url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://brainly.my/tugasan/142738 | 1,485,013,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281151.11/warc/CC-MAIN-20170116095121-00547-ip-10-171-10-70.ec2.internal.warc.gz | 803,381,037 | 10,308 | # Lisa sold 81 magazine subscriptions which is was 27% of her class fundraising goal.how many magazine subscripitons does her class hope to cell
2
dari yatina0genWHI
## Jawapan
2016-03-19T21:29:19+08:00
81 times 100 divide by 27 as we want to know the total of the magazine
2016-03-20T15:59:11+08:00
If 81 represents 27 percent of the whole numbers then you need to take 81 divided by 27 times 100 or 81 times 100 divide by 26 :) | 137 | 432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-04 | longest | en | 0.909642 |
https://keisan.casio.com/exec/system/1244990194 | 1,553,648,473,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912207146.96/warc/CC-MAIN-20190327000624-20190327022624-00253.warc.gz | 523,698,151 | 8,027 | # Incomplete elliptic integral of the 3rd kind Π(x,n,k) Calculator
## Calculates the incomplete elliptic integral of the third kind Π(x,n,k).
$\Pi(x,n,k)={\large\int_{\small 0}^{\hspace{25}\small x}\frac{dt}{(1-nt^2)\sqrt{(1-t^2)(1-k^2t^2)}}}$
x -1≦x≦1 n k -1≦k*x≦1 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit
$\normal Incomplete\ elliptic\ integral\\\hspace{100} of\ the\ 3rd\ kind\ \Pi(x,n,k)\\[10](1)\ \Pi(x,n,k)={\large\int_{\small 0}^{\hspace{25}\small x}\frac{dt}{(1-nt^2)\sqrt{(1-t^2)(1-k^2t^2)}}}\\\hspace{220}-1\le x\le1\\$
Incomplete elliptic integral of the 3rd kind Π(x,n,k)
[1-1] /1 Disp-Num5103050100200
[1] 2014/09/09 03:23 Male / 60 years old level or over / Others / - /
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# Bar Diagram Math
ConceptDraw PRO extended with Divided Bar Diagrams solution from Graphs and Charts area of ConceptDraw Solution Park is the best software for quick and simple drawing the Divided Bar Diagrams and Bar Diagram Math.
## Basic Diagramming
Mathematical Drawing Software - Draw mathematical illustration diagrams easily from examples and templates!
## Mathematics Symbols
ConceptDraw PRO extended with Mathematics solution from the Science and Education area is a powerful diagramming and vector drawing software that offers all needed tools for mathematical diagrams designing.
Mathematics solution provides 3 libraries with predesigned vector mathematics symbols and figures:
Solid Geometry Library, Plane Geometry Library and Trigonometric Functions Library.
## Mathematical Diagrams
ConceptDraw PRO diagramming and vector drawing software extended with Mathematics solution from the Science and Education area is the best for creating: mathematical diagrams, graphics, tape diagrams various mathematical illustrations of any complexity quick and easy.
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## Multi Layer Venn Diagram. Venn Diagram Example
To visualize the relationships between subsets of the universal set you can use Venn diagrams. To construct one, you should divide the plane into a number of cells using n figures. Each figure in the chart represents a single set of, and n is the number of represented sets. Splitting is done in a way that there is one and only one cell for any set of these figures, the points of which belong to all the figures from the set and do not belong to others. The plane on which the figures are represented, is the universal set U. Thus, the point which does not belong to any of the figures, belongs only to U.
## Venn Diagram
Venn diagrams are illustrations used in the branch of mathematics known as set theory. They show the mathematical or logical relationship between different groups of things (sets). A Venn diagram shows all the possible logical relations between the sets.
## Scientific Symbols Chart
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## Circles Venn Diagram
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## Simple Drawing Applications for Mac
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## Venn Diagram Template for Word
This template shows the Venn Diagram. It was created in ConceptDraw PRO diagramming and vector drawing software using the ready-to-use objects from the Venn Diagrams Solution from the "Diagrams" area of ConceptDraw Solution Park.
## 3 Circle Venn. Venn Diagram Example
This example shows the 3 Circle Venn Diagram. The Venn Diagrams visualize all possible logical intersections between several sets. On this example you can see the intersections of 3 sets. Venn Diagrams are widely used in mathematics, logic, statistics, marketing, sociology, etc.
## 3 Circle Venn Diagram. Venn Diagram Example
This template shows the Venn Diagram. It was created in ConceptDraw PRO diagramming and vector drawing software using the ready-to-use objects from the Venn Diagrams Solution from the "Diagrams" area of ConceptDraw Solution Park.
Venn Diagrams visualize all possible logical intersections between several sets and are widely used in mathematics, logic, statistics, marketing, sociology, etc.
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UML state machine's goal is to overcome the main limitations of traditional finite-state machines while retaining their main benefits.
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## Types of Flowchart - Overview
When designing a process or an instruction in clear way, you should consider creating a flowchart. A Process Flow Diagram is the method. You can avoid wasting a lot of time understanding complex concepts as they get clear with different diagrams.
How to Simplify Flow Charting
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## The Best Drawing Program for Mac
ConceptDraw PRO is the professional business graphic software for drawing diagrams and charts with great visual appeal on Mac OS X.
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The solutions from Marketing area of ConceptDraw Solution Park collect templates, samples and libraries of vector stencils for drawing the marketing diagrams and mind maps.
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ConceptDraw PRO creates drawings, diagrams and charts with great visual appeal in Mac OS X.
## Venn Diagrams
Venn Diagrams are the intuitive way to work with sets which allows to visualize all possible logical relations between several sets. They are widely used in mathematics, marketing, logic, statistics, sociology, etc.
ConceptDraw PRO diagramming and vector drawing software extended with Venn Diagrams solution from "Diagrams" area offers set of useful drawing tools for easy creating the Venn Diagrams for any quantity of sets. | 1,235 | 6,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.872041 |
https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map%3A_Organic_Chemistry_(Bruice)/28%3A_The_Chemistry_of_Nucleic_Acids/28.05%3A_The_Biosynthesis_of_DNA_is_Called_Replication | 1,502,920,109,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102663.36/warc/CC-MAIN-20170816212248-20170816232248-00637.warc.gz | 733,807,150 | 17,759 | # 28.5: The Biosynthesis of DNA is Called Replication
The rate of a chemical reaction is the change in concentration over the change in time.
### Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}}$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}}$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Example
For example, in the simple reaction $A + B \rightarrow C + D$ The reaction rate can be defined thusly:
• rate of disappearance of A $\text{rate}=-\dfrac{\Delta[A]}{\Delta{t}}$
• rate of disappearance of B $\text{rate}=-\dfrac{\Delta[B]}{\Delta{t}}$
• rate of formation of C $\text{rate}=\dfrac{\Delta[C]}{\Delta{t}}$
• rate of formation of D) $\text{rate}=\dfrac{\Delta[D]}{\Delta{t}}$
There are many factors that can either slow or speed up the rate of a chemical reaction such as temperature, pressure, concentration, and catalysts. The Rate of a Chemical Reaction is always positive. It can be confusing since the Rate of Disappearance is negative, however when you think about it, a rate should never be negative since the rate is describing how fast the concentration changes with time. The units for the rate is Molarity per Seconds (M/s).
### Reaction Rates from Non-unity Stoichiometric Coefficients
It does not matter whether an experimenter monitors the reagents or products. However, since reagents decrease during reaction, and products increase, there is a sign difference between the two rates. Reagent concentration decreases as the reaction proceeds, giving a negative number for the change in concentration. The products, on the other hand, increase concentration with time, giving a positive number. Since the convention is to express the rate of reaction as a positive number, to solve a problem, set the overall rate of the reaction equal to the negative of a reagent's disappearing rate.
The overall rate also depends on stoichiometric coefficients; consider the more general balanced equation
$$aA+bB \rightarrow cC + dD$$,
where the lower case letters represent the coefficients of the balanced equation and the upper case letters (i.e. A) represent the molecular concentration. As with the example above, the rate of reaction can be defined with respect to loss of reactants or gain of products:
• Rate of Disappearance of reactants: $- \dfrac{1}{a}\dfrac{\Delta [A]}{\Delta t} = - \dfrac{1}{b}\dfrac{\Delta [B]}{\Delta t}$
• Rate of Formation of product: $\dfrac{1}{c}\dfrac{\Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{\Delta [D]}{\Delta t}$
Since Rate of Disappearance and Rate of Formation are equal
$$- \dfrac{1}{a}\dfrac{\Delta [A]}{\Delta t} = - \dfrac{1}{b}\dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{\Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{\Delta [D]}{\Delta t}$$
It is worth noting that the process of measuring the concentration can be greatly simplified by taking advantage of the different physical or chemical properties (i.e.: phase difference, reduction potential, etc.) of the reagents or products involved in the reaction by using the above methods. We have emphasized the importance of taking the sign of the reaction into account in order to get a positive reaction rate. Now, we will turn our attention to the importance of stoichiometric coefficients.
Even though the concentrations of A, B, C and D may all change at different rates, there is only one average rate of reaction. To get this unique rate, choose any one rate and divide it by the stoichiometric coefficient. When the reaction has the formula:
$C_{R1}R_1 + \dots + C_{Rn}R_n \rightarrow C_{P1}P_1 + \dots + C_{Pn}P_n$
The general case of the unique average rate of reaction has the form:
rate of reaction = $$- \dfrac{1}{C_{R1}}\dfrac{\Delta [R_1]}{\Delta t} = \dots = - \dfrac{1}{C_{Rn}}\dfrac{\Delta [R_n]}{\Delta t} = \dfrac{1}{C_{P1}}\dfrac{\Delta [P_1]}{\Delta t} = \dots = \dfrac{1}{C_{Pn}}\dfrac{\Delta [P_n]}{\Delta t}$$
Example
For the reaction: $2A+B \rightarrow 2C + D$
1. find the reaction rate and
2. find the reaction rate
given $$\Delta [A] = 0.002,M$$ and $$\Delta t = 77\, s$$.
SOLUTION
1. rate of reaction = $- \dfrac{1}{2}\dfrac{\Delta [A]}{\Delta t} = - \dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{2}\dfrac{\Delta [C]}{\Delta t} =\dfrac{\Delta [D]}{\Delta t}$
2. rate of disappearance of A = $$- \dfrac{\Delta [A]}{\Delta t} = \dfrac{-0.002M }{\ 77 sec}$$ = -0.000026 M per sec
rate of reaction = $$- \dfrac{1}{a}$$ (rate of disappearance of A)= $$- \dfrac{1}{2}$$ (-0.000026 M per sec) = 0.000013 M per sec
### Average and Instantaneous Reaction Rate
Reaction rates have the general form of (change of concentration / change of time). There are two types of reaction rates. One is called the average rate of reaction, often denoted by (Δ[conc.] / Δt), while the other is referred to as the instantaneous rate of reaction, denoted as either:
$\lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t}$
which is the definition of the derivative
$\dfrac{d [concentration]}{dt}$
The average rate of reaction, as the name suggests, is an average rate, obtained by taking the change in concentration over a time period, for example: -0.3 M / 15 minutes. This is an approximation of the reaction rate in the interval; it does not necessarily mean that the reaction has this specific rate throughout the time interval or even at any instant during that time. The instantaneous rate of reaction, on the other hand, depicts a more accurate value. The instantaneous rate of reaction is defined as the change in concentration of an infinitely small time interval, expressed as the limit or derivative expression above. Instantaneous rate can be obtained from the experimental data by first graphing the concentration of a system as function of time, and then finding the slope of the tangent line at a specific point which corresponds to a time of interest. Alternatively, experimenters can measure the change in concentration over a very small time period two or more times to get an average rate close to that of the instantaneous rate. The reaction rate for that time is determined from the slope of the tangent lines.
### References
1. Petrucci et al. General Chemistry: Principles & Modern Applications, 9th Edition. New Jersey: Prentice-Hall Inc., 2007.
2. Connors, Kenneth. Chemical Kinetics: The Study of Reaction Rates in Solution. New York City: VCH Publishers, Inc., 1990.
### Problems
1. Consider the reaction $$2A + B \longrightarrow C$$. The concentration of [A] is 0.54321M and the rate of reaction is $$3.45 \times 10^{-6} M/s$$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $$A + B \longrightarrow C$$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
### Contributors
• Albert Law, Victoria Blanchard, Donald Le | 2,017 | 7,588 | {"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} | 3.625 | 4 | CC-MAIN-2017-34 | longest | en | 0.879952 |
https://www.convertunits.com/from/trillion+cubic+meter/to/decilitre | 1,607,180,504,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747887.95/warc/CC-MAIN-20201205135106-20201205165106-00008.warc.gz | 624,326,784 | 16,093 | ## ››Convert trillion cubic metre to deciliter
trillion cubic meter decilitre
How many trillion cubic meter in 1 decilitre? The answer is 1.0E-16.
We assume you are converting between trillion cubic metre and deciliter.
You can view more details on each measurement unit:
trillion cubic meter or decilitre
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 1.0E-12 trillion cubic meter, or 10000 decilitre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between trillion cubic meters and deciliters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of trillion cubic meter to decilitre
1 trillion cubic meter to decilitre = 1.0E+16 decilitre
2 trillion cubic meter to decilitre = 2.0E+16 decilitre
3 trillion cubic meter to decilitre = 3.0E+16 decilitre
4 trillion cubic meter to decilitre = 4.0E+16 decilitre
5 trillion cubic meter to decilitre = 5.0E+16 decilitre
6 trillion cubic meter to decilitre = 6.0E+16 decilitre
7 trillion cubic meter to decilitre = 7.0E+16 decilitre
8 trillion cubic meter to decilitre = 8.0E+16 decilitre
9 trillion cubic meter to decilitre = 9.0E+16 decilitre
10 trillion cubic meter to decilitre = 1.0E+17 decilitre
## ››Want other units?
You can do the reverse unit conversion from decilitre to trillion cubic meter, or enter any two units below:
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## ››Definition: Decilitre
A decilitre (or deciliter), abbreviated dL or dl, is one tenth of a litre, or 10^?4 m^3, or 100 millilitre. The SI prefix "deci" stands for one-tenth.
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# Primitive recursive function
Summary
In computability theory, a primitive recursive function is, roughly speaking, a function that can be computed by a computer program whose loops are all "for" loops (that is, an upper bound of the number of iterations of every loop can be determined before entering the loop). Primitive recursive functions form a strict subset of those general recursive functions that are also total functions. The importance of primitive recursive functions lies in the fact that most computable functions that are studied in number theory (and more generally in mathematics) are primitive recursive. For example, addition and division, the factorial and exponential function, and the function which returns the nth prime are all primitive recursive. In fact, for showing that a computable function is primitive recursive, it suffices to show that its time complexity is bounded above by a primitive recursive function of the input size. It is hence not that easy to devise a computable function that is not prim
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Let f(x)=x2 and g(x)=2x+1 be two real functions. Find (f + g), (f - g) (x), (fg) (x) and (fg)(x)
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Solution
(f+g):R→[0,∞) defined by (f+g)(x)=x2+2x+1=(x+1)2 (f−g):R→R defined by (f−g)(x)=x2−2x−1 (fg):R→R defined by (fg)(x)=2x3+x2 (fg):R→R defined by (fg)(x)=x22x+1
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Convert 1-DOF PID controller to 2-DOF controller
Description
example
C2 = make2DOF(C1) converts the one-degree-of-freedom PID controller C1 to two degrees of freedom. The setpoint weights b and c of the 2-DOF controller are 1, and the remaining PID coefficients do not change.
C2 = make2DOF(C1,b) specifies the setpoint weight for the proportional term.
example
C2 = make2DOF(C1,b,c) specifies the setpoint weights for both the proportional and derivative terms.
Examples
collapse all
Design a 1-DOF PID controller for a plant.
G = tf(1,[1 0.5 0.1]);
C1 = pidtune(G,'pidf',1.5)
C1 =
1 s
Kp + Ki * --- + Kd * --------
s Tf*s+1
with Kp = 1.12, Ki = 0.23, Kd = 1.3, Tf = 0.122
Continuous-time PIDF controller in parallel form.
Convert the controller to two degrees of freedom.
C2 = make2DOF(C1)
C2 =
1 s
u = Kp (b*r-y) + Ki --- (r-y) + Kd -------- (c*r-y)
s Tf*s+1
with Kp = 1.12, Ki = 0.23, Kd = 1.3, Tf = 0.122, b = 1, c = 1
Continuous-time 2-DOF PIDF controller in parallel form.
The new controller has the same PID gains and filter constant. It also contains new terms involving the setpoint weights b and c. By default, b = c = 1. Therefore, in a closed loop with the plant G, the 2-DOF controller C2 yields the same response as C1.
T1 = feedback(G*C1,1);
CM = tf(C2);
T2 = CM(1)*feedback(G,-CM(2));
stepplot(T1,T2,'r--')
Convert C1 to a 2-DOF controller with different b and c values.
C2_2 = make2DOF(C1,0.5,0.75)
C2_2 =
1 s
u = Kp (b*r-y) + Ki --- (r-y) + Kd -------- (c*r-y)
s Tf*s+1
with Kp = 1.12, Ki = 0.23, Kd = 1.3, Tf = 0.122, b = 0.5, c = 0.75
Continuous-time 2-DOF PIDF controller in parallel form.
The PID gains and filter constant are still unchanged, but the setpoint weights now change the closed-loop response.
CM_2 = tf(C2_2);
T2_2 = CM_2(1)*feedback(G,-CM_2(2));
stepplot(T1,T2_2,'r--')
Input Arguments
collapse all
1-DOF PID controller, specified as a pid object or a pidstd object.
Setpoint weight on proportional term, specified as a real, nonnegative, finite value. If you do not specify b, then C2 has b = 1.
Setpoint weight on derivative term, specified as a real, nonnegative, finite value. If you do not specify c, then C2 has c = 1.
Output Arguments
collapse all
2-DOF PID controller, returned as a pid2 object or pidstd2 object. C2 is in parallel form if C1 is in parallel form, and standard form if C1 is in standard form.
For example, suppose C1 is a continuous-time, parallel-form pid controller of the form:
${C}_{1}={K}_{p}+\frac{{K}_{i}}{s}+\frac{{K}_{d}s}{{T}_{f}s+1}.$
Then C2 is a parallel-form 2-DOF pid2 controller, which has two inputs and one output. The relationship between the inputs, r and y, and the output u of C2 is given by:
$u={K}_{p}\left(br-y\right)+\frac{{K}_{i}}{s}\left(r-y\right)+\frac{{K}_{d}s}{{T}_{f}s+1}\left(cr-y\right).$
The PID gains Kp, Ki, and Kd, and the filter time constant Tf are unchanged. The setpoint weights b and c are specified by the input arguments b and c, or 1 by default. For more information about 2-DOF PID controllers, see Two-Degree-of-Freedom PID Controllers.
The conversion also preserves the values of the properties Ts, TimeUnit, Sampling Grid, IFormula, and DFormula.
Version History
Introduced in R2015b | 1,058 | 3,324 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.832155 |
https://scsynth.org/t/bounded-loop-power-operator-in-synthdefs/8866 | 1,726,744,170,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652028.28/warc/CC-MAIN-20240919093719-20240919123719-00001.warc.gz | 461,156,228 | 12,112 | # Bounded Loop / "Power Operator" in SynthDefs?
Do we have a “Power Operator” that works in a SynthDef, somewhat equivalent to `⍣` in APL? For example as in `({1+÷⍵}⍣3)1`, applying one function (`{1+÷⍵}` in this case) n times (3 in this case) to one initial argument?
I don’t remember seeing anything other than n.do{}
``````4.do({ source = AllpassN.ar(source, 0.050, [Rand(0, 0.05), Rand(0, 0.05)], 1) })
``````
Not the same thing as it is (it would return a function if valid); something along those lines:
``````{AllpassN.ar(_,0.050,[Rand(0,0.05),Rand(0,0.05)],1)}⍣4
``````
I’ve seen some people playing with operators inside the SD, you guys might know that. (Or even cooler things)
I think we don’t.
As an aside, we (SC community) have a love-hate relationship with syntax sugar. Proposals for new syntax sugar are often met with, “oh yeah, that would be cool.” Then, some time down the road, the same syntax sugar may be held up as an example of “why SC is so confusing to learn – too many ways to do the same thing.”
So for me, whether a new syntax sugary thing makes the cut or not depends on how much of a usability improvement the new way is. If the existing way is painful and the new way is significantly easier (and if it’s a frequently-encountered bottleneck), then maybe go for it. If the existing way isn’t that bad and the new way is an incremental improvement (or a side case that doesn’t happen very often), maybe it isn’t necessary.
For me, writing the do loop isn’t so awful. Others’ mileage may vary – so I wouldn’t argue too strenuously against it. (Sure, this thread isn’t a proposal as such, fwiw.)
hjh
1 Like
This one is pretty easy to add actually because it is just a left–fold.
This is would be the long version using `inject`.
``````( {_ + 4}!n).inject(inital_value, {|acc, f| f.(acc) })
``````
You can then add an operator (or adverb) to Object as a shorthand (I’ll use `⍣` as an example here) , where the right-hand side is expected to be an event with the args in. These args can be validated by the member function.
``````inital_value ⍣ (n: 10, func: (_ + 4))
``````
You could invert this by extending `Function` too
``````(_ + 1) ⍣ (n: 10, init: 2)
``````
Just occasionally, the pure object oriented nature of smalltalk is quite a joy.
Personally, do think that these kinds of operators (piping and the like) should be included because they occur a lot in signal processing.
Here is an implementation for ` 1 |>.doN (\n: 10, \func: (_ + 0.1))`
``````
+Object {
^case
if(f.isKindOf(Event).not, {Error("doN requires an Event").throw});
if(f.size != 2, {Error("doN requires an Event of size 2").throw});
if(f.includesKey(\n).not, {Error("doN requires an Event with the key 'n' ").throw});
if(f[\n].isKindOf(Number).not, {Error("doN requires an Event with the key 'n' which is a Number").throw});
if(f.includesKey(\func).not, {Error("doN requires an Event with the key 'func'").throw});
if(f[\func].isKindOf(Function).not, {Error("doN requires an Event with the key 'func' which is a Function").throw});
({f[\func]}.dup(f[\n])).inject( this, {|acc, f| f.(acc) });
}
}
}
``````
2 Likes
Have a look at miSCellaneous_lib’s GFIS class. I wrote it for implementing Di Scipio’s FIS synthesis, but you can use it also for nested allpass filters. By the way, that hadn’t come to my mind yet, so thanks for the nudge.
In the concrete example there’s no real benefit in terms of typing. However GFIS has some options – e.g. returning different iteration levels – that might be interesting to explore.
``````
// variant with do loop
(
{
var source = Decay2.ar(Impulse.ar(1)) * Saw.ar(100, 0.1);
4.do({ source = AllpassN.ar(source, 0.05, [Rand(0, 0.05), Rand(0, 0.05)], 1) });
source
}.play
)
// same with GFIS
// different tone colours come from the random components
(
{
var source = Decay2.ar(Impulse.ar(1)) * Saw.ar(100, 0.1);
GFIS.ar({ |x| AllpassN.ar(x, 0.05, [Rand(0, 0.05), Rand(0, 0.05)], 1) }, source, 4)
}.play
)
// proof of concept: difference gives silence
// GFIS by default leaks DC (as it was designed with Agostino Di Scipio's FIS synthesis in mind)
// hence turn it off for check
(
{
var source = Decay2.ar(Impulse.ar(1)) * Saw.ar([100, 150], 0.1);
var gfis = GFIS.ar(
{ |x| AllpassN.ar(x, 0.05, [0.01, 0.011], 1) },
source,
4,
leakDC: false
);
4.do({ source = AllpassN.ar(source, 0.05, [0.01, 0.011], 1) });
source - gfis
}.play
)
``````
1 Like
I reckon this particular form of ⍣:
``````(({⍵*2}⍣3)2)≡256
``````
perhaps written as:
``````{|x|x**2}.iterate(3,2)==256
``````
is rather useful, particularly in the middle of a sequence of bindings, where one cannot easily write a do loop.
1 Like
Maybe this?
``````+ AbstractFunction {
iterate { |n, i|
var r = i;
n.do { r = this.(r)};
^r
}
}
/*
// (({⍵*2}⍣3)2)≡256
{|x|x**2}.iterate(3,2) == 256
-> true
*/
``````
It could also work with a implementation of a LazyList (I think a quark tried something like this already)
I’ll take a look at Di Scipio’s Iterated Nonlinear Functions, it sounds quite interesting. Thank you!
Nested allpass filters are not uncommon. It could be interesting if part of something more general.
FreeVerb, for example:
Yea, sure.
As you’ve noted, my suggestions do not involve altering the language itself.Given that SuperCollider’s sclang is currently experiencing conservative cycle, it hadn’t even crossed my mind.
I don’t have a love/hate relationship with this historical aspect of sclang. I just appreciate it. That’s how it is.
On a broader note, it’s important to recognize that what might initially seem like ‘syntactic sugar’ can become eventually a qualitative change. When applied extensively in a certain direction, it can be groundwork for new algebraic approach. See APL as a classic example of that. Another example: applications working with circuits use different ideas to represent processes on signals, which might include local loops etc. Those things are not far away from audio signals, and I’m sure they are used elsewhere.
1 Like
Sure, and FWIW, I wasn’t saying I would absolutely always oppose introducing a new method.
I think what I was driving at is that we have at times tended to add convenience methods in an ad-hoc way, without a broader plan, and this results in messy interfaces. What you’re talking about in this paragraph is a more organized process, which would be good.
hjh
1 Like
Yes, I’m aware of the nested allpass strategies, I just never used GFIS for that.
FWIW, here are some variants derived from the Schroeder ideas:
``````
//////////////////////////
REVERB WITH ALLPASS-FILTER
//////////////////////////
Interesting comment on the 1962 paper by Manfred SCHROEDER from 1962:
1.) Comb and Allpass
// Idea: by chosen params a Comb-Delay can be transformed to an Allpass-Delay:
// frequency response with comb delay -> multiples of 500 Hz 500 Hz
s.freqscope;
(
x = {
var in = WhiteNoise.ar(0.05);
var delayTime = 0.002; // ~500 Hz
var decayTime = 0.1;
var comb = CombL.ar(in, 0.2, delayTime, decayTime);
comb ! 2;
}.play
)
x.release
// flat frequency response ! (= Allpassfilter)
(
x = {
var in = WhiteNoise.ar(0.05);
var delayTime = 0.002; // ~500 Hz
var decayTime = 0.1;
var comb = CombL.ar(in, 0.2, delayTime, decayTime);
// calculate feedback gain
var fbGain = 0.001 ** (delayTime / decayTime).poll(0, fbGain);
// mix with correctl chosen params -> flat frequency response
comb * (1 - (fbGain ** 2)) - (in * fbGain) ! 2;
}.play
)
x.release
// compare with AllpassL -> identical (silence) !
(
x = {
var in = WhiteNoise.ar(0.05);
var delayTime = 0.002; // ~500 Hz
var decayTime = 0.1;
var comb = CombL.ar(in, 0.2, delayTime, decayTime);
// calculate feedback gain
var fbGain = 0.001 ** (delayTime / decayTime).poll(0, fbGain);
// difference = 0 !
comb * (1 - (fbGain ** 2)) - (in * fbGain) - AllpassL.ar(in, 0.2, delayTime, decayTime);
}.play
)
x.release
// allpass + source -> audible frequency (1/delaytime)
(
x = {
var in = WhiteNoise.ar(0.05);
var delayTime = 0.002; // ~500 Hz
var decayTime = 0.1;
var comb = CombL.ar(in, 0.2, delayTime, decayTime);
// calculate feedback gain
var fbGain = 0.001 ** (delayTime / decayTime).poll(0, fbGain);
// allpass + source
comb * (1 - (fbGain ** 2)) - (in * fbGain) + in ! 2;
// identical
// AllpassL.ar(in, 0.2, delayTime, decayTime) + in ! 2;
}.play
)
x.release
/////////////////////////////////////////////
2.) Schroeder Reverb I: serial allpass (P.221, 222)
(
s.options.blockSize = 1;
s.reboot;
)
(
// number of allpas filters in sequence
~num = 5;
~fxBus = Bus.audio(s, 2);
// force nicely distributed random numbers for deviation of allpassDelayFactor
// this seed worked ok for me, try others
~minCombDelayTime = 30;
~allpassDelayFactorDeviationMax = 0.1;
~allpassDelayFactorDeviations = { rand2(~allpassDelayFactorDeviationMax) } ! ~num + 1;
"~allpassDelayFactorDeviations: ".postln;
~allpassDelayFactorDeviations.do(_.postln);
SynthDef(\schroeder_I, { |outBus, amp = 1, initDelay = 30, cutoff = 7000, overallGain = 0.89|
var out, in = In.ar(~fxBus, 2);
var sig = in;
var maxDelay = 0.2;
// Schroeder suggestions
var allpassGain = 0.7;
var firstAllpassDelay = 30;
var allpassDelayFactor = 1/3;
var allpassDelay = firstAllpassDelay * 0.001;
var fb = LocalIn.ar(2);
sig = DelayL.ar(fb + sig, 0.2, initDelay * 0.001 - ControlDur.ir);
{ |i|
var decay;
(i != 0).if { allpassDelay = allpassDelay * allpassDelayFactor * ~allpassDelayFactorDeviations[i] };
decay = allpassDelay * log(0.001) / log(allpassGain);
sig = AllpassL.ar(sig, maxDelay, allpassDelay, decay);
} ! ~num;
// variant: damping of high frequencies
LocalOut.ar(overallGain * BHiShelf.ar(sig, cutoff, 1, -18));
out = (1 - (overallGain ** 2)) * sig - (in * overallGain);
Out.ar(outBus, out * amp)
specs: (
amp: [0, 1, \db, 0, 0.5],
initDelay: [2, 1000, 3, 0, 30], // in ms
overallGain: [0, 1, \lin, 0, 0.89],
cutoff: [200, 16000, \exp, 0, 7000]
)
)
SynthDef(\sawPerc, { |out, freq = 400, att = 0.01, rel = 0.1, amp = 0.1|
var env = EnvGen.ar(Env.perc(att, rel), doneAction: 2);
Out.ar(out, Saw.ar(freq * [1, 1.02], amp) * env)
)
// needs miSCellaneous_lib
\schroeder_I.sVarGui.gui(sliderWidth: 320, labelWidth: 120)
// alternative:
SynthDescLib.global[\schroeder_I].makeGui
(
x = Pbind(
\instrument, \sawPerc,
\dur, 0.2,
\note, Prand([0, 2, 4, 7, 9], inf),
\octave, 4,
\amp, 0.3,
\out, ~fxBus
).play
)
x.stop
// stop reverb in GUI
/////////////////////////////////////////////
2.) Schroeder Reverb II: 4 parallele Combs + 2 Allpass in Serie (S. 223)
(
~combNum = 4;
~allpassNum = 2; // if you change this, you must adapt allpassDelayTimes and allpassGains
~fxBus = Bus.audio(s, 2);
// force nicely distributed random numbers for comb delaytimes
// this seed worked ok for me, try others
~combDelayTimeFactors = { |i|
((~combDelayTimeSpread - 1 * i) + rand(~combDelayTimeSpread - 1)) / ~combNum + 1
} ! ~combNum;
"minCombDelayTime: ".post;
~minCombDelayTime.post;
" ms".postln;
"combDelayTimeFactors: ".postln;
~combDelayTimeFactors.do(_.postln);
SynthDef(\schroeder_II_vs_1, { |outBus, amp = 0.7, mix = 0.12, minCombDelay = 30, revTime = 1|
var minCombDelayTime = 30;
var combDelayTimes = ~combDelayTimeFactors * minCombDelay * 0.001;
// calculate from desired reverb time according to Schroeder
var combGains = 10 ** (-3 * combDelayTimes / revTime);
// fix allpass delaytime choices by Schroeder
var allpassDelayTimes = [5, 1.7] * 0.001;
var allpassGains = [0.7, 0.7];
var sig, out;
var in = In.ar(~fxBus, 2);
var maxDelay = 0.2;
var combDecayTimes = combDelayTimes * log(0.001) / log(combGains);
var allpassDecayTimes = allpassDelayTimes * log(0.001) / log(allpassGains);
// core:
// ~combNum parallel Combs for L and R ...
sig = { |i| CombL.ar(in[i], maxDelay, combDelayTimes, combDecayTimes).sum } ! 2;
// ... followed by 2 sequential Allpasses
{ |i| sig = AllpassL.ar(sig, maxDelay, allpassDelayTimes[i], allpassDecayTimes[i]) } ! ~allpassNum;
out = mix * sig - ((1 - mix) * in);
Out.ar(outBus, out * amp)
specs: (
amp: [0, 1, \db, 0, 0.7],
mix: [0, 1, 3, 0, 0.12],
minCombDelay: [1, 150, \lin, 0, 30],
revTime: [0, 15, 3, 0, 1]
)
)
SynthDef(\sawPerc, { |out, freq = 400, att = 0.01, rel = 0.1, amp = 0.1|
var env = EnvGen.ar(Env.perc(att, rel), doneAction: 2);
Out.ar(out, Saw.ar(freq * [1, 1.02], amp) * env)
)
// needs miSCellaneous_lib
\schroeder_II_vs_1.sVarGui.gui(sliderWidth: 320, labelWidth: 120)
// alternative:
SynthDescLib.global[\schroeder_II_vs_1].makeGui
// minCombDelay should be 30 (recommended by S.)
(
x = Pbind(
\instrument, \sawPerc,
\dur, 0.2,
\note, Prand([0, 2, 4, 7, 9], inf),
\octave, 4,
\amp, 0.3,
\out, ~fxBus
).play
)
x.stop
// stop reverb in GUI
(
~combNum = 4;
~fxBus = Bus.audio(s, 2);
// force nicely distributed random numbers for comb delaytimes
// this seed worked ok for me, try others
~minCombDelayTime = 30;
~combDelayTimeFactors = { |i|
((~combDelayTimeSpread - 1 * i) + rand(~combDelayTimeSpread - 1)) / ~combNum + 1
} ! ~combNum;
"minCombDelayTime: ".post;
~minCombDelayTime.post;
" ms".postln;
"combDelayTimeFactors: ".postln;
~combDelayTimeFactors.do(_.postln);
SynthDef(\schroeder_II_vs_2, { |outBus, amp = 0.7, mix = 0.12, revTime = 1, revCorr = 0.5, srcCorr = 0.2, cutoff = 3500|
var minCombDelayTime = 30;
var combDelayTimes = ~combDelayTimeFactors * minCombDelayTime * 0.001;
// calculate from desired reverb time according to Schroeder
var combGains = 10 ** (-3 * combDelayTimes / revTime);
// fix allpass delaytime choices by Schroeder
var allpassDelayTimes = [5, 1.7] * 0.001;
var allpassGains = [0.7, 0.7];
var sig, out;
var in = In.ar(~fxBus, 2);
var maxDelay = 0.2;
var combDecayTimes = combDelayTimes * log(0.001) / log(combGains);
var allpassDecayTimes = allpassDelayTimes * log(0.001) / log(allpassGains);
// core:
// dampened input, 4 parallel Combs for L and R ...
sig = { |i| CombL.ar(BHiShelf.ar(in[i], cutoff, 1, -18), maxDelay, combDelayTimes, combDecayTimes).sum } ! 2;
// ... followed by 2 sequential Allpasses
{ |i| sig = AllpassL.ar(sig, maxDelay, allpassDelayTimes[i], allpassDecayTimes[i]) } ! ~allpassNum;
// correlations for src and reverb
in = [
];
// better: SelectX.ar(srcCorr, [in, in.sum ! 2])
sig = [
];
// better: SelectX.ar(revCorr, [sig, sig.sum ! 2])
out = mix * sig - ((1 - mix) * in);
Out.ar(outBus, out * amp)
specs: (
amp: [0, 1, \db, 0, 0.7],
mix: [0, 1, 3, 0, 0.12],
revTime: [0, 15, 3, 0, 1],
revCorr: [0, 1, \lin, 0, 0.5],
srcCorr: [0, 1, \lin, 0, 0.2],
cutoff: [200, 16000, \exp, 0, 3500]
)
)
SynthDef(\sawPerc_2, { |out, freq = 400, att = 0.01, rel = 0.1, pan = 0, amp = 0.1|
var env = EnvGen.ar(Env.perc(att, rel), doneAction: 2);
Out.ar(out, Pan2.ar(Saw.ar(freq, amp), pan) * env)
)
// needs miSCellaneous_lib
\schroeder_II_vs_2.sVarGui.gui(sliderWidth: 320, labelWidth: 120)
// alternative:
SynthDescLib.global[\schroeder_II_vs_2].makeGui
(
x = Pbind(
\instrument, \sawPerc_2,
\dur, 0.2,
\note, Prand([0, 2, 4, 7, 9], inf) + [0, 16],
\octave, 4,
\pan, [-1, 1],
\amp, 0.3,
\out, ~fxBus
).play
)
x.stop
// stop reverb in GUI
``````
1 Like
Yes, a stream form is nice too, perhaps as below?
Which suggests placing the count second, so the “switch on nil” idiom works nicely:
``````{ |x| x ** 2 }.iterate(2, 3) == 256
{ |x| x ** 2 }.iterate(2).nextN(4) == [2, 4, 16, 256]
``````
Functon>>iterate:
`````` iterate { |anObject|
var state = anObject;
^FuncStream {
var next = state;
state = this.value(state);
next
} {
state = anObject
}
}
``````
Ps. You don’t need the `var r` above, you’re allowed to mutate `i`!
1 Like
Rohan @rdd , all of these implementations are very idiomatic, and fit perfectly into the sclang style.
I like them.
I know. It just didn’t feel right to mutate an argument named “initial”, or start with an argument named “result” as the initial value. That was just a naming thing here.
EDIT, OT: I was trying out your hsc3 project, and it seems you’re reworking the syntax for some things. Is it under revision right now?
Thanks for the reply, quite cool.
Actually it could be also an `unfoldr`, couldn’t it? Foldr reduces a list to a value, while unfoldr builds a list from a seed value. If you want to have the intermediate signals/number/values too.
``````unfoldr (\x -> if x == 0 then Nothing else Just (x, x-1)) 10
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
OR:
iterate :: (a -> a) -> a -> [a]
iterate f x = x : iterate f (f x)
``````
Something too would be a replicate, a map, and a reduce.
Something useful too would be a replicate, a map, and a reduce.
I’m sure you know this already, but replicate is !, map is collect, foldl1 is reduce?
Ie.
``````1!5 == [1,1,1,1,1]
[1,2,3].collect(_+1) == [2,3,4]
[1,2,3].reduce(_-_) == -4
``````
Cf.
``````replicate 5 1 == [1,1,1,1,1]
map (+ 1) [1,2,3] == [2,3,4]
foldl1 (-) [1,2,3] == -4
``````
But in any case, the first part of the diagram above would, in Sc, just be:
``````LBCF.ar(input, 0.84, 0.2, [1557, 1617, 1491, 1422, &etc.]).sum
``````
Ps. I don’t think there’s been much change at hsc3? Write me off list if anything isn’t working?
1 Like
Yes, of course. I was thinking as one operation. All that is simple programming, I’m just wondering about a simple “mapreduce” pattern, no claims of novelty no far))))
Going a bit further with the “mapReduce” idea, map can process in more different ways, or even analyzing. There could be a shuffle intermediate data, and reduce could also do a bit more, like adjusting levels or applying crossfades etc. Theoretically those kind of things could be done in parallel, but that’s not the case inside a SD> | 5,797 | 17,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.92914 |
http://fuchu-navi.com/6-4-puzzle-time-answer-key/ | 1,624,370,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00197.warc.gz | 17,686,014 | 15,027 | 6.4 Puzzle Time Answer Key. Write the letter of each answer in the box containing the exercise number. Note that the time in time piece is also 9 in this equation). For use before activity 6.4 36 points; She says that it does seem that those involved in spiritual practices are often the most resistant to the daily she likes the fact that students enjoy themselves, help each other, and learn at the same time. 0606 puzzletime linear quadratic and square roots. 100 ap w = substitute 45 for w and 80 for p. Check for the answer with the explanation below. Can you solve this simple math puzzle? 9 + 9 + 9 = 27 (so, time piece = 9 ; 6.3 puzzle time holdup in the yard when two clothespins held up a pair of pants 6.4 start thinking! Solve each system on a separate sheet of paper. Solution to maths puzzle that requires you to make 6 with 3 identical digits using different mathematical operations. Solve for a.you need to get 36 points. Time piece table penguin puzzle. For example how to make 8 8 8 into 6?
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Honors Geometry Chapter 6 Review Name ***Questions bec. For example how to make 8 8 8 into 6? Time piece table penguin puzzle. Solve for a.you need to get 36 points. Solution to maths puzzle that requires you to make 6 with 3 identical digits using different mathematical operations. 100 ap w = substitute 45 for w and 80 for p. Solve each system on a separate sheet of paper. Can you solve this simple math puzzle? Check for the answer with the explanation below. 6.3 puzzle time holdup in the yard when two clothespins held up a pair of pants 6.4 start thinking! Write the letter of each answer in the box containing the exercise number. 0606 puzzletime linear quadratic and square roots. For use before activity 6.4 36 points; 9 + 9 + 9 = 27 (so, time piece = 9 ; She says that it does seem that those involved in spiritual practices are often the most resistant to the daily she likes the fact that students enjoy themselves, help each other, and learn at the same time. Note that the time in time piece is also 9 in this equation).
