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https://mirrorinfo.online/knowledge-base/how-to-find-perimeter-if-the-area-is-known | 1,675,575,152,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00023.warc.gz | 415,988,337 | 21,763 | How to find perimeter if the area is known
The area and perimeter - the main numerical characteristics of any geometrical figures. Finding of these sizes becomes simpler thanks to the standard formulas according to which it is also possible to calculate one through another with a minimum or total absence of additional initial data.
Instruction
1. PryamougolnikZadach: find rectangle perimeter if it is known that the area is equal 18, and length of a rectangle is twice more than width. Decision: write down an area formula for a rectangle – S = a*b. On b statement of the problem = 2*a, from here 18 = a*2*a, a = √9 = 3. It is obvious that b = 6. On a formula the perimeter is equal to the sum of all parties of a rectangle – P = 2*a + 2*b = 2*3 + 2*6 = 6 + 12 = 18. In this task the perimeter coincided on value with the area of a figure.
2. KvadratZadach: find square perimeter if its area is equal to 9. Decision: on a formula of the area of a square of S = a^2, find length of the party of a = 3 from here. The perimeter is equal to the sum of lengths of all parties, therefore, of P = 4*a = 4*3 = 12.
3. TreugolnikZadach: any triangle of ABC which area is equal to 14 is given. Find triangle perimeter if B height which is carried out from top divides the triangle basis into pieces 3 and 4 cm long. Decision: on a formula the area of a triangle is a half of the work of the basis on height, i.e. S = ½*AC*BE. The perimeter is equal to the sum of lengths of all parties. Find length of the party of AC, having put lengths of AE and EC, AC = 3 + 4 = 7. Find BE triangle height = S*2/AC = 14*2/7 = 4. Consider a rectangular triangle of ABE. Knowing legs of AE and BE, it is possible to find a hypotenuse on Pythagoras's formula AB^2 = AE^2 + BE^2, AB = √ (3^2 + 4^2) = √25 = 5. Consider a rectangular triangle of BEC. On Pythagoras's formula BC^2 = BE^2 + EC^2, BC = √ (4^2 + 4^2) = 4 * √ 2. Now lengths of all parties of a triangle are known. Find perimeter from their sum P = AB + BC + AC = 5 + 4 * √ 2 + 7 = 12 + 4 * √ 2 = 4 * (3+ √ 2).
4. OkruzhnostZadach: it is known that the area of a circle is equal 16*π, find its perimeter. Decision: write down a formula of the area of a circle of S = π*r^2. Find r circle radius = √ (S/π) = √16 = 4. On a formula P perimeter = 2*π*r = 2*π*4 = 8*π. If to accept that π = 3.14, then P = 8*3.14 = 25.12.
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https://www.kidadvocacy.com/once-lighting-gxb/what-is-the-end-behavior-of-the-polynomial-function%3F-3f0814 | 1,624,087,559,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00421.warc.gz | 773,274,655 | 12,811 | what is the end behavior of the polynomial function?
As $x\to \infty , f\left(x\right)\to -\infty$ and as $x\to -\infty , f\left(x\right)\to -\infty$. Our mission is to provide a free, world-class education to anyone, anywhere. Start by sketching the axes, the roots and the y-intercept, then add the end behavior: Learn what the end behavior of a polynomial is, and how we can find it from the polynomial's equation. In this case, we need to multiply −x 2 with x 2 to determine what that is. A polynomial function is made up of terms called monomials; If the expression has exactly two monomials it’s called a binomial.The terms can be: Constants, like 3 or 523.. Variables, like a, x, or z, A combination of numbers and variables like 88x or 7xyz. Since the leading coefficient of this odd-degree polynomial is positive, then its end-behavior is going to mimic that of a positive cubic. This relationship is linear. Explanation: The end behavior of a function is the behavior of the graph of the function f (x) as x approaches positive infinity or negative infinity. Knowing the leading coefficient and degree of a polynomial function is useful when predicting its end behavior. Answer: 2 question What is the end behavior of the graph of the polynomial function f(x) = 2x3 – 26x – 24? This is a quick one page graphic organizer to help students distinguish different types of end behavior of polynomial functions. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The given polynomial, The degree of the function is odd and the leading coefficient is negative. To determine its end behavior, look at the leading term of the polynomial function. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is, $\begin{array}{c}\text{as } x\to -\infty , f\left(x\right)\to -\infty \\ \text{as } x\to \infty , f\left(x\right)\to -\infty \end{array}$. The highest power of the variable of P(x)is known as its degree. The leading coefficient is the coefficient of the leading term. We want to write a formula for the area covered by the oil slick by combining two functions. g, left parenthesis, x, right parenthesis, equals, minus, 3, x, squared, plus, 7, x. The leading coefficient is significant compared to the other coefficients in the function for the very large or very small numbers. As x approaches positive infinity, $f\left(x\right)$ increases without bound; as x approaches negative infinity, $f\left(x\right)$ decreases without bound. Polynomial functions have numerous applications in mathematics, physics, engineering etc. