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https://osiris.utwente.nl/student/OnderwijsCatalogusSelect.do?selectie=cursus&cursus=202001045&collegejaar=2022&taal=nl | 1,701,451,177,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00397.warc.gz | 499,597,781 | 6,419 | Sluiten Help Print
Cursus: 202001045
202001045Social Network Structure and Dynamics
Cursus informatie
Cursus202001045
Studiepunten (ECTS)6,5
CursustypeOnderwijseenheid
VoertaalEngels
Contactpersoondr.ir. M.J. van Sinderen
E-mailm.j.vansinderen@utwente.nl
Docenten
Vorige 1-5 van 116-10 van 1111-11 van 11 Volgende 5
Examinator N. Bouali Examinator dr. F.A. Bukhsh Docent dr. D.V. Le Viet Duc Docent J. M. Meylahn Docent C. Morais Fonseca
Collegejaar2022
Aanvangsblok
1B
OpmerkingTCS students register in Osiris. Others contact modulesupport-tcs@utwente.nl . Minor students: register for the minor!
AanmeldingsprocedureZelf aanmelden via OSIRIS Student
Inschrijven via OSIRISJa
Cursusdoelen
body { font-size: 9pt; font-family: Arial } table { font-size: 9pt; font-family: Arial } Learning goal is to be able to recognize and explain network phenomena. Social networks such as Facebook, information networks such as the Web, and institutions such as voting are all IT-enabled. The student will learn: how to recognize and explain structural and dynamic phenomena in these networks, such as cascading behavior and power laws, and how to model and analyze using graph theory and game theory. After following this module, the student is able to: Recognize these phenomena in practice; Apply mathematical models from graph theory, probability, and game theory to describe and analyze them; Explain and predict network phenomena in terms of network structure and behavior; Operationalize and apply these models to existing network data. In this study unit of the module Web Science we focus on the topics: graphs and social networks, information networks, and network dynamics.
Inhoud
body { font-size: 9pt; font-family: Arial } table { font-size: 9pt; font-family: Arial } The study unit Social Network Structure and Dynamics of the module Web Science covers the following topics: Graphs and Social Networks We study basic graph theory concepts such as components, triadic closure, strong and weak ties, homophily (similarity between 'friends') and positive and negative relationships. These concepts are put to work on modeling network data such as collaborations, information linkage, citation, interactions, etc. Students will be able to understand and model network data as graphs, and develop algorithms for analyzing basic graph properties of large volumes of network data (big data). Information Networks Our goal is to understand the structure of information networks on the internet that emerges from citation, liking, commenting, co-authoring, connection with 'friends', hypertext linking, etc. We study properties such as reputation, authority and relevance of web pages and persons. Students will learn to model, understand, and analyze such informational properties in terms of graph theory concepts. Network Dynamics – Population Models and Structural Models We study how people connected in a network influence each other’s behaviour and decisions. First we consider population models which help us to understand informational (or herding) effects and direct-benefit (or network) effects in social processes, and apply this knowledge to analyze the notion of popularity. Then we consider structural models to understand diffusion of information through groups of people, as opposed to a homogeneous population, and explain the small world phenomenon. Students will learn how to model and analyze the processes by which new ideas and innovations are adopted by a population in which groups of people are connected by very short paths.
Voorkennis
Some experience with programming, specifically Python.
Participating study
Bachelor Technical Computer Science
Module
Module 8D
Verplicht materiaal
Book
David Easley and Jon Kleinberg, Networks, Crowds, and Markets: Reasoning About a Highly Connected World. ISBN: 978-0-521-19533-1. The book can be downloaded from http://www.cs.cornell.edu/home/kleinber/networks-book/.
Aanbevolen materiaal
-
Werkvormen
Hoorcollege Project onbegeleid Vragenuur Werkcollege Zelfstudie geen begeleiding
Toetsen
Theoretical Knowledge Practical Knowledge and Skills
Sluiten Help Print | 941 | 4,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-50 | latest | en | 0.706244 |
https://goprep.co/q2b-in-right-angled-triangle-a-is-right-angle-and-b-35-then-i-1nktp7 | 1,611,833,400,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704843561.95/warc/CC-MAIN-20210128102756-20210128132756-00061.warc.gz | 357,084,260 | 31,447 | Q. 2 B3.7( 3 Votes )
# In right angled triangle, ∠A is right angle and ∠B = 35°, then ∠C is _______A. 65°B. 55°C. 75°D. 45°
In the given ∆ABC,
A = 90°
⇒ ∠B = 35°
We know that sum of all the angles of a triangle is 180°.
⇒ ∠A + B + C = 180°
⇒ ∠C = 180° - 90° - 35°
⇒ ∠C = 55°
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# Develop an equity screen - find 25 undervalued stocks #3. Take one of the undervalued stocks from your screen in #2 and prepare a financial analysis...
#2. Develop an equity screen – find 25 undervalued stocks
#3. Take one of the undervalued stocks from your screen in #2 and prepare a financial analysis of this stock. That is, estimate the expected return on the stock and compare it with the required return using the CAPM. What is your estimate of alpha?
#4. Graph the profit/loss for the following options positions
Long at-the-money call with strike 40 and premium 2;
Short at-the-money put with strike 40 and premium 2
Combination of the long call and short put, both with and
and sell call with strike 44 and premium 1
Short straddle: sell call and put, both with strike 40 and premium 3
#5. Suppose the price of a non-dividend paying stock is 40 and can go to 36 or 44 in 1 period
Assume the risk free rate is zero for this period.
Find the value of a call option with strike 40
Find the value of a put option with strike 38
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## Nonfiction » Education & Study Guides » Study guides - Mathematics
High (Secondary) School ‘Grades 9 & 10 - Math – Elementary and Circle Geometry – Ages 14-16’ eBook
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This eBook introduces the student to a variety of rules as they apply to elementary geometry as well as circle geometry.
Secondary School ‘AS-Level – Core 1 & 2 - Maths – Simultaneous Equations and Inequalities – Ages 16-18’ eBook
Price: \$2.99 USD. Words: 920. Language: English. Published: March 21, 2013. Categories: Nonfiction , Nonfiction
This eBook reviews simultaneous equations and inequalities. We introduce simultaneous equations as systems of equations, and consider some relatively simple pairs of simultaneous equations, one pair involving a pair of linear equations, and another pair involving one linear equation and one quadratic equation. We go on to introduce the two methods of solving simultaneous equations, elimim. ...
Secondary School AS-Level - Statistics 1 - Maths - Correlation and Linear Regression - Ages 16-18 - eBook
Price: \$2.99 USD. Words: 850. Language: English. Published: March 27, 2013. Categories: Nonfiction , Nonfiction
This eBook introduces the subjects of correlation and linear regression, presenting the relationship between two variables, reviewing the use of scatter diagrams in its review going on to examine the use of the product moment correlation coefficient in correlation, as well as the concept of regression and linear regression in particular, following up with an introduction of Spearman’s rank corr...
Master Maths: Fractions (add - x divide) HCF, LCM, Rounding
Price: \$2.99 USD. Words: 8,200. Language: English. Published: November 24, 2016. Categories: Nonfiction , Nonfiction
With this chart-topping Smart eBook range, you can tap the answers to the multiple-choice questions and it marks them. This well-illustrated book teaches you about the subjects listed, in a quick, easy and effective way. Ideal for beginners (from 8 years) to GCSE level, students revising, and for work. Part 7 of Master Maths is from Micro Maths PRO which has many good reviews like "Great purchase"
High (Secondary School) Grades 11 & 12 - Math –Integration I – Ages 16-18’ eBook
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This eBook introduces the subject of integration, starting by introducing integration as the inverse of differentiation, with a twist or two. We then introduce the indefinite integral and its constant of integration, and go on to introduce the definite integral. As well as this, we provide ample examples and illustrations to illuminate the prose. We go on to introduce trapeziums linear ..
Middle (Junior High) School ‘Grades 6, 7 & 8 - Math – Fractions, Percentages and Ratio – Ages 11-14’ eBook
Price: \$2.99 USD. Words: 830. Language: English. Published: June 11, 2012. Categories: Nonfiction , Nonfiction
Fractions, percentages and ratio introduces the student to fractions, their representation and their arithmetic, percentages, their representation and their conversion to or from fractions or decimals, as well as ratio, its representation and arithmetic manipulation using ratio.
Master Maths: Properties of Shapes and Lines
Price: \$2.99 USD. Words: 12,530. Language: English. Published: November 23, 2016. Categories: Nonfiction , Nonfiction
With this chart-topping range of Smart eBooks, you can tap the answers to the multiple-choice questions and it marks them. This well-illustrated book teaches you about angles in 2D, 3D shapes, Geometry and more in a quick, easy and effective way. Ideal for beginners (from 8 years) to GCSE level, students revising and for work. Master Maths part 1 from Micro Maths PRO which has reviews like "Great"
Why Is It So Hard To Learn Math?
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More people struggle with math than with any other academic discipline. So much so, that it is commonly excusable to not do well in math but to merely survive. There are reasons why math is so hard to learn. This little book identifies and provides insights into the top 10 reasons why math is so hard to learn. If you know what the bumps are, you can slow down and make it over them in one piece!
Short Tricks of Math
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The book of Short Tricks of Math is prepared for those who want to learn interesting mathematical techniques to boost up the calculating method.
Elementary Algebra Expression Practice Book 1, Grades 4-5
Series: Elementary Algebra Expression Practice · Grade 4 Math · Grade 5 Math. Price: \$0.99 USD. Words: 2,140. Language: English. Published: June 27, 2011. Categories: Nonfiction , Nonfiction
Enhance essential elementary algebra skills with these twenty-one practice problems. Each problem has an expression with mixed operators to reinforce the concept of operator precedence and use of parentheses. Choose a problem from the problem list, and then confirm your answer by easily navigating the link to the complete instructive solution. Most appropriate for 4th and 5th grade students.
The Secret Strategy For Learning Math: The One Thing You Must Understand
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Math is a different kind of subject from all the rest and requires that you understand the differences. If you don't, you will give up too soon and never master math. If you know the secret strategy and apply it from the beginning, you stand a better chance of developing good habits and a good approach to learning math right from the beginning. 2,230 words.
Learn the Multiplication Tables
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Proficiency in multiplication requires a mastery of the multiplication tables. Help your student obtain this ability by using varied repetition that provides immediate positive reinforcement.
Basic Mathematics Review you really need
Series: The Best Math Companion. Price: \$0.99 USD. Words: 910. Language: English. Published: January 4, 2015. Categories: Nonfiction
The important everyday skill is percentage math. We need percentage at home when making big decisions like to buy a new car or house. At work percentage skill is the most important skill because all decisions base of money, discaunt base of percentage and profit will base of percentage. Let me help you to learn more from percentage. | 1,557 | 6,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-35 | latest | en | 0.882776 |
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Unit 3 Section 1 and Section 2 - Unit 3 Exponential and...
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Unit 3 Exponential and Logarithmic Functions In the previous unit, we discussed algebraic functions, their domains, ranges and graphs. For the next two units, we will be looking at functions which are not algebraic, called transcendental functions. For this unit, we will consider a pair of one-to-one transcendental functions, which are inverses of each other. These are the exponential and logarithmic functions. We will also consider their domains, ranges and graphs. In Section 3, we will go back to solving equations. This time, the equations will involve exponential and logarithmic expressions. In the last section, we will also have a glimpse of another set of transcendental functions, called the hyperbolic functions. In this Unit, we aim to do the following: 1. Identify properties and sketch graphs of exponential functions, 2. Identify properties and sketch graphs of logarithmic functions, 3. Solve equations involving exponential and logarithmic functions, 4. Perform operations on hyperbolic and inverse hyperbolic functions, and 5. Solve verbal problems involving exponential and logarithmic functions. 3.1 Exponential Functions: Properties and Graphs Consider this problem: An amount of PhP 5,000 is deposited at 6% interest compounded annually. Find the total amount after 3 years. Let us observe at what happens to the amount invested in the table below. Year Amount invested Interest Earned Total Amount 1 5000 300 5300 2 5300 318 5618 3 5618 337.08 5955.08 Note that every year the amount invested increases because of the interest earned in that year. Thus, every year the interest earned also increases. Now, let us give an expression that will give us the total amount in t years, where t is any natural number. If P is the amount invested and i is the interest rate (in decimal form), then we have . In this expression, the variable for time is the exponent, unlike in the previous expressions that we have seen where we have constants as exponents. An expression like this is called an exponential expression.
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Unit 3. Exponential and Logarithmic Functions page 2 Let us first look at exponential functions and their properties. Definition If and , then the exponential function with base b is defined by . Illustration 1 Let us consider the function defined by . Since the variable is the exponent, the function values will just give us various powers of 2. Observe the following function values when x is an integer shown on the table on the left. 0
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This note was uploaded on 05/13/2011 for the course MATH 17 taught by Professor Dikopaalam during the Spring '11 term at University of the Philippines Los Baños.
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Unit 3 Section 1 and Section 2 - Unit 3 Exponential and...
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Ask a homework question - tutors are online | 713 | 3,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2017-17 | longest | en | 0.914181 |
https://www.nuprl.org/wip/Mathematics/reals/rroot-abs-property.html | 1,696,081,672,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00411.warc.gz | 999,752,181 | 4,613 | ### Nuprl Lemma : rroot-abs-property
`∀i:{2...}. ∀x:ℝ. (rroot-abs(i;x)^i = |x|)`
Proof
Definitions occuring in Statement : rroot-abs: `rroot-abs(i;x)` rabs: `|x|` rnexp: `x^k1` req: `x = y` real: `ℝ` int_upper: `{i...}` all: `∀x:A. B[x]` natural_number: `\$n`
Definitions unfolded in proof : member: `t ∈ T` uall: `∀[x:A]. B[x]` all: `∀x:A. B[x]` subtype_rel: `A ⊆r B` nat: `ℕ` int_upper: `{i...}` nequal: `a ≠ b ∈ T ` decidable: `Dec(P)` or: `P ∨ Q` uimplies: `b supposing a` not: `¬A` implies: `P `` Q` satisfiable_int_formula: `satisfiable_int_formula(fmla)` exists: `∃x:A. B[x]` false: `False` top: `Top` and: `P ∧ Q` prop: `ℙ` nat_plus: `ℕ+` int_nzero: `ℤ-o` true: `True` sq_type: `SQType(T)` guard: `{T}` so_lambda: `λ2x.t[x]` real: `ℝ` so_apply: `x[s]` le: `A ≤ B` less_than': `less_than'(a;b)` uiff: `uiff(P;Q)` rnexp: `x^k1` has-value: `(a)↓` bool: `𝔹` unit: `Unit` it: `⋅` btrue: `tt` ifthenelse: `if b then t else f fi ` bfalse: `ff` bnot: `¬bb` assert: `↑b` reg-seq-nexp: `reg-seq-nexp(x;k)` iff: `P `⇐⇒` Q` rev_implies: `P `` Q` bdd-diff: `bdd-diff(f;g)` ge: `i ≥ j ` rroot-abs: `rroot-abs(i;x)` squash: `↓T` sq_stable: `SqStable(P)` regular-int-seq: `k-regular-seq(f)` less_than: `a < b` label: `...\$L... t` rev_uimplies: `rev_uimplies(P;Q)` subtract: `n - m` cand: `A c∧ B`
Rules used in proof : sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity cut introduction extract_by_obid sqequalHypSubstitution isectElimination thin dependent_functionElimination hypothesisEquality hypothesis applyEquality because_Cache sqequalRule dependent_set_memberEquality_alt setElimination rename natural_numberEquality unionElimination independent_isectElimination approximateComputation independent_functionElimination dependent_pairFormation_alt lambdaEquality_alt int_eqEquality isect_memberEquality_alt voidElimination independent_pairFormation universeIsType addEquality multiplyEquality lambdaFormation_alt instantiate cumulativity intEquality equalityTransitivity equalitySymmetry equalityIstype baseClosed sqequalBase functionEquality inhabitedIsType applyLambdaEquality pointwiseFunctionality promote_hyp baseApply closedConclusion productElimination callbyvalueReduce equalityElimination setEquality functionIsType imageElimination imageMemberEquality universeEquality productIsType minusEquality lessCases isect_memberFormation_alt axiomSqEquality isectIsTypeImplies divideEquality equalityIsType4 remainderEquality hyp_replacement equalityIsType1
Latex:
\mforall{}i:\{2...\}. \mforall{}x:\mBbbR{}. (rroot-abs(i;x)\^{}i = |x|)
Date html generated: 2019_10_30-AM-07_56_17
Last ObjectModification: 2019_10_10-AM-10_22_50
Theory : reals
Home Index | 917 | 2,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.359988 |
https://wikiwap.com/why-first-see-lightning-and-then-hear-the-thunder/ | 1,631,910,906,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00577.warc.gz | 672,563,092 | 29,554 | Home Earth Why first see lightning and then hear the thunder?
Why first see lightning and then hear the thunder?
In summer, there are often lightning and thunder (lightning). When the electric fields between the positive and negative charges in the rain cloud differ significantly, these two types of charges will neutralize and cause lightning. That phenomenon is called lightning discharge. When lightning strikes, there is a flash of intense bright light. Furthermore, the lightning path produces very high temperatures, causing the surrounding air to expand, causing intense explosions suddenly. The flash is lightning, and the explosion is the sound of thunder.
Lightning and thunder arise in the presence of lightning, but why do we see lightning first and then hear the thunder? That’s because the speed of light is much faster than the speed of sound. Light travels through the air at a speed of 300,000 km / s, which equates to 7.5 cycles of equatorial circumference in a second. The sound speed in the air is only 340 m / s, only one-tenth of the speed of light. Since the lightning occurred to the ground, the time was only one-tenth of a second, but with that distance, the thunder had to travel a quite long time. Based on that, we can use the time from seeing lightning to hearing thunder to work out how far away the discharge is.
There are times when only lightning can be heard, but no thunder is heard; it is because the thundercloud is too far away from us, or the sound of thunder cannot be heard enough. As the energy of sound traveling through the air gradually decreases, we cannot hear it in the end.
There will be a corresponding thunder once there is lightning, but why are there times when only a flash of lightning is seen, and thunder’s sound lasts for a while before it stops?
This is because the flash is very long; there are flashes of up to 2 – 3 km, even 10 km. Because lightning is separated from us at different distances, the time it takes for thunder to reach our ears before and after is also different. On the other hand, lightning usually does not arise once and finishes, but it arises several times in a blink of an eye. So when the thunder of the first flash has not ended, the second and third lightning has passed. These thunderstorms are mixed to make thunder resounding forever.
When rain encounters the ground, structures, high mountains, or clouds, the sound is reflected, causing echoes. The time these responses reach our ears also varies greatly, thus causing thunder to resonate. Sometimes due to many reasons, the thunder resounds forever, lasting about a minute.
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Edison's phonograph underwent constant innovation to become a popular electric phonograph. It is made up of the motor, the turntable, the receiver, and the...
What is the principle of the fire detector?
Fire harms human lives and property, damages the ecological environment. Fire prevention is a matter of concern to all humans. If it is possible...
What is the principle of the theft alarm?
In important places such as money, gold, silver, underground record storage, etc., if a fraudster suddenly sneaks in. the burglar alarm will immediately sound... | 798 | 3,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.937404 |
http://www.helpingwithmath.com/by_subject/geometry/geo_quadrilaterals_5g3.htm | 1,417,069,943,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931008105.47/warc/CC-MAIN-20141125155648-00048-ip-10-235-23-156.ec2.internal.warc.gz | 596,933,194 | 18,373 | Please consider helping us with an online survey which is part of an academic research project. The survey should not take longer than 10 minutes. Learn more about it here.
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Home > By Subject > Geometry > Quadrilaterals
The short answer to "what is a quadrilateral?" is that it is a four-sided polygon. In other words it is four-sided, two-dimensional enclosed shape with straight sides.
Your children will learn (typically in 5th Grade) that there are many different types of quadrilaterals and they can be arranged in a hierarchy based on their properties.
Discuss the relationship between the quadrilaterals with your children. Follow the arrows down to see how certain quadrilaterals are also special types of others as they exhibit all their properties. For example, a square is a special type of rectangle. Indeed a square can be said to be a special case of all the other quadrilaterals shown in the table.
The table shows different types of special quadrilaterals arranged in a hierarchy. | 206 | 1,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2014-49 | longest | en | 0.944763 |
https://docs.flojoy.ai/blocks/scipy/signal/kaiser-beta/ | 1,721,653,978,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00283.warc.gz | 181,135,073 | 27,659 | KAISER_BETA
The KAISER_BETA node is based on a numpy or scipy function. The description of that function is as follows: Compute the Kaiser parameter 'beta', given the attenuation 'a'. Params: a : float The desired attenuation in the stopband and maximum ripple in the passband, in dB. This should be a *positive* number. Returns: out : DataContainer type 'ordered pair', 'scalar', or 'matrix'
Python Code
from flojoy import OrderedPair, flojoy, Matrix, Scalar
import numpy as np
import scipy.signal
@flojoy
def KAISER_BETA(
default: OrderedPair | Matrix,
) -> OrderedPair | Matrix | Scalar:
"""The KAISER_BETA node is based on a numpy or scipy function.
The description of that function is as follows:
Compute the Kaiser parameter 'beta', given the attenuation 'a'.
Parameters
----------
a : float
The desired attenuation in the stopband and maximum ripple in
the passband, in dB. This should be a *positive* number.
Returns
-------
DataContainer
type 'ordered pair', 'scalar', or 'matrix'
"""
result = scipy.signal.kaiser_beta(
a=default.y,
)
if isinstance(result, np.ndarray):
result = OrderedPair(x=default.x, y=result)
else:
assert isinstance(
result, np.number | float | int
), f"Expected np.number, float or int for result, got {type(result)}"
result = Scalar(c=float(result))
return result
Find this Flojoy Block on GitHub | 322 | 1,342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.406424 |
https://www.teacherspayteachers.com/Product/Calculus-Integrals-U-substitution-Matching-Game-5203092 | 1,686,057,319,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00308.warc.gz | 1,102,452,684 | 41,780 | EASEL BY TPT
Total:
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# Calculus Integrals: U-substitution Matching Game
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### Description
Integrals U-substitution Matching Game is the perfect activity for your students to sharpen their understanding of U-substitution! Functions include polynomial expression that have the following types of exponents: Positive Integers in Numerator and Denominator, Negative Integers, Rationals, and Radicals.
This product contains 5 Different Matching Card Sets for you and your students. You can choose to use them in any combination to best suite your classroom's needs.
This Set Includes:
5 Different Matching Card Sets
• Each Matching Card Set contains:
• 15 Question cards
Here are some ideas of how you could use this product:
Groups: Pass out a different version of the Matching Cards to each group/student. This promotes individual work seeing as all groups/students will have 15 different problems to solve.
Stations: Set out all 5 Matching Card sets as different stations for students to rotate through. This reinforces the concept with extra practice and allows students to rotate through all 75 problems.
Ultra High Resolution & Gorgeous Typography: All materials are created in ultra high resolution for printing at any size. We use the same mathematical typography used in textbooks and research papers within our products for optimal readability and display.
Look for more Activities & Games on TPT by Qwizy or on qwizy.com! Reviews and Feedback are Always Appreciated. Additionally, if you love this product but wish it covered a different math concept please contact me to let me know. New products can be made quickly to fill any of your needs.
Total Pages
29 pages
Included
Teaching Duration
30 minutes
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Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TPT’s content guidelines. | 724 | 3,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.861393 |
https://www.deriscope.com/products/DayCount__Name__ACT_360.html | 1,709,386,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00043.warc.gz | 727,647,952 | 2,361 | ## ACT 360
Subtype of Name
Actual/360 convention, also known as Act/360, or A/360
Let [T₁,T₂) be the period between two dates T₁ and T₂, of which the length L in annual units must be calculated
Calculation rule:
Let N be the actual number of days of the period [T₁,T₂), formally defined as the difference: T₂ - T₁
Then
L = N/360 | 99 | 330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.939098 |
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# View Sketchup 7.1 For Architectural Visualization: Beginner\'s Guide
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https://tft.brainiac.com/archive/1805/msg00065.html | 1,638,459,399,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00608.warc.gz | 620,503,408 | 4,503 | # Re: Hexagon pattern dungeons
We experimented with a few facing patterns at first, then settled on...
FFF
SXS
SRS
...to keep it simple.
On May 3, 2018, at 8:02 PM, Craig Barber <craigwbar@comcast.net> wrote:
I don't want to answer for him, but I recall one of the early D&D versions had facings defined for squares... if my Jurassic Memory serves.
2.65 huh? Thanks! That's a really USEFUL number, "one that I'll remember for as long as I can." (To quote Moon Unit Zappa). -- Craig B.
On 5/3/2018 5:43 PM, David Bofinger wrote:
I guess that leads to similar effects. The distance between two adjacent megahexes is sqrt(7) = 2.65 times the distance between adjacent hexes. The average distance between two adjacent megasquares is 3 times the distance between adjacent squares. Not too different.
I'd be tempted to say that diagonal moves cost 1.5 but maybe that isn't essential.
A diagonally adjacent character is quite a lot further away than an orthogonally adjacent one.
How did you handle facing? Maybe three front squares, a square where you can attack with a one-handed weapon but not use a shield, a square where you can use a shield but not a weapon, two side hexes and a rear hex?
I think hexes are better than squares in natural environments, but squares are better in artificial ones.
--
David
On 4 May 2018 at 09:20, Rich wrote:
Risking being shunned here. .. I switched to squares: squrares replacing hexes, with a 3 sq. by 3 sq block replacing megahexes. It works well.
Sent from BlueMail
On May 3, 2018, at 6:51 PM, Craig Barber < craigwbar@comcast.net> wrote:
```And if I understood another comment correctly, then yeah, I frequently
laid out hexagon pattern dungeons, enough so it was probably right
around 1/3 , 1/3, 1/3: about 1/3 hexagon, 1/3 square, and 1/3 "other",
like ferinstance Death Test (R) style.
I always liked the hexagonal dungeon layouts because they felt *very*
TFT to me. I always liked the square ones because they felt
traditional "white box" and 1st Ed. D&D, even back in 1985. Just
visceral reactions...
The only problems with squares were that 1) range got messed up on
diagonals to the grain and 2) they didn't really offer an easy use for
megahexes, the only problem with hexes was that villages and towns
didn't usually have rows and rows of lozenge shaped buildings!
Megahexes were a GREAT aid to speed play. "Wait, was that 14 hexes or
15 hexes range? Count again!" TFT could be a light light system
partially because of the megahexes.
After I invented the square megahexes, it was easy to have *both* of my
cakes and eat them both too. Any square grain dungeon could be played
on megahexes from then on. | 733 | 2,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-49 | latest | en | 0.953238 |
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# Pretty Patterns
Students will be able to reproduce and extend a pattern using various objects.
(10 minutes)
• Have students come together as a group.
• To motivate the students, begin by saying that "Today, we will be learning about patterns. Look around the room. Can anyone identify a pattern?"
• If no one knows what a pattern is, then make a simple pattern on the whiteboard, e.g. circle, triangle, circle, triangle, circle, triangle.
• Alternate differently colored markers to make the pattern more easily identifiable.
• Inform the students that this is a pattern of circles and triangles.
(10 minutes)
• Draw five different simple patterns on the board. Use various geometric shapes and objects to create bright, colorful patterns. (This can also be done in advance.)
• At the end of each pattern, one or two spaces should be blank. For example, heart, square, heart, square, heart, ___.
• As a group, have the students call out the shape or object by its name and color.
• When you get to the end ask who knows what the last shape or object should be.
• Have the students give the correct answer and the teacher can draw in the space.
(10 minutes)
• Provide the students with construction paper and glue.
• Ask the students to each create a simple pattern with foam geometric shapes.
(10 minutes)
• Provide each student with a copy of the Cut It Out! worksheet and a pencil. (You can either pre-cut the picture pieces at the bottom of the page for each student or have the students cut out their own pictures.)
• Read the instructions from the worksheet to the students.
• Once students have cut out their picture pieces, have them complete their worksheet by gluing on the correct object were needed.
• Enrichment: Provide advanced students with more complex patterns, such as the ones on the Circle Up! worksheets.
• Support: Cut out struggling students' pictures for them, and allow them to work together to complete their assignments.
(10 minutes)
• Have the students each create a complete pattern with two different shapes.
(10 minutes)
• Have the students listen to the story Busy Bugs: A Book About Patterns.
• Ask students to identify the patterns on the butterflies and ladybugs.
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### New Collection>
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https://discourse.julialang.org/t/matrix-multiplication-with-custom-types/25280 | 1,656,199,127,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00703.warc.gz | 250,777,565 | 6,707 | Matrix multiplication with custom types
Hey there.
So I’m very new to coding with Julia and am trying to get a handle on the “generalizability” of the language, as I’ll call it.
Below I created a Point strcut/class/? and defined the + and * operations of two points. I want to figure out how to take this and make the last line work somewhat like regular matrix multiplication works. Just instead of scalars the entries of the resulting matrix would be Point’s. Could someone point me in the direction to look to do this?
``````import Base.+
import Base.*
struct Point
x::Number
y::Number
end
+(p1::Point, p2::Point) = Point(p1.x + p2.x, p1.y + p2.y)
*(p1::Point, p2::Point) = Point(p1.x * p2.x, p1.y * p2.y)
a = Point(1,1)
b = Point(2,2)
c = a + b
point_matrix = [Point(i, j) for i in 1:3, j in 4:6]
point_matrix
point_vector = [a; b; c]
point_vector
point_matrix * point_vector
``````
Below is the error that is generated
MethodError: no method matching zero(::Point)
Closest candidates are:
zero(!Matched::Type{LibGit2.GitHash}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.1/LibGit2/src/oid.jl:220
zero(!Matched::Type{Pkg.Resolve.VersionWeights.VersionWeight}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.1/Pkg/src/resolve/VersionWeights.jl:19
zero(!Matched::Type{Pkg.Resolve.MaxSum.FieldValues.FieldValue}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.1/Pkg/src/resolve/FieldValues.jl:44
Stacktrace:
[1] generic_matvecmul!(::Array{Point,1}, ::Char, ::Array{Point,2}, ::Array{Point,1}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.1/LinearAlgebra/src/matmul.jl:542
[2] mul! at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.1/LinearAlgebra/src/matmul.jl:76 [inlined]
[3] *(::Array{Point,2}, ::Array{Point,1}) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.1/LinearAlgebra/src/matmul.jl:50
[4] top-level scope at In[12]:1
Also, if anyone could give me formatting tips or a link to a guide on how to make a better post that would be great!
Welcome to Julia!
So the error you’re running into is that it wants a `zero` function defined, which it uses at some point during the matrix multiplication algorithm. So it wants you to define a function `zero(::Point) = ...` that returns a zero object for the `Point` type, i.e. something so that that `x + zero(Point) == x` whenever `x` is a `Point` (in other words, `zero(T)` should return the additive identity for things of type `T`, but it doesn’t know what this is for `Point`s, so it wants you to define the method).
Formatting tips: PSA: how to quote code with backticks
1 Like
Thanks for that! Still getting used to understanding the errors in Julia. I added the lines
``````import Base.zero
zero(x::Point) = Point(0,0)
``````
to the code and it worked.
Is there a place that indicates the types of functions I have to import from Base to get them to work like this? I’ll look through the docs a bit.
The answer is “each of them that you use” (any function from Base that you want to override requires that you do one of two things)
``````import Base: +, *, zero
zero(x::MyType) = ...
``````
or (does not require the import statement)
``````Base.zero(x::MyType) = ...
``````
also, your type will work more quickly if you define it this way
``````struct Point{T<:Real}
x::T
y::T
end
``````
Just to add, with the parametric definition (`struct Point{T <: Real}`), you would want to define `zero` as
``````import Base.zero
zero(::Point{T}) where {T} = Point(zero(T), zero(T))
``````
so that way you get `Point`'s with the right numeric type:
``````julia> zero(Point(1.0, 2.0))
Point{Float64}(0.0, 0.0)
julia> zero(Point(1, 2))
Point{Int64}(0, 0)
``````
This actually illustrates one of the advantages of the parametric definition; without it, you don’t know ahead of time what the numeric types of `p.x` and `p.y` are for a `Point` `p`, but in the parametric case you do since you have access to the parameter `T` to use in `zero`.
(You could define `zero(p::Point) = Point(zero(p.x), zero(p.y))` in the non-parametric case, but this involves a runtime lookup to figure out the type of `p.x` in order to choose the right `zero` method to use when you call `zero(p.x)`, whereas in the parametric case it’s all done at compile time.)
1 Like
Thanks for the extra tidbit Jeffrey!
Thanks for the detailed explanation Eric! I’ll be sure to look this up.
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http://mathhelpforum.com/calculus/153334-estimating-maximum-possible-percentage-error.html | 1,481,063,904,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00357-ip-10-31-129-80.ec2.internal.warc.gz | 164,287,934 | 9,583 | ## Estimating the maximum possible percentage error
1. The problem statement, all variables and given/known data
Hi all! Having a problem with the ending of this question and I was wondering what I needed to do from here...
The coefficient of rigidity, n, of a wire of length L and uniform diameter d is given by
n=AL/d^4
If A is a constant and the parameters L and d are known to within ± 2% of their correct values, estimate the maximum possible percentage error in the calculated value of n.
Here is what I have so far...
dn = dn/dL x dL + dn/dd x dd
dn = d/d^4 dL + L/d^4 dd
n = d/d^4 x ΔL + L/d^4 x Δd
n = ΔL/L + Δd/d
Where do I go from here? Any help will be greatly appreciated | 192 | 694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.913974 |
http://www.usingenglish.com/forum/ask-teacher/28561-finite-vs-infinite-clauses-dummies.html | 1,386,705,835,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164024169/warc/CC-MAIN-20131204133344-00089-ip-10-33-133-15.ec2.internal.warc.gz | 803,962,190 | 12,274 | # Thread: Finite vs. Infinite clauses for dummies
1. Newbie
Join Date
Oct 2006
Posts
3
## Finite vs. Infinite clauses for dummies
Hi,
This is my first time posting. I am taking a course wherein I have to understand the difference between finite and infinite clauses.
On a basic level I understand what the words mean. What I cannot grasp is how to determine if a sentence is finite or infinite.
Can anyone give me a hand...as thought you're talking to a two-year old?
Thanks.
2. ## Re: Finite vs. Infinite clauses for dummies
Non-finite clauses cannot be stand-alone clauses, meaning that they are always subclause of a finite main clause.
1)I asked him to do that.
2) Having done that, I went home.
That doesn't mean, of course, that all subclauses are non-finite.
3) I went to the house that Jack built. (both clauses are finite)
non-finite clauses can be to-clauses (infinitive), like in example 1, -ing clauses (see example 2), or even bear infinitive clauses:
I made him [to] do that
3. Newbie
Join Date
Oct 2006
Posts
3
## Re: Finite vs. Infinite clauses for dummies
Thank you, Mariner.
When you say non-finite clause, is that the same as an infinite clause? Your response talked mostly about non-finite clauses and at the end you said, "non-finite clauses can be to-clauses (infinitive)..." My total brain-block is about infinite and finite clauses. Even after your kind response, I have to admit I am still mostly clueless.
Can you suggest a "method" (or formula, if you will) for picking out a finite clause and an infinite clause? I have done some online exercises and my answers are correct only by chance. I don't truly understand.
Sigh...
Thank you so very much for your time.
Lara
4. ## Re: Finite vs. Infinite clauses for dummies
nonfinite clause=infinite clause
About -to infinitive clause, it's a different thing than "infinite (nonfinite) clause". See Clauses Page 3
As for a method to identify them, the easiest way is to look for the nonfinite clause "signals", such as an -ing ending, a -to+infinitive, or a bare infinitive (bare infinitive means, in essence, a case where "adding" to before the verb does not change the meaning. Here are 3 examples, one for each category. The nonfinite clauses are in bold.
1) Having been there, I don't want to go back.
2) You should tell them to finish immediately.
3) He made me do it.
If there's still something unclear, ask!
5. Newbie
Join Date
Oct 2006
Posts
3
## Re: Finite vs. Infinite clauses for dummies
Dear Mariner,
Thank you for this additional information. Until the information is able to sink below the epidermal level of my head, perhaps your signals will at least get me into the "safe" zone when trying to identify these types of clauses.
Thank you so very much for your time.
Lara
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University of Engineering and Technology Lahore Department of Electrical Engineering Calculus I Problem Set #4 Due date: December 10 , 2010 Announcements Quiz 2 next week on Thursday. It will cover Problem Sets 1-3 and Questions 1-4 of Problem Set 4. That means differentiation will not be a part of the quiz. Office hours: Mondays and Wednesday at 3 p.m. Apart from that I am available every day (except Fridays and Saturday) between 10 a.m.-1 p.m. I may be have drifted downstairs to the Computing Lab on the ground floor. If you do not find me in the office, just ask around in the Software Engineering Centre and you should be able to locate me easily. Reading Thomas’ Calculus: Sections 3.1-3.6. Spivak: Chapter 7 (pages 120-122, and the statements and pictorial “proofs” of Theo- rems 9, 10 and 11), Chapter 8 (pages 131-133), Chapter 9, Chapter 10 (pages 166-168 and 176-179) 1 . 1. Thomas’ Calculus Additional and Advanced Exercises Chapter 2. Question 18, 19.
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Unformatted text preview: 2. Spivak Chapter 7: Question 1 (i), (iv), (vii), (ix), (x) and (xi), 3, 5, 18 (a) and (b). 3. Two nonintersecting roads lead from city A to city B . It is known that two cars traveling from A to B over different roads and joined by a cord of a certain length less than 2 l were able to travel from A to B without breaking the cord. Is it possible for two circular wagons of radius l whose centers move over these roads toward each to pass without touching each other? 2 4. Spivak Chapter 8: Question 1 (i), (ii), (v), (viii), 12. 5. Thomas’ Calculus Exercise 3.1: Question 17(i). 6. Spivak Chapter 9: Question 5, 7, 10, 12, 14 7. Spivak Chapter 10: Questions 1 (i) and (ii), 2 (ii) and (iv), 4 (ii), 5 (i), 7, 11, 33 (i) and (ii). 1 You may feel this is a lot of reading, but essentially you already know most of this material, so we will not cover everything in class. 2 Courtesy V.I. Arnold. 1...
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Get Full Access to Applied Statistics And Probability For Engineers - 3 Edition - Chapter 6-7 - Problem 6-64
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# Solved: Construct a normal probability plot of the insulating fluid breakdown time data
ISBN: 9780471204541 346
## Solution for problem 6-64 Chapter 6-7
Applied Statistics and Probability for Engineers | 3rd Edition
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Problem 6-64
Construct a normal probability plot of the insulating fluid breakdown time data in Exercise 6-2. Does it seem reasonable to assume that breakdown time is normally distributed?
Step-by-Step Solution:
Step 1 of 3
STAT 217: STATISTICAL METHODS In-class Final Exam Review Use JMP to help you answer these questions. Each research question below requires a confidence intervalor hypothesis test since these questions are all about the population, rather than the sample. Decide whether each requires a confidence interval or hypothesis test, then carry out the complete analysis showing all steps. Each question is referring to either the backpack-FinalReview.jmp data or thecaraccidents.jmp data from earlier in the quarter. Steps of a Confidence Interval: Steps of a Hypothesis Test: 1. Describe parameter.
Step 2 of 3
Step 3 of 3
#### Related chapters
Unlock Textbook Solution | 362 | 1,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.798125 |
https://www.enotes.com/homework-help/prove-that-x-x-1-1-1-3x-2-1-x-2-1-x-1-2x-247985 | 1,669,672,552,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710662.60/warc/CC-MAIN-20221128203656-20221128233656-00613.warc.gz | 805,836,987 | 17,613 | # prove that (x/(x+1)+1):(1-3x^2/(1-x^2))=(1-x)/(1-2x)
We have to prove that (x/(x+1)+1):(1-3x^2/(1-x^2))=(1-x)/(1-2x)
Starting with the left hand side
(x/(x+1)+1):(1-3x^2/(1-x^2))
=> [(x/(x+1)+1)]/[(1-3x^2/(1-x^2))]
=> [(x+x+1)/(x+1)]/[(1-x^2-3x^2)/(1-x^2)]
=> [(2x+1)/(x+1)]/[(1-4x^2)/(1-x^2)]
=> [(2x+1)/(x+1)]/[(1-2x)(1+2x)/(1-x)(1+x)]
=> [(2x+1)(1-x)(1+x)/(x+1)(1-2x)(1+2x)]
=> [(1-x)/(1-2x)]
which is the right hand side
This proves that (x/(x+1)+1):(1-3x^2/(1-x^2))=(1-x)/(1-2x)
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We have three 6th grade Science classes and two 8th grade Science classes blogging here from the Pacific Northwest in Chimacum, WA! Sixth graders are learning a bit about Mt Saint Helens, environmental science through fresh water ecology, and physical science this year. Eighth graders are learning about life science this year. Please join us as we learn Science by exploring our world. Mr. G's Blog Mr. G's Class Facebook Page
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WATER CYCLE 12/06/11 qiuck question 11/02/11 Water pollution 10/20/11 Mt.St.Helens 10/03/11 7 random facts!!! 09/30/11
Lesson 6
OK,here is my lesson 6 blog post:
I measured sliding friction force by setting down our surface(course sandpaper) on the table.Then,I
hooked the hook on the Newtons scale to the hook on the wooden block.Then,set the ruler next
to the paper.Pull the newtons scale at a medium speed and the wooden block should follow.Get
to 27 cm and then stop.check the number on the scale.that is the friction force.
I learned,about controlled variables,that Most of the time,you can predict what
happens.Examples are,Surface area,Weight of the load,How far you pull it,etc.
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Copyright (c) 2014 by Conditions of Use Privacy Policy Return to Blogmeister | 856 | 2,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2014-15 | longest | en | 0.842309 |
http://calculator.mathcaptain.com/circumference-calculator.html | 1,532,042,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00510.warc.gz | 57,744,612 | 8,087 | Circumference calculator is an online maths calculator which helps to calculate the circumference of a circle. It is given as 2$\pi$ times of radius. Mathematically it can be represented as,
C=2$\pi$ r
Where r is the radius of curvature of the circle
## Steps for Calculating Circumference
Step1: Pick the given value for radius r.
Step2: Substitute the given value in the equation.
C=2$\pi$ r
### Area of a Circle Calculator
Circumference Calculation Circumference Circle Equation Calculating Arc Length of a Circle A Circle A Chord of a Circle A Sector of a Circle
Calculate the Circumference of a Circle Circumference Diameter Calculator Circumference and Area of a Circle diameter to radius calculator Calculator Calculate Compound Interest Calculator | 168 | 765 | {"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} | 2.9375 | 3 | CC-MAIN-2018-30 | latest | en | 0.716999 |
https://number.academy/1008542 | 1,718,975,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00836.warc.gz | 382,216,977 | 11,176 | # Number 1008542 facts
The even number 1,008,542 is spelled 🔊, and written in words: one million, eight thousand, five hundred and forty-two, approximately 1.0 million. The ordinal number 1008542nd is said 🔊 and written as: one million, eight thousand, five hundred and forty-second. The meaning of the number 1008542 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 1008542. What is 1008542 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,008,542 in other units
The decimal (Arabic) number 1008542 converted to a Roman number is (M)(V)MMMDXLII. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
1008542 seconds equals to 1 week, 4 days, 16 hours, 9 minutes, 2 seconds
1008542 minutes equals to 2 years, 1 month, 9 hours, 2 minutes
### Codes and images of the number 1008542
Number 1008542 morse code: .---- ----- ----- ---.. ..... ....- ..---
Sign language for number 1008542:
Number 1008542 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 1008542 is not a prime number. The closest prime numbers are 1008541, 1008547.
#### Factorization and factors (dividers)
The prime factors of 1008542 are 2 * 17 * 29663
The factors of 1008542 are 1, 2, 17, 34, 29663, 59326, 504271, 1008542.
Total factors 8.
Sum of factors 1601856 (593314).
#### Powers
The second power of 10085422 is 1.017.156.965.764.
The third power of 10085423 is 1.025.845.520.565.556.096.
#### Roots
The square root √1008542 is 1004,261918.
The cube root of 31008542 is 100,283926.
#### Logarithms
The natural logarithm of No. ln 1008542 = loge 1008542 = 13,824016.
The logarithm to base 10 of No. log10 1008542 = 6,003694.
The Napierian logarithm of No. log1/e 1008542 = -13,824016.
### Trigonometric functions
The cosine of 1008542 is -0,94006.
The sine of 1008542 is 0,341008.
The tangent of 1008542 is -0,362752.
## Number 1008542 in Computer Science
Code typeCode value
1008542 Number of bytes984.9KB
Unix timeUnix time 1008542 is equal to Monday Jan. 12, 1970, 4:09:02 p.m. GMT
IPv4, IPv6Number 1008542 internet address in dotted format v4 0.15.99.158, v6 ::f:639e
1008542 Decimal = 11110110001110011110 Binary
1008542 Decimal = 1220020110102 Ternary
1008542 Decimal = 3661636 Octal
1008542 Decimal = F639E Hexadecimal (0xf639e hex)
1008542 BASE64MTAwODU0Mg==
1008542 SHA14639212421b3373a59565b063072269b72927412
1008542 SHA2246839cb665bb659e0b4a4ff2654cef1e31f081bd7d49890202daaed36
1008542 SHA2566052bc03d12b2895575f9587766220281c3e65d8d7ce5007f7552fc81c1347a2
More SHA codes related to the number 1008542 ...
If you know something interesting about the 1008542 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1008542
### Character frequency in the number 1008542
Character (importance) frequency for numerology.
Character: Frequency: 1 1 0 2 8 1 5 1 4 1 2 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1008542, the numbers 1+0+0+8+5+4+2 = 2+0 = 2 are added and the meaning of the number 2 is sought.
## № 1,008,542 in other languages
How to say or write the number one million, eight thousand, five hundred and forty-two in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.008.542) un millón ocho mil quinientos cuarenta y dos German: 🔊 (Nummer 1.008.542) eine Million achttausendfünfhundertzweiundvierzig French: 🔊 (nombre 1 008 542) un million huit mille cinq cent quarante-deux Portuguese: 🔊 (número 1 008 542) um milhão e oito mil, quinhentos e quarenta e dois Hindi: 🔊 (संख्या 1 008 542) दस लाख, आठ हज़ार, पाँच सौ, बयालीस Chinese: 🔊 (数 1 008 542) 一百万八千五百四十二 Arabian: 🔊 (عدد 1,008,542) مليون و ثمانية آلاف و خمسمائة و اثنان و أربعون Czech: 🔊 (číslo 1 008 542) milion osm tisíc pětset čtyřicet dva Korean: 🔊 (번호 1,008,542) 백만 팔천오백사십이 Danish: 🔊 (nummer 1 008 542) en millioner ottetusinde og femhundrede og toogfyrre Hebrew: (מספר 1,008,542) מיליון ושמונת אלפים חמש מאות ארבעים ושתיים Dutch: 🔊 (nummer 1 008 542) een miljoen achtduizendvijfhonderdtweeënveertig Japanese: 🔊 (数 1,008,542) 百万八千五百四十二 Indonesian: 🔊 (jumlah 1.008.542) satu juta delapan ribu lima ratus empat puluh dua Italian: 🔊 (numero 1 008 542) un milione e ottomilacinquecentoquarantadue Norwegian: 🔊 (nummer 1 008 542) en million åtte tusen fem hundre og førtito Polish: 🔊 (liczba 1 008 542) milion osiem tysięcy pięćset czterdzieści dwa Russian: 🔊 (номер 1 008 542) один миллион восемь тысяч пятьсот сорок два Turkish: 🔊 (numara 1,008,542) birmilyonsekizbinbeşyüzkırkiki Thai: 🔊 (จำนวน 1 008 542) หนึ่งล้านแปดพันห้าร้อยสี่สิบสอง Ukrainian: 🔊 (номер 1 008 542) один мільйон вісім тисяч п'ятсот сорок два Vietnamese: 🔊 (con số 1.008.542) một triệu tám nghìn năm trăm bốn mươi hai Other languages ...
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# Abduction as satisfiability
24 Feb 2018, 14:36 • logic
Effectively propositional abductive problems can be cast in terms of satisfiability. This post shows when it's possible and how to do it.
First-order abductive rules consist of an antecedent (on the left) and a consequent (on the right), whereby inference happens by backchaining on a rule, that is, the consequent is given (observed or assumed) and the antecedent can be assumed in order to explain some observables. In symbols
$A \supset C$
Let $$V(\varphi)$$ be the set of all variables occurring in a formula $$\varphi$$. A theory is effectively propositional if the observables are all ground and $$V(A)\subseteq V(C)$$ for all rules. An effectively propositional theory can be straightforwardly converted into a propositional theory so our rules now look like
$p_1 \land \dots \land p_n \supset q_1 \land \dots \land q_m$
where $$p_i,q_i$$ are propositional variables. However these rules are abductive and SAT solvers only work with classical deductive rules. We'll first introduce new propositional variables $$r_i$$ for the ground rules of the theory. We'll now add a few necessary conditions to the theory:
$\begin{array}{c} r_i \rightarrow c \\ r_i \rightarrow a \\ \end{array}$
for all propositional variables $$c$$ in the consequent of $$r_i$$ and $$a$$ in the antecedent of $$r_i$$. We'll also need
$\neg r_i \lor \neg r_j$
for all $$r_i,r_j$$ that share a common variable in their consequents, since a literal can be explained by at most one rule. We'll also need $$o_i$$ for all observables and $$r_i\rightarrow e_a$$ for all variables in the antecedent of $$r_i$$. For every variable representing a literal we'll then add $$o_i \lor e_i$$, that is, every literal in the proof is either observed or assumed by backchaining on a rule. These additional rules ensure that all proofs are correct in the sense that $$I \vdash O$$ and there are no extraneous literals.
We now have a set of classical propositional clauses so we can use a SAT solver to generate abductive proofs (it's easy to see that there's a one-to-one correspondence between abductive proofs and propositional valuations). | 543 | 2,185 | {"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} | 3.53125 | 4 | CC-MAIN-2019-18 | longest | en | 0.899252 |
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# quiz06 - t is given by t-t 2 8 t where you are at(0,0 and...
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Name: Quiz 6 07 Oct 2005 Write the best answer to each question in the box provided. Show your work. For the first two questions, do the following: (a) Give the variable that is changing in the problem to cause a maximum or minimum (this will be the independent variable in part (c)). Tell what this variable means in the problem. (b) Give the domain for the variable that is changing (c) Write down the function you are to maximimize or minimize in terms of the variable that is changing. 1. You toss a ball towards a basket that is located 10 feet in front of you and 10 feet above the ground. The position of the ball at time
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Unformatted text preview: t is given by ( t,-t 2 + 8 t ), where you are at (0,0) and the basket is at (10,10). At what time was the ball closest to the basket? (a) (b) (c) 2. A box is designed with a square base. It needs have a volume of 10 cubic feet. The cost of the material for the top and bottom of the box is \$2 per square foot and the cost of the material for the sides of the box is \$3 per square foot. How much does the most economical box cost? (a) (b) (c) 3. Find the absolute maximum and the absolute minimum points ( x and y-values) of x 3-12 x + 1 when x is in the interval [0 , 3]....
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{[ snackBarMessage ]} | 414 | 1,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.909805 |
https://www.justinmath.com/shaping-stdp-neural-networks-with-periodic-stimulation/ | 1,709,199,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474795.48/warc/CC-MAIN-20240229071243-20240229101243-00072.warc.gz | 835,351,320 | 7,995 | # Shaping STDP Neural Networks with Periodic Stimulation: a Theoretical Analysis for the Case of Tree Networks
by Justin Skycak on
We solve a special case of how to periodically stimulate a biological neural network to obtain a desired connectivity (in theory).
We solve a special case of how to periodically stimulate a biological neural network undergoing spike-timing dependent plasticity to obtain a desired connectivity (in theory).
## Model
We model neuronal networks as weighted directed graphs. Each vertex $i$ is a neuron having activity value $a_i(t) \in \lbrace 0, \delta(0) \rbrace$ at time $t,$ where $\delta$ is the Dirac delta function and represents an action potential or “spike.” Each edge $i \leftarrow j$ is a synapse in which neuron $i$ receives neurotransmitters from neuron $j,$ and the weight of the synapse at time $t$ is given by $w_{ij}(t).$ Neuron $j$ is said to be a neighbor of neuron $i$ if the weight of the synapse $w_{ij}(t)$ is nonzero.