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7come on, ___ you! 8not anonymous. The answer key for tenses exercise which is a dialogue. Complete your quiz offer with 100% accuracy and get credited. If we consider the given numbers 6, 4, 9, 10, 11 to be the month number then we can easily find the killer. Выберете задание unit 0 0.1 in class 0.2 i`m from. In the hardy boys series. Make your own custom word search with our free generator. She says that it does seem that those involved in spiritual practices are often the most resistant to the daily she likes the fact that students enjoy themselves, help each other, and learn at the same time. 4 businessmen 5 faxes 6 salaries 7 mice 8 viruses. Master any subject, one success at a time. What time?7 harry nine o'clock, i think. Seek time is the average time it takes the read/write heads to move and find data; 0606 puzzletime linear quadratic and square roots. Note that the time in time piece is also 9 in this equation). Write the letter of each answer in the box containing the exercise number. C what do you think about people prepare to. A 1 clients 2 keys 3 queries. New questions are added and answers are changed. Workbook 4answer key4 alex we've sold out of those,i'm afraid. 7come on, ___ you! 8not anonymous. Progress review (pages 30 & 31) 1 1 scary 5 lonely 2 naughty 6 cute 3 nervous 2 up 7 became 3 left 8 had 4 become 9 get 5 married workbook 2 answer key photocopiable © oxford university press 5 2 started 6 had 3 did; Transfer rate is the average speed required to transmit data from the disk to the cpu 5 true. Original shows and popular videos in different categories from producers and creators you love. 9 + 9 + 9 = 27 (so, time piece = 9 ; I (know) that i (want) to. One lap = time limit. For exercises, you can reveal the answers first. Check for the answer with the explanation below. Workbook 4answer key3 1 first2 girl3 got4 both5 common6 second / next7 all8 hasn't spoken9 happy4 students' own thatwould be great. These puzzles require you to determine the number and color of dots that belong in the tile with the question mark on it by reducing the possible answers one tile at a time. Leave 2 we definitely won't have time to go to the museum. | 1,848 | 7,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-25 | latest | en | 0.86809 |
https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses | 1,680,099,166,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00073.warc.gz | 527,434,764 | 10,924 | # Group Product Identity therefore Inverses
## Theorem
Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then if either:
$g h = e$
or:
$h g = e$
it follows that:
$g = h^{-1}$
and:
$h = g^{-1}$
### Part 1
$g h = e \implies h = g^{-1}$ and $g = h^{-1}$
### Part 2
$h g = e \implies h = g^{-1}$ and $g = h^{-1}$ | 130 | 346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-14 | latest | en | 0.668513 |
https://metric-calculator.com/convert-cu-yd-to-cl.htm | 1,701,848,710,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00608.warc.gz | 446,933,268 | 8,305 | # Cubic Yards to Centiliters Converter
Select conversion type:
Rounding options:
Convert Centiliters to Cubic Yards (cl to cu yd) ▶
## Conversion Table
cubic yards to centiliters cu yd cl 1 cu yd 76455.4858 cl 2 cu yd 152910.9716 cl 3 cu yd 229366.4574 cl 4 cu yd 305821.9432 cl 5 cu yd 382277.429 cl 6 cu yd 458732.9148 cl 7 cu yd 535188.4006 cl 8 cu yd 611643.8864 cl 9 cu yd 688099.3722 cl 10 cu yd 764554.858 cl 11 cu yd 841010.3438 cl 12 cu yd 917465.8296 cl 13 cu yd 993921.3154 cl 14 cu yd 1070376.8012 cl 15 cu yd 1146832.287 cl 16 cu yd 1223287.7728 cl 17 cu yd 1299743.2586 cl 18 cu yd 1376198.7444 cl 19 cu yd 1452654.2302 cl 20 cu yd 1529109.716 cl
## How to convert
1 cubic yard (cu yd) = 76455.4858 centiliter (cl). Cubic Yard (cu yd) is a unit of Volume used in Standard system. Centiliter (cl) is a unit of Volume used in Metric system.
## Cubic Yards: A Unit of Volume
A cubic yard is a unit of volume that is commonly used in the United States and Canada. It is defined as the volume of a cube with sides of one yard (three feet or 36 inches) in length. One cubic yard is equal to 27 cubic feet or about 0.765 cubic meters. The symbol for cubic yard is yd<sup>3</sup> or cu yd.
## How to Convert Cubic Yards
To convert cubic yard to other units of volume, you need to multiply or divide by the appropriate conversion factor. For example, to convert cubic yard to liters, you need to multiply by 764.555, which is the number of liters in one cubic yard. To convert liters to cubic yard, you need to divide by the same factor.
Here are some common conversion factors for cubic yard:
• 1 cubic yard = 27 cubic feet
• 1 cubic yard = 46,656 cubic inches
• 1 cubic yard = 0.765 cubic meters
• 1 cubic yard = 764.555 liters
• 1 cubic yard = 201.974 US liquid gallons
• 1 cubic yard = 168.179 imperial gallons
## Where Cubic Yard is Useds
Cubic yard are used to measure the volume of various materials, such as soil, sand, gravel, concrete, mulch, compost, etc. They are also used to estimate the amount of space needed for storage or transportation of these materials.
For example, in landscaping and gardening, cubic yard are used to calculate how much soil or mulch is needed to fill a garden bed or cover a lawn. A general rule of thumb is that one cubic yard covers 100 square feet at a depth of three inches.
In construction and engineering, cubic yard are used to measure the volume of concrete or asphalt needed for a project. For example, a driveway that is 10 feet wide and 40 feet long and has a thickness of four inches would require about 4.9 cubic yards of concrete.
Cubic yard are also used in some countries to measure the volume of waste or recyclables collected by garbage trucks or dumpsters. For example, in Canada, a standard garbage truck can hold about 25 cubic yards of waste.
## Definition of the Cubic Yard
A cubic yard is a unit of volume that belongs to the imperial and US customary systems of measurement. It is derived from the unit of length, the yard, which was originally defined as the distance from the tip of the nose to the end of the thumb of King Henry I of England.
A cube is a three-dimensional shape that has six equal square faces. The volume of a cube is calculated by multiplying the length of one side by itself three times. Therefore, the volume of a cube with sides of one yard is one yard times one yard times one yard, or one cubic yard.
## History of Cubic Yards
The origin of the cubic yard can be traced back to the ancient Roman unit of measurement, the amphora, which was a clay vessel used to store liquids such as wine or oil. One amphora was equivalent to about 0.026 cubic meters or 0.035 cubic yards.
The amphora was later replaced by the tun, which was a large wooden barrel used to store wine or beer. One tun was equivalent to about 0.953 cubic meters or 1.28 cubic yards.
The tun was then divided into smaller units, such as the hogshead, the barrel, and the gallon. The gallon was originally defined as the volume of eight pounds of wheat. In 1824, the British Parliament standardized the imperial gallon as the volume of 10 pounds of water at 62 degrees Fahrenheit, which is equal to about 4.546 liters or 0.0012 cubic yards.
In 1836, the US Congress adopted the wine gallon as the standard US liquid gallon, which is equal to about 3.785 liters or 0.001 cubic yards.
The cubic yard was officially adopted as a unit of measurement in both Britain and the United States in the late 19th century.
## Example Conversions of Cubic Yards to Other Units
Here are some examples of how to convert cubic yard to other units of volume:
• To convert 2 cubic yards to cubic feet, multiply by 27: 2 x 27 = 54 cubic feet.
• To convert 3 cubic yards to liters, multiply by 764.555: 3 x 764.555 = 2293.665 liters.
• To convert 4 cubic yards to US liquid gallons, multiply by 201.974: 4 x 201.974 = 807.896 US liquid gallons.
• To convert 5 cubic yards to imperial gallons, multiply by 168.179: 5 x 168.179 = 840.895 imperial gallons.
• To convert 6 cubic yards to cubic meters, multiply by 0.765: 6 x 0.765 = 4.59 cubic meters.
• To convert 7 cubic yards to cubic inches, multiply by 46,656: 7 x 46,656 = 326,592 cubic inches.
• To convert 8 cubic yards to crude barrels, multiply by 4.809: 8 x 4.809 = 38.472 crude barrels.
Cubic yards also can be marked as yd3.
## Definition of the Centiliter
A centiliter (international spelling) or centiliter (American English spelling) (SI symbols cL or cl) is a unit of volume that is used in the metric system. It is defined as one hundredth of a liter, or the volume of a cube with sides of 1 centimeter (0.01 meter) in length.
One centiliter is equal to 10 cubic centimeters, 0.00001 cubic meters, or 0.01 cubic decimeters.
## History of the Centiliter
The centiliter is derived from the liter, which is a unit of volume that was first used by the French chemist Antoine Lavoisier in 1795 as a unit of volume for liquids and gases. The liter was originally defined as the volume of one kilogram of pure water at 4 °C and standard atmospheric pressure. The liter was later redefined several times based on different physical standards, such as a platinum-iridium cylinder and a wavelength of light. The current definition of the liter, adopted in 1964, is based on the cubic meter, which is the SI unit of volume. The liter is not an SI unit, but it is accepted by the CGPM (the standards body that defines the SI) for use with the SI.
The centiliter was introduced as a subunit of the liter in the metric system in 1795, along with other decimal subunits such as the deciliter and the milliliter. The centiliter was used for measuring small volumes of liquids and gases, such as wine, beer, milk, oil, and air. The centiliter was also used for measuring some solids, such as sugar, flour, and salt. The centiliter was adopted as a standard unit of measure in many countries that use the metric system, such as France, Germany, Italy, Spain, and Sweden.
## How to Convert Centiliters
To convert centiliters to other units of volume, you need to multiply or divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert centiliters to cubic centimeters, multiply by 10.
• Example: 2 cL x 10 = 20 cm3
• To convert centiliters to cubic meters, multiply by 0.00001.
• Example: 2 cL x 0.00001 = 0.00002 m3
• To convert centiliters to cubic decimeters, multiply by 0.01.
• Example: 2 cL x 0.01 = 0.02 dm3
• To convert centiliters to gallons (US liquid), multiply by 0.00264.
• Example: 2 cL x 0.00264 = 0.00528 gal
• To convert centiliters to bushels (US), multiply by 0.000374.
• Example: 2 cL x 0.000374 = 0.000748 bu
• To convert centiliters to barrels (oil), multiply by 0.000063.
• Example: 2 cL x 0.000063 = 0.000126 bbl
To convert other units of volume to centiliters, you need to divide by the appropriate conversion factor. Here are some common conversion factors and examples:
• To convert cubic centimeters to centiliters, divide by 10.
• Example: 20 cm3 / 10 = 2 cL
• To convert cubic meters to centiliters, divide by 0.00001.
• Example: 0.00002 m3 / 0.00001 = 2 cL
• To convert cubic decimeters to centiliters, divide by 0.01.
• Example: 0.02 dm3 / 0.01 = 2 cL
• To convert gallons (US liquid) to centiliters, divide by 0.00264.
• Example: 0.00528 gal / 0.00264 = 2 cL
• To convert bushels (US) to centiliters, divide by 0.000374.
• Example: 0.000748 bu / 0.000374 = 2 cL
• To convert barrels (oil) to centiliters, divide by 0.000063.
• Example: 0.000126 bbl / 0.000063 = 2 cL
## Where Centiliters are Used
Centiliters are used for measuring various materials and substances in different countries and applications.
Some examples are:
• In many European countries, such as France, Germany, Italy, Spain, and Sweden, centiliters are used for measuring alcoholic beverages, such as wine, beer, and spirits. For example, a standard glass of wine is usually 10 or 12 centiliters, a bottle of beer is usually 25 or 33 centiliters, and a shot of liquor is usually 2 or 4 centiliters.
• In some Scandinavian countries, such as Norway and Denmark, centiliters are used for measuring milk and cream. For example, a carton of milk is usually 100 or 200 centiliters, and a cup of coffee is usually served with 1 or 2 centiliters of cream.
• In some Asian countries, such as Japan and China, centiliters are used for measuring some traditional medicines and herbal teas. For example, a dose of kampo medicine is usually 10 or 20 centiliters, and a cup of green tea is usually 15 or 20 centiliters.
• In the medical industry, centiliters are used for measuring some fluids and solutions, such as blood plasma, urine, and saline. For example, a unit of blood plasma is usually 20 or 25 centiliters, a urine sample is usually 5 or 10 centiliters, and an intravenous drip is usually set at a rate of 10 or 20 centiliters per hour.
• In the scientific industry, centiliters are used for measuring some chemical reagents and solvents, such as ethanol, acetone, and water. For example, a flask of ethanol is usually 50 or 100 centiliters, a bottle of acetone is usually 25 or 50 centiliters, and a beaker of water is usually 10 or 20 centiliters.
## Example Conversions of Centiliters to Other Units
Here are some example conversions of centiliters to other units of volume:
• 1 cL = 10 cm3
• 1 cL = 0.00001 m3
• 1 cL = 0.01 dm3
• 1 cL = 0.00264 gal
• 1 cL = 0.000374 bu
• 1 cL = 0.000063 bbl
• 1 cL = 0.01 L
Español Russian Français | 2,946 | 10,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-50 | longest | en | 0.772282 |
https://education.lego.com/en-us/lessons/preschool-coding-express/o-shaped-track-looping | 1,632,785,468,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00422.warc.gz | 271,941,847 | 28,494 | Coding Express Set
# O-Shaped Track - Looping
The objective of this lesson is for children to explore and Understand use of the O-shaped track for repeating sequences.
30-45 min.
Intermed.
PreK-K
## Engage
Ask the children if there’s anything they do many times a day or week (e.g., brushing their teeth, showering, cleaning their room).
Tell the children that they’re going to play another game!
Model a sequence of hopping, jumping, running, walking backward, dancing, spinning, or other actions in a circle.
Ask the children to copy what you’ve just done and to repeat (i.e., loop) the sequence at least twice.
Tip: For younger children and beginners, limit your loop to just one or two actions.
## Explore
Ask the children to combine curved and straight track pieces to make an O-shaped train track (six curved and four straight pieces is recommended).
Using the building cards, have the children build two or three places they’d like to visit on the train (see sidebar for an example).
Let’s go on a day trip!
Use some LEGO® DUPLO® figures as passengers.
Tell the children that the passengers would like to have a picnic in the forest and then visit the beautiful castle.
Can you help the passengers take the train to the forest and then to the castle?
## Explain
Tell the children that the passengers enjoyed their trip so much that they’d like to do it again!
Talk with the children about how they could help make this happen.
• Will you be able to help the passengers take the same trip again? How? (The O-shaped track creates loops.)
• Which action bricks will you use and why?
## Elaborate
Encourage the children to build a double-ended track next to the O-shaped track.
Talk about the difference between the two types of tracks.
• What’s the difference between these two types of tracks?
• Will you be able to repeat the same journey on the double-ended track? Why or why not?
## Evaluate
Evaluate the children’s skills development by observing if they’re:
• Observing and describing objects and events
• Identifying cause and effect relationships
## Teacher Support
Children will:
• Understand use of the O-shaped track for repeating sequences
• Be able to compare different train track shapes and their uses
For up to six children
Coding Express set (45025)
Each lesson has been developed using the science, math and technology guidelines from the National Association for the Education of Young Children
(NAEYC), the 21st Century Early Learning framework (P21) and Head Start Early Learning Outcomes Framework.
The learning goals listed at the end of each lesson can be used to determine whether or not each child is developing the relevant early math skills. These bullet points target specific skills or pieces of information that are practiced or presented during each lesson. | 588 | 2,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | longest | en | 0.940097 |
http://physicstasks.eu/3946/work-of-van-der-waals-gas | 1,716,376,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00852.warc.gz | 26,573,161 | 7,866 | ## Work of van der Waals Gas
### Task number: 3946
Determine the work performed by a gas following a van der Waals equation with constants
$a = 0.137\, \mathrm{ J\, m^{3}\, mol^{-2}},$ $b = 38.7\cdot{ 10^{-6}}\, \mathrm{m^{3}\, mol^{-1}}$
during isothermal expansion from volume of 10 l to five times this volume. The initial pressure of the gas was 300 kPa, its amount of substance is 1 mol.
• #### Hint
Van der Waals gas is described by the equation of state
$\left( p+\frac{n^{2}a}{V_{1}^{2}}\right) \left( V_{1}-nb\right) = nRT,$
where V is the volume of the gas, p is the gas pressure, n is the amount of substance, R is the universal molar gas constant, T is thermodynamic temperature, a and b are the given constants (characterizing the gas).
How do you determine the work performed by the gas provided that pressure is the function of a volume?
• #### Analysis
In this task we need to use integral calculus to determine work performed by the gas since the pressure is the function of volume. From the equation of state for van der Waals gas we can determine pressure using volume and integrate the resulting function with respect to volume. The limits of the integral are initial and final volume.
To get a numerical solution we will need to determine temperature from the equation for van der Walls gas.
• #### Given Values
a = 0.137 J m3 mol−2 constant from a van der Waals gas model b = 38.7·10−6 m3 mol−1 constant from a van der Waals gas model V1 = 10 l = 10 dm3 = 0.01 m3 initial volume of the gas V2 = 5V1 = 0.05 m3 volume after isothermal expansion p1 = 300 kPa = 3·105 Pa initial pressure of the gas n = 1 mol amount of substance of the gas W = ? work performed by the gas during isothermal expansion
Table values:
R = 8.31 J K−1 mol−1 universal gas constant
• #### Solution
From the equation of state of van der Waals gas
$\left( p+\frac{n^{2}a}{V^{2}}\right) \left( V-nb\right) = nRT,$
where V is the volume of the gas, p the gas pressure, n is the amount of substance, R is the universal gas constant, T is thermodynamic temperature, a and b are given constants,
we first determine gas temperature T. The expansion is isothermal, the temperature T is therefore constant. We determine it using initial volume V1 and pressure p1.
We obtain:
$T = \frac{\left( p_1+\frac{n^{2}a}{V_{1}^{2}}\right) \left( V_{1}-nb\right) }{nR}.$
The work performed by the gas during a small change in volume dV is determined by
$\mathrm{d}W = p\mathrm{d}V.$
The total performed work is then determined by integration between the limits of initial volume V1 and final volume V2 = 5V1. (We need to use integration calculus because the pressure of the gas changes continuously. Therefore, it is not possible to use a simple equation W =p(V2-V1) that holds for isobaric process with contant pressure!).
From the basic form of van der Waals equation we can determine pressure as a volume function. We will need it in the next step of the calculation:
$\left( p+\frac{n^{2}a}{V^{2}}\right) \left( V-nb\right) = nRT \Rightarrow p+\frac{n^{2}a}{V^{2}} = \frac{nRT}{V-nb}\Rightarrow$ $\Rightarrow p= \frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}.$
Now we can perform the integration:
$W = \int_{V_{1}}^{V_{2}}{p}\, \mathrm{d}V = \int_{V_{1}}^{5V_{1}}{\left( \frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}\right)} \, \mathrm{d}V =$
we factor constants out of the integrals
$=nRT\int_{V_{1}}^{5V_{1}}{\frac{1}{V-nb}}\, \mathrm{d}V - n^{2}a \int_{V_{1}}^{5V_{1}}{\frac{1}{V^{2}}}\, \mathrm{d}V =$
we perform the integration and substitute the limits
$=nRT\left[ \ln \left( V-nb\right) \right] _{V_{1}}^{5V_{1}} + n^{2}a\left[ \frac{1}{V} \right] _{V_{1}}^{5V_{1}} =$ $=nRT\ln \frac{5V_{1}-nb}{V_{1}-nb}+n^{2}a\left( \frac{1}{5V_{1}}-\frac{1}{V_{1}}\right)$ $=nRT\ln \frac{5V_{1}-nb}{V_{1}-nb}- \frac{4n^{2}a}{5V_{1}}.$
• #### Numerical Solution
$T = \frac{\left( p+\frac{n^{2}a}{V_{1}^{2}}\right) \left( V_{1}-nb\right) }{nR} = \frac{\left( 3\cdot{ 10^{5}}+\frac{0.137}{10^{-4}}\right) \cdot \left( 0.01-38.7\cdot{ 10^{-6}}\right) }{8.31}\, \mathrm{K}$ $T \dot{=} 361.3\, \mathrm{K}$
$W=nRT\ln \frac{5V_{1}-nb}{V_{1}-nb}-\frac{4n^{2}a}{5V_{1}}$ $W=\left(8.31\cdot{ 361.3}\cdot \ln \frac{0.05-38.7\cdot{ 10^{-6}}}{0.01-38.7\cdot{ 10^{-6}}}- \frac{4\cdot{0.137}}{0.05}\right) \, \mathrm{J}$ $W \dot{=} 4830.5\, \mathrm{J}\dot{=} 4.8\, \mathrm{kJ}$ | 1,514 | 4,351 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.750569 |
http://chronicle.com/blognetwork/castingoutnines/2006/06/23/what-is-calculus-about-and-does-this-match-what-we-teach/ | 1,371,614,561,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707440693/warc/CC-MAIN-20130516123040-00004-ip-10-60-113-184.ec2.internal.warc.gz | 45,817,848 | 13,043 | Previous Friday random 10 (week 3) Next The lesson of the museum store
What is calculus about? And does this match what we teach?
June 23, 2006, 2:19 pm
I think there is a serious disconnect between what we actually teach in a calculus class, and what students ought to be learning from (and taking away from) a calculus class.
In my syllabus for calculus, and regularly in class, I give students the following definition of calculus:
Calculus is the study of the mathematics of quantities that undergo change.
In addition to that overall slogan, I let them know that everything we do in calculus is targeted at answering one or more of the following basic questions:
1. How can we describe, in a precise way, the manner in which one quantity depends upon another? [We answer this question using the idea of a function.]
2. How can we determine the rate at which a quantity is changing at a given moment and describe the overall manner in which this change occurs? [We answer that question using the idea of a derivative, which needs the idea of a limit.]
3. How can we determine how much change has accumulated in a quantity over a certain period, if the rate at which the quantity changes is not constant? [We answer this question with the idea of the integral and with the Fundamental Theorem of Calculus.]
I think these are all accurate ways to describe the subject which allow students to think about why calculus would be useful — enough so that a four-hour course in it is required — for what they are going on to study. A course in calculus ought to be focused on asking and answering variations on these questions in different kinds of contexts — some purely mathematical, some applied to real questions in applied areas. The focus, in other words, ought to be on the questions and the way in which we answer them.
But after teaching calculus for over a decade, it seems to me that the only goal that calculus textbooks and course desginers really have in mind is to have students get really, really good at algebraic calculations of a certain type, and the students who can do the most complicated and esoteric calculations are the “best” students. Just look at some of the grotesque limit, derivative, and integral calculations students are asked to do in exercises. Some of these are justifiably complicated, because they arise from actual scientific or economic formulas that are themselves complicated. But some are merely contrived in order to present the maximum amount of complication, and for no good reason.
Calculus becomes a course in applied algebra — when in fact, algebra is only one of a whole arsenal of quantitative techniques used to answer the main questions of calculus, and it’s not especially more common or important than the other methods such as numerical or graphical approximation. People who actually use calculus do not use algebra the majority of the time; but we spend a lot more than the majority of time in a calculus class teaching algebra and algebraic techniques.
Why is this the case? Should we really be focusing so much on algebraic techniques in calculus? Could we teach students just as well and have just as effective a course — in terms of actually being able to use the subject matter in a real setting — if we introduced only what was needed to actually answer an important question in front of us, rather than the umpteenth limit calculation involving rationalizing a denominator, or the derivative of ln(ln(ln(ln(x)))), and so on? Might the calculus course be more effective if we did so?
This entry was posted in Calculus, Education, Teaching, Uncategorized. Bookmark the permalink.
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# BUSN6070_Week4Assignment_Ch6,7_Reeves - BUSN 6070 Amber...
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BUSN 6070 Amber Reeves Week 4 Assignment Chapter 6: Pg. 203: #15, 16 Pg. 205/206: #22, 25 Pg. 207/210: # 31, 38 15) 16) 22) Total Revenues \$ 5,000,000.00 Total FC \$ 1,000,000.00 Total Variable \$ 3,000,000.00 Total Prod. & Sol 1,000,000 units a. Selling Price = \$ 5.00 b. VC per unit = \$ 3.00 c. Cont. Margin Per \$ 2.00 \$ 1,000,000.00 d. Break Even Point \$ 500,000.00 e. Income Tax Rate 40% After Tax OP 1,200,000.00 Before Tax Profit 2000000 Unit Sales 1,500,000.00 units 25) Chicken Steak Selling Price per \$4 \$6 VC per unit \$2 \$3 Expected Sales 200,000 300,000 Total FC \$200,000 a. Sales Revenue \$800,000 \$1,800,000 VC \$400,000 \$900,000 Contribution Ma \$400,000 \$900,000 \$1,300,000 Fc 200,000 In order to the make the CVP model work there are three assumptions that are required include separating total costs into fixed and variable components, that cost and revenue throughout the relevant range of activity, and that product mix reaminas constant. Multi a constant product mix so that the break-even point remains unique for multiple produc The sum of the break-even quantities for each company's product line would depend on allocated to that product line. It would depend on what percentage of total fixed costs w product line. If the arbitual allocation changes, then it changes the break-even because mix you need to break-even.
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Profit \$1,100,000.00 31) Basic
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Jill Tulane University ‘16, Course Hero Intern | 696 | 2,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-39 | latest | en | 0.799444 |
https://www.roseindia.net/answers/viewqa/Java-Beginners/15323-recursive-method.html | 1,656,153,948,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00168.warc.gz | 1,047,317,508 | 7,683 | # recursive method
Write a recursive method to add the first n terms of the series
1 + (1/2) - (1/3) + (1/4) - (1/5)...
recursive method
recursive method Write a recursive method to add the first n terms of the series 1 + (1/2) - (1/3) + (1/4) - (1/5
java program on recursive method
java program on recursive method in how many ways can you make change for one dollar(100 cents) using pennies(1-cent coins), nickels(5 cents), dimes(10 cents),and quarter(25 cents)? the coins must add up to the exact total
Fibonacci (Recursive)
to do: Assignment: 1.Write a recursive method that takes in a single integer (x...Fibonacci (Recursive) I need to write a program for my AP CS class... series. 2.Write a method that solves a multiplication problem recursively
ModuleNotFoundError: No module named 'recursive'
ModuleNotFoundError: No module named 'recursive' Hi, My Python... 'recursive' How to remove the ModuleNotFoundError: No module named 'recursive' error? Thanks Hi, In your python environment you
method
method how and where, we can define methods ? can u explain me with full programme and using comments
ModuleNotFoundError: No module named 'djangorestframework-recursive'
ModuleNotFoundError: No module named 'djangorestframework-recursive' ...: No module named 'djangorestframework-recursive' How to remove the ModuleNotFoundError: No module named 'djangorestframework-recursive' error
ModuleNotFoundError: No module named 'djangorestframework-recursive'
ModuleNotFoundError: No module named 'djangorestframework-recursive' ...: No module named 'djangorestframework-recursive' How to remove the ModuleNotFoundError: No module named 'djangorestframework-recursive' error
ModuleNotFoundError: No module named 'recursive-abc'
ModuleNotFoundError: No module named 'recursive-abc' Hi, My... 'recursive-abc' How to remove the ModuleNotFoundError: No module named 'recursive-abc' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive-compression'
ModuleNotFoundError: No module named 'recursive-compression' Hi...: No module named 'recursive-compression' How to remove the ModuleNotFoundError: No module named 'recursive-compression' error? Thanks Hi
ModuleNotFoundError: No module named 'recursive-decorator'
ModuleNotFoundError: No module named 'recursive-decorator' Hi, My... named 'recursive-decorator' How to remove the ModuleNotFoundError: No module named 'recursive-decorator' error? Thanks Hi
ModuleNotFoundError: No module named 'recursive-diff'
ModuleNotFoundError: No module named 'recursive-diff' Hi, My... named 'recursive-diff' How to remove the ModuleNotFoundError: No module named 'recursive-diff' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive_listfunction'
ModuleNotFoundError: No module named 'recursive_listfunction' Hi...: No module named 'recursive_listfunction' How to remove the ModuleNotFoundError: No module named 'recursive_listfunction' error? Thanks Hi
ModuleNotFoundError: No module named 'recursive-parse'
ModuleNotFoundError: No module named 'recursive-parse' Hi, My... named 'recursive-parse' How to remove the ModuleNotFoundError: No module named 'recursive-parse' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive_print'
ModuleNotFoundError: No module named 'recursive_print' Hi, My... named 'recursive_print' How to remove the ModuleNotFoundError: No module named 'recursive_print' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive-sgd'
ModuleNotFoundError: No module named 'recursive-sgd' Hi, My... 'recursive-sgd' How to remove the ModuleNotFoundError: No module named 'recursive-sgd' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive-yaml'
ModuleNotFoundError: No module named 'recursive-yaml' Hi, My... named 'recursive-yaml' How to remove the ModuleNotFoundError: No module named 'recursive-yaml' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'djangorestframework-recursive'
ModuleNotFoundError: No module named 'djangorestframework-recursive' ...: No module named 'djangorestframework-recursive' How to remove the ModuleNotFoundError: No module named 'djangorestframework-recursive' error
ModuleNotFoundError: No module named 'recursive-aronr'
ModuleNotFoundError: No module named 'recursive-aronr' Hi, My... named 'recursive-aronr' How to remove the ModuleNotFoundError: No module named 'recursive-aronr' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive_dict'
ModuleNotFoundError: No module named 'recursive_dict' Hi, My... named 'recursive_dict' How to remove the ModuleNotFoundError: No module named 'recursive_dict' error? Thanks Hi, In your python
ModuleNotFoundError: No module named 'recursive-itertools'
ModuleNotFoundError: No module named 'recursive-itertools' Hi, My... named 'recursive-itertools' How to remove the ModuleNotFoundError: No module named 'recursive-itertools' error? Thanks Hi
ModuleNotFoundError: No module named 'recursive_package'
ModuleNotFoundError: No module named 'recursive_package' Hi, My... named 'recursive_package' How to remove the ModuleNotFoundError: No module named 'recursive_package' error? Thanks Hi, In your
ModuleNotFoundError: No module named 'sympy_recursive'
ModuleNotFoundError: No module named 'sympy_recursive' Hi, My... named 'sympy_recursive' How to remove the ModuleNotFoundError: No module named 'sympy_recursive' error? Thanks Hi, In your python
method
method can you tell me how to write an abstract method called ucapan() for B2 class class A2{ void hello(){ system.out.println("hello from A2"); }} class B2 extends A2{ void hello(){ system.out.println("hello from B2
ModuleNotFoundError: No module named 'colcon-recursive-crawl'
ModuleNotFoundError: No module named 'colcon-recursive-crawl' Hi...: No module named 'colcon-recursive-crawl' How to remove the ModuleNotFoundError: No module named 'colcon-recursive-crawl' error? Thanks Hi
ModuleNotFoundError: No module named 'recursive-monkey-patch'
ModuleNotFoundError: No module named 'recursive-monkey-patch' Hi...: No module named 'recursive-monkey-patch' How to remove the ModuleNotFoundError: No module named 'recursive-monkey-patch' error? Thanks Hi
ModuleNotFoundError: No module named 'colcon-recursive-crawl'
ModuleNotFoundError: No module named 'colcon-recursive-crawl' Hi...: No module named 'colcon-recursive-crawl' How to remove the ModuleNotFoundError: No module named 'colcon-recursive-crawl' error? Thanks Hi
ModuleNotFoundError: No module named 'colcon-recursive-crawl'
ModuleNotFoundError: No module named 'colcon-recursive-crawl' Hi...: No module named 'colcon-recursive-crawl' How to remove the ModuleNotFoundError: No module named 'colcon-recursive-crawl' error? Thanks Hi
ModuleNotFoundError: No module named 'dict-recursive-update'
ModuleNotFoundError: No module named 'dict-recursive-update' Hi...: No module named 'dict-recursive-update' How to remove the ModuleNotFoundError: No module named 'dict-recursive-update' error? Thanks Hi
ModuleNotFoundError: No module named 'recursive-dictionary-update'
ModuleNotFoundError: No module named 'recursive-dictionary-update' ...: No module named 'recursive-dictionary-update' How to remove the ModuleNotFoundError: No module named 'recursive-dictionary-update' error? Thanks
ModuleNotFoundError: No module named 'recursive_print_list'
ModuleNotFoundError: No module named 'recursive_print_list' Hi...: No module named 'recursive_print_list' How to remove the ModuleNotFoundError: No module named 'recursive_print_list' error? Thanks Hi
all combination via recursive function
all combination via recursive function hello i have a php scrpt which generates all possibilities fro a given string recursive function is used in it. i have given its code below \$charset = "abcd"; \$charset_length | 1,857 | 7,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-27 | latest | en | 0.579652 |
https://jeopardylabs.com/play/math-jeopardy11 | 1,526,997,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864790.28/warc/CC-MAIN-20180522131652-20180522151652-00336.warc.gz | 598,895,946 | 8,757 | Theorem
Shapes
Mathematicians
Angles and Lines
Trick Math
### 100
This theorem states the one object is equal to itself.
What is the reflexive property?
### 100
This shape is a quadrilateral with two pairs of opposite parallel lines.
What is a parallelogram?
### 100
This man was an English mathematician and died in 1776. He focused on the Number Theory.
Who is Peter Barlow?
### 100
A triangle has the angle measures of 70 and 30. The third angle is...
What is 80 degrees?
### 100
There are 8 Apples on the table, you take 3. How many do you have?
What is 3 apples?
### 200
This theorem states that if a triangle has two congruent sides then the angles opposite them are congruent.
What is the isosceles triangles base angle theorem?
### 200
This shape is a quadrilateral with two distinct pairs of consecutive sides of the same length.
What is a kite?
### 200
This women was born in Italy in 1718. Became known as one of the greatest female scholars.
Who is Maria Agnesi?
### 200
If one length of a square is 6ft. The area is...
What is 36ft 2?
### 200
I have 2 coins in my hand that add up to 60c. One of the coins isn't a 50c piece. What are the coins?
What is 50c and a 10c (The coin that isn't a 50c piece is a 10c. The other coin is the 50c)?
### 300
This theorem states that if a figure is symmetric then corresponding parts under the symmetry are congruent.
What is the symmetric figures theorem?
### 300
This shape has four sides all of equal lengths.
What is a rhombus?
### 300
This man was born in London in 1791. He once said... "The whole of the developments and operations of analysis are now capable of being executed by machinery. ... As soon as an Analytical Engine exists, it will necessarily guide the future course of science."
Who is Charles Babbage?
### 300
The area of a triangle if the height is 6in and the base is 3cm.
What is 9cm 2?
### 300
A box has nine ears of corn in it. A Squirrel carries out three ears a day, and yet it takes him nine days to carry out all the corn. Explain?
What is he has 2 of his own ears, so he carries out only 1 ear of corn per day?
### 400
This theorem states that corresponding parts of corresponding figures are congruent.
What is CPCF?
### 400
This shape is a quadrilateral with one pair of parallel lines.
What is a trapezoid?
### 400
Born in France, this man was appointed to the Commission set up by the Royal Society to check the claims of Newton.
Who is De Moivre?
### 400
One angle of a triangle is 84 degrees. The other is 59 degrees. The last angle is...
What is 37 degrees?
### 400
If you have 4 melons in one hand, and 7 apples in the other - What do you have?
What is big hands?
### 500
This theorem states that if two angles of a triangle are congruent, then the sides opposite them are congruent.
What is isosceles triangle base angles converse theorem?
### 500
This is a quadrilateral with base angles equal in measure.
What is an isosceles trapezoid?
### 500
This man created the Pythagorean Theorem.
Who is Pythagoras of Samos?
### 500
The base of a triangle is 7in. Another leg is 8in. The hypotenuse is... | 825 | 3,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.955371 |
https://iaeimagazine.org/magazine/2019/2019september/pv-systems-math-sample-calculations/ | 1,591,406,887,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509264.96/warc/CC-MAIN-20200606000537-20200606030537-00210.warc.gz | 382,611,965 | 35,102 | PV Systems Math — Sample Calculations
Looking at the PV array in a PV system, many installers and inspectors are confused by new system voltage calculations that may be required by the Code specific to PV systems. Code Informational Notes also address voltage drop that may be applied to the DC wiring from the array to the inverter. This article will cover both of those subjects.
Example — Module Open-Circuit Voltage
A PV module, or a string of series-connected modules, has a rated open-circuit voltage that is measured (and labeled on the module) at an irradiance of 1000 W/m2 and a cell temperature of 25°C (77°F). This voltage increases from the rated voltage as the temperature drops below 25°C. It is necessary to calculate this voltage at the expected lowest temperature at the installation location to ensure that it is less than the maximum input voltage of the inverter and less than the voltage rating of any connected conductors, switchgear, and overcurrent devices (usually 600 volts). Because parallel connections of strings do not affect open-circuit voltage, the number of strings connected in parallel is not involved with this calculation.
Where module temperature coefficients are available, Section 690.7 of the NEC allows the open-circuit voltage of a PV array to be determined at the lowest expected temperature at the installation location. Alternatively, Table 690.7(A) can be used to determine a multiplier that was applied to either the module- or string- (a series connection of PV modules) rated VOC.
The rated VOC is measured at 25°C (77°F) and is printed on the back of the module and in the module’s technical literature. To use Table 690.7(A), determine the lowest expected temperature, look up the factor from the Table for that temperature (which ranges between 1.02 at 24°C to 1.25 at -40°C), and multiply the factor by the rated VOC.
For example, a module has a VOC of 35 volts (at STC) and is going to be installed where the temperature dips to -17°C.
The factor from Table 690.7(A) is 1.18 and the cold temperature VOC for this module is 35 x 1.18 = 41.3 volts.
If 12 modules were going to be connected in series, the string VOC in cold weather would be 12 x 41.3 = 495.6 volts.
The string voltage at STC-rated conditions could also be calculated first and then apply the temperature factor. In this case, the 12 modules in series would have a string open-circuit voltage of 12 x 35 = 420 volts at 25°C. Then the 1.18 factor is applied to get 1.18 x 420 = 495.6 volts; the same answer as before.
While Table 690.7(A) is still valid and was refined with 5°C increments in the 2011 NEC, new modules may have different technologies than the silicon module technology used to develop the table.
The use of the module temperature coefficients will provide a more accurate calculation of the cold weather maximum system voltage. To add to the confusion, PV module manufacturers present these temperature coefficients in two different ways.
Example — Percentage Coefficients
One way of presenting these data is to specify them as a percentage change, and they are expressed as a percentage change in VOC for a change in temperature measured in degrees Celsius. Note that the temperature used is a change in temperature from the rated 25°C.
For example: The Voc temperature coefficient is given as -0.36% per deg Celsius, or -0.36% / °C.
The module has a VOC of 45 volts at 25°C (77°F) and is going to be installed where the expected lowest temperature is -10°C (14°F). Because the temperature coefficient is given in degrees Celsius, all numbers must be in degrees Celsius. The change in temperature is from 25°C to -10°C. This represents a change in temperature of 35 degrees. The minus sign in the coefficient can be ignored if we remember that the voltage increases as the temperature goes down and vice versa. Of course, if you are an engineer or a mathematician, feel free to use the minus sign in an algebraic equation.
Applying the coefficient shows that the percentage change in VOC resulting from this temperature change is 0.36% / °C x 35°C = 12.6%.
Note that where division and multiplication are involved in a calculation, they are performed from left to right. Where additions and subtractions are combined with multiplications and divisions, the additions and subtractions are performed before any multiplications. Parentheses may be added to clarify the order of the operations, and calculations inside the parentheses should be performed first.
This percentage change can now be applied to the rated VOC of 45 volts. And, at -10°C, the VOC will be 1.126 x 45 = 50.67 V.
For the mathematically oriented person, the equation looks like this:
VOC [cold] = VOC [STC] x (1 + ((-0.36)/100) x (-10-25)))
= 45 x (1 + .0036 x 35)
= 45 x 1.126
= 50.67
Eleven of these modules could be connected in series and the cold-weather voltage would be 11 x 50.67 = 557.37 V, and that voltage is less than a 600-volt equipment limitation.
Example — Millivolt Coefficients
Other PV module manufacturers express the Voc temperature coefficient as a millivolt coefficient. A millivolt is one one-thousandth of a volt, or 0.001 V.
A typical module with an open-circuit voltage (at 25°C) of 65 volts might have a temperature coefficient expressed as:
-240 mV per degree Celsius, or -240 mV/°C
If it is installed where the expected low temperature is -30°C (-22°F), then there is a 55°C degree change in temperature from 25°C to -30°C. Again, the calculations must be accomplished in degrees Celsius, because that is the way the coefficient is presented.
Millivolts are converted to volts by dividing the millivolt number by 1000.
240 mV / 1000 mV/V = 0.24 volts
The module Voc will increase 0.24 V/°C x 55°C = 13.2 volts as the temperature changes from 25°C to -30°C.
The module VOC will increase from 65 volts at 25°C to 65 + 13.2 = 78.2 volts at the -30°C temperature.
In this PV system, the inverter maximum input voltage was listed as 550 volts. How many modules could be connected in series and not exceed this voltage? The maximum inverter voltage of 550 volts is divided by the cold-weather open-circuit voltage for the module of 78.2 volts.
550 / 78.2 = 7.03 modules and the correct answer would be seven modules.
7 x 78.2 V = 547.4 V
Eight modules could not be used because the open-circuit, cold-weather voltage would exceed 550 volts.
8 x 78.2 V = 625.6 V
Example — Expected Lowest Temperature for PV Systems?
What is the source of the expected lowest temperature? Normally, this temperature occurs in the very early morning hours just before sunrise on cold winter mornings. The PV modules are, in many cases, a few degrees colder than the air temperature due to night-sky radiation effects.
The illumination at dawn and dusk are sufficient to produce high VOC, even when the sun is not shining directly on the PV array and has not produced solar heating of the modules. And in many locations, cold temperatures are coupled with high winds and the winds can remove solar heating from the module, even in bright sun.
A conservative approach would be to get weather data showing the record-low temperature of the area and use this as the expected low temperature. Other data show more moderate low temperatures associated with the data used to size heating systems. See the Informational Note on 690.7(A).
The National Renewable Energy Laboratory maintains data that shows the record lows for many locations in the United States (http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/redbook/mon2/state.html).
Local airports and weather stations may have historical data on low temperatures.
Also, weather.com has data on file accessble by ZIP codes (www.weather.com/weather/climatology/monthly/zip code).
Example — Module Short-Circuit Current
In most silicon PV modules, the module short-circuit current does increase very slightly as temperature increases, but the increase is so small as to be negligible at normal module operating temperatures. It is normally ignored.
Example — PV Systems Voltage Drop
In common, utility-interactive PV systems, PV arrays may operate from 50–60 volts up to near 600 volts, depending on the system design. With nominal, peak-power, and open-circuit voltages to deal with, installers and inspectors are sometimes in a quandary as to how to calculate voltage drops from PV arrays to the inverters.
A utility-interactive inverter will normally operate in a manner that keeps the array voltage near the peak-power voltage (also called the maximum power point). While this voltage can vary with temperature—and temperatures vary considerably—using the rated maximum power point voltage and maximum power point current of the modules results in the easiest method of calculating voltage drops.
A typical PV array may have a single string of ten modules in series connected to the inverter 200 feet away with 10 AWG USE-2/RHW-2 conductors. The maximum power point (mpp)numbers for the module are:
Vmpp = 45V
Impp = 5.5 amps
For a single string of 10 modules, the string maximum power point numbers are:
Vmpp = 450V
Impp = 5.5 amps
Table 8 in Chapter 9 of the NEC gives conductor resistance per 1000 feet at 75°C.
For an uncoated, stranded 10 AWG conductor, the resistance is 1.24 ohms per 1000 feet.
The total conductor length (both ways) must be used in the calculation and this is 400 feet.
The resistance for 400 feet of a 10 AWG conductor is 400/1000 x 1.24 = 0.496 ohms.
The current at the maximum power point is 5.5 amps. Voltage drop is found by multiplying this current by the conductor resistance:
5.5 x 0.496 = 2.728 volts.
Expressed as a percentage, 2.278/450 x 100 = 0.606% or about 0.6%, and that is much less than the Informational Note recommendation of 3% for most circuits.
Of course, the losses in the PV DC disconnect were not counted, but they are typically less than 1% on these circuits.
Example — Voltage Drop, Another Perspective
When dealing with the dc input circuits of a utility-interactive inverter, the inverter operates the array at the maximum power point with a maximum power voltage (Vmpp) and a maximum power current (Impp).
Because these parameters are affected by irradiance and temperature, it is difficult to determine what voltage and current should be used in the voltage drop equation.
The procedure above uses module specifications at standard test conditions that are based on the rating conditions of an irradiance of 1000 W/m2 and a cell temperature of 25°C (which is not a very realistic temperature under actual operating conditions).
