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Home Page
# Wednesday 6th January
### Morning Work
Please find today's arithmetic questions and answers attached. Please view the PowerPoint in full screen or 'view slideshow' as the answers are animated to appear for the children to check their work when they have finished.
### Maths
In Maths today we will be looking at using the grid method for multiplication as this will help us when we start to multiply larger numbers.
I have modelled this for the children and the video of this can be found on the following link:
https://st-lukes-church-of-england-notts.primarysite.media/media/grid-method-for-multiplication
Further explanation is given on the document with today's questions. These are grouped into Bronze, Silver, Gold and platinum, the children will be used to this from class. They can choose to start wherever they like but I would encourage all to do some of the Silver and Gold questions as these will be more challenging.
### GPS
For GPS today, we are finishing looking at our work on parenthesis. We have used brackets and dashes and we have already looked at using commas in class.
Yesterday we learnt that the parenthesis can go in the middle of the sentence, or use just one dash and go at the end of the sentence.
For today's tasks, we are going to focus on the parenthesis that go in the middle of the sentence, so you will need either a pair of brackets () a pair of dashes - - or a pair of commas , , correctly placed in each sentence.
Think how well you have understood the work so far and select either the Bronze, Silver or Gold tasks. Each task is explained with examples as you work through. The documents also contain a link to a YouTube video if you need some more help.
### English
In English today I would like you to use the vocabulary you came up with yesterday to create an exciting descriptive paragraph.
Details and a model of how to do this can be found on the PowerPoint attached.
### PE
On Wednesday, we would normally have PE. Why not try one of the following links to PE that you could do at home.
Cosmic Yoga
https://www.youtube.com/user/CosmicKidsYoga
Joe Wicks
https://www.youtube.com/watch?v=BDigyoBrHms&list=PLyCLoPd4VxBsYwx1a3RXMoAuJp1MAVG_O
Dance with Oti Mabuse
https://www.youtube.com/watch?v=GHahd8rQ0hg
### French
In French this half term we will be learning how to describe things in the classroom.
Please follow through the PowerPoint today in full screen mode and practise speaking out loud using the sound clips on each slide. These can be played as many times as you need until you are confident.
Top | 583 | 2,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.940953 |
https://www.justintools.com/unit-conversion/area.php?k1=varas-castellanas-cuads&k2=bunders | 1,627,478,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00419.warc.gz | 876,570,001 | 27,819 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# AREA Units Conversionvaras-castellanas-cuads to bunders
= 6.98737E-5 Bunders
Category: area
Conversion: Varas Castellanas Cuads to Bunders
The base unit for area is square meters (Non-SI/Derived Unit)
[Bunders] symbol/abbrevation: (b)
How to convert Varas Castellanas Cuads to Bunders (varas-castellana to b)?
1 varas-castellana = 6.98737E-5 b.
1 x 6.98737E-5 b = 6.98737E-5 Bunders.
Always check the results; rounding errors may occur.
Definition:
A bunder is a unit of area in the Low Countries (Belgium and the Netherlands). In the Netherlands, it was frequently used in the Achterhoek, Twente, and some parts of Drenthe ..more definition+
In relation to the base unit of [area] => (square meters), 1 Varas Castellanas Cuads (varas-castellana) is equal to 0.698737 square-meters, while 1 Bunders (b) = 10000 square-meters.
1 Varas Castellanas Cuads to common area units
1 varas-castellana = 0.698737 square meters (m2, sq m)
1 varas-castellana = 6987.37 square centimeters (cm2, sq cm)
1 varas-castellana = 6.98737E-7 square kilometers (km2, sq km)
1 varas-castellana = 7.521145711118 square feet (ft2, sq ft)
1 varas-castellana = 1083.044516089 square inches (in2, sq in)
1 varas-castellana = 0.83568249698228 square yards (yd2, sq yd)
1 varas-castellana = 2.6978386398847E-7 square miles (mi2, sq mi)
1 varas-castellana = 1083044516.089 square mils (sq mil)
1 varas-castellana = 6.98737E-5 hectares (ha)
1 varas-castellana = 0.0001726615202898 acres (ac)
Varas Castellanas Cuadsto Bunders (table conversion)
1 varas-castellana = 6.98737E-5 b
2 varas-castellana = 0.0001397474 b
3 varas-castellana = 0.0002096211 b
4 varas-castellana = 0.0002794948 b
5 varas-castellana = 0.0003493685 b
6 varas-castellana = 0.0004192422 b
7 varas-castellana = 0.0004891159 b
8 varas-castellana = 0.0005589896 b
9 varas-castellana = 0.0006288633 b
10 varas-castellana = 0.000698737 b
20 varas-castellana = 0.001397474 b
30 varas-castellana = 0.002096211 b
40 varas-castellana = 0.002794948 b
50 varas-castellana = 0.003493685 b
60 varas-castellana = 0.004192422 b
70 varas-castellana = 0.004891159 b
80 varas-castellana = 0.005589896 b
90 varas-castellana = 0.006288633 b
100 varas-castellana = 0.00698737 b
200 varas-castellana = 0.01397474 b
300 varas-castellana = 0.02096211 b
400 varas-castellana = 0.02794948 b
500 varas-castellana = 0.03493685 b
600 varas-castellana = 0.04192422 b
700 varas-castellana = 0.04891159 b
800 varas-castellana = 0.05589896 b
900 varas-castellana = 0.06288633 b
1000 varas-castellana = 0.0698737 b
2000 varas-castellana = 0.1397474 b
4000 varas-castellana = 0.2794948 b
5000 varas-castellana = 0.3493685 b
7500 varas-castellana = 0.52405275 b
10000 varas-castellana = 0.698737 b
25000 varas-castellana = 1.7468425 b
50000 varas-castellana = 3.493685 b
100000 varas-castellana = 6.98737 b
1000000 varas-castellana = 69.8737 b
1000000000 varas-castellana = 69873.7 b
(Varas Castellanas Cuads) to (Bunders) conversions | 1,245 | 3,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-31 | latest | en | 0.704333 |
http://mizar.uwb.edu.pl/version/current/html/proofs/ortsp_1/36_1_1 | 1,563,215,264,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00268.warc.gz | 111,597,846 | 1,695 | thus ( ex p being Element of S st
( not a _|_ & not x _|_ ) implies ex IT being Element of F st
for q being Element of S st not a _|_ & not x _|_ holds
IT = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y)) ) :: thesis: ( ( for p being Element of S holds
( a _|_ or x _|_ ) ) implies ex b1 being Element of F st b1 = 0. F )
proof
given p being Element of S such that A2: ( not a _|_ & not x _|_ ) ; :: thesis: ex IT being Element of F st
for q being Element of S st not a _|_ & not x _|_ holds
IT = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
take ((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) ; :: thesis: for q being Element of S st not a _|_ & not x _|_ holds
((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
let q be Element of S; :: thesis: ( not a _|_ & not x _|_ implies ((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y)) )
assume ( not a _|_ & not x _|_ ) ; :: thesis: ((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y))
hence ((ProJ (a,b,p)) * (ProJ (p,a,x))) * (ProJ (x,p,y)) = ((ProJ (a,b,q)) * (ProJ (q,a,x))) * (ProJ (x,q,y)) by A1, A2, Th26; :: thesis: verum
end;
thus ( ( for p being Element of S holds
( a _|_ or x _|_ ) ) implies ex b1 being Element of F st b1 = 0. F ) ; :: thesis: verum | 563 | 1,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.762573 |
https://cs.bennington.college/courses/spring2022/gadgets/sys/assignments | 1,723,094,240,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00579.warc.gz | 152,108,959 | 3,731 | Spring 2022
-->
# assignments
## 1. getting started due Tue Feb 22
• Here on this homework page, tell me a bit about yourself - what is your computer background and interest in this course.
• Get the Sparkfun Inventor's Kit from the bookstore if you haven't already.
• After logging into this course site, browse through the links and resource including the syllabus.
• Following the instructions in the SIK Guide, get your kit setup and running, plugged into a laptop, through circuit 1A: blink. (You can do a trial run of this without any wires or components; the code SIK_Circuit_1A-Blink.ino will blink a small blue LED labelled "13" right on the board.)
• Modify the code in the Arduino window on your laptop - perhaps changing the blink length - upload your changes to the board, and see them work.
• If you run into problems, contact the tutors or me for help.
• Finally, tell me here on this homework how all that went.
## 2. light due Tue Mar 8
• Explore the first four projects : 1A (blink), 1B (potentiometer), 1C (photoresistor), 1D (RGB Nightlight) or any variation of those ideas that strikes your fancy.
• Choose one thing that you did and describe it, explaining the circuit, code, and what they do. Taking a picture and attaching it could be helpful.
• Measure the power used by the resistor by measuring both the voltage V and current I with a multimeter and calculating power with P = V * I.
• Find documentation for LEDs like these and see if that is reasonable. About what range of "too little" to "too much" can these LEDs handle? Google "sparkfun LED" or "mouser LED" (one of the major supply stores) to get started. (The idea here is to start to see the many variations of these components, and what their parameters are, and how to read a datasheet like the ones here for example this one.)
## 3. sound due Tue Mar 22
• Explore the next three projects : 2A (buzzer), 2B (digital trumpet), 2C (simon says), or any variation that strikes your fancy.
• Again, choose one thing that you did and describe it, explaining the circuit, code, and what they do. Taking a picture and attaching it could be helpful.
• This time, rather than digging into the hardware with a multimeter, dig into the code : add some debugging statements somewhere that show what's going on, displaying the value of some variables on the serial monitor window, using Serial.print as in experiment C. Explain how this helps understand what the program is doing.
## 4. servo due Tue Apr 12
• Explore the next three projects : 3A (servo), 3B (echo sensor), 3C (motion detector), or any variation that strikes your fancy.
• Again, choose one thing that you did and describe it, explaining the circuit, code, and what they do. Taking a picture and attaching it could be helpful.
• Start thinking about what you want to do for an end of term project ... you may need some lead time to get materials.
## 5. display due Tue Apr 26
• You know the drill: explore the "display" projects: 4A ("hello world"), 4B (temperature sensor), 4C (DIY Who Am I game)
• Choose one thing that you did and describe it, explaining the circuit, code, and what they do.
• Continue to think about a final project. If you'll need stuff beyond your kit, time to start rounding it up.
## 6. robot due Tue May 17
• Last set of experiments: 5A (motor basics), 5B (remote control), 5C (autonomous robot)
• Choose one thing that you did and describe it, explaining the circuit, code, and what they do.
• Describe what you have in mind for a final project.
## 7. final project due Tue May 24
• Last class!
• Design and implement your own gizmo using any of the ideas you've seen this semester, or adding new ones. You can use the components from the kit or add in other stuff.
• Describe what you did, including the code, wiring diagram, and a brief explanation with some photos and/or video.
• Come to class ready for show-n-tell.
## 8. semester evaluation due Mon Jun 6
• A place for the term grade. | 962 | 3,975 | {"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} | 3.046875 | 3 | CC-MAIN-2024-33 | latest | en | 0.912887 |
https://brainmass.com/physics/gauss-law/example-problems-gauss-law-163973 | 1,537,523,546,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157028.10/warc/CC-MAIN-20180921092215-20180921112615-00065.warc.gz | 479,989,151 | 18,300 | Explore BrainMass
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# Gauss' law
This content was STOLEN from BrainMass.com - View the original, and get the solution, here!
A hollow, conducting sphere with an outer radius of 0.242m and an inner radius of 0.197m has a uniform surface charge density of 6.68*10^-6 C/m^2. A charge of -0.830 microC is now introduced into the cavity inside the sphere.
A) What is the new charge density on the outside of the sphere in C/m^2?
B) What is the strength of the electric field just outside the sphere in N/C?
C) What is the electric flux through a spherical surface just inside the inner surface of the sphere in N*m^2/C?
© BrainMass Inc. brainmass.com September 21, 2018, 5:52 am ad1c9bdddf - https://brainmass.com/physics/gauss-law/example-problems-gauss-law-163973
#### Solution Summary
Step by step solution provided.
\$2.19 | 224 | 833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-39 | latest | en | 0.88385 |
http://en.wikipedia.org/wiki/Analytical_expression | 1,417,256,871,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931014369.10/warc/CC-MAIN-20141125155654-00126-ip-10-235-23-156.ec2.internal.warc.gz | 96,314,775 | 17,192 | # Closed-form expression
(Redirected from Analytical expression)
"Closed formula" redirects here. For "closed formula" in the sense of a logic formula with no free variables, see Sentence (mathematical logic).
In mathematics, a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, certain "well-known" operations (e.g., + − × ÷), and functions (e.g., nth root, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions), but e.g. usually no limit. The set of operations and functions admitted in a closed-form expression may vary with author and context.
Problems are said to be tractable if they can be solved in terms of a closed-form expression.
## Example: roots of polynomials
The solutions of any quadratic equation with complex coefficients can be expressed in closed form in terms of addition, subtraction, multiplication, division, and square root extraction, each of which is an elementary function. For example, the quadratic equation:
$ax^2+bx+c=0,\,$
is tractable since its solutions can be expressed as closed-form expression, i.e. in terms of elementary functions:
$x={-b\pm\sqrt{b^2-4ac} \over 2a}$
Similarly solutions of cubic and quartic (third and fourth degree) equations can be expressed using arithmetic, square roots, and cube roots, or alternatively using arithmetic and trigonometric functions. However, there are quintic equations without closed-form solutions using elementary functions, such as x5 − x + 1 = 0.
An area of study in mathematics referred to broadly as Galois theory involves proving that no closed-form expression exists in certain contexts, based on the central example of closed-form solutions to polynomials.
## Alternative definitions
Changing the definition of "well-known" to include additional functions can change the set of equations with closed-form solutions. Many cumulative distribution functions cannot be expressed in closed form, unless one considers special functions such as the error function or gamma function to be well known. It is possible to solve the quintic equation if general hypergeometric functions are included, although the solution is far too complicated algebraically to be useful. For many practical computer applications, it is entirely reasonable to assume that the gamma function and other special functions are well-known, since numerical implementations are widely available.
## Analytic expression
An analytic expression (or expression in analytic form) is a mathematical expression constructed using well-known operations that lend themselves readily to calculation. Similar to closed-form expressions, the set of well-known functions allowed can vary according to context but always includes the basic arithmetic operations (addition, subtraction, multiplication, and division), exponentiation to a real exponent (which includes extraction of the nth root), logarithms, and trigonometric functions.
However, the class of expressions considered to be analytic expressions tends to be wider than that for closed-form expressions. In particular, special functions such as the Bessel functions and the gamma function are usually allowed, and often so are infinite series and continued fractions. On the other hand, limits in general, and integrals in particular, are typically excluded.
If an analytic expression involves only the algebraic operations (addition, subtraction, multiplication, division and exponentiation to a rational exponent) and rational constants then it is more specifically referred to as an algebraic expression.
## Comparison of different classes of expressions
Closed-form expressions are an important sub-class of analytic expressions, which contain a bounded[citation needed] or unbounded number of applications of well-known functions. Unlike the broader analytic expressions, the closed-form expressions do not include infinite series or continued fractions; neither includes integrals or limits. Indeed, by the Stone–Weierstrass theorem, any continuous function on the unit interval can be expressed as a limit of polynomials, so any class of functions containing the polynomials and closed under limits will necessarily include all continuous functions.
Similarly, an equation or system of equations is said to have a closed-form solution if, and only if, at least one solution can be expressed as a closed-form expression; and it is said to have an analytic solution if and only if at least one solution can be expressed as an analytic expression. There is a subtle distinction between a "closed-form function" and a "closed-form number" in the discussion of a "closed-form solution", discussed in (Chow 1999) and below. A closed-form or analytic solution is sometimes referred to as an explicit solution.
## Dealing with non-closed-form expressions
### Transformation into closed-form expressions
The expression:
$f(x) = \sum_{i=0}^\infty {x \over 2^i}$
is not in closed form because the summation entails an infinite number of elementary operations. However, by summing a geometric series this expression can be expressed in the closed-form:[1]
$f(x) = 2x$
### Differential Galois theory
The integral of a closed-form expression may or may not itself be expressible as a closed-form expression. This study is referred to as differential Galois theory, by analogy with algebraic Galois theory.
The basic theorem of differential Galois theory is due to Joseph Liouville in the 1830s and 1840s and hence referred to as Liouville's theorem.
A standard example of an elementary function whose antiderivative does not have a closed-form expression is:
$e^{-x^2}$
whose antiderivative is (up to constants) the error function:
$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2}\,\mathrm dt.$
### Mathematical modelling and computer simulation
Equations or systems too complex for closed-form or analytic solutions can often be analysed by mathematical modelling and computer simulation.
## Closed-form number
Three subfields of the complex numbers C have been suggested as encoding the notion of a "closed-form number"; in increasing order of generality, these are the EL numbers, Liouville numbers, and elementary numbers. The Liouville numbers, denoted L (not to be confused with Liouville numbers in the sense of rational approximation), form the smallest algebraically closed subfield of C closed under exponentiation and logarithm (formally, intersection of all such subfields)—that is, numbers which involve explicit exponentiation and logarithms, but allow explicit and implicit polynomials (roots of polynomials); this is defined in (Ritt 1948, p. 60). L was originally referred to as elementary numbers, but this term is now used more broadly to refer to numbers defined in explicitly or implicitly in terms of algebraic operations, exponentials, and logarithms. A narrower definition proposed in (Chow 1999, pp. 441–442), denoted E, and referred to as EL numbers, is the smallest subfield of C closed under exponentiation and logarithm—this need not be algebraically closed, and correspond to explicit algebraic, exponential, and logarithmic operations. "EL" stands both for "Exponential-Logarithmic" and as an abbreviation for "elementary".
Whether a number is a closed-form number is related to whether a number is transcendental. Formally, Liouville numbers and elementary numbers contain the algebraic numbers, and they include some but not all transcendental numbers. In contrast, EL numbers do not contain all algebraic numbers, but do include some transcendental numbers. Closed-form numbers can be studied via transcendence theory, in which a major result is the Gelfond–Schneider theorem, and a major open question is Schanuel's conjecture.
## Numerical computations
For purposes of numeric computations, being in closed form is not in general necessary, as many limits and integrals can be efficiently computed.
## Conversion from numerical forms
There is software that attempts to find closed-form expressions for numerical values, including RIES,[2] identify in Maple[3] and SymPy,[4] Plouffe's Inverter,[5] and the Inverse Symbolic Calculator.[6] | 1,688 | 8,244 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-49 | latest | en | 0.916402 |
http://www.diyaudio.com/forums/solid-state/174542-name-topology-2.html | 1,501,030,077,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00080.warc.gz | 424,581,176 | 17,923 | What is the name of this topology? - Page 2 - diyAudio
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1st October 2010, 09:04 AM #11 DRC diyAudio Member Join Date: Jan 2003 Location: UK (south west) I still think the analysis will be, more or less, the same as an LTP where the tail resistor is R4/R5 in parellel and Re is R7/2. Assuming C1 is large enough, and of course, the dc analysis is different. dc Last edited by DRC; 1st October 2010 at 09:08 AM.
1st October 2010, 10:35 AM #12 AndrewT diyAudio Member Join Date: Jul 2004 Location: Scottish Borders no. The separate tail sinks defeat the LTP action. An LTP uses the constant current draw of the tail to influence how the other half amplifies it's input signal. __________________ regards Andrew T. Sent from my desktop computer using a keyboard
DRC
diyAudio Member
Join Date: Jan 2003
Location: UK (south west)
Quote:
An LTP uses the constant current draw of the tail to influence how the other half amplifies it's input signal.
thats what R7 + C1 do !
tiefbassuebertr
diyAudio Member
Join Date: Jul 2005
Location: D-55629 Schwarzerden
Quote:
Originally Posted by DF96 This is an emitter-coupled amplifier. The valve analogue is called the cathode-coupled amp. It is a form of differential amp. Differences from a normal LTP are: not necessarily balanced, often quite a 'short' tail, often AC coupling, not necessarily the same transistor type. To a first approximation all BJT behave in exactly the same way: exponential response. That is why changing the transistor had little effect. A normal LTP uses a matched pair in order to maintain DC balance. The ECA will typically have lower even-order distortion than a common-emitter amp. As the output is in phase with the input it might have greater risk of oscillation due to capacitive feedback.
I can also call that emitter coupled cascode (another form of the good known normal cascode or folded cascode).
check out US Patent 6,600,367
http://www.patentstorm.us/patents/66...scription.html
http://www.pat2pdf.org/patents/pat6600367.pdf
and
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This topology I get often automaticly, when I transform a power amp from the non inverted mode to an inverted mode.
But an additional fact is of interest:
After replace R4 and R5 each through current sources (two independend current sources instead commonly used one pcs. for both halves) and R7 through a potentiometer, you get the possibility for easy variation of the open loop gain/damping factor from your whole power amp stage.
Last edited by tiefbassuebertr; 1st October 2010 at 12:19 PM.
DF96
diyAudio Member
Join Date: May 2007
Quote:
no. The separate tail sinks defeat the LTP action.
Get rid of R7 and you have a short-tail pair. The tail is just R4 and R5 in parallel. Obviously this doesn't go down to DC because of the capacitor.
2nd October 2010, 12:06 AM #16 cbdb diyAudio Member Join Date: Oct 2008 Location: Vancouver A diff amp needs a diff input. T2s base is ac grounded. You have 2 outs, the first from a CE amp, the second from a 2 stage amp (cascode) CC then CB (common base). Thats the way it should be analized, not as a diff amp which is possible but thats over complicating things for no reason. (draw the second transistor rotated 90 degrees clockwise, the usual way to draw a CB amp, and maybe it will be easier to understand) Last edited by cbdb; 2nd October 2010 at 12:10 AM.
tiefbassuebertr
diyAudio Member
Join Date: Jul 2005
Location: D-55629 Schwarzerden
Quote:
Originally Posted by cbdb A diff amp needs a diff input. T2s base is ac grounded. You have 2 outs, the first from a CE amp, the second from a 2 stage amp (cascode) CC then CB (common base). Thats the way it should be analized, not as a diff amp which is possible but thats over complicating things for no reason. (draw the second transistor rotated 90 degrees clockwise, the usual way to draw a CB amp, and maybe it will be easier to understand)
Yes, the kind of drawing often determines, how well you understand the right working of a certainly circuit.
Independend of this - despite of the fact of grounding the inverted input it is still a diff amp - then the different between non inverted input and GND.
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All times are GMT. The time now is 12:47 AM. | 1,628 | 6,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-30 | latest | en | 0.899783 |
https://www.cyclingforarmenia.net/answers-to-popular-questions/often-asked-how-to-calculate-w-prime-cycling.html | 1,652,848,033,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521041.0/warc/CC-MAIN-20220518021247-20220518051247-00122.warc.gz | 840,431,500 | 13,876 | # Often asked: How To Calculate W Prime Cycling?
## What is W prime in cycling?
Often referred to as Anaerobic Work Capacity or more currently W ‘ ( W prime ), it is a short term energy source that uses non-aerobic (non oxygen) pathways that fills the gap when you hit the gas.
## How is cycling W calculated?
To work out your power-to-weight ratio figure, simply divide your maximum power output (in watts ) by your body mass in kilograms (kg). For example, an 80kg rider with a maximum sustainable power output of 280 watts has a power-to-weight ratio of 3.5 watts per kilo (commonly abbreviated as 3.5 W /kg or 3.5 W. kg-1).
## What is the W prime?
W Prime is basically a block of energy that the legs can produce when the rider goes over MLSS up to the point where the rider “blows-up” because of the intolerable pain. The W ‘ trace takes a dive at the start of each sprint interval and reflects the amount of energy used from the anaerobic resources.
## What is critical power and W prime?
When exercise tolerance is considered, the power -asymptote is known as CP (or critical speed [CS] when intensity is measured in units of speed rather than power ) and the curvature constant is known as W ′ (i.e., W prime ) and is measured in units of work done, that is, J (or D′ when measured in units of distance, that is
You might be interested: What Is Ftp Cycling?
## What is anaerobic capacity?
Anaerobic capacity is defined as the maximal amount of adenosine triphosphate resynthesized via anaerobic metabolism (by the whole organism) during a specific mode of short-duration maximal exercise.
## What is anaerobic capacity cycling?
Anaerobic capacity is the zone way above your functional threshold. It is a very hard zone which can be kept just up to around 3 minutes. Intervals at this zone are strongly recommended for criterium racing, track racing, and road racing.
## Is 2.5 watts per kg good?
Here are the categories that they recommend you race based on your FTP data and watts per kilogram. The Zwift C category (next from the bottom) says that you should be able to ride at the level between 2.5 to 3.1 w/ kg. If you want to succeed in the A category, you’ll need to be able to hit 4.0 w/ kg or better.
## What is my watts per kg?
Simply put, your watts per kilo (w/ kg ) is your power to weight ratio. Watts per Kilo is your max power output, in watts, divided by your weight in kilos. For example, someone with a weight or mass of 80kg with a sustainable power output of 280 watts will have a power to weight ratio of 3.5 watts per kilo (3.5W/ kg ).
## What is a good cycling FTP?
47% of people have an FTP below 3.2W/kg. 45% of people have an FTP of 3.4W/kg or more. 8% of people have an FTP between 3.2W/kg and 3.4W/kg.
## What is critical power in cycling?
What is Critical Power? Critical Power, or maximum mean power (MMP), is the best average power (measured in watt) you can produce on a bike in a given time frame. A Critical Power for five minutes (denoted as CP5) of 300 watt means that you can drive for five minutes with 300 watt, before you are exhausted.
You might be interested: Quick Answer: How To Properly Wear Cycling Kit?
## How do you find anaerobic capacity?
Anaerobic capacity is expressed as kilogram-Joules (1 kg-m = 9.804 J) and is calculated by adding together each 5-second peak power output over the 30 seconds.
## How can I increase my anaerobic capacity cycling?
To increase anaerobic capacity, you have to perform efforts that burn through the maximum amount of energy you can currently produce through glycolysis. That means repeated short, hard efforts above lactate threshold that rely heavily on fast-twitch muscle fibers.
## How do you increase critical power?
How do I get my Critical Power?
1. After purchasing Stryd, simply go for a few runs with it.
2. Note your Critical Power will change often at first as you do more runs and Stryd gets to know you but will slow down over time with more incremental changes less frequently.
## What is critical power exercise?
The CP may be functionally defined as the highest power output that can be sustained without progressively drawing on W′, where the latter represents, at the onset of exercise, a fixed amount of work that can be done when CP is exceeded.
## Whats a good lactate threshold?
Muscles are producing lactate even at rest, usually about 0.8-1.5 mmol/L. Although the lactate threshold is defined as the point when lactic acid starts to accumulate, some testers approximate this by crossing the lactate threshold and using the point at which lactate reaches a concentration of 4 mmol/L of lactate.
## Often asked: How Can I Keep My Wonder Core From Lifting Up When Cycling?Often asked: How Can I Keep My Wonder Core From Lifting Up When Cycling?
Contents1 How can I improve my core strength for cycling?2 Does the wonder core actually work?3 Does the wonder Core 2 fold away?4 What exercises can you do with a | 1,187 | 4,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-21 | latest | en | 0.940305 |
http://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.208197.html | 1,369,131,658,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699881956/warc/CC-MAIN-20130516102441-00027-ip-10-60-113-184.ec2.internal.warc.gz | 312,349,676 | 4,905 | # SOLUTION: A cellular phone company offers a contract for which the cost , C, in dollars, of t minutes of telephoning is given by C = .25(t – 600) + 65.95 , where it is assumed that t ≥
Algebra -> Algebra -> Graphs -> SOLUTION: A cellular phone company offers a contract for which the cost , C, in dollars, of t minutes of telephoning is given by C = .25(t – 600) + 65.95 , where it is assumed that t ≥ Log On
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Click here to see ALL problems on Graphs Question 208197: A cellular phone company offers a contract for which the cost , C, in dollars, of t minutes of telephoning is given by C = .25(t – 600) + 65.95 , where it is assumed that t ≥ 600 minutes. What times will keep costs between \$110.95 and \$145.95? For the cost to be between \$110.95 and \$145.95, the telephoning time must be between ___ minutes and ___ minutes. Thank you, in advance. Your assistance is vey much appreciated.Answer by ankor@dixie-net.com(15645) (Show Source): You can put this solution on YOUR website!A cellular phone company offers a contract for which the cost , C, in dollars, of t minutes of telephoning is given by C = .25(t – 600) + 65.95 , where it is assumed that t ≥ 600 minutes. What times will keep costs between \$110.95 and \$145.95? : Solve two inequalities .25(t - 600) + 65.95 > 110.95 .25t - 150 + 65.95 > 110.95 .25t - 84.05 > 110.95 .25t > 110.95 + 84.05 .25t > 195 t > t > 780 minutes and .25(t - 600) + 65.95 < 145.95 .25t - 150 + 65.95 < 145.95 .25t - 84.05 < 145.95 .25t < 145.95 + 84.05 .25t < 230 t < t < 920 minutes For the cost to be between \$110.95 and \$145.95, the telephoning time must be between 780 minutes and 920 minutes. | 587 | 1,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2013-20 | latest | en | 0.879954 |
https://scientificnotation.org/34928-in-scientific-notation | 1,603,122,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00187.warc.gz | 520,442,403 | 29,224 | # 34928 in Scientific Notation
Welcome to 34928 in scientific notation, our post about writing the number thirty-four thousand, nine hundred and twenty-eight in scientific notation.
If you have been looking for 34928 in exponential form, 34928 as a power of 10, or what is 34928 in scientific notation, then you are right here, too.
In this post you can find everything about the scientific notation of 34928, including the normalized form and the number written in e-notation.
Read on to learn everything about the scientific notation of thirty-four thousand, nine hundred and twenty-eight.
## What is 34928 in Scientific Notation?
To answer the question what is 34928 in scientific notation let’s recall how to convert 34928 to scientific notation as detailed on our home page.
When you split the number 34928 into a coefficient and a power of 10 you do get 34928 in exponential form, but there is an indefinite number of possibilities.
As you probably want 34928 in normalized scientific notation, the coefficient or significand of 34928 in scientific notation must be in the interval [1,10[.
As there are many ways to express 34928 in scientific notation, in this post we mean 34928 in normalized scientific notation, unless stated otherwise. Therefore:
34928 in scientific notation = 3.4928 × 104.
This can also be expressed as 3.4928 × 10^4, using the caret symbol, or as 3.4928e+4, which is called 34928 in e-notation, further discussed in the section ahead.
Next is our converter. Our tool changes any decimal number including 34928 to standard index form. If your press the swap button, then 34928 is changed back to decimal form.
### Change decimal to scientific
Our tool displays converted numbers in a special form of scientific notation called e-notation, which means that × 104 is replaced by e+4.
## 34928 in Exponential Form
In 34928 in exponential form ab, a is the base, and b is called the exponent, also known as index or power of the exponential form of 34928.
However, it’s important to understand that there is not a unique means to express 34928 in exponential form; in fact there are countless possibilities.
By definition of exponential form, not only the index and coefficient, but also the base can be varied. In contrast to the scientific notation for 34928, which has 10 as base.
While on the subject, here are some more numbers you might be interested in:
In the following paragraph we elaborate on 34928 as a power of 10.
## 34928 as a Power of 10
As outlined above, there is more than one way to write 34928 as a power of 10. The number 34928 as a power of 10 is synonym to 34928 in scientific notation.
If you have found us searching for 34928 in scientific form or, for instance, scientific notation 34928, then you have also learned all you wanted to know about it.
## Conclusion
Note that you can locate many numbers in scientific notation including, but not limited, to 34928 using the search form in the sidebar or at the end of this post.
More about scientific notation can be encountered on our homepage. For questions on thirty-four thousand, nine hundred and twenty-eight in scientific notation, use the comment form at the end of this article.
As another option, you may get in touch with us by email using a meaningful subject line such as, for example, 34928 in scientific notation.
If our content about writing 34928 as a power of 10 has been useful to you, then make sure to bookmark us and to hit the sharing buttons to help spreading the word about us. | 799 | 3,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-45 | latest | en | 0.919105 |
https://www.esaral.com/q/evaluate-the-following-integral-93528 | 1,713,106,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00499.warc.gz | 695,221,392 | 11,152 | # Evaluate the following integral:
Question:
Evaluate the following integral:
$\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)^{2}} d x$
Solution:
Let, $\mathrm{I}=\int \frac{2 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}+2\right)^{2}} \mathrm{dx}$
Let $x^{2}+2=t \Rightarrow 2 x d x=d t$
$\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{(\mathrm{t}-1) \mathrm{t}^{2}}$
Now, let, $\frac{1}{(t-1) t^{2}}=\frac{A}{t-1}+\frac{B}{t}+\frac{C}{t^{2}}$
$\Rightarrow 1=A t^{2}+B t(t-1)+C(t-1)$
For $t=1, A=1$
For $t=0, C=-1$
For $t=-1, B=-1$
$\therefore I=\int \frac{d t}{t-1}-\int \frac{d t}{t}-\int \frac{d t}{t^{2}}$
$\Rightarrow I=\log |t-1|-\log |t|+\frac{1}{t}+c$
So, $\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)^{2}} d x=\log |t-1|-\log |t|+\frac{1}{t}+c$ | 375 | 806 | {"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} | 4.40625 | 4 | CC-MAIN-2024-18 | latest | en | 0.250432 |
http://www.pearltrees.com/u/16759191-wikipedia-encyclopedia | 1,542,826,798,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039749562.99/warc/CC-MAIN-20181121173523-20181121194633-00017.warc.gz | 485,977,242 | 26,172 | # Planck constant
Plaque at the Humboldt University of Berlin: "Max Planck, discoverer of the elementary quantum of action h, taught in this building from 1889 to 1928." In 1905 the value (E), the energy of a charged atomic oscillator, was theoretically associated with the energy of the electromagnetic wave itself, representing the minimum amount of energy required to form an electromagnetic field (a "quantum"). Further investigation of quanta revealed behaviour associated with an independent unit ("particle") as opposed to an electromagnetic wave and was eventually given the term photon. The Planck relation now describes the energy of each photon in terms of the photon's frequency. This energy is extremely small in terms of ordinary experience. Since the frequency , wavelength λ, and speed of light c are related by λν = c, the Planck relation for a photon can also be expressed as The above equation leads to another relationship involving the Planck constant. Value Significance of the value
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Quantum entanglement Quantum entanglement is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the quantum state of each particle cannot be described independently – instead, a quantum state may be given for the system as a whole. Such phenomena were the subject of a 1935 paper by Albert Einstein, Boris Podolsky and Nathan Rosen,[1] describing what came to be known as the EPR paradox, and several papers by Erwin Schrödinger shortly thereafter.[2][3] Einstein and others considered such behavior to be impossible, as it violated the local realist view of causality (Einstein referred to it as "spooky action at a distance"),[4] and argued that the accepted formulation of quantum mechanics must therefore be incomplete. History However, they did not coin the word entanglement, nor did they generalize the special properties of the state they considered.
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Amplituhedron An amplituhedron is a geometric structure that enables simplified calculation of particle interactions in some quantum field theories. In planar N = 4 supersymmetric Yang–Mills theory, an amplituhedron is defined as a mathematical space known as the positive Grassmannian. The connection between the amplituhedron and scattering amplitudes is at present a conjecture that has passed many non-trivial checks, including an understanding of how locality and unitarity arise as consequences of positivity. Research has been led by Nima Arkani-Hamed. Edward Witten described the work as “very unexpected" and said that "it is difficult to guess what will happen or what the lessons will turn out to be.
Research Reveals New Man-Made Ozone-Depleting Gases in the Atmosphere - SciTech Daily False-color view of total ozone over the Antarctic pole. The purple and blue colors are where there is the least ozone, and the yellows and reds are where there is more ozone. Credit: NASA New research from scientists at the University of East Anglia reveals four new man-made gases in the atmosphere that are contributing to the destruction of the ozone layer. New research published in the journal Nature Geoscience reveals that more than 74,000 tonnes of three new chlorofluorocarbons (CFCs) and one new hydrochlorofluorocarbon (HCFC) have been released into the atmosphere. Scientists made the discovery by comparing today’s air samples with air trapped in polar firn snow – which provides a century-old natural archive of the atmosphere.
Intelligent design Intelligent design (ID) is the pseudoscientific view[1][2] that "certain features of the universe and of living things are best explained by an intelligent cause, not an undirected process such as natural selection."[3] Educators, philosophers, and the scientific community have demonstrated that ID is a religious argument, a form of creationism which lacks empirical support and offers no tenable hypotheses.[4][5][6] Proponents argue that it is "an evidence-based scientific theory about life's origins" that challenges the methodological naturalism inherent in modern science,[7][8] while conceding that they have yet to produce a scientific theory.[9] The leading proponents of ID are associated with the Discovery Institute, a politically conservative think tank based in the United States.[n 1] Although they state that ID is not creationism and deliberately avoid assigning a personality to the designer, many of these proponents express belief that the designer is the Christian deity.[n 2]
Related: | 1,946 | 9,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | longest | en | 0.938131 |
http://teaching.csse.uwa.edu.au/units/CITS2002/workshops/workshop4.php | 1,586,016,063,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370524043.56/warc/CC-MAIN-20200404134723-20200404164723-00098.warc.gz | 176,905,142 | 6,211 | CITS2002 Systems Programming
# Department of Computer Science and Software Engineering
## CITS2002 Systems Programming
### Workshop 4: Functions and 2-D arrays
Consider a simple game often seen on mobile phones, with names such as Jawbreaker and Bejeweled. The game consists of cells arranged in a square; each cell initially holds an object of one of (say) 4 colours. The goal is to remove objects by clicking on them; removing an object also removes each of its neighbours that have the same colour (in the directions North, South, East, West).
Every time n neighbouring like-coloured objects are removed, you score n2 points.
Consider the game at left - a 6x6 board employing 4 colours. The top-left cell has the coordinates (col=0,row=0), the bottom-right cell has the coordinates (col=5,row=5). You are NOT expected to develop a graphical interface!
• How, and where, would you declare and define a 2-dimensional array to hold each cell's character values of 'R', 'G', 'B', 'Y', or ' ' (for empty)?
• Consider the function void reset(void);
which initializes each cell's colour. Note - the expression (rand() % 4) produces one of 4 random integers: 0, 1, 2, or 3.
• Consider the function bool valid(int col, int row);
which determines if the indicated coordinates are 'within' the array (on the board).
• Consider the function int click(int col, int row);
which simulates someone clicking in the indicated cell.
1. If the cell is non-empty, we remove its object, and all neighbouring objects (N,S,E,W) of the same colour (yes, this part is getting hard!).
2. When all like-coloured neighbours have been removed, objects "fall down" if there is an empty cell below them.
3. If n like-coloured objects are removed, the function should return the value of n2.
4. If ever a complete column becomes empty, adjacent columns should "move toward the centre of the board" to keep all objects together.
BEFORE the workshop session you're strongly encouraged to think how you would do it. You are not expected to have developed and tested a perfect solution before the workshop, but you should at least scratch out your approach on paper.
From the information recently presented in lectures, we know:
• our programs can have functions, which identify a single task or calculation to be performed,
• functions may receive parameters, which direct how the function performs its task,
• functions may return a value, often a calculated result, to the code which called our function,
• we may combine multiple pieces of data, and refer to each using a single name, into an array,
• arrays may have 1, 2, 3, or more dimensions, which may be considered similar to a vector, rectangle, cube, ... of data
• we employ integer indices to access individual array elements.
So, before the workshop session, think how you can combine the (above) information that we already know to develop a solution.
# The University of Western Australia
## University information
CRICOS Code: 00126G | 685 | 3,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-16 | latest | en | 0.874559 |
http://www.wiibrew.org/w/index.php?title=Senet&action=history | 1,558,820,468,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258451.88/warc/CC-MAIN-20190525204936-20190525230936-00343.warc.gz | 357,123,263 | 10,103 | # Senet
General Senet pedrocrespo Board game 0.05b Creative Commons by-nc-nd 3.0 Download Downloadable via the Homebrew Browser
Senet, a board game from predynastic and ancient Egypt, is the oldest board game whose ancient existence has been confirmed, dating to circa 3500 BC. Soon you'll be able to play it on your Wii, and understand why Tutankhamen has two senet gameboards inside his tomb.
## Rules (based on Spanish Wikipedia's)
The objective in Senet is to remove your pieces from the board before your opponent, by following a few rules, advancing your own pieces and catching or blocking your adversary's ones.
It is a game for two players and consists of a board with three parallel rows of ten squares each as well as ten small playing pieces. One player controls the five conical pieces and the other plays the cylindrical ones.
The order of the squares is 1 through 10 from left to right on the first row, 11 through 20 from right to left on the second row and 21 through 30 from left to right on the third row. At the start of the game, the first row must be filled by the gamers' pieces, arranged alternatively (ABABABABAB). Several special squares exist. They are 15, 26, 27, 28 and 29 (these special squares can contain hieroglyphic drawings or symbols).
If a player falls in square 27 he must begin from square 15 (as happens in some squares of the 'Game of the Oca') but if square 15 is engaged, the piece that fell into square 27 must move to the nearest free square to square 1. Squares 26, 28, and 29 are squares where pieces are protected.
Moving the piece out of the board (to a supposed square 31) is counted as a move.
Dice are not used to determine the advancement of pieces, rather four small sticks with two white faces and two black faces are rolled. The amount moved is decided by how many white faces are rolled (1, 2, 3 or 4). If all black faces are rolled, move six. There does not exist a roll resulting in a move of 5.
Every time a player rolls a 1, 3 or 6, he gets another turn. After moving the piece he wanted or was able to move, he rolls and goes again until he rolls a 2 or a 4.
When two pieces of the same player are in consecutive squares, they are both protected and cannot be captured by the adversary; when they are three pieces of the same player, they form a barrier that the opponent cannot jump, but the barrier owner can jump.
The capture of an opponent's piece consists of interchanging the position of the capturing piece by the position of the captured piece. This can only be done when the captured piece is not protected and on the last square of capturing piece's movement.
When it is not possible to move forwards (protected pieces or barrier) but moving backwards is possible, it is obligatory to move backwards.
Game start. Starting setup is shown, with special squares filled: green are protection squares, blue is the square that makes you move to the red one if you land on it. Gameplay direction. In this picture you can see the direction pieces must be moved while playing, except if they can't move this way, and have to move backwards instead.
Pieces protecting themselves. If it is the conic piece's turn and he rolls a 2 or 3, he won't be able to move forwards, because the cylindrical pieces are next to each other and protected from being captured. So the conic piece would have to move backwards because it can. In this case, if he rolled a 2 he would capture the cylindrical piece two squares behind him, but this move wouldn't be beneficial to this player, because he would now have three opponent's pieces ahead. The conic piece could jump the opponent's pieces with a 4 or a 6, because there's no barrier. Of course, if he rolls a 1 he can move to the next square, because it's empty.
Barrier. If it is the conic piece's turn, he won't be able to move forwards unless he scores a 1, because of the barrier of three cylindrical pieces preventing him from jumping over them but still allowing their owner to jump them. In this case (a roll other than 1), the conic piece would have to move backwards, but it only could do so by rolling a 4 or 6, because the cylindrical rear pieces are protecting themselves. If backward moving is not possible and this player hasn't any other movable piece on the board, he must pass his turn.
## Controls
The menu is controlled using Wiimote 1
Action
Move pointer
Select item
Move through menu. In the tutorial, continue reading.
### Game
The game is controlled using Wiimotes 1 and 2
Action
Move pointer
Roll the sticks
Select piece to move
+ If a player can't move, he can pass the turn by holding PLUS, while the adversary presses MINUS.
## Screenshots
Main Menu Tut-ankh-orial Ingame
## Development stat
### v0.05b (10-07-09)
Passing turn function has been fixed because it wasn't working properly. Please, let me know any bug you'd find. Suggestions are welcome too.
### v0.05 (09-07-09)
It includes all the rules shown above but two:
• jumping multiple barriers isn't allowed.
• if advancing forward is not possible, move backwards if it is.
A basic howtoplay tutorial (tut-ankh-orial) has been included too, but it's just this same page for now. --Pedrocrespo 23:09, 9 July 2009 (UTC) | 1,231 | 5,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-22 | latest | en | 0.958709 |
https://www.encyclopediaofmath.org/index.php/Polynomial_of_best_approximation | 1,558,247,142,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254253.31/warc/CC-MAIN-20190519061520-20190519083520-00175.warc.gz | 777,232,027 | 6,780 | # Polynomial of best approximation
A polynomial furnishing the best approximation of a function in some metric, relative to all polynomials constructed from a given (finite) system of functions. If is a normed linear function space (such as or , ), and if
is a system of linearly independent functions in , then for any the (generalized) polynomial of best approximation
(*)
defined by the relation
exists. The polynomial of best approximation is unique for all if is a space with a strictly convex norm (i.e. if and , then ). This is the case for , . In , which has a norm that is not strictly convex, the polynomial of best approximation for any is unique if is a Chebyshev system on , i.e. if each polynomial
has at most zeros on . In particular, one has uniqueness in the case of the (usual) algebraic polynomials in , and also for the trigonometric polynomials in the space of continuous -periodic functions on the real line, with the uniform metric. If the polynomial of best approximation exists and is unique for any , it is a continuous function of .
Necessary and sufficient conditions for a polynomial to be a best approximation in the spaces and are known. For example, one has Chebyshev's theorem: If is a Chebyshev system, then the polynomial (*) is a polynomial of best approximation for a function in the metric of if and only if there exists a system of points , , at which the difference
assumes values
and, moreover,
The polynomial (*) is a polynomial of best approximation for a function , , in the metric of that space, if and only if for ,
In , the conditions
are sufficient for to be a polynomial of best approximation for , and if the measure of the set of all points at which is zero, they are also necessary; see also Markov criterion.
There exist algorithms for the approximate construction of polynomials of best uniform approximation (see e.g. [3], [5]).
#### References
[1] N.I. [N.I. Akhiezer] Achiezer, "Theory of approximation" , F. Ungar (1956) (Translated from Russian) [2] N.P. Korneichuk, "Extremal problems in approximation theory" , Moscow (1976) (In Russian) [3] V.K. Dzyadyk, "Introduction to the theory of uniform approximation of functions by polynomials" , Moscow (1977) (In Russian) [4] V.M. Tikhomirov, "Some problems in approximation theory" , Moscow (1976) (In Russian) [5] P.J. Laurent, "Approximation et optimisation" , Hermann (1972) [6] E.Ya. Remez, "Foundations of numerical methods of Chebyshev approximation" , Kiev (1969) (In Russian) | 600 | 2,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-22 | latest | en | 0.903306 |
https://www.teacherspayteachers.com/Product/Sector-Area-Guided-Notes-3181304 | 1,547,612,767,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583656665.34/warc/CC-MAIN-20190116031807-20190116053807-00291.warc.gz | 943,178,039 | 19,074 | # Sector Area Guided Notes
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1. Resources to teach circle vocabulary along with calculating circumference, area, arc length, and sector area. I created all of these resources when I realized my tenth grade geometry students did not have a good foundation in circle knowledge. The entire bundle could be used to get students from a
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Product Description
Use these interactive notes to introduce students to finding areas of sectors in circles.
The 2-page notes include basic circle vocabulary, review of area, steps for finding sector area, practice, and challenge problems involving shaded regions.
These notes do not use the standard sector area formula. Instead, I try to get students to understand the concept of sector area by calculating the area of a circle, finding a fraction or decimal for the sector’s portion, and then multiplying these two figures.
The pages are meant to be filled in as a class with the teacher aiding some portions and then allowing students to try some practice problems on their own before checking as a group.
Check out the preview!
Thank you for your interest in this resource from Rise over Run.