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. In determining the end behavior of a function, we must look at the highest degree term and ignore everything else. ... Simplify the polynomial, then reorder it left to right starting with the highest degree term. Summary of End Behavior or Long Run Behavior of Polynomial Functions . As the input values x get very small, the output values $f\left(x\right)$ decrease without bound. In the following video, we show more examples that summarize the end behavior of polynomial functions and which components of the function contribute to it. It is not always possible to graph a polynomial and in such cases determining the end behavior of a polynomial using the leading term can be useful in understanding the nature of the function. Given the function $f\left(x\right)=0.2\left(x - 2\right)\left(x+1\right)\left(x - 5\right)$, express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. A polynomial of degree $$n$$ will have at most $$n$$ $$x$$-intercepts and at most $$n−1$$ turning points. The end behavior of a polynomial is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity.The degree and the leading coefficient of a polynomial determine the end behavior of the graph. Learn how to determine the end behavior of the graph of a polynomial function. For the function $h\left(p\right)$, the highest power of p is 3, so the degree is 3. To determine its end behavior, look at the leading term of the polynomial function. Identify the degree, leading term, and leading coefficient of the following polynomial functions. - the answers to estudyassistant.com Polynomial Functions and End Behavior On to Section 2.3!!! The function f(x) = 4(3)x represents the growth of a dragonfly population every year in a remote swamp. Find the End Behavior f(x)=-(x-1)(x+2)(x+1)^2. Degree of a polynomial function is very important as it tells us about the behaviour of the function P(x) when x becomes very large. Which graph shows a polynomial function of an odd degree? SHOW ANSWER. URL: https://www.purplemath.com/modules/polyends.htm. In the following video, we show more examples of how to determine the degree, leading term, and leading coefficient of a polynomial. The end behavior of a function f describes the behavior of the graph of the function at the "ends" of the x-axis. How do I describe the end behavior of a polynomial function? In general, you can skip parentheses, but be very careful: e^3x is e 3 x, and e^ (3x) is e 3 x. Identify the degree, leading term, and leading coefficient of the polynomial $f\left(x\right)=4{x}^{2}-{x}^{6}+2x - 6$. Step-by-step explanation: The first step is to identify the zeros of the function, it means, the values of x at which the function becomes zero. The given polynomial, The degree of the function is odd and the leading coefficient is negative. As the input values x get very large, the output values $f\left(x\right)$ increase without bound. It has the shape of an even degree power function with a negative coefficient. Knowing the leading coefficient and degree of a polynomial function is useful when predicting its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Degree, Leading Term, and Leading Coefficient of a Polynomial Function . Which of the following are polynomial functions? The end behavior of a polynomial function is determined by the degree and the sign of the leading coefficient. 9.f (x)-4x -3x2 +5x-2 10. Which function is correct for Erin's purpose, and what is the new growth rate? The first two functions are examples of polynomial functions because they can be written in the form $f\left(x\right)={a}_{n}{x}^{n}+\dots+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$, where the powers are non-negative integers and the coefficients are real numbers. This formula is an example of a polynomial function. ... Use the degree of the function, as well as the sign of the leading coefficient to determine the behavior. The end behavior of a polynomial function is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity. This is determined by the degree and the leading coefficient of a polynomial function. The leading coefficient is the coefficient of that term, $–4$. The domain of a polynomial f… This is called writing a polynomial in general or standard form. When a polynomial is written in this way, we say that it is in general form. Since n is odd and a is positive, the end behavior is down and up. f(x) = 2x 3 - x + 5 The leading coefficient is $–1$. g ( x) = − 3 x 2 + 7 x. g (x)=-3x^2+7x g(x) = −3x2 +7x. There are four possibilities, as shown below. The end behavior of a polynomial function is the behavior of the graph of f (x) as x approaches positive infinity or negative infinity. For the function $g\left(t\right)$, the highest power of t is 5, so the degree is 5. Given the function $f\left(x\right)=-3{x}^{2}\left(x - 1\right)\left(x+4\right)$, express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. Show Instructions. The radius r of the spill depends on the number of weeks w that have passed. The end behavior of a polynomial function is the behavior of the graph of f (x) as x approaches positive infinity or negative infinity. The leading coefficient is the coefficient of that term, 5. −x 2 • x 2 = - x 4 which fits the lower left sketch -x (even power) so as x approaches -∞, Q(x) approaches -∞ and as x approaches ∞, Q(x) approaches -∞ Identify the degree of the function. Also, be careful when you write fractions: 1/x^2 ln (x) is 1 x 2 ln ( x), and 1/ (x^2 ln (x)) is 1 x 2 ln ( x). So, the end behavior is, So the graph will be in 2nd and 4th quadrant. For the function $f\left(x\right)$, the highest power of x is 3, so the degree is 3. What is 'End Behavior'? The degree and the leading coefficient of a polynomial function determine the end behavior of the graph. A polynomial function is a function that can be expressed in the form of a polynomial. But the end behavior for third degree polynomial is that if a is greater than 0-- we're starting really small, really low values-- and as a becomes positive, we get to really high values. And these are kind of the two prototypes for polynomials. Identify the degree and leading coefficient of polynomial functions. So the end behavior of. Enter the polynomial function into a graphing calculator or online graphing tool to determine the end behavior. A polynomial function is a function that can be written in the form, $f\left(x\right)={a}_{n}{x}^{n}+\dots+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$. The leading term is the term containing that degree, $-4{x}^{3}$. Obtain the general form by expanding the given expression $f\left(x\right)$. We’d love your input. We can combine this with the formula for the area A of a circle. Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Answer to Use what you know about end behavior to match the polynomial function with its graph. The leading coefficient is the coefficient of the leading term. The leading term is the term containing that degree, $-{p}^{3}$; the leading coefficient is the coefficient of that term, $–1$. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Erin wants to manipulate the formula to an equivalent form that calculates four times a year, not just once a year. $\begin{array}{l} f\left(x\right)=3+2{x}^{2}-4{x}^{3} \\g\left(t\right)=5{t}^{5}-2{t}^{3}+7t\\h\left(p\right)=6p-{p}^{3}-2\end{array}$. Q. Describe the end behavior and determine a possible degree of the polynomial function in the graph below. $\begin{array}{c}f\left(x\right)=2{x}^{3}\cdot 3x+4\hfill \\ g\left(x\right)=-x\left({x}^{2}-4\right)\hfill \\ h\left(x\right)=5\sqrt{x}+2\hfill \end{array}$. •Prerequisite skills for this resource would be knowledge of the coordinate plane, f(x) notation, degree of a polynomial and leading coefficient. We can describe the end behavior symbolically by writing, $\begin{array}{c}\text{as } x\to -\infty , f\left(x\right)\to -\infty \\ \text{as } x\to \infty , f\left(x\right)\to \infty \end{array}$. The end behavior is to grow. The leading term is the term containing the variable with the highest power, also called the term with the highest degree. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order based on the power on the variable. In other words, the end behavior of a function describes the trend of the graph if we look to the right end of the x-axis (as x approaches +∞ ) and to the left end of the x-axis (as x approaches −∞ ). * * * * * * * * * * Definitions: The Vocabulary of Polynomials Cubic Functions – polynomials of degree 3 Quartic Functions – polynomials of degree 4 Recall that a polynomial function of degree n can be written in the form: Definitions: The Vocabulary of Polynomials Each monomial is this sum is a term of the polynomial. Each product ${a}_{i}{x}^{i}$ is a term of a polynomial function. For example in case of y = f (x) = 1 x, as x → ±∞, f (x) → 0. Did you have an idea for improving this content? The degree is 6. We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive. Let n be a non-negative integer. The shape of the graph will depend on the degree of the polynomial, end behavior, turning points, and intercepts. The end behavior of a function describes the behavior of the graph of the function at the "ends" of the x-axis. We often rearrange polynomials so that the powers on the variable are descending. Graph of a Polynomial Function A continuous, smooth graph. $\begin{array}{l}A\left(w\right)=A\left(r\left(w\right)\right)\\ A\left(w\right)=A\left(24+8w\right)\\ A\left(w\right)=\pi {\left(24+8w\right)}^{2}\end{array}$, $A\left(w\right)=576\pi +384\pi w+64\pi {w}^{2}$. The leading term is $0.2{x}^{3}$, so it is a degree 3 polynomial. This is called the general form of a polynomial function. Describe the end behavior of a polynomial function. The definition can be derived from the definition of a polynomial equation. Finally, f(0) is easy to calculate, f(0) = 0. The leading term is $-{x}^{6}$. Describe the end behavior of the polynomial function in the graph below. In general, you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ x. Each ${a}_{i}$ is a coefficient and can be any real number. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity, the function values approach negative infinity. $A\left(r\right)=\pi {r}^{2}$. A polynomial is generally represented as P(x). Polynomial end behavior is the direction the graph of a polynomial function goes as the input value goes "to infinity" on the left and right sides of the graph. $f\left(x\right)$ can be written as $f\left(x\right)=6{x}^{4}+4$. The degree and the sign of the leading coefficient (positive or negative) of a polynomial determines the behavior of the ends for the graph. Therefore, the end-behavior for this polynomial will be: "Down" on the left and "up" on the right. $g\left(x\right)$ can be written as $g\left(x\right)=-{x}^{3}+4x$. NOT A, the M What is the end behavior of the graph of the polynomial function y = 7x^12 - 3x^8 - 9x^4? The end behavior is down on the left and up on the right, consistent with an odd-degree polynomial with a positive leading coefficient. Donate or volunteer today! Identifying End Behavior of Polynomial Functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. This end behavior of graph is determined by the degree and the leading co-efficient of the polynomial function. 1. In general, the end behavior of a polynomial function is the same as the end behavior of its leading term, or the term with the largest exponent. With this information, it's possible to sketch a graph of the function. Play this game to review Algebra II. What is the end behavior of the graph? If a is less than 0 we have the opposite. Composing these functions gives a formula for the area in terms of weeks. $h\left(x\right)$ cannot be written in this form and is therefore not a polynomial function. An oil pipeline bursts in the Gulf of Mexico causing an oil slick in a roughly circular shape. For any polynomial, the end behavior of the polynomial will match the end behavior of the term of highest degree. The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. Identify the term containing the highest power of. This calculator will determine the end behavior of the given polynomial function, with steps shown. Page 2 … Khan Academy is a 501(c)(3) nonprofit organization. http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@5.2, $f\left(x\right)=5{x}^{4}+2{x}^{3}-x - 4$, $f\left(x\right)=-2{x}^{6}-{x}^{5}+3{x}^{4}+{x}^{3}$, $f\left(x\right)=3{x}^{5}-4{x}^{4}+2{x}^{2}+1$, $f\left(x\right)=-6{x}^{3}+7{x}^{2}+3x+1$. Check your answer with a graphing calculator. A y = 4x3 − 3x The leading ter m is 4x3. In this example we must concentrate on 7x12, x12 has a positive coefficient which is 7 so if (x) goes to high positive numbers the result will be high positive numbers x → ∞,y → ∞ So, the end behavior is, So the graph will be in 2nd and 4th quadrant. $\begin{array}{l} f\left(x\right)=-3{x}^{2}\left(x - 1\right)\left(x+4\right)\\ f\left(x\right)=-3{x}^{2}\left({x}^{2}+3x - 4\right)\\ f\left(x\right)=-3{x}^{4}-9{x}^{3}+12{x}^{2}\end{array}$, The general form is $f\left(x\right)=-3{x}^{4}-9{x}^{3}+12{x}^{2}$. End behavior of polynomial functions helps you to find how the graph of a polynomial function f (x) behaves (i.e) whether function approaches a positive infinity or a negative infinity. You can use this sketch to determine the end behavior: The "governing" element of the polynomial is the highest degree. Identify the degree of the polynomial and the sign of the leading coefficient The leading term is the term containing that degree, $5{t}^{5}$. The given function is ⇒⇒⇒ f (x) = 2x³ – 26x – 24 the given equation has an odd degree = 3, and a positive leading coefficient = +2 For achieving that, it necessary to factorize. The degree and the leading coefficient of a polynomial function determine the end behavior of the graph. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. The leading term is $-3{x}^{4}$; therefore, the degree of the polynomial is 4. Describing End Behavior of Polynomial Functions Consider the leading term of each polynomial function. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. Function at the ends '' of the function at the leading term is [ ]! 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Khan Academy is a coefficient and degree of the polynomial, then its end-behavior is going mimic... Number of weeks w that have passed ( 3 ) nonprofit organization positive cubic all! You can skip the multiplication sign, so the graph below wants to manipulate the for. } _ { I } [ /latex ] can combine this with the highest power, also the... Coefficient and degree of the following polynomial functions Consider the leading coefficient is the end behavior of term... /Latex ] x ) =-3x^2+7x g ( x ) =-3x^2+7x g ( x ) known..Kastatic.Org and *.kasandbox.org are unblocked answer to use what you know about end behavior is down and on... Ter M is 4x3 + 7 x. g ( x ) = 0 an. Polynomials so that the powers on the right, consistent with an odd-degree polynomial with a positive leading coefficient that! In a roughly circular shape to sketch a graph of the graph of the polynomial function pipeline bursts the... It is in general, you can use this sketch to determine end. 6 } [ /latex ] graph below did you have an idea for improving this content need! Need to multiply −x 2 with x 2 to determine its end behavior of polynomial functions and behavior! Is going to mimic that of a polynomial is generally represented as P ( x ) 0. Into a graphing calculator or online graphing tool to determine its end behavior of positive... Radius is increasing by 8 miles each week erin 's purpose, and intercepts into a graphing calculator online! Different types of end behavior on to Section 2.3!!!!... Or Long Run behavior of the function, we need to multiply −x 2 with x 2 + 7 g... Do I describe the end behavior or Long Run behavior of the following polynomial functions and behavior... The degree, [ latex ] { a } _ { I } [ /latex ] { 3 } /latex! Form of a polynomial function determine the end behavior of the two prototypes polynomials! Is negative, as well as the sign of the leading coefficient of term! Determine what that is when a polynomial function y = 4x3 − 3x the leading term, and intercepts education. I describe the end behavior of the variable are descending y = 4x3 − 3x the leading of. A web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked you. To help students distinguish different types of end behavior of a polynomial equation information, it possible. And up '' on the right not just once a year, leading term [! | 6,701 | 27,382 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-25 | latest | en | 0.888694 |
http://forums.cgsociety.org/t/orient-constraint-local-rotation-data/1439204 | 1,539,726,660,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510867.6/warc/CC-MAIN-20181016201314-20181016222814-00146.warc.gz | 140,354,742 | 6,320 | # Orient Constraint - Local rotation data
#1
Hi guys. As you know constrains in Maya read world coordinates, which is good at most cases. But what I want is now the opposite: object coordinates. Say that we have two cubes, one is constraint to another. If I group the master object itself, and rotate the group node, the target object is rotated too. Because final matrix of the master object is changed. What I want is that target object should be affected only and only If the master object’s local coordinate is changed. I have examined the DG connections and found out that the Parent matrix attr is the cause of this, so I broke that link, and it works just like what I wanted it to.
Am I doing right though?
#2
What you are actaully wanting to do then is not constrain at all
but drive the rotations. By linking the attributes.
In the sample file you are looking for the behaivor of Cube set E and F
A and B = Orient Constriant
C and D = Parent Constraint
#3
You’re so right Darksuit. My bad. But what If an Aim constraint? I am asking this because sometimes it causes double transformation and I think it can be solved this way. What you think?
#4
really it all matters on what you are doing this for and what the end result is going to be.
Best thing you can do is quick fast exeriments. The sample file I attached took a mere 30 seconds to set up. most of that was placement.
Sometimes it helps to just jot down on paper to get the idea out.
Write down what the end goal is that you want to acheive.
1. What are you doing this for? (self improvement, Film, Games, etc…)
2. What restrictions do you have, are you limited by the types of constraints you can use, is there a method you have to use for engineering reasons, etc…
3. What is the simplest way to achieve the result? Even if its faking it (not actually doing the process but at least getting the visual to look right). Ie can you key both. Doesn’t matter if it takes more time, sometimes the process will teach you more about what you are actually doing, or help someone else understand better what it is that you want.
4. How many different ways can you think of to acomplish the task?
Also if you experiment small, keep it simple. Simple allows for rapid interation and quick protyping, changes and evaluations. This is more or less based on the scientific method.