One conservative approach is to use the module manufacturer’s specified normal cell operating temperature from the module data when it is available. Normal cell operating temperatures are typically in the range of 42°C to 52°C and is a measured temperature of the module when the irradiance is 800 W/m2, ambient temperature is 20°C and the wind is blowing across the module at 1m/sec.
This temperature is used to adjust the Vmpp and Impp numbers presented at standard test conditions to values associated with the higher normal cell operating temperature. T
The maximum power voltage will be reduced and the maximum power current will be increased very little if at all. Voltage drop will be reduced.
Excerpted from IAEI’s Photovoltaic Power Systems, 2017. Purchase online at iaei.org. | 2,629 | 11,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-24 | latest | en | 0.895163 |
https://www.biotite-python.org/apidoc/biotite.sequence.align.align_banded.html | 1,709,352,649,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00567.warc.gz | 670,817,698 | 6,288 | # biotite.sequence.align.align_banded¶
biotite.sequence.align.align_banded(seq1, seq2, matrix, band, gap_penalty=- 10, local=False, max_number=1000)[source]
Perform a local or semi-global alignment within a defined diagonal band. 1
The function requires two diagonals that defines the lower and upper limit of the alignment band. A diagonal is an integer defined as $$D = j - i$$, where i and j are sequence positions in the first and second sequence, respectively. This means that two symbols at position i and j can only be aligned to each other, if $$D_L \leq j - i \leq D_U$$. With increasing width of the diagonal band, the probability to find the optimal alignment, but also the computation time increases.
Parameters
seq1, seq2Sequence
The sequences to be aligned.
matrixSubstitutionMatrix
The substitution matrix used for scoring.
bandtuple(int, int)
The diagonals that represent the lower and upper limit of the search space. A diagonal $$D$$ is defined as $$D = j-i$$, where $$i$$ and $$j$$ are positions in seq1 and seq2, respectively. An alignment of sequence positions where $$D$$ is lower than the lower limit or greater than the upper limit is not explored by the algorithm.
gap_penaltyint or tuple(int, int), optional
If an integer is provided, the value will be interpreted as linear gap penalty. If a tuple is provided, an affine gap penalty is used. The first integer in the tuple is the gap opening penalty, the second integer is the gap extension penalty. The values need to be negative. (Default: -10)
localbool, optional
If set to true, a local alignment is performed. Otherwise (default) a semi-global alignment is performed.
max_numberint, optional
The maximum number of alignments returned. When the number of branches exceeds this value in the traceback step, no further branches are created.
Returns
alignmentslist of Alignment
The generated alignments. Each alignment in the list has the same similarity score, which is the maximum score possible within the defined band.
align_optimal
Guarantees to find the optimal alignment at the cost of greater compuation time and memory requirements.
Notes
The diagonals give the maximum difference between the number of inserted gaps. This means for any position in the alignment, the algorithm will not consider inserting a gap into a sequence, if the first sequence has already -band[0] more gaps than the second sequence or if the second sequence has already band[1] more gaps than the first sequence, even if inserting additional gaps would yield a more optimal alignment. Considerations on how to find a suitable band width are discussed in 2.
The restriction to a limited band is the central difference between the banded alignment heuristic and the optimal alignment algorithms 34. Those classical algorithms require $$O(m \cdot n)$$ memory space and computation time for aligning two sequences with lengths $$m$$ and $$n$$, respectively. The banded alignment algorithm reduces both requirements to $$O(\min(m,n) \cdot (D_U - D_L))$$.
Implementation details
The implementation is very similar to align_optimal(). The most significant difference is that not the complete alignment table is filled, but only the cells that lie within the diagonal band. Furthermore, to reduce also the space requirements the diagnoal band is ‘straightened’, i.e. the table’s rows are indented to the left. Hence, this table
. . x x x . . . . . . . . x x x . . . . . . . . x x x . . . . . . . . x x x . . . . . . . . x x x .
is transformed into this table:
x x x x x x x x x x x x x x x
Filled cells, i.e. cells within the band, are indicated by x. The shorter sequence is always represented by the first dimension of the table in this implementation.
References
1
W. R. Pearson, D. J. Lipman, “Improved tools for biological sequence comparison.,” Proceedings of the National Academy of Sciences of the United States of America, vol. 85, pp. 2444–2448, April 1988. doi: 10.1073/pnas.85.8.2444
2
J. Gibrat, “A short note on dynamic programming in a band,” BMC Bioinformatics, vol. 19, pp. 226, June 2018. doi: 10.1186/s12859-018-2228-9
3
S. B. Needleman, C. D. Wunsch, “A general method applicable to the search for similarities in the amino acid sequence of two proteins,” Journal of Molecular Biology, vol. 48, pp. 443–453, March 1970. doi: 10.1016/0022-2836(70)90057-4
4
T. F. Smith, M. S. Waterman, “Identification of common molecular subsequences,” Journal of Molecular Biology, vol. 147, pp. 195–197, March 1981. doi: 10.1016/0022-2836(81)90087-5
Examples
Find a matching diagonal for two sequences:
>>> sequence1 = NucleotideSequence("GCGCGCTATATTATGCGCGC")
>>> sequence2 = NucleotideSequence("TATAAT")
>>> table = KmerTable.from_sequences(k=4, sequences=[sequence1])
>>> match = table.match(sequence2)[0]
>>> diagonal = match[0] - match[2]
>>> print(diagonal)
-6
Align the sequences centered on the diagonal with buffer in both directions:
>>> BUFFER = 5
>>> matrix = SubstitutionMatrix.std_nucleotide_matrix()
>>> alignment = align_banded(
... sequence1, sequence2, matrix,
... band=(diagonal - BUFFER, diagonal + BUFFER), gap_penalty=(-6, -1)
... )[0]
>>> print(alignment)
TATATTAT
TATA--AT | 1,291 | 5,209 | {"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": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.843745 |
https://www.glassdoor.com/Interview/You-and-the-interviewer-play-a-game-There-are-piles-of-stones-on-a-table-each-with-a-different-number-of-stones-On-your-QTN_242121.htm | 1,563,646,385,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526560.40/warc/CC-MAIN-20190720173623-20190720195623-00191.warc.gz | 719,682,485 | 50,097 | Pocket Gems Interview Question: You and the interviewer play ... | Glassdoor
## Interview Question
Software Engineer Interview San Francisco, CA
# You and the interviewer play a game. There are piles of
stones on a table, each with a different number of stones. On your turn, you can take from the table any positive number of stones from any single pile. The person who takes the last stone wins. What's your move?
1
Suppose We have 3 piles
Case 1: When 0 stones left implies (Loosing Postion 'L')
XOR of non of stones in piles is 0^0^0 =0
Case 2: When 1 stone remains in only one pile and rest piles are empty(Winning Pos W)
XOR of stones 1^0^0=1
So in-general if we play first we should always leave that no.of stones is pile such that XOR of remaining stones is that pile with other piles is 0.
Pragnya Sagar Choudhury on Aug 15, 2012
1
Suppose We have 3 piles
Case 1: When 0 stones left implies (Loosing Postion 'L')
XOR of non of stones in piles is 0^0^0 =0
Case 2: When 1 stone remains in only one pile and rest piles are empty(Winning Pos W)
XOR of stones 1^0^0=1
So in-general if we play first we should always leave that no.of stones is pile such that XOR of remaining stones is that pile with other piles is 0.
Pragnya Sagar Choudhury on Aug 15, 2012
0
You want to make sure that all piles have an even number of stones: the XOR answer works only assuming that the amount in each pile is the same.
Anonymous on Sep 3, 2012
2
This problem is well explained in this video | 408 | 1,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.912336 |
https://oeis.org/A013340 | 1,627,136,130,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00053.warc.gz | 459,300,004 | 3,735 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A013340 arctanh(exp(x)-sech(x))=x+2/2!*x^2+3/3!*x^3+20/4!*x^4+165/5!*x^5... 0
0, 1, 2, 3, 20, 165, 1142, 10983, 136040, 1753545, 25376042, 422373963, 7641560060, 149590274925, 3195495898142, 73536613981743, 1808159717424080, 47465557573756305, 1324783497842287442 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS FORMULA a(n) ~ (n-1)! / (2 * r^n), where r = log((1 + (19 - 3*sqrt(33))^(1/3) + (19 + 3*sqrt(33))^(1/3))/3) = 0.60937786343600623153680337116839869542853927931... is the root of the equation exp(r)-sech(r) = 1. - Vaclav Kotesovec, Feb 05 2015 MATHEMATICA CoefficientList[Series[ArcTanh[Sinh[x]*(1 + Tanh[x])], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Feb 05 2015 *) CROSSREFS Sequence in context: A128977 A069323 A009721 * A218873 A012416 A261317 Adjacent sequences: A013337 A013338 A013339 * A013341 A013342 A013343 KEYWORD nonn AUTHOR Patrick Demichel (patrick.demichel(AT)hp.com) EXTENSIONS Prepended missing a(0)=0 from Vaclav Kotesovec, Feb 05 2015 STATUS approved
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Last modified July 24 09:52 EDT 2021. Contains 346273 sequences. (Running on oeis4.) | 564 | 1,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.556547 |
https://de.mathworks.com/matlabcentral/answers/1844538-use-matlab-function-block-to-plot-into-figure-created-by-matlab-script | 1,675,212,413,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00805.warc.gz | 216,148,251 | 28,481 | # Use Matlab Function Block to plot into figure created by Matlab Script
1 view (last 30 days)
Jannis on 6 Nov 2022
Commented: Jannis on 15 Nov 2022
Hello,
my plan is to create a figure with a Matlab script and then plot the results from my Simlink Simulation in this specific plot, using a Matlab function block.
Here is the script where the figure to plot the results is created:
ax1 = axes;
imshow(map_scales, 'Parent',ax1)
hold on
My original idea was to use the ax1 in the matlab function block to plot the results in this figure. But this does not seem to work:
Here is a picture of the maltabl function block and the code inside it:
function path_plotting(x_pos, x_pos_prev, y_pos, y_pos_prev)
x_coodinates = [x_pos_prev, x_pos];
y_coodinates = [y_pos_prev, y_pos];
plot(x_coodinates, y_coodinates, 'LineWidth', 2, 'Color', 'b', 'Parent', ax1)
end
Does anyone have an idea how to solve this?
Simon Chan on 6 Nov 2022
Try this in the function path_plotting
function path_plotting(x_pos, x_pos_prev, y_pos, y_pos_prev, ax1)
x_coodinates = [x_pos_prev, x_pos];
y_coodinates = [y_pos_prev, y_pos];
plot(ax1,x_coodinates, y_coodinates, 'LineWidth', 2, 'Color', 'b')
end
Jannis on 6 Nov 2022
HI, thanks for you answer, but this does not seem to work. I think there might be no solution for my problem because simulink does not support the data type of an axes vairable.
I changed ax1 to a parameter data type and I received this error
See screenshot below
Paul on 13 Nov 2022
Edited: Paul on 13 Nov 2022
Hi Jannis,
I got this model to work
The code in the Matlab Function block is
function y = fcn(u,t)
persistent ax1
if isempty(ax1)
ax1 = gca;
end
plot(ax1,t,u,'r.')
y = u;
end
In your use case, maybe you can replace the ax1 = gca; with either your code, or a function that runs your code to create the image and return the axis handle. Anyway, I think that whatever you're trying to do is feasible.
##### 2 CommentsShowHide 1 older comment
Jannis on 15 Nov 2022
Thank you very much this solves my problem
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Translated by | 623 | 2,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.841365 |
https://oeis.org/A122414 | 1,618,640,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00446.warc.gz | 504,919,316 | 3,649 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A122414 Triangle T(n,k) for 1 <= k <= n read by rows, where T(n,k) = 1 if gcd(n,k) is prime, 0 otherwise. 2
0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS First column is all zeros, main diagonal is A010051 (characteristic function of primes). LINKS PROG (PARI) {m=14; v=vector(m, x, vector(x)); for(n=1, m, for(k=1, n, if(isprime(gcd(n, k)), v[n][k]=1))); for(n=1, m, for(k=1, n, print1(v[n][k], ", ")))} CROSSREFS Cf. A010051, A122415. Row sums are in A117494. [From Klaus Brockhaus, May 29 2009] Sequence in context: A272664 A074937 A143518 * A288216 A189628 A289239 Adjacent sequences: A122411 A122412 A122413 * A122415 A122416 A122417 KEYWORD nonn,tabl AUTHOR Klaus Brockhaus, Sep 03 2006 STATUS approved
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Last modified April 17 02:14 EDT 2021. Contains 343059 sequences. (Running on oeis4.) | 706 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-17 | latest | en | 0.615633 |
https://www.slideserve.com/synnove/the-29th-annual-acm-icpc-world-finals | 1,580,007,643,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00048.warc.gz | 1,094,043,850 | 17,301 | # The 29th Annual ACM-ICPC World Finals - PowerPoint PPT Presentation
The 29th Annual ACM-ICPC World Finals
1 / 49
The 29th Annual ACM-ICPC World Finals
## The 29th Annual ACM-ICPC World Finals
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. The 29th AnnualACM-ICPC World Finals 1. Shanghai Jiaotong University 2. Moscow State University 3. St. Petersburg Institute of Fine Mechanics and Optics 4. University of Waterloo … 17. St. Petersburg State University
2. Zero Knowledge Proofs and Protocols A proof is whatever convinces me. Shimon Even, 1978 Nikolay Vyahhi St. Petersburg State University Joint Advanced Student[s] School 2005
3. Example (graph 3-coloring) Problem (G3C):Given a graph, color its vertices with red, green, blue such that if any two vertices are joined by an edge then they receive different colors. (13/14 ) = 0,929 (13/14)10 = 0,477 (13/14)100 = 6,047*10-4 (13/14)1000 = 6,536*10-33 Probability, that A can cheat (when B opened n2 edges) at most: (1-1/n)n2 e-n
4. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
5. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
6. Introduction Conception of Zero-Knowledge Proofs
7. Introduction • Applications: • authentication // user proves to system, that he is valid user Weakness: Adversary E can prove to B, that she is A, just by asking A to prove it to her and simulating this protocol with B. • protecting against chosen message attack by augmenting the ciphertext by a zero-knowledge proof of knowledge of the cleartext. • non-oblivious commitment schemes • …
8. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
9. Interactive Proof Systems • Intuitively, what should we require from an efficient theorem-proving procedure? • That it should be possible to “prove” a true theorem. • That it should be impossible to “prove” a false theorem. • That communicating the “proof” should be efficient. Namely regardless of how much time it takes to come up with the proof, its correctness should be efficiently verified. • More formal. An interactive Turing machine (ITM) is a Turing machine equipped with read-only input tape, a work tape, a random tape, one read-only and one write-only communication tapes. The random tape contains an infinite sequence of random bits, and can be scanned only from left to right.
10. Interactive Proof Systems Interactive Turing Machine
11. Interactive Protocol • An interactive protocol is an ordered pair of ITM’s A (prover) and B (verifier) such that A and B share the same input tape, B’s write-only communication tape is A’s read-only communication tape and vice versa. • Machine A is not computationally bounded, while B is bounded by a polynomial in the length of common input. • The two machines take turns in being active, with B being active first. During an active stage A(B) first perform some internal computation using its tapes; and, second, it writes a string (for B(A)) on its write-only communication tape. Then it deactivates and machine B(A) becomes active. • Machine Baccepts (or rejects) the input by outputting “accept” (or “reject”) and terminating the protocol.
12. Interactive Protocol Interactive Turing Machines
13. Interactive Proof Systems • An interactive protocol (A,B) is called an interactive proof system for language L over {0,1}* if we have the following: • For each k, for sufficiently large x in L given as input to (A,B), B halts and accepts with probability at least 1-|x|-k. • For each k, for sufficiently large x NOT in L, for any ITM A’, on input x to (A’,B), B accepts with probability at most |x|-k. The probabilities here are taken over the readings of random bits of A and B. • Interactive Polynomial time (IP) is the class of languages for which there exists interactive proof system.
14. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
15. Zero-Knowledge For every polynomial time B’, the distribution that B’ “sees” on all its tapes, when interacting with A on input x∈L, is “indistinguishable” from a distribution that can be computed from x in polynomial time.
16. Example (QNR) Problem (QNR): QNR = { (x,y) | y is quadratic nonresidue mod x }. There is no such z, that y = z2 mod x. • So let’s try to prove with zero-knowledge for some y, that it is from QNR. With prover A, verifier B, input (x,y) and |x|=n. • B begins by flipping coins to obtain random bits b1,b2,…,bn. • Then B flips additional coins for obtaining random z1,z2…zn (0<zi<x and gcd(zi,x)=1 for each zi). • B computes w1,w2,…,wn as follows: • wi = (zi2) mod x, if bi=0 • wi = (zi2y) mod x, otherwise, if bi=1 • B sends w1,w2,…,wn to A. • A computes (somehow) for each i whether or not wi is quadratic residue mod x, and sends this information (c1,c2,…,cn) to B. • B checks if bi=ci for every i, and if so is “convinced” that (x,y)∈QNR.
17. Example (QNR) Is it zero-knowledge? NO! Why?
18. Example (QNR) What if B were to cheat? B could begin by setting w1=42 for example, and then behave correctly. So, B can compute whether or not 42 is a quadratic residue x, given x and a quadratic nonresidue y. At this time it is not known how compute this in polynomial time, so this proof system may not be zero-knowledge!
19. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
20. Indistinguishability of Random Variables • Consider families of random variables U = {U(x)}, where x∈L, a particular subset of {0,1}*, and all random variables take values in {0,1}*. • Let U(x) and V(x) be two families of random variables. • We want to express the fact that, when the length of x increases, U(x) essentially becomes “replaceable” by V(x). • So, a random sample is selected form U(x) or from V(x) and it is handed to a “judge”. After studying the sample, he proclaims, from which families our sample is.
21. Indistinguishability of Random Variables Two families of random variables {U(x)} and {V(x)} are: • Equal if the judge’s verdict will be meaningless even if he is given samples of arbitrary size and he can study them for an arbitrary amount of time. • Statically indistinguishable if the judge’s verdict became meaningless when he is given an infinite amount of time but only random, polynomial (in |x|) size samples to work on. • Computationally indistinguishable if the judge’s verdict become meaningless when he is only given polynomial (|x|)-size samples and polynomial (|x|) time.
22. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
23. Approximability of Random Variables • Let M be a probabilistic Turing machine that on input x always halts. We denote by M(x) the random variable that, for each string, which is equal to α, have the same probability that M on input x outputs α. • U is perfectly approximable on L if there exist a probabilistic Turing machine M, running expected polynomial time, such that for all x∈L, M(x) is equal to U(x). • U is statically (computationally) approximable on L if there exist a probabilistic Turing machine M, running expected polynomial time, such that for families of random variables {M(x)} and {U(x)} are statically (computationally) indistinguishable on L.
24. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
25. Zero-Knowledge • ITM B’ has an extra input tape H, which length is bounded above be a polynomial in the length of x. • When B’ interacts with A, A sees only x on its tape, whereas B’ sees (x,H). • So H is a some knowledge about x that the cheating B’ already possess. Or H can be considered as the history of previous interactions that B’ is trying to use to get knowledge from A. • Let ViewA,B’(x,H) be the random variables whose value is view of B’ (random tape, messages between parties, private tape). For convenience, we consider each view to be a string from {0,1}* of length |x|c for some fixed c>0.
26. Zero-Knowledge Interactive Turing Machines
27. Zero-Knowledge • Let L be a language and (A,B) a protocol. Let B’ be as above. We say that (A,B) is perfectly (statically) (computationally) zero-knowledge on L for B’ if the family of random variables ViewA,B is perfectly (statically) (computationally) approximable on L’ = { (x,H) | x∈L and |H|=|x|c} • We say that interactive protocol (A,B) if perfectly (statically) (computationally) zero-knowledge on L if it is perfectly (statically) (computationally) zero-knowledge on L for all probabilistic polynomial time ITM B’. Note, that this definition only depends on A and not at all on B. • Usually, only computationally zero-knowledge is consided.
28. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
29. Known Facts and Open Problems • Every language in NP has a perfect zero knowledge proof (if one-way permutations exists). • Every language in IP has a zero knowledge proof. • It’s known that (obvious) • Goldreich’s belief is that • The relationship of PZK and SZK remains an open problem (with no evidence either way).
30. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
31. Examples (GI) Problem (GI – Graph Isomorphism): You have two graphs (G0,G1), are they isomorphic? Exercise 0: Think out zero-knowledge proof for this problem. A knows, that G0 and G1 are isomorphic (and how its are) and tries to prove this fact to B. • A chooses one graph (G0 or G1), and transform it to any another isomorphic one G2 (anyhow). • A sends this graph G2 to B. • B flips a coin, and sends this bit b (0 or 1) to A. • A mustshow isomorphism of G2 and Gb to B, otherwise B can not accept.
32. Examples (GI) • If A cheating, she can’t show isomorphism of those two graphs with probability ½. But A can cheat with ½ probability also. • If B repeats this protocol n times, so A can cheat with probability only ½n=2-n (at most). • B can’t get some additional information from this interaction.
33. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
34. Examples (GNI) Problem (GNI - Graph NonIsomorphism): You have two graphs (G0,G1), are they nonisomorphic? • B chooses one graph (G0 or G1), and transform it to any another isomorphic one G2 (anyway). • B sends this graph G2 to A. • A must say, which graph was chosen by B. • If A cheating, so graphs G0 and G1 are isomorphic, and she can not say exactly, to which one G2 is isomorphic. Probability of being caught is 1-½n. • B can not get some additional information from this interaction. • Are you surein the last point?
35. Examples (GNI) It is not zero-knowledge! The same situation as with QNR earlier.
36. Examples (GNI) Problem (GNI - Graph NonIsomorphism): You have two graphs (G0,G1), are they nonisomorphic? • We must modify verifier B, so that he’ll prove to the prover A, that he (B) knows the answer to his query graph (i.e. he knows an isomorphism to the appropriate input graph), and the prover answers the query only if she is convinced of this claim. • Of course, that B’s proof must be zero-knowledge.
37. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
38. Example (QNR) Problem (QNR): QNR = { (x,y) | y is quadratic nonresidue mod x }. There is no such z, that y = z2 mod x. • Bpicks at random integer r and one bit. • if bit=0 then B sets w = r2 mod x, • otherwise w = r2y mod x. B sends w to A. • For some 1<=j<=m, B picks random integer rj1,rj2 and random bitj. B sets • aj=r2j1 mod x • bj=yr2j2 mod x If bitj=1, B sends A the ordered pair (aj,bj), else (bj,aj). • A sends B an m-long random bit vector i=i1,i2,…,im.
39. Example (QNR) • B sends A the sequence v=v1,v2,…,vm. • if ij=0 then vj = (rj1,rj2) • if ij=1 then • if bit=0 then vj = rrj1 mod x • else vj = yrrj2 mod x. The intuition behind this step is as follows: if ij=0, then B is convincing A that pair was chosen correctly; if ij=1 then B is convincing that if pair was chosen correctly, then w was chosen correctly. • A verifies that the sequence v was properly constructed, If not, A sends terminate to B and halts. Otherwise. A sets answer = 0 if w is a quadratic residue mod x and 1 otherwise, A sends answer to B.
40. Example (QNR) • B checks whether answer = bit. If so B continues the protocol, otherwise B rejects and halts. • After m repetition of this protocol, if B did not reject thus far, B accepts and halts. • Conclusion: So, we force B to prove, that he is not cheating. And now he can not obtain any other information from this protocol (only is y a quadratic nonredisue or not). => It’s a (statically) zero-knowledge proof.
41. Non-Interactive ZK Proofs General Idea: Using one-way function instead of verifier B. • A generates n random numbers, and so generates n different isomorphic (to initial) problems. • A publish all this new problems. • A uses one-way functions, to generate “random” bit string b from definitions of that new problems, which was published (it’ll be like B’s random tape). • If bi=0 then A proves isomorphism of initial and i-th new problem, otherwise she opens solution of i-th new problem. Then A publish this information. • Anyone can verify this proof without interaction.
42. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
43. Related papers • S. Goldwasser, S. Micali, C. Rackoff. “The knowledge complexity of interactive proof systems”, 1989 (1986). • U. Fiege, A. Fiat, A. Shamir. “Zero-Knowledge Proofs of Identity”, 1988. • B. Schneier. “Applied Cryptography”, 1996. • O. Goldreich. “Foundation of Cryptography”, 2001.
44. Thank you!
45. Questions?
46. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
47. Exercises • ZK proof for G3C by using a phone/email (you can’t see, what your opponent do, so you can’t believe in something sometimes).
48. Agenda • Introduction • Theory: • Interactive Proof Systems, Interactive Protocol • Zero-Knowledge, QNR example • Indistinguishability of Random Variables • Approximability of Random Variables • Zero-Knowledge • Known Facts and Open Problems • Examples: • GI • GNI • QNR • Related papers • Exercises
49. Thank you again! | 4,308 | 17,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-05 | latest | en | 0.693037 |
https://liusson.com/homework-solution-use-mathematical-induction-to-prove-the-following-be-sure-to-a-clearly-state-the-property-you-are-trying-to/ | 1,606,258,407,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00526.warc.gz | 398,290,262 | 18,937 | # Homework Solution: Use mathematical induction to prove the following. Be sure to (a) clearly state the property you are trying to…
Use mathematical induction to prove the following. Be sure to (a) clearly state the property you are trying to prove: (b) state and prove the basis case: (c) state the inductive hypothesis: (d) state and prove the inductive step, clearly indicating where you used the inductive hypothesis. sigma^n_i = 1 n*n! = 1 + 1! + 2 * 2! + .... + n*n! = (n + 1)! - 1for all n greaterthanorequalto 1
To prove 1.1!+2.2!+3.3!+....................+n.n!=(n+1)!-1 using induction
Use unversified gathering to establish the subjoined. Be trusting to (a) obviously recite the possessions you are obscure to establish: (b) recite and establish the cause case: (c) recite the inductive theory: (d) recite and establish the inductive plod, obviously indicating where you used the inductive theory. sigma^n_i = 1 n*n! = 1 + 1! + 2 * 2! + …. + n*n! = (n + 1)! – 1ce entire n greaterthanorequalto 1
## Expert Exculpation
To establish 1.1!+2.2!+3.3!+………………..+n.n!=(n+1)!-1 using gathering
Basic plod :
Let n=1
then 1.1!=1.1=1
and (1+1)!-1=2!-1=2-1=1
Hereafter the loving office is penny ce n=1
Gathering Plod:
Assume that office is penny ce F(k)
i.e., 1.1!+2.2!+3.3!+………+k.k!=(k+1)!-1;
Now cogitate F(k+1) = 1.1!+2.2!+3.3!+………+k.k!+(k+1)(k+1)!
= (k+1)!-1 + (k+1)(k+1)! past by overhead asumption
= (k+1)! (k+1+1) -1
= (k+1)!(k+2) -1
= (k+2)! -1 As (n+1).n = (n+1)!
Therefore hereafter establishd by F(k+1)
Hereafter by unversified gathering it is establishd that
1.1!+2.2!+3.3!+4.4!+………….+n.n! = (n+1)!-1 | 612 | 1,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-50 | latest | en | 0.798081 |
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# test1fa00v1 - AB = Find the instantaneous power absorbed by...
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1 EE2120 TEST No. 1.v1 September 14, 2000 NAME (please print) SSN INSTRUCTIONS Failure to follow these instructions may result in your work NOT being accepted for credit. DO NOT OPEN UNTIL THE INSTRUCTOR TELLS YOU All work is individual. Concentrate on your own paper at all times STOP WORKING AS SOON AS THE INSTRUCTOR TELLS YOU When instructed, walk to the front and leave your work on the instruc- tor’s desk If you finish ahead of time, walk to the front and leave your work on the instructor’s desk SHOW ALL YOUR WORK IN EACH PROBLEM
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Prepared by Dr. Jorge L. Aravena. Last revised on September 14, 2000 2 (15 pts) A B The source delivers 100mW of power. If I AB = 20mA V AB = ________ Point ____ has higher voltage than point _____ (A/B) In the source, positive charges move from ____ to ____ I S (20pts) A B R (sec) 0 ]; [ ) ( ³ = - t mC e Q t q t o The positive charges crossing point A are given by Find I
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Unformatted text preview: AB = ________ Find the instantaneous power absorbed by the resistor at t = 1 sec P R (1) = __________ The total energy absorbed by the resistor is E R = _________ R = 10k Q o = 10mC Terminal ____ has higher voltage than ____ Prepared by Dr. Jorge L. Aravena. Last revised on September 14, 2000 3 (20pts) +-I S 4k 8k I x aI x 2k 6k A B I S = 12mA , a=3[V/ mA ] V AB = _______ Prepared by Dr. Jorge L. Aravena. Last revised on September 14, 2000 4 (25pts) I S I 1 6k 6k 6k 1k 4k I 2 I 5 I 4 I 3 I S = 12mA , I 1 = 4mA I 2 = _______ I 3 =_______ I 4 = _______ I 5 = _______ V AB = _______ A B Prepared by Dr. Jorge L. Aravena. Last revised on September 14, 2000 5 • (25pts) _ + 8k 4k 12k R V S V S = 24V , R = 6k I o I 1 I o = ________ I 1 = ________ P R = ________ (P R is the power dissipated in the resistor R)...
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Ask a homework question - tutors are online | 752 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.845653 |
https://patriciabarber.com/money-train-otkdh/isosceles-triangle-theorems-cb07ab | 1,632,739,338,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00465.warc.gz | 485,202,651 | 10,017 | Since this is an isosceles triangle, by definition we have two equal sides. If two angles of a triangle are congruent the sides opposite them are congruent. Contact Person: Donna Roberts. See the section called AA on the page How To Find if Triangles are Similar.) Lines Containing Altitudes of a Triangle (V1) Orthocenter (& Questions) 2. If two sides of a triangle are congruent, then angles opposite to those sides are congruent. The isosceles triangle theorem states the following: Isosceles Triangle Theorem. Compare the isosceles triangle on the left . 4 lessons in Pythagoras Theorem 2: Use Pythagoras' theorem to show that a triangle is right-angled; Use Pythagoras’ theorem to find the length of a line segment; Use Pythagoras’ theorem with Isosceles Triangles; Apply Pythagoras' theorem to two triangles AB = AC To Prove :- ∠B = ∠C Construction:- Draw a bisector of ∠A intersecting BC at D. Proof:- In BAD and CAD AB = AC ∠BAD = ∠CAD AD = AD BAD ≅ CAD Thus, ∠ABD = ∠ACD ⇒ ∠B = ∠C Hence, angles opposite to equal sides are equal. If the line from an angle of a triangle Triangle Congruence Theorems (SSS, SAS, & ASA Postulates) Triangles can be similar or congruent. A triangle with two equal sides is an isosceles triangle. Theorem 2: The base angles of an isosceles triangle are congruent. Theorem 7.2 :- Angle opposite to equal sides of an isosceles triangle are equal. And using the base angles theorem, we also have two congruent angles. In such spaces, it takes a form that says of vectors x, y, and z that if. Consider isosceles triangle A B C \triangle ABC A B C with A B = A C, AB=AC, A B = A C, and suppose the internal bisector of ∠ B A C \angle BAC … If two sides of a triangle are congruent the angles opposite them are congruent. 1. If the bisector of an angle in a triangle An isosceles triangle is generally drawn so it is sitting on its base. is perpendicular to the opposite side, the triangle is isosceles. Theorems about Isosceles Triangles Dr. Wilson. The converse of the Isosceles Triangle Theorem is also true. Given :- Isosceles triangle ABC i.e. MathBitsNotebook.com About this website. Similar triangles will have congruent angles but sides of different lengths. Note: The definition of an isosceles triangle states that the triangle has two congruent "sides". So AB/BD = AC/CE With the use of CPCTC, the theorems stated above can be proven true. The line segment bisects the vertex angle. Sometimes it is specified as having exactly two sides of equal length, and sometimes as having at least two sides of equal length, the latter version thus including the equilateral triangle as a special case. An isosceles triangle is known for its two equal sides. An isosceles triangle is one of the many varieties of triangle differentiated by the length of their sides. Theorems included:Isosceles triangle base angle theorems.An Equilateral triangle is also equiangular.An Equiangular triangle is also equilateral.There are 4 practice problems that consist of 2 part answers in the foldable for st The peak or the apex of the triangle can point in any direction. Incenter Exploration (A) Incenter Exploration (B) Incenter & Incircle Action! MathBits' Teacher Resources (The Isosceles DecompositionTheorem) In an If an "inclusive" isosceles trapezoid is defined to be "a trapezoid with congruent legs", a parallelogram will be an isosceles trapezoid. If ∠ A ≅ ∠ B, then A C ¯ ≅ B C ¯. But BF = CE 4. Isosceles Triangle Theorem - Displaying top 8 worksheets found for this concept.. If two sides in a triangle are congruent, then the angles opposite the congruent sides are congruent angles 2. 1. A point is on the perpendicular bisector 3. Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal. The altitude creates the needed right triangles, the congruent legs of the triangle become the congruent hypotenuses, and the altitude becomes the shared leg, satisfying HL. same as that 90 degrees. In geometry, an isosceles triangle is a triangle that has two sides of equal length. The altitude to the base of an isosceles triangle bisects the base. Suppose a triangle ABC is an isosceles triangle, such that; AB = AC [Two sides of the triangle are equal] Hence, as per the theorem 2; ∠B = ∠C. Conversely, if the base angles of a triangle are equal, then the triangle is isosceles. Some of the worksheets for this concept are 4 isosceles and equilateral triangles, Isosceles triangle theorem 1a, , 4 angles in a triangle, Section 4 6 isosceles triangles, Isosceles triangle theorem 1b, Do now lesson presentation exit ticket, Isosceles and equilateral triangles name practice work. 3. congruent, then the sides opposite the congruent angles are congruent from this site to the Internet then the angles opposite the congruent sides are congruent angles. Isosceles Triangle Theorems and Proofs. So here once again is the Isosceles Triangle Theorem: If two sides of a triangle are congruent, then angles opposite those sides are congruent. If two sides in a triangle are congruent, Topical Outline | Geometry Outline | MathBitsNotebook.com | MathBits' Teacher Resources Isosceles Triangle Theorems. Congruent triangles will have completely matching angles and sides. Side AB corresponds to side BD and side AC corresponds to side BF. We then take the given line – in this case, the apex angle bisector – as a common side, and use one additional property or given fact to show that the triangles formed by this line are congruent. is, and is not considered "fair use" for educators. These can be tricky little triangles, If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle. the base, the following conditions are equivalent: 4. at its midpoint, then the triangle is isosceles. Terms of Use Contact Person: Donna Roberts. Slider. (Extra Credit): If the bisector of an Hypotenuse Leg Theorem-If the hypotenuse and a pair of … 5. so beware! Angles opposite to equal sides is equal (Isosceles Triangle Property) SSS (Side Side Side) congruence rule with proof (Theorem 7.4) RHS (Right angle Hypotenuse Side) congruence rule with proof (Theorem 7.5) Angle opposite to longer side is … The line segment is perpendicular to the base. In this video I will take you through the two Isosceles Triangle Theorems, as well as two proofs which make use of these theorems. The altitude to the base of an isosceles triangle bisects the vertex angle. 6. But the definition of isosceles trapezoid stated above, mentions congruent base "angles", not sides (or legs).Why? Today we will learn more about the isosceles triangle and its theorem. Topical Outline | Geometry Outline | To show this is true, draw the line BF parallel to AE to complete a parallelogram BCEF:Triangles ABC and BDF have exactly the same angles and so are similar (Why? When the altitude to the base of an isosceles triangle is drawn, two congruent triangles are formed, proven by Hypotenuse - Leg. If two sides of a triangle are congruent, the angles opposite them are congruent. Two sides of this triangle are the radii of the circle and the same lengths. (Difficult to see might be the Pythagorean theorem, and perhaps that is why so many proofs have been offered.) 2. Transcript. The base angles theorem suggests that if you have two sides of a triangle that are congruent, then the angles opposite to them are also congruent. If two angles in a triangle are ‖ x − z ‖ = ‖ y − z ‖ . sides. If two angles of a triangle are congruent, then the sides opposite those angles are congruent. Isosceles Triangle TheoremCorresponding SidesTranslationFormRight Angles. Incenter + Incircle Action (V2)! Isosceles Triangles We are now ready to prove the well-known theorem about isosceles triangles, namely that the angles at the base are equal. TERMS IN THIS SET (10) Triangles A Q R and A K P share point A. Triangle A Q R is rotated up and to the right for form triangle A Q R. If two angles in a triangle are congruent, then the sides opposite the congruent angles are congruent sides. The altitude to the base of an isosceles triangle bisects the vertex angle. Or. The altitude to the base of an isosceles triangle bisects the base. The angles opposite to equal sides of an isosceles triangle are also equal in measure. Isosceles Triangle Theorem. And so the third angle So that is going to be the same as that right over there. A triangle can be drawn by joining the ends of the two radii together. Please read the ". \[\begin{align} \angle \text{ABC} &= \angle \text{ACB} \\ Theorem: If two angles of a triangle are congruent, then the sides opposite the angles are congruent The altitude to the base of an isosceles triangle bisects the vertex angle. So AB/BD = AC/BF 3. Which fact helps you prove the isosceles triangle theorem, which states that the base angles of any isosceles triangle have equal measure? This may not, however, be the case in all drawings. The line segment meets the base at its midpoint. When the altitude to the base of an isosceles triangle is drawn, two congruent triangles are formed, proven by Hypotenuse - Leg. of a line segment if and only if it lies the same distance from the Each angle of an equilateral triangle is the same and measures 60 degrees each. If a triangle is isosceles, the triangle formed by its base and the angle bisectors of its base angle is also isosceles-If 2 sides of a triangle are congruent then the angle bisector/altitude/median/ high perpendicular bisector of the vertex angle is also an angle bisector/ altitude/ median/ perpendicular bisector. Their interior angles and … If two sides of a triangle are congruent, then the angles opposite those sides are congruent. two endpoints. In an isosceles triangle, the angles opposite to the equal sides are equal. Terms of Use Concepts Covered: Isosceles and Equilateral theorems practice foldable. is an isosceles triangle, we're going to have two This angle, is the same as that angle. Isosceles Triangle Theorem: Discovery Lab; Geometric Mean Illustration; Points of Concurrency. These two isosceles theorems are the Base Angles Theorem and the Converse of the Base Angles Theorem. with the scalene triangle on the right. Isosceles Triangle Theorem: A triangle is said to be equilateral if and only if it is equiangular. The above figure shows you how this works. The base angles of an isosceles triangle are congruent. The isosceles triangle theorem holds in inner product spaces over the real or complex numbers. triangle is isosceles. And we can see that. Triangle Congruence: SAS. x + y + z = 0 and ‖ x ‖ = ‖ y ‖ , {\displaystyle x+y+z=0 {\text { and }}\|x\|=\|y\|,} then. which is perpendicular to the opposite side meets the opposite side Examples of isosceles triangles include the isosceles right triangle, the golden triangle, and the faces of bipyramids and certain Catalan solids. The following corollaries of equilateral triangles are derived from the properties of equilateral triangle and Isosceles triangle theorem. The converse of the base angles theorem, states that if two angles of a triangle are congruent, then sides opposite those angles are congruent. Proof: Consider an isosceles triangle ABC where AC = BC. isosceles triangle, if a line segment goes from the vertex angle to To make its converse, we could exactly swap the parts, getting a bit of a mish-mash: If angles opposite those sides are congruent, then two sides of a triangle are congruent. 7. The Isosceles triangle Theorem and its converse as a single biconditional statement can be written as - According to the isosceles triangle theorem if the two sides of a triangle … The altitude to the base of an isosceles triangle bisects the base. Conversely, if the two angles of a triangle are congruent, the corresponding sides are also congruent. 1. 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Triangle ABC where AC = BC triangle have equal measure, which states that the base of an isosceles bisects. ) triangles can be similar or congruent z that if are also equal z that.. | 5,171 | 21,426 | {"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} | 4.71875 | 5 | CC-MAIN-2021-39 | latest | en | 0.855476 |
http://smallbusiness.chron.com/expensetosales-ratio-20429.html | 1,477,390,753,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720026.81/warc/CC-MAIN-20161020183840-00451-ip-10-171-6-4.ec2.internal.warc.gz | 237,238,377 | 20,579 | # Expense-To-Sales Ratio
by Kelly Bowlin, studioD
Ratios are used in accounting to measure performance in areas such as sales, expenses, cost of goods, assets and liabilities. Ratios help management make decisions by comparing a ratio from one period to the next and making appropriate changes based on the differences. The expense-to-sales ratio measures the operating expenses of a business shown on the profit and loss statement, with the gross sales of the business that are also shown on the profit and loss statement. This ratio is a quick indicator of rising or decreasing costs or rising or declining sales.
## Determining the Expense-to-Sales Ratio
To determine the expense-to-sales ratio, refer to the profit and loss statement. Take the total expenses total shown near the bottom of the statement and divide this number into the gross sales figure shown at the top of the statement. For example, if the expenses total \$15,000 and the gross sales are \$30,000, the expense-to-sales ratio would be 50 percent (\$15,000 divided by \$30,000).
## Factors in Expense-to-Sales Ratio
In a period of inflation, when costs are rising, the expense-to-sales ratio might increase because it is more expensive to produce a sale. Increased efficiency in the production process can also affect the expense-to-sales ratio. Over time, as efficiency improves, it could take less time to produce a certain good. When this happens, the expense-to-sales ratio should decrease.
## Management Decisions
Because the expense-to-sales ratio measures efficiency of the operation from one period to the next, managers should pay close attention to it. If the expense-to-sales ratio is rising, this means either expenses are on the rise or sales have declined based on the expenses. Management might have to cut costs or find out why sales have declined.
## Considerations
Management should routinely perform a ratio analysis to measure the expense and income trends of their business. Most managers look at ratios such as the expense-to-sales ratio on a monthly basis and make recommendations to upper management based on those observations. Obtaining timely financial statements is key to utilizing the expense-to-sales ratio. | 437 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.949477 |
https://www.backyardchickens.com/threads/4-sq-ft-of-floor-space-seriously.435757/ | 1,506,466,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696696.79/warc/CC-MAIN-20170926212817-20170926232817-00646.warc.gz | 732,354,032 | 15,864 | # 4 sq. ft. of floor space?! Seriously?!
Discussion in 'Turkeys' started by kitchwitch, Jan 3, 2011.
1. ### kitchwitchChillin' With My Peeps
Feb 3, 2009
Greensburg, Pa
Hello! I was debating on getting a few Bourbon Reds this spring and in preparation for the happy event I did what every expectant poultry-mother does: I bought Storey's Guide to Raising Turkeys.
They don't really go into great detail about how much room the birds should be given, just a small blurb stating the following:
For Heavy varieties, provide 1 square foot of floor space per poult up to 6 weeks of age. From 6 to 12 weeks, increase the floor space to 2 square feet per poult; from 12 weeks to 16 weeks, allow a minimum of 3 square feet. Mixed sexes grown in confinement need 4 square feet for floor space per bird from 16weeks to market.
To be clear I give my chickens 4 square foot per bird in their coop so I was struck a bit dumb when reading that Turkeys could apparently be kept like that as well.