Suggested Resources:
Circle Vocabulary Posters
Geometry Bundle
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\$1.99 | 312 | 1,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-04 | latest | en | 0.896841 |
https://electronics.stackexchange.com/questions/539708/cant-for-the-life-of-me-figure-out-voltage-of-batteries-in-series | 1,725,958,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00402.warc.gz | 214,613,071 | 43,607 | # Can't for the life of me figure out voltage of batteries in series
Imagine a 1.5V battery.
1. Before the battery is connected to anything, is there any voltage on either end? That is, is there already a build up of negative charge on the anode? OR is it that when the battery is connected to a circuit electrons easily moved from the anode enters the conductor and push charges to the cathode creating a voltage difference escalating the process to a potential difference of 1.5V?
I assume the end already has a charge buildup creating an E-field but because of the material(e.g. air) it's too weak to displace the electrons.
As a conductor is connected the E-field in the conducting material is strong enough to Force the electrons through.
1. If there are 2 1.5V batteries and the anode of one is pressed against the cathode of the other why doesn't current flow? One end is positive and the other is negative and the ends are conducting?
2. If there are 2 1.5V batteries connected in SERIES and connected to a circuit the potential doubles. Now I can kind of understand why as the second cathode has an additional set of positive charges increasing the E-field. But then why do the two batteries even have to be connected, wouldn't it be enough for them to be close to each other and just connect the cathode of one to the anode of the other?
So then I assume there is no charge until the ends are connected to a conductor. But that leads back to question 2. The anode of one connected to the cathode of the other doesn't create a current?
1. What happens if the two batteries are not connected in SERIES but one cathode is connected to the others anode, is there a current but with still 1.5V potential? It shouldn't have to be a closed loop as electrons don't go round and round, just from lower potential to higher potential.
It's all very confusing for me.
• Just to clarify. Two batteries are IN SERIES if the anode of one is in contact with the cathode of the other. Period. Two batteries are in parallel if their cathodes are connected to each other and their anodes are connected to each other. If two batteries, A and B, are connected so that Cathode A touches Anode B and Cathode B touches Anode A, then the batteries are IN SERIES and also SHORT CIRCUITED. Commented Dec 29, 2020 at 7:49
• When two elements are in series, they have the same current flowing through them. When two elements are in parallel, they have the same voltage or potential across them. By definition. Commented Dec 29, 2020 at 7:50
• You really need to do some reading and then post a set of refined questions which include diagrams that illustrate the point you are trying to make. The questions and terminology (used in the wrong context) in this posting do not make sense. Commented Dec 29, 2020 at 8:04
• I think that a lot of your misunderstanding comes from the way you seem to be looking at 'voltage' as if it's a single-point measurement. Voltage is also referred to as 'potential difference' because it's always a measurement between 2 points. Whenever someone says that 'the voltage a node X in a circuit is Y volts', there's an implied reference node that's being measured against (the 'ground' or 0V node). So you asking "is there any voltage on either end" doesn't really make any sense because voltage is a measurement from one end to the other. Commented Dec 29, 2020 at 14:57
1. The voltage (aka "potential difference") exists already when the battery is disconnected. Your following assumption is correct: the reason why there is no current is that the resistance of air is too big to allow any electron movement at 1.5V. "As a conductor is connected the E-field in the conducting material is strong enough to Force the electrons through" - just to be clear: the E-field does not become stronger, but the resistance of the wire is small enough for current to flow.
2. The circuit is not closed, so there can't be any current. The voltage adds up to 3V accross both batteries.
3. "If there are 2 1.5V batteries conected in paralell and connected to a circuit the potential doubles" - I don't know if you are aware of the meaning of the word "potential". The electrical potential difference (aka voltage) does not double, there will still be 1.5V. But the capability of delivering current doubles (do you mean that with "potential"?). As the E-field is proportional to the voltage, it is the same as for one battery. "[...] and just connect the cathode of one to the anode of the other?" that descibes a series connection...
4. No, it's the other way around: no current will flow (it still isn't a closed circuit) and the voltage remains 1.5V
I think you need a better understanding of voltage and current.
A good analogy for current is water flowing down outside a volcano in a pipe system, where water can never leak or exit the pipe.
Inside the volcano, water gets heated up and vaporates to the top. Only if you connect a pipe from top of the volcano to its bottom, water (current) will circulate.
The voltage analogy is the height of the volcano. A higher volcano does not mean that more water flows, because it also depends on the pipe diameter (resistance).
If you stack a volcano on top of another one, height (voltage) will be doubled. But the bottom volcano tip must match the top volcano base, i.e. they must connect e.g. with a pipe (conductor).
E-field is the steepness of the volcano and is not needed to be considered as it is already expressed by the height (voltage) of the volcano.
I hope this helps a bit for you to answer the questions by yourself.
• I like the air pump analogy. Run a vacuum cleaner (the battery) and air enters in one end and blows out the other. Plug either end with your hand (resistance) and the airflow (current) stops, but you can still feel the pressure (voltage) either sucking or pushing on your hand. Commented Dec 29, 2020 at 10:27
• @DarylK Pneumatic systems (your air pump) are not good analogs for electrical systems because air is compressible and electrons are not compressible (in common electrical situations). That's why hydraulic analogies using water or other incompressible fluids are preferred. Commented Dec 29, 2020 at 21:27
• Heh, by who? Are O2 atoms compressible? No, but in aggregate they are. So are electrons in aggregate. What better analog for a capacitor could there be than an air tank? Also, I was providing an example the OP could try without making a mess on his carpet. :) Commented Dec 31, 2020 at 11:45 | 1,510 | 6,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-38 | latest | en | 0.963598 |
http://quadibloc.com/crypto/co040501.htm | 1,571,726,326,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987803441.95/warc/CC-MAIN-20191022053647-20191022081147-00442.warc.gz | 163,572,828 | 2,375 | [Next] [Up/Previous] [Index]
# Differential and Linear Cryptanalysis
Multi-round ciphers such as DES are clearly very difficult to crack. One property they have is that even if one has some corresponding plaintext and ciphertext, it is not at all easy to determine what key has been used.
### Differential Cryptanalysis
However, if one is fortunate enough to have a large quantity of corresponding plaintext and ciphertext blocks for a particular unknown key, a technique called differential cryptanalysis, developed by Eli Biham and Adi Shamir, is available to obtain clues about some bits of the key, thereby shortening an exhaustive search.
After two rounds of DES, knowing both the input and output, it is trivial to determine the two subkeys used, since the outputs of both f-functions are known. For each S-box, there are four possible inputs to produce the known output. Since each subkey is 48 bits long, but the key is only 56 bits long, finding which of the four possibilities is true for each group of six bits in the subkeys is a bit like solving a crossword puzzle.
Once the number of rounds increases to four, the problem becomes much harder. However, it is still true that the output depends on the input and the key. For a limited number of rounds, it is inevitable, without the need for any flaws in the S-boxes, that there will be some cases where a bit or a combination of bits in the output will have some correlation with a simple combination of some input bits and some key bits. Ideally, that correlation should be absolute with respect to the key bits, since there is only one key to solve for, but it can be probabilistic with respect to the input and output bits, since there need to be many pairs to test.
As the number of rounds increases, though, the simple correlations disappear. Differential cryptanalysis represents an approach to finding more subtle correlations.
Instead of saying "if this bit is 1 in the input, then that bit will be 0 (or 1) in the output", we say "changing this bit in the input changes (or does not change) that bit in the output".
In fact, however, a complete pattern of which bits change and do not change in the input and in the output is the subject of differential cryptanalysis. The basic principle of differential cryptanalysis, in its classic form, is this: the cipher being attacked has a characteristic if there exists a constant X such that given many pairs of plaintexts A, B, such that B = A xor X, if a certain statement is true about the key, E(B,k) = E(A,k) xor Y for some constant Y will be true with a probability somewhat above that given by random chance.
### Linear Cryptanalysis
Linear cryptanalysis, invented by Mitsuru Matsui, is a different, but related technique. Instead of looking for isolated points at which a block cipher behaves like something simpler, it involves trying to create a simpler approximation to the block cipher as a whole.
For a great many plaintext-ciphertext pairs, the key that would produce that pair from the simplified cipher is found, and key bits which tend to be favored are likely to have the value of the corresponding bit of the key for the real cipher. The principle is a bit like the summation of many one-dimensional scans to produce a two-dimensional slice through an object in computer-assisted tomography.
[Next] [Up/Previous] [Index] | 707 | 3,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | longest | en | 0.949655 |
https://www.stat.math.ethz.ch/pipermail/r-devel/2018-July/076456.html | 1,680,222,958,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00082.warc.gz | 1,097,719,020 | 3,911 | # [Rd] problem with display of complex number
Martin Maechler m@echler @ending from @t@t@m@th@ethz@ch
Thu Jul 19 11:27:19 CEST 2018
```TL;DR : It's more complicated and needs more discussion
(which I start below)
> Hi,
> > 1e10+5i
> [1] 1e+10+0e+00i
> > Im(1e10+5i)
> [1] 5
>
> maybe little better...
>
> --- R-3.5.1.orig/src/main/complex.c 2018-03-26 07:02:25.000000000 +0900
> +++ R-3.5.1/src/main/complex.c 2018-07-10 12:50:42.523874767 +0900
> @@ -381,6 +381,7 @@
> r->i = fround(pow10 * x->i, digits)/pow10;
> } else {
> digits = (double)(dig);
> + if(digits < 1) digits=1; /* a little better */
> r->r = fround(x->r, digits);
> r->i = fround(x->i, digits);
> }
>
>
> --
> Best Regards,
> --
> Eiji NAKAMA <nakama (a) ki.rim.or.jp>
> "\u4e2d\u9593\u6804\u6cbb" <nakama (a) ki.rim.or.jp>
Thanks a lot, Eiji!
Yes, your proposed change does "prevent the worst", in this
case, however it is more complicated.
I do think the current display {i.e., format(), as.character(),..},
of such complex numbers is far from optimal, not only in the example
you show above but also more generally, when the real and
complex parts of the complex vector(s) are of different magnitude.
If only the Mod(z) of the complex number z was of importance, one
could argue that indeed it makes sense to use the same format
for the real and imaginary parts; this is however not always the
case, and rather the Re(z) and/or Im(z) are of interest in
themselves. For such cases, the current formatting of complex
numbers is unfortunate also in my view.
I've checked other systems:
1) octave, matlab, python
2) mathematica, maple
and they all format Re() and Im() separately, and I think we
should do.
In R/tests/reg-tests-2.Rout.save, line 5641 ff, we have had,
from svn c39481 | 2006-09-22 ), log 'tweak printing of complex numbers' ,
> ## printing of complex numbers of very different magnitudes
> 1e100 + 1e44i
[1] 1e+100+0e+00i
> 1e100 + pi*1i*10^(c(-100,0,1,40,100))
[1] 1e+100+ 0.000000e+00i 1e+100+ 0.000000e+00i 1e+100+ 0.000000e+00i
[4] 1e+100+ 0.000000e+00i 1e+100+3.141593e+100i
> ## first was silly, second not rounded correctly in 2.2.0 - 2.3.1
> ## We don't get them lining up, but that is a printf issue
> ## that only happens for very large complex nos.
>
>
> ### end of tests added in 2.4.0 ###
and with your change the output of the above example also
changes.
I now found more history:
The fundamental changes is indeed from 2005, for R 2.2.0 :
------------------------------------------------------------------------
r35253 | ripley | 2005-08-11 18:34:24 +0200 (Thu, 11 Aug 2005) | 2 lines
Changed paths:
M NEWS
M src/main/complex.c
M src/main/format.c
M tests/arith.R
M tests/arith.Rout.save
M tests/complex.R
M tests/complex.Rout.save
M tests/print-tests.Rout.save
enhance printing of complex numbers to use pairs together.
---------------------------------------------------------------
where the NEWS entry (now in <R>/doc/NEWS.2 ) has been
o The printing of complex numbers has changed, handling numbers
as a whole rather than in two parts. So both real and
imaginary parts are shown to the same accuracy, with the
'digits' parameter referring to the accuracy of the larger
component, and both components are shown in fixed or
scientific notation (unless one is entirely zero when it is
always shown in fixed notation).
There was a bug in the implementation on which Robin Hankin
https://stat.ethz.ch/pipermail/r-devel/2006-September/042792.html
where Brian Ripley mentioned the (above, not the buggy one)
behavior as "by design", and Robin then also found and mentioned
the above entry. ... and notably the above regression test was
added after the bug fix also in Sept. 2006.
Apart from that NEWS entry (now basically only visible in the
one sentence in the ?signif help page
"Complex numbers are rounded to retain the specified number
of digits in the larger of the components"
And where this does make (some) sense for
signif(<complex>, digits),
it seems wrong to me (and you) for format() and print() and
I think we should reconsider using it when format()ing complex
numbers.
Note that your proposed change does change the z_prec_r()
function which in fact *is* used for signif(<complex>), and so
your change also does affect signif(<complex>) which I think is
more than I'd want here.
For instance, it would change the following (regression test in
tests/complex.R )
> signif(1.678932e80+0i, 5)
[1] 1.678932e+80+0i
>
instead of really showing 5 digits for the real part as it does
now :
[1] 1.6789e+80+0i
---------------------
I'm proposing to reconsider and change such that
1) print() {and auto-print} and format() of complex numbers should
treat Re() and Im() separately, and identically to how double()
are print()ed / format()ted in such case, notably, using
'digits' to both parts "separately".
2) We can and probably should keep signif(<complex>)'s behavior
as current, as that has been documented to behave as it does
for almost 13 years.
Martin
``` | 1,515 | 5,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.774617 |
http://mathhelpforum.com/trigonometry/44144-solve-after-sum-product-print.html | 1,526,925,884,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00402.warc.gz | 186,957,338 | 3,327 | solve after sum to product
• Jul 20th 2008, 02:36 PM
dashreeve
solve after sum to product
need values after using sum to product formula.
I used sum to product and get 2sin4xcosx=0
setting each factor to 0, 2sin4x=0 and cosx=0,
I have the answers in the book, but need to know how to get to the answers. x=0, pi/4, pi/2 etc. (adding [(npi)/4] each time.
• Jul 20th 2008, 03:10 PM
Soroban
Hello, dashreeve!
Are you really having trouble solving elementary trig equations?
Quote:
Solve: . $\displaystyle 2\sin4x\:=\:0\:\text{ and }\;\cos x\:=\:0$
$\displaystyle 2\sin4x \:=\:0 \quad\Rightarrow\quad \sin4x \:=\:0 \quad\Rightarrow\quad 4x \:=\:0 + \pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{4}n \;\;{\color{blue}[1]}$
$\displaystyle \cos x \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{\pi}{2} + \pi n \;\;{\color{blue}[2]}$
But all the solutions of [2] are included in [1].
Therefore, the solution is: .$\displaystyle x \:=\:\frac{\pi}{4}n\;\;\;\text{ for }n \in I$
• Jul 20th 2008, 03:21 PM
dashreeve
thanks for the encouraging remarks! I'm having a lot of trouble understanding this stuff actually. And half way through the course now, and haven't seen the technique you use there (which is much easier than what I've been doing). Did I miss the rule about sin4x=0 can move to 4x=0+pi(n)??? When was I supposed to learn that? I know my professor has us jumping around the book, but I can't imagine I would forget that little rule..
• Jul 21st 2008, 08:19 AM
Soroban
Hello, dashreeve!
Suppose we have: .$\displaystyle 2\sin 3x \:=\:1$
Then we have: .$\displaystyle \sin3x \:=\:\frac{1}{2}$
We know that: .$\displaystyle \sin\frac{\pi}{6} \:=\:\frac{1}{2}\:\text{ and }\:\sin\frac{5\pi}{6} \:=\:\frac{1}{2}$
So the angle is either $\displaystyle \frac{\pi}{6}\,\text{ or }\,\frac{5\pi}{6}$ . . . plus some multiple of $\displaystyle 2\pi.$
So we have: .$\displaystyle 3x \;=\;\begin{Bmatrix}\dfrac{\pi}{6} + 2\pi n \\ \\[-3mm]\dfrac{5\pi}{6} + 2\pi n \end{Bmatrix}$
Therefore: . $\displaystyle x \;=\;\begin{Bmatrix}\dfrac{\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{5\pi}{18} + \dfrac{2\pi}{3}n \end{Bmatrix}$
• Jul 23rd 2008, 08:25 AM
dashreeve
I see. Thanks for your help on several questions. It's really teaching me a lot. | 813 | 2,243 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-22 | latest | en | 0.789053 |
https://math.stackexchange.com/questions/432419/definiteness-of-omega | 1,579,402,872,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00164.warc.gz | 562,427,077 | 31,192 | # Definiteness of omega
A homework(ish) problem from models of set theory:
Define $\varphi(x) :\leftrightarrow Lim(x) \land \forall y\in x \, (Lim(y)\rightarrow y=0)$ where $Lim(x)$ means that $x$ is a limit ordinal. $\varphi$ says that $x=\omega$. I can prove that $\varphi(x)$ is a definite formula : For any transitive model $M\subseteq V$ of $ZF$ and $x\in M$, $\varphi(x)$ is true iff the relativation $\varphi(x)^M$ is true. (because all quantifiers used in the definition of $\varphi(x)$ are bounded)
Now, being asked to prove that the term $\omega= \bigcap \{ x \, |\, Ind(x) \,\}$ is definite (where $Ind(x)\leftrightarrow 0\in x\land \forall y\in x : y\cup \{y\}\in x$), I want to argue as follows.
ZF proves $\exists x \varphi(x)$ and $\forall x (\varphi (x) \leftrightarrow x=\omega)$. So for a model $M$ of ZF, the relativations $\exists x\in M \varphi(x)$ and $\forall x\in M (\varphi(x)\leftrightarrow x=\omega^M)$ should be true. The first relativation implies $\omega\in M$ and the second implies $\omega^M=\omega$.
I have seen a proof that $\omega^M=\omega$ (for transitive models M of ZF) using a more direct argument, but is this also correct?
• Are you talking about transitive models? – Asaf Karagila Jun 29 '13 at 17:45
• Yes, M is supposed to be transitive. Will edit.. – user35359 Jun 29 '13 at 17:57
• Nitpick: Since you wrote $\forall y\in x(\operatorname{Lim}(y)\to y=0)$ instead of $\forall y\in x(\neg\operatorname{Lim}(y))$, it appears that you really intended $\operatorname{Lim}(x)$ to mean that $x$ is not a successor ordinal. – Brian M. Scott Jun 29 '13 at 20:32
• @Brian: I met (on this site) several people who were given definitions that $0$ is a limit ordinal. Not everyone work with the definition that $0$ is the unique non-successor and non-limit ordinal. – Asaf Karagila Jun 29 '13 at 21:11
• @Asaf: They may be stuck using those definitions, but as far as I’m concerned, calling $0$ a limit ordinal is seriously abusing the term limit. At the very least they should be aware that for many of us $0$ is not a limit ordinal. (Mind you, there’s certainly nothing wrong in defining the predicate $mathrm{Lim}$ to include $0$.) – Brian M. Scott Jun 29 '13 at 21:14
Your proof is essentially right. Someone might object that, where you wrote "So for a model $M$ of ZF, the relativations $\dots$ should be true," you should write the actual relativizations, which involve $\varphi^M$, and then invoke the first paragraph of your question to replace $\varphi^M$ with $\varphi$. But this is just a question of how much detail to include.
As long as people are pointing out in the comments that it is customary not to call $0$ a limit ordinal, I might as well also point out the usual terminology "relativizations" (instead of "relativations") and "absolute" (instead of "definite"). | 807 | 2,824 | {"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} | 3.34375 | 3 | CC-MAIN-2020-05 | latest | en | 0.865683 |
http://mathhelpforum.com/algebra/179134-solving-equations-involving-rational-functions-linears.html | 1,481,294,671,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542712.12/warc/CC-MAIN-20161202170902-00480-ip-10-31-129-80.ec2.internal.warc.gz | 172,950,627 | 11,167 | # Thread: Solving equations involving rational functions of linears.
1. ## Solving equations involving rational functions of linears.
i have 4 problems that my dad sent me. /=fraction ^=exponent
1. 2y/y-3=-1/y
so far i got
2y^2=-y+3
2. x/2=3x+8/x
i got
x^2=6x+16
then i dont know what to do next?
3. x-2/x+6= x/2
so far
2x-4=x^2+6
4. y+2/-y=2/y+3
I DONT KNOW WHAT TO DO FOR THIS ONE?!
My questions on proportions.
1. is it possible for a portion to have two different answers for the same variable?
2. Is it true you cant add percents unless same?
Someone help
2. Originally Posted by adit17
i have 4 problems that my dad sent me. /=fraction ^=exponent
1. 2y/y-3=-1/y
so far i got
2y^2=-y+3
What's your dad got to do with this; your math teacher?
And why call there "proportions"?
To start, your equation needs a set of brackets:
2y / (y - 3) = -1 / y
Your 2y^2 = -y + 3 is correct; go 1 step further:
2y^2 + y - 3 = 0
Now solve that for y, by factoring or the quadratic formula.
3. Originally Posted by adit17
i have 4 problems that my dad sent me. /=fraction ^=exponent
1. 2y/y-3=-1/y
so far i got
2y^2=-y+3
2. x/2=3x+8/x
i got
x^2=6x+16
then i dont know what to do next?
3. x-2/x+6= x/2
so far
2x-4=x^2+6
4. y+2/-y=2/y+3
I DONT KNOW WHAT TO DO FOR THIS ONE?!
My questions on proportions.
1. is it possible for a portion to have two different answers for the same variable?
2. Is it true you cant add percents unless same?
Someone help
4. 1. is it possible for a portion to have two different answers for the same variable?
As wilmer said your terminology seems a bit unusual. The equations you posted all involve fractions (portions?), but can be re-arranged into quadratic equations. Quadratic equations can have up to 2 solutions.
for example: x= 1/x has two solutions: x=1, x=-1. You could have got this by rearranging to get $x^2 = 1$
4. y+2/-y=2/y+3
to avoid ambiguity, I assume this is
for this one, try multiplying both sides by (y+3)(y), then expand the brackets on both sides to get a standard quadratic equation | 669 | 2,049 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-50 | longest | en | 0.890876 |
https://kresevski-citrin.com/qa/quick-answer-what-keeps-the-earth-rotating.html | 1,628,091,732,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00139.warc.gz | 356,740,302 | 8,440 | Quick Answer: What Keeps The Earth Rotating?
Is a day actually 24 hours?
Day Length On Earth, a solar day is around 24 hours.
However, Earth’s orbit is elliptical, meaning it’s not a perfect circle.
That means some solar days on Earth are a few minutes longer than 24 hours and some are a few minutes shorter.
On Earth, a sidereal day is almost exactly 23 hours and 56 minutes..
How long does it take for Earth to complete one rotation?
23 hours, 56 minutes and 4.0916 secondsIt only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once its axis.
Does the earth rotate every 24 hours?
Consider the movement of the earth’s surface with respect to the planet’s center. The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers.
Why don’t we feel the earth rotating?
As Earth spins, its surface moves very fast, but so do all the things that are held to its surface by gravity—including you! When you move at the same smooth and steady speed, you don’t usually feel movement. Think of what it’s like when you’re in a moving car.
Why don’t we fly off the earth?
Normally, humans aren’t thrown off the moving Earth because gravity is holding us down. However, because we are rotating with the Earth, a ‘centrifugal force’ pushes us outwards from the centre of the planet. If this centrifugal force were bigger than the force of gravity, then we would be thrown into space.
What would happen if the Earth stopped spinning for 42 seconds?
Assuming that the earth stops suddenly for 42 seconds and then starts spinning again at its normal speed, here’s what would happen: 1. If the earth stops spinning suddenly, the atmosphere will continue to spin. … The winds will also cause erosion to the earth’s crust.
Why does moon not rotate?
A changing orbit. Gravity from Earth pulls on the closest tidal bulge, trying to keep it aligned. This creates tidal friction that slows the moon’s rotation. Over time, the rotation was slowed enough that the moon’s orbit and rotation matched, and the same face became tidally locked, forever pointed toward Earth.
How does the Earth rotate and revolve?
As the Earth rotates, each area of its surface gets a turn to face and be warmed by the sun. … Objects rotate around an axis, but revolve around other objects. So the Earth rotates around its axis as it revolves around the sun. It takes the Earth 365 days, or one year, to complete a revolution.
Why don t the planets fall into the sun?
The reason why planets don’t fall into the sun is that they have a balance between the centripetal acceleration () (falling), due to gravity, and the angular acceleration (), which is due to the angular momentum and the centripetal force. … The sun’s gravity is ~30 times stronger than Earth’s gravity.
Will the earth ever run out of oxygen?
So the short answer is, yes, the earth will be depleted of oxygen. Sic transit gloria mundi. Most of the breathable oxygen in Earth’s atmosphere is supplied by plant life in a process called photosynthesis . We’ll run out of it if we cut down too much of the world’s forests and kill too much plant life in the oceans.
Does Three Gorges Dam slow Earth Rotation?
When the Three Gorges Dam was built, 39 trillion kilograms of water from the Yangtze River built up behind it to 175 meters above sea level. NASA has calculated that the dam only slows the rotation by 0.06 microseconds, which is six hundredths of a millionth of a second. …
What keeps the Earth rotating around the sun?
Just as the Moon orbits the Earth because of the pull of Earth’s gravity, the Earth orbits the Sun because of the pull of the Sun’s gravity. … But the gravity of the Sun alters its course, causing it to travel around the Sun, in a shape very near to a circle.
Will the earth stop rotating?
The Earth will never stop rotating. Earth rotates in the purest, most perfect vacuum in the whole universe—empty space. Space is so empty, so devoid of anything to slow the Earth down, that it just spins and spins, practically without friction.
What Year Will Earth die?
Four billion years from now, the increase in the Earth’s surface temperature will cause a runaway greenhouse effect, heating the surface enough to melt it. By that point, all life on the Earth will be extinct. | 964 | 4,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.927872 |
https://quant.stackexchange.com/questions/30977/what-price-is-used-when-you-exercise-an-option-mid-trading-day | 1,656,905,324,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00102.warc.gz | 514,837,782 | 66,455 | # What price is used when you exercise an option mid-trading-day?
Question says it all, if I exercise an option mid day, is the closing price used to fill the order? In other words, exercise does not lock the price, correct?
• why the down votes? Nov 12, 2016 at 22:07
• When you exercise an American call you pay the amount K and you receive the stock. That's it, that's all there is to Exercising. If you wish you can (simultaneously or later, or never, it is up to you ) sell the stock and receive $S_t$ where $t$ is the time you decide to sell. Then you will have made $S_t-K$. Nov 12, 2016 at 23:56
• your answer assumes the granularity of my question is days. the granularity of my question is minutes. if a stock price is at $10 at 10:45 am, and i exercise my option at 10:45, and at 4:30 the price is at$10.50 and at midnight, during after hours trading, the stock price is at $9.50, which price is used to execute the contract? the key part of my question "mid-trading-day." Nov 13, 2016 at 13:36 • I don't see why @Alex C's answer depends on daily intervals. When you exercise an American equity option you pay$K$and get physical delivery of the stock. The current stock price is irrelevant. Phrased differently - no matter when you exercise during the day (as long as it is before the cutoff time), you trigger the same transactions. Nov 13, 2016 at 14:04 • In the USA, individual stock options are American and exercised via delivery of the stock (with no consideration of its price at some time t). CBOE index options are exercised via cash settlement (no delivery) with reference to the official exchange closing index price BUT THESE OPTIONS ARE EUROPEAN EXERCISE, the exercise takes place at the moment of expiration. Nov 13, 2016 at 15:04 ## 2 Answers It sounds like you want to know the specifics of assignment once an equity option is exercised, and this is really a 'check with your broker' question. Your broker will have a cut-off time to exercise a contact. If you exercise during that day the security will settle on your account after the normal stock settlement period - i.e. as if you had purchased the security on market, something like T+2 (not to be confused with options settlement which is generally T+1). So, it doesn't matter when on a day you exercise, the security will still settle on your account on the same morning. Most brokers will allow you to sell some or all of the assigned stock before the settlement date and lock in a return (e.g. at the same time as you exercise the call option, you sell an equal quantity of the underlying security on-market). This will generally mean you don't need to provide funding on the settlement date. You can reverse the above logic for a put option. Your question is about a situation that never occurs. In the USA, individual stock options are American and exercised via delivery of the stock (with no consideration of its price at some time t). When you exercise an American call you pay the amount K and you receive the stock. That's it, that's all there is to Exercising. If you wish you can (simultaneously or later, or never, it is up to you) sell the stock and receive$S_t$where t is the time you decide to sell, say 10:45am on the 12th of December. Then you will have made$S_t-K$. Cash settled options are a different kettle of fish. CBOE index options are exercised via cash settlement (no delivery) with reference to the official exchange closing index price BUT THESE OPTIONS ARE EUROPEAN EXERCISE, the exercise takes place at the moment of expiration, according to the formula$(S_t-K)^+$using the same official settlement price$S_t\$ for everyone.
In summary then, it is a design issue. If the exchange wants to create an American option they must provide some tradeable instrument (stock, future) which will be physically delivered. If they wish to create a European option they can have physical delivery or cash settlement with respect to an officially determined closing (or sometimes opening) price. There are no other ways to do things AFAIK.
• I realize my question is poorly worded, thanks for your patience. Suppose I sell a put. Someone who owns a put, clicks "exercise" at 10:30am. What happens next, at what time? Does someone get the stock in their account at 10:31? At midnight? What I'm really trying to understand is what is overnight and time risk with a variety of combo positions such as shorting the underlying and buying an OTM put. I have control over when the I open or close the underlying, but it's not clear what kind of time control I have over the exercise of the option. Nov 13, 2016 at 20:56 | 1,082 | 4,616 | {"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} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.95802 |
https://www.physicsforums.com/threads/physics-lab-question-please-help.228813/ | 1,631,949,921,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00498.warc.gz | 948,225,520 | 15,124 | For our physics lab we hung a mass attached to spring above a motion detector. First we took some data with the spring at rest. Using this data, we determined our zo value.
Then we set the spring in motion and took data. Using position, velocity, and time data, we calculated K, U, & K+U = mechanical energy (E)
We were given this info to plot the data on a graph
x= -.5(z-zo)^2 y=.5Mv^2
y=kx+E Plotting this on a graph, we determined the value of k & E
U= .5(z-zo)^2
K= .5mv^2
Now the question: Could this lab be done by hanging the spring mass system from a force probe instead of a distance probe ? How would you do the analysis?
I think this could be done, but I am having trouble visualizing how. Every idea I come up with needs the distance probe also. For example, I thought about how Force= -dU/ds and had this idea: Pull down on the mass-spring system to a measured distance and record the reading of the force probe there. From this you can determine the potential energy and thus the mechanical energy, since there is no KE initially. Then to determine k, hang the mass spring system from a force probe & record the force reading. Then measure the distance the spring stretches from its eq. position using F=-kx you can determine k. But all of these ideas seem to require distance. I have thought long and hard about this but I just can seem to find the right idea. | 335 | 1,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.933038 |
http://www.physicsforums.com/showthread.php?p=2157648 | 1,409,701,287,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535923940.4/warc/CC-MAIN-20140909030338-00021-ip-10-180-136-8.ec2.internal.warc.gz | 1,284,335,508 | 12,540 | # A quick question about gear ratios and how to find them
by Ilyo
Tags: gear, ratios
P: 25 I was not sure where to put this since it is both enginering and math so feel free to move it but to the point I have a few small engines I am going to use on a project and I need to convert the rpm to mph though to do that I need to know the Gear ratio and do not know what that is without tearing the outer caseing of the motors off which is what I would like to avoid if possible All I have is this is it possible to configure the gear ratio from it? And if not is their a way to do so? RPM-180 3.6V DC motor
P: 4,777 you cant convert from RPM to MPH, they are apples and oranges.
PF Gold P: 8,964 The easiest way to check the gear ratio is just to turn the input shaft an even number of rotations and count how many times the output shaft turns.
P: 25
A quick question about gear ratios and how to find them
Quote by Cyrus you cant convert from RPM to MPH, they are apples and oranges.
Though you most likley know more than I do I came across a calculator that does just that
http://www.bgsoflex.com/rpmmph.html
Thank you Danger I shall try that
P: 4,777
Quote by Ilyo Though you most likley know more than I do I came across a calculator that does just that http://www.bgsoflex.com/rpmmph.html Thank you Danger I shall try that
This calculator is not converting RPM to MPH, its doing several other things as well. RPM and MPH cannot be converted.
P: 25 So the site won't help me get an idea of what the mph will be of a 180 rpm motor? And if not is there any way of figureing this out...I could really use this information so I don't need to buy like five motors
P: 4,777 Why don't you explain what you are trying to do. Right now, your question makes no sense. MPH is translational speed, RPM is rotational speed. You can't convert one into another. Apples and Oranges.
Mentor P: 22,305 Cyrus, what he is trying to do requires only the rpm, gear ratio, and tire size (and number of minutes in an hour).
P: 25 Okay then well I know the RPM (180) I know the tire size (3inchs) and I know the amount of mins in a hour (60) though I do not know the gear ratio which is my only real delemia and I am wondering if there is any way of figureing it out without takeing the motor apart to the point where I can see the gears...Since it is encased in a metal cylinder and I highly doubt my skill at putting it back together So mainly I want to know if there is any way to figure it out without takeing apart the motor...Any ideas?
P: 4 The questions is not really clear. You can compute mph from rpm, if you know the distance per revolution. Check any auto with a tachometer and a speedometer. The tire diameter is part of the overall gearing. Calculate the circumference of the tire, and the vehicle will move that distance with each revolution. When you know how many times the tire turns in a minute, or hour, you can calculate the miles per hour. Assume a car in high gear, ie, a 1 to 1 transmission ratio, with a rear end ratio of 4 to 1. The tire will turn over once for every four turns of the engine, whether minutes, seconds, etc. Calculate the circumference of the tire, say 7 feet. At 4 engine rpm, the vehicle will move 7 feet per minute. QED There are formulas for this and you can substitute the unknown for the known quantities, or work backwards from distance over time to calculate the gear ratios. If your motor is turning something direct off the shaft, calculate the circumference of the object being turned times revolutions per minute, second, hour, etc. The calculator referred to in the above post is accurate for what it does. Danger has the easiest solution to calculate gear ratios, simply count the turns of the driving and the driven shaft to figure the gear ratio. I am still not clear on what you want to measure. Pat
P: 25 Okay I get what your saying pat and thank you for that. Seeing that many of you are confused on what this is about and what the question is I simpily wanted to know a way to figure the gear ratio and I belive I can figure it to my knowledge you need the gear ratio wheel diamater and the rpm to get an idea of what the mph would be so that is it in a nut shell and all I really wanted to know was that though I do have another question that goes with this. Does the voltage and wattage of the motor effect the speed or no? (I think it does but I don't know how releveant it is)
P: 4 Ilyo: I am still not sure what you are trying to figure out, but without taking the drive gears apart, Danger's recommendation to count the turns of the driving and driven shaft is the easiest way to figure gear ratio. If your 180 rpm motor drives something at 90 rpm your gear ratio is 2-1. If it drives it at 360 rpm it is 1-2. Diameter and circumference only come into play to determine mph, or distance over time. Regarding voltage and wattage for an electric motor: A motor will turn a certain amount of RPMs depending on its design, meaning how it is wired, for a given voltage. Double the voltage and you will double the RPM, disregarding losses for friction, resistance, etc. (Some motors are designed to accept 110v or 220v, but the connections are different.) Wattage is a different calculation and not is relevant to speed except in an indirect way. It is the product of volts times amps and is a measure of the power used. (Ohms law. Your 100 watt light bulb at 110 volt draws a touch less than one amp. Calculator at http://ourworld.compuserve.com/homep...en/ohmslaw.htm.) For your motor, double the volts and amps will be half, and the wattage will remain the same. To repeat, this depends on the original design of the motor and I am assuming you are not changing the wiring or hookups. (Resistance also gets involved, but, again, I am assuming you are not changing the motor.) Be careful about increasing or decreasing the voltage because the motor may not be mechanically or electrically designed to handle the extra loads imposed. There are also issues of direct and alternating current involved in motors. Don't substitute one for the other. I am really curious about what you want to accomplish. PatC
P: 145 To the original question: if you have the motor's RPM (call it RPMm for RPM motor) and the radius of the tire (R): RPMm * 60min/HR = Rotations per Hour of motor (call it RPHm) You know the speed of the car (in MPH) is related to the rotational speed of the tire (call it RPHt) by: MPH = RPHt * (2*pi*R) This makes sense because for 1 rotation of the tire, the car will move the distance 2*pi*R. Note: this will not be true for a real car, in which the tires undergo slip to move the car, however Im assuming the slip is zero for purposes of simplicity. So now you need the relation between the RPHm and RPHt. This is a direct proportion, i.e.: [B]RPHm = G * RPHt[B] where G is the Gear ratio (in a real car it accounts for the cahnge in angular velocity due to teh torque converter, the transmission, and any other velocity changing items). So, combining all these, you can solve for G by: G = (60 * RPMm) / (MPH/(2*pi*R) ) where G is the Gear ration, RPMm is the RPM of the motor, MPH is the speed if the vehicle, R is the radius of the tire. In response to your second question: in an electric motor changing the supplied voltage will change the rate of acceleration of the motor and the top speed of the motor. The higher the voltage applied to the motor the faster the maximum speed of the motor will be (the Back EMF or counter emf of the motor, which is one thing that impedes a motor's acceleration, increases as the motor spins faster. This back EMF can be larger if the supplied voltage is larger). It will also accelerate faster if you supply a higher voltage since there will be more current traveling through the wire and this, indirectly there is more power delivered to the motor.
P: 25 Thank you both for that and now I understand how to get the gear ratio and how the voltage and wattage account for the speed, I will be doing these things tomarow. Also I have one more question if not to much trouble... How does weight change the speed i'm thinking of between 120-210 lbs would be the weight for most users of my contraption (also one thing I did not understand was the concept of slip could you please elaberate)
P: 4 Ilyo: Absent any other consideration, rotating weight will decrease speed for a given amount of power, or at least delay acceleration to near the speed you want as a practical matter, assuming your application has enough power to attain the desired speed. (The practical application may be sufficient, as opposed to the theoretical loss. I gather this is a practical application to your "contraption.") Slip will result in loss of speed and power (it turns into heat).
P: 145 The top speed should not depend on the mass/weight of the car (im assumong your talking about an automobile here). The weight will affect the acceleration of the car; if it weights more it will accelerate faster but the top speed depends on teh aerodynamic properties of the car and shouldnt be affected by the mass. Longitudinal slip (again im talking about rubber automobile tires) is defined by the "slip ratio", a ratio of how fast the car moves to how "fast" the wheel turns. it is the ratio: [R * (angular velocity of wheel) - (velocity of car)] / (R * (angular velocity of wheel)) during accelerating and [R * (angular velocity of wheel) - (velocity of car)] / (velocity of car) during braking. Now if the car's velocity equalled (wheel's angular velocity) * R, which is a common assumption in physics problems, then this ratio would be one. However in the real world car's tires actually rotate "faster" than the car moves, meaning that where the tire touches the ground, the tire actually is moving in relation to the ground. another way to look at this is an extreme case; a burnout. When someone floors a car and the tires spin like crazy (but the car remains fairly stationary) the slip ratio is extremely large. The same thing happens to all cars at normal speeds and acceleration, just on a smaller scale. A common slip ratio is .1, that is: for a wheel with radius R, for the car to move 2*pi*R distance along the road the tire must rotate 1.1 times (intuitively it would be one time, but because of slip it is 1.1) Slip ratios are necessary for the tire to generate force, and the output force of a tire is a function of the slip ratios. What exactly are you measuring with this? the information im using is more for the automotive industry, although applicable to other vehicles it may be negligible depending on what you are doing.
P: 4 A "heavy" auto on a level surface propelled by an engine, motor, wind, etc, will accelerate slower than an "light" auto propelled by the same force and all other factors being equal.
Sci Advisor PF Gold P: 1,475 see race car physics Dec29-08, 11:31 AM post by ranger mike
Related Discussions Mechanical Engineering 5 Engineering, Comp Sci, & Technology Homework 3 Mechanical Engineering 1 General Physics 2 Introductory Physics Homework 4 | 2,590 | 11,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | latest | en | 0.967687 |
https://www.reddit.com/user/Noddan | 1,495,696,235,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608004.38/warc/CC-MAIN-20170525063740-20170525083740-00203.warc.gz | 927,880,360 | 23,232 | [–][S] 3 points4 points (0 children)
The fill lists are for replacement managers if anyone drops out. If you're not listed then you're in division 4 unless enough spots open up in div 3.
[–][S] 11 points12 points (0 children)
I was bored and didn't feel like studying so I went ahead and put these together. I followed these relegation rules as best I could.
[–] 4 points5 points (0 children)
Earlier this season I was looking into the luck ratings and I believe this is how it's calculated:
If you win your match up in a week, you get a positive luck score equal to the proportion of teams that would have beat you (out of 13 other teams). ie. If you have the 3rd highest PF and win your matchup, your luck score is 2/13, or ~15%.
If you lose, you get a negative luck score equal to the proportion of teams that you would have beat. ie. If you have the 3rd highest PF but lose your matchup, your luck score is -11/13, or ~-85%.
Drop the % and you've got your luck scores.
FF counts any game with a luck score greater than 50 as a lucky win, and any game with a luck score less than -50 as an unlucky loss. This is how you get the luck games stat on the team schedule.
[–] 2 points3 points (0 children)
Really unimpressed with the quality of instruction. I felt like he introduced concepts at an introductory level and proceeded to test them at a much higher level without providing a means for transition. On top of that, the assignments were not well lined up with the concepts taught by the course and varied significantly in difficulty.
Also, I just feel like it needs to be pointed out that the class average on the final was 35 out of 120, not 35%. If that isn't a sign of both poor instruction and assessment, I don't know what is.
[–] 1 point2 points (0 children)
I remember there was talk earlier about promoting regular season as well as playoff winners as up posed to playoff winners and runners up. Is that something we decided against?
[–] 2 points3 points (0 children)
The other trick is goalies never play every game. Only the best even hit 65 games. It's not uncommon for skaters to play all 82 games.
[–] 2 points3 points (0 children)
Do they know something we don't?
[–] 0 points1 point (0 children)
[–] 1 point2 points (0 children)
There's also little room for him on the depth chart with Eberle, Versteeg, Kassian, and possibly Drai if they keep Caggiula at centre. Yakupov's not the kind of player you want sitting in the press box, but he's not going to clear waivers, so I can understand the urgency.
[–] 1 point2 points (0 children)
To be fair, the 5th was a pre-emptive spot start for the STL back to back to start the season.
[–] 1 point2 points (0 children)
The Nailing Yaks are back once again.
[–] 1 point2 points (0 children)
Per season. No real money, everyone is given \$100 to start and you've gotta make it last.
[–] 1 point2 points (0 children)
I'm all for staying in the past too. Just wanted to add a quick thought on +/-:
Since +/- is (almost always) a net 0 stat, adjusting it's value wouldn't change the average points per skater. It would only redistribute some points from players on good teams to players on worse teams. Whether that's valuable or not is up for debate, but it's worth noting.
[–] 2 points3 points (0 children)
From SPRX in the Requests Thread:
"FAAB is that pickups process every day (before games). You can bid an amount of a \$100 budget on players, including \$0, and the player goes to the highest bidder. This should eliminate the advantage to being constantly online, because the player won't go to the first person to login after waivers clear. Also helps create competition when a "good" player gets dropped. Just food for though."
[–] 2 points3 points (0 children)
I'm not sure whether that would be too much or not but I can offer some insight into a few of the classes.
• CMPUT 101 is probably a waste of time. I did 174 with no programming experience and had very little difficulty keeping up. If you have any inclination of doing more than one CMPUT course I would definitely take 174 instead. To be fair, I haven't done 101 so it might be better to hear from someone who has.
• PHYS 124 is essentially just physics 20 again, and only the last chapter or two was new for me. I only had 8 labs over 12 weeks, but they required write ups which were due in 24 hours and took me a while.
• I didn't end up taking it, but I believe the seminars for CHEM 101 are help sessions and therefore attendance isn't mantadory.
Hope that helps a little.
[–] 31 points32 points (0 children)
11/11/11 if I remember correctly.
[–] 0 points1 point (0 children)
If I remember correctly those not taking the course for credit (and thus not paying the tuition for it) were asked to pay something along the lines of \$100 as a participation fee.
For the audition, you sign up for a 10 or so minute time slot and bring something to play that showcases your talents (concert band has no specific requirements, the wind ensemble does I'm pretty sure). I think they also give you a short sight reading passage to read through. It wasn't too bad.
[–] 1 point2 points (0 children)
I can't comment on the Wind Ensemble, but I know for sure you can participate in Concert Band without being enrolled for credits. The marks for Concert Band are something like 80% attendance and participation in classes and concerts, and we had 2 short (1 page) reflection papers for marks as well.
[–] 1 point2 points (0 children)
Individual Game Tiebreakers:
1. Most total starter Points
2. Highest individual starter score
3. Higher bench points total
At least that's what it is in Gretzky so I'm assuming it's default.
[–] 0 points1 point (0 children)
If you could postpone the ruining a week that'd be grand.
[–] -9 points-8 points (0 children)
None of those help against a turn 5 maly double velen mind blast.
Kelley Wentworth impressive batting average by in survivor
[–] 0 points1 point (0 children)
Does this include split votes? ie. Does it count as a "hit" if someone was in on the plan but voted for the person not going home in case of an idol?
[–] 10 points11 points (0 children)
He still has to deal with Jeremy, Tasha, and Spencer all of whom are very capable of winning challenges.
[–] 0 points1 point (0 children)
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# Measuring SLAs of defects (bugs) in a structure
Dear all
I like to share an use case w/ you I like to create in JIRA structure. I like to manage my defects (bugs) during hypercare reg. SLAs and use JIRA structure.
Use case:
The SLA of a bug w/ several priorities = "blocker" have SLAs (like 4 hours for a priority = "blocker"). I like to measure the SLA using creation date & time (create) and today's date and time (NOW() or TODAY()).
The formula should measure if SLA > than defined SLA per priority and post an appropriate text like "critical; roadblock > 4h". If bug is still within SLA then returning the text "within SLA" as a result.
I created the following formula in a column.
IF(priority ="blocker" && ((Today() - created) > 4); "critical: roadblock > 4h";
priority ="critical" && ((Today() - created) > 48); "critical: Delay for > 2 days";
priority ="major" && ((Today() - created) > 96); "SLA of 4 days exceeded";
priority ="minor" OR priority ="trivial" AND ((Today() - created) >
120);"SLA of 5 days exceeded";
"within SLA")
The formula is able to map the correct text per priority but the formula does not take into consideration if SLA is not exceeded (past time < SLA).
I have the impression that the formula does not know the unit of measurement (=hour) of SLA.
How would a solution look like?
Thanks, Marc
1 vote
David Niro
Community Leaders are connectors, ambassadors, and mentors. On the online community, they serve as thought leaders, product experts, and moderators.
Jan 02, 2023
Hello @Marc Dahinden ,
You are correct, the formula is not working due to formatting. The way now() and created are stored in Jira are different than what we see in the UI. Therefore we need to use the HOURS_BETWEEN() function to calculate the duration.
Below is a formula that should work for you, based on your example. I made a couple of changes, saving now() as a variable, as well as the duration calculation itself. These types of calculations can be heavy, so I wanted to make sure we are only doing them once for each issue.
`with _now = now():with _timesince = HOURS_BETWEEN(created,_now):IF priority = "blocker" AND _timesince >4: "critical: roadblock >4h"ELSE IF priority = "critical" AND _timesince >48: "critical: roadblock > 2days"ELSE IF priority = "major" AND _timesince >96: "SLA of 4 days exceeded"ELSE IF (priority = "minor" or priority = "trivial") AND _timesince >120: "SLA of 5 days exceeded"ELSE: "within SLA"`
Please give it a try and let me know if it works for you!
