}+C$$
Example $$\PageIndex{3}$$: Using Substitution with Integrals of Trigonometric Functions
Use substitution to evaluate the integral $$\displaystyle ∫\dfrac{\sin t}{\cos^3t}\,dt.$$
Solution
We know the derivative of $$\cos t$$ is $$−\sin t$$, so we set $$u=\cos t$$. Then $$du=−\sin t\,dt.$$
Substituting into the integral, we have
$∫\dfrac{\sin t}{\cos^3t}\,dt=−∫\dfrac{du}{u^3}.\nonumber$
Evaluating the integral, we get
$−∫\dfrac{du}{u^3}=−∫u^{−3}\,du=−\left(−\dfrac{1}{2}\right)u^{−2}+C.\nonumber$
Putting the answer back in terms of t, we get
$∫\dfrac{\sin t}{\cos^3t}\,dt=\dfrac{1}{2u^2}+C=\dfrac{1}{2\cos^2t}+C.\nonumber$
Exercise $$\PageIndex{3}$$
Use substitution to evaluate the integral $$\displaystyle ∫\dfrac{\cos t}{\sin^2t}\,dt.$$
Hint
Use the process from Example $$\PageIndex{3}$$ to solve the problem.
$$\displaystyle ∫\dfrac{\cos t}{\sin^2t}\,dt = −\dfrac{1}{\sin t}+C$$
Exercise $$\PageIndex{4}$$
Use substitution to evaluate the indefinite integral $$\displaystyle ∫\cos^3t\sin t\,dt.$$
Hint
Use the process from Example $$\PageIndex{3}$$ to solve the problem.
$$\displaystyle ∫\cos^3t\sin t\,dt = −\dfrac{\cos^4t}{4}+C$$
Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, $$u$$ should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of $$u$$. This technique should become clear in the next example.
Example $$\PageIndex{4}$$: Finding an Antiderivative Using u-Substitution
Use substitution to find the antiderivative of $∫\dfrac{x}{\sqrt{x−1}}\,dx. \nonumber$
Solution
If we let $$u=x−1,$$ then $$du=dx$$. But this does not account for the x in the numerator of the integrand. We need to express x in terms of u. If $$u=x−1$$, then $$x=u+1.$$ Now we can rewrite the integral in terms of u:
$∫\dfrac{x}{\sqrt{x−1}}\,dx=∫\dfrac{u+1}{\sqrt{u}}\,du=∫\sqrt{u}+\dfrac{1}{\sqrt{u}}\,du=∫(u^{1/2}+u^{−1/2})\,du.\nonumber$
Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,
\begin{align*} ∫(u^{1/2}+u^{−1/2})\,du &=\dfrac{2}{3}u^{3/2}+2u^{1/2}+C \\[4pt] &= \dfrac{2}{3}(x−1)^{3/2}+2(x−1)^{1/2}+C \\[4pt] &= (x−1)^{1/2}\left[\dfrac{2}{3}(x−1)+2\right]+C \\[4pt] &= (x−1)^{1/2}\left(\dfrac{2}{3}x−\dfrac{2}{3}+\dfrac{6}{3}\right) \\[4pt] &= (x−1)^{1/2}\left(\dfrac{2}{3}x+\dfrac{4}{3}\right) \\[4pt] &= \dfrac{2}{3}(x−1)^{1/2}(x+2)+C. \end{align*}
## Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
Substitution with Definite Integrals
Let $$u=g(x)$$ and let $$g'$$ be continuous over an interval $$[a,b]$$, and let $$f$$ be continuous over the range of $$u=g(x).$$ Then,
$∫^b_af(g(x))g′(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du.$
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $$F(x)$$ is an antiderivative of $$f(x),$$ we have
$∫f(g(x))g′(x)\,dx=F(g(x))+C.$
Then
\begin{align*} ∫^b_af[g(x)]g′(x)\,dx = F(g(x))\bigg|^{x=b}_{x=a} \\[4pt] =F(g(b))−F(g(a)) \\[4pt] = F(u) \bigg|^{u=g(b)}_{u=g(a)} \\[4pt] =∫^{g(b)}_{g(a)}f(u)\,du \end{align*}
and we have the desired result.
Example $$\PageIndex{5}$$: Using Substitution to Evaluate a Definite Integral
Use substitution to evaluate $∫^1_0x^2(1+2x^3)^5\,dx. \nonumber$
Solution
Let $$u=1+2x^3$$, so $$du=6x^2dx$$. Since the original function includes one factor of $$x^2$$ and $$du=6x^2dx$$, multiply both sides of the $$du$$ equation by $$1/6.$$ Then,
\begin{align*} du &=6x^2\,dx \\[4pt] \text{becomes}\quad \dfrac{1}{6}du &=x^2\,dx. \end{align*}
To adjust the limits of integration, note that when $$x=0,u=1+2(0)=1,$$ and when $$x=1,u=1+2(1)=3.$$ Then
$∫^1_0x^2(1+2x^3)^5dx=\dfrac{1}{6}∫^3_1u^5\,du. \nonumber$
Evaluating this expression, we get
\begin{align*} \dfrac{1}{6}∫^3_1u^5\,du &=(\dfrac{1}{6})(\dfrac{u^6}{6})|^3_1 \\[4pt] &=\dfrac{1}{36}[(3)^6−(1)^6] \\[4pt] &=\dfrac{182}{9}. \end{align*}
Exercise $$\PageIndex{5}$$
Use substitution to evaluate the definite integral $$\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy.$$
Hint
Use the steps from Example $$\PageIndex{5}$$ to solve the problem.
$$\displaystyle ∫^0_{−1}y(2y^2−3)^5\,dy = \dfrac{91}{3}$$
Exercise $$\PageIndex{6}$$
Use substitution to evaluate $$\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx.$$
Hint
Use the process from Example $$\PageIndex{5}$$ to solve the problem.
$$\displaystyle ∫^1_0x^2 \cos \left(\dfrac{π}{2}x^3\right)\,dx = \dfrac{2}{3π}≈0.2122$$
Example $$\PageIndex{6}$$: Using Substitution with an Exponential Function
Use substitution to evaluate $∫^1_0xe^{4x^2+3}\,dx. \nonumber$
Solution
Let $$u=4x^3+3.$$ Then, $$du=8x\,dx.$$ To adjust the limits of integration, we note that when $$x=0,\,u=3$$, and when $$x=1,\,u=7$$. So our substitution gives
\begin{align*} ∫^1_0xe^{4x^2+3}\,dx &= \dfrac{1}{8}∫^7_3e^u\,du \\[4pt] &=\dfrac{1}{8}e^u|^7_3 \\[4pt] &=\dfrac{e^7−e^3}{8} \\[4pt] &≈134.568 \end{align*}
Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for $$u$$ after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in Example $$\PageIndex{7}$$.