A neuron becomes “excited” a duration of $\tau \approx 10$ ms [1] after a neighboring neuron spikes ($\tau$ is called the spike propagation latency). If it has been at least $r \approx 5$ ms since the neuron’s previous spike, then the excited neuron spikes ($r$ is called the refractory period). Otherwise, it does not. Formally, we have
\begin{align*} a_i(t) = \begin{cases} \delta(0) & \mbox{if } a_i(t') = 0 \mbox{ for } t-r < t' < t \\[3pt] & \mbox{and } \exists j \in V \mbox{ such that } w_{ij}(t-\tau) > 0 \mbox{ and } a_j(t-\tau) = \delta(0) \\[3pt] 0 & \mbox{otherwise} \end{cases} \end{align*}
The weights evolve over time according to the principles of weight decay and spike-timing dependent plasticity (STDP). Weight decay ensures that the weight of an unused synapse decays to zero. STDP is a biological neural learning rule where the weight is strengthened in a predictive fashion: if neuron $i$ spikes after neuron $j$ and the weight $w_{ij}$ is nonzero, then then the weight $w_{ij}$ increases and the weight $w_{ji}$ decreases. The smaller the time between the spikes of neurons $i$ and $j,$ the greater the magnitude of increase or decrease of the weight.
Let $s_i(t) = t - \min a_i^{-1} (\delta(0))$ be the time since neuron $i$ last spiked. The weights update according to
\begin{align*} \frac{dw_{ij}}{dt} = f_{ij}(t) w_{ij}(t) \delta(0) \end{align*}
where $f_{ij}(t)$ is the STDP function given by
\begin{align*} f_{ij}(t) = \begin{cases} \alpha e^{-ks_j(t)} & \mbox{if } a_i(t) = \delta(0) \\[3pt] - \beta e^{-ks_i(t)} & \mbox{if } a_j(t) = \delta(0) \\[3pt] 0 & \mbox{otherwise}. \end{cases} \end{align*}
In this STDP function, $k$ is on the order of $0.1 \textrm{ ms}^{-1}$ [2] and $\alpha, \beta \ll 1.$
We can also express the weight update as a multiplicative rule
\begin{align*} w_{ij}(t+dt) &= w_{ij}(t) + w_{ij}(t)f_{ij}(t) \\[5pt] &= (1+f_{ij}(t)) w_{ij}(t). \end{align*}
We ensure that there is not a bias towards increase or decrease of weights in the case of symmetric spike-timing differences by choosing $\alpha$ and $\beta$ such that
\begin{align*} \left( 1 + \alpha e^{-ks_j(t)} \right) \left( 1-\beta e^{-ks_i(t)} \right) = 1. \end{align*}
This is accomplished by choosing
\begin{align*} \beta = \frac{\alpha}{1-\alpha \exp(-ks_i(t))}. \end{align*}
With this choice, the STDP rule can be written as
\begin{align*} f_{ij}(t) = \begin{cases} \frac{\alpha}{\exp(ks_j(t))} & \mbox{if } a_i(t) = \delta(0) \\[3pt] -\frac{\alpha}{\exp(ks_i(t))-\alpha} & \mbox{if } a_j(t) = \delta(0) \\[3pt] 0 & \mbox{otherwise}. \end{cases} \end{align*}
In the following sections, we will investigate how periodic stimulation from an external source affects the connectivity of neuronal networks. We initialize weights as finite positive numbers, and we say that a weight
• solidifies if it is asymptotically infinite, i.e. $\lim_{t \to \infty} w_{ij}(t) = \infty$;
• breaks if it is asymptotically zero, i.e. $\lim_{t \to \infty} w_{ij}(t) = 0$; and
• remains fluid if it is neither asymptotically infinite nor asymptotically zero.
## Analysis
We first consider the case where a single neuron in a tree network experiences periodic stimulation from an external source, with period $p$. We consider only cases with $p \geq r = \frac{1}{2}\tau$; other cases with $p < r$ can be analyzed by replacing the period with the effective period:
\begin{align*} p \to \left \lceil \frac{r}{p} \right \rceil p \end{align*}
We find that, by choosing an appropriate period of stimulation, one can solidify or break any subtree by stimulating the root neuron of the subtree.
Proposition. For a network of neurons whose topology is a non-recombining tree, suppose the root neuron is excited by an external pulse every $p$ milliseconds.
• If $p = \frac{1}{2}\tau,$ then the tree will remain fluid.
• If $\frac{1}{2}\tau < p < \tau,$ then the tree will solidify.
• If $p=\tau,$ then the tree will remain fluid.
• If $\tau < p \leq 2\tau,$ then the tree will break.
• If $2\tau < p,$ then the tree will solidify.
Proof. It suffices to prove the result for a single chain of neurons, $\lbrace 1 \to 2 \to \cdots \to N \rbrace$, where neuron 1 receives the pulse. One pulse propagates through the chain every $p$ milliseconds, moving to the next neuron every $\tau$ milliseconds. Neuron $i$ spikes at times
\begin{align*} t=i\tau, i\tau+p, i\tau+2p, \ldots \end{align*}
and neuron $i+1$ spikes at times
\begin{align*} t=i\tau+\tau, i\tau+\tau+p, i\tau+\tau+2p, \ldots. \end{align*}
In other words, neuron $i$ spikes at times
\begin{align*} t-i\tau=0, p, 2p, \ldots \end{align*}
and neuron $i+1$ spikes at times
\begin{align*} t-i\tau = \tau, p+\tau, 2p+\tau, \ldots. \end{align*}
Case 1. If $p=\tau$ then $w_{i+1,i}$ experiences alternating increases and decreases, both with the same spike-time difference. Because our STDP rule has no bias towards increasing or decreasing weights in the case of symmetric spike-timing differences, the weights remain fluid.
Case 2. If $p = \frac{1}{2}\tau$, then $w_{i+1,i}$ again experiences alternating increases and decreases, both with the same spike-time difference. Just as in Case 1, the weights remain fluid.
Case 3. If $p>\tau$ then the order of (shifted) spike-times is as follows:
In this case, $w_{i+1,i}$ experiences alternating increases and decreases, with increases having time difference $\tau$ and decreases having time difference $p-\tau.$ Then we have
\begin{align*} \lim_{t\to\infty} w_{i+1,i}(t) = w_{i+1,i}(0) \lim_{t\to\infty} \left[ \left( 1 + \frac{\alpha}{\exp(k\tau)} \right) \left( 1 - \frac{\alpha}{\exp(k(p-\tau))-\alpha} \right) \right]^{t/p}. \end{align*}
Thus, the chain solidifies if
\begin{align*} \left( 1 + \frac{\alpha}{\exp(k\tau)} \right) \left( 1 - \frac{\alpha}{\exp(k(p-\tau))-\alpha} \right) &> 1 \\[5pt] \frac{1}{\exp(k\tau)} - \frac{\alpha + \exp(k\tau)}{\exp(kp)-\alpha\exp(k\tau)} &> 0 \\[5pt] e^{k(p-\tau)}-2\alpha + e^{kt} &> 0. \end{align*}
Letting $\alpha \to 0,$ we reach $p > 2\tau.$ Thus, if $p>2\tau$, then the chain solidifies, but if $\tau < p \leq 2\tau,$ then the chain breaks.
Case 4. If $\frac{1}{2} \tau < p < \tau$ then the order of (shifted) spike times is as follows:
We see that, $w_{i+1,i}$ experiences alternating increases and decreases, with increases having time difference $(1-\epsilon)\tau = \tau-p$, and decreases having time difference $(2\epsilon-1)\tau = 2p-\tau.$ Then we have
\begin{align*} \lim_{t\to\infty} w_{i+1,i}(t) = w_{i+1,i}(0) \lim_{t\to\infty} \left[ \left( 1 + \frac{\alpha}{\exp(k(\tau-p))} \right) \left( 1 - \frac{\alpha}{\exp(k(2p-\tau))-\alpha} \right) \right]^{t/p}. \end{align*}
Thus, the chain solidifies if
\begin{align*} \left( 1 + \frac{\alpha}{\exp(k(\tau-p))} \right) \left( 1 - \frac{\alpha}{\exp(k(2p-\tau))-\alpha} \right) &> 1 \\[5pt] \frac{1}{\exp(k(\tau-p))} - \frac{\alpha + \exp(k(p-\tau)}{\exp(kp)-\alpha\exp(k(p-\tau))} &> 0 \\[5pt] e^{k(2p-\tau)}-\alpha e^{2k(p-\tau)} - \alpha - e^{k(p-t)} &> 0. \end{align*}
Letting $\alpha \to 0,$ we reach $e^{k(2p-\tau)} > e^{k(p-\tau)},$ which is always true. Thus, the chain solidifies. ■
We are now concerned with breaking or solidifying a particular segment of a tree, since the Proposition gives us a way to break or solidify any subtree as a whole.
Corollary. Suppose that two neurons are stimulated with periods $p$ and $q$, where $q$ is applied to the neuron deeper in the tree. To solidify the connections between $p$ and $q$ while allowing the subtree under $q$ to remain fluid, we can choose
• $q=\frac{1}{2}\tau$ and $p=\frac{n}{2}\tau$, where $n>4$, or
• $q=\tau$ and $p=n\tau$, where $n>2$.
To break the connections between $p$ and $q$ while allowing the subtree under $q$ to remain fluid, we can choose
• $q=\frac{1}{2}\tau$ and $p=\frac{3}{2}\tau$ or $p=2\tau$, or
• $q=\tau$ and $p=2\tau$.
Proof. From the Proposition, we see that to solidify the connections between $p$ and $q,$ we must have $\frac{1}{2}\tau < p < \tau$ or $2\tau < p,$ and to break the connections between $p$ and $q,$ we must have $\tau < p \leq 2\tau.$ To ensure that the subtree under $q$ remains fluid, we can choose $q=\frac{1}{2}\tau$ or $q=\tau$ and choose $p$ such that its pulses are canceled by the stimulation at $q.$ ■
## Future Directions
One direction for future work is in expanding the theoretical analysis to more initial topologies, such as recombining trees, cycles, unions of cycles, and bidirectional networks. Another direction for future work is in making the model more realistic: one could
• include latent weights that can solidify but cannot cause the postsynaptic neuron to spike until solidified, or
• run simulations with e.g. leaky integrate-and-fire neurons.
## References
[1] Tovee, M. J. (1994). Neuronal Processing: How fast is the speed of thought? Current Biology, 4(12), 1125-1127.
[2] Jesper Sjostrom and Wulfram Gerstner (2010) Spike-timing dependent plasticity. Scholarpedia, 5(2):1362.
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http://salted.net/category/computer-game-philosophy/page/2/ | 1,529,887,093,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00095.warc.gz | 269,740,038 | 30,194 | # What is a Computer Game?
Using the Cartesian method we could enumerate this question as:
1. What is a computer game?
1. What is a computer?
1. A system designed to logically process information.
1. System: an causal structure
1. Structure: A consistent arrangement of elements.
2. Causal: Making something happen or not.
1. A causes B if…
2. “If the engine hadn’t stopped there would have been an accident.”
3. Designed: Made with intent for purpose.
4. Logically: principles that flow without contradiction from:
1. The law of The Excluded Middle
1. X is either true or not true.
2. These principles are AND, NOT and OR.
1. There are principles derived from these like XOR.
2. What is a game?:
1. An activity that satisfies the following:
1. It has rules
2. It has objectives
3. It is done for fun
1. What is Fun?
1. A fun activity increases positivity without there necessarily being a corresponding external effect.
2. An activity is fun if someone says they do it for enjoyment.
# The New No Go Game
More than a couple of times in my life I’ve tried to learn the Game Of Go.
This is the ancient Chinese game that is harder to solve (AI or heuristics) than Chess, or, I think, any other game. Only in 2016 was the best player properly beaten by a computer. One with the multiple brains and yadabyte memories driven by the vastness of The Google. It was a big moment for Humanity, I think.
Many good games are easy to learn and hard to master. The Game of Go isn’t like this. It is really hard to learn, and very few master it. They say it takes decades to become a mere “quite good”.
This, to me, has an inevitable subtracting effect on my enjoyment of the game; it is just so hard to be quite good.
It would take so much time and, during that valuable T-Spend, mostly, it would not be fun. There would be few “yeahs!”
It would not have many of those moments of understanding a new tactic or strategy that are common in other exceptional games.
The reason for it’s hardness isn’t like in some games.
The problem isn’t the games complexity, that is , how many rules and parameters and interconnections are involved in it’s realisation, AKA, “playing”.
Nor is it the game’s depth – that meaningful, relative measure of the structure of the hierarchy that represents the emergence of the tactics, strategy and fun experienced by the players.
Rather, perhaps uniquely with the Game Of Go, it is the game’s vast possibility space that seeds the issues for me.
## Possibility Space
A thing’s “possibility space” is the conceptual framework that the thing occupies. A puppy has a possibility space. So does a movie and an omelette.
The concept of “Castling A King” doesn’t exist in the possibility space of Naughts And Crosses. Nor does it exist in Hopscotch. But it does exist in Chess.
When you think about things in terms of their possibility space, you gain a new perspective.
Possibility space is useful to be seen as layered:
If CAK is in Chess and it is possible that Eastender’s might involve Chess (Eg in an episode. Which, of course, it is) then CAK is in the possibility space of Eastenders (Which, it is, isn’t it?).
This does not mean the same as the trivial “CAK is in the possibility space of all things, because, eg, Time-Travel.”
Possibility spaces are much easier to apprehend when they are seen as tied to shared frameworks. Think of how this influences the emergence of “plausibility” as a property of a narrative.
To be meaningful possibility spaces should be internally consistent. It is for this reason that I would say we wouldn’t say that CAK was in the possibility space of Eastenders but not of a historically accurate soap opera set in The Stoneage.
Lastly, possibility space isn’t just about points of property possibilities, but also emergent phenomena that the game originates. The excitement of backgammoning someone is not contained in the necessary and sufficient game description of Backgammon, but it is clearly in the possibility space of the game. It is a foundational property of the gameplay. We can deconstruct the possibilities and we can abstract the possibilities.
The Game Of Go is not deep in its structure. Because the board is nineteen wide, as opposed to eight in chess, there is a computational explosion in the amount of information that needs to be broken down and abstracted in order to significantly understand the game. It is a huge and wide and shallow sea of choices but from this sea emerges game phenomena that I have no idea about. I can see they are going to be there, but I cannot conceive of them. I think if you try to learn Go you will soon understand this, if you don’t already.
Then the realisation is this: until you get good enough to meaningfully understand these emergent game phenomena, then Go is going to remain a relatively shallow game. Shallow tactics and strategies compared to the big picture games of the Go-masters.
In my opinion Go is not very fun to learn because it is so very hard to average.
There is a time/cost/fun/potential/learnability/etc equation with any game.
# Thought Experiment: The Glove Game
Imagine that in the wrist part of the glove is a slit and one side of this is a small red button. Imagine that on the other side of the slit is a small loop that can go around the button, to hold the glove in place on a hand.
You have just imagined a glove. Now imagine this glove floating in a void of nothingness. No other things, no time, no light, no observer. Just the glove. All there is is this glove. Imagine the Glove Universe.
In fact it seems that one cannot imagine a universe that’s just a small red ladie’s glove. It is not conceptually possible for a number of reasons:
You can’t imagine something being “small” if thats’ the only thing there is. Smallness is a relative property, it needs more things to be realised than just one.
You can’t imagine something as being red if there is no light and no observer. Colours don’t make sense in the glove universe. You can imagine the surface of the Glove having properties that, were it on your left hand right now it would look red to you or I.
Perhaps the most interesting reason for why you cannot really imagine the glove universe is because a glove is a special kind of form called an “enatiomorph”. A donut shape is not. Nor is a cube. The letter L is enatiamorphic in two dimensions. A glove is a three dimensional enatiomorph.
It must be right or left hand, it cannot be neither, but it cannot be either without a counterpart. If we had two gloves, and they were incongruent (didn’t fit together) then we would be able to say of one, This is Left and of the other This is Right. But with just one, we cannot.
Things are enatiomorphic in terms of the way they are placed within the world. Back to the glove…
Perhaps, even without the above three issues, we just cannot imagine a glove universe in anything like the same way we can imagine tomorrow’s weather or the things we can imagine.
Perhaps we really can’t imagine the unimaginable. Ponder that.
Luckily, we don’t need need to imagine the unimaginable to be able to think about it. We can discuss idealised worlds that are unimaginable. We can learn from them. They can be tools. This is what thougts experiments are. So now let me guide you through one that I think you will enjoy. I have done this many times face to face.
The Glove Game: Round One
Imagine the glove universe as best you can. It is glove shaped from your perspctive. If you can think of something to loose in the description you can just ditch it. Tru to get to the most idealised thought of a glove.
You are trying to describe something that is logically true of all things that are gloves.
You are trying to describe something that is logically not true in totality of any thing that is not a glove.
We can enumerate:
1. It is a tube that ends in five points at the end of five smaller tubes.
2. One of the tubes is shorter than the others.
1. This tube also is joined to the main tube at a point closer to the main tube entrance and off to one side.
1. It can Extend to the plane of the other four tubes.
What is the most minimal optimal definition of a glove?
When I asked you to imagine the glove at the start it had a small button and a loop etc… Take all that kind of detail out of your imagination. Break it down to the things that are essential to being a glove. Let us call this idealised glove, the simplest glove.
What statements are true of the simplest glove?
What does it mean to say something is True here?
Criticise this definition: “A statement is True about the Glove Universe if what it describes can be found within the Glove Universe.”
• The Glove has four fingers and a thumb.
• The little finger is not longer than the middle finger
• The thumb is not between any fingers.
• It is possible the tip of the thumb could touch the tip of the index finger if the rest of the glove remained the same.
Imagine a list of False statements about the glove:
• The volume of the thumb is greater than the volumes of the other fingers combined.
• The glove has symmetry.
• It is possible to weave the thumb through the other fingers
• The glove has the same topology as a doughnut.
• The Glove Is underneath a Hat.
Is that false? It isn’t true, but it isn’t clear if it is False or meaningless. These kinds of statements are a big issue in the Philosophy of Language.
A statement is Meaningless relative to the Glove Universe Game if it is nether True nor False about the Glove Universe. You might like to think of Meaningless statements as containing things that simply cannot be found in any possible Glove Universe.
• True statements describe things that exist within the Glove Universe.
• By “things” here we mean structures, relations, properties that are contingent upon the stipulation of the universe.
• False statements describe things that do not exist within the Glove Universe but could exist within the Glove Universe.
• Meaningless statements describe things that can not not exist in the Glove Universe.
• These statements are meaningless in the glove universe
• Paris is the Capital of France.
• Mars is often called “The Red Planet”
• The glove is larger than an elephant.
• All gloves are smaller than houses.
• The glove belonged to Audry Hepburn.
• The Glove is left handed.
• We understand this experiment.
• All games are not fun.
This experiment has highlighted a number of things. Perhaps most importantly it’s shown what a Thought Experiment is, in case you didn’t already know. A thought experiment is simply a stipulated possible Universe that is created to be experimented on or questioned about.
We make Thought experiments all the time, “If I won the lottery I would..”, “Imagine all the people, living in Harmony…”
It’s also shown that thought experiments are about what’s relevant to them by stipulation, not by assumption. You can imagine things that are not really possible to exist or imagine and yet, you can see how still we can ask relevant questions about them.
The last thing we saw from this experiment is that all possible statements seem to fit into only one of three categories, True, False or Meaningless and that which list any statement belongs on depends on the stipulated nature of the relevant universe.
# Should I salute magpies?
One of the key advantages of practicing CHE is the ability to quickly sift through life’s mundane choices, enjoying them and knowing that, by and large, you have made what for you were the right choices when it comes to Home Economical issues. How should one clean their clothes, house, self and mind. Is Amazon Prime is justified? Which vitamins should I supplement? How much is optimum salt?
Consider the CHE equation: Should I wear my seatbelt?
It is simple to see on a three-space Risk/Cost/Benefit vector graph that, yes, of course you should wear your seatbelt. It is irrational not to, if you value self preservation. What is interesting is that such indubitable Cartesian conclusions map into the same kind of epistemic grid as things that on the whole seem woo, irrational or nonsensical.
Consider the CHE equation: Should I salute magpies?
This one, when you flesh it out, has a few more nexi than the seatbelt one, but the structure is almost the same; where the two equations differ is in the two driving assumptions.
1. Wearing Seatbelts: It is possible that wearing a seatbelt could save the wearer’s life.
2. Saluting Magpies: It possible that saluting a magpie could increase the saluter’s Luck.
In the case of 2, once we accept the possibility of Luck then it is no difference of kind to move on and reason something like:
1. There is something special called Luck.
2. It is possible this Luck can be increased by agency.
1. I’m assuming that if there is a supernatural (“nonprobabalistic”?) reality to luck then it can be something that can be in some sense accumulated or bestowed on.
1. If this assumption is not accepted then you seem forced to accept that there is Luck but it is distributed stochastically/probabilistically.
1. Luck would be real but its distribution chanced, which seems absurd.
3. It is possible saluting magpies could entail 2 (Luck increase).
4. Saluting magpies is an extremely low risk activity.
5. Saluting magpies is an extremely low cost activity.
6. It is rational to solute magpies.
But if we don’t accept the reality of Luck, we cannot go with Assumption 1 in the CHE reasoning above. It all boils down to the reality of Luck.
With anything abstract and potentially magical in a CHE equation, it needs to be weighted. Is there evidence? Is there mechanism? Is there equivalence? Even then, unless there is a refutation, all we can ultimately say is IDK.
1. I cannot be certain that there is Luck.
2. I cannot be certain that there is no Luck.
The Reality of Luck
I haven’t researched what others have said on Luck, I assume it has been spoken about lots. One thing that seems clear is that people who believe in Luck are believing in something that’s up there with ghosts and deities. For example, for there to be a reality to Luck there needs to be some kind of external agent, some Intelligence, that says “Bob is going to be more likely to win this coin toss.”
That’s a huge new guest to one’s ontological buffet, and I think you cannot have Luck without that. So, if you think your rabbit foot brings you luck, you are tacitly assuming, and please CMV, that there is/might be a deciding and intelligent agent affecting your life.
Luck also has implications to do with temporal logic. The kind of arguments against the logical possibility of changing the past might apply in the case of Luck.
1. At t1 x was not going to happen to P at t3.
2. At t2 P has luck bestowed on them.
3. At t1 x was going to happen to P at t3.
Is that right? I don’t know, it seems so to me.
The point is that accepting Luck is not a small thing, it is a huge thing that brings with it the world being profoundly different to the world without it. But as skeptics, that is no reason to deny the possibility of it.
What about evidence and mechanism? Is there any?
The Physical Argument For Real Luck
We cannot get evidence for Luck. Even if 1000 times out of 1000 I do better with my lucky charm than without it, that could always just be a coincidence.
What about a mechanism for how luck could work? Suppose you were a creator being and you made a universe with individuals in and you wanted to be able to bestow Luck upon them.
How would you do that? What mechanism, in this world, could you use. You would need to use a mechanism that was compatible with this world, or else there would be risk of contradiction. You would need a way to change the outcome of events while the changes being nomologically compatible with reality.
In fact, it seems our universe does have such a mechanism, built in at the bolts, which would allow consistent changes to be made to outcomes – this is quantum indeterminateness. True randomness exists and it could be used to facilitate the bestowing of luck. It is not against the laws of the universe that a bullet could suddenly veer off course. It could happen. If you wanted to bestow Luck upon your creations, you could use the indeterminateness built into your creation.
The reality of Luck has no possible evidence, has a huge ontological payload and has a plausible mechanism in this universe. If I had to choose I would say I do not not belive in the reality of Luck – but I do not have to choose; uncertainty is certain in my world view.
Conclusion: Should I salute magpies?
Real Luck could be real or not. It is fundamentally unknowable which is the case. Luck, if it was real would be something worth having – it would be irrational to think otherwise. Given this, and the minuscule cost and risk of saluting the magpies, in my opinion, the CHE solution to the equation is that yes, I should salute magpies. Why would I not?
# MSM The Miracle Mineral?
My Cousin lives in LA and will only eat food that has been blessed by monks from at least three different Asian religions and then tested, by both mass spectrum analyzer and professional taster, that it is not just Organic but Kosher. He never eats sugar, except in tea and coffee and all other food and drinks.
Even though he is like, really, into healthy living (The last time I stayed with him we spent over eighteen hours in West Hollywood’s biggest health food stores, subsisting just on wheatgrass and zen noodles. These are not a brand name, but a fad that only exists in this particular part of West Holywood and only for one summer in the late nineties. The idea was simple. Gluten was bad. Noodles tasted nice. Instead of being made with carbohydrate, they were made with Zen. They also had a bread made with sourdough) he had never ever once recommended me anything. Not once…
Until two months ago. On the phone he said that he had been taking this supplement and it had changed his life. I didn’t need more of a recommendation, I was all over that stuff within moments of getting off the Skype. I didn’t look into it like I normally do, I just went ahead and took a trip down the Amazon. A family pack for a month was a tenner.
Only then, once the order dispatched email was in, like a really bad scientist and skeptic, did I start to investigate it.
Now over the last decades online I have investigated many things using the power of the internet and books. In the crazy woowoo world of snake oil and superfoods, you have to really get skilled at operating the former from the latter. To think that all claims of benefit are snake oil is just ignorant and unreasonable. Equally to think that just because it had a webpage it has legitimacy is very poor think skills.
This is how I generally do it.
## Firstly, the big question is cui bono, who benefits?
If its something that you cannot make at home or but freely then that’s a redflag to me. This doesn’t mean that that crazy hybrid amino acid transmogifer isn’t going to be amazing, it does mean that while it is proprietary, you should assume there is profit in the promotion, even without benefit.
## Secondly, is it safe?
This is a real tricky one to get through and still today there are a bunch of things I just am not sure what to think of when it comes to their objective safety. MSG, Vitmin E…
## Thirdly, is it worth it?
To me this is the great question that only you can ask, but there are some guidelines and, espeically if many are recomending if for no proffit, it is a good sign.
Forthly, is it open?
There are chemicals in, say, apples. which just seem to do us good. Anyone can access these chamiec
There is also a very proven strategy which is to repeaedly and occasionally stop taking X to see if you iss it. If you do it long enough I think you will get attunes to what is good for you and what is not.
I have been taking it for two months now and, so far, really rate it as a wellbeing optimiser, as something I can imagine I will continue to take; like D3 and Boocha.
• MSM is very very low risk.
• It is quite a low cost.
• It is very high anecdote.
• It has significant scientific evidence.
• It has a plausible and demonstrable explanatory mechanism.
## Have I found it works?
I do feel more energy. I have started running, at about the same time that I started taking it. So its a bit of a mishmash when it comes to isolating the cause. Did I get into running because of MSM or JMR?
I have suddenly started writing poems again like I haven’t for many years, is that MSM? (The point here is the action, not the quality).
Anything else?
Today, for the first time in my life I ran 5k. MSM? It felt like I was going to collapse in a cardiovascular blamache. MSM?
Tonight, it was a consensus that I played the best poker of my life. MSM? I still lost badly, MSM? I dunno!:)
# Chase The Butterflies
I have long been a collector of the various ways we humans have found to express that abstract goodness to life, and the singular, hopeful, response to that goodness; Seize the day. Play The Game. Don’t Worry, Be Happy.
My Uncle Andy died a few years ago, he was a great man; all thought. My older cousin, and Andy’s first male nephew, Yeof , he came to stay, from LA. Just the other day. He told me how Andy had been such a fertile influence on his life; as Andy was to many. He told me of the wise and pristine advise that his uncle had given him, without claim, many long, long years ago… .
“Chase The Butterflies”
Uncle Andy ’47’10
# Thought Experiment Two: The Single Point Universe
This thought experiment is exactly the same as with the Glove Universe, except that it will have less parts. We will simply stipulate a new Universe for the game and then look at the Truth Lists for this Universe
Imagine a universe that is just a circle. No different on the inside or out, but a circle, perfect in its simplicity.
Your Glove Universe and my Glove Universe would have been distinct, it’s very unlikely we could imagine identical gloves, especially not if we started getting really trivial with our Truth List. Perhaps yours had stubbier fingers than mine or yours had thick external seams whereas mine were concealed.
With the Circle universes our Imaginings, and thus stipulations, will be identical. To see this, try to imagine a statement that could be true of your perfect Circle Universe but not True of mine. If you can imagine one in your Universe then you have cheated and not imagined a perfect circle.
Now let’s play the Truth game with this new perfectly circular board.
I will write my Lists using Bullets from now on. I am no mathematician but perhaps I may start my Lists like this:
• True:
• The proportion of the area of the circle is Pi times the radius squared.
• False:
• The circle has exactlly four axis of symmetry.
• Meaningless:
It is a lot harder with Circles than Gloves to fill the Three Lists because the very act of stipulation/imagination/creation limits what’s possible to say about the Circle Universe. Also, and importantly, the fact that all circles in the Game are perfect means they must also be identical. Two things are identical if the totality of their Truth Lists contain all of the same statements on each List, as would be the case with Perfect Circles but nor red gloves. We can imagine the difference in gloves, but not in perfect circles. And therefore, if follows that we can have no differences between our Circle Universes, to show me wrong here you just need to come up with a statement that would be true of your universe but not mine (without changing the rules of the game.)
We are going to move to the even more simple “Board” for our next variant of the Game, in the next part of this Thought Experiment.
Imagine a Single Point universe.
I don’t know what that means in any deep or metaphysical sense. I can’t imagine a Single Point Universe like I think I can a Glove Universe or a cosmic Universe and I certainly can’t imagine it like I can imagine yesterday’s lunch. But, just like with gloves and circles and anything else, I can fill out the Truth List for the Single Point Universe.
So… let’s play, fill out the Truth Lists for a universe that is just a point. No space or time, without change and structure.
When you try to do this, you will soon see that the Meaningless List can be added to easily, but the other two, Truth and False, are much more challenging.
There are only two true statements I can think off about the Single Point universe, and even these I am not sure what they mean. Here is my Truth List for the Single Point Universe:
• True:
• The Point Exists.
• The Point is Identical with Itself.
How can I even be sure that I can have these two True Statement’s on my list? I am not sure that I can, but it strikes me that whatever “existence” is if it is True of the Glove Existing then why would it not be True of the single point existing.
We have stipulated that the universe contains no change. It follows from this stipulation that the point must be identical with itself. If it was not, there would be a change, either in sequence or structure (We shall see what these terms mean in future experiments).
Let’s look to the False List:
• False:
• There is no existence.
• The Point is Identical with Itself.
The Single point universe has two False statements on its List, these are, as you can see, simply the opposite of the Truth List.
If this was the case, if the two lists contained items that couldn’t exist within the same game’s Truth lists, then we would have a problem, the most fundamental of problems, the Contradiction.
Underlying all of these Games we can play is a rule set that contains as its most fundamental rule:
It doesn’t matter what Universe we try to imagine, if we are reasonable then the underlying truths of Logic dictate that things will be consistent. There can be no contradictions. Soon we will see how emergence is a dependency relationship and we will be able to test this with any statement against any possible universe and see that this NonContradiction flows thought reality, possible and actual.In other words, if you are not prepared to accept this most fundamental rule then these Thought Experiments just cannot be for you:)
## Conclusion: The Single Point Philosopher
The Single point universe is logically the most Simple Universe anything could consistently imagine or represent. All we can say are four nontrivial statements, is “that it exists” and “that it is identical with its self” and the negation of these two statements. That’s all we can say, but as we shall see in the next experiments comes, from this most minimal of atoms we can create some amazing things. Before this, I think it would be good to ask the questions we cannot really answer about the Single Point Universe.
So far we haven’t been anywhere near what is traditionally and culturally considered the deeper side of philosophy. In fact, we haven’t really been doing any “philosophizing” at all in these two thought experiments. We have been reporting “the facts” about imaginary universes rather than asking the big Why/What/How? questions common to Philosophy.
As a final exercise, which is ideally suited to the bath, bed or pub, I want you to think about the Single Point Universe in as many ways as you can (or can’t). Try to contemplate the Single Point universe, as we have been; asking questions and suggesting answers. Try to meditate on the single pointed universe, focussing on it with as little distraction as you can (I find this very hard!). Try to visualise the Single Point Universe, even if you never can. Try to doubt it. Try to disprove its possibility.
And when you have tried the above, try, however you can to answer these kind of questions:
1. Could it exist?
2. Can I make sense of it not existing?
3. What is the difference between it existing and not existing?
4. Does it contain Pi?
5. Does it the anything like time or space?
6. Is it true that 4+6=10 in the Single Point Universe?
7. If it exists can it then not exist?
8. Are the Truth Lists of the Single Point Universe contained within the reality we are now in This Universe. Are all things identical with themselves in this universe? Do all things in this universe exist in some sense in this universe?
9. What happens if we add another point exactly like the first?
If you are like me you won’t be able to clearly answer most of the above, but that really doesn’t matter.
# Thought Experiment One: The Glove Game
Close your eyes and imagine an ordinary, small, red ladies glove. Imagine that in the wrist part of the glove is a slit and one side of this is a small red button. Imagine that on the other side of the slit is a small loop that can go around the button, to hold the glove in place on a hand.
You have just imagined a glove. Now I want you to imagine this glove floating in a void of nothingness. No other things, no time, no light, no observer, just the glove. I want you to imagine The Glove Universe.
In fact, you cannot imagine a universe that’s just a small red ladies glove. For a number of reasons:
You can’t imagine something being “small” if thats’ the only thing there is. Smallness is a relative property, it needs more things to be realised.
You can’t imagine something as being red if there is no light and no observer. “Colour’s” don’t make sense in the glove universe (Though you can imagine the surface of the Glove has properties that were it on your hand right now would make it red).
You can’t imagine just a glove because gloves are what’s called “enantiomorphs” (One of my favourite words), this means that they are left or right handed structures that cannot exist without a “counterpart.”
Perhaps, even without the above three issues, we just cannot imagine universes in anything like the same way we can imagine tomorrows weather or the things we can imagine. Perhaps we can’t imagine the unimaginable.
Luckily. We don’t need need to imagine the unimaginable to be able to think about it discuss it and learn from it. Too see this point and too see some other things we are going to play an imaginary game, but one we could all play any time.
The Glove Game: Round One
Start a document that has room for three lists. You can use pen and paper, bullet points, mental notes, whatever, it really doesn’t matter. I will use bullet points for my side of the game.
Label the first List, “True List.”
All you have to do to win Round One is add more True statements about the Glove Universe than I do. When we say “True” in the context of this game we mean:
True: “A statement is True about the Glove Universe if what it describes can be found within the Glove Universes.”
Here is my first Truth List:
• True:
• The Glove has four fingers and a thumb.
When I look at the imaginary universe I see that this is True. The meanings of the words are from outside of the Universe but what they represent can be found inside the Glove universe. If you’re going to try to imagine a six fingered glove, then you lose the game because the game requires an “ordinary, small, red ladies glove.”
It isn’t hard to come up with True statements about the glove as my “True List” shows:
• True
• The Glove has four fingers and a thumb.
• The button is not between the index finger and the thumb.
• The little finger is not longer than the middle finger
• The thumb is not between any fingers.
You can add to your list and I can add to mine and on and on we go. Sometimes we may come up with statements where it isn’t so clear if the statement is True. For example, what do we say about?:
It is possible the tip of the thumb could touch the tip of the index finger if the rest of the glove remained the same.
I don’t know what to say about this. It mentions possibility and conditionals (“if the…”) that don’t seem to belong. We shall discuss these in Round Three, for this round, its pretty clear, none of us can win.
If you want to imagine that the loop is has a tangent that intersects the seam of the thumb at 23% degrees, that’s fine, it’s your thought experiment and so long as your Stipulation of any single fact is consistent with your stipulations of the other facts, you can “imagine” it. And this means the Truth List is just a repository for facts that are consistent with a universe that consists of just an “ordinary, small, red ladies glove.”
The Glove Game: Round Two
For the next round, we have to start on the second List. Label this the “False List”.
The winner of Round Two is the person who comes up with the longest statement list of False statements about the Glove Universe. To see if a statement is False just see if the thing it describes is to be found in the Glove Universe, if it is not, then the statement is False.
Here is the start of my “False List” (Comments are in the lines below):
• False:
• The volume of the thumb is greater than the volumes of the other fingers combined.
• Although we can easily imagine gloves with very big thumbs, that would be outside of the rules of this game which requires “a small red ladies glove…”.
• The glove has symmetry.
• It is possible to weave the thumb through the other fingers
• The glove has the same topology as a doughnut.
It’s pretty easy to come up with False Statements about the Glove Universe. And like with True Statements, when seeking False Statements we also find some statements that don’t seem to be False. For example:
• The Glove Is underneath a Hat.
Seems to be False because the Glove is not underneath a hat. However, it’s not False and yert equally it doesn’t to be a True (That is, Not Flase.). These statements that don’t fit on either list will be discussed in Round Three.
The Glove Game: Round Three
The Third List in the game is the “Meaningless List” and it will take only statements that are meaningless relevant to the Glove Universe. This will take a little bit more to appreciate before we play.
A statement is Meaningless relative to the Glove Universe Game if it is nether True nor False about the Glove Universe. You might like to think of Meaningless statements as containing things that simply cannot be found in any possible Glove Universe.
• True statements describe things that exist within the Glove Universe.
• By “things” here we mean structures, relations, properties that are contingent upon the stipulation of the universe.
• False statements describe things that do not exist within the Glove Universe.
• Meaningless statements describe things cannot exist in the Glove Universe.
• That is, “cannot exist” without cheating and stipulating something other than a “”a small red ladies glove…”.
With an idea of what it means to be “Meaningless,” here is my Meaningless List:
• Meaningless:
• Paris is the Capital of France.
• Mars is often called “The Red Planet”
• The glove is larger than an elephant.
• All gloves are smaller than houses.
• The glove belonged to Audry Hepburn.
• The Glove is left handed.
• We understand this experiment.
• All games are not fun.
This game is rubbish!
Nobody could win Round One, nor Round Two and it now it looks like nobody can win Round Three. In fact, it strikes me that there are always going to be more meaningless statements because most possible statements simply won’t refer to things in the Glove Universe and thus, are meaningless.
Conclusion to the First Thought Experiment
This experiment has highlighted a number of things. Perhaps most importantly it’s shown what a Thought Experiment is, in case you didn’t already know. A thought experiment is simply a stipulated possible Universe that is created to be experimented on or questioned about.
We make Thought experiments all the time, “If I won the lottery I would..”, “Imagine all the people, living in Harmony…”
It’s also shown that thought experiments are about what’s relevant to them by stipulation, not by assumption. You can imagine things that are not really possible to exist or imagine and yet, you can see how still we can ask relevant questions about them.
But with the Glove Game we have employed a useful tool in the Truth Lists that allows us to speak about the possible universe in a pretty precise and useful way. I hope in the next experiment you will have an even more intuitive understanding of the potential.
The last thing we saw from this experiment is that all possible statements seem to fit into only one of three categories, True, False or Meaningless and that which list any statement belongs on depends on the stipulated nature of the relevant universe. This will become very important in future experiments.
The next Experiment will be published here shortly.
# What is Karma?
A twitter post compelled my to quickly try and get down on pixel paper my thoughts on what Karma is.
As with, I think all Dharmic concepts, Karma is best understood as pertaining to systems rather than objects/people etc. So before explaining how I see Karma I’ll outline what a Moral System is to me:
Moral systems have emergent moral properties.
A moral system is a system that can emerge moral properties. I am a moral system. You are. Society is. Religion is. Schools are… and so on. All of these moral systems share the possibility of having moral properties attributed to them. Properties such as right, wrong, fair, cruel and just.
Moral properties are internal, in that they refer to the system or they are external in that they refer to some other system.
In addition these moral systems have the potential of specific attitudes towards other moral properties; my dislike in your unfairness, your compassion for their suffering, a charity’s stance against world debt.
Moral systems have emergent evaluations of moral properties.
Moral systems are able to refer to their own moral systems and in these references they necessarily will value distinctions between moral properties. (I believe it’s these constant valuations that all moral systems have to make that add the core bivalence between right and wrong into our moralities.)
A value within a system is a propensity to pursue or avoid some future state to which the value pertains to. If I value cream-pies then I will pursue those. If I hate cucumbers, I will avoid those. The same is true of moral properties, as moral systems will behave in accordance as to how they value the moral properties they can pursue and avoid.
All of these evaluative properties we can boil down into two abstract notions, the positive and the negative. Loosely, the positive are the things I or you would prefer to be the case and the negative are the contrary.
Karma is the causal interconnectedness between all Moral Systems.
I think that Karma is simply the network of moral causes and effects that radiate from each of our actions out into the world and, importantly, into ourselves.
The moral effects of our actions may be external; you make someone happy, you annoy an entire country . or they may be internal; your pride at your kindness, your guilt at your selfishness. This is all Karma is. We are in a sea Karma. It is our lies and admissions and hopes and fears and all of these motivators that guide not only us as individuals but the institutions around us.
Often people mistake Karma as being a substance or energy or force, and you can see why because all morality is potentially interconnected it is has this substantive aspect but that is just an illusion. The only sense in which Karma can be accurately seen as a Force is by analaogy to physical causation.
The cue ball causes/forces the black ball to move. The bad deed causes/had the karmic effect that Bob was sad.
If you are a good person the chances of you having a good life are increased, not because of some supernatural reward system but because of the “karmic feedback” of your actions and thoughts, internally and externally.
If you are a bad person then your lies and violence and guilt etetcera will increase the chances of you having more negativity in so many ways. Not only the obvious, like punishment etc but also in terms of ones psychology. The Buddha realised this and now so many studies are confirming it.
I think most people, when you take away the “majic” decorations that millennia of Buddhist culture have added to the concept of Karma would see it as the simple and obvious fact: good moral causes have good moral effects.
# Understanding Dependent Origination – The Enlightenment of Dharma
Dependent Origination is the least taught, most misunderstood but most important aspect of all of Buddhism. Dependent Origination is the natural law that the Buddha fully realised when he became enlightened. It is not magic, it is not mystical, it is to do with the grounding truths of all logically possible realities and how they affect the emergent realities we experience as sentient humans.
At first site Dependent Origination, to western analytic minds, seems so obvious, it even has exact analogies in classic logic and metaphysics that have propagated from the pre-Socratic philosophers to modern reason. But what the Buddha did was to not only see it as fundamental but he understood the subsequent effects Dependent Origination has in all aspects of reality and, importantly, the limits it puts on reality.
To understand impermanence, emptiness and negativity is to understand the Dharma but to understand why all systems are impermanent or why there can be no ego, the answer comes back to Dependent Origination.
“If you know Dharma, you know Dependent origination. If you know Dependent Origination, you know dharma.” The Buddha
What Dependent Origination is Not.
If you ask a learned Buddhist what Dependent Origination is, or read in the vast majority of texts concerning it, you will most likely have described a complex wheel of causation. This wheel has twelve stages that encompass rebirth, suffering, ignorance and is termed The Twelve Niddyas.
Historically the Twelve Niddyas came at least four hundred years after the Buddha’s life yet they have become interpreted as being the actual concept of Dependent Origination rather than a mystical (pertaining to the supernatural) interpretation of Dependent Origination as applied to some notion of rebirth. This is a catastrophe to a widespread understanding of Dharma because it obscures the original and crucial meaning of the concept and thus excludes a deep and true understanding of Dharma.
An analogous case would be to say that a constructed, complex economic cycle that nobody but a PHD economist could understand was the law of “supply and demand” and then to expect budding economists to be able to move on in their studies from the confusing and inaccurate set of first principles.
Dependent Origination is a simple and rational principle, when properly understood its about as far from the mystical as one can get.
What Dependent origination Is.
Think for as long as you can about these four statements:
• If I knock at your door, I am on your doorstep.
• If I walk up to your door, I will be on your doorstep.
• If I am never on your doorstep, I will never be knocking on your door.
• If I stop being on your doorstep, I will stop knocking your door.
And then ask yourself these questions:
Are the statements all true in this world? Is there a possible world where the statements might not be true? Are the statements true together, that is, if one is true must they all be true? You can easily come up with trivial examples where they are not true, for example, standing off your doorstep and knocking with a big stick, but without adding any extra information to the statements, it seems that they are absolutely true in all cases.
If, like me, you are forced to conclude that the statements are true in all possible worlds in which they make sense, then you understand the foundational formulation of Dependent Origination. That is it! An anticlimax compared to the esotericism of the Twelve Niddyas, perhaps, but it is this conditionality that is Dependent Origination, nothing more.
In scriptural terms Dependent Origination is most simply expressed as:
• When there is thisthat is.
• With the arising of thisthat arises.
• When this is not, neither is that.
• With the cessation of thisthat ceases.
1.
The first point we need to make about the above expression of Dependent Origination is that the this and that stated will be any this and any that. This is hugely important Dharmically because it referring to all possible things, all possible systems and events and representations. What the Buddha realised was that conditionality isn’t the domain of some isolated syllogism but that it applies to everything, necessarily (The Buddha had no idea about quantum randomness; I am not sure how this would fit in with Dependent Origination.)
When you take this and that as being any possible this and that then it can be seen that Dependent Origination bestows three properties on the causality of reality:
• Transitivity– Conditionality is transitive. If P then Q then R, if not P then not R
o Note that in the world there are countless ways for events to happen, eg, you may open the door for another reason than me knocking on it. But that would be a different event, even though “on paper” it’s the same event.
• Generality –Conditionally applies to any causally related events at any level of abstraction.
• Totality –Conditionality applies to all events/systems/things. There is nothing outside of the conditionality of all things.
Everything is a cause and everything is an effect and the effects of causes are never singular and the causes of effects are never singular. The Buddha realised how reality is the vast, consistent and complex web of change, that there are no distinct unchanging things that can possibly be connected with reality.
Why is Dependent Origination Dharma?
I outlined here about the Three Marks of Existence, Annica, Anataman and Dukka, and how all of Dharma flows from them. I belive when you understand Dependent origination it is possible to see how these Three Marks are necessarily the case. Even now after many years thinking about this it hurts my head to see it all as one body of truth. I think this is why Meditation is considered so important to Dharma practice, as it may offer a way to apprehend these conceptual structures outside of the rigid lingustic structures we are used to relying on. But here goes:
Annica – All things are impermanent
Because all conditioned things originate within this transitive web of causation it follows that there is nothing to remain a constant. There is no thing that is isolated from the changes that flow following the principle of Dependent Origination. This applies from the neurons in my brain that make my knuckles tap on your door to the sound wave you hear and everything that follows from that and leads up to that.
As soon as an event happens, as soon as a thing changes, it is gone and the next change is happening, the effect becomes a new cause. And so on. And so on.
All conditioned things are impermanent. Everything is change.
Anataman – There are no objects/egos/souls
Because all things (systems, events..) are conditioned things it follows that there can be no thing that is isolated from the web of causation and equally no thing that appears from nowhere within the web of causation (QM randomness aside?). If Dependent origination is true in all possible realities then all possible realities are consistent. Consistent interconnectedness and impermanence preclude the possibility of their being something that is excluded from interconnectedness (a distinct thing) or immune to impermanence (an eternal thing.)
Ontologically impermanence and interconnectedness manifest in the impossibility of anomalous things, like miracles and souls. Psychologically impermanence and interconnectedness preclude the possibility of a constant distinct self/ego. There are no distinct objects in mind our world, however the illusions of perspective may make things seem to the contrary.
Dukka –Negativity/Suffering/Decay is Inevitable
Consider a deck of cards arranged neatly in suits and imagine the order changing (shuffling) as all things do. Any change will be a change away from that order, and, in addition, the probability increasingly decreases the more the deck is shuffled that it will to the original neat order. The deck of cards has a finite possibility space and the vastest extent of that space is change away from order not towards
The same is true of all systems that constitute reality; change will take place in a finite possibility space. In physical terms this fact is captured by the Second Law of Thermodynamics. In economic terms by the Law of Diminishing Marginal Returns. In human terms by realisations such as “the more of Great Thing X you have, the less great X becomes.” Change reduces the possibility space and if that space is valued by some other agent (like you) then change tends towards the negative.
These kinds of realisations are the opposite of difficult, if dogs could think like us, they would think the same. It’s not just that all things change but that most change tends towards the negative. The Buddha realised that this fact was self-evident. Moreover, when the inevitable negative was apprehended by sentient beings with qualitative experiences (like you) the realisation, conscious or not, will be bound to create negative experiences in the being.
This is why we suffer, the Buddha thought; because we constantly crave for the inevitable negative to not be the case. Accepting our impermanence and the impermanence of our good experiences is one thing but clinging to the hope that that may not be the case is fundamentally going to only bear bitter fruit. It cannot be any other way.
Dukka applies to solar systems and ecosystems and air-conditioning systems, but where it solidifies into suffering and strain is when the inevitable negative is apprehended by sentient systems, just like me, just like you. The only way to lessen the negativity of the inevitable negative in ourselves is to end attachment to transient things and to remove ignorance about the ontological status of things (ego, object, others…) and see the world as it really is. The method he reasoned from these realisations was to follow the Noble Eightfold Path and embrace the transitive and the inevitable negative.
Conclusion
Dharma is science and reason, to think otherwise is to place the contaminating teachings of scholars and monks who came centuries after the Buddha’s teachings as being more authoritative than the original teachings. It seems this has happened with all religions, but with Buddhism are lucky. We still have the original, wonderful, rational, sceptical discoveries of The Buddha available to us, but in addition, we don’t need them. We could erase all of Buddha’s teachings and start again just from that first principle of Dependent Origination that enlightened the Buddha and arrive using nothing but reason and insight through the Three Marks of Existence, the Four Noble Truths to the practice of The Noble Eightfold Path.
Dharma is simple and rational, not mystical and obscure. It takes us from the core truths of reality to a personal and social morality and understanding where compassion and love are not assumptions but conclusions entailed all the way up from the first principles.
Dharma is truth, to prove it wrong one merely needs to disprove Dependent Origination. | 11,225 | 51,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.94788 |
https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-3-section-3-2-corresponding-parts-of-congruent-triangles-exercises-page-151/19 | 1,721,833,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00565.warc.gz | 685,470,627 | 12,539 | Elementary Geometry for College Students (7th Edition)
- Prove $\angle 1\cong\angle 2$ - Prove $\overline{HF}\cong\overline{HK}$ - Prove $\angle FGH\cong\angle HJK$ - Then $\triangle FHG\cong\triangle HKJ$ according to method AAS - Then $\overline{FG}\cong\overline{HJ}$ according to method CPCTC
*PLANNING: We would prove $\triangle FHG\cong\triangle HKJ$ according to method AAS: - Prove $\angle 1\cong\angle 2$ - Prove $\overline{HF}\cong\overline{HK}$ - Prove $\angle FGH\cong\angle HJK$ 1. $\angle 1$ and $\angle 2$ are right angles. (Given) 2. $\angle 1\cong\angle 2$ (2 corresponding right angles are congruent) 3. H is the midpoint of $\overline{FK}$. (Given) 4. $\overline{HF}\cong\overline{HK}$ (A midpoint divides the line into 2 congruent lines) 5. $\overline{FG}\parallel\overline{HJ}$ (Given) 6. $\angle FGH\cong\angle HJK$ (2 corresponding angles for 2 parallel lines are congruent) So now we have 2 angles and a non-included side of $\triangle FHG$ are congruent with 2 corresponding angles and a non-included side of $\triangle HKJ$. Therefore, 6. $\triangle FHG\cong\triangle HKJ$ (AAS) 7. $\overline{FG}\cong\overline{HJ}$ (CPCTC) | 399 | 1,150 | {"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} | 4.1875 | 4 | CC-MAIN-2024-30 | latest | en | 0.662429 |
https://converter.ninja/volume/imperial-cups-to-metric-teaspoons/423-brcup-to-brteaspoon/ | 1,600,771,584,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00611.warc.gz | 341,859,486 | 5,380 | # 423 imperial cups in metric teaspoons
## Conversion
423 imperial cups is equivalent to 24037.450875 metric teaspoons.[1]
## Conversion formula How to convert 423 imperial cups to metric teaspoons?
We know (by definition) that: $1\mathrm{brcup}\approx 56.826125\mathrm{brteaspoon}$
We can set up a proportion to solve for the number of metric teaspoons.