Best,
David
Thanks David.
Sorry for my basic understanding of formulas.
May I request for a small additional requirement? I hope this is fine.
I like to use colors:
Red bgColor and white text for priority ="blocker" or "critical"
What would be the correct formula for this?
Best, Marc
David Niro
Community Leaders are connectors, ambassadors, and mentors. On the online community, they serve as thought leaders, product experts, and moderators.
Jan 03, 2023
Hello @Marc Dahinden ,
You are very welcome and no apology is necessary! I'm happy to help shed light on some of the nuances.
I'm glad to hear that it is working as expected for you. I modified the formula to include this. You will notice the new local variable (function actually) format_caption(). It defines the color of the background and text. I then used it with the return value for blocker and critical priorities.
`with _now = now():with _timesince = HOURS_BETWEEN(created,_now):with format_caption(caption)= """{panel:borderStyle=solid|borderColor=white|bgColor=red}{color:white}*\$caption*{color}{panel}""":IF priority = "blocker" AND _timesince >4: format_caption("critical: roadblock >4h")ELSE IF priority = "critical" AND _timesince >48: format_caption("critical: roadblock > 2days")ELSE IF priority = "major" AND _timesince >96: "SLA of 4 days exceeded"ELSE IF (priority = "minor" or priority = "trivial") AND _timesince >120: "SLA of 5 days exceeded"ELSE: "within SLA"`
Please let me know if it helps!
Best,
David
Hi David
I get as result in the column:
It seems the code is interpretated as text.
Any idea why?
Best, Marc
David Niro
Community Leaders are connectors, ambassadors, and mentors. On the online community, they serve as thought leaders, product experts, and moderators.
Jan 03, 2023
You are very welcome! I have a very good idea why. It was something I thought to communicate in my previous response, but failed to.
In the Formula window, you will see "Options". Most likely, the drop down is set to "General" at the moment. Changing this to "Wiki Markup" should format the background and text color.
Sorry for neglecting to mention it.
Best,
David
Dave Rosenlund likes this
Thanks, it works
# people like this | 1,399 | 5,542 | {"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} | 2.59375 | 3 | CC-MAIN-2023-50 | longest | en | 0.89813 |
http://mathhelpforum.com/advanced-algebra/217775-lin-alg-proofs-help-print.html | 1,500,651,009,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423785.29/warc/CC-MAIN-20170721142410-20170721162410-00426.warc.gz | 193,606,108 | 3,271 | # Lin alg proofs help
• Apr 19th 2013, 12:21 PM
Tra003
Lin alg proofs help
Hi everyone!
I'm having trouble with these two linear algebra proofs, and help would be greatly appreciated.
1. let V and W be real vector spaces, and let L: V to W be a linear transformation such that Ker(L)=0v and Im(L)=W; let (v1 ,v2 ,v3) be a basis of V; determine whether or not (L(v1) L(v2) L(v3)) is a basis for W
I know how to prove they are lin independent, but how do i prove that they're both lin independent and a generating set?
2. Let A∈Mn(R) be a real n x n matrix (where n is an integer greater than or equal to 2), and assume λ1, λ2 ∈ R are two eigenvalues of A with λ1does not equal λ2 (i.e λ1, λ2 are two distinct eigenvalues of A). let W1 denote the eigenspace of A corresponding with eigenvalue λ1, and let W2 denote the same with λ2 .
Show that we have W1 ∩ W2 = {0Rn}
I know that the W1 is simply the kernel of the matrix (I
λ1-A), but don't know how to get past that.
• Apr 19th 2013, 10:41 PM
chiro
Re: Lin alg proofs help
Hey Tra003.
For a basis, you need N linearly independent vectors where dim(V) = N. Once you have this, you definitely have a basis for any vector space.
For the second one you should show that the two spaces are independent. If you are talking about two vectors then showing they are independent (i.e. not scalar multiples) of each other should suffice and you can do this by setting up a two row matrix and row-reducing it.
• Apr 20th 2013, 05:27 PM
johng
Re: Lin alg proofs help
Hi,
I think even with "easy" problems, it helps to see complete proofs:
Attachment 28053 | 464 | 1,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-30 | longest | en | 0.933442 |
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# One of Krishna Menon's many dreams for his dear homeland was
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One of Krishna Menon's many dreams for his dear homeland was [#permalink]
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12 Jun 2010, 00:29
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46. One of Krishna Menon's many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove all traces of man ever setting foot there.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
B. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and removing
C. many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing
D. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
E. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
OA is
[Reveal] Spoiler:
c
Although OA is c, i feel A is better than C because of more idiomatic to convert to remove...Any comments.
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12 Jun 2010, 11:28
Who the hell is Krishna Menon. LOL
hey gmatcracker2010
I dont know about the history. But I can tell C is best.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and [to] remove ----> to is missing.
A fails to maintain infinitive parallelism. Infinitive parallelism: to …to
You can argue that second "to" is optional. But choice C is sure shot.
Cheers
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12 Jun 2010, 14:27
I narrowed it down to A and C. I'll go with A for sure...
No idea how C is the answer?
looks like option D and E are same
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12 Jun 2010, 14:35
This was an awful question....all the choices seem correct accept the 'dearly' ones.
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16 Jun 2010, 13:56
I think A is better choice than C because the use of gerund in any option has the least possibility for the option to be correct.
In A, [to] remove, the usage is correct as the infinitive usage can be optional when it precedes a verb.
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18 Jun 2010, 13:47
C passes in parallelism. But the use of gerund is a big "?"
Definitely would have gone for A! Let experts have their say on this.
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23 Jun 2010, 08:24
C isn't correct.
Ridiculous. Changes the meaning totally.
One of Menon's many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing
How can the dreams convert something? That is what it means. As if the dreams are converting the islands.
What is the source??? Plz
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23 Jun 2010, 10:26
was sure tht I will be right and picked A ...much to my amazement its C..
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23 Jun 2010, 12:12
IMO C
46. One of Krishna Menon's many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove all traces of man ever setting foot there.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove - incorrect - missing parallelism it should be "to remove" and not only "remove"
B. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and removing
C. many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing - converting and removing - so maintains parallelism, so correct
D. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
E. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
Hope the explanation helps.
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29 Jun 2010, 06:12
amitdesai16 wrote:
IMO C
46. One of Krishna Menon's many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove all traces of man ever setting foot there.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove - incorrect - missing parallelism it should be "to remove" and not only "remove"
B. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and removing
C. many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing - converting and removing - so maintains parallelism, so correct
D. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
E. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
Hope the explanation helps.
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Hi,
This is weird question, first of all D & E are same.
secondly, i don;t agree with OA as C because
whenever there is a tie b/w infinitive (TO verb Form) and ING forms (perhaps called present participle ) then Infinitive is given preference.
So A carry more waightage over C .
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29 Jun 2010, 21:38
onedayill wrote:
amitdesai16 wrote:
IMO C
46. One of Krishna Menon's many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove all traces of man ever setting foot there.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove - incorrect - missing parallelism it should be "to remove" and not only "remove"
B. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and removing
C. many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing - converting and removing - so maintains parallelism, so correct
D. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
E. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
Hope the explanation helps.
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Hi,
This is weird question, first of all D & E are same.
secondly, i don;t agree with OA as C because
whenever there is a tie b/w infinitive (TO verb Form) and ING forms (perhaps called present participle ) then Infinitive is given preference.
So A carry more waightage over C .
I agree with onedayill. Answer has to be A
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30 Jun 2010, 05:47
IMO - C
Either the answer is Option A or C.
Option A: As per my understanding from MGMAT SC, 'to remove' instead of 'remove' is not required: If there are more than one infinitive verbs, then the one after 'and' does not require the 'to' and can be a bare Infinitive.
What does not go in favor of option A is the use of Infinitive itself. As per my understanding, for an infinitive to be used, the sentence structure should be:
- S + Subjunctive Verb + that + Bare Infinitive OR S + Subjunctive Verb + O + Infinitive.
However in the above sentence, there is no subjunctive verb at all. If seen carefully:
One of Krishna Menon's many dreams for his dear homeland[Subject] was[Verb and NOT Subjunctive Verb] to convert.... [Object]. For this reason I believe that the Infinitive itself should not be used and preference for the ING Forms.
Please correct me if I am wrong !!!
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30 Jun 2010, 06:48
Now I need expert's opinion here.
Can some one please explain why A is wrong?
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30 Jun 2010, 10:09
i don't agree with most of the observations here about A.
The only reason why A is wrong is NOT because the sentence is missing the "to" in "to remove," but because the use of the infinitive "to remove" signals (or stresses) an intention. It's not the intention that Krishna that is worried about, it's the [end] result -- (yeah yeah, i know, end result is redundant)!
So when you want to focus on the results, you need to use the -ing (participle form).
Hope that helps.
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30 Jun 2010, 19:18
I went with C also on the basis of parallelism. I felt that answer choice A needed the "to remove" making the sentence correct. where did you find this krishna menon questions? LOL
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30 Jun 2010, 22:14
Go with A...C has correct parallelism but does not carry corret intention.
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01 Jul 2010, 23:01
Everyone's confused...bloody Krishna Menon had such a shitty dream.
IMO: A
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02 Jul 2010, 01:14
gmatcracker2010 wrote:
46. One of Krishna Menon's many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove all traces of man ever setting foot there.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
B. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and removing
C. many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing
D. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
E. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
OA is
[Reveal] Spoiler:
c
Although OA is c, i feel A is better than C because of more idiomatic to convert to remove...Any comments.
Hello Gmatcracker,
Please what is the source and Official Explanation to this question?
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02 Jul 2010, 08:09
gmatcracker2010 wrote:
46. One of Krishna Menon's many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove all traces of man ever setting foot there.
A. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
B. many dreams for his dear homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and removing
C. many dreams for his dear homeland was converting one of Minicoy's islands into a pre-historic nature sanctuary and removing
D. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
E. many dreams for his dearly homeland was to convert one of Minicoy's islands into a pre-historic nature sanctuary and remove
OA is
[Reveal] Spoiler:
c
Although OA is c, i feel A is better than C because of more idiomatic to convert to remove...Any comments.
Here you are talking about dream, so one of krishna dreams was what ? to convert NO but converting x and ...
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31 Jul 2010, 05:35
gauravnagpal wrote:
was sure tht I will be right and picked A ...much to my amazement its C..
Ditto here .........I missed parallel Q....
Re: krishna monon's dreams [#permalink] 31 Jul 2010, 05:35
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# One of Krishna Menon's many dreams for his dear homeland was
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# Overlapping sets
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31 Oct 2017, 11:25
In a Bank 30% of the people speak spanish and 12% of the people who speak French also know Spanish.45 people know both Spanish and French.How many people are there in the bank?
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In a Bank 30% of the people speak spanish and 12% of the people who speak French also know Spanish.45 people know both Spanish and French.How many people are there in the bank?
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Polygons ICSE Class-6th Concise Selina Mathematics Solutions Chapter-28. We provide step by step Solutions of Exercise / lesson-28 Polygons for ICSE Class-6 Concise Selina Mathematics.
Our Solutions contain all type Questions of Exe-28 A and Exe-28 B with Notes on Polygons to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6.
## Polygons ICSE Class-6th Concise Selina Mathematics Solutions Chapter-28
–: Select Topics :–
Exe-28 A,
Exe-28 B,
#### What are Polygons?
• A Polygon is a closed figure made up of lines segments (not curves) in two-dimensions.
• A minimum of three line segments are required for making a closed figure, thus a polygon with a minimum of three sides is known as Triangle.
### Type of Polygons
#### Regular Polygon
If all the sides and interior angles of the polygon are equal, then it is known as a regular polygon.
#### Irregular Polygon
If all the sides and the interior angles of the polygon are of different measure, then it is known as an irregular polygon.
#### Convex Polygon
If all the interior angles of a polygon are strictly less than 180 degrees, then it is known as a convex polygon. The vertex will point outwards from the centre of the shape.
#### Concave Polygon
If one or more interior angles of a polygon are more than 180 degrees, then it is known as a concave polygon. A concave polygon can have at least four sides. The vertex points towards inside of the polygon.
### Exercise – 28 A Polygons ICSE Class-6th Concise Mathematics Selina Solutions
#### Question -1.
State, which of the following are polygons :
Only figure (ii) and (iii) are polygons.
#### Question- 2.
Find the sum of interior angles of a polygon with :
(i) 9 sides
(ii) 13 sides
(iii) 16 sides
(i) 9 sides
No. of sides n = 9
∴Sum of interior angles of polygon = (2n – 4) x 90°
= (2 x 9 – 4) x 90°
= 14 x 90°= 1260°
(ii) 13 sides
No. of sides n = 13
∴ Sum of interior angles of polygon = (2n – 4) x 90° = (2 x 13 – 4) x 90° = 1980°
(iii) 16 sides
No. of sides n = 16
∴ Sum of interior angles of polygon = (2n – 4) x 90°
= (2 x 16 – 4) x 90°
= (32 – 4) x 90° = 28 x 90°
= 2520
#### Question -3.
Find the number of sides of a polygon, if the sum of its interior angles is :
(i) 1440°
(ii) 1620°
(i) 1440°
1440°
Let no. of. sides = n
∴ Sum of interior angles of polygon = 1440°
∴ (2n – 4) × 90° = 1440°
⇒ 2n – 4 = 144090
⇒ 2(n – 2) = 144090
⇒ n – 2 = 144090×2
⇒ n – 2 = 8
⇒ n = 8 + 2
⇒ n = 10
(ii) 1620°
Let no. of sides = n
∴ Sum of angles of polygon = 1620°
∴ (2n – a) × 90° = 1620°
⇒ 2(n – 2) = 162090
⇒ n – 2 = 162090×2
⇒ n – 2 = 9
⇒ n = 9 + 2
⇒ n = 11
#### Question- 4.
Is it possible to have a polygon, whose sum of interior angles is 1030°.
Let no. of. sides be = n
Sum of interior angles of polygon = 1030°
∴ (2n – 4) × 90° = 1030°
⇒ 2(n – 2) = 103090
⇒ (n – 2) = 103090×2
⇒ (n – 2) = 10318
⇒ n = 10318+2
⇒ n = 13918
Which is not a whole number. Hence it is not possible to have a polygon, the sum of whose interior angles is 1030°.
#### Question -5.
(i) If all the angles of a hexagon arc equal, find the measure of each angle.
(ii) If all the angles of an octagon are equal, find the measure of each angle,
(i) If all the angles of a hexagon arc equal, find the measure of each angle.
No. of sides of hexagon, n = 6
Let each angle be = x°
Sum of angles = 6x°
(n – 2) x 180° = Sum of angles
(6 – 2) x 180° = 6x°
4 x 180 = 6x
x = 4×1806
x = 120°
∴ Each angle of hexagon = 120°
(ii) If all the angles of an octagon are equal, find the measure of each angle,
No. of. sides of octagon n = 8
Let each angle be = x°
∴ Sum of angles = 8x°
∴ (2n – 4) × 90° = Sum of angles
(2 × 8 – 4) × 90° = 8x°
12 × 90° = 8x°
⇒ x° =90 x 128
⇒ x° = 135°
∴ Each angle of octagon = 135°
#### Question -6.
One angle of a quadrilateral is 90° and all other angles are equal ; find each equal angle.
Let the angles of a quadrilateral be x°,
x°, x° and 90°
∴ Sum of interior angles of quadrilateral = 360°
⇒ x° + x° + x° + 90° = 360°
⇒ 3x° = 360° – 90°
⇒ x = 2703
⇒ x = 90°
#### Question -7.
If angles of quadrilateral are in the ratio 4 : 5 : 3 : 6 ; find each angle of the quadrilateral.
Let the angles of the quadrilateral be 4x, 5x, 3x and 6x.
∴ 4x + 5x + 3x + 6x = 360°
18x = 360°
x = 36018=20°
∴ First angle = 4x = 4 × 20° = 180°
Second angle = 5x = 5 × 20° = 100°
Third angle = 3x = 3 × 20° = 60°
Fourth angle = 6x = 6 × 20° = 120°
#### Question -8.
If one angle of a pentagon is 120° and each of the remaining four angles is x°, find the magnitude of x.
One angle of a pentagon = 120°
Let remaining four angles be x, x, x and x
Their sum = 4x + 120°
But sum of all the interior angles of a pentagon = (2n – 4) x 90°
= (2 x 5 – 4) x 90° = 540°
= 3 x 180° = 540°
∴ 4x+120o° = 540°
4x = 540° – 120°
4x = 420
x = 4204 ⇒ x = 105°
∴Equal angles are 105° (Each)
#### Question- 9.
The angles of a pentagon are in the ratio 5 : 4 : 5 : 7 : 6 ; find each angle of the pentagon.
Let the angles of the pentagon be 5x, 4x, 5x, 7x, 6x
Their sum = 5x + 4x + 5x + 7x + 6x = 27x
Sum of interior angles of a polygon
= (2n – 4) × 90°
= (2 × 5 – 4) × 90° = 540°
∴ 27x = 540 ⇒ 54027
⇒ x = 20°
∴ Angles are 5 × 20° = 100°
4 × 20° = 80°
5 × 20° = 100°
7 × 20° = 140°
6 × 20° = 120°
#### Question- 10.
Two angles of a hexagon are 90° and 110°. If the remaining four angles arc equal, find each equal angle.
Two angles of a hexagon are 90°, 110°
Let remaining four angles be x, x, x and x
Their sum = 4x + 200°
But sum of all the interior angles of a hexagon
= (2n – 4) × 90°
= (2 × 6 – 4) × 90° = 8 × 90° = 720°
∴ 4x + 200° = 720°
⇒ 4x = 720° – 200° = 520°
⇒ x = 5204=130°
∴ Equal angles are 130° (each)
### Polygons Exe-28 B for ICSE Class-6th Concise Selina Mathematics Solved Questions
#### Question -1.
Fill in the blanks :
In case of regular polygon, with
Number of sides Each exterior angle Each interior angle (i) 6 60° 120° (ii) 8 45° 135° (iii) 10 36° 144° (iv) 18 20° 160° (v) 8 45° 135° (vi) 24 15° 165°
(i) Each exterior angle = 3606=60∘
Each interior angle = 180° – 60° = 120°
(ii) Each exterior angle = 3608 = 45°
Each interior angle = 180° – 45° = 135°
(iii) Since each exterior angles = 36°
∴ Number of sides = 36036=10
Also, interior angle = 180°- 20° = 160°
(iv) Since each exterior angles = 20°
∴ Number of sides = 36020=18
Also, interior angle = 180° – 20° = 160°
(v) Since interior angles = 135°
∴ exterior angle = 180° – 135°
∴ Number of sides = 36045°=8
(vi) Since interior angle = 165°
∴ exterior angle = 180° – 165° = 15°
∴ Number of sides = …36015°..= 24…..
Question -2.
Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 150°
(i) 160°
Let no.of.sides of regular polygon be n.
Each interior angle = 160°
n-2n×180∘=160∘
180n – 360° = 160n
180n – 160n = 360°
20n = 360°
n = 18
(ii) 150°
Let no.of.sides of regular polygon be n.
Each interior angle = 150°
=150∘
180n – 360° = 150n
180n – 150n = 360°
30n = 360°
n = 12
#### Question -3.
Find number of sides in a regular polygon, if its each exterior angle is :
(i) 30°
(ii) 36°
(i) 30°
Let number of sides = n
360n =30∘
n = 36030
n = 12
(ii) 36°
Let no. of. sides = n
360=36°
n = 36036
n = 10
#### Question -4.
Is it possible to have a regular polygon whose each interior angle is :
(i) 135°
(ii) 155°
(i) 135°
No. of. sides = n
Each interior angle = 135°
=135∘
180n – 360° = 135n
180n – 135n = 360°
n = 36045
n = 8
Which is a whole number.
Hence, it is possible to have a regular polygon whose interior angle is 135°.
(ii) 155°
No. of. sides = n
Each interior angle = 155°
=155∘
180n – 360° = 155n
180n – 155n = 360°
25n = 360°
n = 36025
n = 725
Which is not a whole number.
Hence, it is not possible to have a regular polygon whose interior angle is 155°
#### Question -5.
Is it possible to have a regular polygon whose each exterior angle is :
(i) 100°
(ii) 36°
(i) 100°
Let no. of. sides = n
Each exterior angle = 100°
= 360n∘=100∘
∴ n = 360100
n = 185
Which is not a whole number.
Hence, it is not possible to have a regular polygon whose each exterior angle is 100°.
(ii) 36°
et no. of. sides = n
Each exterior angle = 36°
= 360n∘=36∘
∴ n = 36036
n = 10
Which is a whole number.
Hence, it is not possible to have a regular polygon whose each exterior angle is 36°.
#### Question- 6.
The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of this polygon.
(ii) number of sides in the polygon.
(i) each exterior angle of this polygon.
Interior angle : exterior angle = 2 : 1
Let interior angle = 2x° & exterior angle = x°
∴ 2x° + x° = 180°
3x = 180°
x = 60°
∴ Each exterior angle = 60°
(ii) number of sides in the polygon.
Let no.of. sides = n
360n =60°
n = 36060
n = 6
–: End of Polygons ICSE Class-6th Concise Solutions :–
Thanks | 3,379 | 9,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-30 | latest | en | 0.868417 |
https://bitbucket.org/larsyencken/euler/commits/5d051a8305ba | 1,436,309,731,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375102712.76/warc/CC-MAIN-20150627031822-00006-ip-10-179-60-89.ec2.internal.warc.gz | 998,251,097 | 16,778 | # Commits
committed 5d051a8
Julia 06-07.
• Participants
• Parent commits 5b62fbd
# File julia/06-sum-squares.jl
`+#!/usr/bin/env julia`
`+#`
`+# 06-sum-square-difference.jl`
`+# euler`
`+#`
`+# Find the difference between the sum of squares and the square of the sum`
`+# of the first hundred natural numbers.`
`+#`
`+`
`+println(sum(1:100)^2 - sum([i^2 for i = 1:100]))`
# File julia/07-prime.jl
`+#!/usr/bin/env julia`
`+#`
`+# 07-prime.jl`
`+# euler`
`+#`
`+# What is the 10001st prime number?`
`+#`
`+`
`+function primes(n)`
`+ p = Int64[2, 3, 5, 7]`
`+ next = 8`
`+ while length(p) < n`
`+ isprime = true`
`+ for f in p`
`+ if next % f == 0`
`+ isprime = false`
`+ break`
`+ end`
`+ end`
`+ if isprime`
`+ push!(p, next)`
`+ end`
`+ next += 1`
`+ end`
`+`
`+ p`
`+end`
`+`
`+println(primes(10001)[end])` | 353 | 942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.484212 |
http://math.stackexchange.com/questions/232464/how-many-tries-to-get-at-least-k-successes/232532 | 1,419,085,842,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769894.131/warc/CC-MAIN-20141217075249-00103-ip-10-231-17-201.ec2.internal.warc.gz | 177,149,195 | 19,462 | # How many tries to get at least k successes?
The probability $P'$ of getting at least $k$ successes in $n$ independent tries, given probability of a single success $s$, equals one minus the summed probabilities of getting only $0$ to $k-1$ successes:
$P'(k, n, s) = 1 - \sum\limits_{i=0}^{k-1} P(i, n, s)$
where the probability $P$ of getting exactly $k$ successes is:
$P(k, n, s) = \,_nC_k \cdot s^k \cdot (1 - s)^{n - k}$
Now suppose I want to know how many tries I need to achieve a given probability $P'$. How do I solve for $n$?
This question is a lot like the question here:
On the Total Number of Tries Required to have $n$ Successes
But in that question each trial is not independent, since it's about selecting stones from a bag without replacement. In my question, each trial is independent.
-
In general, or if you are given $i$ and $s$? – glebovg Nov 7 '12 at 22:49
You can assume constant $k$ and $s$. ($i$ is just for the sum.) – Paul A Jungwirth Nov 7 '12 at 22:50
Try doing it numerically, or by trial and error. – glebovg Nov 7 '12 at 22:54
Sure, it's easy enough to graph it, but what does that teach you? :-) – Paul A Jungwirth Nov 7 '12 at 23:03
@PaulAJungwirth Could you clarify what you mean by "Now suppose I want to know how many tries I need to achieve a given probability P′. How do I solve for n?" Do you want to know the probability that it takes $n$ trials to get $k$ succes? – Jean-Sébastien Nov 7 '12 at 23:06
There is no exact closed form solution. If $k$ is moderately large, the normal approximation to the binomial will give you an approximate answer, and you can then do a numerical search around there for the exact answer. For the normal approximation we want $$\Phi\left(\frac{k-ns}{\sqrt{ns(1-s)}}\right) \geq .9,$$ where $\Phi$ is the normal cdf. You can take the inverse cdf, simplify this to a quadratic in $\sqrt n$, and find a value for $n$. Since this is not exact, you can then check other nearby values of $n$ to find the exact solution.
Here's an R function that solves it for you using this approach, though not particularly efficiently:
binomial.size <- function(k,s,p) {
# Find the smallest sample size n such that a binomial(n,s) has
# probability at least p of having k sucesses.
# our first approximation comes from solving the quadratic:
n <- ceiling(((-qnorm(p)+sqrt(qnorm(p)^2+4*k/(1-s)))/(2*sqrt(s/(1-s))))^2)
while(pbinom(k-1,n,s,lower.tail=FALSE)<p) { # our n might be too small...
n <- n+1
}
while(pbinom(k-1,n-1,s,lower.tail=FALSE)>p) { # or too large...
n <- n-1
}
return(n)
}
binomial.size(15,0.4,0.5)
## 37
pbinom(14,37,0.4,lower.tail=FALSE)
## 0.54
pbinom(14,36,0.4,lower.tail=FALSE)
## 0.48--too small!
-
Thank you! I'll need to digest the math a bit more, but the R function gives correct results for a few examples I tried. – Paul A Jungwirth Nov 8 '12 at 18:39
If I want to read more about the mathematics of binomial distributions, where is a good place to look? I've encountered them in statistics, but what would be the math course that introduces them? Probability? – Paul A Jungwirth Nov 8 '12 at 18:45
A probability class will probably use the binomial for examples, but you won't spend a lot of time on it. A combinatorics class will probably give you more exposure to its use in various types of problems. I'm not sure what exactly you're looking for, though. – Jonathan Christensen Nov 9 '12 at 2:35
This is a partial solution
For the case exactly $k$ success:
The probability that a sequence of independant Bernoulli trials takes $n$ steps before getting the $k^{th}$ success is given by (an adaptation of) the binomial negative distribution. Think of it this way:
You need your $n^{th}$ trial to be a success. You also need $k-1$ successes in the $n-1$ remaining spots. The rest are failures. Let $X$ be the number of trials it takes to get $k$ success. The probability of $X$ being $n$ is then given by $$\Pr(X=n)={n-1\choose k-1}s^{k}(1-s)^{n-k}$$ Its median seems to be somewhat complicated, see for example this.
-
Thank you for your help! I can't tell how this differs from what I wrote about finding $P$ for exactly $k$ successes, so how does it get me closer to a solution for at least $k$ successes? Perhaps you could expand on it a bit to help me understand? – Paul A Jungwirth Nov 8 '12 at 18:38 | 1,248 | 4,320 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2014-52 | longest | en | 0.909746 |
https://www.physicsforums.com/threads/on-spinor-representations-and-sl-2-c.549694/ | 1,685,897,194,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00165.warc.gz | 1,035,330,248 | 17,572 | # On spinor representations and SL(2,C)
• teddd
#### teddd
Hi guys!
I still have problem clearing once and for all my doubt on the spinor representation. Sorry, but i just cannot catch it.
1)
-----
Take a left handed spinor, $\chi_L$.
Now, i know it transforms according to the Lorentz group, but why do i have to take the $\Lambda_L$ matrices belonging to $SL(2,\mathbb C)$,
$$\chi'=\Lambda_L\chi$$??
Dimensionally it makes sense, it's like
\left(\begin{align}\chi'_{L1}\\ \chi'_{L2}\end{align}\right)=\left(\begin{align}A &B\\C&D\end{align} \right) \left(\begin{align} \chi_{L1}\\ \chi_{L2} \end{align}\right)
but why exacly $SL(2,\mathbb C)$ matrices and not every other generic 2x2 complex matrix?
-----
2)
----
Is it right to say that $\Lambda_L$ is the representation of the lorentz group which acts on spinors?
I have this doubt becaouse i read everywhere that the spinor is a $\left(0,\frac{1}{2}\right)$ representation of the lorentz group, but i'd say that the spinor os the thing on which the $\Lambda_L$ acts, and it is the $\Lambda_L$ itself to be a represetation of the group!
Last edited:
Is it right to say that ΛL is the representation of the Lorentz group which acts on spinors? I have this doubt because i read everywhere that the spinor is a (0,1/2) representation of the Lorentz group, but I'd say that the spinor is the thing on which the ΛL acts, and it is the ΛL itself to be a representation of the group!
Yes, that's correct.
Yes, that's correct.
OK, that's reassuring, thanks Bill_K, but why then i do read everywhere that the the spinor is a (0,1/2) representation of the Lorentz group?
(I'm reading the pages you suggested me vanhees71...)
There are two two-dimensional inequivalent irreducible representations of the proper orthochronous Lorentz group, labeled (1/2,0) and (0,1/2). The pair of numbers are the spin values of the two pseudo-su(2) subalgebras according to the relation between the Lie algebras $\text{sl}(2,\mathbb{C})$ and $\text{su}(2) \oplus \text{su}(2)$.
In my manuscript I describe two ways to describe the irreps of the Lorentz group, namely the formalism with dotted and undotted tensor indices (Theorem 8) and via the above mentioned Lie algebra equivalence (Theorem 10), which is more convenient for applications.
If you're going to Lie algebras, then you needn't use $\mathfrak{sl}(2,\mathbb{C})$, because of the isomorphism
$$\mathfrak{so}(1,3)^{\text{C}} \simeq \mathfrak{su}(2) \oplus \mathfrak{su}(2)$$
The Lie algebra of SU(2) should be known from angular momentum theory.
Ok, that seems to clarify most of my problem on the reason of using SL(2,C).
But I'm still stuck on the representation thing.
I just cannot solve that.
For example, the group SO(3) can be represented by 3x3 orthgonal matrices, which act on 3-dimensional vectors belonging to a vector space.
Well, i cannot transpose this mechanism on the lorentz (poincarè) group, becaouse everywhere i read that the spinor is a (0,1/2) (or (1/2,0) ) representation of the group; while i understand it as the equivalent of the 3-dimensional vector of the example above.
Can you clarify that to me?
A linear representation of a group is the realization of that group by linear operator on a vector space.
More formally it's a homomorphism $\Phi:G \rightarrow \text{GL}(V)$, where $G$ is the group under consideration and $\text{GL}(V)$ is the group of all invertible linear operators on the vector space $V$. The group product for this is the product of the linear operators (i.e., the composition of linear mappings of the vector space to itself).
Further, a homomorphism between two groups obeys the rule
$$\Phi(g_1 g_2)=\Phi(g_1) \Phi(g_2)$$
for all $g_1,g_2 \in G$.
All these mathematical ideas can be applied also to Lie algebras, leading to the analogous notion of linear representations of Lie algebras.
everywhere i read that the spinor is a (0,1/2) (or (1/2,0) ) representation of the group;
They're being casual with the language. As indicated above, the representation is the set of matrices that correspond to the group elements. A spinor is an element of a vector space whose components transform according to the (0,1/2) representation, but it's just too cumbersome to keep saying that.
...but it's just too cumbersome to keep saying that.
So it's on an improper use of the term representation I've been struggling upon!
But another thing then:
When we compose two right (left) weyl spinor to get something that transforms as a 4-vector, namely by doing $$\left(0,\frac{1}{2}\right)\otimes\left(0,\frac{1}{2}\right)=(0,0)\oplus(0,1)$$
we are actually combining togheter the spinors themselves, but as far as I've understood the notation $\left(0,\frac{1}{2}\right)$ (as well as the others) stands for the linear operators, right? | 1,315 | 4,798 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.881431 |
https://www.socialsecurityguide.net/2022/09/30/how-much-is-my-social-security-taxed/ | 1,713,827,858,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00662.warc.gz | 887,296,448 | 86,570 | HomeFactsHow Much Is My Social Security Taxed
# How Much Is My Social Security Taxed
## How To Calculate Your Social Security Income Taxes
Is Social Security Taxable?
If your Social Security income is taxable, the amount you pay will depend on your total combined retirement income. However, you will never pay taxes on more than 85% of your Social Security income.
Again, if you file as an individual with a total income thats less than \$25,000, you wont have to pay taxes on your Social Security benefits in 2022. For the 2022 tax year , single filers with a combined income of \$25,000 to \$34,000 must pay income taxes on up to 50% of their Social Security benefits. If your combined income is more than \$34,000, you will pay taxes on up to 85% of your Social Security benefits.
For married couples filing jointly, you will pay taxes on up to 50% of your Social Security income if you have a combined income of \$32,000 to \$44,000. If you have a combined income of more than \$44,000, you can expect to pay taxes on up to 85% of your Social Security benefits.
If 50% of your benefits are subject to tax, the exact amount you include in your taxable income will be the lesser of either:
• half of your annual Social Security benefits OR
• half of the difference between your combined income and the IRS base amount
The example above is for someone whos paying taxes on 50% of their Social Security benefits. Things get more complex if youre paying taxes on 85% of your benefits. However, the IRS helps taxpayers by offering software and a worksheet to calculate Social Security tax liability.
## A Look At How It Works
Now that we know how to calculate provisional income and the various taxation thresholds, we can determine how much of a hypothetical Social Security recipient’s benefit will be taxed. Let’s say a married couple takes \$35,000 in distributions from their 401, plus they’re getting \$40,000 in Social Security benefits. They don’t receive any tax-exempt interest.
This means their provisional income would be \$35,000 + which equals \$55,000. Using the taxation chart above, we know the first \$32,000 is completely tax-free.
From there, we can see that 50% of any income over \$32,000 and under \$44,000 is going to be taxed. It’s important to note that this doesn’t mean that income in that range is taxed at 50%, but rather, half of the income in that range is subject to taxation. In other words, this couple will be paying taxes on \$6,000 of their Social Security income within that range .
Finally, they made \$11,000 over \$44,000, which means 85% or \$9,350 will be taxed.
Therefore, out of the couple’s total Social Security income of \$40,000, only \$15,350 is subject to taxation. Once again, this doesn’t mean they’ll be paying \$15,350 in taxes, but rather that this is the only portion of their benefit that will be taxed.
## How To Report Your Social Security Income
Each January, you will receive a Form Social Security 1099, SSA-1099, that shows the total benefits you received for the previous year and the total amount you are required to report to the IRS on your federal tax return.
You will report the amount in Box 5 of Form SSA-1099 and the total amount on line 6a of your Form 1040, U.S. Individual Income Tax Return or Form 1040-SR, U.S. Tax Return for Seniors. The amount considered taxable income depends on what other income you earned during the year, which you should note on line 6b of Form 1040 or Form 1040-SR.
Remember: The taxable amount of your benefits is based on your filing status and combined income. If you need to calculate the taxable amount of Social Security benefits, you can use this worksheet.
Read Also: How To Freeze My Social Security Number
Finally, it is important to remember that you do not have to do anything to receive the COLA increase. All beneficiaries will see an increase in their Social Security without applying for it. Another element to keep in mind in all of this is to remember that after the increase we can reach the reportable minimum, so pay attention to your taxes so you dont make financial mistakes.
Knowing how the Social Security Administrations payment schedule is organized, you will have no trouble getting organized with your finances. That is why the SSA sends out payment dates at the beginning of the year. In this way, all beneficiaries who have an accepted benefit can have complete control over what money will enter their bank accounts and when this will occur.
To contact Social Security, you can call the toll-free number, 1-800-772-1213, or go to your local Social Security office. Our phone lines are open from 7 a.m. to 7 p.m., Monday through Friday. You can also write to them at the following address Social Security Administration, Office of Public Inquiries, Windsor Park Building, 6401 Security Boulevard, Baltimore, MD 21235-6401.
## Social Security Increase In 2023
One of the most expert groups on the subject of the COLA increase is The Senior Citizens League . This group forecasts a 10.1% Social Security increase by 2023. This forecast is quite possible seeing the large increase in overall prices in the past few months. It is still too early to be sure of the total increase, but it is quite possible that it will be something close to that percentage.
We will not have the final figure until October. In any case, all the experts say that this increase could be negative for some Social Security recipients. This is because it could increase the taxes paid every month. With a higher benefit it is possible to reach the maximum monthly income threshold.
So that would mean we would have a Social Security increase of \$159 at most. Given that this figure is for maximum SSA benefits, most likely not everyone will be able to get it. Normal in such cases, due to the average benefits, is to raise around \$120 more per month for COLA.
As the increase is expressed as a percentage, remember that your Social Security benefit will increase depending on the monthly amount. If you collect a maximum of \$4,194 per month, you will get a higher COLA increase. This increase is not so that pensioners have more money to spend each month, but so as not to lose their purchasing power during inflation.
## Taxing Social Security Benefits Is Sound Policy
Social Security beneficiaries with higher incomes pay income tax on part of their benefits. Those with incomes below \$25,000 pay no tax on benefits, while those with the highest incomes pay tax on as much as 85 percent of their benefits. This arrangement is sound for several reasons:
• The substantial proceeds from taxing Social Security benefits are credited to the Social Security and Medicare trust funds, strengthening the programs financing.
• The taxation of benefits is broadly progressive, since people with low incomes pay nothing and the tax rate on benefits increases with income.
• As an earned benefit, Social Security should be subject to tax, like other earned benefits, such as employer pensions.
• Social Securitys tax treatment is more favorable than that of private defined-benefit pensions, primarily because of the protections for low-income beneficiaries.
## To Find Out If Their Benefits Are Taxable Taxpayers Should:
• Take one half of the Social Security money they collected during the year and add it to their other income.
Other income includes pensions, wages, interest, dividends and capital gains.
• If they are single and that total comes to more than \$25,000, then part of their Social Security benefits may be taxable.
• If they are married filing jointly, they should take half of their Social Security, plus half of their spouse’s Social Security, and add that to all their combined income. If that total is more than \$32,000, then part of their Social Security may be taxable.
Don’t Miss: Social Security Office In Miami
## How Much Of Your Social Security Is Taxable
Its possible and perfectly legal to avoid paying taxes on your Social Security check. In fact, only about 40 percent of recipients pay any federal tax on their benefit.
But heres the caveat: To receive tax-free Social Security, your annual combined, or provisional, income must be under certain thresholds:
• \$25,000, if youre filing as an individual
• \$32,000, if youre married filing jointly
For married filing separately, the Social Security Administration simply says that youll probably pay taxes on your benefits.
Your combined income consists of three parts:
• Tax-exempt interest
• 50 percent of your Social Security income
Add those amounts up and if youre under the threshold for your filing status, you wont be paying federal taxes on your benefit.
Even if youre above this threshold, however, you may not have to pay tax on your full benefit. You may pay taxes on only 50 percent of your benefit or on up to 85 percent of it, depending on your combined income.
• For individual filers:
• Combined income between \$25,000 and \$34,000, up to 50 percent of your benefit is taxable
• Combined income above \$34,000, up to 85 percent of your benefit is taxable
• For married filing jointly:
• Combined income between \$32,000 and \$44,000, up to 50 percent of your benefit is taxable
• Combined income above \$44,000, up to 85 percent of your benefit is taxable
• ## Planning Around The Taxability Of Social Security Benefits
How to pay less in taxes on your Social Security benefits
Planning strategies should be done based on marginal tax rates, which means the leaps in marginal tax rates from including Social Security benefits can and should be a material factor in planning – especially since the rates have the greatest impact on those whose income is relatively modest and may not realize they are exposed to 27.75% marginal tax rates when they “thought” they were in just the 15% or 25% tax brackets.
For many clients, though, the rates are at least partially unavoidable. In many situations, there simply is not enough income flexibility to spread income out to stay below the thresholds. Although notably, for some clients, the best thing to do is to actually accelerate income and lump it together after all, additional income beyond the point that the maximum 85% of Social Security benefits are taxable is subject to only a 15% tax bracket, which is far better than leaving the income until next year when it may be taxed at 27.75% due to the phase-in of Social Security benefits. In fact, in some cases it might even be worthwhile to trigger a bit of additional Social Security benefits taxation just to reach the cap and then add more income beyond it at a current tax bracket!
#### Join over 51,881 fellow advisors now
Read Also: Brockton Ma Social Security Office
## How Much Of My Social Security Benefits Will Be Taxed
Q. I have been collecting Social Security since turning 66 four years ago. This past year in 2021, I collected \$30,671.40. I also have a pension from that was \$23,216.04. I am still working as well as a consultant and made just over \$107,000. I also had to take out \$14,000 from one of my IRAs and I paid \$2,800 in federal taxes. With that said, how much of my Social Security will be taxed?
Still working
But it is different for federal tax purposes.
If your total gross income including otherwise tax-exempt municipal bond interest exceeds \$25,000, you begin to pay federal tax on your Social Security benefits on a sliding scale that eventually reaches a maximum of 85%, said Neil Becourtney, a certified public accountant and tax partner with CohnReznick in Holmdel.
A formula is used where one calculates their provisional income which is defined as the sum of 50% of Social Security benefits, plus tax-exempt interest, plus other items included in your adjusted gross income, he said.
For single taxpayers, Social Security benefits escape federal taxation if provisional income is less than \$25,000. For provisional income between \$25,000 and \$34,000, up to 50% of Social Security benefits are taxable. If provisional income exceeds \$34,000, up to 85% of benefits will be taxable, Becourtney said.
Note to readers: if you purchase something through one of our affiliate links we may earn a commission.
## How The Math Works
The math works like this:
• If your wages were less than \$137,700 in 2020, multiply your earnings by 6.2% to arrive at the amount you and your employer must each pay for a total of 12.4%. If you were self-employed, multiply your earnings up to this limit by 12.4% to calculate the Social Security portion of your self-employment tax.
• If your wages were more than \$137,700 in 2020, multiply \$137,700 by 6.2% to arrive at the amount you and your employer must each pay. Anything you earned over this threshold is exempt from Social Security tax. You would do the same but multiply by 12.4% if you’re self-employed.
For taxes due in 2021, refer to the Social Security income maximum of \$137,700 as you’re filing for the 2020 tax year.
Also Check: Social Security Offices In Colorado
## Are Social Security Benefits Taxable
Up to 50% or even 85% of your Social security benefits are taxable if your provisional or total income, as defined by tax law, is above a certain base amount. Your Social Security income may not be taxable at all if your total income is below the base amount.
If youre married and filing jointly with your spouse, your combined incomes and social security benefits are used to figure your total income.
## What Home Improvements Are Tax Deductible 2021
Until December 31, 2021, you may claim a tax credit for energy-efficient modifications to your house, such as energy-efficient windows, doors, skylights, roofs, and insulation,Washington adds. Air-source heat pumps, central air conditioning, hot water heaters, and circulating fans are among the other changes.
Read Also: Social Security Office In Victoria Texas
## How Much Social Security Is Taxable
If you are planning to retire, its important to take into account all costs you will incur including how much Social Security is taxable.
Taxes from Social Security are a surprise to many retirees who are unaware of the factors that determine whether your monthly benefit will be taxable.
Fortunately, it is not hard to understand how much Social Security is taxable.
## How We Make Money
The offers that appear on this site are from companies that compensate us. This compensation may impact how and where products appear on this site, including, for example, the order in which they may appear within the listing categories. But this compensation does not influence the information we publish, or the reviews that you see on this site. We do not include the universe of companies or financial offers that may be available to you.
Also Check: Social Security Ticket To Work Program
## How To Minimize Taxes On Your Social Security
If your Social Security benefit is relatively fixed, albeit with small annual increases, you really have only two avenues left to get into that tax-free zone: reducing tax-exempt interest or adjusted gross income. And since most people dont have tax-exempt interest, youre left with one option.
Therefore, the secret is to reduce your adjusted gross income in order to prevent provisional income from triggering a tax on Social Security, says Kelly Crane, president and chief investment officer at Napa Valley Wealth Management in St. Helena, California.
Here are a few ways to reduce your adjusted gross income to get into the tax-free zone:
## How Much Of Your Social Security Income Is Taxable
How much your Social Security benefits will be if you make \$30,000, \$35,000 or \$40,000
Social Security payments have been subject to taxation above certain income limits since 1983. No inflation adjustments have been made to those limits since then, so most people who receive Social Security benefits and have other sources of income pay some taxes on the benefits.
However, regardless of income, no taxpayer has all their Social Security benefits taxed. The top level is 85% of the total benefit. Heres how the Internal Revenue Service calculates how much is taxable:
• The calculation begins with your adjusted gross income from Social Security and all other sources. That may include wages, self-employed earnings, interest, dividends, required minimum distributions from qualified retirement accounts, and other taxable income.
• Tax-exempt interest is then added.
• If that total exceeds the minimum taxable levels, then at least half of your Social Security benefits will be considered taxable income. You must then take the standard or itemize deductions to arrive at your net income. The amount you owe depends on precisely where that number lands in the federal income tax tables.
Combined Income = Adjusted Gross Income + Nontaxable Interest + Half of Your Social Security Benefits
The key to reducing taxes on your Social Security benefit is to reduce the amount of taxable income you have when you retire, but not to reduce your total income.
You May Like: Social Security Office Opelousas La
## How To Plan For Social Security Taxes
A balanced approach to distribution is the best way to plan for Social Security taxes, says Freitag. Keep in mind that too much emphasis on one type of distribution or another is not the way to go.
It is better to have a mix of income streams in retirement, he says. As an example, if all your income is taxable, then adding Social Security just makes it worse across the board.
## Just Started Collecting Social Security Here’s How To Know Whether You’ll Owe Taxes On It
Roughly 1 in every 2 older adults will pay federal income taxes on a portion of their Social Security benefits for the 2020 tax year.
To be sure, this usually happens only if you have other substantial income in addition to your Social Security benefits, such as wages, self-employment, interest, dividends and other taxable income that must be reported on your tax return, according to Uncle Sam.