For housing I was planning on using semi-permanent hoop houses (moving them for 3 seasons and keeping them stationary during winter) and doing the math, if I have an 8x12 structure I could have 24 turkeys (and 16 turkeys in an 8x8)?! It just doesn't seem right to me.
Personally I was figuring on keeping 3 hens and a tom in an 8x8 hoop house (only 1 hen and 1 tom would be wintered over), is that appropriate?
2. ### SEDChillin' With My Peeps
Aug 4, 2009
North West Alabama
What you are doing should work. The 4sf per turkey is for meat birds. If you were planning on raising them to butcher then you may go by the Storey's Guide to Raising Turkeys suggestion because they would be butchered between 4 and 6 months. But just remember, the more room, the better.
3. ### Steve_of_sandspoultryOverrun With Chickens
SED is correct that square footage is for meat birds, Broad Breasted types. They don't really move around alot especially once they get a bunch of weight on them. Bourbon Reds will be alot more active so the more room the better but if you are growing them out the freezer it would work. Remember you are going to need at least 6 months grow out time for a heritage variety. I'm thinking that you wantint to raise your own next year is the reason for wintering over pair? By only keeping a pair if you loose one you are out of luck for breeding your own unless you can find a replacement localy. Just something to keep in mind.
Steve
4. ### SEDChillin' With My Peeps
Aug 4, 2009
North West Alabama
I agree with Steve. I would at least keep a trio and probably keep on the lookout for who may have an extra Tom just in case something happens to him. | 665 | 2,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-39 | longest | en | 0.965311 |
https://www.coursehero.com/file/6659842/elemprob-fall2010-page39/ | 1,498,555,209,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321306.65/warc/CC-MAIN-20170627083142-20170627103142-00286.warc.gz | 843,531,862 | 66,802 | elemprob-fall2010-page39
# elemprob-fall2010-page39 - E XY = E X E Y Proof By the...
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15 Expectations As in the one variable case, we have E g ( X,Y ) = XX g ( x,y ) p ( x,y ) in the discrete case and E g ( X,Y ) = Z Z g ( x,y ) f ( x,y ) dxdy in the continuous case. If we set g ( x,y ) = x + y , then E ( X + Y ) = Z Z ( x + y ) f ( x,y ) dxdy = Z Z xf ( x,y ) dxdy + Z Z yf ( x,y ) dxdy. If we now set g ( x,y ) = x , we see the first integral on the right is E X , and similarly the second is E Y . Therefore E ( X + Y ) = E X + E Y. Proposition 15.1 If X and Y are independent, then E [ h ( X ) k ( Y )] = E h ( X ) E k ( Y ) . In particular,
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Unformatted text preview: E ( XY ) = ( E X )( E Y ) . Proof. By the above with g ( x,y ) = h ( x ) k ( y ), E [ h ( X ) k ( Y )] = Z Z h ( x ) k ( y ) f ( x,y ) dxdy = Z Z h ( x ) k ( y ) f X ( x ) f Y ( y ) dxdy = Z h ( x ) f X ( x ) Z k ( y ) f Y ( y ) dy dx = Z h ( x ) f X ( x )( E k ( Y )) dx = E h ( X ) E k ( Y ) . 39...
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## This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.
Ask a homework question - tutors are online | 470 | 1,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.809436 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Mathematical_Methods_in_Chemistry_(Levitus)/15%3A_Matrices | 1,721,764,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00090.warc.gz | 136,665,559 | 31,124 | # 15: Matrices
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##### Chapter Objectives
• Learn the nomenclature used in linear algebra to describe matrices (rows, columns, triangular matrices, diagonal matrices, trace, transpose, singularity, etc).
• Learn how to add, subtract and multiply matrices.
• Learn the concept of inverse.
• Understand the use of matrices as symmetry operators.
• Understand the concept of orthogonality.
• Understand how to calculate the eigenvalues and normalized eigenvectors of a 2 × 2 matrix.
• Understand the concept of Hermitian matrix
• 15.1: Definitions
Some types of matrices have special names.
The sum of two matrices A and B (of the same dimensions) is a new matrix of the same dimensions, C = A+ B. The sum is defined by adding entries with the same indices.
• 15.3: Matrix Multiplication
If A has dimensions m×n and B has dimensions n×p , then the product AB is defined, and has dimensions m×p .
• 15.4: Symmetry Operators
A symmetry operation, such as a rotation around a symmetry axis or a reflection through a plane, is an operation that, when performed on an object, results in a new orientation of the object that is indistinguishable from the original.
• 15.5: Matrix Inversion
The inverse of a square matrix A , sometimes called a reciprocal matrix, is a matrix A−1 such that AA−1=I , where I is the identity matrix.
• 15.6: Orthogonal Matrices
A nonsingular matrix is called orthogonal when its inverse is equal to its transpose.
• 15.7: Eigenvalues and Eigenvectors
Since square matrices are operators, it should not surprise you that we can determine its eigenvalues and eigenvectors. The eigenvectors are analogous to the eigenfunctions we discussed for quantum mechanics.
• 15.8: Hermitian Matrices
A Hermitian matrix (or self-adjoint matrix) is a square matrix with complex entries that is equal to its own conjugate transpose. Hermitian matrices are a generalization of the symmetric real matrices we just talked about, and they also have real eigenvalues, and eigenvectors that form a mutually orthogonal set.
• 15.9: Problems
This page titled 15: Matrices is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marcia Levitus via source content that was edited to the style and standards of the LibreTexts platform. | 2,288 | 6,613 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.193333 |
https://cmtext.indiana.edu/acoustics/chapter1_standing.php | 1,611,035,837,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00110.warc.gz | 276,587,773 | 9,990 | # Chapter One: An Acoustics Primer
#### 13. What are standing waves?
A particular pattern of constructive and destructive interference is called a standing wave, which is essential to the way most musical instruments produce sound, but very undesirable in the listening environment of an electronic or recording studio.
The prior examination of constructive and destructive interference, along with concepts of reflected phase come into play. When two traveling sound waves are propagating in opposite directions, as a result of reflections inside a bounded system, be it a tube, string or room, the interference of these waves combine to create a resultant wave that appears to stand still at certain resonant frequencies called modes. These modes generally correspond with harmonic partials, with the first mode producing the fundamental. In a listening environment, these are called room modes, and the result is they unduly enhance or suppress certain frequencies at different spots in the room.
String Modes: The fundamental resonating frequency of a string is determined by its tension, mass and length (see formula). An ideal string* will produce the fundamental and all harmonic partials, each as a standing wave force that will be combined with the others to produce the complex fluid shape and motion of the string (for a fascinating view of a bowed violin string demonstrating the modified Helmholtz "kink wave" motion, view this video).
Click video image to play/pause
### First Six String Modes
Simulation created with free Falstad applets at www.falstad.com. Try them!
Nodes and antinodes on the resultant vibrating string correspond to points of minimum (node) and maximum (antinode) displacement of the string, as illustrated in the video example below. String modes always have the mode + 1 number of nodes, and the mode number of antinodes. So a string vibrating in the fundamental mode has a node at each fixed end point, and one antinode in the middle, the point of maximum displacement. The second mode (2nd partial) has three nodes and two antinodes, and so forth. As you will note from the video below, the non-end nodes of the partials divide the string up into segments which are the inverse of the partial number, so the 2nd partial node is at 1/2 the string length, the 3rd partial nodes divide at 1/3 and 2/3, etc.. The location of the first node on either end of a string will also produce that partial as a harmonic if the string is touched lightly at those spots.
Click video image to play/pause
### String Standing Wave Example
Simulation created with free Falstad applets at www.falstad.com. Try them!
The video above demonstrates the 7th mode of vibration for a string bound on both ends. Note the 8 nodes and 7 antinodes. Because a bound string reflects 180° out of phase at its boundaries, play and pause the video to see the effect as well as the result created by the constructive and destructive interference of the two oppositely-traveling waves underneath.
*An ideal string has the optimal balance of mass, length and tension to produce perfect harmonic partials, even when struck or plucked. However, many strings, particularly on guitars and piano are not ideal and produce slightly inharmonic partials. Windings on lower strings attempt to increase mass to mitigate some of the non-ideal characteristics.
Standing Waves in Wind Instruments:
Woodwind instruments are examples of half- or quarter-wave resonators that produce multiple standing wave modes. The differences are whether the tube is open at both ends (flute, including the embouchure hole) or closed at one end, such as the oboe or clarinet—or cylindrical or conical. As discussed in the prior pages, air columns at boundaries reflect differently from string reflections (for air, at a closed end, there is no phase change, at an open end there is a 180° change). In the case of air columns, pressure nodes and antinodes refer to the minimal (node) pressure change and maximum (antinode) pressure change (either above or below equilibrium) in the tube. At the closed end, where there is no phase shift, there is maximum pressure change from the constructive interference that occurs and we would call that a pressure antinode. That is the location that maximum compression or rarification takes place. At an open end (for example, an oboe), the 180° flip creates a pressure node of minimum pressure change due to the destructive interference, so a pressure node represents a location of equilibrium. A flute, which is considered open on both ends has pressure nodes on both ends. If you return to an earlier graphic, you will notice that pressure and displacement (how far an air molecule has been pushed off its origin) are in a cosine-sine relationship. Therefore displacement nodes and antinodes are in this same offset relationship. An air pressure node is an air displacement antinode and visa versa.
Click video image to play/pause
### Pressure Nodes and Antinodes in Cylinder Open on One End
LEGEND: N = pressure node, A = pressure antinode, green = maximum compression, red = maximum rarefaction, black = equilibrium
Simulation created with free Falstad applets at www.falstad.com. Try them!
[GRAPHIC OF AIR COLUMN STANDING MODES HERE, similar to these]
Standing Waves in a Fully Closed Tube:
We have not yet mentioned tubes which are closed on both ends. Below is a picture of a fully enclosed tube with a loudspeaker at one end (some examples use a pulled rosin-coated rod to excite the sound inside the tube). As certain resonant frequencies are played, the pellets in the tube form the nodes and antinodes of a standing wave pattern for that mode. Each end of the tube forms a pressure antinode, with the pellets visually responding to create the standing wave pressure nodes (flat spots) and antinodes (ridges) across the tube. This is an example of a Kundt's Tube experiment.
Exhibit built and designed by the WonderLab Exhibit Team led by Don Marvel, Bloomington, Indiana. | 1,258 | 6,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.887931 |
https://greenemath.com/Pre-Algebra_Practice/Decimals_to_Fractions_Practice/Decimals-to-Fractions.html | 1,726,014,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00633.warc.gz | 240,898,343 | 4,008 | Practice Objectives
• Demonstrate the ability to convert a decimal into a fraction
• Demonstrate the ability to convert a decimal into a mixed number
## Practice Converting Decimals to Fractions
Instructions:
Answer 7/10 questions correctly to pass.
Problem:
Correct!
Not Correct!
Converting a Decimal Number into a Fraction or Mixed Number:
1. Write any whole number
2. Count the number of decimal places
3. Write the decimal part as the numerator
4. Write the denominator as a 1 followed by the number of zeros from step 2
5. Simplify the fraction if the GCF between the numerator and denominator is not 1
Step-by-Step:
You Have Missed 4 Questions... | 152 | 662 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.766154 |
https://proxieslive.com/tag/filters/ | 1,590,974,789,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413901.34/warc/CC-MAIN-20200601005011-20200601035011-00149.warc.gz | 500,483,704 | 13,436 | ## work with sets, filters and cycles
Im trying to represented a set of data with a mathematical expression but I don’t know what is the best method for this.
In programming you have data set like
``var x = [1,2,3,4] ``
and you can apply filter like:
``x.filter( element => element > 2 ) ``
or loops like:
``x.foreach( element => { some filter conditions } ) ``
exists a mathematical expression for this cases?
An apology if the question is very ambiguous, but I would like your guidance to begin to understand programming processes at a more abstract level and to be able to represent cases with clear and universal mathematical expressions that can later be translated into programming languages.
## Optimization of a query to retrieve records randomly with multiple joins and filters
I have the following schema:
This question was posted also in StackOverflow, but I want to consult also to specialists more focused on DB administration because the nature of my project. Sorry if this is a mistake
Right now, the table `Property` hold more than 70K records. I’m developing an update to support more than 500 concurrent sessions. The application will support a map a to make the searches, that’s why `GeoLocation` declares `Coordinate` as `geography` data type. Now we have a big problem, because the response time for some queries (the most important ones) is very slow. I mean, the application has to return around 1000 records at once if there are that quantity of results for the specified parameters.
The parameters are distributed on all the tables of the schema (actually, it’s a portion of the schema). Being `Features` a table which holds all the principal “characteristics” of the properties (# of bedrooms, # of garages, etc).
With that on mind, the query that is taking so much time right now is the following:
``DECLARE @cols NVARCHAR(MAX), @query NVARCHAR(MAX); DECLARE @properties TABLE( [ID] INT ) INSERT INTO @properties SELECT p.[Id] FROM[Property] p INNER JOIN[GeoLocation] AS[g] ON[p].[Id] = [g].[PropertyId] INNER JOIN[PropertyFeature] AS[pf] ON[pf].[PropertyId] = [p].[Id] INNER JOIN[Feature] AS[f] ON[pf].[FeatureId] = [f].[Id] WHERE[g].[Address] IS NOT NULL AND(([g].[Address] <> N'') OR[g].[Address] IS NULL) AND[pf].[FeatureId] IN( Select ID from feature where featuretype = 1) GROUP BY p.Id, p.ModificationDate ORDER BY [p].ModificationDate DESC, newid() OFFSET 0 ROWS FETCH NEXT 1000 ROWS ONLY DECLARE @features TABLE( [Name] NVARCHAR(80) ) INSERT INTO @features select Name from feature where FeatureType = 1 CREATE TABLE #temptable ( Id INT, Url NVARCHAR(200), Title NVARCHAR(300), Address NVARCHAR(200), Domain Tinyint, Price Real, Image NVARCHAR(150), Name NVARCHAR(80), Value NVARCHAR(150) ) INSERT INTO #temptable SELECT [t].[Id], [t].[Url], [t].[GeneratedTitle] AS[Title], [t].[Address], [t].[Domain], [t].[Price], (SELECT TOP(1) ISNULL([m].[Resize1200x1200], [m].Resize730x532) FROM [Multimedia] AS[m] WHERE [t].[Id] = [m].[PropertyId] and m.MultimediaType = 1 ORDER BY [m].[Order]) AS[Image], [t].[Name], [t].[Value] FROM (SELECT [p].[Id], [p].[Url], [p].[GeneratedTitle], [g].[Address], [p].[Domain], [pr].[Amount] AS Price, [p].[ModificationDate], [f].[Name], [pf].[Value] FROM [Property] AS [p] INNER JOIN [GeoLocation] AS[g] ON [p].[Id] = [g].[PropertyId] INNER JOIN [PropertyFeature] AS[pf] ON [pf].[PropertyId] = [p].[Id] INNER JOIN [Feature] AS[f] ON [pf].[FeatureId] = [f].[Id] INNER JOIN [Operation] AS [o] ON [p].[Id] = [o].[PropertyId] INNER JOIN [OperationType] AS [o0] ON [o].[OperationTypeId] = [o0].[Id] INNER JOIN [Price] AS [pr] ON [pr].[OperationId] = [o].[Id] WHERE p.Id in (Select Id from @properties) GROUP BY [p].[Id], [p].[Url], [p].[GeneratedTitle], [g].[Address], [p].[Domain], [pr].[Amount], [p].[ModificationDate], [f].[Name], [pf].[Value]) AS[t] ORDER BY[t].[ModificationDate] DESC SET @cols = STUFF( ( SELECT DISTINCT ','+QUOTENAME(c.[Name]) FROM @features c FOR XML PATH(''), TYPE ).value('.', 'nvarchar(max)'), 1, 1, ''); SET @query = 'SELECT [Id], [Url], [Title], [Address], [Domain], [Price], [Image], ' + @cols + ' FROM (SELECT [Id], [Url], [Title], [Address], [Domain], [Price], [Image], [Value] AS [value], [Name] AS[name] FROM #temptable)x PIVOT(max(value) for name in ('+@cols+')) p'; EXECUTE(@query); DROP TABLE #temptable ``
The execution plan and Live query statistics show me the following:
Query Execution Plan
The previous query tries to obtain randomly a X number of records IDs, holding all the filter criteria to obtain only the IDs of the records which meet that criteria. The time right now is up to 15 seconds. It’s a lot if we talk about more than 400 users using concurrently the application.
Please, help me with this. I’m three weeks trying to solve this problem with no success, but a lot of advances has been made (before it was consuming 2 minutes in avg).
If it helps, I can give you access to a “dummy” deployed version of the DB with the same quantity of records to test and see directly the problem.
===================================================================================================== INDEXES:
the indexes that are currently on the tables are the following:
``GO CREATE UNIQUE NONCLUSTERED INDEX IX_Property_ModificationDate ON [dbo].[Property] (ModificationDate DESC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE NONCLUSTERED INDEX [IX_Property_ParentId_StatusCode] ON [dbo].[Property] ([ParentId] ASC, [StatusCode] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Property_ParentId_StatusCode_Id_ModificationDate] ON [dbo].[Property] ([ParentId] ASC, [StatusCode] ASC, [Id] ASC, [ModificationDate] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Property_ParentId] ON [dbo].[Property]([ParentId] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Property_Identity_Domain_StatusCode] ON [dbo].[Property]([Identity] ASC, [Domain] ASC, [StatusCode] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Property_Id_ModificationDate] ON [dbo].[Property] (Id ASC, ModificationDate ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Property_PublisherId] ON [dbo].[Property]([PublisherId] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Property_RealEstateTypeId] ON [dbo].[Property]([RealEstateTypeId] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE INDEX FIX_Property_StatusCode_Online ON [dbo].[Property](StatusCode) WHERE StatusCode = 1 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE INDEX FIX_Property_StatusCode_Offline ON [dbo].[Property](StatusCode) WHERE StatusCode = 0 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE INDEX FIX_Property_Domain_Urbania ON [dbo].[Property](Domain) WHERE Domain = 1 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE INDEX FIX_Property_Domain_Adondevivir ON [dbo].[Property](Domain) WHERE Domain = 2 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO GO CREATE NONCLUSTERED INDEX [IX_GeoLocation_PropertyId_ModificationDate] ON [dbo].[GeoLocation] (PropertyId ASC, [ModificationDate] DESC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_GeoLocation_PropertyId_Address] ON [dbo].[GeoLocation] (PropertyId ASC, [Address] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE UNIQUE NONCLUSTERED INDEX IX_GeoLocation_ModificationDate ON [dbo].[GeoLocation] (ModificationDate DESC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE NONCLUSTERED INDEX [IX_GeoLocation_Ubigeo] ON [dbo].[GeoLocation]([Ubigeo] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE UNIQUE NONCLUSTERED INDEX [IX_GeoLocation_PropertyId] ON [dbo].[GeoLocation]([PropertyId] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE SPATIAL INDEX SIX_GeoLocation_Coordinate ON [dbo].[GeoLocation](Coordinate) GO CREATE INDEX FIX_GeoLocation_Domain_Urbania ON [dbo].[GeoLocation](Domain) WHERE Domain = 1 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO CREATE INDEX FIX_GeoLocation_Domain_Adondevivir ON [dbo].[GeoLocation](Domain) WHERE Domain = 2 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO GO CREATE NONCLUSTERED INDEX [IX_Multimedia_PropertyId_Order] ON [dbo].[Multimedia] (PropertyId ASC, [Order] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Multimedia_PropertyId] ON [dbo].[Multimedia]([PropertyId] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_Multimedia_Order] ON [dbo].[Multimedia]([Order] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [PK_Multimedia_Property] ON [dbo].[Multimedia]([Id] ASC, [PropertyId] ASC); GO CREATE INDEX FIX_Multimedia_MultimediaType_Image ON [dbo].[Multimedia](MultimediaType) WHERE MultimediaType = 1 WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON) GO GO CREATE NONCLUSTERED INDEX [IX_PropertyFeature_PropertyId_FeatureId] ON [dbo].[PropertyFeature] (PropertyId ASC, [FeatureId] ASC) WITH( SORT_IN_TEMPDB = OFF, DROP_EXISTING = OFF, FILLFACTOR = 90, ONLINE = ON); GO CREATE NONCLUSTERED INDEX [IX_PropertyFeature_FeatureId] ON [dbo].[PropertyFeature]([FeatureId] ASC); GO CREATE NONCLUSTERED INDEX [IX_PropertyFeature_PropertyId] ON [dbo].[PropertyFeature]([PropertyId] ASC); GO CREATE NONCLUSTERED INDEX [IX_PropertyFeature-FeatureId] ON [dbo].[PropertyFeature]([Id] ASC, [FeatureId] ASC); GO CREATE NONCLUSTERED INDEX [IX_PropertyFeature_Property] ON [dbo].[PropertyFeature]([Id] ASC, [PropertyId] ASC); GO CREATE NONCLUSTERED INDEX [IX_Operation_PropertyId] ON [dbo].[Operation]([PropertyId] ASC); GO CREATE NONCLUSTERED INDEX [IX_Operation_OperationTypeId] ON [dbo].[Operation]([OperationTypeId] ASC); GO CREATE NONCLUSTERED INDEX [IX_Price_OperationId] ON [dbo].[Price]([OperationId] ASC); GO CREATE NONCLUSTERED INDEX [IX_Price_Operation] ON [dbo].[Price]([Id] ASC, [OperationId] ASC); ``
## Best Practices for Filters
I have several domains that I want to filter from my projects so they never attempt to post to them.
I have 2 questions.
1. Is it better to create a .txt file with a list of all the domains I want to filter out and enter in the URL of that .txt file in the “Options” —> “Filter” section and GSA will pull the list daily, -OR- is it better to list each domain individually within the Filter list itself?
2. If I want to filter/block a domain and ALL of it’s subdomains, what format do I use? I have seen *domain.com suggested?
For example, if I want to block all art.blog subdomains, and I put in *art.blog would that also block bestart.blog (I don’t want that, I only want to blog art.blog subdomains).
## Single and double Encoding of parameters to bypass security filters
I was practising some labs to get better at XSS and SSRF. I found that sometimes I just encode characters once and the security filter is bypassed and occasionally I have to encode it two times.
How does the filter work in the backend? Can someone explain with a scenario?
## Country Select with region filters
I’m tinkering with a design for user to select a single country, or multiple countries that are historically/culturally grouped (eg United Kingdom).
Just looking for feedback, and wondering if anyone has seen something along these lines.
Dropdown 1: Region (whole world, Europe, Nth America, Sth America, SE Asia etc)
Drowdown 2: Country (all, then list of appropriate countries based on dropdown 1)
The user first selects a region (eg Europe) then an Country (eg Germany).
If Country=All, then the search results will return all countries in the selected region (or all countries if region=whole world).
Changing region changes the options in the Country list.
One issue I have is how to deal with regions like United Kingdom. Ignoring Brexit, the UK countries are also in the Europe region. One country being in multiple regions is not a problem. The problem is there are quite a few regions like this (UK, British Isles, Balkans, Caribbean etc). I can absolutely see users wanting to have results returned for “all countries in the UK”.
Should I just have lots of regions? All world, continents (EU, Africa etc), then these special subsets (UK etc)? Or would you go All world, then Continents+SpecialRegions [sorted a-z]?
## Implementing Filters as Tabs in a Web Dashboard
We are trying to provide users with a strong filter mechanism on a Dashboard that collects visit data. It feels like we have arrived at a point where only a few pieces of the puzzle are missing and we don’t know how to place them.
The user should be able to:
1. Define the filter parameters.
2. Save them for future use along with a name.
3. Apply them for one-time use to quickly check or search something without the need for saving them.
4. Modify the parameters of an existing saved filter and save it again.
5. Open tabs in the software itself and apply different filters (saved and unsaved) to easily switch from one type of data to another.
6. Close tabs.
Constraints:
There is a live view tab that is static and can’t be modified or removed by the user. (Mentioned from now on as the default tab)
Solution: (So far) 1. A filter button on the dashboard that opens a side panel for filters that contains all the parameters and allows the user to save a filter, save as existing or just apply.
1. A Tab System that works like you’d expect. It allows the user to change the filter applied in a tab by clicking on a drop-down icon on the right side of the tab itself and selecting from the saved filters list.
2. A plus button at the end of the tab list which works exactly like the drop-down above, prompting the user to select which filter view to apply on the new tab avoiding any empty states or empty tabs open.
Points we still can’t seem to get a grip on:
1. How to make the experience of applying a filter to any open tab as seamless as possible? For example, if the user is on a tab and he clicks on the Filters button and changes some parameters and clicks Apply. Should we update the view to Unsaved View? Should we keep the name of the tab same with an asterisk to denote unsaved state with a reset button? Or should we open a new tab (potentially leading to a lot of unsaved tabs open?)
2. If subsequently, he refreshes his page? Should the unsaved view disappear (being unsaved) or should we keep it in memory (potentially not really creating a need to use the ‘Save Filter’ feature)
3. How to treat the default tab in all this? Should we disable all filtering options in the default tab to avoid any confusion leading to some loss in discoverability of filter features that are available as they will not show in the default tab or should we show filtering options on the default tab and open any changes made on the default tab in a new tab leading to a little bit of inconsistency (changing any filters from the default tab always opens a new tab while changing any filters from other tabs only makes changes to that specific tab and then implementing a fail safe or an error state if the user tries to make changes to the default tab with the maximum limit reached)
4. Creating a responsive design around the solution.
## how to use more than ten filters in conditions of list view
Can we create more than ten filters in list . SharePoint 2010 is providing only 10 filters OOTB and in my scenario i am having 35 filters how can i make it with. can we do it with JQuery ?
## ‘Select All’ option in SharePoint 2013 list column filters
I have SharePoint 2013 list. Need to get ‘Select All’ option in list filter (on the columns’ filters). Currently SP only has ‘Clear Filters’. The feauture I need is the same as in Excel Autofilter:
## String filtering – process hundreds to millions of filters
What would be the most efficient way (whether with algorithims, cpu(s), DBs & SQL, distributed computing, etc) to process many strings, say ~1000/minute, and filter each string over 100s to potentially millions of different filter parameters. A parameter can be a simple statement such as including or not including the word “cat” or including “dog” not “cat” and as “complicated” as including multiple boolean gates with added timestamp ranges (logs). Each individual filter that marks true would be collected and some operation would run for each.
## ‘select All’ option in SharePoint list column filters
I need ‘Select All’ option in SharePoint list column filter. So that if I click ‘Select All’ option than all filters of check boxes has to select. | 4,615 | 18,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.909677 |
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Author Message
fidvorum.vze.cam
Registered: 12.11.2006
From:
Posted: Saturday 30th of Dec 07:35 Hi all, I just got started with my logarithmic formula with variables class. Boy! This thing is really awful ! I just never seem to understand the point behind any topic . The result? My grades suffer. Is there any expert who can lend me a helping hand?
nxu
Registered: 25.10.2006
From: Siberia, Russian Federation
Posted: Sunday 31st of Dec 10:29 Although I understand what your situation is , but if you could elaborate a bit on the areas in which you are facing struggling, then I might be in a better position to help you out . Anyhow I have some advice for you, try Algebrator. It can solve a wide range of questions, and it can do so within minutes. And that’s not it , it also gives a detailed step-by-step description of how it arrived at a particular answer. That way you don’t just find a solution to your problem but also get to understand how to go about solving it. I found this program to be particularly useful for solving questions on logarithmic formula with variables. But that’s just my experience, I’m sure it’ll be good no matter what the topic is .
Momepi
Registered: 22.07.2004
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Posted: Monday 01st of Jan 10:24 I agree. Algebrator not only gets your assignment done faster, it actually improves your understanding of the subject by providing useful tips on how to solve similar problems . It is a very popular online help tool among students so you should try it out.
Jrobhic
Registered: 09.08.2002
From: Chattanooga, TN
Posted: Tuesday 02nd of Jan 08:11 adding matrices, subtracting fractions and least common measure were a nightmare for me until I found Algebrator, which is truly the best math program that I have ever come across. I have used it frequently through several math classes – Basic Math, Intermediate algebra and Intermediate algebra. Just typing in the math problem and clicking on Solve, Algebrator generates step-by-step solution to the problem, and my math homework would be ready. I really recommend the program. | 720 | 2,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-13 | latest | en | 0.896218 |
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Get started with the Help page
# Schedule for Photovoltaic Efficiency in Simple mode
How can I create an schedule for energy efficiency to add in the "PhotovoltaicPerformanceSimple" using the Conversion mode scheduled depend on the temperature of the surface of painel? I want to relate the surface temperature (or at last the temperature of wheather file) to the effiency of the painel. Could be with a equation or a table of temperature x = efficiency y. It is possible? I want to compare some different tecnologies production.
Thank You.
Henrique Lauffer
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Sort by ยป oldest newest most voted
If you want to use the PhotovoltaicPerformance:Simple option, then I would suggest to run two simulations:
• The first one is to obtain the surface temperature, by reporting the Surface Outside Face Temperature on a Timestep level. Make sure, that the modeled surface has similar construction materials to a PV panel's materials. Once you have this, you can calculate the efficiency of the PV system in excel for every timestep and save it in a .csv file. In the documentation of SAM equation 9.32 can help you determine, how change in temperature changes the efficiency. This equation is using cell temperature, which is usually 0-3 degrees higher, than the surface temperature.
• Then you can import the .csv to EnergyPlus with the Schedule:File object and run the simulation again, with the new efficiency schedule to get the PV output.
An other option is, to go for a more detailed PV model, like PhotovoltaicPerformance:EquivalentOne-Diode or PhotovoltaicPerformance:Sandia which have the temperature dependency already built-in. I would only suggest to go for this option, if you now exactly what type of PV modules you have, and you can get their exact parameters from the Sandia or CEC database.
more
Thanks, this input schedule with file helped to get the efficiency hourly. The photovoltaics are simulated by shadow surface, instead of modeled the surface with similar construction materials, can I get this temperature of the shadow surface? Thanks
( 2019-01-27 09:30:35 -0500 )edit
e+ doesn't calculate the temperature of shading surfaces. For the first simulation to get the temperatures, in Sketchup-Euclid you can make a new zone and copy-paste the shadowing surfaces there, and delete the original shade surfaces. e+ might throw a warning due to non-enclosed zones, but the simulation should be fine.
( 2019-01-28 02:57:52 -0500 )edit | 556 | 2,560 | {"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} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.898082 |
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# 2 Marks
Unit-I
1. What is meant by finite element?
A small units having definite shape of geometry and nodes is called finite element.
2. What is meant by node or joint?
Each kind of finite element has a specific structural shape and is inter- connected with the
adjacent element by nodal point or nodes. At the nodes, degrees of freedom are located. The
forces will act only at nodes at any others place in the element.
3. What is the basic of finite element method?
Discretization is the basis of finite element method. The art of subdividing a structure in
to convenient number of smaller components is known as discretization.
4. What are the types of boundary conditions?
Primary boundary conditions
Secondary boundary conditions
5. State the methods of engineering analysis?
Experimental methods
Analytical methods
Numerical methods or approximate methods
6. What are the types of element?
1D element
2D element
3D element
7. State the three phases of finite element method.
Preprocessing
Analysis
Post Processing
8. What is structural problem?
Displacement at each nodal point is obtained. By these displacements solution stress and
strain in each element can be calculated.
9. What is non structural problem?
Temperature or fluid pressure at each nodal point is obtained. By using these values
properties such as heat flow fluid flow for each element can be calculated.
i
2 Marks
10. What are the methods are generally associated with the finite element analysis?
Force method
Displacement or stiffness method.
11. Explain stiffness method.
Displacement or stiffness method, displacement of the nodes is considered as the
unknown of the problem. Among them two approaches, displacement method is desirable.
12. What is meant by post processing?
Analysis and evaluation of the solution result is referred to as post processing.
Postprocessor computer program help the user to interpret the result by displaying them in
graphical form.
13. Name the variation methods.
Ritz method.
Ray-Leigh Ritz method.
14. What is meant by degrees of freedom?
When the force or reaction act at nodal point node is subjected to deformation. The
deformation includes displacement rotation, and or strains. These are collectively known as
degrees of freedom
15. What is meant by discretization and assemblage?
The art of subdividing a structure in to convenient number of smaller components is
known as discretization. These smaller components are then put together. The process of uniting
the various elements together is called assemblage.
16. What is Rayleigh-Ritz method?
It is integral approach method which is useful for solving complex structural problem,
encountered in finite element analysis. This method is possible only if a suitable function is
available.
17. What is Aspect ratio?
It is defined as the ratio of the largest dimension of the element to the smallest dimension.
In many cases, as the aspect ratio increases the in accuracy of the solution increases. The
conclusion of many researches is that the aspect ratio should be close to unity as possible.
ii
2 Marks
## 18. What is truss element?
The truss elements are the part of a truss structure linked together by point joint which
transmits only axial force to the element.
19. What are the h and p versions of finite element method?
It is used to improve the accuracy of the finite element method. In h version, the order of
polynomial approximation for all elements is kept constant and the numbers of elements are
increased. In p version, the numbers of elements are maintained constant and the order of
polynomial approximation of element is increased.
20. Name the weighted residual method.
Point collocation method
Sub domain collocation method
Lest squares method
Galerkins method.
21. List the two advantages of post processing.
Required result can be obtained in graphical form. Contour diagrams can be used to
understand the solution easily and quickly.
22. During discretization, mention the places where it is necessary to place a node?
Cross-section changing point
Different material interjections point
## 23. What is the difference between static and dynamic analysis?
Static analysis: The solution of the problem does not vary with time is known as static
analysis
Example: stress analysis on a beam
Dynamic analysis: The solution of the problem varies with time is known as dynamic
analysis
Example: vibration analysis problem.
24. Name any four FEA softwares.
ANSYS
NASTRAN
COSMOS
DYANA
iii
2 Marks
## 25. Differentiate between global and local axes.
Local axes are established in an element. Since it is in the element level, they change with
the change in orientation of the element. The direction differs from element to element.
Global axes are defined for the entire system. They are same in direction for all the
elements even though the elements are differently oriented.
26. Distinguish between potential energy function and potential energy functional
If a system has finite number of degree of freedom (q1, q2, and q3), then the potential
energy expressed as,
= f (q1, q2, and q3)
It is known as function. If a system has infinite degrees of freedom then the potential
energy is expressed as
## 27. What are the types of loading acting on the structure?
Body force (f)
Traction force (T)
28. Define the body force
A body force is distributed force acting on every elemental volume of the body
Unit: Force per unit volume.
Example: Self weight due to gravity
29. Define traction force
Traction force is defined as distributed force acting on the surface of the body.
Unit: Force per unit area.
Example: Frictional resistance, viscous drag, surface shear
Point load is force acting at a particular point which causes displacement.
iv
2 Marks
UNIT 2
1. What are the basic steps involved in the finite element modeling?
Discretization of structure.
Numbering of nodes.
2. Write down the general finite element equation.
FKu
where
FForce vector
KStiffness Matrix
uDisplacement Matrix
3. What is discretization?
The art of subdividing a structure in to a convenient number of smaller components is
known as discretization.
4. What are the classifications of coordinates?
Global coordinates
Local coordinates
Natural coordinates
5. What is Global coordinates?
The points in the entire structure are defined using coordinates system is known as global
coordinate system.
6. What is natural coordinates?
A natural coordinate system is used to define any point inside the element by a set of
dimensionless number whose magnitude never exceeds unity. This system is very useful in
assembling of stiffness matrices.
7. Define shape function.
Approximate relation (x,y) = N1 (x,y) 1 + N2 (x,y) 2 + N3 (x,y) 3
Where 1, 2, and 3 are the values of the field variable at the nodes N 1, N2, and N3 are
the interpolation functions.N1, N2, and N3 are also called shape functions because they are used to
express the geometry or shape of the element.
2 Marks
## 8. What are the characteristic of shape function?
It has unit value at one nodal point and zero value at other nodal points. The sum of shape
function is equal to one.
9. Why polynomials are generally used as shape function?
Differentiation and integration of polynomial are quite easy. The accuracy of the result
can be improved by increasing the order of the polynomial. It is easy to formulate and
computerize the finite element equations
10. How do you calculate the size of the global stiffness matrix?
Global stiffness matrix size = Number of nodes X Degrees of freedom per node.
11. Write down the expression of stiffness matrix for one dimensional bar element.
## 12. State the properties of stiffness matrix
It is a symmetric matrix
The sum of elements in any column must be equal to zero
It is an unstable element. So the determinant is equal to zero.
13. Write down the expression of stiffness matrix for a truss element.
14. Write down the expression of shape function N and displacement u for one dimensional
bar element.
U = N1u1+N2u2
N1 = 1-X / l
N2 = X / l
15. Define total potential energy.
Total potential energy, = Strain energy (U) + potential energy of the external forces(W)
vi
2 Marks
## 16. State the principle of minimum potential energy.
Among all the displacement equations that satisfied internal compatibility and the
boundary condition those that also satisfy the equation of equilibrium make the potential energy
a minimum is a stable system.
17. Write down the finite element equation for one dimensional two noded bar element.
## 18. What is truss?
A truss is defined as a structure made up of several bars, riveted or welded together.
19. State the assumption are made while finding the forces in a truss
All the members are pin jointed.
The truss is loaded only at the joint
The self weight of the members is neglected unless stated.
20. State the principles of virtual energy?
A body is in equilibrium if the internal virtual work equals the external virtual work for
the every kinematically admissible displacement field.
21. What is essential boundary condition
Primary boundary condition or EBC Boundary condition which in terms of field variable
is known as Primary boundary condition.
22. Natural boundary conditions
Secondary boundary natural boundary conditions which are in the differential form of
field variable is known as secondary boundary condition.
vii
2 Marks
UNIT - 3
1. How do you define two dimensional elements?
Two dimensional elements are define by three or more nodes in a two dimensional plane.
The basic element useful for two dimensional analysis is the triangular element.
2.What is CST element?
Three noded triangular elements are known as CST. It has six unknown displacement
degrees of freedom (u1, v1, u2, v2, u3, v3). The element is called CST because it has a constant
strain throughout it.
3. What is LST element?
Six nodded triangular elements are known as LST. It has twelve unknown displacement
degrees of freedom. The displacement function for the elements are quadratic instead of linear as
in the CST.
4. What is QST element?
Ten nodded triangular elements are known as Quadratic strain triangle. It is also called as
cubic displacement triangle.
5. Write down the stiffness matrix equation for two dimensional CST elements.
Stiffness matrix KBT DBAt
BT -Strain displacement
D -Stress strain matrix
6. Draw the shape functions of a CST element.
viii
2 Marks
7. What are the differences between 2 Dimensional scalar variable and vector variable
elements?
Two dimensional scalar variable elements have only one direction independent variable
per node. Two dimensional triangular element stiffness matrix size is 3 x 3. Two dimensional
vector variable elements have direction dependent variable at each node. The element stiffness
matrix of a triangular element is of size 6x6.
8. Write down the governing differential equation for a two dimensional steady-state heat
transfer problem.
## 9. What is Area coordinates?
L1 = A1/A
L2 = A2/A
L3 = A3/A
10. Write the applications of CST element?
Use in areas where the strain gradient is small.
Use in mesh transition areas (fine mesh to coarse mesh).
Avoid using CST in stress concentration or other crucial areas in the structure, such as
edges of holes and corners.
Recommended for quick and preliminary FE analysis of 2-D problems.
11. Write down the strain displacement matrix for CST element.
## 12. Explain the important properties of CST element.
1. The strain components are constant throughout the volume of the element.
2. It has six unknown displacement degrees of freedom.
13. What are the ways in which a 3D problem can be reduced to a 2D problem?
1. Plane Stress: One dimension is too small when compared to other dimensions
ix
2 Marks
## Example: Gear thickness is small
2. Plane strain: One dimsensional problem is too large when compared to other two
dimensions
Example: Long Pipe
3. Axisymmetric: Geometry is symmetric about axis
Example: Cooling tower
## 14. Distinguish between Lagrange and Hermition interpolation functions.
At the inter element boundary the displacements are continuous in the case of Lagrange
interpolation function whereas in addition to displacements, slopes are continuous in the case of
Hermition interpolation functions
15. Define, super parametic element.
If the order of the shape functions for describing the geometry is more than that for
describing field variable, then the element is said to be super parametric.
16. Write down Jacobian matrix for 4 noded quadrilateral elements. | 2,716 | 12,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-26 | latest | en | 0.901565 |
https://www.physicsforums.com/threads/sketch-a-graph-of-the-number-of-candies-for-a-few-hours.291204/ | 1,722,804,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00521.warc.gz | 726,730,317 | 18,462 | # Sketch a graph of the number of candies for a few hours
• vampire2008
In summary, the conversation discusses how to sketch a graph of the number of candies in a jar, taking into account the removal of 10% or 20% of the candies each time. The speaker suggests rounding the decimal to get the correct number of remaining candies, and clarifies that the graph will be a line graph.
vampire2008
these days i was doing my lab homework , but i am confused one question ,
1. Consider a candy jar, initially with 1000 candies. You walk past it once each hour. Since you don’t want anyone to notice that you’re taking candy, each time you take 10% of the candies remaining in the jar. Sketch a graph of the number of candies for a few hours.
2. How would the graph change if instead of removing 10% of the candies, you removed 20%? Sketch your new graph.
i compute them , then i got the decimal , but i think it is not correct , because the number of candies should be integer , for example , if there are 729 candies remaining in the jar , 729-729*10%= 656.1 , but i think should be 729-720*10% =657, which one is correct ? any hint for that ? how about 20% ?
the other question is if this graph is point or a line ?
Eventually you will get a decimal, you have to round. Simply plot that graph. So start with 1,000...900...810...729...657...etc... as you've stated there can only be whole candies in the jar. Graph is a line graph.
thanks , Dje, but i still have a question about that, you mean i should do this 1000-1000*10%=900, 900-900*10%=810, 810-810*10%=729 , 729-729*10%=656.1, round 656.1 should be 656 , then 656-656*10%=590.4 , round is 590 , ... right ?
what about 20%, same above ,right ?
thank you !
vampire2008 said:
thanks , Dje, but i still have a question about that, you mean i should do this 1000-1000*10%=900, 900-900*10%=810, 810-810*10%=729 , 729-729*10%=656.1, round 656.1 should be 656 , then 656-656*10%=590.4 , round is 590 , ... right ?
what about 20%, same above ,right ?
thank you !
Yes that is correct. Simply take the whole part of the number to do your calculations. And indeed the 20% is the same as above
omg... why don't you just use calculate for each point and get over with it?
## 1. How do I sketch a graph of the number of candies for a few hours?
To sketch a graph of the number of candies for a few hours, you will need to use a graphing tool or software. Start by labeling the x-axis as "Time (in hours)" and the y-axis as "Number of candies." Then, plot the data points for each hour and connect them with a line. Make sure to include a legend and title for your graph.
## 2. What data do I need to sketch a graph of the number of candies for a few hours?
To sketch a graph of the number of candies for a few hours, you will need to have the data points for the number of candies at each hour. This can be obtained by counting the candies at regular intervals or by using a data set provided to you.