$1 brcup 423 brcup ≈ 56.826125 brteaspoon x brteaspoon$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{brteaspoon}\approx \frac{423\mathrm{brcup}}{1\mathrm{brcup}}*56.826125\mathrm{brteaspoon}\to x\mathrm{brteaspoon}\approx 24037.450875\mathrm{brteaspoon}$
Conclusion: $423 brcup ≈ 24037.450875 brteaspoon$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 metric teaspoon is equal to 4.16017490872979e-05 times 423 imperial cups.
It can also be expressed as: 423 imperial cups is equal to $\frac{1}{\mathrm{4.16017490872979e-05}}$ metric teaspoons.
## Approximation
An approximate numerical result would be: four hundred and twenty-three imperial cups is about twenty-four thousand and thirty-seven point four five metric teaspoons, or alternatively, a metric teaspoon is about zero times four hundred and twenty-three imperial cups.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 397 | 1,458 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-40 | latest | en | 0.695519 |
http://magoosh.com/hs/act/act-math/2015/act-challenge-question-12-what-is-s/ | 1,477,119,652,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718840.18/warc/CC-MAIN-20161020183838-00474-ip-10-171-6-4.ec2.internal.warc.gz | 158,960,343 | 11,529 | The Algebra Song is currently stuck in my head, so I thought I’d share a tough ACT Math algebra problem with you this week. (If you clicked on the link–you’re welcome.)
Can you solve it? We’ll post the answer later this week, but if you think you have the perfect approach to this problem, post it in the comments below–your solution may just get featured!
What would s have to be so that is divisible by (x + 2)?
1. 9
2. 5
3. 2
4. -6
5. -13 | 122 | 444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | latest | en | 0.9608 |
http://www.docstoc.com/docs/9647038/Construction-Estimating | 1,409,354,084,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500833461.95/warc/CC-MAIN-20140820021353-00004-ip-10-180-136-8.ec2.internal.warc.gz | 342,454,399 | 40,968 | Construction Estimating
Document Sample
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Construction Estimating
1.1 1.2 1.3 1.4 Introduction Estimating Defined Estimating Terminology Types of Estimates
Conceptual Estimates • Time and Location Adjustments • Detailed Estimates
1.5
Contracts
Method of Award • Method of Bidding/Payment
James E. Rowings, Jr.
Peter Kiewit Sons’, Inc.
1.6
Computer-Assisted Estimating
1.1 Introduction
The preparation of estimates represents one of the most important functions performed in any business enterprise. In the construction industry, the quality of performance of this function is paramount to the success of the parties engaged in the overall management of capital expenditures for construction projects. The estimating process, in some form, is used as soon as the idea for a project is conceived. Estimates are prepared and updated continually as the project scope and definition develops and, in many cases, throughout construction of the project or facility. The parties engaged in delivering the project continually ask themselves “What will it cost?” To answer this question, some type of estimate must be developed. Obviously, the precise answer to this question cannot be determined until the project is completed. Posing this type of question elicits a finite answer from the estimator. This answer, or estimate, represents only an approximation or expected value for the cost. The eventual accuracy of this approximation depends on how closely the actual conditions and specific details of the project match the expectations of the estimator. Extreme care must be exercised by the estimator in the preparation of the estimate to subjectively weigh the potential variations in future conditions. The estimate should convey an assessment of the accuracy and risks.
1.2 Estimating Defined
Estimating is a complex process involving collection of available and pertinent information relating to the scope of a project, expected resource consumption, and future changes in resource costs. The process involves synthesis of this information through a mental process of visualization of the constructing process for the project. This visualization is mentally translated into an approximation of the final cost.
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At the outset of a project, the estimate cannot be expected to carry a high degree of accuracy, because little information is known. As the design progresses, more information is known, and accuracy should improve. Estimating at any stage of the project cycle involves considerable effort to gather information. The estimator must collect and review all of the detailed plans, specifications, available site data, available resource data (labor, materials, and equipment), contract documents, resource cost information, pertinent government regulations, and applicable owner requirements. Information gathering is a continual process by estimators due to the uniqueness of each project and constant changes in the industry environment. Unlike the production from a manufacturing facility, each product of a construction firm represents a prototype. Considerable effort in planning is required before a cost estimate can be established. Most of the effort in establishing the estimate revolves around determining the approximation of the cost to produce the one-time product. The estimator must systematically convert information into a forecast of the component and collective costs that will be incurred in delivering the project or facility. This synthesis of information is accomplished by mentally building the project from the ground up. Each step of the building process should be accounted for along with the necessary support activities and embedded temporary work items required for completion. The estimator must have some form of systematic approach to ensure that all cost items have been incorporated and that none have been duplicated. Later in this chapter is a discussion of alternate systematic approaches that are used. The quality of an estimate depends on the qualifications and abilities of the estimator. In general, an estimator must demonstrate the following capabilities and qualifications: • • • • • • • • • Extensive knowledge of construction Knowledge of construction materials and methods Knowledge of construction practices and contracts Ability to read and write construction documents Ability to sketch construction details Ability to communicate graphically and verbally Strong background in business and economics Ability to visualize work items Broad background in design and code requirements
Obviously, from the qualifications cited, estimators are not born but are developed through years of formal or informal education and experience in the industry. The breadth and depth of the requirements for an estimator lend testimony to the importance and value of the individual in the firm.
1.3 Estimating Terminology
There are a number of terms used in the estimating process that should be understood. AACE International (formerly the American Association of Cost Engineers) developed a glossary of terms and definitions in order to have a uniform technical vocabulary. Several of the more common terms and definitions are given below.
1.4 Types of Estimates
There are two broad categories for estimates: conceptual (or approximate) estimates and detailed estimates. Classification of an estimate into one of these types depends on the available information, the extent of effort dedicated to preparation, and the use for the estimate. The classification of an estimate into one of these two categories is an expression of the relative confidence in the accuracy of the estimate.
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Construction Estimating
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Conceptual Estimates
At the outset of the project, when the scope and definition are in the early stages of development, little information is available, yet there is often a need for some assessment of the potential cost. The owner needs to have a rough or approximate value for the project’s cost for purposes of determining the economic desirability of proceeding with design and construction. Special quick techniques are usually employed, utilizing minimal available information at this point to prepare a conceptual estimate. Little effort is expended to prepare this type of estimate, which often utilizes only a single project parameter, such as square feet of floor area, span length of a bridge, or barrels per day of output. Using available, historical cost information and applying like parameters, a quick and simple estimate can be prepared. These types of estimates are valuable in determining the order of magnitude of the cost for very rough comparisons and analysis but are not appropriate for critical decision making and commitment. Many situations exist that do not warrant or allow expenditure of the time and effort required to produce a detailed estimate. Feasibility studies involve elimination of many alternatives prior to any detailed design work. Obviously, if detailed design were pursued prior to estimating, the cost of the feasibility study would be enormous. Time constraints may also limit the level of detail that can be employed. If an answer is required in a few minutes or a few hours, then the method must be a conceptual one, even if detailed design information is available. Conceptual estimates have value, but they have many limitations as well. Care must be exercised to choose the appropriate method for conceptual estimating based on the available information. The estimator must be aware of the limitations of his estimate and communicate these limitations so that the estimate is not misused. Conceptual estimating relies heavily on past cost data, which is adjusted to reflect current trends and actual project economic conditions. The accuracy of an estimate is a function of time spent in its preparation, the quantity of design data utilized in the evaluation, and the accuracy of the information used. In general, more effort and more money produce a better estimate, one in which the estimator has more confidence regarding the accuracy of his or her prediction. To achieve significant improvement in accuracy requires a larger-than-proportional increase in effort. Each of the three conceptual levels of estimating has several methods that are utilized, depending on the project type and the availability of time and information. Order of Magnitude The order-of-magnitude estimate is by far the most uncertain estimate level used. As the name implies, the objective is to establish the order of magnitude of the cost, or more precisely, the cost within a range of +30 to –50%. Various techniques can be employed to develop an order-of-magnitude estimate for a project or portion of a project. Presented below are some examples and explanations of various methods used. Rough Weight Check When the object of the estimate is a single criterion, such as a piece of equipment, the order-of-magnitude cost can be estimated quickly based on the weight of the object. For the cost determination, equipment can be grouped into three broad categories: 1. Precision/computerized/electronic 2. Mechanical/electrical 3. Functional Precision equipment includes electronic or optical equipment such as computers and surveying instruments. Mechanical/electrical equipment includes pumps and motors. Functional equipment might include heavy construction equipment, automobiles, and large power tools. Precision equipment tends to cost ten times more per pound than mechanical/electrical equipment, which in turn costs ten times per pound more than functional equipment. Obviously, if you know the average cost per pound for a particular class of equipment (e.g., pumps), this information is more useful than a broad category estimate. In any case, the estimator should have a feel for the approximate cost per pound for the three
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categories so that quick checks can be made and order-of-magnitude estimates performed with minimal information available. Similar approaches using the capacity of equipment, such as flow rate, can be used for order-of-magnitude estimates. Cost Capacity Factor This quick method is tailored to the process industry. It represents a quick shortcut to establish an orderof-magnitude estimate of the cost. Application of the method involves four basic steps: 1. Obtain information concerning the cost (C1 or C2) and the input/output/throughput or holding capacity (Q1 or Q2) for a project similar in design or characteristics to the one being estimated. 2. Define the relative size of the two projects in the most appropriate common units of input, output, throughput, or holding capacity. As an example, a power plant is usually rated in kilowatts of output, a refinery in barrels per day of output, a sewage treatment plant in tons per day of input, and a storage tank in gallons or barrels of holding capacity. 3. Using the three known quantities (the sizes of the two similar plants in common units and the cost of the previously constructed plant), the following relationship can be developed: C1 C 2 = (Q1 Q2 )
x
where x is the appropriate cost capacity factor. With this relationship, the estimate of the cost of the new plant can be determined. 4. The cost determined in the third step is adjusted for time and location by applying the appropriate construction cost indices. (The use of indices is discussed later in this chapter.) The cost capacity factor approach is also called the six-tenths rule, because in the original application of the exponential relationship, x was determined to be equal to about 0.6. In reality, the factors for various processes vary from 0.33 to 1.02 with the bulk of the values for x around 0.6. Example 1 Assume that we have information on an old process plant that has the capacity to produce 10,000 gallons per day of a particular chemical. The cost today to build the plant would be \$1,000,000. The appropriate cost factor for this type of plant is 0.6. An order-of-magnitude estimate of the cost is required for a plant with a capacity of 30,000 gallons per day. C = \$1, 000, 000(30, 000 10, 000)
0.6
= \$1, 930, 000
Comparative Cost of Structure This method is readily adaptable to virtually every type of structure, including bridges, stadiums, schools, hospitals, and offices. Very little information is required about the planned structure except that the following general characteristics should be known: 1. 2. 3. 4. 5. Use — school, office, hospital, and so on Kind of construction — wood, steel, concrete, and so on Quality of construction — cheap, moderate, top grade Locality — labor and material supply market area Time of construction — year
By identifying a similar completed structure with nearly the same characteristics, an order-of-magnitude estimate can be determined by proportioning cost according to the appropriate unit for the structure. These units might be as follows: 1. Bridges — span in feet (adjustment for number of lands) 2. Schools — pupils
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3. 4. 5. 6.
Stadium — seats Hospital — beds Offices — square feet Warehouses — cubic feet
Example 2 Assume that the current cost for a 120-pupil school constructed of wood frame for a city is \$1,800,000. We are asked to develop an order-of-magnitude estimate for a 90-pupil school. Solution. The first step is to separate the per-pupil cost. \$1,800,000 120 = \$15, 000 pupil Apply the unit cost to the new school. \$15,000 pupil ¥ 90 pupils = \$1, 350, 000 Feasibility Estimates This level of conceptual estimate is more refined than the order-of-magnitude estimate and should provide a narrower range for the estimate. These estimates, if performed carefully, should be within ±20 to 30%. To achieve this increase in accuracy over the order-of-magnitude estimate requires substantially more effort and more knowledge about the project. Plant Cost Ratio This method utilizes the concept that the equipment proportion of the total cost of a process facility is about the same, regardless of the size or capacity of the plant, for the same basic process. Therefore, if the major fixed equipment cost can be estimated, the total plant cost can be determined by factor multiplication. The plant cost factor or multiplier is sometimes called the Lang factor (after the man who developed the concept for process plants). Example 3 Assume that a historical plant with the same process cost \$2.5 million, with the equipment portion of the plant costing \$1 million. Determine the cost of a new plant if the equipment has been determined to cost \$2.4 million. C = 2.4 (1.0 2.5) C = 6 million dollars Floor Area This method is most appropriate for hospitals, stores, shopping centers, and residences. Floor area must be the dominant attribute of cost (or at least it is assumed to be by the estimator). There are several variations of this method, a few of which are explained below. Total Horizontal Area For this variation, it is assumed that cost is directly proportional to the development of horizontal surfaces. It is assumed that the cost of developing a square foot of ground-floor space will be the same as a square foot of third-floor space or a square foot of roof space. From historical data, a cost per square foot is determined and applied uniformly to the horizontal area that must be developed to arrive at the total cost. Example 4 Assume that a historical file contains a warehouse building that cost \$2.4 million that was 50 ft ¥ 80 ft with a basement, three floors, and an attic. Determine the cost for a 60 ft ¥ 30 ft warehouse building with no basement, two floors, and an attic.
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Solution. Determine the historical cost per square foot. Basement area 1st floor 2nd floor 3rd floor Attic Roof TOTAL 4000 4000 4000 4000 4000 4000 24,000
\$2, 400,000 24,000 = \$100 ft 2 Next, calculate the total cost for the new project. 1st floor 2nd floor Attic Roof TOTAL 1800 1800 1800 1800 7200
7200 ft 2 ¥ \$100 ft 2 = \$720, 000 Finished Floor Area This method is by far the most widely used approach for buildings. With this approach, only those floors that are finished are counted when developing the historical base cost and when applying the historical data to the new project area. With this method, the estimator must exercise extreme care to have the same relative proportions of area to height to avoid large errors. Example 5 Same as the preceding example. Solution. Determine historical base cost. 1st floor 2nd floor 3rd floor TOTAL 4000 4000 4000 12,000 ft2fa
\$2, 400,000 12,000 = \$200 ft 2fa where ft2fa is square feet of finished floor area. Next, determine the total cost for the new project. 1st floor 2nd floor TOTAL 1800 1800 3600 ft2fa
3600 ft 2fa ¥ \$200 ft 2fa = \$720, 000
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As can be seen, little difference exists between the finished floor area and total horizontal area methods; however, if a gross variation in overall dimensions had existed between the historical structure and the new project, a wider discrepancy between the methods would have appeared. Cubic Foot of Volume Method This method accounts for an additional parameter that affects cost: floor-to-ceiling height. Example 6 The same as the preceding two examples, except that the following ceiling heights are given: Old Structure 1st floor 2nd floor 3rd floor Solution. Determine the historical base cost. 14 ¥ 4000 = 56,000 ft3 10 ¥ 4000 = 40,000 ft3 10 ¥ 4000 = 40,000 ft3 TOTAL = 136,000 ft3 14 10 10 New Structure 12 12 —
\$2, 400,000 136,000 ft 3 = \$17.65 ft 3 Next, determine the total cost for the new warehouse structure. 1st floor 2nd floor TOTAL 1800 ft2 ¥ 12 ft = 21,600 ft3 1800 ft2 ¥ 12 ft = 21,600 ft3 = 43,200 ft3
43,200 ft 3 ¥ \$17.65 ft 3 = \$762, 500 Appropriation Estimates As a project scope is developed and refined, it progresses to a point where it is budgeted into a corporate capital building program budget. Assuming the potential benefits are greater than the estimated costs, a sum of money is set aside to cover the project expenses. From this process of appropriation comes the name of the most refined level of conceptual estimate. This level of estimate requires more knowledge and effort than the previously discussed estimates. These estimating methods reflect a greater degree of accuracy. Appropriation estimates should be between ±10 to 20%. As with the other forms of conceptual estimates, several methods are available for preparing appropriation estimates. Parametric Estimating/Panel Method This method employs a database in which key project parameters, project systems, or panels (as in the case of buildings) that are priced from past projects using appropriate units are recorded. The costs of each parameter or panel are computed separately and multiplied by the number of panels of each kind. Major unique features are priced separately and included as separate line items. Numerous parametric systems exist for different types of projects. For process plants, the process systems and piping are the
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parameters. For buildings, various approaches have been used, but one approach to illustrate the method is as follows:
Parameter Site work Foundations and columns Floor system Structural system Roof system Exterior walls Interior walls HVAC Electrical Conveying systems Plumbing Finishes Unit of Measure Square feet of site area Building square feet Building square feet Building square feet Roof square feet Wall square feet minus exterior windows Wall square feet (interior) Tons or Btu Building square feet Number of floor stops Number of fixture units Building square feet
Each of these items would be estimated separately by applying the historical cost for the appropriate unit for similar construction and multiplying by the number of units for the current project. This same approach is used on projects such as roads. The units or parameters used are often the same as the bid items, and the historical prices are the average of the low-bid unit prices received in the last few contracts. Bay Method This method is appropriate for buildings or projects that consist of a number of repetitive or similar units. In the plan view of a warehouse building shown in Fig. 1.1, the building is made up of three types of bays. The only difference between them is the number of outside walls. By performing a definitive estimate of the cost of each of these bay types, an appropriation estimate can be made by multiplying this bay cost times the number of similar bays and totaling for the three bay types. Example 7 We know from a definitive estimate that the cost of the three bay types is as follows: Type I = \$90,000 Type II = \$120,000 Type III = \$150,000
III
II
III
II 4@60' II
I
II
I
II
III
II
III
3@60'
FIGURE 1.1 Plan view — warehouse building.
Determine the cost for the building structure and skin (outer surface). Solution. 2 Type I @ 90,000 = 6 Type II @ 120,000 = 4 Type III @ 150,000 = TOTAL \$180,000 \$720,000 \$600,000
= \$1,500,000
After applying the bay method for the overall project, the estimate is modified by making special allowances (add-ons) for end walls, entrances, stairs, elevators, and mechanical and electrical equipment.
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Plant Component Ratio This method requires a great deal more information than other methods used in the process industry. Definitive costs of the major pieces of equipment are needed. These can be determined from historical records or published data sources. Historical records also provide the data that identifies the relative percentage of all other items. The total project cost is then estimated as follows: TPC = ET 1 – PT
where TPC = total plant cost ET = total estimated equipment cost PT = sum of percentages of other items or phases (major account divisions). Example 8 The total equipment cost for a plant is estimated to be \$500,000. The following percentages represent the average expenditures in other cost phases:
Engineering, overhead, and fees Warehousing Services Utilities Piping Instrumentation Electrical Buildings TOTAL 22% 5% 2% 6% 20% 5% 6% 4% 70% = PT
(1.0 – 0.70)
500, 000
= \$1, 670, 000
While the solution here appears simple, in fact, the majority of time and effort is spent collecting the equipment cost and choosing the appropriate percentages for application.
It is often desirable when preparing conceptual estimates to utilize cost data from a different period of time or from a different location. Costs vary with time and location, and it is, therefore, necessary to adjust the conceptual estimate for the differences of time and location from the historical base. A construction cost-indexing system is used to identify the relative differences and permit adjustment. Cost Indexing A cost index is a dimensionless number associated with a point in time and/or location that illustrates the cost at that time or location relative to a base point in time or base location. The cost index provides a comparison of cost or cost change from year to year and/or location to location for a fixed quantity of services and commodities. The concept is to establish cost indices to avoid having to estimate all of the unique features of every project, when it is reasonable to assume that the application of relative quantities of resources is constant or will follow the use of historical data on a proportional basis without knowledge of all of the design details. If the cost index is developed correctly, the following simple relationship will exist: New cost New index = Historical cost Historical index
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An example of the way in which a cost index might be computed is given below. The cost elements used for developing a cost index for concrete in 1982 are as follows: = C1 = four hours for a carpenter C2 = one cubic yard concrete = C3 = three hours for laborer = C4 = 100 fbm lumber (2 ¥ 10) = C5 = 100 # rebar = C6 = one hour from an ironworker = \$240 \$60 \$66 \$49 \$35 \$50
Ca = 240 + 60 + 66 + 49 + 35 + 50 = 500 Calculating Cb similarly for another time or location involves the following steps: C1 = four hours for a carpenter = C2 = one cubic yard concrete = C3 = three hours for laborer = C4 = 100 fbm lumber (2 ¥ 10) = C5 = 100 # rebar = C6 = one hour from an ironworker = \$200 \$58 \$90 \$42 \$36 \$44
Ca = 200 + 58 + 90 + 42 + 36 + 44 = 470 Using the CIa as the base with an index equal to 100, the CIb index can be calculated as follows: CI b = (Cb Ca ) ¥ 100 = ( 470 500) ¥ 100 = 94 The key to creating an accurate and valid cost index is not the computational approach but the correct selection of the cost elements. If the index will be used for highway estimating, the cost elements should include items such as asphalt, fuel oil, paving equipment, and equipment operators. Appropriately, a housing cost index would include timber, concrete, carpenters, shingles, and other materials common to residential construction. Most of the cost indices are normalized periodically to a base of 100. This is done by setting the base calculation of the cost for a location or time equal to 100 and converting all other indices to this base with the same divisor or multiplier. While it is possible to develop specialized indices for special purposes, numerous indices have been published. These include several popular indices, such as the Engineering News Record building cost index and construction cost index and the Means Building Construction Cost Data construction cost index and historical cost index. These indices are developed using a wide range of cost elements. For example, the Means’ construction cost index is composed of 84 construction materials, 24 building crafts’ labor hours, and 9 different equipment rental charges that correspond to the labor and material items. These cost indices are tabulated for the major metropolitan areas four times each year and for the 16 major UCI construction divisions. Additionally, indices dating back to 1913 can be found to adjust costs from different periods of time. These are referred to as historical cost indices. Application of Cost Indices These cost indices can have several uses: • Comparing costs from city to city (construction cost indices) • Comparing costs from time to time (historical cost indices)
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• Modifying costs for various cities and times (both) • Estimating replacement costs (both) • Forecasting construction costs (historical cost indices) The cost index is only a tool and must be applied with sound judgment and common sense. Comparing Costs from City to City The construction cost indices can be used to compare costs between cities, because the index is developed identically for each city. The index is an indicator of the relative difference. The cost difference between cities for identical buildings or projects in a different city can be found by using the appropriate construction cost indices (CCI). The procedure is as follows: Cost, city A = CCI for city A (Known cost, city B) CCI for city B
(Known cost, city B) - (Cost, city A) = Cost difference
Comparing Costs from Time to Time The cost indices can be used to compare costs for the same facility at different points in time. Using the historical cost indices of two points in time, one can calculate the difference in costs between the two points in time. It is necessary to know the cost and the historical index for time B and the historical cost index for time A. Cost, time A = HCI for time A (Cost, time B) HCI for time B
(Known cost, time B) - (Cost, time A) = Cost difference
Modifying Costs for Various Cities and Times The two prior uses can be accomplished simultaneously, when it is desired to use cost information from another city and time for a second city and time estimate. Care must be exercised to establish the correct relationships. The following example illustrates the principle. Example 9 A building cost \$2,000,000 in 2000 in South Bend. How much will it cost to build in Boston in 2002? Given: HCI, 2002 = HCI, 2000 = CCI, S. Bend = CCI, Boston = 114.3 102.2 123.4 134.3
(HCI, 2002) (CCI, Boston) (Cost, S. Bend) = (Cost, Boston) (HCI, 2000) (CCI, S. Bend) (114.3) (134.3) (2, 000, 000) = \$2, 430, 000 (102.2) (123.4)
Estimating Replacement Costs The historical cost index can be used to determine replacement cost for a facility built a number of years ago or one that was constructed in stages.
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Example 10 A building was constructed in stages over the last 25 years. It is desired to know the 2002 replacement cost for insurance purposes. The building has had two additions since the original 1981, \$300,000 portion was built. The first addition was in 1990 at a cost of \$200,000, and the second addition came in 1994 at a cost of \$300,000. The historical cost indices are as follows: 2002 1994 1990 1981 Solution. The cost of the original building is 100 \$300, 000 = \$1, 255, 000 23.9 The cost of addition A is 100 \$200, 000 = \$578, 000 34.6 The cost of addition B is 100 \$300, 000 = \$602, 000 49.8 So, Total replacement cost = \$2, 435,000 Construction Cost Forecasting If it is assumed that the future changes in cost will be similar to the past changes, the indices can be used to predict future construction costs. By using these past indices, future indices can be forecast and, in turn, used to predict future costs. Several approaches are available for developing the future index. Only one will be presented here. The simplest method is to examine the change in the last several historical cost indices and use an average value for the annual change in the future. This averaging process can be accomplished by determining the difference between historical indices each year and finding the average change by dividing by the number of years. 100.0 = 49.8 = 34.6 = 23.9 = HCl HCl HCl HCl
Detailed Estimates
Estimates classified as detailed estimates are prepared after the scope and definition of a project are essentially complete. To prepare a detailed estimate requires considerable effort in gathering information and systematically forecasting costs. These estimates are usually prepared for bid purposes or definitive budgeting. Because of the information available and the effort expended, detailed estimates are usually fairly accurate projections of the costs of construction. A much higher level of confidence in the accuracy of the estimate is gained through this increased effort and knowledge. These types of estimates are used for decision making and commitment. The Estimating Process Estimating to produce a detailed construction cost estimate follows a rigorous process made up of several key steps. These key steps are explained below.
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Familiarization with Project Characteristics The estimator must be familiar with the project and evaluate the project from three primary avenues: scope, constructibility, and risk. Having evaluated these three areas in a general way, the estimator will decide whether the effort to estimate and bid the work has a potential profit or other corporate goal potential (long-term business objective or client relations). In many cases, investigation of these three areas may lead to the conclusion that the project is not right for the contractor. The contractor must be convinced that the firm’s competitive advantage will provide the needed margin to secure the work away from competitors. Scope — Just because a project is available for bidding does not mean that the contractor should invest the time and expense required for the preparation of an estimate. The contractor must carefully scrutinize several issues of scope for the project in relation to the company’s ability to perform. These scope issues include the following: 1. 2. 3. 4. 5. 6. Technological requirements of the project Stated milestone deadlines for the project Required material and equipment availability Staffing requirements Stated contract terms and associated risk transfer Nature of the competition and likelihood of an acceptable rate-of-return
The contractor must honestly assess the technological requirements of the project to be competitive and the internal or subcontractor technological capabilities that can be employed. This is especially true on projects requiring fleets of sophisticated or specialized equipment or on projects with duration times that dictate employment of particular techniques such as slipforming. On these types of projects, the contractor must have access to the fleet, as in the case of an interstate highway project, or access to a knowledgeable subcontractor, as in the case of high-rise slipforming. The contractor must examine closely the completion date for the project as well as any intermediate contractual milestone dates for portions of the project. The contractor must feel comfortable that these dates are achievable and that there exists some degree of time allowance for contingencies that might arise. Failure to complete a project on time can seriously damage the reputation of a contractor and has the potential to inhibit future bidding opportunities with the client. If the contract time requirements are not reasonable in the contractor’s mind after having estimated the required time by mentally sequencing the controlling work activities, two choices exist. The obvious first choice is to not bid the project. Alternatively, the contractor may choose to reexamine the project for other methods or sequences which will allow earlier completion. The contractor should not proceed with the estimate without a plan for timely completion of the project. A third issue that must be examined in relation to the project’s scope is availability to satisfy the requirements for major material commodities and equipment to support the project plan. Problems in obtaining structural steel, timber, quality concrete, or other materials can have pronounced effects on both the cost and schedule of a project. If these problems can be foreseen, solutions should be sought, or the project should not be considered for bidding. Staffing requirements, including staffing qualifications as well as required numbers, must be evaluated to determine if sufficient levels of qualified manpower will be available when required to support project needs. This staffing evaluation must include supervisory and professional support and the various crafts that will be required. While the internal staffing (supervisory and professional support) is relatively simple to analyze, the craft availability is extremely uncertain and to some degree uncontrollable. With the craft labor in much of the construction industry (union sector) having no direct tie to any one construction company, it is difficult to predict how many workers of a particular craft will be available during a particular month or week. The ability to predict craft labor availability today is a function of construction economy prediction. When there is a booming construction market, some shortfalls in craft labor supply can be expected with a result of higher labor costs or longer project durations.
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Constructibility — A knowledgeable contractor, having made a preliminary review of the project documents, can assess the constructibility of the project. Constructibility evaluations include examination of construction quality requirements, allowable tolerances, and the overall complexity of the project. The construction industry has general norms of quality requirements and tolerances for the various types of projects. Contractors tend to avoid bidding for projects for which the quality or tolerances specified are outside those norms. The alternative for the contractor is to overcompensate for the risk associated with achieving the requirements by increasing their expectation of cost. Complexity of a project is viewed in terms of the relative technology requirement for the project execution compared with the technology in common practice in the given area. Where the project documents indicate an unusual method to the contractor, the contractor must choose to either accept the new technology or not bid. The complexity may also come about because of dictated logistical or scheduling requirements that must be met. Where the schedule does not allow flexibility in sequence or pace, the contractor may deem the project unsuitable to pursue through bidding. The flexibility left to the contractor in choosing methods creates interest in bidding the project. The means and methods of work are the primary ways that contractors achieve competitive advantage. This flexibility challenges the contractor to develop a plan and estimate for the work that will be different and cheaper than the competition’s. Risk — The contractor must also evaluate the myriad of potential problems that might be encountered on the project. These risks can include the following: • • • • • • • • • • • • • • • • Material and workmanship requirements not specified Contradictory clauses interpreted incorrectly Impossible specifications Unknown or undiscovered site conditions Judgment error during the bidding process Assumption of timely performance of approvals and decisions by the owners Interpretation and compliance requirements with the contract documents Changes in cost Changes in sequence Subcontractor failure Suspension of work Weather variations Environmental issues Labor and craft availability Strikes and labor disputes Utility availability
This list represents a sample of the risks, rather than an inclusive listing. In general, a construction firm faces business risks, project risks, and operational risks, which must be offset in some way. Contract terms that transfer unmanageable risk or categories of risk that are not easily estimated discourage participation in bidding. Contractors assess the likelihood of success in the bidding process by the number of potential competitors. Typically, more competition means lower markups. Lower markup reduces the probability for earning acceptable margins and rates of return associated with the project risks. Examine the Project Design Another aspect of the information important to the individual preparing the estimate is the specific design information that has been prepared. The estimator must be able to read, interpret, and understand the technical specifications, the referenced standards and any project drawings, and documents. The
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estimator must closely examine material specifications so that an appropriate price for the quality and characteristics specified can be obtained. The estimator must use sound judgment when pricing substitute materials for providing an assumption of “or-equal” quality for a material to be used. A thorough familiarity and technical understanding is required for this judgment. The same is also true for equipment and furnishings that will be purchased. The estimator must have an understanding of referenced documents that are commonly identified in specifications. Standards of testing and performance are made a part of the specifications by a simple reference. These standards may be client standards or more universal standards, such as State Highway Specifications or ASTM (American Society for Testing and Materials) documents. If a specification is referenced that the estimator is not familiar with, he or she must make the effort to locate and examine it prior to bid submittal. In some cases, the specifications will identify prescribed practices to be followed. The estimator must assess the degree to which these will be rigidly enforced and where allowances will be made or performance criteria will be substituted. Use of prescriptive specifications can choke innovation by the contractor but may also protect the contractor from performance risks. Where rigid enforcement can be expected, the estimator should follow the prescription precisely. The drawings contain the physical elements, their location, and their relative orientation. These items and the specifications communicate the designer’s concept. The estimator must be able to examine the drawings and mentally visualize the project as it will be constructed to completion. The estimator relies heavily on the information provided in the drawings for determination of the quantity of work required. The drawings provide the dimensions so that lengths, widths, heights, areas, volumes, and numbers of items can be developed for pricing the work. The drawings show the physical features that will be part of the completed project, but they do not show the items that may be required to achieve completion (such as formwork). It is also common that certain details are not shown on the drawings for the contractor but are developed by shop fabricators at a later time as shop drawings. The estimator must keep a watchful eye for errors and omissions in the specifications and drawings. Discrepancies are often identified between drawings, between specifications, or between drawings and specifications. The discrepancies must be resolved either by acceptance of a risk or through communication with the designer. The best choice of solution depends on the specifics of the discrepancy and the process or the method for award of contract. Structuring the Estimate The estimator either reviews a plan or develops a plan for completing the project. This plan must be visualized during the estimating process; it provides the logical flow of the project from raw materials to a completed facility. Together with the technical specifications, the plan provides a structure for the preparation of the detailed estimate. Most estimators develop the estimate around the structure of the technical specifications. This increases the likelihood that items of work are covered without duplication in the estimate. Determine the Elements of Cost This step involves the development of the quantities of work (a quantity survey) to be performed and their translation into expected costs. Translating a design on paper into a functioning, completed project involves the transformation and consumption of a multitude of resources. These basic ingredients or resources utilized and incorporated in a project during construction can be classified into one of the following categories: 1. 2. 3. 4. 5. Labor Material Equipment Capital Time
Associated with the use or consumption of each of these resources is a cost. It is the objective of the estimator performing a detailed estimate to identify the specific types of resources that will be used, the
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Cost of Labor For a detailed estimate, it is imperative that the cost of labor resources be determined with precision. This is accomplished through a three-part process from data in the construction bidding documents that identify the nature of work and the physical quantity of work. The first step in the process involves identifying the craft that will be assigned the work and determining the hourly cost for that labor resource. This is termed the labor rate. The second part of the process involves estimation of the expected rate of work accomplishment by the chosen labor resource. This is termed the labor productivity. The third step involves combining this information by dividing the labor rate by the labor productivity to determine the labor resource cost per physical unit of work. The labor cost can be determined by multiplying the quantity of work by the unit labor resource cost. This entire process will be illustrated later in this chapter; however, an understanding of labor rate and labor productivity measurement must first be developed. Labor Rate — The labor rate is the total hourly expense or cost to the contractor for providing the particular craft or labor resource for the project. This labor rate includes direct costs and indirect costs. Direct labor costs include all payments made directly to the craftsworkers. The following is a brief listing of direct labor cost components: 1. 2. 3. 4. 5. 6. Wage rate Overtime premium Travel time allowance Subsistence allowance Show-up time allowance Other work or performance premiums
The sum of these direct labor costs is sometimes referred to as the effective wage rate. Indirect labor costs include those costs incurred as a result of use of labor resources but which are not paid directly to the craftsworker. The components of indirect labor cost include the following: 1. 2. 3. 4. 5. 6. 7. 8. 9. Vacation fund contributions Pension fund contributions Group insurance premiums Health and welfare contributions Apprenticeship and training programs Workers’ compensation premiums Unemployment insurance premiums Social security contribution Other voluntary contribution or payroll tax
It is the summation of direct and indirect labor costs that is termed the labor rate — the total hourly cost of providing a particular craft labor resource. Where a collective bargaining agreement is in force, most of these items can be readily determined on an hourly basis. Others are readily available from insurance companies or from local, state, and federal statutes. Several of the direct cost components must be estimated based on past records to determine the appropriate allowance to be included. These more difficult items include overtime, show-up time, and performance premiums. A percentage allowance is usually used to estimate the expected cost impact of such items. Labor Productivity — Of all the cost elements that contribute to the total project construction cost, labor productivity ranks at the top for variability. Because labor costs represent a significant proportion of the total cost of construction, it is vital that good estimates of productivity be made relative to the productivity that will be experienced on the project. Productivity assessment is a complex process and not yet fully understood for the construction industry. The following example illustrates the calculation of a unit price from productivity data.
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Example 11 To form 100 square feet of wall requires 6 hours of carpenter time and 5 hours of common laborer time. This assumption is based on standards calculated as averages from historical data. The wage rate with burdens for carpenters is \$60.00/h. The wage rate with burdens for common laborers is \$22.00/h. Solution. The unit cost may be calculated as follows: Carpenter — 6 h at \$60.00/h = \$360.00 Laborer — 5 h at \$22.00/h = \$110.00 Total labor cost for 100 ft2 = \$470.00
Labor cost per ft 2 = \$470.00 100 ft 2 = \$4.70 ft 2 This labor cost is adjusted for the following conditions: Weather adjustment Job complexity Crew experience Management 1.05 1.04 0.95 1.00
Adjusted unit cost = 4.70 ¥ 1.05 ¥ 1.04 ¥ 0.95 ¥ 1.00 = \$4.88 ft 2 Equipment Resources One of the most important decisions a contractor makes involves the selection of construction equipment. Beyond simple construction projects, a significant number of the activities require some utilization of major pieces of equipment. This equipment may either be purchased by the contractor or leased for the particular project at hand. The decision for selection of a particular type of equipment may be the result of an optimization process or may be based solely on the fact that the contractor already owns a particular piece of equipment that should be put to use. This decision must be anticipated or made by the estimator, in most cases, to forecast the expected costs for equipment on a project being estimated. Equipment Selection Criteria It is important for the estimator to have a solid background in and understanding of various types of construction equipment. This understanding is most important when making decisions about equipment. The estimator, having recognized the work to be performed, must identify the most economical choice for equipment. There are four important criteria that must be examined to arrive at the best choice: 1. 2. 3. 4. Functional performance Project flexibility Companywide operations Economics
Functional performance is only one criterion, but an important one, for the selection of construction equipment. For each activity, there is usually a clear choice based on the most appropriate piece of equipment to perform the task. Functional performance is usually examined solely from the perspective of functional performance. The usual measures are capacity and speed. These two parameters also give rise to the calculation of production rates. A second criterion that must be used is project flexibility. Although each task has an associated, appropriate piece of equipment based on functional performance, it would not be prudent to mobilize a different piece of equipment for each activity. Equipment selection decisions should consider the multiple uses the item of equipment possesses for the particular project. The trade-off between mobilization
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expense and duration versus efficiency of the operation must be explored to select the best fleet of equipment for the project. Companywide usage of equipment becomes an important factor when determining whether to purchase a particular piece of equipment for a project application. If the investment in the equipment cannot be fully justified for the particular project, then an assessment of future or concurrent usage of the equipment is necessary. This whole process necessarily influences selection decisions by the estimator because the project cost impacts must be evaluated. Equipment that can be utilized on many of the company projects will be favored over highly specialized single-project oriented equipment. The fourth, and probably most important, criterion the estimator considers is the pure economics of the equipment selection choices. Production or hourly costs of the various equipment alternatives should be compared to determine the most economical choice for the major work tasks involving equipment. A later section in this chapter explains and illustrates the process of determining equipment costs that the estimator should follow. Production Rates Equipment production rates can be determined in a relatively simple fashion for the purposes of the estimator. Most manufacturers produce handbooks for their equipment that provide production rates for tasks under stated conditions. Equipment Costs Equipment costs represent a large percentage of the total cost for many construction projects. Equipment represents a major investment for contractors, and it is necessary that the investment generate a return to the contractor. The contractor must not only pay for the equipment purchased but also pay the many costs associated with the operation and maintenance of the equipment. Beyond the initial purchase price, taxes, and setup costs, the contractor has costs for fuel, lubricants, repairs, and so on, which must be properly estimated when preparing an estimate. A system must be established to measure equipment costs of various types to provide the estimator with a data source to use when establishing equipment costs. The cost associated with equipment can be broadly classified as direct equipment costs and indirect equipment costs. Direct equipment costs include the ownership costs and operating expenses, while indirect equipment costs are the costs that occur in support of the overall fleet of equipment but which cannot be specifically assigned to a particular piece of equipment. Each of the broad cost categories will be discussed in greater detail in the following sections. Direct Equipment Costs Direct equipment expenses are costs that can be assigned to a particular piece of equipment and are usually divided into ownership and operating expenses for accounting and estimating. The concept behind this separation is that the ownership costs occur regardless of whether the equipment is used on a project. Ownership Costs Ownership costs include depreciation, interest, insurance, taxes, setup costs, and equipment enhancements. There are several views taken of ownership costs relating to loss in value or depreciation. One view is that income must be generated to build a sufficient reserve to replace the equipment at the new price, when it becomes obsolete or worn out. A second view is that ownership of a piece of equipment is an investment, and, as such, must generate a monetary return on that investment equal to or larger than the investment made. A third view is that the equipment ownership charge should represent the loss in value of the equipment from the original value due purely to ownership, assuming some arbitrary standard loss in value due to use. These three views can lead to substantially different ownership costs for the same piece of equipment, depending on the circumstances. For simplicity, ownership will be viewed as in the third view. The depreciation component of ownership cost will be discussed separately in the following section.
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Depreciation Costs Depreciation is the loss in value of the equipment due to use and/or obsolescence. There are several different approaches for calculating depreciation, based on hours of operation or on real-time years of ownership. In both cases, some arbitrary useful life is assumed for the particular piece of equipment based on experience with similar equipment under similar use conditions. The simplest approach for calculating depreciation is the straight-line method. Using the useful life, either hours of operation or years, the equipment is assumed to lose value uniformly over the useful life from its original value down to its salvage value. The salvage value is the expected market value of the equipment at the end of its useful life. Operating Expenses Operating costs are items of cost directly attributable to the use of the equipment. Operating costs include such items as fuel, lubricants, filters, repairs, tires, and sometimes operator’s wages. Obviously, the specific project conditions will greatly influence the magnitude of the operating costs. It is, therefore, important that on projects where the equipment is a significant cost item, such as large civil works projects like dams or new highway projects, attention must be given to the job conditions and operating characteristics of the major pieces of equipment. Equipment Rates The equipment rates used in an estimate represent an attempt to combine the elements of equipment cost that have been explained above. The pricing of equipment in an estimate is also influenced by market conditions. On very competitive projects, the contractor will often discount the actual costs to win the project. In other cases, even though the equipment has been fully depreciated, a contractor may still include an ownership charge in the estimate, because the market conditions will allow the cost to be included in the estimate. Materials Costs Materials costs can represent the major portion of a construction estimate. The estimator must be able to read and interpret the drawings and specifications and develop a complete list of the materials required for the project. With this quantity takeoff, the estimator then identifies the cost of these materials. The materials costs include several components: the purchase price, shipping and packaging, handling, and taxes. There are two types of materials: bulk materials and engineered materials. Bulk materials are materials that have been processed or manufactured to industry standards. Engineered materials have been processed or manufactured to project standards. Examples of bulk materials are sand backfill, pipe, and concrete. Examples of engineered materials are compressors, handrailing, and structural steel framing. The estimator must get unit price quotes on bulk materials and must get quotes on the engineered materials that include design costs as well as processing and other materials costs. Subcontractor Costs The construction industry continues to become more specialized. The building sector relies almost entirely on the use of specialty contractors to perform different trade work. The heavy/highway construction industry subcontracts a smaller percentage of work. The estimator must communicate clearly with the various subcontractors to define the scope of intended work. Each subcontractor furnishes the estimator with a quote for the defined scope of work with exceptions noted. The estimator must then adjust the numbers received for items that must be added in and items that will be deleted from their scope. The knowledge of the subcontractor and any associated risk on performance by the subcontractor must also be assessed by the estimator. The estimator often receives the subcontractor’s best estimate only a few minutes before the overall bid is due. The estimator must have an organized method of adjusting the overall bid up to the last minute for changes in the subcontractor’s prices.
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Example 12 For use as structural fill, 15,000 cubic yards of material must be hauled onto a job site. As the material is excavated, it is expected to swell. The swell factor is 0.85. The material will be hauled by four 12-yd3 capacity trucks. The trucks will be loaded by a 1.5-yd3 excavator. Each cycle of the excavator will take about 30 sec. The hauling time will be 9 min, the dumping time 2 min, the return time 7 min, and the spotting time 1 min. The whole operation can be expected to operate 50 min out of every hour. The cost of the trucks is \$66/h and the excavator will cost about \$75/h. What is the cost per cubic yard for this operation? Solution. Excavator capacity = 1.5 yd 3 ¥ 0.85 = 1.28 yd 3 cycle Hauler capacity = 12 ¥ 0.85 = 10.2 yd 3 cycle Number of loading cycles = 10.2 1.28 = 8 cycle Truck cycle time: Load 8 cycles ¥ 0.5 min = Haul Dump Return Spot TOTAL Fleet production: 4 ¥ (50 23) ¥ 10.2 = 89.75 yd 3 h 15, 000 89.75 = 168 h Cost: 168 ¥ 66 ¥ 4 = \$44,352 168 ¥ 75 = \$12,600 TOTAL \$56,952 4 min 9 min 2 min 7 min 1 min 23 min
56,952 15,000 = \$3.80 yd 3 Example 13 It is necessary to place 90 cubic yards of concrete. Site conditions dictate that the safest and best method of placement is to use a crane and a 2-cubic-yard bucket. It is determined that to perform the task efficiently, five laborers are needed — one at the concrete truck, three at the point of placement, and one on the vibrator. It is assumed that supervision is done by the superintendent. The wage rate for laborers is \$22.00/h. Time needed: Setup Cycle: Load Swing, dump, and return TOTAL
30 min 3 min 6 min 9 min
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No. of cycles Total cycle time Disassembly subtotal Inefficiency (labor, delays, etc.) 10% of cycle time Total operation time Amount of time needed (adjusted to workday) Laborers — five for 8 hours at \$22.00/h Cost per 90 yd3 Cost per cubic yard Example 14
90/2 = 45 ¥ 9 = = = 405 + 15 + 41 = = = = \$880/90 yd3 =
45 cycles 405 min 15 min 41 min 461 min 8h \$880.00 \$880.00 \$9.78/yd3
A small steel-frame structure is to be erected, and you are to prepare an estimate of the cost based on the data given below and the assumptions provided. The unloading, erection, temporary bolting, and plumbing will be done by a crew of 1 foreman, 1 crane operator, and 4 structural steel workers with a 55-ton crawler crane. The bolting will be done by two structural-steel workers using power tools. The painting will be done by a crew of three painters (structural-steel) with spray equipment. For unloading at site, erection, temporary bolting, and plumbing, allow 7 labor-hours per ton for the roof trusses, and allow 5.6 labor-hours per ton for the remaining steel. Assume 60 crew hours will be required for bolting. Allow 1.11 labor-hours per ton for painting. Materials: A 36 structural Costs: Structural steel supply: Fabrication: Freight cost: Field bolts: Paint: Labor costs:
Steel trusses Columns, etc.
15 tons 50 tons
Equipment costs:
44¢/lb \$800/ton — trusses \$410/ton — other steel \$2.65/100 lb 250 @ \$1.10 each 41 gallons @ \$30.00/gallon Assume payroll taxes and insurance are 80% of labor wage; use the following wages: Foreman \$24.10 Crane Operator \$21.20 Structural steel worker \$22.10 Painter \$20.20 Crane \$915.00/day Power tools \$23.40/day Paint equipment \$68.00/day \$300.00 40% of field labor cost 12% of all costs
Materials: Structural steel: 65 ¥ 2000 ¥.44 Freight: 65 ¥ 2000/100 ¥ 2.65 Field bolts: 250 ¥ \$1.10 Paint: 41 ¥ 30
= \$57,200 = 3445 = 275 = 1230 \$62,150
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Fabrication: Truss: 15 ¥ 800 Frame: 50 ¥ 410
= \$12,000 = 20,500 \$32,500
Labor crew costs: Erection: 1 foreman: 1 crane operator: 4 structural steel workers: Paint: 3 painters: Bolting: 2 structural steel workers: Erection: Frame: (50 ¥ 5.6)/6 Trusses: (15 ¥ 7)/6 Paint: (65 ¥ 1.11)/3 Bolting:
\$24.10 21.20 88.40 \$133.70 \$60.60 \$44.20
= 46.7 crew hours — 6 days 46.7 ¥ \$133.40 = \$6239 = 17.5 crew hours — 2 days 17.5 ¥ 133.40 = \$2340 = 24 crew hours — 3 days 24 ¥ 60.60 = \$1455 60 ¥ 44.20 = \$2652 Total labor = \$12,686 \$7320 187 204 300 \$8011 Summary
Equipment: Crane: 8 days ¥ 915/day = Power tools: 8 days ¥ 23.40/day = Paint equipment: 3 days ¥ 68/day = Move in/out = TOTAL
Materials: Fabrication: Labor: Equipment: Payroll taxes and insurance: 80% of 12,686 Overhead: 40% of (12,686 + 10,149) TOTAL Profit: 12% of 90,781 Bid
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\$62,150 32,500 12,686 8011 10,149 9134 \$134,630 16,156 = \$150,786
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Project Overhead Each project requires certain items of cost that cannot be identified with a single item of work. These items are referred to as project overhead and are normally described in the general conditions of the contract. The items that are part of the project overhead include but are not limited to the following: • • • • • • • • • • • Bonds Permits Mobilization Professional services (such as scheduling) Safety equipment Small tools Supervision Temporary facilities Travel and lodging Miscellaneous costs (e.g., cleanup, punch list) Demobilization
Each of these types of items should be estimated and included in the cost breakdown for a project. Markup Once the direct project costs are known, the estimator adds a sum of money to cover a portion of the general overhead for the firm and an allowance for the risk and investment made in the project — the profit. Each of these elements of markup is in large part determined by the competitive environment for bidding the project. The more competition, the less the markup. General Overhead Each business has certain expenses that are not variable with the amount of work they have under contract. These expenses must be spread across the projects. The typical method for spreading general overhead is to assign it proportionally according to the size of the project in relation to the expected total volume of work for the year. General overhead costs typically include the following: • • • • • • • • • • Salaries (home office) Employee benefits Professional fees Insurance Office lease or rent Office stationery and supplies Maintenance Job procurement and marketing Home office travel and entertainment Advertising
The only restriction on the items of general overhead is that they must have a legitimate business purpose. The estimator typically will start with the proportional amount and then add a percentage for profit. Profit The profit assigned to a project should recognize the nature of risk that the company is facing in the project and an appropriate return on the investment being made in the project. The reality is that the profit is limited by the competition. A larger number of bidders requires that a smaller profit be assigned to have a chance at having the low bid. This process of assigning profit is usually performed at the last minute by the senior management for the company submitting the bid.
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1.5 Contracts
The estimator prepares the estimate in accordance with the instructions to bidders. There are numerous approaches for buying construction services that the estimator must respond to. These various approaches can be classified by three characteristics: the method of award, the method of bidding/payment, and incentives/disincentives that may be attached.
Method of Award
There are three ways in which construction contracts are awarded: competitive awards, negotiated awards, and combination competitive-negotiated awards. With a purely competitive award, the decision is made solely on the basis of price. The lowest bidder will be awarded the project. Usually, public work is awarded in this manner, and all who meet the minimum qualifications (financial) are allowed to compete. In private work, the competitive method of award is used extensively; however, more care is taken to screen potential selective bidders. The term selective bid process describes this method of competitive award. At the opposite extreme from competitive awards are the negotiated awards. In a purely negotiated contract, the contractor is the only party asked to perform the work. Where a price is required prior to initiating work, this price is negotiated between the contractor and the client. Obviously, this lack of competition relieves some of the tension developed in the estimator through the competitive bid process because there is no need to be concerned with the price another contractor might submit. The contractor must still, if asked, provide a firm price that is acceptable to the client and may have to submit evidence of cost or allow an audit. As the purely competitive and purely negotiated method of contract awards represent the extremes, the combination competitive-negotiated award may fall anywhere in between. A common practice for relatively large jobs is to competitively evaluate the qualifications of several potential constructors and then select and negotiate with a single contractor a price for the work.
Method of Bidding/Payment
Several methods of payment are used to reimburse contractors for the construction services they provide. These methods of payment include lump sum or firm price, unit-price, and cost-plus. Each of these methods of payment requires an appropriate form of bidding that recognizes the unique incentive and risk associated with the method. The requirements for completeness of design and scope definition vary for the various types. The lump-sum or firm-price contract is widely used for well-defined projects with completed designs. This method allows purely competitive bidding. The contractor assumes nearly all of the risk, for quantity and quality. The comparison for bidding is based entirely on the total price submitted by the contractors, and payment for the work is limited to the agreed-upon contract price with some allowance for negotiated changes. The lump sum is the predominant form used for most building projects. The unit-price contract is employed on highway projects, civil works projects, and pipelines. For these projects, the quality of the work is defined, but the exact quantity is not known at the time of bidding. The price per unit is agreed upon at the time of bidding, but the quantity is determined as work progresses and is completed. The contractor, therefore, assumes a risk for quality performance, but the quantity risk is borne by the owner. There is a strong tendency, by contractors, to overprice or front-load those bid items that will be accomplished first and compensate with lower pricing on items of work that will be performed later. This allows contractors to improve their cash flow and match their income closer to their expenses. Each unit-price given must include a portion of the indirect costs and profits that are part of the job. Usually, quantities are specified for bidding purposes so that the prices can be compared for competitive analysis. If contractors “unbalance” or front-load certain bid items to an extreme, they risk being excluded from consideration. The unit-price approach is appropriate for projects where the quantity of work is not known, yet where competitive bidding is desirable.