Also Check: Greenville South Carolina Social Security Office | 3,833 | 18,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.955811 |
http://www.cram.com/flashcards/statistics-exam-2-398207 | 1,503,325,730,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108709.89/warc/CC-MAIN-20170821133645-20170821153645-00201.warc.gz | 524,139,133 | 27,785 | • Shuffle
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### 31 Cards in this Set
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paired t-test technique for determining if differences exist between two dependent-based means What does the one sample z-test and the one sample t-test determine? if our sample mean is significantly different from the population mean What type of test would be used when pairs of participants are matched on one or more characteristics and randomly placed in two different groups (e.g. experimental and control) paired t-test What type of test would be used when twin studies (or siblings) where one twin is in Group 1 and the other is in Group 2 paired t-test What is the null hypothesis for the paired t-test? u1=u2 There is no significant difference in performance within the two conditions What does the r in the paired t-test formula mean? the correlation between the scores from the 2 dependent samples. What should the magnitude of the r be for the scores to be considered dependent (i.e. related) .4 or greater What is the criteria for rejecting the null hypothesis? p value is less than or equal to an alpha of .05 When is the z-test used? when the sample data has a perfect normal distribution What does the one sample z-test and the one sample t-test determine? if the sample mean is significantly different from the population mean When could the z-test be used? when all sample data had perfect normal distributions t distribution series of approximations of the normal curve for different sample sizes independent t-test, two-sample t-test, Student's t-test comparing 2 sample means rather than a sample mean to a population mean What makes up the independent variable in a independent t-test? there is 1 IV with 2 levels: the 2 different groups that constitute you samples What are the assumptions of the independent t-test. randomness, normality, interval/ratio data, 2 independent groups, 10 or more participants per group, homogeneity of variance What independent t-test assumption do you not have to worry about if the group sizes are equal? homogeneity of variance What do you do if the group sizes are not equal in an independent t-test? test for the assumption of equal variance Levene's test used for testing for equal variance what do you do if the variances between the gruops are equal in an independent t-test? You can use a different formula for t that uses a "pooled" or average variance for the groups. This usually results in a lower error term and thus greater power. When is the effect size used? before a study to estimate sample size- based on an average effect size from the literature What does effect sizehelp estimage? the meaningfulness of your treatment What can you use to get a better picture of the magnitude of the difference between two means after the study? effect size How is the effect size most often used? to describe the difference between the mean of the experimental group and the control group omega squared another way of evaluating effectiveness of treatment When should omega squared be calculated? after a t-test if a significant difference is found What does omega squared indicate? the proportion of the variance in the dependent variable that is explained by the independent variable normal distribution scores on the dependent variable can be expressed as a standard score as a basis of comparison What is the most common standard score used with a normal distribution? z-score z-scores the standardized score used to produce a normal curve with mean=0 and s=1. It essentially represents a raw score expressed in standard deviation units. Can convert raw scores from a distribution into z-scores. What must be known before raw scores can be reported as z-scores? mean and standard deviation The equations for calculating what is used by most stats programs based on z scores? skewness | 884 | 4,274 | {"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} | 2.96875 | 3 | CC-MAIN-2017-34 | longest | en | 0.881541 |
http://www.uff.br/trianglecenters/X1319.html | 1,508,812,282,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827853.86/warc/CC-MAIN-20171024014937-20171024034937-00586.warc.gz | 569,987,888 | 3,531 | ## X(1319) (BEVAN-SCHRÖDER POINT)
Interactive Applet
You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon , select the center name from the list and, then, click on the vertices A, B and C successively.
The JRE (Java Runtime Environment) is not enabled in your browser!
This applet was built with the free and multiplatform dynamic geometry software C.a.R..
Information from Kimberling's Encyclopedia of Triangle Centers
Trilinears f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = (b + c - 2a)/ (b + c - a)
= g(A,B,C) : g(B,C,A) : g(C,A,B), where g(A,B,C) = 2 - 2 cos A - cos B - cos C (Peter J. C. Moses)
Barycentrics af(a,b,c) : bf(b,c,a) : cf(c,a,b)
Let X'Y'Z' be the pedal triangle of the Bevan point, W = X(40); then X(1319) is the point, other than W, in which the circles AWX', BWY', CWZ' concur. (Floor van Lamoen, Hyacinthos #6321, 6352).
X(1319) lies on these lines:
1,3 11,515 12,1125 37,604 44,1317 48,1108 59,518 73,1104 77,1122 106,1168 108,953 210,956 214,519 226,535 355,499 392,993 513,663 529,908 840,934 910,1055 961,1255
X(1319) is the {X(1),X(56)}-harmonic conjugate of X(65).
X(1319) = midpoint of X(1) and X(36)
X(1319) = reflection of X(1155) in X(36)
X(1319) = isogonal conjugate of X(1320)
X(1319) = inverse-in-circumcircle of X(56)
X(1319) = inverse-in-incircle of X(65)
X(1319) = cevapoint of X(902) and X(1404)
X(1319) = crosspoint of X(1) and X(104)
X(1319) = crosssum of X(1) and X(517)
X(1319) = crossdifference of any two points on line X(9)X(650)
This is a joint work of
Humberto José Bortolossi, Lis Ingrid Roque Lopes Custódio and Suely Machado Meireles Dias. | 698 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-43 | latest | en | 0.650436 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=15&t=31059&p=97912 | 1,597,374,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00188.warc.gz | 382,112,575 | 11,271 | ## Homework number 1.25a [ENDORSED]
$E=hv$
paulacamara1E
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am
### Homework number 1.25a
Why is the power of the final answer -19? I keep getting a different result every time I repeat the calculation.
Isabelle De Rego 1A
Posts: 40
Joined: Fri Apr 06, 2018 11:02 am
### Re: Homework number 1.25a [ENDORSED]
Too find the energy emitted by the photon you use the formula: E=hc/wavelength
First: convert wavelength from nanometers to meters (589*10^-9)
Then plug it into the equation: (6.626*10-34)(3*10^8)/(589*10^-9)
Energy= 3.37*10^-19
Posts: 41
Joined: Wed Nov 15, 2017 3:01 am
### Re: Homework number 1.25a
What units would the final answer be in?
princessturner1G
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am | 268 | 760 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-34 | latest | en | 0.794915 |
https://jp.mathworks.com/matlabcentral/answers/412491-sum-of-n-matrices?s_tid=prof_contriblnk | 1,579,722,954,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607407.48/warc/CC-MAIN-20200122191620-20200122220620-00111.warc.gz | 510,341,807 | 24,333 | # Sum of N matrices
11 ビュー (過去 30 日間)
Pedro Guevara 2018 年 7 月 28 日
Edited: Pedro Guevara 2018 年 7 月 30 日
The reason for this new message was to request a new favor.
Is there a code or function in matlab that allows me to add an undefined number (N) of matrices, which I already have a program? As an example I put the following situation to be understood better:
Aux = Mat (1) + Mat (2) + Mat (3) + .... + Mat (n)
I thank you for your attention and I hope you can help me with my predicament.
#### 8 件のコメント
Matt J 2018 年 7 月 28 日
The program that I already started has been pretty much taken advantage of using the eval function and I do not want to lose all that work.
Why not? That work hasn't helped you. It has hurt you, and will continue to hurt you. I'm willing to bet money that whatever you did with eval can be remade in 1/10th the time using the better practices described at Stephen's link.
Pedro Guevara 2018 年 7 月 28 日
OK, I will try to make my code implementing indexing. Could you ask for help if you need it? I appreciate your attention.
Stephen Cobeldick 2018 年 7 月 29 日
@Pedro Guevara: the approach using eval is specifically advised against by all experienced MATLAB users and by the MATLAB documentation, which has a whole page explaining why to avoid doing this. It states "A frequent use of the eval function is to create sets of variables such as A1, A2, ..., An, but this approach does not the array processing power of MATLAB and is not recommended. The preferred method is to store related data in a single array."
You have already already wasted more than one day trying to work with this inefficiently designed code, in contrast per isakson gave a much better solution in just a few seconds, which is simpler, neater, more efficient, less buggy, and easier to work with. This is one reason why you should rewrite your code: it will save you time! Do you see the pattern here? If you continue to design your code like this, then you will continue to fight your code for no real benefit. Or you could use arrays and indexing, which is what the MATLAB documentation recommends, and make developing, debugging, and running your code much faster.
サインイン to comment.
### 採用された回答
per isakson 2018 年 7 月 29 日
I've modified the example of my comment.
n = 12;
sz = [3,3];
MKG = nan( sz(1),sz(2), n );
for ix = 1 : n
MKG(:,:,ix) = ones(sz); % inv(M_Trans)*M_Kele*M_Trans;
end
sum( MKG, 3 )
outputs
ans =
12 12 12
12 12 12
12 12 12
>>
#### 1 件のコメント
Pedro Guevara 2018 年 7 月 30 日
I think I love you. XD I am trying to understand your code by making it run step by step and apparently, with some modifications, you can help me with the purpose of my work. Thank you very much.
サインイン to comment. | 733 | 2,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-05 | latest | en | 0.926378 |
http://algebra-expression.com/simplifying-expressions.html | 1,506,145,328,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00304.warc.gz | 16,132,420 | 8,903 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
# Simplifying Expressions
Overview
• Section 1.8 in the textbook
– Like Terms & Combining Like Terms
– Simplifying Expressions
Like Terms & Combining Like
Terms
Combining Like Terms
• Two terms are said to be like terms if they
satisfy BOTH of the following conditions:
– The variables must be the same
– The variables must be raised to the same
power
(numbers) and retain the variable
Ex 1: Simplify by combining the like terms:
a) 2x + 7 – 5 + x – 1
b) 5a – 2b + 2 – 6b – 3
c) mn + 4m – mn – 2n + m + 9n
d) -x2 + 4x – 3 + 2x2 – x
Simplifying Expressions
• Recall the Distributive Property:
a(b + c) = a · b + a · c
• Usually no operation next to the number
outside of the parentheses
– Implied to be multiplication
• To simplify an expression:
– Apply the Distributive Property if necessary
– Combine any like terms
Ex 2: Simplify:
a) 4(2x – 5) + 3(x + 1)
b) 8(2 – y) – (3 – y)
c) 5(2x2 – 5x + 1) – 3(x3 + x2 – 2) + 2x
Ex 3: Translate and simplify:
a) The difference of (5x – 9) and (x + 2)
b) (9y2 – 5) less (2y2 + y – 3)
c) Subtract (18 – 7z) from (z + 7)
Summary
• After studying these slides, you should
know how to do the following:
– Identify and combine like terms
– Simplify expressions | 600 | 1,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-39 | longest | en | 0.766971 |
https://socratic.org/questions/what-is-the-standard-form-of-f-x-x-1-x-3-x-5-2 | 1,653,776,732,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00501.warc.gz | 586,566,754 | 6,022 | # What is the standard form of f(x)=(x+1)(x+3)+(x+5)^2 ?
Mar 8, 2016
$f \left(x\right) = \textcolor{p u r p \le}{2 {x}^{2} + 14 x + 28}$
#### Explanation:
$f \left(x\right) = \left(x + 1\right) \left(x + 3\right) + {\left(x + 5\right)}^{2}$
Simplify.
$f \left(x\right) = \textcolor{red}{\left(x + 1\right) \left(x + 3\right)} + \textcolor{b l u e}{\left(x + 5\right) \left(x + 5\right)}$
FOIL each pair of binomials.
$f \left(x\right) = \textcolor{red}{\left(x \cdot x + 3 \cdot x + 1 \cdot x + 1 \cdot 3\right)} + \textcolor{b l u e}{\left(x \cdot x + 5 \cdot x + 5 \cdot x + 5 \cdot 5\right)}$
Simplify.
$f \left(x\right) = \textcolor{red}{{x}^{2} + 3 x + x + 3} + \textcolor{b l u e}{{x}^{2} + 5 x + 5 x + 25}$
Gather like terms.
$f \left(x\right) = \textcolor{red}{{x}^{2}} + \textcolor{b l u e}{{x}^{2}} + \textcolor{red}{3 x} + \textcolor{red}{x} + \textcolor{b l u e}{5 x} + \textcolor{b l u e}{5 x} + \textcolor{red}{3} + \textcolor{b l u e}{25}$
Simplify.
$f \left(x\right) = \textcolor{p u r p \le}{2 {x}^{2} + 14 x + 28}$ | 472 | 1,046 | {"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} | 4.59375 | 5 | CC-MAIN-2022-21 | latest | en | 0.431377 |
https://www.oreilly.com/library/view/interactive-data-visualization/9781449340223/ch11.html | 1,695,869,009,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00693.warc.gz | 1,010,047,360 | 26,367 | ## Chapter 11. Layouts
Contrary to what the name implies, D3 layouts do not, in fact, lay anything out for you on the screen. The layout methods have no direct visual output. Rather, D3 layouts take data that you provide and remap or otherwise transform it, thereby generating new data that is more convenient for a specific visual task. It’s still up to you to take that new data and generate visuals from it.
Here is a complete list of all D3 layouts:
• Bundle
• Chord
• Cluster
• Force
• Histogram
• Pack
• Partition
• Pie
• Stack
• Tree
• Treemap
In this chapter, I introduce three of the most common: pie, stack, and force. Each layout performs a different function, and each has its own idiosyncrasies.
If you are curious about any of the other D3 layouts, check out the many examples on the D3 website, and be sure to reference the official API documentation on layouts.
## Pie Layout
`d3.layout.pie()` might not be as delicious as it sounds, but it’s still worthy of your attention. Obviously, its typical use is for creating pie charts, like the example in Figure 11-1.
Feel free to open up the sample code for this in 01_pie.html and poke around.
To draw those pretty wedges, we need to know a number of measurements, including an inner and outer radius for each wedge, plus the starting and ending angles. The purpose of the pie layout is to take your data and calculate all those messy angles for you, sparing you from ever having to think about radians.
Remember radians? In case you don’t, here’s a quick refresher. Just as there are 360° in a circle, there are 2π radians. So π radians equals 180°, or half a circle. Most people find it easier to think in terms of degrees; computers prefer radians.
For this pie chart, let’s start, as usual, with a very simple dataset:
``var` `dataset` `=` `[` `5``,` `10``,` `20``,` `45``,` `6``,` `25` `];``
We can define a default pie layout very simply as:
``var` `pie` `=` `d3``.``layout``.``pie``();``
Then, all that remains is to hand off our data to the new `pie()` function, as in `pie(dataset)`. Compare the datasets before and after in Figure 11-2.
The pie layout takes our simple array of numbers and generates an array of objects, one object for each value. Each of those objects now has a few new values—most important, `startAngle` and `endAngle`. Wow, that was easy!
Now, to actually draw the wedges, we turn to `d3.svg.arc()`, a handy built-in function for drawing arcs as SVG `path` elements. We haven’t talked about `path`s yet, but they are SVG’s answer to drawing irregular forms. Anything that’s not a `rect`, `circle`, or another basic shape can be drawn as a `path`. The catch is, the syntax for defining `path` values is not particularly human-friendly. For example, here’s the code for the big, red wedge in Figure 11-1:
````<path` `fill=``"#d62728"` `d=``"M9.184850993605149e-15,-150A150,150 0 0,1`
` 83.99621792063931,124.27644738657631L0,0Z"``></path>````
If you can understand that, then you don’t need this book.
The bottom line is, it’s best to let functions like `d3.svg.arc()` handle generating `path`s programatically. You don’t want to try writing this stuff out by hand.
Arcs are defined as custom functions, and they require inner and outer radius values:
````var` `w` `=` `300``;`
`var` `h` `=` `300``;`
`var` `outerRadius` `=` `w` `/` `2``;`
`var` `innerRadius` `=` `0``;`
`var` `arc` `=` `d3``.``svg``.``arc``()`
`.``innerRadius``(``innerRadius``)`
`.``outerRadius``(``outerRadius``);````
Here I’m setting the size of the whole chart to be 300 by 300 square. Then I’m setting the `outerRadius` to half of that, or 150. The `innerRadius` is zero. We’ll revisit `innerRadius` in a moment.
We’re ready to draw some wedges! First, we create the SVG element, per usual:
````//Create SVG element`
`var` `svg` `=` `d3``.``select``(``"body"``)`
`.``append``(``"svg"``)`
`.``attr``(``"width"``,` `w``)`
`.``attr``(``"height"``,` `h``);````
Then we can create new groups for each incoming wedge, binding the pie-ified data to the new elements, and translating each group into the center of the chart, so the `path`s will appear in the right place:
````//Set up groups`
`var` `arcs` `=` `svg``.``selectAll``(``"g.arc"``)`
`.``data``(``pie``(``dataset``))`
`.``enter``()`
`.``append``(``"g"``)`
`.``attr``(``"class"``,` `"arc"``)`
`.``attr``(``"transform"``,` `"translate("` `+` `outerRadius` `+` `", "` `+` `outerRadius` `+` `")"``);````
Note that we’re saving a reference to each newly created `g` in a variable called `arcs`.
Finally, within each new `g`, we append a `path`. A `path`s path description is defined in the `d` attribute. So here we call the `arc` generator, which generates the path information based on the data already bound to this group:
````//Draw arc paths`
`arcs``.``append``(``"path"``)`
`.``attr``(``"fill"``,` `function``(``d``,` `i``)` `{`
`return` `color``(``i``);`
`})`
`.``attr``(``"d"``,` `arc``);````
Oh, and you might be wondering where those colors are coming from. If you check out 01_pie.html, you’ll note this line:
``var` `color` `=` `d3``.``scale``.``category10``();``
D3 has a number of handy ways to generate categorical colors. These might not be your favorite colors, but they are quick to drop into any visualization while you’re in the prototyping stage. `d3.scale.category10()` creates an ordinal scale with an output range of 10 different colors. (See the wiki for more information on these color scales as well as perceptually calibrated color palettes, based on research by Cynthia Brewer, that are included with D3.)
Lastly, we can generate text labels for each wedge:
````arcs``.``append``(``"text"``)`
`.``attr``(``"transform"``,` `function``(``d``)` `{`
`return` `"translate("` `+` `arc``.``centroid``(``d``)` `+` `")"``;`
`})`
`.``attr``(``"text-anchor"``,` `"middle"``)`
`.``text``(``function``(``d``)` `{`
`return` `d``.``value``;`
`});````
Note that in `text()`, we reference the value with `d.value` instead of just `d`. This is because we bound the pie-ified data, so instead of referencing our original array (`d`), we have to reference the array of objects (`d.value`).
The only thing new here is `arc.centroid(d)`. WTH? A centroid is the calculated center point of any shape, whether that shape is regular (like a square) or highly irregular (like an outline of the state of Maryland). `arc.centroid()` is a super-helpful function that calculates and returns the center point of any arc. We translate each `text` label element to each arc’s centroid, and that’s how we get the text labels to float right in the middle of each wedge.
Bonus tip: Remember how `arc()` required an `innerRadius` value? We can expand that to anything greater than zero, and our pie chart becomes a ring chart like the one shown in Figure 11-3:
``var` `innerRadius` `=` `w` `/` `3``;``
Check it out in 02_ring.html.
One more thing: The pie layout automatically reordered our data values from largest to smallest. Remember, we started with `[ 5, 10, 20, 45, 6, 25 ]`, so the small value of 6 should have appeared between 45 and 25, but no—the layout sorted our values in descending order, so the chart began with 45 at the 12 o’clock position, and everything just goes clockwise from there.
## Stack Layout
`d3.layout.stack()` converts two-dimensional data into “stacked” data; it calculates a baseline value for each datum, so you can “stack” layers of data on top of one another. This can be used to generate stacked bar charts, stacked area charts, and even streamgraphs (which are just stacked area charts but without the rigid starting baseline value of zero).
For example, we’ll start with the stacked bar chart in Figure 11-4.
Let’s say you had some data like this:
````var` `dataset` `=` `[`
`{` `apples``:` `5``,` `oranges``:` `10``,` `grapes``:` `22` `},`
`{` `apples``:` `4``,` `oranges``:` `12``,` `grapes``:` `28` `},`
`{` `apples``:` `2``,` `oranges``:` `19``,` `grapes``:` `32` `},`
`{` `apples``:` `7``,` `oranges``:` `23``,` `grapes``:` `35` `},`
`{` `apples``:` `23``,` `oranges``:` `17``,` `grapes``:` `43` `}`
`];````
The first step is to rearrange that data into an array of arrays. Each array represents the data for one category (e.g., apples, oranges, or grapes). Within each category array, you’ll need an object for each data value, which itself must contain an `x` and a `y` value. The `x` in our case is just an ID number. The `y` is the actual data value:
````var` `dataset` `=` `[`
`[`
`{` `x``:` `0``,` `y``:` `5` `},`
`{` `x``:` `1``,` `y``:` `4` `},`
`{` `x``:` `2``,` `y``:` `2` `},`
`{` `x``:` `3``,` `y``:` `7` `},`
`{` `x``:` `4``,` `y``:` `23` `}`
`],`
`[`
`{` `x``:` `0``,` `y``:` `10` `},`
`{` `x``:` `1``,` `y``:` `12` `},`
`{` `x``:` `2``,` `y``:` `19` `},`
`{` `x``:` `3``,` `y``:` `23` `},`
`{` `x``:` `4``,` `y``:` `17` `}`
`],`
`[`
`{` `x``:` `0``,` `y``:` `22` `},`
`{` `x``:` `1``,` `y``:` `28` `},`
`{` `x``:` `2``,` `y``:` `32` `},`
`{` `x``:` `3``,` `y``:` `35` `},`
`{` `x``:` `4``,` `y``:` `43` `}`
`]`
`];````
Our original `dataset` looks like Figure 11-5 in the console.
Then we initialize our stack layout function, and call it on `dataset`:
````var` `stack` `=` `d3``.``layout``.``stack``();`
`stack``(``dataset``);````
Now the stacked data is shown in Figure 11-6.
Can you spot the difference? In the stacked data, each object has been given a `y0` value. This is the baseline value. Notice that the `y0` baseline value is equal to the sum of all the `y` values in the preceding categories. For example, reading left to right across the top, the first object’s `y` value is 5, and its `y0` is zero. To the right (in the oranges column), the first object has a `y` value of 10, and a `y0` of 5. (Aha! That’s just the value of the first object’s `y`!) On the far right (in the grapes column), we see a `y` value of 22, and a `y0` of 15 (which is just 5 + 10).
To “stack” elements visually, now we can reference each data object’s baseline value as well as its height. See all the code in 03_stacked_bar.html. (I leave it as an exercise to you to align the stacks against the bottom x-axis.) Here’s the critical excerpt:
````var` `rects` `=` `groups``.``selectAll``(``"rect"``)`
`.``data``(``function``(``d``)` `{` `return` `d``;` `})`
`.``enter``()`
`.``append``(``"rect"``)`
`.``attr``(``"x"``,` `function``(``d``,` `i``)` `{`
`return` `xScale``(``i``);`
`})`
`.``attr``(``"y"``,` `function``(``d``)` `{`
`return` `yScale``(``d``.``y0``);`
`})`
`.``attr``(``"height"``,` `function``(``d``)` `{`
`return` `yScale``(``d``.``y``);`
`})`
`.``attr``(``"width"``,` `xScale``.``rangeBand``());````
Note how for `y` and `height` we reference `d.y0` and `d.y`, respectively.
## Force Layout
Force-directed layouts are so-called because they use simulations of physical forces to arrange elements on the screen. Arguably, they are a bit overused, yet they make a great demo and just look so darn cool. Everyone wants to learn how to make one, so let’s talk about it.
Force layouts are typically used with network data. In computer science, this kind of dataset is called a graph. A simple graph is a list of nodes and edges. The nodes are entities in the dataset, and the edges are the connections between nodes. Some nodes will be connected by edges, and others won’t. Nodes are commonly represented as circles, and edges as lines. But of course the visual representation is up to you—D3 just helps manage all the mechanics behind the scenes.
The physical metaphor here is of particles that repel each other, yet are also connected by springs. The repelling forces push particles away from each other, preventing visual overlap, and the springs prevent them from just flying out into space, thereby keeping them on the screen where we can see them.
Figure 11-7 provides a visual preview of what we’re coding toward.
D3’s force layout expects us to provide nodes and edges separately, as arrays of objects. Here we have one `dataset` object that contains two elements, `nodes` and `edges`, each of which is itself an array of objects:
````var` `dataset` `=` `{`
`nodes``:` `[`
`{` `name``:` `"Adam"` `},`
`{` `name``:` `"Bob"` `},`
`{` `name``:` `"Carrie"` `},`
`{` `name``:` `"Donovan"` `},`
`{` `name``:` `"Edward"` `},`
`{` `name``:` `"Felicity"` `},`
`{` `name``:` `"George"` `},`
`{` `name``:` `"Hannah"` `},`
`{` `name``:` `"Iris"` `},`
`{` `name``:` `"Jerry"` `}`
`],`
`edges``:` `[`
`{` `source``:` `0``,` `target``:` `1` `},`
`{` `source``:` `0``,` `target``:` `2` `},`
`{` `source``:` `0``,` `target``:` `3` `},`
`{` `source``:` `0``,` `target``:` `4` `},`
`{` `source``:` `1``,` `target``:` `5` `},`
`{` `source``:` `2``,` `target``:` `5` `},`
`{` `source``:` `2``,` `target``:` `5` `},`
`{` `source``:` `3``,` `target``:` `4` `},`
`{` `source``:` `5``,` `target``:` `8` `},`
`{` `source``:` `5``,` `target``:` `9` `},`
`{` `source``:` `6``,` `target``:` `7` `},`
`{` `source``:` `7``,` `target``:` `8` `},`
`{` `source``:` `8``,` `target``:` `9` `}`
`]`
`};````
As usual for D3, you can store whatever data you like within these objects. Our nodes are simple—just names of people. The edges contain two values each: a source ID and a target ID. These IDs correspond to the nodes above, so ID number 3 is Donovan, for example. If 3 is connected to 4, then Donovan is connected to Edward.
The data shown is a bare minimum for using the force layout. You can add more information, and, in fact, D3 will itself add a lot more data to what we’ve provided, as we’ll see in a moment.
Here’s how to initialize a force layout:
````var` `force` `=` `d3``.``layout``.``force``()`
`.``nodes``(``dataset``.``nodes``)`
`.``links``(``dataset``.``edges``)`
`.``size``([``w``,` `h``])`
`.``start``();````
Again, this is just the bare minimum. We specify the nodes and links to be used, the size of the available space, and then call `start()` when we’re ready to go.
This generates a default force layout, but the default won’t be ideal for every dataset. As you would expect, there are all kinds of options for customization. See the API wiki for all the gory details. Here, I’ve increased the `linkDistance` (the length of the edges between connected nodes) as well as the negative `charge` between nodes, so they will repel each other more. (It’s not personal; I just want them to spread out a bit.)
````var` `force` `=` `d3``.``layout``.``force``()`
`.``nodes``(``dataset``.``nodes``)`
`.``links``(``dataset``.``edges``)`
`.``size``([``w``,` `h``])`
`.``linkDistance``([``50``])` `// <-- New!`
`.``charge``([``-``100``])` `// <-- New!`
`.``start``();````
Next, we create an SVG line for each edge:
````var` `edges` `=` `svg``.``selectAll``(``"line"``)`
`.``data``(``dataset``.``edges``)`
`.``enter``()`
`.``append``(``"line"``)`
`.``style``(``"stroke"``,` `"#ccc"``)`
`.``style``(``"stroke-width"``,` `1``);````
Note that I set all the lines to have the same stroke color and weight, but of course you could set this dynamically based on data (say, thicker or darker lines for “stronger” connections, or some other value).
Then, we create an SVG circle for each node:
````var` `nodes` `=` `svg``.``selectAll``(``"circle"``)`
`.``data``(``dataset``.``nodes``)`
`.``enter``()`
`.``append``(``"circle"``)`
`.``attr``(``"r"``,` `10``)`
`.``style``(``"fill"``,` `function``(``d``,` `i``)` `{`
`return` `colors``(``i``);`
`})`
`.``call``(``force``.``drag``);````
I set all circles to have the same radius, but each gets a different color fill, just because it’s prettier that way. Of course, these values could be set dynamically, too, for a more useful visualization.
You’ll notice the last line of code, which enabled drag-and-drop interaction. (Comment that out, and the user won’t be able to move nodes around.)
Lastly, we have to specify what happens when the force layout “ticks.” Yes, these ticks are different from the axis ticks addressed earlier, and definitely different from those little blood-sucking insects. Physics simulations use the word “tick” to refer to the passage of some amount of time, like the ticking second hand of a clock. For example, if an animation were running at 30 frames per second, you could have one tick represent 1/30th of a second. Then each time the simulation ticked, you’d see the calculations of motion update in real time. In some applications, it’s useful to run ticks faster than actual time. For example, if you were trying to model the effects of climate change on the planet 50 years from now, you wouldn’t want to wait 50 years to see the results, so you’d program the system to tick on ahead, faster than real time.
For our purposes, what you need to know is that D3’s force layout “ticks” forward through time, just like every other physics simulation. With each tick, the force layout adjusts the position values for each node and edge according to the rules we specified when the layout was first intialized. To see this progress visually, we need to update the associated elements—the lines and circles:
````force``.``on``(``"tick"``,` `function``()` `{`
`edges``.``attr``(``"x1"``,` `function``(``d``)` `{` `return` `d``.``source``.``x``;` `})`
`.``attr``(``"y1"``,` `function``(``d``)` `{` `return` `d``.``source``.``y``;` `})`
`.``attr``(``"x2"``,` `function``(``d``)` `{` `return` `d``.``target``.``x``;` `})`
`.``attr``(``"y2"``,` `function``(``d``)` `{` `return` `d``.``target``.``y``;` `});`
`nodes``.``attr``(``"cx"``,` `function``(``d``)` `{` `return` `d``.``x``;` `})`
`.``attr``(``"cy"``,` `function``(``d``)` `{` `return` `d``.``y``;` `});`
`});````
This tells D3, “Okay, every time you tick, take the new x/y values for each line and circle and update them in the DOM.”
Wait a minute. Where did these x/y values come from? We had only specified `name`s, `source`s, and `target`s!
D3 calculated those x/y values and appended them to the existing objects in our original `dataset` (see Figure 11-8). Go ahead and open up 04_force.html, and type `dataset` into the console. Expand any node or edge, and you’ll see lots of additional data that we didn’t provide. That’s where D3 stores the information it needs to continue running the physics simulation. With `on("tick", …)`, we are just specifying how to take those updated coordinates and map them on to the visual elements in the DOM.
The final result, then, is the lovely entanglement displayed in Figure 11-9.
Again, you can run the final code youself in 04_force.html.
Note that each time you reload the page, the circles and lines spring to life, eventually resting in a state of equilibrium. But that final state of rest is different every time because there is an element of randomness to how the circles enter the stage. This illustrates the unpredictable nature of force-based layouts: they might be quite different each time you use them, and they rely on the data to provide structure. If your data is more structured, you might get a better visual result.
Interactivity can be useful for improving our view of the data. I don’t like how the edges overlap here, so I’ll drag the pink circle out and around (see Figure 11-10), then I’ll move the red one up and to the left (see Figure 11-11).
As a bonus, interactivity + physics simulation = irresistable demo. I can’t explain it, but it’s true. For some reason, we humans just love seeing real-world things replicated on screens.
Get Interactive Data Visualization for the Web now with the O’Reilly learning platform.
O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers. | 5,808 | 19,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-40 | latest | en | 0.872414 |
https://itecnotes.com/electrical/electrical-distribution-of-current-in-kirchoffs-junction-law/ | 1,701,647,058,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00709.warc.gz | 363,292,761 | 6,317 | # Electrical – Distribution of Current in Kirchoff’s Junction Law
circuit analysiscurrentkirchhoffs-laws
I know that Kirchoff's law says that the incoming current is equal to outgoing current.
But I want to understand his: suppose we have 4 wires like the one in diagram.
So let us consider current i1, comes to the junction and gets split into three direction towards i2, i3, i4 and similar is the case with current i2 but then why we imagine only the current i3 and i4 as outgoing current whereas current is also going to i1 and i2.
May be I am missing something but I am stuck on this problem for very long. | 145 | 615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | longest | en | 0.950518 |
https://www.experts-exchange.com/questions/21824384/monkeys.html | 1,531,810,794,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589573.27/warc/CC-MAIN-20180717051134-20180717071134-00580.warc.gz | 860,346,141 | 16,897 | # monkeys
suppose 8 monkeys take 8 minutes to eat 8 bananas,
how many monkeys would it take to eat 48 bananas in 48 minutes
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Commented:
Hi shilpi84,
1min 1 monkey 1/8 banana
48 mins 1 monkey 48/8 = 6 banana
48 mins 8 monkey 48 banana
Cheers!
Sunnycoder
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Commented:
Don't accept my answer !!! Accept sunnycoder's :)
0
EngineerCommented:
Although it seems to be very simple in this case, it won't be the same everytime.
So, you must consider this:
More the monkeys, more the bananas they can eat in same time.
More the bananas, more the time required.
Total "work" to be done = 8 bananas
Rate of "work" = 8 bananas/(8 monkeys * 8 minutes) = 1/8
So, the total work required to be done = 48 bananas.
Since rate of doing work is the same, 48/(x * 48) = 1/8 | 360 | 1,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.933385 |
http://www.rapidtables.com/calc/light/how-candela-to-lumen.htm | 1,369,504,334,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706009988/warc/CC-MAIN-20130516120649-00058-ip-10-60-113-184.ec2.internal.warc.gz | 675,879,050 | 3,466 | # How to convert candela to lumens
How to convert luminous intensity in candela (cd) to luminous flux in lumens (lm).
You can calculate but not convert candela to lumens, since lumens and candela do not represent the same quantity.
### Candela to lumens calculation
For uniform, isotropic light source, the luminous flux Φv in lumens (lm) is equal to the luminous intensity Iv in candela (cd),
times the solid angle Ω in steradians (sr):
Φv(lm) = Iv(cd) × Ω(sr)
The solid angle Ω in steradians (sr) is equal to 2 times pi times 1 minus cosine of half the apex angle θ in degrees (º):
Ω(sr) = 2π(1 - cos(θ/2))
The luminous flux Φv in lumens (lm) is equal to the luminous intensity Iv in candela (cd),
times 2 times pi times 1 minus cosine of half the apex angle θ in degrees (º):
Φv(lm) = Iv(cd) × ( 2π(1 - cos(θ/2)) )
So
lumens = candela × ( 2π(1 - cos(degrees/2)) )
Or
lm = cd × ( 2π(1 - cos(º/2)) )
#### Example
Find the luminous flux Φv in lumens (lm) when the luminous intensity Iv in candela (cd) is 400cd and the apex angle is 60°:
Φv(lm) = 400cd × ( 2π(1 - cos(60°/2)) ) = 336.7lm
Lumens to candela calculation | 366 | 1,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.664926 |
https://chemistry.stackexchange.com/questions/32264/what-is-the-difference-between-these-equations-for-the-contact-process | 1,652,859,539,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00190.warc.gz | 230,727,680 | 62,807 | # What is the difference between these equations for the Contact process?
Some webpages describe the contact process as this (I left out $\ce{SO2}$, $\ce{SO3}$ production):
$\ce{SO3 + H2SO4 -> H2S2O7}$
After this, it is controllable to add water to the oleum therefore making liquid sulfuric acid. However, some other sites describe this:
$\ce{SO3 + H2O -> H2SO4}$
I am confused about what really happens in the absorption tower. I've read about the wet acid, but it states that adding water causes gaseous $\ce{H2SO4}$ to form, which is condensed.
EDIT: In the absorption tower, trioxide sulfur enters as gas, and sulfuric acid comes down as liquid. As a result the liquid comes out as oleum (first eq.). The other contact process states different: the liquid entering is 98% sulfuric acid, and 2% water, this water reacts with the sulfur trioxide (this little water avoids the heat problem), giving strengthened sulfuric acid (let's say 99.5% sulfuric acid, and 0.5% water). Still confused.
In order to avoid the formation of problematic sulfuric acid mist, sufur trioxide is not directly absorbed into water. Instead, it is absorbed in sulfuric acid of 98 %–99 % concentration, which approximately corresponds to an azeotrope where the partial pressures of sufur trioxide, sulfuric acid, and water are relatively low.
Sulfuric acid reacts with the absorbed sulfur trioxide to form disulfuric acid:
$$\ce{H2SO4 + SO3 <=> H2S2O7}$$
The disulfuric acid can be converted to sulfuric acid by reaction with added water:
$$\ce{H2S2O7 + H2O <=> 2 H2SO4}$$
Thus, the overall net reaction is as follows:
\begin{align} \ce{H2SO4 + SO3 \;&<=> H2S2O7}\\ \ce{H2S2O7 + H2O \;&<=> 2 H2SO4}\\ \hline\ce{SO3 + H2O \;&<=> H2SO4} \end{align}
• Thank you, but sources state either one of the equations as the contact process, which is what happens in the absorption tower. And adding the water occurs later, outside the tower.
– Dave
May 31, 2015 at 21:38
The answer to your question is thoroughly explained in Chemistry of the Elements (Second Edition), A. Earnshaw and Norman Greenwood, 1997, p. 708:
(...) The $\ce{SO3}$ gas cannot be absorbed directly in water because it would first come into contact with the water-vapour above the absorbant and so produce a stable mist of fine droplets of $\ce{H2SO4}$ which would then pass right through the absorber and out into the atmosphere. Instead, absorption is effected by 98% $\ce{H2SO4}$ in ceramic packed towers and sufficient water is added to the circulating acid to maintain the required concentration. (...)
Two pages following, the process is represented as:
$$\ce{SO3 + H2O\ (\text{in 98% }H2SO4) -> H2SO4}$$
In the Industrial Manufacture of Sulfuric Acid box (p. 708-710), the $\ce{H2S2O7}$ species is not mentioned. However, the discussion about concentrated Sulfuric Acid states:
It is clear that "pure" anhydrous sulfuric acid, far from being a single substance in the bulk liquid phase, comprises a dynamic equilibrium involving at least seven well-defined species. The concentration of the self-dissociation products in $\ce{H2SO4}$ at 25º (expressed in millimoles of solute per kg solvent) are:
$$\begin{array}{lcr} \ce{HSO4-} & \ce{H3SO4+} & \ce{H3O+} & \ce{HS2O7-} & \ce{H2S2O7} & \ce{H2O} & \text{Total}\\ \hline 15.0 & 11.3 & 8.0 & 4.4 & 3.6 & 0.1 & 42.4 \end{array}$$
Considering the above, you can assume that the absorption of $\ce{SO3}$ in 98% $\ce{H2SO4}$ will lead to the formation of $\ce{H2S2O7}$ but also several other species. If you are really interested in industrial production details and properties of sulfuric acid, this reference is really valuable.
• Thanks. This is what I read how they add water to the sulfur trioxide in the absorption tower. Which is different to the other contact process equation, hence my confusion.
– Dave
May 31, 2015 at 21:36 | 1,103 | 3,845 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-21 | longest | en | 0.908548 |
https://www.coursehero.com/file/p1unl2k7/cr-ea-tur-es-contain-the-sa-me-fi-xe-d-prop-ortions-of-14-to-nonradioactive-12/ | 1,627,328,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.92/warc/CC-MAIN-20210726183622-20210726213622-00514.warc.gz | 715,105,875 | 67,817 | cr ea tur es contain the sa me fi xe d prop ortions of 14 to nonradioactive 12
# Cr ea tur es contain the sa me fi xe d prop ortions
• 68
This preview shows page 33 - 36 out of 68 pages.
creatures contain the same fixed proportions of 14( to nonradioactive 12C as the atmosphere. After an organism dies, it stops assimilating 14(, and the amount of 14( in it begins to decay exponentially. We can then determine the time elapsed since the death of the organism by measuring the amount of 14( left in it. For example, if a donkey bone contains 73% as much 14( as a living donkey and it died t years ago, then by the formula for radioactive de-cay (Section 4.6), 0.73 = (l.OO)e-(tln2)/5730 We solve this exponential equation to find t = 2600, so the bone is about 2600 years old.
334 CHAPTER 4 I Exponential and Logarithmic Functions CHECK YOUR ANSWER lfx = 17, we get log2(25-17) = log28 = 3 V l' Logarithmic Equations A lugarirlunic equation is one in which a logarithm of the variable occurs. For example, logl r + 2) = 5 To solve for x. we write the equation in exponential form. E\ponential rorm X= 32-2 = 30 Solve for x Another way of looking at the first step is to raise the base. 2, to each side of the equation. Rui-.e 2 to each -,ide Propert) or logarithms X= 32-2 = 30 Solve for .I The method used to solve this simple problem is typical. We summarize the steps as follows. GUIDELINES FOR SOLVING LOGARITHMIC EQUATIONS 1. Isolate the logarithmic term on one side of the equation; you might lirst need to combine the logarithmic terms. 2. Write the equation in exponential form (or raise the base to each side of the equation). 3. Solve for the variable. EXAMPLE 6 I Solving Logarithmic Equations Solve each equation for x. (a) In x = 8 (b) log2(25 -x) = 3 SOLUTION (a) In .r = 8 Therefore, .r = e8 = 298 1. Gi,·en equation Exponential form We can also solve this problem another way: In .r = 8 Gi1cn equation Rai~e e to each -,ide Propcrt) of In (b) The first step is to rewrite the equation in exponential form. Gi1cn equation 25 -.r = 2' c\poncntial form (or raise 2 to each -.ide) 25 -.r = 8 X= 25-8 = 17 , NOW TRY EXERCISES 37 AND 41
CHECK YOUR ANSWER x = - 4: log( - 4 + 2) + log( - 4 -I ) = log( -2) + log( - 5) undefined X .r = 3: log(3 + 2) + log(3 -I) 3 = log 5 + log 2 = log(5 · 2) = log 10 = I V' 0 ----+r----+-=--1--==;=::--:;:=:==t----+---6 -3 FIGURE 2 SECTION 4.5 I Exponential and Logarithmic Equations 335 EXAMPLE 7 I Solving a Logarithmic Equation Solve the equation 4 + 3 log(2x) = 16. S 0 LUTI 0 N We flrst isolate the logarithmic term. This allows us to write the equation in exponential form. 4 + 3 log(2x) = 16 3 log(2x) = 12 log(2x) = 4 2x = 10~ X= 5000 CHECK YOUR ANSWER If x = 5000, we get Gt1·en equation Subtract -1 Divid..: by J E\pon..:ntial form (or raise I0 10 ea<.:h ' ide) Di1 ide hy:! 4 + 3 log 2(5000) = 4 + 3 log 10,000 = 4 + 3(4) = 16 V' . NOW TRY EXERCISE 43 EXAMPLE 8 I Solving a Logarithmic Equation Algebraically and Graphically Solve the equation log(.r + 2) + log(x-I ) = I algebraically and graphically. SOLUTION 1: Algebraic We flrst combine the logarithmic terms, using the Laws of Logarithms. | 1,039 | 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} | 4.34375 | 4 | CC-MAIN-2021-31 | latest | en | 0.863593 |
https://www.usingenglish.com/forum/threads/116113-larger-than-or-as-large-as | 1,477,518,103,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720973.64/warc/CC-MAIN-20161020183840-00160-ip-10-171-6-4.ec2.internal.warc.gz | 1,006,355,008 | 15,102 | # Thread: larger than or as large as
1. ## larger than or as large as
1. A is three times larger than B.
2. B is three times as large as B.
What's the difference in meaning between the above sentences?
3. A is one-sixth larger than B.
4. A is one-sixth as large as B.
What's the difference in meaning between #3 and #4?
2. ## Re: larger than or as large as
If B is 100 pounds in mass, then
1. A = 3*100 = 300 lbs
2. B (I think you mean A) = 3*100 = 300 lbs
3. A = 1.17 * 100 = 117 lbs
4. A = 100/6 = 16.67 lbs
3. ## Re: larger than or as large as
Originally Posted by jlinger
If B is 100 pounds in mass, then
1. A = 3*100 = 300 lbs
2. B (I think you mean A) = 3*100 = 300 lbs
I disagree with 1.
If:
i. A is three times larger than B, and
ii. B = 100 pounds, then
iii. A is 300 pounds larger than B, i.e. 400 lbs.
'Three times larger' should not be used to mean 'three times as large'.
This can easily be understood by comparing "half as large" (0.5 times) with "a half larger" (1.5 times).
I think terms like 'three times larger' should not be used at all, because there will always be misunderstandings between those who understand the term differently.
With 'three times as large', there is rarely disagreement about meaning.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 414 | 1,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.978982 |
http://openstudy.com/updates/508df681e4b078e4677b6a1e | 1,448,690,535,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398451648.66/warc/CC-MAIN-20151124205411-00274-ip-10-71-132-137.ec2.internal.warc.gz | 172,055,959 | 9,862 | ## Bubblez33 3 years ago Find the GCF of the list of terms 44m^2n^2, 66mn
1. gaara438125
GCF of just 66 and 44 is 22 and the 44m^2n^2 and 66mn both have M and N as variables. lol i'm trying to explain without giving the answer, it's in the rules :P
2. Bubblez33
22mn
3. gaara438125
bingo
4. Bubblez33
lol...thanks | 119 | 321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-48 | longest | en | 0.882771 |
https://www.xszz.org/faq-2/question-2018101938930.html | 1,558,292,758,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255092.55/warc/CC-MAIN-20190519181530-20190519203530-00320.warc.gz | 1,024,223,658 | 12,625 | # Calculating weights in a NN
So I am currently trying to implement my first NN with a genetic algorithm for training and a sigmoid activation function. It's all good but I'm not quite sure in what ranges the weights must be. I've searched some about the question but with no luck. How does one choose the ranges of the weights in a NN? What does it depend on?
The weights can be seen as an intrinsic property of the problem you're trying to solve using the GA/NN approach; there's no general best value fo these, so you're best off studying different weight spans (w.r.t. training sets) with other parameters fixed.
E.g., study different settings for parameter weightSpan in
weights \in [-weightSpan/2, weightSpan/2],
and let your initial chromosomes describe weights with randomized values in this range. Your squashing function (sigmoid) is used to grade the NN response to the range [0, 1].
Finding an appropriate weight span is, much like setting a value of number of hidden layer, a process if problem-specific testing. ("There is no free lunch").
<hr>
Edit:
I thought I'd add that the easiest way to study different weight spans is probably to set a fixed weight span, say [-1, 1], and study the squashing constant in your squashing function (sigmoid). I.e., study different (non-negative) values of constant c in your sigmoid
σ(s) = 1 / (1 + e^(-c*s))
人点赞 | 316 | 1,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-22 | latest | en | 0.900691 |
https://www.thenational.academy/teachers/lessons/what-are-magnets-cgvkee | 1,713,638,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00142.warc.gz | 929,196,620 | 36,766 | # What are magnets?
In this lesson, we will learn what magnets are and test putting magnets together in different ways to see whether the ends are attracted or repelled.
# What are magnets?
In this lesson, we will learn what magnets are and test putting magnets together in different ways to see whether the ends are attracted or repelled.
## Lesson details
### Key learning points
1. Know what magnets are
2. Describe when magnets attract and repel
3. Describe how to test the strength of a magnet
### Licence
This content is made available by Oak National Academy Limited and its partners and licensed under Oak’s terms & conditions (Collection 1), except where otherwise stated.
## Video
Share with pupils
## Worksheet
Share with pupils
## Starter quiz
Share with pupils
### 3 Questions
Q1.
Is friction a contact or non-contact force?
non-contact
Q2.
In magnetism, 'attract' means...
push apart
Q3.
Which of the following is NOT a non-contact force?
Gravitational force
Magnetic force
## Exit quiz
Share with pupils
### 3 Questions
Q1.
What are the two poles of a magnet called?
North and East poles
Correct answer: North and South poles
North and West poles
Q2.
What does 'repel' mean?
When objects are drawn together
When objects are on top of each other
Correct answer: When the objects push each other apart
Q3.
What's the sticky way to remember the sides of a magnet that attract?
Opposites are good
Opposites repel | 332 | 1,443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | latest | en | 0.88925 |
https://math.dartmouth.edu/archive/m22f06/public_html/m22syllabus.html | 1,566,470,223,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317037.24/warc/CC-MAIN-20190822084513-20190822110513-00179.warc.gz | 548,984,180 | 2,474 | Last updated September 11, 2006
Syllabus
The following is a tentative syllabus for the course. The weekly syllabus contained in the HW Assignments page will always be accurate.
