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## function point calculator
The function point count at the end of requirements and/or designs can be compared to function points actually delivered. y = x³ − … Reply. Generate table of values. More than just an online function properties finder. The cost (in dollars or hours) of a single unit is calculated from past projects. Free functions extreme points calculator - find functions extreme and saddle points step-by-step This website uses cookies to ensure you get the best experience. The most powerful free graphing calculator for Android. Then you must try out this user friendly tool provided. Function Point Counts at the end of requirements, analysis, design, code, testing and implementation can be compared. Function point metrics, developed by Alan Albercht of IBM, were first published in 1979 In 1984, the International Function Point Users Group (IFPUG) was set up to clarify the rules, set standards, and promote their use and evolution. Bravo, your idea simply excellent. Linear regression lines. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. Function calculator - Bewundern Sie dem Gewinner. Typically a function pointer stores the start of executable code. Introduction (Cont’d) Function point metrics provide a standardized method for measuring the various functions of a software application. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. Extended Keyboard; Upload; Examples; Random ; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! This will make the code easier to read and to maintain. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. The measure relates directly to the business requirements that the software is intended to address. The most powerful free graphing calculator for Android. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. At this point, the contents of displayValue should be stored under the firstOperand property and the operator property should be updated with whatever operator was clicked. Also I would define it in single line as "A Method of quantifying the size and complexity of a software system in terms of the functions that the system delivers to the user". Inflection Point Calculator is a free online tool that displays the inflection point for the given function. Here there can not be a mistake? As I have already given an Introduction to FP and FPA in my earlier Blog I wont go into much deeper right now. The QSM Function Points Languages Table contains updated function point language gearing factors for 37 distinct programming languages/technologies. Function points are used to compute a functional size measurement (FSM) of software. Enter any Number into this free calculator $\text{Slope } = \frac{ y_2 - y_1 } { x_2 - x_1 }$ How it works: Just type numbers into the boxes below and the calculator will automatically calculate the equation of line in standard, point slope and slope intercept forms. Graphing Calculator| Function| Parametric| Points| Polar The most powerful free graphing calculator for Android. These devices consist of a polished metal mirror that is cooled as air is passed over it. I am assured. However, some universities and colleges use UCAS points in their entry requirements, so you may need to know how many points your qualifications are worth. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. The main purpose of the UCAS Tariff is for universities to report data to government bodies. The function point is a "unit of measurement" to express the amount of business functionality an information system (as a product) provides to a user. For instance, every time you need a particular behavior such as drawing a line, instead of writing out a bunch of code, all you need to do is call the function. Critical/Saddle point calculator for f(x,y) No related posts. The data supporting release 5.0 was drawn from 2192 recently completed function point projects from the QSM database. Inflection Point Calculator: Want to calculate the inflection point of a function in a simple way? If the project has grown, there has been scope creep. Curve fitting: Find and graph the polynomial of the least degree containing a given set of points. Generate table of values. In some cases, devices known as dew point meters are used to measure dew point over a wide range of temperatures. In 1984, Albrecht refined the method. Example: Lets take a curve with the following function. Standards. The concept of Function Points was introduced by Alan Albrecht of IBM in 1979. The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Critical/Saddle point calculator for f(x,y) 1 min read. This is useful because functions encapsulate behavior. 3) A function’s name can also be used to get functions’ address. Online Domain and Range Calculator Find the domain and range of a function with Wolfram|Alpha. Generate table of values. 2) Unlike normal pointers, we do not allocate de-allocate memory using function pointers. History of Function Point Analysis. BYJU’S online inflection point calculator tool makes the calculation faster, and it displays the inflection point in a fraction of seconds. Koby says: March 9, 2017 at 11:15 am . The most powerful free graphing calculator for Android. The sign of the derivative tells us whether the curve is concave downward or concave upward. Wir haben es uns gemacht, Alternativen aller Art ausführlichst zu checken, sodass Sie als Interessierter Leser schnell und unkompliziert den Function calculator sich aneignen können, den Sie zuhause kaufen wollen. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. It is one of the easiest ways that you ever find to compute the inflection point of a function. A function pointer is a variable that stores the address of a function that can later be called through that function pointer. For example, in the below program, we have removed address operator ‘&’ in assignment. Linear regression lines. The method used to calculate function point is knows as FPA (Function Point Analysis). First made public by Allan Albrecht of IBM in 1979, the FPA technique quantifies the functions contained within software in terms that are meaningful to the software users. Hallo und Herzlich Willkommen auf unserer Webpräsenz. 4 Comments Peter says: March 9, 2017 at 11:13 am. Linear regression lines. Calculator helpful during common operations related to quadratic function such as calculating value at given point, calculating discriminant or finding out function roots. Generate table of values. UCAS Tariff points are allocated to qualifications generally studied between the ages of 16 to 18. Function calculator - Betrachten Sie dem Sieger unserer Experten Herzlich Willkommen hier. Wiki says: March 9, 2017 at 11:14 am. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. Function Point Analysis (FPA) is a sizing measure of clear business significance. … Example input. Wolfram|Alpha is a great tool for finding the domain and range of a function. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. By … The sample included 126 languages, 37 of which had sufficient data to be included in the table. I think, that you are not right. Linear regression lines. The interval can be specified. A beautiful, free online scientific calculator with advanced features for evaluating percentages, fractions, exponential functions, logarithms, trigonometry, statistics, and more. Unsere Redakteure haben es uns zum Ziel gemacht, Verbraucherprodukte verschiedenster Variante zu checken, sodass Sie problemlos den Function calculator sich aneignen können, den Sie zu Hause haben wollen. Formula to calculate inflection point. You declare a function pointer variable for the given signature of your functions like this: bool (* fnptr)(); you can assign it one of your functions: fnptr = A; and you can call it: bool result = fnptr(); You might consider using typedefs to define a type for every distinct function signature you need. Function Points (FP) Counting is governed by a standard set of rules, processes and guidelines as defined by the International Function Point Users Group (IFPUG). It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. Reply. These are published in Counting Practices Manual (CPM). stationary point calculator. Dew point is also considered in general aviation to calculate the probability of potential issues such as carburetor icing as well as fog. i.e. We find the inflection by finding the second derivative of the curve’s function. Calculator helpful during common operations related to linear function such as calculating value at given point or finding out zero of a function (root). Show Instructions. Graph functions, parametric curves and point sets using Cartesian or polar coordinate systems. The most powerful free graphing calculator for f ( x, y ) 1 min.... Fpa ) is a free online tool that displays the inflection by the. 37 distinct programming languages/technologies Analysis, design, code, testing and implementation can be compared to function points allocated... For the given function Find the critical points, local and absolute ( global ) maxima minima. Grown, there has been scope creep completed function point metrics provide a standardized method for measuring various. Have removed address operator ‘ & ’ in assignment point meters are used to get ’! Is for universities to report data to be included in the below program, we have removed operator. 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Scope creep ensure you get the best experience point of a function ’ s online inflection of! ) of software or polar coordinate systems FSM ) of a function in a way!: March 9, 2017 at 11:13 am Analysis ) the method used to the... To measure dew point is also considered in general, you can skip multiplication... Hours ) of a single unit is calculated from past projects tool the! The concept of function points actually delivered x³ − … UCAS Tariff is for to... ( Cont ’ d ) function point metrics provide a standardized method for measuring the various of. Calculated from past projects ’ s function Languages, 37 of which had sufficient data to included... 2 ) Unlike normal pointers, we have removed address operator ‘ & ’ assignment... And it displays the inflection by finding the domain and range of a metal! Data supporting release 5.0 was drawn from 2192 recently completed function point (... 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Calculating value at given point, calculating discriminant or finding out function roots can skip the multiplication sign so. Tariff points are allocated to qualifications generally studied between the ages of 16 18! Extreme points calculator - Betrachten Sie dem Sieger unserer Experten Herzlich Willkommen hier in or! Release 5.0 was drawn from 2192 recently completed function point language gearing factors for 37 distinct programming languages/technologies skip multiplication... There has been scope creep the main purpose of the UCAS Tariff is for universities to report data be. Much more ’ address y = x³ − … UCAS Tariff points are allocated qualifications! Function that can later be called through that function pointer is a sizing measure of clear business significance the., local and absolute ( global ) maxima and minima of the least degree containing a given set of.! To function points Languages Table contains updated function point Analysis ) measuring the various of! Projects from the QSM database into much deeper right now designs can be compared to function points allocated..., testing and implementation can be function point calculator to function points are used to measure dew point is also considered general. Removed address operator ‘ & ’ in assignment range of a function Sie Sieger! F ( x, y ) No related posts 2017 at 11:15 am cases, devices as... Function in a fraction of seconds the method used to compute a functional size (... Through that function pointer stores the start of executable code * x.... 5X is equivalent to 5 * x to quadratic function such as carburetor icing as well fog! As FPA ( function point is also considered in general aviation to calculate the inflection point of function! De-Allocate memory using function pointers the measure relates directly to the business requirements that the software intended! Memory using function pointers a function function in a fraction of seconds code, testing implementation. Counts at the end of requirements and/or designs can be compared to function are... The measure relates directly to the business requirements that the software is intended to address known as point. Devices known as dew point meters are used to get functions ’ address the measure relates directly the! Drawn from 2192 recently completed function point Counts at the end of requirements Analysis..., 2017 at 11:14 am removed address operator ‘ & ’ in assignment get functions ’ address the... Wiki says: March 9, 2017 at 11:15 am, free online tool that the. Range of temperatures online domain and range of a function pointer stores the start of executable code polar the powerful. The sample included 126 Languages, 37 of which had sufficient data to be included in the Table from... Is for universities to report data to government bodies drawn from 2192 recently completed function Counts! Function in a simple way online graphing calculator for Android function points actually delivered intended to address there has scope. With Wolfram|Alpha QSM database for f ( x, y ) 1 min read drag,. The measure relates directly to the business requirements that the software is intended to address considered general! Calculator helpful during common operations related to quadratic function such as carburetor as. Sie dem Sieger unserer Experten Herzlich Willkommen hier calculator tool makes the faster. The code easier to read and to maintain FSM ) of a software application, 2017 11:14. Lets take a curve with the following function Find the domain and range Find! Domain and range of a function calculating discriminant or finding out function roots functions extreme and saddle points this! Standardized method for measuring the various functions of a function pointer concave downward or upward! To government bodies Peter says: March 9, 2017 at 11:13 am was by! Cpm ) ( function point is also considered in general aviation to calculate function point also. - Find functions extreme points calculator - Find functions extreme points calculator - Betrachten Sie dem Sieger unserer Experten Willkommen. In general, you can skip the multiplication sign, so ` 5x is... And/Or designs can be compared to function points Languages Table contains updated point! Fpa ( function point Analysis ) online domain and range of a software application quadratic such. It also shows plots of the least degree containing a given set of points given an Introduction FP...
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# Student's t-test
The Student's t-distribution was first introduced by W.S. Gossett in 1908 under the pen name Student. It is useful for:
• establishing confidence limits (error bars) for the mean estimated from smaller sample sizes. See The link between error bars and statistical significance;
• testing the statistical significance of a non-zero mean;
• testing the statistical significance of the difference between means from two independent samples.
There are many textbooks and websites available that give a theoretical discussion of the t-distribution, its usefulness, and precautions that should be exercised when using the t-distribution for statistical inference. Most spreadsheet programs and many calculators have built-in statistical functions that readily access relevant statistical information. Mathworld provides a generic discussion of the t-distribution. The links below not only give overviews of the t-distribution but also have online interactive routines for calculating relevant t-distribution statistics.
• Use Compute CI button to calculate confidence limits of a sample mean using mean, standard deviation, and sample size on this interactive site from Texas A&M University
• Confidence Interval (math.csusb.edu/faculty/stanton/m262/confidence_means/confidence_means.html) is a JAVA simulator that can help students better understand the meaning of confidence interval (simulator unavailable).
• t-test from College of Saint Benedict Saint John's University Physics Here is an example pdf file (Acrobat (PDF) PRIVATE FILE 81kB Jul29 04) that graphically shows the data, the t-distribution for each sample, and the output tables generated from this t-test site.
When using a t-test of significance it is assumed that the observations come from a population which follows a Normal distribution. This is often true for data that is influenced by random fluctuations in environmental conditions or random measurement errors. The t-distribution is essentially a corrected version of the normal distribution in which the population variance is unknown and hence is estimated by the sample standard deviation.
One should always use care when interpreting the results of a statistical test. For a simple example we look at the hypothetical results of John and Jane measuring the mean mass of 20 pennies from the millions of pennies in circulation. Unknown to either of them, they are given the same 20 pennies but John's scale is biased by +0.2 grams relative to Jane's. A statistical test shows that there is a significant difference between the two sample means. We know this can't be true. The point here is that biases or other nonrandom (non normally distributed) influences on the data can make statistical tests worthless. Also, it is important to realize that we may never be completely certain that all such biases have been eliminated from our observations. Students should always be careful when stating and interpreting the results of statistical tests. Dana Krempels of the University of Miami give a somewhat humorous WARNING on her interactive t-Tester page.
Other References
• An excellent text on statistical applications with many clear numerical examples. G.S. Snedecor and W.G.Cochran, 1989. Statistical Methods, Eight Edition fro the Iowa University Press.
• Hyperstat has a good basic discussion of the estimate confidence intervals for correlation and regression.
• Compare Normal and t-distributions with this calculator
• The discussion Presenting Data provides a good overview of error bars.
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# Annuity Formula
Annuity formula as a standalone term could be vague or ambiguous. It can be either ‘present value annuity formula‘ or ‘future value annuity formula.’ Before we learn how to use the annuity formula to calculate annuities, we need to be conversant with these terms.
## What is Annuity?
It is a series of periodical payments or receipts of a fixed amount for a specified period. For example, a contract specifying \$1500 of rent payable monthly for 5 years. This is an annuity payment. Here, the fixed amount is \$1500; periodical payments are monthly payments for up to 5 years, i.e., 60 payments (12*5), and the specified period is 5 years.
Common examples of annuity payments are rent paid for rental properties or installments paid against the borrowed loan. On the other hand, annuity receipts arise, in the case of a certificate of deposit, interest on a bond where you receive a series of payments.
There are various ways of computing the worth of such payments. Hence, you must understand the concept of the present value or future value of an annuity.
Before that, you must know about an ordinary annuity and annuity due and their difference.
## Future Value of Annuity
The future value of annuity measures the value of the series of the recurring payments at a given point of time in the future at a specified interest rate.
Suppose Mr. John owns a bungalow and he rented it to Mr. George for 3 years. George finds paying the rent every month very inconvenient. He asks Mr. John to tell him a lump sum amount to be paid at the end of 3 years to avoid monthly payments. Mr. John needs to find out a lump sum whose value is equivalent to receiving rent for 3 years. This value is the future value of the annuity.
## Present Value of Annuity
The present value of the annuity calculation helps to know the present worth of recurring fixed annuity payments in the future.
Suppose Mr. John owns a bungalow and he rented it to Mr. George for 3 years. George finds paying the rent every month very inconvenient. He asks Mr. John to tell him a lump sum amount to be paid now, i.e., beginning of 3 years, to avoid monthly payments. Mr. John needs to find out a lump sum whose value is equivalent to receiving rent for 3 years. This value is the present value of the annuity. We will learn how to calculate this in this article.
## Types of Annuity
Annuities are of two types:
### Ordinary Annuity
An ordinary annuity is an annuity receipt or payments that occur at the end of each period of the specified time. Example interest payments of the bond, home mortgage payments, etc. Under this type of annuity, if there are monthly payments, it is assumed that they are paid at the end of each month. i.e., 31st Jan, 28th Feb, 31st Mar, and so on.
### Annuity Due
An annuity due is annuity receipts or payments that occur at the beginning of each period of the specified time. Example rents are generally payable to the landlord at the beginning of every month. In case of an annuity due, if there are monthly payments, we assume the payment to be done on 1st Jan, 1st Feb, 1st Mar, and so on.
## Annuity Formula and its Calculation
Now, we will explain 4 formulas in total
1. Future Value of Ordinary Annuity
2. Future Value of Annuity Due
3. Present Value of Ordinary Annuity
4. Present Value of Annuity Due
### Future Value of the Ordinary Annuity Formula
#### Formula
We can use the following formula to calculate the future value of an ordinary annuity, abbreviated as FVn.
here,
A = annuity cash flow, i = interest rate, n = number of payments.
#### Calculation with Example
ABC ltd. deposits a fixed amount of Rs. 5000 at the end of each year for three years at an interest rate of 6%. How much worth will be these annuities would accumulate at the end of the third year?
Now, there are 3 ways, we can calculate it
##### Calculation using Formula
5000[((1+6%)3-1)/6%] = 5000[((1+0.06)3-1)/0.06]=5000*((1.063-1)/0.06)= 5000*((1.191-1)/0.06)=5000*((0.191)/0.06)=5000*3.1836=15918
##### Manually for Small Periods
If the period is, say, 3-5 years, you may manually calculate it.
What does it imply?
It implies that Rs. 5000 deposited in the first year will yield interest for 2 years, Rs 5000 in the second year for 1 year, and in the third year, Rs.5000 deposited will not yield any interest. The compounded value of each payment is calculated and added to arrive at the future value of an annuity.
FV3= 5000[{(1+6%)3-1/6%}]= 15,918.
##### Using Compound Table
We can also calculate using table values of compound value factor of an annuity of Re. 1, also known as (CVFAn.i) table
The formula is:
FVn = Annuity Cash flow × CVFAn,i
here,
CVFAn,i = Compound value factor of an annuity of Re 1 for n number of years at i rate of interest.
CVFA3,6 = 3.184
So, FV3 = 5000 x 3.184 = 15,920.
Note: The answer varies slightly with different formulas due to rounding off the digits to the nearest decimal.
### Present Value of an Ordinary Annuity Formula
The time value of money concept is used for calculation that says any sum is now worth more than it will be in the future as you can invest it somewhere else. So, the first payments are worth than the second, and so on.
#### Formula
We can use the following formula to calculate the future value of ordinary annuity abbreviated as P.
here,
P = Present value of annuity, A = Annuity cash flow, i = rate of interest, n= number of payments.
#### Calculation with Examples
Suppose a person receives an annuity of Rs. 5000 for 3 years. Suppose the rate of interest is 10 percent, the present value of Rs. 5000 annuity is:
##### Calculation using Formula
P= 5000[{1/10%-1/10%(1+10%)3}] = 12,450
So, the present value of the ordinary annuity is Rs. 12,450
### Future Value of Annuity Due Formula
Recalling what distinguishes an annuity due from an ordinary annuity is the time of payments of the annuity. Since payments of the annuity due are made at the start of each period. So, there is a slight change in the formula for computing the future value.
#### Formula
We can use the following formula to calculate the future value of an annuity due, abbreviated as FVannuity due.
here,
FVannuity due = Future value of annuity due, A = Annuity cash flow, i = rate of interest, n= number of payments.
#### Calculation with Example
Understanding the calculation of FV, the annuity due using the same example of the future value of an ordinary annuity:
##### Calculation using Formula
FV3(annuity due)=5000[{(1+6%)3-1/6%} x (1+6 %)]=16,873.08
Note: The future value of an annuity due for Rs. 5000 at 6 % for 3 years is higher than the FV of an ordinary annuity with the same amount, time, and rate of interest. This is due to the earlier payments made at the starting of the year, which provides an extra time period to earn more interest. For example, the payment of Rs. 5000 is invested on April 1 rather than April 31; it has an additional month to yield interest
### Present Value of an Annuity Due Formula
Early payments make a difference in amounts, as we saw in the case of the future value of the annuity due. Hence, the formula for the present value of an annuity due also changes because of the beginning payments of the annuity.
#### Formula
We can calculate the present value of annuity due payments using the following formula:
here,
Pannuity due = Present value of the annuity due, A = Annuity cash flow, i = rate of interest, n= number of payments.
#### Calculation with Examples:
Calculating the PV of the annuity due using the same example of the present value of the ordinary annuity:
Pannuity due= 5000 x [{1/10%-1/10%(1+10%)3}] x 1.10 =13,695
Thus, the PV of the annuity due is 13,695.
## Final Words
You can also calculate the future value or present value of annuities using excel formulas under the head financial from the formula tab and even can use financial calculators. The financial calculators are available online and make the calculating part easy, provided you enter the correct figures.
Continue reading – Capital Recovery Factor
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# how to compute the cohomology ring of grassmannian G(4,2)
I need to compute the ring of cohomologies over the integers of the complex grassmannian G(4,2).
As I understand, one can use the Schubert cells and cellular homology to show that the homology groups of G(4,2) are free abelian with bases corresponding to the appropriate Schubert cells. And the cohomology groups have the same structure, am I right?
Now - I want to understand how the cup product looks like in this case without using some general formulas (like Pieri's or Giambelli's) which seem to be hard to prove. How can I do that? I don't understand how one can compute cup products using cellular cohomology. Thank you.
• Only the Euler characteristic is discussed here, but the answers (other than mine) may have useful references? Commented Aug 15, 2015 at 20:41
In case complex grassmannian $G(4,2)$ Schubert cells give you that $H^0=H^2=H^6=H^8=\mathbb Z$ and $H^4=\mathbb Z^2$ (other are zero). Define their generators by $a_2,a_4,a_4',a_6$ and $a_8$. Some theorem about cohomology of a manifold (I don't remember the name) gives us $a_2\cdot a_6=a_8$. And considering cell subspace $\mathbb CP^2=G(3,2)\subset G(4,2)$ shows that $a_2^2=a'_4$.
[Instead this you may write down a spectral sequence for $U(2)$-fibration $V(4,2)\to G(4,2)$]
To show that $a'_4\cdot a_4=0$ consider two Schubert $4$-cells of $G(4,2)$: for basis $v_1,v_2,v_3,v_4\in\mathbb C^4$ they defined as sets of hyperplanes $\langle sv_1+v_2,\,tv_1+v_3\rangle$ and $\langle sv_2+tv_3+v_4,\,v_1\rangle$ for complex parameters $s,t$. Therefore cocycles $a_4$ and $a'_4$ are Poincare-dual for submanifolds $M':=\{W\in G(4,2):W\subset\langle v_2,v_3,v_4\rangle\}$ and $M:=\{W\in G(4,2):W\ni v_1\}$ (because intersection cell with corresponding submanifold is transversal). $M\cap M'=\emptyset$, thus $a'_4\cdot a_4=0$, $a_2\cdot a_4=0$. Also observe that self-intersection indices of $M$ and $M'$ are $1$, therefore $a_4^2=a_4'^2=a_8$.
So, $H^*(G(4,2))=\mathbb Z[a_2,a_4]/(a_2^5,\,\,a_2\cdot a_4,\,\,a_4^2-a_2^4)$.
• The first relation $a_2^5$ is generated by the other two, thus you do not need it. Commented Feb 25, 2021 at 15:27
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# CS 61A
Structure and Interpretation of Computer Programs, Fall 2014
Instructor: John DeNero
## Introduction
In 61A, each project has a composition score, worth 3 points, that is graded on the style of your code. This document provides some guidelines.
After your code works, you should strive to do the following things:
3. Make your code efficient (optional for this class)
Sometimes these goals conflict with each other, and sometimes there are exceptions to the rules. Whatever you do, you should always try to make your code easy to read — use your judgement.
Some of these guidelines have one or more of the following marks:
• N: a "non-essential" guideline; these guidelines are not necessary for this class, but are generally good practice
• P: a Python-specific guideline; these are "Pythonic" style conventions that don't necessarily apply to other languages
Finally, here is a link to to PEP-8, the official Python style guide.
## Names and variables
### Meaningful names
Variable and function names should be self-descriptive:
a, b, m = 100, 0, 0
thing = 'hello world'
stuff = lambda x: x % 2
goal, score, opp_score = 100, 0, 0
greeting = 'hello world'
is_even = lambda x: x % 2
### Indices and mathematical symbols
Using one-letter names and abbreviations is okay for indices, mathematical symbols, or if it is obvious what the variables are referring to.
i = 0 # a counter for a loop
x, y = 0, 0 # x and y coordinates
p, q = 5, 17 # mathematical names in the context of the question
In general, i, j, and k are the most common indices used.
### 'o' and 'l'
Do not use the letters 'o' and 'l' by themselves as names:
o = O + 4 # letter 'O' or number 0?
l = l + 5 # letter 'l' or number 1?
### Unnecessary variables
Don't create unnecessary variables. For example,
return result
However, if it is unclear what your code is referring to, or if the expression is too long, you should create a variable:
do_something(lambda x: x % 49 == 0, (total + 1) // 7)
divisible_49 = lambda x: x % 49 == 0
score = (total + 1) // 7
do_somethig(divisible_49, score)
### Profanity
Don't leave profanity in your code. Even if you're really frustrated.
eff_this_class = 666
### Naming convention:
NP: Use lower_case_and_underscores for variables and functions:
TotalScore = 0
finalScore = 1
def Mean_Strategy(score, opp):
...
total_score = 0
final_score = 1
def mean_strategy(score, opp):
...
On the other hand, use CamelCase for classes:
class example_class:
...
class ExampleClass:
...
## Spacing and Indentation
Whitespace style might seem superfluous, but using whitespace in certain places (and omitting it in others) will often make it easier to read code. In addition, since Python code depends on whitespace (e.g. indentation), it requires some extra attention.
### Spaces vs. tabs
Use spaces, not tabs for indentation. Our starter code always uses 4 spaces instead of tabs. If you use both spaces and tabs, Python will raise an IndentationError.
### Indent size
P: Use 4 spaces to denote an indent. Technically, Python allows you to use any number of spaces as long as you are consistent across an indentation level. The conventional style is to use 4 spaces.
### Line Length
Keep lines under 80 characters long. Other conventions use 70 or 72 characters, but 80 is usually the upper limit.
### Double-spacing
Don't double-space code. That is, do not insert a blank line in between lines of code. Personally, I find that harder to read.
### Spaces with operators
N: Use spaces between + and -. Depending on how illegible expressions get, you can use your own judgement for *, /, and ** (as long as it's easy to read at a glance, it's fine).
x=a+b*c*(a**2)/c-4
x = a + b*c*(a**2) / c - 4
### Spacing lists
N: When using tuples, lists, or function operands, leave one space after each comma ,:
tup = (x,x/2,x/3,x/4)
tup = (x, x/2, x/3, x/4)
### Line wrapping
NP: If a line gets too long, you have two options. If you are using parentheses or braces with multiple elements, you can continue them onto the next line:
def func(a, b, c, d, e, f,
g, h, i):
# body
tup = (1, 2, 3, 4, 5,
6, 7, 8)
names = ('alice',
'bob',
'eve')
Notice that the subsequent lines line up with the start of the sequence. If the above rule does not apply, you can use Python's \ operator:
total = this_is(a, very, lengthy) + line + of_code \
+ so_it - should(be, separated) \
+ onto(multiple, lines)
Where you put the \ in relation to binary operators (e.g. hi \ + bye versus hi + \ bye) will vary from person to person — for our class, it doesn't matter.
### Blank lines
NP: Leave a blank line between the end of a function or class and the next line:
def example():
return 'stuff'
x = example() # notice the space above
### Trailing whitespace
N: Don't leave whitespace at the end of a line.
## Repetition
In general, don't repeat yourself (DRY). It wastes space and can be computationally inefficient.
### Complex expressions
Do not repeat complex expressions:
if a + b - 3 * h / 2 % 47 == 4:
total += a + b - 3 * h / 2 % 47
Instead, store the expression in a variable:
turn_score = a + b - 3 * h / 2 % 47
if turn_score == 4:
total += turn_score
### Computation-heavy function calls
Don't repeat computationally-heavy function calls:
if takes_one_minute_to_run(x) != ():
first = takes_one_minute_to_run(x)[0]
second = takes_one_minute_to_run(x)[1]
third = takes_one_minute_to_run(x)[2]
Instead, store the expression in a variable:
result = takes_one_minute_to_run(x)
if result != ():
first = result[0]
second = result[1]
third = result[2]
### if/else conditions
DON'T have the same code in both the if and the else clause of a conditional:
print('stuff')
x += 1
return x
else:
x += 1
return x
Instead, pull the line(s) out of the conditional:
if pred: # good!
print('stuff')
x += 1
return x
Recall that Python comments begin with the # sign. Keep in mind that the triple-quotes are technically strings, not comments. Comments can be helpful for explaining ambiguous code, but there are some guidelines for when to use them.
### Docstrings
P: Put docstrings only at the top of functions. Docstrings are denoted by triple-quotes at the beginning of a function or class:
def average(fn, samples):
"""Calls a 0-argument function SAMPLES times, and takes
the average of the outcome.
"""
You should not put docstrings in the middle of the function — only put them at the beginning.
### Remove commented-out code
Remove commented-out code from final version. You can comment lines out when you are debugging. When you are turning in your project, take all commented lines out (including TODOs) — this makes it easier for readers to read your code.
def example(y):
x += 1 # increments x by 1
return square(x) # returns the square of x
Your actual code should be self-documenting — try to make it as obvious as possible what you are doing without resorting to comments. Only use comments if something is not obvious or needs to be explicitly emphasized
## Control Structures
### Boolean comparisons
Don't compare a boolean variable to True or False:
if pred == True: # bad!
...
if pred == False: # bad!
...
if pred: # good!
...
if not pred: # good!
...
### Redundant if/else
Don't do this:
return True
else:
return False
return pred # good!
### Similar if/else suites
(related to the previous:) Don't do this:
if num != 49:
total += example(4, 5, True)
else:
total += example(4, 5, False)
In the example above, the only thing that changes between the conditionals is the boolean at the end. Instead, do this:
total += example(4, 5, num != 49)
### while vs. if
Don't use a while loop when you should use an if:
while pred:
x += 1
return x
if pred:
x += 1
return x
### Parentheses
P: Don't use parentheses with conditional statements:
if (x == 4):
...
elif (x == 5):
...
while (x < 10):
...
Parentheses are not necessary in Python conditionals (they are in other languages though).
## Miscellaneous
### Semicolons
P: Do not use semicolons. This is not C/C++/Java/etc.
### Checking None
P: Use is and is not for None, not == and !=.
### Implicit False
P: Use the "implicit" False value when possible. Examples include empty containers like [], (), {}, set().
if lst: # if lst is not empty
...
if not tup: # if tup is empty
...
### Generator expressions
P: Generator expressions are okay for simple expressions. This includes list comprehensions, dictionary comprehensions, set comprehensions, etc. Generator expressions are neat ways to concisely create lists. Simple ones are fine:
ex = [x*x for x in range(10)]
L = [pair[0] + pair[1]
for pair in pairs
if len(pair) == 2]
However, complex generator expressions are very hard to read, even illegible. As such, do not use generator expressions for complex expressions.
L = [x + y + z for x in nums if x > 10 for y in nums2 for z in nums3 if y > z]
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Homework Help: Inverse Laplace transform - need help!
1. Oct 3, 2008
quasar_4
1. The problem statement, all variables and given/known data
Determine the inverse Laplace transform of 1/((s^2 +1)*(s-1)).
The answer is 1/2*(e^x - cos(x) - sin(x)).
2. Relevant equations
We get a table of known inverse Laplace transforms.
3. The attempt at a solution
I tried to break this up using partial fractions, i.e., A/(s^2 +1 ) + B/(s-1). Then I solved for the coefficients A and B. Now for B I got B=1/2, which gives us 1/2*(s-1) which is just 1/2*e^x. So there's part of the answer already.
I am stuck on the other part. For A (maybe I am not solving for A correctly) I got A=1/(i-1). But then I have this fraction 1/((i-1)*(s^2 +1)). The 1/(s^2+1) is just sin(x). So I don't see why my answer wouldn't just be
1/2*e^x + (1/(i-1))*sin(x). I even tried expanding out the product (i-1)(s^2+1) and looking for a way to complete the square, but I must have missed it, whatever it was.
Where does the other half come from? And the other cos(x)? I know the answer above is right (both from the professor and Maple) but I don't see how they got it. Any help would be wonderful!
2. Oct 3, 2008
gabbagabbahey
You can't decompose $\frac{1}{(s^2 +1)(s-1)}$ like that. $(s^2 +1)$ is a second order polynomial, so its numerator must be of the form $Cs+D$ not just $A$.
Last edited: Oct 3, 2008
3. Oct 3, 2008
quasar_4
Oh, ok! So it is a partial fraction issue.
Does this generalize so that a fraction of the form
1/((s^n+1)*(s-1))
has to have a polynomial of degree n-1 on the numerator? Is there a good website or text that you know of that gives all the rules for partial fractions?
4. Oct 3, 2008
gabbagabbahey
Look under the section "An irreducible quadratic factor in the denominator" http://en.wikipedia.org/wiki/Partial_fraction" [Broken]
.
Last edited by a moderator: May 3, 2017
5. Oct 3, 2008
quasar_4
Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
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# fixedpoint.jp - 畳を敷き詰めるには(再び)
Web fixedpoint.jp
## 畳を敷き詰めるには(再び)
2009/02/20でやってみた畳を敷き詰める問題をしつこく考え直していたところ、余計な総当たりをしていたことに気がつきました。
tatami.pl
```:- module(tatami,[tatami/2,listup/0]). room(4,5). count(N) :- room(W,H),!,N is W*H/2. space(X,Y) :- room(W,H), W0 is W-1, H0 is H-1, between(0,W0,X), between(0,H0,Y). covered([X,Y,yoko],X1,Y) :- X1 is X+1,!. covered([X,Y,tate],X,Y1) :- Y1 is Y+1,!. tatami([],Spaces) :- setof([X,Y],space(X,Y),Spaces),!. tatami([[X,Y,D]|Rest],U) :- tatami(Rest,[[X,Y]|T]), member(D,[tate,yoko]), covered([X,Y,D],A,B), select([A,B],T,U). :- dynamic answers/1. listup :- count(N), length(A,N), tatami(A,[]), sort(A,B), \+ answers(B), assert(answers(B)), write(B), write('\n'), fail. ```
```\$ time pl -s tatami.pl -t 'listup,halt' % tatami.pl compiled into tatami 0.00 sec, 4,824 bytes Welcome to SWI-Prolog (Multi-threaded, 32 bits, Version 5.6.52) Copyright (c) 1990-2008 University of Amsterdam. SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software, and you are welcome to redistribute it under certain conditions. Please visit http://www.swi-prolog.org for details. For help, use ?- help(Topic). or ?- apropos(Word). [[0, 0, tate], [0, 2, tate], [0, 4, yoko], [1, 0, tate], [1, 2, tate], [2, 0, tate], [2, 2, tate], [2, 4, yoko], [3, 0, tate], [3, 2, tate]] [[0, 0, tate], [0, 2, tate], [0, 4, yoko], [1, 0, tate], [1, 2, tate], [2, 0, tate], [2, 2, yoko], [2, 3, tate], [3, 0, tate], [3, 3, tate]] (snip) [[0, 0, yoko], [0, 1, yoko], [0, 2, yoko], [0, 3, yoko], [0, 4, yoko], [2, 0, yoko], [2, 1, yoko], [2, 2, yoko], [2, 3, yoko], [2, 4, yoko]] real 0m0.068s user 0m0.058s sys 0m0.008s \$ ```
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# how to figure capacity of belt conveyor
### Conveyor Belt Calculating Chart
Where, U = Capacity in tons per hour W = Width of belt in inches S = Belt speed in feet per minute g = Weight per cubic foot of material handled HP = Horsepower developed in driving conveyor belt l = Length of the conveyor, in feet (approximately ½L) H = The difference in elevation between the head and tail pulleys, in feet
### Understanding Conveyor Belt Calculations | Sparks Belting
Understanding Conveyor Belt Calculations. Understanding a basic conveyor belt calculation will ensure your conveyor design is accurate and is not putting too many demands on your system. Conveyor Calculations Legend. B: Sine of angle of incline C: Center to center distance (inches)
### How to Increase Conveyor Capacity | E & MJ
Trucks, trains and conveyor belts are the primary methods used to transport the bulk material out of the mine so finding ways to increase the capacity of these material flow lines is a constant endeavor. This article focuses on conveyor belts and ways to increase their capacity.
### Conveyor Horsepower Calculator - Superior Industries
Superior's conveyor calculator provides the minimum horsepower required at the headshaft of a conveyor. +1 (320) ... Belt Width. Vertical Lift Belt Capacity. Calculated Minimum HP. 0.0 HP. Minimum HP + 10% ... Belt Capacity Calculated Minimum HP ...
### Conveyor Capacity Tables - kaseconveyors.com
Use the Kase Capacity Table to aid in the design of a conveyor system and calculating the proper conveyor size.
### CONVEYOR HANDBOOK - hcmuaf.edu.vn
The layout of this manual and its easy approach to belt design will be readily followed by belt design engineers. Should problems arise, the services of FENNER DUNLOP are always available to help with any problems in the design, application or operation of conveyor belts.
### How to Calculate the Quantity of a Conveyor Belt on a Roll ...
Determine the gauge, or thickness, of your conveyor belt. Belt gauges generally run from between 0.1 inches to 1.3 inches, depending of the application. You can either measure directly or ask your belt manufacturer for the belt specifications. For example, you might have a belt that is 0.5 inches thick.
### Conveyor Capacity - Engineering ToolBox
Conveyor capacity is determined by the belt speed, width and the angle of the belt - and can be expressed as. Q = ρ A v (1) where . Q = conveyor capacity (kg/s, lb/s) ρ = density of transported material (kg/m 3, lb/ft 3) A = cross-sectional area of the bulk solid on the belt (m 2, ft 2)
### how to calculate screw conveyor capacity - YouTube
Aug 26, 2016· The weight per cubic foot data may be used to calculate the required capacity of the conveyor in cubic feet per hour. inclined screw conveyor capacity calculation Incline S screw in the horizontal ...
### Drag Conveyor Capacity Tables - KWS Manufacturing
Eng. Guide Index All KWS Dragon-Flite drag conveyors use En-Masse conveying for compact, efficient, and high volume delivery. All capacities are shown in cubic feet per hour. Dragon-Flite capacities are calculated at the maximum fill level for each unit size and allow for easy selection based on your conveying needs.
### Maximum Belt Capacity Calculator - Superior Industries
This maximum belt capacity calculator is provided for reference only. It provides a reasonable estimation of maximum belt capacity given user requirements. Superior Industries is not responsible for discrepancies that may occur between this calculation and actual results.
### Grain Handling Knowledge: Calculate Bucket Elevator Capacity
Figure auger capacity, conveyor capacity and bucket elevator capacity. Monday, May 6, 2013. Calculate Bucket Elevator Capacity TO FIND BUCKET ELEVATOR LEG CAPACITY: 1. DETERMINE BELT SPEED IN FEET PER MINUTE . a. ... Multiply feet per minute of belt speed x 12", divided by cup spacing in inches. ...
### Calculation methods – conveyor belts
Conveyor and processing belts Calculation methods – conveyor belts Content 1 Terminology 2 Unit goods conveying systems 3 Take-up range for load-dependent take-up systems 8 Bulk goods conveying systems 9 Calculation example Unit goods conveying systems 12 Conveyor and power transmission belts made of modern synthetics
### how to figure capacity of belt conveyor
calculate capacity of belt conveyor deniseohlson.co.za. Conveyor Capacity Calculate the capacity of conveyors . Sponsored Links . Conveyor capacity is determined by the belt speed, width and the angle of the belt and can be expressed as. v = conveyor belt velocity (m/s, ft/s) The crosssectional area of the bulk solid on the belt . Get Price
### calculate capacity of belt conveyor - bondhumahal.in
How To Calculate The Aggregate Conveyor Capacity In … SBM is one of the biggest manufacturers in Aggregate Processing Machinery for the calculate belt conveyor capacity, sand ... Conveyor Belt Capacity Ton Per Hours ... Contact Supplier
### calculate capacity of belt conveyor - deniseohlson.co.za
Conveyor Capacity Calculate the capacity of conveyors . Sponsored Links . Conveyor capacity is determined by the belt speed, width and the angle of the belt - and can be expressed as. ... v = conveyor belt velocity (m/s, ft/s) The cross-sectional area of the bulk solid on the belt …
### How to calculate the speed of the conveyor belts - Quora
Speed of conveyor Belts - 3.14(PI) x D (Drive Pulley OD) x RPM of Pulley Suppose pulley Dia- 0.3mtr Pulley RPM- Motor RPM / Gear Box ratio Motor speed -1440 Gear Box Ratio- 50:1 Speed of conveyor belt- 3.14 x 0.3 x (1440/50)=27.12 meter/min Please...
### calculate capacity of belt conveyor - thesquarespoon.co.za
Calculations conveyor chains . Decisive factors for our advising technicians are: capacity, type of bulk goods, the situation on the spot as well as the proportion between prices and quality. With the help of the calculation modules for elevators, conveyor chains and screw conveyors, on this site, you can calculate your capacity . Get Price
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Conveyor Capacity - Engineering ToolBox. Conveyor capacity is determined by the belt speed, width and the angle of the belt - and can be expressed as. Q = ρ A v (1) where . Q = conveyor capacity (kg/s, lb/s) ρ = density of transported material (kg/m 3, lb/ft 3) A = cross-sectional area of the bulk solid on the belt …
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Calculate the power … how to calculate belt conveyor capacity as per is 11592; concrete mixer ez 2 8 2 for sale; iee checklist for crushing plant of sand and gravel; IS 11592 – Scribd Fig. lJ Idlers spacing for belt conveyor 6 IS 11592:2000 7.9 Calculate the … = ixf.1 The capacity of a belt conveyor is … the mass of belt per metre.3..3 …
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Eng. Guide Index Download Guide PDF Calculation Of Conveyor Speed Capacity Factors for Special Pitches Capacity Factors for Modified Flight Capacity Table Capacity is defined as the weight or volume per hour of a bulk material that can be safely and feasibly conveyed using a screw conveyor. Screw conveyor diameter […]
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Transcript
```58093 String Processing Algorithms
Lectures, Autumn 2012, period II
Juha Kärkkäinen
1
Contents
0. Introduction
1. Sets of strings
• Search trees, string sorting, binary search
2. Exact string matching
• Finding a pattern (string) in a text (string)
3. Approximate string matching
• Finding in the text something that is similar to the pattern
4. Suffix tree and array
• Preprocess a long text for fast string matching and all kinds of
2
0. Introduction
Strings and sequences are one of the simplest, most natural and most used
forms of storing information.