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A third method, with many variations, is the cost-plus method of bidding/payment. With this method, the contractor is assured of being reimbursed for the costs involved with the project plus an additional amount to cover the cost of doing business and an allowance for profit. This additional amount may be calculated as a fixed fee, a percent of specified reimbursable costs, or a sliding-scale amount. The cost to the owner with this method of bidding/payment is open-ended; thus, the risks lie predominantly with the owner. This method is used in instances where it is desired to get the construction work underway prior to completion of design, or where it is desired to protect a proprietary process or production technology and design. Many of the major power plant projects, process facilities, and other long-term megaprojects have used this method in an attempt to shorten the overall design/construct time frame and realize earlier income from the project. Of the several variations used, most relate to the method of compensation for the “plus” portion of the cost and the ceiling placed on the expenditures by the owner. One of the variations is the cost plus a fixed fee. With this approach, it is in the contractor’s best interest to complete the project in the least time with the minimum nonreimbursable costs so that his profits during a given time period will be maximized. Where the scope, although not defined specifically, is generally understood, this method works well. The owner must still control and closely monitor actual direct costs. A second variation is the cost plus a percentage. This method offers little protection for the owner on the cost of the project or the length of performance. This method, in fact, may tempt the contractor to prolong project completion to continue a revenue stream at a set return. The sliding-scale approach is a third approach. This method of compensation is a combination of the two approaches described above. With this approach, a target amount for the project cost is identified. As costs exceed this amount, the fee portion decreases as a percentage of the reimbursable portion. If the costs are less than this target figure, there may be a sliding scale that offers the contractor an increased fee for good cost containment and management. In addition to the method of calculations of the plus portion for a cost-plus method, there may be a number of incentives attached to the method. These typically take the form of bonuses and penalties for better time or cost performance. These incentives may be related to the calendar or working day allowed for completion in the form of an amount per day for early completion. Similarly, there may be a penalty for late completion. The owner may also impose or require submittal of a guaranteed maximum figure for a contract to protect the owner from excessive costs.
1.6 Computer-Assisted Estimating
The process of estimating has not changed, but the tools of the estimator are constantly evolving. The computer has become an important tool for estimators, allowing them to produce more estimates in the same amount of time and with improved accuracy. Today, the computer is functioning as an aid to the estimator by using software and digitizers to read the architect/engineer’s plans, by retrieving and sorting historical cost databases, by analyzing information and developing comparisons, and by performing numerous calculations without error and presenting the information in a variety of graphical and tabular ways. The microcomputer is only as good as the programmer and data entry person. The estimator must still use imagination to create a competitive plan for accomplishing the work. The computer estimating tools assist and speed the estimator in accomplishing many of the more routine tasks. Many commercially available programs and spreadsheets are used by estimators for developing their final estimates of cost. These are tools that calculate, sort, factor, and present data and information. The selection of a software program or system is a function of the approach used by the contractor and the particular work processes and cost elements encountered. The most widely used tool is still the spreadsheet because it gives the estimator a tool for flexible organization of data and information and the capacity to make quick and accurate calculations.
© 2003 by CRC Press LLC
Construction Estimating
1-27
Defining Terms1
Bid — To submit a price for services; a proposition either verbal or written, for doing work and for
supplying materials and/or equipment.
Bulk materials — Material bought in lots. These items can be purchased from a standard catalog
description and are bought in quantity for distribution as required.
Cost — The amount measured in money, cash expended, or liability incurred, in consideration of goods
Direct cost — The cost of installed equipment, material, and labor directly involved in the physical
construction of the permanent facility.
Indirect cost — All costs that do not become part of the final installation but which are required for
the orderly completion of the installation. Markup — Includes the percentage applications, such as general overhead, profit, and other indirect costs. Productivity — Relative measure of labor efficiency, either good or bad, when compared to an established base or norm. Quantity survey — Using standard methods to measure all labor and material required for a specific building or structure and itemizing these detailed quantities in a book or bill of quantities. Scope — Defines the materials and equipment to be provided and the work to be done.
References
Adrian, J.J. 1982. Construction Estimating. Reston Publishing, Reston, VA. American Association of Cost Engineers, Cost Engineer’s Notebook, Morgantown, WV. Bauman, H.C. 1964. Fundamentals of Cost Engineering in the Chemical Industry. Reinhold Publishing, Florence, KY. Collier, K.F. 1974. Fundamentals of Construction Estimating and Cost Accounting. Prentice-Hall, Englewood Cliffs, NJ. Gooch, K.O. and Caroline, J. 1980. Construction for Profit. Reston Publishing, Reston, VA. Hanscomb, R. et al. 1983. Yardsticks for Costing. Southam Business Publications, Ltd., Toronto. Helyar, F.W. 1978. Construction Estimating and Costing. McGraw-Hill Ryerson Ltd., Scarborough, Ontario. Humphreys, ed. 1984. Project and Cost Engineers’ Handbook. Marcel Dekker, New York. Hunt, W.D. 1967. Creative Control of Building Costs. McGraw-Hill, New York. Landsdowne, D.K. 1983. Construction Cost Handbook. McGraw-Hill Ryerson Ltd., Scarborough, Ontario. Neil, J.M. 1982. Construction Cost Estimating for Project Control. Prentice-Hall, Englewood Cliffs, NJ. Peurifoy, R.L. 1975. Estimating Construction Costs. McGraw-Hill, New York. Seeley, I.H. 1978. Building Economics. The Macmillan Press, Ltd., London. Vance, M.A. 1979. Selected List of Books on Building Cost Estimating. Vance Bibliographies, Monticello, IL. Walker, F.R. 1980. The Building Estimator’s Reference Book. Frank R. Walker Publishing, Chicago, IL.
Source: American Association of Cost Engineers (AACE, Inc.), AACE Recommended Practices and Standards, November 1991.
1
© 2003 by CRC Press LLC
1-28
The Civil Engineering Handbook, Second Edition
Further Information
For more information on the subject of cost estimating, one should contact the following professional organizations that have additional information and recommended practices. AACE, International (formerly the American Association of Cost Engineers), 209 Prairie Ave., Suite 100, Morgantown, WV 26507, 800–858-COST. American Society of Professional Estimators, 11141 Georgia Ave., Suite 412, Wheaton, MD 20902, 301–929–8848. There are numerous textbooks on the subject of cost estimating and construction cost estimating. Cost engineering texts usually have a large portion devoted to both conceptual estimating and detailed estimating. The following reference materials are recommended: Process Plant Construction Estimating Standards. Richardson Engineering Services, Mesa, AZ. Contractor’s Equipment Cost Guide. Data quest — The Associated General Contractors of America (AGC). The Building Estimator’s Reference Book. Frank R. Walker, Lisle, IL. Means Building Construction Cost Data. R.S. Means, Duxbury, MA. Estimating Earthwork Quantities. Norseman Publishing, Lubbock, TX. Caterpillar Performance Handbook, 24th ed. Caterpillar, Peoria, IL. Means Man-Hour Standards. R.S. Means, Duxbury, MA. Rental Rates and Specifications. Associated Equipment Distributors. Rental Rate Blue Book. Data quest — The Dun & Bradstreet Corporation, New York. Historical Local Cost Indexes. AACE — Cost Engineers Notebook, Vol. 1. Engineering News Record. McGraw-Hill, New York. U.S. Army Engineer’s Contract Unit Price Index. U.S. Army Corps of Engineers. Chemical Engineering Plant Cost Index. McGraw-Hill, New York. Bureau of Labor Statistics. U.S. Department of Labor.
© 2003 by CRC Press LLC
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sofda free Junior Manager | 16,642 | 78,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2014-35 | longest | en | 0.875731 |
https://mathematica.stackexchange.com/questions/156635/how-to-find-the-limit-of-an-infinite-recursion-relation | 1,576,101,934,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00109.warc.gz | 458,382,984 | 32,178 | # How to find the limit of an infinite recursion relation
The recursive relation is $f(n+1) = 2 + \frac{1}{f(n)}$, I would like to find $$\lim_{n\to\infty} f(n)$$ given $f(1)=3$.
I tried to use RSolve but it does not give me a definite value.
RSolve[{f[n + 1] - 1/f[n] == 2}, f[1] == 3, n]
• Try this: FixedPoint[2 + 1/# &, 3.]. It applies the function repeatedly until the output doesn't change. – Anjan Kumar Sep 27 '17 at 16:22
• Thank you, FixedPoint worked! – Henry Wang Sep 27 '17 at 16:26
• @AnjanKumar - FixedPoint[2 + 1/# &, 3.] // RootApproximant – Bob Hanlon Sep 27 '17 at 16:45
• @BobHanlon That's much better. – Anjan Kumar Sep 27 '17 at 16:50
• You have incorrect syntax for RSolve. – Edmund Sep 27 '17 at 22:13
In Version 11.2, you can use RSolveValue to obtain the answer as shown below (unfortunately, earlier versions return Indeterminate for this input).
=================
RSolveValue[{f[n + 1] - 1/f[n] == 2, f[1] == 3},
f[Infinity], n]
(* 1 + Sqrt[2] *)
N[%]
(* 2.41421 *)
FixedPoint[2 + 1/# &, 3.]
(* 2.41421 *)
=====================
Hope this helps.
It should be RSolve rather than Rsolve and the boundary condition must be inside the list of equations.
Clear[f];
eqns = {f[n + 1] - 1/f[n] == 2, f[1] == 3};
soln = RSolve[eqns, f, n][[1]];
Verifying that the solution satisfies the equations
eqns /. soln // Simplify
(* {True, True} *)
f[n] /. soln // FullSimplify
(* 1 + Sqrt[2] - (2 Sqrt[2] (1 - Sqrt[2])^n)/((1 - Sqrt[2])^n + (1 + Sqrt[2])^n) *)
Limit[f[n] /. soln, n -> Infinity]
(* 1 + Sqrt[2] *)
Limit[f[n] /. soln, n -> -Infinity]
(* 1 - Sqrt[2] *)
Just for fun:
cp = {x, x} /. Solve[x == 2 + 1/x, x];
fp = cp[[2, 1]]
f[x0_, n_] :=
Sequence @@ {{##}, {{##}[[2]], {##}[[2]]}} & @@@
Partition[NestList[2 + 1/# &, x0, n], 2, 1]
Plot[{2 + 1/x, x}, {x, -5, 5},
Epilog -> {Red, PointSize[0.02], Point@cp}]
ListAnimate@
Table[Plot[{2 + 1/x, x}, {x, 2.3, 3},
Epilog -> {Red, PointSize[0.02], Point[cp], Line[f[3, j]]},
PlotRange -> {2.3, 2.45}], {j, 1, 5}] | 788 | 2,014 | {"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-2019-51 | latest | en | 0.640499 |
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# Find the answer if you are genius; How many legs are present in the bus?
You entered into a bus. Inside the bus you saw 7 girls. Each girl has 7 bags. In each bag, there are 7 big cats Each big cat has 7 little cats. Each cat has 4 legs. Question: How many legs are present in the bus?
This question is closed to new answers.
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# Can anybody give a good solution to this one? An eight-card
Author Message
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Can anybody give a good solution to this one? An eight-card [#permalink]
### Show Tags
28 Jul 2003, 19:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Can anybody give a good solution to this one?
An eight-card deck consists of 4 pairs of each suit. 2 cards are taken at random without replacement. What is the probability to have two cards of the same suit?
тАв 1/2
тАв 1/3
тАв 1/4
тАв 1/7
тАв 1/8
Last edited by bb on 28 Jul 2003, 19:53, edited 1 time in total.
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### Show Tags
28 Jul 2003, 21:05
bb wrote:
Can anybody give a good solution to this one?
An eight-card deck consists of 4 pairs of each suit. 2 cards are taken at random without replacement. What is the probability to have two cards of the same suit?
тАв 1/2
тАв 1/3
тАв 1/4
тАв 1/7
тАв 1/8
The absolute quickest way to solve this is to say:
Whatever card I pick as the first card, there is only one card of the seven remaining to match its suit. Hence, the answer is 1/7
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Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
Kudos [?]: 235 [0], given: 0
Founder
Joined: 04 Dec 2002
Posts: 15585
Kudos [?]: 28510 [0], given: 5114
Location: United States (WA)
GMAT 1: 750 Q49 V42
### Show Tags
28 Jul 2003, 22:37
AkamaiBrah wrote:
bb wrote:
Can anybody give a good solution to this one?
An eight-card deck consists of 4 pairs of each suit. 2 cards are taken at random without replacement. What is the probability to have two cards of the same suit?
тАв 1/2
тАв 1/3
тАв 1/4
тАв 1/7
тАв 1/8
The absolute quickest way to solve this is to say:
Whatever card I pick as the first card, there is only one card of the seven remaining to match its suit. Hence, the answer is 1/7
Very good reasoining!!!
You are the man!
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### Show Tags
29 Jul 2003, 07:58
Another way would be:
Favorable outcome = 4 (because there are only 4 cases where both the cards will be of same suite)
Total outcome = 8C2 = 28
P = 4/28 = 1/7
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29 Jul 2003, 13:02
great soluton akamai ..
you are a star
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Discussion in 'Homework Help' started by Whalum, Apr 3, 2009.
1. ### Whalum Thread Starter Member
Mar 10, 2009
11
0
Graph the voltage as a function of time in point P. The voltage E is given in the picture.
I'm so lost I don't even know where to start. All I know is, that there's no reflection between R1 and Z01.
Any tips on how to start are appreciated!
2. ### beenthere Retired Moderator
Apr 20, 2004
15,815
282
Transmission lines tend to be sensitive to frequency in terms of length and so on. But think of the duration of the pulse and the propagation time through the TL - recalling the speed of light. That may clarify your problem.
3. ### Whalum Thread Starter Member
Mar 10, 2009
11
0
Nope, not getting it. This whole concept of transmission line is somehow so different from anything I've done before. Let's try baby steps. So it takes 2.5 microseconds for the pulse to reach point P. But how do I determine the voltage drop in R1? Can I use normal voltage division? So it would be $U_{R1}= \frac{R_1}{R_1+Z_{01}+Z_{02}+R_L}E=2V$
4. ### KL7AJ Senior Member
Nov 4, 2008
2,013
273
Alas...our ultraparanoid server at work has excised your drawings....or I'd be able to help. Could you attach the files to an email to me?
eric.nichols@eielson.af.mil
Eric
5. ### KL7AJ Senior Member
Nov 4, 2008
2,013
273
I'm currently teaching a comprehensive section on the Smith Chart in my A.C. theory class.
Mar 10, 2009
11
0
7. ### wr8y Active Member
Sep 16, 2008
232
1
I wish I was there.
8. ### KL7AJ Senior Member
Nov 4, 2008
2,013
273
You can enroll in the Fall semester.
eric
Nov 25, 2008
394
2
Lefty
10. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
There are two points where reflections will occur. Where the two transmission lines meet (with unequal characteristic impedances) and at the load - due to load mismatch with the second line.
Work out the reflection coefficients at each point and draw a time graph with a superposition of the forward and backward traveling pulses, having regard to the reflection coefficient signs.
11. ### wr8y Active Member
Sep 16, 2008
232
1
Alaska? I moved to Georiga from Michigan 8 years ago 'cause 40 years of winter was enough! No thanks, I'll stay here and enjoy the warm weather and year around hiking. Besides, "Georgia HOPE" pays for me to go to school here.
But I WOULD like to sit in your class...
12. ### Whalum Thread Starter Member
Mar 10, 2009
11
0
Will the reflected signals turn to heat completely at R1? And will the signal that reaches the load turn to heat as well?
Last edited: Apr 4, 2009
13. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
Yes.
You may have deduced by now that owing to the match discontinuity at the junction of the two lines and the load mismatch at the end of the second line you will end up with a pulse of ever decreasing magnitude and alternating sign "bouncing" back and forth between the two reflection points in the second line at light speed. Depending on where you "stand" on the second line as an observer, determines what you record as the time plot of the pulses as they come and go. In your original question I'm assuming the diagram intends to indicate the observer's position as mid-point on the second line.
14. ### Whalum Thread Starter Member
Mar 10, 2009
11
0
So what I get now, is: voltage drop in the first resistor and TL is 4V, so the voltage goin into the second TL is 8V. Reflection coefficient between first and second TL is .5, so the magnitude of signal that gets transmitted, is 1.5 times greater, 12V. And the pulse that gets "trapped" in the second TL gets half the magnitude and opposite sign every reflection.
There are two things bothering me. My logics are against my course material, which says that the transmission coefficient can be calculated using $\tau=1+ \rho$, where rho is the reflection coefficient. Are we getting free energy? And the second thing: is the trapped pulse really going to be there for ever?
Last edited: Apr 5, 2009
15. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
Not sure where the 4V came from.
Since the first line has Z0=100 and R1=100 I would expect the pulse amplitude at the line input would be 6V.
Also not sure about $\tau=1+\rho$ .... never used it myself.
If I wanted to find the pulse amplitude when it first enters the second line at the line junction, I would do an energy balance for the incident, reflected & forward values.
I think it would go something like ...
W_incident = W_reflected +W_forward [W being the energy]
Why?: The sum of reflected and forward energies must equate to the incident energy.
So..
$Wi=(E^2/Z01)*\delta t$
where $E=6V \ and\ \delta t = 1\mu sec$
I'd need to keep in mind that there is a different impedance on the second line - Z02=300Ω.
As to the pulse trapped on the second line, I think it would gradually loose energy each time it hits the mismatch at each end. Some energy leaks back to the source and some goes into the load. Does it go on forever? I guess it stops when all the original energy is exhausted - which would be at time=∞. This is an ideal situation in which no line losses are included.
I'm happy to be corrected on any of these assumptions.
Do you actually have access to a solution for the problem? Say from the source text ...
16. ### Whalum Thread Starter Member
Mar 10, 2009
11
0
Unfortunately, no. This is an assignment for extra points in the next mid-term.
I got the 4V from voltage division: $U=\frac{R_1+Z_{01}}{R_1+Z_{01}+Z_{02}+R_L}E=4V$. I'm with you, though, that the transmission coefficient, $\tau$, can't be right. It would make more sense to subtract the reflection coefficient from 1. Maybe a typo in the material?
I'll try the energy method and see what I'll get.
17. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
I'm thinking the transmission coefficient should be something like ....
TC=1-ρ^2
So if ρ=0.5
TC=0.75
Hence a pulse of E volts magnitude traversing the line intersection would have a reflected component of 0.5*E and a transmitted component of 0.75*E.
18. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
Doubt if that's the way to go ....
Just consider the first transmission line as having 100Ω impedance as "viewed" from the source feed point.
I know this all sounds crazy - especially when you first start on transmission lines. We get so used to solving problems with Kirchoff's laws we often balk at these new concepts at the first (second ...third ...) encounter.
The problem is we are dealing with guided waves traveling along these lines with distributed parameters - not discrete circuits. I remember an RF engineer saying once - "50 ohm coax has 50 ohms impedance whether its 100 miles long or 1 inch long". Knock me over with a feather .....!
19. ### Whalum Thread Starter Member
Mar 10, 2009
11
0
This sounds a bit more logical to me, but aren't we still getting free energy?
Does this really change something? The pulse goes through the whole line, so isn't it same, if we replace the line with resistor with the same resistance. If not, how do you get the 6V?
20. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
782
I would view the pulse source of 12V amplitude as being imposed on a series combination comprising R1 and Z01 - much like a voltage divider.
Equal values of R1 and Z01 would give equal voltage drops - 6V each.
Free energy - not really.
Remember energy is proportional to the square of the voltage. I realize it is confusing if you just add the two voltages algebraically.
You ask what appears to be a logical question - "Isn't 0.5*E+0.75*E greater than the original E value - that means more energy! ...?"
You need to think it through carefully. | 2,043 | 7,693 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 10, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-44 | longest | en | 0.909784 |
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# Early in the development of a new product line, the critical
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Early in the development of a new product line, the critical [#permalink]
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08 Apr 2011, 04:05
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Early in the development of a new product line, the critical resource is talent. New marketing ventures require a degree of managerial skill disproportionate to their short-term revenue prospects. Usually, however, talented managers are assigned only to established high-revenue product lines and, as a result, most new marketing ventures fail. Contrary to current practice, the best managers in a company should be assigned to development projects.
Which one of the following, if true, most strengthens the author’s argument?
(A) On average, new ventures under the direction of managers at executive level survive no longer than those managed by lower-ranking managers.
(B) For most established companies, the development of new product lines is a relatively small part of the company’s total expenditure.
(C) The more talented a manager is, the less likely he or she is to be interested in undertaking the development of a new product line.
(D) The current revenue and profitability of an established product line can be maintained even if the company’s best managers are assigned elsewhere.
(E) Early short-term revenue prospects of a new product line are usually a good predictor of how successful a product line will ultimately be.
Guys i know the answer so there is no point replying to this Message with
IMO A,B,C,D,E
What i wanna discuss is about how B and C fail to strengthen the argument.
(B) For most established companies, the development of new product lines is a relatively small part of the company’s total expenditure.
Cant this be the reason why Talented managers are not assigned to new development projects.The company does not want to spend much on the new projects and therefore it decides to assign mediocre managers to the new development projects in order to keep the Project cost as low as possible.
(C) The more talented a manager is, the less likely he or she is to be interested in undertaking the development of a new product line.
Cant this be 1 of the reasons for the company not to assign taleneted managers to the new dev. projects
[Reveal] Spoiler: OA
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Re: LSAT Repeated quest [#permalink]
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08 Apr 2011, 04:21
What is the argument? That since more talent is required during development phase, best managers should be assigned to development projects.
C is saying that talented/ best managers are not interested in development projects mostly. How can this strengthen author's argument? In fact it is sort of weakening only - that best managers should NOT be assigned to development projects, since mostly they will not be interested in these projects, and this arguable can affect adversely their perfomance and that of the project.
B says that development projects are a small part of the expenditure. Assuming that best managers will mean more expenditure on development projects, this is contadictory to the author's conclusion, hence it cant strengthen the argument.
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Re: LSAT Repeated quest [#permalink]
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08 Apr 2011, 04:36
vivesomnium wrote:
What is the argument? That since more talent is required during development phase, best managers should be assigned to development projects.
C is saying that talented/ best managers are not interested in development projects mostly. How can this strengthen author's argument? In fact it is sort of weakening only - that best managers should NOT be assigned to development projects, since mostly they will not be interested in these projects, and this arguable can affect adversely their perfomance and that of the project.
B says that development projects are a small part of the expenditure. Assuming that best managers will mean more expenditure on development projects, this is contadictory to the author's conclusion, hence it cant strengthen the argument.
I was saying C reinforces the premise that the current practice is not to assign talented managers to new projects.
Anything that reinforces the premise strengthens the argument, doesnt it ?
I guess you are only looking at the conclusion for both the answer choices.Need Expert opinion to confirm whether 1 can do it your way.
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Re: LSAT Repeated quest [#permalink]
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08 Apr 2011, 05:35
mundasingh123 wrote:
vivesomnium wrote:
What is the argument? That since more talent is required during development phase, best managers should be assigned to development projects.
C is saying that talented/ best managers are not interested in development projects mostly. How can this strengthen author's argument? In fact it is sort of weakening only - that best managers should NOT be assigned to development projects, since mostly they will not be interested in these projects, and this arguable can affect adversely their perfomance and that of the project.
B says that development projects are a small part of the expenditure. Assuming that best managers will mean more expenditure on development projects, this is contadictory to the author's conclusion, hence it cant strengthen the argument.
I was saying C reinforces the premise that the current practice is not to assign talented managers to new projects.
Anything that reinforces the premise strengthens the argument, doesnt it ?
I guess you are only looking at the conclusion for both the answer choices.Need Expert opinion to confirm whether 1 can do it your way.
First of all mundasingh123 I am not an expert still would like to share my thoughts on this.
I have my apprehensions about your view that "Anything that reinforces the premise strengthens the argument".Here it means actually that reinforcing the premises actually reinforces the arguments's conclusion contrary to your view of looking at it I assume.Because when you want to strengthen an argument you support/strengthen the conclusion by giving additional premises to strengthen it.All the premises contribute towards the conclusion not towards other premises.
So both option B and C are weakening the argument(conclusion per se) as very well explained by vivesomnium.
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Re: LSAT Repeated quest [#permalink]
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08 Apr 2011, 07:57
D is clearly the winner
now why B and C fail -
(B) For most established companies, the development of new product lines is a relatively small part of the company’s total expenditure. and hence there is no need to assign a experienced manager to it. This weakens the argument
(C) The more talented a manager is, the less likely he or she is to be interested in undertaking the development of a new product line. and hence he will not show much interest in making the product a success. This also weakens the argument.
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Re: LSAT Repeated quest [#permalink]
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08 Apr 2011, 15:02
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Hey Mundasingh:
Thanks for the invite to weigh in! This is a fantastic example of why it's so critical to identify the conclusion of an argument before attacking the answer choices - a task that I don't think is typically taught all that well. While the author definitely does talk about talented managers typically NOT being assigned to development projects, his argument is that they SHOULD BE assigned to them. Ultimately, there are three ways to find a conclusion in one of these arguments:
THREE WAYS TO IDENTIFY THE CONCLUSION
1) Conclusion language (such as "therefore", "thus", "in conclusion", "so", etc.)
2) A call for action ("We should", "they must", etc.)
3) The effect of a cause-effect relationship (It is raining, so the parade will be cancelled. The rain causes the cancellation, so "the parade will be cancelled" is the conclusion)
3a) To better exemplify this, try the "Why Test". Facts in these arguments don't have a reason why. Why is it raining? It doesn't say. But it does give a reason "why" the parade will be cancelled - because it's raining. So "the parade will be canceled", because the argument provides a reason why, is a conclusion based upon existing facts.
So...given all that, the conclusion of this argument is "the best managers in a company should be assigned to development projects" - it has a call for action "SHOULD BE", and it passes the why test. Why should they be assigned to those projects? The passage says because "early in projects the critical resource is talent"; because "new ventures require a disproportionate degree of managerial skill"; etc. So the conclusion is "the best managers should be assigned to development projects".
B weakens that, giving a reason that they shouldn't be (it's a small portion of the company's operations).
C weakens that, giving a reason that they shouldn't be (they probably don't want to)
D strengthens it, giving another reason why they should be (we already know that they'll significantly improve the odds of the new project's success; D says that they'll also not really be missed in the regular operations)
I hope that helps - keep in mind that the conclusion is by far the most important part of any of these arguments, so take care in making sure that you identify and embrace it.
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Re: Early in the development of a new product line, the critical [#permalink]
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Re: Early in the development of a new product line, the critical [#permalink] 11 Jun 2016, 14:39
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# Early in the development of a new product line, the critical
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,040 | 13,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-22 | latest | en | 0.888348 |
https://www.bartleby.com/essay/Apple-s-Iphone-Price-On-Australia-Jump-F3UTWTQ3FTDX | 1,716,105,425,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00710.warc.gz | 598,682,060 | 9,812 | # Apple 's Iphone Price On Australia Jump 15 Per Cents
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1.0 Article Summary The article of “APPLE’S IPHONE PRICES IN AUSTRALIA JUMP 15 PER CENT OVERNIGHT”will discuss increasing the price of apple product which is a few of iPhones in Australia. According to Rayner, Rising the price of iPhones, it is because of the rate Australian dollar falls down. When first time apple introduced the new iPhone 6 and iPhone 6 plus on September 2014, AU\$1 equals to US\$0,89 (Rayner 2015). However, last year on March 2015, Apple company announced that AU\$1 equals to US\$0,77 (Rayner 2015). It means Australian dollar falls down 15 per cent that affects iPhones price rises. For instance, previously, the price of iPhone 6 in 16 GB storage is AU\$869, however starting March 2015, the price is rised from AU\$869 to …show more content…
2013, 3). What is the meaning of demand ? “demand is a relationship between price of the good and the quantity demanded of the good (Curtin University 2015).” The law of demand is, “Holding everything else constant, [ceteris paribus] when the price of a product increases the quantity demanded will fall, and when the price of a product decreases, the quantity demanded will rise (Curtin University 2015).” According to Curtin University, “Price and quantity demanded are negatively related, as the law of demand.” The demand curve In the figure 1, previously, customers who want to buy iPhone 6 in 16 GB storage must pay AU\$869, however, they must pay more which the cost is AU\$999 now and the quatity demanded from point X to point Y automatically decreases. It means that “there is a movement along from point A to point B because the price rises (Curtin University 2015).” From this diagram, it shows that the price of iPhone 6 in 16 GB storage rises from AU\$869 to AU\$999. Because the price rises, the quantity demanded falls from 1,5 to 1,3. “A definition of supply is a relationship between quantity supplied and the price of good (Curtin University 2015) .” What is supply ? Supply curve has relationship among the price of good and quantity supplied, when the sales of producers are affected remain the same (Curtin | 528 | 2,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.917464 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-find-the-indefinite-integral-of-int-2x-3-10x-5-dx | 1,725,814,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00085.warc.gz | 83,395,257 | 5,876 | # How do you find the indefinite integral of int (-2x^-3+10x^-5) dx?
Aug 24, 2016
$\frac{1}{x} ^ 2 - \frac{5}{2 {x}^{4}} + c$
#### Explanation:
integrate each term using the $\textcolor{b l u e}{\text{power rule for integration}}$
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\int \left(a {x}^{n}\right) \mathrm{dx} = \frac{a}{n + 1} {x}^{n + 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\Rightarrow \frac{- 2}{- 2} {x}^{-} 2 + \frac{10}{- 4} {x}^{-} 4 + c$
$\Rightarrow \int \left(- 2 {x}^{-} 3 + 10 {x}^{-} 5\right) \mathrm{dx} = \frac{1}{x} ^ 2 - \frac{5}{2 {x}^{4}} + c$ | 283 | 634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.326419 |
https://benvitalenum3ers.wordpress.com/2016/05/14/squarefree-semiprimes-n-%CF%83n-%CF%86n-and-%CF%83n-%CF%86n/ | 1,501,159,085,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428257.45/warc/CC-MAIN-20170727122407-20170727142407-00578.warc.gz | 508,407,130 | 36,120 | ## SquareFree semiprimes n, σ(n), φ(n), and σ(n) * φ(n)
$\sigma (n)$ is the sum of divisors function
Euler’s totient function $\phi (n)$ is the number of positive integers not exceeding $n$ that have no common divisors with $n$ (other than the common divisor 1).
In other words, $\phi (n)$ is the number of integers $m$ coprime to $n$ such that $1 \; \leq \; m \; \leq \; n$
Product = $\phi (n) \cdot \sigma (n)$
Squarefree semiprimes are numbers $n$ such that $\phi \,(n) \; + \; \sigma \,(n) \; = \; 2 \, (n+1)$
These results show a pattern.
Prove that a positive integer $n$ is the product of two primes differing by 2 iff
$\sigma (n) \cdot \phi (n) \; = \; (n + 1) \,(n - 3)$
Also, prove that a positive integer $n$ is the product of two primes differing by 4 iff
$\sigma (n) \cdot \phi (n) \; = \; (n + 3) \,(n - 5)$
and a positive integer $n$ is the product of two primes differing by 6 iff
$\sigma (n) \cdot \phi (n) \; = \; (n + 5) \,(n - 7)$
and a positive integer $n$ is the product of two primes differing by 8 iff
$\sigma (n) \cdot \phi (n) \; = \; (n + 7) \,(n - 9)$
and a positive integer $n$ is the product of two primes differing by 10 iff
$\sigma (n) \cdot \phi (n) \; = \; (n + 9) \,(n - 11)$
and a positive integer $n$ is the product of two primes differing by 12 iff
$\sigma (n) \cdot \phi (n) \; = \; (n + 11) \,(n - 13)$ | 524 | 1,420 | {"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": 23, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-30 | longest | en | 0.706802 |
https://www.pokecommunity.com/archive/index.php/t-111497.html | 1,513,286,413,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550986.47/warc/CC-MAIN-20171214202730-20171214222730-00780.warc.gz | 815,669,928 | 2,860 | PDA
View Full Version : [Tutorial] Hex Editing Legendary Pokemon
Elementologist
October 16th, 2007, 9:36 AM
Hex Editing Legendary Pokemon
Well this tutorial is for beginners so if you think you are one then read on.
What You Need
1. A Pokemon Rom (preferably FR or LG)
2. Hex Workshop (Get it from download.com)
3. Advance Map (Unless you have another way to find the right offsets)
What You Do
Step 1
Open Advance Map and load you rom.
Step 2
Go to the third level of Cerulian Cave (We Are Going to use Mewtwo) and right click + copy Mewtwo's Offset without the "\$"
Step 3
Open Hex Workshop and click "Edit" and then "Goto".
Paste the offset (in this case for FR 1624F5) and select "hex" instead of "decimal" and then "beginning of file".
So then you will see this:
022A 8100 0325 8701 210D 8002 0006 01E0
7A1A 0825 8801 6A5A 30A1 9600 0200 679F
7F17 0866 C528 1400 3356 0100 6DB6 9600
4600 0029 0708 2538 0127 2A07 0826 0D80
So The Red Colored One Is The Pokemon (9600, well 96 in decimal is 150 which is Mewtwo's position in the Pokedex).
And The Blue Colored One Is The Level (4600, 46 in decimal is 70 which is Mewtwo's level).
So change these two values however you like.
To find out what hex number the pokemon you want has, open calculator click "view" and then "scientific". After that select the Decimal box and write the Pokemon's Pokedex Number.
Then Select the hex box and whatever you see just add to zeros behind it. Same goes with the level, so write the level in the Decimal box.
Here are some examples:
Entei F400
Suicune F500
Raikou F300
Ho-Oh FA00
10 Level: A000
20 Level: 1400
30 Level: 1E00
40 Level: 2800
50 Level: 3200
60 Level: 3C00
70 Level: 4600
80 Level: 5000
90 Level: 5A00
100 Level: 6400
Final Step
In Hex workshop click "file" and then "save".
You're Done!
NOTE
REMEMBER TO MAKE A BACKUP OF YOUR ROM BEFORE DOING ANY OF THIS
Elementologist
October 26th, 2007, 1:54 AM
Come on guys. 130 views and none of you has anything to say.
thethethethe
October 26th, 2007, 2:26 AM
Well, this is pretty pointless now, since scripting a wildbattle, has become an extremely basic thing.
cooley
October 26th, 2007, 1:23 PM
Yeah, Mostly everyone knows how to do it now, And we now have a lot of tools,
that can help us to do so. (e.g. poketronic, etc.)
mbgbben
December 24th, 2007, 9:47 AM
Everytime i script a pokemon with the hex editor or hex workshop the pokemon always comes back after the battle so I can always battle it again.Help?
ZodiacDaGreat
December 24th, 2007, 1:48 PM
I tutorial helps mostly noobs, who donot script. The way you're suggesting will cause glitches later on in the game, due to flags and stuff. So its better to create your original wild battle.
mbgbben
December 25th, 2007, 1:09 AM
So there is no way or what? | 864 | 2,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-51 | latest | en | 0.750769 |
https://sonichours.com/how-many-hours-are-in-50-years/ | 1,702,088,054,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00146.warc.gz | 582,805,731 | 20,570 | General
# How Many Hours Are In 50 Years
Knowing how many hours are in 50 years is important because you’ll need the answer for a calculation. For instance, one year has 365 days and 50 weeks have 168 days. In the same way, you’ll need to know how many hours are in a month. Then, you can use this formula to calculate how many hours are in a year. It’s very useful because you’ll be able to convert different kinds of times.
To convert hours into days, divide five years by 59 minutes. This way, you get an accurate answer for five years. The same goes for ten years, fifteen months, and twenty-five days. Each of these times is converted into a day, or an hour. A year that’s fifty years old is equal to 450 minutes. A year with fewer days will be equal to 650 hours.
If you’re interested in learning more about time and how much it is worth, you can take the age of a baby. It’s a good idea to keep a journal of your life, since that will make it easier to refer to when you’re older. You can also convert years into hours by using a calculator. If you’re wondering how long you’ll be able to live in 50 years, you’ll need to calculate your average number of days.
A lifetime of years has six months, which is equivalent to 18250 days. A year has 438000 hours, or 26080 thousand minutes. It’s a great way to look at time and how much it’s worth. The millennium is a 1,000-year-old millennium. The only way to find the answer is to get yourself a time converter for your needs. It’s very easy!
A decade is made up of a hundred days. A century is divided into a hundred days. A hundred years are four thousand. A lifetime is the same as the previous century. The millennium is a thousand days, and the millennium is five thousand years. A day is one minute, and a year is eight minutes. So, you can see how many hours are in fifty-two decades.
How many hours are in fifty-two years? A century is a thousand months. So, a century is a hundred and a year is fifty. A millennium is 100 centuries. A decade is 500 years. And it’s the same time. However, you can find out how many hours are in fifty-two years. There are no leap-years. So, you can even see if it’s a century or more!
Likewise, the millennium is five years. Similarly, a year is a thousand months. Then you can count on a year to see how many hours are in a day. In fact, a decade is a thousand days. Then you’ll be able to count on just one year, or you can add up each month until you reach a century. Then you’ll be able and feel the difference!
How many hours are in 50 years? A year is the same as a month, so a decade is a month. Then, a year is a week. A day is a year. And a year is a year. If you’re thinking about your life in terms of months, consider the millennium to be a thousand years. This will allow you to calculate the same amount of hours as a century.
Visit the rest of the site for more useful articles! | 696 | 2,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.960987 |
https://www.nagwa.com/en/videos/592197201608/ | 1,680,263,992,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949642.35/warc/CC-MAIN-20230331113819-20230331143819-00173.warc.gz | 999,038,782 | 33,006 | # Lesson Video: Solving Quadratic Equations with Complex Roots Mathematics • 10th Grade
In this video, we will learn how to solve quadratic equations whose roots are complex numbers.
17:27
### Video Transcript
In this video, we will learn how to solve quadratic equations whose roots are complex numbers. We will begin by learning how to solve simple equations which have complex solutions and then look at what the introduction of the complex number set does to our understanding of the quadratic formula and the discriminant. Finally, weβll learn how to reconstruct a quadratic equation given a complex root.
During our exploration of the concept of numbers, we will have come across equations that have no solutions or at least those that we presume to have no real solutions. However, the introduction of the set of complex numbers opens up a whole new world when it comes to these kinds of equations.
Letβs consider the equation two π₯ squared equals negative eight. To solve this equation, we begin by dividing through by two. That tells us that π₯ squared is equal to negative four. We will then find the square root of both sides of the equation.
In the past, we might have said that this equation, whose solution is found by finding the square root of a negative number, doesnβt actually have any real solutions. And of course, this statement would be entirely correct. However, weβre no longer just dealing with real numbers. We now have the imaginary number set. And we recall that these revolve around the single letter π, where π is defined as the solution to the equation π₯ squared equals negative one. But we often just say that π is equal to the square root of negative one. And this means we can now solve our equation by finding the square root of negative four.
To find the square root of a negative number, we split it up. So, for example, the square root of negative π is the same as the square root of π multiplied by negative one, which in turn is equal to the square root of π multiplied by the square root of negative one. And since the square root of negative one is π, we see that the square root of negative π is the same as π root π. In this example, π₯ is equal to plus or minus π root four. And of course, the square root of four is two. And we see that the equation two π₯ squared equals negative eight has two solutions: two π and negative two π. Letβs consider another equation that we would have previously been unable to solve.
Solve the equation five π₯ squared plus one equals negative 319.
We can begin by solving this equation just as we would any other, by performing a series of inverse operations. Weβll start by subtracting one from both sides of the equation. Five π₯ squared plus one minus one is simply five π₯ squared. And negative 319 minus one is negative 320.
Next, we divide through by five. And we see that π₯ squared is equal to negative 64. Our final step is to find the square root of both sides of the equation. The square root of π₯ squared is π₯. And remember, we can take both the positive and negative roots of negative 64. And we see that π₯ is equal to plus or minus the square root of negative 64.
At this point, we choose to rewrite negative 64 as 64 multiplied by negative one. And we then see that the square root of negative 64 is the same as the square root of 64 multiplied by the square root of negative one. The square root of negative one though is π, and the square root of 64 is eight. And we finished solving our equation. π₯ has two solutions. Itβs eight π and negative eight π.
Now in fact we can apply the usual methods for solving equations to help us solve any equation with nonreal roots. In the case of a quadratic equation, we might struggle to factorise a quadratic expression. But we can apply the other two methods weβre confident in. These are the quadratic formula and completing the square. And there are advantages and disadvantages to both.
The quadratic formula can be a little nicer to work with when the coefficient of π₯ squared is not equal to one. And weβll see in a moment that we can use part of the formula to help us find the nature and number of roots of the equation. The completing the square method though can be fairly efficient when the coefficient of π₯ squared is equal to one.
Now itβs very much personal preference. Weβll use the quadratic formula mainly throughout this video. Then we will look at both methods during our next example. Letβs remind ourselves of the quadratic formula and the discriminant. For a quadratic equation given as ππ₯ squared plus ππ₯ plus π equals zero, where π is not equal to zero, the roots are given by π₯ equals negative π plus or minus the square root of π squared minus four ππ all over two π. And we use the discriminant to find the nature of the roots of the equation. Itβs the part of the formula that sits inside the square root: π squared minus four ππ.
It makes sense that if the discriminant is greater than zero, the square root of the discriminant will be a real number. And this means that there will be two real roots for our equation. If the value of the discriminant is equal to zero, there will be exactly one solution. This is known as a repeated root. It occurs when the turning point of the curve touches the π₯-axis exactly once. And what about if the value is less than zero? Well, weβve seen that the square root of a negative number is not a real number. So there are no real roots. This means that the curve doesnβt actually intersect the π₯-axis. Letβs look at an example of a quadratic equation which has nonreal roots.
Solve the quadratic equation π₯ squared minus four π₯ plus eight equals zero. Before we solve this equation, we can, if we want, double-check the nature of the roots of the equation by finding the value of the discriminant. Remember, the discriminant of an equation of the form ππ₯ squared plus ππ₯ plus π equals zero is given by the formula π squared minus four ππ. And itβs sometimes denoted by this little triangle.
In our equation, π is the coefficient of π₯ squared. Itβs one. π is the coefficient of π₯. Itβs negative four. And π is the constant. Itβs eight. The discriminant of our equation is therefore negative four squared minus four multiplied by one multiplied by eight, which is negative 16.
We know that if the discriminant is greater than zero, the equation has two real roots. If itβs equal to zero, it has exactly one real root. And if itβs less than zero, it has no real roots. Our discriminant is less than zero. So the equation π₯ squared minus four π₯ plus eight equals zero has no real roots.
Knowing that weβre going to get two complex roots, letβs solve this equation by first looking at the quadratic formula. The solutions to the equation are found by negative π plus or minus the square root of π squared minus four ππ all over two π.
We already saw that π squared minus four ππ in our example is equal to negative 16. So the solutions to our quadratic equation are given by negative negative four plus or minus the square root of negative 16 all over two multiplied by one. That simplifies to four plus or minus the square root of negative 16 all over two.
And at this stage, weβre going to rewrite the square root of negative 16 as the square root of 16 multiplied by the square root of negative one. And this is useful because we know that the square root of 16 is four and we know that the square root of negative one is π. So π₯ is equal to four plus or minus four π all over two. And we can simplify. And we see that the solutions to the quadratic equation are π₯ equals two plus two π and π₯ equals two minus two π.
Now in fact this isnβt the only method for solving this equation. We could have completed the square. And this is very much personal preference in an example like this. Letβs see what that wouldβve looked like.
The first thing we do is halve the coefficient of π₯. Half of negative four is negative two. So we write π₯ minus two all squared. Now negative two squared is four. So we subtract this four and then add on that value of eight. And of course, all this is equal to zero.
We can simplify our equation somewhat, and we get π₯ minus two all squared plus four equals zero. Weβre going to solve this by subtracting four from both sides. And that gives us π₯ minus two all squared equals negative four. Weβre then going to find the square root of both sides of this equation. The square root of π₯ minus two all squared is π₯ minus two. And remember, we can take both the positive and negative roots of negative four. So we see that π₯ minus two is equal to plus or minus the square root of negative four.
Now using the same method as earlier, we can see that the square root of negative four is actually the same as two π. And we can complete this solution by adding two to both sides of the equation. And once again, we see the solutions to our equation to be two plus two π and two minus two π.
In fact, itβs no accident that the roots of the equation are complex conjugates of one another. It really makes a lot of sense, especially given our second method of solving, that this would be true for any quadratic equation with complex roots.
We can say that the nonreal roots of a quadratic equation with real coefficients occur in complex conjugate pairs. And there is actually a lovely little proof of this. Letβs say we have a quadratic equation of the form ππ₯ squared plus ππ₯ plus π. Weβre going to let πΌ be a solution to this equation. And we say that πΌ star is the complex conjugate to πΌ.
Weβre going to substitute this complex conjugate into our equation. And when we do, we get π multiplied by πΌ star all squared plus π multiplied by πΌ star plus π. And here we recall the fact that, for any two complex numbers, the conjugate of their product is equal to the product of their conjugates. This means that the square of the conjugate of our solution is equal to the conjugate of the square.
And we see the first part becomes π multiplied by πΌ squared star. Since π, π, and π are real numbers β remember, our quadratic equation has real coefficients and we also know that the conjugate of a real number is just that number β this can be further rewritten as shown. And finally, we recall the fact that, for two complex numbers π§ one and π§ two, the conjugate of their sum is equal to the sum of their conjugates. And we see that π of πΌ star is equal to π multiplied by πΌ squared plus π multiplied by πΌ plus π star.
Now we already said that πΌ is a solution to the equation. This means that ππΌ star plus ππΌ plus π must be equal to zero. And we, of course, know that the conjugate of zero is simply zero. And weβve seen that since π of πΌ star is equal to zero, πΌ star must also be a solution to this equation. And in fact, this is called the complex conjugate root theorem, and it can be extended into solving polynomials. Letβs have a look at a number of examples of where this theorem can be used to solve problems involving quadratics.
The complex numbers π plus ππ and π plus ππ, where π, π, π, and π are real numbers, are the roots of a quadratic equation with real coefficients.
Given that π is not equal to zero, what conditions, if any, must π, π, π, and π satisfy?
In this question, we are told that π plus ππ and π plus ππ are roots to our quadratic equation with real coefficients. This equation would usually be of the form ππ₯ squared plus ππ₯ plus π equals zero, though π, π, and π are not to be confused with the letters π, π, and π in our complex numbers. So weβll rewrite this as ππ₯ squared plus ππ₯ plus π equals zero.
Now we know that the nonreal roots of a quadratic equation with real coefficients occur in complex conjugate pairs. And remember, to find the conjugate, we change the sign of the imaginary part. So the conjugate of π plus ππ is π minus ππ. And π plus ππ and π plus ππ must be complex conjugates of one another by this theorem. This means that the conjugate of π plus ππ must be equal to π plus ππ. So we say that π minus ππ equals π plus ππ.
And for two complex numbers to be equal, their real parts must be equal. So here we equate π and π. But their imaginary parts must also be equal. So we equate the imaginary parts. And we see that negative π equals π. So the conditions that π, π, π, and π must satisfy here is that π must be equal to π and negative π must be equal to π. This time, weβre going to use our knowledge of the nature of complex roots of quadratic equations to reconstruct an equation given one of its roots.
Find the quadratic equation with real coefficients which has five plus π as one of its roots.
Weβre told that five plus π is a root of the quadratic equation. And remember, we know that the nonreal roots of a quadratic equation which has real coefficients occur in complex conjugate pairs. To find the complex conjugate, we change the sign of the imaginary part. And we can therefore see that the roots of our equation are five plus π and five minus π. And this means that our quadratic equation is of the form π₯ minus five plus π multiplied by π₯ minus five minus π is equal to zero. And this comes from the fact that when we solve a quadratic equation by factoring, we equate each expression inside the parentheses to zero.
So in this case, we would have π₯ minus five plus π equals zero and π₯ minus five minus π equals zero. We would solve this first equation by adding five plus π to both sides. And we see that π₯ equals five plus π. And we solve the second equation by adding five minus π to both sides. And we get that second root π₯ equals five minus π.
Weβre going to need to distribute these parentheses. Letβs use the grid method here since thereβs a number of bits and pieces that could trip us up. π₯ multiplied by π₯ is π₯ squared. π₯ multiplied by negative five plus π is negative π₯ five plus π. Similarly, we get negative π₯ multiplied by five minus π. And negative five plus π multiplied by negative five minus π gives us a positive five minus π multiplied by five plus π. And weβll distribute these brackets using the FOIL method.
Multiplying the first term in the first bracket by the first term in the second bracket gives us 25. We multiply the outer two terms β thatβs five π β and the inner two terms β thatβs negative five π. And five π minus five π is zero. So these cancel each other out. And then we multiply the last terms. Negative π multiplied by π is negative π squared. And since π squared is equal to negative one, we see that these brackets distribute to be 25 minus negative one, which is equal to 26. So our quadratic equation is currently of the form π₯ squared minus π₯ multiplied by five plus π minus π₯ multiplied by five minus π plus 26.
We collect like terms. And we get π₯ squared minus five plus π plus five minus π multiplied by π₯ plus 26. π minus π is zero. And weβre left with π₯ squared minus 10π₯ plus 26 equals zero.
Now actually there is a formula that we can use that will save us some time. If we have a quadratic equation with real roots and a compact solution π plus ππ, the equation of that quadratic is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero. π is the real part of the solution. Here thatβs five. And π is the imaginary part. In our solution, thatβs one.
We can substitute what we know about our complex number into the formula. And we get π₯ squared minus two times five times π₯ plus five squared plus one squared. Two times five is 10, and five squared plus one squared is 26. And we see once again we have the same quadratic equation. And we should be able to see now why this method can be a bit of a time saver.
In this video, weβve learned that we can use the quadratic formula or completing the square to solve equations with no real roots by giving our answers as complex numbers. Weβve also seen that these solutions occur in complex conjugate pairs. And weβve also learned that we can reconstruct a quadratic equation given one of its complex solutions. If the solution is π plus ππ, the quadratic equation is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero. | 4,529 | 16,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2023-14 | latest | en | 0.948092 |
https://then24.com/2023/03/18/gross-salary-vs-net-salary-what-are-their-differences-and-how-are-they-calculated/ | 1,680,075,640,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00582.warc.gz | 610,579,503 | 23,549 | If you are looking for your first job or just got your first opportunity then it is important that you know the difference between gross and net salary. This will not only allow you to compare options and choose the best one, but also to know the taxes and contributions that are discounted from the total amount you earn.
## What is net salary?
According to the Federal Labor Law, the wage or salary is the economic compensation that the employer must pay the worker for his work. This can be set per unit of time, per unit of work, per commission, at a fixed price or in any other way.
Regarding the amount that the worker receives, there are some factors that intervene, for example, it is not the same to talk about net and gross salary. The first refers to the monetary amount that the worker receives, that is, the money that he receives in his account once minus taxes and social security contributions.
To calculate it, the IMSS quotas, ISR withholding, Infonavit Credits, among some others, must be subtracted from the gross salary. This amount depends on the type of employment contract of each worker.
## What is gross salary?
While the gross salary is the total amount before those withholdings are applied. This represents the money that a worker receives for the work that he performs in the company, but that is subject to tax withholdings on each payroll that will be deducted, obtaining the net salary as a result.
Therefore, the gross salary is the sum of all the amounts received by a worker, that is, the Base salary plus bonuses such as seniority, overtime, night work, residence, improvements, toxicity, etc.
### How to calculate gross salary withholdings?
The deductions that apply to gross salary include both tax payments and payments destined to Social Security such as medical expenses or retirement.
### Income Tax (ISR)
As the site explains Factorial, to calculate this deduction you must review the Income Tax (ISR) tables for the current year that are available on the website of the Tax Administration System (SAT). Once there you must continue to do the following:
• Locate the gross monthly salary within the table corresponding to the “lower limit” and “upper limit”.
• Subtract the “lower limit” value from the total income. The result you get will be the base.
• Locate the rate corresponding to the gross salary range in the table.
• Calculate the percentage of the rate to base that you obtained previously.
• Add the fixed fee located in the table to the result.
### IMSS fees
The employer worker quotas They are the total sum of the amount that is contributed to the IMSS by workers, employers and the quota that the Mexican government contributes. These quotas make up a patrimony that is destined to the retirement, unemployment and old age of the worker or worker, in addition to guaranteeing their social security and medical attention.