Date Section(s) in Text Brief Description
9/20 1.1 Systems of Linear Equations
9/21 (x-hour) x-hour proof workshop
9/22 1.2 Row Reduction and Echelon Forms
9/25 1.3 Vector Equations
9/27 1.4, 1.5 The Matrix Equation Ax = b, Solution Sets of Linear Systems
9/28 (x-hour) x-hour proof workshop
9/29 1.7 Linear Independence
10/2 1.8 Introduction to Linear Transformations
10/4 1.9 The Matrix of a Linear Transformation
10/5 (x-hour) 2.1, 2.2 Matrix Operations, The Inverse of a Matrix
10/6 2.2, 2.3 The Inverse of a Matrix, Characterizations of Invertible Matrices
10/9 catch-up day; section 2.6 if time permits
10/11 4.1 Vector Spaces and Subspaces
10/12 (x-hour) in-class part of exam 1 (covers sections 1.1 – 1.9, 2.1 – 2.3 minus section 1.6)
10/13 4.2 Null Spaces, Column Spaces, and Linear Transformations
10/16 4.3 Linearly Independent Sets; Bases
10/18 4.4 Coordinate Systems
10/19 (x-hour) x-hour proof workshop
10/20 4.5 The Dimension of a Vector Space
10/23 4.6 Rank
10/25 4.7 Change of Basis
10/26 (x-hour) x-hour proof workshop
10/27 3.1, 3.2 Introduction to Determinants, Properties of Determinants
10/30 3.2, 3.3 Properties of Determinants, Determinants as Area or Volume, Linear Transformations
11/1 5.1 Eigenvectors and Eigenvalues
11/2 (x-hour) in-class part of exam 2 (covers sections 4.1 – 4.7, 3.1 – 3.3)
11/3 5.2 The Characteristic Equation
11/6 5.3 Diagonalization
11/8 5.4 Eigenvectors and Linear Transformations
11/9 (x-hour) x-hour proof workshop
11/10 6.1 Inner Product, Length, and Orthogonality
11/13 6.2 Orthogonal Sets
11/15 6.3 Orthogonal Projections
11/16 (x-hour) x-hour proof workshop
11/17 6.4 The Gram-Schmidt Process
11/20 no class
11/22 no class (Thanksgiving break)
11/23 (x-hour) no class (Thanksgiving break)
11/24 no class (Thanksgiving break)
11/27 7.1 Diagonalization of Symmetric Matrices
11/29 wrap up; section 7.4 (the singular values of an m × n matrix, end of the invertible matrix theorem) if time permits
12/3 Final Exam, 15:00 – 18:00 (3:00 – 6:00 pm) | 737 | 2,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.689566 |
http://blog.csdn.net/twobqn123/article/details/17542217 | 1,519,165,587,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813109.36/warc/CC-MAIN-20180220204917-20180220224917-00332.warc.gz | 47,416,110 | 13,513 | # codeforces #221(div2)B. I.O.U.
B. I.O.U.
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Imagine that there is a group of three friends: A, B and С. A owes B 20 rubles and B owes C 20 rubles. The total sum of the debts is 40 rubles. You can see that the debts are not organized in a very optimal manner. Let's rearrange them like that: assume that A owes C 20 rubles and B doesn't owe anything to anybody. The debts still mean the same but the total sum of the debts now equals 20 rubles.
This task is a generalisation of a described example. Imagine that your group of friends has n people and you know the debts between the people. Optimize the given debts without changing their meaning. In other words, finally for each friend the difference between the total money he should give and the total money he should take must be the same. Print the minimum sum of all debts in the optimal rearrangement of the debts. See the notes to the test samples to better understand the problem.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100; 0 ≤ m ≤ 104). The next m lines contain the debts. The i-th line contains three integers ai, bi, ci (1 ≤ ai, bi ≤ nai ≠ bi; 1 ≤ ci ≤ 100), which mean that person ai owes person bi ci rubles.
Assume that the people are numbered by integers from 1 to n.
It is guaranteed that the same pair of people occurs at most once in the input. The input doesn't simultaneously contain pair of people (x, y) and pair of people (y, x).
Output
Print a single integer — the minimum sum of debts in the optimal rearrangement.
Sample test(s)
Input
5 3
1 2 10
2 3 1
2 4 1
Output
10
Input
3 0
Output
0
Input
4 3
1 2 1
2 3 1
3 1 1
Output
0
Note
In the first sample, you can assume that person number 1 owes 8 rubles to person number 2, 1 ruble to person number 3 and 1 ruble to person number 4. He doesn't owe anybody else anything. In the end, the total debt equals 10.
In the second sample, there are no debts.
In the third sample, you can annul all the debts.
ps:我擦，这么简单的题目我在比赛的时候都没想出来，真不知道我到底怎么了，每天都在训练，但是感觉还是这么弱，我到底怎样提高啊。但是我绝对不会放弃的，我热爱它。
/*
@author : liuwen
*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#include <cmath>
using namespace std;
const int maxn=200;
int G[maxn],n,m;
int main()
{
//freopen("in.txt","r",stdin);
while(cin>>n>>m){
int u,v,w;
memset(G,0,sizeof(G));
while(m--){
cin>>u>>v>>w;
G[u]-=w;
G[v]+=w;
}
int ans=0;
for(int i=1;i<=n;i++){
if(G[i]>0) ans+=G[i];
}
cout<<ans<<endl;
}
return 0;
}
• 本文已收录于以下专栏：
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• 89
## Codeforces 456 Div2
（写到一半突然断电。。。。。无力，原谅我不写题意了） B: 很显然，如果k=1k=1，答案就是n 若k≥2k≥2，答案就是n的第一个1开始，将后面所有位全部转换为1后的值。 #...
• qq_34454069
• 2018年01月06日 15:12
• 49
## Codeforces Round #214div2题解
A 直接判min(a, b) + min(c, d) B 按%k分类求和 或者暴力扫 C 把每个食物当成重量a - k * b, 价值a的物品 然后就是做背包问题求重量和为0的最优解 ...
• ztxz16
• 2013年11月25日 02:00
• 308
## codeforces #Round354-div2-B
• West___wind
• 2016年05月26日 14:12
• 192
## codeforces round #214 div2 CD
C. Dima and Salad time limit per test 1 second memory limit per test 256 megabytes input stand...
• u012350533
• 2013年11月25日 17:57
• 874
## 【解题报告】Codeforces Round #374 (Div. 2)
• TRiddle
• 2016年10月02日 00:50
• 416
## Codeforces 456 A. Laptops
• qingshui23
• 2015年07月07日 09:43
• 828
举报原因： 您举报文章：codeforces #221(div2)B. I.O.U. 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 1,473 | 4,073 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-09 | latest | en | 0.896942 |
https://www.studysmarter.us/textbooks/physics/college-physics-urone-1st-edition/fluid-dynamics-and-its-biological-and-medical-applications/q1cq-what-is-the-difference-between-flow-rate-and-fluid-velo/ | 1,680,072,377,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00244.warc.gz | 1,109,227,267 | 21,731 | Suggested languages for you:
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Found in: Page 425
### College Physics (Urone)
Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000
# What is the difference between flow rate and fluid velocity? How are they related?
The fluid velocity is variation in position with time while the flow rate is the quantity of water passed per second from an area and the flow rate is equal to the product of cross section area and fluid velocity.
See the step by step solution
## Step 1: Difference between flow rate and fluid velocity
The fluid velocity of a fluid is the variation in the position of the fluid with time and position. The volume of moving fluid that passes through a certain area of cross section in unit time is called the flow rate.
## Step 2: Relation between fluid velocity and flow rate
Suppose a fluid is moving with fluid velocity $${\rm{v}}$$ through a cross-section of the area $${\rm{A}}$$ with diameter $${\rm{d}}$$.
The cross-section area for fluid flow is given as:
$${\rm{A}} = \frac{{\rm{\pi }}}{{\rm{4}}}{{\rm{d}}^{\rm{2}}}$$
The flow rate of fluid is given as:
$${\rm{Q}} = {\rm{Av}}$$
Here, $${\rm{Q}}$$ is the flow rate and $${\rm{v}}$$is the fluid velocity.
Substitute all the values in the above equation.
$$\begin{array}{l}{\rm{Q}} = \left( {\frac{{\rm{\pi }}}{{\rm{4}}}{{\rm{d}}^{\rm{2}}}} \right){\rm{v}}\\{\rm{Q}} = \frac{{\rm{\pi }}}{{\rm{4}}}{{\rm{d}}^{\rm{2}}}{\rm{v}}\end{array}$$ | 447 | 1,500 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-14 | latest | en | 0.782884 |
http://www.numbersaplenty.com/12322203021113 | 1,591,284,879,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347441088.63/warc/CC-MAIN-20200604125947-20200604155947-00466.warc.gz | 191,359,396 | 3,524 | Search a number
12322203021113 = 19164514151943
BaseRepresentation
bin1011001101001111110010…
…0010001010111100111001
31121121222202111220110121212
42303103330202022330321
53103341332433133423
642112424131012505
72411151451343006
oct263237442127471
947558674813555
1012322203021113
113a208a9403527
121470160a3a135
136b4c954448bb
1430857d82c8ad
151657de962578
hexb34fc88af39
12322203021113 has 4 divisors (see below), whose sum is σ = 12386717173248. Its totient is φ = 12257688868980.
The previous prime is 12322203021101. The next prime is 12322203021179. The reversal of 12322203021113 is 31112030222321.
Adding to 12322203021113 its reverse (31112030222321), we get a palindrome (43434233243434).
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 12322203021113 - 210 = 12322203020089 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (12322203021313) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 32257075781 + ... + 32257076162.
It is an arithmetic number, because the mean of its divisors is an integer number (3096679293312).
Almost surely, 212322203021113 is an apocalyptic number.
It is an amenable number.
12322203021113 is a deficient number, since it is larger than the sum of its proper divisors (64514152135).
12322203021113 is an equidigital number, since it uses as much as digits as its factorization.
12322203021113 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 64514152134.
The product of its (nonzero) digits is 864, while the sum is 23.
The spelling of 12322203021113 in words is "twelve trillion, three hundred twenty-two billion, two hundred three million, twenty-one thousand, one hundred thirteen".
Divisors: 1 191 64514151943 12322203021113 | 612 | 2,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.83493 |
http://www.unilad.co.uk/news/can-you-crack-the-code-on-britains-top-spy-agencys-christmas-card/ | 1,495,961,146,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609610.87/warc/CC-MAIN-20170528082102-20170528102102-00301.warc.gz | 844,316,938 | 10,925 | # Can You Crack The Code On Britain’s Top Spy Agency’s Christmas Card?
By :
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The Telegraph
Britain’s top spy agency has just sent out its Christmas card and, as you’d expect, it comes with an catch. Don’t worry though, it’s not timed to self destruct.
The card doesn’t have a nice picture of a snowy village or a shitty Christmas joke, instead the spies have hidden their Christmas message in a secret code.
Clever participants have to solve a puzzle to reveal an image.
To solve the puzzle you have to know that each square is either black or white. Each row or column is labelled with a string of numbers which tell you the length of consecutive runs of black squares, and are displayed in the order that the runs appear in that line.
The GCHQ website gives the example that a label “2 1 6” indicates sets of two, one and six black squares, each of which will have at least one white square separating them.
As it’s Christmas, the spies have lent a hand, some of the squares are already coloured to make it a bit easier. Apparently if you can solve the puzzle it then leads to a series of more complex challenges.
GCHQ director Robert Hannigan sent the puzzle out to his Christmas card list. If participants can solve all the puzzles then they can email their answer to GCHQ by January 31.
If you fancy the challenge but aren’t on the spies mailing list you can have a go here. Those who enjoyed the brainteaser have been asked to make a donation to the National Society for the Prevention of Cruelty to Children. | 338 | 1,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.925415 |
https://gmatclub.com/forum/if-2-y-2-3-y-1-14-0-which-of-the-following-could-be-the-265102.html | 1,576,384,772,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00093.warc.gz | 382,063,145 | 151,481 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the
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If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
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08 May 2018, 01:51
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If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
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08 May 2018, 02:21
2
The given equation can be changed to
$$14y^2-3y-2=0$$
$$14y^2 - 7y + 4y -2 =0$$
7y (2y-1) + 2 (2y-1) =0
(7y+2) (2y -1) =0
y can be either -2/7 or 1/2.
Ans :- B
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Posts: 3158
Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
### Show Tags
08 May 2018, 03:59
Solution
Given:
• 2∗$$y^{−2}+3∗y^{−1}$$−14=0
To find:
• The possible value of y
Approach and Working:
• If we assume the value of $$y^{-1}$$ to be x, we can rewrite the equation as
o 2$$x^2$$ + 3x – 14 = 0
Or, 2$$x^2$$ + 7x – 4x – 14 = 0
Or, (x – 2) (2x + 7) = 0
Or, x = 2 , $$\frac{-7}{2}$$
• Replacing the value of x with y^-1, we can write:
o $$\frac{1}{y}$$ = 2
Or, y = $$\frac{1}{2}$$
o Similarly, $$\frac{1}{y} = \frac{-7}{2}$$
Or, y =$$\frac{-2}{7}$$
As per the options, the correct answer is option B.
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
### Show Tags
08 May 2018, 04:04
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
I think back solving will be the easiest way to deal with this question. First simplify the equation the plug in values. B is the correct answer. Here it is clear that fraction will be the correct answer and no negative value won't work. Thus u can't A and E immediately.
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
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09 May 2018, 01:38
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
For such questions, back solving is the best and the fastest way.
First simplify the given equation and then try each answer choice.
B is the correct choice.
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
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09 May 2018, 16:45
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
Simplifying we have:
2/y^2 + 3/y - 14 = 0
Multiplying by y^2 we have:
2 + 3y - 14y^2 = 0
14y^2 - 3y - 2 = 0
(7y + 2)(2y - 1) = 0
7y + 2 = 0 → y = -2/7
2y - 1 = 0 → y = 1/2
Since only 1/2 is given as one of the choices. Choice B is the correct answer.
Alternate solution:
If we let x = y^-1, then the equation can be rewritten as 2x^2 + 3x - 14 = 0. Let’s solve it:
(2x + 7)(x - 2) = 0
2x + 7 = 0 → x = -7/2
x - 2 = 0 → x = 2
Since x = y^-1 = 1/y, y = 1/x. Therefore, y is either 1/(-7/2) = -2/7 or 1/2. Since only 1/2 is given as one of the choices. Choice B is the correct answer.
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Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink]
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16 Mar 2019, 19:48
ScottTargetTestPrep wrote:
Bunuel wrote:
If $$2*y^{-2} + 3*y^{-1} - 14 = 0$$, which of the following could be the value of y?
A. 2
B. 1/2
C. 2/7
D. -1/2
E. -7
Simplifying we have:
2/y^2 + 3/y - 14 = 0
Multiplying by y^2 we have:
2 + 3y - 14y^2 = 0
14y^2 - 3y - 2 = 0
(7y + 2)(2y - 1) = 0
7y + 2 = 0 → y = -2/7
2y - 1 = 0 → y = 1/2
Since only 1/2 is given as one of the choices. Choice B is the correct answer.
Alternate solution:
If we let x = y^-1, then the equation can be rewritten as 2x^2 + 3x - 14 = 0. Let’s solve it:
(2x + 7)(x - 2) = 0
2x + 7 = 0 → x = -7/2
x - 2 = 0 → x = 2
Since x = y^-1 = 1/y, y = 1/x. Therefore, y is either 1/(-7/2) = -2/7 or 1/2. Since only 1/2 is given as one of the choices. Choice B is the correct answer.
Hello ScottTargetTestPrep!!
Whys is it like the follwoing?
14y^2 - 3y - 2 = 0
(7y + 2)(2y - 1) = 0
Shouldn't the sum of the roots be -3?
Regards!
Re: If 2*y^(-2) + 3*y^(-1) - 14 = 0, which of the following could be the [#permalink] 16 Mar 2019, 19:48
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The very term ‘Physics’ is panicky since it is full of derivations, formulas, equations and problems. Get the help of an Online Physics Tutor for solving problems without panicking.
## 10 steps for solving Physics problems!
1. Don’t panic on seeing the problem
Do not be afraid of the problem given but boost your confidence with a positive attitude for solving the problem successfully.
1. Try to understand the practical situation involved in the problem
Visualize the problem and try to connect it with a situation that is imaginable in our day today life.
1. Read the question again to understand the problem
Now you know the general context of the problem, read the question again to make sure of the requirement of the problem.
1. Sketch out the requirement for clear understanding
Draw an image of the details of the problem. It facilitates your understanding.
1. Organize the information for step by step approach
Organize every detail in order so as to approach the problem without difficulty and go step by step.
1. Verify whether you are using the same units
If you change the units, you will not arrive at answers. Further, your teacher is interested in testing your unit conversion skills many a time. So, take care of the units.
1. Check whether you are using the right formula
You may have to memorize different Physics formulas and choose the right one from your memory. Or, you need to choose from the cheat sheet given to you. Either way, know which formula suits the problem and apply it to solve the problem.
1. Solve the problem.
You have sketched out the problem. You have details and have chosen the formula. Solve with all these things and have confidence while solving. | 357 | 1,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-43 | longest | en | 0.933338 |
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Question
# Let f(x) = √3x - 9 and g(x) = 5xa. State the domain of f. Express your answer in interval
notation.
b. Find (f - g) (3)
c. Find (g/f) (4)
d. Find (fg)(15)
e. Find and simplify: g(x+h) - g (x) / h
If you thank me I will... View the full answer
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A complete walk through of AQA GCSE Maths November 2018 Higher Tier – Paper 3 calculator. Help revise for the 8300 new specification 9-1 mathematics exams and your mock. This walkthrough tutorial has full solutions to each question, so you can use it like a mark scheme.
Choose to watch the whole walkthrough, or use the handy timestamps below to jump straight to the questions you want help with:
0:00:00 – Intro
0:01:45 – Q 1 – Translations with Column Vectors
0:02:34 – Q 2 – Metric unit conversion with expressing one number as a fraction of another
0:03:41 – Q 3 – Find the nth term of a linear sequence
0:04:54 – Q 4 – Lowest Common Multiples
0:05:17 – Q 5 – Error Intervals
0:06:23 – Q 6 – Interpreting a Time series graph
0:11:00 – Q 7 – Proofs
0:15:26 – Q 8 – Compound interest
0:17:35 – Q 9 – Perimeter of compound shapes (circle and rectangle)
0:22:37 – Q10 – Solving linear inequalities
0:23:50 – Q11 – Trigonometry in right angled triangles
0:25:28 – Q12 – Plotting cubic equations
0:26:15 – Q13 – Probability and expectation
0:27:38 – Q14 – Mean Average
0:29:50 – Q15 – Surface areas of pyramids and cylinders
0:38:53 – Q16 – Dependant Probability
0:42:18 – Q17 – Indices
0:43:05 – Q18 – Histograms
0:46:52 – Q19 – Limits of Accuracy
0:50:02 – Q20 – Transformation of curves
0:52:55 – Q21 – Speed distance time
1:00:29 – Q22 – Iterative formula
1:06:09 – Q23 – Similar triangles with Pythagoras theorem
1:10:30 – Q24 – Volumes of spheres and cones with ratio
1:17:06 – Q25 – Transformaions – Invariant points after reflection
1:18:29 – Q26 – Function notation – Inverse functions
1:21:52 – Q27 -Coordinate geometry – Intersection of circle and straight line
1:31:45 – Q28 – Properties of a parabola
1:31:52 – Outro
Corrections: Q25 – I circle the wrong answer. It should of course be B & C! I do clearly say this in the video, but I placed the ring in the wong spot when editing the video together. Doh!
#MathsRulesFools #AQA #GCSEMaths
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——————————————————————————————————– | 841 | 2,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.762633 |
http://kids.britannica.com/students/article/algebra/272805/234083-toc | 1,506,404,933,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695066.99/warc/CC-MAIN-20170926051558-20170926071558-00054.warc.gz | 186,143,298 | 9,588 | ## Basic Principles
### Generalizations
Since the sentence
is the capital of Illinois
is a true statement, we may say that there is a city that is the capital of Illinois.
Similarly, since the equation
is a true statement, we may say that there is a number with such a property that when 5 is subtracted from it the result is 3. More briefly, we say that there exists a number x such that x – 5 = 3.… | 100 | 406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-39 | latest | en | 0.943636 |
https://cpt.hitbullseye.com:443/NIIT-Technologies/NIIT-Technologies-Aptitude-Test.php | 1,716,535,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00727.warc.gz | 161,099,678 | 20,374 | # Sample Aptitude Questions of NIIT-Technologies
Views:61535
1. A number X is 150 more than a second number, Y. If the sum of X and Y is 5 times Y, what is the value of Y ?
1. 50
2. 40
3. 80
4. 60
5. 70
X = 150 + Y
X + Y = 5Y
150 + 2Y = 5Y
150 = 3Y => Y = 50
2. A square field has an area of 50625 m2. Find the cost of fencing around it at Rs. 15 per metre, (in Rs.)
1. 12,500
2. 6750
3. 17,500
4. 13500
5. 16250
Area of square field = 50625 => side of the field = = 225
cost of fencing @ Rs. 15/ meter = 4 × 225 × 15 = Rs. 13500
1. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What will be the weight of the new person ?
1. 76
2. 76.5
3. 85
4. 80
5. 90
Increased wt. of 8 persons = 8 × 2.5 = 20 kg
=>wt. of new person = 65 + 20 = 85 kg
2. 50% of a number is 18 less than two-third of that number. Find the number ?
1. 123
2. 115
3. 119
4. 108
5. 101
Let the no. be x
Given : 50x/100 = 2x/3 – 18
½ x – 2/3 x = - 18
=> x = 108
3. Running at the same constant rate, 6 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
1. 648
2. 1800
3. 2700
4. 2000
5. 3080
Let the no. of bottles be ‘B’. Using chain rule:
4. 652.84 + 482.26 + ? = 1200
1. 62.16
2. 54.18
3. 56.1
4. 64.9
5. 66.1
652.84 + 482.26 + x = 1200,=> 1135.1 + x = 1200 => x = 64.9
5. 60% of 250 -? = 75
1. 25
2. 45
3. 60
4. 75
5. 100
=>60% of 250 - x = 75
= 150 – x = 75 => x = 75
6. 681 + ? * 40 = 1161
1. 14
2. 12
3. 24
4. 16
5. 8
681 + x × 40 = 1161
= 681 + 40x = 1161 => 40x = 1161 – 681 = 480
=> x = 480/40 = 12
7. Different words are formed with the help of the letters of word RELATION. Find the number of ways in which vowels always occupy even places.
1. 292
2. 654
3. 356
4. 576
5. 8! | 820 | 1,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.761316 |
http://math.stackexchange.com/questions/275142/sign-of-roots-of-a-quadratic-equation-with-complex-coefficients | 1,469,556,977,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825048.60/warc/CC-MAIN-20160723071025-00013-ip-10-185-27-174.ec2.internal.warc.gz | 159,525,304 | 17,525 | # sign of roots of a quadratic equation with complex coefficients.
Consider $x^2+ax+b=0$, where $x$ is the variable and $a,b$ are complex coefficients. Is there any condition on $a$ and $b$ which makes sure the roots of the equation have negative real parts?
-
A necessary condition is $\Re a>0$, a sufficient condition is $4|b|<(\Re a)^2$. – Hagen von Eitzen Jan 10 '13 at 13:48
@HagenvonEitzen: How did you obtain this inequalities? Here I have a similler question. Can you help me to solve it? Thank you. math.stackexchange.com/questions/985518/… – Nil Mal Oct 22 '14 at 7:44
@Nilan $-a$ is the sum of the two roots, hence $\Re a>0$ is necessary. – Hagen von Eitzen Oct 22 '14 at 10:30
@HagenvonEitzen: I got that. How did you get the sufficient condition? – Nil Mal Oct 23 '14 at 7:46
Let $\,z_1:=a+bi\,,\,z_2:=c+di\in\Bbb C\,$ be the two roots, then using Viete's formulae:
$$ac-bd+(ad+bc)i=z_1z_2=b:=x+yi$$
$$a+c+(b+d)i=z_1+z_2=-a:=-p-qi$$ | 334 | 948 | {"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} | 3.421875 | 3 | CC-MAIN-2016-30 | latest | en | 0.814777 |
http://vustudents.ning.com/group/cs201introductiontoprogramming/forum/topics/cs201-assignment-2-solution-and-discussion-28-may-2018 | 1,534,526,642,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212639.36/warc/CC-MAIN-20180817163057-20180817183057-00281.warc.gz | 463,936,359 | 24,974 | .... How to Find Your Subject Study Group & Join .... .... Find Your Subject Study Group & Join ....
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CS201 Assignment 2 Solution Spring 2018
Watch Video: https://youtu.be/EPETAxae-dY
Assignment Solution >>>>Enjoy dear friendz
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HERE IS COMPLETE SOLUTION THAT I JUST GOT By M IDREES
UNDERSTAND
PRACTICE
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Coding
#include <iostream>
using namespace std;
void computeUserLevel (char *) ;
int main()
{
char Y [4];
for (int k=0; k<=0; k++)
{
cout"arrays and pointers are same : ";
cin>> Y[0];
cout"switch is a loop : ";
cin>> Y[1];
cout"pointers store memory addresses : ";
cin>>Y[2];
}
computeUserLevel(Y);
return 0;
}
void computeUserLevel (char *ptr)
{
if (*ptr == 'f' && *(ptr+1) == 't' && *(ptr+2) == 't')
{
}
else if (*ptr == 'f' && *(ptr+1) == 't' && *(ptr+2) == 'f')
{
}
else
}
Complete CS201 Assignment 2 Solution Spring 2018
CS201 Assignment#02 Solution
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İzmir, Turkey | 788 | 2,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.694718 |
https://deltaclicks.com/how-to-design-a-game/ | 1,679,407,643,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00590.warc.gz | 247,109,863 | 11,908 | # How to Design a Game
A game is an activity involving skill, knowledge or chance, in which the participants follow a set of rules and attempt to win against another person. Some games involve physical activities; others are based on mental exercises such as logic or reasoning.
#### Agamemnon
There is a broad range of games in the world, from simple card and board games to complex computer simulations. Some games are played for pleasure while others are competitive and involve millions of dollars in prize money.
Most games require multiple players. Some, such as chess and checkers, require strategy and intellectual skills; other games are solely or largely based on chance, like dice.
The first step in designing a game is to decide what tools are necessary to make the gameplay effective. These tools can be in the form of pawns, cards, or other objects; they can also be intangible elements such as a point scored.
Many games are designed for particular environments; a game of hide-and-seek in a school building differs from one in a park, and an auto race on a street course can be different from a road race.
In order to determine if an object is a game, Crawford suggests that the following questions must be answered: what are the goals of the plaything?, what is the opponent’s role in the game and how does that impact the game?, and are there any attacks allowed in the game?
If there are no goals associated with the object, it is a toy; if it has goals, it is a challenge. Similarly, if the opposing player cannot outperform you, but can attack your performance, it is a competition.
The final step in defining a game is to define the rules of the game and how the game is played. These rules can include victory and loss conditions, obstacles, rewards and other incentives.
Game theory is a field of mathematics that studies the interactions between people and their decisions. It uses mathematical models to analyze various types of “games.” The most well-known examples are the Prisoner’s Dilemma and the Bell Curve.
These models can explain how and why players will behave in certain situations, what are the best strategies for each player, and how different decisions affect the outcome of the game. They can be applied to a wide variety of settings, including psychology, economics, politics, war, and business.
It is a relatively new science that is still developing. It is used in many fields, but it has only recently gained popularity and widespread use.
A game is a form of play and an integral part of human nature. It can be an activity for diversion or amusement, and it has been around since the beginning of civilization. | 539 | 2,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-14 | longest | en | 0.97093 |
https://www.physicsforums.com/threads/potential-difference-in-capacitor-help.306853/ | 1,527,260,413,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00071.warc.gz | 829,252,765 | 16,063 | # Homework Help: Potential difference in capacitor help
1. Apr 12, 2009
### rayhan619
1. The problem statement, all variables and given/known data
Initially, the switch in the figure is in position A and capacitors C_2 and C_3 are uncharged. Then the switch is flipped to position B. (Figure attached)
a) Afterward, what is the charge on C_1 capacitor?
b) Afterward, what is the potential difference across C_1 capacitor?
c) Afterward, what is the charge on C_2 capacitor?
d) Afterward, what is the potential difference across C_2 capacitor?
e) Afterward, what is the charge on C_3 capacitor?
f) Afterward, what is the potential difference across C_3 capacitor?
2. Relevant equations
V= IR
3. The attempt at a solution
How do I start this problem?
2. Apr 12, 2009
### LowlyPion
Re: Capacitor
This would be a start.
3. Apr 12, 2009
### rayhan619
Re: Capacitor
and here is the figure...
#### Attached Files:
• ###### 1.jpg
File size:
9.5 KB
Views:
338
4. Apr 12, 2009
### LowlyPion
Re: Capacitor
So what do you think is going to happen?
What is the equivalent capacitance of the additional capacitors that are not initially connected?
5. Apr 12, 2009
### rayhan619
Re: Capacitor
we know C = Q/V
V = 100 V
C1 = 15*10^-6 F
so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C
C2 and C3 is in series. so 1/C = (1/C2 + 1/C3) = (1/30*10^-6)+(1/30*10^-6) = 66666.67
so C = 1.5*10^-5
would the Q be same after switching to point b? If yes, then we can figure out V using above equation.
6. Apr 12, 2009
### LowlyPion
Re: Capacitor
When the connection is broken then there is a fixed charge on the C1. When the switch connects to C2 and C3 then the charge will be shared according to what the equivalent capacitance for all 3 of them.
C2 and C3 = 2*C1
C2 and C3 equivalent C = C1
When you put them in || you have then C1 + C1 = 2 C1
Q = V*C
So with the new C = 2*old C and charge the same, then Vnew = 1/2*Vold
7. Apr 12, 2009
### rayhan619
Re: Capacitor
there are like 6 parts of this problem.
so for part a)
C = Q/V
V = 100 V
C1 = 15*10^-6 F
so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C
and how do I get the potential difference?
8. Apr 13, 2009
### rl.bhat
Re: Capacitor
When uncharged capacitors are connected to the charged capacitors the common potential difference is given by V = (C1V1 + o)/(C1 + C2) | 769 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | longest | en | 0.873435 |
http://math.stackexchange.com/questions/121406/set-of-cluster-points-of-a-bounded-sequence | 1,469,563,818,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.22/warc/CC-MAIN-20160723071025-00256-ip-10-185-27-174.ec2.internal.warc.gz | 156,085,537 | 18,848 | # Set of cluster points of a bounded sequence
Suppose that $\{\alpha_k\}$ is a bounded sequence of real numbers satisfying the condition $\displaystyle\lim_{k\rightarrow\infty}|\alpha_k-\alpha_{k+1}|=0$. Let $\displaystyle m = \varliminf_{k\rightarrow\infty}\alpha_k$ and $\displaystyle M = \varlimsup_{k\rightarrow\infty}\alpha_k$. Prove that the cluster point set of the sequence $\{\alpha_k\}$ is the whole segment $[m;M]$.
-
You can see a proof in the following blog post: mathproblems123.wordpress.com/2009/09/09/property-of-a-sequence – Beni Bogosel Mar 17 '12 at 18:11
Please accept some answers to you questions before asking for more. – Beni Bogosel Mar 17 '12 at 18:51
Thank you for your solution. How about the extension of this result in $\mathbb{R}^n$. If $\{x^k\}\subset \mathbb{R}^n$ is bounded and such that $\displaystyle\|x^k-x^{k+1}\|\rightarrow 0$. Could we conclude that the set of accummulation points of $\{x^k\}$ is nonempty, closed and connected? – impartialmale Mar 17 '12 at 21:22
No, we can't. There could be only two accumulation points and the sequence could go back and forth between them along infinitely many different paths. The result in one dimension only comes about because there's only a single path connecting two accumulation points. – joriki Mar 17 '12 at 23:18
Here is an example showing that in higher dimension the cluster point set of an unbounded sequence $(x_n)$ such that $\|x_{n+1}-x_n\|\to0$ may be disconnected, for the reason explained by @joriki (naturally, this cluster point set is still closed).
In words, consider the curves $\gamma_n$, each making a vee joining the points $a=(0,0)$ and $b=(2,0)$ through the bottom point $(1,-2^n)$, and the curve $\gamma$ which concatenates them, joining $a$ to $b$ through $\gamma_{2n}$, then $b$ to $a$ through $\gamma_{2n+1}$, and so on, at speed roughly $1$.
In maths, for each $n\geqslant0$, $\gamma_n:[0,2^n]\to\mathbb R^2$ is defined by $$\gamma_n(t)=(2^{1-n}t,|2^n-2t|-2^n),$$ and the curve $\gamma:[0,+\infty)\to\mathbb R$ is defined by $\gamma(2^{2n}+t)=\gamma_{2n}(t)$ for every $t$ in $[0,2^{2n}]$ and by $\gamma(2^{2n+1}+t)=\gamma_{2n+1}(2^{2n+1}-t)$ for every $t$ in $[0,2^{2n+1}]$.
Now, consider $x_n=\gamma(\sqrt{n})$ for every $n\geqslant0$. Since $\sqrt{n+1}-\sqrt{n}\to0$ and $\|\gamma'\|$ is bounded, $\|x_{n+1}-x_n\|\to0$. Since $\sqrt{n}\to\infty$, the sequence $(x_n)$ passes near $a$ and near $b$ infinitely often (in fact $x_{2^{4n}}$ is exactly $a$ and $x_{2^{4n+2}}$ is exactly $b$, for every $n\geqslant0$).
The set of limit points of $(x_n)$ is $\{a,b\}$, which is not connected. Since the union of the paths $\gamma_n$ is unbounded, $(x_n)$ is unbounded.
-
The fact that the set of cluster points of such sequence is connected for sequences in compact metric spaces (and, consequently, for bounded sequences in $\mathbb R^n$) is shown, for example in a paper by Asic and Adamovic; see here for the exact reference. – Martin Sleziak Jun 7 '13 at 16:14
@Martin Oops. Unsurprisingly, the sequence constructed here is unbounded... I modified my post. Thanks. – Did Jun 7 '13 at 17:13 | 997 | 3,111 | {"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} | 3.59375 | 4 | CC-MAIN-2016-30 | latest | en | 0.850727 |
https://goprep.co/ex-13.b-q14-a-cylindrical-bucket-28-cm-in-diameter-and-72-cm-i-1njpyt | 1,638,067,175,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00101.warc.gz | 382,684,587 | 38,146 | # <span lang="EN-US
Given,
Diameter of cylindrical bucket = 28 cm
Height of bucket = 72 cm
Volume of water in bucket = πr2h =
Length of rectangular tank = 66 cm
Width of tank = 28 cm
Let rise in water level in rectangular tank = h cm
Volume of cylinder = Volume of rectangular tank
= 66 × 28 × h
h =
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The question consRS Aggarwal & V Aggarwal - Mathematics
<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics | 255 | 977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-49 | latest | en | 0.805414 |
http://www.sceneadvisor.com/Virginia/margin-of-error-random-sample.html | 1,561,075,712,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999291.1/warc/CC-MAIN-20190620230326-20190621012326-00283.warc.gz | 290,616,434 | 8,041 | Website Design, Redesign, Maintenance & Support, Hosting, Computer Virus Removal and OS Reloads.
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# margin of error random sample Cana, Virginia
To find the critical value, follow these steps. Bush/Dick Cheney, and 2% would vote for Ralph Nader/Peter Camejo. If we did have some idea about this number , possibly through previous polling data, we would end up with a smaller margin of error.The formula we will use is: E The larger the margin of error, the less confidence one should have that the poll's reported results are close to the true figures; that is, the figures for the whole population.
ISBN 0-87589-546-8 Wonnacott, T.H. Bush/Dick Cheney, and 2% would vote for Ralph Nader/Peter Camejo. This would mean a margin of error of plus or minus 8 percentage points for individual candidates and a margin of error of plus or minus 16 percentage points for the Good as-is Could be even better © 2004 by Raosoft, Inc..
Since we don't know the population standard deviation, we'll express the critical value as a t statistic. Recommended allowance for sampling error of a percentage * In Percentage Points (at 95 in 100 confidence level)** Sample Size 9 n/a 1,000 750 500 250 100 Percentage near 10 2% If you create a sample of this many people and get responses from everyone, you're more likely to get a correct answer than you would from a large sample where only Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.
The amount of precision that can be expected for comparisons between two polls will depend on the details of the specific polls being compared. Some polling organizations, including Pew Research Center, report margins of error for subgroups or make them available upon request. 5What determines the amount of error in survey estimates? But taking into account sampling variability, the margin of error for that 3-point shift is plus or minus 8 percentage points. Mahwah, NJ: Lawrence Erlbaum Associates. ^ Drum, Kevin.
It can be calculated as a multiple of the standard error, with the factor depending of the level of confidence desired; a margin of one standard error gives a 68% confidence If a poll has a margin of error of 2.5 percent, that means that if you ran that poll 100 times -- asking a different sample of people each time -- But polls often report on subgroups, such as young people, white men or Hispanics. It asserts a likelihood (not a certainty) that the result from a sample is close to the number one would get if the whole population had been queried.
What is coverage error? But there are other factors that also affect the variability of estimates. Margin of error = Critical value x Standard deviation of the statistic Margin of error = Critical value x Standard error of the statistic If you know the standard deviation of ISBN0-471-61518-8.
For example, in the accompanying graphic, a hypothetical Poll A shows the Republican candidate with 48% support. If You Loved This Article, You Might Also Love Sample Correctly to Measure True Improvement Levels Eliminating the Fear About Using Confidence Intervals How to Determine Sample Size, Determining Sample Size For example, if the true value is 50 percentage points, and the statistic has a confidence interval radius of 5 percentage points, then we say the margin of error is 5 As an example of the above, a random sample of size 400 will give a margin of error, at a 95% confidence level, of 0.98/20 or 0.049—just under 5%.
How to Compute the Margin of Error The margin of error can be defined by either of the following equations. But they are present nonetheless, and polling consumers should keep them in mind when interpreting survey results. It should be: "These terms simply mean that if the survey were conducted 100 times, the actual percentages of the larger population would be within a certain number of percentage points The estimated percentage plus or minus its margin of error is a confidence interval for the percentage.
This level is the percentage of polls, if repeated with the same design and procedure, whose margin of error around the reported percentage would include the "true" percentage. For tolerance in engineering, see Tolerance (engineering). The math behind it is much like the math behind the standard deviation. That's because pollsters often want to break down their poll results by the gender, age, race or income of the people in the sample.
But, if the sample size is increased from 750 to 1,000, the statistical error drops from 4 to 3%. You can also use a graphing calculator or standard statistical tables (found in the appendix of most introductory statistics texts). Along with the confidence level, the sample design for a survey, and in particular its sample size, determines the magnitude of the margin of error. Your recommended sample size is 377
This is the minimum recommended size of your survey.
Register iSixSigmawww.iSixSigma.comiSixSigmaJobShopiSixSigmaMarketplace Create an iSixSigma Account Login Sample size calculator . For tolerance in engineering, see Tolerance (engineering). You might also enjoy: Sign up There was an error. Anonymous • 1 month ago Mr.
Please try again. The margin of error for the difference between two percentages is larger than the margins of error for each of these percentages, and may even be larger than the maximum margin Anonymous • 1 month ago I find one thing troubling. The margin of error is a measure of how close the results are likely to be. | 1,193 | 5,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-26 | latest | en | 0.866618 |
https://walkccc.me/LeetCode/problems/0298/ | 1,685,608,796,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00506.warc.gz | 674,411,030 | 65,119 | # 298. Binary Tree Longest Consecutive Sequence
• Time: $O(n)$
• Space: $O(h)$
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 class Solution { public: int longestConsecutive(TreeNode* root) { if (root == nullptr) return 0; return dfs(root, root->val, 0, 0); } private: int dfs(TreeNode* root, int target, int length, int maxLength) { if (root == nullptr) return maxLength; if (root->val == target) maxLength = max(maxLength, ++length); else length = 1; return max(dfs(root->left, root->val + 1, length, maxLength), dfs(root->right, root->val + 1, length, maxLength)); } };
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 class Solution { public int longestConsecutive(TreeNode root) { if (root == null) return 0; return dfs(root, -1, 0, 1); } private int dfs(TreeNode root, int target, int length, int max) { if (root == null) return max; if (root.val == target) max = Math.max(max, ++length); else length = 1; return Math.max(dfs(root.left, root.val + 1, length, max), dfs(root.right, root.val + 1, length, max)); } }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 class Solution: def longestConsecutive(self, root: Optional[TreeNode]) -> int: if not root: return 0 def dfs(root: Optional[TreeNode], target: int, length: int, maxLength: int) -> int: if not root: return maxLength if root.val == target: length += 1 maxLength = max(maxLength, length) else: length = 1 return max(dfs(root.left, root.val + 1, length, maxLength), dfs(root.right, root.val + 1, length, maxLength)) return dfs(root, root.val, 0, 0) | 509 | 1,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 3.328125 | 3 | CC-MAIN-2023-23 | latest | en | 0.314714 |
http://www.ck12.org/geometry/Design-Problems-in-Three-Dimensions/lesson/Design-Problems-in-Three-Dimensions/r8/ | 1,454,960,221,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153998.27/warc/CC-MAIN-20160205193913-00183-ip-10-236-182-209.ec2.internal.warc.gz | 335,128,442 | 32,416 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Design Problems in Three Dimensions
## Optimization of surface area and volume.
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Practice Design Problems in Three Dimensions
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Design Problems in Three Dimensions
An ice cream cone is made from a thin wafer cookie that is rolled to make the cone shape. The radius of the ice cream cone is 2.5 centimeters and the slant height is 11.5 centimeters. How many square centimeters of wafer cookie are needed to make each ice cream cone?
#### Guidance
When modeling in three dimensions, you are often interested in maximizing or minimizing some aspect of the situation. Perhaps you want to minimize materials used in order to minimize cost. Perhaps you want to maximize the volume in order to maximize profit.
When working with these types of problems, first create your model. Then, test different possibilities or create an equation that shows the relationship between the variable you are looking to maximize or minimize and another variable in the problem.
To find the maximum or minimum value when you know the equation, you can graph the equation and look for where the graph is at a maximum or a minimum.
Example A
A very thin glass container is being designed to hold a candle (it will cover the candle on all sides except the top). The volume of the candle will be 60 cubic inches. If the candle is in the shape of a cylinder, what's an equation that models the amount of glass needed given the radius of the candle?
Solution: Because the glass is very thin, the amount of glass needed is essentially the surface area of the side and bottom of the cylinder.
• The volume of the cylinder is 60 cubic inches, so . This means that .
• The bottom of the cylinder has a surface area of .
• The lateral face of the cylinder is a rectangle whose dimensions are the height of the cylinder and the circumference of the base. The surface area of the lateral face is .
The overall surface area is therefore . This is the approximate amount of glass needed for a given radius.
Example B
Using the information from Example A, what should the approximate radius and height of the candle be so that the least amount of glass is used for the container?
Solution: Use a calculator or graphing program to make a graph of the function . Look for where there is a minimum in the graph.
The minimum looks to be a radius of approximately 2.6 inches, producing a surface area of approximately . When the radius is 2.6 inches, the height will be .
Example C
A new aquarium is designing a central tank. It will be either a cylinder or a square based prism that is in the center of the building, visible from all floors. If the volume of the tank must be 12,000 cubic feet and the height must be 30 feet, should the tank be a cylinder or a square based prism to minimize the glass needed to build the tank? Note that the top of the tank will be open and the bottom will be concrete.
Solution: The glass needed has to do with the surface area of the tank, not including the top or the bottom (which will not be glass).
For the cylinder, the surface area of the lateral face is . Since the height must be 30 feet, the surface area is . To find the radius, use the fact that the volume must be 12000 cubic feet.
The surface area of the cylinder is
For the square based prism, the surface area is the sum of the rectangular faces. If the side of the square base is , the surface area is . Since the height must be 30 feet, the surface area is . To find the side length, use the fact that the volume must be 12000 cubic feet.
The surface area of the prism is .
The tank should be a cylinder in order to minimize the amount of glass needed.
Concept Problem Revisited
An ice cream cone is made from a thin wafer cookie that is rolled to make the cone shape. The radius of the ice cream cone is 2.5 centimeters and the slant height is 11.5 centimeters. How many square centimeters of wafer cookie are needed to make each ice cream cone?
When unwrapped, the cone is a sector of a circle. The arc length is the circumference of the base of the cone and the radius is the slant height (11.5 cm).
To find what fraction of the circle this is, note that the circumference of the full circle would have been . Therefore, this is of the circle. To find the area needed for the ice cream cone, find the area of the sector.
#### Vocabulary
Modeling in three dimensions is choosing a solid to represent a real life object and analyzing the solid to help answer questions about the real life object.
Optimization is maximizing or minimizing a particular value.
#### Guided Practice
Use the information from the concept problem for these questions.
1. To save money, you decide you could either make the cones skinnier or shorter. Which change will cause the biggest change in the surface area: shortening the radius to 2 centimeters or shortening the slant height to 11 centimeters?
2. Waffle cones have a radius of 4 centimeters and a slant height of 16 centimeters. How much bigger is the surface area of the waffle cone compared to the surface area of the regular ice cream cone?
3. Come up with a function that outputs the area of wafer cookie needed per cone given the radius and slant height of the cone.
1. If the radius is 2 cm, the new surface area will be . If the slant height is 11 cm, the new surface area will be . The biggest change in surface area is caused by shortening the radius.
2. The surface area of the waffle cone is . This surface area is 110.74 square centimeters bigger than a regular ice cream cone.
3. Let and . .
#### Practice
1. Come up with at least 5 rectangles with a perimeter of 24 inches. Which rectangle has the biggest area?
2. If you did not consider a square in #1, compare the area of a square with perimeter 24 inches to the other rectangles that you came up with. What do you notice?
3. A pentagon has a given perimeter. Make a conjecture about what type of pentagon with this perimeter will have the biggest area.
4. A rectangular prism has a given surface area. Make a conjecture about what type of rectangular prism will have the biggest volume.
A new peanut butter company wants to stand out compared with the competition, so decides to design a container that is a truncated pyramid. The top of the container is a square that is 4 inches by 4 inches. The bottom of the container is a square that is 2 inches by 2 inches. The container is 4 inches tall.
5. If the pyramid hadn't been truncated, how tall would it have been?
6. Find the volume of the truncated pyramid.
7. Find the surface area of the truncated pyramid.
8. One cubic inch holds 0.45 ounces of peanut butter. How much peanut butter can this container hold?
9. A typical peanut butter jar is a cylinder with a height of 4 inches and a diameter of 3.9 inches. Compare and contrast this type of jar with the new truncated pyramid container.
An open faced box is being made from a square piece of paper that measures 10 inches by 10 inches. The box will be made by cutting small congruent by sized squares out of each corner.
10. What's an equation that relates to the volume of the box?
11. Graph the equation from #10 and explain what it shows.
12. What size squares should you remove from each corner to maximize the volume?
13. Why does it not make sense to try to minimize the volume of the box?
14. The length of the side of the original piece of square paper is . Come up with an equation that relates , , and the volume of the box.
15. Use your equation to verify that a 15 inch by 15 inch piece of paper with 2 inch by 2 inch square corners removed will produce a box with a volume of . | 1,768 | 7,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2016-07 | latest | en | 0.875455 |
http://slidegur.com/doc/198818/oscilloscope | 1,521,374,586,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645613.9/warc/CC-MAIN-20180318110736-20180318130736-00704.warc.gz | 270,475,964 | 11,304 | Oscilloscope
```Oscilloscope Fundamentals
For Electrical Engineering and Physics Undergraduate Students
Agenda
What is an oscilloscope?
Probing basics (low-frequency model)
Making voltage and timing measurements
Properly scaling waveforms on-screen
Understanding oscilloscope triggering
Oscilloscope theory of operation and performance
specifications
What is an oscilloscope?
os·cil·lo·scope (ə-sĭl'ə-skōp')
Oscilloscopes convert electrical input signals into a visible trace on a screen
- i.e. they convert electricity into light.
Oscilloscopes dynamically graph time-varying electrical signals in two
dimensions (typically voltage vs. time).
Oscilloscopes are used by engineers and technicians to test, verify, and
debug electronic designs.
Oscilloscopes will be the primary instrument that you will use in your
EE/Physics labs to test assigned experiments.
Terms of Endearment (what they are called)
Scope – Most commonly used terminology
DSO – Digital Storage Oscilloscope
Digital Scope
Digitizing Scope
Analog Scope – Older technology oscilloscope, but still around today.