Example $$\PageIndex{7}$$: Using Substitution to Evaluate a Trigonometric Integral
Use substitution to evaluate $∫^{π/2}_0\cos^2θ\,dθ. \nonumber$
Solution
Let us first use a trigonometric identity to rewrite the integral. The trig identity $$\cos^2θ=\dfrac{1+\cos 2θ}{2}$$ allows us to rewrite the integral as
$∫^{π/2}_0\cos^2θ\,dθ=∫^{π/2}_0\dfrac{1+\cos2θ}{2}\,dθ. \nonumber$
Then,
\begin{align*} ∫^{π/2}_0\left(\dfrac{1+\cos2θ}{2}\right)\,dθ =∫^{π/2}_0\left(\dfrac{1}{2}+\dfrac{1}{2}\cos 2θ\right)\,dθ \\[4pt] =\dfrac{1}{2}∫^{π/2}_0\,dθ+∫^{π/2}_0\cos2θ\,dθ. \end{align*}
We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let $$u=2θ.$$ Then, $$du=2\,dθ,$$ or $$\dfrac{1}{2}\,du=dθ$$. Also, when $$θ=0,\,u=0,$$ and when $$θ=π/2,\,u=π.$$ Expressing the second integral in terms of $$u$$, we have
\begin{align*}\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}∫^{π/2}_0 \cos 2θ\,dθ &=\dfrac{1}{2}∫^{π/2}_0\,dθ+\dfrac{1}{2}\left(\dfrac{1}{2}\right)∫^π_0 \cos u \,du \\[4pt] &=\dfrac{θ}{2}\,\bigg|^{θ=π/2}_{θ=0}+\dfrac{1}{4}\sin u\,\bigg|^{u=θ}_{u=0} \\[4pt] &=\left(\dfrac{π}{4}−0\right)+(0−0)=\dfrac{π}{4} \end{align*}
## Key Concepts
• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable $$u$$ and $$du$$ for appropriate expressions in the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.
## Key Equations
• Substitution with Indefinite Integrals $∫f[g(x)]g′(x)\,dx=∫f(u)\,du=F(u)+C=F(g(x))+C \nonumber$
• Substitution with Definite Integrals $∫^b_af(g(x))g'(x)\,dx=∫^{g(b)}_{g(a)}f(u)\,du \nonumber$
## Glossary
change of variables
the substitution of a variable, such as $$u$$, for an expression in the integrand
integration by substitution
a technique for integration that allows integration of functions that are the result of a chain-rule derivative | 5,523 | 15,210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-24 | latest | en | 0.403109 |
http://powerpointpresentationon.blogspot.com/2013/08/ppt-on-magnetic-forces-materials-and.html | 1,524,247,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944677.39/warc/CC-MAIN-20180420174802-20180420194802-00221.warc.gz | 259,875,172 | 20,242 | ## Monday, August 5, 2013
### PPT On Magnetic Forces Materials And Inductance
Magnetic Forces Materials And Inductance Presentation Transcript:
1.Magnetic Forces, Materials, & Inductance
2.Magnetic Field
A bar magnet has a magnetic field around it. This field is 3D in nature and often represented by lines LEAVING north and ENTERING south
To define a magnetic field you need to understand the MAGNITUDE and DIRECTION
We sometimes call the magnetic field a B-Field as the letter “B” is the SYMBOL for a magnetic field with the TESLA (T) as the unit.
3.Magnetic Force on a moving charge
If a MOVING CHARGE moves into a magnetic field it will experience a MAGNETIC FORCE. This deflection is 3D in nature.
4.Lorentz Force Equation
The force on a moving particle arising from both Electric and Magnetic fields.
C is the speed of light.
5.Lorentz Force Equation
Electron Orbits in Magnetron.
Proton Paths in cyclotron.
Plasma Characteristics in MHD generator.
Charge particle motion in both combined Electric and Magnetic Fields.
6.Force on a Differential Current Element
Assume two current loops
7.Force on a Differential Current Element
8.Force and Torque on a closed circuit
9.Torque
10.Force and Torque on a closed circuit
If B is uniform on the loop current , then the net force is zero (no displacement)
If B is not uniform, the net force is not always zero
The magnetic field causes a Torque on the loop
This torque will try to make the magnetic field due to the loop current in the same direction of the external magnetic field B
11.Force and Torque on a closed circuit
12.The Nature of Magnetic Materials
Materials have a different behavior in magnetic fields
Accurate description requires quantum theory
However, simple atomic model (central nucleus surrounded by electrons) is enough for us
There are 3 magnetic dipole moments:
1. Moment due to rotation of the electrons
2. Moment due to the spin of the electrons
3. Moment due to the spin of the nucleus
13.We will study 6 types:
Diamagnetic
Paramagnetic
Ferromagnetic
Ferrimagnetic
Anti-ferromagnetic
Super paramagnetic
14.Diamagnetic Materials
Without an external magnetic field, diamagnetic materials have no net magnetic field
With an external magnetic field, they generate a small magnetic field in the opposite direction
The value of this opposite field depends on the external field and the diamagnetic material
15.Atoms have a small net magnetic dipole moment
The random orientation of atoms make the average dipole moment in the material zero
Without an external field, there is no magnetic property | 582 | 2,581 | {"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} | 3.296875 | 3 | CC-MAIN-2018-17 | latest | en | 0.861431 |
https://sbainvent.com/fluid-mechanics/acceleration-along-a-streamline/ | 1,590,584,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00533.warc.gz | 531,567,123 | 21,118 | # Acceleration along a Streamline
As a particle moves along a streamline and its velocity changes an acceleration along a streamline will occur. To understand acceleration along streamline it is helpful to use a coordinate system to define the streamline. Two examples of coordinate systems are the Cartesian and the polar coordinate system. First, the Cartesian coordinate system considers x, y, and z directions. On the other hand, the polar coordinate system considers the radius r, and angle θ from a specific origin.