Employer worker contributions are calculated based on the following factors:
• The worker’s base salary (SBC).
• The Measurement and Update Unit (UMA).
• Percentages set within the Social Security Law.
• Insurance and social benefits provided by the IMSS.
#### Work risk insurance
To obtain the percentage to be paid, the employer takes as a reference the worker’s base contribution salary and the classes of work risk specified in the Regulation of the Social Security Law on affiliation, company classification, collection and inspection:
• Class I: 0.54355 percent.
• Class II: 1.13065 percent.
• Class III: 2.59840 percent.
• Class IV: 4.65325 percent.
• Class V: 7.58875 percent.
#### Health and maternity insurance
The sickness and maternity insurance provides coverage for medical care, hospital care and benefits in kind.
According to the Factorial site, to carry out the insurance calculation, the employee contribution base salary (SBC) and do the following:
• Obtain the amount of the daily salary and subtract the discount corresponding to the difference of the SBC and three times the UMA which is 0.40 percent.
• Multiply the gross salary by the deduction for medical expenses which corresponds to 0.375 percent.
• Carry out the same process with the percentage corresponding to cash benefits which is 0.25 percent.
• Add up all the results to find out how much you allocate to the IMSS.
#### Disability and life insurance
This type of insurance is used when the worker suffers an accident due to work and is disabled or dies. Said insurance is covered by both the employer and the collaborator, corresponding to the first 1.75 percent of the SBC and the second 0.625 percent.
#### Retirement insurance, unemployment in old age and old age
In this case, only the employer is the one who must provide the 2 percent of SBC. While for unemployment at advanced age and old age, the employer must provide the 3,150 percent of the SBCwhile the worker corresponds to the 1.125 percent of his SBC.
#### Nurseries and social benefits
This insurance grants the insured and their beneficiaries childcare services for their children, that is, it covers the risk of not being able to provide care for their children during the working day.
In this case, the employer must cover the additional 1 percent to the SBC that the worker has
## Are taxes and contributions deducted from the minimum wage?
No, the withholding of taxes or contributions does not apply to the minimum salary, the same happens when the net monthly salary is less than the minimum salary.
The minimum wage in the North Border Free Zone It is 312 pesos a dayequivalent to 9 thousand 372 pesos for 30 working days. While in the rest of the country it is 207 pesos dailyequivalent to 6 thousand 223 pesos for 30 days worked.
With information from Leticia Hernández. | 1,192 | 5,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.966324 |
http://www.algebra.com/algebra/homework/Expressions-with-variables/Expressions-with-variables.faq.question.21171.html | 1,369,196,490,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701314683/warc/CC-MAIN-20130516104834-00070-ip-10-60-113-184.ec2.internal.warc.gz | 312,388,217 | 4,607 | # SOLUTION: {{{(x^2-64)/(x^2-36)*(x-6)/(x+8)}}}
Algebra -> Algebra -> Expressions-with-variables -> SOLUTION: {{{(x^2-64)/(x^2-36)*(x-6)/(x+8)}}} Log On
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Algebra: Expressions involving variables, substitution Solvers Lessons Answers archive Quiz In Depth
Question 21171: Answer by rahman(247) (Show Source): You can put this solution on YOUR website! = = = | 157 | 553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-20 | latest | en | 0.743444 |
http://oeis.org/A124280 | 1,534,623,717,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213737.64/warc/CC-MAIN-20180818193409-20180818213409-00432.warc.gz | 298,346,650 | 3,630 | This site is supported by donations to The OEIS Foundation.
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A124280 Expansion of 1/(1-x-x^2+x^3-x^4). 2
1, 1, 2, 2, 4, 5, 9, 12, 20, 28, 45, 65, 102, 150, 232, 345, 529, 792, 1208, 1816, 2761, 4161, 6314, 9530, 14444, 21821, 33049, 49956, 75628, 114356, 173077, 261761, 396110, 599150, 906576, 1371377, 2074913, 3138864 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Diagonal sums of number triangle A124279. LINKS Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1). FORMULA a(n) = sum{k=0..floor(n/2), sum{j=0..n-2k, C(j,n-2k-j)C(k,n-2k-j)}}. a(n) = a(n-1)+a(n-2)-a(n-3)+a(n-4), a(0)=1, a(1)=1, a(2)=2, a(3)=2. - Carl Najafi, May 06 2014 CROSSREFS Sequence in context: A089935 A038000 A204856 * A088518 A001224 A102526 Adjacent sequences: A124277 A124278 A124279 * A124281 A124282 A124283 KEYWORD easy,nonn AUTHOR Paul Barry, Oct 24 2006 STATUS approved
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Last modified August 18 16:13 EDT 2018. Contains 313833 sequences. (Running on oeis4.) | 503 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.548911 |
https://hoven-in.appspot.com/Home/Aptitude/Decimal-Numbers/maths-aptitude-mock-test-on-decimals-and-floating-point-numbers-008.html | 1,638,402,747,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00079.warc.gz | 367,073,971 | 6,328 | # Decimal Numbers Quiz Set 008
### Question 1
What is \${31 × 8.8 - 27 × 3.2}/4.2\$?
A
\$44{8/21}\$.
B
\$44{5/21}\$.
C
\$44{1/21}\$.
D
\$44{11/21}\$.
Soln.
Ans: a
Observe that all fractional numbers are having one digit to the right of the dot. So we can cancel all the dots and get a plain expression: \${31 × 88 - 27 × 32}/42\$ which equals \${932/21}\$, which is same as: \$44{8/21}\$
### Question 2
For what value of x would this expression be a perfect square: \$6.39 × 6.39 + x × 6.39 + 0.5 × 0.5\$?
A
1.
B
10.
C
0.1.
D
0.01.
Soln.
Ans: a
Comparing the given expression with \$a^2 + 2 × a ×b + b^2 = (a + b)^2\$, we see that if a = 6.39, and b = 0.5 then by inspection, x should be equal to 2b = 2 × 0.5 = 1.
### Question 3
What is x in \$0.001/x = 0.1\$?
A
0.01.
B
0.1.
C
1.0.
D
0.001.
Soln.
Ans: a
The expression becomes \$x = {1/1000} × {10}\$ = 0.01.
### Question 4
If \$7.06/0.9\$ = 7.844444, then \$0.0706/9\$ = ?
A
0.007844444.
B
0.07844444.
C
0.7844444.
D
7.844444.
Soln.
Ans: a
Numerator is divided by 100, and denominator multiplied by 10, so the answer should get divided by 1000.
### Question 5
What is x in \${437/0.0437} = {4.37/x}\$?
A
\$4.37 × 10^{-4}\$.
B
\$4.37 × 10^{-3}\$.
C
\$4.37 × 10^{-2}\$.
D
\$4.37 × 10^{-5}\$.
Soln.
Ans: a
Transposing, x = \$4.37 × {0.0437/437}\$ which gives 0.000437, which, in scientific form is \$4.37 × 10^{-4}\$.
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# For a satellite escape speed is 11kms If the satellite is
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For a satellite escape speed is 11kms If the satellite is launched at an angle of 60 with the vertical what will be the escape speed 11 km sec because C escape speed R Radius of earth 2GM X No chan Ru Lox No change of earth Constant change | 87 | 336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.789777 |
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What is the overall effect of a change in resistance on the power? According to P = V^2/R, increasing the resistance decreases the power. This intuitively makes sense. If we hold voltage constant and increase the resistance, less current is supplied to the load and power decreases. But, according to P = I^2R, increases the resistance increases power. Funny thing that this also can be rationalized. We hold current constant and increase the resistance. This increase the voltage, and thus power. W/o knowing what to hold constant, can we identify whether power increases or decreases?
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15.1
INTRODUCTION
The basic mode superposition method, which is restricted to linearly elastic analysis, produces the complete time history response of joint displacements and member forces. In the past there have been two major disadvantages in the use of this approach. First, the method produces a large amount of output information that can require a significant amount of computational effort to conduct all possible design checks as a function of time. Second, the analysis must be repeated for several different earthquake motions in order to assure that all frequencies are excited, since a response spectrum for one earthquake in a specified direction is not a smooth function. There are computational advantages in using the response spectrum method of seismic analysis for prediction of displacements and member forces in structural systems. The method involves the calculation of only the maximum values of the displacements and member forces in each mode using smooth design spectra that are the average of several earthquake motions.
2
STATIC AND DYNAMIC ANALYSIS
The purpose of this chapter is to summarize the fundamental equations used in the response spectrum method and to point out the many approximations and limitations of the method. For, example it cannot be used to approximate the nonlinear response of a complex three-dimensional structural system. The recent increase in the speed of computers has made it practical to run many time history analyses in a short period of time. In addition, it is now possible to run design checks as a function of time, which produces superior results, since each member is not designed for maximum peak values as required by the response spectrum method.
15.2
DEFINITION OF A RESPONSE SPECTRUM
For three dimensional seismic motion, the typical modal Equation (13.6) is rewritten as
2 & n + 2 ζ n ω n y(t)n + ω n y(t)n = & & y(t)
(15.1)
& & & & & & pnx u(t)gx + pny u(t)gy + pnz u(t)gz
where the three Mode Participation Factors are defined by pni = - φ n Mi in
T
which i is equal to x, y or z. Two major problems must be solved in order to obtain an approximate response spectrum solution to this equation. First, for each direction of ground motion maximum peak forces and displacements must be estimated. Second, after the response for the three orthogonal directions is solved it is necessary to estimate the maximum response due to the three components of earthquake motion acting at the same time. This section will address the modal combination problem due to one component of motion only. The separate problem of combining the results from motion in three orthogonal directions will be discussed later in this chapter. For input in one direction only, Equation (15.1) is written as
& n & y(t) & & & + 2 ζ n ω n y(t)n + ω 2 y(t)n = pni u(t)g n
(15.2)
2). a period T in seconds. For this acceleration input. However. After . A A plot of ω y (ω ) MAX is defined as the pseudo-velocity spectrum and a plot of ω 2 y (ω ) MAX is defined as the pseudo-acceleration spectrum. period which has some physical significance for zero damping only.5b) The pseudo-acceleration spectrum.2) can be solved for &&(t ) and substituted into Equation (15. where S (ω ) a = ω 2 y (ω ) MAX and T= 2π ω (15. single degree-of-freedom system.3) Equation (15. The true values for maximum velocity and acceleration must be calculated from the solution of Equation (15.RESPONSE SPECTRUM ANALYSIS 3 & & Given a specified ground motion u(t)g . For this reason. different curve will exist for each different value of damping. It is standard to present the curve in terms of S(ω ) vs. It is apparent that all response spectrum curves represent the properties of the earthquake at a specific site and are not a function of the properties of the structural system. the total acceleration of the system is equal to ω 2 y (t ) .2) at various values of ω and plot a curve of the maximum peak response y (ω ) MAX . S(ω ) a .2). governed by Equation (15. between the pseudo-acceleration spectrum and the total acceleration spectrum. damping value and assuming pni = −10 it . the curve is by definition the displacement response spectrum for the earthquake motion.5a) and (15. however. There is a mathematical relationship. is given by & & & & u(t ) T = &&(t ) + u(t ) g y (15.3) which y yields & & & u(t ) T = −ω 2 y (t ) − 2ξωy (t ) (15. The total acceleration of the unit mass. for the special case of zero damping. These three curves are normally plotted as one curve on special log paper. the displacement response spectrum curve is normally not plotted as modal displacement y (ω ) MAX vs ω . curve has the units of acceleration vs.4) Therefore. is possible to solve Equation (15. these pseudovalues have minimum physical significance and are not an essential part of a response spectrum analysis.
f kn . for a structural model. it would require considerable space to clearly define terms such as peak ground velocity. . Furthermore. is shown in Figure 15. For the earthquake motions given in Figure 15.4 TYPICAL RESPONSE SPECTRUM CURVES A ten second segment of the Loma Prieta earthquake motions. can now be calculated for a typical mode n with period Tn and corresponding spectrum response value S (ω n ) .2b The velocity curves have been intentionally omitted since they are not an essential part of the response spectrum method.4 STATIC AND DYNAMIC ANALYSIS an estimation is made of the linear viscous damping properties of the structure.1a. relative velocity spectrum and absolute velocity spectrum.7) The corresponding internal modal forces. a specific response spectrum curve is selected. pseudo velocity spectrum.3 CALCULATION OF MODAL RESPONSE The maximum modal displacement. by use of an iterative algorithm.2a and 15. The record has been corrected.1. for zero displacement.6) The maximum modal displacement response of the structural model is calculated from u n = y (Tn ) MAX φ n (15. 15. velocity and acceleration at the beginning and end of the ten second record. The maximum modal response associated with period Tn is given by y (Tn ) MAX = S (ω n ) ω2 n (15. 15. the response spectrum curves for displacement and pseudo-acceleration are summarized in Figure 15. recorded on a soft site in the San Francisco Bay Area. are calculated from standard matrix structural analysis using the same equations as required in static analysis.
seconds Figure 15.1a. Typical Earthquake Ground Acceleration .Percent of Gravity 2 0 -2 -4 -6 -8 -10 -12 0 1 2 3 4 5 6 7 8 9 10 TIME .1b. Typical Earthquake Ground Displacements .RESPONSE SPECTRUM ANALYSIS 5 25 20 15 10 5 0 -5 -10 -15 -20 -25 0 1 2 3 4 5 6 7 8 9 10 TIME .Inches .seconds Figure 15.
2b. Relative Displacement Spectrum y (ω ) MAX .Seconds Figure 15.6 20 18 16 14 12 10 8 6 4 2 0 0 1 2 3 STATIC AND DYNAMIC ANALYSIS 1. Sa = ω 2 y (ω ) MAX .Seconds Figure 15.0 Perent Damping 5.Inches 100 90 80 70 60 50 40 30 20 10 0 0 1 2 3 4 5 1.0 Percent Damping 4 5 PERIOD .0 Percent Damping PERIOD . Pseudo Acceleration Spectrum.2a.0 Percent Damping 5.Percent of Gravity .
97 seconds. perhaps we should consider using the relative displacement spectrum as the fundamental form for selecting a design earthquake.1a. the absolute acceleration spectrum. Figure 15. will converge to 11. Therefore.2a. Figure 15. The high frequency. the behavior of a rigid structure is not a function of the viscous damping value.2a. It is important to note the significant difference between one and five percent damping for this typical soft site record.6 seconds for one percent damping and 16. for the earthquake defined by Figure 15.8) & & where ug MAX is the peak ground acceleration. the multiplication by ω 2 tends to completely eliminate the information contained in the long period range.01 percent of gravity at 2. indicates maximum values at a period of 0.2b. has the same value for a very short period system. the maximum relative displacement is directly proportional to the maximum forces developed in the structure. However. shown in Figure 15. part of the curve should always be defined by & & y (ω ) MAX = ug MAX / ω 2 or & & y (T ) MAX = ug MAX T2 4π 2 (15.2b. have been associated with soft sites. short period. Since most structural failures. shown in Figure 15. The maximum ground displacement shown in Figure 15.9 inches at a period of 1.92 seconds. Also. Figure 15. For long period systems. This type of real physical behavior is fundamental to the design of base isolated structures.RESPONSE SPECTRUM ANALYSIS 7 The maximum ground acceleration.62 inches for long periods and all values of damping. have physical significance. the mass of the one-degree-of-freedom structure does not move significantly and has approximately zero absolute displacement.2a.62 inches at 1.0 inches at a period of four seconds for five percent damping. This is due to the physical fact that a very rigid structure moves as a rigid body and the relative displacements within the structure are equal to zero as indicated by Figure 15. is 20. . and the absolute acceleration spectrum. the relative displacement spectrum curves. Also.2b. For this earthquake the maximum relative displacement is 18.1b is -11. The relative displacement spectrum. during recent earthquakes.64 seconds for both values of damping. It is important to note that the pseudo acceleration spectrum.
Because many engineers and building codes are not requiring the use of the CQC method. The cross-modal coefficients.5 THE CQC METHOD OF MODAL COMBINATION The most conservative method that is used to estimate a peak value of displacement or force within a structure is to use the sum of the absolute of the modal response values. from the maximum modal values. in which a large number of frequencies are almost identical. The peak value of a typical force can now be estimated. The double summation is conducted over all modes. method [2] that was first published in 1981. occur at the same point in time. CQC. Another very common approach is to use the Square Root of the Sum of the Squares.9) where f n is the modal force associated with mode n . SRSS. Similar equations can be applied to node displacements.8 STATIC AND DYNAMIC ANALYSIS 15. It is based on random vibration theories and has found wide acceptance by most engineers and has been incorporated as an option in most modern computer programs for seismic analysis. for the CQC method with constant damping are ρnm 8ζ 2 (1 + r ) r 3/ 2 = (1 − r 2 ) 2 + 4ζ 2r (1 + r ) 2 (15. this assumption is not justified.10) . This approach assumes that the maximum modal values. for all modes. The relatively new method of modal combination is the Complete Quadratic Combination. relative displacements and base shears and overturning moments. by the CQC method with the application of the following double summation equation: F= ∑∑ f ρ n n m nm fm (15. For three dimensional structures. on the maximum modal values in order to estimate the values of displacement or forces. one purpose of this chapter is to explain by example the advantages of using the CQC method and illustrate the potential problems in the use of the SRSS method of modal combination. ρ nm . The SRSS method assumes that all of the maximum modal values are statistically independent.
It is important to note that the cross-modal coefficient array is symmetric and all terms are positive. 15. the center of mass.0. of all floors. Figure 15. A Simple Three Dimensional Building Example .6 NUMERICAL EXAMPLE OF MODAL COMBINATION The problems associated with the use of the absolute sum and the SRSS of modal combination can be illustrated by their application to the four story building shown in Figure 15.3. The building is symmetrical. however.3.RESPONSE SPECTRUM ANALYSIS 9 where r = ω n / ω m and must be equal to or less than 1. is located 25 inches from the geometric center of the building.
and a response spectrum analysis were conducted. y.10 STATIC AND DYNAMIC ANALYSIS Figure 15. produces base shears which under-estimate the exact values in the direction of the loads by approximately 30 percent and over-estimate the base shears normal to the loads by a factor of ten. as well as torsion components.6b. Therefore. The building was subjected to one component of the Taft. Because of the small mass eccentricity. earthquake. The SRSS method. are exact. The maximum modal base shears in the four frames for the first five modes are shown in Figure 15.5. note that there is not a mode shape in a particular given direction as implied in many building codes and some text books on elementary dynamics.6 summarizes the maximum base shears.4. the model represents a very common three dimensional building system. An exact time history analysis. using all 12 modes. the fundamental mode shape has x. Figure 15. The time history base shears. Figure 15.6a. in each of the four frames. 1952. which is normal in real structures.4 Frequencies and Approximate Directions of Mode Shapes The direction of the applied earthquake motion. a table of natural frequencies and the principal direction of the mode shape are summarized in Figure 15. Also. using different methods. The sum of the absolute . One notes the closeness of the frequencies which is typical of most three dimensional building structures that are designed to equally resist earthquakes from both directions. Figure 15.
grossly over-estimates all results. Base Shears in Each Frame for First Five Modes Figure 15. Figure 15. The CQC method.RESPONSE SPECTRUM ANALYSIS 11 values.5.6.6d.6c. produces very realistic values that are close to the exact time history solution. Figure 15. Figure 15. Comparison of Modal Combination Methods .
998 1. It is of importance to note the existence of the relatively large off-diagonal terms that indicate which modes are coupled.006 0. Table 15. normal to the external motion.99 44. In addition.42 If one notes the signs of the modal base shears.000 0.180 0.05 Mode 1 2 3 4 5 ω n rad/sec 1 1.000 0.006 0.006 0.3.998 0.000 0.004 0.87 13.998 1.186 5 0.006 1.ζ = 0.006 0.006 0.19 54.12 STATIC AND DYNAMIC ANALYSIS The modal cross-correlation coefficients for this building are summarized in Table 15.000 13. .006 0.93 43.1.004 3 0.004 0. it is apparent how the application of the CQC method allows the sum of the base shears in the direction of the external motion to be added directly. the sum of the base shears.006 0.998 0. The ability of the CQC method to recognize the relative sign of the terms in the modal response is the key to the elimination of errors in the SRSS method.180 4 0.000 0.004 2 0. tend to cancel.186 1. shown in Figure 15. Modal Cross-Correlation Coefficients .1.
7.5 1 0.5 2 1. At the present time. 3 Normallized Pseudo Acceleration 2.Seconds Figure 15. many building codes specify design spectra in the form shown in Figure 15.7 DESIGN SPECTRA Design spectra are not uneven curves as shown in Figure 15.RESPONSE SPECTRUM ANALYSIS 13 15.2 since they are intended to be the average of many earthquakes. .5 0 0 2 4 6 8 10 PERIOD .7 Typical Design Spectrum The Uniform Building Code has defined specific equations for each range of the spectrum curve for four different soil types. For major structures it is now common practice to develop a site-dependent design spectrum which includes the effect of local soil conditions and distance to the nearest faults.
the only rational earthquake design criterion is that the structure must resist an earthquake of a given magnitude from any possible direction. curved bridges. One option in existing design codes for buildings and bridges requires that members be designed for "100 percent of the prescribed seismic forces in one direction plus 30 percent of the prescribed forces in the perpendicular direction". For time history input. Therefore.1. arch dams or piping systems. resist earthquake motions of magnitude S2 at 90o to the angle θ ". it is valid to assume that these normal motions are statistically independent. when maximum ground acceleration occurs. for most geographical locations. Based on these assumptions. it is possible to perform a large number of dynamic analyses at various angles of input in order to check all points for the critical earthquake directions. However. they give no indication on how the directions are to be determined for complex structures. the direction of the earthquake which produces the maximum stresses. In addition. because of the complex nature of three dimensional wave propagation. However. during a finite period of time. a principal direction exists. these "percentage" rules yield approximately the same results as the SRSS method.8 ORTHOGONAL EFFECTS IN SPECTRAL ANALYSIS A well-designed structure should be capable of equally resisting earthquake motions from all possible directions. For most structures this direction is not known and. cannot be estimated. Or. in a particular member or at a specified point. Other codes and organizations require the use of 40 percent rather than 30 percent. For structures that are rectangular and have clearly defined principal directions. In addition to the motion in the principal direction. a statement of the design criterion is "a structure must resist a major earthquake motion of magnitude S1 for all possible angles θ and. These motions are shown schematically in Figure 15. is not apparent. For complex three dimensional structures such as non-rectangular buildings. It is reasonable to assume that motions that take place during an earthquake have one principal direction [1]. the cost of such a study would be prohibitive. at the same point in time. a probability exists that motions normal to that direction will occur simultaneously. Such an elaborate study could conceivably produce a different critical input direction for each stress evaluated.14 STATIC AND DYNAMIC ANALYSIS 15. .
(15. It will be shown.7 indicates that the basic input spectra S1 and S2 are applied at an arbitrary angle θ . the maximum member forces calculated are invariant with respect to the selection system. stress or displacement F is produced by this input. In order to simplify the analysis. Definition of Earthquake Spectra Input Figure 15. . that maximum values for all members can be exactly evaluated from one computer run in which two global dynamic motions are applied.RESPONSE SPECTRUM ANALYSIS 15 15. Or. it will be assumed that the minor input spectrum is some fraction of the major input spectrum. 90 S2 90 S1 θ Plan View 0 Figure 15. S2 = a S1 where a is a number between 0 and 1. Furthermore.1 Basic Equations For Calculation Of Spectral Forces The stated design criterion implies that a large number of different analyses must be conducted in order to determine the maximum design forces and stresses. in this section. a force. At some typical point within the structure.11) Recently.0. Menun and Der Kiureghian [3] presented the CQC3 method for the combination of the effects of orthogonal spectrum.7.8.
This method is acceptable by most building codes. It is important to note that for equal spectra a = 1 .16) in which f 0 n and f 90 n are the modal values produced by 100 percent of the lateral spectrum applied at 0 and 90 degrees respectively and f z n is the modal response from the vertical spectrum which can be different from the lateral spectrum.13) (15.15) FZ2 = ∑ ∑ f z n ρ nm f z m n m (15. with any reference system. the value F is not a function of θ and the selection of the analysis reference system is arbitrary. Or.14) F0− 90 = ∑ ∑ f 0 n ρ nm f 90 m n m (15.12) where F02 = ∑ ∑ f 0 n ρ nm f 0 m n m 2 F90 = ∑ ∑ f 90 n ρ nm f 90 m n m (15. FMAX = 2 2 F02 + F90 + Fz (15.17) This indicates that it is possible to conduct only one analysis. . and the resulting structure will have all members that are designed to equally resist earthquake motions from all possible directions.16 STATIC AND DYNAMIC ANALYSIS The fundamental CQC3 equation for the estimation of a peak value is 2 2 F = [ F02 + a 2 F90 − (1 − a 2 ) ( F02 − F90 ) sin 2 θ + 2 (1 − a ) F0−90 sin θ cosθ + F ] 2 1 2 2 z (15.
shown in Figure 15. therefore.85. the value of θ in Equation (15. it is necessary to calculate the critical angle that produces the maximum response.50 and 0.2 The General CQC3 Method For a=1 the CQC3 method reduces to the SRSS method. 15. are pinned at the top where they are connected to an in-plane rigid diaphragm.18) At the present time no specific guidelines have been suggested for the value of a . with a equal 1.8. However. which are not a function of the reference system used by the engineer.8. is identical to the SRSS method and produces results. this can be over conservative since real ground motions of equal value in all directions have not been recorded. Normally. Differentiation of Equation (15. One example will be presented in order to show the advantages of the method.3 Examples Of Three Dimensional Spectra Analyses The previously presented theory clearly indicates that the CQC3 combination rule.8. Reference [3] presented an example with values a between 0.12) and setting the results to zero yields θcr = 1 −1 2 F0−90 tan [ 2 2 ] 2 F0 − F90 (15.12) is not known. . Note that the masses are not at the geometric center of the structure. for all structural systems.0. The columns. which are subjected to bending about the local 2 and 3 axes.17) Two roots exist for Equation (15. A very simple one-story structure. was selected to compare the results of the 100/30 and 100/40 percentage rules with the SRSS rule.17) that must be checked in order that the following equation is maximum: 2 2 FMAX = [ F02 + a 2 F90 − (1 − a 2 ) ( F02 − F90 ) sin 2 θ cr − 2 (1 − a 2 ) F0− 90 sin θ cr cosθ cr + Fz2 ]2 1 (15. The structure has two translations and one rotational degrees-of-freedom located at the center of mass.RESPONSE SPECTRUM ANALYSIS 17 15.
I 22 = 100 ft 4 I 33 = 200ft 4 E = 30 k / ft 2 L = 10 ft M TOP = 0. Sym.25k .5 degrees with the x-axis. Periods and Base Reaction Forces Period Seconds 1.924 . .065 ft. 3 2 3 X = Y = 70. Figure 15.2.01029 0. the second mode has no torsion and has a normalized base shear at 22.5 degrees.24 YMoment 3.5 The definition of the mean displacement response spectrum used in the spectra analysis is given in Table 15. 2 2 X X = 150 ft. Since the structure has a plane of symmetry at 22.8.5 0.83 Torsion -115.18 Y 3 2 STATIC AND DYNAMIC ANALYSIS Typical Column: 4 X = Y = 106. it is apparent that columns 1 and 3 (or columns 2 and 4) should be designed for the same forces.83 -9.924 -. Table 15.24 3.383 -.24 3.383 YForce -. Due to this symmetry.383 -.sec 2 / ft 3 3 1 2 0 X = 100 ft.76918 0. Three Dimensional Structure The periods and normalized base shear forces associated with the mode shapes are summarized in Table 15.0 -115.83 9.2.3.717 ft.43102 Mode 1 2 3 XForce .924 XMoment 9.
27 Spectral Displacement Used For Analysis 1.36 0.934 2.901 2.131 1.940 1. However.4 1.743 2.137 1.4 Table 15.750 2.922 2.00 1.652 2.43 85.43102 XMASS 14.922 1.742 1.65 1.4 and 15.RESPONSE SPECTRUM ANALYSIS 19 Table 15.3.137 0.8 -7.455 1.100/30 Rule Member M0 M 90 MSRSS = M0 + M90 2 2 M100/30 Error% 1 2 3 4 0. Table 15.493 2.702 1.22 YMASS 84.794 3.100/30 Rule Member M0 M 90 MSRSS = M0 + M90 2 2 M100/30 Error% 1 2 3 4 2. Participating Masses and Response Spectrum Used Mode Period Seconds 1.5.705 2.904 1.8 For this example.743 2.8 3.705 2. Moments About 2-Axes .797 1.081 14.705 2.76918 0.01029 0. Moments About 3-Axes .705 2. it does illustrate that the 100/30 combination method produces .973 2.493 1.703 1.0 and 90 degrees are summarized in Tables 15.113 0.904 0.00 1 2 3 The moments about the local 2 and 3 axes at the base of each of the four columns for the spectrum applied separately at 0.5 1.4. the maximum forces do not vary significantly between the two methods.703 1.901 2.00 1.8 3.463 1.5 and are compared to the 100/30 rule.702 2.4 -7.
940 1.922 2. Because of symmetry.028 2.455 1.7 Table 15.047 2.750 2. whereas the SRSS combination method produces logical and symmetric moments.705 2.7.684 2.8 -0.7 also illustrate that the 100/40 combination method produces results which are not reasonable.463 1.6 and 15.7 7.757 2.705 2. Both the 100/30 and 100/40 rules fail this simple test.131 1.5 Table 15.901 2.8 The results presented in Tables 15. Moments About 3-Axes .904 0.6.703 2.705 2.2 7.904 1.703 1. Moments About 2-Axes . For example.907 7.901 2.684 1.20 STATIC AND DYNAMIC ANALYSIS moments which are not symmetric.757 2.652 2.6 and 15.100/40 Rule Member M0 M 90 MSRSS = M0 + M90 2 2 M100/40 Error% 1 2 3 4 0.908 2. members 1 and 3 and members 2 and 4 should be designed for the same moments.9 -0.137 0.6 1.705 2.742 1.113 0. The SRSS and 100/40 design moments about the local 2 and 3 axes at the base of each of the four columns are summarized in Tables 15. member 4 would be over-designed by 3.702 2.137 1.8 percent about the local 3-axis using the 100/30 combination rule.9 1.100/40 Rule Member M0 M 90 MSRSS = M0 + M90 2 2 M100/40 Error% 1 2 3 4 2. .922 1.4 percent about the local 2-axis and under-designed by 7.702 1.
9 LIMITATIONS OF THE RESPONSE SPECTRUM METHOD It is apparent that use of the response spectrum method has limitations. These commonly used "percentage combination rules" are empirical and can underestimate the design forces in certain members and produce a member design which is relatively weak in one direction. However. some of which can be removed by additional development. For complex three dimensional structures the use of the 100/40 or 100/30 percentage rule will produce member designs which are not equally resistant to earthquake motions from all possible directions. Some of these additional limitations will be discussed in this section. The author believes that in the future more time history dynamic response analyses will be conducted and the many approximations associated with the use of the response spectrum method will be avoided. Therefore.8. it will never be accurate for nonlinear analysis of multi-degree of freedom structures. in which an SRSS combination of two 100 percent spectra analyses with respect to any user defined orthogonal axes. 15. the results of the SRSS directional combination rule or the input spectra can be multiplied by an additional factor greater than one. One should not try to justify the use of the 100/40 percentage rule because it is conservative in "most cases". the resulting structural design has equal resistance to seismic motions from all directions. it has been shown that the "design of elements for 100 percent of the prescribed seismic forces in one direction plus 30 or 40 percent of the prescribed forces applied in the perpendicular direction" is dependent on the user’s selection of the reference system.0 can be justified. The use of the CQC3 method should be used if a value of a less than 1. . 15. It has been shown that the alternate building code approved method.4 Recommendations On Orthogonal Effects For three dimensional response spectra analyses.RESPONSE SPECTRUM ANALYSIS 21 If a structural engineer wants to be conservative. It will produce realistic results that are not a function of the user selected reference system. will produce design forces that are not a function of the reference system.
22 STATIC AND DYNAMIC ANALYSIS 15. with a shear modulus of unity. 15. in the area where the deformation is to be calculated.9.005 horizontal drift ratio. that stresses calculated from modal stresses can be less than 50 percent of the value calculated using maximum peak values of moments and axial force.19) is to evaluate the equation for each mode of vibration.19) This equation can be evaluated for a specified x and y point in the cross section and for the calculated maximum spectral axial force and moments which are all positive values. the correct and accurate approach for the evaluation of equation (15. with large three dimensional structures. It is apparent that the resulting stress may be conservative since all forces will probably not obtain their peak values at the same time.1 Story Drift Calculations All displacements produced by the response spectrum method are positive numbers.9. It has been the author’s experience. The peak value of shear stress will be a good estimation of the damage index. For response spectrum analysis. .2 Estimation of Spectra Stresses in Beams The fundamental equation for the calculation of the stresses within the cross section of a beam is σ= P My x Mx y + + A Iy Ix (15. The current code suggests a maximum value of 0. An accurate value of the maximum stress can then be calculated from the modal stresses using the CQC double sum method. Therefore. This will take into consideration the relative signs of axial forces and moments in each mode. a plot of a dynamic displaced shape has very little meaning since each displacement is an estimation of the maximum value. A simple method to obtain a probable peak value of shear strain is to place a very thin panel element. Inter-story displacements are used to estimate damage to nonstructural elements and cannot be calculated directly from the probable peak values of displacement. which is the same as panel shear strain if the vertical displacements are neglected.
15.3 Design Checks for Steel and Concrete Beams Unfortunately. if the approach is built into a post processor computer program. A time history analysis may be the only approach that will produce rational design forces.RESPONSE SPECTRUM ANALYSIS 23 15. This approach would improve accuracy and still be conservative. it is not correct to estimate the maximum shear force from a vector summation since the x and y shears do not obtain their peak values at the same time. This would involve first calculating the maximum axial force. . to replace the current state of the art approach of calculating strength ratios based on maximum peak values of member forces. One must check at several angles in order to estimate the maximum and minimum value of the stress at each point in the structure.9. the computational time to calculate the maximum bolt force is trivial. therefore. most design check equations for steel structures are written in terms of "design strength ratios" which are a nonlinear function of the axial force in the member. The design ratios would then be evaluated mode by mode.4 Calculation of Shear Force in Bolts With respect to the interesting problem of calculating the maximum shear force in a bolt.9. The same problem exists if principal stresses are to be calculated from a response spectrum analysis. This would be a tedious approach using hand calculations. A correct method of estimating the maximum shear in a bolt is to check the maximum bolt shear at several different angles about the bolt axis. is proposed by the author. For concrete structures additional development work is required in order to develop a completely rational method for the use of maximum spectral forces in a design check equation because of the nonlinear behavior of concrete members. however. The design ratio for the member would then be estimated by a double-sum modal combination method such as the CQC3 method. assuming the maximum axial force reduction factor remains constant for all modes. A new approximate method. the ratios cannot be calculated in each mode.
"Characteristics of 3-D Earthquake Ground Motions. C.11 REFERENCES 1. For such structures. 1981. true nonlinear time-history response should be used as indicated in Chapter 19. 3. 3. 1975. 9. It is restricted to linear elastic analysis in which the damping properties can only be estimated with a low degree of confidence." Earthquake Engineering and Structural Dynamics. Vol. Watabe. pp. pp. E. Wilson. Number 1. Der Kiureghian. as compared to the SRSS method.24 STATIC AND DYNAMIC ANALYSIS 15. which are commonly used. 15. has very little theoretical background and should not be used for the analysis of complex three dimensional structures. The increase in computational effort. The use of nonlinear spectra. L. Menun and A. should clearly understand that the response spectrum method is an approximate method used to estimate maximum peak values of displacements and forces and that it has significant limitations. The CQC method has a sound theoretical basis and has been accepted by most experts in earthquake engineering. The percentage rule methods have no theoretical basis and are not invariant with respect to the reference system. . J. "A Replacement for the SRSS Method in Seismic Analysis. Der Kiureghian and E. Engineers. February 1998.10 SUMMARY In this chapter it has been illustrated that the response spectrum method of dynamic analysis must be used carefully." Earthquake Engineering and Structural Dynamics. R. Bayo. Earthquake Spectra. A. 2. however. In order for a structure to have equal resistance to earthquake motions from all directions. the CQC3 method should be used to combine the effects of earthquake spectra applied in three dimensions. l87-l92. The CQC method should be used to combine modal maxima in order to minimize the introduction of avoidable errors. Vol. The use of the absolute sum or the SRSS method for modal combination cannot be justified. 365-373. “A Replacement for the 30 % Rule for Muticomponent Excitation”. is small compared to the total computer time for a seismic analysis. Penzien and M. Vol. 13. | 8,072 | 33,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.919099 |
https://ask.sagemath.org/question/34741/how-to-solve-ode-y-4y-y-x-0-using-rk4/ | 1,506,184,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689752.21/warc/CC-MAIN-20170923160736-20170923180736-00492.warc.gz | 644,236,964 | 13,968 | How to solve ODE y'' - 4y' + y - x = 0 using rk4?
This post is a wiki. Anyone with karma >750 is welcome to improve it.
How to solve ODE y''+y''-4y'+x = 0 using rk4?
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This looks like homework. If you want some help, you should ask more precise questions related to your research in solving those exercises, especially where you are locked.
( 2016-09-06 02:32:07 -0500 )edit
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I got a way to solve the differential equation. If wrong please correct;)
To solve this differential equation we must first make a variable substitution to reduce the differential equation for a first order and thus create a system of ODEs. variable change:
w1 = y w2 = y w3 = y '
Thus, deriving the above variables, we have a system of equations of the form:
w1 '= w2 w2 '= f (x, w1, w2) = 4 * w2 + x
desolve_system_rk4([w2, 4*w2 + x ],[w1 , w2], ics = [0,1,0],ivar = x, step = 0.1, end_points = 2)
( 2016-09-06 16:18:07 -0500 )edit
This post is a wiki. Anyone with karma >750 is welcome to improve it.
I tryed
# w = y''
# z = y'
var('x, w, z')
desolve_system_rk4( [w , 4*z + x] , [w,z] , ics = [0,1,0], ivar = x, end_points = 10, step = 0.1)
But does no work correty :(
more | 409 | 1,242 | {"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} | 3.75 | 4 | CC-MAIN-2017-39 | longest | en | 0.871864 |
https://jp.mathworks.com/matlabcentral/cody/problems/1065-make-a-1-hot-vector/solutions/447817 | 1,580,071,512,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00194.warc.gz | 509,349,789 | 15,990 | Cody
# Problem 1065. Make a 1 hot vector
Solution 447817
Submitted on 30 May 2014 by Juan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% assert(isequal(one_hot(1,1),1))
2 Pass
%% assert(isequal(one_hot(2,1),[1 0]))
3 Pass
%% assert(isequal(one_hot(3,3),[0 0 1]))
4 Pass
%% assert(isequal(one_hot(4,1),[1 0 0 0]))
5 Pass
%% assert(isequal(one_hot(5,2),[0 1 0 0 0]))
6 Pass
%% assert(isequal(one_hot(6,1),[1 0 0 0 0 0]))
7 Pass
%% assert(isequal(one_hot(7,1),[1 0 0 0 0 0 0]))
8 Pass
%% assert(isequal(one_hot(8,7),[0 0 0 0 0 0 1 0]))
9 Pass
%% assert(isequal(one_hot(9,7),[0 0 0 0 0 0 1 0 0]))
10 Pass
%% assert(isequal(one_hot(10,4),[0 0 0 1 0 0 0 0 0 0]))
11 Pass
%% assert(isequal(one_hot(11,11),[0 0 0 0 0 0 0 0 0 0 1]))
12 Pass
%% assert(isequal(one_hot(12,1),[1 0 0 0 0 0 0 0 0 0 0 0]))
13 Pass
%% assert(isequal(one_hot(13,6),[0 0 0 0 0 1 0 0 0 0 0 0 0]))
14 Pass
%% assert(isequal(one_hot(14,6),[0 0 0 0 0 1 0 0 0 0 0 0 0 0]))
15 Pass
%% assert(isequal(one_hot(15,12),[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0]))
16 Pass
%% assert(isequal(one_hot(16,13),[0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0]))
17 Pass
%% assert(isequal(one_hot(17,4),[0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]))
18 Pass
%% assert(isequal(one_hot(18,9),[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]))
19 Pass
%% assert(isequal(one_hot(19,9),[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]))
20 Pass
%% assert(isequal(one_hot(20,13),[0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0])) | 826 | 1,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-05 | latest | en | 0.415843 |
https://www.exactlywhatistime.com/days-before-date/july-30/9-days | 1,708,597,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00013.warc.gz | 769,664,698 | 5,634 | # What date is 9 days before Tuesday July 30, 2024?
## Calculating 9 days before Tuesday July 30, 2024 by hand
This page helps you figure out the date that is 9 days before Tuesday July 30, 2024. We've made a calculator to find the date before a certain number of days before a specific date. In this example, we want to know the date 9 days before Tuesday July 30, 2024.
Trying to do this in your head can be really hard and take a long time. An easier way is to use a calendar, either a paper one or an app on your phone or computer, to look at the days before the date you're interested in. But the best and quickest way to find the answer is by using our days before specific date calculator, which you can find here.
If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculator to type in a new question. Remember, figuring out these types of calculations in your head can be really tough, so we made this calculator to help make it much easier for you.
## Sunday July 21, 2024 Stats
• Day of the week: Sunday
• Month: July
• Day of the year: 203
## Counting 9 days backward from Tuesday July 30, 2024
Counting backward from today, Sunday July 21, 2024 is 9 before now using our current calendar. 9 days is equivalent to:
9 days is also 216 hours. Sunday July 21, 2024 is 55% of the year completed.
## Within 9 days there are 216 hours, 12960 minutes, or 777600 seconds
Sunday Sunday July 21, 2024 is day number 203 of the year. At that time, we will be 55% through 2024.
## In 9 days, the Average Person Spent...
• 1933.2 hours Sleeping
• 257.04 hours Eating and drinking
• 421.2 hours Household activities
• 125.28 hours Housework
• 138.24 hours Food preparation and cleanup
• 43.2 hours Lawn and garden care
• 756.0 hours Working and work-related activities
• 695.52 hours Working
• 1138.32 hours Leisure and sports
• 617.76 hours Watching television
## Famous Sporting and Music Events on July 21
• 1904 Sinclair Oil founder Harry Ford Sinclair (28) weds Elizabeth Farrell | 554 | 2,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-10 | latest | en | 0.932553 |
https://www.teacherspayteachers.com/Product/Freebie-Magic-Number-Cards-1192404 | 1,508,516,765,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00563.warc.gz | 981,541,370 | 22,327 | Total:
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# Freebie: Magic Number Cards
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Amaze your students with the "Magic Number Cards" that uses the binary number system in order for the teacher to correctly guess a student's secret number. Students will think it's magic and you'll know the secret. This product will be popular in your classroom and best of all...it's free!
1. Have the student pick a secret number from 1-63.
2. Tell the student to look at the six number cards and to only give you the number cards that has his or her secret number on them.
3. Figure out his or her secret number by adding all the first numbers located in the upper left hand corner of each card that is given to you by the student. For example, if the student gives you two cards and the first numbers on each card are 16 and 8, then 16 + 8 = 24...so their secret number is 24!
Check out my blog post for a more detailed "how-to"!
This freebie has two versions: Number Cards 1-31 for the younger grades and Number Cards 1-63 for the older grades. Both sets come in two sizes.
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FREE | 342 | 1,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-43 | latest | en | 0.919901 |
https://socratic.org/questions/how-do-you-find-the-slope-and-y-intercept-of-y-3 | 1,722,780,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00552.warc.gz | 433,114,915 | 5,952 | # How do you find the slope and y intercept of y=3?
Jan 8, 2016
$s l o p e = 0 \mathmr{and} y -$intercept$= 3$
#### Explanation:
$y = 3$
The general form of slope intercept equation is
$y = m x + c$
Where $m$ is the slope and $c$ is the y-intercept.
The given equation can also be written as
$y = 0 x + 3$$\ldots \ldots \ldots \ldots . \left(i\right)$
From $\left(i\right)$ we can conclude that
$m = 0$ amd $c = 3$
$\implies s l o p e = 0 \mathmr{and} y -$intercept$= 3$ | 179 | 479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.833892 |
https://arc2climate.org/and-pdf/578-analog-and-digital-communication-interview-questions-and-answers-filetype-pdf-174-912.php | 1,642,942,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00030.warc.gz | 170,785,583 | 9,326 | # Analog And Digital Communication Interview Questions And Answers Filetype Pdf
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## Analog And Digital Communication Interview Questions And Answers Filetype
Analog Communication is a data transmitting technique in which information signal is transmitted in analog nature. An Analog signal is a. To produce modulated signal inside the transmitter in Analog.
Communication, analog signal modulates the high carrier frequency. Than this modulated signal is transmitted with the. What is Sampling? What is Sampling Theorem?
Ans: Sampling is defined as the process in which an analog signals are converted into digital signals. It means that a. Sampling Theorem is defined as : The continuous time signal that can be represented in its samples and recovered back. Ans: Pulse Amplitude Modulation is the process by which the amplitude of the regularly spaced pulses varies according to. What is Modulation? What happens in over modulation?
Ans: Modulation is defined as the process in which some characteristics of the signal called carrier is varied according to. In case of over modulation, the modulation index is greater than one and envelope distortion occurs. Ans: In case of Nyquist rate, the sampling frequency is equal to the maximum frequency of the signal and therefore the. Ans: Frequency Modulation can be defined as the frequency of the carrier wc is varied acc. Ans: Under sampling is also known as aliasing effect in which the the sampling frequency is less than the maximum.
Ans: Multiplexing is defined as the process in which a number of message signals are combined together to form. The two types of multiplexing are:. Frequency Division Multiplexing: In this technique, fixed frequency bands are allotted to every user in the complete. Time Division Multiplexing: When the pulse is present for the short time duration and most of the time their is no signal.
Ans: Amplitude Modulation is defined as the process in which the instantaneous value of the amplitude of the carrier is. Sampling frequency must be greater than the frequency of the modulating signal.
Now, Earth has turned into a global village by adopting various processes for the communication. We are dealing with. Thus, with the advancement in the world and technologyCommunication plays a. Communication is the process of connection establishment between two points for. Mainly their are three field of Electronics namely Computers, Communication and Control.
In simplex type, one unit is equipped with a transmitter and the other side is equipped with only one receiver but. Here, the communication system that is capable of transmitting information in both directions but the flow is only one way. In HDX type, one unit is equipped with transmitter as well receiver at.
Full-duplex Ethernet connections work on. Answer: Modulation is the process by which some characteristics of signal known as Carrier Signal is varied in.
Answer: 1. When we want to transmit electrical signal over an antenna, through free space, it must be converted into. Only electro-magnetic waves have the property to travel through space vacuum at the speed of.
To make msg signal or voice signal travel longer distance. To increase the signal to noise ratio. Answer: In analog modulation bandwidth required is low while in digital modulation due to higher bit rate, heigher channel. Answer: Multiplexing is defined as the process in which a number of message signals are combined together to form.
Answer: It suffers from tolerance which occurred in its analog components. In the process of Companding, the weak signals are amplified and strong signals are attenuated before applying them to.
The study and use of electrical devices that operate by controlling the flow of electrons or other electrically charged. Communication means transferring a signal from the transmitter which passes through a medium then the output is.
As a technology, analog is the process of taking an audio or video signal the human voice and translating it into. Digital on the other hand is breaking the signal into a binary format where the audio or video data is.
Digital signals are immune to noise, quality of transmission and reception is good, components used in digital. The application of science to the needs of humanity and a profession in which a knowledge of the mathematical and. If the electronic device is plugged into a standard. Electric devices use line voltage vAC, vAC, etc Electric devices can also be designed to operate on DC. Examples: are incandescent lights, heaters, fridge, stove, etc The process of obtaining a set of samples from a continuous function of time x t is referred to as sampling.
It states that, while taking the samples of a continuous signal, it has to be taken care that the sampling rate is equal to or. What is cut-off frequency? The frequency at which the response is -3dB with respect to the maximum response. Passband is the range of frequencies or wavelengths that can pass through a filter without being attenuated. A stopband is a band of frequencies, between specified limits, in which a circuit, such as a filter or telephone circuit, does.
This range. Since most of. What is modulation? And where it is utilized? Modulation is the process of varying some characteristic of a periodic wave with an external signals. These high frequency carrier signals can be transmitted over the air easily and are capable of travelling long distances.
The characteristics amplitude, frequency, or phase of the carrier signal are varied in accordance with the information. Demodulation is the act of removing the modulation from an analog signal to get the original baseband signal back. Demodulating is necessary because the receiver system receives a modulated signal with specific characteristics and it. Explain AM and FM. AM-Amplitude modulation is a type of modulation where the amplitude of the carrier signal is varied in accordance with.
FM-Frequency modulation is a type of modulation where the frequency of the carrier signal is varied in accordance with. An oscillator is a circuit that creates a waveform output from a direct current input. The two main types of oscillator are. The harmonic oscillators have smooth curved waveforms, while relaxation oscillators have.
What is crosstalk? Crosstalk is a form of interference caused by signals in nearby conductors. The most common example is hearing an. Crosstalk can also occur in radios, televisions, networking equipment, and even. A rectifier changes alternating current into direct current. This process is called rectification. The three main types of.
A rectifier is the opposite of an inverter, which changes direct current into. HWR- The simplest type is the half-wave rectifier, which can be made with just one diode. When the voltage of the. When the voltage is. The result is a clipped copy of the alternating current.
This pulsating. FWR- The full-wave rectifier is essentially two half-wave rectifiers, and can be made with two diodes and an earthed. The positive voltage half of the cycle flows through one diode, and the negative half flows.
The centre tap allows the circuit to be completed because current cannot flow through the other diode. The result is still a pulsating direct current but with just over half the input peak voltage, and double the frequency. A resistor is a two-terminal electronic component that opposes anelectric current by producing a voltage drop between its. The process of storing energy in the capacitor is known as "charging", and involves electric charges of equal.
Capacitors are often used in electric and electronic circuits asenergy-storage devices. They can also be used to.
This property makes them useful in electronic filters. Capacitors are occasionally referred to as condensers. This term is considered archaic in English, but most other. An inductor is a passive electrical device employed in electrical circuits for its property of inductance.
## Analog communication interview questions and answers
Analog Communication is a data transmitting technique in which information signal is transmitted in analog nature. An Analog signal is a. To produce modulated signal inside the transmitter in Analog. Communication, analog signal modulates the high carrier frequency. Than this modulated signal is transmitted with the. What is Sampling? What is Sampling Theorem?
What are the disadvantages of Analog communication? What are the Advantages of Digital Communication? Reliable, Noise effect is very less, power consumption is very less, various Digital ICs are available so circuits not complex , cheap, Error detection and correction is also possible. What are different types of digital modulation? How to convert an analog signal into digital signal? Define the functionality of Sampler, Quantizer?
Routing is the process of finding a path on which data can pass from source to destination. Routing is done by a device called routers, which are network layer devices. The job of the Data Link layer is to check messages are sent to the right device. Another function of this layer is framing. When a switch receives a signal, it creates a frame out of the bits that were from that signal. With this process, it gains access and reads the destination address, after which it forwards that frame to the appropriate port. This is a very efficient means of data transmission, instead of broadcasting it on all ports.
## Analog Communication Interview Questions & Answers
Oct 16, Dear Readers, Welcome to Analog Communication multiple choice questions and answers with explanation. Specially developed for the Electronic Engineering freshers and Today added electrical engineering objective questions and answers in one pdf recently added signal and system 50 most important expected mcq with solution for vizag mt and bel pe exam analog ,digital and power electronics basic level important mcq pdf for upcoming exam electrical machine 25 important mcq pdf for upcoming exam basic electrical itidiploma based mostly Elements of Digital Communication. The elements which form a digital communication system is represented by the following block diagram for the ease of understanding. Following are the sections of the digital communication system.
#### Communication Interview Questions And Answers Pdf
PART 1. PART 2. PART 3. PART 4. PART 5. PART 6. PART 7.
What is sampling? What is Sampling Theorem? Ans Sampling is defined as the process in which an analog signals are converted into digital signals. It means that a continuous time signal is converted into a discrete time signal. You might not require more epoch to spend to go to the ebook start as without difficulty as search for them.