CRO – Cathode Ray Oscilloscope (pronounced “crow”). Even though
most scopes no longer utilize cathode ray tubes to display waveforms,
Aussies and Kiwis still affectionately refer to them as their CROs.
O-Scope
MSO – Mixed Signal Oscilloscope (includes logic analyzer channels of
acquisition)
Probing Basics
Probes are used to transfer the
signal from the device-undertest to the oscilloscope’s BNC
inputs.
There are many different kinds of probes used for
different and special purposes (high frequency
applications, high voltage applications, current, etc.).
The most common type of probe used is called a
“Passive 10:1 Voltage Divider Probe”.
Passive 10:1 Voltage Divider Probe
Passive 10:1 Probe Model
Passive: Includes no active elements such as transistors or amplifiers.
10-to-1: Reduces the amplitude of the signal delivered to the scope’s BNC
input by a factor of 10. Also increases input impedance by 10X.
Note: All measurements must be performed relative to ground!
Low-frequency/DC Model
Passive 10:1 Probe Model
Low-frequency/DC Model: Simplifies to a 9-MΩ resistor in series with the
scope’s 1-MΩ input termination.
Probe Attenuation Factor:
Some scopes such as Agilent’s 3000 X-Series automatically detect 10:1 probes and adjust
all vertical settings and voltage measurements relative to the probe tip.
Some scopes such as Agilent’s 1000 Series require manual entry of a 10:1 probe
attenuation factor.
Dynamic/AC Model: Covered later and during Lab #4.
Understanding the Scope’s Display
Horizontal = 1 µs/div
1 Div
Volts
1 Div
Vertical = 1 V/div
Time
Waveform display area shown with grid lines (or divisions).
Vertical spacing of grid lines relative to Volts/division setting.
Horizontal spacing of grid lines relative to sec/division setting.
Making Measurements – by visual estimation
The most common measurement technique
Ground level (0.0 V)
indicator
Vertical = 1 V/div
V p-p
V max
Horizontal = 1 µs/div
Period
Period (T) = 5 divisions x 1 µs/div = 5 µs, Freq = 1/T = 200 kHz.
V p-p = 6 divisions x 1 V/div = 6 V p-p
V max = +4 divisions x 1 V/div = +4 V, V min = ?
Making Measurements – using cursors
X2 Cursor
X1 Cursor
Y2 Cursor
Y1 Cursor
Manually position A & B cursors to desired measurement points.
Scope automatically multiplies by the vertical and horizontal
scaling factors to provide absolute and delta measurements.
Making Measurements – using the scope’s
automatic parametric measurements
Select automatic parametric measurements with a continuously
Primary Oscilloscope Setup Controls
Horizontal Scaling
(s/div)
Horizontal Position
Trigger Level
Vertical Scaling
(V/div)
Vertical Position
Input BNCs
Agilent’s DSO1000 Series Oscilloscope
Properly Scaling the Waveform
Initial Setup Condition (example)
- Too many cycles displayed.
- Amplitude scaled too low.
Optimum Setup Condition
Trigger Level
Adjust V/div knob until waveform fills most of the screen vertically.
Adjust vertical Position knob until waveform is centered vertically.
Adjust s/div knob until just a few cycles are displayed horizontally.
Adjust Trigger Level knob until level set near middle of waveform vertically.
Setting up the scope’s waveform scaling is an iterative process of making front
panel adjustments until the desired “picture” is displayed on-screen.
Understanding Oscilloscope Triggering
Triggering is often the least understood function of a scope, but is
one of the most important capabilities that you should understand.
Think of oscilloscope “triggering” as
“synchronized picture taking”.
One waveform “picture” consists of many
consecutive digitized samples.
“Picture Taking” must be synchronized to a
unique point on the waveform that repeats.
Most common oscilloscope triggering is based
on synchronizing acquisitions (picture taking)
on a rising or falling edge of a signal at a
specific voltage level.
A photo finish horse race is
analogous to oscilloscope
triggering
Triggering Examples
Trigger level set above waveform
Trigger Point
Trigger Point
Untriggered
(unsynchronized picture taking)
Trigger = Rising edge @ 0.0 V
Negative Time
Positive Time
Trigger = Falling edge @ +2.0 V
Default trigger location (time zero) on DSOs = center-screen (horizontally)
Only trigger location on older analog scopes = left side of screen
Oscilloscope Theory of Operation
Yellow = Channel-specific blocks
Blue = System blocks (supports all channels)
DSO Block Diagram
Oscilloscope Performance Specifications
“Bandwidth” is the most important oscilloscope specification
Oscilloscope “Gaussian” Frequency Response
All oscilloscopes exhibit a low-pass frequency response.
The frequency where an input sine wave is attenuated by 3 dB defines the scope’s
bandwidth.
-3 dB equates to ~ -30% amplitude error (-3 dB = 20 Log
).
Selecting the Right Bandwidth
Input = 100-MHz Digital Clock
Response using a 100-MHz BW scope
Response using a 500-MHz BW scope
Required BW for analog applications: ≥ 3X highest sine wave frequency.
Required BW for digital applications: ≥ 5X highest digital clock rate.
More accurate BW determination based on signal edge speeds (refer to
“Bandwidth” application note listed at end of presentation)
Other Important Oscilloscope Specifications
Sample Rate (in samples/sec) – Should be
≥ 4X BW
Memory Depth – Determines the longest
waveforms that can be captured while still
sampling at the scope’s maximum sample
rate.
Number of Channels – Typically 2 or 4
channels. MSO models add 8 to 32
channels of digital acquisition with 1-bit
resolution (high or low).
Waveform Update Rate – Faster update rates enhance probability of capturing
infrequently occurring circuit problems.
Display Quality – Size, resolution, number of levels of intensity gradation.
Advanced Triggering Modes – Time-qualified pulse widths, Pattern, Video, Serial,
Pulse Violation (edge speed, Setup/Hold time, Runt), etc.
Probing Revisited - Dynamic/AC Probe Model
Passive 10:1 Probe Model
Cscope and Ccable are inherent/parasitic capacitances (not intentionally designed-in)
Ctip and Ccomp are intentionally designed-in to compensate for Cscope and Ccable.
With properly adjusted probe compensation, the dynamic/AC attenuation due to frequencydependant capacitive reactances should match the designed-in resistive voltage-divider
attenuation (10:1).
Where Cparallel is the parallel combination of Ccomp + Ccable + Cscope
Compensating the Probes
Proper Compensation
Channel-1 (yellow) = Over compensated
Channel-2 (green) = Under compensated
Connect Channel-1 and Channel-2 probes to the “Probe Comp” terminal.
Adjust V/div and s/div knobs to display both waveforms on-screen.
Using a small flat-blade screw driver, adjust the variable probe compensation capacitor
(Ccomp) on both probes for a flat (square) response.
The probe and scope input model can be simplified down to a single resistor and
capacitor.
Any instrument (not just scopes) connected to a circuit becomes a part of the
circuit under test and will affect measured results… especially at higher
frequencies.
“Loading” implies the negative affects that the scope/probe may have on the
circuit’s performance.
Assignment
1. Assuming Cscope = 15pF, Ccable = 100pF and Ctip = 15pF, compute Ccomp if
2. Using the computed value of Ccomp, compute CLoad.
3. Using the computed value of CLoad, compute the capacitive reactance of CLoad
at 500 MHz. XC-Load = ______
Using the Oscilloscope Lab Guide and Tutorial
Homework – Read the following sections
before your 1st oscilloscope lab session:
Section 1 – Getting Started
Oscilloscope Probing
Getting Acquainted with the Front Panel
Appendix A – Oscilloscope Block Diagram and
Theory of Operation
Appendix B – Oscilloscope Bandwidth Tutorial
Hands-on Oscilloscope Labs
Section 2 – 7 hands-on labs to learn about the
basics of using an oscilloscope
Oscilloscope Lab Guide and Tutorial
Agilent Technologies
Application Note
Publication #
Evaluating Oscilloscope Fundamentals
5989-8064EN
Evaluating Oscilloscope Bandwidths for your Applications
5989-5733EN
5990-9175EN
Evaluating Oscilloscope Sample Rates vs. Sampling Fidelity
5989-5732EN
Evaluating Oscilloscopes for Best Waveform Update Rates
5989-7885EN
Evaluating Oscilloscopes for Best Display Quality
5989-2003EN
Evaluating Oscilloscope Vertical Noise Characteristics
5989-3020EN
Evaluating Oscilloscopes to Debug Mixed-signal Designs
5989-3702EN
http://cp.literature.agilent.com/litweb/pdf/xxxx-xxxxEN.pdf
Insert pub # in place of “xxxx-xxxx”
Page 25 | 2,393 | 9,605 | {"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} | 2.6875 | 3 | CC-MAIN-2018-13 | latest | en | 0.85482 |
https://mathoverflow.net/questions/150215/who-is-attributed-with-the-conjecture-that-every-multiply-perfect-number-greater | 1,603,227,478,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874135.2/warc/CC-MAIN-20201020192039-20201020222039-00565.warc.gz | 419,767,170 | 29,719 | # Who is attributed with the conjecture that every multiply-perfect number greater than $1$ is even?
I know that Descartes is considered to be the first to ask whether or not odd perfect numbers exist ($n$ such that $\sigma(n)=2n$, where $\sigma(n)$ is the sum of divisors of $n$), and he also discovered several multiply-perfect numbers ($n$ such that $\sigma(n)=kn$ for some integer $k\geq 2$). No odd multiply-perfect numbers greater than one are yet known, and I've heard it called a conjecture that they don't exist several times, but I haven't found an attribution to anyone in particular. Is there an agreed upon origin for the conjecture?
The context of my question (please, feel free to skip this):
I'm looking for someone to attribute in a paper I'm writing that makes a refinement of this form of conjecture for a generalization of the multiply-perfect numbers. Specifically, let $$\dfrac{\sigma(n)}n=\dfrac{k}m$$ where $\gcd(k,m)=1$ and $m$ is a practical number i.e. $m=1$, or $p_i^{a_i+1}\leq 1+\sigma(p_1^{a_1}p_2^{a_2}...p_i^{a_i})$ for every $i\in[1,\omega(m)]$, where $p_1^{a_1}p_2^{a_2}...p_{\omega(m)}^{a_{\omega(m)}}$ is the canonical prime factorization of $m$. I call these pad (practical abundancy denominator) numbers, and conjecture that every such number is practical (i.e. a practical number). $\dfrac{\sigma(n)}n$ is called the abundancy of $n$.
It is easy to see that every even perfect number is practical, and every practical number greater than $1$ is trivially even, so the conjecture that every perfect number is practical is equivalent to the conjecture that they're all even. From this perspective it's natural to ask whether every multiply-perfect number is practical. I've tested essentially every known multiply-perfect number using the factorizations from Achim Flammenkamp's database and all of them so far are practical, so already this seems like a promising lead.
Clearly the multiply-perfect numbers are the special case of the displayed equation where $m=1$, which is trivially practical, so every multiply-perfect number is a pad number. For an idea of their relative size, there are $6484$ pad numbers less than $10^6$, all of them practical, compared with only $10$ multiply-perfect numbers in the same range. An additional fact that makes this conjecture seem quite intuitive is that all small multiples of a practical number are also practical. Specifically, if $a$ is practical and $b\leq \sigma(a)+1$ then $ab$ is practical. Since, $k/m$ is just the abundancy of $n$ in lowest terms, there exists some positive integer $r$ such that $n/m=\sigma(n)/k=r$.
When $r=n$, the multiply perfect numbers, we have no idea why every term would be practical, and yet this is the special case I've tested the furthest, and no counterexamples have yet been found. On the other hand, if $r\leq 1+\sigma(m)$ then $n$ is obviously practical. While in retrospective it seems rather intuitive to me, this was a purely numerical observation. Please let me know if you have ideas about this problem, since I believe it has the potential to remain open for a long time, but someone to attribute is what I'm asking for, since right now I feel I have very few people to attribute in my paper. This is for my first paper, which I think I will try to have published in Experimental Mathematics, since computational experiments were the method by which I reached this generalization. If you have strong opinions about other papers or people who should be attributed then please mention them at least in the comments. I'm not looking to downplay anyone's contributions, and I realize the importance of showing how your work ties in with that of others.
Edit: Note that the same conjecture appears to hold with $\sigma(m)$ replaced by $\sigma^*(m)=\prod_{i=1}^{\omega(m)}(p_i^{a_i}+1)$, the sum of the unitary divisors of $n$, which might be simpler to work with.
The earliest reference seems to be from 1966: E.A. Bugulov, On the question of the existence of odd multiperfect numbers, Kabardino-Balkarskaya State University Učen. Zap. 30 (1966) 9-19. [I could not find this article online.]
Bugulov showed that an odd multiperfect number must have at least 11 distinct prime divisors, more precisely, odd $k$-perfect numbers contain at least $\omega$ distinct prime factors, where $(k,\omega) =$ (3, 11), (4,21), (5, 54),... This result is discussed and improved by Shigeru Nakamura, On k-perfect numbers, Journal of the Tokyo University of Mercantile Marine (Natural Sciences), 33 (1982) 43–50. [Listed here, but not available online.]
• Thank you very much. Are you pretty confident that this was the first reference, of should I keep the question open a little longer in case someone thinks they are aware of an earlier one? Does Bugulov claim the generalized conjecture as his own? I want to make sure I present an accurate history of the problem. – Jaycob Coleman Nov 28 '13 at 22:21
• my source is page 49 of "Unsolved problems in number theory", online here: librarum.org/book/882/65 . I have no earlier reference, but there may well be one, and since I have no access to this obscure paper of Bugulov, I cannot say much more. It seems clear enough that Bugulov spent quite some effort to show that odd multiperfect numbers are rare . That would seem enough to credit him with this conjecture, in the absence of an older source. – Carlo Beenakker Nov 29 '13 at 9:13 | 1,357 | 5,408 | {"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-2020-45 | longest | en | 0.911552 |
https://www.sitepoint.com/boolean-data-type/ | 1,722,898,351,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00357.warc.gz | 784,702,570 | 47,839 | # What Is a Boolean Data Type, and What Are Some Uses?
Learn what the Boolean data type is, and how to use it in programming.
This article details the definition of a Boolean data type and explains its use in programming languages such as Python. There are also examples of Boolean operators that will be useful when understanding Boolean logic and conditional statements.
## What is a Boolean data type?
In computer programs, there are three types of data: text, numbers and Booleans. A Boolean data type is a value that can only be either true or false (these are known as Boolean values).
A true Boolean value might indicate that the object is valid (e.g. an email address has been typed correctly). A false Boolean value indicates that the object is invalid and has not been done correctly (e.g. you’ve forgotten to fill out a required field).
### Boolean values are true and false values
Boolean values have two possible states: true and false. In binary, these are represented by 1 and 0.
Boolean algebra is a type of math that deals with logical operations on logical values, including binary variables. It is the foundation for decisions in programs using logical operators, so it’s important to understand how Booleans work.
### How are Boolean values used in computer programming?
A Boolean value is used to create conditions and control how a program behaves when certain things happen (e.g. if a condition is true, then do something). They can have only two possible values: either 0 or 1. You cannot add or subtract them. A Boolean variable is a special type of memory in a computer that can only store two values: true or false.
### What are the differences between text, numbers and Booleans?
When using Booleans in programming, it’s important to understand the differences between a Boolean variable and other types of data. You need to know how they are stored, and what operations can be performed on them.
#### Booleans
Booleans are usually stored using one byte of memory while text variables use more than one (e.g. two bytes for ASCII, four bytes for Unicode).
#### Text
Some programming languages represent text with an array of characters (e.g. ASCII or Unicode). Text data types have a limited size (typically 256 characters) and do not perform operations on themselves.
#### Numbers
Numbers can be negative, positive, or zero depending on their properties. They are stored as an array of bits which determines if they’re positive (i.e.: 00000000). The same applies to negative numbers (-1, -2, etc).
These types of values can be used in mathematical operations such as addition and subtraction. Booleans cannot be added or subtracted from each other because they only have two states: true and false.
## Truthy and falsy values
There are some special values in programming languages which can be treated as both text and Booleans. These are known as “truthy” or “falsy” values, depending on whether they evaluate to true or false respectively.
For example, 0 is a falsy value because it evaluates to false but “0” is a truthy value as a defined string.
Get to know how various variable states and data types will evaluate to true or false by reading our guide to truthy and falsy in JavaScript.
## Boolean operators
When using Booleans in a program, it’s important to understand the Boolean operators. These are used in conditions and conditional statements which control how the program will behave. Logical operator examples include AND (`&&`), OR (`||`) and NOT (`~`).
### Boolean operator examples
The following are examples of the Boolean operators in programming:
• `>=` – True if a number is greater than or equal to another.
• `<=` – True if a number is less than or equal to another.
• `==` – True if two values are equivalent.
• `!=` – True if two values are not equivalent.
• `&&` – True if both values are true.
• `||` – True if either of the values are true.
• `!` – True if the value is false.
• `~` – Reverses all of the bits in a variable (e.g.: 0000000000000000 becomes 1111111111111111). This can be useful when necessary because it allows you to change Booleans without affecting values or other types of data. It can also be used to revert changes made by a previous statement (e.g.: if (!k) then k = 1).
Boolean operators are used to make decisions in programs and indicate how the program should behave. For example, if p is true AND q is also true, then do something.
### Boolean use-case example
Boolean values are used in conditional tests as discussed below.
• Checks that the email address is valid.
• Checks that the password is at least 6 characters long.
• Checks that both fields are filled out correctly.
### What is an example of a Boolean data type in practice?
The following is an example of a Boolean in C++ code (note that the type of variable is not represented and can be stored using any variable data type):
This function returns true if two numbers add up to zero, otherwise false.
`bool NumberCheck(int x, int y) { return x+y == 0; };`
Boolean data types can also be used as operands in relational (e.g.: “is greater than”) and equality (e.g.: “equals” or “does not equal”) operators to perform such tasks as comparing directory contents and checking file sizes from a website, as shown in JavaScript below:
``````if (document.getElementById("files").value) {
if (parseInt(document.getElementById("files").value) > 1048576) {
alert("You have selected a file larger than 1MB.");
} else {
alert("You have selected a file smaller or equal to 1MB.");
}
} else {
}``````
## History & Origins of the Boolean Data Type
Booleans are named after George Boole, who was a mathematician from the 19th century. He first developed Boolean algebra in 1854.
The Boolean data type was invented in the early 1800s. George Boole created a system of logic that could be used to describe the true values (i.e.: 1) and false values (i.e.: 0) in computers. This developed into what we now call Boolean algebra, which is used as a basis for most computer programming languages.
In 1854, George Boole wrote “An investigation into the Laws of Thought on Which are Founded the Mathematical Theories of Logic and Probabilities”, which laid the mathematical foundation for many computer languages.
Booleans were first used commercially in 1951 by a Bell Labs engineer as a part of AN/FSQ-7, the massive computer that was used by the United States Air Force.
## FAQs
### What are Boolean variables in programming?
The Boolean data type is used to store the logical values true and false. This data type may be used to store information that allows one of two states, on or off, to be stored.
### What is a nullable data type?
A nullable type is used to represent values that may be set to the null (no value) state.
### What is a null value?
null is the data type used to represent objects (e.g., variables, records, etc.) which do not currently exist, are unfinished/unused, or are being discarded so that resources will be freed. It could also be used when a Boolean expression evaluates to nothing and needs to be given a value. It returns false.
What is a non-zero value?
A non-zero value is a value that is not zero. It returns true.
### Can a Boolean data type have the value of null?
No, it cannot, but you can use an undefined Boolean value to represent a null value. If checked against an operator, it returns false.
### What is a Boolean data type in a database?
Boolean data types can be used to store the values true and false in a database.
Booleans are most commonly used in databases to represent yes/no, on/off, or other related states. For example, if an account has been turned off then its status column may contain false. If it’s currently on then true would be stored in the account status Boolean. Another common use is for flags which allow you to store information about what happened (e.g.: ’email sent’, ‘file downloaded’).
Booleans may also be useful to show if a file exists, or if an action has been performed (e.g.: form submitted).
### What are some programming languages that support Boolean data types?
Booleans are available in most programming languages. If you’re using JavaScript, Java, PHP, Python, C, C++ or Swift then you should have access to the Boolean data type.
If null is allowed in your programming language, then both true and false would be allowed as well.
### What are some applications that manipulate Boolean data types?
Booleans can be used by application software to manage flags, yes/no options, on/off states and other related information. They may also be useful in keeping track of resources or performing various tasks (e.g.: deleting files).
## Summary
In this blog post you learned what Boolean values are and how they differ from other types of data. You also learned why it’s important to understand how Booleans work, and how to use Boolean operators in programming.
Joel Falconer
View Author
Joel Falconer is a technical content strategist. He has been managing editor at SitePoint, AppStorm, DesignCrowd, and Envato, and features editor at The Next Web.
booleandata typesfalsytruthy | 1,989 | 9,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.886212 |
http://www.andrew.cmu.edu/course/24-ansys/htm/s4_truss.htm | 1,545,036,706,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00053.warc.gz | 328,566,272 | 7,736 | Carnegie Mellon
Mechanical Engineering
Self-paced learning on the Web
FEM/ANSYS
## S4 3D Truss Structure
Structural #4: Analysis of a 3-D truss structure
Introduction: In this example you will learn to use the 3-D Truss element in ANSYS.
Physical Problem: Analysis of the 3D truss structure shown in the figure below.
Problem Description:
The tower is made up of trusses. You may recall that a truss is a structural element that experiences loading only in the axial direction.
Units: Use S.I. units ONLY
Geometry: the cross sections of each of the truss members is 1.56e-3 sq meter.
Material: Assume the structure is made of aluminum with modulus of elasticity E=75 GPa.
Boundary conditions: The structure is constrained in the X, Y and Z directions at the bottom three corners.
Loading: The tower is loaded at the top tip. The load is in the YZ plane and makes an angle of 75 with the negative Y axis direction. The load value is 2500 N.
Objective:
To determine deflection at each joint. To determine stress in each member. To determine reaction forces at the base. Give three examples where similar 3D trusses are used in practice. Model one of them (with reasonable assumptions of dimensions, material properties and loading) using ANSYS. You don't have to solve it. You can do so to check whether your assumptions were reasonable!!
You are required to hand in print outs for the above.
Figure:
IMPORTANT: Convert all dimensions and forces into SI units.
STARTING ANSYS
Click on ANSYS 6.1 in the programs menu. Select Interactive. The following menu that comes up. Enter the working directory. All your files will be stored in this directory. Also enter 64 for Total Workspace and 32 for Database. Give your file an appropriate job name. Click on Run.
MODELING THE STRUCTURE
Go to ANSYS Utility Menu. Click on Workplane>Change Active CS to..>Global Cartesian. Go to the ANSYS Main Menu.
Click Preprocessor>Modeling>Create>Keypoints>In active CS The following window comes up
Fill in the keypoint number (1,2,3...) and the coordinates. Make sure you get the correct coordinates from the figure. Create all the 10 keypoints. Make sure the numbering of your keypoints matches the numbering of the joints in the figure.
If you cannot see the grid then go to Utility Menu>Display Working Plane
If you cannot see the complete figure then go to Utility Menu>PlotCntrls>Pan Zoom Rotate and zoom out to see the entire figure.
Now create lines connecting the keypoints
Click on Preprocessor>Modeling>Create>Lines>Lines>In Active Coord. Pick the endpoints of each element to create the lines. Rotate the figure for more accessible views.
You can use the Utility Menu>PlotCtrls>Pan Zoom Rotate window to rotate the model and see its 3D nature.
MATERIAL PROPERTIES
Go to the ANSYS Main Menu Click Preprocessor>Material Props>Material Models. In the window that comes up which is shown below, for Material Model 1, choose Structural>Linear>Elastic>Isotropic.
Double click Isotropic for Material Model 1.
Fill in 7.5e10 for the Young's modulus and 0.3 for minor Poisson's Ratio. Click OK Now the material 1 has the properties defined in the above table. We will use this material for the elements of the structure.
ELEMENT PROPERTIES:
SELECTING ELEMENT TYPE:
Click Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types' window that opens click on Add... The following window opens.
Type 1 in the Element type reference number. Click on Structural Link and select 3D spar. Click OK. Close the 'Element types' window. So now we have selected Element type 1 to be a structural Link- 3D spar (cable) element. The trusses will be modeled as elements of type 1, i.e. structural link element. This finishes the selection of element type. Now we need to define the cross sectional area for this element. Go to Preprocessor>Real Constants. In the "Real Constants" dialog box that comes up click on Add In the "Element Type for Real Constants" that comes up click OK. The following window comes up.
Type 1.56e-3 for cross sectional area and click on OK. We have now defined the cross sectional area of the link element.
MESHING:
DIVIDING THE TOWER INTO ELEMENTS: Go to Preprocessor>Meshing>Size Controls>Manual Size>Lines>All Lines. In the menu that comes up type 1 in the field for 'Number of element divisions'.
Click on OK. Now go to Preprocessor>Meshing>Mesh>Lines. Select all the lines and click on OK in the "Mesh Lines" dialog box. Now each line is a truss element (Element 1).
BOUNDARY CONDITIONS AND CONSTRAINTS
APPLYING BOUNDARY CONDITIONS
The tower is constrained in the X, Y and Z directions at the four bottom corners. Go to Main Menu Click on Preprocessor>Loads>Define Loads>Apply>Structural>Displacement>On Keypoints Select the keypoint on which you want to apply displacement constraints. The following window comes up.
Select UX, UY, UZ and click OK.
APPLYING FORCES
First find the components of the force along the Y and Z directions Go to Main Menu Click on Preprocessor>Loads>Define Loads>Apply>Structural>Forces/Moment>On Nodes. Select the top node. Click on OK in the 'Apply F/M on Nodes' window. The following window will appear. Enter the value of the Z-component of the force. Repeat the procedure to apply the Y-component of force.
Now the Modeling of the problem is done
SOLUTION
Go to ANSYS Main Menu>Solution>Analysis Type>New Analysis. Select static and click on OK. Go to Solution>Solve>Current LS. Wait for ANSYS to solve the problem. Click on OK and close the 'Information' window
POST-PROCESSING
Listing the results Go to ANSYS Main Menu Click on General Postprocessing>List Results>Nodal Solution. The following window will come up:
Select DOF solution and All U's. Click on OK. The nodal displacements will be listed as follows.
Similarly you can list the stresses for each element by clicking General Postprocessing>List Results>Element Solution. Now select LineElem Results.
MODIFICATIONS:
You can also plot the displacements and stress. Go to General Postprocessing>Plot Results>Contour Plot>Element Solution. The following window will come up.
Select a stress to be plotted and click OK. The output will be like this. | 1,430 | 6,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-51 | latest | en | 0.881414 |
https://artofproblemsolving.com/wiki/index.php/2020_CIME_I_Problems/Problem_4 | 1,701,415,730,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00182.warc.gz | 136,729,671 | 10,838 | # 2020 CIME I Problems/Problem 4
## Problem 4
There exists a unique positive real number $x$ satisfying $$x=\sqrt{x^2+\frac{1}{x^2}} - \sqrt{x^2-\frac{1}{x^2}}.$$ Given that $x$ can be written in the form $x=2^\frac{m}{n} \cdot 3^\frac{-p}{q}$ for integers $m, n, p, q$ with $\gcd(m, n) = \gcd(p, q) = 1$, find $m+n+p+q$.
## Solution
We simply use the best technique of easy bash.
$x^2 = 2x^2 - 2\sqrt{x^4-\frac{1}{x^4}}$
$x^4 = 4x^4-\frac{4}{x^4}$
$x^8 = \frac{4}{3}$
$x = 2^{\frac{1}{4}}3^\frac{-1}{8}$
The answer is then $14$. | 232 | 538 | {"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": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-50 | latest | en | 0.492059 |
https://www.jiskha.com/questions/543326/can-someone-tell-me-how-to-expand-the-bionomial-d-2-6 | 1,558,786,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258003.30/warc/CC-MAIN-20190525104725-20190525130725-00197.warc.gz | 823,745,818 | 5,633 | # algebra 2
can someone tell me how to expand the bionomial (d-2)^6?
1. 👍 0
2. 👎 0
3. 👁 42
1. (d-2)^6
is the same as (d-2)*(d-2)*.... six times.
You can expand it by multiplication, or you can use the binomial expansion formula, where
(a+b)^n
=a^n + na^(n-1)b + n(n-1)/2!a^(n-2)b^2 +
+ ...b^n
where the general term i is
n(n-1)(n-2)..(n-i)/i! * a^(n-i)b^i
Expanding along these lines, we get for
n=1, (a+b)^1 = a+b
n=2, (a+b)^2 = a^2+2ab+b^2
n=3, (a+b)^3 = a^3+3a^2b+3ab^2+b^3
...
with the coefficients being
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 21 15 6 1
....
Each coefficient is the sum of the one immediately above plus the one to the left.
This pattern is called the pascal's triangle.
http://en.wikipedia.org/wiki/Binomial_theorem
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2. 👎 0
posted by MathMate
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asked by Claire on October 22, 2008 | 900 | 2,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-22 | latest | en | 0.895985 |
https://www.answerspile.com/ExpertAnswers/ensure-you-show-all-formulas-and-state-values-of-variables-substitute-and-solve-provide-a-concludi-pa572 | 1,713,642,351,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817674.12/warc/CC-MAIN-20240420184033-20240420214033-00576.warc.gz | 603,338,307 | 5,725 | Home / Expert Answers / Statistics and Probability / ensure-you-show-all-formulas-and-state-values-of-variables-substitute-and-solve-provide-a-concludi-pa572
# (Solved): Ensure you show all formulas and state values of variables, substitute and solve. Provide a concludi ...
Ensure you show all formulas and state values of variables, substitute and solve. Provide a concluding statement for all parts. Suppose that 31% of families in town own iPads. If sixteen families are surveyed at random, Determine the probability that exactly 4 families own an IPAD. (3 marks) Determine the probability that less than 3 families own iPads. (3 marks) Determine the probability that least 3 families own iPads. (3 marks) Determine the expected number of families with an iPad. (2 marks) There are ten cats and eleven dogs in a pet shop. Nine pets are chosen at random to visit a children’s hospital. Determine the probability that exactly three of the pets will be dogs. (3 marks) Determine the probability that less than 2 of the pets will be dogs. (3 marks) Determine the probability that at least 2 of the pets will be dogs. (3 marks) Determine the expected number of dogs chosen. (2 marks) Determine the probability that less than 3 families own iPads. (3 marks) Determine the probability that least 3 families own iPads. (3 marks) Determine the expected number of families with an iPad. (2 marks)
We have an Answer from Expert | 316 | 1,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-18 | longest | en | 0.888978 |
https://www.polytechforum.com/electrical/how-convert-three-phase-power-to-single-phase-15035-.htm | 1,702,266,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103464.86/warc/CC-MAIN-20231211013452-20231211043452-00166.warc.gz | 1,008,321,559 | 12,283 | # How convert three-phase power to single-phase?
• posted
Hi,
We're just moving into some new office space, and among the things that we inherited was some three-phase power. I haven't actually seen what the outlets look like, because the person who knows where it is isn't in this week, but I was kind of curious if there's any way that we can utilize any of that? We're a typical office environment, and all of our equipment (workstations, servers, etc.) are single-phase.
To go from the three-phase power to single-phase, is it simply a matter of creating three single-phase outputs from the one three-phase (i.e., A+neutral, B+neutral, and C+neutral), and then we can just plug (independent) equipment into each of the 3 single-phase outputs?
Also, if we do that, from a power standpoint, do we end up with each of the single-phase outputs just being 1/3 of the power rating for the original three-phase power?
As often happens, I'm probably asking what might seem like dumb/naive questions, but I'm really glad to have found this NG :)!
Jim
• posted
277-480, 120-208, 120-240, 377-600 these are all pretty common in north America.
• posted
| We're just moving into some new office space, and among the things that we | inherited was some three-phase power. I haven't actually seen what the | outlets look like, because the person who knows where it is isn't in this | week, but I was kind of curious if there's any way that we can utilize any | of that? We're a typical office environment, and all of our equipment | (workstations, servers, etc.) are single-phase.
Larger UPS systems use three-phase power to keep the batteries charged and power the DC to AC conversion. Some can be had that then produce only single-phase, but most of them produce three phase back out. But that's not a big deal as they can all be had with 120 volts.
| To go from the three-phase power to single-phase, is it simply a matter of | creating three single-phase outputs from the one three-phase (i.e., | A+neutral, B+neutral, and C+neutral), and then we can just plug | (independent) equipment into each of the 3 single-phase outputs?
It depends on what kind of three phase power you have. If you have what is known as "208Y/120" then what you describe gets you 120 volts. But if you have "240 delta center tapped", then A+N and C+N gets you 120 volts but B+N gets you 208 volts. With 208Y/120, A+B or B+C or C+A gets you
208 volts, while with 240 delta, A+B or B+C or C+A gets you 240 volts.
You could have another three phase system or voltage, in which case you likely have a transformer or three changing the voltage so you can get something at 120 volts. The power coming in from the electric company could be in "480Y/277", "600Y/347" (primarily in Canada), "480 delta", or some others. You would have multiple fuse boxes or breaker panels to support the high and low voltage circuits in these cases.
If you have no transformers, you're probably getting one of the first two I described. In some locations you might be getting "216Y/125" or "220Y/127".
| Also, if we do that, from a power standpoint, do we end up with each of the | single-phase outputs just being 1/3 of the power rating for the original | three-phase power?
The way you just described, yes. A phase-to-phase connection, which gets a different voltage, could have a higher capacity, since it is using two of the phases and thus gets a higehr voltage with no reduction in current (unless something else uses up some of the capacity).
If you have 208Y/120, you will most likely have circuits distributed over the three phases in a circuit breaker panel designed to interleave the three phases at every 3rd row. A 1-pole breaker gets 120 volts at the phase it is plugged in to. A 2-pole breaker gets 208 volts (this would have been 240 volts if it were single phase), but can be used with 120 volt loads. A 3-pole breaker gets the full three phases.
Whatever the amperage rating of a breaker is, you get that much power at each of the poles/wires coming from it, however many that is. So if you have a 20 amp 1-pole breaker, it can provide 2400 watts to 120 volt loads. If you have a 20 amp 2-pole breaker, it can provide 4800 watts total to
2 sets of 120 volt loads, or 4156 watts to 208 volt loads. A 20-amp 3-pole breaker would let you have a total of 7200 watts for 3 sets of 120 volt loads. 3 sets of 208 volt loads could use up to 4156 watts each as long as the total of 2 does not exceed 4800 watts and the total of all 3 does not exceed 7200 watts. And of course it can supply a real three phase load up to 7200 watts. Electrical codes generally require planned loads to not exceed 80% of capacity, so derate the above figures to determine what you could plan for.
There are PDUs (power distribution units, or power strips) for computer rack cabinets that plug into a three phase receptacle and provide three sets of 120 volt outlets. If everything is rated 20 amps, you'd have that 20 amps three times (16 amps maximum planned usage on each set). You might want to check what you can, or are, doing with a UPS, before jumping into how power is fed to your computers.
| As often happens, I'm probably asking what might seem like dumb/naive | questions, but I'm really glad to have found this NG :)!
You're well ahead of most. The first big step is knowing you needed to ask.
Try to find out what kind of three phase power it is, and the voltages, so we can focus on your real options. Also, if you can, find out what type(s) and size(s) of circuit breaker panel(s) you have.
One important thing to remember is that in a situation outside of an owner-occupied single family house, virtually every jurisdiction does require using the services of a licensed electrician to do any work. People here can give you hints, advice, general direction, and explain your options. But wiring the building is not a do-it-yourself project; not even changing out a receptacle or light switch.
• posted
Phil and SQLit,
I'll have to get more details that you mentioned from someone else who isn't at the office today. Will post back next week sometime.
And Phil, it's not my intention to do any of the actual wiring or whatever, I'm mainly trying to gain an understanding so I can be a little more knowledgeable about what our options might be.
Thanks, and have a great weekend!
Jim
• posted
Depending on your jurisdiction (I don't think you said what country you're in, which sometimes makes a big difference) this could well be illegal. It may happen on construction sites with special splitter boards, but is probably not at all appropriate for an office environment. Systems of supply and protective arrangements (fuses, CBs etc) that may be safe and legal in one environment may not be, in another.
The voltages and the way phase, neutral and earth wiring is arranged is also important. Your best solution is to call an electrician to check this out with certainty. Only then will you be sure that what you want to do is both safe and legal in your particular region.
1/3 of the "power" rating yes, but exactly the same "current" rating. Which of these were you meaning?
Newsgroups are not a safe place to be asking these sorts of questions unless you are already very knowledgeable about electrical distribution and safety.
This is good advice, but you could still end up getting the wrong advice because the information you provide may be incomplete or even wrong, despite your own best efforts. Your comment about asking "someone who isn't at the office today" gives me considerable concern.
If you get the wrong advice because the advisors don't have the full or accurate picture, it could cost you a whole lot more than just getting in an electrician to look at it in the first place. Any advice you get on here can only be based on what you tell people here, which is likely to be incomplete, and may be inaccurate if the people that you're talking to don't know as much as you thought they did.
Electrical safety is too important to leave to chance. Sorry if this sounds negative, but for this sort of thing you need an electrician not a Newsgroup.
My thoughts fwiw.
Moving into a new office is always a great experience, starting from scratch is a great way of getting things organised in ways that you probably couldn't justify changing in an existing office. Good luck with your project. Hope it goes really well for you.
• posted
|> snipped-for-privacy@ipal.net wrote: |> > | To go from the three-phase power to single-phase, is it simply a | matter of |> > | creating three single-phase outputs from the one three-phase (i.e., |> > | A+neutral, B+neutral, and C+neutral), and then we can just plug |> > | (independent) equipment into each of the 3 single-phase outputs? | | Depending on your jurisdiction (I don't think you said what country you're | in, which sometimes makes a big difference) this could well be illegal. It | may happen on construction sites with special splitter boards, but is | probably not at all appropriate for an office environment. Systems of | supply and protective arrangements (fuses, CBs etc) that may be safe and | legal in one environment may not be, in another.
Are you referring to things like rack PDUs?
| Newsgroups are not a safe place to be asking these sorts of questions unless | you are already very knowledgeable about electrical distribution and safety.
In which case he would have no questions to ask.
|> > One important thing to remember is that in a situation outside of an |> > owner-occupied single family house, virtually every jurisdiction does |> > require using the services of a licensed electrician to do any work. |> > People here can give you hints, advice, general direction, and explain |> > your options. But wiring the building is not a do-it-yourself project; |> > not even changing out a receptacle or light switch. | | This is good advice, but you could still end up getting the wrong advice | because the information you provide may be incomplete or even wrong, despite | your own best efforts. Your comment about asking "someone who isn't at the | office today" gives me considerable concern.
The advice here should be on the order of things like what you might be able to do that no one told you about otherwise because they didn't think of it or didn't know about it.
For example, you can probably find as many electricians who will tell you that 240 volt appliances on 208 volts is fine as you can who will tell you that you will have problems with that. It's good to get a 2nd opinion (which in the case of getting it here, would be the 1st opinion).
Of course, never trust anyone here. It's like medical advice. You can find _someone_ that will tell you HOW to do surgery on yourself. There are crackpots in all fields (especially computers).
|> And Phil, it's not my intention to do any of the actual wiring or |> whatever, I'm mainly trying to gain an understanding so I can be a |> little more knowledgeable about what our options might be. | | If you get the wrong advice because the advisors don't have the full or | accurate picture, it could cost you a whole lot more than just getting in an | electrician to look at it in the first place. Any advice you get on here | can only be based on what you tell people here, which is likely to be | incomplete, and may be inaccurate if the people that you're talking to don't | know as much as you thought they did. | | Electrical safety is too important to leave to chance. Sorry if this sounds | negative, but for this sort of thing you need an electrician not a | Newsgroup.
You need an electrician, period. But electricians, despite being licensed, have their faults, too. The best of the advisors on the newsgroup will exceed the median of electricians. What that means is, if the electrician does anything you don't understand, ask for an explanation both from the electrician as well as here. Whenever there is a conflict, get it resolved. I've seen and/or read of cases where licensed electricians did things totally stupid and created very hazardous conditions. One I read about was the guy who put in a three phase panel for 240 delta center tapped, and filled it up with single pole breakers for all the branch circuits (there were no three phase loads). So some of the receptacles or lights had 208 volts. Then there was the electrician who installed grounded outlets in a house and wired all the grounds to the box and neutral because the existing wiring did not have a separate grounding wire. I'm sure many others have some horror stories in this regard.
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners. | 2,903 | 12,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.976644 |
https://direct.physicsclassroom.com/Concept-Builders/Forces-in-2D/Inclined-Plane-Analysis/Teacher-Notes | 1,719,216,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00892.warc.gz | 183,988,358 | 29,871 | ### Notes:
The Solve It! - Inclined Planes Concept Builder is an adjustable-size file that displays nicely on smart phones, on tablets such as the iPad, on Chromebooks, and on laptops and desktops. The size of the Interactive can be scaled to fit the device that it is displayed on. The compatibility with smart phones, iPads, other tablets, and Chromebooks make it a perfect tool for use in a 1:1 classroom.
### Teaching Ideas and Suggestions:
While not all Physics courses discuss inclined plane problems, those that do will appreciate this Concept Builder. This Concept Builder has three variations of inclined plane problems. Each physical situation involves the acceleration of an object along an inclined plane. Students must analyze the situation, determining the parallel- and perpendicular-components of the gravity force (mass is stated), the normal force, the friction force (in Master and Wizard Levels), the net force, and the acceleration. They must also identify that quantity associated with the given information. The numerical values provided to students will vary from session to session and from student to student as they are randomly generated.
This Concept Builder consists of three difficulty levels. Each level includes two situtations that must be analyzed. A free-body diagram and a short description is provided. Two or more numerical values (for example, mass, incline angle, and friction information) are provided. Students must identify the values of up to 9 quantities (7 in Apprentice level and 8 in Master level). A few of the values are stated in the provided description. The others must be calculated from an understanding of the relationships implicit to the situation. It is not necessary to start with the Apprentice Level or to finish with the Wizard Level. The activity was designed such that classrooms jump in and jump out at the locations that are most appropriate to them. We recommend that teachers view the Questions or do the Concept Builder in order to make judgements about what works best for their specific classes.
The three difficulty levels are differentiated as follows:
• Apprentice Difficulty Level Questions 1-2: Analyze a problem involving a force applied at an angle to the horizontal to cause a horizontal acceleration. No friction.
• Master DifficultyLevel Questions 3-4: Analyze a problem involving a force applied at an angle to the horizontal to cause a horizontal acceleration. Friction value is given.
• Wizard Difficulty Level Questions 5-6: Analyze a problem involving a force applied at an angle to the horizontal to cause a horizontal acceleration. Coefficient of friction is given.
In order to complete a difficulty level, a student must successfully analyze both situations of that level. A successful analysis involves determining all the correct information requested - like the parallel- and perpendicular-components of the gravity force, the normal force, the friction force, the net force, and the acceleration. A student can enter values one blank at a time and use the Check Answers button to check. Feedback will be immediate and correct response cannot be changed thereafter. Incorrect responses can be corrected until they are correct. There is no penalty for misses or difficulty (other than the extra time and frustration that can be associated with it). When both analyses are correct, the student is returned to the Main Menu and (if logged in as a Task Tracker user) the Trophy for that difficulty level is displayed.
The most valuable (and most overlooked) aspect of this concept-building activity is the Help Me! feature. Each situation is accompanied by a Help page that discusses in detail the specifics of the situation. Formulas, tips, suggestions, etc. are provided. This Help feature transforms the activity from a question-answering activity into a concept-building activity. The student who takes the time to use the Help pages can be transformed from a guesser to a learner and from an unsure student to a confident student. The "meat and potatoes" of the Help pages are in the sections titled "How to Think About This Situation:" Students need to be encouraged by teachers to use the Help Me! button and to read this section of the page. A student that takes time to reflect upon how they are answering the question and how an expert would think about the situation can transform their naivete into expertise.
### Related Resources
There are numerous resources at The Physics Classroom website that serve as very complementary supports for the Solve It! (with Vectors and Fnet=m•a) Concept Builder. These include:
• Minds On Physics Internet Modules:
The Minds On Physics Internet Modules include a collection of interactive questioning modules that help learners assess their understanding of physics concepts and solidify those understandings by answering questions that require higher-order thinking. Assignments from the Forces in Two Dimensions module make for a great complement to this Concept Builder. They are best used in the middle to later stages of the learning cycle. We recommend mission F2D2 as an accompaniment to this activity. Visit the Minds On Physics Internet Modules.
Users may find that the App version of Minds On Physics works best on their devices. The App Version can be found at the Minds On Physics the App section of our website. The Forces in Two Dimensions module can be found on Part 2 of the six-part App series. Visit Minds On Physics the App.
Additional resources and ideas for incorporating Solve It! (with Vectors and Fnet=m•a) into an instructional unit on Newton's Laws can be found at the Teacher Toolkits section of The Physics Classroom website. Visit Teacher Toolkits. | 1,122 | 5,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-26 | latest | en | 0.920931 |
https://www.burned-calories.com/sport/nordic-walking | 1,563,728,378,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00466.warc.gz | 634,690,001 | 6,114 |
## Calories Burned with Nordic Walking
How many calories do you burn by Nordic Walking? A man weighing 180 Pounds burns approximately 502 kilocalories (kcals) per hour by Nordic Walking. A 170-Pound woman burns about 446 kcals per hour.
Calculate your own calorie usage below or use the table at the bottom of the page for an idea of how many calories you burn by Nordic Walking, depending on your weight and duration of exercise.
### Burned Calories (kcals)
#### How do we calculate the amount of calories burned with Nordic Walking?
This calculation uses the MET value (Metabolic Equivalent of Task) of Nordic Walking. The MET value of Nordic Walking = 5,9 . We multiply the MET value with the person's body weight in kilogram. Thereafter we multiply this with 0.0175 and the time in minutes.
#### Example
A person weighs : 175 pounds
Excercise : Nordic Walking
MET value of Nordic Walking : 5,9
Time : 30 minutes
How many calories are burned during these 30 minutes?
(175/2.2) * 5,9 * 0.0175 * 30 minutes = 251 kcal
Questions or suggestions about burning calories with Nordic Walking? Let us know!
#### Table Calories Burned - Nordic Walking
Use the below table to quickly find out how many calories are burned by Nordic Walking when you are a certain weight and exercise for a certain length of time. For a precise calculation, enter the required information further up the page.
Minutes 120 lbs 140 lbs 160 lbs 180 lbs 200 lbs 220 lbs
15 Minutes 84 kcals 98 kcals 112 kcals 125 kcals 141 kcals 155 kcals
30 Minutes 167 kcals 195 kcals 223 kcals 251 kcals 282 kcals 310 kcals
45 Minutes 251 kcals 293 kcals 335 kcals 376 kcals 423 kcals 465 kcals
60 Minutes 335 kcals 390 kcals 446 kcals 502 kcals 564 kcals 620 kcals
90 Minutes 502 kcals 585 kcals 669 kcals 753 kcals 846 kcals 929 kcals
## Calorie Calculators
Copyright © 2019 / contact : i n f o @ burned-calories.com | 498 | 1,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-30 | latest | en | 0.834684 |
https://ccssmathanswers.com/factors-affecting-interest/ | 1,709,374,639,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00801.warc.gz | 155,774,136 | 56,354 | # Factors Affecting Interest | Main Factors that Influence or Determine the Interest Rates
Factors Affecting Interest: Interest depends on the rate of interest, time, and principal. So, before knowing the factors affecting interest let us know about the interest, principal, rate of interest, and period of time. There are some formulas to calculate the interest. We have provided step-by-step explanations for all the questions in the coming sections. Thus all the students of 5th grade are advised to scroll down this page to know more information on factors affecting interest.