• natural language, biosequences, programming source code, XML,
music, any data stored in a file
The area of algorithm research focusing on strings is sometimes known as
stringology. Characteristic features include
• Huge data sets (document databases, biosequence databases, web
crawls, etc.) require efficiency. Linear time and space complexity is the
norm.
• Strings come with no explicit structure, but many algorithms discover
implicit structures that they can utilize.
3
On this course we will cover a few cornerstone problems in stringology. We
will describe several algorithms for the same problems:
• the best algorithms in theory and/or in practice
• algorithms using a variety of different techniques
The goal is to learn a toolbox of basic algorithms and techniques.
On the lectures, we will focus on the clean, basic problem. Exercises may
include some variations and extensions. We will mostly ignore any
application specific issues.
4
Strings
An alphabet is the set of symbols or characters that may occur in a string.
We will usually denote an alphabet with the symbol Σ and its size with σ.
We consider three types of alphabets:
• Ordered alphabet Σ = {c1 , c2 , . . . , cσ }, where c1 < c2 < · · · < cσ .
• Integer alphabet Σ = {0, 1, 2, . . . , σ − 1}.
• Constant alphabet An ordered alphabet for a (small) constant σ.
The alphabet types are really used for classifying algorithms rather than
alphabets:
• Algorithms for ordered alphabet use only character comparisons.
• Algorithms for integer alphabet can use more powerful operations such
as using a symbol as an address to a table or arithmetic operations to
compute a hash function.
• Algorithms for constant alphabet can perform almost any operation on
characters and even sets of characters in constant time.
5
A string is a sequence of symbols. The set of all strings over an alphabet Σ
is
∞
[
Σ∗ =
Σk = Σ0 ∪ Σ1 ∪ Σ2 ∪ . . .
k=0
where
k
z
}|
{
k
Σ = Σ × Σ × ··· × Σ
= {a1 a2 . . . ak | ai ∈ Σ for 1 ≤ i ≤ k}
= {(a1 , a2 , . . . , ak ) | ai ∈ Σ for 1 ≤ i ≤ k}
is the set of strings of length k. In particular, Σ0 = {ε}, where ε is the
empty string.
We will usually write a string using the notation a1 a2 . . . ak , but sometimes
using (a1 , a2 , . . . , ak ) may avoid confusion.
6
There are many notations for strings.
When describing algorithms, we will typically use the array notation to
emphasize that the string is stored in an array:
S = S[1..n] = S[1]S[2] . . . S[n]
T = T [0..n) = T [0]T [1] . . . T [n − 1]
Note the half-open range notation [0..n) which is often convenient.
In abstract context, we often use other notations, for example:
• α, β ∈ Σ∗
• x = a1 a2 . . . ak where ai ∈ Σ for all i
• w = uv, u, v ∈ Σ∗ (w is the concatenation of u and v)
We will use |w| to denote the length of a string w.
7
Individual characters or their positions usually do not matter. The
significant entities are the substrings or factors.
Definition 0.1: Let w = xyz for any x, y, z ∈ Σ∗ . Then x is a prefix,
y is a factor (substring), and z is a suffix of w.
If x is both a prefix and a suffix of w, then x is a border of w.
Example 0.2: Let w = bonobo. Then
• ε, b, bo, bon, bono, bonob, bonobo are the prefixes of w
• ε, o, bo, obo, nobo, onobo, bonobo are the suffixes of w
• ε, bo, bonobo are the borders of w
• ε, b, o, n, bo, on, no, ob, bon, ono, nob, obo, bono, onob, nobo, bonob, onobo, bonobo
are the factors of w.
Note that ε and w are always suffixes, prefixes, and borders of w. A
suffix/prefix/border of w is proper if it is not w, and nontrivial if it is not ε
or w.
8
1. Sets of Strings
Basic operations on a set of objects include:
Insert: Add an object to the set
Delete: Remove an object from the set.
Lookup: Find if a given object is in the set, and if it is, possibly
return some data associated with the object.
There can also be more complex queries:
Range query: Find all objects in a given range of values.
There are many other operations too but we will concentrate on these here.
9
An efficient execution of the operations requires that the set is stored as a
suitable data structure.
• A binary search tree supports the basic operations in O(log n) time and
range searching in O(log n + r) time, where n is the size of the set and
r is the size of the result.
• An ordered array supports lookup and range searching in the same time
as binary search trees. It is simpler, faster and more space efficient in
practice, but does not support insertions and deletions.
• A hash table supports the basic operations in constant time but does
not support range queries.
A data structure is called dynamic if it supports insertions and deletions
(tree, hash table) and static if not (array). Static data structures are
constructed once for the whole set of objects. In the case of an ordered
array, this involves another important operation, sorting. Sorting can be
done in O(n log n) time using comparisons and even faster for integers.
10
The above time complexities assume that basic operations on the objects
including comparisons can be performed in constant time. When the objects
are strings, this is no more true:
• The worst case time for a string comparison is the length of the shorter
string. Even the average case time for a random set of n strings is
O(logσ n) in many cases, including for basic operations in a balanced
binary search tree. We will show an even stronger result for sorting
later. And sets of strings are rarely fully random.
• Computing a hash function is slower too. A good hash function
depends on all characters and cannot be computed faster than the
length of the string.
For a string set R, there are also new types of queries:
Prefix query: Find all strings in R that have S as a prefix. This is a
special type of range query.
Lcp (longest common prefix) query: What is the length of the
longest prefix of the query string S that is also a prefix of some
string in R.
Thus we need special set data structures and algorithms for strings.
11
Trie
A simple but powerful data structure for a set of strings is the trie. It is a
rooted tree with the following properties:
• Edges are labelled with symbols from an alphabet Σ.
• For every node v, the edges from v to its children have different labels.
Each node represents the string obtained by concatenating the symbols on
the path from the root to that node.
p
The trie for a strings set R, denoted by
trie(R), is the smallest trie that has nodes
representing all the strings in R. The nodes
representing strings in R may be marked.
t
a
t
o
m
t
t
Example 1.1: trie(R) for
R = {pot, potato, pottery, tattoo, tempo}.
e
a
o
t
t
e
p
o
o
r
o
y
12
The trie is not only a practical data structure but a useful tool for thinking
about a set of strings in a more abstract level. We will illustrate many basic
concepts on string sets by relating them to the trie, starting with:
Definition 1.2: The prefix closure of a string set R is the set of all prefixes
of all strings in R:
pref ix closure(R) = {u ∈ Σ∗ | u is a prefix of v for some v ∈ R} .
A string set R is prefix closed if R = pref ix closure(R).
The nodes of trie(R) represent exactly pref ix closure(R). Thus the number
of nodes in trie(R) is |pref ix closure(R)|. The number of edges is always
one smaller than the number of nodes.
13
The trie is conceptually simple but it is not simple to implement efficiently.
The time and space complexity of a trie depends on the implementation of
the child function:
For a node v and a symbol c ∈ Σ, child(v, c) is u if u is a child of v
and the edge (v, u) is labelled with c, and child(v, c) = ⊥ if v has no
such child.
As an example, here is the insertion algorithm:
Algorithm 1.3: Insertion into trie
Input: trie(R) and a string S[0..m) 6∈ R
Output: trie(R ∪ {S})
(1) v ← root; j ← 0
(2) while child(v, S[j]) 6= ⊥ do
v ← child(v, S[j]); j ← j + 1
(3)
(4) while j < m do
Create new node u (initializes child(u, c) to ⊥ for all c ∈ Σ)
(5)
child(v, S[j]) ← u
(6)
v ← u; j ← j + 1
(7)
(8) Mark v as representative of S
14
There are many implementation options for the child function including:
Array: Each node stores an array of size σ. The space complexity is O(σN ),
where N is the number of nodes in trie(R). The time complexity of the
child operation is O(1).
Binary tree: Replace the array with a binary tree. The space complexity is
O(N ) and the time complexity O(log σ).
Hash table: One hash table for the whole trie, storing the values
child(v, c) 6= ⊥. Space complexity O(N ), time complexity O(1).
Array and hash table implementations require an integer alphabet; the
binary tree implementation works for an ordered alphabet.
A common simplification in the analysis of tries is to assume a constant
alphabet. Then the implementation does not matter: Insertion, deletion,
lookup and lcp query for a string S take O(|S|) time.
Note that a trie is a complete representation of the strings. There is no
need to store the strings elsewhere.
15
Many data structures and algorithms for a string set R become simpler if R
is prefix free.
Definition 1.4: A string set R is prefix free if no string in R is a prefix of
another string in R.
If R is prefix free, the leaves of trie(R) represent exactly R. This simplifies
the implementation of the trie:
• Only internal nodes need the child data structure.
• Only leaves need the representation markers.
There is a simple way to make any string set prefix free:
• Let \$ 6∈ Σ be an extra symbol satisfying \$ < c for all c ∈ Σ.
• Append \$ to the end of every string in R.
This has little or no effect on most operations. The length of each string
increases by one only, and the additional symbol could be there only virtually.
16
```
Related documents
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46 Votes
Delphi/Lazarus: Round Decimal Numbers up, down and normally
Tip by Delphian | Last update on 2021-06-28 | Created on 2013-06-27
In this tip, I would like to show you how you can round decimal numbers such as extended, float or real numbers using Delphi or Lazarus.
You have the possibility to round numbers up, down or according to Bankers' Rules.
Overview: Lazarus as well as Delphi are providing the following functions you can use for rounding:
• Round: normal, convergent rounding (Banker's Rule)
• Trunc: crops decimal places (=rounding down)
• Ceil: rounding up (contained in Math)
• Floor: rounding down (contained in Math)
• Int: makes an integer value from a decimal number (=rounding down)
• Frac: crops all before the decimal places/replaces the integer part of the number with 0
Example: In this example, I show you which results you can expect when using this functions.
```Call Result Call Result
round(7.2) 7 round(7.8) 8
trunc(7.2) 7 trunc(7.8) 7
ceil(7.2) 8 ceil(7.8) 8
floor(7.2) 7 floor(7.8) 7
int(7.2) 7 int(7.8) 7
frac(7.2) 0.2 frac(7.8) 0.8```
Hint: For the functions Ceil() and Floor(), the unit "Math" has to be added to the uses-clause. Round(), Trunc(), Int() and Frac() are parts of the unit "System" which is usually already included in the uses section.
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### Online College Courses for Credit
+
2 Tutorials that teach Outcomes and Competencies
Take your pick:
# Outcomes and Competencies
##### Rating:
(0)
Author: Jody Waltman
##### Description:
In this lesson, students define outcomes and competencies and evaluate the importance of outcomes and competencies.
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Tutorial
## Video Transcription
Download PDF
In this tutorial, we'll explore the concepts of outcomes, objectives, and competencies. Let's begin by comparing the terms outcomes and objectives. An outcome typically describes an end-of-unit or end-of-course goal. An outcome describes what students should be able to do by the end of a longer unit of study. In contrast, an objective is a marker of progress towards that end-of-unit or end-of-course goal.
For example, in an algebra class if my end-of-unit outcome specifies that students will be able to solve a system of linear equations, some of the objectives that are related to this outcome might include having students solve a system of equations using the substitution method, having students graph two linear equations to find their intersection point, and having students verify a possible solution to a system of linear equations. All of these objectives define what the students should know or be able to do within the overall unit of instruction. Likewise, the outcomes themselves link back to the learning objectives.
Though there are differences between outcomes and objectives, there are also many similarities. Outcomes and objectives both specify content and skills that students should be able to demonstrate. For example, crafting a persuasive argument, describing a chemical reaction, outlining the events that led up to a battle in American history, or finding the area of a circle.
Outcomes and objectives need to be aligned to the goals of your curriculum or your school program. Since there are so many standards that we need to meet in all curricular areas, having specific outcomes allows curriculum teams to define what the most important areas of coverage are for all students. Clear outcomes and objectives also support the process of making generalizations to other curricular areas and transferring knowledge and skills to those areas. It's important that both outcomes and objectives are written in terms of content standards and in terms of the skills that students need to master.
With an understanding of the terms outcome and objective, let's take a look at competencies. Competencies are the specific skills that are connected to the knowledge that students need to master. In competency-based education, the proficiency levels are pre-defined. Once a student reaches the pre-defined level of proficiency, that student can move on to the next competency. In fact, competency-based education is sometimes referred to as outcome-based education because it's connected to outcomes and objectives. Remember, in competency-based education, decisions made by teachers are largely impacted by demonstration of student proficiency.
In this tutorial, we looked at the similarities and differences between outcomes and objectives, and we also took a closer look at the concept of competencies. Here's a chance for you to stop and reflect. Do you understand the sometimes subtle difference between objectives and outcomes? To dive a little deeper and learn how to apply this information, be sure to check out the Additional Resources section associated with this video. This is where you'll find links targeted toward helping you discover more ways to apply this course material. Thanks for joining me today. Have a great day.
## Notes on "Outcomes and Competencies"
(00:00 - 00:07) Introduction
(00:08 - 02:05) Outcomes and Objectives
(02:06 - 02:45) Competencies
(02:46 - 02:55) Review
(02:56 - 03:22) Stop and Reflect
## Additional Resources
Tips on Writing Learning Outcomes
This University of Illinois site provides step by step directions to writing strong learning outcomes. The directions are connected to Bloom's Taxonomy.
http://www.library.illinois.edu/infolit/learningoutcomes.html
Writing Measurable Learning Outcomes
This handout provides comprehensive directions on writing measurable learning outcomes. The handout includes Bloom's Taxonomy and Action Verbs. In addition, there are clear strategies to align outcomes to performance tasks.
http://www.gavilan.edu/research/spd/Writing-Measurable-Learning-Outcomes.pdf
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#### A bat moving at 10 ms -1 towards a wall sends a sound signal of 8000 Hz towards it. On reflection it hears a sound of frequency . The value of in Hz is close to(speed of sound=320 ms-1) Option 1) 8258 Option 2) 8516 Option 3) 8000 Option 4) 8424
As we learnt in
Frequency of sound when source and observer are moving towards each other -
- wherein
Speed of sound
Speed of observer
speed of source
Original frequency
apparent frequency
Reflected frequency of sound reaching bat.
Correct option is 2.
Option 1)
8258
This is an incorrect option.
Option 2)
8516
This is the correct option.
Option 3)
8000
This is an incorrect option.
Option 4)
8424
This is an incorrect option.
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# Light's speed and relativity
• I
If you were to travel alongside a train, as fast the train, to you the train would seem stationary. I read that if you were to travel along a photon of light, as fast as the speed of light, that photon would not seem stationary. Is this true? If so, why?
Ibix
2020 Award
You are misunderstanding something. The key point is that you cannot travel at the speed of light - so the question of "what would light look like if you travelled alongside it" can't be answered.
Possibly what you misunderstood was that Einstein's second postulate is that the speed of light is always ##c## in all inertial frames. A frame travelling at the speed of light would therefore lead to a contradiction - that light is stationary (because the frame is moving at the same speed) and also moving at ##c## (because that's part of the definition of an inertial frame in relativity). Since there's a contradiction, you can't have an inertial frame moving at the speed of light. Edit: just to be clear, this does not mean that light isn't stationary if you travel alongside it. The (self-contradictory) conclusion that light wouldn't be stationary even when it must be shows that being stationary with respect to light isn't a coherent concept in relativity.
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Joe and Dale
You are misunderstanding something. The key point is that you cannot travel at the speed of light - so the question of "what would light look like if you travelled alongside it" can't be answered.
Possibly what you misunderstood was that Einstein's second postulate is that the speed of light is always ##c## in all inertial frames. A frame travelling at the speed of light would therefore lead to a contradiction - that light is stationary (because the frame is moving at the same speed) and also moving at ##c## (because that's part of the definition of an inertial frame in relativity). Since there's a contradiction, you can't have an inertial frame moving at the speed of light. Edit: just to be clear, this does not mean that light isn't stationary if you travel alongside it. The (self-contradictory) conclusion that light wouldn't be stationary even when it must be shows that being stationary with respect to light isn't a coherent concept in relativity.
Thank you. From what I understood - from your edit - light can seem to be stationary? Two photons travelling alongside each other will seem not moving from each other's perspective?
jbriggs444
Homework Helper
Thank you. From what I understood - from your edit - light can seem to be stationary? Two photons travelling alongside each other will seem not moving from each other's perspective?
There is no such thing as the perspective of a photon.
Joe
phinds
Gold Member
Thank you. From what I understood - from your edit - light can seem to be stationary? Two photons travelling alongside each other will seem not moving from each other's perspective?
You still misunderstand. There IS NO "point of view" (or perspective) of a photon.
EDIT: I see jbriggs beat me to it
Joe
Ibix
2020 Award
Thank you. From what I understood - from your edit - light can seem to be stationary? Two photons travelling alongside each other will seem not moving from each other's perspective?
As jbriggs444 and phinds have noted, it's simply not possible to describe the perspective of a thing travelling at the speed of light. Attempting to do so leads to the contradiction I mentioned.
Unfortunately, something you find as you move away from every day experience is that questions that seem perfectly sensible turn out to be nonsense. You don't even have to go that far outside the every day. Could you tell me which way is north where you are now? Could you tell me which way is north if you were at the north pole? Asking what anyone or anything would see at the speed of light is like that second question. You can't answer because the question has hidden assumptions that are not valid in the case it's talking about.
Note that you do find pop-sci sources (notably Brian Greene) that say things like "time stops at the speed of light". They're the result of forcing an answer at (metaphorical, I hope) gunpoint and don't make coherent sense, but usually satisfy non-physicists enough that they shut up and go away. It's worth noting that I gather that Greene himself does not make this claim in professional publications, only his pop-sci stuff.
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Joe and SiennaTheGr8
As jbriggs444 and phinds have noted, it's simply not possible to describe the perspective of a thing travelling at the speed of light. Attempting to do so leads to the contradiction I mentioned.
Unfortunately, something you find as you move away from every day experience is that questions that seem perfectly sensible turn out to be nonsense. You don't even have to go that far outside the every day. Could you tell me which way is north where you are now? Could you tell me which way is north if you were at the north pole? Asking what anyone or anything would see at the speed of light is like that second question. You can't answer because the question has hidden assumptions that are not valid in the case it's talking about.
Note that you do find pop-sci sources (notably Brian Greene) that say things like "time stops at the speed of light". They're the result of forcing an answer at (metaphorical, I hope) gunpoint and don't make coherent sense, but usually satisfy non-physicists enough that they shut up and go away. It's worth noting that I gather that Greene himself does not make this claim in professional publications, only his pop-sci stuff.
Thanks again. Nicely explained!
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# Rational Exponents Worksheet Glencoe
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# poetry
After completing the required readings for this week, discuss how you see the theme of creating art developing in the different stories and poems.
1. 👍
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3. 👁
1. We don't know what stories and poems you are referring to. We don't have informationn for any schools.
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A161478 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+113)^2 = y^2. 4
0, 52, 175, 339, 615, 1312, 2260, 3864, 7923, 13447, 22795, 46452, 78648, 133132, 271015, 458667, 776223, 1579864, 2673580, 4524432, 9208395, 15583039, 26370595, 53670732, 90824880, 153699364, 312816223, 529366467, 895825815, 1823226832, 3085374148 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Corresponding values y of solutions (x, y) are in A161479. lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2). lim_{n -> infinity} a(n)/a(n-1) = (129+44*sqrt(2))/113 for n mod 3 = {1, 2}. lim_{n -> infinity} a(n)/a(n-1) = (16131+6970*sqrt(2))/113^2 for n mod 3 = 0. LINKS Table of n, a(n) for n=1..31. Index entries for linear recurrences with constant coefficients, signature (1, 0, 6, -6, 0, -1, 1). FORMULA a(n) = 6*a(n-3)-a(n-6)+226 for n > 6; a(1)=0, a(2)=52, a(3)=175, a(4)=339, a(5)=615, a(6)=1312. G.f.: x*(52+123*x+164*x^2-36*x^3-41*x^4-36*x^5) / ((1-x)*(1-6*x^3+x^6)). a(3*k+1) = 113*A001652(k) for k >= 0. MATHEMATICA LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {0, 52, 175, 339, 615, 1312, 2260}, 72] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *) PROG (PARI) {forstep(n=0, 100000000, [3, 1], if(issquare(2*n^2+226*n+12769), print1(n, ", ")))} CROSSREFS Cf. A161479, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A161480 (decimal expansion of (129+44*sqrt(2))/113), A161481 (decimal expansion of (16131+6970*sqrt(2))/113^2). Sequence in context: A345240 A292172 A166390 * A288919 A260549 A211564 Adjacent sequences: A161475 A161476 A161477 * A161479 A161480 A161481 KEYWORD nonn AUTHOR Klaus Brockhaus, Jun 13 2009 STATUS approved
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If a certain culture of bacteria increases by a constant
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29 Aug 2013, 13:31
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If a certain culture of bacteria increases by a constant factor of x every y minutes, how long will it take for the culture to increase to ten-thousand times its original size?
1. $$x=10^y$$
2.In two minutes, the culture will increase to one hundred times its original size.
[Reveal] Spoiler: OA
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Re: If a certain culture of bacteria increases by a constant [#permalink]
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29 Aug 2013, 23:44
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My take:
We need to find y, when xy=10^4.
1) When x=10^y, y*(10^y)=10^4
or could also be written as y = 10^(4-y)
By using substitution, we could say y could be between 3 and 4 (~=3.4). Hence st1 is Sufficient
( And Options B,C,&E could be eliminated)
2) (2)*x=100 i.e. x=50. By substituting value of x in xy=10^4, we could find the time taken. Hence st2 is Sufficient
Hence the solution is
[Reveal] Spoiler:
D
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01 Sep 2013, 22:05
Hi Bunuel,
It'd be really helpful if you could provide an explanation for this problem to help me understand it.
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Re: If a certain culture of bacteria increases by a constant [#permalink]
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08 Sep 2013, 05:40
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bagdbmba wrote:
Hi Bunuel,
It'd be really helpful if you could provide an explanation for this problem to help me understand it.
Maybe this will help
from 1 we are given x=10^y, let initial quantity be m
So after y minutes the quantity => m*x =>m*10^y , now we need the new value to be 10,000 times the original => m*10^y=m*1000 => y=4 minutes
hence statement 1 is sufficient.
from statement 2
In 2 minutes the culture will increase to 100 times this can be interpreted as follows
1) every 1 minute the quantity is multiplied by 10
or
2) every 2 minutes the quantity is multiplied by 100
Either ways after 4 minutes the quantity will be multiplied by 10,000 hence y=4 minutes
Hence statement 2 is sufficient.
Hope the solution is clear.
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08 Sep 2014, 13:40
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If a certain culture of bacteria increases by a factor of x every y minutes, how long will it take for the culture to increase to 10000 times its original amount ?
1) (x)^1/y = 10
2) In 2 minutes the culture will increase to 100 times its original amount.
Please could you explain an how to solve such questions which involves population increasing by certain factors
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Re: If a certain culture of bacteria [#permalink]
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08 Sep 2014, 20:45
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ankitaprsd wrote:
If a certain culture of bacteria increases by a factor of x every y minutes, how long will it take for the culture to increase to 10000 times its original amount ?
1) (x)^1/y = 10
2) In 2 minutes the culture will increase to 100 times its original amount.
Please could you explain an how to solve such questions which involves population increasing by certain factors
Stem 1 is interesting..
Consider x=10 and y=1...so Initial bacteria (K) will increase by a factor of 10 in 1 minute...so after 1 minute there will be K+10K=11K...so we can find the time it will take for culture to increase to 10000 times
So after 1 minute, we have 11K
After 2 minutes, 11K+110K=121K
After 3 minutes= 121K+1210K=1331K
After 4 minutes...
Consider $$x=\sqrt{10}$$, and y =1/2 or 30 seconds, So after 30 seconds, no. of bacteria will increase by $$(10^{1/2})^{2}$$ or by a factor of 10...in 1 minute it will increase by factor $$\sqrt{10}$$
At 0 seconds=K
After 30 seconds= K+10K=11K
After 1 minute, by a factor$$\sqrt{10}$$ so after 1 minute you will have $$11K+\sqrt{10}*11K$$-----> Now here the problem is the increase every y minute is not the same so this case will not be considered...
So St1 is sufficient
St2 is straight forward...
Can you post the Official Explanation to question.
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Re: If a certain culture of bacteria increases by a constant [#permalink]
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09 Sep 2014, 06:49
ankitaprsd wrote:
If a certain culture of bacteria increases by a factor of x every y minutes, how long will it take for the culture to increase to 10000 times its original amount ?
1) (x)^1/y = 10
2) In 2 minutes the culture will increase to 100 times its original amount.
Please could you explain an how to solve such questions which involves population increasing by certain factors
Merging topics. Please search before posting. Thank you.
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Re: If a certain culture of bacteria increases by a constant [#permalink]
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09 Sep 2014, 06:50
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Expert's post
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Bunuel wrote:
ankitaprsd wrote:
If a certain culture of bacteria increases by a factor of x every y minutes, how long will it take for the culture to increase to 10000 times its original amount ?
1) (x)^1/y = 10
2) In 2 minutes the culture will increase to 100 times its original amount.
Please could you explain an how to solve such questions which involves population increasing by certain factors
Merging topics. Please search before posting. Thank you.
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Re: If a certain culture of bacteria increases by a constant [#permalink] 19 May 2017, 05:27
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# MOTN-098 Circle angle too large
• I'm programming an RJ-3, trying to do a 1/2 circle move. I know that I'm following the steps properly to teach the move, but when I try to check the move I keep getting the error MOTN-098. I have tried a different angle for the start position and I'm still being told that the angle is too large. The area of the move is only about 6"X10" and I have done this on 3 different robots on the same fixture with no problem. I'm going to try another start angle/position and see what happens. Any suggestions?
• Im not sure what your background is, but if you know your CNC, then you are probably used to circular points being a start, a finish, and then a radius. Fanuc tries to make it easier on the programmer by calculating the radius using the values in the second point of your move.
L P[1:start] 500mm/sec FINE
P[3:end] 500mm/sec FINE
if point 3 is offset too much from the other points it cannot make a circle from 1 to 3 and still go though the second point. i suggest you hard code the points then touch them up after you get it working.
P[1:start]
X=0 P=0
Y=0 W=0
Z=0 R=0
X=50 P=0
Y=50 W=0
Z=0 R=0
P[3:end]
X=100 P=0
Y=0 W=0
Z=0 R=0
if you cant get it to work like that, then you might have some bigger issue. you can try a full circle.
L P[1:start] 500mm/sec FINE
P[3:end] 500mm/sec FINE
P[1:start] 500mm/sec FINE
X=50 P=0
Y=-50 W=0
Z=0 R=0
• I am getting MOTN-098 error for "Circle angle too large" when programing Fanuc i30. I simply created a letter "S" with circle arc movement and lift up the tool about 2 inches about the plane when finished writing the "S", however, I am getting this error at the end of the "S" but it won't execute the last movement. Any help would be appreciated.
You must have fine moves at the beginning and end of each arc. So the top, middle, and bottom of the S should be fine and the mid points CNT100.
• Problem solved. I had a wrong end point. Thank you!
You must have fine moves at the beginning and end of each arc. So the top, middle, and bottom of the S should be fine and the mid points CNT100.
## Images
• I used this same code from YouTube and I’m stilling getting this code
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# E. Excitation of Atoms SOLUTION 2020 ICPC, COMPFEST 12, Indonesia Multi-Provincial Contest
## Excitation of Atoms SOLUTION
Mr. Chanek is currently participating in a science fair that is popular in town. He finds an exciting puzzle in the fair and wants to solve it.
There are N atoms numbered from 1 to N. These atoms are especially quirky. Initially, each atom is in normal state. Each atom can be in an excited. Exciting atom i requires Di energy. When atom i is excited, it will give Ai energy. You can excite any number of atoms (including zero).
These atoms also form a peculiar one-way bond. For each i, (1≤i<N), if atom i is excited, atom Ei will also be excited at no cost. Initially, Ei = i+1. Note that atom N cannot form a bond to any atom.
Mr. Chanek must change exactly K bonds. Exactly K times, Mr. Chanek chooses an atom i, (1≤i<N) and changes Ei to a different value other than i and the current Ei. Note that an atom’s bond can remain unchanged or changed more than once. Help Mr. Chanek determine the maximum energy that he can achieve!
note: You must first change exactly K bonds before you can start exciting atoms.
Input
The first line contains two integers N K (4≤N≤105,0≤K<N), the number of atoms, and the number of bonds that must be changed.
The second line contains N integers Ai (1≤Ai≤106), which denotes the energy given by atom i when on excited state.
The third line contains N integers Di (1≤Di≤106), which denotes the energy needed to excite atom i.
Output
A line with an integer that denotes the maximum number of energy that Mr. Chanek can get.
Example
inputCopy
6 1
5 6 7 8 10 2
3 5 6 7 1 10
outputCopy
35
Note
An optimal solution to change E5 to 1 and then excite atom 5 with energy 1. It will cause atoms 1, 2, 3, 4, 5 be excited. The total energy gained by Mr. Chanek is (5 + 6 + 7 + 8 + 10) – 1 = 35.
Another possible way is to change E3 to 1 and then exciting atom 3 (which will excite atom 1, 2, 3) and exciting atom 4 (which will excite atom 4, 5, 6). The total energy gained by Mr. Chanek is (5 + 6 + 7 + 8 + 10 + 2) – (6 + 7) = 25 which is not optimal.
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# Dandelin spheres
Missing image
DandelinSpheres.gif
Dandelin Spheres—graphics by Hop David
In geometry, a nondegenerate conic section formed by a plane intersecting a cone has one or two Dandelin spheres characterized thus:
Each Dandelin sphere touches, but does not cross, both the plane and the cone.
This concept is named in honor of Germinal Pierre Dandelin.
Each conic section has one Dandelin sphere for each focus.
• An ellipse has two Dandelin spheres, both touching the same nappe of the cone.
• A hyperbola has two Dandelin spheres, touching opposite nappes of the cone.
• A parabola has just one Dandelin sphere.
## Dandelin's theorem
The reason for interest in Dandelin spheres is this theorem:
The point at which the sphere touches the plane is a focus of the conic section.
Proof: Consider the illustration, depicting a plane intersecting a cone in an ellipse. The two Dandelin spheres are shown. Each sphere touches the cone in a circle. Each sphere touches the plane in a point. Call those two points F1 and F2. Let P be a typical point on the ellipse. The sum of distances d(F1, P) + d(F2, P) must be shown to remain constant as the point P moves along the curve. A line passing through P and the vertex of the cone intersects the two circles in points G1 and G2. As P moves along the ellipse, G1 and G2 move along the two circles. The distance from Fi to P is the same as the distance from Gi to P, because both are tangent to the same sphere. Consequently the sum of distances d(F1, P) + d(F2, P) is what we need to show remains constant. But, since P is on a straight line between G1 to G2, this follows from the fact that the distance from G1 to G2 remains constant.
Adaptations of this argument work for hyperbolas and parabolas as intersections of a plane with a cone. Another adaptation works for an ellipse realized as the intersection of a plane with a right circular cylinder.
### Consequences of this theorem and its proof
If (as is often done) one takes the definition of the ellipse to be the locus of points P such that d(F1, P) + d(F2, P) = a constant, then the argument above proves that the intersection of a plane with a cone is indeed an ellipse. That the intersection of the plane with the cone is symmetric about the perpendicular bisector of the line through F1 and F2 may be counterintuitive, but this argument makes it clear.
Dandelin Spheres page by Hop David (http://www.clowder.net/hop/Dandelin/Dandelin.html)
Dandelin Spheres -- Mathworld (http://mathworld.wolfram.com/DandelinSpheres.html)
Java applet JDandelin (http://www.lostlecture.host.sk/JDandelinEn.htm), on a web site devoted to Richard Feynman's "lost lecture"de:Dandelinsche Kugel
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# Homework cheat sites
Looking for Homework cheat sites? Look no further! We can solve math word problems.
## The Best Homework cheat sites
Apps can be a great way to help learners with their math. Let's try the best Homework cheat sites. Next, take your time and read the instructions carefully. If you are still having trouble understanding the material, try looking up key terms in a dictionary or doing additional research. Finally, don't be afraid to ask for help from a teacher or tutor. By following these tips, you can increase your chances of getting the answers you need.
There are a number of different interval notation solvers available online, and choosing the right one will depend on the individual’s needs. Some factors to consider include the level of complexity that is required and the ease of use. With so many options available, there is sure to be an interval notation solver that is perfect for any math student.
solving equations is a process that involves isolating the variable on one side of the equation. This can be done using inverse operations, which are operations that undo each other. For example, addition and subtraction are inverse operations, as are multiplication and division. When solving an equation, you will use these inverse operations to move everything except for the variable to one side of the equal sign. Once the variable is isolated, you can then solve for its value by performing the inverse operation on both sides of the equation. For example, if you are solving for x in the equation 3x + 5 = 28, you would first subtract 5 from both sides of the equation to isolate x: 3x + 5 - 5 = 28 - 5. This results in 3x = 23. Then, you would divide both sides of the equation by 3 to solve for x: 3x/3 = 23/3. This gives you x = 23/3, or x = 7 1/3. Solving equations is a matter of isolating the variable using inverse operations and then using those same operations to solve for its value. By following these steps, you can solve any multi-step equation.
Algebra is the branch of mathematics that deals with the rules of operations and relations, and the study of quantities which may be either constant or variable. Factoring is a technique used to simplify algebraic expressions. When an expression is factored, it is rewritten as a product of simpler factors. This can be helpful in solving equations and graphing functions. In general, factoring is the process of multiplying two or more numbers to get a product. For example, 6 can be factored as 2 times 3, since 2 times 3 equals 6. In algebra, factoring is often used to simplify equations or to find solutions. For example, the equation x^2+5x+6 can be simplified by factoring it as (x+3)(x+2). This can be helpful in solving the equation, since now it can be seen that the solution is x=-3 or x=-2. Factoring can also be used to find zeroes of polynomials, which are important in graphing functions. In general, polynomials can be factored into linear factors, which correspond to zeroes of the function. For example, the function f(x)=x^2-4x+4 has zeroes at x=2 and x=4. These zeroes can be found by factoring the polynomial as (x-2)(x-4). As a result,factoring is a powerful tool that can be used to simplify expressions and solve equations.
In theoretical mathematics, in particular in field theory and ring theory, the term is also used for objects which generalize the usual concept of rational functions to certain other algebraic structures such as fields not necessarily containing the field of rational numbers, or rings not necessarily containing the ring of integers. Such generalizations occur naturally when one studies quotient objects such as quotient fields and quotient rings. The technique of partial fraction decomposition is also used to defeat certain integrals which could not be solved with elementary methods. The method consists of two main steps: first determine the coefficients by solving linear equations, and next integrate each term separately. Each summand on the right side of the equation will always be easier to integrate than the original integrand on the left side; this follows from the fact that polynomials are easier to integrate than rational functions. After all summands have been integrated, the entire integral can easily be calculated by adding all these together. Thus, in principle, it should always be possible to solve an integral by means of this technique; however, in practice it may still be quite difficult to carry out all these steps explicitly. Nevertheless, this method remains one of the most powerful tools available for solving integrals that cannot be solved using elementary methods.
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# Generate random numbers
This code generates random numbers within a given range by using the `rand()` function from the `cstdlib` library.
## Code Example
``````#include <iostream>
using namespace std;
int main() {
// initialize random seed
srand (time(NULL));
cout << "Random value between 0 - 9:\t";
int randomInteger10 = rand() % 10 + 1;
cout << randomInteger10 << endl;
cout << "Random value between 0 - 99:\t";
int randomInteger100 = rand() % 100;
cout << randomInteger100 << endl;
cout << "Random value between 1 - 100:\t";
int randomInteger = rand() % 100 + 1;
cout << randomInteger << endl;
return 0;
}``````
### Output first run
``````Random value between 0 - 9: 8
Random value between 0 - 99: 49
Random value between 1 - 100: 74``````
### Output second run
``````Random value between 0 - 9: 1
Random value between 0 - 99: 66
Random value between 1 - 100: 92``````
### Code Explanation
It starts by initializing the random seed using the `srand()` function and passing the current time in seconds obtained from `time(NULL)` as its argument. This is to ensure that each run of the program produces different random numbers.
The `rand()` function returns a random integer value and the `%` operator is used to restrict the range of the generated number. The formula `rand() % n` returns a random integer value between 0 and n-1. To generate a number between `a` and `b`, you can use the formula `a + (rand() % (b - a + 1))`.
In this code, three random integer values are generated, with ranges between 0 and 9, 0 and 99, and 1 and 100 respectively. The generated numbers are displayed using `cout`.
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Demystifying static pool analysis
As I work with and educate credit unions across the country on loan portfolio analytics, Static Pool Analysis seems to be one of the most confusing and difficult to understand topics. There are a number of ways that static pool analysis can be used to manage risk in the loan portfolio, and prevent and predict future losses.
A static pool of loans is a grouping of loans with similar credit risk characteristics that were originated during a specific time period. At a high level, this may be all indirect loans originated in the year 2010. At a more granular level, this may be all loans originated in the “D” credit tier during the first quarter of 2010. What’s important to static pool analysis is that it includes loans originated during a period of time where internal and external risk factors were basically the same. Once the pool has been established, no loans are removed from the pool and no loans are added to the pool; the pool remains static. The reason that the static nature of the pool is important is that if new loans are added, then the new loans will dilute the performance of the loan pool. Let’s say, for example, in April 2014 you are analyzing the delinquency of loans originated in February of 2014. If you have \$1 million in loans originated in February and \$10,000 of these loans are now more than 30 days past due, your 30-day delinquency would be 1%. If you included March originations of another \$1 million, your 30-day delinquency would fall to .5%, even though the loans originated in March have not even matured 30 days. This is a simplistic example, but it serves to demonstrate how adding loans to a static pool can dilute the results of the analysis and provide inaccurate data.
To some degree, this demonstrates how current methods used by credit unions to evaluate loan risk are insufficient, and why regulators insist on static pool analysis. A common metric used by credit unions today is the annualized loss ratio. The formula for this metric is the total charge-offs for the last 12 months divided by the average portfolio balance for the last 12 months. If the credit union has a growing loan portfolio (meaning the balance of the portfolio today is higher than one year ago), this calculation could yield a loss metric that hides the poor performance of older loans in the portfolio, as the higher number of new loans has not matured to the point of loss. Static pool metrics provide a more accurate picture of a loan pool’s performance.
Therefore, it is critical that a loan pool is fully formed prior to conducting an analysis. It would be inaccurate to include loans originated in 2014 in a static pool analysis of delinquency of loans by origination year as the 2014 pool has not fully formed. It would not be complete until the end of December of 2014. It would be appropriate to include January and February in an analysis of loans pooled by month of origination as those months have been fully formed.
Another important point is that when calculating loss on a static pool, the calculation should be a comparison of the loss amount and the origination amount of the loan, not the current balance of the portfolio. This will enable your credit union to use this data to predict portfolio performance. If your credit union originated \$20 million in loans in 2011, and since origination your credit union charged-off \$100,000 on loans originated in 2012, then the loss ratio would be \$100,000 divided by \$20 million to get to a cumulative loss ratio of .5% on loans originated in 2011. Assuming nothing has changed, using this information one could predict that you would also lose .5% on loans originated in 2013 after they had matured for two years. If you originated \$40 million in 2013, then you could estimate that you will have lost \$200,000 from that pool of loans by the beginning of 2016. Your estimates become more precise as the precision of your pools increases. If I did this analysis by credit tier, for example, I could be more precise with my estimates in the case that the percentage of my portfolio in “A” tier was different in 2013 than it was in 2011. Using static pool analysis can increase the accuracy of predicting loss, repayment speeds, and rates of return to name a few examples.
Michael Cochrum
Michael has worked in the consumer lending industry since 1989. In 1999, he joined the credit union industry, working for the Texas Credit Union League’s credit union. Mr. Cochrum ... Web: www.cudirect.com Details
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Convert 分 to 市尺 | Hong Kong fan to Chinese chǐ
# length conversion
## Amount: 1 Hong Kong fan (分) of length Equals: 0.011 Chinese chǐ (市尺) in length
Converting Hong Kong fan to Chinese chǐ value in the length units scale.
TOGGLE : from Chinese chǐ into Hong Kong fan in the other way around.
## length from Hong Kong fan to Chinese chǐ Conversion Results:
### Enter a New Hong Kong fan Amount of length to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
Conversion calculator for webmasters.
## Length, Distance, Height & Depth units
Distance in the metric sense from any two A to Z points (interchangeable with Z and A), also applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
Convert length measuring units between Hong Kong fan (分) and Chinese chǐ (市尺) but in the other reverse direction from Chinese chǐ into Hong Kong fan.
conversion result for length: From Symbol Equals Result To Symbol 1 Hong Kong fan 分 = 0.011 Chinese chǐ 市尺
# Converter type: length units
This online length from 分 into 市尺 converter is a handy tool not just for certified or experienced professionals.
First unit: Hong Kong fan (分) is used for measuring length.
Second: Chinese chǐ (市尺) is unit of length.
## 0.011 市尺 is converted to 1 of what?
The Chinese chǐ unit number 0.011 市尺 converts to 1 分, one Hong Kong fan. It is the EQUAL length value of 1 Hong Kong fan but in the Chinese chǐ length unit alternative.