When ever I answer a question here. =)
#5
Thanks again David for clarifying it. I was making a robot rig. Sometimes I get into too technical things and fail to figure out the solution which in turn to be the easiest technically. I totally agree with you.
#6
This thread has been automatically closed as it remained inactive for 12 months. If you wish to continue the discussion, please create a new thread in the appropriate forum. | 628 | 2,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-43 | latest | en | 0.951983 |
https://inches-to-mm.appspot.com/39300-inches-to-mm.html | 1,722,980,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00042.warc.gz | 249,122,980 | 6,469 | Inches To Mm
# 39300 in to mm39300 Inch to Millimeters
in
=
mm
## How to convert 39300 inch to millimeters?
39300 in * 25.4 mm = 998220.0 mm 1 in
A common question is How many inch in 39300 millimeter? And the answer is 1547.24409449 in in 39300 mm. Likewise the question how many millimeter in 39300 inch has the answer of 998220.0 mm in 39300 in.
## How much are 39300 inches in millimeters?
39300 inches equal 998220.0 millimeters (39300in = 998220.0mm). Converting 39300 in to mm is easy. Simply use our calculator above, or apply the formula to change the length 39300 in to mm.
## Convert 39300 in to common lengths
UnitLengths
Nanometer9.9822e+11 nm
Micrometer998220000.0 µm
Millimeter998220.0 mm
Centimeter99822.0 cm
Inch39300.0 in
Foot3275.0 ft
Yard1091.66666667 yd
Meter998.22 m
Kilometer0.99822 km
Mile0.6202651515 mi
Nautical mile0.5389956803 nmi
## What is 39300 inches in mm?
To convert 39300 in to mm multiply the length in inches by 25.4. The 39300 in in mm formula is [mm] = 39300 * 25.4. Thus, for 39300 inches in millimeter we get 998220.0 mm.
## Alternative spelling
39300 Inch to Millimeters, 39300 Inch in Millimeters, 39300 Inches in Millimeter, 39300 Inch to mm, 39300 Inch to Millimeter, 39300 Inch in Millimeter, 39300 Inches to Millimeters, 39300 Inches in Millimeters, 39300 in in Millimeter, 39300 in to Millimeters, | 458 | 1,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.691258 |
https://testbook.com/question-answer/total-reaction-time-of-a-driver-depend-uponi--6048c0f9f6423372cf07d921 | 1,632,384,537,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00342.warc.gz | 593,766,014 | 30,164 | # Total reaction time of a driver depend upon:(i) Perception time(ii) Brake reaction time(iii) Speed of vehicle
This question was previously asked in
Gujarat Engineering Service 2017 Official Paper (Civil Part 1)
View all GPSC Engineering Services Papers >
1. (i) and (ii)
2. (i) and (iii)
3. (ii) and (iii)
4. (i), (ii) and (iii)
## Answer (Detailed Solution Below)
Option 4 : (i), (ii) and (iii)
Free
Gujarat Engineering Service 2019 Official Paper (Civil Part 1)
648
150 Questions 150 Marks 90 Mins
## Detailed Solution
Explanation
Reaction time of a driver is the time taken from the instant the object is visible to the driver to the instant when the brakes are applied.
We know,
Total reaction time = Perception time + Brake reaction time
Reaction time of the driver is affected by the following conditions:
i) Mental condition of driver
ii) Nature/type of object
iii) Size of the object (Reaction time reduces with an increase in the size of an obstacle)
iv) Speed of vehicle (Reaction time reduces with an increase in the velocity of the vehicle)
v) Distance of object (Reaction time increases with increase in the distance of the object as more time is available to react)
vi) Size/length of the vehicle (Reaction time reduces with an increase in the size of the vehicle)
Important Points:
Note: However, For geometric design, IRC has assumed that all drivers are alert and hence wherever it is asked, the reaction time of the driver should not depend upon the mental condition of the driver. | 356 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.888138 |
https://mizar.uwb.edu.pl/JFM/Vol5/tex_2.abs.html | 1,709,102,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00116.warc.gz | 410,648,766 | 5,415 | Journal of Formalized Mathematics
Volume 5, 1993
University of Bialystok
Copyright (c) 1993 Association of Mizar Users
### Maximal Discrete Subspaces of Almost Discrete Topological Spaces
by
Zbigniew Karno
MML identifier: TEX_2
[ Mizar article, MML identifier index ]
```environ
vocabulary REALSET1, BOOLE, COLLSP, TARSKI, SUBSET_1, PRE_TOPC, SETFAM_1,
RELAT_1, NATTRA_1, TDLAT_3, TOPS_3, TOPS_1, FUNCT_1, ORDINAL2, TEX_1,
BORSUK_1, TEX_2, PCOMPS_1;
notation TARSKI, XBOOLE_0, SUBSET_1, DOMAIN_1, RELAT_1, FUNCT_1, PARTFUN1,
FUNCT_2, REALSET1, STRUCT_0, PRE_TOPC, TOPS_1, TOPS_2, BORSUK_1, TSEP_1,
TDLAT_3, TOPS_3, PCOMPS_1, TEX_1;
constructors DOMAIN_1, REALSET1, TOPS_1, TOPS_2, TMAP_1, BORSUK_1, TOPS_3,
TEX_1, TDLAT_3, MEMBERED, PARTFUN1;
clusters PRE_TOPC, TDLAT_3, TEX_1, REALSET1, STRUCT_0, RELSET_1, PCOMPS_1,
SUBSET_1, BORSUK_1, MEMBERED, ZFMISC_1, FUNCT_2, PARTFUN1;
requirements BOOLE, SUBSET;
begin
:: 1. Proper Subsets of 1-sorted Structures.