In order to define a streamline, a streamline coordinate system is used to show the motion of a fluid particle. It does this by using unit vectors in two directions. One direction is along the streamline $\hat{s}$, while the other is normal to the streamline $\hat{n}$. Since a streamline coordinate system is expressed this way, the direction of these two vectors will change in respect to the Cartesian Coordinate system. This is because of the fact that the two vectors must either be tangent to the streamline or normal to the streamline, while the direction of the streamline can change itself.
### Acceleration along a Streamline
To determine the acceleration along a streamline you need to think about Dynamics. What happens when a particle accelerates around a curve? If a particle is accelerating around a curve there will be a normal ($\hat{n}$) and possibly a tangential ($\hat{s}$) acceleration. As a result, as a fluid particle moves along a streamline a tangential and normal acceleration could develop. This would result in the following equation.
(Eq 1) $a = \frac{Dv}{Dt} = a_s\hat{s}+a_n\hat{n}$
It is important to realize that there is a difference, besides their direction, between normal and tangential acceleration. This difference is how the two acceleration occur. First, let’s talk about tangential acceleration. A tangential acceleration occurs the way any other acceleration would occur. There has to be a change in velocity of the fluid particle along the streamline in relation to time. As a result a tangential acceleration will be produced. To determine a tangential acceleration the following equation would be used.
(Eq 2) $a_s=v\frac{∂v}{∂s}$
Now we have to look at normal acceleration. Normal acceleration behaves differently than other types of acceleration. It will occur regardless of if the velocity is changing with time or if the velocity is constant. Instead it occurs because the particle is changing direction. As a result a normal acceleration will occur when there is a curve in a streamline. Its value will be based off of the current particle velocity and radius (r) of the streamlines curve.
(Eq 3) $a_n = \frac{v^2}{r}$
Finally, the total acceleration of the fluid particle can be found by finding the magnitude of the tangential and normal acceleration.
(Eq 4) $a=\sqrt{a_s^2+a_n^2}$
| | 632 | 2,893 | {"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} | 4.3125 | 4 | CC-MAIN-2020-24 | latest | en | 0.877855 |
https://www.studypool.com/discuss/1296091/Gauss-Jordan-elimination-method-show-work-?free | 1,505,892,978,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686705.10/warc/CC-MAIN-20170920071017-20170920091017-00352.warc.gz | 825,686,015 | 14,322 | ##### Gauss-Jordan elimination method (show work)
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Solve the system of linear equations using the Gauss-Jordan elimination method.
2x + 10y = −1 −6x + 8y = 22
Find x and y
Nov 28th, 2015
IMG_3189[1].JPG IMG_3190[1].JPG
Nov 28th, 2015
What is the final x & y answer?
Nov 28th, 2015
...
Nov 28th, 2015
...
Nov 28th, 2015
Sep 20th, 2017
check_circle | 158 | 440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.662684 |
https://math.answers.com/other-math/Where_is_the_center_of_the_circle_given_by_the_equation_x_-_3_squared_plus_y_plus_2_squared_equals_9 | 1,708,819,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00809.warc.gz | 389,543,632 | 47,224 | 0
Where is the center of the circle given by the equation x - 3 squared plus y plus 2 squared equals 9?
Updated: 4/28/2022
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The center of the circle given by the equation (x - 3)2 plus (y + 2)2 = 9 is (3,-2).
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Exactly as it's stated, that equation describes a straight line, not a circle. If you take out the phrase "times 2" from both places where it's used and replace it with "squared", then the equation describes a circle, centered at (-5, 3), with a radius of 5.
What is the equation of x squared plus y squared equals 16 with a horizontal compression of a half?
In order to take your circle and squash it horizontally to 1/2 of its original width (2),change the equation to4x2 + y2 = 16 | 601 | 2,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.909764 |
https://klu.ai/glossary/brute-force-search | 1,717,109,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00793.warc.gz | 288,503,091 | 13,132 | # What is brute-force search?
by Stephen M. Walker II, Co-Founder / CEO
## What is brute-force search?
Brute-force search, also known as exhaustive search or generate and test, is a general problem-solving technique and algorithmic paradigm that systematically enumerates all possible candidates for a solution and checks each one for validity. This approach is straightforward and relies on sheer computing power to solve a problem.
In the context of computer science, a brute-force algorithm might find the divisors of a natural number by enumerating all integers and checking whether each divides the number without remainder. Another example is the traveling salesman problem, where the brute-force solution calculates the total distance for every possible route.
While a brute-force search is simple to implement and will always find a solution if it exists, it can be inefficient due to the potentially large number of candidate solutions. Therefore, it's typically used when the problem size is limited, or when there are problem-specific heuristics that can reduce the set of candidate solutions. It's also used when the simplicity of implementation is more important than speed, such as in critical applications where any errors in the algorithm would have serious consequences.
Despite its simplicity and guaranteed solution finding, the brute-force approach has several disadvantages. It's inefficient, often with a time complexity above the O(N!) order of growth, and relies more on the power of a computer system than on good algorithm design. These algorithms run slowly and are not constructive or creative compared to other algorithms.
In the context of cybersecurity, a brute-force attack involves systematically checking all possible keys until the correct key is found. This strategy can theoretically be used against any encrypted data, with longer keys exponentially more difficult to crack than shorter ones.
## How does brute-force search compare to other AI search methods?
Brute-force search, in comparison to other AI search methods, is the most straightforward approach that does not rely on any domain-specific knowledge or heuristics to guide the search. It systematically explores all possible states or solutions until it finds one that satisfies the problem's criteria. This method is guaranteed to find a solution if one exists, but it can be highly inefficient, especially as the size of the problem space increases.
Other AI search methods aim to improve efficiency by using various strategies:
• Divide and conquer — This technique breaks down a problem into smaller, more manageable subproblems, solves each one, and combines the results. Algorithms like merge sort and quick sort use this approach to achieve better performance.
• Dynamic programming — This method stores the results of overlapping subproblems to avoid redundant calculations, significantly reducing the time complexity from exponential to polynomial in many cases.
• Greedy algorithms — These make the locally optimal choice at each step with the hope of finding a global optimum. However, they do not always guarantee an optimal solution.