## 3. How do I interpret the graph of the number of candies for a few hours?
The graph of the number of candies for a few hours shows the relationship between time and the number of candies. The x-axis represents time and the y-axis represents the number of candies. The higher the line on the graph, the more candies there are at that specific time. The slope of the line also indicates the rate at which the number of candies is changing over time.
## 4. Can I use a different scale for my graph of the number of candies for a few hours?
Yes, you can use a different scale for your graph of the number of candies for a few hours. However, it is important to keep the scale consistent on both the x-axis and y-axis to accurately represent the data. For example, if the scale on the x-axis is in hours, the scale on the y-axis should also be in whole numbers to maintain accuracy.
## 5. How can I make my graph of the number of candies for a few hours more visually appealing?
To make your graph of the number of candies for a few hours more visually appealing, you can add color, use different shapes for data points, and include a title and legend. You can also adjust the size and font of the text to make it easier to read. Additionally, make sure to label all axes and include units for each axis to make the graph easier to understand.
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2K | 1,243 | 4,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-33 | latest | en | 0.892358 |
http://planetmath.org/Tractrix | 1,521,490,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647146.41/warc/CC-MAIN-20180319194922-20180319214922-00117.warc.gz | 229,457,441 | 5,733 | # tractrix
Tractrix (from the Latin verb trahere ‘pull, drag’) is the curve along which a small object (tractens) moves when pulled on a horizontal plane with a piece of thread by a puller (tractendus) which moves rectilinearly.
Let the object initially be in the $xy$-plane on the $x$-axis in the point $(a,\,0)$ and the puller in the origin; $a$ is the of the pulling thread. Then the puller begins to move along the $y$-axis in the positive direction. The object follows drawing the path curve $y=y(x)$ so that the line determined by the thread is at every the tangent of the curve. This condition gives in the point $(x,\,y)$ the differential equation
$\frac{dy}{dx}=-\frac{\sqrt{a^{2}-x^{2}}}{x}$
with the initial condition$y(a)=0$. The solution is
$y=\int_{x}^{a}\frac{\sqrt{a^{2}-x^{2}}}{x}\,dx$
or
$y=\pm(a\ln{\frac{a+\sqrt{a^{2}-x^{2}}}{x}}-\sqrt{a^{2}-x^{2}}).$
Here the minus alternative is for the case that the puller moves in the negative direction from the origin. In fact, both branches, corresponding to both signs, belong to the tractrix. The branches meet in the cusp point $(a,\,0)$.
The substitution $x:=a\cos{t}$ gives for the tractrix the parametric
$x=a\cos{t},\quad y=\pm a(\ln\frac{1+\sin{t}}{\cos{t}}-\sin{t}).$
Another one is
$x=\frac{a}{\cosh{u}},\quad y=\pm a(u-\tanh{u}),$
where $\cosh$ and $\tanh$ are the hyperbolic functions cosinus hyperbolicus and tangens hyperbolica.
Remarks
1. 1.
It is obvious that the line, on which the puller goes, is the asymptote of the tractrix. The curve thus has the property that its tangent, between the asymptote and the point of tangency, has the ($=a$).
2. 2.
The differential equation of the orthogonal curves of the tractrix is
$\frac{dy}{dx}=\frac{x}{\sqrt{a^{2}-x^{2}}},$
whence they are the circles $x^{2}+(y-C)^{2}=a^{2}$.
3. 3.
The arc length of one branch on the interval$[b,\,a]$ is simply
$\int_{b}^{a}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx=a\int_{b}^{a}\frac{dx}% {x}=a\ln\frac{a}{b}.$
4. 4.
The area $A$ between the tractrix and its asymptote is $\frac{\pi a^{2}}{2}$. This may be calculated ordinarily as
$A=2\int_{0}^{a}(a\ln{\frac{a+\sqrt{a^{2}-x^{2}}}{x}}-\sqrt{a^{2}-x^{2}})\,dx;$
integrating by parts and using the area of a quarter-circle yield
$A=2\left[a\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{\,\quad a}x\ln\frac{a+% \sqrt{a^{2}-x^{2}}}{x}-a\int_{0}^{a}x\frac{d}{dx}\left(\ln\frac{a+\sqrt{a^{2}-% x^{2}}}{x}\right)\,dx-\frac{\pi a^{2}}{4}\right]$
and moreover
$A=2a\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{\,\quad a}\left[x\ln(a+\sqrt% {a^{2}-x^{2}})-x\ln{x}+a\arcsin\frac{x}{a}\right]-\frac{\pi a^{2}}{2}=2a\left(% 0-0+a\cdot\frac{\pi}{2}\right)-\frac{\pi a^{2}}{2}=\frac{\pi a^{2}}{2}$
(see this entry (http://planetmath.org/GrowthOfExponentialFunction) for $\lim_{x\to 0+}x\ln{x}=0$). Another way to determine $A$ is differential-geometric: as the object draws the tractrix from above to down, the thread turns $180^{\mathrm{o}}$ and thus sweeps an area equal to a half-circle.
5. 5.
The envelope of the normal lines of the tractrix, i.e. the evolute of the tractrix is the catenary (or “chain curve”) $x=a\cosh{\frac{y}{a}}$.
Title tractrix Tractrix 2013-03-22 15:18:32 2013-03-22 15:18:32 pahio (2872) pahio (2872) 26 pahio (2872) Derivation msc 51N05 SubstitutionNotation Catenary EulersSubstitutionsForIntegration tractrix | 1,229 | 3,402 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 31, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-13 | latest | en | 0.813758 |
http://mathhelpforum.com/calculus/75015-polar-coordinated-two-variable-calculus-print.html | 1,529,709,159,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864822.44/warc/CC-MAIN-20180622220911-20180623000911-00310.warc.gz | 203,683,999 | 2,854 | polar coordinated two variable calculus
• Feb 22nd 2009, 05:30 AM
James0502
polar coordinated two variable calculus
Using polar coordinates evaluate
int int over D 3(x+y) and D is x^2 + y^2 <=9, x>=0
Ok, I have
int (between -pi/2 and pi/2) d theta . int(between 3 and 0) 3r^2(costheta + sintheta)
giving 18
is this right?
it seems i'm missing something I think
many thanks
• Feb 22nd 2009, 09:44 AM
HallsofIvy
Quote:
Originally Posted by James0502
Using polar coordinates evaluate
int int over D 3(x+y) and D is x^2 + y^2 <=9, x>=0
Ok, I have
int (between -pi/2 and pi/2) d theta . int(between 3 and 0) 3r^2(costheta + sintheta)
giving 18
is this right?
it seems i'm missing something I think
many thanks
Although the integral, $\displaystyle 3\int_{\theta= -\pi/2}^{\pi/2} \int_{r= 0}^3 r^2(cos(\theta)+ sin(\theta) dr d\theta$ is correct, I do NOT get 18 for that integral. Could you show how you got that?
• Feb 22nd 2009, 09:57 AM
James0502
erm.. ok
integrate first wrt r - to give 9 (cos theta + sin theta) then wrt theta to give
9(sin theta - cos theta)
putting limits in gives 18
oh wait
i need to multiply by 3? | 398 | 1,140 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-26 | latest | en | 0.857198 |
https://digitaloptionswina.web.app/mottern15834kic/present-value-of-future-monthly-cash-flows-excel-1470.html | 1,624,539,559,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00450.warc.gz | 200,464,075 | 6,505 | ## Present value of future monthly cash flows excel
How to Discount Cash Flow, Calculate PV, FV and Net Present Value Present value (PV) is what the future cash flow is worth today. cash flow calculations and more in-depth coverage of DCF usage, see the Excel-based ebook Financial Calculate Present Value of Future Cash Flows For example, a court settlement might entitle the recipient to \$2,000 per month for 30 years, but the receiving
25 Nov 2007 Note the distinction between the PV of a single sum and the future value (For additional assistance reading a cash flow diagram, click here.) For example, the PV of \$100 in 3 years at 5% under monthly compounding is \$86.10. If all we want is the PV of a single sum, we can use Excel's PV function as 24 Aug 2014 If we don't do this, then the cash flows will be discounted far too aggressively because Excel will think that each column represents 12 months, not 4 Oct 2015 However, the best way to get rid of this time period issue is using exact dates for the current and future cash flows. We can use XNPV formula and A simple cash flow is a single cash flow in a specified future time period; it can be Discounting a cash flow converts it into present value dollars and enables the user of compounding periods during the year (2 = semi-annual; 12 = monthly). The correct NPV formula in Excel uses the NPV function to calculate the present value of a series of future cash flows and subtracts the initial investment. I.e. the future value of the investment (rounded to 2 decimal places) is \$12,047.32. Future Value of a Series of Cash Flows (An Annuity) If you want to calculate the future value of an annuity (a series of periodic constant cash flows that earn a fixed interest rate over a specified number of periods), this can be done using the Excel FV function. Present Value of a Series of Cash Flows (An Annuity) If you want to calculate the present value of an annuity (a series of periodic constant cash flows that earn a fixed interest rate over a specified number of periods), this can be done using the Excel PV function. The syntax of the PV function is:
## The Excel function to calculate the NPV is “ NPV ”. The NPV, or Net Present Value, is the present value, or actual value, of a future flow of funds. The present value of a future cash flow is the current worth of it. To know the current value, you must use a discount rate.
A simple cash flow is a single cash flow in a specified future time period; it can be Discounting a cash flow converts it into present value dollars and enables the user of compounding periods during the year (2 = semi-annual; 12 = monthly). The correct NPV formula in Excel uses the NPV function to calculate the present value of a series of future cash flows and subtracts the initial investment. I.e. the future value of the investment (rounded to 2 decimal places) is \$12,047.32. Future Value of a Series of Cash Flows (An Annuity) If you want to calculate the future value of an annuity (a series of periodic constant cash flows that earn a fixed interest rate over a specified number of periods), this can be done using the Excel FV function. Present Value of a Series of Cash Flows (An Annuity) If you want to calculate the present value of an annuity (a series of periodic constant cash flows that earn a fixed interest rate over a specified number of periods), this can be done using the Excel PV function. The syntax of the PV function is:
### Use Excel Formulas to Calculate the Present Value of a Single Cash Flow or a Series fv is the future value of the investment;; rate is the interest rate per period (as a How To Calculate Present Value When Interest is Compounded Monthly.
Unlike most of finance courses, in this course, you are going to learn how to use excel to find present value of future cash flows. In addition to the present value, An even cash flow of regularly scheduled payments defines an annuity. If you borrow money to start your business, the monthly payments are calculated using an annuity formula. in this case, the annuity formula discounts a series of future payments to calculate their present value. Kinds of Functions Used in Excel. Here's how to use Excel to generate a term-loan amortization schedule. With this approach, a large percentage of your monthly payment is applied to interest in The present value, which is the original loan amount, or \$100,000 in this example. When you work with periodic cash flows, and you want to derive a general B. The time value of money means that cash flows at different points of time differ in value Note: All Excel functionalities described in this handbook are based on Microsoft Excel the analysis is done by monthly, quarterly or annually, for example. The higher the discount rate, the lower is the present value of future cash 25 Nov 2007 Note the distinction between the PV of a single sum and the future value (For additional assistance reading a cash flow diagram, click here.) For example, the PV of \$100 in 3 years at 5% under monthly compounding is \$86.10. If all we want is the PV of a single sum, we can use Excel's PV function as
### NPV = F / [ (1 + r)^n ] where, PV = Present Value, F = Future payment (cash flow), r = Discount rate, n = the number of periods in the future A guide to the NPV formula in Excel when performing financial analysis.
This article teaches you how to calculate the NPV (Net Present Value) using Excel. The Excel function to calculate the NPV is "NPV". The NPV, or Net Present Value, is the present value, or actual value, of a future flow of funds. The present value of a future cash flow is the current worth of it. To know the current value, you must use a discount rate. Excel Financial Functions Find Future and Present Values from Scheduled Cash Flows in Excel Here's how to set up a Future Value formula that allows compounding by using an interest rate and referencing cash flows and their dates.
## The PV (Present Value), NPV (Net Present Value), and FV (Future Value) functions in Excel 2016 all found on the Financial button’s drop-down menu on the Ribbon’s Formulas tab (Alt+MI) enable you to determine the profitability of an investment. Calculating the Present Value The PV, or Present Value, function returns the present value of an […]
Here's how to use Excel to generate a term-loan amortization schedule. With this approach, a large percentage of your monthly payment is applied to interest in The present value, which is the original loan amount, or \$100,000 in this example. When you work with periodic cash flows, and you want to derive a general B. The time value of money means that cash flows at different points of time differ in value Note: All Excel functionalities described in this handbook are based on Microsoft Excel the analysis is done by monthly, quarterly or annually, for example. The higher the discount rate, the lower is the present value of future cash 25 Nov 2007 Note the distinction between the PV of a single sum and the future value (For additional assistance reading a cash flow diagram, click here.) For example, the PV of \$100 in 3 years at 5% under monthly compounding is \$86.10. If all we want is the PV of a single sum, we can use Excel's PV function as 24 Aug 2014 If we don't do this, then the cash flows will be discounted far too aggressively because Excel will think that each column represents 12 months, not
How to Discount Cash Flow, Calculate PV, FV and Net Present Value Present value (PV) is what the future cash flow is worth today. cash flow calculations and more in-depth coverage of DCF usage, see the Excel-based ebook Financial Calculate Present Value of Future Cash Flows For example, a court settlement might entitle the recipient to \$2,000 per month for 30 years, but the receiving 1 Mar 2018 Calculating the future value of a present single sum for in the future or to calculate monthly payments for a loan, among other The NPV function can be used when calculating the present value of unequal future cash flows. 1 Mar 2017 Can Excel calculate the Net Present Value (NPV) of cash flows over The higher the discount rate, the less the future cash flows are worth today. Excel's NPV function requires regularly spaced cash flows – each month, Present Value of Future Cash Flow is nothing but the intrinsic value of the Cash Flow an annual interest rate of 10%, and the monthly EMI Installment for 30 years. In order to calculate present value, using the PV Factor formula in excel, we Free financial calculator to find the present value of a future amount, or a can be used to calculate the present value of a certain amount of money in the future or or cash flow, NPV represents the net of all cash inflows and all cash outflows , 22 Mar 2011 I have a client who is to receive £4300 a month for 97 months. A cash flow is much more flexible than Excel's PV function and can be used to | 1,989 | 8,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.87167 |
https://joningram.org/questions/Trigonometry/1083834 | 1,695,325,525,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00629.warc.gz | 382,141,833 | 4,635 | # Find the Tangent of the Angle (5pi)/3
Find the Tangent of the Angle (5pi)/3
Take the tangent of .
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the fourth quadrant.
The exact value of is .
The result can be shown in multiple forms.
Exact Form:
Decimal Form:
Do you know how to Find the Tangent of the Angle (5pi)/3? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name three hundred sixteen million three hundred ninety-four thousand five hundred sixty-nine
### Interesting facts
• 316394569 has 8 divisors, whose sum is 321412608
• The reverse of 316394569 is 965493613
• Previous prime number is 2237
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 9
• Sum of Digits 46
• Digital Root 1
### Name
Name one billion five hundred twenty-six million two hundred thirty-six thousand seven hundred sixty-one
### Interesting facts
• 1526236761 has 32 divisors, whose sum is 2814289920
• The reverse of 1526236761 is 1676326251
• Previous prime number is 17
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 39
• Digital Root 3
### Name
Name three hundred twelve million two hundred thirteen thousand two hundred ninety-eight
### Interesting facts
• 312213298 has 8 divisors, whose sum is 473962608
• The reverse of 312213298 is 892312213
• Previous prime number is 83
### Basic properties
• Is Prime? no
• Number parity even
• Number length 9
• Sum of Digits 31
• Digital Root 4 | 430 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.730064 |
https://studyalgorithms.com/string/length-of-longest-palindrome-that-can-be-built-from-a-string/?shared=email&msg=fail | 1,638,123,093,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00325.warc.gz | 630,532,188 | 30,414 | Home Strings Length of longest palindrome that can be built from a string
Length of longest palindrome that can be built from a string
Question: Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
Input: “abccccdd”
Output: 7
In the above example, the longest palindrome in the given string is “dccaccd” whose length is 7.
A palindrome consists of letters with equal partners, plus possibly a unique center (without a partner). The letter i from the left has its partner i from the right. For example in ‘abcba’, ‘aa’ and ‘bb’ are partners, and ‘c’ is a unique center.
Imagine we built our palindrome. It consists of as many partnered letters as possible, plus a unique center if possible. This motivates a greedy approach.
Algorithm:
We use a HashSet to find the number of partnered characters. Since we are only interested in the length of the longest palindrome possible, we can then have the result.
public int longestPalindrome(String s) { if(s==null || s.length()==0) return 0; HashSet<Character> hs = new HashSet<>(); int count = 0; for(int i = 0; i < s.length(); i++) { if(hs.contains(s.charAt(i))) { hs.remove(s.charAt(i)); count++; } else { hs.add(s.charAt(i)); } } if(!hs.isEmpty()) return count * 2 + 1; return count * 2; }
Code language: Java (java)
A working solution can be found at:- https://ideone.com/8CNFQG
Time Complexity: O(N), where N is the length of s. We need to count each letter.
Space Complexity: O(1), the space for our count, as the alphabet size of s is fixed.
We should also consider that in a bit complexity model, technically we need O(log N) bits to store the count values. | 414 | 1,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.833271 |
http://m.blog.csdn.net/samuex/article/details/40841 | 1,511,343,091,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00153.warc.gz | 181,183,893 | 12,922 | ## CSDN博客
### A Taste of AOP from Solving Problems with OOP and Design Patterns (Part I)
public class Calculator
{
public int Add(int x, int y) { return x + y; }
}
public void Test()
{
Calculator calculator = new Calculator();
Assert.IsNotNull(calculator);
}
public class Calculator
{
public int Add(int x, int y)
{
int result = x + y;
Console.WriteLine(" = {0}", result);
return result;
}
}
public class Calculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result = x + y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
}
public class Calculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result = x + y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
public int Subtract(int x, int y)
{
Console.Write("Subtract({0},{1})", x, y);
DateTime startTime = PreciseTimer.Now;
int result = x - y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
}
public void Test()
{
ICalculator calculator = new Calculator();
Assert.IsNotNull(calculator);
Assert.AreEqual(5, calculator.Subtract(8,3));
}
public interface ICalculator
{
int Subtract(int x, int y);
}
public class Calculator: ICalculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result = x + y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
public int Subtract(int x, int y)
{
Console.Write("Subtract({0},{1})", x, y);
DateTime startTime = PreciseTimer.Now;
int result = x - y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
}
public void Test()
{
ICalculator calculator = new Calculator();
Assert.IsNotNull(calculator);
Assert.AreEqual(5, calculator.Subtract(8,3));
}
public class Calculator: ICalculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result = x + y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
public int Subtract(int x, int y)
{
Console.Write("Subtract({0},{1})", x, y);
DateTime startTime = PreciseTimer.Now;
int result = x - y;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Console.Write(" [{0}] ", elapsedTime);
Console.WriteLine(" = {0}", result);
return result;
}
}
public class CalculatorTimer: ICalculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result =
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
return result;
}
public int Subtract(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result =
Decoratee.Subtract(x, y);
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
return result;
}
}
public interface ICalculatorDecorator
{
void InitDecorator(ICalculator decoratee);
ICalculator Decoratee { get; }
}
public abstract class CalculatorDecorator: ICalculatorDecorator
{
private ICalculator decoratee;
void ICalculatorDecorator.InitDecorator(ICalculator decoratee)
{
this.decoratee = decoratee;
}
// FIXED: to use implicit interface implementation instead explicit way
// so that derived classes could access this property. Thanks
greatqn!
ICalculator ICalculatorDecorator.Decoratee
public ICalculator Decoratee
{
get { return this.decoratee; }
}
}
public class CalculatorTimer: CalculatorDecorator, ICalculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
return result;
}
public int Subtract(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result = Decoratee.Subtract(x, y);
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
return result;
}
}
public class CalculatorFactory
{
private CalculatorFactory() {}
private static Type[] GetObjectGraph()
{
ArrayList objectGraphTypes = new ArrayList();
// Add decoratee as 1st type to be created
// and then add all decorators
return (Type[])objectGraphTypes.ToArray(typeof(Type));
}
public static ICalculator CreateInstance()
{
Type[] objectGraphTypes = GetObjectGraph();
ICalculator result = null;
foreach (Type calcType in objectGraphTypes)
{
ICalculator calcImpl = (ICalculator)Activator.CreateInstance(calcType);
if (calcImpl is ICalculatorDecorator)
{
((ICalculatorDecorator)calcImpl).InitDecorator(result);
}
result = calcImpl;
}
return result;
}
}
public void Test()
{
ICalculator calculator = CalculatorFactory.CreateInstance();
Assert.IsNotNull(calculator);
Assert.AreEqual(5, calculator.Subtract(8,3));
}
public class CalculatorLogger: CalculatorDecorator, ICalculator
{
public int Add(int x, int y)
{
Console.WriteLine(" = {0}", result);
return result;
}
public int Subtract(int x, int y)
{
Console.Write("Subtract({0}, {1})", x, y);
int result = Decoratee.Subtract(x, y);
Console.WriteLine(" = {0}", result);
return result;
}
}
public class CalculatorFactory
{
private static Type[] GetObjectGraph()
{
ArrayList objectGraphTypes = new ArrayList();
// Add decoratee as 1st type to be created
// and then add all decorators
return (Type[])objectGraphTypes.ToArray(typeof(Type));
}
public static ICalculator CreateInstance() …
}
public abstract class CalculatorDecorator: ICalculatorDecorator, ICalculator
{
private ICalculator decoratee;
void ICalculatorDecorator.InitDecorator(ICalculator decoratee)
{
this.decoratee = decoratee;
}
public ICalculator Decoratee
{
get { return this.decoratee; }
}
{
return result;
}
int ICalculator.Subtract(int x, int y)
{
DoPreSubtract(ref x, ref y);
int result = decoratee.Subtract(x, y);
DoPostSubtract(x, y, ref result);
return result;
}
protected virtual void DoPreAdd(ref int x, ref int y) {}
protected virtual void DoPostAdd(int x, int y, ref result) {}
protected virtual void DoPreSubtract(ref int x, ref int y) {}
protected virtual void DePostSubtract(int x, int y, ref result) {}
}
public interface ICalculatorDecorator
{
void InitDecorator(ICalculator decoratee, IDictionary context);
ICalculator Decoratee { get; }
IDictionary Context { get; }
}
public abstract class CalcuatorDecorator: ICalculatorDecorator
{
private ICalculator decoratee;
private IDictionary context;
public void InitDecorator(ICalculator decoratee, IDictionary context)
{
this.decoratee = decoratee;
this.context = context;
}
public ICalculator Decoratee
{
get { return this.decoratee; }
}
public IDictionary Context
{
get { return this.context; }
}
}
public class CalculatorFactory
{
public static ICalculator CreateInstance()
{
Type[] objectGraphTypes = GetObjectGraph();
ICalculator result = null;
IDictionary context = new Hashtable();
foreach (Type calcType in objectGraphTypes)
{
ICalculator calcImpl = (ICalculator)Activator.CreateInstance(calcType);
if (calcImpl is ICalculatorDecorator)
{
((ICalculatorDecorator)calcImpl).InitDecorator(result, context);
}
result = calcImpl;
}
return result;
}
}
public class CalculatorTimer: CalculatorDecorator, ICalculator
{
public int Add(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
return result;
}
public int Subtract(int x, int y)
{
DateTime startTime = PreciseTimer.Now;
int result = Decoratee.Subtract(x, y);
TimeSpan elapsedTime = PreciseTimer.ElapsedSince(start);
Context["Subtract.ElapsedTime"] = elapsedTime;
return result;
}
}
public class CalculatorLogger: CalculatorDecorator, ICalculator
{
public int Add(int x, int y)
{
Console.WriteLine(" = {0} {1}", result, Context["Add.ElapsedTime"]);
return result;
}
public int Subtract(int x, int y)
{
Console.Write("Subtract({0}, {1})", x, y);
int result = Decoratee.Subtract(x, y);
Console.WriteLine(" = {0} {1}", result, Context["Subtract.ElapsedTime"]);
return result;
}
}
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Author Message
butchichay
New User
Joined: 04 Sep 2006
Posts: 2
Posted: Mon Jun 18, 2007 6:58 pm Post subject: How to compute for an exponential of a number Hi, Would anyone know how to compute for an exponential of a number? For example, 4^2 = 16 What should i do with 4 to get 16? Any help would be greatly appreciated. Thanks!
Craq Giegerich
Senior Member
Joined: 19 May 2007
Posts: 1512
Location: Virginia, USA
Posted: Mon Jun 18, 2007 7:35 pm Post subject: Re: EXPONENTIAL Syntax
butchichay wrote: Hi, Would anyone know how to compute for an exponential of a number? For example, 4^2 = 16 What should i do with 4 to get 16? Any help would be greatly appreciated. Thanks!
4 ** 2
Did you try looking in the manual.
Phrzby Phil
Active Member
Joined: 31 Oct 2006
Posts: 971
Location: Richmond, Virginia
Posted: Mon Jun 18, 2007 9:01 pm Post subject: ** is also fairly conventional, so always a good guess for your test.
butchichay
New User
Joined: 04 Sep 2006
Posts: 2
Posted: Tue Jun 19, 2007 11:34 am Post subject: Re: How to compute for an exponential of a number Thanks for the reply! It was really of big help!
arpita soni
New User
Joined: 19 Jun 2007
Posts: 3
Location: mumbai
Posted: Tue Jun 19, 2007 12:27 pm Post subject: Re: How to compute for an exponential of a number Hi, Kindly let me know the way how to compute the exponential of a number: e.g: 5 ^ 4 = 32 combinations. answer to the above query has not been provided.
dick scherrer
Site Director
Joined: 23 Nov 2006
Posts: 19270
Location: Inside the Matrix
Posted: Tue Jun 19, 2007 2:08 pm Post subject:
Hello,
How to calculate an "exponential" is posted already in the topic
What does this mean?
Quote: 5 ^ 4 = 32 combinations.
5 raised to the 4th is not 32 - 2 raised to the 4th is 32 and that is what is shown in the previous info.
Please show what you mean by 32 combinations. . .
tosaurabh20
New User
Joined: 08 Jun 2007
Posts: 26
Location: Noida
Posted: Tue Jun 19, 2007 5:34 pm Post subject: Re: How to compute for an exponential of a number 2 raise to power of 4 is not 32. it is 2 raise to power of 5 which comes out to be 32. which can be interpreted as 2**5 in cobol. Thanks Saurabh
dick scherrer
Site Director
Joined: 23 Nov 2006
Posts: 19270
Location: Inside the Matrix
Posted: Tue Jun 19, 2007 6:31 pm Post subject:
Hello,
My bad - didn't check well when keying this
Quote: Quote: 5 ^ 4 = 32 combinations. 5 raised to the 4th is not 32 - 2 raised to the 4th is 32
As has been pointed out, 2 to the 5th is 32. What i was trying to point out is that 5 to the 4th isn't 32 (unless that also was a typo. . .
Thanx for the correction
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Deskargatu Txertatu Itxi Simulazio honen uneko kopia txertatu Erabili ezazu HTML hau simulazio honen uneko kopia txertatzeko. Alda dezakezu txertatutako simulazioaren zabalera eta altuera "width" eta "height" ezaugarriak aldatuz. Irudi bat txertatu eta bertan klik eginez simulazioa abiaraziko da
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Build rectangles of various sizes and relate multiplication to area. Partition a rectangle into two areas to discover the distributive property.
• Recognize that area represents the product of two numbers.
• Develop and justify a strategy that uses the area model to simplify a multiplication problem.
• Represent a multiplication problem as the proportional area of a rectangle.
• Looks for patterns in the total area calculation.
Ohizko alineamenduak
Common Core - Math
2.NBT.A.2
Count within 1000; skip-count by 5s, 10s, and 100s.
2.OA.C.4
Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends.
3.MD.C.7
Relate area to the operations of multiplication and addition.
3.MD.C.7c
Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning.
3.MD.C.7d
Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems.
3.OA.A.1
Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.
3.OA.A.3
Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.1
3.OA.B.5
Apply properties of operations as strategies to multiply and divide.2 Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
4.OA.A.1
Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations.
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• TextEncoderLite-3c9f6f0.js | 2,175 | 6,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-45 | latest | en | 0.559102 |
https://www.physicsforums.com/threads/how-do-you-know-the-vector-potential-inside-a-solenoid-only-depends-on-r.340908/ | 1,519,242,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813712.73/warc/CC-MAIN-20180221182824-20180221202824-00381.warc.gz | 925,901,562 | 15,441 | # How do you know the vector potential inside a solenoid only depends on r?
1. Sep 27, 2009
### AxiomOfChoice
How can you argue, by symmetry, that the vector potential inside a solenoid depends only on $\rho$, the perpendicular distance from the axis of the solenoid? And how can you argue that there is only a $\hat{\phi}$ (azimuthal) component of the vector potential, such that $\vec{A}$ takes the form
$$\vec{A}(\vec{r}) = A_\phi(\rho) \hat{\phi}$$
My E&M teacher is fond of talking about how we can "use symmetry arguments" to reach these conclusions without actually providing the arguments for us. It's a little frustrating...what symmetry arguments is he talking about?
Last edited: Sep 27, 2009
2. Sep 27, 2009
### AxiomOfChoice
Ok, I think I understand why $A$ has only an azimuthal component. The equation for the vector potential reads
$$\vec{A}(\vec{r}) = \frac{\mu_0}{4 \pi} \int\int \frac{\vec{K}(\vec{r}') dA}{|\vec{R}|},$$
and since $\vec{K}(\vec{r}) = \frac{N}{L} I \hat{\phi}$ (where N is the # of turns and L is the length), and we can't obtain other components from performing the surface integral, we're stuck with only a $\phi$ component for the vector potential.
Is that right?!
3. Sep 27, 2009
### Born2bwire
Basically, the geometry of the problem is invariant in the z hat and phi hat dimensions. There is no change in the problem with respect to these variables and dimensions. If the entire system is invariant in a dimension, then any other properties of the system must also be invariant.
4. Sep 28, 2009
### diazona
To put it another way: the vector potential has to be determined by the physical charge and current configuration, right? And in a cylindrical solenoid, that configuration exhibits radial symmetry - that is, you can rotate it around its axis, and the charge/current configuration looks just the same as it did before.
Now suppose the vector potential did depend on $\phi$. Whatever kind of angular dependence you can imagine, it provides some way to distinguish one direction from another... there has to be some "feature" of the vector potential that points in some particular direction, for it to have angular dependence. Let's say that you start by orienting the solenoid so that this feature points upwards (i.e. in a chosen direction which we'll designate as "up"). Then rotate the solenoid around by 90 degrees. Now the feature of the vector potential points to the left. But the charge and current distribution generating the vector potential still looks the same! How can the same distribution create two different vector potentials, one with this "feature" pointing up and one with it pointing to the left?
(The answer, of course, is that it can't, which is why we know that the vector potential must be radially symmetric) | 685 | 2,796 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.828125 | 4 | CC-MAIN-2018-09 | latest | en | 0.903789 |
https://brainly.com/question/305375 | 1,484,960,338,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280891.90/warc/CC-MAIN-20170116095120-00247-ip-10-171-10-70.ec2.internal.warc.gz | 802,562,226 | 8,922 | 2015-02-13T20:16:46-05:00
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X-4 ≤ 4-x
x+x ≤ 4+4
2x ≤ 8
x ≤ 4
x ∈ ( +∞: 4 >
-x-2 > 2(3-x)
-x-2 > 6 -2x
-x+2x > 6+2
x>8
x∈(8 ; +∞)
yw
gianrs, if you wouldn't mind, could you solve three more problems for me? Thanks if you do | 182 | 529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.882188 |
https://www.physicsforums.com/threads/some-physics-help-needed.6188/ | 1,670,483,294,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00072.warc.gz | 974,128,588 | 14,606 | # Some physics help needed
Skyblitz
Hi, I'm having a bit of trouble with a problem..
basically the question is as follows:
A particle falling covers 64% of the entire height in its last second of the fall. What is the total height of the fall?
I've tried equating two formulas, and a few other things but I can't seem to go anywhere. Any help would be appreciated.
I also need clarification on something.. My teacher said that on a position-time graph, that if there is a portion that has constant velocity (ie, no curve but a straight line), that you can't have instantaneous velocity since you can't draw a tangent to a point.. but I was under the impression that instantaneous velocity was just the velocity and any given point of time?
Thanks!
Staff Emeritus
Gold Member
Hi Skyblitz,
Post your equations which you tried, so we can see if/where there is a problem.
EDIT: Re the instantaneous velocity, you can have one. The subject which covers it is (dum, dum, dum, da!) calculus. Basically, you get the instantaneous velocity by taking two close points, and bringing them closer and closer together, watching what the intersecting line does. As the distance between the points approaches zero, the intersecting line becomes the tangent.
Last edited: | 274 | 1,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-49 | latest | en | 0.972392 |
https://justaaa.com/statistics-and-probability/470395-the-data-below-are-the-final-exam-scores-of-10 | 1,722,700,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00624.warc.gz | 275,341,026 | 9,964 | Question
# The data below are the final exam scores of 10 randomly selected history students and the...
The data below are the final exam scores of 10 randomly selected history students and the number of hours they slept the night before the exam. Find the equation of the regression line for the given data. What would be the predicted score for a history student who spent 15 hours the previous night? Is this a reasonable question? Round your predicted score to the nearest whole number. Round the regression line values to the nearest hundredth.
Hours, X, 3 5 2 8 2 4 4 5 6 3
Scores, Y, 65 80 60 88 66 78 85 90 90 71
for x=15 , predicted score =56.11+5.04*15 =132
as x=15, falls outside the range of values on which this regression is based upon, therefore we should not predict scores from this regression equation for a student who spent 15 hours the previous night
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 232 | 939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.869751 |
http://cprogrammingcodes.blogspot.com/2012/01/diagonal-sum-of-matrix.html | 1,394,176,644,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999636575/warc/CC-MAIN-20140305060716-00017-ip-10-183-142-35.ec2.internal.warc.gz | 43,910,712 | 31,109 | ## 1.10.2012
### Diagonal sum of matrix
Q. Write a C program to find the sum of diagonal elements in matrix.
Hint: If you have following typical square 3X3 matrix, then diagonal elements will be:
1 2 3
5 6
7 8 9
In above 3X3 matrix there are two types diagonal element as:
first diagonal elements are 1, 5, 9 and
second diagonal elements are 3, 5,7
Ans.
/*program to find the sum of diagonal elements of the matrix*/
#include<stdio.h>
#include<conio.h>
#define MAX 5
int main()
{
int mat[MAX][MAX],row,col;
int i,j,d1=0,d2=0;
printf("Enter no. of rows and columns : ");
scanf("%d%d",&row,&col);
printf("Enter the elements of matrix:\n");
if(row==col)
{
for(i=0; i<row; i++)
{
for(j=0; j<col; j++)
scanf("%d",&mat[i][j]);
}
for(i=0,j=col-1; i<row || j>=0; i++,j--)
{
d1=d1+mat[i][i];
d2=d2+mat[i][j];
}
printf("\nThe sum of first diagonal elements : %d",d1);
printf("\nThe sum of second diagonal elements : %d",d2);
}
else
{
printf("Rows and columns are not equal!");
printf("\nTry again!");
}
getch();
return 0;
}
output:-
Enter no. of rows and columns : 3 2
Enter the elements of matrix:
Rows and columns are not equal!
Try again!
Enter no. of rows and columns : 3 3
Enter the elements of matrix:
1 2 3
4 5 6
7 8 9
The sum of first diagonal elements : 15
The sum of second diagonal elements : 15
You might also like:
1. This comment has been removed by the author.
2. its very useful
3. hats off | 465 | 1,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2014-10 | longest | en | 0.504416 |
https://academicwritersden.com/mgt324-chapter-7-question-7-page-161-of-operations-supply-management-14th-ed-management-homework-help/ | 1,652,772,715,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00794.warc.gz | 139,671,439 | 13,731 | # mgt324 Chapter 7: Question #7 – Page 161 of Operations & Supply Management 14th Ed, management homework help
1. Chapter 7: Question #7 – Page 161 of Operations & Supply Management 14th Ed (2 pts)
For each of the following variables,explain the differences (in general) as one moves from a workcenter to an assembly line environment.
• a. Throughput time (time to convert raw materials into product)
• b. Capital/labor intensity
• c. Bottlenecks
2. Chapter 7: Question #8 – Page 161 of Operations & Supply Management 14th ed (2 pts) ( Reference the example 7.1 on page 152)
A book publisher has fixed costs of \$300,000 and variable costs per book of \$8.00. The book sells for \$23.00 per copy
• a. What is the break even point?
FACILITY LAYOUT PROBLEM
ABC Corporation is planning to assemble a lawnmower using an assembly line. The following table provides the job elements and their precedence elements and the element times.
ELEMENT PRECEDENCE ELEMENT ELEMENT TIME A none 30 seconds B none 30 seconds C A 20 seconds D B 20 seconds E C,D 30 seconds F E 40 seconds G F 10 seconds
Draw the precedence diagram of this assembly job. (2PTS) ( see page 176-178 of your text book, Exhibit 8.9&8.10).
If the production requirement for the lawnmower is 500 units per 8 hour shift, what will be the speed of this assembly line? (2PTS)
(Line speed = Time allotted per shift in seconds/no. of units produced per shift)
SERVICE PROBLEMS
Chapter 9: Question #1&3 – Page 217 of Operations & Supply Management 14th ed (2 pts)
• 1.What is the term used for a bundle of goods and services that are provided in some environment by every service operation?
• 2. List the 3 significant ways in which service systems differ from manufacturing systems. | 439 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.834846 |
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# What is the tens digit of positive integer x ?
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What is the tens digit of positive integer x ? [#permalink]
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19 Mar 2012, 09:59
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What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30.
(2) x divided by 110 has a remainder of 30.
Math Expert
Joined: 02 Sep 2009
Posts: 52437
Re: What is the tens digit of positive integer x ? [#permalink]
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19 Mar 2012, 10:10
2
2
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30 --> x=100q+30: 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient.
(2) x divided by 110 has a remainder of 30 --> x=110p+30: 30, 140, 250, 360, ... so, there are more than 1 value of the tens digit possible. Not sufficient.
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Re: What is the tens digit of positive integer x ? [#permalink]
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19 Jun 2014, 08:48
Bunuel wrote:
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30 --> x=100q+30: 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient.
(2) x divided by 110 has a remainder of 30 --> x=110p+30: 30, 140, 250, 360, ... so, there are more than 1 value of the tens digit possible. Not sufficient.
Question 1) x divided by 100 has a remainder of 30 --> x=100q+30: 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient. QUESTION :How exactly are 30, 130, 230 computed?
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Posts: 52437
Re: What is the tens digit of positive integer x ? [#permalink]
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19 Jun 2014, 08:53
1
1
sagnik242 wrote:
Bunuel wrote:
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30 --> x=100q+30: 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient.
(2) x divided by 110 has a remainder of 30 --> x=110p+30: 30, 140, 250, 360, ... so, there are more than 1 value of the tens digit possible. Not sufficient.
Question 1) x divided by 100 has a remainder of 30 --> x=100q+30: 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient. QUESTION :How exactly are 30, 130, 230 computed?
If q=0, then x=30;
If q=1, then x=130;
If q=2, then x=230;
If q=3, then x=330;
...
All these numbers when divided by 100 gives the remainder of 30.
Generally, if $$x$$ and $$y$$ are positive integers, there exist unique integers $$q$$ and $$r$$, called the quotient and remainder, respectively, such that $$y =divisor*quotient+remainder= xq + r$$ and $$0\leq{r}<x$$.
For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Notice that $$0\leq{r}<x$$ means that remainder is a non-negative integer and always less than divisor.
This formula can also be written as $$\frac{y}{x} = q + \frac{r}{x}$$.
You really need to brush up fundamentals:
Theory on remainders problems: remainders-144665.html
Units digits, exponents, remainders problems: new-units-digits-exponents-remainders-problems-168569.html
All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199
Hope this helps.
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Re: What is the tens digit of positive integer x ? [#permalink]
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19 Aug 2015, 06:06
This is interesting am benefiting well
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Re: What is the tens digit of positive integer x ? [#permalink]
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23 Apr 2017, 05:53
pratikbais wrote:
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30.
(2) x divided by 110 has a remainder of 30.
Thank you for this question, but what are choices A, B, C , D & E?
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Posts: 52437
Re: What is the tens digit of positive integer x ? [#permalink]
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23 Apr 2017, 21:47
1
rosmann wrote:
pratikbais wrote:
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30.
(2) x divided by 110 has a remainder of 30.
Thank you for this question, but what are choices A, B, C , D & E?
This is a data sufficiency question. Options for DS questions are always the same.
The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
I suggest you to go through the following post ALL YOU NEED FOR QUANT.
Hope this helps.
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Re: What is the tens digit of positive integer x ? [#permalink]
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07 Jul 2018, 08:02
Bunuel wrote:
What is the tens digit of positive integer x ?
(1) x divided by 100 has a remainder of 30 --> x=100q+30: 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient.
(2) x divided by 110 has a remainder of 30 --> x=110p+30: 30, 140, 250, 360, ... so, there are more than 1 value of the tens digit possible. Not sufficient.
Hello,
I am looking for an alternative way to solve this question, as generally advised for gmat prep. Could you please help me with that?
Thank you,
Jetmat
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Re: What is the tens digit of positive integer x ? &nbs [#permalink] 07 Jul 2018, 08:02
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# If x and y are positive integers, which of the following
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If x and y are positive integers, which of the following [#permalink]
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11 Feb 2008, 12:12
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
5
5(x – y)
20x
20y
35x
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### Show Tags
11 Feb 2008, 12:53
bmwhype2 wrote:
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
5
5(x – y)
20x
20y
35x
I think its C
A: obvs can be.
B: x=2 and y=1 5x-5y=5 then and it can be a common a GCF
D: what if 35x=20y
E: what if 35x=20y
Now obvs 20x cannot equal 35x. This is where I gathered that the answer is C. I am not certain here.
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Joined: 02 Feb 2008
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11 Feb 2008, 13:58
i think it's b.
if x and y are 1, 5*(1-1) is not the gcd of 20 and 35
el doktor!
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Joined: 01 Jan 2008
Posts: 622
### Show Tags
11 Feb 2008, 14:24
bmwhype2 wrote:
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
5
5(x – y)
20x
20y
35x
it's C. 20x is not a divisor of 35x because 20 is not a divisor of 35.