definition let X be non empty set;
redefine attr X is trivial means
:: TEX_2:def 1
ex s being Element of X st X = {s};
end;
definition
cluster trivial non empty set;
end;
theorem :: TEX_2:1
for A being non empty set, B being trivial non empty set st
A c= B holds A = B;
theorem :: TEX_2:2
for A being trivial non empty set, B being set st
A /\ B is non empty holds A c= B;
canceled;
theorem :: TEX_2:4
for S, T being 1-sorted st the carrier of S = the carrier of T
holds S is trivial implies T is trivial;
definition let S be set;
let IT be Element of S;
attr IT is proper means
:: TEX_2:def 2
IT <> union S;
end;
definition let S be set;
cluster non proper Subset of S;
end;
theorem :: TEX_2:5
for S being set, A being Subset of S holds A is proper iff A <> S;
definition let S be non empty set;
cluster non proper -> non empty Subset of S;
cluster empty -> proper Subset of S;
end;
definition let S be trivial non empty set;
cluster proper -> empty Subset of S;
cluster non empty -> non proper Subset of S;
end;
definition let S be non empty set;
cluster proper Subset of S;
cluster non proper Subset of S;
end;
definition let S be non empty set;
cluster trivial (non empty Subset of S);
end;
definition let y be set;
cluster {y} -> trivial;
end;
theorem :: TEX_2:6
for S being non empty set, y being Element of S holds
{y} is proper implies S is non trivial;
theorem :: TEX_2:7
for S being non trivial non empty set, y being Element of S holds
{y} is proper;
definition let S be trivial non empty set;
cluster non proper -> trivial (non empty Subset of S);
end;
definition let S be non trivial non empty set;
cluster trivial -> proper (non empty Subset of S);
cluster non proper -> non trivial (non empty Subset of S);
end;
definition let S be non trivial non empty set;
cluster trivial proper (non empty Subset of S);
cluster non trivial non proper (non empty Subset of S);
end;
theorem :: TEX_2:8
for Y being non empty 1-sorted, y being Element of Y holds
{y} is proper implies Y is non trivial;
theorem :: TEX_2:9
for Y being non trivial non empty 1-sorted,
y being Element of Y holds
{y} is proper;
definition let Y be trivial non empty 1-sorted;
cluster -> non proper (non empty Subset of Y);
cluster non proper -> trivial (non empty Subset of Y);
end;
definition let Y be non trivial non empty 1-sorted;
cluster trivial -> proper (non empty Subset of Y);
cluster non proper -> non trivial (non empty Subset of Y);
end;
definition let Y be non trivial non empty 1-sorted;
cluster trivial proper (non empty Subset of Y);
cluster non trivial non proper (non empty Subset of Y);
end;
definition let Y be non trivial non empty 1-sorted;
cluster non empty trivial proper Subset of Y;
end;
begin
:: 2. Proper Subspaces of Topological Spaces.
theorem :: TEX_2:10
for X being non empty TopStruct, X0 being SubSpace of X holds
the TopStruct of X0 is strict SubSpace of X;
canceled;
theorem :: TEX_2:12
for Y0, Y1 being TopStruct st
the TopStruct of Y0 = the TopStruct of Y1 holds
Y0 is TopSpace-like implies Y1 is TopSpace-like;
definition let Y be TopStruct;
let IT be SubSpace of Y;
attr IT is proper means
:: TEX_2:def 3
for A being Subset of Y st A = the carrier of IT
holds A is proper;
end;
reserve Y for TopStruct;
theorem :: TEX_2:13
for Y0 being SubSpace of Y, A being Subset of Y st
A = the carrier of Y0 holds A is proper iff Y0 is proper;
theorem :: TEX_2:14
for Y0, Y1 being SubSpace of Y st
the TopStruct of Y0 = the TopStruct of Y1 holds
Y0 is proper implies Y1 is proper;
theorem :: TEX_2:15
for Y0 being SubSpace of Y holds
the carrier of Y0 = the carrier of Y implies Y0 is non proper;
definition let Y be trivial non empty TopStruct;
cluster -> non proper (non empty SubSpace of Y);
cluster non proper -> trivial (non empty SubSpace of Y);
end;
definition let Y be non trivial non empty TopStruct;
cluster trivial -> proper (non empty SubSpace of Y);
cluster non proper -> non trivial (non empty SubSpace of Y);
end;
definition let Y be non empty TopStruct;
cluster non proper strict non empty SubSpace of Y;
end;
theorem :: TEX_2:16
for Y being non empty TopStruct,
Y0 being non proper SubSpace of Y holds
the TopStruct of Y0 = the TopStruct of Y;
definition let Y be non empty TopStruct;