• Heuristic search methods — These include algorithms like A* Search, which use heuristics to guide the search towards the goal more efficiently than brute-force methods. Heuristics help to prioritize which paths to explore based on an estimate of the cost from the current state to the goal.
• Metaheuristics — These are higher-level procedures designed for finding, generating, or selecting a heuristic that may provide a sufficiently good solution to an optimization problem, especially with incomplete or imperfect information. Examples include genetic algorithms, simulated annealing, and ant colony optimization.
Brute-force search is often used when the problem space is small, when the simplicity of implementation is critical, or when other methods have failed. It is also easier to explain and understand compared to more complex algorithms, which can be advantageous in terms of transparency and trust. However, due to its lack of scalability and inefficiency, brute-force search is generally unsuitable for large-scale or real-time applications where quick responses are necessary.
## What are the benefits, drawbacks, and applications of Brute-force search?
Brute-force search offers a straightforward and easily implementable algorithm in AI, with the definitive advantage of always finding a solution if one exists. Its simplicity makes it an ideal candidate for parallel processing, leveraging modern computing power to improve efficiency. Additionally, it serves as a benchmark for evaluating the performance of more complex algorithms.
However, brute-force search is not without its limitations. It can be time-consuming and resource-heavy, particularly with large search spaces. The lack of heuristics or domain knowledge means it may yield sub-optimal solutions and is prone to getting trapped in local minima.
In practical applications, brute-force search is instrumental in solving constraint satisfaction problems, which are prevalent in planning, scheduling, and resource allocation tasks. It is also employed in pathfinding algorithms, where it exhaustively explores all possible routes to determine the shortest path in a graph-based representation of locations and connections.
## More terms
### What is Intelligence Quotient (IQ)?
Intelligence Quotient (IQ) is a measure of a person's cognitive ability compared to the population at large. It is calculated through standardized tests designed to assess human intelligence. The scores are normalized so that 100 is the average score, with a standard deviation of 15. An IQ score does not measure knowledge or wisdom, but rather the capacity to learn, reason, and solve problems.
### What is an embodied agent?
An embodied agent in the field of artificial intelligence (AI) is an intelligent agent that interacts with its environment through a physical or virtual body. This interaction can be with a real-world environment, in the case of physically embodied agents like mobile robots, or with a digital environment, in the case of graphically embodied agents like Ananova and Microsoft Agent. | 1,128 | 6,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | longest | en | 0.910555 |
https://www.homeworkminutes.com/q/post-university-fin-201-unit-2-case-problem-2-4347/ | 1,716,222,051,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00743.warc.gz | 736,714,267 | 10,855 | # post university fin 201 unit 2 case problem 2
Question # 00004347 Posted By: spqr Updated on: 12/02/2013 12:31 AM Due on: 12/28/2013
Subject Finance Topic Finance Tutorials:
Question
Answer Question 25 and 26 which you will find on Page 46 of your textbook. (10th edition)
For Q25, draw up anincome statement andbalance sheet. Be sure to follow the format as outlined in the text and on the lecture.
For Q26, cite the formula foroperating cash flow and then plug-in the correct numbers. ForQ26, you need to submit only Operating Cash Flow,not all three cash flow measures.
After you have completed your income statement and balance sheet, compute the following financial ratios for both fiscal years: 1) current ratio; 2)quick ratio; 3)accounts receivable turnover; 4)inventory turnover; 5)debt to equity; 6)profit margin; 7)return on equity. For all ratios, cite the formula and then plug-in numbers. Show all your work, not just the final answer. Remember to compute ratios for both years!
Does this company appear to be a healthy company? Comment, and include financial numbers and ratios from your work, above, to support your answer which should be appx 2-3 paragraphs, single-spaced.
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1. ## Solution: post university fin 201 unit 2 case problem 2
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2. ## Solution: Post University FIN 201 (Principles of Finance) Unit 2 submission and feedback. 9.6/10 Grade
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Great! We have found the solution of this question! | 564 | 2,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.809189 |
https://ncalculators.com/algebra/vector-dot-product-calculator.htm | 1,718,411,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00817.warc.gz | 379,930,454 | 14,701 | # Vector Dot Product Calculator
Vector A
i + j + k
Vector B
i + j + k
A . B = 64
CALCULATE
Vector Dot Product Calculator to find the resultant vector by multiplying two vectors. The concept of the vector dot product is used to describe the product of physical quantities which have both a magnitude and a direction associated with them.
## Scalar or Dot Product
The Dot Product also known as Scalar Product, of two vectors is the sum of the component wise products. The dot product of two vectors in the same direction is equal to the product of their magnitudes. The scalar product of two perpendicular vectors is zero. The Scalar or Dot product of the vectors properties are
1. The commutative Law A . B = B . A
2. The Distributive Law A . (B + C) = A. B + A . C
3. A (B . C) = B . (AC)
4. A . A >= 0; and A. A = 0 if and only if A = 0
The scalar or dot products of vectors is used in many applications of mathematics, physics and other engineering operations. When it comes to calculate the dot product of two vectors, this vector dot product calculator can help you to find out the resulting vector. | 272 | 1,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.910935 |
https://www.teacherspayteachers.com/Product/Imaginary-Numbers-2139744 | 1,495,642,173,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607848.9/warc/CC-MAIN-20170524152539-20170524172539-00486.warc.gz | 949,442,972 | 25,921 | Total:
\$0.00
# Imaginary Numbers
Subjects
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### PRODUCT DESCRIPTION
Imaginary Numbers
resource includes:
question cards
worksheet (with same questions/ use for students that miss out on class activity)
recording sheets (3 options)
This activity provides a great opportunity to practice and assess student understanding of imaginary numbers
Students answer 12 questions such as; simplify expressions with imaginary numbers, find the sum or product of imaginary numbers, solve a quadratic with a non-real solution.
How to Use:
The 12 questions can be individually presented to students on cards. For each question there is a matching answer card. Students can match the question with the answer or start with the answer and find the question.
OR use the worksheet with all twelve questions and have students respond to each question without an answer choice. The worksheet also provides an alternate assignment for absent students or those students that are out of class when you do the activity.
H.O.T.*Challenge; Give students an answer card and ask for a question to match the answer. This requires a complete understanding of the concept and moves the student to think on a higher level than is normally assessed.
3 options for recording answers. Multiple copies are on each page, print only what is needed for your class size.