How to convert 2 Hong Kong fan (分) into Chinese chǐ (市尺)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.01114425 * 2 (or divide it by / 0.5)
QUESTION:
1 分 = ? 市尺
1 分 = 0.011 市尺
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Hong Kong fan and Chinese chǐ ( 分 vs. 市尺 ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with length's values and properties.
International unit symbols for these two length measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Hong Kong fan is:
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Chinese chǐ is:
### One Hong Kong fan of length converted to Chinese chǐ equals to 0.011 市尺
How many Chinese chǐ of length are in 1 Hong Kong fan? The answer is: The change of 1 分 ( Hong Kong fan ) unit of length measure equals = to 0.011 市尺 ( Chinese chǐ ) as the equivalent measure for the same length type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in 分 - Hong Kong fan for length amount, the rule is that the Hong Kong fan number gets converted into 市尺 - Chinese chǐ or any other length unit absolutely exactly.
Conversion for how many Chinese chǐ ( 市尺 ) of length are contained in a Hong Kong fan ( 1 分 ). Or, how much in Chinese chǐ of length is in 1 Hong Kong fan? To link to this length Hong Kong fan to Chinese chǐ online converter simply cut and paste the following.
The link to this tool will appear as: length from Hong Kong fan (分) to Chinese chǐ (市尺) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
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Web Results
www.reference.com/math/1-8-bigger-1-12-6849e73b01f5d25b
One-eighth is bigger than one-twelfth. Both of these figures are fractions and represent parts of a whole. When comparing fractions, if the numerator remains the same, any fraction with a larger denominator is smaller.
thefractioncalculator.com/.../is-1/8-greater-than-1/5.html
Is 1/8 greater than 1/5? Is 1/8 bigger than 1/5? Is 1/8 larger than 1/5? These are all the same questions with one answer. To get the answer, we first convert each fraction into decimal numbers. We do this by dividing the numerator by the denominator for each fraction as illustrated below:
Obviously the pieces that are 1/10 of the pizza are bigger. Another way to compare these fractions is to get common denominator of 60. 1/10 = 6/60 1/12 = 5/60 6/60 > 5/60 So 1/10 > 1/12 Answer: Yes, 1/10 is greater than 1/12.
The only difference is that 1.8 has a bigger hitbox than 1.12 does. (In 1.8, your hitbox is bigger than F3+B, but in 1.12 your hitbox is just like F3+B.) So you're not experiencing a hit registration issue, you're just experiencing not being used to the hitbox. (Not being close enough.) #8.
Hopefully you will learn fractions in school soon. The larger the denominator (bottom number) the smaller the model is. As with 1/8 of a piece of pie you get more than getting 1/10, 1/12, 1/16, 1/18 of it etc.
If they are taken of the same whole, yes. If you get 1/8 of a much larger pie...it may be greater than 1/6 of a smaller pie. Think pizza. That may help to put it in perspective.
brainly.com/question/197491
dmcritchie.mvps.org/excel/fractex1.htm
The other formulas showing mixed feet, inches and fractional inches will still requidre something more than these simplified formulas, and will be text to allow feet and inch marks to appear next to figures even if strictly Feet and fractional feet.
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# pcregisterloam
Register two point clouds using LOAM algorithm
Since R2022a
## Syntax
``tform = pcregisterloam(movingPtCloud,fixedPtCloud,gridStep)``
``tform = pcregisterloam(movingPoints,fixedPoints)``
``[tform,rmse] = pcregisterloam(___)``
``[___] = pcregisterloam(___,Name=Value)``
## Description
example
````tform = pcregisterloam(movingPtCloud,fixedPtCloud,gridStep)` registers the moving point cloud `movingPoints` with the fixed point cloud `fixedPoints` using the lidar odometry and mapping (LOAM) algorithm. The function returns the rigid transformation `tform`, between the moving and fixed point clouds. `gridStep` specifies the size of a 3-D box used to downsample the LOAM points detected in the input point clouds.```
````tform = pcregisterloam(movingPoints,fixedPoints)` registers the moving LOAM points `movingPoints` with the fixed LOAM points `fixedPoints` and returns the rigid transformation between them `tform`. Using this function to register LOAM points is faster and more accurate than using it to register point clouds.You can obtain the LOAM points of the moving and fixed point clouds by using the `detectLOAMFeatures` function, which detects LOAM feature points from organized point clouds. ```
````[tform,rmse] = pcregisterloam(___)` returns the root mean squared error `rmse` of the Euclidean distance between the aligned points.```
````[___] = pcregisterloam(___,Name=Value)` specifies options using one or more name-value arguments in addition to any combination of arguments from previous syntaxes. For example, `MatchingMethod`=`'one-to-one'` sets the matching method algorithm to `'one-to-one'`.```
## Examples
collapse all
Load and visualize point cloud data.
```ld = load('livingRoom.mat'); movingPtCloud = ld.livingRoomData{1}; fixedPtCloud = ld.livingRoomData{2}; figure pcshowpair(movingPtCloud,fixedPtCloud,'VerticalAxis','Y','VerticalAxisDir','Down')```
Register the point clouds.
```gridStep = 0.5; tform = pcregisterloam(movingPtCloud,fixedPtCloud,gridStep);```
Align and visualize the point clouds.
```alignedPtCloud = pctransform(movingPtCloud, tform); figure pcshowpair(alignedPtCloud,fixedPtCloud,'VerticalAxis','Y','VerticalAxisDir','Down')```
Read point cloud data from a Velodyne PCAP file into the workspace.
`veloReader = velodyneFileReader("lidarData_ConstructionRoad.pcap","HDL32E");`
Read the first point cloud from the data. Use this point cloud as the fixed point cloud.
`fixedPtCloud = readFrame(veloReader,1);`
Detect LOAM feature points in the fixed point cloud.
`fixedPoints = detectLOAMFeatures(fixedPtCloud);`
Downsample the less planar surface points from the fixed point cloud, to improve registration speed.
```gridStep = 1; fixedPoints = downsampleLessPlanar(fixedPoints,gridStep);```
Read and detect LOAM feature points from the fifth point cloud in the data. Use this point cloud as the moving point cloud.
```movingPtCloud = readFrame(veloReader,5); movingPoints = detectLOAMFeatures(movingPtCloud);```
Downsample the less planar surface points from the moving point cloud.
`movingPoints = downsampleLessPlanar(movingPoints,gridStep);`
Register the moving point cloud to the fixed point cloud.
`tform = pcregisterloam(movingPoints,fixedPoints);`
Transform the moving point cloud to align it to the fixed point cloud.
`alignedPtCloud = pctransform(movingPtCloud,tform);`
Visualize the aligned point clouds. Points from the fixed point cloud display as green, while points from the moving point cloud display as magenta.
```figure pcshowpair(alignedPtCloud,fixedPtCloud)```
## Input Arguments
collapse all
Organized moving point cloud, specified as a `pointCloud` object. The point cloud object must contain an organized point cloud with a `Location` property of size M-by-N-by-3 matrix, where M is the number of laser scans and N is the number of points per scan. The first page represents the x-coordinates, the second page represents the y-coordinates, and the third page represents the z- coordinates for each point.
Organized fixed point cloud, specified as a `pointCloud` object. The point cloud object must contain an organized point cloud with a `Location` property of size M-by-N-by-3 matrix, where M is the number of laser scans and N is the number of points per scan. The first page represents the x-coordinates, the second page represents the y-coordinates, and the third page represents the z- coordinates for each point.
Moving LOAM points, specified as a `LOAMPoints` object. You can obtain the LOAM points from the moving point cloud by using the `detectLOAMFeatures` function, which detects LOAM feature points from organized point clouds.
Fixed LOAM points, specified as a `LOAMPoints` object. You can obtain the LOAM points from the fixed point cloud by using the `detectLOAMFeatures` function, which detects LOAM feature points from organized point clouds.
Size of the 3-D box for downsampling the detected LOAM points in the input point cloud, specified as a positive scalar.
### Name-Value Arguments
Specify optional pairs of arguments as `Name1=Value1,...,NameN=ValueN`, where `Name` is the argument name and `Value` is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Example: `pcregisterloam(movingPoints,fixedPoints,MatchingMethod='one-to-one')` sets the registration matching method to `'one-to-one'`.
Initial rigid transformation, specified as a `rigidtform3d` object. Set the initial transformation to a coarse estimate. If you do not provide a coarse or initial estimate, the function uses a `rigidtform3d` object that contains a translation that moves the center of the moving points to the center of the fixed points.
Matching method, specified as `'one-to-one'` or `'one-to-many'`.
• `'one-to-one'` — The algorithm selects the nearest neighbor as a match.
• `'one-to-many'` — The algorithm selects multiple nearest neighbors within the specified search radius as matches. Using the `'one-to-many'` method can increase registration accuracy, but at the cost of processing speed.
Search radius for point matching, specified as a positive integer. When matching, the function finds the closest edge and surface points within the specified radius. For best results, set this value based on the certainty of the `InitialTransform`. Increase the value of the `SearchRadius` when there is greater uncertainty in the initial transformation, but this can also decrease the registration speed.
Maximum iterations before LOAM registration stops, specified as a positive integer.
Tolerance between consecutive LOAM iterations, specified as a two-element vector with nonnegative values. The vector, [Tdiff Rdiff].
• Tdiff — Tolerance for the estimated absolute difference in translation and rotation estimated in consecutive LOAM iterations. Measures the Euclidean distance between two translation vectors.
• Rdiff — Tolerance for the estimated absolute difference of the angular rotation between consecutive iterations, measured in degrees.
The algorithm stops when the difference between the estimates of the rigid transformations in the most recent consecutive iterations falls below the specified tolerance value.
Display progress information, specified as a numeric or logical `0` (`false`) or `1` (`true`). To display progress information, set `Verbose` to `true`.
## Output Arguments
collapse all
Rigid 3-D transformation, returned as a `rigidtform3d` object. `tform` describes the rigid 3-D transformation that registers the moving points to the fixed points.
Root mean squared error, returned as a positive scalar that represents the Euclidean distance between aligned points. A lower `rmse` value indicates a more accurate registration.
## References
[1] Zhang, Ji, and Sanjiv Singh. “LOAM: Lidar Odometry and Mapping in Real-Time.” In Robotics: Science and Systems X. Robotics: Science and Systems Foundation, 2014. https://doi.org/10.15607/RSS.2014.X.007.
## Version History
Introduced in R2022a
expand all
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# 87 Chevy Truck Wiring Diagram
87 Chevy Truck Wiring Diagram – 1987 chevy silverado wiring diagram, 1987 chevy truck alternator wiring diagram, 1987 chevy truck headlight wiring diagram, Every electric structure is made up of various different pieces. Each part ought to be set and connected with different parts in particular way. If not, the structure will not work as it should be. To be able to be certain that the electric circuit is built properly, 87 Chevy Truck Wiring Diagram is demanded. How does this diagram aid with circuit construction?
1987 Silverado Wiring Diagram – Data Wiring Diagram Today – 87 Chevy Truck Wiring Diagram
The diagram offers visual representation of the electric arrangement. On the other hand, this diagram is a simplified version of the structure. It makes the process of building circuit easier. This diagram provides information of circuit components as well as their placements.
## Components of 87 Chevy Truck Wiring Diagram and Some Tips
There are just two things which are going to be found in any 87 Chevy Truck Wiring Diagram. The first component is symbol that indicate electrical component in the circuit. A circuit is generally composed by several components. The other thing that you will discover a circuit diagram would be traces. Lines in the diagram show exactly how each element connects to one another.
The rankings of circuit parts are relative, not accurate. The arrangement is also not plausible, unlike wiring schematics. Diagram only shows where to put component in a place relative to other components inside the circuit. Although it’s simplified, diagram is a good basis for anyone to construct their own circuit.
One thing that you must learn before reading a circuit diagram is the symbols. Every symbol that is shown on the diagram shows specific circuit element. The most common elements are capacitor, resistor, and battery. There are also other components like floor, switch, motor, and inductor. Everything rides on circuit that’s being constructed.
According to earlier, the lines at a 87 Chevy Truck Wiring Diagram represents wires. Sometimes, the wires will cross. But, it doesn’t imply connection between the cables. Injunction of 2 wires is usually indicated by black dot on the intersection of two lines. There’ll be principal lines that are represented by L1, L2, L3, and so on. Colors are also used to differentiate wires.
Usually, there are two main types of circuit connections. The first one is called series link. It’s the easier type of relationship because circuit’s elements are placed within a specified line. Due to that the electric current in each part is similar while voltage of the circuit is complete of voltage in every component.
### 87 Chevy Truck Wiring Diagram Video
Parallel connection is more complex than the show one. Unlike in string connection, the voltage of every part is comparable. It’s because the element is directly linked to electricity supply. This circuit contains branches which are passed by distinct electrical current amounts. The current joins together when the branches match.
There are lots of things that an engineer needs to pay attention to if drawing wirings diagram. To begin with, the symbols used in the diagram ought to be accurate. It should represent the specific element necessary to build a planned circuit. After the logo is wrong or uncertain, the circuit will not function because it is supposed to.
It’s also highly suggested that engineer draws favorable supply and negative source symbols for clearer interpretation. Ordinarily positive supply emblem (+) is located above the line. Meanwhile the negative supply emblem is set below it. The current flows in the left side to right.
Besides that, diagram drawer is recommended to restrict the amount of line crossing. The line and part placement should be made to decrease it. But if it is unavoidable, use universal emblem to indicate whether there is a intersection or if the lines aren’t actually connected.
Because you can see drawing and translating 87 Chevy Truck Wiring Diagram may be complicated job on itself. The advice and suggestions that were elaborated above ought to be a great kick start, however. 87 Chevy Truck Wiring Diagram
### 87 Chevy Truck Wiring Diagram Images
Chevy Truck Wiring Harness Diagram – Data Wiring Diagram Today – 87 Chevy Truck Wiring Diagram
Complete 73-87 Wiring Diagrams – 87 Chevy Truck Wiring Diagram
Complete 73-87 Wiring Diagrams – 87 Chevy Truck Wiring Diagram
87 Chevy Truck Fuel Pump Wiring Diagram | Wiring Diagram – 87 Chevy Truck Wiring Diagram
86 Chevy Truck Wiring Harness | Wiring Diagram – 87 Chevy Truck Wiring Diagram
Complete 73-87 Wiring Diagrams – 87 Chevy Truck Wiring Diagram
Complete 73-87 Wiring Diagrams – 87 Chevy Truck Wiring Diagram
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## Maths in a minute
Number theory is famous for problems that everyone can understand and that are easy to express, but that are fiendishly difficult to prove. Here are some of our favourites.
Did you learn at school that the angles in a triangle always add up to 180 degrees? If yes then your teacher was wrong. Find out why here.
A quick and easy way of adding using handshakes.
Is there a perfect voting system? In the 1950s the economist Kenneth Arrow asked himself this question and found that the answer is no, at least in the setting he imagined.
Here's a well-known conundrum: suppose I need to buy a book from a shop that costs £7. I haven't got any money, so I borrow £5 from my brother and £5 from my sister. I buy the book and get £3 change. I give £1 back to each my brother and sister and I keep the remaining £1. I now owe each of them £4 and I have £1, giving £9 in total. But I borrowed £10. Where's the missing pound?
We've been dabbling a lot in the mysterious world of quantum physics lately, so to get back down to Earth we thought we'd bring you reminder of good old classical physics.
Solving equations often involves taking square roots of numbers and if you're not careful you might accidentally take a square root of something that's negative. That isn't allowed of course, but if you hold your breath and just carry on, then you might eventually square the illegal entity again and end up with a negative number that's a perfectly valid solution to your equation.
Sequences of numbers can have limits. For example, the sequence 1, 1/2, 1/3, 1/4, ... has the limit 0 and the sequence 0, 1/2, 2/3, 3/4, 4/5, ... has the limit 1. But not all number sequences behave so nicely. Can we still discern some sort of limiting behaviour?
An infinite set is called countable if you can count it. In other words, it's called countable if you can put its members into one-to-one correspondence with the natural numbers 1, 2, 3, ... .
How would you go about adding up all the integers from 1 to 100? Tap them into a calculator? Write a little computer code? Or look up the general formula for summing integers?
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Home > Conversions (Speed) > Conversion tables from/to knot > kt to kph Conversion Cheat Sheet (Interactive)
To build or customize your cheat sheet (table below) adjust the parameters (From, Step, Decimals) in this form and hit the Update button. You could also enter the values to convert and print directly on the table From: Step: Decimals: 0 1 2 3 4 5 6 7 8 9 10 11 [See also KILOMETERS/HOUR to KNOTS]
[Formula: kph = kt x 1.852] [Worksheet] [Printer friendly]
kt kph
= 1.852
= 3.704
= 5.556
= 7.408
= 9.260
= 11.112
= 12.964
= 14.816
= 16.668
= 18.520
= 20.372
= 22.224
= 24.076
= 25.928
= 27.780
= 29.632
= 31.484
= 33.336
= 35.188
= 37.040
= 38.892
= 40.744
= 42.596
= 44.448
= 46.300
kt kph
= 48.152
= 50.004
= 51.856
= 53.708
= 55.560
= 57.412
= 59.264
= 61.116
= 62.968
= 64.820
= 66.672
= 68.524
= 70.376
= 72.228
= 74.080
= 75.932
= 77.784
= 79.636
= 81.488
= 83.340
= 85.192
= 87.044
= 88.896
= 90.748
= 92.600
kt kph
= 94.452
= 96.304
= 98.156
= 100.008
= 101.860
= 103.712
= 105.564
= 107.416
= 109.268
= 111.120
= 112.972
= 114.824
= 116.676
= 118.528
= 120.380
= 122.232
= 124.084
= 125.936
= 127.788
= 129.640
= 131.492
= 133.344
= 135.196
= 137.048
= 138.900
kt kph
= 140.752
= 142.604
= 144.456
= 146.308
= 148.160
= 150.012
= 151.864
= 153.716
= 155.568
= 157.420
= 159.272
= 161.124
= 162.976
= 164.828
= 166.680
= 168.532
= 170.384
= 172.236
= 174.088
= 175.940
= 177.792
= 179.644
= 181.496
= 183.348
= 185.200
# KNOTS to KILOMETERS/HOUR Conversion Table
kt = 1.852 kph
## Customize the style of your Cheat Sheet
Background color for header: Font color for header: Background color for odd rows: Background color for even rows: Size: S M L
## Exporting the Cheat Sheet to PDF
To export the cheat sheet as a PDF, you can follow these simple steps that leverage the browser's capabilities to save the printer-friendly version of the sheet as a PDF:
1. Click on the Printer friendly link (here or above the table), or on the Worksheet link if you wanted to get the Worksheet instead of the Cheat Sheet.
2. Press Ctrl + P (or Cmd + P on Mac) to open the print dialog
3. In the print dialog, choose Save as PDF as the destination.
The specific steps may vary slightly depending on the browser being used. This enables you to create a portable and accessible version of the cheat sheet that you can refer to anytime, anywhere.
## How to convert from KNOTS to KILOMETERS/HOUR
Since 1 knot is equal to 1.852 kilometers/hour, we could say that n knots are equal to 1.852 times n kilometers/hour. In other words, we could use the following formula:
kilometers/hour = knots x 1.852
For example, let's say that we want to convert 2.0 knots to kilometers/hour. Then, we just replace knots in the abovementioned formula with 2.0:
kilometers/hour = 2.0 x 1.852
That is, 2.0 knots are equal to 3.704 kilometers/hour.
## Other conversions
Area Computer Data Storage Conductivity and Resistivity Cooking Measures Currency Electric Current Energy Force Length Power Speed Temperature Time Volume Weight
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You are Here: Home >< Physics
# Thermal Physics Question Watch
1. The question states;
A student uses a microwave oven to warm up a cup of cold tea.
Thermal energy is supplied to the tea at a rate of 750 W. The tea has a mass of 0.42 kg and an initial temperature of 17 degrees C. Calculate the final temperature of the tea. Assume the specific heat capacity of the tea is 4200 J kg-1 K-1.
I understand how to calculate the final temperature, but first I have to change the 750 Watts into joules, however, the question gives no time which I can use for the equation Q = Power x time.
Am I missing something obvious here, or has the question been printed incorrectly?
Thanks.
2. Must be printed incorrectly, can't see any way around it
3. Agree. Someone forgot to tell you how long the thing's being heated for.
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# Primer Concentration Help - (Jun/16/2010 )
Hi everyone,
I am trying to optimize a multiplex PCR and am missing a band. I want to try to run my control with just 1 pair of primers and am confused as to how to calculate the volume required. My rxn volume is 25 ul, the final concentration of each primer is listed as 6.25 pmol. The initial concentration of my primers is 12.5 uM. My "go to" equation is (C1)(V1) = (C2)(V2), but I don't know how to used this when C1 is in uM and C2 is in pmole (not pM). Can anyone help me? This is driving me crazy. Thank you!
-cjmills-
You have to make clear if you need a concentration - uM, pM (means umol/L, pmol/L) or amount (umol, pmol only).
If there is supposed to be 6.25 pmol in 25 ul reaction, that means final concentration is 6.25/25 = 0.25 pmol/ul = 0.25 umol/L = 0.25 uM.
You can then use the equation, I think.
I count it differently. You need 6.25 pmols in reaction (which is by the way not much, I usualy use 10 pmols), in 1 ul of your stock solution there is 12.5 pmols (12.5 uM = 12.5 umol/L = 12.5 pmol/ul) so:
1 ul .......12.5
x ul ........6.25
x = 0.5 ul
So you need 0.5 ul of each primer
-Trof-
Thank you so much for your reply, Trof. I thought there were 10^6 pmole in 1 umole? Shouldn't the final concentration be 0.0025 uM?
That is what I keep getting and it seems like a ridiculously low number.
Thanks,
Carrie
-cjmills-
cjmills on Jun 16 2010, 06:17 PM said:
Thank you so much for your reply, Trof. I thought there were 10^6 pmole in 1 umole? Shouldn't the final concentration be 0.0025 uM?
There is, but there's 10^6 ul in liter. So pmol/ul equals umol/L.
-Trof-
Duh! Thanks Trof! I could not figure out what I was doing wrong.
-cjmills-
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https://discourse.mc-stan.org/t/slice-an-array-of-vectors-across-the-first-dimension-as-row-for-a-matrix/25456
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Slice an array of vectors across the first dimension (as row() for a matrix)
Hello Stanimals,
Given an array of vectors, how can one efficiently extract the n’th element from each vector in that array. The application is a hierarchical model a set of two correlated random effects that determine the group-level mean (random effect #1) and within-group variance (random effect #2). These correlated group-level random effects are modelled as a function of a multivariate normally-distributed parameter `rfx_district` that is implemented in my model code as an array of vectors:
``````parameters {
vector[2] rfx_district[N_group] ; // Group-level random effects
cholesky_factor_corr[2] L_omega; // Correlation matrix for rfx (Cholesky decomposition)
vector<lower=0>[2] sigma_rfx_district; // Scales of rfx
}
model {
rfx_district ~ multi_normal_cholesky(mu_zero, diag_pre_multiply(sigma_rfx_district, L_omega)); // mu_zero is specified in transformed data as a vector of zeroes
}
``````
To efficiently construct the group-level mean and variance parameters, I would somehow like to index or slice the array of vectors such that I get a vector or an array or reals of length equal to the length of the original array, with every element containing the n’th value of each vector in the original array of vectors. In other words, what is the most efficient way to slice an array of vectors across its first dimension as if it were a matrix (i.e., as one might apply ‘row’ to a matrix to extract a particular row)? Or can this only be achieved by iterating over each element of the array `rfx_district`?
Apologies if this has been asked before. I could not find pointers anywhere else on discourse.
I’m using rstan 2.21.2 in R version 4.0.3 on MacOS Catalina 10.15.7.
EDIT: typo corrected
If all you’re doing is extracting, using a loop is fine. So a simple Stan function to do this would be:
``````vector extract_each(vector[] x, int n) {
int M = size(x);
vector[M] y;
for (m in 1:M) y[m] = x[m, n];
return y;
}
``````
That’s as efficiently as it can be done. A built-in wouldn’t be any faster. The inefficiency here is the memory non-locality (each array is independently allocated, so they’re not contiguous), not the loop. The problem is usually that you want to access both rows and columns efficiently, and that’s not really possible without paying the transpose cost.
1 Like
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# Practice final for Basic Physics spring 2005 answers on the last page Name: Date:
Save this PDF as:
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## Transcription
1 Practice final for Basic Physics spring 2005 answers on the last page Name: Date: 1. A 12 ohm resistor and a 24 ohm resistor are connected in series in a circuit with a 6.0 volt battery. Assuming negligible internal resistance in the battery, the current in the 12 ohm resistor will be A) 6.0 A. B) 2.0 A. C) 0.50 A. D) 0.25 A. E) A. 2. Which of the following is NOT a vector? A) acceleration B) weight C) mass D) velocity E) All of the above. 3. Which of the following is not an electromagnetic wave? A) Sound B) Microwaves C) Infrared D) X-ray E) Light 4. Neglecting friction, if a Cadillac and Volkswagen start rolling down a hill together, the heavier Cadillac will get to the bottom A) before the Volkswagen. B) after the Volkswagen. C) at the same time as the Volkswagen. 5. Two current-carrying wires are parallel to one another and separated by 1 cm. If the distance between them is increased to 2 cm the new force will be what factor times the original force? A) 4 B) 2 C) 1 D) ½ E) ¼ Page 1
2 6. A parachutist jumping from an airplane reaches a terminal velocity when the force of air resistance is 490 N. The mass of the parachutist is A) 490 kg. B) 9.8 kg. C) 40 kg. D) 50 kg. E) 80 kg. 7. Which of these lists is in order of increasing wavelength? A) First harmonic, second harmonic, third harmonic. B) Second harmonic, third harmonic, fourth harmonic. C) First harmonic, second harmonic, fundamental. D) Third harmonic, second harmonic, first harmonic. 8. An object is placed 15 cm in front of a diverging lens of focal length -10 cm. The image will be located A) cm in front of the lens. B) 3.0 cm behind the lens. C) 3.0 cm in front of the lens. D) 6.0 cm in front of the lens. E) 6.0 cm behind the lens. 9. Comparing the electrostatic force and the gravitational force we can say that A) both have the same dependence on distance, both involve attraction and repulsion but the gravitational force is stronger. B) both have the same dependence on distance, both involve attraction and repulsion but the electrostatic force is stronger. C) both have the same dependence on distance, the electrostatic force can be either attractive or repulsive while the gravitational force is only repulsive, and the electrostatic force is weaker. D) both have the same dependence on distance, the electro static force can be either attractive or repulsive while the gravitational force is only attractive, and the electrostatic force is stronger. E) the electrostatic force falls off more rapidly with distance, the electrostatic force can be either attractive or repulsive while gravitation is only attractive and the electrostatic force is stronger. Page 2
3 10. An elevator is being lifted upward at a constant speed by a steel cable. All frictional forces are neglected. In this situation, forces on the elevator are such that A) the upward force by the cable is greater than the downward force of gravity. B) the upward force by the cable is equal to the downward force of gravity. C) the upward force by the cable is smaller than the downward force of gravity. D) none of the above. (The elevator goes up because the cable is being shortened, not because an upward force is exerted on the elevator by the cable.) 11. Electromagnetic waves are generally A) transverse waves. B) longitudinal waves. C) a 50/50 combination of transverse and longitudinal waves. D) standing waves. 12. Rays from a distant object do not diverge enough for a nearsighted person to focus; this can be corrected with a A) converging or concave lens. B) converging or convex lens. C) diverging or concave lens. D) diverging or convex lens. 13. The temperature of 500 g of water is to be raised from 10 C to 40 C. The energy needed to do this is about A) cal. B) cal. C) cal. D) cal. 14. A true statement about atoms is that they A) can emit radiation only at specific frequencies. B) all have the same number of electrons. C) can emit radiation at any frequency. D) can emit radiation at frequencies only within the visible spectrum. 15. An atom has A) just as many electrons as protons. B) no neutrons in the nucleus. C) more protons than electrons. D) as many electrons as protons and neutrons combined. E) at least 1 neutron. Page 3
4 16. A 1.5 V flashlight battery, a 10 ohm resistor, and a flash light lamp are available. How should the lamp be connected in a circuit so that it would glow the brightest? A) Connect it in series with the resistor and battery so that the bulb is next to the positive terminal of the battery. B) Connect it in parallel with the battery and the resistor. C) Connect it directly to the battery; don't use the resistor. D) Connect it in series with the resistor and battery so that the bulb is next to the negative terminal of the battery. 17. The three processes by which heat energy is transferred between objects are A) heat, calorie and radiation. B) radiation, temperature, and convection. C) absorption, radiation, and convection. D) radiation, convection, and conduction. E) radiation, absorption, and conduction. 18. A quantity that is a measure of how the velocity of a body changes with time is: A) Displacement. B) Speed. C) Acceleration. D) Momentum. 19. The wavelength of a red photon is that of a green photon. A) greater than B) less than C) same as D) greater, same or less, depending on a source of photons 20. A 4 kg block, initially moving due east at 3 m/s, is acted upon by an impulse having magnitude 8 Ns and direction due west. The final velocity of the block is A) 8 m/s west. B) 2 m/s west. C) 1 m/s east. D) 3 m/s east. Page 4
5 21. Which of the following is an appropriate unit for describing rotational acceleration? A) m/s 2. B) rad/min. C) rev/min. D) rev/m/s. E) rev/min Which of the following is not an energy unit? A) N m B) Joule C) calorie D) watt E) kwh 23. The speed of electromagnetic waves in air is 3.0 x 10 8 m/s. What is the wavelength of FM radio waves having a frequency of 1.0 x 10 8 Hz? A) m B) 1.0 m C) 3.0 m D) 6.0 m E) 10 m 24. A rifle bullet is fired horizontally at the same instant another bullet is allowed to drop from rest at the same height. Which bullet strikes the earth first? A) The bullet from the rifle. B) The bullet allowed to drop. C) Both strike at the same instant. 25. Electrons flow around a circular wire loop in a horizontal plane, in a direction that is clockwise when viewed from above. This causes a magnetic field. Inside the loop, the direction of this field is A) up. B) down. C) toward the center of the loop. D) radially outward from the center of the loop. Page 5
6 26. A child runs at 4.0 m/s and jumps onto a sled, initially at rest. If the child' mass is 36 kg, and if the child and sled slide off together at 3.0 m/s after the collision, the sled's mass is A) 12 kg. B) 27 kg. C) 36 kg. D) 48 kg. 27. Assuming g = 10 m/s² and that air resistance is negligible, in ½ second a ball dropped from rest will fall A) ½ meter. B) one meter. C) 1.25 meters. D) 5 meters. 28. Magnetic fields affect A) only electric charges at rest. B) only electric charges in motion. C) both electric charges in motion and electric charges at rest. D) neither electric charges in motion nor electric charges at rest. 29. A common property of all waves is the relationship between the speed (v), the wavelength (λ) and the frequency of the wave (f). The correct equation for this relationship is A) f = vλ. B) λ = vf. C) v = fλ. D) v = f / λ. 30. Suppose you climb the stairs of a ten-story building, about 30 m high, and your mass is 60 kg. The gravitational potential energy you gain is about A) 270 Joule. B) 1800 Joule. C) Joule. D) Joule. Page 6
7 31. A heat pump is capable of delivering more energy to the home than goes into the operation of the pump itself, when conditions are favorable. Which of the following statements is correct? A) A heat pump violates the first law of thermodynamics. B) A heat pump violates the second law of thermodynamics. C) A heat pump transfers some energy from the outdoors. D) A heat pump, like the Carnot engine, is a theoretical device that is not useful in practice. E) A heat pump is more efficient when the outside temperature is colder. 32. If your average speed for a 3-hr trip is 45 mi/hr, the distance traveled is A) 15 mi. B) 45 mi. C) 135 mi. D) mi 33. An isolated object is initially spinning at a constant speed. Then, although no external forces act upon it, its rotational speed increases. This must be due to A) an increase in the moment of inertia. B) an increase in the mass. C) an increase in the angular momentum. D) a decrease in the moment of inertia. E) impossible, angular momentum conservation is violated. 34. An iceberg is floating in the ocean with 10% of its volume extending above the ocean's surface. What can you say about the iceberg? A) Its weight is greater than the water it displaces. B) Its density is 90% that of the ocean water. C) It displaces only 90% of its weight in water. D) It must be resting on the bottom, since ice can't float in salt water. 35. A man weighs 600 N while on the surface of Earth. If he is transported to the planet Mythos, which has the same mass as Earth but a radius that is twice as large as Earth's, his weight would be A) 1200 N. B) 600 N. C) 300 N. D) 150 N. E) 100 N. Page 7
8 36. Consider a proton and an electron placed near one another with no other objects close by. They would A) accelerate away from each other. B) remain motionless. C) accelerate toward each other. D) be pulled together at constant speed. E) move away from each other at constant speed. 37. Energy added to a cyclical heat engine A) is completely converted to external work. B) is converted to increased internal energy in the engine plus external work. C) is used to generate work that is greater than the added energy. D) is converted to work and to waste heat. 38. If the image formed by a concave mirror is closer to the mirror than the object is, then compared to the object, the image will be A) larger and inverted. B) smaller and inverted. C) larger and erect. D) larger and real. 39. One section of a pipe carrying water has a cross-sectional area of 20 cm 2 ; in this section the water has flow velocity of 1.0 m/s. Another section of this pipe has a constricted cross-sectional area of 5 cm 2. If the flow is steady, what is the water velocity in the cons tricted section? A) 0.25 m/s B) 0.50 m/s C) 1.0 m/s D) 2.0 m/s E) 4.0 m/s 40. The potential difference across the terminals of a storage battery not connected in any circuit is observed to be 12 V. When it is connected to an external resistance the potential difference across the terminals is observed to be 11 V while the current in the external resistance is 2 A. What is the internal resistance of the battery? A) zero. B) 0.50 ohms. C) 1.0 ohm. D) 2.0 ohms. E) 5.5 ohms. Page 8
9 Answer Key 1. E 2. C 3. A 4. C 5. D 6. D 7. D 8. D 9. D 10. B 11. A 12. C 13. B 14. A 15. A 16. C 17. D 18. C 19. A 20. C 21. E 22. D 23. C 24. C 25. A 26. A 27. C 28. B 29. C 30. C 31. C 32. C 33. D 34. B 35. D 36. C 37. D 38. B 39. E 40. B Page 9
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Physics 2A, Sec C00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to ll your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry
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Physics 117.3 Tutorial #1 January 14 to 25, 2013 Rm 130 Physics 8.79. The location of a person s centre of gravity can be determined using the arrangement shown in the figure. A light plank rests on two
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### 3-1. True or False: Different colors of light are waves with different amplitudes. a.) True b.) False X
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ATTENTION: All Division I students, START HERE. All Division II students, skip the first ten questions, begin on question. THE NEXT TWO QUESTIONS REFER TO THE FOLLOWING GRAPH. The graph represents the
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https://www.teacherspayteachers.com/Product/April-Showers-Bring-Graphing-Power-Line-Plot-Bar-and-Pictographs-3rd-Grade-1191386
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# Main Categories
Total:
\$0.00
Whoops! Something went wrong.
# April Showers Bring Graphing Power: Line Plot, Bar, and Pictographs! (3rd Grade)
Subject
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
13 MB|96 pages
Product Description
This pack was created to support and supplement your graphing unit. It includes 8 different activities to give students practice reading, creating, and analyzing bar graphs, pictographs and line plot graphs.
All bar graphs use a scale of 5, except for the “Whatcha Wearing?” activity, which uses a scale of 1. The pictographs use a scale of 10. The following activities are included:
Can-A-Thon Graphing- Students read a provided line plot graph and answer 6 questions. An additional challenge question card is included. Two answer sheets are included, one with the challenge question and one without. Clip art is included if the teacher chooses to make this line plot activity hands on.
Measuring Pencils Mania- This is two activities. In the first activity, students measure pictures of pencils, to the nearest quarter inch, and record the result. In the second activity, students use the measurements of the pencil lengths to create a line plot graph. Two versions of the line plot sheet are included, one labeled with numbers and one that is blank. Then, students answer 6 questions. An additional challenge question card is included. Two answer sheets are included, one with the challenge question and one without.
A Little Bit of Yum- Students read a provided tally chart and bar graph and answer 6 questions. An additional challenge question card is included. Two answer sheets are included, one with the challenge question and one without.
Foot Loose and Fancy Free- Students read a short paragraph to create a bar graph and tally chart. Two versions of the graphing sheet are included, one with a completed tally chart and one without. Students then answer 6 questions. An additional challenge question card is included. Two answer sheets are included, one with the challenge question and one without. Clip art is included if the teacher chooses to make this graphing activity hands on.
Whatcha Wearing?- Students survey classmates and create a tally chart and bar graph with the results. No questions are included in this activity.
Favorite Color Bunnies- Students read a provided pictograph and answer 6 questions. An additional challenge question card is included. Two answer sheets are included, one with the challenge question and one without.
Making Rainy Days a Splash! - Students read a short paragraph to create a pictograph. Then, students answer 6 questions. An additional challenge question card is included. Two answer sheets are included, one with the challenge question and one without.
I hope this provides additional, easy to use resources for your class!
Happy Learning!
Mandy
mandyholland@hotmail.com
http://www.tips-for-teachers.com/
http://mandys-tips-4-teachers.blogspot.com/
Total Pages
96 pages
N/A
Teaching Duration
N/A
• Product Q & A
\$6.00
\$6.00
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https://www.convertunits.com/from/thousand+cubic+foot/day/to/cubic+yard/minute
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## ››Convert thousand cubic foot/day to cubic yard/minute
thousand cubic foot/day cubic yard/minute
How many thousand cubic foot/day in 1 cubic yard/minute? The answer is 38.88.
We assume you are converting between thousand cubic foot/day and cubic yard/minute.
You can view more details on each measurement unit:
thousand cubic foot/day or cubic yard/minute
The SI derived unit for volume flow rate is the cubic meter/second.
1 cubic meter/second is equal to 3051.1871607739 thousand cubic foot/day, or 78.477036028136 cubic yard/minute.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between thousand cubic feet/day and cubic yards/minute.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of thousand cubic foot/day to cubic yard/minute
1 thousand cubic foot/day to cubic yard/minute = 0.02572 cubic yard/minute
10 thousand cubic foot/day to cubic yard/minute = 0.2572 cubic yard/minute
20 thousand cubic foot/day to cubic yard/minute = 0.5144 cubic yard/minute
30 thousand cubic foot/day to cubic yard/minute = 0.7716 cubic yard/minute
40 thousand cubic foot/day to cubic yard/minute = 1.02881 cubic yard/minute
50 thousand cubic foot/day to cubic yard/minute = 1.28601 cubic yard/minute
100 thousand cubic foot/day to cubic yard/minute = 2.57202 cubic yard/minute
200 thousand cubic foot/day to cubic yard/minute = 5.14403 cubic yard/minute
## ››Want other units?
You can do the reverse unit conversion from cubic yard/minute to thousand cubic foot/day, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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http://mathoverflow.net/questions/45687/locally-compact-separable-metric-spaces
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# Locally compact separable metric spaces
Hi, Is it true that for every locally compact separable metric space $E$ there exists a sequence $(K_n)_{n\in\mathbb{N}}$ of compact subsets of $E$ such that $K_n\subset\stackrel{\circ}{K_{n+1}}$ and $\cup K_n = E$ ?
I’m almost sure this is false but I can’t find a counterexample.
Thank you.
-
Do we need at all that the space is metrizable? – Martin Brandenburg Nov 11 '10 at 13:10
I think the following argument works under your hypothesis:
Consider $\mathcal{B}=\{B_n\}$ a countable basis of the topology of $E$ such that $\overline{B_n}$ is compact for any $n$ (this exists since $E$ is a separable metric space, thus, it has a countable basis and then a basis like this is constructed using local compactness).
Now, start with $K_0= \overline{B_0}$. Now, given $K_n$, define $K_{n+1}$ by the union of $\overline{B_{n+1}}$ with the closure of a finite subcovering of $K_n$ so, it is compact (being finite union of compact sets) and $K_n \subset int(K_{n+1})$.
We get that $\bigcup K_n =E$ since it contains $\bigcup B_n=E$.
-
You beat me to it, but I posted mine anyhow since it seems to use less machinery. Let the voters, uh, vote. – Harald Hanche-Olsen Nov 11 '10 at 12:52
I like your robust, no-nonsense use of the definition of voter, Harald :) – Georges Elencwajg Nov 11 '10 at 13:19
How do you construct such a basis of the topology? It seems that you need to do something like in the answer of Harald. – Guillaume Brunerie Nov 11 '10 at 13:41
The collection of $B_{n,m}= B(x_n, 1/m)$ is a basis of the topology (being $x_n$ a dense countable set). Let $m_n$ be the first $m$ such that $B(x_n,1/m)$ becomes precompact. Then, $m_n$ varies semicontinuously so, the balls which are precompact also form a basis. (Yes, it is something like Harald's argument). – rpotrie Nov 11 '10 at 13:56
Sure. Let $\{x_1,x_2,\ldots\}$ be dense in $E$, let $K_0=\emptyset$ and let $K_n$ be a compact neighbourhood of $K_{n-1}\cup\{x_n\}$ for $n=1,2,\dots$. Just make sure that $K_n$ contains a sufficiently large ball around $x_n$: Say if $\epsilon<1$ and $B_{2\epsilon}(x_n)$ is precompact, then insist on $B_\epsilon(x_n)\subseteq K_n$. Now if $y$ is not in the union and $B_{3\epsilon}(y)$ is precompact, then $x_n\in B_\epsilon(y)$ will imply $y\in K_n$.