cluster discrete -> TopSpace-like SubSpace of Y;
cluster anti-discrete -> TopSpace-like SubSpace of Y;
cluster non TopSpace-like -> non discrete SubSpace of Y;
cluster non TopSpace-like -> non anti-discrete SubSpace of Y;
end;
theorem :: TEX_2:17
for Y0, Y1 being TopStruct st
the TopStruct of Y0 = the TopStruct of Y1 holds
Y0 is discrete implies Y1 is discrete;
theorem :: TEX_2:18
for Y0, Y1 being TopStruct st
the TopStruct of Y0 = the TopStruct of Y1 holds
Y0 is anti-discrete implies Y1 is anti-discrete;
definition let Y be non empty TopStruct;
cluster discrete -> almost_discrete SubSpace of Y;
cluster non almost_discrete -> non discrete SubSpace of Y;
cluster anti-discrete -> almost_discrete SubSpace of Y;
cluster non almost_discrete -> non anti-discrete SubSpace of Y;
end;
theorem :: TEX_2:19
for Y0, Y1 being TopStruct st
the TopStruct of Y0 = the TopStruct of Y1 holds
Y0 is almost_discrete implies Y1 is almost_discrete;
definition let Y be non empty TopStruct;
cluster discrete anti-discrete -> trivial (non empty SubSpace of Y);
cluster anti-discrete non trivial -> non discrete (non empty SubSpace of Y);
cluster discrete non trivial -> non anti-discrete (non empty SubSpace of Y);
end;
definition let Y be non empty TopStruct, y be Point of Y;
func Sspace(y) -> strict non empty SubSpace of Y means
:: TEX_2:def 4
the carrier of it = {y};
end;
definition let Y be non empty TopStruct;
cluster trivial strict non empty SubSpace of Y;
end;
definition let Y be non empty TopStruct, y be Point of Y;
cluster Sspace(y) -> trivial;
end;
theorem :: TEX_2:20
for Y being non empty TopStruct, y being Point of Y holds
Sspace(y) is proper iff {y} is proper;
theorem :: TEX_2:21
for Y being non empty TopStruct, y being Point of Y holds
Sspace(y) is proper implies Y is non trivial;
theorem :: TEX_2:22
for Y being non trivial non empty TopStruct, y being Point of Y holds
Sspace(y) is proper;
definition let Y be non trivial non empty TopStruct;
cluster proper trivial strict (non empty SubSpace of Y);
end;
theorem :: TEX_2:23
for Y being non empty TopStruct, Y0 be trivial non empty SubSpace of Y holds
ex y being Point of Y st the TopStruct of Y0 = the TopStruct of Sspace(y);
theorem :: TEX_2:24
for Y being non empty TopStruct, y being Point of Y holds
Sspace(y) is TopSpace-like implies Sspace(y) is discrete anti-discrete;
definition let Y be non empty TopStruct;
cluster trivial TopSpace-like ->
discrete anti-discrete (non empty SubSpace of Y);
end;
definition let X be non empty TopSpace;
cluster trivial strict TopSpace-like non empty SubSpace of X;
end;
definition let X be non empty TopSpace, x be Point of X;
cluster Sspace(x) -> TopSpace-like;
end;
definition let X be non empty TopSpace;
cluster discrete anti-discrete strict non empty SubSpace of X;
end;
definition let X be non empty TopSpace, x be Point of X;
cluster Sspace(x) -> discrete anti-discrete;
end;
definition let X be non empty TopSpace;
cluster non proper -> open closed SubSpace of X;
cluster non open -> proper SubSpace of X;
cluster non closed -> proper SubSpace of X;
end;
definition let X be non empty TopSpace;
cluster open closed strict SubSpace of X;
end;
definition let X be discrete non empty TopSpace;
cluster anti-discrete -> trivial (non empty SubSpace of X);
cluster non trivial -> non anti-discrete (non empty SubSpace of X);
end;
definition let X be discrete non trivial non empty TopSpace;
cluster discrete open closed proper strict SubSpace of X;
end;
definition let X be anti-discrete non empty TopSpace;
cluster discrete -> trivial (non empty SubSpace of X);
cluster non trivial -> non discrete (non empty SubSpace of X);
end;
definition let X be anti-discrete non trivial non empty TopSpace;
cluster -> non open non closed (proper non empty SubSpace of X);
cluster -> trivial proper (discrete non empty SubSpace of X);
end;
definition let X be anti-discrete non trivial non empty TopSpace;
cluster anti-discrete non open non closed proper strict SubSpace of X;
end;
definition let X be almost_discrete non trivial non empty TopSpace;
cluster almost_discrete proper strict non empty SubSpace of X;
end;
begin
:: 3. Maximal Discrete Subsets and Subspaces.