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Joined: 01 May 2007
Posts: 793
### Show Tags
11 Feb 2008, 15:47
I agree C. for same reason. I
Re: MGMAT - GCF [#permalink] 11 Feb 2008, 15:47
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A004520 Generalized nim sum n + n in base 10. 20
0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88, 80, 82, 84, 86, 88, 0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 40, 42 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS a(n) = n + n in carryless arithmetic mod 10. - N. J. A. Sloane, Jul 23 2010. REFERENCES E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982. J. H. Conway, On Numbers and Games. Academic Press, NY, 1976. LINKS David Applegate, Marc LeBrun and N. J. A. Sloane, Carryless Arithmetic (I): The Mod 10 Version. R. Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) 463-535, doi, Section 4.4. FORMULA Generalized nim sum m + n in base q: write m and n in base q and add mod q with no carries, e.g. 5 + 8 in base 3 = "21" + "22" = "10" = 1. MATHEMATICA carrylessAdd[m_, n_, b_] := Block[{lm = IntegerLength[m, b], ln = IntegerLength[n, b]}, mx = Max[lm, ln]; idm = IntegerDigits[m, b, mx]; idn = IntegerDigits[n, b, mx]; FromDigits[ Mod[ idm + idn, b], b]]; Table[ carrylessAdd[n, n, 10], {n, 0, 76}] (* Robert G. Wilson v, Aug 23 2010 *) PROG (Python) def A004520(n): return int(''.join(str(2*int(d) % 10) for d in str(n))) # Chai Wah Wu, Jun 29 2020 CROSSREFS When sorted and duplicates removed, gives A014263. - N. J. A. Sloane, Aug 03 2010 Sequence in context: A243590 A169933 A113603 * A169918 A169916 A073909 Adjacent sequences: A004517 A004518 A004519 * A004521 A004522 A004523 KEYWORD nonn,base AUTHOR EXTENSIONS More terms from Robert G. Wilson v, Aug 23 2010 STATUS approved
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Last modified July 31 02:33 EDT 2021. Contains 346367 sequences. (Running on oeis4.) | 883 | 2,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-31 | latest | en | 0.618254 |
https://ivypanda.com/essays/statistical-thinking-and-collection-of-information/ | 1,606,210,803,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176049.8/warc/CC-MAIN-20201124082900-20201124112900-00064.warc.gz | 345,715,670 | 16,823 | Cite this
# Statistical Thinking and Collection of Information Report
## Introduction to Statistics
This chapter seeks to provide a clear understanding of concepts such as statistical thinking, critical thinking, types of data, and collection of sample data in a bid to gain more statistical literacy.
The journal starts by providing the definitions of some commonly used terms in statistics as follows.
Data refer to collections of facts and observations, such as measurements, genders, or survey responses from statistical inquiries.
Statistics refer to an organized set of scientific concepts and processes that are used to obtain, organize, analyze, present, and interpret data in an attempt to make informed decisions that influence human lives.
A Population is a collection of all measurements or data that are being considered.
Census refers to collection of information about every unit in a population by the means of a complete count approach. It provides a true measure of the population.
Sample refers to representation of the entire population.
Statistical thinking is a philosophical tool that is used in the process of data analysis. It involves the creation of a relationship between processes and statistics. Critical thinking is a process that is used in statistical thinking to enable the researcher to derive sense out of the results of a study. The process that is involved in conducting a statistical study consists of several steps that include preparation, analysis, and drawing of conclusions.
Various types of data such as sample statistic and population parameters are used in statistics to make inferences and/or conclusions about a population from which the data were obtained. A population parameter is a statistical quantity that describes the features of a given population whereas a sample statistic refers to a statistical measure that represents a limited size of the population.
Collection of Sample Data is a critical process in statistical methods that requires consideration of appropriate methods of obtaining sample data. A simple random sample refers to a small representation of the entire population that offers a fair means of selection to each population unit. The basics of collecting sample data significantly drive the statistical methods. Largely, collection of sample data is done through observational studies and experiments.
## Example 1
45-percent of 200 employees at a certain motor vehicle factory earn less than 25,000 USD annually.
Relevance: This example presents a population parameter since it deals with a small group of people.
## Summarizing and Graphing
This chapter mainly focuses on the distribution of data by the use of summarizing and graphing methods such as frequency distributions, histograms, graphs that enlighten, and graphs that deceive.
Frequency Distribution is a summarizing and graphing technique that represents the outcomes that are obtained from a sample study.
A histogram is usually a graph of a frequency distribution that is used in statistical summarizing and graphing because it is simpler to interpret than a distribution table.
Graphs that Enlighten
• Scatterplots
• Time-Series Graph
• Dot plots
Graphs that Deceive
• Pictographs
• Nonzero axis graphs
## Example 2
In a study conducted at an anonymous livestock firm, in each of the firm’s 20 sections, representatives were asked to reveal how many animals were vaccinated against a certain disease outbreak. The results that were obtained are as presented below:
3, 2, 1, 0, 1, 4, 0, 1, 1, 1, 2, 2, 3, 2, 3, 2, 0, 4, 0, 1
Steps:
1. Divide the results (x) into intervals, and then count the number of results in each interval.
2. Create a table and divide it into columns three columns. Tag these columns number of animals, tally, and frequency.
3. Enter corresponding tally values of the number of animals as provided in the data above.
4. Obtain a summation of the tally values for each animal category and enter them in the corresponding Frequency column entries.
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1. IvyPanda. "Statistical Thinking and Collection of Information." May 14, 2020. https://ivypanda.com/essays/statistical-thinking-and-collection-of-information/.
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References
IvyPanda. (2020) 'Statistical Thinking and Collection of Information'. 14 May. | 1,121 | 5,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-50 | latest | en | 0.938504 |
https://market.subwiki.org/w/index.php?title=Law_of_joint_demand_for_two_goods&mobileaction=toggle_view_desktop | 1,585,835,188,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00544.warc.gz | 581,644,135 | 7,244 | # Law of joint demand for two goods
The law of joint demand for two goods is a variant for two goods of the law of demand.
## Statement
The law of joint demand for two goods says that if we have:
$p_{A,1} < p_{A,2}, p_{B,1} < p_{B,2}$
then at least one of these must hold:
$q_A(p_{A,1}, p_{B,1}) \ge q_A(p_{A,2}, p_{B,2})$
$q_B(p_{A,1}, p_{B,1}) \ge q_B(p_{A,2}, p_{B,2})$
In other words, it cannot happen that the quantity demanded increases for both goods when both their increases.
When the goods are complementary goods, then in fact, both the above inequalities hold (i.e., both quantities demanded decrease or stay the same). However, when the goods are substitute goods, it is possible for the quantity demanded to increase for one of them when both prices increase. This is because the increase in price of one good pushes some of the demand for it to the other good.
The law of joint demand places a constraint on the nature of the demand surface. | 259 | 966 | {"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": 3, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-16 | latest | en | 0.935965 |
https://socratic.org/questions/what-is-4-65-divided-by-0-31 | 1,624,220,114,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488253106.51/warc/CC-MAIN-20210620175043-20210620205043-00263.warc.gz | 475,731,952 | 5,629 | # What is 4.65 divided by 0.31?
May 14, 2016
can be written as:
• $\frac{465}{100} : \frac{31}{100}$
by rule of math the second number is reversed and multiplied
we get:
$\frac{465}{100} \cdot \frac{100}{31}$
here hundreds cancel out and we have:
$\frac{465}{31} = 15$ | 96 | 274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-25 | latest | en | 0.805075 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=Israeli+cubit&To=wavelength | 1,539,872,184,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00501.warc.gz | 443,584,819 | 3,884 | Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Israeli cubit = 0.55372 length (length) ``` Related Measurements: Try converting from "Israeli cubit" to actus (Roman actus), astronomical unit, barleycorn, Biblical cubit, digitus (Roman digitus), en (typography en), engineers chain, finger, gradus (Roman gradus), Greek cubit, Greek span, li (Chinese li), line, link (surveyors link), marathon, pica (typography pica), sazhen (Russian sazhen), spindle, stadia (Greek stadia), verst (Russian verst), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: Israeli cubit = .0156071 actus (Roman actus), .77857143 archin (Russian archin), 65.4 barleycorn, .00252315 cable length, 2,180 caliber (gun barrel caliber), .02752525 chain (surveyors chain), .00605556 city block (informal), 3,150.97 en (typography en), .00605556 football field, 1,661.16 French, .00275253 furlong (surveyors furlong), 21.8 inch, 553,720 micron, .0000983 parasang, 1.25 Roman cubit, 1.83 shaku (Japanese shaku), .00605556 soccer field, 2.42 span (cloth span), 18.27 sun (Japanese sun), .6608063 vara (Mexican vara).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 486 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-43 | longest | en | 0.650758 |
https://freethoughtblogs.com/atrivialknot/category/origami/page/3/ | 1,591,200,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00343.warc.gz | 339,902,062 | 14,402 | # Origami: Life inside
Life Inside, designed by Ekaterina Lukasheva
Ah, here’s a lovely origami model that I have not yet shared. Not much to say about this one. It’s a 30-unit model with icosahedral symmetry. I used 3 distinct colors, and the petals are curled with a toothpick.
# Origami: Spiral creases
Four 45 degree spiral creases
This is going to be one of those origami posts where I talk way too much about math. But before I get to the math, I will explain how you can make one of these things entirely with ordinary arts and craft tools.
“Ordinary tools” is the relevant bit here, since my understanding is that experts in curved-crease origami don’t use ordinary tools, they use things like vinyl cutters. When I first tried making these, I could not find any instructions for how to make these models using ordinary tools (I later found an article by Ekaterina Lukasheva), so when I finally figured out a method, I wanted to share it.
#### Making a template
Before we draw the creases directly on the paper, we need to make a template. The template ensures that each of the four curves are identical to each other.
# Origami: Star with spirals
Star with Spirals, designed by Meenakshi Mukerji
Meenakshi Mukerji featured some of my origami on her guest gallery page, including this model. Then I realized I hadn’t posted this one on my blog yet, so here it is.
Regular readers have probably noticed that I feature an awful lot of models designed by Meenakshi Mukerji. That’s because I have four of her books! The Star with Spirals is featured on the cover of Origami Inspirations, which I would say is the best of the four. Although, if you’re reluctant to commit, you might consider one of the Exquisite Origami books, which are cheaper because they’re in grayscale.
# Origami: Three-form
Three-Form, a model designed by me. Some of the component cubes are taken directly from Meenakshi Mukerji.
I was looking through my photos, and I realized that there are several large models that I never got around to sharing. This is one of them. The Three-Form consists of 24 little cubes, assembled into a larger mathematical design.
This one is inspired by General Relativity, Einstein’s theory of gravity. At the time I was reading Sean Carroll’s textbook on the subject, and I was lamenting how difficult it was to visualize the mathematical concepts therein.
# Origami: Arrow Illusion, v2
Announcement: I now have a flickr account for my origami photos! Now the images are bigger, and you can find them all in one place. There are also lots of photos that I haven’t (yet) posted on my blogs.
Arrow Illusion, version 2. An original design. My hand is in the photo to show that this is a real mirror.
Over a year ago, I designed an origami arrow that points in the opposite direction when looked at in a mirror. I wanted to revisit the design, and make it easier to fold, because this is basically the most popular origami thing I have ever done. Below, I show detailed diagrams, and a sneak peek behind the curtain. [Read more…]
# Origami: Toroidal Cube
Toroidal Cube, a design by me
In case it’s not clear what’s going on in the photo, the model is in the shape of a cube, and we are looking at it directly onto one of the vertices. However, the vertex has been cut out, and there is a triangular hole in its place. There is also a triangular hole on the opposite end, so that you can see directly through the cube. The triangles aren’t really triangles, but they look triangular from just the right perspective.
Today’s model was inspired by the regular toroid. You may have heard of a torus, which is the shape of a donut, or a mug. A toroidal polyhedron (aka a toroid) is a polyhedron which is also a torus. A regular toroid is a toroidal polyhedron where each face has the same number of sides, and each vertex connects the same number of edges.
# Origami: Cube Plus Alpha
Tomoko Fuse’s Simple Sonobe 12-unit Assembly Plus Alpha, from Unit Origami: Multidimensional Transformations
Fuse has a series of models that consist of basic polyhedra, with extra pieces of paper attached as embellishments. This model is a cube (made of 12 pieces of paper) with a pyramid added to each face, and 3 spikes to each vertex. All in all, that’s 42 pieces of paper. This is a pretty neat idea. Since it is made of three distinct types of units, it defies the usual convention of making modular origami from many identical units.
This one’s a fairly old model, apparently made in 2013. I gave it away as a gift so I don’t know if it’s still living, or deceased. | 1,108 | 4,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-24 | latest | en | 0.941954 |
https://www.emmamichaels.com/9859/prime-factorization-of-52.html | 1,685,255,054,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00171.warc.gz | 824,362,313 | 14,670 | Breaking News
# Prime Factorization Of 52
Prime Factorization Of 52. For the first 5000 prime numbers, this calculator indicates the index of the prime number. 4 rows the following steps help to find the factors of 52 using the prime factorization method. The prime factorization is where a positive integer is written as a product of its prime factors. 2 x 2 x 13.the prime factorization of 21 is:
Prime factorization of 52 = 2 x 2 x. If we gather all the prime factors: The prime factorization is where a positive integer is written as a product of its prime factors.
## If it is not a prime factor it;
To get the prime factorisation of 52, we have to start with dividing it by primes 52 ÷ 26 = 2 26 ÷ 13 = 2 13 ÷ 1 = 13 so here he prime factorisation of 52 = 2 x 2 x 13 = 2 2 x 13 1 we can check it in. If it is not a prime factor it; The nth prime number is denoted as prime [n], so prime [1] = 2, prime [2].
## 52 ÷ 2 = 26 26 ÷ 2 = 13 13 ÷ 13 = 1 So The Prime Factorization Of 52 Can Be Mathematically Written As Follows:
Find 2 factors of the number; Let’s use (2, 13) 2 and 13 are both prime factors. To get the prime factorisation of 52, we have to start with dividing it by primes 52 ÷ 26 = 2 26 ÷ 13 = 2 13 ÷ 1 = 13 so here he prime factorisation of 52 = 2 x 2 x 13 = 2 2 x 13 1 we can check it in. 52 is not a prime number.
### The Nth Prime Number Is Denoted As Prime [N], So Prime [1] = 2, Prime [2].
All the prime numbers that are used to divide in the prime factor tree are the prime factors of 52.
### Kesimpulan dari Prime Factorization Of 52.
Prime factorization by the ladder method. In other words, a composite number is any integer greater than one that is not a prime number. The prime factorization of 52 is shown below: The prime factors of 52 are 2 and 13. | 534 | 1,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-23 | longest | en | 0.907908 |
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### Chapter 9.6
Work-Energy Theorem
http://www.veoh.com/browse/videos/category/entertainment/watch/v70742516sqFsJS2#
http://www.gcse.com/fm/braking_distance_answer.htm
9.6Work-Energy Theorem
The work-energy theorem states that whenever work is done, energy changes.
Work = ∆KE
Work equals the change in kinetic energy.
Calculating Stopping Distance
• Work = change in KE, can be rewritten…
• Fd = ½ mv2
• What is the stopping distance for a 650 kg car that is traveling 5 m/s if 4,500 N of braking force is applied?
• d = ½ mv2
F
d = 1.8 m
• How fast is 5 m/s?
• 11 mph
• Multiply 5 m/s by 2.2
• This stopping distance does not take into account the reaction time.
Calculating Stopping Distance
• Calculate the stopping distance for the same car that travels at 10 m/s.
• 7.2 m.
• How does this stopping distance compare with the stopping distance at 5 m/s?
• It is four times greater!
• Double the speed, quadruple the stopping distance.
Calculate Stopping Distance
• Fd = ½ mv2
-Calculate the difference in stopping distance for a car that travels at 30 mph and the same car that travels 60 mph. Assume that the mass of the car is 800 kg and the braking force is 5000 N. Show your work and analyze your results.
How does speed influence stopping distance?
9.6Work-Energy Theorem
For moving objects such as cars:
The more kinetic energy it has, the more work is required to stop it.
Twice as much kinetic energy means twice as much work.
Brakes do work on wheels (you do work by pushing the brake pedal). When a car brakes, the work is the friction force (supplied by the brakes) multiplied by the distance over which the friction force acts.
KE is transformed by work (friction) into thermal energy, sound energy and larger-scale vibrations.
9.6Work-Energy Theorem
• Kinetic energy often appears hidden in different forms of energy, such as heat, sound, light, electricity and large-scale vibrations. All require work, and the energy required to do this work is supplied by the kinetic energy of the moving object.
• Random molecular motion is sensed as heat.
• Sound consists of molecules vibrating in rhythmic patterns.
• Light energy originates in the motion of electrons within atoms.
• Electrons in motion make electric currents.
• Larger-scale vibrations require energy to occur
9.6Work-Energy Theorem
• Due to friction, energy is transferred both into the floor and into the tire when the bicycle skids to a stop.
• An infrared camera reveals the heated tire track on the floor.
9.6Work-Energy Theorem
• Due to friction, energy is transferred both into the floor and into the tire when the bicycle skids to a stop.
• An infrared camera reveals the heated tire track on the floor.
• The warmth of the tire is also revealed.
kinetic energy is transformed into thermal energy, sound and vibrations, which represent work done to slow the bike (Fd)
9.6Work-Energy Theorem
A car moving at twice the speed of another has four times as much kinetic energy, and will require four times as much work to stop.
The frictional force is nearly the same for both cars, so the faster one takes four times as much distance to stop.
Kinetic energy depends on speed squared.
9.6Work-Energy Theorem
Typical stopping distances for cars equipped with antilock brakes traveling at various speeds. The work done to stop the car is friction force × distance of slide.
9.6Work-Energy Theorem
Typical stopping distances for cars equipped with antilock brakes traveling at various speeds. The work done to stop the car is friction force × distance of slide.
9.6Work-Energy Theorem
Typical stopping distances for cars equipped with antilock brakes traveling at various speeds. The work done to stop the car is friction force × distance of slide.
9.6Work-Energy Theorem
think!
A friend says that if you do 100 J of work on a moving cart, the cart will gain 100 J of KE. Another friend says this depends on whether or not there is friction. What is your opinion of these statements?
9.6Work-Energy Theorem
think!
A friend says that if you do 100 J of work on a moving cart, the cart will gain 100 J of KE. Another friend says this depends on whether or not there is friction. What is your opinion of these statements?
Answer:
Careful. Although you do 100 J of work on the cart, this may not mean the cart gains 100 J of KE. How much KE the cart gains depends on the net work done on it.
9.6Work-Energy Theorem
think!
When the brakes of a car are locked, the car skids to a stop. How much farther will the car skid if it’s moving 3 times as fast?
9.6Work-Energy Theorem
think!
When the brakes of a car are locked, the car skids to a stop. How much farther will the car skid if it’s moving 3 times as fast?
Answer:
Nine times farther. The car has nine times as much kinetic energy when it travels three times as fast:
9.6Work-Energy Theorem
What is the work-energy theorem? | 1,481 | 6,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.819497 |
https://www.aqua-calc.com/one-to-all/volume/preset/cubic-micrometer/41 | 1,643,297,680,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00269.warc.gz | 691,574,589 | 7,341 | # Convert cubic micrometers [µm³] to other units of volume
## micrometers³ [µm³] volume conversions
41 µm³ = 3.32 × 10-20 acre-foot µm³ to ac-ft 41 µm³ = 6.03 × 10-11 capsule 0 µm³ to cap 0 41 µm³ = 4.56 × 10-11 capsule 00 µm³ to cap 00 41 µm³ = 2.99 × 10-11 capsule 000 µm³ to cap 000 41 µm³ = 4.1 × 10-11 capsule 00E µm³ to cap 00E 41 µm³ = 5.26 × 10-11 capsule 0E µm³ to cap 0E 41 µm³ = 8.54 × 10-11 capsule 1 µm³ to cap 1 41 µm³ = 2.28 × 10-12 capsule 10 µm³ to cap 10 41 µm³ = 4.1 × 10-12 capsule 11 µm³ to cap 11 41 µm³ = 8.2 × 10-12 capsule 12 µm³ to cap 12 41 µm³ = 5.47 × 10-12 capsule 12el µm³ to cap 12el 41 µm³ = 1.28 × 10-11 capsule 13 µm³ to cap 13 41 µm³ = 1.14 × 10-10 capsule 2 µm³ to cap 2 41 µm³ = 1.52 × 10-10 capsule 3 µm³ to cap 3 41 µm³ = 2.05 × 10-10 capsule 4 µm³ to cap 4 41 µm³ = 3.15 × 10-10 capsule 5 µm³ to cap 5 41 µm³ = 1.71 × 10-12 capsule 7 µm³ to cap 7 41 µm³ = 1.46 × 10-12 capsule Su7 µm³ to cap Su7 41 µm³ = 41 000 000 000 000 cubic angstroms µm³ to ų 41 µm³ = 1.22 × 10-50 cubic astronomical unit µm³ to au³ 41 µm³ = 4.1 × 10-11 cubic centimeter µm³ to cm³ 41 µm³ = 5.04 × 10-21 cubic chain µm³ to ch³ 41 µm³ = 4.1 × 10-14 cubic decimeter µm³ to dm³ 41 µm³ = 4.1 × 10-20 cubic dekameter µm³ to dam³ 41 µm³ = 6.7 × 10-18 cubic fathom µm³ to ftm³ 41 µm³ = 1.45 × 10-15 cubic foot µm³ to ft³ 41 µm³ = 5.04 × 10-24 cubic furlong µm³ to fur³ 41 µm³ = 4.1 × 10-23 cubic hectometer µm³ to hm³ 41 µm³ = 2.5 × 10-12 cubic inch µm³ to in³ 41 µm³ = 4.1 × 10-26 cubic kilometer µm³ to km³ 41 µm³ = 4.84 × 10-65 cubic light year µm³ to ly³ 41 µm³ = 4.1 × 10-17 cubic meter µm³ to m³ 41 µm³ = 2 501 973.51 cubic microinches µm³ to µin³ 41 µm³ = 41 cubic microns µm³ to µ³ 41 µm³ = 0.003 cubic mil µm³ to mil³ 41 µm³ = 9.84 × 10-27 cubic mile µm³ to mi³ 41 µm³ = 4.1 × 10-8 cubic millimeter µm³ to mm³ 41 µm³ = 41 000 000 000 cubic nanometers µm³ to nm³ 41 µm³ = 6.45 × 10-27 cubic nautical mile µm³ to nmi³ 41 µm³ = 1.4 × 10-66 cubic parsec µm³ to pc³ 41 µm³ = 4.1 × 10+19 cubic picometers µm³ to pm³ 41 µm³ = 0.003 cubic thou µm³ to thou³ 41 µm³ = 5.36 × 10-17 cubic yard µm³ to yd³ 41 µm³ = 4.1 × 10-13 deciliter µm³ to dl 41 µm³ = 4.1 × 10-15 dekaliter µm³ to dal 41 µm³ = 8.2 × 10-10 drop µm³ to gt 41 µm³ = 4.1 × 10-23 gigaliter µm³ to Gl 41 µm³ = 1.44 × 10-13 Imperial cup µm³ to imperial c 41 µm³ = 1.44 × 10-12 Imperial fluid ounce µm³ to imperial fl.oz 41 µm³ = 9.02 × 10-15 Imperial gallon µm³ to UK gal 41 µm³ = 7.21 × 10-14 Imperial pint µm³ to imperial pt 41 µm³ = 3.61 × 10-14 Imperial quart µm³ to UK qt 41 µm³ = 4.1 × 10-17 kiloliter µm³ to kl 41 µm³ = 4.1 × 10-14 liter µm³ to l 41 µm³ = 4.1 × 10-20 megaliter µm³ to Ml 41 µm³ = 1.64 × 10-13 metric cup µm³ to metric c 41 µm³ = 4.1 × 10-12 metric dessertspoon µm³ to metric dstspn 41 µm³ = 2.73 × 10-12 metric tablespoon µm³ to metric tbsp 41 µm³ = 8.2 × 10-12 metric teaspoon µm³ to metric tsp 41 µm³ = 4.1 × 10-11 milliliter µm³ to ml 41 µm³ = 2.58 × 10-13 oil barrel µm³ to bbl 41 µm³ = 4.1 × 10-26 teraliter µm³ to Tl 41 µm³ = 1.73 × 10-13 US cup µm³ to US c 41 µm³ = 5.55 × 10-12 US dessertspoon µm³ to US dstspn 41 µm³ = 1.39 × 10-12 US fluid ounce µm³ to fl.oz 41 µm³ = 1.08 × 10-14 US gallon µm³ to US gal 41 µm³ = 8.66 × 10-14 US pint µm³ to pt 41 µm³ = 4.33 × 10-14 US quart µm³ to US qt 41 µm³ = 2.77 × 10-12 US tablespoon µm³ to US tbsp 41 µm³ = 8.32 × 10-12 US teaspoon µm³ to US tsp
#### Foods, Nutrients and Calories
ENERGY DRINK, UPC: 611269716467 contain(s) 3 calories per 100 grams (≈3.53 ounces) [ price ]
15 foods that contain Choline, free. List of these foods starting with the highest contents of Choline, free and the lowest contents of Choline, free
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Sunset Gold weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Nickel dichloride [NiCl2] weighs 3 550 kg/m³ (221.61926 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-500, liquid (R500) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
gram centimeter per second is a centimeter-gram-second (CGS) measurement unit of momentum.
A force that acts upon an object can cause the acceleration of the object.
oz t/dm³ to sl/metric tsp conversion table, oz t/dm³ to sl/metric tsp unit converter or convert between all units of density measurement.
#### Calculators
Online Food Calculator. Food Weight to Volume Conversions | 2,131 | 4,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.127202 |
https://byjus.com/question-answer/if-we-take-a-circle-and-a-hemisphere-of-the-same-radius-then-the-total-41/ | 1,638,495,368,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00061.warc.gz | 224,434,635 | 18,438 | Question
# If we take a circle and a hemisphere of the same radius, then the total surface area of the hemisphere is___ times the area of the circle.
Solution
## Let the radius of hemisphere and circle be r. Then, area of a circle = πr2 Total surface area of hemisphere = 3πr2 Hence, total surface area of the hemisphere is three times area of the circle.
Suggest corrections | 89 | 380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-49 | latest | en | 0.752816 |
https://www.worksheeto.com/post_writing-linear-equations-point-slope-form-worksheet_697848/ | 1,656,494,887,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103626162.35/warc/CC-MAIN-20220629084939-20220629114939-00222.warc.gz | 1,132,821,496 | 6,604 | # Writing Linear Equations Point-Slope Form Worksheet
📆 1 Jan 1970
🔖 Line Category
📂 Gallery Type
The New Testament
### What is the point-slope form?
A point-slope form is used. An equation is written in a point-slope form of the line that passes through the point. m is -4 m is -2 m is 6 y. m is 7 and m is 2. The equations are graphed.
### What is the name of the worksheets that are designed to help 8th grade and high school?
The point-slope form is the equation of a line. This exclusive ensemble ofprintables was designed to help 8th grade and high school learners comprehend the basics of converting a to.
### What is the color of a line in slope intercept form?
A step by step guide to writing equations. The equation of a line in a slope intercept form is: The slope should be identified. You can find the (y)u 2013 Intercept.
### What is the point-slope form of the equation of the line?
Write the point-slope form of the equation of the line through the given point. Through: (2, -1), through: (1, 3), through: (-2, 5), through: (4, 3) The equation can be graphed. y + 3 is -3/4. 2 4 6 8 6 y - 1 is 7/2. 2 4 6 8 10 is a combination of the letters x +.
### What is the point-slope form used to find a line?
The point-slope form is used to find the equation of a line. To practice finding the equations, download and print the worksheets. The equation can be obtained by plugging in the value of the line.
### What is the y and y given in the question?
Plug it into the slope intercept formula. Plug the x and y given in the question into the point slope formula. Write equation with only and y-Intercept.
### What is the name of the worksheet that Graphs Linear Equations?
The equation of a line given a point was found on 10/01/2019. Slope color is used in the writing of linear equations. Writing standard form equations.
### What equation is parallel to the given line?
The equation of the line that is parallel to the given line is written in point-slope.
### What type of problem type is listed below?
A generator for graphs and slopes. Pick at least one problem type. The equation in the form y is used to graph a line.
### What is the name of the point-slope form written as?
The point-slope form is written as eq. m(x - x) The slope is known as /eq, where m is the line.
### What is the name of the Worksheet?
The name of the form is the Algebra 1 Point Slope Form. Period U00a9L r2]0b2P0P KCut_aK SwofDtew_aSrfew ZAclvlK Jr JiXgFhMtas_ OrfeFseorVvxed. The equation of the line through the given.
### What is the name of the equation that is used to write linear equations?
To graph and write linear equations using data. In example 5 6 Part 1 you can use the deufb01nition to write an equation about altitude and the boiling point of water.
### What is the name of the letter?
Linear equations are written on a date. Period Write the point-slope form of the equation. You can create your own with Infinite Algebra 1. There is a trial available.
### What is the point-slope form used to find a line?
The point-slope form is used to find the equation of a line. To practice finding the equations, download and print the worksheets. The equation can be obtained by plugging in the value of the line.
### What equation is parallel to the given line?
The equation of the line that is parallel to the given line is written in point-slope.
### What is the set of a compulsive print?
The slope-Intercept form has an equation in it. This set of exercises for students to practice writing linear equations in slope-Intercept form, forming an equation when the slope and y-Intercept of a line are given, and identifying the and y-Intercept of an equation from the graph is a compulsive print.
### What is the Zip Zip?
\$2.00 Zip. A hands-on way for your students to practice converting between the three different forms of linear equations. There are 24 puzzle pieces in this resource.
### What is Point-Slope Form?
This form can be used to write equations that give slope and a point, but can also be used to write equations that give two points. What is the point-slope form? The point-slope form is 1.
### What is the name of the linear equations worksheets?
On January 1, 2022, U00b7 The linear equations are covered in these linear equations sheets, which include finding linear equations from two and graphing equations on the coordinate plane from either y intercept form or point slope form. There are 21 posts about word problems. From word problems.
### What is the name of the comics that slope from point as the card?
Point Slope Form And Writing Linear Equations will be done on 10/08/2021. Everyone would agree that the guidelines slope from point to point.
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Submit | 1,192 | 4,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-27 | latest | en | 0.938981 |
https://www.onlinemath4all.com/area-and-polygons.html | 1,579,393,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00551.warc.gz | 1,028,371,779 | 14,932 | # AREA AND POLYGONS
Area and polygons :
A polygon is a plane shape with straight sides.The area of a polygon measures the size of the region enclosed by the polygon. It is measured in units squared.
Here you can find area of different polygons.
## Area of polygons - Formula
A triangle with all three sides of equal length. All the angles are 60°.Area of equilateral triangle= (√3/4) a² A scalene triangle is a triangle that has three unequal sides.Area of scalene triangle = √s(s-a)(s-b)(s-c) A parallelogram is a quadrilateral with opposite sides parallel.Area of parallelogram = b x h A shape which is having four sides is generally called quadrilateral. Area of quadrilateral= (1/2) x d x (h₁+h₂) A plane figure with four straight sides and four right angles, especially one with unequal adjacent sides, in contrast to a square.Area of rectangle = length x breadth A plane figure with four equal straight sides and four right angles.Area of square = 4 a A rhombus is a parallelogram with four equal sides and opposite equal angles.Area of rhombus = (1/2) x (d₁ x d₂) A quadrilateral with one pair of sides parallel.Area of trapezium = (1/2) x h (a+ b)
## Area and polygons examples
Example 1 :
Find the area of the parallelogram
Solution :
Area of the parallelogram = b x h
here base (b) = 14 cm and
height (h) = 6 cm
= 14 x 6
= 84 cm²
Let us see the next example problem on "Area and polygons"
Example 2 :
What is the area of a parallelogram that has a base of 12 ¾ in and a height of 2 ½ in.?
Solution :
Area of the parallelogram = b x h
here base (b) = 12 ¾ inches => 51/4 inches
height (h) = ½ => 5/2 inches
= (51/4) x (5/2)
= (255/8)
⅞ square inches
Let us see the next example problem on "Area and polygons"
Example 3 :
Find the area of trapezoid
Solution :
Area of the trapezoid = (1/2) h (a + b)
a = 36 inches, b = 42 inches and h = 24 inches
= (1/2) x 24 (36 + 42)
= (1/2) x 24 x 78
= 936 in²
Example 4 :
The bases of a trapezoid are 11 meters and 14 meters. Its height is 10 meters. What is the area of the trapezoid?
Solution :
Area of the trapezoid = (1/2) h (a + b)
a = 11 m, b = 14 m and h = 10 m
= (1/2) x 10 (11 + 14)
= 5 x 25
= 125 m²
Example 5 :
The diagonals of a rhombus are 21 m and 32 m. What is the area of the rhombus?
Solution :
Area of rhombus = (1/2) x (d₁ x d₂)
d₁ = 21 m and d₂ = 32 m
= (1/2) x (21 x 32)
= 21 x 16
= 336 m²
Example 6 :
Find the area of the given figure. Explain how you found your answer.
Solution :
In the given figure we can find two shapes trapezium and rectangle.
ABCD is a trapezium
DCEF is a rectangle
Area of the given figure
= area of trapezium + area of rectangle
Area of trapezium (ABCD) = (1/2) x h (a+ b)
DC = EF = 18 ft
a = 10 ft, b = 18ft and h = 6 ft
= (1/2) x 6 x (10 + 18) ==> 84 square ft
Area of rectangle (DCEF) = length x breadth
length = 18 ft and breadth = 12 ft
= 18 x 12 ==> 216 square ft
Area of given figure = 84 + 216
= 300 square ft
## Related topics
After having gone through the stuff given above, we hope that the students would have understood "Area and polygons".
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Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,553 | 5,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2020-05 | latest | en | 0.733399 |
https://www.jagranjosh.com/general-knowledge/gk-questions-and-answers-on-physics-reflection-of-light-set-1-1481775740-1 | 1,537,881,357,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00079.warc.gz | 768,442,988 | 38,390 | GK Questions & Answers on Physics: Reflection of Light Set 1
15-DEC-2016 09:52
Light is a form of energy without it we cannot see things around us. Light enables us to see objects from which it comes or from which it is reflected. We detect light with our eyes. When light falls on the surface of an object, some of it sent back and the process of sending back the light rays which fall on the surface of an object, is called the reflection of light.
1. On which theory nature of light depends upon:
A. Wave theory
B. Particle theory
C. Both A and B
D. Only A
Ans. C
2. Why light is said to have a dual nature?
A. It exhibits the properties of wave and particles.
B. It exhibits the properties of reflection and diffraction.
C. It has both interference and polarisation effect.
D. None of the above
Ans. A
3. Which object does not reflect more light?
A. Polished Surface
B. Shining Surface
C. Unpolished
D. Both A and B
Ans. C
4. Name a metal which is the best reflector of light?
A. Gold
B. Silver
C. Iron
D. Magnesium
Ans. B
5. An incident ray is:
A. The point at which incident ray falls on the mirror.
B. The ray of light which is sent back by the mirror.
C. The ray of light which falls on the mirror surface.
D. The ray which makes right angle to the mirror surface.
Ans. C
6. Which statement is correct about the laws of reflection:
(i) The incident ray, normal ray and the reflected ray all lie in the same plane.
(ii) The angle of reflection is always equal to the angle of incidence.
(iii) The angle of incidence is equal to the angle formed by normal ray.
(iv) The angle of reflection is equal to 90o.
Options are:
A. Both (i) and (ii)
B. Both (ii) and (iii)
C. Both (iii) and (iv)
D. (i), (ii) and (iv) are correct
Ans. A
7. When a parallel beam of incident light is reflected as a parallel beam in one direction, this reflection is known as:
A. Diffuse reflection
B. Interference
C. Diffraction
D. Regular reflection
Ans. D
8. Anything which gives out light rays is called :
A. Real Image
B. Virtual Image
C. Object
D. Image
Ans. C
9. We see the image of our face when we look into the mirror. It is due to:
A. Interference
B. Diffraction
C. Polarisation
D. Reflection
Ans D
10. The image formed on a cinema screen is an example of:
A. Real Image
B. Virtual Image
C. Point Image
D. Plane Image
Ans. A
Hence, we can say that light travels in a straight line and consists of dual nature of both wave and particle. We see objects clearly due to the light which have been emitted by the object itself or it may have been reflected by the object.
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##### Anonymous
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Date Posted: 5 months ago
Views: 28
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1st
I saw your writing '8', '4' and '2' on the temperature axis. For the 24.0 deg C, the point is around halfway between 2 and 4, so we can take it as 3 (after all, the 'days' in this question can be taken to be whole numbers).
Date Posted: 5 months ago
Eric Nicholas K
5 months ago
(a) 24.0 deg C. There are 30 days in September and the total frequency is 30 (this seen in 3 + 5 + 8 + 7 + 4 + 2 + 1 = 30) and the lowest temperature seen is 24.0 deg C.
(b) 25.0 deg C. This is the most common temperature, seen in 8 of the 30 days.
(c) This is similar in working to questions such as calculating average speed given different journey speeds, distances and travel timings.
(d) 14 out of the 30 days have temperatures above 25 deg C, so we just express this in percentage form. | 283 | 987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-26 | latest | en | 0.913461 |
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# Alternating Current (AC) vs. Direct Current (DC)
An electric current is the movement of electric charge. In electric circuits, this charge is predominantly conveyed through the motion of electrons within a conductive wire. However, it can also be transported by ions within an electrolyte or by a combination of ions and electrons, as observed in ionized gases, commonly referred to as plasma.
In scientific notation, the customary symbol for current is the uppercase letter "I". The internationally recognized unit for measuring current is the ampere, denoted by the symbol "A". Specifically, one ampere of current corresponds to the passage of one coulomb of electrical charge across a specific point within a circuit over the span of one second.
Electricity exhibits two distinct modes of flow:
1. Alternating Current (AC)
2. Direct current (DC)
The dissimilarity between AC and DC arises from the specific direction in which electrons move. In the case of DC, electrons maintain a consistent and unidirectional flow, commonly referred to as "forward." Conversely, AC involves electrons constantly changing their direction, alternating between "forward" and "backward" movements. While a DC current has the capability to increase or decrease in intensity, it remains confined to a single direction without ever reversing its course. Once the flow of current undergoes a reversal in direction, it transforms into an AC current.
## Key Differences:
Following are the key differences between AC and DC:
1. Direction of Current Flow: In AC, the flow of electric charge alternates periodically, changing direction in a cyclical manner. In DC, the flow of electric charge remains in a single, constant direction.
2. Voltage Polarity: In AC, the voltage polarity reverses along with the changing current direction. In DC, the voltage polarity remains constant.
3. Generation: AC is generated using devices such as generators and alternators, which produce a continuously changing current. DC can be generated from sources like batteries, power supplies, solar cells, or dynamos.
4. Transmission and Distribution: AC is primarily used for long-distance power transmission and distribution due to its ability to easily change voltage levels using transformers. DC transmission requires specialized equipment and is more commonly used for shorter distances or specific applications.
5. Power Loss: AC experiences power losses due to factors like resistance and inductance. DC power transmission has lower power losses compared to AC transmission over long distances.
6. Conversion: AC can be converted into DC using rectifiers, while DC can be converted into AC using inverters or motor-generator sets.
7. Power Supply: AC is the standard power supply provided by utility companies in many regions. DC is used in specific applications like powering electronic devices, batteries, and renewable energy systems.
8. Frequency and Voltage Levels: AC frequency and voltage levels vary depending on the region. In North America, the standard is 60 Hz and 120 volts, while in Europe, it is typically 50 Hz and 220 to 240 volts. DC voltage levels can be adjusted based on the application requirements.
9. Safety Considerations: AC has a higher risk of causing electric shocks due to its changing polarity and higher voltage levels. DC is generally considered safer for low-voltage applications.
10. Applications: AC is commonly used for residential and commercial power supply, electric motors, large appliances, and industrial machinery. DC is prevalent in electronic devices, telecommunications, battery-powered systems, and automotive applications.
## Alternating Current (AC)
Alternating current (AC) characterizes the movement of electric charge that cyclically changes its direction. Consequently, the voltage level accompanying the current also undergoes reversals in polarity. AC can be visualized as a sinusoidal waveform, often referred to as a sine wave, which manifests as a smooth curved line. However, it is worth noting that in certain applications, alternative waveforms such as triangular or square waves may be employed.
Notably, audio and radio signals transmitted through electrical wires exemplify instances of alternating current. These specific types of AC carry encoded information, or modulation, within the signal itself, such as audio for sound reproduction or video for visual content. It is pertinent to mention that these currents typically oscillate at higher frequencies compared to those utilized in power transmission systems.
AC (alternating current) is widely adopted and highly favored as the primary form of electric power for various applications, including household equipment, offices, and buildings. In North America, the standard AC voltage is 120 volts, and it oscillates at a frequency of 60 hertz, signifying a change in direction 60 times per second. Conversely, in Europe, the prevalent AC frequency is typically 50 hertz, accompanied by voltage levels ranging from 220 to 240 volts.
## Direct Current (DC)
Direct current (DC) refers to the type of electric current where the flow of electric charge occurs in a single, constant direction. It is generated by various sources such as batteries, power supplies, solar cells, thermocouples, or dynamos. While DC primarily flows through conductors like wires, it can also traverse insulators, semiconductors, or even in vacuum environments, such as in electron or ion beams.
To convert alternating current (AC) into direct current, a rectifier is employed. This device, historically consisting of electromechanical elements and nowadays typically employing electronic components, ensures that current flows in only one direction, allowing for the extraction of direct current from an AC supply.
Indeed, direct current (DC) can be transformed into alternating current (AC) through the utilization of devices like inverters or motor-generator sets. This conversion enables the generation of AC power from a DC source.
DC finds widespread usage in powering digital electronics, including devices such as cell phones and computers. Even desktop computers rely on a power supply unit (PSU) to convert the alternating current from the household power source into direct current, which is then utilized to power the electronic components.
Furthermore, DC is suitable for energy storage applications, as it can be stored in batteries and power cells for later use. This feature allows for the efficient utilization of energy, especially in situations where a continuous power supply is not readily available or during periods of high demand.
### Conclusion
When encountering a device labeled as "AC/DC," it signifies that the device is compatible with both Alternating Current (AC) and Direct Current (DC) power sources. This versatile characteristic enables the device to operate seamlessly regardless of the type of electrical current available. Whether the device is connected to an AC power outlet or powered by a DC source, it can function efficiently and effectively. The "AC/DC" designation grants users the flexibility to utilize the device in various settings, whether they have access to AC power from a standard electrical grid or rely on DC power from batteries, power supplies, or other DC sources. | 1,386 | 7,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | latest | en | 0.928374 |
http://www.troublefreepool.com/threads/92753-Calculating-Volume-in-a-customr-shaped-pool | 1,526,930,246,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00243.warc.gz | 480,815,519 | 13,304 | # Thread: Calculating Volume in a customr shaped pool
1. ## Calculating Volume in a customr shaped pool
I am in need of help calculating the volume in our custom shaped pool here at the hotel I work for. I've calculated it myself but it was a real pain to do so and I am not supremely confident in my math. The pool & hotel are going on 25 years old and past stated volumes I've found in the records seem suspect and there are several different #s so the only way to know for sure is to recalculate it. I've attached a drawing I did showing the shape and measurements I made myself. Can anyone calculate it and tell me what you get so I can compare it to what I got?
Mark
Pool Measurements Pic.jpg
2. ## Re: Calculating Volume in a customr shaped pool
The easiest way to calculate is by using a fas-dpd test. Using a 25ml sample you can test to within .2 ppm. Make a chlorine addition using pool math here, estimating your gallons, from what you got to a target number.............retest...........and see if it is high or low. You can get within 500-1000 gallons or so which is close enough.