An answer sheet is included and a solutions page for the worksheet (same problems as those on the cards).
Suggested activities:
Scavenger Hunt: Given a question, find the solution.
Match-Up Challenge; Working on teams, students ‘race’ to find all the matches.
Speed-Math - Set-up like speed-dating where each student is an expert on one card and sits across from an ‘intern.’ They swap cards and share answers, instructing when necessary.
Scoot or some version of it
Quick Review or Warm-up to keep skills alive.
The questions in this series cover skills for Algebra 2
They are great for differentiation, diagnostics, review and assessment.
Want to save on a BUNDLE?
This resource is included in the following bundle.
COMPLEX NUMBERS Resource Packet
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10+
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Teaching Duration
40 Minutes
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\$2.25 | 552 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.924559 |
https://runestone.academy/ns/books/published/boelkins-ACS/sec-5-4-parts.html | 1,708,837,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00644.warc.gz | 507,016,361 | 21,684 | # Active Calculus
## Section5.4Integration by Parts
In Section 5.3, we learned the technique of $$u$$-substitution for evaluating indefinite integrals. For example, the indefinite integral $$\int x^3 \sin(x^4) \, dx$$ is perfectly suited to $$u$$-substitution, because one factor is a composite function and the other factor is the derivative (up to a constant) of the inner function. Recognizing the algebraic structure of a function can help us to find its antiderivative.
Next we consider integrands with a different elementary algebraic structure: a product of basic functions. For instance, suppose we are interested in evaluating the indefinite integral
\begin{equation*} \int x \sin(x) \, dx\text{.} \end{equation*}
The integrand is the product of the basic functions $$f(x) = x$$ and $$g(x) = \sin(x)\text{.}$$ We know that it is relatively complicated to compute the derivative of the product of two functions, so we should expect that antidifferentiating a product should be similarly involved. Intuitively, we expect that evaluating $$\int x \sin(x) \, dx$$ will involve somehow reversing the Product Rule.
To that end, in Preview Activity 5.4.1 we refresh our understanding of the Product Rule and then investigate some indefinite integrals that involve products of basic functions.
### Preview Activity5.4.1.
In Section 2.3, we developed the Product Rule and studied how it is employed to differentiate a product of two functions. In particular, recall that if $$f$$ and $$g$$ are differentiable functions of $$x\text{,}$$ then
\begin{equation*} \frac{d}{dx} \left[ f(x) \cdot g(x) \right] = f(x) \cdot g'(x) + g(x) \cdot f'(x)\text{.} \end{equation*}
1. For each of the following functions, use the Product Rule to find the function’s derivative. Be sure to label each derivative by name (e.g., the derivative of $$g(x)$$ should be labeled $$g'(x)$$).
1. $$\displaystyle g(x) = x\sin(x)$$
2. $$\displaystyle h(x) = xe^x$$
3. $$\displaystyle p(x) = x\ln(x)$$
4. $$\displaystyle q(x) = x^2 \cos(x)$$
5. $$\displaystyle r(x) = e^x \sin(x)$$
1. $$\displaystyle \int xe^x + e^x \, dx$$
2. $$\displaystyle \int e^x(\sin(x) + \cos(x)) \, dx$$
3. $$\displaystyle \int 2x\cos(x) - x^2 \sin(x) \, dx$$
4. $$\displaystyle \int x\cos(x) + \sin(x) \, dx$$
5. $$\displaystyle \int 1 + \ln(x) \, dx$$
3. Observe that the examples in (b) work nicely because of the derivatives you were asked to calculate in (a). Each integrand in (b) is precisely the result of differentiating one of the products of basic functions found in (a). To see what happens when an integrand is still a product but not necessarily the result of differentiating an elementary product, we consider how to evaluate
\begin{equation*} \int x\cos(x) \, dx\text{.} \end{equation*}
1. First, observe that
\begin{equation*} \frac{d}{dx} \left[ x\sin(x) \right] = x\cos(x) + \sin(x)\text{.} \end{equation*}
Integrating both sides indefinitely and using the fact that the integral of a sum is the sum of the integrals, we find that
\begin{equation*} \int \left(\frac{d}{dx} \left[ x\sin(x) \right] \right) \, dx = \int x\cos(x) \, dx + \int \sin(x) \, dx\text{.} \end{equation*}
In this last equation, evaluate the indefinite integral on the left side as well as the rightmost indefinite integral on the right.
2. In the most recent equation from (i.), solve the equation for the expression $$\int x \cos(x) \, dx\text{.}$$
3. For which product of basic functions have you now found the antiderivative?
### Subsection5.4.1Reversing the Product Rule: Integration by Parts
Problem (c) in Preview Activity 5.4.1 provides a clue to the general technique known as Integration by Parts, which comes from reversing the Product Rule. Recall that the Product Rule states that
\begin{equation*} \frac{d}{dx} \left[ f(x) g(x) \right] = f(x) g'(x) + g(x) f'(x)\text{.} \end{equation*}
Integrating both sides of this equation indefinitely with respect to $$x\text{,}$$ we find
$$\int \frac{d}{dx} \left[ f(x) g(x) \right] \, dx = \int f(x) g'(x) \, dx + \int g(x) f'(x) \, dx\text{.}\tag{5.4.1}$$
On the left side of Equation (5.4.1), we have the indefinite integral of the derivative of a function. Temporarily omitting the constant that may arise, we have
$$f(x) g(x) = \int f(x) g'(x) \, dx + \int g(x) f'(x) \, dx\text{.}\tag{5.4.2}$$
We solve for the first indefinite integral on the left to generate the rule
$$\int f(x) g'(x) \, dx = f(x) g(x) - \int g(x) f'(x) \, dx\text{.}\tag{5.4.3}$$
Often we express Equation (5.4.3) in terms of the variables $$u$$ and $$v\text{,}$$ where $$u = f(x)$$ and $$v = g(x)\text{.}$$ In differential notation, $$du = f'(x) \, dx$$ and $$dv = g'(x) \, dx\text{,}$$ so we can state the rule for Integration by Parts in its most common form as follows:
\begin{equation*} \int u \, dv = uv - \int v \, du\text{.} \end{equation*}
To apply integration by parts, we look for a product of basic functions that we can identify as $$u$$ and $$dv\text{.}$$ If we can antidifferentiate $$dv$$ to find $$v\text{,}$$ and evaluating $$\int v \, du$$ is not more difficult than evaluating $$\int u \, dv\text{,}$$ then this substitution usually proves to be fruitful. To demonstrate, we consider the following example.