What is Electronic? The study and use of electrical devices that operate by controlling the flow of electrons or other electrically charged particles. What is communication? Communication means transferring a signal from the transmitter which passes through a medium then the output is obtained at the receiver. Different types of communications? Analog and digital communication.
Я рисковал всю свою жизнь. Хотите меня испытать. Что ж, попробуйте! - Он начал нажимать кнопки мобильника. - Ты меня недооценил, сынок.
Да, сэр. У нас все это записано на пленку, и если вы хотите… - Исчезает фильтр Х-одиннадцать! - послышался возглас техника. - Червь преодолел уже половину пути. - Забудьте про пленку, - сказал Бринкерхофф.
Трюк, старый как мир. Никуда я не звонил. ГЛАВА 83 Беккеровская веспа, без сомнения, была самым миниатюрным транспортным средством, когда-либо передвигавшимся по шоссе, ведущему в севильский аэропорт.
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Circle - the Chord Lengths when Divided in to Equal Segments
Chord Length Calculator
The length - L - of a chord when dividing a circumference of a circle into equal number of segments can be calculated from the table below. The chord length - L - in the table is for a "unit circle" with radius = 1.
To calculate the actual length of a chord - multiply the "unit circle" length - L - with the radius for the the actual circle.
.
Example - Chord Length
A circle with radius 3 m is divided in 24 segments. From the table below: the length - L - of a single chord in a "unit circle" with 24 segments is 0.2611 units.
The length of the chord for a circle with radius 3 m can be calculated as
0.2611 (3 m) = 0.7833 m
The summarized length of all chords in the circle can be calculated as
(0.7833 m) 24
= 6.2653 (3m)
= 18.7959 m
The circumference of the circle can be calculated as
C = 2 π r
= 2 π (3 m)
= 18.8496 m
Circle - Chord Lengths
Number of Segments
- n -
Center Angle - θ -Length of Single Chord
- L -
Summarized Lengths of Chords
2 180.0000 3.1416 2.0000 4.0000
4 90.0000 1.5708 1.4142 5.6569
6 60.0000 1.0472 1.0000 6.0000
8 45.0000 0.7854 0.7654 6.1229
10 36.0000 0.6283 0.6180 6.1803
12 30.0000 0.5236 0.5176 6.2117
14 25.7143 0.4488 0.4450 6.2306
16 22.5000 0.3927 0.3902 6.2429
18 20.0000 0.3491 0.3473 6.2513
20 18.0000 0.3142 0.3129 6.2574
22 16.3636 0.2856 0.2846 6.2619
24 15.0000 0.2618 0.2611 6.2653
26 13.8462 0.2417 0.2411 6.2679
28 12.8571 0.2244 0.2239 6.2700
30 12.0000 0.2094 0.2091 6.2717
32 11.2500 0.1963 0.1960 6.2731
34 10.5882 0.1848 0.1845 6.2742
36 10.0000 0.1745 0.1743 6.2752
38 9.4737 0.1653 0.1652 6.2760
40 9.0000 0.1571 0.1569 6.2767
42 8.5714 0.1496 0.1495 6.2773
44 8.1818 0.1428 0.1427 6.2778
46 7.8261 0.1366 0.1365 6.2783
48 7.5000 0.1309 0.1308 6.2787
50 7.2000 0.1257 0.1256 6.2791
52 6.9231 0.1208 0.1208 6.2794
54 6.6667 0.1164 0.1163 6.2796
56 6.4286 0.1122 0.1121 6.2799
58 6.2069 0.1083 0.1083 6.2801
60 6.0000 0.1047 0.1047 6.2803
62 5.8065 0.1013 0.1013 6.2805
64 5.6250 0.0982 0.0981 6.2807
66 5.4545 0.0952 0.0952 6.2808
68 5.2941 0.0924 0.0924 6.2810
70 5.1429 0.0898 0.0897 6.2811
72 5.0000 0.0873 0.0872 6.2812
74 4.8649 0.0849 0.0849 6.2813
76 4.7368 0.0827 0.0826 6.2814
78 4.6154 0.0806 0.0805 6.2815
80 4.5000 0.0785 0.0785 6.2816
82 4.3902 0.0766 0.0766 6.2816
84 4.2857 0.0748 0.0748 6.2817
86 4.1860 0.0731 0.0730 6.2818
88 4.0909 0.0714 0.0714 6.2819
90 4.0000 0.0698 0.0698 6.2819
92 3.9130 0.0683 0.0683 6.2820
94 3.8298 0.0668 0.0668 6.2820
96 3.7500 0.0654 0.0654 6.2821
98 3.6735 0.0641 0.0641 6.2821
100 3.6000 0.0628 0.0628 6.2822
102 3.5294 0.0616 0.0616 6.2822
104 3.4615 0.0604 0.0604 6.2822
106 3.3962 0.0593 0.0593 6.2823
108 3.3333 0.0582 0.0582 6.2823
110 3.2727 0.0571 0.0571 6.2823
112 3.2143 0.0561 0.0561 6.2824
114 3.1579 0.0551 0.0551 6.2824
116 3.1034 0.0542 0.0542 6.2824
118 3.0508 0.0532 0.0532 6.2824
120 3.0000 0.0524 0.0524 6.2825
122 2.9508 0.0515 0.0515 6.2825
124 2.9032 0.0507 0.0507 6.2825
126 2.8571 0.0499 0.0499 6.2825
128 2.8125 0.0491 0.0491 6.2826
130 2.7692 0.0483 0.0483 6.2826
132 2.7273 0.0476 0.0476 6.2826
134 2.6866 0.0469 0.0469 6.2826
136 2.6471 0.0462 0.0462 6.2826
138 2.6087 0.0455 0.0455 6.2826
140 2.5714 0.0449 0.0449 6.2827
142 2.5352 0.0442 0.0442 6.2827
144 2.5000 0.0436 0.0436 6.2827
146 2.4658 0.0430 0.0430 6.2827
148 2.4324 0.0425 0.0425 6.2827
150 2.4000 0.0419 0.0419 6.2827
152 2.3684 0.0413 0.0413 6.2827
154 2.3377 0.0408 0.0408 6.2827
156 2.3077 0.0403 0.0403 6.2828
158 2.2785 0.0398 0.0398 6.2828
160 2.2500 0.0393 0.0393 6.2828
162 2.2222 0.0388 0.0388 6.2828
164 2.1951 0.0383 0.0383 6.2828
166 2.1687 0.0379 0.0378 6.2828
168 2.1429 0.0374 0.0374 6.2828
170 2.1176 0.0370 0.0370 6.2828
172 2.0930 0.0365 0.0365 6.2828
174 2.0690 0.0361 0.0361 6.2828
176 2.0455 0.0357 0.0357 6.2829
178 2.0225 0.0353 0.0353 6.2829
180 2.0000 0.0349 0.0349 6.2829
182 1.9780 0.0345 0.0345 6.2829
184 1.9565 0.0341 0.0341 6.2829
186 1.9355 0.0338 0.0338 6.2829
188 1.9149 0.0334 0.0334 6.2829
190 1.8947 0.0331 0.0331 6.2829
192 1.8750 0.0327 0.0327 6.2829
194 1.8557 0.0324 0.0324 6.2829
196 1.8367 0.0321 0.0321 6.2829
198 1.8182 0.0317 0.0317 6.2829
200 1.8000 0.0314 0.0314 6.2829
202 1.7822 0.0311 0.0311 6.2829
204 1.7647 0.0308 0.0308 6.2829
206 1.7476 0.0305 0.0305 6.2829
208 1.7308 0.0302 0.0302 6.2829
210 1.7143 0.0299 0.0299 6.2830
212 1.6981 0.0296 0.0296 6.2830
214 1.6822 0.0294 0.0294 6.2830
216 1.6667 0.0291 0.0291 6.2830
218 1.6514 0.0288 0.0288 6.2830
220 1.6364 0.0286 0.0286 6.2830
222 1.6216 0.0283 0.0283 6.2830
224 1.6071 0.0280 0.0280 6.2830
226 1.5929 0.0278 0.0278 6.2830
228 1.5789 0.0276 0.0276 6.2830
230 1.5652 0.0273 0.0273 6.2830
232 1.5517 0.0271 0.0271 6.2830
234 1.5385 0.0269 0.0269 6.2830
236 1.5254 0.0266 0.0266 6.2830
238 1.5126 0.0264 0.0264 6.2830
240 1.5000 0.0262 0.0262 6.2830
242 1.4876 0.0260 0.0260 6.2830
244 1.4754 0.0258 0.0258 6.2830
246 1.4634 0.0255 0.0255 6.2830
248 1.4516 0.0253 0.0253 6.2830
250 1.4400 0.0251 0.0251 6.2830
252 1.4286 0.0249 0.0249 6.2830
254 1.4173 0.0247 0.0247 6.2830
256 1.4063 0.0245 0.0245 6.2830
258 1.3953 0.0244 0.0244 6.2830
260 1.3846 0.0242 0.0242 6.2830
262 1.3740 0.0240 0.0240 6.2830
264 1.3636 0.0238 0.0238 6.2830
266 1.3534 0.0236 0.0236 6.2830
268 1.3433 0.0234 0.0234 6.2830
270 1.3333 0.0233 0.0233 6.2830
272 1.3235 0.0231 0.0231 6.2830
274 1.3139 0.0229 0.0229 6.2830
276 1.3043 0.0228 0.0228 6.2830
278 1.2950 0.0226 0.0226 6.2831
280 1.2857 0.0224 0.0224 6.2831
282 1.2766 0.0223 0.0223 6.2831
284 1.2676 0.0221 0.0221 6.2831
286 1.2587 0.0220 0.0220 6.2831
288 1.2500 0.0218 0.0218 6.2831
290 1.2414 0.0217 0.0217 6.2831
292 1.2329 0.0215 0.0215 6.2831
294 1.2245 0.0214 0.0214 6.2831
296 1.2162 0.0212 0.0212 6.2831
298 1.2081 0.0211 0.0211 6.2831
300 1.2000 0.0209 0.0209 6.2831
302 1.1921 0.0208 0.0208 6.2831
304 1.1842 0.0207 0.0207 6.2831
306 1.1765 0.0205 0.0205 6.2831
308 1.1688 0.0204 0.0204 6.2831
310 1.1613 0.0203 0.0203 6.2831
312 1.1538 0.0201 0.0201 6.2831
314 1.1465 0.0200 0.0200 6.2831
316 1.1392 0.0199 0.0199 6.2831
318 1.1321 0.0198 0.0198 6.2831
320 1.1250 0.0196 0.0196 6.2831
322 1.1180 0.0195 0.0195 6.2831
324 1.1111 0.0194 0.0194 6.2831
326 1.1043 0.0193 0.0193 6.2831
328 1.0976 0.0192 0.0192 6.2831
330 1.0909 0.0190 0.0190 6.2831
332 1.0843 0.0189 0.0189 6.2831
334 1.0778 0.0188 0.0188 6.2831
336 1.0714 0.0187 0.0187 6.2831
338 1.0651 0.0186 0.0186 6.2831
340 1.0588 0.0185 0.0185 6.2831
342 1.0526 0.0184 0.0184 6.2831
344 1.0465 0.0183 0.0183 6.2831
346 1.0405 0.0182 0.0182 6.2831
348 1.0345 0.0181 0.0181 6.2831
350 1.0286 0.0180 0.0180 6.2831
352 1.0227 0.0178 0.0178 6.2831
354 1.0169 0.0177 0.0177 6.2831
356 1.0112 0.0176 0.0176 6.2831
358 1.0056 0.0176 0.0176 6.2831
360 1.0000 0.0175 0.0175 6.2831
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Dedication iii
Preface v
Acknowledgement vii
1 Preliminaries 1
1.1 Basic notions of functional analysis . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Differentiability in Banach spaces . . . . . . . . . . . . . . . . . . . . 3
1.1.2 Duality mapping in Banach spaces . . . . . . . . . . . . . . . . . . . 5
1.1.3 The signum function . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.4 Convex functions and sub-differentials . . . . . . . . . . . . . . . . . 8
2 Characteristic Inequalities 11
2.1 Uniformly convex spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1.1 Strictly convex spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.1.2 Inequalities in uniformly convex spaces . . . . . . . . . . . . . . . . . 14
2.2 Uniformly smooth spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2.1 Inequalities in uniformly smooth spaces . . . . . . . . . . . . . . . . . 19
2.2.2 Characterization of uniformly smooth spaces by the duality maps . . 21
3 Sunny Nonexpansive Retraction 23
3.1 Construction of sunny nonexpansive retraction in Banach spaces . . . . . . . 23
4 An Application of Sunny Nonexpansive Retraction 33
Bibliography 37
ix
CHAPTER ONE
Preliminaries
1.1 Basic notions of functional analysis
In this chapter, we recall some definitions and results from linear functional analysis.
Proposition 1.1.1 (The Parallelogram Law) Let X be an inner product space. Then for
arbitrary x; y 2 X,
kx + yk2 + kx yk2 = 2
kxk2 + kyk2
:
Theorem 1.1.1 (The Riesz Representation Theorem) Let H be a Hilbert space and let f be
a bounded linear functional on H. Then there exists a unique vector y0 2 H such that
f(x) = hx; y0i for each x 2 H and ky0k = kfk:
Theorem 1.1.2 Let X be a reflexive and strictly convex Banach space, K be a nonempty,
closed, and convex subset of X. Then for any fixed x 2 X there exists a unique m 2 K
such that
kx mk = inf
k2K
kx kk:
Proof. Let x 2 X be fixed, and define Px : X ! R [ f1g by
Px(k) =
kx kk; if k 2 K,
1; if k =2 K.
Clearly Px is convex. Indeed, let 2 (0; 1) and k1; k2 2 X. If any of k1 or k2 is not in K,
then Px(k1 + (1 )k2) Px(k1) + (1 )Px(k2) since the right handside is 1. Now
1
2 CHAPTER 1. PRELIMINARIES
suppose both elements are in K. Then
Px(k1 + (1 )k2) = k(k1 x) + (1 )(k2 xk
k(k1 x)k + k(1 )(k2 xk
= Px(k1) + (1 )Px(k2):
We next show that Px is lower semicontinuous. By the continuity of the map
k 7! kx kk; k 2 K;
we have Px is lower semicontinuous on K. We now show Px is lower semicontinuous on
Kc. Let x0 2 Kc and 2 R such that < Px(x0). Since K is closed, Kc is an open
neighbourhood of x0 and < Px(y) 8y 2 Kc . Hence Px is lower semicontinuous on Kc and
therefore on the whole X. Obviously Px is proper. Next, we show that Px is coercive. Let
y 2 X. Then
Px(y) kyk kxk
kyk
kyk
2
=
kyk
2
provided kyk 2kxk:
This implies that Px(y) ! 1 as kyk ! 1. Thus, Px is lower semicontinuous, convex,
proper, and coersive. Hence there exists m 2 X such that
Px(m) Px(m) 8m 2 X:
Since Px(y) = 1 for all y 2 Kc and K 6= ;, we must have m 2 K. Furthermore kxmk =
Px(m) Px(k) = kx kk; 8k 2 K. This completes the proof. We now show that m 2 K
is unique. Indeed, if x 2 K then m = x and hence it is unique. Suppose x 2 Kc and m 6= n
such that kxmk = kxnk kxkk 8k 2 K, then 1
kx mk
k
1
2
((xm)+(xn))k < 1.
This implies that kx 1
2 (m + n)k < kx mk and this contradict the fact that m is a
minimizing vector in K. Therefore m 2 K is unique.
Corollary 1.1.1 Let X be a uniformly convex Banach space and K be any nonempty, closed
and convex subset of X. Then for arbitrary x 2 X there exists a unique k 2 K such that
kx kk = inf
k2K
kx kk:
Remark If H is a real Hilbert space and M is any nonempty, closed, and convex subset of
H then in view of the above corollary, then there exists a unique map PM : H ! M defined
by x 7! PMx; where kx PMxk = inf
m2M
kx mk. This map is called the projection map.
The following properties of projection map PM of H onto M are well known.
(1) z = PMx , hx z;m zi 0 8m 2 M.
(2) kPMxPMyk2 hxy; PMxPMyi 8x; y 2 H, which implies that kPMxPMyk
kx yk 8x; y 2 H, i.e., PM is nonexpansive.
(3) PM(PMx + t(x PMx)) = PMx 8t 0.
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 3
1.1.1 Differentiability in Banach spaces
Let X and Y be two real normed linear spaces and U be a nonempty open subset of X.
Definition 1.1.1 (Directional Differentiability) Let f : U ! Y be a map. Let x0 2 U and
v 2 Xnf0g. We say that f has directional derivative at x0 in the direction of v if
lim
t!0
f(x0 + tv) f(x0)
t
;
exists in the normed linear space Y . We denote by f0(x0; v) to be the directional derivative
of f at x0 in the direction of v.
Example Let f be the function defined from R2 into R by
f(x1; x2) =
(
x1x22
x21
+x22
; if (x1; x2) 6= (0; 0)
0; otherwise.
Then f has directional derivative at (0; 0) in any direction.
To see this, let v = (v1; v2) 2 R2nf0; 0g, t 6= 0; then
f(0 + tv) f(0)
t
=
f(tv)
t
=
v1v2
2
v2
1 + v2
2
:
Thus,
lim
t!0
f(0 + tv) f(0)
t
=
v1v2
2
v2
1 + v2
2
= f
0
(0; v):
So f has directional derivative at (0; 0) in any direction.
Example Let f be the function defined from R2 into R by
f(x1; x2) =
x1x2
x21
+x22
; if (x1; x2) 6= (0; 0)
0; otherwise.
Then, this function f has no directional derivative at (0; 0) in any direction.
To see this, let v = (v1; v2) 2 R2nf0; 0g; and t 6= 0; then
f(0 + tv) f(0)
t
=
f(tv)
t
=
v1v2
t(v2
1 + v2
2)
and so the limit does not exists in R. Therefore, the directional derivative of the function f
does not exists at (0; 0) in any direction.
Definition 1.1.2 (Gateaux Differentiability) Let f : U ! Y be a map . Let x0 2 U. The
function f is said to be Gâteaux Differentiable at x0 if :
4 CHAPTER 1. PRELIMINARIES
1. f has directional derivative at x0 in every direction v 2 X n f0g and
2. there exists a bounded linear map A 2 B(X; Y ) (depending on x0) such that f0(x0; v) =
A(v) for all v element of X n f0g.
In this case the map f0(x0; 🙂 is called the Gâteaux differential of f at x0 and is denoted by
DGf(x0) or f0G
(x0).
In other words, f is Gâteaux differentiable at x0 if there exists a bounded linear map A 2
B(X; Y ) such that
lim
t!0
f(x0 + tv) f(x0)
t
= A(v); 8v 2 X n f0g
Remark From the above definition, it is obvious that if a function is Gâteaux differentiable
at a point, then it has a directional derivative in all directions at that point.
However, the converse is not true in general. We refer to the above example, it is clear
that the directional derivative of the function f exists at (0,0) in any direction but f0(0; )
is not linear.
Definition 1.1.3 A Banach space X is said to have Gateaux differentiable norm if the limit
lim
t!0
kx + tyk kxk
t
exists for each x; y 2 X with kxk = kyk = 1:
Definition 1.1.4 A Banach space X is said to have uniformly Gateaux differentiable norm
if for each y 2 X with kyk = 1; the limit
lim
t!0
kx + tyk kxk
t
is attained uniformly in x 2 X with kxk = 1.
Definition 1.1.5 Let (M; ) be a metric space. A mapping T : M ! M is called a contraction
if there exists k 2 [0; 1) such that (Tx; Ty) k(x; y) for all x; y 2 M. If k = 1, then
T is called non-expansive.
Theorem 1.1.3 (Banach Contraction Mapping Principle). Let (M; ) be a complete metric
space and T : M ! M be a contraction. Then T has a unique fixed point, i.e., there exists
a unique x 2 M such that Tx = x.
Definition 1.1.6 A mapping L : l1 ! R is called a Banach limit if
(a) L is linear and continuous.
(b) L(x) 0 if x 0, where x = (xn)n with xn 0 8n:
(c) L(x) = L(x); where denotes the shift operator defined by
(x1; x2; x3; :::) = (x2; x3; x4; :::) i:e: (xn) = (xn+1):
(d) L(1) = 1
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 5
1.1.2 Duality mapping in Banach spaces
In order to find the analogue of the identities (1) and (2) in Banach spaces, we need to have
a suitable replacement for the inner product. In this section, we give the definition and
properties of duality mapping in an arbitrary normed space.
Definition 1.1.7 A continuous and strictly increasing function : R+ ! R+ such that
(0) = 0 and lim
t!1
(t) = 1 is called a gauge function.
Definition 1.1.8 Given a gauge function , the mapping J : X ! 2X defined by
J(x) = fx 2 X : hx; xi = kxkkxk; kxk = (kxk)g
is called the duality map with gauge function where X is any normed space. The normalized
duality map J is a particular case of J where (t) = t; t 2 R+.
Lemma 1.1.1 Let be a gauge function and
‘(t) =
Z t
0
(s)ds;
then ‘ is a convex function on R+.
Proposition 1.1.2 In a normed linear space X, for every gauge function , J(x) is not
empty for any x 2 X.
Proof: If x = 0 then trivially we have J(x) 6= ; by taking x = 0.
For x 6= 0 in X, Hahn-Banach theorem guarantee the existence of f 2 X such that
kfk = 1 and hx; fi = kxk. Take x = (kxk)f then clearly kxk = (kxk) and hx; xi =
(kxk)hx; fi = (kxk)kxk. Hence the proof.
Proposition 1.1.3 In a real Hilbert space H, the normalized duality map is the identity map.
Proof: Since H is a Hilbert space, we identify H = H. Let x 2 H, since hx; xi = kxk2,
then x 2 J(x). If y 2 J(x), then hx; yi = kxkkyk and kxk = kyk.
So that kx yk2 = hx y; x yi = hx; xi hy; xi hx; yi + hy; yi = 0. Therefore y = x.
Hence the proof.
Proposition 1.1.4 In a Banach space X, let J be a duality map of gauge function . Then
for every x in X, x 6= 0 and every in R, we have
J(x) = sign()
(jjkxk)
(kxk)
J(x):
Proof: Let x 2 X such that x 6= 0 and 2 R. We need to show that
J(x) sign()
(jjkxk)
kxk
J(x) and
6 CHAPTER 1. PRELIMINARIES
sign()
(jjkxk)
(kxk)
J(x) J(x):
Now, let u 2 J(x). Then
x; sign()
(jjkxk)
(kxk)
u
= jj
(jjkxk)
(kxk)
hx; ui
= jjkxk(jjkxk)
= kxk(kxk) and
ksign()(kxk)
(kxk) uk = (kxk). Thus we have, sign()(kxk)
(kxk) J(x) J(x).
To see the other inclusion, we take y = x, = 1
and use the above result, we get
sign()(kyk)
(kyk) J(y) J(y). This implies sign( 1
) (kxk)
(jjkxk)J(x) J(x), ie., J(x)
sign()jjkxk
kxk J(x). Hence, we conclude that J(x) = sign()(jjkxk)
(kxk) J(x).
Corollary 1.1.2 Let X be a real Banach space and J be the normalized duality map on X.
Then J(x) = J(x), 8 2 R, 8x 2 X.
Proof: For the normalized duality map, (t) = t, t 2 R+. It follows from the above proposition
that
J(x) = J(x) = sign() jjkxk
kxk J(x) = J(x) = J(x).
Lemma 1.1.2 Let X be a normed linear space and J be a normalized duality map on X.
Then for any x 2 X
J(x) = fx 2 X : hx; xi = kxk2; kxk kxkg:
Proof: For x = 0, the results holds trivially. For x 6= 0, it is immediate that J(x) fx 2
X : hx; xi = kxk2; kxk kxkg.
Suppose that u 2 X such that hx; ui = kxk2 and kuk kxk. We need to show
kxk kuk.
kuk = sup
kxk6=0
hx;ui
kxk which implies that kxk = hx;ui
kxk kuk. Hence kuk = kxk and so
u 2 J(x).
Proposition 1.1.5 Let X be a real Banach space and J be the duality map on X, then
(a) For every x 2 X, the set J(x) is convex and weak closed in X.
(b) J is monotone in the sense that hxy; xyi 0 8x; y 2 X and x 2 J(x); y 2 J(y).
Proof:(a) Let x 2 X; x; y 2 J(x) and 2 (0; 1). We need to show that x +(1)y 2
J(x).
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 7
hx; x + (1 )yi = hx; xi + (1 )hx; yi
= kxk2 + (1 )kxk2
= kxk2:
Also,kx +(1)yk kxk+(1)kyk = kxk. Therfore by lemma (1.1.2) x +(1
)y 2 J(x). Hence J(x) is convex.
To show J(x) is weak closed, define for each x 2 X, a map x : X ! R by x(f) = hx; fi.
Then J(x) = 1
x (kxk2) \ B(0; kxk) and so is weak closed in X since x is continuous if
we put the weak topology on X.
(b) Let x; y 2 X and x 2 J(x); y 2 J(y). Then
hx y; x yi = hx; xi hx; yi hy; xi + hy; yi
= kxk2 + kyk2 hx; yi hy; xi
kxk2 2kxkkyk + kyk2
= (kxk + kyk)2 0:
Hence, J is monotone.
1.1.3 The signum function
Definition 1.1.9 The signum function denoted by sgn, is the function sgn : R ! R defined
as
sgn(x) =
8<
:
1 if x > 0
0 if x = 0
1 if x < 0
We now state and prove the following properties:
(a) For all x 2 R, sgn(x) = sgn(x).
(b) For all x 2 R, jxj = sgn(x)x.
(c) For all x 2 R, x 6= 0 d
dx jxj = sgn(x).
Proof.
(a) Let x 2 R.
Case 1. x > 0. In this case sgn(x) = 1, therefore sgn(x) = 1 = sgn(x).
Case 2. x < 0. In this case sgn(x) = 1, therefore sgn(x) = 1 = (1) = sgn(x).
Case 3. x = 0, then sgn(x) = 0 = sgn(x).
(b) Let x 2 R.
If x > 0, then jxj = x and sgn(x) = 1 so that jxj = x = sgn(x)x.
If x < 0, then jxj = x and sgn(x) = 1 so that we obtain jxj = x = sgn(x)x.
If x = 0. Then jxj = 0 and sgn(x) = 0, so we have jxj = 0 = sgn(x)x.
8 CHAPTER 1. PRELIMINARIES
(c) Let x 2 R such that x 6= 0.
Case 1. x > 0. In this case jxj = x and sgn(x) = 1. Thus d
dx jxj = 1 = sgn(x).
Case 2. x < 0. In this case jxj = x and sgn(x) = 1. So d
dx jxj = 1 = sgn(x).
Definition 1.1.10 Let C X be nonempty subset of a Banach space X and D C be
non-empty. A retraction Q from C to D is a mapping Q : C ! D such that Qx = x for
x 2 D. Q is nonexpansive if kQx Qyk kx yk; x; y 2 C.
Definition 1.1.11 A retraction Q from C to D is sunny if Q satisfies the property:
Q(Qx + t(x Qx)) = Qx for x 2 C and t > 0 whenever Qx + t(x Qx) 2 C. A retraction
Q from C to D is sunny nonexpansive if Q is both sunny and nonexpansive.
Lemma 1.1.3 [8] Let C be a nonempty closed convex subset of a smooth Banach space E,
D C nonempty, j : E ! E the normalized duality mapping of E, and Q : C ! D a
retraction. Then the following are equivalent:
(i) hx Qx; j(y Qx)i 0 for all x 2 C and y 2 D;
(ii) Q is both sunny and nonexpansive.
1.1.4 Convex functions and sub-differentials
In this section, we present the basic notion of convex functions and sub-differential of a
convex function.
Definition 1.1.12 Let D be a non-empty subset of a normed linear space X. The set D is
called convex if for each x; y 2 D and for any 2 (0; 1) we have x + (1 )y 2 D.
Definition 1.1.13 Let f : X ! R be a map. Then D(f) = fx 2 X : f(x) < +1g is called
the effective domain of f. The function f is proper if D(f) 6= ; i.e 9×0 2 X : f(x0) 2 R.
Definition 1.1.14 Let D be a non-empty convex subset of X. Let f : D ! R [ f+1g,
then f is said to be convex if for any 2 (0; 1) and for all x; y 2 D we have
f(x + (1 )y) f(x) + (1 )f(y):
Definition 1.1.15 A convex function f on a convex domain D X is said to be uniformly
convex on D if there exists a function : R+ ! R+ with (t) = 0 , t = 0 such that
f(x + (1 )y) f(x) + (1 )f(y) (1 )(kx yk); 8 2 [0; 1]:
If whenever = 1
2 then f is uniformly convex at centre on D.
Definition 1.1.16 The sub-differential of a convex function f is a map @ : X ! 2X
defined by
@f(x) = fx 2 X : f(y) f(x) hy x; xi; 8y 2 Xg:
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 9
Definition 1.1.17 For p > 1, let (t) = tp1 be the gauge function. Then we define the
generalized duality map Jp : X ! 2X by
Jp(x) = fx 2 X : hx; xi = kxkkxk; kxk = (kxk) = kxkp1g:
Observe that for p = 2, we have Jp = J2 = J which is the normalized duality map defined
in the previous section.
Proposition 1.1.6 For every x 6= 0 in a Banach space X,
@kxk = fu 2 X : hx; ui = kxk = kuk; kuk = 1g:
Proof. We note that
@kxk = fu 2 X : hy x; ui kyk kxk 8y 2 Xg:
Now, let u 2 X : hx; ui = kxk = kuk; kuk = 1: Then, for arbitrary y 2 X we have
hy x; ui = hy; ui hx; ui kyk kxk:
This implies that u 2 @kxk:
Conversely, if u is in @kxk then
hy; ui = h(y + x) x; ui kx + yk kxk kyk:
From this we get that kuk 1 and with y = 0 in the definition of @kxk; we have kxk
hx; ui kxkkuk. Therefore kuk = 1; kxk = hx; ui and the result holds.
Lemma 1.1.4 J(x) = @ (kxk) for each x in a Banach space X, where (kxk) =
R kxk
0 (s)ds.
Theorem 1.1.4 For p > 1, JP is the sub-differential of the functional 1
p
kxkp.
Proof From the definition of Jp, we note that the gauge function is given by (t) = tp1; p >
1. By the theorem above, we get
Jp = J(t)(x) = @
Z kxk
0
(t)dt = @
Z kxk
0
tp1dt = @
1
p
kxkp
;
completing the proof.
GET THE FULL WORK | 6,756 | 15,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-16 | latest | en | 0.443157 |
https://getrevising.co.uk/revision-cards/physics_11_1 | 1,716,912,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00074.warc.gz | 227,443,178 | 14,608 | # P1.1 Energy Transfer by Heating
?
• Created by: Cornetto
• Created on: 27-05-15 16:20
• Thermal radiation is emitted by all objects as electromagnetic waves, and the hotter the object the more infrared radiation emitted.
• Matt, black surfaces absorb and emit more infrared radiation, but light, shiny surfaces are better reflectors.
• Radiation can happen even through a vacuum, unlike convection and conduction.
• The temperature of an object is determined by the average thermal energy of each of the particles in the object.
• When infrared radiation from the sun gets absorbed by the earth, some gets reflected, but the infrared light cannot then penetrate the atmosphere, because of greenhouse gases.
1 of 7
## Conduction
• Conduction occurs in solids, where thermal energy provided to one part of an object is spread along the whole object until the whole object has the same temperature.
• Vibrations are passed on through the atoms, when particles with more kinetic energy pass on their kinetic energy to other particles.
• Temperature is proportional to the average vibrational energy of particles.
• Metals like copper are better conductors because they have delocalised electrons that can easily travel through the object and pass on their kinetic energy.
• Plastic and wood are good insulators, as well as fluids.
2 of 7
## Convection
• Convection is a heat transfer that occurs in liquids and gases.
• A convection current:
• When the fluid is heated, it is supplied with more thermal and kinetic energy, causing it to vibrate more and then spread out.
• This expansion means that the hotter fluid has a lower density.
• The number of particles or mass of particles does not change when the substance is heated, but the volume does.
• density = mass / volume shows how an increase in volume results in a decrease in density.
• The hotter fluid close to a heat source rises above the colder fluid, and colder fluid takes up the place of the hot fluid.
• Once the hotter fluid is far enough from the heat source, it becomes cool once again and sinks because of its high density.
• Convection results in the rising and falling of hot air on Earth, which causes tropical storms and hurricanes.
• Sea breezes are formed when cooler air from the sea takes the place of the hot air that has risen from the land during warm weather.
3 of 7
## Heat and Temperature
• There are differences between heat and temperature and the two should not be mixed up.
• Heat is thermal energy moving from one place to another, and is measured in Joules (J).
• Temperature is the average kinetic energy in the particles that make up an object, and can be measured in Degrees Celcius, Degrees Fahrenheit or Kelvin.
• One object has a higher temperature than another object with the same mass, we can say that it has more thermal energy.
• The net movement of thermal energy is always from a hot object to a cold object.
• If a beaker of water has the same thermal energy as another beaker of water, then both beakers of water poured into a larger beaker will have double the amount of energy as one beaker, but the temperature will stay the same because temperature tells us the average kinetic energy in the beaker.
• Objects at absolute zero have no thermal energy at all, so they are regarded as stationary.
• An object with a lower surface to volume ratio will lose heat energy at a slower rate.
4 of 7
## Evaporation and Condensation
• Evaporation occurs when molecules on the surface of a liquid are able to gain enough kinetic energy to escape the liquid and become a gas.
• Factors speeding up rate of evaporation include:
• Temperature difference between the liquid and surroundings.
• Surface area of the liquid.
• If there is a draught to take away molecules that have escaped.
• The effects of evaporation can be used to numb a person before they receive an injection:
• A liquid that easily evaporates, like an alcohol, is placed on the skin.
• In order to have enough energy to evaporate, it takes kinetic energy from the skin it is touching.
• The average thermal energy on the skin is decreased because of thermal energy being passed on to the fluid, so the skin is much colder and more numb.
• Condensation occurs when water molecules in a gas become cool enough to turn back into a liquid.
• This is what happens when some of the water molecules in the air hits a cold surface like a mirror in a steamy bathroom. A film of water covers the surface.
5 of 7
## Specific Heat Capacity
• The specific heat capacity of a substance tells us how much energy is needed to increase the temperature of 1kg of the substance by 1°C.
• For example, in order to increase 1kg of water by 1°C, 4200J of energy will be needed, but only 130J will be needed to increase 1kg of lead by 1°C.
• The equation for specific heat capacity is:
• E = m × c × θ
• E stands for energy in Joules (J)
• m stands for mass in kilograms (kg)
• c stands for specific heat capacity of the substance in J/kg°C
• θ stands for the temperature change in °C
• The equation is helpful in telling us how much energy will be needed to heat a specfic substance by a certain amount, or even what the temperature change is based on the energy needed, mass and specific heat capacity of the substance.
• Storage heaters use electricity at night to heat special bricks or concrete, which have a high specific heat capacity. This means that, during the day, they can slowly warm up the room, after having used electricity when it was cheaper.
6 of 7
## Heatng and Insulating Buildings
• There are many ways that buildings can be insulated, so that the least amount of heat is lost to surroundings, and less money is wasted on heating:
• Loft insulation uses fibreglass to reduce rate of heat transfer though the roof, while air between fibres helps reduce loss of heat through convection because the air is trapped and unable to move.
• Cavity wall insulation is when the space between the outside wall and inside wall is filled with a special insulator that traps air in small pockets to minimise convection currents.
• A draught excluder is fitted on doors so that air cannot escape through the doors easily because of convection.
• Double glazing is when two panes of glass are fitted on a window pane, with a small space for air or a vacuum, so that conduction is reduced, as well as convection if a vacuum is in place.
• The U-value tells us the energy per second passing through one square metre of a material, when the temperature difference is 1°C. A lower U-value means the material is a better insulator of heat.
• Payback time tells us how cost-effective a solution is, with a lower payback meaning more cost-effective:
• payback time (yrs) = cost of installation (£) ÷ savings per year in fuel costs (£)
7 of 7 | 1,473 | 6,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.925547 |
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## how to graph fractions
You will be shown the link option for the pdf file related to the page you are opening. In addition, you can look at the pages in Chrome or Firefox because they have to be displayed correctly in the hottest version of the browser without further steps on your part. You can easily find the link for that page. The Sitemap page for the website will be available for download. Here are some additional topics. Here too we give you the design of your geometric intuition. This will be very close to the x axis, but will never be touched.
If there are fractions with different denominators, it may be necessary to estimate their position or to search for frequent denominators among all the fractions involved. When using the models, the denominator will remain in place and prevent problems. This is a simple method to see if this equation describes y as a function of x. You can click on any equation to get a larger view of the equation.
You can use ” in your formula and use the cursor to change the value ” to look at the ways that influence the chart.
If you still have problems, then you’re already in the baseball stadium where you want to be. You can choose the difficulty from the graphic sector. Unfortunately.The point of view in drinking a point is at stake. Find this track point, make sure the steps are set to 1 and what are we really looking for? It’s easy to use and simple to understand. Be sure to ask why you are satisfied. In our analytical era, having the ability to interpret and create graphics is a very useful skill. They are intended for use by instructors to determine what homework they want.
Download If you know exactly which file you want to download or if you need to convert File to a fraction it could be useful, where you will be able to find the type of value x that could be more useful. Start by trying to reproduce the things that give the same result. You can choose different variables to adjust the type of graph paper that will be generated. You can choose a unique variable to adjust this chart worksheet to suit your needs. Almost all functions will have a number of parts in this way. This is a special quality that you can store in your work in the URL (link to the website).
By : study.com
## how to graph fractions
By : www.showme.com
## how to graph fractions
By : www.mathcaptain.com
When using the model, the denominator will remain in place and prevent the problem. These actions could be a real struggle to teach. Whether there are fractions with different denominators, you will have to move closer to their position or locate a common denominator among all the fractions concerned.
Spreadsheets are generated randomly, which means you get a different one at any time. Mathematics works like anything else, so if you need to gain a good knowledge, then you will have to practice. It is a simple and thorough method. Equation Write an equation that can be used to symbolize the relationship. When the equations are written in standard form, it is very easy to get the intercepts. There is another way to graph standard equations, and it looks like x and y intersections.
The software algebra tutorial section provides easy-to-understand explanations for each step in solving algebra problems. Students can use the drawer symbol. Most students do not have the opportunity to sit long enough in the first two phases. Students study creativity and visualization in different ways. I found that the simple and detailed instructions, explanations on how the formula worked were of invaluable help. When they work for leaders or for their own business, they must correctly analyze a situation, for example with a strategy to solve a problem, using the appropriate tools. communicate the results. It is not only the basic students, but also the students who practice advanced algebra.
You will need another decision mode. Look at this diagram because it will help you understand the procedure. This procedure continues until everything is correctly labeled. So see if you can adapt to get the result that interests you.
Mixed numbers can be entered directly. All natural numbers are also called counting numbers. The approach has led to many different numbers that have the exact nomenclature, the context provides further clues to the actual number. Don’t forget that there are an infinite number of solutions.
If you still have difficulty, chances are you already know where you’ll need it. If you had this problem alone, you may have found three unique points using the value table. There are different solutions, which are all the other points involved: every point of this line is a cure for the equation. For the slope of the line to be useful, it requires the coordinates of at least one point in play: in our analytical era, the ability to interpret and create graphics is an extremely useful skill. The first thing I did was discover the interception. | 990 | 4,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.933889 |
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[{:(0,-(1)/(3),-(1)/(2)) ,((1)/(3),0,-(1)/(5)),(1)/(2) ,((1)/(5),0):}]
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# If x and n are integers and x^(n + 2) = 16x^2, what is the least possi
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If x and n are integers and x^(n + 2) = 16x^2, what is the least possi [#permalink]
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19 Feb 2019, 23:29
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If x and n are integers and $$x^{(n + 2)} = 16x^2$$, what is the least possible value of 16 + x ?
A. -16
B. -4
C. 0
D. 12
E. 16
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Re: If x and n are integers and x^(n + 2) = 16x^2, what is the least possi [#permalink]
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19 Feb 2019, 23:50
1
$$x^{(n+2)}=16x^2$$
$$x^n.x^2=16x^2$$
$$x=0$$ or $$x^n=16$$
$$x^n=16$$ has the following solutions
$$4^2=16,(-4)^2=16,2^4=16,(-2)^4=16,16^1=16$$
Hence $$x$$ can be $$-4,-2,0,2,4,16$$ .
Least value of $$16+x=16-4=12$$
IMO C
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If x and n are integers and x^(n + 2) = 16x^2, what is the least possi [#permalink]
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Updated on: 20 Feb 2019, 01:41
Bunuel wrote:
If x and n are integers and $$x^{(n + 2)} = 16x^2$$, what is the least possible value of 16 + x ?
A. -16
B. -4
C. 0
D. 12
E. 16
x^n*x^2=16*x^2
so
x^n=16
x=(+/-4)^2 or (+/-2)^2 , or ( 16)^1
so least value of 16+x would be at x=-4
16-4 = 12
IMO D
Originally posted by Archit3110 on 20 Feb 2019, 01:00.
Last edited by Archit3110 on 20 Feb 2019, 01:41, edited 1 time in total.
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Re: If x and n are integers and x^(n + 2) = 16x^2, what is the least possi [#permalink]
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20 Feb 2019, 01:35
How you wrote x^2=16
when x^n*x^2=16*x^2
Posted from my mobile device
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Re: If x and n are integers and x^(n + 2) = 16x^2, what is the least possi [#permalink]
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20 Feb 2019, 01:41
mohitranjan05 wrote:
How you wrote x^2=16
when x^n*x^2=16*x^2
Posted from my mobile device
mohitranjan05 edited thanks
Re: If x and n are integers and x^(n + 2) = 16x^2, what is the least possi [#permalink] 20 Feb 2019, 01:41
Display posts from previous: Sort by | 1,111 | 3,083 | {"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} | 4.1875 | 4 | CC-MAIN-2019-43 | latest | en | 0.812978 |
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Aug 15 2006 | 4:36 pm
Okay, I'm determined to learn Jitter. I'm inspired by the jitter recipes. It seems that the tutorials are the way to learn, so for the 2nd time I'm staring them again. This time, however, I'm doing it in an interactive way, making sure I don't move on without trying to creatively process the information.
I was a little intimidated by y'all at first, but I thought I would jump right in and start asking questions. I realize that some of the info. may be covered in later chapters so feel free to just tell me to keep working through and I'll get to my answer.
Here is a patch that I made that randomly fills a 2x2 matrix. I'm looking for an easier way to fill a larger matrix without putting a "random" module on each coordinate. I'd like to have all the coordinates continuously changing as well. I'm not so good with equations, so an answer that is more visual would help. I'm willing to work on my math though if that really is the best way.
Patch is below:
thanks!
max v2;
• Aug 15 2006 | 5:37 pm
if you make the matrix 8 X 8 and change the two randoms that were 4 to
16, it works fine. If you double the randoms over the matrix size, it
works. Is this not what you wanted?
Keith
On 8/15/06, Tyson Rogers wrote:
>
> Okay, I'm determined to learn Jitter. I'm inspired by the jitter recipes. It seems that the tutorials are the way to learn, so for the 2nd time I'm staring them again. This time, however, I'm doing it in an interactive way, making sure I don't move on without trying to creatively process the information.
>
> I was a little intimidated by y'all at first, but I thought I would jump right in and start asking questions. I realize that some of the info. may be covered in later chapters so feel free to just tell me to keep working through and I'll get to my answer.
> Here is a patch that I made that randomly fills a 2x2 matrix. I'm looking for an easier way to fill a larger matrix without putting a "random" module on each coordinate. I'm not so good with equations, so an answer that is more visual would help. I'm willing to work on my math though if that really is the best way.
>
> Patch is below:
>
> thanks!
>
>
> max v2;
> #N vpatcher 345 433 945 833;
> #P origin 0 30;
> #P toggle 36 52 15 0;
> #P toggle 294 20 15 0;
> #P window setfont "Sans Serif" 9.;
> #P window linecount 1;
> #P newex 36 86 52 196617 metro 11;
> #P newex 294 56 52 196617 metro 10;
> #P button 196 63 15 0;
> #P newex 247 156 57 196617 pack 0 0 0;
> #P message 252 185 102 196617 setcell \$2 \$3 val \$1;
> #P newex 312 100 52 196617 random 4;
> #P window linecount 2;
> #P newex 194 100 53 196617 random 257;
> #P user jit.pwindow 411 324 82 62 0 1 0 0 1 0;
> #P newex 254 100 52 196617 random 4;
> #P button 126 126 15 0;
> #P window linecount 1;
> #P newex 139 265 46 196617 jit.print;
> #P newex 131 237 105 196617 jit.matrix 1 char 2 2;
> #P window linecount 8;
> #P comment 437 79 100 196617 okay , randomly filling a 2x2 matrix. now , what's an easier way to fill a larger matrix without using a random module for each coordinate?;
> #P connect 14 0 12 0;
> #P connect 12 0 3 0;
> #P connect 8 0 1 0;
> #P connect 3 0 1 0;
> #P connect 1 0 2 0;
> #P connect 10 0 6 0;
> #P connect 11 0 10 0;
> #P connect 6 0 9 0;
> #P connect 9 0 8 0;
> #P connect 10 0 4 0;
> #P connect 4 0 9 1;
> #P connect 7 0 9 2;
> #P connect 13 0 11 0;
> #P connect 10 0 7 0;
> #P connect 1 0 5 0;
> #P pop;
>
>
• Aug 15 2006 | 5:38 pm
also, you could try urn and group, if you want to fill all at the same time.
On 8/15/06, keith manlove wrote:
> if you make the matrix 8 X 8 and change the two randoms that were 4 to
> 16, it works fine. If you double the randoms over the matrix size, it
> works. Is this not what you wanted?
>
> Keith
>
> On 8/15/06, Tyson Rogers wrote:
> >
> > Okay, I'm determined to learn Jitter. I'm inspired by the jitter recipes. It seems that the tutorials are the way to learn, so for the 2nd time I'm staring them again. This time, however, I'm doing it in an interactive way, making sure I don't move on without trying to creatively process the information.
> >
> > I was a little intimidated by y'all at first, but I thought I would jump right in and start asking questions. I realize that some of the info. may be covered in later chapters so feel free to just tell me to keep working through and I'll get to my answer.
> > Here is a patch that I made that randomly fills a 2x2 matrix. I'm looking for an easier way to fill a larger matrix without putting a "random" module on each coordinate. I'm not so good with equations, so an answer that is more visual would help. I'm willing to work on my math though if that really is the best way.
> >
> > Patch is below:
> >
> > thanks!
> >
> >
> > max v2;
> > #N vpatcher 345 433 945 833;
> > #P origin 0 30;
> > #P toggle 36 52 15 0;
> > #P toggle 294 20 15 0;
> > #P window setfont "Sans Serif" 9.;
> > #P window linecount 1;
> > #P newex 36 86 52 196617 metro 11;
> > #P newex 294 56 52 196617 metro 10;
> > #P button 196 63 15 0;
> > #P newex 247 156 57 196617 pack 0 0 0;
> > #P message 252 185 102 196617 setcell \$2 \$3 val \$1;
> > #P newex 312 100 52 196617 random 4;
> > #P window linecount 2;
> > #P newex 194 100 53 196617 random 257;
> > #P user jit.pwindow 411 324 82 62 0 1 0 0 1 0;
> > #P newex 254 100 52 196617 random 4;
> > #P button 126 126 15 0;
> > #P window linecount 1;
> > #P newex 139 265 46 196617 jit.print;
> > #P newex 131 237 105 196617 jit.matrix 1 char 2 2;
> > #P window linecount 8;
> > #P comment 437 79 100 196617 okay , randomly filling a 2x2 matrix. now , what's an easier way to fill a larger matrix without using a random module for each coordinate?;
> > #P connect 14 0 12 0;
> > #P connect 12 0 3 0;
> > #P connect 8 0 1 0;
> > #P connect 3 0 1 0;
> > #P connect 1 0 2 0;
> > #P connect 10 0 6 0;
> > #P connect 11 0 10 0;
> > #P connect 6 0 9 0;
> > #P connect 9 0 8 0;
> > #P connect 10 0 4 0;
> > #P connect 4 0 9 1;
> > #P connect 7 0 9 2;
> > #P connect 13 0 11 0;
> > #P connect 10 0 7 0;
> > #P connect 1 0 5 0;
> > #P pop;
> >
> >
>
• Aug 15 2006 | 5:45 pm
jit.noise ?
On 8/15/06, keith manlove wrote:
> if you make the matrix 8 X 8 and change the two randoms that were 4 to
> 16, it works fine. If you double the randoms over the matrix size, it
> works. Is this not what you wanted?
>
> Keith
>
> On 8/15/06, Tyson Rogers wrote:
> >
> > Here is a patch that I made that randomly fills a 2x2 matrix. I'm looking for an easier way to fill a larger matrix without putting a "random" module on each coordinate. I'm not so good with equations, so an answer that is more visual would help. I'm willing to work on my math though if that really is the best way.
> >
• Aug 15 2006 | 6:18 pm
Thanks!
Keith, what does doubling the randoms over the matrix size do?
Scott, I think jit.noise is doing exactly what I was trying to do.
I'm going to keep working through the tutorials, I think I have the jist of what I needed.
• Aug 15 2006 | 8:36 pm
Hi Tyson,
jit.noise i nice and fast but is only one kind of noise. There are
some really beautiful noise options in jit.bfg which runs slower but
is vast in possiblities.
wes
On 8/15/06, Tyson Rogers wrote:
>
> Thanks!
>
> Keith, what does doubling the randoms over the matrix size do?
>
> Scott, I think jit.noise is doing exactly what I was trying to do.
>
> I'm going to keep working through the tutorials, I think I have the jist of what I needed.
>
>
>
• Aug 17 2006 | 7:29 am
On 15-Aug-2006, at 22:36, Wesley Smith wrote:
> jit.noise i nice and fast but is only one kind of noise. There are
> some really beautiful noise options in jit.bfg which runs slower but
> is vast in possiblities.
There are also vast possibilities in the new Litter Bundle for Jitter
package: Gaussian noise, exponential noise, Poisson noise, beta and
arc-sine noise, band-limited white noise, and much more.
And they're fast.
A public beta will be coming out very soon. Stay tuned.
-- Peter
-------------- http://www.bek.no/~pcastine/Litter/ -------------
Peter Castine | +--> Litter Power & Litter Bundle for Jitter
| Industry-strength math for everyday use
|..................................................
Composer | iCE: Essential Tools for Sequencing, Recording
Sonic Artist | & Interface Building in Max/MSP
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p@castine.de | http://www.castine.de/ home|chez nous|wir | 2,745 | 8,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-17 | longest | en | 0.975996 |
https://maslinandco.com/6083256 | 1,675,032,150,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499768.15/warc/CC-MAIN-20230129211612-20230130001612-00387.warc.gz | 412,629,912 | 3,616 | # Calculate: 3x+7=25
## Expression: \$3x+7=25\$
Move the constant to the right-hand side and change its sign
\$3x=25-7\$
Subtract the numbers
\$3x=18\$
Divide both sides of the equation by \$3\$
\$x=6\$
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http://www.bankrate.com/loans/auto-loans/rule-of-78-watch-out-for-this-auto-loan-trick/ | 1,511,266,373,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00552.warc.gz | 356,660,188 | 13,558 | This is one rebate auto shoppers should avoid.
Some auto lenders still use the archaic and costly “Rule of 78s” formula to calculate a rebate of finance charges when a customer pays off a loan early. This rebate is actually a sneaky prepayment penalty.
“The Rule of 78s is a historical anachronism,” says David Rubinstein, vice president of the Virginia Citizens Consumer Council. “It’s simply another way of padding a loan.”
The Rule of 78s is a mathematical formula that was devised in the days before modern calculators. The formula was a quick way for lenders in the 1920s and 1930s to estimate payoff amounts when a customer paid ahead on an installment loan. It’s still around today.
Also known as the sum-of-the-digits method, the Rule of 78s gets its name from the sum of the digits one through 12 — the number of months in a year.
Wrong way
For a borrower looking to end an auto loan early, there isn’t a worse way a lender could calculate your payoff amount. The Rule of 78s formula packs extra interest charges into the early months of a loan. Using Rule of 78s, a lender typically collects three-quarters of a loan’s interest in the first half of a loan term.