3. ## Re: Calculating Volume in a customr shaped pool
Originally Posted by woodyp
The easiest way to calculate is by using a fas-dpd test. Using a 25ml sample you can test to within .2 ppm. Make a chlorine addition using pool math here, estimating your gallons, from what you got to a target number.............retest...........and see if it is high or low. You can get within 500-1000 gallons or so which is close enough.
You can calculate pool volume(total gallons) with a DPD test? I've never heard about that. Are we talking about the same thing or is it possible you misunderstood my question?
4. ## Re: Calculating Volume in a customr shaped pool
I see about 68000 gallons breaking it up into circles and rectangles and half-circles. You did an excellent job with the scale. I used paintshop to draw the circles and the relative dimensions were darn close!
What Woody is talking about is using poolmath. If you plug in 68000 and go about treating the chemistry that way, it shouldn't take long to see if you're overshooting or undershooting consistently. That applies to everything. Not just chlorine, but also TA & pH. Between the calculator and Effects of Adding Chemicals down at the bottom, within a few additions you should know roughly how much you're over or underdosing. So you adjust the volume at the top of poolmath up or down. Just keep raising or lowering the volume until you get it bracketed and then narrow the gap. It will take a while, but the point comes when you hit your target every single time you add something. That's when you know the volume.
5. ## Re: Calculating Volume in a customr shaped pool
What ^^^^Richard^^^^ said.
FAS-dpd test. Drop based.
6. ## Re: Calculating Volume in a customr shaped pool
Or you can calculate the volume by using the TA or Muriatic Acid method.
Have fun with this one!
http://www.troublefreepool.com/threa...ng-Pool-Volume
7. ## Re: Calculating Volume in a customr shaped pool
Wow Zep! That is a LOT of pool.
Welcome to the forum...we really want to see this one for sure.
8. ## Re: Calculating Volume in a customr shaped pool
Better not let Jerry Brown see that thing. He might try to use it for as a new found water source.
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• | 815 | 3,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-22 | latest | en | 0.938762 |
https://kr.mathworks.com/matlabcentral/cody/problems/2202-flip-the-bit/solutions/1947645 | 1,606,397,256,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188146.22/warc/CC-MAIN-20201126113736-20201126143736-00183.warc.gz | 367,850,979 | 16,927 | Cody
Problem 2202. Flip the bit
Solution 1947645
Submitted on 24 Sep 2019 by Sam McPheeters
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
s = '1001'; y_correct = '0110'; assert(all(isequal(flipbit(s),y_correct)&ischar(flipbit(s))))
2 Pass
s = '11'; y_correct = '00'; assert(all(isequal(flipbit(s),y_correct)&ischar(flipbit(s))))
3 Pass
s = '1'; y_correct = '0'; assert(all(isequal(flipbit(s),y_correct)&ischar(flipbit(s))))
4 Pass
s = '100000001'; y_correct = '011111110'; assert(all(isequal(flipbit(s),y_correct)&ischar(flipbit(s))))
5 Pass
s = '00001'; y_correct = '11110'; assert(all(isequal(flipbit(s),y_correct)&ischar(flipbit(s))))
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Start Hunting! | 272 | 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} | 2.625 | 3 | CC-MAIN-2020-50 | latest | en | 0.51318 |
https://math.stackexchange.com/questions/3108591/compact-imbedding-on-weighted-sobolev-spaces | 1,560,861,590,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998724.57/warc/CC-MAIN-20190618123355-20190618145355-00171.warc.gz | 508,675,559 | 34,275 | # Compact Imbedding on weighted Sobolev Spaces
I am trying to solve the simple Eigenvalue problem
$$-\Delta{u}=\lambda u, \; u = 0$$ on $$\partial{\Omega}$$
in Sobolev spaces on a smooth exterior (and thus unbounded) domain $$\Omega \subset \mathbb{R}^{n}$$. Unfortunately, the Rellich-Kondrachov compactness theorem cannot be used on $$\Omega$$.
I read something online about using weighted Sobolev Spaces, but didn´t find any concrete references and now tried to do it on my own.
I am trying to prove the compactness of the embedding of $$H^{1}_{0}(\Omega,\omega)$$ in $$L^{2}(\Omega,\omega)$$, where $$H^{1}_{0}(\Omega,\omega)$$ is the completion of $$C^{\infty}_{c}(\Omega)$$ in regards to the norm $$||u||^{2}:=\int_{\Omega}|u|^{2}\omega + |Du|^{2}$$, where $$\omega(x)=1/|x|^{2+\delta}$$ for some small $$\delta > 0$$.
My idea: Let $$(u_{n}) \subset H^{1}_{0}(\Omega,\omega)$$ be a bounded set. Using the Hardy inequality i get $$\int_{\Omega/B_{R}(0)}|u_{n}|^{2}\omega \le 1/|R|^{\delta} \cdot C \cdot \int_{\Omega/B_{R}(0)}|Du_{n}|^{2}$$, which can be made $$< \epsilon$$ for sufficiently large $$R$$.
On $$B_{R}(0)\cap \Omega$$ I use the classic Rellich-Kondrachov theorem to find a finite number of functions in $$L^{2}(\Omega\cap B_{R}(0),\omega)$$ which cover the $$u_{n}$$ up to $$\epsilon$$. Thus, the $$u_{n}$$ are totally bounded.
Does this make sense?
• Do you assume that $0\in \Omega$ so that $L^2(\Omega\cap B_R)=L^2(\Omega\cap B_R,w)$ with equivalent norms? Otherwise I don't see how you use Rellich-Kondrachov for the bounded part. The rest seems fine to me. – MaoWao Feb 11 at 15:19
• Thanks! I will adjust $\omega$ slightly to make it work in the general case of $0 \notin \Omega$. One more question: Since the dual $(H^{1}_{0}(\Omega,\omega))´$ is way smaller than the weighted space itself, I am having trouble applying the methods one usually uses with compact operators. Could you briefly explain how I use the spectral theorem in this case? – Falc14 Feb 12 at 10:52 | 632 | 2,003 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-26 | latest | en | 0.776366 |
http://www.boards.ie/vbulletin/showthread.php?t=2056650270 | 1,448,579,448,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447860.26/warc/CC-MAIN-20151124205407-00008-ip-10-71-132-137.ec2.internal.warc.gz | 329,751,205 | 16,932 | How to combat the bothersome incircle construction inaccuracy? - boards.ie
24-05-2012, 14:03 #1 reznov Registered User Join Date: Jan 2012 Posts: 920 How to combat the bothersome incircle construction inaccuracy? I have trouble constructing a perfect incircle which doesn't go past the three sides of a triangle. Always a part of the circle will jut out and it's an eyesore as i strive to develop a professional representation of my mathematical ability. Will I lose many marks? How do you construct a sexy incircle?
24-05-2012, 14:10 #2 JeaicMaG Join Date: Feb 2012 Posts: 787 You should be able to judge it by eye.
24-05-2012, 14:11 #3 reznov Registered User Join Date: Jan 2012 Posts: 920 Just a little bit always goes astray. I'm aiming for that Maths certificate and don't want the chief examiner to get the wrong impression. I do it by eye! Maybe I need an eye replacement?
24-05-2012, 14:16 #4 cutty9 Closed Account Join Date: May 2012 Posts: 61 you need optilase. if you can't get the incircle construction fully exact, i would kill myself reznov.
24-05-2012, 14:18 #5
reznov
Registered User
Join Date: Jan 2012
Posts: 920
Quote:
Originally Posted by cutty9 you need optilase. if you can't get the incircle construction fully exact, i would kill myself reznov.
Wait so if I don't get it right, you will kill yourself? A hefty sacrifice indeed comrade.
Anyone? Any suggestions? Or should I hope for the best as there's no technique?
(2) thanks from:
24-05-2012, 16:31 #6 Canard J'ai une araignée au plafond Join Date: Dec 2008 Posts: 8,114 Mod: Leaving Cert Draw the circle first, draw a triangle around it, add in the construction lines.
(4) thanks from:
24-05-2012, 16:42 #7
JeaicMaG
Join Date: Feb 2012
Posts: 787
Quote:
Originally Posted by Patchy~ Draw the circle first, draw a triangle around it, add in the construction lines.
That's... That's brilliant!
24-05-2012, 16:48 #8 Canard J'ai une araignée au plafond Join Date: Dec 2008 Posts: 8,114 Mod: Leaving Cert Works every time, FUQ DA SYSTEM
24-05-2012, 18:16 #9
reznov
Registered User
Join Date: Jan 2012
Posts: 920
Quote:
Originally Posted by Patchy~ Draw the circle first, draw a triangle around it, add in the construction lines.
If I get the certificate in Maths, I'm dedicating it to YahFaoiSin and you.
Thanks from:
24-05-2012, 18:23 #10
finality
Registered User
Join Date: Feb 2012
Posts: 1,763
Quote:
Originally Posted by Patchy~ Draw the circle first, draw a triangle around it, add in the construction lines.
What if they give you specific measurements for the triangle?
24-05-2012, 18:30 #11 Togepi Closed Account Join Date: Oct 2011 Posts: 4,807 They usually just give you the triangle don't they?
24-05-2012, 19:11 #12 Canard J'ai une araignée au plafond Join Date: Dec 2008 Posts: 8,114 Mod: Leaving Cert I thought that too. I always thought it was kind of accepted the incircles just dont work, my godly JC maths teacher taught us that sneaky trick. But I guess youd just make the radius around the length of the side or something like that, might take a few tries but if all else fails just do trial and error or do it the conventional way even if it messes up a bit.
24-05-2012, 19:41 #13
ChemHickey
Closed Account
Join Date: Jun 2011
Posts: 3,477
Quote:
Originally Posted by Patchy~ I thought that too. I always thought it was kind of accepted the incircles just dont work, my godly JC maths teacher taught us that sneaky trick. But I guess youd just make the radius around the length of the side or something like that, might take a few tries but if all else fails just do trial and error or do it the conventional way even if it messes up a bit.
Use your protractor first. then draw the lines after words
24-05-2012, 19:51 #14 finality Registered User Join Date: Feb 2012 Posts: 1,763 I actually don't have trouble with the incircle. Finding the incentre accurately is easy, then just try to sort of line your compass up with the part of the triangle that's closest to the centre, then draw a circle. If it's out by like half a millimetre it's not going to be a problem anyway.
24-05-2012, 22:51 #15 m4r0s Registered User Join Date: May 2012 Posts: 1 assume everyone knows it...but you've to draw a 90 degree line to one of the sides from the incentre, works every time for me | 1,229 | 4,337 | {"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} | 2.59375 | 3 | CC-MAIN-2015-48 | latest | en | 0.862848 |
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lecture 36
# lecture 36 - 1 Summary of Lecture#36 • Ideal transformers...
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Unformatted text preview: 1 Summary of Lecture #36 11/17/2010 • Ideal transformers • Unity coupling coefficient ( k = 1 ) • Infinite self and mutual inductance • Basic properties of ideal transformers • Voltage ratio • Current ratio • Impedance Transformation • Examples • Transformers with unity coupling coefficient • Examples • Equivalent T and Π circuits 1 Coupled coils and transformers L 1 , L 2 , M are finite. M =k √ L 1 L 2 ≤ √ L 1 L 2 Ideal transformers M = √ L 1 L 2 , a = √ L 1 /L 2 k = 1 Infinite L 1 , L 2 and M a = N 1 :N 2 , turn ratio Ideal Ideal 0 ≤ k ≤ 1 Two isolated coils Finite L 1 and L 2 M = 0 k = 0 A transf. is ideal if k=1 and L 1 , L 2 and M → ∞ . Basic Properties of ideal transformers : (1) Voltage ratio 2 1 2 1 2 1 2 1 2 1 2 1 N N a ) j ( V ) j ( V N N a ) s ( V ) s ( V N N a ) t ( v ) t ( v = = ϖ ϖ = = = = 2 1 2 1 2 1 2 1 2 1 2 1 N N a ) j ( V ) j ( V N N a ) s ( V ) s ( V N N a ) t ( v ) t ( v- =- = ϖ ϖ- =- =- =- = Basic Properties of ideal transformers : (2) Current ratio 1 2 2 1 1 2 2 1 1 2 2 1 N N a 1 ) j ( I ) j ( I N N a 1 ) s ( I ) s ( I N N a 1 ) t ( i ) t ( i- =- = ϖ ϖ- =- =- =- = 1 2 2 1 1 2 2 1 1 2 2 1 N N a 1 ) j ( I ) j ( I N N a 1 ) s ( I ) s ( I N N a 1 ) t ( i ) t ( i = = ϖ ϖ = = = = i 1 and i 2 go into the transformer terminals. (3) Impedance transformation Valid for dots on the same side or opposite sides ) s ( Z a ) s ( Z 2 in = 2 1 2 1 2 1 N N L L L M M L a = = = = Turn ratio: N 1 /N 2 Z(s) ) s ( Z N N ) s ( Z a ) s ( I ) s ( V a ) s ( I ) s ( V a ) s ( I a 1 ) s ( aV ) s ( I ) s ( V ) s ( Z 2...
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# What is the greatest common factor of 9 8?
Updated: 4/28/2022
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11y ago
The greatest common factor of 9 , 8 = 1
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The Greatest Common Factor (GCF) is: 9 | 288 | 874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.904786 |
https://www.physicsforums.com/threads/settle-a-bet-on-time-and-distance.644788/ | 1,516,104,091,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00631.warc.gz | 952,693,360 | 16,188 | # Settle a bet on time and distance.
1. Oct 17, 2012
### hazzatori
Hear is my first post and go easy Hawkins i aint! but its to settle a little bet at work.
Lets say i am stood in a Universe that obeys all of the laws of physics as our own, no differences at all,
but all that is in this Universe is
1 Platform with a 12 hour clock face on it
and
1 Giant 12 hour clock face
both are synced to same time.
(oh and and some light so we can see, O2 so we can breath etc)
when i stand on the platform for the first time time starts, when i step off time stops
The distance from the platform to the clock face is exactly 1 Light Year, but the clock is so large all i can see in my peripheral vision is the clock face.
Am i right in believing that firstly, it would take me one year from stepping on the platform to even see the clock face?
then as i do see it and as it is precisely 1 light year away, i would see the time on both clock faces as exactly the same time in fact they should even tick over precisely.
then if the platform moves closer to the large clockface the times would not stay synced or would they?
2. Oct 17, 2012
### jbriggs444
This probably belongs in the Special Relativity forum.
So you have two clocks. They are one light-year apart. The far clock is big enough to be seen from that distance. The two clocks are synchronized.
By convention, we take this to mean that the clocks are synchronized in the frame of reference within which they are both at rest. Also, by convention, the synchronization is such that the measured speed of light, timed with the use of the two clocks is judged to be the same going from near clock to far as from far clock to near.
You step on the platform with the near clock. No, it does not take you a year to see the far clock. You see it immediately. It had been illuminated a year ago and light reflected from past illumination is continuously arriving at the near platform. The light is all ready for someone to step up and start seeing it.
You would see the time displayed by the far clock showing a date and time as of one year ago. That's when the light from that year-ago reading left from the far clock's face.
The fact that the clocks are synchronized does not mean that you _see_ both displays showing the same time. It means that _once you account for speed of light delays_, both displays would show the same time. That's part of the standard convention for synchronization.
If the one clock is moved _slowly_ to the other you would find that the two clocks show the same time when they meet and that it does not matter which is moved to the other.
If one clock is moved _rapidly_ to the other you would find that the clock that was moved shows a time that is one year stale as compared to the clock that was not moved.
"slowly" is in the limit as the clock transport speed decreases toward zero.
"rapidly" is in the limit as the clock transport speed increases toward light speed.
In between the high speed and low speed limits you would find that the clock that moved is stale, but not a whole year stale.
3. Oct 17, 2012
### Naty1
What does this mean:
4. Oct 17, 2012
### mrspeedybob
This is a little ambiguous, and the different ways it should be interpreted affect the answer.
It could be interpreted to mean that each clock starts when it's detector sees you step onto the platform. In this case you will see the distant clock 2 years behind. It takes 1 year for the signal that you stepped onto the platform to reach the clock, and another year for the light from the clock to reach you after it starts ticking.
Now suppose there is a big "go" light on the far clock. The clock starts ticking at the same time the "go" light turns on. You step onto the plate when you see the go light turn on. In this case you see both clocks tick in sync.
Now you did leave the option to interpret the whole thing as a trick question. You stated the distance to be exactly 1 light year and you specified a 12 hour clock face. Based on a year that is exactly 365 days long you would see the clocks tick in sync regardless. It would depend on the specific interpretation weather you see the far clock start ticking at the same time, 730 hour hand revolutions later, or 1460 hour hand revolutions later. In any case, once you see them both start ticking they should read the same time and tic in sync. | 1,000 | 4,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2018-05 | longest | en | 0.976193 |
https://www.nag.co.uk/numeric/R/html23/g02ab.html | 1,560,769,018,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998473.44/warc/CC-MAIN-20190617103006-20190617125006-00027.warc.gz | 843,532,047 | 2,334 | g02ab {NAGFWrappers} R Documentation
## g02ab: Computes the nearest correlation matrix to a real square matrix, augmented g02aa to incorporate weights and bounds
### Description
g02ab computes the nearest correlation matrix, in the Frobenius norm or weighted Frobenius norm, and optionally with bounds on the eigenvalues, to a given square, input matrix.
### Usage
g02ab(g, opt, alpha, w,
n = nrow(w),
errtol = 0.0,
maxits = 0,
maxit = 0)
### Arguments
g double array G, the initial matrix. opt string Indicates the problem to be solved. opt='A': The lower bound problem is solved. opt='W': The weighted norm problem is solved. opt='B': Both problems are solved. alpha double The value of α. w double array The square roots of the diagonal elements of W, that is the diagonal of W^(1)/(2). n integer: default = nrow(w) The size of the matrix G. errtol double: default = 0.0 The termination tolerance for the Newton iteration. If errtol <= 0.0 then n \times sqrt(machine precision) is used. maxits integer: default = 0 Specifies the maximum number of iterations to be used by the iterative scheme to solve the linear algebraic equations at each Newton step. maxit integer: default = 0 Specifies the maximum number of Newton iterations.
### Details
R interface to the NAG Fortran routine G02ABF.
### Value
G double array A symmetric matrix (1)/(2)(G + G^T) with the diagonal set to I. W double array If opt='W', 'B', the array is scaled so max(W_i) = 1 for i=1 . . . n. X double array Contains the nearest correlation matrix. ITER integer The number of Newton steps taken. FEVAL integer The number of function evaluations of the dual problem. NRMGRD double The norm of the gradient of the last Newton step. IFAIL integer ifail =0 unless the function detects an error or a warning has been flagged (see the Errors section in Fortran library documentation).
NAG
### Examples
ifail <- 0
opt <- "b"
alpha <- 0.02
g <- matrix(c(2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2,
-1, 0, 0, -1, 2), nrow = 4, ncol = 4, byrow = TRUE)
w <- matrix(c(100, 20, 20, 20), nrow = 4, ncol = 1,
byrow = TRUE)
errtol <- 1e-07
maxits <- 200
maxit <- 10
ans <- g02ab(g, opt, alpha, w)
if (ifail == 0) {
writeLines(sprintf("\n Nearest Correlation Matrix\n",
"\n"))
x <- ans$X print(x) iter <- ans$ITER
writeLines(sprintf("\n Number of Newton steps taken: %d\n",
iter))
feval <- ans$FEVAL writeLines(sprintf(" Number of function evaluations: %d\n", feval)) alpha <- ans$ALPHA
writeLines(sprintf(" \n\n Alpha: %30.3f\n",
alpha))
}
[Package NAGFWrappers version 24.0 Index] | 738 | 2,569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-26 | latest | en | 0.632594 |
https://www.mindmeister.com/464263473/circuits | 1,548,082,653,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583792784.64/warc/CC-MAIN-20190121131658-20190121153658-00220.warc.gz | 879,513,257 | 13,089 | # Circuits
Get Started. It's Free
Circuits
1.2.1. .
## 2. What are circuit elements?
### 2.1. principles
2.1.1. RC circuit
2.1.2. Ohm's law
2.1.2.1. emf
2.1.3. Kirchhoff's laws
2.1.3.1. Sum(V) = 0
2.1.3.1.1. loop rule
2.1.3.2. Sum(I) = 0
2.1.3.2.1. nodal rule
### 2.2. application of principles
2.2.1. Two loop circuits
2.2.1.1. .
2.2.1.2. ...thinking of node rule systems as a system of simultaneous eqns
### 2.3. the elements
2.3.1. resistors
2.3.2. capacitors
2.3.3. inductors
2.3.4. batteries | 235 | 517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-04 | latest | en | 0.529033 |
https://web2.0calc.com/questions/i-need-help-with-the-last-two-columns | 1,596,985,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738555.33/warc/CC-MAIN-20200809132747-20200809162747-00230.warc.gz | 544,022,864 | 7,491 | +0
# I need help with the last two columns
0
805
3
+176
1. For each of the professions in the left column, calculate the annual pay based on full-time, year-round employment consisting of 2,000 hours a year (40 hours per week for 50 weeks each year). Record your calculations under "Annual income" in the table. Then, find the difference between each annual wage figure and both the poverty threshold and the median household income. If the difference is a negative number, record it as such.
Hourly wage Annual income Difference between annual wage and federal poverty line Difference between annual wage and median household income Federal minimum wage \$7.25 Oregon’s minimum wage \$8.95 Average for all occupations \$23.87 Marketing managers \$51.90 Family-practice doctors \$82.70 Veterinary assistants \$11.12 Police officers \$26.57 Child-care workers \$9.38 Restaurant cooks \$10.59 Air-traffic controllers \$58.91
Data source: U.S. Bureau of Labor Statistics (2013)
Oct 10, 2019
#1
+26006
0
For the last two columns you will need the poverty threshold and median household income......you'll have to use what was given to you for these values or you will have to look them up if they weren't given to you in this or another question....Sorry !
Oct 10, 2019
#2
+26006
0
I found (for 2013) that poverty line = 12119 for a single person
and MHI = 51939 Though I do not know if these are the values they want you to use, here is an example
first line 7.25 x 2000 = annul income = 14500 14500 - 12119 = \$2381 ABOVE poverty line 14500 - 51939 = 37439 BELOW MHI
etc
etc
Oct 10, 2019
#3
+176
0
Thank you so much. I makes more sense now.
Jazminfun0128182 Oct 11, 2019 | 453 | 1,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-34 | latest | en | 0.901579 |
https://gist.github.com/chelsyx/ebc8ff5d7125a79d4297 | 1,652,743,224,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00290.warc.gz | 338,708,497 | 24,002 | {{ message }}
Instantly share code, notes, and snippets.
# chelsyx/donutchart.R
Forked from mages/doughnut.R
Last active Jan 21, 2016
Donut Charts in R
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#From http://stackoverflow.com/questions/13615562/ggplot-donut-chart library(ggplot2) # Create test data. dat = data.frame(count=c(10, 60, 30), category=c("A", "B", "C")) # Add addition columns, needed for drawing with geom_rect. dat$fraction = dat$count / sum(dat$count) dat = dat[order(dat$fraction), ] dat$ymax = cumsum(dat$fraction) dat$ymin = c(0, head(dat$ymax, n=-1)) p1 = ggplot(dat, aes(fill=category, ymax=ymax, ymin=ymin, xmax=4, xmin=3)) + geom_rect() + coord_polar(theta="y") + xlim(c(0, 4)) + labs(title="Basic ring plot") p2 = ggplot(dat, aes(fill=category, ymax=ymax, ymin=ymin, xmax=4, xmin=3)) + geom_rect(colour="grey30") + coord_polar(theta="y") + xlim(c(0, 4)) + theme_bw() + theme(panel.grid=element_blank()) + theme(axis.text=element_blank()) + theme(axis.ticks=element_blank()) + labs(title="Customized ring plot") library(gridExtra) png("ring_plots_1.png", height=4, width=8, units="in", res=120) grid.arrange(p1, p2, nrow=1) dev.off()
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#' Doughnut plot #' #' Draw a doughnut plot #' #' This function draws a pie chart with a hole; a doughnut plot #' The doughnut plot is based on the pie plot, see \code{\link{pie}}. #' #' @param x a vector of non-negative numerical quantities. #' The values in \code{x} are displayed as the areas of #' doughnut slices. #' @param labels one or more expressions or character strings giving #' names for the slices. Other objects are coerced by #' \code{\link{as.graphicsAnnot}}. For empty or \code{NA} #' (after coercion to character) labels, no label nor #' pointing line is drawn. #' @param edges the circular outline of the doughnut is approximated #' by a polygon with this many edges. #' @param outer.radius the doughnut is drawn centered in a square box #' whose sides range from \eqn{-1} to \eqn{1}. If the #' character strings labeling the slices are long it may be #' necessary to use a smaller radius. #' @param inner.radius Size of the inner radius to be whited #' @param clockwise logical indicating if slices are drawn clockwise #' or counter clockwise (i.e., mathematically positive #' direction), the latter is default. #' @param init.angle number specifying the \emph{starting angle} (in #' degrees) for the slices. Defaults to 0 (i.e., #' \sQuote{3 o'clock}) unless \code{clockwise} is true where #' \code{init.angle} defaults to 90 (degrees), (i.e., #' \sQuote{12 o'clock}). #' @param density the density of shading lines, in lines per inch. #' The default value of \code{NULL} means that no shading #' lines are drawn. Non-positive values of \code{density} #' also inhibit the drawing of shading lines. #' @param angle the slope of shading lines, given as an angle in #' degrees (counter-clockwise). #' @param col a vector of colors to be used in filling or shading #' the slices. If missing a set of 6 pastel colours is #' used, unless \code{density} is specified when #' \code{par("fg")} is used. #' @param border, lty (possibly vectors) arguments passed to #' \code{\link{polygon}} which draws each slice. #' @param main an overall title for the plot. #' @param \dots graphical parameters can be given as arguments to #' \code{pie}. They will affect the main title and #' labels only. #' #' @author Original pie plot by R Core, amended by Markus Gesmann #' for a doughnut plot #' @keywords hplot #' @seealso \code{\link{pie}} #' @export #' @examples #' x <- c(2,4,3,2,4) #' doughnut(x) #' ## Add lables #' doughnut(x, labels=LETTERS[1:5]) doughnut <- function (x, labels = names(x), edges = 200, outer.radius = 0.8, inner.radius=0.6, clockwise = FALSE, init.angle = if (clockwise) 90 else 0, density = NULL, angle = 45, col = NULL, border = FALSE, lty = NULL, main = NULL, ...) { if (!is.numeric(x) || any(is.na(x) | x < 0)) stop("'x' values must be positive.") if (is.null(labels)) labels <- as.character(seq_along(x)) else labels <- as.graphicsAnnot(labels) x <- c(0, cumsum(x)/sum(x)) dx <- diff(x) nx <- length(dx) plot.new() pin <- par("pin") xlim <- ylim <- c(-1, 1) if (pin[1L] > pin[2L]) xlim <- (pin[1L]/pin[2L]) * xlim else ylim <- (pin[2L]/pin[1L]) * ylim plot.window(xlim, ylim, "", asp = 1) if (is.null(col)) col <- if (is.null(density)) palette() else par("fg") col <- rep(col, length.out = nx) border <- rep(border, length.out = nx) lty <- rep(lty, length.out = nx) angle <- rep(angle, length.out = nx) density <- rep(density, length.out = nx) twopi <- if (clockwise) -2 * pi else 2 * pi t2xy <- function(t, radius) { t2p <- twopi * t + init.angle * pi/180 list(x = radius * cos(t2p), y = radius * sin(t2p)) } for (i in 1L:nx) { n <- max(2, floor(edges * dx[i])) P <- t2xy(seq.int(x[i], x[i + 1], length.out = n), outer.radius) polygon(c(P$x, 0), c(P$y, 0), density = density[i], angle = angle[i], border = border[i], col = col[i], lty = lty[i]) Pout <- t2xy(mean(x[i + 0:1]), outer.radius) lab <- as.character(labels[i]) if (!is.na(lab) && nzchar(lab)) { lines(c(1, 1.05) * Pout$x, c(1, 1.05) * Pout$y) text(1.1 * Pout$x, 1.1 * Pout$y, labels[i], xpd = TRUE, adj = ifelse(Pout$x < 0, 1, 0), ...) } ## Add white disc Pin <- t2xy(seq.int(0, 1, length.out = n*nx), inner.radius) polygon(Pin$x, Pin\$y, density = density[i], angle = angle[i], border = border[i], col = "white", lty = lty[i]) } title(main = main, ...) invisible(NULL) } | 1,713 | 5,879 | {"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} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.586721 |
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# Sleeping Drivers
Author Message
Manager
Joined: 11 May 2010
Posts: 216
Kudos [?]: 141 [0], given: 11
### Show Tags
11 May 2010, 02:21
I would go with C
Pondering between C and D though.
Kudos [?]: 141 [0], given: 11
Manager
Joined: 21 Jan 2010
Posts: 220
Kudos [?]: 104 [0], given: 38
### Show Tags
11 May 2010, 05:27
Another vote for B.
Try to rephrase each choice according to the structure of "Most drivers think they can tell when they are about to fall asleep, but they cannot."
You can rephrase for choice A, C, D and E. But not B.
A: Most people think they can tell when they are too drunk to drive, but they cannot
C: Most industrial workers think they know when they should stop working, but they cannot.
D: Most people who have just donated blood think they can tell when they are ready to walk out, but they cannot.
E: Most people who are being treated for schizophrenia think they can tell when they can halt their medical treatments, but they cannot.
Can you do that with B?
Kudos [?]: 104 [0], given: 38
Intern
Joined: 21 Feb 2010
Posts: 33
Kudos [?]: 6 [0], given: 9
Location: Ukraine
### Show Tags
12 May 2010, 04:31
I`ll go with B
what is the OA?
Kudos [?]: 6 [0], given: 9
Manager
Joined: 16 Feb 2010
Posts: 179
Kudos [?]: 34 [0], given: 17
### Show Tags
15 May 2010, 07:42
yet another vote for B
Kudos [?]: 34 [0], given: 17
Intern
Joined: 06 Sep 2009
Posts: 20
Kudos [?]: 2 [0], given: 9
Location: Mumbai
### Show Tags
19 May 2010, 00:07
ykaiim wrote:
First, thanks Ravi for this post. IMO B.
Second, it is definitly a Main Point Question and not a FLAW one. Read the question stem - Each of the following illustrates the principle that the passage illustrates.
When ppl are having some effect of some cause then they show behavioral changes in stimuly and concentration. ppl are poor in this selfawreness.
So, it follows some action is taken on/by ppl and some judgement on doing something is then asked. We need to check this logic. Now, lets check the choices:
(A) People who have been drinking alcohol are not good judges of whether they are too drunk to drive. [Show behavioral changes in stimuly and concentration - Yes]
(B) Elementary school students who dislike arithmetic are not good judges of whether multiplication tables should be included in the school’s curriculum. [Show behavioral changes in stimuly and concentration - No]
(C) Industrial workers who have just been exposed to noxious fumes are not good judges of whether they should keep working. [Show behavioral changes in stimuly and concentration - Yes]
(D) People who have just donated blood and have become faint are not good judges of whether they are ready to walk out of the facility. [Show behavioral changes in stimuly and concentration - Yes]
(E) People who are being treated for schizophrenia are not good judges of whether they should continue their medical treatments. [Show behavioral changes in stimuly and concentration - Yes]
thanks for your explanation. i was convinved with C till i came across your explanation.
Kudos [?]: 2 [0], given: 9
Manager
Joined: 21 Jan 2010
Posts: 220
Kudos [?]: 104 [0], given: 38
### Show Tags
19 May 2010, 10:21
Very good question. Thanks for the explanation!
Kudos [?]: 104 [0], given: 38
Re: Sleeping Drivers [#permalink] 19 May 2010, 10:21
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Additional info for A Discourse Concerning Algebra - English Algebra Since 1685
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Most, for example, those of the (modern) form ax 3 ± bx 2 ± cx = 0 were essentially quadratics (the solution x = 0 would not have counted); others, of the form ax 3 = bx + c;ax 3 = bx 2 + c or ax 3 = bx 2 + cx+d. were solved wrongly by simply applying the standard rules for quadratics. 99 Where Gerardi had treated fifteen equation types, Dardi treated no fewer than 194, including some with fourth powers, all, however, reducible by a simple substitution to one of al-Khwārizmī's six basic forms. 100 Later authors repeated his rules for these equations as though they applied generally, without stating the special conditions under which they applied.
He also wrote a computus, a text on the reckoning of date and time, also in verse. 26 HOW ALGEBRA WAS ENTERTAINED AND CULTIVATED IN EUROPE Fig. 1 The final verse of the Carmen de algorismo of Alexander Ville Dieu, and a multiplication table with a list of squares and cubes (Bodleian Library MS Savile 17, f. 108v). Usually known as the Massa computi it is internally dated at 1200,44 so his algorism may be supposed to date from roughly the same period. Even more influential than Ville Dieu's Carmen was the Algorismus of Johannes Sacrobosco (c.
8 Aldhelm (640–709), Abbot of Malmesbury and Bishop of Sherborne, was the author of Liber de septenario, a treatise on the number seven, but it was a mystical rather than mathematical work. 15 In England, Bede's learning was never more than a fragile candle in a vast surrounding darkness, and was all but extinguished in the invasions and instability of the three following centuries. Only early in the twelfth century did scholars in England and elsewhere across Europe become aware of the knowledge that all this time had been accumulating in Islamic Spain, and some of the more adventurous travelled south and brought back texts that were to set intellectual life in northern Europe on a new course. | 1,127 | 4,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-04 | longest | en | 0.924606 |
https://lifethroughamathematicianseyes.wordpress.com/2014/07/15/first-day/ | 1,532,180,231,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00020.warc.gz | 696,561,807 | 45,767 | As I was telling you in my previous post (link) I was at Stuttgart, and the time-table was more intense than I had expected at the beginning, though I enjoyed it a lot… It was really interesting, and it will take time to share everything I was doing there, but I promise that I will post as much as possible.
I will start with my first day, obviously. As a normal 1st day, it was more about presenting the university and the city, than math, but we have also done some ‘Mathematics Problem Solving’. The lecture-presentation was made by Prof. Dr. Fischer and it was extremely interesting.
We started with some general ideas about mathematics:
• ‘Mathematics is the art of problem solving.’ (by G. Polya)
• ‘I do believe that problems are the heart of mathematics.’ (by P.Halmos)
• Mathematics is a living, breathing, changing organism with many facets… It is creative, powerful, and even artistic.
• Mathematics is not a set of formulas to be applied to a list of problems at the end of textbook chapters.
• Mathematics is thinking not computing.
• Mathematical problem solving requires creativity (have an idea, see something) and intuition (feeling), but also precise and logical thinking.
Then we discussed in groups some easy math problems; trying to use some of our basic knowledge and not complicated things. Here are some of the problems (I will post the proofs later to give you some thinking time):
• Is √2+√3 rational?
• There is no polynomial p(x) with integer coefficients such that p(1)=4 and p(3)=5 hold.
• Find a formula for the value of the sum 1*1!+2*2!+…+n*n!
• p(x) and q(x) are polynomials with equal coefficients, but in reverse order. Is there any relation between the roots of p and q?
• Is there in every year a Friday the 13th?
Then we did some harder questions in which he was explaining some important concepts that are used a lot in mathematics and also in industry or finance/insurance, such as the pigeonhole principle, divide and conquer or extremum principle. Here are some of the problems I liked the most (again I will post the solutions later to let you digest these harder questions):
• Let there be given 5 lattice points (points with integer coordinates) in the plane. Show that there is a lattice point on the interior of one of the line segments joining 2 of these points. How many points do we need if we consider the 3D space instead of the plane?
• On a certain island we have: 13 green chameleons; 17 red ones and 24 brown. Whenever 2 chameleons of different colors meet, they both change their color into the 3rd color. Is it possible that at one time all chameleons have the same color?
• In the plane a set of 2n points is given, n points are colored red and the remaining n blue. Every red point is connected with a blue point by a straight line segment. Show that it is always possible to do this so that these line segments do not intersect.
• On a circle points are selected and the chords joining them in pairs are drawn. Assuming that no 3 of these chords are concurrent (except at the endpoints), how many points of intersection are there?
In the end he also recommended some books that are now on my must read list. Here are some:
• How to solve it: A new aspect of mathematical method by G. Polya (1945) [this is the 1st book printed about this subject]
• Mathematical discovery: on understanding, learning and teaching problem solving by G. Polya (1962)
• Problem-solving through problems by L.C.Larson (1983)
• Mathematical thinking: problem-solving and proof by J.P.D’Angelo, D.B.West (1997)
• Problem-solving strategies by A.Engel (1998)
• Principles of mathematical problem solving by M.J.Erickson, J.Flowers (1999)
It was an incredible fun presentation, and I enjoyed the way he was explaining everything to us. He changed the way in which I think about math problems, but also at life problems in general. Because if we think more about this, we use the same thinking procedures to solve our everyday problems. And this kind of lectures really help us understand life in a different way, and help us on our thinking process ( I am jealous that the students there in Stuttgart have this kind of lectures regularly). I totally recommend to check anything related with problem solving (math ones, especially) you will not regret it.
Check my Facebook page, my Tumblr, my just started Google+ page and also my new Twitter (I am really new to the last 2 things, so bare with me if you can see stupid mistakes there) and Instagram. | 1,037 | 4,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-30 | latest | en | 0.975029 |
https://matheducators.stackexchange.com/questions/9962/what-is-a-good-method-for-drawing-a-m%C3%B6bius-band-on-the-blackboard | 1,719,278,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00515.warc.gz | 339,241,503 | 44,358 | # What is a good method for drawing a Möbius band on the blackboard?
This week I'm going to give a talk on fiber bundles, and I found myself with an unexpected problem. Since I'm not using slides, I'll need to draw a Möbius band on the blackboard. Usually what I do is simply draw a rectangle with some arrows to indicate the identification, but I think I could do better.
Is there a method that can help me to draw a good Möbius band on the blackboard? I'm looking for something that looks good and is easy to draw.
• I would show one made of paper, and I would practice drawing from good images online. Commented Nov 9, 2015 at 19:34
• If fact, it is so easy to make paper ones everyone in the class should be encouraged to make a copy. Commented Nov 9, 2015 at 20:22
• By talk do you mean a class you are teaching, or a research seminar, or something else? Paper ones will not be entirely appropriate in some contexts. Commented Nov 10, 2015 at 7:04
• @JosephMalkevitch Very good idea! In my class we actually made our copies, and that was in high school. As a very beautiful and meaningful mathematical concept I wear a Möbius strip as my wedding ring, but as an object it is unfortunately too small for demonstration in front from a whole class. Commented Nov 10, 2015 at 17:11
• Also look at: davidparker.com/janine/mathpage/topology.html#MOBIUS Commented Nov 11, 2015 at 18:44
Draw the bottom three-quarters of an oval:
Flesh that out to make the bottom half of the strip:
Connect one of the open ends at the top to the bottom on the other side:
Now draw a straight line across the top:
Finally fill in the last edge at the back:
An advantage to this approach is that it highlights the fact that the mobius strip starts out as an ordinary strip, but has a twist in one spot.
Note I figured out these steps by making various physical mobius strips with very thin width relative to their length, and manipulating them so that they were as loop-like as possible. One of the key features is the straight line at the top. That is due to how a curve in paper looks from the side -- same as at the sides of the picture.
Finally, I echo everyone else's sentiments: you should bring strips of paper so that everyone can make a real physical mobius strip of their own. There's nothing like seeing the physical real object to support the imagination!
• I was wondering something similar, but I also thought it might be too broad. This question matheducators.stackexchange.com/questions/7864/… was on a similar theme, but didn't have a "how can I actually do it by hand" answer (yet) Commented Nov 10, 2015 at 1:49
• I'm using yours. It's amazing. Commented Nov 10, 2015 at 11:04
• @Newman If you are using this one, you might want to accept his answer by clicking the checkmark at the upper lefthand corner of the post. Commented Nov 10, 2015 at 15:19
• @BenjaminDickman I've put an answer to that other question on quadric surfaces, if you're interested matheducators.stackexchange.com/questions/7864/… Commented Nov 13, 2015 at 4:50
• UPDATE: I used this drawing in my talk. It went great. This drawing is specially good to show that there isn't a natural homeomorphism between the fibers and the typical fiber, but two such homeomorphisms , and they are related by "reflect in the midpoint". Commented Nov 15, 2015 at 17:20
I searched google images and found many nice renditions. Here one that you may prefer from this link
(source: umich.edu) .
I especially like that it shows the width of the paper and doesn't draw the strip as a line or piece of string. Having the students make them is always a hit.
• I really like this one--per chance are there other figures you draw nicely? Commented Nov 10, 2015 at 4:17
• @Addem I wish I could draw anything. I got this from a link as explained in the beginning of the post. Commented Nov 10, 2015 at 12:29
Since I use Seifert's algorithm to construct surfaces a lot, I tend to go for the flat approach:
Instead of drawing on the blackboard, use a 3D model and pass it around when you're finished using it for demonstration.
If the class is small enough, you can bring in a strip of paper for each student to make his/her own.
The demonstration will be much more powerful this way and the lesson more likely to stick.
• This allows people to actually vote up this particular answer as an independent idea instead of just always having it in the comments.
– WBT
Commented Nov 10, 2015 at 16:34
A Möbius band can be constructed by cutting a cylinder and then rejoining its ends after a twist. So draw it the same way.
Draw a cylinder. This should be easy enough to do. Cut it by rubbing out parts of it. And then link it back with a twist.
This is in the same spirit as David's answer, with the added benefit that your audience can follow its construction with their imagination.
However, the use of blackboard may present a problem, as erasing tends to make an area less readable. Perhaps a different colour for the twist would help. | 1,222 | 5,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.962254 |
http://algs4.cs.princeton.edu/home/ | 1,412,209,303,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663612.31/warc/CC-MAIN-20140930004103-00271-ip-10-234-18-248.ec2.internal.warc.gz | 8,102,630 | 5,784 | # Algorithms, 4th Edition
essential information that
every serious programmer
algorithms and data structures
## Textbook.
The textbook Algorithms, 4th Edition by Robert Sedgewick and Kevin Wayne [ Amazon · Pearson ] surveys the most important algorithms and data structures in use today. We motivate each algorithm that we address by examining its impact on applications to science, engineering, and industry. The textbook is organized into six chapters:
• Chapter 1: Fundamentals introduces a scientific and engineering basis for comparing algorithms and making predictions. It also includes our programming model.
• Chapter 2: Sorting considers several classic sorting algorithms, including insertion sort, mergesort, and quicksort. It also includes a binary heap implementation of a priority queue.
• Chapter 3: Searching describes several classic symbol table implementations, including binary search trees, red-black trees, and hash tables.
• Chapter 4: Graphs surveys the most important graph processing problems, including depth-first search, breadth-first search, minimum spanning trees, and shortest paths.
• Chapter 5: Strings investigates specialized algorithms for string processing, including radix sorting, substring search, tries, regular expressions, and data compression.