#### Example5.4.1.
Evaluate the indefinite integral
\begin{equation*} \int x\cos(x) \, dx \end{equation*}
using integration by parts.
Solution.
When we use integration by parts, we have a choice for $$u$$ and $$dv\text{.}$$ In this problem, we can either let $$u = x$$ and $$dv = \cos(x) \, dx\text{,}$$ or let $$u = \cos(x)$$ and $$dv = x \, dx\text{.}$$ While there is not a universal rule for how to choose $$u$$ and $$dv\text{,}$$ a good guideline is this: do so in a way that $$\int v \, du$$ is at least as simple as the original problem $$\int u \, dv\text{.}$$
This leads us to choose 1 $$u = x$$ and $$dv = \cos(x) \, dx\text{,}$$ from which it follows that $$du = 1 \, dx$$ and $$v = \sin(x)\text{.}$$ With this substitution, the rule for integration by parts tells us that
\begin{equation*} \int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \cdot 1 \, dx\text{.} \end{equation*}
All that remains to do is evaluate the (simpler) integral $$\int \sin(x) \cdot 1 \, dx\text{.}$$ Doing so, we find
\begin{equation*} \int x \cos(x) \, dx = x \sin(x) - (-\cos(x)) + C = x\sin(x) + \cos(x) + C\text{.} \end{equation*}
Observe that when we get to the final stage of evaluating the last remaining antiderivative, it is at this step that we include the integration constant, $$+C\text{.}$$
The general technique of integration by parts involves trading the problem of integrating the product of two functions for the problem of integrating the product of two related functions. That is, we convert the problem of evaluating $$\int u \, dv$$ to that of evaluating $$\int v \, du\text{.}$$ This clearly shapes our choice of $$u$$ and $$v\text{.}$$ In Example 5.4.1, the original integral to evaluate was $$\int x \cos(x) \,dx\text{,}$$ and through the substitution provided by integration by parts, we were instead able to evaluate $$\int \sin(x) \cdot 1 \, dx\text{.}$$ Note that the original function $$x$$ was replaced by its derivative, while $$\cos(x)$$ was replaced by its antiderivative.
#### Activity5.4.2.
Evaluate each of the following indefinite integrals. Check each antiderivative that you find by differentiating.
1. $$\displaystyle \int te^{-t} \, dt$$
2. $$\displaystyle \int 4x \sin(3x) \, dx$$
3. $$\displaystyle \int z \sec^2(z) \,dz$$
4. $$\displaystyle \int x \ln(x) \, dx$$
Hint.
1. Try $$u=t\text{.}$$
2. Let $$dv=\sin(3x)dx$$
3. Remember that $$\int \sec^2(z) \,dz = \tan(z)\text{.}$$
4. Note that $$\ln(x) \, dx$$ has a simple derivative to work with.
### Subsection5.4.2Some Subtleties with Integration by Parts
Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating
\begin{equation*} \int \arctan(x) \, dx\text{.} \end{equation*}
Initially, this problem seems ill-suited to integration by parts, since there does not appear to be a product of functions present. But if we note that $$\arctan(x) = \arctan(x) \cdot 1\text{,}$$ and realize that we know the derivative of $$\arctan(x)$$ as well as the antiderivative of $$1\text{,}$$ we see the possibility for the substitution $$u = \arctan(x)$$ and $$dv = 1 \, dx\text{.}$$ We explore this substitution further in Activity 5.4.3.
In a related problem, consider $$\int t^3 \sin(t^2) \, dt\text{.}$$ Observe that there is a composite function present in $$\sin(t^2)\text{,}$$ but there is not an obvious function-derivative pair, as we have $$t^3$$ (rather than simply $$t$$) multiplying $$\sin(t^2)\text{.}$$ In this problem we use both $$u$$-substitution and integration by parts. First we write $$t^3 = t \cdot t^2$$ and consider the indefinite integral
\begin{equation*} \int t \cdot t^2 \cdot \sin(t^2) \, dt\text{.} \end{equation*}
We let $$z = t^2$$ so that $$dz = 2t \, dt\text{,}$$ and thus $$t \, dt = \frac{1}{2} \, dz\text{.}$$ (We are using the variable $$z$$ to perform a “$$z$$-substitution” first so that we may then apply integration by parts.) Under this $$z$$-substitution, we now have
\begin{equation*} \int t \cdot t^2 \cdot \sin(t^2) \, dt = \int z \cdot \sin(z) \cdot \frac{1}{2} \, dz\text{.} \end{equation*}
The resulting integral can be evaluated by parts. This, too, is explored further in Activity 5.4.3.
These problems show that we sometimes must think creatively in choosing the variables for substitution in integration by parts, and that we may need to use substitution for an additional change of variables.
#### Activity5.4.3.
Evaluate each of the following indefinite integrals, using the provided hints.
1. Evaluate $$\int \arctan(x) \, dx$$ by using Integration by Parts with the substitution $$u = \arctan(x)$$ and $$dv = 1 \, dx\text{.}$$
2. Evaluate $$\int \ln(z) \,dz\text{.}$$ Consider a similar substitution to the one in (a).
3. Use the substitution $$z = t^2$$ to transform the integral $$\int t^3 \sin(t^2) \, dt$$ to a new integral in the variable $$z\text{,}$$ and evaluate that new integral by parts.
4. Evaluate $$\int s^5 e^{s^3} \, ds$$ using an approach similar to that described in (c).
5. Evaluate $$\int e^{2t} \cos(e^t) \, dt\text{.}$$ You will find it helpful to note that $$e^{2t} = e^t \cdot e^t\text{.}$$
### Subsection5.4.3Using Integration by Parts Multiple Times
Integration by parts is well suited to integrating the product of basic functions, allowing us to trade a given integrand for a new one where one function in the product is replaced by its derivative, and the other is replaced by its antiderivative. The goal in this trade of $$\int u \, dv$$ for $$\int v \, du$$ is that the new integral be simpler to evaluate than the original one. Sometimes it is necessary to apply integration by parts more than once in order to evaluate a given integral.
#### Example5.4.2.
Evaluate $$\int t^2 e^t \, dt\text{.}$$
Solution.