There are two basic types of auto loans: simple interest loans and pre-computed loans. The Rule of 78s can only be applied to pre-computed loans that are paid ahead of schedule. To understand why this is such a lousy deal for consumers, you have to understand how a pre-computed loan works.
With a pre-computed loan, the interest owed over the life of the loan is calculated using a standard amortization table. Once you sign on the dotted line for this type of loan, you’re obligated to pay back principal plus the full amount of interest that will accrue over the entire term of the loan.
To sum up, interest on a pre-computed loan is calculated in advance and you’re on the hook for every penny of it when you sign.
In contrast, with a simple-interest loan you’re charged interest each day based on the balance you owe. So the quicker you pay down your balance the less interest you pay. A simple interest loan with no prepayment penalties rewards customers who pay ahead.
Pay ahead with a pre-computed loan that applies the “Rule of 78s” method to prepayments and you’ll be slammed with a penalty, disguised as a rebate.
Caution: Interest padding ahead
Let’s say you’re ready to pay off your 48-month auto loan a year early. Because you signed on for a pre-computed loan, you’re on the hook for 48 months worth of interest even though you’re paying off the loan in 36 months.
But your lender is going to do you a “favor.” You don’t have to pay 48 months worth of interest. Instead, he’s going to determine your payout amount including a “rebate” for those 12 months worth of finance charges you won’t have to pay.
But your payout amount won’t be what you deserve. The reason? Using the “Rule of 78s” method, your lender applies more of your previous payments toward interest and less of your previous payments toward principal.
Since less is applied toward principal, the amount you owe will be higher than expected. The earlier you try to pay off one of these loans the more you’ll have to pay. The higher the interest rate, the more that payoff amount is going to hurt.
“If it had overcharged the lender and undercharged the consumer, it would have disappeared decades ago,” says Jean Ann Fox, director of consumer protection for Consumer Federation of America.
“It’s a dirty little secret.”
Turning on the warning lights
In 1992, the U.S. Congress outlawed the use of the “Rule of 78s” formula in closed-end loans longer than 61 months.
“It just gets very egregious with a longer-term loan,” says Elizabeth Renuart, staff attorney at the National Consumer Law Center.
States outlawing use of the Rule of 78s formula in installment loans of five years and less: Arizona Michigan Delaware Minnesota Idaho Nebraska Iowa Nevada Kansas New Hampshire Maine New York Maryland Oregon Massachusetts South Dakota Vermont Source: CARLAW, a monthly legal reporting service for legal compliance specialists in the automobile industry.
Whether a lender can apply the “Rule of 78s” method to installment loans of five years or less is a matter of state law. Currently, 17 states prohibit the practice.
Earlier last year, U.S. Rep. John LaFalce, D-N.Y., introduced a bill (H.R. 1054) that would eliminate the use of the Rule of 78s formula in credit transactions.
Fortunately for consumers, simple interest loans are now the norm in the auto financing business. The vast majority of auto lenders do not use pre-computed auto loans and they do not use the Rule of 78s method to calculate prepayments.
“The Rule of 78s as it applies to installment auto sales is a relic of the past,” says David Robertson, executive director of the Association of Finance and Insurance Professionals.
“In today’s mainstream market, that would be an absolute rarity.”
The pre-computed Rule of 78s auto loans that do exist today tend to be found in the subprime market. Folks with less-than-perfect credit should be on the lookout.
“Buy here, pay here” auto lots and lenders that specialize in offering loans to borrowers with badly damaged credit may offer these consumer-unfriendly loans.
“All the ones I’ve seen have had really high interest rates,” says Mark Eskeldson, an auto expert and author of CarInfo.com, a consumer information and advocacy Web site.
“If a car dealer is trying to put you into a rule of 78s loan it’s fairly safe to assume that the dealer has packed your interest rate — he’s inflated it.”
Watch out for ‘interest rebates’
Don’t let this happen to you. Be leery of signing any financing contract that mentions a refund or rebate of interest. That’s a sure sign you’re about to sign on for a pre-computed loan and not a simple interest loan. Be sure to ask.
“If you see that there may be a refund of interest, that’s the first red flag that you don’t want this loan,” Eskeldson says.
And because it puts the most bucks in his pocket, there’s a good chance that a lender offering a pre-computed loan will apply the Rule of 78s formula to all prepayments.
Check the front of a loan contract to see whether it allows a refund or rebate of interest. Flip over to the back of the contract and look under the section on prepayments for further details. Some contracts even mention Rule of 78s.
“You’re more likely to find it in subprime, but you can’t assume it wasn’t used in the contract you signed,” Fox says. “You have to look.”
Avoid signing on to loans that apply the Rule of 78s formula to prepayments. If you’ve already signed on the dotted line, you’re best bet is to make your payments as scheduled. Because of the penalties, there’s really no point in paying ahead.
“You’re stuck,” says Jack Gillis, author of The Car Book. “You have no leverage. They’re not going to let you out of the deal. If you refinance you just end up paying more.”
— Updated: Jan. 1, 2002 | 1,543 | 6,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.956123 |
https://socratic.org/questions/how-do-you-graph-y-3cos-2x-pi-2 | 1,631,831,129,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00303.warc.gz | 587,451,064 | 5,972 | # How do you graph y = 3cos(-2x + pi/2)?
Jun 7, 2018
As below.
#### Explanation:
$y = A \cos \left(B x - C\right) + D$ is the standard form
$y = 3 \cos \left(- 2 x + \frac{\pi}{2}\right)$
$A = 3 , B = - 2 , C = \frac{\pi}{2} , D = 0$
$A m p l i t u \mathrm{de} = | A | = 3$
$\text{Period } = \frac{2 \pi}{|} B | = \frac{2 \pi}{2} = \pi$
$\text{Phase Shift } = - \frac{C}{B} = - \frac{\frac{\pi}{2}}{-} 2 = \frac{\pi}{4}$, color(red)(pi/4 " to the right"
$\text{Vertical Shift } = 0$
graph{3 cos(pi/2 - 2x) [-10, 10, -5, 5]} | 243 | 534 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "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} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.472461 |
https://jp.mathworks.com/matlabcentral/profile/authors/18516014 | 1,721,902,155,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00518.warc.gz | 279,256,251 | 23,908 | # Niklas Kurz
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# Thanks to anyone who helps with this question!
0
137
5
+41
A triangle T is separated into two congruent triangles by a straight cut. Which of the following statements are true?
A) T must have two sides with equal length.
B) The cut must be perpendicular to one side of T.
C) T must have two interior angles with the same measure.
D) The cut must go through the midpoint of one side of T.
Sep 16, 2021
#1
+115899
+1
This is a good question.
One choice can be dismissed easily enough, which would that be, and why?
Sep 17, 2021
#2
+41
+1
My guess is D. is it right?
CCeed Sep 18, 2021
#3
+115899
0
Do you mean D is the only one that is right or the only one that can easily be dismissed?
Melody Sep 18, 2021
#4
+115899
+1
Actually, looking again, I do not know.
Certainly all of them can be true whether or not they need to be true is a different matter.
I am withdrawing my earlier assertion that one can be easily removed as the answer. I misread it.
Thinking about it now one at a time.
I think all of them have to be true.
Sep 18, 2021
#5
+41
+1
got it! thank you!
CCeed Sep 18, 2021 | 332 | 1,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-05 | longest | en | 0.969964 |
https://freeboson.org/teaching/?dispnote=exp14 | 1,695,850,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00128.warc.gz | 292,516,733 | 4,380 | Home » Teaching
# Lecture Notes
Below are the lecture notes from my course on multivariable calculus, as well as some handouts I give to students for some of the experiments in the introductory physics lab courses. Note that I update my notes each term, but this website may not link to the latest editions.
### Notes on IPL/CPS Physics Labs
$\newcommand{\vect}[1]{{\mathbf{#1}} }$ $\renewcommand{\vec}[1]{{\overrightarrow{#1}} }$ $\newcommand{\norm}[1]{{\left|\left|#1\right|\right|} }$ $\newcommand{\D}{{{\mathcal D}} }$ $\newcommand{\C}{{{\mathcal C}} }$ $\newcommand{\E}{{{\mathcal E}} }$ $\newcommand{\R}{{\mathbb{R}} }$ $\newcommand{\Rtwo}{{\mathbb{R}^2} }$ $\newcommand{\Rthree}{{\mathbb{R}^3} }$ $\newcommand{\V}{{\mathbb{V}} }$ $\newcommand{\T}{{\mathbb{T}} }$ $\newcommand{\nhat}{{\vect{\hat{n}}} }$ $\newcommand{\vi}{{\vect{i}} }$ $\newcommand{\vj}{{\vect{j}} }$ $\newcommand{\vk}{{\vect{k}} }$ $\newcommand{\f}{{\vect{f}} }$ $\newcommand{\g}{{\vect{g}} }$ $\newcommand{\vz}{{\vect{0}} }$ $\newcommand{\Du}{{D_{\vect{u}}} }$ $\DeclareMathOperator{\sgn}{sgn}$ $\DeclareMathOperator{\dist}{dist}$ $\DeclareMathOperator{\area}{Area}$ $\DeclareMathOperator{\vol}{Volume}$ $\newcommand{\xcm}{{x_\mathrm{CM}} }$ $\newcommand{\ycm}{{y_\mathrm{CM}} }$ $\DeclareMathOperator{\grad}{\bf grad}$ $\DeclareMathOperator{\ave}{ave}$ $\DeclareMathOperator{\std}{std}$
Notes on Experiment 14: Standing Waves
Sujeet Akula
Error Analysis
Standing Waves in a String
For a given configuration with $n$ nodes, you have $n-1$ measurements of half-wavelengths, say, $\{\ell_i\}_{i=1}^{i=n-1}$. From here, you calculate the average, $\ell$, and the standard deviation $\sigma_\ell$. This gives the error that you propagate to error bars, $\delta \ell = \sqrt{\sigma_\ell^2 + \Delta_\ell^2}$, where $\Delta_\ell$ is the systematic error. A good estimate of the system error is to measure the width of a node in your standing waves. You can either measure the widths of each and average for each standing wave, or simply take the largest width (which will be the width of a node in the lowest $n$ standing wave). Since you will make a plot with $v_{\mathrm{string}}^2$, we will propagate the error to this. Note that the frequency is fixed at $f=120 \mathrm{Hz}$. \begin{align} \delta\lambda &= 2\delta \ell \\ \delta v_{\mathrm{string}} &= f\delta\lambda \\ \delta (v_{\mathrm{string}}^2) &= 2v_{\mathrm{string}}\delta v_{\mathrm{string}} \end{align} The uncertainty in tension is much simpler. Since $F_t = mg$, and we will not make multiple measurements of the mass, take $\delta m = \Delta_m$, and let $\Delta_m = 1\,\mathrm{gram}$. Then, $\delta F_t = g\times(1\,\mathrm{gram})$. (Of course you will adjust $\Delta_m$ numerically to suit your units.) Thus, you plot the tension $F_t = mg$ on the $y$-axis, and on the $x$-axis, you have $v_{\mathrm{string}}^2$, with $x$ error bars, $\delta (v_{\mathrm{string}}^2)$, and $y$ error bars, $\delta F_t$. Your $y$ error bars will be fixed for each point, while your $x$ error bars will typically grow with $v_{\mathrm{string}}$. Be sure to calculate the slope and error in slope using http://freeboson.org/slope; the slope is the linear mass density of the string.
Standing Waves in the Air
The analysis here is similar. You will have written down the water-levels that correspond to the maxima, which you find on the way up and the way down. So, for $n$ such maxima, say you have $\{(x_i^u,x_i^d)\}_{i=1}^{i=n}$. Since there is a time-delay associated with hearing the resonance, we will immediately average these quantities, without considering statistical error to produce $\{x_i : x_i = (x_i^u + x_i^d)/2\}$. From here, we should be able to produce a set of $n-1$ simple differences that give the half-wavelengths: $\{\ell_i : \ell_i = x_{i+1}-x_i\}_{i=1}^{i=n-1}$. This means that for each tuning fork, you should have the given frequency $f$, measurements of the half-wavelengths $\{\ell_i\}_{i=1}^{i=n-1}$, where $n$ varies with the frequency. This time, we will be plotting $\lambda$ against $1/f$, so the analysis is simple. From your data ${\ell_i}$, calculate the average $\ell$, and the standard deviation $\sigma_\ell$. Then, $\delta \ell = \sqrt{\sigma_\ell^2 + \Delta_\ell^2}$, and we can set $\Delta_\ell = 0.5\,\mathrm{cm}$. From here, $\lambda = 2\ell$ and $\delta \lambda = 2 \delta \ell$. Assume no uncertainty in the frequency $f$, and therefore no uncertainty in $1/f$, so that you can plot $\lambda$ on the $y$-axis and $1/f$ on the $x$-axis. Your error bars in $y$ will be $\delta \lambda$. Again, calculate the slope and slope uncertainty. The slope here is the speed of sound. | 1,420 | 4,647 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-40 | latest | en | 0.575499 |
https://www.equationsworksheets.net/systems-of-equations-elimination-worksheet-pdf/ | 1,716,453,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00004.warc.gz | 682,934,251 | 15,095 | # Systems Of Equations Elimination Worksheet Pdf
Systems Of Equations Elimination Worksheet Pdf – Expressions and Equations Worksheets are designed to help kids learn faster and more efficiently. These worksheets include interactive exercises and challenges based on sequence of operations. These worksheets will help children can master both basic and complex concepts in a brief amount of duration. The PDF worksheets are free to download and could be used by your child to practise maths equations. These materials are great to students in the 5th-8th grades.
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These worksheets are geared towards students in fifth and eighth grade. These worksheets are ideal for students struggling to calculate percentages. There are three types of problems that you can pick from. You have the choice to solve single-step issues that include whole numbers or decimal numbers or to use words-based methods to solve fractions and decimals. Each page contains 10 equations. The Equations Worksheets are utilized by students in the 5th-8th grades.
These worksheets are a great way to learn fraction calculation and other algebraic concepts. The majority of these worksheets allow users to select from three different types of problems. It is possible to select the one that is numerical, word-based, or a mixture of both. It is essential to pick the problem type, because every problem is different. Each page contains ten problems and is a wonderful resource for students from 5th to 8th grade.
These worksheets aid students in understanding the relationship between variables and numbers. The worksheets let students work on solving polynomial problems and discover how to use equations in their daily lives. If you’re looking for an excellent educational tool to discover the basics of equations and expressions, you can start by exploring these worksheets. These worksheets will teach you about the various types of mathematical problems along with the different symbols utilized to represent them.
These worksheets can be extremely beneficial to students in the beginning grades. These worksheets will help them master the art of graphing and solving equations. These worksheets are perfect to get used to working with polynomial variable. They can also help you learn how to factor and simplify them. It is possible to find a wonderful set of expressions and equations worksheets for children at any grade level. Doing the work yourself is the best method to master equations.
There are plenty of worksheets to study quadratic equations. Each level comes with their own worksheet. These worksheets can be used to solve problems to the fourth degree. Once you’ve finished a level, you’ll be able to proceed to solve other types equations. Once you have completed that, you are able to work to solve the same problems. For instance, you could, find the same problem as an extended one. | 611 | 3,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.943218 |
https://mathematica.stackexchange.com/questions/187847/finding-the-minimum-points-of-a-discrete-trigonometric-equation/188080 | 1,582,132,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00235.warc.gz | 457,649,115 | 33,359 | # Finding the minimum points of a discrete trigonometric equation
Basically I'm trying to solve the equation
$$A\Big(\sin(\theta_a-\theta_{a+1}) + \sin(\theta_a-\theta_{a-1})\Big) + B\Big(n_a^z \sin(\theta_a) - n_a^x \cos(\theta_a)\Big) = 0$$ subjected to the constraint that $$A\Big(\cos(\theta_a-\theta_{a+1}) + \cos(\theta_a-\theta_{a-1})\Big) + B\Big(n_a^x \sin(\theta_a) + n_a^z \cos(\theta_a) \Big) > 0.$$ In addition I would like all of the angles $$\theta_a$$ to lie on the interval $$(-\pi,\pi)$$.
Here $$A$$ and $$B$$ are some numerical constants and $$a$$ labels the sites of a lattice. For concreteness let's say that $$a=1,2,3,\dots,N$$ where $$N$$ is some finite even number. Furthermore the numbers $$n_a^z = 1$$ and $$n_a^x = 0$$ for all sites except $$a = N/2$$. At the site $$a = N/2$$ the two numbers can be chosen arbitrarily. For instance $$n_{N/2}^z = -1/\sqrt{2}$$ and $$n_{N/2}^x = -1\sqrt{2}$$.
Furthermore I'm using the periodic boundary conditions $$\theta_1 = \theta_N = 0$$.
To solve this numerically I have tried using the function FindRoot on the first equation. However, this results in that some of the angles I find are not satisfying the second constraint. Is there a nice way of solving the above two equations numerically for a finite lattice in Mathematica?
The problem is basically that my initial guess $$(\theta_a = \pi/4)$$ in FindRoot ends up giving me an angle that satisfies the first equation, but does not satisfy the second equation.
• Should $A$ multiply both of the first two terms in the first equation? Also, is $\theta_{N+1}$ the same thing as $\theta_1$ (i.e., periodic boundary conditions)? – Michael Seifert Dec 13 '18 at 18:26
• $A$ Should multiply both of the first terms. Yes, I'm using periodic boundary conditions – MOOSE Dec 13 '18 at 18:28
• You could always generate a large number of roots of the first system using randomized initial guesses and then select all those for which the second conditions hold. It's a bit kludgy, but it might work. – Michael Seifert Dec 13 '18 at 18:50
• This may be a naive suggestion, but can you rephrase the problem as a constrained optimization problem? E.g. minimize the square of the LHS subject to the given constraints. If so, check out Mathematica's overview of Constrained Optimization: reference.wolfram.com/language/tutorial/…. – Robert Jacobson Dec 13 '18 at 19:12
• @RobertJacobson it might be! Do you know if it is easy to minimize this problem using for instance NMinimize? I'm having some difficulty due to that the equations are coupled. – MOOSE Dec 14 '18 at 10:21
# Rephrase the problem as a nonlinear optimization problem
Since $$N$$ is the name of a built-in function in Mathematica, we use $$M$$. We do the simplest nontrivial case of $$M=4$$. Since $$A$$ and $$B$$ are just constants, I'll set them both to 1. I'll use DiscreteDelta to take care of the $$n_a^x$$ and $$n_a^z$$ variables. We will "index" $$\theta$$ in the Mathematica way, as θ[a], setting θ[1]=θ[M]=0.
M=4;
A=1;
B=1;
nx[a_, M_]:=-1/Sqrt[2] DiscreteDelta[M/2-a];
nz[a_, M_]:=(1-DiscreteDelta[M/2-a]) + -1/Sqrt[2] DiscreteDelta[M/2-a];
θ[1]=0;
θ[M]=0;
Let's write the objective function to match the question but use Replace to make the $$\theta_a$$ periodic with $$\theta_{M+1}=\theta_1$$. The objective function is the square of the original function. The square has a minimum of zero where the original function is zero.
objf[a_, M_, A_, B_]:=(A(Sin[θ[a]-θ[a+1]] + Sin[θ[a]-θ[a-1]]) +
B(nz[a, M] Sin[θ[a]] - nx[a, M]Cos[θ[a]]))^2 /.θ[x_]->θ[(Mod[x, M]+1)]
Actually, we have $$M$$ equations to solve, so we need to minimize $$M$$ quadratics, or, equivalently, minimize their sum.
f = Sum[objf[a, M, A, B], {a, 1, M}];
We repeat the above for the constraints, except we make a list instead of a sum.
constr[a_, M_, A_, B_]:=A(Cos[θ[a]-θ[a+1]] + Cos[θ[a]-θ[a-1]]) +
B(nx[a, M] Sin[θ[a]] + nz[a, M]Cos[θ[a]])>0 /.θ[x_]->θ[(Mod[x, M]+1)]
c = Table[constr[a, M, A, B], {a, 1, M}];
Since $$\theta_1=\theta_M=0$$, our unknowns are $$\theta_2, \theta_3, \theta_4, \ldots, \theta_{M-1}$$. (Of course, for $$M=4$$, that's just $$\theta_2$$ and $$\theta_3$$.) We put the objective function, constratins, and variables into NMinimize and cross our fingers.
v=Table[θ[a], {a, 2, M-1}];
NMinimize[{f, c}, v]
{0.255348, {θ[2] -> 4.63221*10^-23, θ[3] -> -0.261482}}
A quick Plot3D confirms this is at least reasonable. | 1,412 | 4,416 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-10 | latest | en | 0.848299 |
https://gmatclub.com/forum/three-bodies-a-b-and-c-start-moving-around-a-circular-track-131449.html?kudos=1 | 1,488,008,946,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171670.52/warc/CC-MAIN-20170219104611-00588-ip-10-171-10-108.ec2.internal.warc.gz | 713,649,727 | 65,114 | Three bodies A, B and C start moving around a circular track : GMAT Problem Solving (PS)
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# Three bodies A, B and C start moving around a circular track
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Three bodies A, B and C start moving around a circular track [#permalink]
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27 Apr 2012, 13:07
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Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?
A. 30 seconds
B. 60 seconds
C. 15 seconds
D. 10 seconds
E. 25 seconds
[Reveal] Spoiler: OA
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Re: Three bodies A, B and C start moving around a circular track [#permalink]
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29 Aug 2012, 07:34
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Speed of A,B and C are 3, 5, 9 m/s respectively.
Considering A&B:
Speed of B is (5-3)=2 m/s more than that of A. So with this relative speed it will take 60/2= 30 sec to cover the full length.
Considering B&C:
Relative speed is (9-5)=4 m/s. So, B&C will meet after every 60/4=15 sec.
Considering A&C:
Relative speed is (9-3)=6 m/s. So, A&C will meet after every 60/6=10 sec.
The time when all three will meet together is the LCM of values 30, 15 and 10.
That is 30. Because
30=30*1 ( So A,B meet)
30=15*2 (So, B,C meet)
30=10*3 (So, A,C meet)
So, after 30 sec they will meet again.
A Follow-up Question: When will A,B and C meet together at the start point?
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Re: Three bodies A, B and C start moving around a circular track [#permalink]
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27 Apr 2012, 18:25
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Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?
A. 30 seconds
B. 60 seconds
C. 15 seconds
D. 10 seconds
E. 25 seconds
A - 3 m/s, B - 5 m/s, C - 9 m/s
When will they meet if they are moving in the same direction?
When B covers one (or multiple) complete circle more than A and C also covers one (or multiple) complete circle more than A.
B's speed is 2 m/s more than A so he will take 60/2 = 30 s to complete one full circle more than A. In 60 secs he will cover 2 circles more than A and so on...
C's speed is 6 m/s more than A so he will take 60/6 = 10 s to complete one full circle more than A. In 20 secs he will cover 2 circles more than A and in 30 sec he will cover 3 circles more than A.
So in 30 s, all A, B and C will be at the same point. Answer A
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 101 Kudos [?]: 912 [2] , given: 43 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 25 Aug 2012, 11:23 2 This post received KUDOS 1 This post was BOOKMARKED vdadwal wrote: Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving? A. 30 seconds B. 60 seconds C. 15 seconds D. 10 seconds E. 25 seconds If they all meet after T seconds, it means they covered the distances 3T, 5T, and 9T respectively. Since they all arrive to the same spot, it means that the differences taken pairwise between the distances must be positive integer multiples of the length of the track, which is 60m. So, 2T, 4T, and 6T must all be multiples of 60. 2T multiple of 60 means T multiple of 30. The smallest T with this property is 30 and is on the list of answers. Answer A. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Director Joined: 24 Aug 2009 Posts: 507 Schools: Harvard, Columbia, Stern, Booth, LSB, Followers: 17 Kudos [?]: 693 [1] , given: 276 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 25 Aug 2012, 10:37 1 This post received KUDOS Hi Karishma, After calculating Relative speed of B & C over A. We can take LCM of time taken to complete one round by B & C to find out when all three will meet. This shortcut is preferred once anyone has mastered the logic as suggested by karishma _________________ If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth -Game Theory If you have any question regarding my post, kindly pm me or else I won't be able to reply Intern Joined: 22 Jan 2012 Posts: 35 Followers: 0 Kudos [?]: 21 [0], given: 0 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 27 Apr 2012, 14:07 For this question , we will start with the body who is the slowest i.e. 3m/sec. Since the questions asks us when all the three bodies are going to meet , so assume the time t is reqduired to do this Distance travelled by I body: 3t Distance travelled by II body: 5t Distance travelled by III body: 9t the distances should be equal to meet , and that is possible because of a circular track , length of the track: 60m Keep this is mind , for circular track The point on the circular track = n * length of the track + remaining distance where n is a positive integer Like if someone travels from point Z on the track 200 m then actually he is far from point Z by 20m . As ,200 = 3*60 +20 Now i inserted the values : A: 30 secs I=3*30=90 = one length of track +30 ; II=5*30=150 = 2 lenth of track + 30; III= 9*30 =270 = 4 lenth of track +30; so everyone is at 30 m after 30 secs. Hence A is the answer. Senior Manager Joined: 21 Jan 2010 Posts: 344 Followers: 3 Kudos [?]: 179 [0], given: 12 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 27 Apr 2012, 17:45 thanks , probably this is the quickest way for this , but any algebric way to handle this? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2169 Kudos [?]: 14027 [0], given: 222 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 27 Aug 2012, 21:41 fameatop wrote: Hi Karishma, After calculating Relative speed of B & C over A. We can take LCM of time taken to complete one round by B & C to find out when all three will meet. This shortcut is preferred once anyone has mastered the logic as suggested by karishma Yes, you are right. You get that the time taken by B to complete one extra circle is 30 secs and time taken by C to complete one extra circle is 10 secs. You take their LCM which is 30 secs. The theory explains why you should take the LCM. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Three bodies A, B and C start moving around a circular track [#permalink]
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18 Aug 2013, 00:34
VeritasPrepKarishma wrote:
Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?
A. 30 seconds
B. 60 seconds
C. 15 seconds
D. 10 seconds
E. 25 seconds
A - 3 m/s, B - 5 m/s, C - 9 m/s
When will they meet if they are moving in the same direction?
When B covers one (or multiple) complete circle more than A and C also covers one (or multiple) complete circle more than A.
B's speed is 2 m/s more than A so he will take 60/2 = 30 s to complete one full circle more than A. In 60 secs he will cover 2 circles more than A and so on...
C's speed is 6 m/s more than A so he will take 60/6 = 10 s to complete one full circle more than A. In 20 secs he will cover 2 circles more than A and in 30 sec he will cover 3 circles more than A.
So in 30 s, all A, B and C will be at the same point. Answer A
Hi Karishma
I have lost control over my understanding though u mentioned very clearly.
Requesting you to again depicts the same for me.
Also the theory behind the logic and any other question of similar kind.
Rgds
Prasannajeet
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Re: Three bodies A, B and C start moving around a circular track [#permalink]
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18 Aug 2013, 02:11
Bluelagoon wrote:
Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?
A. 30 seconds
B. 60 seconds
C. 15 seconds
D. 10 seconds
E. 25 seconds
i like to back solve it: (it takes at most 40 seconds) i explained my way below:
For circular distance of 60 meter, 60,120,180,240 all end in the same point
for body A , 30 * 3 = 90 m = 60 + 30 m (so 30m ahead from the starting point)
for body, B, 30 * 5 = 150 m = 120 + 30 m (so 30m ahead from the starting point)
for body C, 30 * 9 = 270 m = 240 + 30 m (so 30m ahead from the starting point)
Everyone 30m ahead of the starting point after 30 sec.
i am lucky enough that the 1st answer satisfies my findings.
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Re: Three bodies A, B and C start moving around a circular track [#permalink]
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19 Aug 2013, 04:12
prasannajeet wrote:
VeritasPrepKarishma wrote:
Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?
A. 30 seconds
B. 60 seconds
C. 15 seconds
D. 10 seconds
E. 25 seconds
A - 3 m/s, B - 5 m/s, C - 9 m/s
When will they meet if they are moving in the same direction?
When B covers one (or multiple) complete circle more than A and C also covers one (or multiple) complete circle more than A.
B's speed is 2 m/s more than A so he will take 60/2 = 30 s to complete one full circle more than A. In 60 secs he will cover 2 circles more than A and so on...
C's speed is 6 m/s more than A so he will take 60/6 = 10 s to complete one full circle more than A. In 20 secs he will cover 2 circles more than A and in 30 sec he will cover 3 circles more than A.
So in 30 s, all A, B and C will be at the same point. Answer A
Hi Karishma
I have lost control over my understanding though u mentioned very clearly.
Requesting you to again depicts the same for me.
Also the theory behind the logic and any other question of similar kind.
Rgds
Prasannajeet
Check out this post on circular motion: http://www.veritasprep.com/blog/2012/08 ... n-circles/
See if this helps else I will try to explain more in detail.
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Get started with Veritas Prep GMAT On Demand for 199 Veritas Prep Reviews Senior Manager Joined: 13 May 2013 Posts: 472 Followers: 3 Kudos [?]: 165 [0], given: 134 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 19 Aug 2013, 13:01 Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving? A. 30 seconds B. 60 seconds C. 15 seconds D. 10 seconds E. 25 seconds From the onset I kind of figured you could solve with LCM's but I wasn't entirely sure why. I tried solving by figuring out how long it would take each of them to make one complete revolution and keep counting until their times aligned but I think that is incorrect because we're trying to figure out how long it takes for them to "meet up" we cannot solve that way. Is this correct? If that is the case, then we need to figure out their relative speeds to determine when each body (let's call them A, B, C for the slow, medium and fast objects respectively) reaches the other. B's relative rate to A is 5-3 = 2m/second so it takes B 30 seconds to move 60 meters away from A. In other words, at the 30 second mark, A and B are next to one another. A has traveled 90 meters and B has traveled 150 meters. C's relative rate to A is 9-3 = 6m/second so it takes C 10 seconds to move 60 meters away from A. Every 10 seconds, it moves 60 meters (one full revolution) away from A. In 30 seconds (the time it takes A and B to meet up) it is 3 full revolutions ahead of A but is also next to it on the circle. ANSWER A. 30 seconds. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2169 Kudos [?]: 14027 [0], given: 222 Re: Three bodies A, B and C start moving around a circular track [#permalink] ### Show Tags 20 Aug 2013, 20:37 Expert's post 2 This post was BOOKMARKED WholeLottaLove wrote: Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving? A. 30 seconds B. 60 seconds C. 15 seconds D. 10 seconds E. 25 seconds From the onset I kind of figured you could solve with LCM's but I wasn't entirely sure why. I tried solving by figuring out how long it would take each of them to make one complete revolution and keep counting until their times aligned but I think that is incorrect because we're trying to figure out how long it takes for them to "meet up" we cannot solve that way. Is this correct? If that is the case, then we need to figure out their relative speeds to determine when each body (let's call them A, B, C for the slow, medium and fast objects respectively) reaches the other. B's relative rate to A is 5-3 = 2m/second so it takes B 30 seconds to move 60 meters away from A. In other words, at the 30 second mark, A and B are next to one another. A has traveled 90 meters and B has traveled 150 meters. C's relative rate to A is 9-3 = 6m/second so it takes C 10 seconds to move 60 meters away from A. Every 10 seconds, it moves 60 meters (one full revolution) away from A. In 30 seconds (the time it takes A and B to meet up) it is 3 full revolutions ahead of A but is also next to it on the circle. ANSWER A. 30 seconds. LCM of time works for a question of a different type: When will they meet for the first time AT THE STARTING POINT after they started moving? Time take by A to cover a circle = 60/3 = 20 sec Time taken by B to cover a circle = 60/5 = 12 sec Time taken by C to cover a circle = 60/9 sec So every 20 sec, A will be at the starting point. Every 12 secs B will be at the starting point. Every 60/9 sec, C will be at the starting point. Taking their LCM, we get 60. So every 60 sec, all three will be at the starting point. All meet for the first time at the starting point after they start moving after 60 sec. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for199
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Re: Three bodies A, B and C start moving around a circular track [#permalink] 23 Dec 2016, 06:54
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https://calculat.io/en/length/inches-to-mm/548 | 1,701,804,848,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00392.warc.gz | 174,958,528 | 21,712 | # Convert 548 inches to mm
## How many mm is 548 inches?
548 Inches is equal to 13919.2 Millimeters
## Explanation of 548 Inches to Millimeters Conversion
Inches to Millimeters Conversion Formula: mm = in × 25.4
According to 'inches to mm' conversion formula if you want to convert 548 (five hundred forty-eight) Inches to Millimeters you have to multiply 548 by 25.4.
Here is the complete solution:
548″ × 25.4
=
13919.2 mm
(thirteen thousand nine hundred nineteen point two millimeters)
## About "Inches to Millimeters" Calculator
This converter will help you to convert Inches to Millimeters (in to mm). For example, it can help you find out how many mm is 548 inches? (The answer is: 13919.2). Enter the number of inches (e.g. '548') and hit the 'Convert' button.
## Inches to Millimeters Conversion Table
InchesMillimeters
13538.2 mm
13563.6 mm
13589 mm
13614.4 mm
13639.8 mm
13665.2 mm
13690.6 mm
13716 mm
13741.4 mm
13766.8 mm
13792.2 mm
13817.6 mm
13843 mm
13868.4 mm
13893.8 mm
13919.2 mm
13944.6 mm
13970 mm
13995.4 mm
14020.8 mm
14046.2 mm
14071.6 mm
14097 mm
14122.4 mm
14147.8 mm
14173.2 mm
14198.6 mm
14224 mm
14249.4 mm
14274.8 mm
## FAQ
### How many mm is 548 inches?
548″ = 13919.2 mm | 413 | 1,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.629991 |
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# Lesson 2-5
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### Lesson 2-5
1. 1. Exponential Models Recall that an exponential function is of the form f ( x ) = ab x , where a ≠ 0, b >0, and b ≠ 1. When using an exponential function as a model for a real-life situation, x often represents time. Since f (0)= a , a is referred to as the initial value of the dependent variable. b , on the other hand, is known as the growth factor . Initial Value Growth Factor
2. 2. How do exponential models compare with linear models? The parameter that controls the “growth” of each type of functions works differently in each case. A linear model exhibits constant increase/decrease , which is determined by m . An exponential model exhibits constant percentage change , which is determined by b . Rate of increase/decrease Growth factor (% change)
3. 3. How do exponential models compare with linear models? To find an equation of a line, we need two points. This is related to the fact that there are two parameters in the equation: m and b . How many points do you think we need to determine an exponential equation?
4. 4. We can use the points to write two equations: If we divide the second equation by the first: Finally, we can find a : Let’s find an exponential equation that includes the points (2, 6) and (4,10): We know the equation has this form:
5. 5. We can use the points to write two equations: If we divide the second equation by the first: Finally, we can find a : Let’s find an exponential equation that includes the points (2, 6) and (4,10): We know the equation has this form:
6. 6. We can use the points to write two equations: If we divide the second equation by the first: Finally, we can find a : Let’s find an exponential equation that includes the points (2, 6) and (4,10): We know the equation has this form: Can’t my calculator do this for me??? | 553 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2016-50 | latest | en | 0.903353 |
https://www.coursehero.com/file/5692021/10-23-09/ | 1,529,784,213,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865181.83/warc/CC-MAIN-20180623190945-20180623210945-00475.warc.gz | 783,358,346 | 126,568 | 10-23-09
# 10-23-09 - Economics 448 Class Notes GET A BETTER...
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Economics 448 – Class Notes – 10/23/09 GET A BETTER DEFINITION OF WAGE CONTRACT Reservation level of utility – the level utility needed to get the worker to accept the contract Wage will be a function of the work effort MODEL #1: In the example on the handout: How much would the worker need to be paid in order to ACCEPT THE CONTRACT and also PUT IN THE LOW LEVEL OF EFFORT, and: How much would the worker need to be paid in order to ACCEPT THE CONTRACT and also PUT IN THE HIGH LEVEL OF EFFORT, and: o Asking for the participation constraint LOW LEVEL OF EFFORT: A = 0 U(w, aL) = w^(1/2) – 0 = 9 I have to construct my contract so that the utility received from putting in the low level of effort is GREATER than the reservation level of utility Reservation level of utility = 9 Thus, w >= 81 If I want the worker to put in the low level of effort, you have to pay the worker \$81 This is the level of effort where it doesn’t matter if the worker comes to work and puts in low effort or not come to work at all This is a simple contract: worket is paid \$81 to come to work and do the low level of effort PROFIT TO FIRM (low level) Profit = TR – TC Profit = 70 – 81 = (-11) NOT WORTH IT TO THE FIRM if you only get the worker to put in a low level of effort How much do I have to pay the agent (worker) to put in the high level of effort? U(w, aH) = w^(1/2) – 5 >= 9 w^(1/2) >= 14 W(H) = \$196 (high level of effort wage) Firm’s profit: Profit = \$270 - \$196 = \$74
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We want the worker to put in the high level of effort, so I need to structure the contract so that
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Jill Tulane University ‘16, Course Hero Intern | 697 | 2,839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.930864 |
https://programs.programmingoneonone.com/2021/09/leetcode-summary-ranges-problem-solution.html | 1,701,250,458,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00777.warc.gz | 556,841,164 | 32,708 | # Leetcode Summary Ranges problem solution
In this Leetcode Summary Ranges problem solution, You are given a sorted unique integer array nums.
Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.
Each range [a,b] in the list should be output as:
1. "a->b" if a != b
2. "a" if a == b
## Problem solution in Python.
```class Solution(object):
def summaryRanges(self, nums):
summary = []
if len(nums) > 1:
for i in range(len(nums)-1):
if i == 0:
start = nums[0]
end = nums[0]
if nums[i+1] == end+1:
end = end+1
else:
if start!=end:
summary.append(str(start)+"->"+str(end))
else:
summary.append(str(start))
start = nums[i+1]
end = nums[i+1]
if start!=end:
summary.append(str(start)+"->"+str(end))
else:
summary.append(str(start))
if len(nums) == 1:
summary.append(str(nums[0]))
return summary
```
## Problem solution in Java.
```import java.util.*;
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> ranges = new ArrayList<>();
int startI = 0;
int endI = 0;
for(int i=0; i<nums.length; i++) {
startI = i;
while(i < nums.length-1 && nums[i+1] == nums[i] + 1) {
i++;
}
endI = i;
if (startI == endI) {
} else {
}
}
return ranges;
}
}
```
## Problem solution in C++.
```class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> result;
for (int i = 0; i < nums.size(); ++i) {
for (int j = i; j < nums.size(); ++j) {
if (j == nums.size() - 1 || (long long)nums[j + 1] - nums[j] > 1) {
string s = to_string(nums[i]);
if (j > i) {
s += "->" + to_string(nums[j]);
}
result.push_back(s);
i = j;
break;
}
}
}
return result;
}
};
```
## Problem solution in C.
```char ** summaryRanges(int* nums, int numsSize, int* returnSize){
uint i,start,capacity = 8;
char **output;
char *str;
*returnSize = 0;
if(!numsSize)
return output;
output = (char**)malloc(sizeof(char*)*capacity);
for(i = 0,start = 0; i < numsSize; i++)
{
if(nums[i] - (i - start) != nums[start])
{
str = (char*)malloc(sizeof(char)*25);
if(i-start == 1)
sprintf(str,"%d",nums[start]);
else
sprintf(str,"%d->%d",nums[start],nums[i-1]);
if(*returnSize == capacity)
{
capacity *=2;
output = (char**)realloc(output,sizeof(char*)*capacity);
}
output[(*returnSize)++] = str;
start = i;
}
}
str = (char*)malloc(sizeof(char)*23);
if(i-start == 1)
sprintf(str,"%d",nums[start]);
else
sprintf(str,"%d->%d",nums[start],nums[i-1]);
if(*returnSize == capacity)
{
capacity += 1;
output = (char**)realloc(output,sizeof(char*)*capacity);
}
output[(*returnSize)++] = str;
start = i;
return output;
}
``` | 806 | 2,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.362497 |
http://math.stackexchange.com/questions/42746/summation-of-complex-numbers/42749 | 1,469,265,458,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00250-ip-10-185-27-174.ec2.internal.warc.gz | 158,288,534 | 18,388 | # Summation of complex numbers
This is a series problem where the terms are complex numbers. I am looking for a better approach to solving this problem.
If $\displaystyle z = \frac{1+i}{\sqrt2}$, Evaluate $1 + z + z^2 + ... + z^{20}$
The way I solved this was to evaluate the terms upto the 8th term. like below,
\begin{align} z^2 &= i \\ z^3 &= -\frac{1}{\sqrt2} +\frac{1}{\sqrt2}i \\ z^4 &= -1 \\ z^5 &= -\frac{1}{\sqrt2} -\frac{1}{\sqrt2}i \\ z^6 &= -i \\ z^7 &= -\frac{1}{\sqrt2} +\frac{1}{\sqrt2}i \\ z^8 &= 1 \\ \end{align}
Then evaluating the first 8 terms, $$S_7 = 1 + z + ... + z^7 = 0$$
Which implies that, $$S_20 = 0 + 0 + z^{17} + z^{18} + z^{19} + z^{20} = 1 + z + z^2 + z^3$$
Thus giving the solution, $$S_20 = 1 + (1 + \sqrt2)i$$
1. Just looking at the series itself($1 + z + z^2 + ...$) makes me think that the series is similar to a binomial series. The way the terms cancelled out made me think of a telescopic series. Can you guys shed some light on this?
2. While this solution works it took me a while to get there. Is there a more elegant/less tedious way of doing this?
-
@mathguy: the value of $z_{7}$ is wrong? there should be a $"+"$ sgin – user9413 Jun 2 '11 at 10:49
@mathguy, note that as soon as you got $z^2=i$, you could have gone right to $z^8=(z^2)^4=i^4=1$, so $z^8-1=0$, so $(z-1)(z^7+z^6+\dots+z+1)=0$, so $z^7+z^6+\dots+z+1=0$. – Gerry Myerson Jun 2 '11 at 11:24
@Gerry Myerson, very elegant and concise! Thank you. – mathguy80 Jun 3 '11 at 9:19
@Chandru, yup fixed that. – mathguy80 Jun 3 '11 at 9:20
@mathguy80: Thanks. PLease accept an answer :) – user9413 Jun 3 '11 at 9:21
1. If $z \ne 1$, we have
$$1 + z + \ldots z^{n-1} = \frac{z^n-1}{z-1}$$
So it greatly reduces the number of computation
1. Note that in your case $z = e^{\frac{i \pi}{4}}$, which makes it faster to compute the powers of $z$.
Applying those two ideas gets you the result.
-
Thanks. So it's a binomial series. Does $z$ being complex have any bearing on the series? – mathguy80 Jun 3 '11 at 9:23
Use the formula for the sum of a geometric series and the fact that $$\bigl(\cos\theta + i\sin\theta)^{n} = \cos{n\theta} + i \sin{n \theta}$$
-
I considered that but converting to polar form would have been more tedious. :) – mathguy80 Jun 3 '11 at 9:24 | 836 | 2,281 | {"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} | 4.3125 | 4 | CC-MAIN-2016-30 | latest | en | 0.835534 |
http://impresospopulares.colsan.edu.mx/index.php/read/elements-of-logic-and-formal-science | 1,544,966,147,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827727.65/warc/CC-MAIN-20181216121406-20181216143406-00579.warc.gz | 142,878,780 | 9,517 | # Get Elements of Logic and Formal Science PDF
By C. West Churchman, None
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Extra resources for Elements of Logic and Formal Science
Sample text
Thus, if one thing not stronger than another (p doesn't imply r) then something of equal strength or weaker than the former (something p implies) is not stronger than the latter (does not imply r) Or, if one thing is not weaker than another, then something of equal strength or stronger than the former is not weaker than the is , . latter. The results may be manner analogous to the For suppose we have two premises. generalized in a generalization of Principle 2. : ELEMENTS OF LOGIC 30 q, which imply r; then we may replace p or q (or both) by stronger forms, or we may replace r by a weaker form, and p and the resulting implication is That valid.
THE LOGIC OF PROPOSITIONS 25 may also be applied in the case where the implinot valid. " In Principle 2b. then q^ ("q follows: "If a 2, as does not follow that he is man is not honest general, then: If the is by Principle it statement p does not imply the statement does not imply p' ("p is false") q, false") its principles. That what the old principles which express is, it finds new principles generalization of Principle 2 expressed and more as well. The illustration: Suppose you were can best be seen by means of an presenting a debate in which you had four points admittedly substantiating your argument.
Implication is the principal relation of logic or the science of the laws of reasoning. When we say that a certain statement p "implies" another statement q, we mean that there is a relation between the former and the latter, a relation which is such that whenever the former is true, the latter is true also. " The word "implies," or one of its synonyms, is used constantly, but usually subconsciously. " That is, one method of denying that "p implies ^" is true consists in showing that there are cases where p is true and q is not true. | 684 | 3,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-51 | longest | en | 0.890479 |
https://blogs.ubc.ca/infiniteseriesmodule/units/unit-1/infinite-series/koch-snowflake-example/ | 1,669,526,375,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00442.warc.gz | 169,805,683 | 16,112 | # Koch Snowflake Example
Previous: A Geometric Series Problem with Shifting Indicies
Next: Videos on the Introduction to Infinite Series
## Problem
Suppose we would like to calculate the area of the "Koch Snowflake". The Koch Snowflake is an object that can be created from the union of infinitely many equilateral triangles (see figure below).
Starting with the equilateral triangle, this diagram gives the first three iterations of the Koch Snowflake (Creative Commons, Wikimedia Commons, 2007).
We construct the Koch Snowflake in an iterative process. Starting with an equilateral triangle, each iteration consists of altering each line segment as follows:
• divide the line segment into three segments of equal length
• draw an equilateral triangle that has the middle segment from step 1 as its base and points outward
• remove the line segment that is the base of the triangle from step 2
The Koch Snowflake is the limit approached as the number of iterations goes to infinity.
## Solution
The Koch Snowflake after 5 iterations. The first iteration is blue, the second green, the third yellow, the fourth is red, and the fifth is black (Creative Commons, image from Wikimedia Commons).
Now, to derive an expression for the area of our construction at the iteration, let's start with the fifth iteration. The fifth iteration of the snowflake is shown below, with its iterations in different colours.
Blue and Green Triangles
Assume that the one blue triangle as unit area. Each side of the green triangle is exactly 1/3 the length of a side of the blue triangle, and therefore has exactly 1/9 the area of the blue triangle. There are three green triangles, so the green and blue triangles have an area of
Blue, Green, and Yellow Triangles
Each of the 12 = 3•4 yellow triangles has 1/9 the area of a green triangle, or the area of a blue triangle. The area of the blue, green, and yellow triangles is
Blue, Green, Yellow, and Red Triangles
Each of the 48 = 3•4•4 red triangles has 1/9 the area of a yellow triangles, or the area of a blue triangle. The area of the blue, green, yellow, and red triangles is
Total Area
The total area of the snowflake uses the infinite sequence
.
We will add all the terms of the series together, and add 1, to produce the following sum
Seeing that this is a geometric series with a = 1/3 and r = 4/9, we immediately conclude that this series converges and is equal to
Previous: A Geometric Series Problem with Shifting Indicies
Next: Videos on the Introduction to Infinite Series
### 3 Responses to Koch Snowflake Example
1. Ryan Singley says:
This is all wrong. The number of yellow triangles is 12 not 9, and that makes the infinite sequence wrong as well.
2. yea says:
“each yellow triangle has 1/9 the area of a green triangle, or 1/27 the area of a blue triangle. ” But if a yellow triangle is 1/9 of a green triangle which is in turn 1/9 of a blue triangle, wouldn’t that make a yellow triangle equivalent to 1/81 of a blue triangle? Also- you say “The area of the blue, green, and yellow triangles is 1 + 3(1 / 9) + 9(1 / 27) = 5 / 3.”, you multiply 1/27 by 9 when it should be multiplied by 12 because there are 12 yellow triangles. Am I just really confused or is your math all wrong?
3. Geoffrey says:
This is wrong!
The r = 4/9 not 1/9. | 803 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.907494 |
https://www.extendoffice.com/excel/formulas/excel-subtotal-by-age.html | 1,721,211,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00020.warc.gz | 673,801,304 | 58,125 | ## Subtotal invoice amounts by age in Excel
To sum the invoice amounts based on age as below screenshot shown may be a common task in Excel, this tutorial will show how to subtotal invoice amounts by age with a normal SUMIF function.
#### Subtotal invoice amount based on age with SUMIF function
Dealing with this task, you can use a simple SUMIF function in Excel, the generic syntax is:
=SUMIF(range, criteria, sum_range)
• range: The range of cells which criteria is applied.
• criteria: The criteria to sum data based on;
• sum_range: The range of cells that you want to sum.
For example, I want to sum all the invoice amounts whose age is less than or equal to 30, or whose age is greater than 30, please apply the following formula:
1. Enter or copy the below formula into a blank cell:
=SUMIF(\$B\$2:\$B\$10,F2,\$D\$2:\$D\$10)
2. And then, drag the fill handle down to copy the formula to other cells, and the amount subtotals will be calculated, see screenshot:
#### Relative function used:
• SUMIF:
• The SUMIF function can help to sum cells based on one criterion.
#### More articles:
• Sum Values By Group In Excel
• Sometimes, you may have the requirement to sum values based on group in a table. For example, I have a list of products with their corresponding amounts in another column, now, I want to get the subtotal amount for each product as below screenshot shown. This tutorial will introduce some formulas for solving this task in Excel.
• Sum Smallest Or Bottom N Values
• In Excel, it is easy for us to sum a range of cells by using the SUM function. Sometimes, you may need to sum the smallest or bottom 3, 5 or n numbers in a data range as below screenshot shown. In this case, the SUMPRODUCT together with the SMALL function can help you to solve this problem in Excel.
• Sum Smallest Or Bottom N Values Based On Criteria
• In previous tutorial, we have discussed how to sum the smallest n values in a data range. In this article, we will perform a further advanced operation – to sum the lowest n values based on one or more criteria in Excel.
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• Brings Efficient Tabs to Office (include Excel), Just Like Chrome, Edge and Firefox. | 1,006 | 4,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.876173 |
https://sts-math.com/post_32.html | 1,547,824,767,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00004.warc.gz | 660,365,799 | 4,748 | Determine whether there is a maximum or minimum vale for the given function, and find that value. f(x)= x^2+6x+4
Every second degree function has either a maximum (if a is negative) or a minimum (if a is positive). Our function, thus, has a minimum. The formula for it is:
$$(x, y)=(\frac{-b}{2a}, \frac{-\Delta}{4a})$$
a is 1, b is 6, c is 4, $$\Delta=b^2-4ac=36-16=20$$, so the minimum is at coordinates (-3,5), that is the function doesn’t ever get below -5 and it gets there only when the argument is -3.
If you take the first derivative of f(x) you get:
f’(x) = 2x + 6
The max or min is where the derivative is 0.
So. 0 = 2x + 6
Do some algebra and x = -3
Substitute that back into the original equation:
f(-3) = (-3)^2 + 6(-3) + 4
And you get -5
So your value is at the point (-3,5)
RELATED: | 269 | 808 | {"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} | 4.15625 | 4 | CC-MAIN-2019-04 | longest | en | 0.837861 |
https://www.jiskha.com/questions/7901/a-link-to-a-how-to-page-please-since-this-is-not-my-area-of-expertise-i-searched-google | 1,597,098,638,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738699.68/warc/CC-MAIN-20200810205824-20200810235824-00455.warc.gz | 709,977,395 | 6,018 | # Resume
Since this is not my area of expertise, I searched Google under the key words "resume write" to get these possible sources:
http://jobstar.org/tools/resume/index.php
http://www.how-to-write-a-resume.org/resume_writing.htm
http://www.soyouwanna.com/site/syws/resume/resume.html
I hope this helps. Thanks for asking.
thank you :)
1. 👍 0
2. 👎 0
3. 👁 326
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asked by Selena on September 4, 2014 | 964 | 3,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.937059 |
http://mathhelpforum.com/pre-calculus/177448-remainder-theorem.html | 1,524,237,794,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00229.warc.gz | 205,815,492 | 9,918 | # Thread: Remainder theorem
1. ## Remainder theorem
The expression $\displaystyle f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $\displaystyle x^2 - 2x$. When $\displaystyle f(x)$ is divided by $\displaystyle x - 1$ , the remainder is 8 more than when it is divided by $\displaystyle x + 1$. Factorise $\displaystyle f(x)$ completely.