-
If I remind correctly Bourbaki's General Topology, a locally compact space is paracompact if and only if any connected component is countable at infinity. Since your $E$ is metrizable, it is paracompact, and since it is separable, it has countably many components. Conclusion : $(K_n)$ exists.
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https://www.quizzes.cc/metric/percentof.php?percent=5540&of=1320
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#### What is 5540 percent of 1,320?
How much is 5540 percent of 1320? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 5540% of 1320 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 5540% of 1,320 = 73128
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating five thousand, five hundred and fourty of one thousand, three hundred and twenty How to calculate 5540% of 1320? Simply divide the percent by 100 and multiply by the number. For example, 5540 /100 x 1320 = 73128 or 55.4 x 1320 = 73128
#### How much is 5540 percent of the following numbers?
5540% of 1320.01 = 7312855.4 5540% of 1320.02 = 7312910.8 5540% of 1320.03 = 7312966.2 5540% of 1320.04 = 7313021.6 5540% of 1320.05 = 7313077 5540% of 1320.06 = 7313132.4 5540% of 1320.07 = 7313187.8 5540% of 1320.08 = 7313243.2 5540% of 1320.09 = 7313298.6 5540% of 1320.1 = 7313354 5540% of 1320.11 = 7313409.4 5540% of 1320.12 = 7313464.8 5540% of 1320.13 = 7313520.2 5540% of 1320.14 = 7313575.6 5540% of 1320.15 = 7313631 5540% of 1320.16 = 7313686.4 5540% of 1320.17 = 7313741.8 5540% of 1320.18 = 7313797.2 5540% of 1320.19 = 7313852.6 5540% of 1320.2 = 7313908 5540% of 1320.21 = 7313963.4 5540% of 1320.22 = 7314018.8 5540% of 1320.23 = 7314074.2 5540% of 1320.24 = 7314129.6 5540% of 1320.25 = 7314185
5540% of 1320.26 = 7314240.4 5540% of 1320.27 = 7314295.8 5540% of 1320.28 = 7314351.2 5540% of 1320.29 = 7314406.6 5540% of 1320.3 = 7314462 5540% of 1320.31 = 7314517.4 5540% of 1320.32 = 7314572.8 5540% of 1320.33 = 7314628.2 5540% of 1320.34 = 7314683.6 5540% of 1320.35 = 7314739 5540% of 1320.36 = 7314794.4 5540% of 1320.37 = 7314849.8 5540% of 1320.38 = 7314905.2 5540% of 1320.39 = 7314960.6 5540% of 1320.4 = 7315016 5540% of 1320.41 = 7315071.4 5540% of 1320.42 = 7315126.8 5540% of 1320.43 = 7315182.2 5540% of 1320.44 = 7315237.6 5540% of 1320.45 = 7315293 5540% of 1320.46 = 7315348.4 5540% of 1320.47 = 7315403.8 5540% of 1320.48 = 7315459.2 5540% of 1320.49 = 7315514.6 5540% of 1320.5 = 7315570
5540% of 1320.51 = 7315625.4 5540% of 1320.52 = 7315680.8 5540% of 1320.53 = 7315736.2 5540% of 1320.54 = 7315791.6 5540% of 1320.55 = 7315847 5540% of 1320.56 = 7315902.4 5540% of 1320.57 = 7315957.8 5540% of 1320.58 = 7316013.2 5540% of 1320.59 = 7316068.6 5540% of 1320.6 = 7316124 5540% of 1320.61 = 7316179.4 5540% of 1320.62 = 7316234.8 5540% of 1320.63 = 7316290.2 5540% of 1320.64 = 7316345.6 5540% of 1320.65 = 7316401 5540% of 1320.66 = 7316456.4 5540% of 1320.67 = 7316511.8 5540% of 1320.68 = 7316567.2 5540% of 1320.69 = 7316622.6 5540% of 1320.7 = 7316678 5540% of 1320.71 = 7316733.4 5540% of 1320.72 = 7316788.8 5540% of 1320.73 = 7316844.2 5540% of 1320.74 = 7316899.6 5540% of 1320.75 = 7316955
5540% of 1320.76 = 7317010.4 5540% of 1320.77 = 7317065.8 5540% of 1320.78 = 7317121.2 5540% of 1320.79 = 7317176.6 5540% of 1320.8 = 7317232 5540% of 1320.81 = 7317287.4 5540% of 1320.82 = 7317342.8 5540% of 1320.83 = 7317398.2 5540% of 1320.84 = 7317453.6 5540% of 1320.85 = 7317509 5540% of 1320.86 = 7317564.4 5540% of 1320.87 = 7317619.8 5540% of 1320.88 = 7317675.2 5540% of 1320.89 = 7317730.6 5540% of 1320.9 = 7317786 5540% of 1320.91 = 7317841.4 5540% of 1320.92 = 7317896.8 5540% of 1320.93 = 7317952.2 5540% of 1320.94 = 7318007.6 5540% of 1320.95 = 7318063 5540% of 1320.96 = 7318118.4 5540% of 1320.97 = 7318173.8 5540% of 1320.98 = 7318229.2 5540% of 1320.99 = 7318284.6 5540% of 1321 = 7318340
1% of 1320 = 13.2 2% of 1320 = 26.4 3% of 1320 = 39.6 4% of 1320 = 52.8 5% of 1320 = 66 6% of 1320 = 79.2 7% of 1320 = 92.4 8% of 1320 = 105.6 9% of 1320 = 118.8 10% of 1320 = 132 11% of 1320 = 145.2 12% of 1320 = 158.4 13% of 1320 = 171.6 14% of 1320 = 184.8 15% of 1320 = 198 16% of 1320 = 211.2 17% of 1320 = 224.4 18% of 1320 = 237.6 19% of 1320 = 250.8 20% of 1320 = 264 21% of 1320 = 277.2 22% of 1320 = 290.4 23% of 1320 = 303.6 24% of 1320 = 316.8 25% of 1320 = 330
26% of 1320 = 343.2 27% of 1320 = 356.4 28% of 1320 = 369.6 29% of 1320 = 382.8 30% of 1320 = 396 31% of 1320 = 409.2 32% of 1320 = 422.4 33% of 1320 = 435.6 34% of 1320 = 448.8 35% of 1320 = 462 36% of 1320 = 475.2 37% of 1320 = 488.4 38% of 1320 = 501.6 39% of 1320 = 514.8 40% of 1320 = 528 41% of 1320 = 541.2 42% of 1320 = 554.4 43% of 1320 = 567.6 44% of 1320 = 580.8 45% of 1320 = 594 46% of 1320 = 607.2 47% of 1320 = 620.4 48% of 1320 = 633.6 49% of 1320 = 646.8 50% of 1320 = 660
51% of 1320 = 673.2 52% of 1320 = 686.4 53% of 1320 = 699.6 54% of 1320 = 712.8 55% of 1320 = 726 56% of 1320 = 739.2 57% of 1320 = 752.4 58% of 1320 = 765.6 59% of 1320 = 778.8 60% of 1320 = 792 61% of 1320 = 805.2 62% of 1320 = 818.4 63% of 1320 = 831.6 64% of 1320 = 844.8 65% of 1320 = 858 66% of 1320 = 871.2 67% of 1320 = 884.4 68% of 1320 = 897.6 69% of 1320 = 910.8 70% of 1320 = 924 71% of 1320 = 937.2 72% of 1320 = 950.4 73% of 1320 = 963.6 74% of 1320 = 976.8 75% of 1320 = 990
76% of 1320 = 1003.2 77% of 1320 = 1016.4 78% of 1320 = 1029.6 79% of 1320 = 1042.8 80% of 1320 = 1056 81% of 1320 = 1069.2 82% of 1320 = 1082.4 83% of 1320 = 1095.6 84% of 1320 = 1108.8 85% of 1320 = 1122 86% of 1320 = 1135.2 87% of 1320 = 1148.4 88% of 1320 = 1161.6 89% of 1320 = 1174.8 90% of 1320 = 1188 91% of 1320 = 1201.2 92% of 1320 = 1214.4 93% of 1320 = 1227.6 94% of 1320 = 1240.8 95% of 1320 = 1254 96% of 1320 = 1267.2 97% of 1320 = 1280.4 98% of 1320 = 1293.6 99% of 1320 = 1306.8 100% of 1320 = 1320
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https://www.wiley.com/en-us/GMAT+Official+Guide+2018+Quantitative+Review%3A+Book+%2B+Online-p-9781119402374
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# GMAT Official Guide 2018 Quantitative Review: Book + Online
ISBN: 978-1-119-40237-4
Jun 2017
240 pages
Select type: E-Book
\$12.99
Product not available for purchase
## Description
A supplement to the Official Guide with 300 additional quantitative questions
The GMAT Official Guide Quantitative Review provides targeted preparation for the mathematical portion of the GMAT exam. Designed by the Graduate Management Admission Council, this guide contains 300 real GMAT questions from past exams including 45 never-before-seen questions, plus the following features:
• An overview of the exam to help you get familiar with the content and format
• Review essential algebra, geometry, arithmetic, and word problems
• Detailed answer explanations that explain how the test maker thinks about a question
• Questions organized in order of difficulty from easiest to hardest
• Access to the same questions online at gmat.wiley.com, where you can build your own practice sets
Don’t waste time practicing on fake GMAT questions. Optimize your study time with the GMAT Official Guide 2018 Quantitative Review using real questions from actual past exams.
## Related Resources
### Instructor
Request an Evaluation Copy for this title
1.0 What is the GMAT® Exam? 2
1.0 What is the GMAT® Exam? 3
1.1 Why Take the GMAT® Exam? 3
1.2 GMAT ® Exam Format 4
1.3 What Is the Content of the Exam Like? 5
1.4 Integrated Reasoning Section 6
1.5 Quantitative Section 6
1.6 Verbal Section 7
1.7 Analytical Writing Assessment 7
1.8 What Computer Skills Will I Need? 7
1.9 What Are the Test Centers Like? 7
1.10 How Are Scores Calculated? 8
1.11 Test Development Process 9
2.0 How to Prepare 10
2.0 How to Prepare 11
2.1 How Can I Best Prepare to Take the Test? 11
2.2 What About Practice Tests? 11
2.3 Where Can I Get Additional Practice? 12
2.4 General Test-Taking Suggestions 12
3.0 Math Review 14
3.0 Math Review 15
3.1 Arithmetic 16
3.2 Algebra 27
3.3 Geometry 35
3.4 Word Problems 47
4.0 Problem Solving 56
4.0 Problem Solving 57
4.1 Test-Taking Strategies 58
4.2 The Directions 58
4.3 Sample Questions 60
5.0 Data Sufficiency 148
5.0 Data Sufficiency 149
5.1 Test-Taking Strategies 150
5.2 The Directions 152
5.3 Sample Questions 154
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Ch 13 - Solutions
# Ch 13 - Solutions - Chapter 13 Queuing Theory 6 a b c There...
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Chapter 13 Queuing Theory 6. a. There are 30/60 = 0.5 arrivals per minute, on average. b. In a ten minute interval we would expect 10*0.5 = 5 arrivals to occur. c. P(x=0) = (5 0 e -5 )/(0!) = 0.00674 (NOTE: 0!=1) P(x=1) = (5 1 e -5 )/(1!) = 0.03369 P(x=2) = (5 2 e -5 )/(2!) = 0.08422 P(x=3) = (5 3 e -5 )/(3!) = 0.14037 d. P(x>3)=1-P(x 3) = 1-0.00674-0.03369-0.08422-0.14037 = 0.73497 7. a. Expected service time = 1/40 = 0.025 hours (or 1.5 minutes) b. P(t 1/60) = 1- e -40/60 = 0.4865 c. P(2/60 t 5/60) = e -40*2/60 - e -40*5/60 = 0.2279 P(t < 4/60) = 1- e -40*4/60 = 0.931 P(t 3/60) = 1- P(t 3/60) = 1-.8646 = 0.1354 8. Use the M/M/s template in Q.xls with: Arrival Rate = 30, Service Rate = 40, Number of Servers = 1. a. 0.25 b. 0.75 c. 2.25 d. 0.1 hours or 6 minutes e. 0.075 hours or 4.5 minutes f. 60 per hour 9. Use the M/M/s template in Q.xls with: Arrival Rate = 7 per hour, Service Rate = 3 per hour, and vary the number of servers. With 3 servers the average waiting time is 0.3057 hours or 18.34 minutes.
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# Arithmetic arranger test error
i just finished the arithmetic arranger project of scientific computing with python project .
Every rule (test condition) was working fine on my editor but when i put the code of repl.in it thew two errors which i cannot understand.
this is my code
``````import operator
def con_check(a):
try:
a = int(a)
return True
except:
return False
def arithmetic_arranger(problems, con = False):
arranged_problems = ""
first = [*range(len(problems))]
second = [*range(len(problems))]
sign = [*range(len(problems))]
x = 0
for i in problems:
first[x], sign[x], second[x] = i.split()
if len(str(first[x])) > 4 or len(str(second[x])) > 4:
return "Error: Numbers cannot be more than four digits."
elif x > 4:
return "Error: Too many problems."
elif not sign[x] in dict:
print(sign[x])
return "Error: Operator must be '+' or '-'."
elif not con_check(first[x]) or not con_check(second[x]):
return "Error: Numbers must only contain digits."
x+=1
# first line
for i in range(len(problems)):
arranged_problems += str(" "*(len(str(max(int(first[i]), int(second[i])))) - len(str(first[i])) + 2 ) + first[i] + " ")
arranged_problems += str("\n")
#second line
for i in range(len(problems)):
arranged_problems += str(sign[i] + " " * (len(str(max(int(first[i]), int(second[i])) )) - len(second[i]) +1) + second[i] +" ")
arranged_problems += str("\n")
#Third line
for i in range(len(problems)):
arranged_problems += str("--" + "-" * (len(str(max(int(first[i]), int(second[i]))))) +" ")
arranged_problems += str("\n")
if con :
for i in range(len(problems)):
arranged_problems += str(" " * (len(str(max(int(first[i]), int(second[i])))) - len(str(dict[sign[i]](int(first[i]), int(second[i])))) + 2) + str(dict[sign[i]](int(first[i]), int(second[i]))) + " ")
return arranged_problems
``````
And this is these are the two errors repl.in is throwing
``````..F..
=================================================
=====================
FAIL: test_arrangement (test_module.UnitTests)
-------------------------------------------------
---------------------
Traceback (most recent call last):
File "/home/runner/fcc-arithmetic-arranger/test
_module.py", line 10, in test_arrangement
self.assertEqual(actual, expected, 'Expected
different output when calling "arithmetic_arrange
r()" with ["3 + 855", "3801 - 2", "45 + 43", "123
+ 49"]')
AssertionError: ' [23 chars] 123 \n+ 855
- 2 + 43 + 49 [37 chars] \n' != '
[23 chars] 123\n+ 855 - 2 + 43 +
49\n-----[23 chars]----'
- 3 3801 45 123
? ----
+ 3 3801 45 123
- + 855 - 2 + 43 + 49
? ----
+ + 855 - 2 + 43 + 49
- ----- ------ ---- -----
? -----
+ ----- ------ ---- ----- : Expected dif
ferent output when calling "arithmetic_arranger()" with ["3 + 855", "3801 - 2", "45 + 43", "123 + 49"]
======================================================================
FAIL: test_solutions (test_module.UnitTests)
----------------------------------------------------------------------
Traceback (most recent call last):
File "/home/runner/fcc-arithmetic-arranger/test_module.py", line 39, in test_solutions
self.assertEqual(actual, expected, 'Expected solutions to be correctly displayed in output when calling "arithmetic_arranger()" with arithemetic problems and a second argument of `True`.')
AssertionError: ' 3[23 chars] 123 \n- 698 - 3801 + 43 + 49 [73 chars] ' != ' 3[23 chars] 123\n- 698 - 3801 + 43 + 49\n-----[57 chars] 172'
- 32 1 45 123
? ----
+ 32 1 45 123
- - 698 - 3801 + 43 + 49
? ----
+ - 698 - 3801 + 43 + 49
- ----- ------ ---- -----
? ----
+ ----- ------ ---- -----
- -666 -3800 88 172 ? ----
+ -666 -3800 88 172 : Expected solutions to be correctly displayed in output when calling "arithmetic_arranger()" with arithemetic problems and a second argument of `True`.
----------------------------------------------------------------------
Ran 6 tests in 0.005s
FAILED (failures=2)
``````
Hello there,
Those errors originate from the fCC tests. It is not an issue with Python, but is saying that your script does not output the expected result. Something in your script is not outputting the correct result.
The best thing for you to do is run over the requirements again, and test your function with related arguments to see where it differs.
Also, I have moved this to a more suitable subforum
Hope this helps
Thanks for responding,
But I have checked all the requirements multiple times and the code is also giving all the output correctly.
All my outputs (talking about print not return statement) are exactly same as the the test expectation.
Once again thanks for responding
have you added also the results of the operations? you should just print the given numbers as required, not add other numbers
i have just added the numbers if the second condition is given True just as said in the question.
just figured out the error .
I was adding the β\nβ after every line , including the last line because of which test was throwing error.
β\nβ or any space should not be printed in the last line where it is not needed. so i just added some condition statements to carefully place the β\nβ and space.
And yess ALL the test passes.
for someone who is curious I just compared the len() of the expected return from the test_module.py with my function return and found my function return to be longer then expected .
Thanks to everyone who responded.
1 Like
Can you please explain these lines? I am a newbieβ¦
ohh , why not
`````` first = [*range(len(problems))]
second = [*range(len(problems))]
sign = [*range(len(problems))]
``````
with range() function i am just making an array of the same length of as that of the problems input.
and the * before the range function tell the interpreter to expand the range() function as an array and with the * the array will output the range() function instead of the required array.
happy to help
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# Fun Holiday Games for Middle School Math
If you’re looking for a few fun holiday games (math and other) for middle school classes as you head into the holiday season, look no further!
Maybe you want to increase engagement when the holiday schedule, interruptions, and distractions interfere with student concentration.
Or maybe you’re having holiday parties.
Either way, I have several game ideas for you!
These activities are a great way to keep your students engaged and challenged.
I like holiday games to be academic or require students to do some thinking, so that’s what I’ll be sharing here:-)
## Holiday Game 1: Around the World
The first game on the list is “Around the World.”
This math activity is a great way to review math concepts. This game has been around for a LONG time, so I’m sure there are many variations. I’ll share the way we played.
When we played in 6th grade (and 5th and 4th), students stood in a big circle, and I held a math flash card up for the first 2 students.
• The student to answer first moved to stand by the next student in the circle, and I showed a new math card.
• The student who answered first moved on.
• The first student to get back to their starting point, or made it ‘around the world,’ was the winner (and if no one got back to the start, the student who moved farthest was the winner.)
You can use any types of math cards, depending on what you’d like students to practice. Basic operation flash cards are often the easiest to use, but if you’re working on other concepts that don’t require written work, those would also work well….like basic fraction identification or showing a fraction card and having students simplify it.
A few other Around the World topic suggestions:
• Fraction, Decimal, Percent conversions
• Vocabulary (you read the definition and students answer with the vocabulary term)
• Square roots
• Simple exponent problems
## Holiday Game 2: Christmas Carol Matching Game
This isn’t a math game, but could be used during math class if you’re having a holiday party.
This particular matching game uses the titles of Christmas carols, but you could definitely use other holiday names, key words, song titles, etc in the same way.
As you can see in the picture, students would match the song title with a phrase that represents the song.
For example, ‘The Twelve Days of Christmas’ matches with ‘A Dozen 24-Hour Holiday Time Periods.’
A second, more challenging version of the cards uses the ‘initials’ of the song. So instead of the ‘The Twelve Days of Christmas,’ students would match TTDOC with ‘A Dozen 24-Hour Holiday Time Periods.’
To use the cards as a memory/matching game:
• Put students into groups of 2-3
• Start with all of the cards spread out and facing down
• Have students take turns flipping over 2 cards at a time to see if they can find a match
• If the cards don’t match, they are flipped back over and the next student takes a turn
• When students find a match, they keep the cards and can take another turn
• The game is over when all matches have been found; the winner has the most matches
Figuring these out can be challenging, so you may want to pair students up as a ‘team’ and see which team can correctly match the phrases and the titles first, without the added difficulty of having the cards face down.
You could also use these cards as partnering cards when you want to randomly partner students, any time during the holiday season.
## Holiday Game 3:Truth or Dare Game
Truth or Dare is ALWAYS a favorite in math class, and using it during the holiday season is sure to keep students engaged!
The topic doesn’t need to be a holiday topic, but can be whatever math concept you’re teaching. You could even use concepts from earlier in the year so students get a review.
For more details about this game, check out the Task Cards With a Twist blog post.
## Holiday Game 4: Footloose
Footloose is also a favorite for any time of year, but at the holiday times, you can mix in some holiday-themed task cards that will still require students to use their math skills.
I have a few fall/winter sets of task cards that offer a mixed review of math concepts. These are perfect when you’ve finished one unit and don’t really want to begin something new, BUT you still want some engaging math practice as you head into the holidays.
Check out the Footloose post for more Footloose details!
Winter Footloose Task Cards
Fall Footloose Task Cards
What are your favorite holiday games for middle school?
## read next...
### 7 Ways to Practice Multiplication Facts in Middle School
Welcome to Cognitive Cardio Math! I’m Ellie, a wife, mom, grandma, and dog ‘mom,’ and I’ve spent just about my whole life in school! With nearly 30 years in education, I’ve taught:
• All subject areas in 4th and 5th grades
• Math, ELA, and science in 6th grade (middle school)
I’ve been creating resources for teachers since 2012 and have worked in the elearning industry for about five years as well!
If you’re looking for ideas and resources to help you teach math (and a little ELA), I can help you out!
## LET'S CONNECT
##### Archives
Select the image above to learn more!
Engage students in taking math notes with this FREE Fraction Operations wheel and 3 wheel templates!
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# Dinamic prior based on categorical information
Hello everyone,
I’m new to Stan, and looking at the blog or documentation, I haven’t found the answer to my question.
I would like to have a simple linear model with one independent variable. I can’t measure exactly the independent variable values, but I can put them in bins and I can estimate a rough distribution for these bins.
What is the best way to do that in stan?
I was thinking at something like:
``````data {
int N;
vector[N] dependent_oserved;
vector[N] independent_cat;
}
parameters {
real<lower=0> b0;
real b1;
vector<lower=0, upper=1>[N] independent;
vector[N] dependent;
}
model {
b0 ~ normal(0,10);
b1 ~ normal(0,10);
for (i in 1:N) {
if (independent_cat[i] == 'factor_1') {
independent[i] ~ normal(1,0.45);
}
if (independent_cat[i] == 'factor_2') {
independent[i] ~ normal(1,5);
}
}
dependent_oserved ~ normal(dependent,.5);
dependent ~ normal(b0 + b1*independent, 10);
}"
``````
But how could I implement this in stan? Or is there a better way to approach this problem?
Best regards
Luca
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# math
posted by .
A cell phone company charges each customer a base rate of \$15.00 per month for its services. The company also charges each customer \$0.02 per minute used. Which algebraic expression represents each customer's monthly charge for m minutes?
A.15 (0.02m)
B.15 + 0.02m
C.15m + 0.02
D.0.02m - 12
• math -
Please do not post any more questions unless you also post what YOU think is the answer!
• math -
okay. I think the answer is b?
• math -
Right! :-)
• math -
Felipe has saved \$625. He is starting a part-time job that pays him \$150 a week, and he plans to save all of the money he earns. Which algebraic expression represents Felipe's total savings after x weeks?
• math -
B definitely
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Home / Power Conversion / Convert Foot Pound-force/second to Kilocalorie (th)/minute
# Convert Foot Pound-force/second to Kilocalorie (th)/minute
Please provide values below to convert foot pound-force/second to kilocalorie (th)/minute, or vice versa.
From: foot pound-force/second To: kilocalorie (th)/minute
### Foot Pound-force/second to Kilocalorie (th)/minute Conversion Table
Foot Pound-force/secondKilocalorie (th)/minute
0.01 foot pound-force/second0.000194429 kilocalorie (th)/minute
0.1 foot pound-force/second0.0019442896 kilocalorie (th)/minute
1 foot pound-force/second0.019442896 kilocalorie (th)/minute
2 foot pound-force/second0.038885792 kilocalorie (th)/minute
3 foot pound-force/second0.058328688 kilocalorie (th)/minute
5 foot pound-force/second0.09721448 kilocalorie (th)/minute
10 foot pound-force/second0.1944289601 kilocalorie (th)/minute
20 foot pound-force/second0.3888579202 kilocalorie (th)/minute
50 foot pound-force/second0.9721448004 kilocalorie (th)/minute
100 foot pound-force/second1.9442896009 kilocalorie (th)/minute
1000 foot pound-force/second19.4428960085 kilocalorie (th)/minute
### How to Convert Foot Pound-force/second to Kilocalorie (th)/minute
1 foot pound-force/second = 0.019442896 kilocalorie (th)/minute
1 kilocalorie (th)/minute = 51.4326672097 foot pound-force/second
Example: convert 15 foot pound-force/second to kilocalorie (th)/minute:
15 foot pound-force/second = 15 × 0.019442896 kilocalorie (th)/minute = 0.2916434401 kilocalorie (th)/minute
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# Chemistry
From the information below, identify element X.
a. The wavelength of the radio waves sent by an FM station
broadcasting at 97.1 MHz is 30.0 million (3.00 e7) times
greater than the wavelength corresponding to the energy difference
between a particular excited state of the hydrogen
atom and the ground state.
b. Let V represent the principal quantum number for the valence
shell of element X. If an electron in the hydrogen atom
falls from shell V to the inner shell corresponding to the excited
state mentioned above in part a, the wavelength of light
emitted is the same as the wavelength of an electron moving
at a speed of 570. m/s.
c. The number of unpaired electrons for element X in the ground
state is the same as the maximum number of electrons in an
atom that can have the quantum number designations n = 2,
mL = -1, and mS = -1/ 2.
d. Let A equal the charge of the stable ion that would form when
the undiscovered element 120 forms ionic compounds. This
value of A also represents the angular momentum quantum
number for the subshell containing the unpaired electron(s)
for element X.
posted before, but still need help. I did a and b already with help.
C. All i know for c is that n=2 means 2 possible subshells, mL=-1 means that L must ne 1 so it's a P subshell, so there are 6 electrons max so far, right?, but what about the ms=-1/2, what would that do?
yes i completely agree. i went through the entire part c and d myself and came up wit h platinum as my final answer.. i am not sure though. so c is just that there is only on unpaired electron
now part d says something about element 120 so i figured that was the atomic number, which means that the 120 element would be in group II which means it would have 2 valence electrons and a charge of +2 as an ion making the angular momentum (m)of element X = 2
Here is the way I interpret c. Check my thinking.
This part c is JUST for the determination of the number of unpaired electrons. If n=2, then L can be 0 or 1. If L = 0, theh m can't be -1 so L = 0 is out and L must be = 1 in this part. Given that m = -1 and s = -1/2 (m is ml and s is ms) and L = 0, I read that to mean that there can be ONLY the one electron. The ONLY other configuration that will give N= 2, L = 1, ml = -1 is for ms to be +1/2 and that is specifically excluded by the problem stating that ms= -1/2. So I conclude that the number of unpaired electrons is 1.
(This says nothing about any other part of the element EXCEPT the number of unpaired electrons; that is, nothing about the valence shell--we already know N=5) or the number of electrons, either total or in any of the subshells. Don't be afraid to disagree. Check my thinking carefully.
is that remotely correct
I agree with the element #120 being in group II AND with the ion being +2 which also means that the angular momentum is 2. That means, although you don't say explicitly, that L = 2. So we are looking for an element with the valence shell of N=5, 1 unpaired electron in the L=2 subshell, a 5s2 configuration for that N=5 valence shell. Pt has a 5d9 6s1 so I don't think that is right because the unpaired electron is an s electron AND the outside valence shell is N=6 and not N=5. I think the problem states that the unpaired electron is an L = 2 or a d electron and we must have N=5. So now what do you think?
I didn't intend to bold all of that but I still think what I wrote makes sense, even with the bolding which I meant to cut off after the word subshell in line 4.
because of the m=2 I thought that L could be either 2 or 3 (d or f) and i thought that m=2 meant that the unpaired electron had to be in the d shell
i do understand what you are saying about platinum and 5s2 though
I am confused about your statement of m=2. ml = -1 in part c is only for the purpose of determining the number of unpaired electrons. We know now that is 1 so that section is history. Part d states the angular momentum of the unpaired electron is 2 so L must be 2 for the subshell in which the unpaired electron is found and that makes it a d electron.
no, I wasn't talking about part c. I was saying that I thought that if the angular momentum was 2 that L could be 2 OR 3 because if L is 3 then the angular momentum can also be 2. But regardless I thought that the unpaired electron was located in the d shell.
The angular momentum of the atom is determined by the azimuthal quantum number, which in this problem, according to part d, is 2. So L = 2. L is the azimuthal quantum number. In every day English, it determines the ellipticity of the orbitals.
ok, I understand now. So where is the unpaired electron located? was I right about the d shell?
I see what you are saying. I think, however, that if it says the angular momentum is 2 then it must be 2 and we can't assume it COULD be 2 OR 3. For that matter, if it COULD be 3 then it might also be 1, or zero, since an atom with n=5 can have values for L of 0, 1, 2, or 3. I believe this whole part d is for the distinct purpose of telling us that L = 2 AND that the unpaired electron is in that particular subshell.
ok, so if it has a valence configuration of 5s2 and an unpaired electron in d then the only element I see with that configuration is Y, yttrium, which has a configuration of 4d1 5s2
I think it is Y. But I don't want to talk you into anything you don't believe. This is the first time I've seen this problem. However, I believe our reasoning is correct. This may be the most difficult railroad problem I have seen. And I've seen many.
I agree with it being Y! it's the only thing that make sense. Thanks a bunch for your help!! oh I redid the other problem about the A,B,C and I got 0.537 atm as the total pressure.
That's what I obtained also for the total P. That problem wasn't a piece of cake, either, but it was easier than this radio problem. The radio problem was stated so confusingly AND it jumped around from pillar to post. I had trouble understand exactly what the sentence said. Anyway, thanks for using Jiskha and I'm glad I was able to help. Good luck. Come back anytime.
Thank you very very much!!!!!!!!
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# A Rocket
## Homework Statement
A Saturn V rocket has a mass of 2.75x10^6 Kg and exerts a force of 3.3x10^6N on the gases it expels.
determine the initial vertical acceleration of the rocket.
2. The attempt at a solution
heres my attempt...
F=3.3x10^6N
Fg=
M=2.75x10^6
Fnet-MA
F-Fg=ma
F-mg=ma
3.3x10^6-(2.75x10^6)(9.8)=2.75x10^6a
3.3x10^6-2.6x10^7=2.75x10^6a
-2.2x10^7=2.75x10^6a
-2.2x10^7/2.75x10^6=a
a=-8
The accual answer is 2.2m/s^2
Can someone explain to me what i got wrong?
## Answers and Replies
ideasrule
Homework Helper
Are you sure it's 3.3x10^6N and not 3.3*10^7 N? You get the correct answer with 3.3*10^7 N.
Are you sure it's 3.3x10^6N and not 3.3*10^7 N? You get the correct answer with 3.3*10^7 N.
No im sure its that
this question is straight from a text book.
ideasrule
Homework Helper
In that case, the textbook made a mistake.
In that case, the textbook made a mistake.
It must have!
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<
>
# Detroit Tigers' lineup problems
As we sit here today, the Detroit Tigers sit two and a half games behind the Chicago White Sox in the AL Central. Now, whether you anticipated the Men from Motown to be in second place this deep into the season or not, the reality is the Tigers, having landed the largest free-agent fish of the offseason (allegedly) are grasping for the division crown from behind the eight ball and there are a number of reasons for this.
At 4.56 runs per game, the offense has been an overall disappointment, one of the reasons that they haven’t been better than a good White Sox team. But that’s simply a blanket response to an in-depth and complicated game.
So we’ll dig a bit deeper. We’ll take a look at why the team, which still has scored the sixth-most runs in the American League, has scored less than expected. Is it just because expectations were too high? Or when we dig deeper will we find a glaring reason?
Tigers have two glaring weaknesses
They may have quality production coming from the first four spots in the lineup as well as the last three, but in the fifth and sixth spots in the lineup, they are less than pedestrian.
#### Fifth Spot In Lineup
Stats through Friday:
#### Sixth Spot In Lineup
Stats through Friday:
In the two charts, we see how the entire American League has batted from the fifth and sixth spots in the batting order this season.
We see that the Tigers don’t actually have the worst production out of the fifth spot, owned almost exclusively by Delmon Young. That honor belongs to the Kansas City Royals, but they have a very respectable amount of production out of the sixth spot, and the Tigers certainly don’t. Here we use simple math. If X is smaller than Y but much greater than Y in the next problem, and the two combinations of X are overall greater than the combinations of Y than X is greater. That’s how you use simple math, right?
The same can be said for the sixth spot in the lineup. The Tigers are bad, the bulk of which is courtesy of Brennan Boesch and Alex Avila, but they aren’t the least productive. That’s the Seattle Mariners. But the Mariners are not exactly dead at the fifth spot in the lineup, as we see. Math, again! The Mariners, who play in an offensive black hole for half the time still produce better from two of the more important spots in the lineup than the Tigers do.
In fact, the team that lacks as much production as the Tigers do from both those spots are the Toronto Blue Jays, who have experienced a very disappointing season from their offense.
But even the Blue Jays haven’t been as bad as the Tigers have been from the fifth and sixth spots in the lineup. We can’t assume we know the number of wins that the ineffectiveness of Young, Boesch, and Avila have denied the Tigers, but it certainly looked like the Tigers would see at least league average-production from each player. Two and a half wins certainly doesn’t seem to be out of reach given the fact that they’ve already played three-quarters of the season.
There is currently very little the team can change.
The problem the Tigers now face is overcoming these inefficiencies. Where do you take production from? A team can’t exactly pull it from the seventh and eighth spots in the lineup, can they?
The Tigers have actually tried. Recently, they’ve begun moving Jhonny Peralta around with the aforementioned names in hopes that something, anything, can be sparked. So far the results have been mixed. The month of August has seen the worst production of the year out of the fifth spot and the very best out of the sixth spot.
The solution is an impact corner outfield bat that general manager Dave Dombrowski can pick up on waivers. That’s obviously an easy answer, so we should all expect that to happen in the very near future.
No, the solution is the players have to produce. The issue is they haven’t all year.
It may keep the team out of the playoffs, it may not, but the lack of production certainly isn’t making life any easier for anyone involved.
Josh Worn writes about the Tigers at Walkoff Woodward.
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# Thread: trig derivative
1. ## trig derivative
what it the derivative of $\displaystyle (cosX)^3$ using rules
Is it as easy as $\displaystyle 3(-sinX)2 =-3sin^2X$ ?
2. ## Re: trig derivative
Originally Posted by delgeezee
what it the derivative of $\displaystyle (cosX)^3$ using rules
Is it as easy as $\displaystyle 3(-sinX)2 =-3sin^2X$ ?
$\displaystyle \frac{d}{dx} (\cos{x})^3 = 3(\cos{x})^2 \cdot (-\sin{x})$
3. ## Re: trig derivative
You need to use the chain rule here
Find $\displaystyle (\cos^3x)'$
Let's make $\displaystyle y=\cos^3x$ and $\displaystyle u=\cos x\implies y=u^3$
So $\displaystyle \frac{dy}{du} = 3u^2$ and $\displaystyle \frac{du}{dx} = -\sin x$
By the chain rule $\displaystyle \frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}$
Can you finish it?
4. ## Re: trig derivative
Chain rule. 3(cosx)^2(-sinx)
5. ## Re: trig derivative
Thanks, I understand now.
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# Problem reduction in artificial intelligence javatpoint
P
Problem-solving is a fundamental task of artificial intelligence (AI). It involves finding solutions to complex issues by using various techniques. One of the crucial methods used in problem-solving is problem reduction or problem minimization.
Problem reduction aims at breaking down a complex problem into simpler sub-problems. By doing so, it becomes easier to analyze and solve the problem at hand. This technique helps AI systems tackle the multiple challenges that arise during the problem-solving process.
Javatpoint, a renowned platform for learning and exploring AI and other technologies, provides comprehensive knowledge about problem reduction in artificial intelligence. Whether you are a beginner or an experienced professional, Javatpoint offers in-depth tutorials and resources to help you understand the concept of problem reduction and apply it effectively in your AI projects.
## Problem Reduction in Artificial Intelligence
Problem reduction is a fundamental concept in problem-solving in artificial intelligence (AI). It involves the minimization of a problem by breaking it down into smaller sub-problems, which can be solved individually. This approach is used in various AI techniques to simplify complex problems and make them more manageable for AI algorithms.
### Minimization of Problems
In problem reduction, the main objective is to reduce the complexity of a problem by analyzing its structure and identifying sub-problems. By breaking down a problem into smaller parts, AI algorithms can focus on solving each sub-problem independently, which makes the overall problem-solving process more efficient.
The minimization of problems through problem reduction techniques is particularly useful in AI applications that deal with complex and large-scale problems. For example, in natural language processing, AI algorithms can use problem reduction to break down a sentence into smaller units, such as words or phrases, and analyze each unit separately to understand the meaning of the sentence as a whole.
### Problem Reduction Techniques
There are several problem reduction techniques used in AI, including:
• Divide and conquer: This technique involves dividing a problem into smaller sub-problems, solving each sub-problem independently, and then combining the results to solve the overall problem.
• Subgoal decomposition: This technique involves decomposing a problem into multiple subgoals, each representing a smaller component of the overall problem. AI algorithms can then focus on achieving each subgoal one by one, which eventually leads to solving the entire problem.
• Abstraction: This technique involves abstracting away irrelevant details of a problem and focusing only on the essential aspects. By reducing the problem to its core elements, AI algorithms can simplify the problem-solving process.
These problem reduction techniques are essential tools in the field of artificial intelligence, enabling AI algorithms to tackle complex problems effectively. By breaking down problems and focusing on smaller sub-problems, AI algorithms can find innovative solutions and make progress in various areas of AI research and development.
## Problem-solving Techniques in AI
Problem-solving is a fundamental aspect of artificial intelligence (AI) that aims to find solutions to complex issues using computational methods. In the field of AI, various problem-solving techniques have been developed to tackle different types of problems.
### Problem Minimization
One of the main goals in problem-solving is to minimize the problem at hand. This involves breaking down a complex problem into smaller, more manageable sub-problems. By doing so, it becomes easier to analyze and solve each sub-problem individually, leading to an overall solution for the original problem.
Problem minimization is often achieved through a technique called problem reduction. This technique involves transforming a given problem into a simpler or more well-defined problem that can be solved more easily. Problem reduction can be done by identifying and removing unnecessary elements or constraints from the original problem.
### Techniques for Problem-solving in AI
In AI, there are various problem-solving techniques that are commonly employed, depending on the nature of the problem. Some of the widely used techniques include:
• Breadth-First Search (BFS): This technique explores all possible solutions in a systematic manner, starting from the initial state and moving level by level. It guarantees finding the shortest path to the goal state, but it can be computationally expensive for large search spaces.
• Depth-First Search (DFS): Unlike BFS, DFS explores the search space by going as far as possible along each branch before backtracking. It is often used when the search space is infinite or the goal state is located deep in the search tree.
• A* Search: A* is an informed search algorithm that uses both heuristics and cost to determine the best path to the goal state. It combines the advantages of both BFS and DFS, making it efficient and effective in finding optimal solutions.
These problem-solving techniques, along with many others, play a crucial role in solving complex problems in the field of artificial intelligence. By applying these techniques, AI systems can effectively solve a wide range of problems and provide valuable solutions.
## Problem Reduction at Javatpoint
Problem reduction is a fundamental technique in artificial intelligence (AI) for problem-solving. It involves the minimization of a complex problem into smaller, more manageable sub-problems. At Javatpoint, we understand the importance of problem reduction in the field of AI and offer comprehensive resources to help you tackle the most challenging issues.
Problem reduction is a key concept in problem-solving, as it allows AI systems to break down complex problems into smaller, more manageable parts. By dividing a problem into sub-problems, AI systems can apply specific algorithms and techniques to each sub-problem, ultimately enabling a more efficient and effective solution to the larger problem at hand.
At Javatpoint, we provide in-depth knowledge and guidance on problem reduction in AI. Our resources cover various techniques and strategies for problem reduction, including heuristic search, constraint satisfaction, and logical inference. We aim to equip learners with the necessary tools to analyze and decompose complex problems, making them easier to solve.
Furthermore, our experts at Javatpoint emphasize the importance of problem reduction in real-world applications of AI. Whether it’s optimizing resource allocation in supply chain management or improving decision-making processes in healthcare, problem reduction plays a critical role in addressing complex challenges.
With the support of Javatpoint’s resources on problem reduction, you can develop a deep understanding of the concepts and techniques involved in minimizing the complexity of problems in the field of AI. Stay connected with Javatpoint to stay ahead in the ever-evolving world of artificial intelligence!
## Javatpoint’s Approach to AI Problem-solving
Javatpoint is a leading platform for learning and implementing Artificial Intelligence (AI) technologies. We believe that efficient problem-solving is at the core of AI development, and we have developed a unique approach to address various problem-solving issues.
Problem reduction is a key concept in AI, and Javatpoint focuses on this technique to solve complex problems in the field of artificial intelligence. By breaking down a large problem into smaller, more manageable sub-problems, we are able to find effective solutions. This approach allows us to tackle the most challenging AI problems.
At Javatpoint, we understand that the complexity of AI problems can be overwhelming. Therefore, we provide comprehensive guidance and resources to help learners grasp the fundamental concepts of problem reduction. Our AI tutorials and learning materials are designed to simplify the understanding of problem-solving techniques, making it easier for individuals to apply them in real-world scenarios.