Refer Articles:
### What is meant by Interest?
Interest is nothing but the extra amount of money you get after depositing the money. Interest is a payment from a borrower to a depositor of an amount above repayment of the amount that is borrowed at a particular rate.
### Factors Affecting interest
There are two types of interest. They are Simple Interest and Compound Interest. The three important factors that affect interest are
1. Principal
2. Rate of interest
3. Period of time
Principal: It is the amount of money borrowed from the lender.
Period of time: It is meant by the time or number of years for which the money is debited.
Rate of interest: The interest that is calculated at a rate percent which is finalized by the bank every year.
### Formula for Simple Interest
The simple interest depends on the rate of interest, principal, and period of time. S.I remains the same every year. Here is the formula to calculate the Simple Interest (S.I)
S.I = P × T × R/100
Where,
P is Principal
T is a period of time
R is the rate of interest
Principal + Interest = Amount
P + I = A
### Formula for Compound Interest
Compound Interest means the interest that is earned on prior interest added to the principal. Due to compounding, the total amount that is borrowed grows exponentially. The interest is calculated on a daily, weekly, monthly, or yearly basis.
A = P(1 + r/n)nt
Where,
A = Amount
P = Principal
n = number of times interest is compounded per unit ‘t’
t = time
r = Rate of interest
Also, Refer:
### Factors Affecting Interest Rate Examples
Example 1.
Find the interest for 25000 at 7 percentage per annum for 4 years ?
Solution:
Given,
Principal = 25000
Rate = 7 per annum
Time = 4 years
Interest = P × T × R/100
P = principal
R = rate
T = time
Interest = principal × time × rate/100
= 25000 × 4 × 7/100
=7000
Thus the interest is $7000. Example 2. Find the interest to be paid at the end of the year on$700 at 3 percent per annum?
Solution:
Given that,
Principal = $700 Rate = 3 per annum Time = 1 years Interest = P × T × R/100 P = principal R = rate T = time Interest = principal × time × rate/100 = 700 × 3 × 1/100 = 21 Thus the interest is$21.
Example 3.
Find the interest to be paid at the end of the year on $8000 at 6 1/2 percentage per annum ? Solution: Given that, Principal =$8000
Rate = 6 1/2 per annum
Time = 1 years
Interest = P × T × R/100
P = principal
R = rate
T = time
Interest = principal × time × rate/100
= $8000 × 1 × 6 1/2/100 =$8000 × 1 × 6.5/100
= $520 Therefore the interest is$520.
Example 4.
Find the interest for 12370 at 2 percentage per annum for 4 years ?
Solution:
Given,
Principal = $12370 Rate = 2 per annum Time = 4 years Interest = P × T × R/100 P = principal R = rate T = time Interest = principal × time × rate/100$12370 × 4 × 2/100
= $989.6 Thus the interest is$989.6
Example 5.
Find the interest for Rs 2000 at 4% per annum for 6 years ?
Solution:
Given,
Principal = 2000
Rate of interest = 4 per annum
Time = 6 years
Interest = P × T × R/100
P = principal
R = rate
T = time
Interest = principal × time × rate/100
= Rs. 2000 × 6 × 4/100
= Rs. 480
Thus the interest is Rs. 480
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Scroll to Top | 1,039 | 3,796 | {"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} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.952246 |
https://www.naalaa.com/forum/post-364.html | 1,726,241,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00360.warc.gz | 833,622,466 | 11,702 | How to create a board game aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 02-11-2024, 09:09 PM I was thinking about how I could create a game similar to the game of checkers but I have no idea how to start. Does anyone know how to create this type of game or something similar to have a base and be able to create more games of that style. I put a link to a video in case I was not understood correctly. https://www.youtube.com/watch?v=r-7R2sCW3Ro Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 02-13-2024, 06:56 PM I implemented Connect Four (https://en.m.wikipedia.org/wiki/Connect_Four) and Sudoku in some early version of naalaa, but that's about it. The biggest difficulty with most of these games is creating a good computer opponent (well, not in Sudoku). aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 02-13-2024, 07:54 PM My question is not very specific, I start from scratch with something simpler. How do I move the pieces diagonally on a board? johnno56 Senior Member Posts: 304 Threads: 31 Joined: Nov 2023 Reputation: 3 02-13-2024, 09:12 PM My first guess... Imagine the player is at coordinates 0,0 on an x,y grid. Move x to the left: negative. Right: positive Move y up: negative. Down: positive. up-left would be x-1,y-1 up-right would be x+1,y-1 down-left would be x-1,y+1 down-right would be x+1,y+1 I hope this helps... Logic is the beginning of wisdom. aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 02-14-2024, 04:27 PM (02-13-2024, 09:12 PM)johnno56 Wrote: My first guess... Imagine the player is at coordinates 0,0 on an x,y grid. Move x to the left: negative. Right: positive Move y up: negative. Down: positive. up-left would be x-1,y-1 up-right would be x+1,y-1 down-left would be x-1,y+1 down-right would be x+1,y+1 I hope this helps... It might work yes, thanks johnno56 johnno56 Senior Member Posts: 304 Threads: 31 Joined: Nov 2023 Reputation: 3 02-14-2024, 07:10 PM We are both learning on this particular topic... I have never "created" a board game (programmed from scratch) before. But, since you have raised the subject, I have been researching board games and how they are coded. I can see how this could be quite addictive! Making games like Checkers and Chess etc. have always been on my "to do" list, but have never really summoned up enough courage, to actually make a commitment... lol I am hoping this will change... I had a weird idea about board design and player movement... N7 has a "TileMap Editor" that could be used to both create the board and move the player(s)... Not exactly sure how to do that, but I imagine that it could be done... Could save a LOT of work... more research!! Cool... Yes. Quite addictive... Logic is the beginning of wisdom. aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 02-14-2024, 07:49 PM (02-14-2024, 07:10 PM)johnno56 Wrote: We are both learning on this particular topic... I have never "created" a board game (programmed from scratch) before. But, since you have raised the subject, I have been researching board games and how they are coded. I can see how this could be quite addictive! Making games like Checkers and Chess etc. have always been on my "to do" list, but have never really summoned up enough courage, to actually make a commitment... lol I am hoping this will change... I had a weird idea about board design and player movement... N7 has a "TileMap Editor" that could be used to both create the board and move the player(s)... Not exactly sure how to do that, but I imagine that it could be done... Could save a LOT of work... more research!! Cool... Yes. Quite addictive... Right now I'm trying to make a platform game, I have about half of the first scene. I just need to touch up the collisions, add enemies and the camera, I do everything manually without using the editor, which is something I haven't tried yet. I like doing it this way better. I will 100% get into the board game when I finish the platform game, my idea is to make something similar to the checkers game and then with that base I can create other games with similar gameplay. I am subscribed to a YouTube channel where there are many abstract board games with a base similar to checkers and since these games are not available for PC, it seemed like a good idea to design them myself. I don't show a link because the channel is in Spanish. johnno56 Senior Member Posts: 304 Threads: 31 Joined: Nov 2023 Reputation: 3 02-14-2024, 09:57 PM Platformers? The TileMap Editor is ideal for creating platformers but I am not that familiar with it. I still use the old school method of creating a text-based "map". The game "reads" the map and "translates" each text character to the desired "tile" image. That's the fun part... the "real" work is coding movement, collisions, scores etc... I tried using this method to create a clone of Mario brothers, which include "camera" movement, but that was a long time ago using sdlbasic. Not even sure if I still have it... (the game only had Mario moving across the game area - no enemies or collectables etc) - no promises... I will see if I can find it and try to reproduce it using N7 - assuming I can of course... lol Looking forward to seeing your game. Gotta love platformers... Logic is the beginning of wisdom. aliensoldier Member Posts: 85 Threads: 16 Joined: Nov 2023 Reputation: 2 02-23-2024, 04:34 PM I'm going to share the code I've made because I'm at an impasse and I don't know how to continue. I have three problems: 1-there is a cursor that moves around the board and if you stand on top of a player's piece and press "z" it picks it up and you can drop it on another square with the "x" key, but the problem is that it drops it anywhere box and I want it to only be able to be dropped in the dark boxes and I don't know how to do it. 2-when I grab a player token you can grab other player tokens at the same time and I only want it to grab one token at a time but I haven't been able to do it. 3-When I have a player token I should only be able to move one square per turn but it can be moved to any square, as could be done. Attached Files damas.zip (Size: 467.34 KB / Downloads: 6) Marcus Administrator Posts: 316 Threads: 39 Joined: Nov 2023 Reputation: 3 02-23-2024, 09:06 PM (This post was last modified: 02-23-2024, 11:16 PM by Marcus.) (02-23-2024, 04:34 PM)aliensoldier Wrote: I'm going to share the code I've made because I'm at an impasse and I don't know how to continue. I have three problems: 1-there is a cursor that moves around the board and if you stand on top of a player's piece and press "z" it picks it up and you can drop it on another square with the "x" key, but the problem is that it drops it anywhere box and I want it to only be able to be dropped in the dark boxes and I don't know how to do it. 2-when I grab a player token you can grab other player tokens at the same time and I only want it to grab one token at a time but I haven't been able to do it. 3-When I have a player token I should only be able to move one square per turn but it can be moved to any square, as could be done. I made two posts but deleted them now, after looking up the rules of Checkers. Hopefully this code can help you: Code:```constant GRID = 8 ' number of rows and columns. constant SQUARE = 53 ' size of each square. set window "test", 640, 480 set redraw off ' create 8x8 array filled with zeroes. board = fill(0, GRID, GRID) ' place pieces. for y = 0 to 2 for x = 0 to GRID - 1 if (y + x)%2 = 1 board[x][y] = 1 if (GRID - 1 - y + x)%2 = 1 board[x][GRID - 1 - y] = 2 next ' center board in window. boardX = (width(primary) - GRID*SQUARE)/2 boardY = (height(primary) - GRID*SQUARE)/2 ' set to 1 or 2 when a piece has been picked up. holdingPiece = 0 ' original position of picked up piece. originalPieceX = 0 originalPieceY = 0 ' loop until user presses escape. while not keydown(KEY_ESCAPE, true) ' check if mouse is inside board. mx = mousex() my = mousey() gridX = mx - boardX gridY = my - boardY if gridX >= 0 and gridX < GRID*SQUARE and gridY >= 0 and gridY < GRID*SQUARE ' convert to grid coordinates. gridX = int(gridX/SQUARE) gridY = int(gridY/SQUARE) ' click? if mousebutton(0, true) ' not holding a piece? if not holdingPiece ' any piece at that position? if board[gridX][gridY] ' store information about the piece. holdingPiece = board[gridX][gridY] originalPieceX = gridX originalPieceY = gridY ' remove piece from board. board[gridX][gridY] = 0 endif ' already holding piece. else ' spot empty? if not board[gridX][gridY] ' same position or valid distance for move? here both players can move up and ' down. only "kings" should be able to do that. dx = |gridX - originalPieceX| dy = |gridY - originalPieceY| if dx = 0 and dy = 0 or dx = 1 and dy = 1 ' put piece down. board[gridX][gridY] = holdingPiece holdingPiece = 0 endif endif endif endif endif ' draw. set color 0, 0, 0 cls ' draw board. for y = 0 to GRID - 1 for x = 0 to GRID - 1 ' this formula gives a chess pattern. if (y + x)%2 set color 64, 64, 64 else set color 208, 208, 208 draw rect boardX + x*SQUARE, boardY + y*SQUARE, SQUARE, SQUARE, true ' any piece? if board[x][y] DrawPiece(board[x][y], boardX + (x + 0.5)*SQUARE, boardY + (y + 0.5)*SQUARE) endif next next ' Draw picked up piece. if holdingPiece DrawPiece(holdingPiece, mx, my) redraw fwait 60 wend function DrawPiece(player, x, y) if player = 1 set color 240, 240, 240 else set color 32, 32, 32 draw ellipse x, y, SQUARE/2 - 4, SQUARE/2 - 4, true set color 0, 0, 0 draw ellipse x, y, SQUARE/2 - 4, SQUARE/2 - 4, false endfunc``` As it is now, both players can always move backwards or forwards diagonally. But according to the rules, only "kings" can move backwards. board[x][y] is now either: 0 = empty, 1 = player 1 piece or 2 = player 2 piece. I guess you can simply add two more values: 3 = player 1 king piece and 4 = player 2 king piece, and make it so that only kings can move backwards. Edit And I'm sorry for not basing my code on yours. I'm just showing you how I would do it myself, and hopefully it can help you modify your own code « Next Oldest | Next Newest » | 2,905 | 10,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-38 | latest | en | 0.92193 |
http://mathhelpforum.com/calculus/86977-differential-trig-problem.html | 1,480,868,768,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00347-ip-10-31-129-80.ec2.internal.warc.gz | 173,475,848 | 9,802 | 1. differential trig problem
HI, can someone please shed some light on this for me. I spent last night on it without luck. Thank you
show that
d/dx(ln(sqrt(2)-cos x)/(sqrt(2)+cos x) = (2sqrt(2) sin x)/(1+sin^2 x)
2. Carried out step by step, we derive
$f'(x) = \frac{d}{dx}\left(\ln\,\frac{\sqrt{2}-\cos\,x}{\sqrt{2}+\cos\,x}\right)$
$= \frac{d}{dx}(\ln\,(\sqrt{2}-\cos\,x)-\ln\,(\sqrt{2}+\cos\,x))$
$= \frac{d}{dx}(\ln\,(\sqrt{2}-\cos\,x))-\frac{d}{dx}(\ln\,(\sqrt{2}+\cos\,x))$
$= \frac{1}{\sqrt{2}-\cos\,x}\cdot \sin\,x-\frac{1}{\sqrt{2}+\cos\,x}\cdot(-\sin\,x)$
$= \frac{\sin\,x}{\sqrt{2}-\cos\,x}+\frac{\sin\,x}{\sqrt{2}+\cos\,x}$
$= \frac{\sin\,x(\sqrt{2}+\cos\,x)+\sin\,x(\sqrt{2}-\cos\,x)}{2-\cos^2 x}$
$= \frac{\sqrt{2}\sin\,x+\sin\,x\,\cos\,x+\sqrt{2}\si n\,x-\sin\,x\,\cos\,x}{1+1-\cos^2 x}$
$=\frac{2\sqrt{2}\sin\,x}{1+\sin^2 x}.$
3. thank heaps for the effort you put into that. I really appreciate it. | 444 | 920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-50 | longest | en | 0.499189 |
https://www.transtutors.com/questions/in-problem-p7-7-we-demonstrated-how-networks-can-be-trained-using-the-hebb-rule-when-2011102.htm | 1,618,056,885,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056869.3/warc/CC-MAIN-20210410105831-20210410135831-00415.warc.gz | 1,158,089,490 | 15,625 | # In Problem P7.7 we demonstrated how networks can be trained using the Hebb rule when the... 1 answer below »
In Problem P7.7 we demonstrated how networks can be trained using the Hebb rule when the prototype vectors are given in binary (as opposed to bipolar) form. Repeat Exercise E7.1 using the binary representation for the prototype vectors. Show that the response of this binary network is equivalent to the response of the original bipolar network.
Exercise E7.1
Consider the prototype patterns given to the left.
i. Are p1 and P2 orthogonal?
ii. Use the Hebb rule to design an autoassociator network for these patterns.
iii. Test the operation of the network using the test input pattern p1 shown to the left. Does the network perform as you expected? Explain
Problem P7.7
In all of our pattern recognition examples thus far, we have represented patterns as vectors by using “1” and “-1” to represent dark and light pixels (picture elements), respectively. What if we were to use “1” and “0” instead? How should the Hebb rule be changed?
bhavani k
Neural networks can be trained by Hebb rule. It can be done in...
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# GMAT - Quantitative
## Latest Stories About GMAT - Quantitative
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## GMAT Tip: Prime Factors
Today’s GMAT challenge question comes from Kaplan. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution: Good luck on this Challenge Problem involving Prime Factors! Remember, there’s probably a reason that the questions are... Read more »
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## GMAT Tip: Understanding Remainders
Today’s GMAT tip comes from Veritas Prep. In this article, they provide helpful advice on understanding the concept of remainders. Read on to see what they have to say! Unless you teach math, chances are that you haven’t used a remainder in years. Remainders in division are basically placeholders for... Read more »
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## GMAT Tips: Strategic Guessing on the GMAT
Today’s GMAT tip comes from Kaplan. In this article, Kaplan GMAT instructor Bret Ruber provides advice on how to guess strategically on quantitative problems: Almost every GMAT quantitative problem can be solved in more than one way. Some of the various approaches to GMAT math questions include picking numbers, backsolving... Read more »
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## GMAT Tip: Breaking Down a GMATPrep Consecutive Integer Problem
Today’s GMAT tip comes from test prep firm ManhattanGMAT. In this article, they share helpful tips on how to solve consecutive integer problems. Read on to see what they have to say! This week, we’re going to talk about what to know for consecutive integer problems and how to recognize... Read more »
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## GMAT Tip: Solve This Stats Problem, Stat!
Today’s GMAT challenge question comes from Veritas Prep. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution: Set A consists of integers -9, 8, 3, 10, and J; Set B consists of integers -2,... Read more »
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## GMAT Tip: Exponent Manipulation – Tough questions, basic approaches
Today’s GMAT tip comes from Kaplan. In this article, Kaplan GMAT instructor Bret Ruber explains how to tackle tricky quantitative problems involving exponents: Exponents questions are common among advanced quantitative problems and generally fall into two categories, both of which involve a variable in the exponent. First are problems that... Read more »
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## GMAT Tip: Divisibility Mother Lode
Today’s GMAT challenge question comes from ManhattanGMAT. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution: Problem If m is the square of integer n and m is divisible by 98, m must also... Read more »
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## GMAT Tip: GMAT Math Strategy – Think with your pen
Today’s GMAT Tip comes from our friends at Knewton. In this post, they provide helpful hints on how to tackle particularly complex math problems. Read on to see what they have to say! Here is a rather challenging GMAT math problem. Give it a shot: For every positive EVEN integer... Read more »
Join the Clear Admit community for free and conduct unlimited searches of MBA LiveWire, MBA DecisionWire, MBA ApplyWire and the Interview Archive. Register now and you’ll also get 10% off your entire first order. | 797 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.924966 |
https://math.stackexchange.com/questions/117702/orthogonal-projections-with-sum-p-i-i-proving-that-i-ne-j-rightarrow-p | 1,643,158,939,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00152.warc.gz | 428,308,147 | 35,780 | # Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow P_{j}P_{i}=0$
I am reading Introduction to Quantum Computing by Kaye, Laflamme, and Mosca. As an exercise, they write
"Prove that if the operators $P_{i}$ satisfy $P_{i}^{*}=P_{i}$ and $P_{i}^{2}=P_{i}$ , then $P_{i}P_{j}=0$ for $i\ne j$.''
In the context of this problem, it has been assumed that $I=\sum_{i=1}^{n} P_{i}$, where I suppose that $n$ could be infinite. I have shown that this is true in the trivial case $n=2$, but the general case has been eluding me. How should I attack this?
For each $j$, $$P_j=P_jIP_j=P_j\left(\sum_{k=1}^n P_k\right)P_j=\sum_{k=1}^nP_jP_kP_j=P_j+\sum_{k\neq j}P_jP_kP_j,$$ so $\sum\limits_{k\neq j}P_jP_kP_j=0$. For each $i\neq j$, $P_jP_iP_j=(P_iP_j)^*P_iP_j$ is a positive operator, and a sum of positive operators is positive, so $-P_jP_iP_j=\sum\limits_{k\neq i,j}P_jP_kP_j$ is also positive. This is only possible if $P_jP_iP_j=0$. Since $\|P_iP_j\|^2=\|(P_iP_j)^*P_iP_j\|=\|P_jP_iP_j\|$, it follows that $P_iP_j=0$.
Because of the properties you state, $\|P_{j}x\|^{2}=(x,P_{j}x)=(P_{j}x,x)$. Therefore, $$\|x\|^{2} = (\sum_{j}P_{j}x,x)= \sum_{j}\|P_{j}x\|^{2}.$$ Apply this identity to $x=P_{k}y$, and use the fact that $P_{k}^{2}=P_{k}$: $$\|P_{k}y\|^{2} = \sum_{j\ne k}\|P_{j}P_{k}y\|^{2}+\|P_{k}y\|^{2}.$$ The only way this can happen is $P_{j}P_{k}y=0$ for all $j \ne k$.
For all $$i,j$$ you have $$P_i+P_j\le\sum_k P_k\le I$$, hence $$P_i\le I-P_j$$.
Multiplying by $$P_j$$ on left and right on LHS and RHS you then get $$P_j P_i P_j\le P_j(I-P_j)P_j=0,$$ hence $$P_j P_i P_j=0$$, which implies $$P_j P_i P_j=(P_j P_i)(P_j P_i)^\dagger=0$$ and thus $$P_i P_j=P_j P_i=0$$.
You can also prove the other direction: if $$P_i P_j=0$$ for all $$i\neq j$$ then $$\sum_k P_k$$ is a projector, as $$\left(\sum_k P_k\right)^2=\sum_k P_k + \sum_{i
Here is a slight variant of Jonas’ argument.
Assume that $p_{1},\ldots,p_{n}$ are projection elements of a unital $C^{*}$-algebra $A$, where $n \in \Bbb{N}_{\geq 2}$, such that $$\sum_{k = 1}^{n} p_{k} = 1_{A}.$$ Choose distinct $i,j \in [n]$, where $[n] \stackrel{\text{df}}{=} \Bbb{N}_{\leq n}$. Then \begin{align} p_{i} & = p_{i} 1_{A} \\ & = p_{i} \sum_{k \in [n]} p_{k} \\ & = \sum_{k \in [n]} p_{i} p_{k} \\ & = p_{i}^{2} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} \\ & = p_{i} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k}. \end{align} It follows that $$p_{i} p_{j} p_{j}^{*} = p_{i} p_{j} = - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} = - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*},$$ and consequently, $$(\spadesuit) \qquad (p_{i} p_{j}) (p_{i} p_{j})^{*} = p_{i} p_{j} p_{j}^{*} p_{i}^{*} = - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*} p_{i}^{*} = - \sum_{k \in [n] \setminus \{ i,j \}} (p_{i} p_{k}) (p_{i} p_{k})^{*}.$$ On the extreme left of $(\spadesuit)$, we have a positive element, while on the extreme right of $(\spadesuit)$, we have a negative element. This can only mean that both extremes are zero, so $(p_{i} p_{j}) (p_{i} p_{j})^{*} = 0_{A}$. Hence, $$\| p_{i} p_{j} \|_{A}^{2} = \| (p_{i} p_{j}) (p_{i} p_{j})^{*} \|_{A} = \| 0_{A} \|_{A} = 0,$$ or equivalently, $p_{i} p_{j} = 0_{A}$. | 1,460 | 3,265 | {"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": 12, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.732573 |
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Article
Wang X., Pardalos P. M. Journal of Global Optimization. 2015. P. 1-18.
Transportation discrete network design problem (DNDP) is about how to modify an existing network of roads and highways in order to improve its total system travel time, and the candidate road building or expansion plan can only be added as a whole. DNDP can be formulated into a bi-level problem with binary variables. An active set algorithm has been proposed to solve the bi-level discrete network design problem, while it made an assumption that the capacity increase and construction cost of each road are based on the number of lanes. This paper considers a more general case when the capacity increase and construction cost are specified for each candidate plan. This paper also uses numerical methods instead of solvers to solve each step, so it provides a more direct understanding and control of the algorithm and running procedure. By analyzing the differences and getting corresponding solving methods, a modified active set algorithm is proposed in the paper. In the implementation of the algorithm and the validation, we use binary numeral system and ternary numeral system to avoid too many layers of loop and save storage space. Numerical experiments show the correctness and efficiency of the proposed modified active set algorithm.
Added: Dec 24, 2015
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Yuri Evtushenko, Mikhail Posypkin, Turkin A. et al. Journal of Global Optimization. 2018. Vol. 71. No. 1. P. 129-145.
In this paper we propose a method for solving systems of nonlinear inequalities with predefined accuracy based on nonuniform covering concept formerly adopted for global optimization. The method generates inner and outer approximations of the solution set. We describe the general concept and three ways of numerical implementation of the method. The first one is applicable only in a few cases when a minimum and a maximum of the constraints convolution function can be found analytically. The second implementation uses a global optimization method to find extrema of the constraints convolution function numerically. The third one is based on extrema approximation with Lipschitz under- and overestimations. We obtain theoretical bounds on the complexity and the accuracy of the generated approximations as well as compare proposed approaches theoretically and experimentally.
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Chistyakov V., Goldengorin B. I., Pardalos P. M. Journal of Global Optimization. 2012. Vol. 53. No. 3. P. 475-495.
The currently adopted notion of a tolerance in combinatorial optimization is defined referring to an arbitrarily chosen optimal solution, i.e., locally. In this paper we introduce global tolerances with respect to the set of all optimal solutions, and show that the assumption of nonembededdness of the set of feasible solutions in the provided relations between the extremal values of upper and lower global tolerances can be relaxed. The equality between globally and locally defined tolerances provides a new criterion for the multiplicity (uniqueness) of the set of optimal solutions to the problem under consideration.
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Borrero J., Gillen C., Prokopyev O. Journal of Global Optimization. 2017. Vol. 69. No. 1. P. 255-282.
We consider a class of nonlinear integer optimization problems commonly known as fractional 0–1 programming problems (also, often referred to as hyperbolic 0–1 programming problems), where the objective is to optimize the sum of ratios of affine functions subject to a set of linear constraints. Such problems arise in diverse applications across different fields, and have been the subject of study in a number of papers during the past few decades. In this survey we overview the literature on fractional 0–1 programs including their applications, related computational complexity issues and solution methods including exact, approximation and heuristic algorithms.
Added: Feb 9, 2017
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Guan X., He X., Pardalos P. M. et al. Journal of Global Optimization. 2017. P. 1-12.
The inverse max ++ sum spanning tree (MSST) problem is considered by modifying the sum-cost vector under the Hamming Distance. On an undirected network G(VEwc), a weight w(e) and a cost c(e) are prescribed for each edge e∈Ee∈E. The MSST problem is to find a spanning tree T∗T∗ which makes the combined weight maxe∈Tw(e)+∑e∈Tc(e)maxe∈Tw(e)+∑e∈Tc(e) as small as possible. It can be solved in O(mlogn)O(mlogn) time, where m:=|E|m:=|E| and n:=|V|n:=|V|. Whereas, in an inverse MSST problem, a given spanning tree T0T0 of G is not an optimal MSST. The sum-cost vector c is to be modified to c¯c¯ so that T0T0 becomes an optimal MSST of the new network G(V,E,w,c¯)G(V,E,w,c¯) and the cost ∥c¯−c∥‖c¯−c‖ can be minimized under Hamming Distance. First, we present a mathematical model for the inverse MSST problem and a method to check the feasibility. Then, under the weighted bottleneck-type Hamming distance, we design a binary search algorithm whose time complexity is O(mlog2n)O(mlog2n). Next, under the unit sum-type Hamming distance, which is also called l0l0 norm, we show that the inverse MSST problem (denoted by IMSST00) is NP−NP−hard. Assuming NP⊈DTIME(mpolylogm)NP⊈DTIME(mpolylogm), the problem IMSST00 is not approximable within a factor of 2log1−εm2log1−εm, for any ε>0ε>0. Finally, We consider the augmented problem of IMSST00 (denoted by AIMSST00), whose objective function is to multiply the l0l0 norm ∥β∥0‖β‖0 by a sufficiently large number M plus the l1l1 norm ∥β∥1‖β‖1. We show that the augmented problem and the l1l1 norm problem have the same Lagrange dual problems. Therefore, the l1l1 norm problem is the best convex relaxation (in terms of Lagrangian duality) of the augmented problem AIMSST00, which has the same optimal solution as that of the inverse problem IMSST00.
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Du H., Pardalos P. M., Wu W. et al. Journal of Global Optimization. 2013. Vol. 56. No. 2. P. 559-568.
A sensor with two active phases means that active mode has two phases, the full-active phase and the semi-active phase, which require different energy consumptions. A full-active sensor can sense data packets, transmit, receive, and relay the data packets. A semi-active sensor cannot sense data packets, but it can transmit, receive, and relay data packets. Given a set of targets and a set of sensors with two active phases, find a sleep/active schedule of sensors to maximize the time period during which active sensors form a connected coverage set. In this paper, this problem is showed to have polynomial-time (7.875+ε) -approximations for any ε>0 when all targets and sensors lie in the Euclidean plane and all sensors have the same sensing radius R s and the same communication radius R c with R c ≥ 2R s .
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Aleskerov F. T., Subochev A. Journal of Global Optimization. 2013. Vol. 56. No. 2. P. 737-756.
Various Condorcet consistent social choice functions based on majority rule (tournament solutions) are considered in the general case, when ties are allowed: the core, the weak and strong top cycle sets, versions of the uncovered and minimal weakly stable sets, the uncaptured set, the untrapped set, classes of k-stable alternatives and k-stable sets. The main focus of the paper is to construct a unified matrix-vector representation of a tournament solution in order to get a convenient algorithm for its calculation. New versions of some solutions are also proposed.
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In data envelopment analysis, methods for constructing sections of the frontier have been recently proposed to visualize the production possibility set. The aim of this paper is to develop, prove and test the methods for the visualization of production possibility sets using parallel computations. In this paper, a general scheme of the algorithms for constructing sections (visualization) of production possibility set is proposed. In fact, the algorithm breaks the original large-scale problems into parallel threads, working independently, then the piecewise solution is combined into a global solution. An algorithm for constructing a generalized production function is described in detail.
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Chirkov A. Y., Gribanov D., Malyshev D. et al. Journal of Global Optimization. 2019. Vol. 73. No. 4. P. 761-788.
In this paper, we consider the class of quasiconvex functions and its proper subclass of conic functions. The integer minimization problem of these functions is considered, assuming that the optimized function is defined by the comparison oracle. We will show that there is no a polynomial algorithm on log R to optimize quasiconvex functions in the ball of radius R using only the comparison oracle. On the other hand, if the optimized function is conic, then we show that there is a polynomial on log R algorithm (the dimension is fixed). We also present an exponential on the dimension lower bound for the oracle complexity of the conic function integer optimization problem. Additionally, we give examples of known problems that can be polynomially reduced to the minimization problem of functions in our classes.
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Žilinskas J., Goldengorin B. I., Pardalos P. M. Journal of Global Optimization. 2015. Vol. 61. No. 1. P. 91-108.
The earliest approaches to the cell formation problem in group technology, dealing with a binary machine-part incidence matrix, were aimed only at minimizing the number of intercell moves (exceptional elements in the block-diagonalized matrix). Later on this goal was extended to simultaneous minimization of the numbers of exceptions and voids, and minimization of intercell moves and within-cell load variation, respectively. In this paper we design the first exact branch-and-bound algorithm to create a Pareto-optimal front for the bi-criterion cell formation problem.
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Goldengorin B. I., Tijssen G. A., Ghosh D. et al. Journal of Global Optimization. 2003. Vol. 25. No. 4. P. 377-406.
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Evgeny Maslov, Mikhail Batsyn, Panos M. Pardalos. Journal of Global Optimization. 2014. Vol. 59. No. 1. P. 1-21.
In this paper we consider two branch and bound algorithms for the maximum clique problem which demonstrate the best performance on DIMACS instances among the existing methods. These algorithms are MCS algorithm by Tomita et al. (2010) and MAXSAT algorithm by Li and Quan (2010a, b). We suggest a general approach which allows us to speed up considerably these branch and bound algorithms on hard instances. The idea is to apply a powerful heuristic for obtaining an initial solution of high quality. This solution is then used to prune branches in the main branch and bound algorithm. For this purpose we apply ILS heuristic by Andrade et al. (2012). The best results are obtained for p_hat1000-3 instance and gen instances with up to 11,000 times speedup.
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Goldengorin B. I., Ghosh D. Journal of Global Optimization. 2005. Vol. 32. No. 1. P. 65-82.
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Turkensteen M., Malyshev D., Goldengorin B. I. et al. Journal of Global Optimization. 2017. Vol. 68. No. 3. P. 601-622.
The tolerance of an element of a combinatorial optimization problem with respect to its optimal solution is the maximum change of the cost of the element while preserving the optimality of the given optimal solution and keeping all other input data unchanged. Tolerances play an important role in the design of exact and approximation algorithms, but the computation of tolerances requires additional computational time. In this paper, we concentrate on combinatorial optimization problems for which the computation of all tolerances and an optimal solution have almost the same computational complexity as of finding an optimal solution only. We summarize efficient computational methods for computing tolerances for these problems and determine their time complexity experimentally.
Added: Dec 10, 2016 | 2,985 | 12,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-51 | longest | en | 0.890602 |
https://www.mathdoubts.com/derivative-of-csc-inverse-function-proof/ | 1,722,727,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00563.warc.gz | 687,846,443 | 12,590 | Proof of Derivative Rule of Inverse Cosecant function
When $x$ represents a variable, the inverse cosecant function is expressed as $\csc^{-1}{(x)}$ or $\operatorname{arccsc}{(x)}$ in inverse trigonometry. The derivative or differentiation of the inverse co-secant function with respect to $x$ is written in two different forms as follows in differential calculus.
$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\csc^{-1}{(x)}\Big)}$
$(2) \,\,\,$ $\dfrac{d}{dx}{\, \Big(\operatorname{arccsc}{(x)}\Big)}$
In differential calculus, the first principle of differentiation is used for deriving the derivative of inverse cosecant function. In fact, it is used as a formula. Hence, we must learn the proof of the differentiation rule of the inverse cosecant function.
Derivative of Cosecant function in Limit form
According to the principle definition of the derivative, the derivative of inverse cosecant function with respect to $x$ is written in limit form.
$\dfrac{d}{dx}{\, (\csc^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\csc^{-1}{(x+\Delta x)}-\csc^{-1}{x}}{\Delta x}}$
Let $h = \Delta x$, the differential element $\Delta x$ can be simply written as $h$.
$\implies$ $\dfrac{d}{dx}{\, (\csc^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\csc^{-1}{(x+h)}-\csc^{-1}{x}}{h}}$
Now, the differentiation of $\operatorname{arccsc}{(x)}$ function with respect to $x$ can be proved from the first principle of differential calculus.
Evaluate the Limit by Direct Substitution
Firstly, try to evaluate the functionality of rational expression as $h$ approaches $0$ by the direct substitution method.
$= \,\,\,$ $\dfrac{\csc^{-1}{(x+0)}-\csc^{-1}{x}}{0}$
$= \,\,\,$ $\dfrac{\csc^{-1}{x}-\csc^{-1}{x}}{0}$
$= \,\,\,$ $\require{cancel} \dfrac{\cancel{\csc^{-1}{x}}-\cancel{\csc^{-1}{x}}}{0}$
$=\,\,\,$ $\dfrac{0}{0}$
The limit of the inverse trigonometric function in rational form as $h$ tends to $0$ is indeterminate. So, it is impossible to evaluate the function by the direct substitution. Hence, we must think about an alternative method.
Simplify the Inverse trigonometric function
Now, let’s return to the first step of deriving the derivative of inverse cosecant function in limit form.
$\implies$ $\dfrac{d}{dx}{\,(\csc^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\csc^{-1}{(x+h)}-\csc^{-1}{x}}{h}}$
In limits, there is no formula in inverse co-secant function but we have formulas in inverse sine and inverse tan functions. Hence, it is must to convert every inverse cosecant function into either inverse sine or inverse tan function. Unfortunately, there is no law to convert the inverse cosecant into inverse tan function but it is possible to express the inverse cosecant as inverse sine function as per inverse trigonometry.
As per the reciprocal inverse trigonometric identity, $\csc^{-1}{x} \,=\, \sin^{-1}{\Big(\dfrac{1}{x}\Big)}$
$\implies$ $\dfrac{d}{dx}{\, (\csc^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Big(\dfrac{1}{x+h}\Big)}-\sin^{-1}{\Big(\dfrac{1}{x}\Big)}}{h}}$
In numerator, the inverse trigonometric expression represents the difference of inverse sine functions and it can be simplified by the difference identity of sine inverse functions.
$\sin^{-1}{x}-\sin^{-1}{y}$ $\,=\,$ $\sin^{-1}{\Big(x\sqrt{1-y^2}-y\sqrt{1-x^2}\Big)}$
Now, use this formula and simplify the inverse trigonometric expression in the numerator of the rational expression.
$\implies$ $\dfrac{d}{dx}{\, (\csc^{-1}{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\Big(\dfrac{1}{x+h}\Big)\sqrt{1-\Big(\dfrac{1}{x}\Big)^2}}-\Big( \dfrac{1}{x}\Big)\sqrt{1-\Big(\dfrac{1}{x+h}\Big)^2}\Bigg)}{h}}$
Now, concentrate on simplifying the algebraic expression, which is argument of the inverse sine function in the numerator.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x+h}\sqrt{1-\dfrac{1^2}{x^2}}}-\dfrac{1}{x}\sqrt{1-\dfrac{1^2}{(x+h)^2}}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x+h}\sqrt{1-\dfrac{1}{x^2}}}-\dfrac{1}{x}\sqrt{1-\dfrac{1}{(x+h)^2}}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x+h}\sqrt{\dfrac{x^2-1}{x^2}}}-\dfrac{1}{x}\sqrt{\dfrac{(x+h)^2-1}{(x+h)^2}}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1}{x+h} \times \dfrac{\sqrt{x^2-1}}{x}}-\dfrac{1}{x} \times \dfrac{\sqrt{(x+h)^2-1}}{(x+h)}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{1 \times \sqrt{x^2-1}}{(x+h) \times x}}-\dfrac{1 \times \sqrt{(x+h)^2-1}}{x \times (x+h)}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}}{(x+h)x}}-\dfrac{\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}}{x(x+h)}}-\dfrac{\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}}\Bigg)}{h}}$
The function is almost similar to the limit rule of inverse sine function. So, we have to make some adjustments for transforming it into required form.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}}{h}}$ $\times$ $1 \Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}}{h}}$ $\times$ $\dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}\Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}}$ $\times$ $\dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}\Bigg]$
Evaluate the Limits of the functions
Now, use the product rule of limits for evaluating the limit of product of functions by the product of their limits.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}}{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
The first factor in the product is almost similar to the limit rule of inverse sine function but its input must be in the form of argument in inverse sine function or expression in the denominator. Hence, we should evaluate the approaching value of the expression $\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$ firstly.
$(1)\,\,\,$ If $h \,\to\, 0$, then $x+h \,\to\, x+0$. Therefore, $x+h \,\to\, x$
$(2)\,\,\,$ If $x+h \,\to\, x$, then $x \times (x+h) \,\to\, x \times x$. Therefore, $x(x+h) \,\to\, x^2$
$(3)\,\,\,$ We know that $x+h \,\to\, x$ and $x(x+h) \,\to\, x^2$, then $\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$ $\,\to\,$ $\dfrac{\sqrt{x^2-1}-\sqrt{(x)^2-1}}{x^2}$. Now, $\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$ $\,\to\,$ $\dfrac{\sqrt{x^2-1}-\sqrt{x^2-1}}{x^2}$, then $\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$ $\,\to\,$ $\dfrac{0}{x^2}$. Therefore, $\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$ $\,\to\,$ $0$
Let $m \,=\, \dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}$, then $m\,\to\,0$ as $h\,\to\,0$.
Keep the second factor as it is but express the first factor in terms of $m$ for evaluating it.
$=\,\,\,$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{\Big(m\Big)}}{\Big(m\Big)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin^{-1}{m}}{m}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
As per the limit rule of inverse sine function, the limit of ratio of inverse sine function $\sin^{-1}{m}$ to $m$ as $m$ closes to zero is equal to one.
$=\,\,\,$ $1$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{h}}$
Now, we have to evaluate the limit of the remaining function.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)}\Bigg)}{\dfrac{h}{1}}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)} \times \dfrac{1}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\sqrt{x^2-1}-\sqrt{(x+h)^2-1}\Big)}{x(x+h)} \times \dfrac{1}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\sqrt{x^2-1}-\sqrt{(x+h)^2-1}\Big) \times 1}{x(x+h) \times h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{x(x+h)h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{hx(x+h)}}$
The direct substitution will obviously fail if you try it. The expression in the numerator is in radical form. So, the rationalization method is suitable for evaluating it.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{hx(x+h)} \times 1\Bigg)}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{\sqrt{x^2-1}-\sqrt{(x+h)^2-1}}{hx(x+h)}}$ $\times$ $\dfrac{\sqrt{x^2-1}+\sqrt{(x+h)^2-1}}{\sqrt{x^2-1}+\sqrt{(x+h)^2-1}}\Bigg)$
The two fractional functions can be multiplied mathematically. The product of the expressions in numerators is difference of squares of both terms, as per difference of squares algebraic identity.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(\sqrt{x^2-1}\Big)^2-\Big(\sqrt{(x+h)^2-1}\Big)^2}{hx(x+h) \times \Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^2-1-((x+h)^2-1)}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^2-1-(x+h)^2+1}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^2-(x+h)^2-1+1}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^2-(x+h)^2-\cancel{1}+\cancel{1}}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{x^2-(x+h)^2}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
Use the difference of squares formula for factoring the difference of squares in the numerator.
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(x+x+h)(x-(x+h))}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(2x+h)(x-x-h)}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(2x+h)(\cancel{x}-\cancel{x}-h)}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{(2x+h)(-h)}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{-h(2x+h)}{hx(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{-\cancel{h}(2x+h)}{\cancel{h}x(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{-(2x+h)}{x(x+h)\Big(\sqrt{x^2-1}+\sqrt{(x+h)^2-1}\Big)}}$
We can now evaluate the limit of the function as $h$ approaches zero by the direct substitution.
$=\,\,\,$ $\dfrac{-(2x+0)}{x(x+0)\Big(\sqrt{x^2-1}+\sqrt{(x+0)^2-1}\Big)}$
$=\,\,\,$ $\dfrac{-2x}{x(x)\Big(\sqrt{x^2-1}+\sqrt{(x)^2-1}\Big)}$
$=\,\,\,$ $\dfrac{-2x}{x(x)\Big(\sqrt{x^2-1}+\sqrt{x^2-1}\Big)}$
$=\,\,\,$ $\dfrac{-2x}{x(x)\Big(2\sqrt{x^2-1}\Big)}$
$=\,\,\,$ $\dfrac{-2x}{2x(x)\sqrt{x^2-1}}$
$=\,\,\,$ $\dfrac{-\cancel{2x}}{\cancel{2x}(x)\sqrt{x^2-1}}$
$=\,\,\,$ $\dfrac{-1}{x\sqrt{x^2-1}}$
Here, $x \ne 0$ and $x^2-1 > 0$, then $x^2 > 1$. So, $x > \pm 1$. Hence, we take $x$ as $|x|$ for its absolute value.
$=\,\,\,$ $\dfrac{-1}{|x|\sqrt{x^2-1}}$
Thus, the differentiation of the inverse cosecant function rule is derived mathematically by the first principle in differential calculus.
$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (\csc^{-1}{x})}$ $\,=\,$ $-\dfrac{1}{|x|\sqrt{x^2-1}}$
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Get the latest math updates from the Math Doubts by subscribing us. | 5,392 | 13,546 | {"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} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.746627 |
https://search.r-project.org/CRAN/refmans/Compounding/html/pgfInegativebinomial.html | 1,623,597,113,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00564.warc.gz | 465,624,746 | 1,737 | pgfInegativebinomial {Compounding} R Documentation
## Function pgfInegativebinomial
### Description
This function calculates value of the pgf's inverse of the negative binomial distribution.
### Usage
```pgfInegativebinomial(s, params)
```
### Arguments
`s` Value of the parameter of the pgf. It should be from interval [-1,1]. In the opposite pgf diverges. `params` List of the parameters of the negative binomial distribution, such that params<-c(theta,k), where theta is probability, and k is positive number.
### Author(s)
S. Nadarajah, B. V. Popovic, M. M. Ristic
### References
Johnson N, Kotz S, Kemp A (1992) Univariate Discrete Distributions, John Wiley and Sons, New York
http://www.am.qub.ac.uk/users/g.gribakin/sor/Chap3.pdf
### Examples
```params<-c(.9,.7)
pgfInegativebinomial(.5,params)
## The function is currently defined as
pgfDnegativebinomial <- function(s,params) {
k<-s[abs(s)>1]
if (length(k)>0)
warning("Some elements of the vector s are out of interval [-1,1]")
if (length(params)<2)
stop("At least one value in params is missing")
if (length(params)>2)
stop("The length of params is 2")
theta<-params[1]
k<-params[2]
if ((theta>=1)|(theta<=0))
stop ("Parameter theta belongs to the interval (0,1)")
if (k<=0)
stop("Parameter k must be positive")
k*(1-theta)*theta^k/(1-(1-theta)*s)^(k+1)
}
```
[Package Compounding version 1.0.2 Index] | 401 | 1,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2021-25 | latest | en | 0.577304 |
http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-829637-4&title=ct&chapter=11&headerFile=6&state=al&2005 | 1,386,821,591,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164452243/warc/CC-MAIN-20131204134052-00020-ip-10-33-133-15.ec2.internal.warc.gz | 360,612,920 | 4,386 | 1. The coordinates of the vertices of a quadrilateral are (0,1), (3,-3), (5,1), and (-2,-3). Find the area of the quadrilateral. A. 30 units2 B. units2 C. 20 units2 D. 10 units2 Hint 2. Carpeting How much carpeting is needed to cover the floor of an L-shaped room with the dimensions shown in the figure? A. 312 ft2 B. 264 ft2 C. 396 ft2 D. 286 ft2 Hint 3. Find the value of x if the area of the trapezoid is 54. A. 11 B. 8 C. 10 D. 9 Hint 4. Find the area of a circle with a radius of 15 millimeters. Round to the nearest tenth. A. 225.0 mm2 B. 706.9 mm2 C. 1413.7 mm2 D. 94.2 mm2 Hint 5. If a stone is dropped at random on the arrangement of tiles shown, what is the probability the stone lands on a black tile? A. B. C. D. Hint 6. Find the probability of Lose a Turn for the spinner shown. A. B. C. D. Hint 7. This trapezoid is made up of 3 congruent isosceles triangles. Find the exact area of the trapezoid. A. B. 234 C. D. 252 Hint 8. Find the area of a regular octagon with an apothem of 8.5. A. 340 B. 239 C. 182 D. 296 Hint 9. Determine the area of the figure. A. 88 m2 B. 70 m2 C. 136 m2 D. 100 m2 Hint 10. Find the area of the figure. Round to the nearest tenth. A. 261.8 in2 B. 265.9 in2 C. 144.0 in2 D. 222.6 in2 Hint | 445 | 1,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2013-48 | latest | en | 0.726138 |
https://fdocument.org/document/problem-of-the-month-circular-of-the-month-circular-reasoning-page-5-correct.html | 1,716,811,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00874.warc.gz | 205,822,252 | 31,606 | 20
© Noyce Foundation 2012 Problem of the Month: Circular Reasoning Overview: In the Problem of the Month Circular Reasoning, students use geometric reasoning to solve problems involving circles and the number π. The mathematical topics that underlie this POM are the attributes of polygons, circles, the irrational number π, spatial visualization, and angle measurement. In the first level of the POM, students are presented with the task of examining the relationship between the measure of the diameter of a circle and its circumference. Their task involves determining the number of times bigger the circumference is than the diameter (approximation of π). Level B requires students to continue to investigate the number π by investigating the relationship between the measurements of a tennis ball can. They find the height of the can and the circumference of the base of the can. They reason why the two measurements are close to the same size and search for a pattern to determine a rationale. In level C, students investigate different size pizzas made in the form of a circle. The pizzas are divided into fractional slices shaped like a sector of a circle. The students investigate the relationships between the sizes of the pizzas, including the dimensions of the slices (radii and degrees), the costs, and whether the costs are proportional to their sizes. In level D, the students investigate the rings that support wooden barrels. In this problem, the rings are made a foot longer than they were supposed to be. Since the barrels come in a range of sizes from very large to very small, the foot long error must be explored. The goal is to determine the relationship between error in rings to the original size and how the extra length affects a redesigned barrel. In level E, students investigate Archimedes’ approach to calculating an approximation for π. Students are asked to duplicate the drawings and calculations of Archimedes and determine the accuracy of his approximation. They are also asked to justify the formula for the area of a circle. Mathematical Concepts: Geometry and measurement are important real-world experiences. From the most complex structures created by designers, architects, and construction workers to arranging the furniture in a room, geometry and measurement are essential. In this POM, students explore various aspects of circular geometry. This includes understanding the attributes of circles. In particular, students study regular polygons, diameters, circumference, and angles. The attributes studied include the diagonals of polygons as well as angles and their measurements. Students explore the irrational number π. In level E, students use an historical approach to calculate an approximation of π. In addition to the geometric and measurement aspects of this POM, the students are seeking to find patterns, develop functional relationships, make generalizations, and justify their conclusions. The mathematics involves higher-level cognitive skills.