Let $$u = t^2$$ and $$dv = e^t \, dt\text{.}$$ Then $$du = 2t \, dt$$ and $$v = e^t\text{,}$$ and thus
\begin{equation*} \int t^2 e^t \, dt = t^2 e^t - \int 2t e^t \, dt\text{.} \end{equation*}
The integral on the right side is simpler to evaluate than the one on the left, but it still requires integration by parts. Now letting $$u = 2t$$ and $$dv = e^t \, dt\text{,}$$ we have $$du = 2\, dt$$ and $$v = e^t\text{,}$$ so that
\begin{equation*} \int t^2 e^t \, dt = t^2 e^t - \left( 2t e^t - \int 2 e^t \, dt \right)\text{.} \end{equation*}
(Note the parentheses, which remind us to distribute the minus sign to the entire value of the integral $$\int 2t e^t \, dt\text{.}$$) The final integral on the right is a basic one; evaluating that integral and distributing the minus sign, we find
\begin{equation*} \int t^2 e^t \, dt = t^2 e^t - 2t e^t + 2 e^t + C\text{.} \end{equation*}
Of course, even more than two applications of integration by parts may be necessary. In the preceding example, if the integrand had been $$t^3e^t\text{,}$$ we would have had to use integration by parts three times.
Next, we consider the slightly different scenario.
#### Example5.4.3.
Evaluate $$\int e^t \cos(t) \, dt\text{.}$$
Solution.
We can choose to let $$u$$ be either $$e^t$$ or $$\cos(t)\text{;}$$ we pick $$u = \cos(t)\text{,}$$ and thus $$dv = e^t \, dt\text{.}$$ With $$du = -\sin(t) \, dt$$ and $$v = e^t\text{,}$$ integration by parts tells us that
\begin{equation*} \int e^t \cos(t) \, dt = e^t \cos(t) - \int e^t (-\sin(t))\, dt\text{,} \end{equation*}
or equivalently that
$$\int e^t \cos(t) \, dt = e^t \cos(t) + \int e^t \sin(t) \, dt\text{.}\tag{5.4.4}$$
The new integral has the same algebraic structure as the original one. While the overall situation isn’t necessarily better than what we started with, it hasn’t gotten worse. Thus, we proceed to integrate by parts again. This time we let $$u = \sin(t)$$ and $$dv = e^t \, dt\text{,}$$ so that $$du = \cos(t) \, dt$$ and $$v = e^t\text{,}$$ which implies
$$\int e^t \cos(t) \, dt = e^t \cos(t) + \left( e^t \sin(t) - \int e^t \cos(t) \, dt \right)\text{.}\tag{5.4.5}$$
We seem to be back where we started, as two applications of integration by parts has led us back to the original problem, $$\int e^t \cos(t) \, dt\text{.}$$ But if we look closely at Equation (5.4.5), we see that we can use algebra to solve for the value of the desired integral. Adding $$\int e^t \cos(t) \, dt$$ to both sides of the equation, we have
\begin{equation*} 2 \int e^t \cos(t) \, dt = e^t \cos(t) + e^t \sin(t)\text{,} \end{equation*}
and therefore
\begin{equation*} \int e^t \cos(t) \, dt = \frac{1}{2} \left( e^t \cos(t) + e^t \sin(t) \right) + C\text{.} \end{equation*}
Note that since we never actually encountered an integral we could evaluate directly, we didn’t have the opportunity to add the integration constant $$C$$ until the final step.
#### Activity5.4.4.
Evaluate each of the following indefinite integrals.
1. $$\displaystyle \int x^2 \sin(x) \, dx$$
2. $$\displaystyle \int t^3 \ln(t) \, dt$$
3. $$\displaystyle \int e^z \sin(z) \, dz$$
4. $$\displaystyle \int s^2 e^{3s} \, ds$$
5. $$\int t \arctan(t) \,dt$$ (Hint: At a certain point in this problem, it is very helpful to note that $$\frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}\text{.}$$)
Hint.
1. Start with $$u=x^2\text{.}$$
2. Begin with $$u=t^3\text{.}$$
3. You’ll have to integrate by parts twice.
4. Start with $$u=s^2\text{.}$$
5. Try $$u=\arctan(x)\text{.}$$
### Subsection5.4.4Evaluating Definite Integrals Using Integration by Parts
We can use the technique of integration by parts to evaluate a definite integral.
#### Example5.4.4.
Evaluate
\begin{equation*} \int_0^{\pi/2} t\sin(t) \, dt\text{.} \end{equation*}
Solution.
One option is to find an antiderivative (using indefinite integral notation) and then apply the Fundamental Theorem of Calculus to find that
\begin{align*} \int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp \left( -t \cos(t) + \sin(t) \right) \bigg\vert_0^{\pi/2}\\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right)\\ =\mathstrut \amp 1\text{.} \end{align*}
Alternatively, we can apply integration by parts and work with definite integrals throughout. With this method, we must remember to evaluate the product $$uv$$ over the given limits of integration. Using the substitution $$u = t$$ and $$dv = \sin(t) \, dt\text{,}$$ so that $$du = dt$$ and $$v = -\cos(t)\text{,}$$ we write
\begin{align*} \int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp -t \cos(t) \bigg\vert_0^{\pi/2} - \int_0^{\pi/2} (-\cos(t)) \, dt\\ =\mathstrut \amp -t \cos(t) \bigg\vert_0^{\pi/2} + \sin(t) \bigg\vert_0^{\pi/2}\\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right)\\ =\mathstrut \amp 1\text{.} \end{align*}
As with any substitution technique, it is important to use notation carefully and completely, and to ensure that the end result makes sense.
### Subsection5.4.5When $$u$$-substitution and Integration by Parts Fail to Help
Both integration techniques we have discussed apply in relatively limited circumstances. It is not hard to find examples of functions for which neither technique produces an antiderivative; indeed, there are many, many functions that appear elementary but that do not have an elementary algebraic antiderivative. For instance, neither $$u$$-substitution nor integration by parts proves fruitful for the indefinite integrals
\begin{equation*} \int e^{x^2} \, dx \ \ \text{and} \ \ \int x \tan(x) \, dx\text{.} \end{equation*}
While there are other integration techniques, some of which we will consider briefly, none of them enables us to find an algebraic antiderivative for $$e^{x^2}$$ or $$x \tan(x)\text{.}$$ We do know from the Second Fundamental Theorem of Calculus that we can construct an integral antiderivative for each function; $$F(x) = \int_0^x e^{t^2} \, dt$$ is an antiderivative of $$f(x) = e^{x^2}\text{,}$$ and $$G(x) = \int_0^{x} t \tan(t) \, dt$$ is an antiderivative of $$g(x) = x \tan(x)\text{.}$$ But finding an elementary algebraic formula that doesn’t involve integrals for either $$F$$ or $$G$$ turns out not only to be impossible through $$u$$-substitution or integration by parts, but indeed impossible altogether. Antidifferentiation is much harder in general than differentiation.