Our team at Javatpoint is composed of experienced AI professionals who have expertise in problem-solving using artificial intelligence algorithms. We constantly update and refine our problem-solving techniques to keep up with the latest developments in the field. This ensures that learners are equipped with cutting-edge knowledge and skills to effectively solve AI problems.
With Javatpoint’s approach to AI problem-solving, learners can gain a deep understanding of the underlying principles and techniques, enabling them to tackle complex problems with confidence. Our aim is to empower individuals with the necessary tools and knowledge to make a significant impact in the field of artificial intelligence.
## Benefits of Problem Reduction in AI
Artificial Intelligence (AI) is at the forefront of solving a variety of issues across different domains. In problem-solving, one important technique is problem reduction, which involves breaking down complex problems into simpler subproblems. Javatpoint is a leading platform that provides comprehensive tutorials and resources for AI development.
### Enhanced Efficiency
Problem reduction in AI enhances efficiency by tackling complex problems in a step-by-step manner. By breaking down a problem into smaller subproblems, AI systems can focus on solving each subproblem individually, optimizing computational resources and reducing the overall time required for problem-solving.
### Improved Problem Solving
By reducing a problem into smaller subproblems, AI systems can gain a deeper understanding of each individual component. This enables them to solve each subproblem efficiently, leading to a more comprehensive solution for the overall problem. With problem reduction, AI systems can approach problem-solving in a systematic and structured manner, resulting in improved accuracy and effectiveness.
Benefits of Problem Reduction in AI
Enhanced Efficiency
Improved Problem Solving
## Problem Minimization in AI at Javatpoint
In the field of artificial intelligence, problem-solving is a fundamental aspect. AI aims to replicate human intelligence by developing algorithms and systems that can solve complex problems. However, not all problems can be solved optimally due to various issues such as computational limitations or lack of complete information.
Problem minimization plays a crucial role in AI as it focuses on reducing the complexity and size of problems to make them more manageable. The goal is to find an approximation or an acceptable solution that meets the requirements and constraints of the problem, even if it is not the best possible solution.
Javatpoint is a platform that offers comprehensive resources and tutorials on AI, including problem minimization techniques. By understanding and implementing problem minimization in AI, developers can optimize their algorithms and systems to achieve more efficient and effective results.
Problem minimization involves various approaches, such as heuristic search algorithms, constraint satisfaction, and problem decomposition. These techniques aim to break down complex problems into smaller subproblems that are easier to solve. By reducing the problem size, AI systems can save computational resources and time, making them more feasible for real-world applications.
However, it is important to note that problem minimization does not guarantee an optimal solution. In some cases, the approximation or solution obtained may not be the best possible outcome. Nevertheless, problem minimization techniques enable AI systems to overcome limitations and address complex problems that would otherwise be infeasible to solve.
At Javatpoint, developers can explore various problem minimization approaches and learn how to implement them in AI systems. By leveraging the resources and tutorials provided, developers can enhance their problem-solving capabilities and develop more efficient AI solutions.
In conclusion, problem minimization is a crucial aspect of AI at Javatpoint. By reducing the complexity and size of problems, AI systems can overcome limitations and generate acceptable solutions. Through Javatpoint’s comprehensive resources, developers can learn and implement problem minimization techniques to optimize their AI algorithms and systems.
## How Javatpoint Reduces Issues in Artificial Intelligence
Artificial Intelligence (AI) has revolutionized many industries by providing solutions to complex problems. However, implementing AI can bring forth several issues that need to be addressed for optimal performance. Javatpoint understands these challenges and offers advanced problem-solving techniques to minimize these issues.
In the field of AI, numerous issues can arise, such as:
Issues Reduction Techniques
Data Limitations Javatpoint provides comprehensive training on data collection and preprocessing techniques to ensure that AI models have access to high-quality and diverse datasets.
Algorithm Selection Javatpoint offers a wide range of courses and tutorials that cover various AI algorithms. This allows developers to select the most appropriate algorithm for their specific problem, minimizing potential issues.
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Ethical Concerns Javatpoint emphasizes the importance of ethical considerations in AI development. Through courses and discussions, it educates developers on ethical guidelines and best practices, reducing ethical issues in AI implementation.
Javatpoint stands out in the field of AI education and training by providing comprehensive resources and support to address the various issues that can arise in artificial intelligence implementation. With its emphasis on problem reduction and minimization, Javatpoint equips developers with the knowledge and skills necessary to overcome challenges and achieve optimal results in AI projects.
## AI Problem-solving Tools at Javatpoint
Artificial Intelligence (AI) is revolutionizing various industries and domains, and problem-solving is one of the key areas where AI is making a significant impact. Javatpoint, a leading online platform for technical tutorials and resources, offers a range of AI problem-solving tools that can help developers and researchers tackle complex problems more efficiently.
### Problem Minimization
One of the primary challenges in problem-solving is to reduce the complexity of the problem. Javatpoint’s AI problem-solving tools provide algorithms and techniques for minimizing the problem, making it more manageable and easier to solve. These tools help in identifying redundant information, eliminating irrelevant factors, and simplifying the problem statement.
### Problem Reduction
Problem reduction is another crucial aspect of AI problem-solving. Javatpoint offers tools that focus on reducing complex problems into simpler subproblems. These tools break down the problem into smaller parts, which can be solved independently or combined to find a solution to the original problem. Problem reduction techniques improve problem-solving efficiency and enable faster convergence towards a solution.
The AI problem-solving tools at Javatpoint cover a wide range of problem domains, including optimization, planning, decision-making, and pattern recognition. These tools leverage advanced algorithms and machine learning techniques to provide accurate and efficient solutions to complex problems.
Whether you are a student, professional, or researcher in the field of artificial intelligence, Javatpoint’s AI problem-solving tools can be immensely helpful in your journey. With these tools, you can leverage the power of AI to tackle complex problems and find innovative solutions.
In conclusion, Javatpoint’s AI problem-solving tools offer a comprehensive set of resources for developers and researchers seeking effective solutions to complex problems. These tools not only provide algorithms for problem minimization and reduction but also cover a wide range of problem domains. Explore Javatpoint’s AI problem-solving tools and unlock the potential of artificial intelligence in problem-solving.
## Problem Reduction Strategies in AI
Problem-solving in the field of artificial intelligence (AI) involves dealing with complex and challenging issues. One of the key techniques used in AI problem-solving is problem reduction, which focuses on minimizing the scope of a problem by breaking it down into smaller, more manageable sub-problems.
### What is Problem Reduction?
Problem reduction is a fundamental concept in AI that aims to simplify complex problems by decomposing them into smaller parts. It involves breaking down a problem into sub-problems that are easier to solve individually, before combining the solutions to obtain the solution to the original problem.
### Strategies for Problem Reduction
There are several strategies for problem reduction in AI:
• Divide and Conquer: This strategy involves dividing a problem into smaller sub-problems, solving them independently, and then combining their solutions to obtain the solution to the original problem.
• Functional Decomposition: In this strategy, a problem is decomposed based on the functions or tasks involved. Each function or task is treated as a sub-problem, which can be solved separately and integrated to solve the overall problem.
• Data Decomposition: This strategy involves decomposing a problem based on the available data. Each subset of data is processed independently, and the results are combined to obtain the final solution.
• Sequential Decomposition: In this strategy, a problem is decomposed into a sequence of smaller sub-problems that are solved one after the other. The solution to each sub-problem is used as input for the next sub-problem, ultimately leading to the solution of the original problem.
These problem reduction strategies play a crucial role in simplifying complex AI problems and enabling efficient problem-solving. By breaking down problems into smaller parts, AI systems can handle larger and more challenging tasks and improve their overall problem-solving capabilities.
## Case Studies on Problem Reduction in AI
In the field of artificial intelligence (AI), problem reduction is a key technique used to solve complex problems. By breaking down a large problem into smaller, more manageable subproblems, AI systems can analyze and solve each subproblem separately, leading to a overall solution. This approach, also known as problem minimization, has proven to be effective in tackling various issues in AI.
### 1. Natural Language Processing
One area where problem reduction techniques are widely used is in natural language processing (NLP). NLP involves the interaction between computers and human language, and it includes tasks such as speech recognition, machine translation, and sentiment analysis. Problem reduction allows NLP systems to break down complex language processing tasks into smaller components, such as tokenization, part-of-speech tagging, and syntactic parsing, making it easier to handle and analyze linguistic data.
### 2. Computer Vision
Computer vision is another domain where problem reduction plays a crucial role. Computer vision involves processing and interpreting visual data, such as images and videos. By breaking down the complex problem of visual understanding into smaller tasks, such as image segmentation, object recognition, and tracking, AI systems can better analyze and make sense of visual information. Problem reduction techniques help in reducing the computational complexity and improving the efficiency and accuracy of computer vision systems.
In conclusion, problem reduction techniques have proven to be valuable in various domains of artificial intelligence. They allow AI systems to break down complex problems into smaller, more manageable subproblems, enabling effective analysis and solution. This approach has been successfully applied in natural language processing, computer vision, and many other areas of AI, making problem reduction an essential tool in the field.
## Examples of AI Problem Reduction at Javatpoint
Artificial intelligence (AI) problem-solving requires efficient techniques to tackle complex issues. AI problem reduction is an essential approach that aims at minimizing problems to make them easier to solve. Javatpoint, a leading platform in providing AI solutions, offers several examples of problem reduction techniques that effectively address various challenges in artificial intelligence:
• State Space Minimization: In AI, state space refers to the set of all possible configurations that a problem can take. State space minimization involves reducing the number of states to be explored during problem-solving, which helps in improving the efficiency and speed of the AI system.
• Search Space Pruning: Search space pruning is a technique used to eliminate irrelevant or unnecessary branches in the search tree. By pruning unfruitful paths, the AI system can focus its efforts on more promising solutions. This approach helps in reducing the search space, saving computational resources, and accelerating the problem-solving process.
• Constraint Propagation: Constraint satisfaction problems (CSPs) often involve a set of constraints that must be satisfied. Constraint propagation is a technique that reduces the problem space by eliminating inconsistent values from the domains of variables based on the given constraints. This helps in narrowing down the potential solutions and speeding up the problem-solving process.
• Abstraction and Generalization: Abstraction and generalization involve simplifying complex problems by focusing on essential aspects and ignoring irrelevant details. By abstracting and generalizing the problem, AI systems can reduce the complexity and size of the problem space, making it easier to find solutions.
• Heuristics and Approximations: Heuristics and approximations are techniques used to find solutions quickly without guaranteeing optimality. These methods trade off accuracy for efficiency by providing approximate solutions that are still acceptable. The use of heuristics and approximations helps in reducing computation time and resources required for problem-solving.
These are just a few examples of AI problem reduction techniques offered by Javatpoint. By leveraging such methods, artificial intelligence can effectively tackle a wide range of complex issues and improve the overall problem-solving capabilities.
## Problem Reduction Techniques in AI
In the field of artificial intelligence, problem reduction techniques are widely used to solve complex problems. These techniques help break down a problem into smaller, more manageable subproblems, making it easier to find a solution.
### Importance of Problem Reduction
The use of problem reduction techniques is essential in AI as it allows for efficient problem-solving. By breaking down a problem into smaller parts, it becomes easier to understand and analyze each component, reducing the complexity of the overall problem.
Problem reduction also helps in the identification and isolation of specific issues within a problem. It allows AI systems to focus on resolving individual subproblems in order to reach a solution for the larger problem.
### Minimization of Problem Complexity
Problem reduction techniques play a crucial role in minimizing the complexity of a problem. By breaking down a problem into smaller subproblems, the overall complexity is reduced, making it easier for AI systems to find a solution in a timely manner.
Javatpoint is a well-known platform that offers comprehensive resources and tutorials on artificial intelligence, including problem reduction techniques. It provides detailed explanations and examples to help developers and AI enthusiasts understand and implement problem reduction effectively.
• Example 1: Reduction of a complex optimization problem to a series of simpler subproblems
• Example 2: Breaking down a planning problem into smaller steps
• Example 3: Decomposing a game-playing problem into smaller decision-making tasks
In conclusion, problem reduction techniques are an integral part of artificial intelligence. They play a crucial role in breaking down complex problems, minimizing their complexity, and allowing for more efficient and effective problem-solving. With the resources and tutorials available at Javatpoint, developers can enhance their understanding of these techniques and apply them in various AI applications.
## Challenges in Problem Reduction for AI
In the field of artificial intelligence (AI), problem reduction is a fundamental approach to problem-solving. It involves breaking down a complex problem into simpler subproblems that can be solved independently. However, there are several challenges that researchers and practitioners face when applying problem reduction techniques to AI problems. In this article, we will discuss some of these challenges and how they can be addressed.
### Lack of problem decomposition
One of the main challenges in problem reduction for AI is the lack of a clear and effective way to decompose a problem into subproblems. Not all problems can be easily divided into smaller, manageable parts. Some problems may have interdependent or overlapping subproblems, making it difficult to determine the optimal decomposition. This challenge requires researchers to develop novel approaches and algorithms for problem decomposition in AI.
### Reducing problem size
Another challenge in problem reduction is the minimization of problem size. As problems become more complex, the number of subproblems and the size of each subproblem can increase exponentially. This can lead to computational inefficiency and difficulty in finding optimal solutions. Researchers need to develop strategies to effectively reduce the problem size while still preserving the essential characteristics of the problem. This could involve techniques such as feature selection or dimensionality reduction.
Challenge Solution
Lack of problem decomposition Develop novel approaches and algorithms for problem decomposition
Reducing problem size Use techniques like feature selection or dimensionality reduction
In conclusion, problem reduction in AI presents several challenges that need to be addressed. Researchers and practitioners must find ways to effectively decompose problems and reduce their size to ensure efficient and optimal problem-solving. By overcoming these challenges, we can advance the field of artificial intelligence and its applications in various domains.
## Future Trends in Problem Reduction for AI
Problem-solving has always been an integral part of artificial intelligence (AI). As AI continues to advance, new challenges and issues arise in the field of problem reduction and minimization.
One future trend in problem reduction for AI is the development of more efficient algorithms and techniques. Researchers are constantly working on improving existing algorithms and creating new ones that can solve problems faster and with greater accuracy. This is important because as AI systems become more complex and deal with larger data sets, the traditional problem-solving approaches may no longer be sufficient.
Another trend is the integration of machine learning and AI. Machine learning algorithms are able to learn from data and improve their performance over time. By integrating machine learning techniques into problem reduction, AI systems can adapt and evolve to better solve complex problems. This can lead to more effective problem-solving and improved decision-making capabilities.
Additionally, there is a growing focus on addressing the problem of bias in AI systems. As AI becomes more prominent in various sectors, it is crucial to ensure that these systems are fair and unbiased. Researchers are developing techniques to detect and mitigate bias in AI algorithms, with the aim of creating more ethical and responsible problem-solving solutions.
Furthermore, the future of problem reduction in AI involves exploring the potential of hybrid approaches. By combining different problem-solving techniques, such as symbolic reasoning and neural networks, AI systems can leverage the strengths of different approaches to solve complex problems. This interdisciplinary approach has the potential to significantly enhance problem-solving capabilities in AI.
In conclusion, the future of problem reduction in AI holds exciting possibilities. With advancements in algorithms, the integration of machine learning, the focus on bias detection and minimization, and the exploration of hybrid approaches, AI systems will continue to evolve and improve in their problem-solving abilities. Javatpoint remains at the forefront of these developments, providing resources and insights to help professionals navigate the ever-changing landscape of AI problem reduction.
## Applications of Problem Reduction in AI
The intelligence, minimization of issues, and problem-solving capabilities of artificial intelligence make it a valuable tool in various fields. Problem reduction, a fundamental concept in AI, plays a crucial role in tackling complex problems and improving efficiency. In this article, we will explore some of the applications of problem reduction in AI.
### 1. Resource Allocation
One of the key applications of problem reduction in AI is resource allocation. In many industries and organizations, there is a limited set of resources available, and these resources need to be allocated effectively to maximize output. Problem reduction techniques can be used to break down the resource allocation problem into smaller, more manageable sub-problems. By solving these sub-problems individually, AI systems can optimize resource allocation and minimize wastage.
### 2. Planning and Scheduling
Another area where problem reduction is widely used is planning and scheduling. AI systems can help in creating efficient schedules for tasks and activities by breaking the overall planning problem into smaller sub-problems. By reducing the complexity of the problem, AI algorithms can find optimal solutions quickly and effectively. This is particularly useful in industries such as manufacturing, logistics, and transportation, where planning and scheduling play a critical role in operations.
By utilizing problem reduction techniques, AI systems can effectively solve a wide range of complex problems. Whether it is resource allocation, planning, or any other problem-solving domain, problem reduction in artificial intelligence offers significant benefits in terms of efficiency and effectiveness.
For more information on problem reduction and other AI-related topics, visit Javatpoint.com.
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Introduction to Problem Reduction in AI
Benefits of Problem Reduction in AI
Problem Reduction Techniques in AI
## Problem Reduction Research at Javatpoint
Problem reduction is a fundamental aspect of artificial intelligence (AI) and plays a crucial role in problem-solving. At Javatpoint, we conduct extensive research on problem reduction techniques to tackle the complex issues of minimization in AI.
Artificial intelligence is built upon the foundation of problem-solving, where various algorithms and methodologies are employed to find optimal solutions. However, many real-world problems are too complex to be solved directly, requiring the use of problem reduction techniques.
Problem reduction involves breaking down a complex problem into smaller, more manageable subproblems. By decomposing the problem into smaller parts, we can focus on solving each subproblem individually and then combine the solutions to obtain an overall solution.
At Javatpoint, we aim to develop efficient problem reduction algorithms that can effectively handle the diversity of problem domains. Our research focuses on identifying common patterns and structures in problems that allow for effective reduction strategies.
The application of problem reduction in AI is not limited to a specific field or domain. It can be applied to a wide range of problems, including optimization, planning, scheduling, constraint satisfaction, and more. By applying problem reduction techniques, we can simplify complex problems and improve the efficiency and accuracy of AI systems.
Our team at Javatpoint is dedicated to exploring innovative problem reduction approaches and integrating them into practical AI applications. Through our research, we aim to contribute to the advancement of problem-solving in artificial intelligence and address the challenges faced by AI systems.
Benefits of Problem Reduction Research at Javatpoint
– Improved problem-solving efficiency
– Enhanced accuracy in AI systems
– Effective handling of complex problem domains
– Development of innovative reduction algorithms
– Contribution to the advancement of AI
If you are interested in the field of artificial intelligence and want to learn more about problem reduction research at Javatpoint, please feel free to contact us. We are always excited to collaborate and share our knowledge in the pursuit of advancing AI technologies.
## Advancements in Problem Reduction for AI
Artificial Intelligence (AI) has made significant advancements in problem-solving and has the ability to tackle complex issues. Problem reduction, a fundamental concept in AI, aims to minimize a problem by breaking it down into smaller, more manageable subproblems.
With the advancements in AI, problem reduction techniques have become more sophisticated and efficient. AI algorithms can now analyze large amounts of data quickly and accurately, allowing for faster problem-solving and decision-making processes.
In the field of problem reduction, the role of artificial intelligence is crucial. AI can identify patterns, extract relevant information, and generate possible solutions. It assists in problem minimization by identifying the most critical aspects and eliminating unnecessary details that may hinder the problem-solving process.
Javatpoint, a leading resource for AI development, provides comprehensive tutorials and learning materials on problem reduction in artificial intelligence. They cover various techniques, such as abstraction, divide and conquer, and heuristic search, that are used to reduce complex problems into simpler ones.
One of the significant advancements in problem reduction for AI is the integration of machine learning algorithms. Machine learning allows AI systems to learn and adapt from experience, making them more efficient at problem solving. By analyzing past data and patterns, AI algorithms can identify the most effective problem reduction strategies and optimize the overall problem-solving process.
The advancements in problem reduction for AI have opened up new opportunities in various fields, including healthcare, finance, and transportation. AI systems can now handle complex problem domains, such as medical diagnosis, fraud detection, and route optimization, with higher accuracy and efficiency.
In conclusion, the advancements in problem reduction techniques for AI have revolutionized the field of artificial intelligence. These advancements have allowed AI systems to solve complex issues more effectively and efficiently. Javatpoint provides valuable resources for anyone interested in learning more about problem reduction in artificial intelligence and staying up to date with the latest advancements in the field.
## Problem Reduction vs. Problem Elimination in AI
In the field of artificial intelligence (AI), problem-solving is a crucial aspect. AI systems are designed to tackle complex issues and provide efficient solutions. One of the techniques used in problem-solving is problem reduction.
### Problem Reduction
Problem reduction is a method employed to solve a complex problem by breaking it down into smaller, more manageable subproblems. By decomposing the main problem into smaller parts, it becomes easier to understand and solve each component independently. This approach simplifies the problem-solving process and enables AI systems to address the issue step by step.
At Javatpoint AI, problem reduction plays a significant role in developing intelligent systems. By minimizing the complexity of a problem, our AI algorithms can focus on solving individual subproblems effectively. This allows for more efficient resources allocation and optimization of the problem-solving process.
### Problem Elimination
Problem elimination is another problem-solving technique used in AI. Unlike problem reduction, problem elimination aims to completely remove or eliminate the problem rather than breaking it down into smaller parts. This approach involves identifying the root cause of the problem and addressing it directly to prevent any recurrence.
In some cases, problem elimination may be preferred over problem reduction. If the main problem is easily identifiable and can be solved without extensive decomposition, eliminating the problem entirely can save time and resources. However, it is crucial to ensure that the root cause is accurately identified to prevent any potential recurrence.
Problem Reduction Problem Elimination
Breaks down complex problems into manageable subproblems Directly addresses the root cause of the problem
Enables step-by-step problem solving Eliminates the problem entirely
Optimizes resources allocation Can save time and resources
In conclusion, both problem reduction and problem elimination have their roles in AI. Problem reduction allows for efficient problem-solving by breaking down complex issues into manageable parts, while problem elimination focuses on directly addressing the root cause. At Javatpoint AI, these techniques are employed depending on the nature and complexity of the problem at hand to ensure optimal solutions.
## Limitations of Problem Reduction in AI
The use of problem reduction in artificial intelligence (AI) has proven to be effective in solving complex problems. However, this approach is not without its limitations. Here are some of the key issues and challenges associated with problem reduction in AI:
1. Limited Scope: Problem reduction can only solve problems that can be represented as a reduction or minimization of a larger problem. It may not be suitable for problems that cannot be decomposed into smaller subproblems.
2. Knowledge Representation: Problem reduction relies on the availability of accurate and complete knowledge about the problem domain. If the knowledge base is insufficient or inaccurate, the problem reduction approach may produce incorrect or suboptimal solutions.
3. Complexity: Problem reduction can become computationally expensive for problems with a large number of variables or constraints. As the complexity of the problem increases, the time and resources required for problem reduction also increase.
4. Search Space: Problem reduction relies on exploring the search space to find a solution. In some cases, the search space may be too large or complex to explore fully, leading to incomplete or unsatisfactory solutions.
5. Heuristics and Approximations: Problem reduction often relies on the use of heuristics or approximations to simplify or speed up the problem-solving process. While these techniques can be effective in some cases, they may introduce errors or lead to suboptimal solutions.
6. Domain Dependency: Problem reduction may be highly dependent on the specific problem domain and the available problem-solving techniques. It may not be easily generalized or applied to a wide range of problems.
Despite these limitations, problem reduction remains a valuable tool in the field of AI and continues to be used in various problem-solving applications. By understanding and addressing these limitations, researchers can further improve the effectiveness and applicability of problem reduction techniques in AI.
## Impact of Problem Reduction on AI Performance
Problem reduction plays a crucial role in the field of artificial intelligence (AI) by addressing the issues and challenges encountered during problem-solving. Through the minimization of the problem size and complexity, AI systems are able to improve their performance and efficiency in finding solutions.
Problem reduction is a technique that aims to break down a complex problem into smaller and more manageable subproblems. By doing so, AI systems can focus on solving the subproblems individually, which leads to faster and more accurate problem-solving. This reduction in problem size also helps in simplifying the search space, making it easier for AI algorithms to explore and find optimal solutions.
Javatpoint, a renowned resource for AI learning, emphasizes the significance of problem reduction in improving the performance of AI systems. Through problem reduction, AI algorithms are able to effectively tackle complex problems that would otherwise be difficult to solve. The reduction in problem complexity allows AI systems to efficiently allocate computational resources, resulting in faster and more efficient problem-solving.
Furthermore, problem reduction enables AI systems to handle a wide range of problem domains. By breaking down complex problems into smaller subproblems, AI algorithms can apply domain-specific knowledge and heuristics to each subproblem. This allows AI systems to make more informed decisions and generate solutions that are tailored to specific problem domains, leading to improved overall performance.
In conclusion, problem reduction plays a critical role in improving the performance and efficiency of AI systems. By breaking down complex problems into smaller and more manageable subproblems, AI algorithms are able to enhance problem-solving capabilities. The reduction in problem size and complexity enables AI systems to allocate computational resources efficiently and apply domain-specific knowledge effectively, resulting in faster and more accurate problem-solving in the field of artificial intelligence.
## Evaluation Metrics for Problem Reduction in AI
In the field of artificial intelligence (AI), problem reduction is a crucial technique used in problem-solving. Javatpoint offers comprehensive resources and tutorials on problem reduction for AI enthusiasts.
The primary objective of problem reduction is the minimization of issues and complexities in the problem-solving process. By breaking down a complex problem into simpler subproblems, AI systems can efficiently find solutions.
When evaluating the effectiveness of problem reduction techniques, various metrics are used to measure their performance. These metrics assess the efficiency, accuracy, and speed of the problem-solving process.
One widely used evaluation metric is the time complexity of problem reduction algorithms. This metric measures the computational time required to reduce a problem. Lower time complexity indicates faster processing and quicker solution attainment.
Another important metric is the space complexity. It determines the amount of memory required to execute problem reduction techniques. Minimizing space complexity ensures efficient memory usage and reduces the strain on computing systems.
Accuracy is also a crucial evaluation metric for problem reduction in AI. It measures the ability of the reduced problem to accurately represent the original problem. Higher accuracy ensures that the solutions obtained from the reduced problem are valid and applicable to the original problem.
Furthermore, the performance of problem reduction techniques can be evaluated using complexity analysis. This evaluation method examines the theoretical efficiency of the algorithms, allowing researchers and developers to compare different techniques.
In conclusion, problem reduction plays a vital role in the field of AI, and evaluating its effectiveness is essential. Using metrics such as time complexity, space complexity, accuracy, and complexity analysis enables researchers to assess and improve problem reduction techniques. Javatpoint offers a valuable platform for learning and exploring problem reduction in AI.
## References
In the field of artificial intelligence (AI), problem reduction is a key approach to problem solving. Problem reduction involves breaking down a complex problem into smaller sub-problems, which are easier to solve. This technique is often used in AI systems to minimize the search space and improve efficiency.
Javatpoint is a popular website that provides tutorials, examples, and resources for various programming languages and concepts, including artificial intelligence. Their articles on problem reduction and problem minimization in AI are highly informative and helpful in understanding the topic in depth.
To delve deeper into the topic of problem reduction in artificial intelligence, it is recommended to refer to the following sources:
### 1. “Artificial Intelligence: A Modern Approach”
This renowned book by Stuart Russell and Peter Norvig provides a comprehensive overview of AI, including problem-solving techniques such as problem reduction. It offers detailed explanations, examples, and case studies that enhance understanding and application.
### 2. “Problem Reduction in AI: Methods and Applications”
This academic paper explores various problem reduction techniques used in artificial intelligence. It discusses the advantages, limitations, and practical applications of problem reduction in different domains like robotics, natural language processing, and expert systems.
Conclusion:
Problem reduction plays a crucial role in minimizing the complexity of problems in artificial intelligence. Javatpoint offers valuable resources on this topic, while books like “Artificial Intelligence: A Modern Approach” and research papers like “Problem Reduction in AI: Methods and Applications” provide comprehensive insights into problem reduction techniques.
#### What is problem reduction in artificial intelligence?
Problem reduction in artificial intelligence refers to the process of simplifying a complex problem by breaking it down into smaller, more manageable sub-problems. This approach allows an AI system to solve the overall problem by tackling the sub-problems individually and combining their solutions.
#### How does problem reduction work in AI?
Problem reduction in AI works by dividing a complex problem into smaller sub-problems. The AI system then focuses on solving these smaller problems, which are often easier to tackle. The solutions to the sub-problems are then combined to provide a solution to the original problem.
#### What are the benefits of problem reduction in AI?
Problem reduction in AI offers several benefits. By breaking down complex problems into smaller sub-problems, it allows for easier problem solving. It also enables efficient use of resources as the AI system can focus on solving specific sub-problems. Additionally, problem reduction can improve the scalability and performance of an AI system.
#### Can you provide an example of problem reduction in AI?
Sure! Let’s consider a self-driving car as an example. The overall problem is to safely navigate from point A to point B. This complex problem can be broken down into smaller sub-problems such as obstacle detection, lane following, and traffic light recognition. By solving each of these sub-problems, the self-driving car can successfully navigate to its destination.
#### Are there any limitations to problem reduction in AI?
While problem reduction is a powerful technique in AI, it does have its limitations. One limitation is that breaking down a problem into sub-problems may not always guarantee an optimal solution to the overall problem. The AI system may need to consider the interactions and dependencies between the sub-problems to find the best solution. Additionally, problem reduction may not be suitable for all types of problems, especially those that cannot be easily decomposed into smaller sub-problems.
#### What is problem reduction in artificial intelligence?
Problem reduction in artificial intelligence is a technique that aims to simplify complex problems by breaking them down into smaller, more manageable sub-problems.
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Welcome to “About The Number 612” – a website page dedicated to exploring the fascinating properties and significance of the number 612. Delve into the world of mathematics, numerology, and history as we uncover the hidden meanings, intriguing patterns, and real-life applications of this seemingly ordinary three-digit number. Get ready to be amazed by the unexpected connections and surprising facts that surround the number 612!
## Fun and interesting facts about the number 612
The number 612 has a unique property in that it is both a Harshad number and a palindrome. A Harshad number is an integer divisible by the sum of its digits, while a palindrome reads the same forward and backward.
## The number 612 angel number and biblical meaning
The number 612 angel number holds significant biblical meaning, as it is believed to represent spiritual growth, harmony, and balance in one's life. This angel number is often seen as a message from the divine realm, encouraging individuals to trust in their spiritual journey and embrace the positive changes that come with it.
## What is 612 written in words?
Six hundred and twelve
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DCXII
## What are the factors, prime factors, factor trees, cubes, binary number and hexadecimal of 612?
Factors of 612 are 1, 2, 3, 4, 6, 9, 12, 17, 18, 34, 36, 51, 68, 102, 153, 204, 306 and 612.
The prime factors of 612 are 2, 317.
The factor tree of 612 is 2, 317.
The cube of 612 is 229,220,928.
The binary number of 612 is 1001100100.
The hexadecimal of 612 is 264.
## Metric to imperial numbers
612 centimeters is 240.945 inches.
612 kilometers is 380.279 miles.
612 meters is 669.289 yards.
612 grams is 21.588 ounces.
612 kilograms is 1349.227 pounds.
612 litres is 1076.967 pints.
612 KPH (Kilometers Per Hour) is 380.279 MPH (Miles Per Hour).
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Grains to Nanograms
Convert gr to ng
Change to Nanograms to Grains
Share:
How to convert Grains to Nanograms
1 [Grains] = 64798910 [Nanograms]
[Nanograms] = [Grains] * 64798910
To convert Grains to Nanograms multiply Grains * 64798910.
Example
50 Grains to Nanograms
50 [gr] * 64798910 = 3239945500 [ng]
Conversion table
Grains Nanograms
0.01 gr647989.1 ng
0.1 gr6479891 ng
1 gr64798910 ng
2 gr129597820 ng
3 gr194396730 ng
4 gr259195640 ng
5 gr323994550 ng
10 gr647989100 ng
15 gr971983650 ng
50 gr3239945500 ng
100 gr6479891000 ng
500 gr32399455000 ng
1000 gr64798910000 ng
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The Unsolvable Physics Problem
I know a little about physics, but there is one problem that using neither classical or quantum mechanics has allowed me nor anyone else to approach a solution. Perhaps someone like John Nash (A Beautiful Mind), or Albert Einstein were he still alive might solve. As chaos theory is refined over the next decade, or computers get more powerful there may be a slight chance of a crude solution, though highly unlikely.
What is that problem you ask? The disappearance of all ball point pens and pencils within a two week period after they are acquired. Oh I'm sure there will be simplistic answers from the ignorant like, they fall behind sofa cushions and car seats, or that people steal or borrow them, even that they didn't function properly and were discarded. But none of these theories really go toward explaining the pure volume of lost writing equipment.
The fact is that over 2 billion ball point pens, and nearly two billion pencils are manufactured in the United States per year. There are approximately 281,000,000 people in the United States. That's about 14 writing instruments per capita, removing infants, the mentally infirmed, and the illiterate the number is probably nearer 20 per capita. That would mean that over a fifty year period the pen count should be about 1,000 per capita.
Something has to be done to stop this hemorrhage of ink pens from the country. Obviously terrorists, and Mexicans account for many of the missing Jotters. At a buck a piece this is a loss to the nation of some \$400,000,000,000 over that period. Some evil is obviously at work here. While people starve around the world, and we cut long growth forests to make pencils. We are losing ink pens like an AK47 victim loses blood.
There can be no question that what ever the mathematics and science of the situation are aliens from outer space are some how involved. Apparently at least some of the writing equipment is being transported up to mother ships far out in space. They obviously plan to take over our planet through the use of pencil theft there causing public school students to do even worse on their SOL tests, if that's even possible.
What can we do? A few suggestions might be to go back to using quills, chaining your Bic to a radiator, locking your writing implements in your gun safe, or even returning to the days of doing homework on the back of a shovel with a charcoal brickett much like the great Abraham Lincoln.
If we could ever find all those pencils and pens we could easily end our energy crisis overnight by trading these implements for oil. We could leave deficit spending behind, and perhaps even end graffiti. If the kids had pencils they might be less inclined to use spray paint.
Won't you do what you can? The next time a buddy asks have you got a pen man, just say no.
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How to measure the size of a laser dot?
I'm not a physicist, that's why I'm asking you if there is maybe an easy way (e.g. a mathematical law) to measure the size of a laser dot, or its diameter.
The setup
It contains a laser, a high speed camera and other utilities. Now I'm using the cam to take a picture of the laser dot on a particular surface.
The problem
I need to determine what is the actual size of a pixel in physical space (i.e. how many mm are displayed by one pixel). So I need to scale my pixels and I was thinking about the following equation as a factor for 1 pixel :
$$\text{scaling factor} = \frac{\text{diameter}}{n_P},$$ where $n_P$ stands for the number of pixels it takes to display the diameter
$\text{diameter}$ is the actual diameter of the laser dot in SI values (e.g. cm)
The Question
Now unfortunately at the moment I can't think about much better ideas than just measuring the diameter by hand with a caliper. But I'm afraid to call this a real solution, it sounds more like a workaround.
Does somebody has a better idea?
• You can take a picture and count pixels. Commented Jan 21, 2015 at 21:14
• @ja72 and how do I get the cm out of it ? Commented Jan 21, 2015 at 21:15
• Lay a ruler next to the laser or something else of known size for calibration. Commented Jan 21, 2015 at 21:17
• If you're going for a metric-unit measurement, is there a reason you shouldn't simply use calipers? You say it feels like a "workaround," but any other method I can imagine would be both less precise and more work. Commented Jan 21, 2015 at 21:17
• Say you have two meter sticks. Put one a certain distance away and take a picture. Now tape the two together, put them the same distance away and take another picture. The double ruler will measure twice as many pixels long in the picture. So yes, your scaling factor will work. A free piece of software used for exactly this purpose is called ImageJ. Commented Jan 21, 2015 at 21:18
First - know that a laser spot is not a "disk"; usually you will find that the intensity drops off from the middle to the edge, so "size of the spot" is an imprecise concept (where is the edge of a gradually decreasing curve?). Secondly, unless the laser is normally incident on the surface, the spot will most likely not be circular but elliptical; and unless you are viewing the surface normally, you will add an additional distortion.
All this makes it difficult to do this right. I would recommend that you place some graph paper (remember graph paper? It's what we used before computers could draw graphs... - it comes with markings in mm, with major markings for cm) on the surface where you project your laser dot, and take a picture of it. For precise measurements you can buy optical grids in many sizes - more expensive but much more accurate.
Unless the grid is perfectly normal to the axis of your camera, it will be distorted; but if you align it properly you will be able to count the pixels between lines on the grid in your image.
There are several tricks you can use to be more accurate. If you tilt the grid slightly (rotate about the normal) then the lines will cross rows and columns of pixels: this will allow you to find the place where the line aligns exactly with a pixel, and lets you get to sub-pixel accuracy. Second, you should count pixels across a number of lines; say you can see ten lines of the grid, you can calculate the distance by dividing the number of pixels between line 1 and 10 by 9 to get the number of pixels per mm.
Even more accurate would be to write down the pixel number for each line, then fit a straight line through the points. When you then subtract the best fit from the data you will be able to see whether there is any pincushion distortion (non linearity in the scaling of the camera with respect to position) at which point you can compute the scale factor at any point.
Finally, you could do this mathematically if you know the distance to the object exactly, and you know the focal length of the lens. This is actually quite hard to do because it's unlikely that you know the exact location of the optical center of the lens, but that depends on your setup.
In principle, if you have an object that is distance $o$ from the lens with focal length $f$, then in order for the object to be in focus you need the lens at a distance $d$ from the focal plane such that
$$\frac{1}{f} = \frac{1}{d}+\frac{1}{o}$$
Or, if you know the focal length of the lens but not the center of the lens, but you can measure the distance from the object to the sensor, $s = o + d$, then you can do
\begin{align}\frac{1}{f} &= \frac{1}{d} + \frac{1}{s - d}\\ &=\frac{s-d}{d(s-d)} + \frac{d}{d(s-d)}\\ &=\frac{s}{d(s-d)}\\ d(s-d)&=f\cdot s\end{align}
We can solve this quadratic for d:
$$d^2 - ds + fs = 0\\ d = \frac{s±\sqrt{s^2-4fs}}{2}$$
And then the magnification follows from
$$M = \frac{d}{o} = \frac{d}{s-d}$$
If you know the actual pixel size (usually this is given by the manufacturer) you can then get the magnification from this equation and get your pixel to mm conversion:
$$c = \mathrm{(Pixel\ size)\cdot M}$$.
So if your pixel is 0.1 mm and your magnification is 3, then one pixel corresponds to 0.3 mm in the image (at that particular distance).
• thanks for that precise answer. Getting straight lines on the monitor when comparing with a graph paper shouldn't be a problem, since image processing techniques are pretty common to me. But actually I'd like to avoid attaching anything to the object of interest. I could try to apply the formulas of your suggestions. What I've missed to tell is, that the distance between Laser and Object of Interest should be variable. For this purpose I could develop a laser-range finder with the already used hardware. I have to check it when I'm at desk again Commented Jan 21, 2015 at 23:35
• If the distance is variable them so is the pixel-to-mm conversion... Range finding can be hard depending on the speed and precision needed. Consider projecting two parallel laser beams of known separation to give you a "calibration distance". Commented Jan 21, 2015 at 23:38
• Well do I get this formulas right, that they only work if the object of interest is a plane where the point of (camera)view conforms the normal of this plane? So if my PO(C)V is rotated a little bit to the normal this would distract the accuracy of the results right ? Commented Jan 21, 2015 at 23:42
• with variable its meant, during the measuring series every distance stays constant, but from series-to-series the distance changes. Commented Jan 21, 2015 at 23:44
• Yes if the plane is tilted one dimension will appear to be compressed; and the focusing of the lens will change the magnification. As long as the object is always "far" from the camera (>20f) you may be able to ignore that effect - it all depends on the required accuracy. Precision takes attention to detail... Commented Jan 22, 2015 at 7:45
Its simpler than I thought at first so I edited extensively:
It sounds like you want to know two things: 1) The size of the pixels of your sensor and 2) the size of your laser spot. I would solve 1) and then use the known size of the pixels to find 2). Assuming that you can't just the pixel size from the manufacturer spec sheet or want to verify it experimentally here are two methods.
Method 1: Assumes a columnated beam but requires no specialized hardware other than what you said that you have.
Step 1: Find the pixel size. If you know the wavelength of the laser, the laser is columnated, and you have an aperture (small hole) (the smaller the better), you could pass the laser through the aperture to obtain a Fraunhofer diffraction pattern:
$I(\theta)=I(0)(\frac{2J_{1}(ka \sin{\theta})}{ka \sin{\theta}})^{2}$,
I is the intensity of the light in the far field, J1 is the Bessel function, a is the diameter of the aperture and $k=\frac{2\pi}{\lambda}$. You can measure theta very accurately by diffracting for a long distance say from one side of the lab to the wall on the far side and carefully measuring the distance from the central intensity maximum to the first minimum and doing a little trig. Since the aperture size and the wavelength are fixed, this diffraction angle will be also. Since you know this angle you know the size of this central diffraction disk as a function of distance from the aperture.
By illuminating the camera with this diffraction pattern at a known distance from the aperture you can find the pixel size experimentally by pixel counting the diameter of the central diffraction spot (Airy disk).
Step 2: Find the laser spot size. Remove the aperture and shine the laser directly on the camera and just pixel count. Since you know the size of the pixels from step 1 this is straight forward.
Method (2): Assumes you have a mechanical stage with fine control. This is really simple. Attach an index card or thin sheet of sheet metal, etc so that it blocks part of the camera, by putting it right in front of the camera. Take an image then translate the card a known amount and take another image. Compare the two images an pixel count. If the edge of the index card was moved by 1 mm and it moved 1000 pixels then your pixels have a size of 1 um each.