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Problem of the Month: Circular Reasoning
Overview: In the Problem of the Month Circular Reasoning, students use geometric reasoning to solve problems involving circles and the number π. The mathematical topics that underlie this POM are the attributes of polygons, circles, the irrational number π, spatial visualization, and angle measurement. In the first level of the POM, students are presented with the task of examining the relationship between the measure of the diameter of a circle and its circumference. Their task involves determining the number of times bigger the circumference is than the diameter (approximation of π). Level B requires students to continue to investigate the number π by investigating the relationship between the measurements of a tennis ball can. They find the height of the can and the circumference of the base of the can. They reason why the two measurements are close to the same size and search for a pattern to determine a rationale. In level C, students investigate different size pizzas made in the form of a circle. The pizzas are divided into fractional slices shaped like a sector of a circle. The students investigate the relationships between the sizes of the pizzas, including the dimensions of the slices (radii and degrees), the costs, and whether the costs are proportional to their sizes. In level D, the students investigate the rings that support wooden barrels. In this problem, the rings are made a foot longer than they were supposed to be. Since the barrels come in a range of sizes from very large to very small, the foot long error must be explored. The goal is to determine the relationship between error in rings to the original size and how the extra length affects a redesigned barrel. In level E, students investigate Archimedes’ approach to calculating an approximation for π. Students are asked to duplicate the drawings and calculations of Archimedes and determine the accuracy of his approximation. They are also asked to justify the formula for the area of a circle. Mathematical Concepts: Geometry and measurement are important real-world experiences. From the most complex structures created by designers, architects, and construction workers to arranging the furniture in a room, geometry and measurement are essential. In this POM, students explore various aspects of circular geometry. This includes understanding the attributes of circles. In particular, students study regular polygons, diameters, circumference, and angles. The attributes studied include the diagonals of polygons as well as angles and their measurements. Students explore the irrational number π. In level E, students use an historical approach to calculate an approximation of π. In addition to the geometric and measurement aspects of this POM, the students are seeking to find patterns, develop functional relationships, make generalizations, and justify their conclusions. The mathematics involves higher-level cognitive skills.
Problem of the Month Circular Reasoning Page 1 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Problem of the Month
Circular Reasoning Level A Janet and Lydia want to learn more about circles. They decide to measure different size circles that they can find. They measure the circles in two ways. One way is across a centerline of the circle. The centerline is called the diameter. Another way is around the outside of the circle. The distance around the outside of a circle is called the circumference. Measure various size circles using either physical objects or the included page of circles. Measure both the circumference and diameter of each circle with a tape measure (paper or cloth tape). Create a table that compares the length of the diameter with the circumference of the circle. Measure at least five different size circles.
Diameter Circumference
Compare the numbers between the two columns. What patterns do you see? Explain any relationship you see between the length of the diameter and the circumference.
π π
Problem of the Month Circular Reasoning Page 2 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Level B Measure a tennis ball can. First measure the height of the can, from the top of the can to the bottom. Record the measurement.
Next, measure the circumference around the can. Be accurate with your measurements.
Compare the measurements. Explain the relationship between the two measurements. How does the number of balls in the can relate to the measurements and their relationship? What other parts of the can or tennis balls can you measure and compare? Describe the relationship of those measurements? How are they related?
Problem of the Month Circular Reasoning Page 3 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Level C You work for a pizza parlor. You currently have three sizes of pizza: regular (area 25 square inches), large (area 36 square inches) and giant (area 64 square inches). The storeowner has asked you to determine some information about the amount of pizza that is made. How much bigger is the giant pizza than the large pizza in terms of area? How much bigger is the large pizza than the regular pizza in terms of area? The owner sells a regular cheese pizza for \$8.25. He wants to know what to charge for the large and giant sized pizzas. He wants to cover the cost of the ingredients and make a proportional amount of profit on the other two pizzas. What should he charge for a large and giant cheese pizza? The owner sells pizza by the slice. He has been selling a large pizza by the slice for \$1.35. Each pizza is cut into eight equal slices. The owner wants to know whether he has priced it right to make more money selling by the slice than the whole pizza. Explain to him, using mathematics, how his prices for the slices and the whole pizza compare. Now the owner wants to cut each pizza into nine slices, instead of eight. How much smaller will the slices be? He is interested in comparing both the width (angle in numbers of degrees) and the area of each slice. The owner wants to sell the nine slices at the same price as he did when the pizza was cut into eight slices. He wants to know how much more he will make cutting the pizzas into smaller slices. He wants to compare the prices of selling the nine slices individually with the price of selling an entire large pizza. He asks you to calculate and explain the difference in prices and amounts. The owner also thinks he can save more money if he sells slices from regular sized pizzas instead of large sized pizzas. You cut both the regular and large pizzas into nine equal slices. How much larger is a single slice from a large pizza than that of a slice from a regular pizza? Explain your answer in terms of degrees of the angle and area of the slice.
Regular 25 sq. in. Large
36 sq. in. Giant 64 sq. in.
Problem of the Month Circular Reasoning Page 4 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Level D You are a quality controller for a barrel manufacturing company. Your company makes barrels of all different sizes. The company makes very large storage vats for vineyards. They even make very small barrels as earrings that are sold in the gift shop next to the factory. Without exception, the company makes every barrel the same way. They use wood supported with metal rings that go around the barrels. A computer controls the ring-making machine. The computer malfunctioned and made every ring, from the smallest for the earrings to the largest for the wine vats, larger than normal by exactly one foot. In other words, every ring your factory has turned out recently has a circumference that is twelve inches longer than the right size. Your boss has asked you to investigate this matter.
How does the altered rings compare to the original rings? How much bigger do the altered rings look in comparison to the rings on the barrels that they were designed to fit? How will the altered rings compare in size to the rings on the small barrels, like the earrings? How will the altered rings compare to the rings on the large �barrels, like the wine vats? What should the company do with all those altered rings? It would be very costly to dispose of them. Your boss has asked you to tell him what size barrels need to be made to use those rings. Are the barrels usable? What size would you have to make the new barrels in relationship to the original size barrels? What conclusion can you draw from your findings? Can you explain your findings using mathematics? Why are the results what they are?
Problem of the Month Circular Reasoning Page 5 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Level E Archimedes is considered one of the greatest mathematicians of all time. He lived in the Greek city of Syracuse, Sicily between 287 BC and 212 BC and is known for his work in many areas of mathematics and science. Archimedes mathematical work includes inventing methods of calculating the area and volume of geometric objects, such as the circles, parabolas, cylinders, cones, spheres, hyperbolas and ellipses. He also spent time investigating a rational number approximation of π. The method he used to approximate π involved examining the size of many-sided polygons. Archimedes, as the story is told, drew a large circle on the floor of his apartment. He measured the diameter of the circle and called that distance one unit in length. Next he began to inscribe and circumscribe different size polygons, tangent to the circle. Archimedes would measure the perimeters of the polygons and average the two to approximate π. As the numbers of sides of the polygons were increased, the shaped of the polygon looked more like the circle. The polygons got closer and closer to being the size of the circle and perimeters became ever closer to the value of π. Using that method, he was able to calculate the approximation of π, correct to a value between 3 1/7 and 3 10/71. How precise was his approximation of π? Give your answer in terms of the number of correct decimal places. Estimate how many sides his final polygon needed to be to get a number that accurate? Explain your estimate. Illustrate the method Archimedes used by drawing an interior regular polygon inscribed in a circle and an exterior regular polygon circumscribed about the same circle. Determine the perimeter of both polygons and show how π might be approximated. Find the area of both polygons and compare the areas with that of the circle. Explain how the area formula is derived from the area formula of the polygon.
Problem of the Month Circular Reasoning Page 6 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Problem of the Month Circular Reasoning
Primary Version Level A
Materials: A picture of a circle, circular objects (i.e. counters, coins), the π-day picture, paper and pencil. Discussion on the rug: (Teacher holds up a picture of a circle) “What is the name of this shape?” (Students respond to the question). (Teacher hands out circular objects) “These objects are shaped like a circle. Tell us about the shapes?” (Students explore the objects and describe the circles). (Teacher holds up a circle) “A circle is a special shape. Long ago people discovered the circle and tried to measure it. They found a special number called pi. Some people who like math celebrate circles on a special day, March 14. It is called pi day. Here is a picture of people having a picnic on pi day.” (Teacher hands-out the picture of the π-day picnic). In small groups: (Each student has the picture of the π-day picnic). “Look at the picture of the picnic. Find all the things in the picture that are shaped like a circle. Circle each shape you find.” (Students answer the following questions). How many things did you find that are shaped like a circle? Name the things in the picture that are shaped like a circle. (At the end of the investigation, have students either discuss or dictate a response to this summary question below.) Explain how you decided which ones to circle.
π π
Problem of the Month Circular Reasoning Page 7 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
π day picnic
Problem of the Month Circular Reasoning Page 8 (c) Noyce Foundation 2008. To reproduce this document, permission must be granted by the Noyce Foundation: [email protected].
Diameter Circumference
Problem of the Month Circular Reasoning
Task Description – Level A This task challenges a student to explore various aspects of circular geometry. Students are asked to examine the relationship between the measure of the diameter of a circle and its circumference. Their task involves determining the number of times bigger the circumference is than the diameter (approximation of π).
Common Core State Standards Math - Content Standards Geometry Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres.) K.G.1 Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2 Correctly name shapes regardless of their orientations or overall size. Measurement and Data Describe and compare measurable attributes. K.MD.1 Describe measurable attributes of objects, such as length or weight. Describe measurable attributes of a single object. Measure and estimate lengths in standard units. 2.MD.1 Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. 2.MD.4 Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. Operations and Algebraic Thinking Generate and analyze patterns. 4.OA.5 Generate a number or shape pattern that follows a given rule. … Expressions and Equations Solve real-life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. Functions Define, evaluate, and compare functions. 8.F.1 Understand that a function is a rule that assigns to each input exactly one output. … Use functions to model relationships between quantities. 8.F.4 Construct a function to model a linear relationship between two quantities. … High School – Functions - Building Functions Build a function that models a relationship between two quantities. F-BF.1 Write a function that describes a relationship between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context.
Common Core State Standards Math – Standards of Mathematical Practice MP.2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem
situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. MP.4 Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
Problem of the Month
Circular Reasoning Task Description – Level B
This task challenges a student to explore various aspects of circular geometry. A student is required to continue to investigate the number π by investigating the relationship between the measurements of a tennis ball can, such as the height of the can and the circumference of the base of the can. They are to reason why the two measurements are close to the same size and to search for a pattern to determine a rationale for this relationship. Additionally, a student is asked to explore and describe the relationship between the number of balls in a can and the height of a can.
Common Core State Standards Math - Content Standards Geometry Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). K.G.1 Describe objects in the environment using names of shapes, and describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, and next to. K.G.2 Correctly name shapes regardless of their orientations or overall size. Measurement and Data Describe and compare measurable attributes. K.MD.1 Describe measurable attributes of objects, such as length or weight. Describe measurable attributes of a single object. Measure and estimate lengths in standard units. 2.MD.1 Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. 2.MD.4 Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. Operations and Algebraic Thinking Generate and analyze patterns. 4.OA.5 Generate a number or shape pattern that follows a given rule. … Expressions and Equations Solve real-life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. Functions Define, evaluate, and compare functions. 8.F.1 Understand that a function is a rule that assigns to each input exactly one output. … Use functions to model relationships between quantities. 8.F.4 Construct a function to model a linear relationship between two quantities. … High School – Functions - Building Functions Build a function that models a relationship between two quantities. F-BF.1 Write a function that describes a relationship between two quantities. a. Determine an explicit expression, a recursive process, or steps for calculation from a context.
Common Core State Standards Math – Standards of Mathematical Practice MP.2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. MP.4 Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
Problem of the Month
Circular Reasoning Task Description – Level C
This task challenges a student to explore various aspects of circular geometry. A student is asked to investigate different size pizzas made in the form of a circle. The pizzas are divided into fractional slices shaped like a sector of a circle. The students investigate the relationships between the sizes of the pizzas, including the dimensions of the slices (radii and degrees), the costs, and whether the costs are proportional to their sizes.
Common Core State Standards Math - Content Standards Operations and Algebraic Thinking Represent and solve problems involving addition and subtraction. 2.OA.1 Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, … Ratio and Proportional Relationships Understand ratio concepts and use ratio reasoning to solve problems. 6.RP.3 c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times
the quantity); solve problems involving finding the whole, given a part and the percent. Analyze proportional relationships and use them to solve real-world and mathematical problems. 7.RP.1 Recognize and represent proportional relationships between quantities. 7.RP.3 Use proportional relationships to solve multistep ratio and percent problems. … Geometry Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. 7.G.4 Know the formulas for the area and circumference of a circle and use them to solve problems… 7.G.5 Use facts about …angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure.
Common Core State Standards Math – Standards of Mathematical Practice MP.2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. MP.4 Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are
comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
Problem of the Month: Circular Reasoning
Task Description – Level D This task challenges a student to explore various aspects of circular geometry. A student is asked to investigate the rings that support wooden barrels. In this problem, the rings are made a foot longer than they were supposed to be. Since the barrels come in a range of sizes from very large to very small, the foot long error must be explored. The goal is to determine the relationship between error in rings to the original size and how the extra length affects a redesigned barrel.
Common Core State Standards Math - Content Standards Geometry Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. 7.G.4 Know the formulas for the area and circumference of a circle and use them to solve problems… 7.G.6 Solve real-world and mathematical problems involving area, volume and surface area of two- and three- dimensional objects… Ratio and Proportional Relationships Analyze proportional relationships and use them to solve real-world and mathematical problems. 7.RP.2 Recognize and represent proportional relationships between quantities. 7.RP.3 Use proportional relationships to solve multistep ratio and percent problems. … Expressions and Equations Solve real-life and mathematical problems using numerical and algebraic expressions and equations. 7.EE.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. a. Solve word problems leading to equations of the form px+q=r and p(x+q)=r, where p, q, and r
are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width?
Common Core State Standards Math – Standards of Mathematical Practice
MP.2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. MP.4 Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of
interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose.
Problem of the Month
Circular Reasoning Task Description – Level E
This task challenges a student to explore various aspects of circular geometry. A student is asked to investigate Archimedes’ approach to calculating an approximation for π. Students are asked to duplicate the drawings and calculations of Archimedes and determine the accuracy of his approximation. They are also asked to justify the formula for the area of a circle.
Common Core State Standards Math - Content Standards Geometry Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. 7.G.4 Know the formulas for the area and circumference of a circle and use them to solve problems… 7.G.6 Solve real-world and mathematical problems involving area, volume and surface area of two- and three- dimensional objects… High School – Geometry - Congruence Make geometric constructions. G-CO.12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line. G-CO.13 Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle. High School – Geometry - Similarity, Right Triangles, and Trigonometry Apply trigonometry to general triangles. G-SRT.11 (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).
Common Core State Standards Math – Standards of Mathematical Practice MP.2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. MP.4 Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical
situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. | 7,173 | 37,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.943693 |
http://oeis.org/A209422 | 1,571,228,664,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00514.warc.gz | 143,424,564 | 4,265 | This site is supported by donations to The OEIS Foundation.
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A209422 Triangle of coefficients of polynomials v(n,x) jointly generated with A209415; see the Formula section. 3
1, 3, 5, 1, 9, 2, 1, 15, 6, 2, 1, 25, 13, 7, 2, 1, 41, 28, 16, 8, 2, 1, 67, 56, 38, 19, 9, 2, 1, 109, 109, 82, 49, 22, 10, 2, 1, 177, 206, 173, 112, 61, 25, 11, 2, 1, 287, 382, 352, 252, 146, 74, 28, 12, 2, 1, 465, 697, 701, 543, 347, 184, 88, 31, 13, 2, 1, 753, 1256, 1368, 1144, 784, 459, 226, 103, 34, 14, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS For a discussion and guide to related arrays, see A208510. LINKS G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened FORMULA u(n,x) = x*u(n-1,x) + v(n-1,x), v(n,x) = u(n-1,x) + v(n-1,x) + 1, where u(1,x)=1, v(1,x)=1. G.f.: (1 + (1 - x)*t - t^2)/((1 - t)*(1 - (x + 1)*t + (x - 1)*t^2)) = 1 + 3*t + (5 + x)*t^2 + ... . - G. C. Greubel, Jan 03 2018 EXAMPLE First five rows: 1; 3; 5, 1; 9, 2, 1; 15, 6, 2, 1; First three polynomials v(n,x): 1, 3, 5 + x. MATHEMATICA u[1, x_] := 1; v[1, x_] := 1; z = 16; u[n_, x_] := x*u[n - 1, x] + v[n - 1, x]; v[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1; Table[Expand[u[n, x]], {n, 1, z/2}] Table[Expand[v[n, x]], {n, 1, z/2}] cu = Table[CoefficientList[u[n, x], x], {n, 1, z}]; TableForm[cu] Flatten[%] (* A209421 *) Table[Expand[v[n, x]], {n, 1, z}] cv = Table[CoefficientList[v[n, x], x], {n, 1, z}]; TableForm[cv] Flatten[%] (* A209422 *) CoefficientList[CoefficientList[Series[(1 + (1 - x)*t - t^2)/((1 - t)*(1 - (x + 1)*t + (x - 1)*t^2)), {t, 0, 10}], t], x]// Flatten (* G. C. Greubel, Jan 03 2018 *) CROSSREFS Cf. A209421, A208510. Sequence in context: A197326 A235605 A212695 * A320386 A112411 A283838 Adjacent sequences: A209419 A209420 A209421 * A209423 A209424 A209425 KEYWORD nonn,tabf AUTHOR Clark Kimberling, Mar 09 2012 STATUS approved
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Last modified October 16 08:15 EDT 2019. Contains 328051 sequences. (Running on oeis4.) | 1,005 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-43 | latest | en | 0.452532 |
https://community.filemaker.com/thread/72133 | 1,503,503,572,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.75/warc/CC-MAIN-20170823152006-20170823172006-00154.warc.gz | 744,039,895 | 23,878 | 2 Replies Latest reply on Sep 4, 2013 11:33 PM by bradleyboggs
# Newb Question on Logical Calculations
Hello all,
Pardon me if this is ridiculously basic, but I'm just getting going with Filemaker Pro after a few classes with The Support Group and have hit a wall on my calculated field...
I have a "Total On Hand" field which is a sum of the in/out transactions in a separate Inventory Transactions table (this is used to track incoming materials for inventory purposes, and materials as they're pulled, to decrement the On Hand amount and also have a record of what material lots are used in each order. With these basics, the calculation works just fine.
One wrench in that plan: Sometimes we use left-overs or scraps to fill an order. When this is the case, I need to still keep track of the material lot used for an order, but don't want my calculation to count those records. I've created a "Scap" Field with a Checkbox on the layout - if it's checked it has a 1 in the field, if it's not checked the field is empty.
I only want this "Total On Hand" Calculation to count when the "Scrap" Field is empty, but no matter what I do with IF and CASE and CHOOSE functions I can't seem to get it right.
Here's the working calculation for the on-handbefore considering the scap:
If
(IsEmpty (Sum (InvTrans::UnitsIn) = Sum (InvTrans::UnitsOut) ); 0; Sum (InvTrans::UnitsIn) - Sum(InvTrans::UnitsOut)
Just need to modify that not to count the records where InvTrans::Scrap is Not Empty (or where InvTrans::Scrap = "1").
Any help is greatly appreciated!
Thanks!
• ###### 1. Re: Newb Question on Logical Calculations
If you make a calc field in the InvTrans table with the formula: if(Scrap = 1; UnitsOut ; "")
and use a Sum to count that field should work
Hope that helps,
Best regards,
Ruben van den Boogaard
Infomatics Software
ruben@infomatics.nl
• ###### 2. Re: Newb Question on Logical Calculations
Thanks a bunch! I knew the solution would be something ridiculously simple, and indeed that turned out to be the case.
Your tip also gave me a better understanding of these calculations as well, so thanks for that!
On to the next hurdle. | 529 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-34 | longest | en | 0.91879 |
https://it.mathworks.com/matlabcentral/fileexchange/50608-counting-the-floating-point-operations-flops | 1,550,500,218,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247486936.35/warc/CC-MAIN-20190218135032-20190218161032-00503.warc.gz | 585,610,740 | 19,724 | File Exchange
## Counting the Floating Point Operations (FLOPS)
version 1.0.0.0 (576 KB) by Hang Qian
### Hang Qian (view profile)
Scan and parse each line of MATLAB codes, and infer FLOPS based on matrix sizes
Updated 23 Apr 2015
The program counts FLOPS of a MATLAB file, either as a script or function. By scanning and parsing each line of the MATLAB codes, we infer the floating point operations based on matrix sizes. Arithmetic operations, matrix decompositions, elementary functions and common statistics functions are counted. The program supports user-defined new rules, which can override our predetermined rules by a spreadsheet. For details of software usage, refer to the enclosed PDF documentation ‘User Guide for FLOPS’.
Usage:
Step 1: Prepare your MATLAB codes in a script or function, say fileName.m.
Step 2: Save all the variables in a MAT file. For example:
save MATfileName.mat
Step 3: Profile the MATLAB codes.
profile on
(run your MATLAB codes here, say filename(A,B,C,D))
profileStruct = profile('info');
Step 4: Count FLOPS by calling the function
FLOPS(fileName,MATfileName,profileStruct);
The parsing and counting results will be displayed on the screen and sent to output arguments.
Hang Qian
### Hang Qian (view profile)
Hi Mohamed,
Thank you very much for the bug report. The flop count software does not know how MATLAB parses code, and the software only recognizes simple code patterns. Users are encouraged to simplify code so that the software can identify the operators and the corresponding flops. I try my best to revise the software so that it will recognize more complicated code patterns.
mohnait
### mohnait (view profile)
Hi Hang,
Thank you for this useful tool.
I noticed a bug related to Matlab comments.
I tested your tool putting only: "x = magic(3);%+magic(3);" in a script and I got: "FLOPs = 9 * 1 = 9", where it should be zero.
I got the correct "FLOPs = 0 * 1 = 0" After adding a space between the semicolon and the % symbol for comments.
Cheers,
Mohamed
Hang Qian
### Hang Qian (view profile)
Hi Darshan,
Thank you for your interest in the FLOPS counter. If we are aware of the theoretic FLOPS for a function, we could add a rule to the spreadsheet ExtendedRules.xlsx. For example, suppose that A is a m-by-p matrix and we believe SVD requires 2np^2+2n^3 FLOPS, we could add a row to the spreadsheet (with three columns), say
First column: svd
Second column: 2*nrow*ncol^2+2*nrow^3
Third column: some arbitrary text for comments
After running the program, the second output argument “Details” should show how much FLOPS are counted for each line of the code. If the extended rule is recognized by the software, we should see the additional FLOPS shown in “Details”.
Regards,
Hang Qian
Darshan Patil
### Darshan Patil (view profile)
Hi Hang,
Since you mentioned this code does not compute flops for nested functions, I was wondering how one can have a work around in order to compute flops for built in matlab functions like null() or svd() etc.. Thanks for your contribution.
Hang Qian
### Hang Qian (view profile)
Hi Gillani,
Thank you for your interests in the FLOP counting tool.
If the arithmetic operations are based on complex numbers, you may consider overriding the internal rules by editing the spreadsheet ExtendedRules.xlsx. Basically, overload +, -, *,/, ect in the first and second columns of the spreadsheet.
Alternatively, you can modify the codes directly by rewriting the internal rules. From about Line 1130, there is a switch...case... block, in which FLOPS of arithmetic operations + - * / are defined.
GILLANI
### GILLANI (view profile)
Hi Hang Qiang,
Thanks for the nice tool. When I multiply two complex numbers, it shows 1 FLOP, which is actually 6. Any idea that how to fix this? Actually, my application is based on matrices with complex values.
jeyakarthikeyan P V
### jeyakarthikeyan P V (view profile)
thank you for valuable and important contribution. It is very much useful to my research.
Ahvand
### Ahvand (view profile)
Thanks Hang for your useful codes. I will cite your codes if I succeed to publish a paper.
Hang Qian
### Hang Qian (view profile)
Hi Alireza,
Thank you for your interest in my codes. I developed that tool when I was working on a research paper. I tried my best to make the FLOPS counting as accurate as possible. Comments and bug reports are welcome and sincerely appreciated.
As for the citation, I think the file exchange ID and the version number is unique to each MATLAB central submission, and other researchers can retrieve the original file to reproduce the results. So I would suggest citation like
Qian, Hang (2015). Counting the Floating Point Operations (FLOPS), MATLAB Central File Exchange, No. 50608, Ver. 1.0, Retrieved June 30, 2015.
Hi Hang,
Thanks for the file. It's very useful in my research. How can I cite your work in my papers?
Thanks,
Alireza
##### MATLAB Release Compatibility
Created with R2014a
Compatible with any release
##### Platform Compatibility
Windows macOS Linux
### Discover Live Editor
Create scripts with code, output, and formatted text in a single executable document. | 1,223 | 5,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-09 | latest | en | 0.830204 |
https://jasmcole.com/2014/09/14/only-a-fool/?like_comment=343&_wpnonce=db72d5ccc8 | 1,653,668,153,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00176.warc.gz | 380,827,970 | 31,507 | # Only a Fool
I’ve spend an unreasonably significant fraction of my life sitting in traffic, particularly along the infamous M4 corridor in England. Now, living in London I’m well acquainted with the various ways traffic jams can form. The ones which result from stuffing too many cars into too tiny streets have an obvious cause. Somewhat more irritating are those forming for seemingly no reason out on the motorway – queueing for ages at a standstill only to suddenly drive away like nothing happened. I’ve spent enough time in this unenviable state that it’s almost (though inevitably not quite) worth dedicating some brain time to thinking about why.
Now the science and analysis of traffic flow is apparently so dull to most people that spies can use the job as cover. Undeterred, we can take a peek at some exposition on the subject and realise there’s much more going on under the surface.
We’d like to construct a model of traffic flow to try and understand where this ghost traffic comes from. For some basic definitions, lets assume each car is labelled $i$, with 1D position $x_i$ and velocity $v_i$. Further assume a given car will accelerate to match the one in front of it (up to the speed limit), and slow down to avoid a collision. The equation of motion for a car is then
$\frac{dv_i}{dt} = \lambda(v_{i+1} - v_i)$
where $\lambda$ represents how hard the driver likes to accelerate. We can integrate this equation once to give
$\frac{dx_i}{dt} = \lambda(x_{i+1} - x_i) + C$
where $C$ is some integration constant. The term inside the brackets is the distance between the cars, which can be related to the density of cars
$\rho = \frac{1}{x_{i+1} - x_i}$.
Now cars have a certain length $L$, so there is a maximum peak density the traffic can reach, $\rho_{max} = 1/L$. At this point we have a traffic jam, so
$v_i = \frac{dx_i}{dt} = 0 \Rightarrow C = -\frac{\lambda}{\rho_{max}}$.
Now for the other limit, at the moment our cars will keep on accelerating as they get more spread out. At some point this becomes illegal, so we need to impose a maximum speed limit $v_{max}$. We then know the car speed as a function of traffic density
$v_i = \left\{ \begin{array}{ll} v_{max} & \rho < \rho_{min} \\ \frac{\lambda}{\rho} - \frac{\lambda}{\rho_{max}} & \rho_{min} < \rho < \rho_{max} \end{array} \right.$
where the minimum density $\rho_{min} = \lambda/(L\lambda + v_{max})$. This is plotted below for different $\lambda$, and it is obvious that as drivers get dopier ($\lambda$ getting smaller), they start to balk at dense traffic and slow down more quickly.
This is interesting to know (and confirms a privately-held hypothesis that all of this bloody traffic is caused by stupid drivers), but we can steal a bit of intuition from other areas of physics. For a simple 1-lane road with no junctions or exits, cars coming in = cars going out, or the flux of cars is conserved. This implies that the car density and velocity must together satisfy the continuity equation, used in physics wherever a flux is conserved. If there is some source of cars $s(x,t)$, then we have
$\frac{\partial\rho}{\partial t} + \frac{\partial(v\rho)}{\partial x} = s(x,t)$.
We also know from the ‘microscopic’ analysis above of individual cars that the mean velocity is dependent on density such that
$v\rho = \frac{\lambda}{\rho_{max}}(\rho_{max} - \rho)$
and
$\frac{\partial(v\rho)}{\partial x} = \frac{d(v\rho)}{d\rho}\frac{\partial \rho}{\partial x} = -\frac{\lambda}{\rho_{max}}\frac{\partial \rho}{\partial x}$.
Putting everything together then, the car density distribution satisfies
$\left(\frac{\partial}{\partial t} - \frac{\lambda}{\rho_{max}}\frac{\partial}{\partial x}\right)\rho = s(x,t)$.
This should be recognisable as a wave equation, for backwards-travelling waves with velocity $\lambda/\rho_{max}$. Traffic disturbances then travel backwards towards the incoming traffic. If the drivers are alert or the traffic density is low the traffic information propagates quickly. If the traffic is dense and the drivers dopy then traffic disturbances take a long time to travel back to the unwary commuter. I should note that there are many empirical and observational models relating traffic density and velocity, and this is only the simplest choice. The general form of the equation above is known rather tastily as the inviscid form of Burgers’ equation.
This is nice to know, so let’s try some numerical experiments to see if we can reproduce this behaviour. Specifically I model a simple straight road full of cars. Each car reacts to the one ahead of it by matching speed, as above. If the distance between cars gets too low, the driver panics and slams on the brakes to slow down much more quickly. The cars are ‘injected’ into the road with a small uniform random spread in speeds and distances, and get deleted as they exit the road.
To simulate a traffic jam, I insert a short section of the road with a lower speed limit where the drivers try to slow down to match the speed limit. I’ll plot these results as distance-speed plots, with dots representing cars, as in the example animation below. Cars enter from the left and begin reacting to the car ahead of them. The horizontal line represents the lower speed limit.
Now if we scale this up a bit we can simulate a longer section of road to see how the traffic pattern develops:
We can note a few things: the initial random velocity distribution smooths itself out quickly as the drivers react to one another. The traffic flows fairly smoothly despite the reduced speed limit, and cars accelerate smoothly out of the reduced speed zone.
We can have a look what happens when the speed limit is switched off. Annoyingly, the low-speed region persists despite there being no reason for it – the speed limit is maximum! This is the first hint that traffic can stick around for a while even after the original reason for it has gone.
What if the drivers are more alert? We can adjust the $\lambda$ parameter to make the drivers react more aggressively to changes in velocity. As plotted below, we can see that the region of traffic now slides smoothly back towards the incoming cars. Now not only is there traffic for no reason, it’s not even in the right place!
This is easy to see in a space-time diagram too. The low-speed region is created by the restriction, then propagates backwards and flattens as it goes – the traffic gradually speeds up as cars are able to spread out.
For a proper traffic jam though, we’ll need a bit of gridlock. Lets suppose the drivers aren’t being sensible and obeying the 2-second rule, but are bunched up more closely than that. Using the same speed limits and driver awareness as above, the following animation shows what happens if instead the drivers are instead obeying a 1-second rule.
In this case a traffic jam appears immediately as cars begin to pile up the start of the speed limit (coloured red), and persists long after the restriction is lifted.
Don’t be a fool then, and let us all enjoy our dreary M4 drives in peace.
## 10 thoughts on “Only a Fool”
1. Olivier says:
Those graphs are pretty interesting and fun to look at!
Any chance you could put the code to generate them somewhere, so that we can play and create our own traffic jams :-)?
Like
1. Yes of course, once the code is cleaned up a bit! I’ll comment back here once it’s uploaded.
Like
1. steve says:
Hey Jason,
Any chance that code can be made available?
The plots(and your blog) are great, awesome work keep it up!!!
Like
2. Steve says:
I have recently started following your blog after your article on WiFi.. and this article has made me feel so justified in doing so. I have noticed these standing waves of traffic and been so incredibly frustrated with them. It is amazing that you’ve found some basic physics principles to model them as such. Kudos.
Like
1. Hi Steve, I’m glad you like the post, thanks for the kind words. I didn’t invent any of this, just quoted some lecture notes I found on Google. I do think a little animation goes a long way though, so I hope I’m adding some tangible value to what’s already out there.
Like
3. Buttfranklin says:
Very interesting!
Like
4. Hello Jason! It’s so interesting to see a simulation of your models and how easily you define them. When I was at the school of eng. we used some models to predict traffic behaviour in complex networks (containing roads of different capacities and diferent nodes) and used Wardrop model. But never examined this efect of “stupid drivers” who accelerate-brake.
BTW I’d be very honoured if you could check out my blog and this humble contribution to another blog, modelling air resistance in 1-D:
cheers from Barcelona!
Like
1. Hi from London! I’m glad you like the post, I don’t really know anything about traffic modelling but it seems quite interesting – I’ll have a look at the model you mentioned. I had a look at (a translated version) of your post, I like the ‘phase space’ diagrams of your ODE solutions. It is also possible to get an analytic solution for constant quadratic drag, see e.g. here http://www.physics.udel.edu/~szalewic/teach/419/cm08ln_quad-drag.pdf
Liked by 1 person
1. Ok, I’ll take a look!! thanks for reading the translated version (it was written in catalan and it was aimed at general public)
TC 😉
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http://www.doitpoms.ac.uk/tlplib/index.php?tag=2,28,37,40,45,51 | 1,369,034,158,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698493317/warc/CC-MAIN-20130516100133-00083-ip-10-60-113-184.ec2.internal.warc.gz | 418,434,283 | 8,816 | DoITPoMS
TLP Library
Teaching and learning packages (TLPs) are self-contained, interactive resources, each focusing on one area of Materials Science.
Introduction To Anisotropy
It is common in basic analysis to treat bulk materials as isotropic - their properties are independent of the direction in which they are measured. However the atomic scale structure can result in properties that vary with direction. This teaching and learning package (TLP) looks into typical examples of such anisotropy and gives a brief mathematical look into modelling the behaviour.
Atomic Force Microscopy
Provides a brief introduction to atomic force microscopy (AFM), some of the ways it is commonly used and some of the problems faced.
Atomic Scale Structure of Materials
This teaching and learning package provides an introduction to crystalline, polycrystalline and amorphous solids, and how the atomic-level structure has radical consequences for some of the properties of the material. It introduces the use of polarised light to examine the optical properties of materials, and shows how a variety of simple models can be used to visualise important features of the microstructure of materials.
Bending and Torsion of Beams
This teaching and learning package provides an introduction to the mechanics of beam bending and torsion, looking particularly at the bending of cantilever and free-standing beams and the torsion of cylindrical bars.
Brillouin Zones
This teaching and learning package provides an introduction to Brillouin zones in two and three dimensions and is aimed at developing familiarity with Brillouin Zones. It will not cover any specific applications. Brillouin Zones are particularly useful in understanding the electronic and thermal properties of crystalline solids.
Crystallinity in Polymers
An understanding of polymer crystallinity is important because the mechanical properties of crystalline polymers are different from those of amorphous polymers. Polymer crystals are much stiffer and stronger than amorphous regions of polymer.
Crystallographic Texture
This teaching and learning package (TLP) introduces the concept of texture in crystalline materials such as common metals and metallic alloys.
Crystallography
Crystalline materials are characterised by a regular atomic structure that repeats itself in all three dimensions. In other words the structure displays translational symmetry.
Deformation of Honeycombs and Foams
Highly porous materials, such as honeycombs, foams and fibrous structures, are an important class of material in both synthetic and biological systems. They are used in many different ways, but their mechanical behaviour is often of great importance as they are pressed, bent, sat on or chewed. An important class of these materials can be considered as made up of cells, so-called cellular structures. Here we describe how these materials deform, elastically and irreversibly.
Introduction To Dislocations
Dislocations are crucially important in determining the mechanical behaviour of materials. This teaching and learning package provides an introduction to dislocations and their motion through a crystal. A 'bubble raft' model is used to demonstrate some of the features of dislocations and other lattice defects. Some methods for observing real dislocations in materials are examined.
Elasticity in Biological Materials
This teaching and learning package (TLP) discusses the elasticity of biological materials. Whilst some show Hookean elasticity, the vast majority do not. Non-linear elasticity is considered, in particular J-shaped and S-shaped curves. Viscoelasticity is also discussed, using hair and spiders' silk as examples.
Epitaxial Growth
This TLP enables you to explore the way in which perfect thin crystalline layers are deposited epitaxially (i.e. in the same crystal orientation) on semiconductor substrates. This is the way many electronic and opto-electronic devices are now fabricated using techniques such as molecular beam epitaxy (MBE).
Examination of a Manufactured Article
This TLP provides an introduction to the deconstruction and investigation of the materials and processes used in an everyday item or article.
The Glass Transition in Polymers
This teaching and learning package is based on a lecture demonstrations used within the Department of Materials Science and Metallurgy at the University of Cambridge. The package is aimed at first year undergraduate Materials Science students and focuses on the glass transition in polymers.
Lattice Planes and Miller Indices
This teaching and learning package provides an introduction to the method used to describe planes of atoms in a crystalline material. The practical uses of describing planes of atoms are also addressed.
Liquid Crystals
This Teaching and Learning Package provides an introduction to liquid crystals, their physical properties and their modern-day applications.
Mechanics of Fibre-reinforced Composites
This teaching and learning package (TLP) gives an introduction to the nature of fibre-reinforced composite materials and their basic mechanical characteristics.
Microstructural Examination
This teaching and learning package (TLP) looks at how what we see in micrographs relates to equilibrium phase diagrams and cooling routes for alloy systems.
Phase Diagrams and Solidification
Phase diagrams are a useful tool in metallurgy and other branches of materials science. They show the mixture of phases present in thermodynamic equilibrium. This teaching and learning package looks at the theory behind phase diagrams, and ways of constructing them, before running through an experimental procedure, and presenting the results which can be obtained.
Introduction To Photoelasticity
This tutorial is based on lab work within the Department of Materials Science and Metallurgy at the University of Cambridge. The tutorial provides an introduction to the topic of photoelasticity and preparation for lab work. Photographs illustrate many features of birefringence in polymers under polarised light.
Polymer Basics
This teaching and learning package is an introduction to the basic concepts of polymer science. It includes molecular structure, synthesis and tests for identification.
Raman Spectroscopy
An introduction to the analysis of materials and chemicals by the Raman scattering of light.
Reciprocal Space
This TLP shows the construction of reciprocal lattices from real ones, use of the Ewald sphere for diffraction experiments and some other applications of reciprocal space.
Introduction To Semiconductors
This teaching and learning package provides a very basic introduction to semiconductors. These materials are essential to the operation of solid state electronic devices.
Slip in Single Crystals
This teaching and learning package explains how plastic deformation of materials occurs through the mechanism of slip. Slip involves dislocation glide on particular slip planes. The geometry of slip is explained, and electron microscopy techniques are used to show slip occurring in single crystals of cadmium.
Solid Solutions
This teaching and learning package is based on a practical used within the Department of Materials Science and Metallurgy at the University of Cambridge. The package is aimed at first year undergraduate Materials Science students and focuses on the different types of solid solution and the thermodynamic principles involved in understanding them.
The Stereographic Projection
This TLP covers the use of the Stereographic projection and Wulff nets.
The Stiffness of Rubber
This teaching and learning package is based on two experiments which demonstrate the behaviour of rubber under tension. The first displays the unusual behaviour of a rubber strip when heated under tension; the second considers the behaviour of a rubber membrane under tension. In both cases the behaviour is considered theoretically in terms of the molecular structure of rubber and the thermodynamic entropy changes involved.
Structure of Bone and Implant Materials
This teaching and learning package (TLP) describes the structure of bone from the macro-scale to the micro-scale and considers its description as a biological composite. The structure of hip replacements is described and common implant materials are discussed in relation to the mechanical properties of bone.
Superelasticity and Shape Memory Alloys
This teaching and learning package (TLP) introduces the phenomena of superelasticity and the shape memory effect.
Thermal Expansion and the Bi-material Strip
This teaching and learning package (TLP) is based on lab work in the Department of Materials Science and Metallurgy at the University of Cambridge. The TLP provides an introduction to the topic of thermal expansion, and its application, together with the different stiffness of materials, in the bi-material strip. The TLP leads you through experiments to measure Young's Modulus from the deflection of a cantilever beam, and to estimate the boiling temperature of nitrogen and the expansivity of a polycarbonate material from the curvature of a bi-material strip immersed in liquid nitrogen.
Transmission Electron Microscopy
Transmission electron microscopy is a very important tool in materials science for investigating the fine-scale structure of materials. This TLP serves as an introduction to the basic concepts and structure of the transmission electron microscope.
X-ray Diffraction Techniques
This teaching & learning package provides an introduction to X-ray diffraction. It describes the main crystallographic information that can be obtained and experimental methods most commonly used. | 1,796 | 9,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.898611 |
http://priceandwilloughby.com/index.php?main_page=index&cPath=47_49&page=1&sort=2a | 1,539,930,176,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00529.warc.gz | 296,463,262 | 7,975 | # Middle School
Product Image Item Name+ Price
### Add and Subtract Multi Decimals
Item number 140 Add and Subtract Multi Decimals Add and Subtract Multi Decimals: In these activities, the students will learn to add and subtract...
\$54.00
Max: 1
### Add and Subtract Rational Numbers
Item number 154 Add and Subtract Rational Numbers Add and Subtract Rational Numbers: In these activities the students learn about to add and subtract...
\$54.00
### Area of Polygons
Item number 189 Area of Polygons Area of Polygons: In these activities the students learn to determine the area of various polygons, when given the...
\$54.00
### Cones, Cylinders, and Spheres
Item number 158 Cones, Cylinders, and Spheres Cones, Cylinders, and Spheres: In these activities the students learn to determine the volume of cones,...
\$54.00
### Congruence and Similarity
Item number 143 Common Core Content Activity: Congruence and Similarity Congruence and Similarity: In these activities, we learn about the congruent...
\$54.00
### Creating a Table
Common Core Content Activity: Creating a Table Creating a Table comes with a pretest/posttest plus six days of activities to teach the skills. In...
\$54.00
Max: 1
### Cubes and Right Prisms
Item number 184 Cubes and Right Prisms Cubes and Right Prisms: In these activities the students learn to determine the volume and surface area of...
\$54.00
### Do I Have Enough? (money)
Item #317 Do I Have Enough? These activities work on understanding money. There are 30 pages of determining if you have enough...
\$4.50
### Equivalent Expressions
Item number 190 Equivalent Expressions Equivalent Expressions: In these activities the students learn to solve expressions, determine if they are...
\$54.00
### Evaluate Numerical Expressions
Item number 153 Evaluate Numerical Expressions Evaluate Numerical Expressions with Whole Number Exponents: In these activities the students learn...
\$54.00 | 449 | 1,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-43 | latest | en | 0.805291 |
https://www.mathlearnit.com/fraction-as-decimal/what-is-201-92-as-a-decimal | 1,722,966,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00092.warc.gz | 670,564,617 | 7,250 | # What is 201/92 as a decimal?
## Solution and how to convert 201 / 92 into a decimal
201 / 92 = 2.185
Fraction conversions explained:
• 201 divided by 92
• Numerator: 201
• Denominator: 92
• Decimal: 2.185
• Percentage: 2.185%
201/92 converted into 2.185 begins with understanding long division and which variation brings more clarity to a situation. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
201 / 92 as a percentage 201 / 92 as a fraction 201 / 92 as a decimal
2.185% - Convert percentages 201 / 92 201 / 92 = 2.185
## 201/92 is 201 divided by 92
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (201) by the denominator (92). Here's 201/92 as our equation:
### Numerator: 201
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. Overall, 201 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Large two-digit conversions are tough. Especially without a calculator. Now let's explore the denominator of the fraction.
### Denominator: 92
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 92 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Ultimately, don't be afraid of double-digit denominators. So without a calculator, let's convert 201/92 from a fraction to a decimal.
## How to convert 201/92 to 2.185
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 92 \enclose{longdiv}{ 201 }$$
We will be using the left-to-right method of calculation. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Solve for how many whole groups you can divide 92 into 201
$$\require{enclose} 00.2 \\ 92 \enclose{longdiv}{ 201.0 }$$
We can now pull 184 whole groups from the equation. Multiply this number by 92, the denominator to get the first part of your answer!
### Step 3: Subtract the remainder
$$\require{enclose} 00.2 \\ 92 \enclose{longdiv}{ 201.0 } \\ \underline{ 184 \phantom{00} } \\ 1826 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, continue to the next step.
### Step 4: Repeat step 3 until you have no remainder
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 201/92 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/92 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.218 per hour and not$20 and 201/92.
### When to convert 2.185 to 201/92 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 201/92 = 2.185 what would it be as a percentage?
• What is 1 + 201/92 in decimal form?
• What is 1 - 201/92 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 2.185 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 92 Denominator What is 201/82 as a decimal? What is 191/92 as a decimal? What is 201/83 as a decimal? What is 192/92 as a decimal? What is 201/84 as a decimal? What is 193/92 as a decimal? What is 201/85 as a decimal? What is 194/92 as a decimal? What is 201/86 as a decimal? What is 195/92 as a decimal? What is 201/87 as a decimal? What is 196/92 as a decimal? What is 201/88 as a decimal? What is 197/92 as a decimal? What is 201/89 as a decimal? What is 198/92 as a decimal? What is 201/90 as a decimal? What is 199/92 as a decimal? What is 201/91 as a decimal? What is 200/92 as a decimal? What is 201/92 as a decimal? What is 201/92 as a decimal? What is 201/93 as a decimal? What is 202/92 as a decimal? What is 201/94 as a decimal? What is 203/92 as a decimal? What is 201/95 as a decimal? What is 204/92 as a decimal? What is 201/96 as a decimal? What is 205/92 as a decimal? What is 201/97 as a decimal? What is 206/92 as a decimal? What is 201/98 as a decimal? What is 207/92 as a decimal? What is 201/99 as a decimal? What is 208/92 as a decimal? What is 201/100 as a decimal? What is 209/92 as a decimal? What is 201/101 as a decimal? What is 210/92 as a decimal? What is 201/102 as a decimal? What is 211/92 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 92 Denominator 202/92 as a percentage 201/93 as a percentage 203/92 as a percentage 201/94 as a percentage 204/92 as a percentage 201/95 as a percentage 205/92 as a percentage 201/96 as a percentage 206/92 as a percentage 201/97 as a percentage 207/92 as a percentage 201/98 as a percentage 208/92 as a percentage 201/99 as a percentage 209/92 as a percentage 201/100 as a percentage 210/92 as a percentage 201/101 as a percentage 211/92 as a percentage 201/102 as a percentage | 1,783 | 6,800 | {"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} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.919948 |
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