• Chapter 6: Context highlights connections to systems programming, scientific computing, commercial applications, operations research, and intractability.
## Online courses.
You can take our free Coursera MOOCs. They include a full set of lecture videos and assessments. Algorithms, Part I covers Chapters 1 through 3; Algorithms, Part II covers Chapters 4 through 6.
## Booksite.
Reading a book and surfing the web are two different activities: This booksite is intended for your use while online (for example, while programming and while browsing the web); the textbook is for your use when initially learning new material and when reinforcing your understanding of that material (for example, when reviewing for an exam). The booksite consists of the following elements:
• Excerpts. A condensed version of the text narrative, for reference while online.
• Java code. The algorithms and clients in this textbook.
• Exercise solutions. Solutions to selected exercises.
## To get started.
Here are instructions for setting up a simple Java programming environment [ Mac OS X · Windows · Linux ]. | 455 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2014-41 | longest | en | 0.895921 |
https://ttlc.intuit.com/community/tax-credits-deductions/discussion/recovery-rebate-credit-turbotax-calculation-error/00/2189998 | 1,660,064,672,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571056.58/warc/CC-MAIN-20220809155137-20220809185137-00055.warc.gz | 528,271,062 | 28,089 | • Submit a question
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Level 2
# Recovery Rebate Credit TurboTax calculation error
There appears to be a calculation error in TurboTax regarding the Recovery Rebate Credit. I'll cut right to it, I'm filing married filing jointly, and the totals of EIP1 and EIP2 received was \$4769.08. As married with two dependents, I was eligible for \$3400 in the first-round (EIP1) and \$2400 in the second-round (EIP2), for a total eligibility of \$5800. The Recovery Rebate Credit is reduced by 5% of the amount by which my AGI exceeds the married filing jointly threshold (\$150,000). According to TurboTax, my 2020 AGI was \$169,412. I exceeded the threshold by \$19,412. 5% of \$19,412 is \$970.60. So taking the \$5800 total I was eligible for and subtracting \$970.60 from it, I get \$4829.40. That's the total amount I should have received in total from EIP1 and EIP2. Again the total we received for EIP1 and EIP2 was \$4769.08. Subtracting \$4769.08 from \$4829.40 is \$60.32. So I think I should be receiving a Recovery Rebate Credit of \$60.32. It's small but it's significant. There is no way in the program to manually calculate the credit, so I'm stuck with what appears to be a wrong calculation. So unless my understanding of the Credit is wrong, there is a bug in TurboTax potentially not giving taxpayers their full refund.
23 Replies
Level 15
## Recovery Rebate Credit TurboTax calculation error
suggest walking your math through the official IRS worksheet; please post back if you think it is still in error.
The \$970.60 gets substracted from the eligibility of each payment and the eligibility of the two payments is assessed separately; you appear to be combining everything.
see page 58 and then see lines 14 and 18
https://www.irs.gov/pub/irs-pdf/i1040gi.pdf
Returning Member
## Recovery Rebate Credit TurboTax calculation error
Turbotax says based on my AGI of \$79,805 I'm eligible for the full \$1200 first stimulus and the full \$600 second stimulus. I received \$698.50 from the first stimulus already. When I plug that in, it says I'm only due a \$621 credit. That doesn't add up to the expected \$1,105.50. What gives?
Further I don't understand where Turbotax is getting my AGI from. It looks like it excluded \$10,800 out of \$17,892 from my AGI. But, didn't the IRS come out and say all unemployment is excluded from AGI for purposes of calculating stimulus payments?
Expert Alumni
## Recovery Rebate Credit TurboTax calculation error
The IRS clarified on Tuesday that the unemployment benefits will not count toward adjusted gross income, the figure used to determine whether people are eligible for the \$1,400 stimulus payment. That means more people will be eligible for the stimulus checks.
With the new stimulus bill, up to \$10,200 in last year's unemployment payments can be exempt from taxes if your adjusted gross income (AGI) is less than \$150,000, according to new exclusions from the IRS. If your AGI is higher, you can't exclude any unemployment compensation
When figuring any of the following deductions or exclusions, include the full amount of your unemployment benefits reported on Schedule 1, line 7 (unreduced by any exclusion amount): taxable social security benefits, IRA deduction, student loan interest deduction, nontaxable amount of Olympic or Paralympic medals and USOC prize money, the exclusion of interest from Series EE and I U.S. Savings Bonds issued after 1989, the exclusion of employer-provided adoption benefits, the tuition and fees deduction, and the deduction of up to \$25,000 for active participation in a passive rental real estate activity. See the specific form or instructions for more information. If you file Form 1040-NR, you aren’t eligible for all of these deductions. See the Instructions for Form 1040-NR for details.
Returning Member
## Recovery Rebate Credit TurboTax calculation error
Yeah, I read that exact language in the same news article, and that's where the confusion comes from. The question is whether ALL unemployment is excluded from determining whether you qualify for ANY of the stimulus checks, or whether just the first \$10,200 is excluded. TurboTax seems to believe it is just the first \$10,200. The author of that articled also emailed me back to say \$10,200 and they should update their article. Can anyone definitively answer where ALL unemployment is excluded from determining eligibility for any of the three stimulus checks, or just the first \$10,200. (Also, FYI, the question I posted only related to the first two stimulus checks.)
Expert Alumni
## Recovery Rebate Credit TurboTax calculation error
The IRS clarified on Tuesday that the unemployment benefits will not count toward adjusted gross income, the figure used to determine whether people are eligible for stimulus payments. That means more people will be eligible for the stimulus checks.Mar 25, 2021
Returning Member
## Recovery Rebate Credit TurboTax calculation error
On the "Let's double-check the amount you received" page, it reads "based on the info you entered and your combined adjusted gross income of \$185,575, you and 'wife's name' are eligible for the following stimulus payments:". Then it shows First-Round: \$2400 and Second-Round: \$1200. I then checked that was NOT what we received and entered amount we received of \$898. Doing the math, we should receive a Recovery Rebate Credit of \$2702. However, this amount is not included on 1040 line 30 Recovery Rebate Credit. Is there a problem with Turbotax CD?
Level 15
## Recovery Rebate Credit TurboTax calculation error
@JPM20211 - TT is correct, but the wording is terrible.
the Recovery Credit is dependent on a) your filing status, b) your dependents under the age of 17 and c) your income for 2020 LESS what you have received for the first two stimulus payments. That statement in TT fails to recognize either the haircut for having an AGI greater than \$150,000 or the EIPs received to date
Since your income exceeds \$150,000, the amount due you is reduced by 5% for the amount over \$150,000, or \$1779 reduction in this case. I assume you received \$898 for the 1st payment and nothing for the 2nd payment
walk through the worksheet on page 59
https://www.irs.gov/pub/irs-pdf/i1040gi.pdf
from the \$2400 you would subtract \$1779 which is what you are eligible for LESS what you received (\$898) and since the answer cannot be negative, you are eligible for \$0 for EIP1
for the \$1200 you would subtract \$1779 and since the answer cannot be negative, you are eligible for \$0 for EIP2
Note that you cannot be asked to return any money under this formula, even if the result is negative. Line 30 can never be negative.
hence Line 30 is zero and is correct.
Level 15
## Recovery Rebate Credit TurboTax calculation error
Note that you should not receive the 3rd stimulus as your income exceeds \$160,000, which was the new limit set by Congress.
Level 1
## Recovery Rebate Credit TurboTax calculation error
TT showed a \$3,600 Rebate Credit as my "Refund Due". But IRS web says we are due \$0.00. TT advertising "guarantees accuracy" so when is TT going to send me the \$3,600 mistake it made?
Expert Alumni
## Recovery Rebate Credit TurboTax calculation error
When TurboTax in the Review asked you if you received the first and second stimulus payment, how did you answer? Did you tell TurboTax that you had received these payments and told TurboTax how much for each one?
TurboTax is not liable for errors in input.
**Say "Thanks" by clicking the thumb icon in a post
New Member
## Recovery Rebate Credit TurboTax calculation error
I haven't had a chance to figure out why, but my refund was cut in half from \$1200 -\$600 by the IRS with the message there was an error in calculations, no details given. Since u used TurboTax and have a pretty straight forward return I can only assume there is a glitch in TurboTax.
Level 2
## Recovery Rebate Credit TurboTax calculation error
Like kathya, we filed with TT only to find that the glitch messed things up. So instead of a refund, we now owe.
Level 15
## Recovery Rebate Credit TurboTax calculation error
@ssarfati Lots of people gave incorrect answers when they went through the recovery rebate credit section of their tax returns, resulting in a bigger refund showing on the Form 1040 than they were really entitled to get. The IRS cross checks the recovery rebate credit on line 30 and when it sees that you already got money you said you did not receive, they recalculate and reduce your refund. They will send a detailed letter in several weeks.
**Disclaimer: Every effort has been made to offer the most correct information possible. The poster disclaims any legal responsibility for the accuracy of the information that is contained in this post.**
Level 1
## Recovery Rebate Credit TurboTax calculation error
Bill, turbotax may not be liable for the input, but they are responsible for how they pose the questions. They are also responsible for how it handles the input. So I ran across this issue today. 6/7/2021 Turbotax presents: "Lets double-check the amount received" (in regards to stimulus). Then it presents the amounts for first and second round payments. It asks; "Is this what XXX received?" If one answers yes, then it processes that as you received no stimulus payments. This may have been corrected now, but on 6/7/2021 that's how it processed (I screenshot it for documentation purposes) that input. That to me clearly indicates an issue with how it displayed data and incorrectly processed it. Does turbotax offer rebates/refunds for incorrectly processing input?
v | 2,305 | 9,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | latest | en | 0.949527 |
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Algebra Pre-Algebra and Basic Algebra Math Forum
April 17th, 2014, 03:43 PM #11
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Quote:
Originally Posted by the john Also, if the circle doesn't require having such tangent lines then . . .
With the smaller radius, (x - 1)² + (y - 4)² = 1 or x² + y² - 2x - 8y + 16 = 0. Note that the center (1, 4) is the midpoint between the other two circles' centers.
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# show that a circle can be drawn touching the line
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32
By using some auxiliary equations, an under determined systems of equations
1
can
be found. Then, by choosing the values of a few of the constants or , the other con-
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particular, by using different combinations of
a
i
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page 40, this has certain advantages.
The error term for Runge-Kutta's method can be hard, but not impossible to deter-
mine. It can be shown that the order of the error term is always going to be less than or
equal to the number of
k
terms and if there are more than four
k
terms, the order must be
strictly less than the number of
k
terms. So, using five
k
terms can not increase the order of
the method, although it can reduce the constant in the error term. Thus, using four
k
terms
is often the most efficient of the Runge-Kutta methods.
The most popular constants for the 4
th
order Runge-Kutta method yield:
where:
The Runge-Kutta method uses a series of Euler like steps to create trial points into
the interval that is being integrated. As is shown in FIGURE 22., the first evaluation of
f()
gets the slope of the path at time
t
and this slope is used to get a trial evaluation at the mid
point of the interval. The second evaluation of
f()
is used to get a different evaluation at the
mid point. The trial points are always taken at the location that is the result of starting at
x(t)
and moving along the given slope for the desired length of time. Even though
k
2
and
k
3
are both taken at the mid point,
k
2
will have (in this example) a worse approximation of
1. A list of equations that has more variables (unknowns) than equations. When the system of equations is
under determined you will always be free to pick (or be forced to pick depending on how you look at it) the
values for some of the variables in order to find the value of the rest of the variables.
a
i
b
j
x t h
+
(
)
x t
( )
16
k
1
2
k
2
2
k
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k
4
+
+
+
(
)
+
=
k
1
hf x t
( )
(
)
=
k
2
hf x t h
2
+
(
)
12
k
1
+
(
)
=
k
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hf x t h
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+
(
)
12
k
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=
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hf x t h
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= | 656 | 2,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.937502 |
https://engineering-math.org/2017/03/13/mortgage-project/ | 1,553,502,603,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203842.71/warc/CC-MAIN-20190325072024-20190325094024-00290.warc.gz | 478,754,538 | 39,676 | Mortgage Project
A home will likely be the biggest purchase a person ever makes, along with being the most intimidating purchase. But the suitable news is most of the problems homebuyers face have a quick solution, if accomplished before trying to get a mortgage.
Also the purpose of this project is help you to overcome the imitating home buying process, definitely it will gave a chance to get familiar with the process:
• Check your credit report so you are mindful of what your current credit score is before for a loan. Credit reporting agencies must give you one free report annually.
• Try to work closely with your banker to figure out how much you can borrow and which loan exactly fits you.
• Acquire what current mortgage rates are. Bankers are there to help you understand how that translates into monthly mortgage payments and that’s the purpose of this project.
The following formula is used for figuring out a monthly home mortgage payment:
$M=\frac{Lr\left[1+\frac{r}{12}\right]^{12t}}{12\left[\left(1+\frac{r}{12}\right)^{12t}-1\right]}$
where: L=the loan amount in dollars
r = the annual interest rate expressed as decimal
t = the number of years of the loan
M = the monthly payment in dollars
You are looking to buy a $325,000 home in Haverhill. If Bank of America will give them a 30-year mortgage at 6% annual interest rate for the cost of the house after they receive a 10% down payment. A. Determine the loan amount? B. How much their monthly payment will be? C. At the end of the 30-years, how much total money will you have paid to Bank of America for your home? In another word how much did the$325,000 house really cost the couple?
D. How much interest will they have paid?
E. How many of her monthly payment go toward the interest?
F. What percent increase over the cost of the home does this interest represent?
G. Redo and re-answer all questions, but this time for 15 years?
Do analysis comparison between 30 and 15 years mortgage (at least one page not double spacing).
You can purchase the solution to this problem now! Just click the button below!
If you are looking for help in your online class or homework and projects, you can send me an email at jonaspaulbdelacruz@gmail.com with the following information:
Course:
Budget:
Name:
Course URL:
User ID: | 532 | 2,305 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-13 | longest | en | 0.960921 |
http://www.instructables.com/topics/curious-about-Capacitors-/ | 1,519,101,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00034.warc.gz | 488,649,368 | 9,053 | 1128Views12Replies
### Author Options:
M just curious about capacitors that the voltage they store when they get charged
i have two capacitors laying around for test
on one is written 400v 22uf
other is 420v 100uf
so how much voltage can they hold when they get charged fully ???
and any simple way to charge them ???
i want make something for shocking pranks.... :D
Tags:
## 12 Replies
The voltage rating printed on the side of a capacitor tells you the
maximum voltage you can place across it, without breaking it. There may
also be some markings indicating polarity. Its sort of the same deal with polarity. If you reverse the polarity there's the possibility of breaking the capacitor, sometimes spectacularly.
So there's your answer. A capacitor, rated at 400 VDC, is fully charged when it has 400 volts DC across it. A capacitor, rated at 420 VDC, is fully charged when it has 420 volts DC across it.
That's why the question, "How much voltage can they hold when they get charged fully?", the question is not all that meaningful.
It is sort of like asking, "How much stretch can a rubber band hold?"
For example, I've got a rubber band here that can hold 10 centimeters of stretch.
So you've got a mental picture of this now: A rubber band. It started out in a roughly circular shape, maybe one centimeter in diameter, but now I have stretched into a taught, thin line, 10 cm in length.
What happens if I try to stretch it some more? For example, lets say I want to stretch it out to 11 cm. So I pull on the ends a little harder, but before I can get it to stretch to 11cm, the rubber band breaks.
Why did it break? Because it was only rated for 10 cm of stretch.
A question that might be more meaningful is to ask how much energy can a capacitor hold. The amount of energy stored in a capacitor, ideally, is
U = (1/2) *C*V^2
where V is the voltage measured across the terminals of the capacitor, C is capacitance of the capacitor in farads, and U is in joules of stored energy.
The amount of energy a capacitor stores is equal to the work required to to move charge into it and increase its voltage. Similarly, the energy stored in a rubber band is equal the work required to stretch it. Actually the formula for energy stored in an ideal rubber band, or metal spring, is similar, something like
U = (1/2) *k*x^2
where k is a "spring constant" with units of force divided by distance, and x is displacement, the length, or distance the rubber band, or spring, gets stretched.
It is sort of the same story with capacitors and rubber bands. Both can store energy, and for both you can break the darned thing if you try to store more energy than the materials can physically withstand.
Anyway, that concludes the physics lesson. Hope that brings you closer to enlightenment. If not, you can always read the rest, here:
http://en.wikipedia.org/wiki/Capacitor
Regarding the thing you want to build, the capacitors you've mentioned are roughly the same size, voltage and capacitance, as those found in a flash for a camera, be it disposable film camera, or obsolete film camera. Both of those are pretty cheap and easy to find, at the time of this writing. The circuit that charges the big capacitor, "big" in this case meaning around 300 uF or so, this circuit produces around 300 volts DC, so that's not going to break the capacitors you mentioned (since those are rated for voltage higher than 300 VDC)
So I'm going to recommend looking into camera flash charging circuits. Here's a search for instructables about "camera flash shocker" :
https://www.instructables.com/tag/type-id/?sort=non...
and I'm guessing that some of those are, pretty much the same thing you want to build.
I'm not sure how safe toys like this are, but at the same time I'm not very good at lecturing on safely. Just be careful, because those goddamn monkeys bite, I tell ya!
http://m.imdb.com/title/tt0078788/quotes?qt=qt0324...
D3zire (author)2014-02-24
thanks for the pretty good explaining ... :) but lets say i cant get a disposable camera circuit... any other method to charge them ??
i have other circuits that has capacitors .. cant i use them ??
Jack A Lopez (author)2014-02-25
Well, the camera flash charger pretty much has everything you want; i.e.
It runs on batteries, is small enough to fit in one hand, and the complete circuit is cheap and easy to find.
If you want to build your own, the part that's hard to find, if you try to build it part by part, is the transformer.
I think that's why the circuit Iceng just uppped is using a 8ohm-to-1000ohm impedance matching transformer, because that's a pretty common part. If you live in the former US, I think RadioShack(r) has that part.
But if you live in the former US, then you could just buy a disposable camera from some brick-and-mortar store. Or buy an obsolete film camera from a thrift store. Or buy the used circuit board from place like this:
http://www.goldmine-elec-products.com/prodinfo.asp...
Anyway, going back to this idea of building such a thing from discrete components, this page:
http://www.talkingelectronics.com/projects/XenonFl...
has a bunch of circuit diagrams which the author copied from found flash charger circuits.
Also the circuit Iceng upped, here
https://cdn.instructables.com/FCL/1722/HS1RRYEY/FCL...
looks to me like it would work. The part of that circuit on the output side of the transformer is a network of diodes and capacitors, called a voltage multiplier.
http://en.wikipedia.org/wiki/Voltage_multiplier
and you can sort of increase or decrease the voltage it produces by adding or removing stages. If I remember correctly the trade off in adding more stages, is that current available at the last stage gets smaller and smaller, so the time required to charge a capacitor in the last stage gets longer and longer, until it becomes an inconveniently looooooooooooooong time.
D3zire (author)2014-02-25
that is the problem i live in india where even disposable camera are pretty costly and hard to find... i have this mosquito zapper which has capacitor attached to it.. so can i replace it with bigger capacitor and charge it how much volt needs to charge the capacitor.. and i am not ee not even close i cant understand that diagram thats what i was trying to tell iceng too..
iceng (author)2014-02-25
+1
Great pointers, especially the
http://www.talkingelectronics.com
A
iceng (author)2014-02-24
As an older EE ... I can tell you that four hundred volts will provide a good shock and probably kill the subject of your foolish desire for attention.
Here is a circuit that can charge to 400 VDC if you simply take out 3 identical sections.
Be aware if you exceed the capacitor voltage you have, they do have a tendency to steam explode and fling its chemicals at your eyes and face.
D3zire (author)2014-02-25
well that is scary.. anyway to make it safe i have some reduced volt capacitors too.. like 240v .. i don't want to kill anyone :( and i i am not a ee not even close.. so can you simplify that diagram please if possible
iceng (author)2014-02-25
The process of creating the pain of electric shock penetrating the skin without killing is to limit the DC current.
The 10nF/400v capacitors are 0.01uF/400v compared to your 22uF they are 2200 less current storage capacity.
Also the stun gun puts eight of these in series further diminishing the capacitance to .00125uF which is 17,600 less then your 22uF cap.
And ... that circuit is as simple as it comes !
steveastrouk (author)2014-02-25
If someone tried to "prank" me with an electric shock, they would find themselves on the deck, nursing broken teeth.
Just sayin'
steveastrouk (author)2014-02-25
If someone tried to "prank" me with an electric shock, they would find themselves on the deck, nursing broken teeth.
mpilchfamily (author)2014-02-24
Like the caps say on there side they can hold 400V and 420V respectively. To charge them give them any voltage between 0 and 400v/420v and your set. If you want to play around with shocking pranks then get a hold of some disposable camera flash cicuits. They only need slight modification to make some fun little toys.
D3zire (author)2014-02-24
thanx for explaining :) | 1,975 | 8,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-09 | longest | en | 0.958523 |
http://mymathforum.com/elementary-math/346798-why-2nd-formula-c-i-used.html | 1,566,178,445,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314638.49/warc/CC-MAIN-20190819011034-20190819033034-00231.warc.gz | 139,134,198 | 9,269 | My Math Forum Why 2nd formula of C.I is used?
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July 22nd, 2019, 06:33 AM #1 Senior Member Joined: Aug 2014 From: India Posts: 458 Thanks: 1 Why 2nd formula of C.I is used? Value of X, Ram invested Rs 6400 on compound interest at the rate of X% and gets an amount of Rs 8464 as interest after 2 yrs. Quantity II → A/P = (1+ R/100)^n → √(8464/6400) = (1+ R/100)^n → 23/20 = 1 + R/100 → R = 15% There are C.I formulas: C.I = P([1 + R/100]^n – 1); A = P (1 + R/100)^n Why 2nd formula of C.I is used? Last edited by skipjack; July 22nd, 2019 at 03:55 PM. Reason: to correct use of parentheses
July 22nd, 2019, 09:43 AM #2 Senior Member Joined: Jun 2019 From: USA Posts: 120 Thanks: 40 Both formulas say the same thing. You could use either one and get the same answer. Thanks from Ganesh Ujwal
July 22nd, 2019, 03:55 PM #3 Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 The first formula had misplaced parentheses, which I've corrected, so A = P + C.I.
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# Problem 44343. Pair Primes
Solution 1745607
Submitted on 8 Mar 2019 by Jean-Marie Sainthillier
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 = 51; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
2 Pass
x = 3; y_correct = 2485; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
3 Pass
x = 4; y_correct = 136162; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match'))))
4 Pass
x = 5; y_correct = 8578934; assert(isequal(pairPrimes(x),y_correct)) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[51,2485,136162,8578934]),regexp(fileread('pairPrimes.m'),'[\d\.\+\-\*\/]+','match')))) | 404 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-05 | latest | en | 0.308203 |
https://www.belmontday.org/news/learning-updates-week-april-6-10/ | 1,708,913,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00190.warc.gz | 659,176,386 | 39,870 | Search
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# Learning Updates for Week of April 6 to 10
April 13, 2020
With the switch to offsite learning due to the coronavirus outbreak, our faculty will be sharing various curriculum highlights from their “classrooms” over the coming weeks.
In theater arts, first graders were given an assignment to come up with three different characters. These characters could be real, like a police officer, teacher, surfer, or wildlife enthusiast, or they could be make-believe, like Bugs Bunny, a dragon, or a genie. The students were asked to think about each character in terms of physicality, facial expression, and setting. Students used items found around their home to create costumes for their characters. Finally, students had someone at home take a picture of their characters to share.
– Christopher Parsons, theater arts teacher
Sixth graders just finished a unit on fractions in math class. They learned how to add, subtract, multiply, and divide with fractions using models and applying the strategies to solve real-life problems (see the photo for an example). The students will now take these strategies to support their learning in our new unit on decimals as they learn operations with decimals, their relationship to fractions, and the importance of place value.
– Elinor Hannum, middle school math teacher
Second Grade Absorbs A Lesson On Water
Second graders have been exploring the properties of water in the science lessons. Recently, students did a challenge called “climbing water.” They challenge was to find a way to move water from one cup to another using a paper towel without touching or pouring the water. The challenge allowed students to see capillary action at work. After following an instructional video, students tried out the challenge and shared their work with their teachers.
– Nancy Fell and Sunny Lee, grade 2 teachers
Students Team Up to Investigate Biomes
Students in seventh grade science groups one and two have been learning about the characteristics of the six major biomes of the world—rainforests, deserts, grasslands, deciduous forests, boreal forests, and tundras. With a partner, students split up the biomes so that they each had three to study. Students created a climate graph for each of their three biomes using Google Sheets and then researched information such as climate, world location, longitude, latitude, soil structure, flora, and fauna.
The pairs of students then combined their work into a final submission. This is a project that students usually do in the regular class setting working closely to share information and feedback about their work with each other over the course of about eight class periods. It was great to see each partnership work together online to complete this assignment to the same standards that they would be required to meet in the classroom in less time! Check out an example of their work here.
– Sandra Trentowsky, grades 7 & 8 science teacher
Eighth Grade Scientists Cook Up a Lesson on Bonds
On Wednesday, the eighth grade science gold group had its very first distance learning chemistry lab. In this lab, students found materials in their homes that contained either ionic bonds or covalent bonds. They then investigated properties of these ionic compounds and covalent compounds by melting them, boiling them, and dissolving them in water. Students performed the experiments in their kitchens, applying their knowledge of chemical bonding to better understand some important cooking principles; for example, why sugar will melt easily but salt does not. Our first distance lab was a great success, and we are looking forward to doing more soon!
– Leal Carter, grades 7 & 8 science teacher
## February 16, 2024
Coach Tzelnic took the helm of the boys’ varsity basketball team three years ago. That team started a pair of sixth graders, lost Friday Night Hoops by 20 points, and finished their season with a 2-6 record. Fast forward to…
## February 16, 2024
Eighth Graders Share Original Poems After a week of learning about and experimenting with the Shakespearean sonnet, the modern English ghazal, and contemporary free verse poetry, eighth graders worked hard this week on crafting their poetry portfolios that touch on…
## February 2, 2024
Pre-k Welcomes Guests for Lunar New Year Celebration On Wednesday, our pre-kindergarten classes were joined by the parents of Averie, Oliver, William, and Zachary for a celebration of the Lunar New Year! We learned all about foods with special meaning…
## February 2, 2024
Belmont Day’s girls’ basketball program recorded a pair of wins against Nashoba Brooks this week, improving their combined record to an impressive 11-0. The junior varsity team continued its torrid pace by doubling up Nashoba Brooks. Relying on strong defense… | 1,001 | 4,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | longest | en | 0.961559 |
https://adc.bmj.com/highwire/markup/183005/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed | 1,716,760,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00364.warc.gz | 68,768,731 | 3,296 | Table 1
Rounding rules for summary statistics
Summary statisticReportingExamples (where useful)
MeanUse enough decimal places to give either the SD to two significant digits,7 or the SE to one significant digit3320 g
3.32 kg
PercentageIntegers, or one decimal place for values under 10%. Values over 90% may need one decimal place if their complement is informative. Use two or more decimal places only if the range of values is less than 0.1%0.1%
5.3%
27%
89%
99.6%
Mean differenceUse enough decimal places to give the SE to one or two significant digits. For a standardised mean difference use one or two decimal places
Regression coefficientAs with the mean difference.
Correlation coefficientOne or two decimal places, or more when very close to ±10.03
−0.7
0.89
0.999
Risk ratioRound to two significant digits if the leading non-zero digit is four or more, otherwise round to three (the rule of four11). Alternatively use one/two significant digits rather than two/three. For ORs very close to 1 (eg, in logistic regression with a continuous variable) use three decimal places or else report the log OR×100 as the percentage odds to one decimal place130.0321
0.062
0.76
1.05
4.2
11.3
55
1.042
4.1%
SDOne or two significant digits7570 g
0.57 kg
9 mm Hg
2.5 mL
SEOne or two significant digits
CIUse the same rule as for the corresponding effect size (be it mean, percentage, mean difference, regression coefficient, correlation coefficient or risk ratio), perhaps with one less significant digit
Test statistics: t, F, χ2, etcUp to one decimal place and up to two significant digits
t=−1.3
F=11
χs=4.1
p valueRound up to one significant digit, within the limits shown in the examples. The lower limit may be smaller than 0.001, but never 0.000. For genome-wide association studies use the power of 10 format>0.9
0.4
0.1
0.08
0.05
0.003
<0.001
6.10−9 | 506 | 1,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-22 | latest | en | 0.779503 |
https://www.geteasysolution.com/12x+2(x+8)+6=8-(-4x+6) | 1,575,881,411,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00318.warc.gz | 707,391,803 | 5,391 | # 12x+2(x+8)+6=8-(-4x+6)
## Simple and best practice solution for 12x+2(x+8)+6=8-(-4x+6) equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the equation solver your own equation and let us solve it.
## Solution for 12x+2(x+8)+6=8-(-4x+6) equation:
12x+2(x+8)+6=8-(-4x+6)
We move all terms to the left:
12x+2(x+8)+6-(8-(-4x+6))=0
We multiply parentheses
12x+2x-(8-(-4x+6))+16+6=0
We calculate terms in parentheses: -(8-(-4x+6)), so:
8-(-4x+6)
determiningTheFunctionDomain
-(-4x+6)+8
We get rid of parentheses
4x-6+8
We add all the numbers together, and all the variables
4x+2
Back to the equation:
-(4x+2)
We add all the numbers together, and all the variables
14x-(4x+2)+22=0
We get rid of parentheses
14x-4x-2+22=0
We add all the numbers together, and all the variables
10x+20=0
We move all terms containing x to the left, all other terms to the right
10x=-20
x=-20/10
x=-2
` | 379 | 1,045 | {"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} | 4.53125 | 5 | CC-MAIN-2019-51 | longest | en | 0.703494 |
http://easy-ciphers.com/hypoglossal | 1,571,853,537,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00023.warc.gz | 65,804,743 | 18,798 | Easy Ciphers Tools:
Caesar cipher
Caesar cipher, is one of the simplest and most widely known encryption techniques. The transformation can be represented by aligning two alphabets, the cipher alphabet is the plain alphabet rotated left or right by some number of positions.
When encrypting, a person looks up each letter of the message in the 'plain' line and writes down the corresponding letter in the 'cipher' line. Deciphering is done in reverse.
The encryption can also be represented using modular arithmetic by first transforming the letters into numbers, according to the scheme, A = 0, B = 1,..., Z = 25. Encryption of a letter x by a shift n can be described mathematically as
Plaintext: hypoglossal
cipher variations: izqphmpttbm jarqinquucn kbsrjorvvdo lctskpswwep mdutlqtxxfq nevumruyygr ofwvnsvzzhs pgxwotwaait qhyxpuxbbju rizyqvycckv sjazrwzddlw tkbasxaeemx ulcbtybffny vmdcuzcggoz wnedvadhhpa xofewbeiiqb ypgfxcfjjrc zqhgydgkksd arihzehllte bsjiafimmuf ctkjbgjnnvg dulkchkoowh evmldilppxi fwnmejmqqyj gxonfknrrzk
Decryption is performed similarly,
(There are different definitions for the modulo operation. In the above, the result is in the range 0...25. I.e., if x+n or x-n are not in the range 0...25, we have to subtract or add 26.)
Atbash Cipher
Atbash is an ancient encryption system created in the Middle East. It was originally used in the Hebrew language.
The Atbash cipher is a simple substitution cipher that relies on transposing all the letters in the alphabet such that the resulting alphabet is backwards.
The first letter is replaced with the last letter, the second with the second-last, and so on.
An example plaintext to ciphertext using Atbash:
Plain: hypoglossal Cipher: sbkltolhhzo
Baconian Cipher
To encode a message, each letter of the plaintext is replaced by a group of five of the letters 'A' or 'B'. This replacement is done according to the alphabet of the Baconian cipher, shown below.
```a AAAAA g AABBA m ABABB s BAAAB y BABBA
b AAAAB h AABBB n ABBAA t BAABA z BABBB
c AAABA i ABAAA o ABBAB u BAABB
d AAABB j BBBAA p ABBBA v BBBAB
e AABAA k ABAAB q ABBBB w BABAA
f AABAB l ABABA r BAAAA x BABAB
```
Plain: hypoglossal Cipher: AABBB BABBA ABBBA ABBAB AABBA ABABA ABBAB BAAAB BAAAB AAAAA ABABA
Affine Cipher
In the affine cipher the letters of an alphabet of size m are first mapped to the integers in the range 0..m - 1. It then uses modular arithmetic to transform the integer that each plaintext letter corresponds to into another integer that correspond to a ciphertext letter. The encryption function for a single letter is
where modulus m is the size of the alphabet and a and b are the key of the cipher. The value a must be chosen such that a and m are coprime.
Considering the specific case of encrypting messages in English (i.e. m = 26), there are a total of 286 non-trivial affine ciphers, not counting the 26 trivial Caesar ciphers. This number comes from the fact there are 12 numbers that are coprime with 26 that are less than 26 (these are the possible values of a). Each value of a can have 26 different addition shifts (the b value) ; therefore, there are 12*26 or 312 possible keys.
Plaintext: hypoglossal
cipher variations:
izqphmpttbm
wvurtirddbi
krytfetnnbe
yncvravxxba
mjgxdwxhhbw
afkzpszrrbs
cxsdnkdllbk
qtwfzgfvvbg
epahlchffbc
slejxyjppby
ghiljulzzbu
udmnvqnjjbq
jarqinquucn
xwvsujseecj
lszugfuoocf
zodwsbwyycb
nkhyexyiicx
bglaqtassct
dyteolemmcl
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fqbimdiggcd
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venowrokkcr
kbsrjorvvdo
yxwtvktffdk
mtavhgvppdg
apextcxzzdc
olizfyzjjdy
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ezufpmfnndm
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pmjagzakkez
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hsdkofkiief
vohmabmsseb
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mdutlqtxxfq
azyvxmvhhfm
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ejodtwdvvfw
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uxajdkjzzfk
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yhqrzurnnfu
nevumruyygr
bazwynwiign
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dshawfaccgf
rolcibcmmgb
fkpeuxewwgx
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vybkelkaagl
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lmnqozqeegz
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etibxgbddhg
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jezkurkssir
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ybenhonddjo
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fedacrammkr
tahconcwwkn
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jotiybiaakb
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jcvaopaggsp
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arihzehllte
onmjlajvvta
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qfunjsnppts
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upkvfcvddtc
iloxryxnnty
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rgvoktoqqut
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tydsilskkul
vqlwgdweeud
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zabecnessun
nwfgojgccuj
ctkjbgjnnvg
qpolnclxxvc
elsnzynhhvy
shwpluprrvu
gdarxqrbbvq
uzetjmtllvm
wrmxhexffve
knqztazppva
yjubfwbzzvw
mfydrsdjjvs
abcfdofttvo
oxghpkhddvk
dulkchkoowh
rqpmodmyywd
fmtoazoiiwz
tixqmvqsswv
hebsyrsccwr
vafuknummwn
xsnyifyggwf
loraubaqqwb
zkvcgxcaawx
ngzestekkwt
bcdgepguuwp
pyhiqlieewl
evmldilppxi
srqnpenzzxe
gnupbapjjxa
ujyrnwrttxw
ifctzstddxs
wbgvlovnnxo
ytozjgzhhxg
mpsbvcbrrxc
alwdhydbbxy
ohaftufllxu
cdehfqhvvxq
qzijrmjffxm
fwnmejmqqyj
tsroqfoaayf
hovqcbqkkyb
vkzsoxsuuyx
jgduatueeyt
xchwmpwooyp
zupakhaiiyh
nqtcwdcssyd
bmxeizeccyz
pibguvgmmyv
defigriwwyr
rajksnkggyn
gxonfknrrzk
utsprgpbbzg
ipwrdcrllzc
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khevbuvffzu
ydixnqxppzq
avqblibjjzi
orudxedttze
cnyfjafddza
qjchvwhnnzw
efgjhsjxxzs
sbkltolhhzo
hypoglossal
vutqshqccah
xmbuqzuwwaz
lifwcvwggav
zejyoryqqar
bwrcmjckkaj
psveyfeuuaf
dozgkbgeeab
rkdiwxiooax
fghkitkyyat
tclmupmiiap
The decryption function is
where a - 1 is the modular multiplicative inverse of a modulo m. I.e., it satisfies the equation
The multiplicative inverse of a only exists if a and m are coprime. Hence without the restriction on a decryption might not be possible. It can be shown as follows that decryption function is the inverse of the encryption function,
ROT13 Cipher
Applying ROT13 to a piece of text merely requires examining its alphabetic characters and replacing each one by the letter 13 places further along in the alphabet, wrapping back to the beginning if necessary. A becomes N, B becomes O, and so on up to M, which becomes Z, then the sequence continues at the beginning of the alphabet: N becomes A, O becomes B, and so on to Z, which becomes M. Only those letters which occur in the English alphabet are affected; numbers, symbols, whitespace, and all other characters are left unchanged. Because there are 26 letters in the English alphabet and 26 = 2 * 13, the ROT13 function is its own inverse:
ROT13(ROT13(x)) = x for any basic Latin-alphabet text x
An example plaintext to ciphertext using ROT13:
Plain: hypoglossal Cipher: ulcbtybffny
Polybius Square
A Polybius Square is a table that allows someone to translate letters into numbers. To give a small level of encryption, this table can be randomized and shared with the recipient. In order to fit the 26 letters of the alphabet into the 25 spots created by the table, the letters i and j are usually combined.
1 2 3 4 5
1 A B C D E
2 F G H I/J K
3 L M N O P
4 Q R S T U
5 V W X Y Z
Basic Form:
Plain: hypoglossal Cipher: 3245534322134334341113
Extended Methods:
Method #1
Plaintext: hypoglossal
method variations: ndutmqtxxfq sizyrvycclv xoedwadhhqa ctkibfinnvf
Method #2
Bifid cipher
The message is converted to its coordinates in the usual manner, but they are written vertically beneath:
```h y p o g l o s s a l
3 4 5 4 2 1 4 3 3 1 1
2 5 3 3 2 3 3 4 4 1 3 ```
They are then read out in rows:
3454214331125332334413
Then divided up into pairs again, and the pairs turned back into letters using the square:
Plain: hypoglossal Cipher: subocfphntl
Method #3
Plaintext: hypoglossal
method variations: rzshbsnodan zshbsnodanr shbsnodanrz hbsnodanrzs bsnodanrzsh snodanrzshb nodanrzshbs odanrzshbsn danrzshbsno anrzshbsnod nrzshbsnoda
Permutation Cipher
In classical cryptography, a permutation cipher is a transposition cipher in which the key is a permutation. To apply a cipher, a random permutation of size E is generated (the larger the value of E the more secure the cipher). The plaintext is then broken into segments of size E and the letters within that segment are permuted according to this key.
In theory, any transposition cipher can be viewed as a permutation cipher where E is equal to the length of the plaintext; this is too cumbersome a generalisation to use in actual practice, however.
The idea behind a permutation cipher is to keep the plaintext characters unchanged, butalter their positions by rearrangement using a permutation
This cipher is defined as:
Let m be a positive integer, and K consist of all permutations of {1,...,m}
For a key (permutation) , define:
The encryption function
The decryption function
A small example, assuming m = 6, and the key is the permutation :
The first row is the value of i, and the second row is the corresponding value of (i)
The inverse permutation, is constructed by interchanging the two rows, andrearranging the columns so that the first row is in increasing order, Therefore, is:
Total variation formula:
e = 2,718281828 , n - plaintext length
Plaintext: hypoglossal
first 5040 cipher variations(39615626 total)
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# Institute of Banking Personnel Selection (IBPS) Exam pattern & Syllabus
• Institute of Banking Personnel Selection (IBPS) Exam pattern & Syllabus
• ### Institute of Banking Personnel Selection (IBPS) Exam pattern & Syllabus
WR1334 > 30-12-2012, 09:45 AM
EXAM PATTERN
1. Reasoning: 50 questions of 50 marks
2. English Language: 50 questions of 50 marks
3. Numerical Ability: 50 questions of 50 marks
4. General Awareness: 50 questions of 50 marks
5. Computer Knowledge: 50 questions of 50 marks
Total: 250 marks
EXAM SYLLABUS
1. ENGLISH LANGUAGE
Vocabulary: Synonyms, Antonyms, Homonyms, Word Formation, Sentence Completion
Comprehension: Theme detection, Deriving Conclusion, Passage Completion
Error Detection and Rearrangement: Passage Correction, Sentence Correction, Spelling
Grammar: Active Passive Voice, Direct Indirect Speech
General Usage: Idioms and Phrases
2. LOGICAL REASONING
Verbal Reasoning: Analogy, Coding Decoding, Blood Relation, Sitting Arrangement, Series Completion, Syllogism, Decision Making, Statement Reasoning
Non-verbal Reasoning: Series Completion, Analogy, and Classification
3. QUANTITATIVE APTITUDE
Arithmetic: Numbers, Simplification, Roots, Average, Surds & Indices, Percentage, Profit & Loss, Ratio & Proportion, Partnership, Chain Rule, Time & Work, Pipes & Cisterns, Time & Distance, Problems on Trains, Boats & Streams, Alligation, Simple Interest, Compound Interest, Stocks & Shares, Clocks, Logarithms, Mensuration, Volume & Surface Area, Permutation & combination, Probability, Heights & Distances
Data Interpretation: Tabulation, Bar Graphs, Line Graphs, Pie Charts
4. BANKING AWARENESS
Questions in this component will be aimed at testing the candidate’s general awareness of the environments around him and its application to society. These questions will be such that they do not require a specific study of any discipline.
Knowledge of current events
Everyday observations and experience in the scientific aspect as may be expected of any educated person
India and its neighboring countries
Sports, History, Culture, Geography, Economy, General Polity, Indian Constitution, Scientific Research
Over 70% of the questions will be related to Economy, Banking and Finance
5. COMPUTER AWARENESS
General Knowledge on Computer usage and application.
• ### RE: Institute of Banking Personnel Selection (IBPS) Exam pattern & Syllabus
Preetha > 30-12-2012, 02:29 PM
Useful info!!
• ### RE: Institute of Banking Personnel Selection (IBPS) Exam pattern & Syllabus
rajsathya > 30-12-2012, 06:57 PM
good info for banking dude........ | 632 | 2,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | latest | en | 0.761557 |
https://www.thestudentroom.co.uk/showthread.php?t=4387518 | 1,534,527,295,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212639.36/warc/CC-MAIN-20180817163057-20180817183057-00704.warc.gz | 1,022,190,934 | 39,677 | You are Here: Home >< A-levels
# M1 force diagrams help watch
1. Hi, I've never come across this sort of question before so I was wondering how I would approach it? I'm finding mechanics quite difficult in general too.
The angle on the incline is 40.1 and the reaction force between the boat and the plane is 825N.
b) The rope starts to fray. A horizontal force P is applied to the front of the boat so that the
rope does not have to hold as much weight. The frayed rope remains taut. If the tension in the
rope reaches 550N the rope will snap. We do not want the rope to snap. What is the minimum
value that the force P can take?
2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.
Just quoting in Danny Dorito so she can move the thread if needed
Spoiler:
Show
(Original post by Danny Dorito)
x
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