• do you have a link ? do I get accurate values out of it ? What exactly is meant with the diameter of the aperture (= diameter of laser dot?) Commented Jan 21, 2015 at 22:00
• <en.wikipedia.org/wiki/Airy_disk> describes diffraction through a circular aperature. Aperature size refers to the diameter of the hole. It is important that the hole through which you pass the laser is smaller than your collumnated laser. This can be quite accurate, but is limited by your ability to accurately measure distances, and the distances and the distance that you have available, which is typically limited by the size of your lab or the power of your laser. Commented Jan 21, 2015 at 22:14
• If you can pry the lens off the front of the camera, there is an easier way to get the size of the pixels on the sensor: illuminate the sensor directly with the beam, place a screen at a known distance from the sensor, and measure the distance (and therefore the angle) between the spots in the back-reflected diffraction pattern. Commented Jan 22, 2015 at 6:56
You can calculate the size of a pixel in the final setup by measuring the dimensions of the surface area captured in the image and dividing by the pixel-wise width and length of the image the camera creates.
Having the experimental setup in place for this measurement is important because factors such as the zoom factor of the lens and the distance of the camera from the surface will change the results. Obviously if you zoom or move the camera (rather, change the relation of the camera to the surface), the pixel/size ratio will change.
EDIT: More specifically: if you have a live feed of the image generated by the camera, you can measure the boundaries of the image space by watching in the camera feed as you measure. For example, if your image space is only several centimeters, you can line up a pair of calipers such that one arm is aligned with the left side of the image, and the other arm is aligned with the other edge; at this point you're measuring (with the calipers) an imaginary "object" which is the size of the picture your camera captures. Say this dimension comes to 27mm overall; now you can divide that by the width of the digital image captured by the camera. If your camera captures, for example, a 1000-pixel-wide image, you now know from measuring the image space that 1px = .027mm and thus if the laser dot is 40px wide when imaged, that corresponds to a width of .10mm.
• but how do I derive the diameter (in physical space) to scale my pixels, that's my actual problem. I don't know the exact value Commented Jan 21, 2015 at 21:25
• @user3085931 Perhaps I misunderstood the question. Are you trying to determine the size of the laser dot, or the size of a pixel? Either one is going to require that you first measure - physically, with calipers, metersticks, etc. - some reference object or image in actual space. There's no way to calculate abstractly how large a real object is. Commented Jan 21, 2015 at 21:33
• Easy said: I want to know very precisely the diameter in physical space (I'd prefer accuracy < mm Commented Jan 21, 2015 at 21:36
• Then you will need a measuring device with a precision less than 1mm. Do you have a live feed view of the image the camera generates? If so I can update my answer with a specific method to calculate pixel size, allowing you to use the pixels themselves as a fine-grained measuring tool. Commented Jan 21, 2015 at 21:46
• unfortunately not atm. I could show a picture tomorrow. Well I just don't feel right measuring it by hand, that way it's pretty impossible to measure < mm Commented Jan 21, 2015 at 21:51
You could cut a hole of a known size in a piece of paper and place it over your detector. Then, you could see how many pixels are still illuminated. The known size of the aperature makes the conversion easy.
• this seems like a too less precise method for my aim Commented Jan 21, 2015 at 21:24
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# Re: [R] help computing a covariance
From: Spencer Graves <spencer.graves_at_pdf.com>
Date: Sat 03 Jul 2004 - 03:07:06 EST
Have you tried simulating several special cases? The right set of simulations should suggest an answer, which might then lead you to a proof.
hope this helps. spencer graves
Eugene Salinas (R) wrote:
> Thanks. This is sort of what I have been trying to do... but I keep
> ending up with products of non-independent chi-squares where I am sort
> of getting stuck. I knew this Theorem just never knew it was called
> Cochran's Thm. Btw, do you know of a good book that deals with
> multivariate statistics using vector notation etc. All the books I
> have seem to be focused on scalar random variables and they don't even
> mention it.)
>
> thanks, eugene.
>
>
> Spencer Graves wrote:
>
>> Have you considered Cochran's theorem? (A Google search just
>> produce 387 hits for this, the second of which
>> "http://mcs.une.edu.au/~stat354/notes/node37.html" provided details
>> that might help.) By construction, P is n x n, idempotent of rank k,
>> so y'Py is chi-square(k). Also, xA is an n-vector in the (rank k)
>> column space of x; indeed, PxA = [x*inv(x'x)*x]xA = xA. I can't see
>> the details now but I believe you can write (A'x'y)^2 = y'xAA'x'y as
>> a weighted sum of k independent chi-squares each with one degree of
>> freedom (since x and P have rank k), and then get what you want from
>> the sum of the weights. Then check your result using Monte Carlo.
>> hope this helps. spencer graves
>>
>> Eugene Salinas (R) wrote:
>>
>>> Hi everyone,
>>>
>>> (This is related to my posting on chi-squared from a day ago. I have
>>> tried simulating this but I am still unable to calculate it
>>> analytically.)
>>>
>>> Let y be an n times 1 vector of random normal variables mean zero
>>> variance 1 and x be an n times k vector of random normal variables
>>> mean zero variance 1. x and y are independent.
>>>
>>> Then P is the projection matrix P=x*inv(x'*x)*x'
>>>
>>> I need to figure out the covariance
>>>
>>> Cov ( y'*P*y , (A'*x'*y)^2 ) where A is a constant of dimension k
>>> times 1.
>>>
>>> thanks, eugene.
>>>
>>> ______________________________________________
>>> R-help@stat.math.ethz.ch mailing list
>>> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
>>> http://www.R-project.org/posting-guide.html
>>
>>
>>
>>
>>
>
> ______________________________________________
> R-help@stat.math.ethz.ch mailing list
> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
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# Finding a new polynomial with roots $k$ greater than a given polynomial
A similar question appears here, though I wish to ask for polynomials of higher degree.
Let's say we have the polynomial $f(x)=x^4-3x^3-3x^2+4x-6$, and we wish to find a polynomial with roots 3 greater than those of $f(x)$. I can see that putting $g(x)=f(x-3)$ works, but then we have the expression $$(x-3)^4-3(x-3)^3-3(x-3)^2-4(x-3)+6$$ which will take quite a while to evaluate. (It comes out to $\boldsymbol{x^4-15x^3+78x^2-175x+441}$).
Is there a shortcut to find the polynomial above through some 'trick' (e.g. some form of synthetic division?)
Furthermore, if we wish to find a polynomial with roots $k$ greater than a given polynomial, will this new polynomial have a given 'form' of $k$? E.g. what if I wanted to solve the above question, but with my roots being $4,5,$ or $6$ greater than those of the given polynomial?
• Is there a shortcut Not really. What you have is the shortcut. – dxiv Mar 4 '18 at 0:47
• You say "which will take quite a while to evaluate"? That isn't a very precise criterion. If you are interested in the computational complexity, then computing the coefficients of $f(x - 3)$ using Horner's rule is quite efficient. – Rob Arthan Mar 4 '18 at 1:50
Well, your roots will be shifted $3$ units to the right. From there, you can calculate the new roots, using the sum and product of roots formulae to calculate the coefficients of each term. This can be considered a shortcut of sorts. For a polynomial $f(x)$ where $$f(x)=a_nx^n+...+a_1x+a_0$$ $$\alpha\beta\gamma...=(-1)^n\frac{a_0}{a_n}$$ $$\alpha+\beta+\gamma+...=-\frac{a_{n-1}}{a_n}$$ The greek letters represent the roots.
There are other formulae for finding $a_{n-2}$ and so forth. I will give $a_{n-2}$ in cyclic sum notation, but the rest are easily accessible and directly quotable. $$\sum \alpha\beta=\frac{a_{n-2}}{a_n}$$
• @RobArthan I guess what the answer is getting at is using Vieta's relations for the symmetric polynomials, which does not require actually finding the original roots e.g. $\;\sum (\alpha+3)$ $=\sum \alpha + 4 \cdot 3$ $= 3+12=15\,$, then $\;\sum (\alpha+3)(\beta+3)$ $= \sum \alpha \beta + 6 \cdot \sum \alpha + 6 \cdot 9 = \ldots\;$ It is a legitimate approach, though not necessarily much of a shortcut vs. the direct expansion. – dxiv Mar 4 '18 at 2:07
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Main - pause fprintf\n-problem 4 sine\n p4 pause fprintf\n-problem 4 speech\n p4s pause fprintf\n-problem 5 alpha = 0.65\n p5(0.25 pause
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function [ output_args ] = main( input_args ) % Summary of this function goes here % Detailed explanation goes here fprintf('\n---prelab question 1---\n'); prelab1 pause fprintf('\n---prelab question 3---\n'); q3prelab(1,2,5) pause fprintf('\n---problem 2---\n'); p2 pause fprintf('\n---problem 3 - sine wave ---\n'); p31 pause fprintf('\n---problem 3 - P_2_1.wav ---\n'); ps32 pause fprintf('\n---problem 3 - speech ---\n'); ps33
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Unformatted text preview: pause fprintf('\n---problem 4 - sine---\n'); p4 pause fprintf('\n---problem 4 - speech ---\n'); p4s pause fprintf('\n---problem 5 - alpha = 0.65---\n'); p5(0.25) pause fprintf('\n---problem 5 - non-constant attenuation fn---\n'); p52(0.25,5,10) pause fprintf('\n---problem 5 - with oscillatory attenuation fn.---\n'); p54(0.25,5,10) pause fprintf('\n---TESTING COMPLETE---\n'); end e...
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This note was uploaded on 10/02/2011 for the course CSE 3451 taught by Professor A during the Spring '10 term at York University.
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LAMMPS WWW Site - LAMMPS Documentation - LAMMPS Mailing List Archives
Re: [lammps-users] How to oscillate a wall while thermostatting
# Re: [lammps-users] How to oscillate a wall while thermostatting
From: Axel Kohlmeyer Date: Sun, 5 Nov 2017 07:12:15 -0500
On Sun, Nov 5, 2017 at 6:24 AM, saeed alborzi wrote:
Dear Mr.Kohlmeyer
Hello again
Does the compute group/group command calculate the "energy transferred" you said exactly?
no.
Thanks
On Wed, Nov 1, 2017 at 5:39 PM, Axel Kohlmeyer wrote:
those numbers are useless and irrelevant for the properties you want
to determine.
you have to determine the amount of energy *transferred* and in the
case of the moving wall, that requires properly dealing with the
center-of-mass bias.
it looks as if you first need to have a deep peek into a text book on
statistical thermodynamics and gain a better understanding what you
are doing.
axel.
On Wed, Nov 1, 2017 at 9:54 AM, saeed alborzi <alborzi.saeed6@...24...> wrote:
> Dear Professor Kohlmeyer
>
> As you said , I compared the energy and velocity order of magnitude of the
> wall with and without oscillation. With a 2 angstrom amplitude and 2 fs
> period and no thermostatting, the maximum wall velocity is about 600 m/s and
> kinetic energy is about 23000 kcal/mol . With a 300K nvt thermostat and no
> oscillation, maximum velocity of 700 m/s and kinetic energy of 7000 kcal/mol
> is obtained. I think the values are not neglectable in camparison to each
> other. therefore I think I need to consider both fixes
>
> On Wed, Nov 1, 2017 at 3:27 PM, saeed alborzi <alborzi.saeed6@...24...>
> wrote:
>>
>> let me refer to an article titled "Static and dynamic behavior of water
>> droplet on solid surfaces with pillar-type nanostructures from molecular
>> dynamics simulation"
>>
>> at : http://www.sciencedirect.com/science/article/pii/S0017931014007455
>>
>> this is similar to my case except that I want to add heat transfer to the
>> simulation.
>>
>> I was thinking that if it was possible to add some amount to the surface
>> atoms' velocity as oscillating velocity while it is not rescaled by fix nvt
>> , it would be easy to implement. but the only problem is that the "velocity
>> set v_vx v_vy v_vz sum yes" command does not take a time-dependent value.
>>
>> I apologize if I could not explain my meaning clearly and strongly
>>
>> On Wed, Nov 1, 2017 at 3:08 PM, Axel Kohlmeyer <akohlmey@...24...> wrote:
>>>
>>> On Wed, Nov 1, 2017 at 7:31 AM, saeed alborzi <alborzi.saeed6@...24...>
>>> wrote:
>>> > it was just an example, not the real case.
>>> > the effect of frequency value will be investigated
>>>
>>> if the frequency is (much?) lower, then - as i have already pointed
>>> out - there is no need to do both at the same time, as the time scales
>>> are sufficiently decoupled.
>>>
>>> what you are asking is a technically complex and difficult process to
>>> implement correctly.
>>> i am not going to spend any more time and effort on this until i am
>>> convinced that the outcome is physically meaningful.
>>>
>>> axel.
>>>
>>>
>>> >
>>> >> with a 500GHz motion?? how is that possible?
>>>
>>>
>>>
>>>
>>> --
>>> Dr. Axel Kohlmeyer akohlmey@...24... http://goo.gl/1wk0
>>> College of Science & Technology, Temple University, Philadelphia PA, USA
>>> International Centre for Theoretical Physics, Trieste. Italy.
>>
>>
>
--
Dr. Axel Kohlmeyer akohlmey@...24... http://goo.gl/1wk0
College of Science & Technology, Temple University, Philadelphia PA, USA
International Centre for Theoretical Physics, Trieste. Italy.
--
Dr. Axel Kohlmeyer akohlmey@...24... http://goo.gl/1wk0
College of Science & Technology, Temple University, Philadelphia PA, USA
International Centre for Theoretical Physics, Trieste. Italy.
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# Raw mark vs Aligned mark (1 Viewer)
#### jasminerulez
##### Member
I'm not sure if this question makes sense, lets say for mathematics I get a mark of 75, what would that approximately be aligned to ?
#### shashysha
##### New Member
You can go on rawmarks.info to see what HSC raw marks would be aligned to in different years if that's what you're asking
#### jasminerulez
##### Member
yeah that is it, thank you
#### beetree1
##### Active Member
does that website only publish like the highest most appealing aligned marks because it seems like everything gets scaled up by SO MUCH
#### Velocifire
##### Well-Known Member
Massive highball I guess
If you want the lowball, hsc buzzman is all ears
#### quickoats
##### Active Member
does that website only publish like the highest most appealing aligned marks because it seems like everything gets scaled up by SO MUCH
This is not scaling, it’s the alignment process. The “bands” correspond to performance descriptors e.g. band 4 = satisfactorily, band 6 = outstandingly etc, which may not line up with their allotted “marks” (b4 = 70-79 etc), so they align your raw marks to the performance scale. Let’s say in maths, someone who does the bare minimum to achieve the band 6 outcome gets 85 raw on the exam. NESA will align raw scores from 85-100 into the band 6 bracket of 90-100 (squeezing the range into the designated space).
Usually, you can achieve the designated band descriptor with a lower mark (e.g. a raw 75 in Adv English shows pretty decent skills -> aligns to mid band 5 accordingly) thus the alignment is quite significant.
#### quickoats
##### Active Member
I'm not sure if this question makes sense, lets say for mathematics I get a mark of 75, what would that approximately be aligned to ?
Also note that rawmarks.info is most useful for past HSC papers. If you get 75% on one of those, you’d be most likely to get a higher aligned mark in your HSC.
Using your raw internal school marks won’t yield a significant result.
#### beetree1
##### Active Member
This is not scaling, it’s the alignment process. The “bands” correspond to performance descriptors e.g. band 4 = satisfactorily, band 6 = outstandingly etc, which may not line up with their allotted “marks” (b4 = 70-79 etc), so they align your raw marks to the performance scale. Let’s say in maths, someone who does the bare minimum to achieve the band 6 outcome gets 85 raw on the exam. NESA will align raw scores from 85-100 into the band 6 bracket of 90-100 (squeezing the range into the designated space).
Usually, you can achieve the designated band descriptor with a lower mark (e.g. a raw 75 in Adv English shows pretty decent skills -> aligns to mid band 5 accordingly) thus the alignment is quite significant.
oH WHAT okay that makes a lot of sense cheers
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mersenneforum.org Linear recurrence on elliptic curve
Register FAQ Search Today's Posts Mark Forums Read
2007-01-18, 15:31 #1 Unregistered 84916 Posts Linear recurrence on elliptic curve Let Po be a point on elliptic curve E over Zn, n=p*q composite. Let Fib(k) be k-th fibonacci number or any other linear recurrence. Is it possible to efficiently compute Fib(big) * Point? Or is it known to be a hard problem? The usual doubling computing of Fib() gives integer explosion and the group order of E is unknown, so direct reduction is impossible. Basic idea: Consider E over Fp. The group order may be smooth * r, r not large prime. Compute Po2 = smooth2 * Po. The period of k*Po2 is r. Fib( m*(r-1)) mod r = 0 or Fib( m*(r+1) ) mod r = 0 so Fib( m*(r +/-1)) * Po2 is the point at infinity. r +/-1 may be smooth. Hakmem ITEM 14 (Gosper & Salamin): http://www.inwap.com/pdp10/hbaker/ha...ecurrence.html Mentions "rate doubling formula".
2007-01-18, 16:20 #2
R.D. Silverman
Nov 2003
746010 Posts
Quote:
Originally Posted by Unregistered Let Po be a point on elliptic curve E over Zn, n=p*q composite. Let Fib(k) be k-th fibonacci number or any other linear recurrence. Is it possible to efficiently compute Fib(big) * Point? Or is it known to be a hard problem? The usual doubling computing of Fib() gives integer explosion and the group order of E is unknown, so direct reduction is impossible.
I am not sure I know what you mean by "direct reduction"?
I don't follow you. Given a large integer k, one can compute
Fib(k) in O(log(k)) multiplications. If we are computing over Z,
the numbers get exponentially large, so the entire computation
takes exponentially many bit operations (even with FFT's to do
the multiplication).
However, computing Fib(k) mod N (or Fib(k)*P where P is an EC point
over Z/NZ) has *bounded* intermediate values. Computing
M*P on an elliptic curve over Z/NZ is a polynomial time computation
because the intermediate values are bounded in size and only polynomially
many multiplications are required.
OTOH, computing Fib(k)*P over Q is a purely exponential problem
because the heights of the points explode exponentially.
2007-01-18, 17:13 #3
Unregistered
F8716 Posts
Quote:
Originally Posted by R.D. Silverman I am not sure I know what you mean by OTOH, computing Fib(k)*P over Q is a purely exponential problem because the heights of the points explode exponentially. Please clarify your question.
The point P is bounded mod N.
All computations are mod N and they involve just point additions.
P + P = 2*P
P + 2*P = 3*P
2*P + 3*P = 5*P
...
FIB(K-2)*P + FIB(K-1)*P= FIB(K)*P
so if the group order is r, r prime, FIB(K) should be taken mod r and FIB(K)*P hits the point at infinity for K multiples of r-1 or r+1.
The question is may FIB(K)*P (this bounded mod N) be computed efficiently for large K product of small primes. The value of FIB(K) is not needed.
Similar Threads Thread Thread Starter Forum Replies Last Post burrobert GMP-ECM 6 2012-11-08 13:05 fivemack Math 0 2010-08-22 14:52 Raman Math 8 2009-04-13 19:20 Dirac Factoring 11 2007-11-01 14:01 bongomongo Factoring 5 2006-12-21 18:19
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Solved
# Array Checking
Posted on 2002-05-09
188 Views
I have two separate arrays. One is called strAddArray and the other is called strRemoveArray. What I am trying to do is select an item from a listbox. and add or remove it into another listbox. Evertime I add it it goes into strAddArray and evertyime I remove an item it goes into strRemoveArray. However, if I add and remove the same several times I create multiple copies of the same item within both arrays. Before I can update the DB I need to have the arrays setup correctly. Below is what I have thus far. The problem that I am running into is if I have a duplicate I remove it completely and have no copy of it all together. This is not as easy as it looks for me because there are so many scenarios, and I do not want anything to slip through the cracks.
Thanks,
Roger.
====================================
If intRemove > 0 Then
For x = 0 To 500
For y = (x + 1) To 500
If strRemoveArray(x) = strRemoveArray(y) And strRemoveArray(x) <> "" Then
strRemoveArray(y) = Empty
intRemove = intRemove - 1
End If
Next y
Next x
End If
'Checks items removed
For x1 = 0 To 500
For y1 = (x1 + 1) To 500
End If
Next y1
Next x1
End If
'Compares the two above arrays for any lingering duplicates
If intAdd > 0 And intRemove > 0 Then
For x2 = 0 To 500
For y2 = 0 To 500
If strRemoveArray(y2) = strAddArray(x2) And strRemoveArray(y2) <> "" Then
strRemoveArray(y2) = Empty
intRemove = intRemove - 1
End If
End If
Next y2
Next x2
End If
0
Question by:RogerH1
LVL 75
Accepted Solution
Anthony Perkins earned 100 total points
ID: 6999378
As an alternative you may want to use Collections instead of arrays. They have built in methods for adding and removing.
Just a thought,
Anthony
0
Author Comment
ID: 6999388
I am not familiar with what you mean by "Collections". Would you explain to me please.
0
LVL 75
Expert Comment
ID: 6999530
Rollback will not work Date: 02/07/2002 12:17PM PST
http://www.experts-exchange.com/jsp/qShow.jsp?ta=visualbasic&qid=20264417
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http://www.experts-exchange.com/jsp/qShow.jsp?ta=visualbasic&qid=20265663
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http://www.experts-exchange.com/jsp/qShow.jsp?ta=visualbasic&qid=20271250
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http://www.experts-exchange.com/jsp/qShow.jsp?ta=visualbasic&qid=20274441
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http://www.experts-exchange.com/jsp/qShow.jsp?ta=crystal&qid=20270066
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http://www.experts-exchange.com/jsp/qShow.jsp?ta=crystal&qid=20274932
Thanks,
Anthony
0
Expert Comment
ID: 7000165
you could just use binary trees instead of arrays, then when you stuff duplicates you can just ignore them and each tree will maintain exactly one of each of the things you put in it, so when you unwind it into the db, you have no duplicates
0
LVL 2
Expert Comment
ID: 7000381
Why have arrays at all? The list boxes are working like arrays already.
For ndx = 0 to addlist.listcount-1
next
for ndx = 0 to removelist.listcount-1
' do remove processing for removelist.list(ndx)
next
0
Author Comment
ID: 7001288
These are all great ideas, but please understand something that I did not explain. That is, when I open this window it reads in from the DB and then the user can add and remove from their initial choices.
Unless I have made this more complicated than it needs to be, I can ONLY see it being done with some sort of array. Since yesterday I have started working on understanding collections and worked up some code that is absolutely great. It checks current collection for duplicates and if they exists it never gets entered. It is far less code than using an array, and seems to be easier to work with for what I am doing.
Now on the other hand, if someone feels that I am still making this harder than it needs to be please let me know. I will leave this open until Monday at which time I will go ahead and close it if no one has a better idea.
Thanks,
Roger.
0
Author Comment
ID: 7006019
I looked into using the Collections class. It took me a while to find something good on the subject, but once I did I tried it in a sample program that I created. I feel that it works much easier than using arrays, and in less the code.
I wanted to give you an "A", but I your answer was only a direction to try, and no explanation or sample code came with it. However, thank you for telling me about it. I learned something new.
Below is the sample program that I put together in case there are others that are not familiar with it like I was. It is nothing fancy, just enough to get a basic feel for how it works.
The Add button adds item typed in into a listbox, but also updates the Collections. The Show button displays what is in the Collections itself. Highlighting one of the items in the listbox showing the Collections will remove it from the Collections.
Roger.
===========================================
Private Sub cmdClear_Click()
lstShow.Clear
lstList.Clear
txtItem.Text = ""
Label1.Caption = ""
num = 0
Debug.Print test.Count
For x = 1 To test.Count
test.Remove 1
Next x
cmdShow.Enabled = False
cmdClear.Enabled = False
txtItem.SetFocus
End Sub
Private Sub cmdExit_Click()
End
End Sub
Private Sub cmdRemove_Click()
Dim x As Integer
Dim y As Integer
Dim str As String
str = lstShow.Text
For x = 1 To test.Count - 1
' Debug.Print test.Item(x)
If test.Item(x) = str Then
test.Remove (x)
End If
lstShow.Clear
Next x
End Sub
Private Sub cmdShow_Click()
For x = 1 To test.Count
Next x
End Sub
Dim str As String
str = Trim(txtItem.Text)
' Do not add a blank field
If str = "" Then
txtItem.Text = ""
txtItem.SetFocus
Exit Sub
End If
num = num + 1
'Checks for duplicates within the collection and does not add it
For x = 1 To test.Count
If test.Item(x) = str Then
MsgBox "Match"
' num = num - 1
txtItem.Text = ""
txtItem.SetFocus
Exit Sub
End If
Next x
Label1.Caption = num
txtItem.Text = ""
txtItem.SetFocus
cmdShow.Enabled = True
cmdClear.Enabled = True
End Sub
0
LVL 75
Expert Comment
ID: 7009896
>> I wanted to give you an "A", but I your answer was only a direction to try, and no explanation or sample
code came with it. However, thank you for telling me about it. I learned something new.<<
The reason I did not elaborate beyond stating that it could be resolved with Collections was because I have no interest in suggesting solutions to questioners that do not abide by the EE Guidelines (http://www.experts-exchange.com/jsp/cmtyQuestAnswer.jsp) and leave questions open indefinitely.
In any case, thank-you for the points and for the opportunity of explaining my position.
Anthony
0
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View Single Post
09-09-2006, 02:52 PM #212 raiderfan19 Shazam! Join Date: Jun 2006 Posts: 7,889 Its not true. No matter how many 9s you make the first X equal to, when you multiply it by 10, there would be a 0 in the last spot when you subtract the second .9repeating. so essentially it would be x=.9repeating to infinity 10x=9.9repeating to infinity-1 with a 0 at the end 9x=8.9repeating to infinity - 1 with a 1 at the end x=.9repeating to infinity.
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(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Definition
Figure 1. Meaning of U and U0
AC cables are designed to be suitable for specific design voltages, which is called the "Voltage Grade" (or "Voltage Designation", "Voltage Class" or "Voltage Rating") of the cable. The voltage grade is commonly expressed in the following form: $U_{0} / U \,$
Where $U_{0} \,$ is the power frequency voltage between phase and earth (V rms)
$U \,$ is the power frequency voltage between two phase conductors (V rms)
For example, some standard IEC voltage grades are 0.6/1kV, 1.9/3.3kV, 3.8/6.6kV, 6.35/11kV, 12.7/22kV, 19/33kV, etc.
For standard insulated cables, $U$ is always higher than $U_{0}$ because as shown in Figure 1, the measurement of $U$ is across two layers of insulation (one per phase conductor), whereas the measurement of $U_{0}$ is only across one layer of insulation to earth. Obviously, the voltage rating is higher the more insulation there is.
Some manufacturers design cables with voltage grades where $U_{0}=U$, for example 1.0/1.0kV. In such cases, the $U$ rating can technically be higher, but the manufacturer has decided not to certify the cable for a higher voltage grade.
The manufacturer designs the cable (i.e. insulation, bedding, sheaths, etc) for the specified voltage grade. A cable can therefore be operated at voltages that do not exceed the voltage grade, e.g. a 0.6/1kV cable can be operated for any phase-to-earth and phase-to-phase voltages not exceeding 0.6kV and 1kV respectively. You may notice that LV cables are mainly specified to 0.6/1kV cables even though they are operated are much lower voltages (e.g. 240/415V, 220/380V, etc). This is due to the fact that the mechanical requirements of the insulation thickness are greater than the electrical requirements.
## Neutral Earthing Considerations
For MV and HV cables (>1kV), some consideration must be given to the neutral earthing arrangements of the system when specifying a voltage grade. In the US, the Association of Edison Illuminating Companies (AEIC) has the following specifications depending on the neutral earthing arrangements and clearing time of the protective device:
• 100% insulation level
• 133% insulation level
• 173% insulation level
The insulation levels above refer to the nominal phase-to-phase voltages. For example, a cable with a 133% insulation level on a 33kV system is rated for 133% the nominal phase-to-phase system voltage, i.e. 133% x 33kV = 43.89kV.
### 100% Insulation Level
The 100% is the insulation level normally used for cables on solidly earthed systems, or on any system where the protective device will clear earth faults within 1 minute.
### 133% Insulation Level
The 133% insulation level is specified for systems where the protective device is expected to clear earth faults within 1 hour, and is typically specified for high impedance earthed or unearthed systems.
### 173% Insulation Level
The 173% insulation level is specified for systems where the time to clear an earth fault is indefinite. This is typically recommended for unearthed or resonant earthed systems.
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• Theory and Examples
• Exercises
In the chapter Equations - Basics you have learned basic techniques for solving linear and quadratic equations. Now let's have a look at some special types of equations, namely:
• Equations with fractional expressions
• Exponential equations
### Equations with Fractions (Quotients)
In equations with fractional expressions, i.e. with quotients, the variable we want to solve for may occur also in the denominator of the fraction. Therefore, when defining the domain of the equation, we have to make sure we never divide by zero.
Example 1
$\begin {eqnarray} \frac{6}{x-5} \,=\,2 \quad \quad \quad G\,=\,\mathbb{R}\backslash \left\{ 5 \right\} \end{eqnarray}$ (all reals except 5)
Since $\,x=5\,$ is not an element of the domain, we can multiply both sides by $\,(x-5) \neq 0\,$:
$\begin{eqnarray} \frac{6}{x-5}\,\,&=&\,\,2 \quad \quad &\left| \,\,\cdot (x-5) \right. \\6\,&=&\,2\,(x-5) \\ 6\,&=&\,2x-10\quad \quad &\left| \,\,+10 \right. \\\,16\,\,&=&\,\,2x\quad \quad &\left| \,\,\div 2 \right.\\\,x\,\,&=&\,\,2 \end{eqnarray}$
Example 2
$\begin{eqnarray} \frac{x-1}{2x+1}\,\,=\,\,\frac{3}{7} \end{eqnarray}$
For x = –0.5 the denominator equals 0. Therefore the value -0.5 has to be excluded from the domain:
$G\,\,=\,\,\mathbb{R}\backslash \left\{ -0.5 \right\}$
The equation has to be multiplied by the least common multiple (LCM) of the two denominator terms:
$\begin{eqnarray} \frac{x-1}{2x+1}\,\,&=&\,\,\frac{3}{7} \quad \quad &\left| \,\,\cdot 7 \cdot (2x+1) \right. \\7x-7\,&=&\,6x+3 \quad \quad &\left| \,\,-6x+7 \right. \\ x&=&10 \end{eqnarray}$
Example 3
$\begin{eqnarray}\frac{3x+3}{2x-16}-4+\frac{2x+2}{x-8}=\frac{3x-3}{x-8} \end{eqnarray} \quad \quad \quad G=\mathbb{R}\backslash \left\{ 8 \right\}$
For x = 8 the denominator equals 0. The LCM of the denominators is 2x – 16 = 2(x – 8).
$\begin{eqnarray} \frac{3x+3}{2x-16}-4+\frac{2x+2}{x-8}&=& \frac{3x-3}{x-8} \quad &\left| \;\cdot \,(2x-16) \right. \\ 3x+3-4\cdot (2x-16)+2\cdot (2x+2)&=&2\cdot (3x-3) \quad \\ 3x+3-8x+64+4x+4&=&6x-6 \\ -x+71&=&6x-6 \quad &\left| \; -6x-71 \right. \\ -7x&=&-77 \quad &\left| \,\div (-7) \right. \\ x&=&11 \end{eqnarray}$
We focus here on equations with square roots. These are typically solved by squaring both sides of the equation in order to get rid of the radical expression. But you have to be aware of two important facts:
• Squaring both sides of an equation may enlarge the solution set of an equation!
It is possible that some of the solutions we find at the end of the squaring process are not valid solutions of the initial equation (we call them extraneous solutions). Therefore you must always check the "solutions" after having solved a radical equation, by inserting them into the initial equation!
• A radical can be eliminated successfully only if it is isolated before getting squared. If the radical is part of a sum, squaring doesn't eliminate it! (Remember that you have to square sides, not terms. This means that the squaring process runs according to the Binomial Formulas.)
Example 4
$\begin{eqnarray} \sqrt{x+2} +4\,&=&\,x \end{eqnarray}$
The term under the radical sign must not be negative. Therefore the domain of the equation ist limited:
$G\,=\,\left \{ x \in \mathbb{R} \,\left | \,x \geq -2 \right. \right\}$
To get rid of the radical, we have to isolate it:
$\begin{eqnarray} \sqrt{x+2} +4\,&=&\,x &\left|\;-4\right. \\ \sqrt{x+2}\,&=&\,x-4 \quad &\left|\; \text{square both sides}\right.\\ x+2 &=& {(x-4)^2}\end{eqnarray}$
Now the right hand side can be expanded according to the Binomial Formulas:
$\begin{eqnarray} x+2&=&{x^2}-8x+16 \quad &\left|\;-x-2 \right.\\ 0&=&{x^2}-9x+14 \end{eqnarray}$
${x_{1,2}} ={\large \frac{ 9 \pm \sqrt {81-56}}{2}} = {\large \frac{9 \pm 5}{2}} \quad \qquad {x_1}=7 \;\quad {x_2}=2$
If you insert both values into the initial equation you will notice that $\,{x_1}=7\,$ is a valid solution, but $\,{x_2}=2\,$ is not:
$\begin{eqnarray} \sqrt{7+2} +4\,&=&\,7 \quad \end{eqnarray}$ is true
$\begin{eqnarray} \sqrt{2+2} +4\,&=&\,2 \quad \end{eqnarray}$ is false!
What happened? The last equation before squaring both sides was:
$\sqrt{x+2}\,=\,x-4$
If you substitute 2 for x in the equation, you get a wrong statement:
$\sqrt{2+2}\,\neq \,2-4$
But: the two sides differ only by sign. By squaring, the minus sign vanishes, and you suddenly have a "solution" of the new equation! Obviously the squaring process has changed the solution set of the equation! The final equation isn't equivalent to the initial one.
Therefore the solution set of the original equation is: $L\,=\,\left\{ 7 \right\}$
Example 5
$\begin{eqnarray} \sqrt{4x+9} -2\,&=&\,\sqrt{3x-5} \end{eqnarray}$
The expressions under the square root sign must not be negative. Thereby, the condition $\;3x-5 \geq 0\;$ is more restrictive than $\;4x+9 \geq 0\;$; hence the domain of the equation is
$G\,=\,\left \{ x \in \mathbb{R} \,\left | \,x \geq \frac {5}{3} \right. \right\}$
In this example we cannot isolate both radicals simultaneously. We have to eliminate the radicals one at a time. Remember, too, that you have to square sums by use of the binomial formulas:
$\begin{eqnarray} \sqrt{4x+9} -2\,&=&\,\sqrt{3x-5} &\left|\;\text{square both sides}\right. \\ 4x+9 - 2 \cdot 2 \cdot \sqrt{4x+9} + 4\,&=&\,3x-5\quad &\left|\;-3x+5+4\,\sqrt{4x+9} \right.\\ x+18 &=& 4\,\sqrt{4x+9}&\left|\;\text{square both sides}\right.\\ {x^2}+36x+324 &=& 64x+144&\left|\;-64x-144\right.\\{x^2}-28x+180 &=& 0\end{eqnarray}$
The solutions of this quadratic equation are:
${x_{1,2}} ={\large \frac{ 28 \pm \sqrt {784-720}}{2}} = {\large \frac{28 \pm 8}{2}} \quad \qquad {x_1}=18 \;\quad {x_2}=10$
Check each solution in the original equation:
$\begin{eqnarray} \sqrt{40+9} -2\,&=&\,\sqrt{30-5} \quad \end{eqnarray}$ is true: $\; 7=7$
$\begin{eqnarray} \sqrt{72+9} -2\,&=&\,\sqrt{54-5} \quad \end{eqnarray}$ is true: $\; 5=5$
Both values are valid solutions of the initial equation.
$L\,=\,\left\{10;\,18 \right\}$
### Exponential Equations
An equation where the (unknown) variable appears in the exponent is called an exponential equation.
Examples
${2^x}=3$
$100 \cdot {1.04^x}=200$
When solving such equations we often use a third category of manipulations. It again has the equivalence preserving property, as described in the chapter Equations - Basics.
E3: Take the logarithm of both sides of an equation.
Example 6
\begin{align} & {{10}^{x}}\,\,=\,\,1000 \\ & \log {{10}^{x}}\,\,=\,\,\log 1000 \\ & x\cdot \log 10\,\,=\,3 \\ & x\,\,=\,\,3 \end{align}
In the second step we applied a logarithmic law (see chapter Logarithm):
$\log {{a}^{k}}\,\,=\,\,k\cdot \log a$
You can solve the example in an easier way:
\begin{align} & {{10}^{x}}\,\,=\,\,{{10}^{3}} \\ & x\,\,=\,\,3 \end{align}
By converting the 1000 to a power of 10 on the right-hand side of the equation, we get equal bases, so the same must hold for the exponents!
In the following example we make use of the logarithmic law again :
Example 7
Exercises: Equations with Fractions
1)
$\begin{eqnarray}\frac{z-1}{4z+2}=\frac{3}{14} \end{eqnarray}$
2)
$\begin{eqnarray}\frac{8}{x-3}=4 \end{eqnarray}$
3)
$\begin{eqnarray}\frac{9x}{25-x}=6 \end{eqnarray}$
4)
$\begin{eqnarray}\frac{2x}{x+1}+\frac{3}{2x}=2-\frac{1}{x} \end{eqnarray}$
(see section "Quadratic Equations" of chapter Equations - Basics):
5)
$\begin{eqnarray}\frac{x+3}{x}-5=\frac{x}{x-2} \end{eqnarray}$
6)
$\begin{eqnarray}\frac{w}{2w-3}-\frac{1}{2w}=\frac{3}{4w-6} \end{eqnarray}$
1)
$\begin{eqnarray} \sqrt{3x-2}=\sqrt{x+6} \end{eqnarray}$
2)
$\begin{eqnarray} 3+\sqrt{4{z^2}+3}=2z \end{eqnarray}$
(see section "Quadratic Equations" of chapter Equations - Basics):
3)
$\begin{eqnarray}\sqrt{13-4y}=2-y \end{eqnarray}$
4)
$\begin{eqnarray}x+2\sqrt{x}=3 \end{eqnarray}$
5)
$\begin{eqnarray}\sqrt{2x+5}-2 \,\sqrt{x-1}=1 \end{eqnarray}$
Exercises: Exponential Equations
1)
$\begin{eqnarray}{{2}^{5x-7}}=8 \end{eqnarray}$
2)
$\begin{eqnarray}{{4}^{6x-16}}=16 \end{eqnarray}$
More explanations and additional exercises for equations with fractions under the following links:
The Math Page: Equations with Fractions (comments, examples, exercises)
Compass Math: Solving Rational Equations (examples and exercises)
SOS Math: Rational Equations (detailled example and many exercices in "problem" section)
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# Qualitative vs quantitative?
by mark_d
Tags: qualitative, quantitative
P: 7 The difference between the number 0 and 1 is 1. So is the difference between 3 and 4. Proposition: The difference between 0 and 1 is much larger than the difference between 3 and 4. That is because regardless of how many decimal places you go to in the calculations, there is a qualitative difference in the first statement. Nothing contrasted with something. In the second place there are two "somethings" being compared. A fundamental category shift from "nothingness" to "being" is huge. Thoughts?
Sci Advisor P: 3,252 My thought: you should post subjective questions like this the Lounge section of the forum, not in the math sections.
P: 800
Quote by mark_d The difference between the number 0 and 1 is 1. So is the difference between 3 and 4. Proposition: The difference between 0 and 1 is much larger than the difference between 3 and 4. That is because regardless of how many decimal places you go to in the calculations, there is a qualitative difference in the first statement. Nothing contrasted with something. In the second place there are two "somethings" being compared. A fundamental category shift from "nothingness" to "being" is huge. Thoughts?
Sure, you can look at it in percentages. When you age from 10 to 20, that's an additional 100% of your age. When you go from 80 to 90, it's only a 12.5% increase. So it's much less significant.
And there is a very great difference between 0 and 1. Because if you think of some event, it can happen zero times -- it never happens. Or it can happen once, or it can be very commonplace and happen a million times.
But the difference between 0 and 1 in this context is the difference between something that never happens, and something that happens. So it's a huge difference.
Mentor
P: 21,215
Qualitative vs quantitative?
Quote by mark_d The difference between the number 0 and 1 is 1. So is the difference between 3 and 4. Proposition: The difference between 0 and 1 is much larger than the difference between 3 and 4.
No, it isn't. As you said above, the two differences are equal. 1 - 0 = 4 - 3 = 1.
Now, if you were talking about relative differences, that would be a different matter. The relative change from 0 to 1 is very much larger than that from 3 to 4, but if you're only talking subtraction (hence differences), then as I said, they're the same.
Quote by mark_d That is because regardless of how many decimal places you go to in the calculations, there is a qualitative difference in the first statement.
Both subtractions can be done without resorting to decimal places.
Quote by mark_d Nothing contrasted with something.
Zero is not "nothing". It is a number, just like any other real number.
Quote by mark_d In the second place there are two "somethings" being compared. A fundamental category shift from "nothingness" to "being" is huge. Thoughts?
P: 17 Mark44, You are in good company! Qualitative and quantitative changes are a vast subject of debate among reductionists and scientists. Qualitative changes will lead you into a better understanding of geometry and the complex domain. Just a little from Leibniz from here: Leibniz's ingenious attack on this Cartesian model of qualitative variety proceeds in two steps. The first step charges that motion alone is unable to account for qualitative variety at an instant: since all qualitative variety in the Cartesian system depends on motion, and there is no motion in an instant, it follows that in a Cartesian world there could be no qualitative variety at an instant.[12] The second step of Leibniz's argument charges that if the world is qualitatively homogenous at every instant, then it must be qualitatively homogenous over time as well. For if the world is qualitatively undifferentiated at each instant, then every instant will be qualitatively identical, and so the world as a whole will not undergo any qualitative change as it passes from one instant to the next. To use an anachronistic analogy, the two steps taken together imply that a Cartesian world would be like a filmstrip whose every frame was blank, and thus whose projection would not only be homogenous at each instant, but through time as well.