### Subsection5.4.6Summary
• Through the method of integration by parts, we can evaluate indefinite integrals that involve products of basic functions such as $$\int x \sin(x) \, dx$$ and $$\int x \ln(x) \, dx\text{.}$$ Using a substitution enables us to trade one of the functions in the product for its derivative, and the other for its antiderivative, in an effort to find a different product of functions that is easier to integrate.
• If the algebraic structure of an integrand is a product of basic functions in the form $$\int f(x) g'(x) \, dx\text{,}$$ we can use the substitution $$u = f(x)$$ and $$dv = g'(x) \,dx$$ and apply the rule
\begin{equation*} \int u \, dv = uv - \int v \, du \end{equation*}
to evaluate the original integral $$\int f(x) g'(x) \, dx$$ by instead evaluating
\begin{equation*} \int v \, du = \int f'(x) g(x) \, dx\text{.} \end{equation*}
• When deciding to integrate by parts, we have to select both $$u$$ and $$dv\text{.}$$ That selection is guided by the overall principle that the new integral $$\int v \, du$$ not be more difficult than the original integral $$\int u \, dv\text{.}$$ In addition, it is often helpful to recognize if one of the functions present is much easier to differentiate than antidifferentiate (such as $$\ln(x)$$), in which case that function often is best assigned the variable $$u\text{.}$$ In addition, $$dv$$ must be a function that we can antidifferentiate.
### Exercises5.4.7Exercises
#### 1.Choose which method to use.
For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate, or if neither method is appropriate. Do not evaluate the integrals.
1. $$\int\,x\sin x\,dx$$
• integration by parts
• substitution
• neither
2. $$\int\,{x\over 1 + x^{2}}\,dx$$
• substitution
• neither
• integration by parts
3. $$\int\,x e^{x^{2}}\,dx$$
• neither
• integration by parts
• substitution
4. $$\int\,x \cos(x^{2})\,dx$$
• integration by parts
• neither
• substitution
5. $$\int\,{1\over\sqrt{4 x + 1}}\,dx$$
• integration by parts
• neither
• substitution
(Note that because this is multiple choice, you will not be able to see which parts of the problem you got correct.)
#### 2.Product involving $$\cos(5 x)$$.
Use integration by parts to evaluate the integral.
$$\displaystyle\int 3 x \cos (4 x)\, dx =$$ $$+C$$
#### 3.Product involving $$e^{8 z}$$.
Find the integral
$$\int\, \left(z+1\right)e^{5z} dz =$$
#### 4.Definite integral of $$t e^{-t}$$.
Evaluate the definite integral.
$$\displaystyle\int_{0}^{5} t e^{-t} dt =$$
#### 5.
Let $$f(t) = te^{-2t}$$ and $$F(x) = \int_0^x f(t) \, dt\text{.}$$
1. Determine $$F'(x)\text{.}$$
2. Use the First FTC to find a formula for $$F$$ that does not involve an integral.
3. Is $$F$$ an increasing or decreasing function for $$x \gt 0\text{?}$$ Why?
#### 6.
Consider the indefinite integral given by $$\int e^{2x} \cos(e^x) \, dx\text{.}$$
1. Noting that $$e^{2x} = e^x \cdot e^x\text{,}$$ use the substitution $$z = e^{x}$$ to determine a new, equivalent integral in the variable $$z\text{.}$$
2. Evaluate the integral you found in (a) using an appropriate technique.
3. How is the problem of evaluating $$\int e^{2x} \cos(e^{2x}) \, dx$$ different from evaluating the integral in (a)? Do so.
4. Evaluate each of the following integrals as well, keeping in mind the approach(es) used earlier in this problem:
• $$\displaystyle \int e^{2x} \sin(e^x) \, dx$$
• $$\displaystyle \int e^{3x} \sin(e^{3x}) \, dx$$
• $$\displaystyle \int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx$$
#### 7.
For each of the following indefinite integrals, determine whether you would use $$u$$-substitution, integration by parts, neither*, or both to evaluate the integral. In each case, write one sentence to explain your reasoning, and include a statement of any substitutions used. (That is, if you decide in a problem to let $$u = e^{3x}\text{,}$$ you should state that, as well as that $$du = 3e^{3x} \, dx\text{.}$$) Finally, use your chosen approach to evaluate each integral. (* one of the following problems does not have an elementary antiderivative and you are not expected to actually evaluate this integral; this will correspond with a choice of “neither” among those given.)
1. $$\displaystyle \int x^2 \cos(x^3) \, dx$$
2. $$\int x^5 \cos(x^3) \, dx$$ (Hint: $$x^5 = x^2 \cdot x^3$$)
3. $$\displaystyle \int x\ln(x^2) \, dx$$
4. $$\displaystyle \int \sin(x^4) \, dx$$
5. $$\displaystyle \int x^3 \sin(x^4) \, dx$$
6. $$\displaystyle \int x^7 \sin(x^4) \, dx$$
Observe that if we considered the alternate choice, and let $$u = \cos(x)$$ and $$dv = x \, dx\text{,}$$ then $$du = -\sin(x) \, dx$$ and $$v = \frac{1}{2}x^2\text{,}$$ from which we would write $$\int x\cos(x) \, dx = \frac{1}{2}x^2 \cos(x) - \int \frac{1}{2}x^2 (-\sin(x)) \, dx\text{.}$$ Thus we have replaced the problem of integrating $$x \cos(x)$$ with that of integrating $$\frac{1}{2}x^2 \sin(x)\text{;}$$ the latter is clearly more complicated, which shows that this alternate choice is not as helpful as the first choice. | 7,235 | 22,590 | {"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": 5, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-10 | latest | en | 0.828791 |
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