2. Originally Posted by Ilsa
The expression $\displaystyle f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $\displaystyle x^2 - 2x$. When $\displaystyle f(x)$ is divided by $\displaystyle x - 1$ , the remainder is 8 more than when it is divided by $\displaystyle x + 1$. Factorise $\displaystyle f(x)$ completely.
Well, f(x) is divisible by $\displaystyle x^2 - 2x = x(x - 2)$ so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
$\displaystyle f(x) = ax^3 - (a + 3b)x^2 + 2bx$
From the original statement (divisible by x(x - 2)), we get
$\displaystyle f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0$
Call
$\displaystyle f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d$
Then we know that
$\displaystyle f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8$
Three equations, three unknowns.
-Dan
3. Originally Posted by topsquark
Well, f(x) is divisible by $\displaystyle x^2 - 2x = x(x - 2)$ so we know f(x) is divisible by x. Thus c = 0 immediately. Consider then
$\displaystyle f(x) = ax^3 - (a + 3b)x^2 + 2bx$
From the original statement (divisible by x(x - 2)), we get
$\displaystyle f(2) = a(2)^3 - (a + 3b)(2)^2 + 2b(2) = 0$
Call
$\displaystyle f(1) = a(1)^3 - (a + 3b)(1)^2 + 2b(1) = d$
Then we know that
$\displaystyle f(-1) = a(-1)^3 - (a + 3b)(-1)^2 + 2b(-1) = d - 8$
Three equations, three unknowns.
-Dan
If f(x) is divisible by x, how is c=0?
4. $\displaystyle f(x) = ax^3 - ( a + 3b )x^2 + 2bx + c$ is excatly divisible by $\displaystyle x^2 - 2x$.
If x divides f(x) then
$\displaystyle \displaystyle \frac{f(x)}{x}$ leaves no remainder. Thus
$\displaystyle \displaystyle \frac{ax^3 - (a + 3b)x^2 + 2bx + c}{x} = ax^2 - (a + 3b)x + 2b + \frac{c}{x}$
This gives a remainder of c, but f(x) is divisible by x, thus c = 0.
You can do this quicker by using the fact that if f(x) is divisible by x - r then f(r) = 0. In this case r = 0, so f(0) = 0.
-Dan | 865 | 2,253 | {"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} | 4.5625 | 5 | CC-MAIN-2018-17 | latest | en | 0.781069 |
https://jp.mathworks.com/matlabcentral/cody/problems/44743-pizza-order/solutions/1644119 | 1,603,155,691,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00260.warc.gz | 386,549,338 | 16,637 | Cody
Problem 44743. Pizza order
Solution 1644119
Submitted on 12 Oct 2018 by Edgar Guevara
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
d1 = 2;d2 = 1; n_correct = 4; assert(isequal(Pizza2(d1,d2),n_correct))
p = 4
2 Pass
d1 = 1;d2 = 2; n_correct = 0.25; assert(isequal(Pizza2(d1,d2),n_correct))
p = 0.2500
3 Pass
d1 = 5;d2 = 7; n_correct = 0.75; assert(isequal(Pizza2(d1,d2),n_correct))
p = 0.7500
4 Pass
d1 = 7;d2 = 5; n_correct = 2; assert(isequal(Pizza2(d1,d2),n_correct))
p = 2
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Start Hunting! | 261 | 739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.604291 |
https://help.libreoffice.org/latest/ko/text/swriter/01/06100000.html | 1,721,007,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00892.warc.gz | 273,515,901 | 4,921 | # Sort
์ ํํ ๋จ๋ฝ ๋๋ ํ ํ์ ์ฌ์ ์์ด๋ ์ซ์์์ผ๋ก ์ ๋ ฌํฉ๋๋ค. ์ ๋ ฌ ํค๋ฅผ ์ธ ๊ฐ๊น์ง ์ง์ ํ์ฌ ์ฌ์ ์ ๋ ฌ ํค์ ์ซ์ ์ ๋ ฌ ํค๋ฅผ ๊ฒฐํฉํ ์ ์์ต๋๋ค.
์ด ๋ช
๋ น์ ์ฌ์ฉํ๋ ค๋ฉด...
Choose Table - Sort.
#### From the tabbed interface:
Choose Table - Sort.
On the Table menu of the Table tab, choose Sort.
Sort
## ํค 1 - 3
Specifies additional sorting criteria. You can also combine sort keys.
## ์ด 1 - 3
Enter the number of the table column that you want to use as a basis for sorting.
## ํค ์ ํ 1 - 3
Select the sorting option that you want to use.
## ์์
### ์ค๋ฆ์ฐจ์
Sorts in ascending order, (for example, 1, 2, 3 or a, b, c).
## ๋ด๋ฆผ์ฐจ์
Sorts in descending order (for example, 9, 8, 7 or z, y, x).
## ์ด
Sorts the columns in the table according to the current sort options.
## ํ
ํ์ฌ ์ ๋ ฌ ์ต์
์ ๋ฐ๋ผ ํ์ ํ์ด๋ ์ ํ ์์ญ์ ๋จ๋ฝ์ ์ ๋ ฌํฉ๋๋ค.
## ๊ตฌ๋ถ์
์ธ์ํ ์ ์๋ ๋จ๋ฝ ๊ธฐํธ๋ฅผ ์ฌ์ฉํ์ฌ ๋จ๋ฝ์ ๊ตฌ๋ถํฉ๋๋ค. ๋จ๋ฝ์ ์ ๋ ฌํ ๋ ํญ์ด๋ ๋ฌธ์๊ฐ ๊ตฌ๋ถ ๊ธฐํธ๋ก ์๋ํ๋๋ก ์ง์ ํ ์๋ ์์ต๋๋ค.
## ํญ
์ ํ๋ ๋จ๋ฝ์ด ํญ์ผ๋ก ๊ตฌ๋ถ๋์ด ์๋ ๋ชฉ๋ก๊ณผ ์ผ์นํ๋ฉด ์ด ์ต์
ํ๋๋ฅผ ์ ํํฉ๋๋ค.
## ๊ธ๊ผด
์ ํ๋ ์์ญ์์ ๊ตฌ๋ถ ๊ธฐํธ๋ก ์ฌ์ฉํ ๋ฌธ์๋ฅผ ์
๋ ฅํฉ๋๋ค. ์ฌ๊ธฐ์ ์ ์๋ ๊ตฌ๋ถ ๊ธฐํธ๋ฅผ ์ฌ์ฉํ์ฌ LibreOffice์ ์ ํ๋ ๋จ๋ฝ์์ ์ ๋ ฌ ํค์ ์์น๋ฅผ ์ง์ ํ ์ ์์ต๋๋ค.
## ์ ํ
๊ตฌ๋ถ ๊ธฐํธ๋ก ์ฌ์ฉํ ๋ฌธ์๋ฅผ ์ ํํ ์ ์๋ ํน์ ๋ฌธ์ ๋ํ ์์๋ฅผ ์ฝ๋๋ค.
## ์ธ์ด
์ ๋ ฌ ๊ท์น์ ์ ์ํ๋ ์ธ์ด๋ฅผ ์ ํํฉ๋๋ค. ์ผ๋ถ ์ธ์ด๋ ํน์ ๋ฌธ์๋ฅผ ๋ค๋ฅธ ์ธ์ด์ ๋ค๋ฅด๊ฒ ์ ๋ ฌํฉ๋๋ค.
## ๋/์๋ฌธ์ ๊ตฌ๋ถ
ํ๋ฅผ ์ ๋ ฌํ ๋ ๋๋ฌธ์์ ์๋ฌธ์๋ฅผ ๊ตฌ๋ณํฉ๋๋ค. ํ๊ตญ์ด์ ๊ฒฝ์ฐ ํน์ํ ์ฒ๋ฆฌ๊ฐ ์ ์ฉ๋ฉ๋๋ค.
ํ๊ตญ์ด์ ๊ฒฝ์ฐ ๋๋ฌธ์ ๋ฐ ์๋ฌธ์๋ฅผ ๊ตฌ๋ถ์ ์ ํํ์ฌ ๋ค์ค ๋์กฐ๋ฅผ ์ ์ฉํฉ๋๋ค. ๋ค์ค ๋์กฐ๋ฅผ ์ฌ์ฉํ๋ฉด ์์์ ๋/์๋ฌธ์์ ๋ถ์์ ๋ฌด์ํ๊ณ ํญ๋ชฉ์ ์๋ ํํ๋ฅผ ๋จผ์ ๋น๊ตํฉ๋๋ค. ๋์ผํ ํญ๋ชฉ์ผ๋ก ํ๊ฐ๋๋ฉด ์์์ ๋ถ์์ ๋น๊ตํฉ๋๋ค. ์ฌ๊ธฐ์๋ ์์์ด ๋์ผํ ํญ๋ชฉ์ผ๋ก ํ๊ฐ๋๋ฉด ์์์ ๋/์๋ฌธ์, ๋ฌธ์ ๋๋น ๋ฐ ์ผ๋ณธ์ด ๊ฐ๋ ์ฐจ์ด๋ฅผ ๋น๊ตํฉ๋๋ค. | 1,988 | 2,383 | {"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} | 2.59375 | 3 | CC-MAIN-2024-30 | latest | en | 0.173434 |
https://ditaxml.org/bloomfield/university-physics-solution-manual-chapter-11.php | 1,643,064,134,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00158.warc.gz | 267,165,865 | 8,403 | chapter_11 University Physics 13e(Young/Freedman Chapter Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Oct 11, 2019. Mahander Oad …
## Chapter 11 Solutions University Physics With Modern
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11. a. 139 cm 2.3 cm 320 cm2 or 3.2 102 cm2 b. 3.2145 km 4.23 km 13.6 km2 12. a. 13.78 g 11.3 mL 1.22 g/mL b. 18.21 g 4.4 cm3 4.1 g/cm3 Section Review 1.1 Mathematics and Physics pages 3–10 page 10 13. Math Why are concepts in physics described with formulas? The formulas are concise and can be used to predict new data. 14. Magnetism The University Physics Solution Manual 14th Ed Chapter 1, Problem 1 Starting with the definition 1 in Newton's 3rd Law Lab Procedure for 11 University Physics by Troy Tennant.
Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet.
Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who
Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN
Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit 11. a. 139 cm 2.3 cm 320 cm2 or 3.2 102 cm2 b. 3.2145 km 4.23 km 13.6 km2 12. a. 13.78 g 11.3 mL 1.22 g/mL b. 18.21 g 4.4 cm3 4.1 g/cm3 Section Review 1.1 Mathematics and Physics pages 3–10 page 10 13. Math Why are concepts in physics described with formulas? The formulas are concise and can be used to predict new data. 14. Magnetism The
Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet.
Sep 11, 2013 · Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m …
Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN
Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
exam physics formulas chapter 11 Flashcards Quizlet. Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use, Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m ….
### Physics 7502
exam physics formulas chapter 11 Flashcards Quizlet. Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m …, 11. a. 139 cm 2.3 cm 320 cm2 or 3.2 102 cm2 b. 3.2145 km 4.23 km 13.6 km2 12. a. 13.78 g 11.3 mL 1.22 g/mL b. 18.21 g 4.4 cm3 4.1 g/cm3 Section Review 1.1 Mathematics and Physics pages 3–10 page 10 13. Math Why are concepts in physics described with formulas? The formulas are concise and can be used to predict new data. 14. Magnetism The.
### Popular Videos University Physics - YouTube
exam physics formulas chapter 11 Flashcards Quizlet. Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning..
Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m … Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet.
Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m … Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet. Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality! Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who
Sep 11, 2013 · Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
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Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning. Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
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Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Oct 11, 2019. Mahander Oad … Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
## chapter_11 University Physics 13e(Young/Freedman Chapter
exam physics formulas chapter 11 Flashcards Quizlet. Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit, Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!.
### Chapter 11 Solutions University Physics With Modern
exam physics formulas chapter 11 Flashcards Quizlet. Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use, Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN.
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Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning. Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
Oct 15, 2015В В· Access solution manuals of Physics textbooks at CrazyforStudy.com, every solution manual is prepared by expert and highly qualified Physics tutors. ШЁШ§ШШ« ЩЃЩЉ University of Kufa at ISBN: 9780130352569 Solution Manual of PHYSICS PRINCIPLES WITH APPLICATIONS (6th Edition) Textbook written by вЂDouglas C Giancoli’. 17. ISBN Mar 15, 2018В В· Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m …
Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!
Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit 11. a. 139 cm 2.3 cm 320 cm2 or 3.2 102 cm2 b. 3.2145 km 4.23 km 13.6 km2 12. a. 13.78 g 11.3 mL 1.22 g/mL b. 18.21 g 4.4 cm3 4.1 g/cm3 Section Review 1.1 Mathematics and Physics pages 3–10 page 10 13. Math Why are concepts in physics described with formulas? The formulas are concise and can be used to predict new data. 14. Magnetism The
Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m … Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!
Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who
Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning. Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use
Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Oct 11, 2019. Mahander Oad … Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!
### chapter_11 University Physics 13e(Young/Freedman Chapter
Physics 7502. Oct 11, 2015 · University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning., Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit.
Popular Videos University Physics - YouTube. Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet., Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!.
### Physics 7502
Popular Videos University Physics - YouTube. University Physics Solution Manual 14th Ed Chapter 1, Problem 1 Starting with the definition 1 in Newton's 3rd Law Lab Procedure for 11 University Physics by Troy Tennant. Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning..
Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning. Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!
Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet. Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m … Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Oct 11, 2019. Mahander Oad …
Sep 11, 2013 · Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit University Physics Solution Manual 14th Ed Chapter 1, Problem 1 Starting with the definition 1 in Newton's 3rd Law Lab Procedure for 11 University Physics by Troy Tennant.
Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m … Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit
Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning. Learn exam physics formulas chapter 11 with free interactive flashcards. Choose from 500 different sets of exam physics formulas chapter 11 flashcards on Quizlet.
Access University Physics with Modern Physics 14th Edition Chapter 11 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality! Sep 11, 2013В В· Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who
Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane Paul Stanley Beloit College encouraged to refer students to the Student’s Solution Manual for these exercises and problems. E1-11 Lunar months are based on the Earth’s position, and as the Earth moves around the orbit Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Student Solutions Manual for University Physics, Volume 1 book. Read reviews from world’s largest community for readers. Oct 11, 2019. Mahander Oad …
Inha University Department of Physics Chapter 1. Problem Solutions 1. Inha University Department of Physics 11. A galaxy in the constellation UrsaMajor is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the at the beginning of this chapter, we can use Oct 11, 2015В В· University Physics, 13e (Young/Freedman) Chapter 11 Equilibrium and Elasticity 11.1 Conceptual Questions 1) If the torque on an object adds up to zero A) the forces on it also add up to zero. B) the object is at rest. C) the object cannot be turning. D) the object could be accelerating linearly but it could not be turning.
Mar 15, 2018 · Q1.11 Displacement is the distance between the starting point and ending point. When the bicyclist has made a half-circle she is a distance 2R 1000 m … Sep 11, 2013 · Contribute to sonhuytran/MIT8.01SC.2010F development by creating an account on GitHub. MIT8.01SC.2010F / References / University Physics with Modern Physics, 13th Edition Solutions Manual / M13_YOUN7066_13_ISM_C13.pdf. Find file Copy path unknown Lecture Notes for Chap 01, Exercises 1.1 -> 1.3 169ff4b Sep 11, 2013. 0 contributors. Users who
AbeBooks.com: Applied Petroleum Reservoir Engineering (3rd Edition) (9780133155587) by Terry, Ronald E.; Rogers, J. Brandon and a great selection of similar New, Used and Collectible Books available now at great prices. Applied petroleum reservoir engineering 3rd edition solution manual Torrance Title: Applied Petroleum Reservoir Engineering 3rd Edition Terry Solutions Manual Author: Terry Subject | 7,736 | 29,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.773043 |
https://philoid.com/question/15503-add-and-express-the-sum-as-a-mixed-fraction-i-and-ii-and-iii-and-iv-and | 1,686,060,806,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00230.warc.gz | 504,003,640 | 11,910 | ## Book: RD Sharma - Mathematics
### Chapter: 1. Rational Numbers
#### Subject: Maths - Class 8th
##### Q. No. 4 of Exercise 1.1
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
4
##### Add and express the sum as a mixed fraction:(i) and(ii) and(iii) and(iv) and
(i) The denominators of the given rational numbers 5 and 10 respectively.
The L.C.M of 5 and 10 is 10
Now,
We write the given rational numbers into forms in which both of them have the same denominator
=
And,
=
Therefore,
+ =
=
= 1
(ii) The denominators of the given rational numbers 7 and 4 respectively.
The L.C.M of 7 and 4 is 28
Now,
We write the given rational numbers into forms in which both of them have the same denominator
=
And,
=
Therefore,
- =
=
= 1
(iii) The denominators of the given rational numbers 6 and 8 respectively.
The L.C.M of 6 and 8 is 24
Now,
We write the given rational numbers into forms in which both of them have the same denominator
=
And,
=
Therefore,
- =
=
= -8
(iv) The denominators of the given rational numbers 6 and 8 respectively.
The L.C.M of 6 and 8 is 24
Now,
We write the given rational numbers into forms in which both of them have the same denominator
=
And,
=
Therefore,
+ =
=
= 17
1
2
3
4 | 350 | 1,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.938783 |
https://www.fatiando.org/verde/v1.0.0/tutorials/decimation.html | 1,721,537,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00413.warc.gz | 662,402,168 | 5,945 | # Data Decimation¶
Often times, raw spatial data can be highly oversampled in a direction. In these cases, we need to decimate the data before interpolation to avoid aliasing effects.
import matplotlib.pyplot as plt
import cartopy.crs as ccrs
import verde as vd
For example, our sample shipborne bathymetry data has a higher sampling frequency along the tracks than between tracks:
# Load the data as a pandas.DataFrame
data = vd.datasets.fetch_baja_bathymetry()
# Plot it using matplotlib and Cartopy
crs = ccrs.PlateCarree()
plt.figure(figsize=(7, 7))
ax = plt.axes(projection=ccrs.Mercator())
ax.set_title("Locations of bathymetry measurements from Baja California")
# Plot the bathymetry data locations as black dots
plt.plot(data.longitude, data.latitude, ".k", markersize=1, transform=crs)
vd.datasets.setup_baja_bathymetry_map(ax)
plt.tight_layout()
plt.show()
Class verde.BlockReduce can be used to apply a reduction/aggregation operation (mean, median, standard deviation, etc) to the data in regular blocks. All data inside each block will be replaced by their aggregated value. BlockReduce takes an aggregation function as input. It can be any function that receives a numpy array as input and returns a single scalar value. The numpy.mean or numpy.median functions are usually what we want.
import numpy as np
Blocked means and medians are good ways to decimate data for interpolation. Let’s use a blocked median on our data to decimate it to our desired grid interval of 5 arc-minutes. The reason for using a median over a mean is because bathymetry data can vary abruptly and a mean would smooth the data too much. For data varies more smoothly (like gravity and magnetic data), a mean would be a better option.
reducer = vd.BlockReduce(reduction=np.median, spacing=5 / 60)
print(reducer)
Out:
BlockReduce(adjust='spacing', center_coordinates=False,
reduction=<function median at 0x150b196bc2f0>, region=None,
spacing=0.08333333333333333)
Use the filter method to apply the reduction:
coordinates, bathymetry = reducer.filter(
coordinates=(data.longitude, data.latitude), data=data.bathymetry_m
)
plt.figure(figsize=(7, 7))
ax = plt.axes(projection=ccrs.Mercator())
ax.set_title("Locations of decimated data")
# Plot the bathymetry data locations as black dots
plt.plot(*coordinates, ".k", markersize=1, transform=crs)
vd.datasets.setup_baja_bathymetry_map(ax)
plt.tight_layout()
plt.show()
By default, the coordinates of the decimated data are obtained by applying the same reduction operation to the coordinates of the original data. Alternatively, we can tell BlockReduce to return the coordinates of the center of each block:
reducer_center = vd.BlockReduce(
reduction=np.median, spacing=5 / 60, center_coordinates=True
)
coordinates_center, bathymetry = reducer_center.filter(
coordinates=(data.longitude, data.latitude), data=data.bathymetry_m
)
plt.figure(figsize=(7, 7))
ax = plt.axes(projection=ccrs.Mercator())
ax.set_title("Locations of decimated data using block centers")
# Plot the bathymetry data locations as black dots
plt.plot(*coordinates_center, ".k", markersize=1, transform=crs)
vd.datasets.setup_baja_bathymetry_map(ax)
plt.tight_layout()
plt.show()
Now the data are ready for interpolation.
Total running time of the script: ( 0 minutes 2.447 seconds)
Gallery generated by Sphinx-Gallery | 788 | 3,352 | {"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} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.74234 |
https://climate-cms.org/posts/2018-11-27-Value-when-other-is-max.html | 1,721,465,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00679.warc.gz | 137,258,529 | 13,053 | # Find the value of a variable at the times when another variable is maximum#
Claire Carouge, CLEX CMS
Let’s say you have two variables that vary in space and time. You can easily calculate the maximum of one variable for each spatial point across time. Now, you would like to know what are the values of the second variable at the same time the first reaches a maximum. And this, for each spatial point.
As we shall see, it is relatively easy to do with two variables of the same rank, and that is what the where() function is designed for. But, depending on which Python package you are using, this problem can be more complicated with variables of different ranks, or not! In this blog, we will look at both scenarios.
We first need to import some packages, read in some data and calculate the maximum in time at all spatial points. As usual, I’m going to use CMIP data as it’s easily accessible.
%matplotlib inline
import xarray as xr
import numpy as np
# Let's get the 2m temperature and the sensible heat flux.
# We do not want to decode the time unit.
# as datetime objects can't be plotted.
ds = xr.open_dataset('/g/data/rr3/publications/CMIP5/output1/CSIRO-BOM/ACCESS1-0/historical/mon/atmos/Amon/r1i1p1/latest/tas/tas_Amon_ACCESS1-0_historical_r1i1p1_185001-200512.nc',
decode_times=False)
ds1 = xr.open_dataset('/g/data/rr3/publications/CMIP5/output1/CSIRO-BOM/ACCESS1-0/historical/mon/atmos/Amon/r1i1p1/latest/hfss/hfss_Amon_ACCESS1-0_historical_r1i1p1_185001-200512.nc',
decode_times=False)
tas = ds.tas
hfss = ds1.hfss
tas
<xarray.DataArray 'tas' (time: 1872, lat: 145, lon: 192)>
[52116480 values with dtype=float32]
Coordinates:
* time (time) float64 6.753e+05 6.754e+05 ... 7.323e+05 7.323e+05
* lat (lat) float64 -90.0 -88.75 -87.5 -86.25 ... 86.25 87.5 88.75 90.0
* lon (lon) float64 0.0 1.875 3.75 5.625 7.5 ... 352.5 354.4 356.2 358.1
height float64 ...
Attributes:
standard_name: air_temperature
long_name: Near-Surface Air Temperature
units: K
cell_methods: time: mean
cell_measures: area: areacella
history: 2012-01-18T23:37:46Z altered by CMOR: Treated scalar d...
associated_files: baseURL: http://cmip-pcmdi.llnl.gov/CMIP5/dataLocation...
hfss
<xarray.DataArray 'hfss' (time: 1872, lat: 145, lon: 192)>
[52116480 values with dtype=float32]
Coordinates:
* time (time) float64 6.753e+05 6.754e+05 ... 7.323e+05 7.323e+05
* lat (lat) float64 -90.0 -88.75 -87.5 -86.25 ... 86.25 87.5 88.75 90.0
* lon (lon) float64 0.0 1.875 3.75 5.625 7.5 ... 352.5 354.4 356.2 358.1
Attributes:
standard_name: surface_upward_sensible_heat_flux
long_name: Surface Upward Sensible Heat Flux
units: W m-2
cell_methods: time: mean
cell_measures: area: areacella
history: 2012-01-15T11:36:06Z altered by CMOR: replaced missing...
associated_files: baseURL: http://cmip-pcmdi.llnl.gov/CMIP5/dataLocation...
# Calculate the maximum temperature at each spatial point along the time axis:
tas_max=tas.max('time')
tas_max
<xarray.DataArray 'tas' (lat: 145, lon: 192)>
array([[258.12787, 258.12787, 258.12787, ..., 258.12787, 258.12787, 258.12787],
[259.0565 , 259.0456 , 259.02908, ..., 259.0787 , 259.07742, 259.07217],
[259.79144, 259.71857, 259.64832, ..., 260.0159 , 259.9404 , 259.86655],
...,
[273.00952, 273.0114 , 273.0609 , ..., 273.01083, 273.00726, 273.0072 ],
[273.2563 , 273.26328, 273.27258, ..., 273.25238, 273.25394, 273.25162],
[273.18732, 273.18732, 273.18732, ..., 273.18732, 273.18732, 273.18732]],
dtype=float32)
Coordinates:
* lat (lat) float64 -90.0 -88.75 -87.5 -86.25 ... 86.25 87.5 88.75 90.0
* lon (lon) float64 0.0 1.875 3.75 5.625 7.5 ... 352.5 354.4 356.2 358.1
height float64 ...
## Variables of the same dimensionality#
Now we want to find the values of the sensible heat flux when the 2m temperature is maximum.
# Get the values of the sensible heat flux when temperature is maximum
hfss_at_max= hfss.where(tas == tas_max)
hfss_at_max.mean('time').plot(size=8)
hfss_at_max
CPU times: user 942 ms, sys: 3.44 s, total: 4.38 s
Wall time: 4.78 s
As you can see, hfss_at_max has the same dimensions as the initial arrays (hfss and tas) but with a lot of missing values. Only the times and spatial points where tas is maximum will have values. So you need to then sample in some way that makes sense for your analysis to plot it. Here we arbitrarily chose to plot the mean in time.
## Variables of different dimensionality#
It really depends a lot on what object you are using. It is really easy with xarray arrays and not so much with numpy arrays
To illustrate this, we are going to look at the problem of finding at what times the 2m temperature is maximum at each point
### Solution with xarray#
In this case, it is as simple as the previous case with variables of the same rank. Note that the result is a full 3D array with lots of missing values (NaN). There are only values when tas is at a maximum for that point in space.
time_at_max=tas.time.where(tas == tas_max)
time_at_max.mean('time').plot(size=8)
time_at_max
<xarray.DataArray (time: 1872, lat: 145, lon: 192)>
array([[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
...,
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]]])
Coordinates:
* time (time) float64 6.753e+05 6.754e+05 ... 7.323e+05 7.323e+05
height float64 1.5
* lat (lat) float64 -90.0 -88.75 -87.5 -86.25 ... 86.25 87.5 88.75 90.0
* lon (lon) float64 0.0 1.875 3.75 5.625 7.5 ... 352.5 354.4 356.2 358.1
Attributes:
bounds: time_bnds
units: days since 0001-01-01
calendar: proleptic_gregorian
axis: T
long_name: time
standard_name: time
Note: The time unit is number of days since 0001-01-01, so one would have to convert to a more usable format for scientific usage
## Solution for numpy arrays#
We’ll use tas and tas.time again but without using the xarray built-in methods. If I try the numpy equivalent solution:
nptime_at_max = np.where(tas == tas_max, tas.time, np.nan)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-8-027b2651c8a9> in <module>()
----> 1 nptime_at_max = np.where(tas == tas_max, tas.time, np.nan)
ValueError: operands could not be broadcast together with shapes (1872,145,192) (1872,) ()
This does not work as numpy (and xarray) can only perform a where() operation on arrays of the same shape. When provided with arrays of different shapes (like in this case), both numpy and xarray will try to make them conform with each other by expanding the smallest arrays to the shape of the biggest array. The values of the smallest arrays are copied across all missing dimensions. This process is called broadcasting.
The problem is numpy is quite conservative in its broadcasting rules and can not perform it in the case above. xarray is much better at broadcasting as it uses all the metadata stored in the DataArray to identify the dimensions in each array. That is the reason why DataArray.where() worked above.
If you have numpy arrays, the best solution is to quickly transform your numpy arrays into xarray DataArrays. You only need to name the dimensions. Any name will do, as long as you give the same name for the common dimensions in the :
#DataArray.values will return only the numpy array with the values and none of the metadata stored in the DataArray
new_tas = xr.DataArray(tas.values, dims=('t','l','L'))
new_time = xr.DataArray(tas.time.values, dims='t')
new_max = new_tas.max(dim='t')
new_time_at_max = new_time.where(new_tas == new_max)
new_time_at_max
<xarray.DataArray (t: 1872, l: 145, L: 192)>
array([[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
...,
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]]])
Dimensions without coordinates: t, l, L
As you see, you can keep your values and the time in separate arrays (new_tas and new_time). You don’t need to add the time array as a coordinate to the 3D array. Although it can be a good idea to do it in general as it keeps the data self-describing.
## Extension#
The other advantage of using xarray is it allows you to extend the functionalities beyond the built-in functions a lot more easily. The idea here is to use the groupby().apply() workflow to apply a user-defined function. You could obviously use it as a solution to the current problem:
def check_max(data):
return np.where(data == tas_max, data.time, np.nan)
tasmax_dates = tas.groupby('time').apply(check_max)
tasmax_dates.mean('time').plot(size=8)
tasmax_dates
<xarray.DataArray 'tas' (time: 1872, lat: 145, lon: 192)>
array([[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
...,
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]],
[[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan],
...,
[nan, nan, ..., nan, nan],
[nan, nan, ..., nan, nan]]])
Coordinates:
* lat (lat) float64 -90.0 -88.75 -87.5 -86.25 ... 86.25 87.5 88.75 90.0
* lon (lon) float64 0.0 1.875 3.75 5.625 7.5 ... 352.5 354.4 356.2 358.1
height float64 ...
* time (time) float64 6.753e+05 6.754e+05 ... 7.323e+05 7.323e+05
But this is slower than using the built-in functions directly. And it doesn’t keep the attributes (like time_bnds, units, calendar, etc)! It is then best to keep this approach for more complex problems that can not be easily solved otherwise. | 3,239 | 10,386 | {"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} | 3.015625 | 3 | CC-MAIN-2024-30 | latest | en | 0.758323 |
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Question From class 12 Chapter JEE MAINS
# Let and let , be circles with radii , respectively. Assume that and touch externally for . It is also given that the x-axis and the line are tangential to each of the circle . Then are in (A) an arithmetic progression with common difference (B) a geometric progression with common ratio (C) an arithmetic progression with common difference (D) a geometric progression with common ratio
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If a,b,c are in geometric progression and a,2b,3c are in arithmetic progression, then what is the common ratio r such that ?
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Let P be the principal and the rate of interest be for first year for second year for third year and so on and in the last for the nth year. Then the amount A and the compound interest C.I. at the end of n years are given by respectively.
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Microconcepts | 845 | 3,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-16 | latest | en | 0.900714 |
https://stats.stackexchange.com/questions/44194/definition-of-stationary-distribution-in-continuous-time-markov-chains/171109 | 1,582,986,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00364.warc.gz | 563,521,472 | 32,376 | # Definition of stationary distribution in continuous time markov chains
I found the following definition:
"A probabilitly distribution $\pi = \{\pi_x\}_{x \in S}$ on the state space $S$ is called a stationary distribution for the Markov chain if for every $t > 0$,
$$\pi^T P_t = \pi^T$$
What does $P_t$ mean? I thought it was the t'th step matrix of the transition matrix P but then this would be for discrete time markov chains and not continuous, right?
Oh wait, is it the transition matrix at time t?
You can always get a continuous time version of a discrete one by simply "Poissonizing" it. For example, if you have a discrete time Markov chain with transition matrix $T$ you get a continuous time version by considering $$P_t = \sum_{n\geq 0} \frac{t^n}{n!}\exp(-t)T^n$$ Hence the above definition makes sense in the context of continuous time Markov chains.
• Yeah, but what does $P_t$ mean? Is it the t'th step transition matric or the transition matrix at time t? Or are these two things the same? – Kaish Nov 22 '12 at 15:10
• $T^n$ is the $n$-th step transition matrix, i.e. the probability of finding the walker in state $y$ after $n$ steps, given it has started in state $x$ is $(y,T^n x)$. Similarly, the probability of being in state $y$ after time $t$ and starting at $x$ is $(y,P_t x)$. – Sven Stodtmann Nov 22 '12 at 15:53
I am answering rather than commenting due to lack of reputation:
Sven, your claim is incorrect: in your expression, T must be an infinitesimal rate matrix whose rows sum to 0, not a transition matrix whose rows sum to 1.
With CTMCs, different things happen to those who wait longer. $P_t$ denotes a transition matrix between observations at time $t_0$ and time $t_0 + t$. When $t$ goes to $0$, it approaches the identity matrix, and when $t$ goes to infinity, it approaches a matrix where every row in $\pi$. Those claim may require regularity conditions such as irreducibility. My entire answer also assumes the process is time-homogenous, i.e. $P_t$ doesn't depend on $t_0$. | 546 | 2,028 | {"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} | 3.703125 | 4 | CC-MAIN-2020-10 | latest | en | 0.916011 |
https://www.thepoorcoder.com/hackerrank-lazy-evaluation-solution/ | 1,717,061,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00598.warc.gz | 898,697,630 | 14,578 | # Hackerrank Lazy Evaluation Solution
Lazy evaluation is an evaluation strategy that delays the assessment of an expression until its value is needed.
Ruby introduced a lazy enumeration feature. Lazy evaluation increases performance by avoiding needless calculations, and it has the ability to create potentially infinite data structures.
Example:
power_array = -> (power, array_size) do
1.upto(Float::INFINITY).lazy.map { |x| x**power }.first(array_size)
end
puts power_array.(2 , 4) #[1, 4, 9, 16]
puts power_array.(2 , 10) #[1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
puts power_array.(3, 5) #[1, 8, 27, 64, 125]
In this example, lazy avoids needless calculations to compute power_array.
If we remove lazy from the above code, then our code would try to compute all ranging from to Float::INFINITY.
To avoid timeouts and memory allocation exceptions, we use lazy. Now, our code will only compute up to first(array_size).
Your task is to print an array of the first palindromic prime numbers.
For example, the first palindromic prime numbers are .
Input Format.
A single line of input containing the integer .
Constraints
You are not given how big is.
Output Format.
Print an array of the first palindromic primes.
Sample Input.
5
Sample Output.
[2, 3, 5, 7, 11]
### Solution in ruby
Approach 1.
# Enter your code here. Read input from STDIN. Print output to STDOUT
require 'prime'
n = gets.to_i
p Prime.each.lazy.select {|i| i.to_s == i.to_s.reverse}.first(n)
Approach 2.
# Enter your code here. Read input from STDIN. Print output to STDOUT
require 'prime'
def yield_var(var_num)
sum = (2..Float::INFINITY).lazy.select{
|x| x.to_s.reverse.to_i == x && Prime.prime?(x) }.first(var_num.to_i)
return sum
end
num = gets.chomp
print yield_var(num)
Approach 3.
# Enter your code here. Read input from STDIN. Print output to STDOUT
def prime?(num)
num >= 2 && (2..Math.sqrt(num)).none? { |div| (num % div).zero? } ? true : false
end
def palindrome?(num)
num.to_s == num.to_s.reverse
end
p 2.upto(Float::INFINITY).lazy.select { |num| palindrome?(num) && prime?(num) }
.first(gets.to_i) | 601 | 2,124 | {"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} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.619531 |
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A097841 First differences of Chebyshev polynomials S(n,83) = A097839(n) with Diophantine property. 5
1, 82, 6805, 564733, 46866034, 3889316089, 322766369353, 26785719340210, 2222891938868077, 184473245206710181, 15309056460218076946, 1270467212952893676337, 105433469618629957059025 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS (9*b(n))^2 - 85*a(n)^2 = -4 with b(n)=A097840(n) give all positive solutions of this Pell equation. LINKS Indranil Ghosh, Table of n, a(n) for n = 0..520 Tanya Khovanova, Recursive Sequences Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16. Index entries for linear recurrences with constant coefficients, signature (83, -1). Index entries for sequences related to Chebyshev polynomials. FORMULA a(n) = ((-1)^n)*S(2*n, 9*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials. G.f.: (1-x)/(1 - 83*x + x^2). a(n) = S(n, 83) - S(n-1, 83) = T(2*n+1, sqrt(85)/2)/(sqrt(85)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120. a(n) = 83*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=82. - Philippe Deléham, Nov 18 2008 EXAMPLE All positive solutions of Pell equation x^2 - 85*y^2 = -4 are (9=9*1,1), (756=9*84,82), (62739=9*6971,6805), (5206581=9*578509,564733), ... MATHEMATICA CoefficientList[Series[(1-x)/(1-83x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *) LinearRecurrence[{83, -1}, {1, 82}, 20] (* G. C. Greubel, Jan 13 2019 *) PROG (PARI) my(x='x+O('x^20)); Vec((1-x)/(1-83*x+x^2)) \\ G. C. Greubel, Jan 13 2019 (Magma) m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-83*x+x^2) )); // G. C. Greubel, Jan 13 2019 (Sage) ((1-x)/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019 (GAP) a:=[1, 82];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019 CROSSREFS Sequence in context: A252705 A239670 A292423 * A116123 A116142 A054214 Adjacent sequences: A097838 A097839 A097840 * A097842 A097843 A097844 KEYWORD nonn,easy AUTHOR Wolfdieter Lang, Sep 10 2004 STATUS approved
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Last modified August 12 04:50 EDT 2024. Contains 375085 sequences. (Running on oeis4.) | 1,016 | 2,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.53964 |
http://www.jiskha.com/display.cgi?id=1286830925 | 1,498,598,380,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321553.70/warc/CC-MAIN-20170627203405-20170627223405-00142.warc.gz | 551,340,764 | 3,844 | # college physics
posted by .
Imagine that your car can travel 25 mi on 1 gal of gas when traveling on a level road at 55mi/h. Estimate your gas mileage while traveling at 55 mi/h up an incline that gains 3m in elevation for every 100 m one goes forward along the incline. Assume that your car has a mass of 1500 kg.
• college physics -
At a playground, a 19 kg child plays on a slide that drops through a height of 2.3. The child starts at rest at the top of the slide. On the way down, the slide does a nonconervative work of -361 J on the child. What is the child's speed at the bottom of the slide? | 157 | 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} | 3.21875 | 3 | CC-MAIN-2017-26 | latest | en | 0.911256 |
https://nrich.maths.org/10424/note | 1,656,696,905,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00652.warc.gz | 472,313,424 | 5,616 | ### Pebbles
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Bracelets
Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads?
### Sweets in a Box
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
# Multiply Multiples 2
## Multiply Multiples 2
In the equation below each square represents a missing digit:
One possible solution is:
Can you work out some different ways to balance the equation?
#### Why do this activity?
This problem provides the children with an opportunity to practise multiplying a single digit number by a multiple of 100. It also reinforces learning about equations being balanced and may lead to conversations about common factors. It encourages children to record their results, notice patterns and make predictions.
#### Possible approach
It might be good to let the class explore the problem first before trying to encourage systematic working. The class will hopefully discover issues such as 600 x 1 = 600 x 1 (does this count?!) and 200 x 6 = 300 x 4 = 400 x 3 = 600 x 2 (where there is more than one option). It may be appropriate to then discuss how the class is going to record solutions.
Working systematically might involve starting with the lowest values possible. 100 x 1 = 100 x 1. With 100 on each side of the equals sign, can this be made in any other way? Decide as a class if this example actually counts as both sides of the equation are identical. Then move on to 100 x 2 = 200 x 1. Is there another way of making 200? Then try 100 x 3, 100 x 4 and so on.
Alternatively, you might encourage the class to work through all of the options for 100x and then move on to 200 x 1, 200 x 2, 200 x 3 ”¦
Organising results according to how many ways in which the equation can be balanced will support the children to notice patterns and make conjectures.
Some children may move onto the extension tasks (below).
#### Key questions
Is there another way of balancing the equation? How many ways are there?
#### Possible extensions
Extension 1: Can you explain why there is sometimes just one way of balancing the equation and on other occasions there are lots of ways of doing it? What do you notice about the numbers?
Extension 2: Have a go at Multiply Multiples 3 which involves multiplying a multiple of 10 by another multiple of 10.
#### Possible support
Learners might like to have a go at Multiply Multiples 1 before tackling this activity.
Some children may benefit from using Diennes (base 10) apparatus and physically making the amounts. For example, 1200 (12 hundreds) can be organised as 600 x 2, 400 x 3, 300 x 4 and 200 x 6.
Children may be provided with multiplication grids or calculators as appropriate. | 688 | 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} | 4.28125 | 4 | CC-MAIN-2022-27 | latest | en | 0.922249 |
https://studysoup.com/tsg/24459/college-physics-1-edition-chapter-13-problem-4cq | 1,611,454,428,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00732.warc.gz | 579,232,505 | 11,798 | ×
Get Full Access to College Physics - 1 Edition - Chapter 13 - Problem 4cq
Get Full Access to College Physics - 1 Edition - Chapter 13 - Problem 4cq
×
# If you add boiling water to a cup at room temperature,
ISBN: 9781938168000 42
## Solution for problem 4CQ Chapter 13
College Physics | 1st Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
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College Physics | 1st Edition
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2
Problem 4CQ
If you add boiling water to a cup at room temperature, what would you expect the final equilibrium temperature of the unit to be? You will need to include the surroundings as part of the system. Consider the zeroth law of thermodynamics.
Step-by-Step Solution:
Step 1 of 3
Solution 4CQ
According to the zeroth law of thermodynamics, when two objects with different temperatures are in thermal contact, the heat flows from hot object to the colder object till the temperature of both objects become same.
Since the temperature of surrounding is less than the temperature of the boiling water, the heat energy flows from hot water to the surrounding air so that they reaches thermal equilibrium with each other if place for long time.
Step 2 of 3
Step 3 of 3
#### Related chapters
Unlock Textbook Solution | 319 | 1,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.850488 |
https://infinitylearn.com/surge/question/mathematics/area-ofparallelogrambcef--50-cm2-then-find-the-area-of-p/ | 1,685,758,636,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00113.warc.gz | 350,321,393 | 15,319 | Area of parallelogram BCEF = 50 cm2, then find the area of parallelogram BCDA.
# Area of parallelogram BCEF = 50 cm2, then find the area of parallelogram BCDA.
1. A
25 cm2
2. B
50 cm2
3. C
100 cm2
4. D
75 cm2
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### Solution:
Given, the area of parallelogram BCEF = 50 cm2
Both the parallelograms BCEF and BCDA are on the same base and between the same parallels. Hence, both the parallelograms will have the same area.
Therefore, the area of parallelogram BCDA = 50 cm2
## Related content
Distance Formula Perimeter of Rectangle Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formulae Volume of Cylinder Perimeter of Triangle Formula Area Formulae Volume Formulae | 224 | 838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-23 | latest | en | 0.68235 |
https://ch.mathworks.com/matlabcentral/cody/problems/1035-generate-a-vector-like-1-2-2-3-3-3-4-4-4-4/solutions/1710874 | 1,582,607,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146033.50/warc/CC-MAIN-20200225045438-20200225075438-00306.warc.gz | 308,374,592 | 15,867 | Cody
# Problem 1035. Generate a vector like 1,2,2,3,3,3,4,4,4,4
Solution 1710874
Submitted on 20 Jan 2019 by Daniel Turizo
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 2; y_correct = [1 2 2]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 5; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5]; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 10; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10]; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = 12; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ... 11 11 11 11 11 11 11 11 11 11 11 ... 12 12 12 12 12 12 12 12 12 12 12 12]; assert(isequal(your_fcn_name(x),y_correct))
5 Pass
x = 9; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9]; assert(isequal(your_fcn_name(x),y_correct))
6 Pass
x = 7; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7]; assert(isequal(your_fcn_name(x),y_correct))
7 Pass
x = 15; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ... 11 11 11 11 11 11 11 11 11 11 11 ... 12 12 12 12 12 12 12 12 12 12 12 12 ... 13 13 13 13 13 13 13 13 13 13 13 13 13 ... 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ... 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15]; assert(isequal(your_fcn_name(x),y_correct)) | 981 | 1,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-10 | latest | en | 0.320275 |
https://tawasul-med.com/find-936 | 1,674,958,523,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00066.warc.gz | 583,160,401 | 3,794 | # Math solver
Math solver can be a helpful tool for these students. So let's get started!
## The Best Math solver
Math solver can help students to understand the material and improve their grades. Math homework can be challenging, but it can also be rewarding. It can be a great way to learn new concepts and practice problem-solving skills. Of course, it is also important to get help when needed and to ask questions when you're stuck. With a little effort and perseverance, you can be successful in math homework.
Assuming you want a word problem to be solved: A plumber charges \$25 for a service call plus \$50 per hour of service. What will be the total cost for a 4 hour job? The total cost for the 4 hour job would be \$200.
When calculating a circle’s radius, you need to take into account both the radius of the circle’s circumference and the radius of its diameter. You can use this formula to solve for either or both: With these formulas, all you have to do is find the radius of each side in relation to the other one. You should also remember that the radius increases as your circle gets larger. If a circle has a radius of 1 unit, then its radius will double (or triple) as it grows from 1 unit in size. Once you know how much bigger a circle is than another one, you can calculate its diameter. Divide the first circle’s circumference by the second one’s diameter and multiply by pi to get the answer.
Algebra is one of the most important and valuable subjects any student can learn. But it can also be one of the hardest. That’s why it’s so important to have a good understanding of basic algebraic concepts before you even step foot into your first math class. When you’re first learning algebra, the best way to go about it is to break down each problem into smaller, more manageable pieces. This will make it easier to understand how each piece fits together and how they relate to each other. Another great way to make sure you understand what you’re doing when you’re solving algebra problems is to use a math calculator. They allow you to break down your problems into easy-to-understand steps and will save you time and frustration in the long run. If you find that you’re having a hard time with algebra, don’t hesitate to ask for help. It might just be that you need a refresher on some of the basics before diving into more complex problems.
A differential equation is an equation that relates a function with one or more of its derivatives. In order to solve a differential equation, we must first find the general solution, which is a function that satisfies the equation for all values of the variable. The general solution will usually contain one or more arbitrary constants, which can be determined by using boundary conditions. A boundary condition is a condition that must be satisfied by the solution at a particular point. Once we have found the general solution and determined the values of the arbitrary constants, we can substitute these values back into the solution to get the particular solution. Differential equations are used in many different areas of science, such as physics, engineering, and economics. In each case, they can help us to model and understand complicated phenomena.
## We cover all types of math problems
Very educational and helpful for people having a hard time with math, but let’s all be honest everyone who downloaded this app was bad at math lol. And very useful than a calculator. But the calculator is occasional so nice!
### Lilliana Barnes
This is a really good app, I have been struggling in math, and whenever I have late work, this app helps me! Plus, there is barely any ads! But you should really add a Eureka Math book thing for 1st, 2nd, 3rd, 4th, 5th, 6th grade, and more! | 792 | 3,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.941221 |
http://stackoverflow.com/questions/11341913/truth-tables-with-in-j | 1,433,277,377,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036266.4/warc/CC-MAIN-20150601214356-00098-ip-10-180-206-219.ec2.internal.warc.gz | 155,424,913 | 16,867 | # truth tables with /~ in J
I was experimenting with generating truth tables in J:
nand =: *:
nand /~ 0 1
1 1
1 0
bxor =: 22 b. NB. Built-in bitwise XOR
bxor /~ 0 1
0 1
1 0
Now I want to define my own logical xor, which I did like so:
xor =: 3 : 0
]y NB. monadic case is just the identity
:
(x*.-.y)+.(y*.-.x) NB. dyadic case is (x AND NOT y) OR (y AND NOT x)
)
This works as I expect when I call it directly.
0 xor 0 1
0 1
1 xor 0 1
1 0
But it doesn't generate a truth table:
xor /~ 0 1
0 0
Why not?
I thought maybe the problem was that ]/~ 0 1 itself produced a 1 x 2 array, so I changed the monadic part to use nand (*:y) because it produces the 2x2 array:
*:/~ 0 1
1 1
1 0
xor =: 3 : 0
*:y NB. certainly wrong, but at least has 2x2 shape.
:
(x*.-.y)+.(y*.-.x)
)
But I still get the same behavior:
xor /~ 0 1
0 0
Can someone help me understand the flaw in my thinking?
-
Update: I found ~:, which implements logical XOR, and ~:/~ 0 1 builds the truth table correctly... But I still don't understand why my hand-coded one doesn't play nice with /~. – tangentstorm Jul 5 '12 at 10:14
## 1 Answer
Your xor has infinite rank, while *:,~: have rank 0. You can verify that by using b.: v b. 0 like so:
~: b. 0
_ 0 0
*: b. 0
0 0 0
xor b. 0
_ _ _
What this means is that xor operates on the list 0 1 rather than on each individual atom 0, 1.
You will get the result you expect if you use xor with rank 0:
xor"0 /~ 0 1
0 1
1 0
Or if you define xor to be of rank 0.
-
See here and here for ranks. The v b. 0 above gives the rank for monadic, dyadic-left, dyadic-right case of the verb v, respectively. – Eelvex Jul 5 '12 at 11:28
Thanks! I'd been trying to understand what " did, too. Now I'm starting to see the pieces fit together. :) – tangentstorm Jul 5 '12 at 11:44 | 640 | 1,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | latest | en | 0.865095 |
https://stats.stackexchange.com/questions/369704/how-do-i-interpret-the-result-of-individual-survival-probabilities-in-survival-a | 1,718,863,088,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00562.warc.gz | 473,520,802 | 39,957 | # How do I interpret the result of individual survival probabilities in Survival Analysis in R?
I have developed a survival model using Cox Regression. The aim is to obtain individual probabilities of survival on a given date or time. Now the steps used are:
fit<-coxph(Surv(time,status=0)~$$X_1+X_2+X_3+X_4+.....+X_p$$)
summary(fit)
cox_fit<-survfit(fit)
cox_lp_predict<-predict(fit,type='lp')
basehaz_cox<-basehaz(fit)
individual_probability_at_t1<-as.data.frame(exp(-b_1)^(exp(cox_lp_predict)))
individual_probability_at_t2<-as.data.frame(exp(-b_2)^(exp(cox_lp_predict)))
fit<-coxph(Surv(time,status=0)~$$X_1+X_2+X_3+X_4+.....+X_p$$)Here status =0 because we want to analyse those observations where status=0, infact in my case status is churn flag=0$$=>$$ No churn,1$$=>$$Churn
cox_lp_predict<-predict(fit,type='lp')#This step is to find $$X_1\beta_1+X_2\beta_2+.....+x_p\beta_p$$ for each observation
basehaz_cox<-basehaz(fit)#Now let's say after obtaining the base hazard we get the cumulative base hazard i.e $$H_0(t)$$ at $$t=t_1$$ is $$=b_1$$
And let's say after obtaining the base hazard we get the cumulative base hazard using basehaz_cox<-basehaz(fit) i.e $$H_0(t)$$ at $$t=t_2$$ is $$=b_2$$
individual_probability_at_t1<-as.data.frame(exp(-b_1)^(exp(cox_lp_predict)))This is in line with the formula $$S(t|X)=exp(-\int_0^t h_0(t)dt)^{exp(\beta_1x_1+...\beta_px_p)}=S(t|X)=exp(-H_0(t))^{exp(\beta_1x_1+...\beta_px_p)}$$
Now my question is that when we obtain an appended dataframe i.e
Merged_individual_prob_at_t1_t2<-merge(individual_probability_at_t1$$Probability_at_t1,individual_probability_at_t2$$Probability_at_t2)
We get a table like this :
Serial No. Prob(T>230) Prob(T>360) Prob(T>500)
1 0.1125 0.3455 0.3221
2 0.2344 0.3244 0.2877
3 0.2556 0.4456 0.3211
And so on.
My question exactly here for certain observations like the ones I show above the survival probability increases for a certain duration of time and then it reduces like the above table. Mind it most of the observations show a decreasing survival probability over the three time instances calculated but for some of them this increasing trend for a certain period comes and then it again starts reducing. My question is why does this happen? Is my approach correct? If not rectify me anywhere,please help.
• The question as it is presented now is pretty messy. You should run diagnostics and figure out exactly where the logic breaks down and ask a specific question for that specific step. Commented Oct 2, 2018 at 7:45
• First how to obtain individual probabilities of each individual at time t?Second, how do I interpret the increase and then decrease in survival probabilities of certain individuals over time? Commented Oct 2, 2018 at 7:49
If you want to get individual probabilities from a fitted Cox model from the survival package in R, you can use the survfit() method, e.g.,
fm <- coxph(Surv(time, status == 0) ~ x1 + x2 + x3, data = your_data) | 875 | 3,036 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.619535 |
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