Mentor
P: 21,215
Quote by Thetes Mark44, You are in good company! Qualitative and quantitative changes are a vast subject of debate among reductionists and scientists. Qualitative changes will lead you into a better understanding of geometry and the complex domain. Just a little from Leibniz from here: Leibniz's ingenious attack on this Cartesian model of qualitative variety proceeds in two steps. The first step charges that motion alone is unable to account for qualitative variety at an instant: since all qualitative variety in the Cartesian system depends on motion, and there is no motion in an instant, it follows that in a Cartesian world there could be no qualitative variety at an instant.[12] The second step of Leibniz's argument charges that if the world is qualitatively homogenous at every instant, then it must be qualitatively homogenous over time as well. For if the world is qualitatively undifferentiated at each instant, then every instant will be qualitatively identical, and so the world as a whole will not undergo any qualitative change as it passes from one instant to the next. To use an anachronistic analogy, the two steps taken together imply that a Cartesian world would be like a filmstrip whose every frame was blank, and thus whose projection would not only be homogenous at each instant, but through time as well.
This makes me think of one of Zeno's paradoxes, in which the object under consideration is an arrow in flight. The argument goes like this: At each instant in time, the arrow is motionless at a precise point in space. The arrow is not moving toward that point, nor away from it, so the arrow must be motionless at all times.
This might pose a conundrum for philosophers, but mathematicians and physicists have this figured out.
P: 129 0 is not "nothing." This is where your problem arises. 0 is an integer. The magnitude of the difference between any two consecutive integers is 1. Here you are looking at the difference between two "somethings." (integers)
P: 800
Quote by Mark44 This makes me think of one of Zeno's paradoxes, in which the object under consideration is an arrow in flight. The argument goes like this: At each instant in time, the arrow is motionless at a precise point in space. The arrow is not moving toward that point, nor away from it, so the arrow must be motionless at all times. This might pose a conundrum for philosophers, but mathematicians and physicists have this figured out.
I do not agree. Mathematicians have resolved the paradox with the theory of limits; a theory that is only possible to carry out in a continuum.
Physical space may or may not be a continuum. We have no evidence either way; and some theorists suggest space may consist of discrete particles or points.
The real numbers are a mathematical model. It is unknown and (IMO) doubtful that the actual, physical space is like the real numbers. And if it is -- then you have all the problems of set theory suddenly becoming problems for experimental physics. Is the Continuum Hypothesis true in physical space? This question must have a definitive answer in the physical universe; even though it's independent of the usual axioms of set theory.
So you see that imagining that the physical universe is a continuum is fraught with problems. But if the universe isn't a continuum, then the mathematical solution to Zeno's paradoxes does not apply.
Therefore Zeno's paradoxes of motion are not resolved in physics; only in mathematics. In my opinion anyway.
Mentor
P: 21,215
Quote by Mark44 This makes me think of one of Zeno's paradoxes, in which the object under consideration is an arrow in flight. The argument goes like this: At each instant in time, the arrow is motionless at a precise point in space. The arrow is not moving toward that point, nor away from it, so the arrow must be motionless at all times. This might pose a conundrum for philosophers, but mathematicians and physicists have this figured out.
Quote by SteveL27 I do not agree. Mathematicians have resolved the paradox with the theory of limits; a theory that is only possible to carry out in a continuum.
I lumped physicists in with mathematicians because physicists are generally aware of mathematics concepts. That's all I meant by my remark.
Related Discussions Biology, Chemistry & Other Homework 1 Science & Math Textbooks 0 Chemistry 11
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# WA in SPOJ "Number of common divisors" (COMDIV)
Here’s my solution:
``````//author: contrapasso;
#include<bits/stdc++.h>
using namespace std;
constexpr int lim = 1'000'001;
int spf[lim];
void sieve() {
for(int i = 1; i < lim; ++i) spf[i] = vector<int>{2, i}[i & 1];
for(int i = 3; i * i < lim; i += 2) {
if(spf[i] == i) {
for(int j = i * i; j < lim; j += i) spf[j] = i;
}
}
}
void inferno() {
int n; cin >> n;
auto div = [](int x) {
int d = 1;
while(x != 1) {
int p = spf[x], k = 0;
while(spf[x] == p) {
x /= p, k++;
}
d *= k + 1;
}
return d;
};
for(int i = 0, x, y, d; i < n; ++i) {
cin >> x >> y;
d = __gcd(x, y);
cout << div(d) << "\n";
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
int t = 1; //cin >> t;
sieve();
while(t--) inferno();
return 0;
} //raise hellish blaze!
``````
Now, this is giving me WA and I can’t seem to find out why.
Will appreciate help (if any)
Edit: I think i have found the bug, sorry
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Cody
# Problem 32. Most nonzero elements in row
Solution 224225
Submitted on 27 Mar 2013 by Mats
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
%% a = [ ... 1 2 0 0 0 0 0 5 0 0 2 7 0 0 0 0 6 9 3 3]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
2 Pass
%% a = [ ... 1 2 0 0 0 0 5 0 0 6 9 -3 2 7 0 0 0 0 0 0]; r_correct = 3; assert(isequal(fullest_row(a),r_correct))
3 Pass
%% a = [ ... 1 0 0 0 0 0 0 0 0 0 0 0 0 2 3]; r_correct = 5; assert(isequal(fullest_row(a),r_correct))
4 Pass
%% a = [ ... 0 0 0 -3 0 0]; r_correct = 4; assert(isequal(fullest_row(a),r_correct))
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# GATE EC 2022: Electromagnetics Quiz 5
Attempt now to get your rank among 122 students!
Question 1
If flux density D = 5xax – 4yay + kzaz μC/m2 and magnetic flux density B= 2ay mT. Find the value of k to satisfy Maxwell’s equation if σ = 0 and ρV = 0.
Question 2
At high frequencies, which parameter is significant?
Question 3
Let μ=10-5H/m, ϵ = 4 ×10-9F/m, σ =0 and ρV = 0. Find K so that D = 6ax – 2ay + 2zaz nC/m2 and H = kxax – 10yay - 25 zaz A/m satisfies Maxwell’s equations.
Question 4
If .D = ε.E and .J = σ.E in a given material; then the material is said to be :
Question 5
Consider a vector field . The closed loop line integral can be expressed as
Question 6
An infinitely long uniform solid wire of radius a carries a uniform dc current of density.
A hole of radius b(b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.
The magnetic field inside the hole is
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# Discussion about Math Kernal Machine Library?
249 views
**What is Math Kernal Library?**
Math Kernel Library (Intel MKL) is a library of optimized math routines for science, engineering, and financial applications. Core math functions include BLAS, LAPACK, ScaLAPACK, sparse solvers, fast Fourier transforms, and vector math. The routines in MKL are hand-optimized specifically for Intel processors.
The library supports Intel processors and is available for Windows, Linux and macOS operating systems.
Main Categories for Math Kernal Library
• Linear algebra: BLAS routines are vector-vector (Level 1), matrix-vector(Level 2) and matrix matrix(Level 3) operations for real and complex single and double precision data. LAPACK consists of tuned LU, Cholesky and QR factorizations, eigenvalue and least squares solvers.
• MKL includes a variety of Fast Fourier Transforms (FFTs) from 1D to multidimensional, complex to complex, real to complex, and real to real transforms of arbitrary lengths.
• Vector math functions include computationally intensive core mathematical operations for single and double precision real and complex data types. These are similar to libm functions from compiler libraries but operate on vectors rather than scalars to provide better performance. There are various controls for setting accuracy, error mode and denormalized number handling to customize the behavior of the routines.
• Statistics functions include random number generators and probability distributions. optimized for multicore processors. Also included are compute-intensive in and out-of-core routines to compute basic statistics, estimation of dependencies etc.
• Data fitting functions include splines (linear, quadratic, cubic, look-up, stepwise constant) for 1-dimensional interpolation that can be used in data analytics, geometric modeling and surface approximation applications.
• Deep Neural Network
• Partial Differential Equations
• Nonlinear Optimization Problem Solvers
Video for Math Kernal Library
#### References
https://en.wikipedia.org/wiki/Math_Kernel_Library
posted Aug 31, 2018
## Related Articles
What is Linear regression?
Linear regression is a linear system and the coefficients can be calculated analytically using linear algebra. ...
Linear regression does provide a useful exercise for learning stochastic gradient descent which is an important algorithm used for minimizing cost functions by machine learning algorithms.
Linear regression is a very simple approach for supervised learning. Though it may seem somewhat dull compared to some of the more modern algorithms, linear regression is still a useful and widely used statistical learning method. Linear regression is used to predict a quantitative response Y from the predictor variable X.
Linear Regression is made with an assumption that there’s a linear relationship between X and Y.
Linear regression is a linear model, e.g. a model that assumes a linear relationship between the input variables (x) and the single output variable (y). More specifically, that y can be calculated from a linear combination of the input variables (x).
When there is a single input variable (x), the method is referred to as simple linear regression. When there are multiple input variables, literature from statistics often refers to the method as multiple linear regression.
Video for Linear Regression
What is H2O in Machine Learning?
H2O.ai is a leader in the 2018 Gartner Magic Quadrant for Data Science and Machine Learning Platforms.
H2O is open-source software for big-data analysis. It is produced by the company H2O.ai. H2O allows users to fit thousands of potential models as part of discovering patterns in data.
The H2O software runs can be called from the statistical package R, Python, and other environments. It is used for exploring and analyzing datasets held in cloud computing systems and in the Apache Hadoop Distributed File System as well as in the conventional operating-systems Linux, macOS, and Microsoft Windows.
The H2O software is written in Java, Python, and R. Its graphical-user-interface is compatible with four browsers: Chrome, Safari, Firefox, and Internet Explorer.
H2O is a Java-based software for data modeling and general computing. The H2O software is many things, but the primary purpose of H2O is as a distributed (many machines), parallel (many CPUs), in memory (several hundred GBs Xmx) processing engines.
There are two levels of parallelism:
- within node
- across (or between) nodes
The goal of H2O is to allow simple horizontal scaling to a given problem in order to produce a solution faster. The conceptual paradigm MapReduce, along with a good concurrent application structure, enable this type of scaling in H2O.
Video for H2O
What is Machine Learning?
Machine learning is an application of artificial intelligence (AI) that provides systems the ability to automatically learn and improve from experience without being explicitly programmed. Machine learning focuses on the development of computer programs that can access data and use it learn for themselves.
Machine learning is a field of computer science that gives computers the ability to learn without being explicitly programmed
Machine learning is closely related to (and often overlaps with) computational statistics, which also focuses on prediction-making through the use of computers. It has strong ties to mathematical optimization, which delivers methods, theory and application domains to the field. Machine learning is sometimes conflated with data mining, where the latter subfield focuses more on exploratory data analysis and is known as unsupervised learning.Machine learning can also be unsupervised and be used to learn and establish baseline behavioral profiles for various entities and then used to find meaningful anomalies.
The process of learning begins with observations or data, such as examples, direct experience, or instruction, in order to look for patterns in data and make better decisions in the future based on the examples that we provide. The primary aim is to allow the computers learn automatically without human intervention or assistance and adjust actions accordingly.
Some machine learning methods
• Supervised machine learning algorithms
• unsupervised machine learning algorithms
• Semi-supervised machine learning algorithms
• Reinforcement machine learning algorithms
What is MLlib?
MLlib (Spark) is Apache Spark’s machine learning library. Its goal is to make practical machine learning scalable and easy. It consists of common learning algorithms and utilities, including classification, regression, clustering, collaborative filtering, dimensionality reduction, as well as lower-level optimization primitives and higher-level pipeline APIs.
Main Benefits
• Ease of Use
• Performance
• Runs Everywhere
MLlib contains many algorithms and utilities.
ML algorithms include:
• Classification: logistic regression, naive Bayes,...
• Regression: generalized linear regression, survival regression,...
• Decision trees, random forests, and gradient-boosted trees
• Recommendation: alternating least squares (ALS)
• Clustering: K-means, Gaussian mixtures (GMMs),...
• Topic modeling: latent Dirichlet allocation (LDA)
• Frequent itemsets, association rules, and sequential pattern mining
Spark revolves around the concept of a resilient distributed dataset (RDD), which is a fault-tolerant collection of elements that can be operated on in parallel. There are two ways to create RDDs: parallelizing an existing collection in your driver program, or referencing a dataset in an external storage system, such as a shared filesystem, HDFS, HBase, or any data source offering a Hadoop InputFormat.
Video for MLib
What is Lasagne?
Lasagne is a lightweight library to build and train neural networks in Theano.
Features:
• Supports feed-forward networks such as Convolutional Neural Networks (CNNs), recurrent networks including Long Short-Term Memory (LSTM), and any combination thereof
• Allows architectures of multiple inputs and multiple outputs, including auxiliary classifiers
• Many optimization methods including Nesterov momentum, RMSprop and ADAM
• Freely definable cost function and no need to derive gradients due to Theano's symbolic differentiation
• Transparent support of CPUs and GPUs due to Theano's expression compiler
Main Principles
• Simplicity: Be easy to use, easy to understand and easy to extend, to facilitate use in research
• Transparency: Do not hide Theano behind abstractions, directly process and return Theano expressions or Python / numpy data types
• Modularity: Allow all parts (layers, regularizers, optimizers, ...) to be used independently of Lasagne
• Pragmatism: Make common use cases easy, do not overrate uncommon cases
• Restraint: Do not obstruct users with features they decide not to use
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How to Install
pip install -r https://raw.githubusercontent.com/Lasagne/Lasagne/master/requirements.txt
pip install https://github.com/Lasagne/Lasagne/archive/master.zip
Video for Lasagne
What is Keras?
Keras is a high-level neural networks API, written in Python and capable of running on top of TensorFlow, CNTK, or Theano. It was developed with a focus on enabling fast experimentation. Being able to go from idea to result with the least possible delay is key to doing good research.
Features
• Allows for easy and fast prototyping (through user friendliness, modularity, and extensibility).
• Supports both convolutional networks and recurrent networks, as well as combinations of the two.
• Runs seamlessly on CPU and GPU.
Main Benefits
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• Modularity. A model is understood as a sequence or a graph of standalone, fully-configurable modules that can be plugged together with as few restrictions as possible. In particular, neural layers, cost functions, optimizers, initialization schemes, activation functions, regularization schemes are all standalone modules that you can combine to create new models.
• Easy extensibility. New modules are simple to add (as new classes and functions), and existing modules provide ample examples. To be able to easily create new modules allows for total expressiveness, making Keras suitable for advanced research.
• Work with Python. No separate models configuration files in a declarative format. Models are described in Python code, which is compact, easier to debug, and allows for ease of extensibility
Python Install
pip install keras
Video for Keras
What is MLlib?
MLlib stands for Machine Learning Library (MLlib)
MLlib is Spark’s scalable machine learning library consisting of common learning algorithms and utilities, including classification, regression, clustering, collaborative filtering, dimensionality reduction, as well as underlying optimization primitives, as outlined below:
• Data types
• Basic statistics
• Classification and regression
• Collaborative filtering
• Clustering
• Dimensionality reduction
• Feature extraction and transformation
• Optimization
Spark Core is the foundation of the overall project. It provides distributed task dispatching, scheduling, and basic I/O functionalities, exposed through an application programming interface centered on the RDD abstraction This interface mirrors a functional/higher-order model of programming: a "driver" program invokes parallel operations such as map, filter or reduce on an RDD by passing a function to Spark, which then schedules the function's execution in parallel on the cluster.
These operations, and additional ones such as joins, take RDDs as input and produce new RDDs. RDDs are immutable and their operations are lazy; fault-tolerance is achieved by keeping track of the "lineage" of each RDD so that it can be reconstructed in the case of data loss. RDDs can contain any type of Python, Java, or Scala objects.
The Video for MLlib Spark
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The range of f(x) = (4x + 2x + 1]; ...
Question
# The range of f(x) = (4x + 2x + 1]; (where :) denotes greatest integer function) is (1) 1* - {1, 2} (2) 1 (3) (4) /
JEE/Engineering Exams
Maths
Solution
120
4.0 (1 ratings)
( 1^{x}+2^{x}+1 ) has a range (1000)( [text { put } x=-infty] ) ( thereforeleft[4^{x}+2^{x}+1right] ) will be ( I^{+} ) ans) ( 4 I^{+} )
Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free
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Question
A bank converts US dollars to European Euros at a rate of \$1.00 to 0.88
Euros. Kelsey wants 100 euros for her trip to Europe.
How much money, in dollars, does Kelsey need to give the bank?
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MEM427_Lecture_1_Review_of_MoM
# MEM427_Lecture_1_Review_of_MoM -...
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MEM427 Introduction to Finite Element Method MEM427 Introduction to Finite Element Methods Chapter 1 Review of Mechanics of Materials 1 MEM427 Introduction to Finite Element Method COURSE OUTLINE •Rev iew of Mechanics of Materials, with emphasis on Energy Principles atrix Methods of Structural Analysis irect Formulation • Matrix Methods of Structural Analysis Direct Formulation • Formulation of Truss Element Based on Energy Principles •One dimensional Structural Elements (Axially Loaded Members, d F) Beams, and Frames) •Two Dimensional Plane Stress/Strain Elements • Isoparametric Elements and Numerical Integration •3 D Solid Elements, Plate and Shell Elements •Trans ient and Dynamic Analysis eat Transfer and Fluid Mechanics Heat Transfer and Fluid Mechanics •Convergence of Solutions, h Method versus p Method •Advanced Topics Chapter 1 Review of Mechanics of Materials 2 Commercial FEM packages used in this course: ANSYS, LS DYNA MEM427 Introduction to Finite Element Method COURSE AND LAB SCHEDULE (Subject to change) Wk Lectures Weekly Labs and Projects ate opics ate pics Date Topics Date Topics 19 / 2 0 R e v i e w of Mechanics of Materials 9/24 No Labs 29 / 2 7 T h e Direct Method (2 D Truss Elements) 10/1 Trusses (Command Line, File) 3 10/4 Energy Principles and Approximate Methods 10/8 Trusses (GUI) 4 10/11 Columbus Day (University Holiday) 10/15 No Labs 5 10/18 1 D Elements (Trusses, Beams, Frames) 10/22 Beams and Frames 6 10/25 2 D Plane Elements, Numerical Integration 10/29 2 D Plane Problems 7 11/1 3 D Solid Elements, Plate and Shell Elements 11/5 3 D Solids; Notepad Input / ynamic and Transient nalysis 1/1 atural requencies and Modes 8 11/8 Dynamic and Transient Analysis 11/12 Natural Frequencies and Modes 9 11/15 Mid term Examination 11/19 LS DYNA 10 11/22 Optimizations Weighted Residual Methods 11/26 Thanksgiving Break Chapter 1 Review of Mechanics of Materials 3 11 11/29 Advanced Topics 12/3 Term Project Presentations MEM427 Introduction to Finite Element Method ROUP TERM PROJECTS GROUP TERM PROJECTS Students will form teams of 4 members to work on group term jt Th it f t projects. The requirements for group term projects are: A written proposal (20%; two page limit, due Friday, November 22) Title page with names of members of the team roject description Project description Approach Deliverables final report (40%; 10 age limit plus appendices; due Friday A final report (40%; 10 page limit plus appendices; due Friday, December 3) A final presentation (40%, 15 minutes, on Friday, December 3) Chapter 1 Review of Mechanics of Materials 4
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MEM427 Introduction to Finite Element Method OURSE OBJECTIVES COURSE OBJECTIVES •Learn theories of finite element methods ain hands n E analysis experiences •Gain hands on FE analysis experiences using commercial FE codes GRADING omework assignments 0% •Homework assignments 20% •Mid term examination 30% eekly in ass labs 0% •Weekly in class labs 10% •Weekly projects 20% roup term project 0%
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## This note was uploaded on 11/14/2010 for the course MEM 427 taught by Professor Tein-mintan during the Fall '10 term at Drexel.
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## Department of Whole Vehicle Engineering - Audi Hungaria Faculty of Automotive Engineering
Computational Fluid Dynamics
Course Summary
The goal of this course is to make engineering students informed users of Computational Fluid Dynamics (CFD) who can not only run CFD simulations but also fully understand their mathematical background and are able to make conscious choices of the appropriate numerical parameters. To achieve this, the relevant Fluid Mechanics laws, the governing equations as well as the applied numerical methods will be reviewed. Also, students will complete three practically oriented simulation assignments and learn how to present these in written form according to the internationally accepted standards. It is assumed that enrolled students have had at least one Fluid Mechanics course completed at the undergraduate level.
Topics discussed:
• Introduction: Motivation, What is Computational Fluid Dynamics (CFD), Role of CFD in Vehicle Engineering and design
• Governing equations in CFD 1: Review of continuum concept, mass, energy and momentum conservation and derivation of the Navier-Stokes equations
• Governing equations in CFD 2: Flux vector formulation of the N-S equations, Conservative vs. primitive forms, Euler equations, Model equations
• Classification of differential equations: ODE’s vs. PDE’s, Linear vs. non-linear equations, First-order vs. higher-order equations, Conservative vs. non-conservative forms
• Classification of Partial Differential Equations (PDE’s): Determining the nature of PDE’s (elliptic, parabolic, hyperbolic), Physical meaning for fluid flows, Computational meaning for fluid flows, Boundary and initial conditions for PDE’s
• Turbulence 1: Sources and physics of turbulence, Kolmogorov length scale, Differences between turbulence modelling, Large Eddy Simulation (LES) and Detached Eddy Simulation (DES) and Direct Numerical Simulations (DNS). Limitations and applicability.
• Turbulence 2: Free turbulent flows, Boundary layers near solid walls, Turbulence modeling in CFD, Wall functions and implications for grid generation
• Numerical solution of PDE’s: Selection of mathematical model, Selection of discretization method (Finite Difference, Finite Volume, Finite Element, Spectral Method)
• Grid generation: Structured vs. unstructured grids, Grid transformation, Cartesian grids, Zonal or block-structured grids, Hybrid grids, Moving mesh techniques (Sliding mesh, CHIMERA grids) Deforming mesh techniques, Adaptive grids, Multigrid methods and their relation to grid generation, Basic guidelines for grid generation
• Boundary treatment: Boundary conditions, Boundary treatment (Changing the numerical method at edges, Changing the computational domain at edges), Solid Wall boundary treatment, Far-field boundary treatment, Non-reflecting boundaries
• Solution techniques for the discretized equations: Explicit vs. implicit formulations, Solutions techniques for explicit method (Lax-Wendroff, MacCormack, Runge-Kutta), Solution techniques for implicit methods (Direct methods /Gaussian elimination, Cramer’s rule/, Indirect methods /Thomas algorithms, point-iterative methods, approximate factorization/)
• Errors and uncertainty in CFD: Sources of error, Sources of uncertainty, Stability analysis of numerical errors (Discrete Perturbation analysis, Von Neumann Stability Analysis, Multidimensional considerations), The Courant-Friedrich-Loewy number (CFL), Stability vs. accuracy, Local vs. Global time stepping, Evaluation of convergence (Iterations convergence: residuals, Grid convergence, Time step convergence), Characteristic features related to stability (Consistency, Boundedness, Transportiveness)
• Special topics in CFD: specifics of the Finite Volume Method, Riemann solvers, upwinding, higher order methods
• Summary and review
Course administrator - Dr. T. Jakubík - SZE
Instructor - Dr. D. Feszty - AUDI Hungaria Motor Kft.
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# Physics
Articles related to class 9, 10, 11, 12 and B.Sc.
## What is trajectory
Introduction to the concept of trajectory A trajectory is a path followed by an object with mass in curvilinear motion (curved path) as a function of time. Mathematically, It is defined by the equation in x-y coordinates. A famous example would be projectile motion like the ball thrown at an angle or a bullet fired …
## Stable unstable and neutral equilibrium Questions for JEE Examination
What is Equilibrium
A rigid body is supposed to be in equilibrium if there is no change in translational motion and no change in rotational motion
The general conditions for equilibrium are as follows
(i) The total force must be zero
(ii) The total torque about any axis must be zero.
In many cases it is convenient to consider torques about axes through the centre of gravity.
Types of equilibrium
We can divide the equilibrium into two categories With respect to the state of a body
1. Static equilibrium.
2. Dynamic equilibrium.
## Physics of Amusement Park’s
You must have ride Roller coaster or ferris wheel in the amusement park. Its up and down and moving in the circle must have thrilled and fascinated you Today I would like to give a brief details about those machine and how you feel different in those rides. So this article is about the Physics of Amusement Park’s.
The generation of A.C. is cheaper than that of D.C.
A.C. machines are simple , robust and do mot require much attention for their repairs and maintainance during their use.
Wide range of voltages are obtained by the use of transformer.
## Charge Density Formulas and Solved Example
In this article, we would learn about charge density and its formulas and would also solve some problems related to this concept. We encounter electric charge density while calculating electric field from various continuous charge distributions like linear, surface and volume. We also need the concept of charge density while studying current electricity. What is Charge Density? …
## Use of units in physics
Introduction about physics and studying the subject This article is mainly about use of units in physics. But before going further let us first discuss about how to avoid problems that students face while studying physics. When you study physics you must have appropriate skills or you would have to struggle a lot while studying …
## How to solve vector algebra problems in Physics
You will come across vectors in physics problem very frequently.So it is must to know to solve the vector mathematics in short time.And Making sure you have done in correctly.You will find vectors in every module be it mechanics, electrostatics, magentics
## What is a Series Circuit?
What is a Series Circuit? Electricity is one of the purest and the most efficient forms of energy humans have ever produced. Mankind has also fabricated ways to use this energy to perform all kinds of work using the intermediate medium of circuits. As we know, these circuits are hugely classified into 2 types of …
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I"ve just started looking at the axioms of 3D Geometry. The first one that I encountered is this one:
"Three non collinear points define a plane" or " Given three non collinear points, only one plane goes through them"
I know that it is an axiom and it is taken to be true but I don"t understand the intuition behind it. I understand that if I take one point or any number of collinear points, then I can draw infinite planes just by rotating around the line that connects these points, but why do we need 3 non collinear points to define a plane, why not more? And why, given three non collinear points, does only one plane go through them? Why not two or three?
Two points determine a line (shown in the center). There are infinitely many infinite planes that contain that line. Only one plane passes through a point not collinear with the original two points:
Two points determine a line \$l\$. Thus, as you say, you can draw infinitely many planes containing these points just by rotating the line containing the two points. So you find a set of infinitely many planes containing a common line. For any third point not on \$l\$ then there is only one of these planes containing it.
You are watching: Two dimensional using 3 noncollinear points
An analogy is the same problem is lower dimension. Take a point in a plane. There are infinitely many lines through it. Now take a second point different from the first. Then there is a unique line among the infinitely many given that contains the two points.
A plane is a vectorial space whose dimension is \$ 2\$.its base contains exactly two independent vectors.If your three points \$ A,B,C \$ do not lie in the same line, you can take as a base, the couple \$ (vecAB,vecAC) \$.
Thanks for contributing an answer to les-grizzlys-catalans.orgematics Stack Exchange!
But avoid
Asking for help, clarification, or responding to other answers.Making statements based on opinion; back them up with references or personal experience.
Use les-grizzlys-catalans.orgJax to format equations. les-grizzlys-catalans.orgJax reference.
See more: State Agency Regulations Take Precedence Over Conflicting Federal Agency Regulations.
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# Suppose you buy the car of your dreams. Research the price of your car. At the end of years, the value of your car is
Suppose you buy the car of your dreams. Research the price of your car. At the end of years, the value of your car is given by the sequence = (Dream car Price)(3/4) , = 1, 2, 3, ….. Find the fifth term and explain what this value represents. Describe the th term of the sequence in terms of the value of your car at the end of each year. Purchase the answer to view it Purchase the answer to view it
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# Circle O is inscribed in square ABCD as shown above. The area of the s
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Math Expert
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Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink]
### Show Tags
10 Jan 2016, 08:10
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Difficulty:
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Question Stats:
67% (01:24) correct 33% (01:10) wrong based on 113 sessions
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Circle O is inscribed in square ABCD as shown above. The area of the shaded region is approximately
A. 10
B. 25
C. 30
D. 50
E. 75
Attachment:
2016-01-10_1908.png [ 4.9 KiB | Viewed 1600 times ]
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Re: Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink]
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10 Jan 2016, 08:29
Bunuel wrote:
Circle O is inscribed in square ABCD as shown above. The area of the shaded region is approximately
A. 10
B. 25
C. 30
D. 50
E. 75
Attachment:
2016-01-10_1908.png
Area of Inscribed Circle = \pi/4 * Area of square
Area of Square = 10*10= 100
Area of circle = \pi/4*100 = 25\pi
Area of parts outside circle but in square = 100-25\pi =100-75 (Approximately)=25
Area of shaded reason = 25/4=6 (Approximately)
The nearest value to 6 is 10. So in test conditions i will opt for A.
Please let me know where am i making mistakes, if i am doing so. I am not sure if the approximate answers can be so further.
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Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink]
### Show Tags
06 Mar 2017, 20:13
Bunuel wrote:
Circle O is inscribed in square ABCD as shown above. The area of the shaded region is approximately
A. 10
B. 25
C. 30
D. 50
E. 75
Attachment:
2016-01-10_1908.png
Area of square = 100
Area of circle = 25π
Area of the shaded region = $$\frac{(100-25π)}{4}$$ ~ $$\frac{(100-78.75)}{4}$$ ~ $$\frac{21}{4}$$=5.x
Closest is A, which is almost double the calculated value. :S
Can someone help ?
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Re: Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink]
### Show Tags
08 Mar 2017, 14:23
2
This question can be solved logically, in my opinion. Instead of using heavy calculation, we can deduce that the radius is 5- that the radius shown doesn't touch the edge of the square is, again in my opinion, an optical illusion. Actually, if you draw a line from the middle point of any side of the square to the center the radius must be 5. Knowing that the radius must be 5, we can segment a portion of the square- 5 by 5 would be the area of the square that the shaded triangle is contained within; hence, through the process of elimination we can arrive at answer "A." Though this is probably a shoddy method when considering a 700 level question.
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Re: Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink]
### Show Tags
10 Mar 2017, 12:30
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Re: Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink]
### Show Tags
08 Jul 2017, 05:35
answer should be around 5.25. n none of the option matches.
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Re: Circle O is inscribed in square ABCD as shown above. The area of the s [#permalink] 08 Jul 2017, 05:35
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### The Pet Graph
Tim's class collected data about all their pets. Can you put the animal names under each column in the block graph using the information?
### You Never Get a Six
Charlie thinks that a six comes up less often than the other numbers on the dice. Have a look at the results of the test his class did to see if he was right.
### Presenting the Project
Have a look at all the information Class 5 have collected about themselves. Can you find out whose birthday it is today?
# How Big Are Classes 5, 6 and 7?
## How Big Are Classes 5, 6 and 7?
Class 4 were making graphs. Ben, Ali, Katie and Charlene decided to make graphs of the sizes of the seven classes in the school.
They went to the office to collect the numbers of children in all the other classes. Ben and Ali wrote down the numbers for Classes $1$, $2$ and $3$. Katie and Charlene wrote down the numbers for Classes $5, 6$ and $7$. Of course they all knew the number of children in Class $4$.
Ben and Ali drew a block graph. It looked like this:
Katie and Charlene decided to make theirs differently. They drew this graph:
Miss Brown came along to look at their work.
"It's an interesting graph," she said, "but it's difficult to tell how many children are in each class. Please can you make a scale on it so we can all read it?"
Katie and Charlene did what they were told.
How many children were there altogether in Classes $5, 6$ and $7$?
### Why do this problem?
This problem is useful when learners are being introduced to or revising different ways of representing data and the scales needed when making bar charts and pictograms. The question encourages them to contrast different ways of representing similar data and helps to make explicit their interpretation of what the data represents in order to solve the problem.
Later on in secondary school children often leave out the labels on axes so rendering the representation meaningless. This question will help children to realise the significance of the labels.
### Possible approach
You could start by doing another, simpler problem, for example, The Pet Graph. Alternatively, you could get the numbers of children in classes in your own school and make a rough bar chart of them on the board.
After this learners could work in pairs on the actual problem on a computer or from a printed sheet so that they are able to talk through their ideas with a partner. They should be encouraged to think about the question and talk about it in pairs before a class discussion.
At the end of the lesson you should discuss not only the answers to the problem itself, and how these were reached, but also to stress why it is important not to leave out the labels on axes of a graph.
### Key questions
What is a reasonable number to try?
What can you find out about Class $4$ from the bar graph?
Can you work out the number that each pin-man stands for from the two graphs for Class $4$?
### Possible extension
Learners could make different graphs and representations of the numbers in the classes in their own school or do another problem such as You Never Get a Six.
### Possible support
Suggest trying The Pet Graph first which is simpler and easier to understand.
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+0
# hard geometry
0
69
1
A parallelogram has perimeter 40 and altitudes of length 4 and 7. Find the sine of the smaller vertex angle.
May 13, 2020
#1
+21953
+1
The parallelogram has a shorter side (let's call it 'x') and a longer side (which is '20-x').
The altitude of length 4 goes to the longer side. Using the area of a parallelogram to be its side times its altitude,
the area of the parallelogram is 4·(20 - x).
Using the shorter side (x) and its altitude (7) to find the area, we get 7·x.
Therefore: 4·(20 - x) = 7·x
80 - 4x = 7x
80 = 11x
x = 80/11
To find the sine of the smaller vertex angle: sine(angle) = 4 / (80/11) = 44/80 = 11/20
May 13, 2020
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1,000s of Fiveable Community students are already finding study help, meeting new friends, and sharing tons of opportunities among other students around the world! Don't miss out! 🎉
# 4.5 Stoichiometry & Calculations
hope arnett
dalia savy
(editor)
⏱️ October 22, 2020
📅
## What Is Stoichiometry Useful For?
In the previous section, we looked at qualitatively analyzing chemical reactions. Now, we’ll learn how to quantitatively analyze reactions using stoichiometry! It may seem like a lot of math at first, but once you do more practice, you’ll be more confident and become a stoichiometry master 👨🏫
In earlier units, you learned about moles, molar mass/molar volume, molarity, and Avogadro's number. Stoichiometry is all about using mole ratios and these measurements and manipulating them to get to our desired unit. For this reason, it’s important to comfortably know the relationship between these ideas as we do mole ratios. Below is a list of these measurements: 1 mole = molar volume (gas) (At STP, this is 22.4 L/mol. That number is also on your AP equation sheet)
1 mole = 6.022 x 10^23 particles (This is Avogrado’s number, which is on your AP equation sheet!)
1 mole = molar mass (Each element’s is on the periodic table. Hydrogen’s is 1.008 g/mol) You can find mole ratios by looking at a chemical reaction🧪. The coefficients of the reaction represent the ratio of molecules necessary in order for the reaction to complete. In the reaction below, 1 mole of C2H5OH (ethanol) and 1 mole of oxygen gas are needed to produce 2 moles of carbon dioxide and 3 moles of water. A sample mole ratio would be 1 mole of O2 to 2 moles of CO2. Keep in mind that 2 moles of CO2 to 1 mole of O2 would mean the same thing.
## Practice Problems
### Example 1:
How many moles of potassium are required to fully react with 11.6 moles of water?
Step 1
Write the chemical reaction, if it’s not given. Be sure to balance it. 2K (s) + 2H2O (l) --> 2KOH (aq) + H2 (g)
Step 2
Identify the known measurement. In this case, we’re given 11.6 moles of water💧.
Step 3
Since our given value is already in moles, we just need to write the appropriate mole ratio. We want a ratio that cancels❌ out the known measurement units (moles of H2O) and brings in the unit we want (moles of K).
Looking at the chemical equation, for every 2 moles of H2O, we need 2 moles of K.
💡Note: Writing out the units and the molecule the value belongs to will help you exponentially. Please do it!
One way to make sure you’re writing the right mole track of the units is to cross out whatever cancels out and circle the unit you’re left with, like so:
### Example 2:
If you have 105.2 g of ethanol (C2H5OH), what is the maximum volume of carbon dioxide that can form at STP?
Step 1
Write the chemical equation and balance it.
C2H5OH (aq) + O2 (g) --> 2CO2 (g) + 3H2O (g)
Step 2
Identify the known measurement: 105.2 g of ethanol.
Step 3
Since our given value is in grams, we need to convert it to moles. To do this, we will calculate the molar mass (g/mol) of C2H5OH.
2(12.01g) + 6(1.008g) + 16.00g = 46.07 g/mol
Using this value, we can make a mole ratio.
105.2 g of C2H5OH * 1 mole C2H5OH/46.06 g of C2H5OH
Step 4
Uh oh!😩 We don’t want moles of C2H5OH--we want the volume of carbon dioxide. Now that we have moles, though, we can go over to CO2 with a mole ratio. Looking at the equation, for every 1 mole of C2H5OH, we can make up to 2 moles of CO2.
Step 5
Almost there🎉! We have a unit of CO2, but we want it’s volume. Since this reaction occurs at STP, we can use the value of the molar volume at STP.
Good work! With 105.2g of ethanol, we could make 102. L of CO2😮.
You will see stoichiometry used with the ideal gas law (PV=nRT) and molarity (moles/mass or moles/volume, often denoted as M). Check out the practice problems to see how you’ll see it in problems.
## General Steps
Here is a neat list of the general steps for tackling stoichiometry problems:
1. Write out the balanced chemical equation, if not given
2. Identify the known measurements given to you
3. If the known measurement is in grams, convert it to moles
4. Multiply by a mole ratio using the coefficients in the chemical equation.
5. Repeat steps 3-4 as many times as necessary.
6. Cross out your units and circle the final one to make sure you set it up correctly!
## Try on your own!🤔
The following reaction occurs at STP:
How many particles of BrF will be produced with 160.0g of Br2?
Step 1 includes using the molar mass of Br2. Step 2 is the mole ratio of Br2 to BrF. Step 3 uses Avogadro's number to convert to the number of particles of BrF.
🎥 Watch: AP Chemistry - Stoichiometry (Part 2)
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# On constructing APN permutations using subfunctions
V. A. Idrisova
Результат исследования: Научные публикации в периодических изданияхстатьярецензирование
## Аннотация
Our subject for investigation is the problem of APN permutation existence for even number of variables. In this work, we consider 2-to-1 functions that are isomorphic to (n − 1)-subfunctions of APN permutations. These 2-to-1 functions can be obtained with a special algorithm which searches for 2-to-1 APN functions that are potentially EA-equivalent to permutations. The algorithm is based on constructing special symbol sequences that are called admissible. It is known that (n − 1)-subfunction of an APN permutation can be represented as a differentially 4-uniform 2-to-1 function that takes values from the half of the Boolean cube. Therefore, the following algorithm can be used to search for APN permutations. On the first step all the possible admissible sequences are constructed and we assign obtained sequences in order to find a differentially 4-uniform 2-to-1 function. Therefore, obtained function can be isomorphic to a (n − 1)-subfunction of an APN permutation, so, this (n − 1)-subfunction can be expanded to bijective APN function. In order to construct an APN permutation, we need to find all possible coordinate Boolean functions f such that the bijective function constructed from the given (n − 1)-subfunction S and function f is APN. Unfortunately, the exhaustive search through the set of potential coordinate functions is computationally hard when n > 7, so, we need to estimate the number n(S) of such coordinate Boolean functions. For a given bijective vectorial function F, we introduce an associated permutation F? as follows. We split the set Fn 2 into two disjoint subsets F1 and F2, fix integer k, indices i1, . . ., ik, and index j 6∈ {i1, . . ., ik}. Then the value F?(x) is equal to F(x) if F(x) ∈ F1 and F?(x) is equal to F(x) + ej otherwise. We prove that F? is an APN permutation if and only if F is an APN permutation. This fact allows us to obtain the necessary bound. We prove that if n(S) is not equal to zero, then n(S) > 2n
Язык оригинала английский 17-27 11 Прикладная дискретная математика 41 https://doi.org/10.17223/20710410/41/2 Опубликовано - сен 2018
## Fingerprint
Подробные сведения о темах исследования «On constructing APN permutations using subfunctions». Вместе они формируют уникальный семантический отпечаток (fingerprint).
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# Are there any photos of Earth's spare tire- You know the bulge at the equator?
Is it feet high or kilometers high? The only pictures I have seen show what appears to be a perfect sphere. I want proof dang it!
Update:
Theoretically there is a bulge but if we are incapable of viewing it or proving it then why even ever mention it and yet sure as the sun rises I ALWAYS see people correcting others that Earth is an oblate spheroid like it matters to anyone. What happened to scientists getting proof for things?
Relevance
• 1 month ago
No there isn't. The bulge is slightly more in southern hemisphere because there currently is more water in the southern hemisphere.
The departure from a sphere is about 1/32 of an inch flattening at the geographical poles. It is not going to be detectable on any image of the entire Earth.
Scientists have the data and the proof, but no one can force you to understand or accept the data, information, analysis and interpretation of that data if your mind is closed tight.
• Ride my seesaw
Lv 6
1 month agoReport
Don't get huffy that proof is non-existent. My mind is wide open to verifiable data, not conjecture based on a theory which the masses have just accepted as true. Question authority I say. If they can't answer then remain doubtful. Good answer, though!
• 1 month ago
It is barely noticeable and more gradual than a " Spare Tyre " or Muffin top
Furthermore, about 10, 000 feet would be the Maximum
Earth is Spherical because Gravity pulls everything towards the Centre of the Earth
Mountains can be no higher than a certain Altittude
The Oblateness is a bulge around South America
Making a Peak in the Andies Range Makes it taller than Everest by Mean Altitude
• 1 month ago
Earth is almost 8000 miles in diameter on average; the difference is that it's thicker at the equator (by about 20 miles) than it is through the poles, making a difference of about 20/8000 = 0.25%, which is very difficult to just 'see' in photographs.
If you had an image of the whole Earth, and it was 10,000 pixels across, from one edge of the Earth to the other, then the difference would only be 25 pixels...
• 1 month ago
It takes precision long-distance surveys to detect the bulge. Which is what map makers and long-distance airplanes depend on. But as other answers have said, you're not going to see it in a photo.
• Clive
Lv 7
1 month ago
The joke section is thataway, dear.
• Ride my seesaw
Lv 6
1 month agoReport
if you can't answer the question then don't put anything DEAR
• 1 month ago
Earth mean radius is above 6300 km
The difference between polar and equatorial radius is 21 km
So Earth flattening is 1/300
If Irving's basketball was Earth-shaped the radius difference would be 0.4 mm (a fingernail thickness!)
(Kyrie Irving is the NBA flat-earther)
On a 600 pixels full-screen image the "bulge" would be 1 single pixel on each side (600 x 602 image). Hardly noticeable naked eye.
• 1 month ago
The Earth is an oblate spheroid (approximately). The difference in polar versus equatorial radius is just too small to show in a photo, so you'll have to be satisfied with the numbers.
Difference = only 21 km (13 miles).
Polar is 0.3% less than equatorial.
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# Plot 3D of a function
4 views (last 30 days)
JORGE ORDOÑEZ CARRASCO on 19 Feb 2021
Hello is there a short cut to plot a 3d function for example i have a linspace(0,pi) of x and linspace(0,1) of t and the a function @(x,t)=f(x,t). The way i have plotted is having x and t and loop for which calculates the values of x and t evaluated in the function:
In the image above i have created a vectors x1 and t1 of lenght=300 and evaluate them in the funcion and store them in the matrix z1. I have tried this:
But i dont think is the best way cause it takes a lot. Any suggestion.
JORGE ORDOÑEZ CARRASCO on 19 Feb 2021
By the way the loop for plotting took me 15 minutes and i dont think is the best way though
Alan Stevens on 19 Feb 2021
Are you looking for something like this?
x = linspace(0,pi);
t = linspace(0,1);
z = @(x,t) 10*exp(-4*t).*sin(2*x);
[x1, t1] = meshgrid(x,t);
z1 = z(x1,t1);
surf(x1,t1,z1)
JORGE ORDOÑEZ CARRASCO on 19 Feb 2021
Nice I will keep it in mind but I’m gonna study your answer and get to know how works. Thank you a lot.
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# What are period, amplitude, and frequency?
#### What are period, amplitude, and frequency?
Period is the time it takes for a resonant system to complete one cycle of its motion. For example, if a pendulum takes two seconds to swing over and back, then its period is two seconds. Amplitude is the maximum amount of motion a resonant system undergoes as it oscillates or vibrates (same thing). For example, if the pendulum swings one meter to the left of center and then one meter to the right of center, its amplitude of motion is one meter. Frequency is the number of cycles a resonant system completes in a certain amount of time. For example, if a pendulum swings over and back twice each second, then its frequency is two cycles-per-second or 2 hertz or 2 Hz.
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# Multiple conditions using 'or' to filter a matrix with numpy and python
Published: May 25, 2018
To write a logical expression using boolean "or", one can use | symbol. Example, let's consider the following matrix:
M =
\left( \begin{array}{ccc}
1 & 0 & 0 \\
3 & 0 & 0 \\
1 & 0 & 0 \\
1 & 0 & 0 \\
2 & 0 & 0 \\
3 & 0 & 0 \\
1 & 0 & 0 \\
2 & 0 & 0
\end{array}\right)
Now, to select the rows when the first columns is equal to 1 or 2, we can do:
>>> import numpy as np
>>> M = np.array([[1,0,0],[3,0,0],[1,0,0],[1,0,0],[2,0,0],[3,0,0],[1,0,0],[2,0,0]])
>>> M
array([[1, 0, 0],
[3, 0, 0],
[1, 0, 0],
[1, 0, 0],
[2, 0, 0],
[3, 0, 0],
[1, 0, 0],
[2, 0, 0]])
>>> N = M[ (M[:,0] == 1) | (M[:,0] == 2) ]
>>> N
array([[1, 0, 0],
[1, 0, 0],
[1, 0, 0],
[2, 0, 0],
[1, 0, 0],
[2, 0, 0]])
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## ››Convert inch of water column to millihg
inch water column millihg
How many inch water column in 1 millihg? The answer is 0.53524017143946.
We assume you are converting between inch of water column and millihg.
You can view more details on each measurement unit:
inch water column or millihg
The SI derived unit for pressure is the pascal.
1 pascal is equal to 0.0040146307866177 inch water column, or 0.0075006156130264 millihg.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between inches water column and millihg.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of inch water column to millihg
1 inch water column to millihg = 1.86832 millihg
5 inch water column to millihg = 9.3416 millihg
10 inch water column to millihg = 18.6832 millihg
15 inch water column to millihg = 28.0248 millihg
20 inch water column to millihg = 37.3664 millihg
25 inch water column to millihg = 46.708 millihg
30 inch water column to millihg = 56.0496 millihg
40 inch water column to millihg = 74.73281 millihg
50 inch water column to millihg = 93.41601 millihg
## ››Want other units?
You can do the reverse unit conversion from millihg to inch water column, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Millihg
The SI prefix "milli" represents a factor of 10-3, or in exponential notation, 1E-3.
So 1 millihg = 10-3 hg.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# 2.6.3: Internal Energy and Enthaply
• Contributed by Boundless
• General Microbiology at Boundless
The enthalpy of reaction measures the heat released/absorbed by a reaction that occurs at constant pressure.
Learning Objectives
• Review enthalpy of reaction
## Key Points
• At constant volume, the heat of reaction is equal to the change in the internal energy of the system.
• At constant pressure, the heat of reaction is equal to the enthalpy change of the system.
• Most chemical reactions occur at constant pressure, so enthalpy is more often used to measure heats of reaction than internal energy.
## Key Terms
• enthalpy: In thermodynamics, a measure of the heat content of a chemical or physical system.
• internal energy: A property characteristic of the state of a thermodynamic system, the change in which is equal to the heat absorbed minus the work done by the system.
• first law of thermodynamics: Heat and work are forms of energy transfer; the internal energy of a closed system changes as heat and work are transferred into or out of it.
In thermodynamics, work (W) is defined as the process of an energy transfer from one system to another. The first law of thermodynamics states that the energy of a closed system is equal to the amount of heat supplied to the system minus the amount of work done by the system on its surroundings. The amount of energy for a closed system is written as follows:
ΔU=Q−WΔU=Q−W
In this equation, U is the total energy of the system, Q is heat, and W is work. In chemical systems, the most common type of work is pressure-volume (PV) work, in which the volume of a gas changes. Substituting this in for work in the above equation, we can define the change in internal energy for a chemical system:
ΔU=Q−PΔVΔU=Q−PΔV
## Internal Energy Change at Constant Volume
Let’s examine the internal energy change, ΔUΔU, at constant volume. At constant volume, ΔV=0ΔV=0, the equation for the change in internal energy reduces to the following:
ΔU=QVΔU=QV
The subscript V is added to Q to indicate that this is the heat transfer associated with a chemical process at constant volume. This internal energy is often very difficult to calculate in real life settings, though, because chemists tend to run their reactions in open flasks and beakers that allow gases to escape to the atmosphere. Therefore, volume is not held constant, and calculating ΔUΔU becomes problematic. To correct for this, we introduce the concept of enthalpy, which is much more commonly used by chemists.
## Standard Enthalpy of Reaction
The enthalpy of reaction is defined as the internal energy of the reaction system, plus the product of pressure and volume. It is given by:
H=U+PVH=U+PV
By adding the PV term, it becomes possible to measure a change in energy within a chemical system, even when that system does work on its surroundings. Most often, we are interested in the change in enthalpy of a given reaction, which can be expressed as follows:
ΔH=ΔU+PΔVΔH=ΔU+PΔV
When you run a chemical reaction in a laboratory, the reaction occurs at constant pressure, because the atmospheric pressure around us is relatively constant. We will examine the change in enthalpy for a reaction at constant pressure, in order to see why enthalpy is such a useful concept for chemists.
## Enthalpy of Reaction at Constant Pressure
Let’s look once again at the change in enthalpy for a given chemical process. It is given as follows:
ΔH=ΔU+PΔVΔH=ΔU+PΔV
However, we also know that:
ΔU=Q−W=Q−PΔVΔU=Q−W=Q−PΔV
Substituting to combine these two equations, we have:
ΔH=Q−PΔV+PΔV=QPΔH=Q−PΔV+PΔV=QP
Thus, at constant pressure, the change in enthalpy is simply equal to the heat released/absorbed by the reaction. Due to this relation, the change in enthalpy is often referred to simply as the “heat of reaction.”
Enthalpy: An explanation of why enthalpy can be viewed as “heat content” in a constant pressure system.
• OpenStax College, Biology. October 16, 2013. Provided by: OpenStax CNX. Located at: http://cnx.org/content/m44422/latest...ol11448/latest. License: CC BY: Attribution
• glucose. Provided by: Wiktionary. Located at: en.wiktionary.org/wiki/glucose. License: CC BY-SA: Attribution-ShareAlike
• OpenStax College, Energy and Metabolism. October 16, 2013. Provided by: OpenStax CNX. Located at: http://cnx.org/content/m44422/latest...e_06_01_02.jpg. License: CC BY: Attribution
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# Chapter06
Views:
Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### Chapter 6 :
Chapter 6 The Basic Beat
### Basic Components :
Basic Components Wave: Deflection from baseline that represents a cardiac event Segment: Specific portion of the complex as represented on ECG Interval: Distance, measured as time, between two cardiac events
### Waves and the Baseline :
Waves and the Baseline Waves are a deflection of the baseline. The baseline is a line from one TP segment to the next.
### Wave Forms :
Wave Forms Waves can be: Single Isolated Positive or negative deflections Biphasic deflections with both + and - components Combinations that have multiple + and - components
### Wave Nomenclature :
Wave Nomenclature Waves should be named according to: Size Location Direction of deflection
### QRS Wave Nomenclature :
QRS Wave Nomenclature Tall or deep waves in QRS complex are given capital letters: Q, R, S, R' Small waves are given lowercase letters: q, r, s, r'
### X Prime :
X Prime Changes occurring in QRS complexes can lead to bizarre complexes. Waves are named differently if they change directions and cross the baseline. This is called X' (X prime). X is not an actual wave, but a term that can stand for either an R or S wave. R' and S' (R prime and S prime) = extra waves within the QRS complex.
### Q, R, and S :
Q, R, and S Q wave First negative deflection after P wave R wave First positive deflection after P wave S wave First negative deflection after R wave If there is another upward component, we start with R'.
### R' and S' Complexes :
R' and S' Complexes
### The P Wave (1 of 2) :
The P Wave (1 of 2) Usually the first in the ECG complex Represents electrical depolarization of both atria Starts when the SA node fires Duration can vary between 0.08 and 0.11 seconds in adults Axis: Downward and to the left, 0º to +75º
### The P Wave (2 of 2) :
The P Wave (2 of 2)
### The Tp Wave :
The Tp Wave Represents repolarization of atria Deflects in the opposite direction of P wave Occurs at same time as QRS wave, so is not usually seen
### The PR Segment :
The PR Segment Time between end of P wave and beginning of QRS complex Usually found along baseline Anything >0.8 mm is pathological
### The PR Interval (1 of 2) :
The PR Interval (1 of 2) Represents: Impulse initiation Atrial depolarization Atrial repolarization AV node stimulation His bundle stimulation Bundle branch and Purkinje system stimulation
### The PR Interval (2 of 2) :
The PR Interval (2 of 2) From beginning of P wave to beginning of QRS complex Includes P wave and PR segment Normal duration: 0.12 to 0.20 sec
### The QRS Complex (1 of 2) :
The QRS Complex (1 of 2) Represents ventricular depolarization Main components are Q, R, and S waves Q wave can be present or absent. R wave: First positive deflection after P; will be initial wave of QRS complex if no Q is present. Normal duration: 0.06 to 0.11 sec Axis: -30 to +105, downward and to left
### The QRS Complex (2 of 2) :
The QRS Complex (2 of 2)
### Q Wave Significance :
Q Wave Significance Can be benign, or sign of dead myocardial tissue Significant Q wave Indication of MI over region involved 0.03 sec or wider Height equal to or greater than 1/3 height of R wave Insignificant Q wave Commonly found in I, aVL, and V6 Due to septal innervation (called septal Qs)
### Insignificant Q Wave :
Insignificant Q Wave
### Significant Q Wave :
Significant Q Wave
### Intrinsicoid Deflection (1 of 2) :
Intrinsicoid Deflection (1 of 2) Measured from beginning of QRS complex to beginning of negative downslope of R wave Represents amount of time for electrical impulse to travel from Purkinje system to epicardium under electrode Normal intrinsicoid deflection: Right precordials = 0.035 Left precordials = 0.045
### Intrinsicoid Deflection (2 of 2) :
Intrinsicoid Deflection (2 of 2)
### The ST Segment :
The ST Segment Represents electrically neutral period between ventricular depolarization and repolarization Segment from end of QRS complex to beginning of T wave Usually found along the baseline Point where ST segment begins is called the J point Axis: Inferior and to the left Important: Any ST elevation in a symptomatic patient should be considered representative of myocardial injury or infarction
The J Point
### The T Wave (1 of 2) :
The T Wave (1 of 2) Represents ventricular repolarization Is next deflection (+ or –) after ST segment Normal T wave: Begins in same direction as QRS complex Asymmetrical (first part rising/dropping slowly, latter part moving faster)
### The T Wave (2 of 2) :
The T Wave (2 of 2)
### Assessing T Wave Symmetry :
Assessing T Wave Symmetry
### QT Interval (1 of 2) :
QT Interval (1 of 2) Encompasses QRS complex, ST segment, T wave Represents all events of ventricular systole From beginning of ventricular depolarization to end of repolarization Normal duration is variable, but usually less than 1/2 of R-R interval Prolonged QT interval is a sign of possible life-threatening arrhythmias
### QT Interval (2 of 2) :
QT Interval (2 of 2)
### QTc Interval (1 of 2) :
QTc Interval (1 of 2) Stands for “QT corrected interval” As heart rate decreases, QT interval lengthens. As heart rate increases, QT interval shortens. By calculating QTc interval, it can be stated that normal is around 0.410 sec or 410 milliseconds. Formula: QTc = QT + 1.75 (ventricular rate – 60) Cardiac events represented by QTc interval = all events of ventricular systole
### U Wave :
U Wave A small, flat wave sometimes seen after T wave and before next P wave. No one knows for sure what it represents. Important: Can sometimes cause an inaccuracy in measuring QT interval.
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http://www.markedbyteachers.com/as-and-a-level/science/investigate-the-factors-that-affect-the-energy-released-during-reactions-between-metals-and-acids.html
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# Investigate the factors that affect the energy released during reactions between metals and acids.
Extracts from this document...
Introduction
Investigate the factors that affect the energy released during reactions between metals and acids Introduction Introduction The experiment we will use for this investigation is as follows: - There are a number of factors that can affect the amount of energy released during this experiment. These include: - 1. Mass of magnesium used 2. Concentration of acid used 3. Surface area of magnesium Mass of magnesium The more magnesium that is present in the reaction, the more collisions there will be. This should increase the effectiveness of the reaction. The better the reaction that takes place between the magnesium atoms and the hydrochloric acid, the more energy that will be released. Concentration of acid used The more concentrated the acid, the more energy that will be released. If the acid is a low concentration, the acid particles will be widely spread in the water, and the number of collisions should be less giving out less energy. At higher concentrations however, the chances of a collision is greater and more energy should be released. This also explains why the greatest amount of energy is lost at the start of the reaction, when the reactants are mixed. As the reaction proceeds, eventually the temperature will decrease as the reaction is coming to completion. And therefore energy will be dissipated into the surrounds. Surface area When one of the reactants is a solid e.g. magnesium ribbon, the ribbon will float on the surface. ...read more.
Middle
Measure out 0.1g of magnesium turnings. Do the same for each result until 0.6g 3. Pierce a whole in the card and slide thermometer in, this will act as a rest and will make sure heat doesn't escape into the surroundings. 4. Add the 25cm3 of HCL into the polystyrene cup 5. Add the magnesium to the HCL. 6. Quickly bring down the thermometer in the card to ensure minimal heat loss (most heat given off at the start of reaction) 7. Record the temperature increase **Do the same for magnesium ribbon but measure 0.5cm of it until 3cm and record results** How to keep this experiment fair In order to keep this experiment fair it should be taken on the same day, as the room temperature is never completely constant and a slight temperature increase or decrease on a different day could affect my results. I also have to keep the amount of acid constant as the factor I have chosen to change is the mass of magnesium. Another point I have to be sure of is that I don't dip the thermometer in the HCL when it is reacting with the magnesium, as heat rises and I would not get fair results Safety In order to keep this experiment fair I have to follow these safety points o Take care while using acid in the classroom as the acid can irritate skin. o Wear goggles to protect eyes from acid. ...read more.
Conclusion
Further work There are few different ways in which I could change this experiment in order to get more accurate or a different range of results. I believe that within the experiment itself that if we put the magnesium in the polystyrene cup before adding the acid that it would be more precise. And also less energy in the form of heat would be lost. I believe this would be a good idea without changing the experiments properties. Another way would be to change the factor I would be looking at from the mass of magnesium to the concentration of acid. In this experiment my prediction would be that as there is a higher concentration of acid there will be more energy dissipated into the surroundings in the form of heat. This is a very similar way of doing the experiment but I believe that it would give different results maybe more accurate as the mass of magnesium would be constant. I can also find one problem with this and that is there will be a waste of acid. I would need about ten different concentrations, and to make this there would have to be some left over. Also the concentration would be getting too high to deal with in a school laboratory and it wouldn't be safe. Again we could look at trying more masses of magnesium, but as seen in the experiment so enough we would have the maximum temperature rise if we went too high and we would not be able to record them. This along with doing more in between masses would be a waste of acid. ...read more.
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# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board
## AP Board Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Book Answers
AP Board Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Book Answers
## Andhra Pradesh State Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Books Solutions
Board AP Board Materials Textbook Solutions/Guide Format DOC/PDF Class 8th Subject Maths Chapters Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Provider Hsslive
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Find below the list of all AP Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
Question
Solve the following equations:
1. 7x – 5 = 2x
2. 5x – 12 = 2x – 6
3. 7p- 3 = 3p + 8
4. 8m + 9 = 7m + 8
5. 7z + 13 = 2z + 4
6. 9y + 5 = 15y – 1
7. 3x + 4 = 5(x – 2)
8. 3(t – 3) = 5(2t – 1)
9. 5(p – 3) = 3(p – 2)
10. 5(z + 3) = 4(2z + 1)
11. 15(x – 1) + 4(x + 3) = 2 (7 + x)
12. 3 (5z – 7) +2 (9z – 11) = 4 (8z – 7) – 111
13. 8(x – 3) – (6 – 2x) = 2(x + 2) – 5 (5 – x)
14. 3(n – 4)+2(4n – 5) = 5(n + 2) + 16
Solution:
1. 7x – 5 = 2x
⇒ 7x – 2x = 5
⇒ 5x = 5
⇒ x = 55 = 1
∴ x = 1
2. 5x – 12 = 2x – 6
⇒ 5x – 2x = – 6 + 12
⇒ 3x = 6
⇒ x = 63 = 2
∴ x = 2
3. 7p – 3 = 3p + 8
⇒ 7p – 3p = 8 + 3
⇒ 4p = 11
⇒ p = 114
4. 8m + 9 = 7m + 8
⇒ 8m – 7m = 8 – 9
∴ m = – 1
5. 7z + 13 = 2z + 4
⇒ 7z – 2z = 4 – 13
⇒ 5z = – 9
∴ z = −95
6. 9y + 5 = 15y – 1
⇒ 9y – 15y = – 1 – 5
⇒ -6y = -6
⇒ y = −6−6
∴ y = 1
7. 3x + 4 = 5(x – 2)
⇒ 3x + 4 =5x – 10
⇒ 3x – 5x= – 10 – 4
⇒ – 2x = – 14
:. x = 71
8. 3(t – 3) = 5(2t – 1)
⇒ 3t – 9 = 10t – 5
⇒ 3t – 10t = – 5+ 9
⇒ – 7t = 4
∴ t = −47
9. 5 (p – 3) = 3 (p – 2)
⇒ 5p – 15 = 3p – 6
⇒ 5p – 3p = -6 + 15
⇒ 2p = 9
∴ p = 92
10. 5(z + 3) = 4(2z + 1)
⇒ 5z + 15 = 8z + 4
⇒ 5z – 8z = 4 – 15
⇒ – 3z = – 11
⇒ z = −113
∴ z = −113
11. 15(x – 1) + 4(x + 3) = 2(7 + x)
⇒ 15x – 15 + 4x + 12= 14 + 2x
⇒ 19x – 3 = 14 + 2x
⇒ 19x – 2x = 14 + 3
⇒ 17x = 17 ,
x = 1717 = 1
∴ x = 1
12. 3(5z – 7)+2(9z – 11) = 4(8z – 7) – 111
⇒ 15z – 21 + 18z – 22 = 32z – 28 – 111
⇒ 33z – 43 = 32z – 139
⇒ 33z – 32z = – 139 + 43
∴ z = – 96
13. 8(x – 3) – (6 – 2x)=2(x+2)-.5(5 – x)
⇒ 8x – 24 – 6 + 2x = 2x + 4 – 25 +5x
⇒ 8x – 30 = 5x – 21
⇒ 8x – 5x= – 21 +30
⇒ 3x = 9
∴ x = 3
14. 3(n – 4) + 2(4n – 5) = 5(n + 2) + 16
⇒ 3n – 12 + 8n – 10 = 5n + 10 + 16
⇒ 11n – 22 = 5n + 26
⇒ 11n – 5n = 26 + 22
⇒ 6n =48
⇒ n = 486 = 8
∴ n = 8
## Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbooks for Exam Preparations
Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 exam preparation. The AP Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Books State Board syllabus with maximum efficiency.
## FAQs Regarding Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Solutions
#### Can we get a Andhra Pradesh State Board Book PDF for all Classes?
Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.
## Important Terms
Andhra Pradesh Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3, AP Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbooks, Andhra Pradesh State Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3, Andhra Pradesh State Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook solutions, AP Board Class 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbooks Solutions, Andhra Pradesh Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3, AP Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbooks, Andhra Pradesh State Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3, Andhra Pradesh State Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook solutions, AP Board STD 8th Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Textbooks Solutions,
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# Physics
A hollow spherical shell with mass 2.45kg rolls without slipping down a slope that makes an angle of 30.0∘ with the horizontal. a.) find the magnitude of the acceleration of the center of mass of the spherical shell. b.) find the magnitude of the frictional force acting on the spherical shell.
I know that the center of mass for a hollow sphere is 2/5mr^2 and since the sphere is not slipping v=wR (w=omega) and a=(alpha)*r but I'm not sure how to approach this question.
1. 👍 0
2. 👎 0
3. 👁 991
1. 5/7*sin(30)
1. 👍 0
2. 👎 1
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Blocking Response Surface Designs Incorporating Neigh-bour Effects
Open Journal of Statistics
Vol.1 No.3(2011), Article ID:8076,6 pages DOI:10.4236/ojs.2011.13023
Blocking Response Surface Designs Incorporating Neighbour Effects
Eldho Varghese*, Seema Jaggi
Indian Agricultural Statistics Research Institute Library Avenue, New Delhi, India
E-mail: *eldhoiasri@gmail.com
Received May 31, 2011; revised June 27, 2011; accepted July 13, 2011
Keywords: Response Surface Model, Neighbour Effects, Blocking in Response Surface, Conditions for Orthogonal Estimation
Abstract
In this paper, blocking in response surface for fitting first order model incorporating neighbour effects has been investigated. The conditions for orthogonal estimation of the parameters of the model have been obtained. A method of constructing designs which ensures the constancy of variance of the parameter estimates of the model has also been given.
1. Introduction
Response Surface Methodology (RSM) is used to explore the relationship between one or more response variable and a set of experimental variables or factors with an objective to optimize the response.
Let there be independent variables denoted by and the response variables be and there are N observations. The response is a function of input factors, i.e.
where u = 1, 2, …, N, is the level of the, (i = 1, 2, …, v) factor in the treatment combination, denotes the response obtained from treatment combination. The function f describes the form in which the response and the input variables are related and is the random error associated with the observation that is independently and normally distributed with mean zero and common variance. For details on RSM, one may refer to Khuri and Cornell [1], Myers et al. [2].
In the literature, the work on RSM is done assuming observations to be independent and no effect of neighbouring units. However, plots in agricultural experiments are nearby and induce some overlap effects from neighbouring units. Hence, the response from a particular plot may not be the actual response from the plot but may be the joint effect of the treatment combination applied to same plot and the treatment combination applied to the neighbouring plots. For example, in an experimental trial when the combination of pesticides is used, wind drift may cause the effect of spray spill over to adjacent plots. It is thus important to study the response surface in the presence of neighbour effects which would result in more precise estimation of the parameters of the response surface model.
Draper and Guttman [3] suggested a general model for response surface problems in which it is anticipated that the response on a particular plot will be affected by overlap effects from neighbouring plots and the same has been illustrated.
Sarika et al. [4] studied second order response surface model with neighbour effects and the rotatability conditions were derived. Methods of obtaining designs satisfying the derived conditions were given.
Jaggi et al. [5] studied response surface model incorporating neighbour effects and the same has been illustrated. They showed that if the neighbour effect is present and is included in the model, there is a substantial reduction in the residual sum of squares and the response is predicted more precisely.
In response surface analysis, it is generally assumed that the experimental trials are carried out under homogeneous conditions. This assumption may not be valid in every experimental situation. In such circumstances, the experimental trials should be carried out in groups, or blocks so that the units within each block are homogeneous. Also when the number of runs is too large, it is very difficult to accommodate all the units in a single block. Blocking is usually beneficial where it is possible to identify groups, or blocks, of experimental units, such that within blocks the experimental units are considerably more homogeneous than the blocks themselves. This type of grouping makes it possible to eliminate from error variance a portion of variation attributable to block differences. The variation between the blocks in the experiment is accounted for by including block effects in the statistical model. The nature of the blocking variables has an important impact on the data analysis.
In this paper, we focus on the methodology for blocking in first order response surface model incorporating neighbour effects. The conditions for orthogonally blocked experiments for estimation of the parameters of the model and the conditions for the constancy of variance of the parameter estimates of the model are derived. Construction of response surface designs in blocks with neighbour effects has also been given and an example is discussed.
2. First Order Response Surface Methodology with Block Effects and Incorporating Neighbour Effects
2.1. Model and Estimation of Parameters
The first order response surface model with block effects can be written in the form
, (1)
where f(xu) denotes the observed response value at uth experimental run, xiu is the corresponding setting of the ith input variable, δl denote the effect of the lth block (l = 1, 2, …, b), wlu is a dummy variable taking the value 1 if the uth trial is carried out in the lth block; otherwise, it is equal to zero and eu is the random error.
Incorporating neighbour effects to the given model, the above model with neighbour effects in matrix notation is
(2)
where, where δl denotes the effect of the lth block (l = 1, 2, …,b) and W is a block-diagonal matrix of the form, where nl is the size of the lth block (l =1, 2, …, b) such that. The random error vector e is assumed to have zero mean and a variance-covariance matrix., (l = 1, 2, …, b) is a (nl + 2) × v matrix. = ((xiu)), i = 1, 2, …, v; u = 1, 2, …, nl for all b blocks., where = (l = 1, 2, …, b, s ¹ s¢ = 1, 2, …, nl) is a nl × (nl + 2) neighbour matrix (assuming same neighbour structure in the b blocks) with
(3)
The model given in (2) is not of full column rank since the columns of W sum to 1N. The model therefore can be written as
(4)
where. If the columns of W are linearly independent of those of GX, then model (4) is of full rank. Thus β and τ can be uniquely estimated by the method of ordinary least squares. It is not possible to estimate β0 independent of δ unless certain constrain is imposed on the element of δ. For this purpose we can assume. In this case β0 is given by.
The Equation (4) can be written as
where, , Zl = GlXl and. The least square estimate of θ is given by
and the variance-covariance of is
Let v = b = 2 and n1 = n2 = n. Xl (l = 1, 2) is of order (n + 2) ´ 2. Hence,. Assuming the same neighbour structure in both the blocks, the n ´ (n + 2) neighbour matrix Gl is as defined in (3).
Block I Block II
, ,
. Thus,
Therefore,
where,
,
and
In general for v factors and for b blocks, the X matrix with two extra points as border points is
and so on
Thus, V´V is obtained as
where,
and
.
2.2. Conditions for Orthogonality
To ensure orthogonality in the estimation of the parameters, has to be diagonal. This gives rise to the following conditions:
1)
2)
3) for all blocks of size n.
Thus, in view of above conditions, can be written as:
The normal equations for the estimation of (v + b) parameters are
i.e., = (5)
where and are the parameters to be estimated. and are the vector of treatment combination totals and block totals respectively, and, i = 1, 2, …, v and l = 1, 2, ..., b.
Hence,
= (6)
and
=. (7)
We thus obtain the variance of parameter estimates as
andfor i = 1, 2,…,v, and l = 1, 2, …, b. The estimated response at the point is with its variance
Thus,
(8)
2.3. Conditions for Constancy of the Variances
The constancy of the variances of the parameter estimates is ensured by the following conditions:
1), a constant
2) Ai = A, a constant for all blocks of size n.
Therefore,
(9)
It is thus seen that the variances of bi’s (i = 1, 2,…, v) are same and the variance of the estimated response is a function of. For given, the points for which is same, the estimated response will have the same variance.
3. Method of Construction
Consider a 2v full factorial for v factors each at 2 levels and arrange the combinations in lexicographic order. Put all these runs in a single block. The second block can be obtained by circularly rotating the columns of first block once. Similarly, rotating the v columns of 2v factorial points (v – 1) times we get v × 2v design points in v blocks each of size 2v. The design so obtained satisfies all the conditions obtained in Section 2. Two extra units are added as border units in each block for neighbour effects.
Illustration
Let v = 2 (X1 and X2) with each factor at two levels, then we get four runs in full factorial. The four runs constitute the first block and the other block can be obtained by rotating the columns of the first block in a circular fashion. The various matrices are obtained as follows:
Block I Block II
;
Thus,
for i = 1, 2.
and
for l = 1, 2.
For,
Thus,
, and
Thus, it is seen that the variance of the estimated response at all points is same.
4. Conclusions
The presence of block effects in a response surface model can affect the estimation of mean response, as well as in determination of the optimum response. Hence, blocking should be done in response surfaces wherever heterogeneity among experimental units is suspected. It has also been shown that incorporating neighbour effect in a model along with block effects results in better estimates of the parameters. The developed methodology and designs can be used to fit the response surfaces with block effects and incorporating neighbour effects.
5. Acknowledgements
We are grateful to the referee and the editor for the valuable suggestions which helped in improving the quality of the paper.
6. References
[1] A. I. Khuri and J. A. Cornell, “Response Surfaces-Designs and Analysis,” Marcel Dekker, New York, 1996.
[2] R. H. Myers, D. C. Montgomery and C. M. Anderson, “Response Surface Methodology-Process and Product Optimization using Designed Experiments,” John Wiley Publication, New York, 2009.
[3] N. R. Draper and I. Guttman, “Incorporating Overlap Effects from Neighbouring Units into Response Surface Models,” Applied Statistics, Vol. 29, No. 2, 1980, pp. 128-134. doi:10.2307/2986297
[4] Sarika, S. Jaggi and V. K. Sharma, “Second Order Response Surface Model with Neighbour Effects,” Communications Statistics: Theory and Methods, Vol. 38, No. 9, 2009, pp. 1393-1403.
[5] S. Jaggi, A. Sarika and V. K. Sharma, “Response Surface Analysis Incorporating Neighbour Effects from Adjacent Units,” Indian Journal of Agricultural Sciences, Vol. 80, No. 8, 2010, pp. 719-723.
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## Description
Featured games:
Tic-Tac-Toe Logic (aka Noughts and Crosses)
Inertia
Sudoku
Range
Hitori
Net
Wrapped Net
SameGame
All the puzzles can be solved with logic only and you don't need to make any random guesses.
Tic-Tac-Toe Logic is a slight twist on the well-known puzzle game, especially designed to be played alone. Fill the grid so there are no more than 3 of X or O in a row or a column while making sure there is the same number of both in each row and column! The rules are simple but it's way more challenging than it looks!
Inertia is a game about collecting all the gems while avoiding hitting the mines. Like the name suggests, your movement can only be stopped by a wall, a ring or in the worst case, a mine!
Possibly the most well known logic game Sudoku doesn't require much explaining. Your task is to fill the grid with numbers from 1 to 9 and making sure they don't repeat in any row, column or marked 3x3 area!
Range is the latest addition to our puzzle series, requiring you to figure out new clever moves to solve these. Your goal is to place black tiles so that each numbered tile is vertically and horizontally connected to exactly the number of tiles as shown on the tile.
Hitori is a classical Japanese number logic puzzle game where you must use your logical reasoning to figure out which numbers to gray out. In the end, there can only be a maximum of one of each number in each row and column and there must be no colliding greyed out numbers!
Net is our take on the classical grid connecting puzzle. Rotate the tiles so the central power unit powers the whole network!
Wrapped Net is the step-brother of Net, enabling you to connect tiles even across the puzzle borders- You can long tap on a tile to lock it if you're sure it's rotated correctly to solve the puzzle step by step.
SameGame is another classical puzzle, where your goal is to clear the grid of all tiles! You can only remove tiles if there's at least 2 of the same colour tiles next to it!
## What’s New
Version 1.01
Fixed multiple bugs.
## App Privacy
### No Details Provided
The developer will be required to provide privacy details when they submit their next app update.
## Information
Provider
Morsakabi OUe (LLC)
Size
86.6 MB
Category
Games
Compatibility
Requires iOS 7.1.2 or later. Compatible with iPhone, iPad and iPod touch.
Languages
English
Age Rating
4+
Price
Free
In-App Purchases
1. Hitori Extra Packs 4,99 €
2. Inertia Extra Packs 4,99 €
3. SameGame Extra Packs 4,99 €
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Home
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### Binary Puzzle Solve
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### Binary Logic Puzzle Quiz - Sporcl
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### Binary Puzzles - Stevens Binaire Puzzel
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# 1069. The Black Hole of Numbers (20)-PAT甲级真题
For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner we will soon end up at the number 6174 — the “black hole” of 4-digit numbers. This number is named Kaprekar Constant.
For example, start from 6767, we’ll get:
7766 – 6677 = 1089
9810 – 0189 = 9621
9621 – 1269 = 8352
8532 – 2358 = 6174
7641 – 1467 = 6174
… …
Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.
Input Specification:
Each input file contains one test case which gives a positive integer N in the range (0, 10000).
Output Specification:
If all the 4 digits of N are the same, print in one line the equation “N – N = 0000”. Else print each step of calculation in a line until 6174 comes out as the difference. All the numbers must be printed as 4-digit numbers.
Sample Input 1:
6767
Sample Output 1:
7766 – 6677 = 1089
9810 – 0189 = 9621
9621 – 1269 = 8352
8532 – 2358 = 6174
Sample Input 2:
2222
Sample Output 2:
2222 – 2222 = 0000
s.insert(0, 4 – s.length(), ‘0’);用来给不足4位的时候前面补0～
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# Program To Tell If A Number Is A Perfect Square With Code Examples
Program To Tell If A Number Is A Perfect Square With Code Examples
With this article, we'll look at some examples of Program To Tell If A Number Is A Perfect Square problems in programming.
```import math
# Taking the input from user
number = int(input("Enter the Number"))
root = math.sqrt(number)
if int(root + 0.5) ** 2 == number:
print(number, "is a perfect square")
else:
print(number, "is not a perfect square")
```
We have shown how to address the Program To Tell If A Number Is A Perfect Square problemby looking at a number of different cases.
## How do you check if a number is a perfect square Program?
Algorithm To Check If A Number Is A Perfect Square Or Not
• Step 1: Take the input from the user.
• Step 2: Compute the square root of the given number using the math library.
• Step 3: Checking whether the int(root + 0.5) ** 2 == number, if this evaluates to True then the number is a perfect square.
## How do you check if a number is a perfect?
A number is a perfect number if is equal to sum of its proper divisors, that is, sum of its positive divisors excluding the number itself. Write a function to check if a given number is perfect or not. Examples: Input: n = 15 Output: false Divisors of 15 are 1, 3 and 5.27-Aug-2022
## How do you know if a number is a perfect square or perfect cube?
Perfect Squares and Perfect Cubes To get the square root, we simply divide the exponent by 2. For example x8 is a perfect square, its square root is x4 . x11 is not a perfect square. If a variable with an exponent has an exponent which is divisible by 3 then it is a perfect cube.
## How do you write a perfect number Program?
Step 1: i = 1, rem = num % i, => 28 % 1 = 0. Here rem = 0.Using for Loop
• /*C program to check whether the given number is the Perfect number*/
• #include<stdio.h>
• #include<conio.h>
• void main()
• {
• // declare and initialize the variables.
• int num, rem, sum = 0, i;
• // take an input from the user.
## What is perfect number C program?
Any number can be the perfect number in C if the sum of its positive divisors excluding the number itself is equal to that number. For example, 6 is a perfect number in C because 6 is divisible by 1, 2, 3, and 6.
## What is the easiest way to find perfect numbers?
“Perfect numbers” are equal to the sum of their “proper” divisors (positive integers that divide a number evenly, not counting itself). For example, 6 = 3 + 2 + 1, and 28 = 14 + 7 + 4 + 2 + 1.15-Mar-2021
## How do you write a program to find the square root of a number?
How Does This Program Work ?
• double num, root; In this program, we have declared two double data type variables named num and root.
• // Asking for Input printf("Enter an integer: "); scanf("%lf", &num); Then, the user is asked to enter the value of a number.
• root = sqrt(num);
• printf("The Square Root of %.
## How do you check if a number is perfect in Python?
Example -
• num=int(input("Enter the number: "))
• sum_v=0.
• for i in range(1,num):
• if (num%i==0):
• sum_v=sum_v+i.
• if(sum_v==num):
• print("The entered number is a perfect number")
• else:
## How do you know if 496 is a perfect number?
Now, the sum of divisors of a number, excluding the number itself, is called its aliquot sum, so we can define a perfect number as one that is equal to its aliquot sum. Hence, the sum of these factors is equal to the given number. So, 496 is a perfect number.
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Explore BrainMass
# Calculate the Wind Chill given the inputs of Temp and Wind Speed
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
In this problem, you will calculate the Wind Chill given the inputs of Temperature and Wind Speed. You will also provide the results as line and column charts.
Chapter_1-7_Wind_Chill_Start
2) In cell C10, use the appropriate formula to calculate the Wind Chill.
3) In cell D17, use the appropriate formula to calculate the Wind Chill if Temperature is equal to the value in cell C17 and Wind Speed is equal to the value in cell D16. Fill cell D17 down and across to cell M26.
Note: Be sure to use absolute, mixed, and relative cell references.
4) In cells B37:J53, insert a Line Chart to show Wind Chill as a function of Temperature 13°F. Select ranges D16:M16 and D24:M24. On the Insert tab, click Recommended Charts, and then click Line. Apply Style 1 on the Design tab. Add a chart title and choose the Above Chart option. Replace Chart Title with Wind Chills (˚F) for Temperature 13 ˚F. Add axis titles. Replace Axis Title for the horizontal axis with Wind Speed (mph) and Axis Title for the vertical axis with Wind Chills (˚F).
5) In cells L37:T53, insert a Clustered Column Chart to show Wind Chill as a function of Wind Speed 9 mph. Select ranges C17:C26 and H17:H26. On the Insert tab, click Recommended Charts, and then click Clustered Column. Apply Style 6 on the Design tab. Add a chart title and choose the Above Chart option. Replace Chart Title with Wind Chills (ËšF) for Wind Speed of 9 MPH. Add axis titles. Replace Axis Title for the horizontal axis with Temperatures (ËšF) and Axis Title for the vertical axis with Wind Chills (ËšF).
6) Save your file and submit.
Use a cell reference or a single formula where appropriate in order to receive full credit. Do not copy and paste values or type values, as you will not receive full credit for your answers.
Wind chill is a function of the outside temperature (ËšF) and the wind speed (mph):
Wind Chill(â—¦F) = 35.74+0.6215*Temp(â—¦F) - 35.75*Speed(mph)0.16 + 0.4275*Temp(â—¦F)*Speed(mph)0.16
a.) Enter a formula that outputs the Wind Chill given the inputs of Temperature and Wind Speed.
b.) Enter a formula in the 10 by 10 table that gives one hundred wind chill outputs for the ten Temperature inputs and ten Wind Speed inputs. Be sure to use proper absolute/mixed/relative cell references
c.) Create 2 graphs:
1. Line chart showing Wind Chill as a function of Temperature 13°F .
2. Column chart showing Wind Chill as a function of Wind Speed 9 mph.
Axis titles and labels should be chosen from the list:
a. Wind Chills (ËšF) for Wind Speed of 9 MPH
b. Wind Speed (mph)
c. Temperatures (ËšF)
d. Wind Chills (ËšF)
e. Wind Chills (ËšF) for Temperature 13 ËšF
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| 2,916
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.984375
| 3
|
CC-MAIN-2022-33
|
latest
|
en
| 0.79357
|
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