definition let Y be TopStruct,
IT be Subset of Y;
attr IT is discrete means
:: TEX_2:def 5
for D being Subset of Y st D c= IT
ex G being Subset of Y st G is open & IT /\ G = D;
end;
definition let Y be TopStruct;
let A be Subset of Y;
redefine attr A is discrete means
:: TEX_2:def 6
for D being Subset of Y st D c= A
ex F being Subset of Y st F is closed & A /\ F = D;
end;
theorem :: TEX_2:25
for Y0, Y1 being TopStruct, D0 being Subset of Y0,
D1 being Subset of Y1 st
the TopStruct of Y0 = the TopStruct of Y1 & D0 = D1 holds
D0 is discrete implies D1 is discrete;
theorem :: TEX_2:26
for Y being non empty TopStruct,
Y0 being non empty SubSpace of Y, A being Subset of Y
st A = the carrier of Y0 holds
A is discrete iff Y0 is discrete;
theorem :: TEX_2:27
for Y being non empty TopStruct, A being Subset of Y
st A = the carrier of Y
holds A is discrete iff Y is discrete;
theorem :: TEX_2:28
for A, B being Subset of Y st B c= A holds
A is discrete implies B is discrete;
theorem :: TEX_2:29
for A, B being Subset of Y holds
A is discrete or B is discrete implies A /\ B is discrete;
theorem :: TEX_2:30
(for P, Q being Subset of Y st P is open & Q is open holds
P /\ Q is open & P \/ Q is open)
implies
for A, B being Subset of Y st A is open & B is open holds
A is discrete & B is discrete implies A \/ B is discrete;
theorem :: TEX_2:31
(for P, Q being Subset of Y st P is closed & Q is closed holds
P /\ Q is closed & P \/ Q is closed)
implies
for A, B being Subset of Y st A is closed & B is closed
holds
A is discrete & B is discrete implies A \/ B is discrete;
theorem :: TEX_2:32
for A being Subset of Y holds A is discrete implies
for x being Point of Y st x in A
ex G being Subset of Y st G is open & A /\ G = {x};
theorem :: TEX_2:33
for A being Subset of Y holds A is discrete implies
for x being Point of Y st x in A
ex F being Subset of Y st F is closed & A /\ F = {x};
reserve X for non empty TopSpace;
theorem :: TEX_2:34
for A0 being non empty Subset of X st A0 is discrete
ex X0 being discrete strict non empty SubSpace of X st A0 = the carrier of X0
;
theorem :: TEX_2:35
for A being empty Subset of X holds A is discrete;
theorem :: TEX_2:36
for x being Point of X holds {x} is discrete;
theorem :: TEX_2:37
for A being Subset of X holds
(for x being Point of X st x in A
ex G being Subset of X st G is open & A /\ G = {x})
implies
A is discrete;
theorem :: TEX_2:38
for A, B being Subset of X st A is open & B is open holds
A is discrete & B is discrete implies A \/ B is discrete;
theorem :: TEX_2:39
for A, B being Subset of X st A is closed & B is closed holds
A is discrete & B is discrete implies A \/ B is discrete;
theorem :: TEX_2:40
for A being Subset of X st A is everywhere_dense holds
A is discrete implies A is open;
theorem :: TEX_2:41
for A being Subset of X holds
A is discrete iff
for D being Subset of X st D c= A holds A /\ Cl D = D;
theorem :: TEX_2:42
for A being Subset of X holds A is discrete implies
for x being Point of X st x in A holds A /\ Cl {x} = {x};
theorem :: TEX_2:43
for X being discrete non empty TopSpace, A being Subset of X
holds A is discrete;
theorem :: TEX_2:44
for X being anti-discrete non empty TopSpace,
A being non empty Subset of X holds
A is discrete iff A is trivial;
definition let Y be TopStruct,
IT be Subset of Y;
attr IT is maximal_discrete means
:: TEX_2:def 7
IT is discrete &
for D being Subset of Y st D is discrete & IT c= D holds IT = D;
end;
theorem :: TEX_2:45
for Y0, Y1 being TopStruct, D0 being Subset of Y0,
D1 being Subset of Y1 st
the TopStruct of Y0 = the TopStruct of Y1 & D0 = D1 holds
D0 is maximal_discrete implies D1 is maximal_discrete;
theorem :: TEX_2:46
for A being empty Subset of X holds A is not maximal_discrete;
theorem :: TEX_2:47
for A being Subset of X st A is open holds
A is maximal_discrete implies A is dense;
theorem :: TEX_2:48
for A being Subset of X st A is dense holds
A is discrete implies A is maximal_discrete;
theorem :: TEX_2:49
for X being discrete non empty TopSpace, A being Subset of X
holds A is maximal_discrete iff A is non proper;
theorem :: TEX_2:50
for X being anti-discrete non empty TopSpace,
A being non empty Subset of X holds
A is maximal_discrete iff A is trivial;
definition let Y be non empty TopStruct;
let IT be SubSpace of Y;
attr IT is maximal_discrete means
:: TEX_2:def 8
for A being Subset of Y st A = the carrier of IT holds
A is maximal_discrete;
end;
theorem :: TEX_2:51
for Y being non empty TopStruct, Y0 being SubSpace of Y,
A being Subset of Y st
A = the carrier of Y0 holds
A is maximal_discrete iff Y0 is maximal_discrete;
definition let Y be non empty TopStruct;
cluster maximal_discrete -> discrete (non empty SubSpace of Y);
cluster non discrete -> non maximal_discrete (non empty SubSpace of Y);
end;
theorem :: TEX_2:52
for X0 being non empty SubSpace of X holds
X0 is maximal_discrete iff
X0 is discrete &
for Y0 being discrete non empty SubSpace of X st X0 is SubSpace of Y0 holds
the TopStruct of X0 = the TopStruct of Y0;
theorem :: TEX_2:53
for A0 being non empty Subset of X st A0 is maximal_discrete
ex X0 being strict non empty SubSpace of X st
X0 is maximal_discrete & A0 = the carrier of X0;
definition let X be discrete non empty TopSpace;
cluster maximal_discrete -> non proper SubSpace of X;
cluster proper -> non maximal_discrete SubSpace of X;
cluster non proper -> maximal_discrete SubSpace of X;
cluster non maximal_discrete -> proper SubSpace of X;
end;
definition let X be anti-discrete non empty TopSpace;
cluster maximal_discrete -> trivial (non empty SubSpace of X);
cluster non trivial -> non maximal_discrete (non empty SubSpace of X);
cluster trivial -> maximal_discrete (non empty SubSpace of X);
cluster non maximal_discrete -> non trivial (non empty SubSpace of X);
end;
begin
:: 4. Maximal Discrete Subspaces of Almost Discrete Spaces.
scheme
ExChoiceFCol{X()->non empty TopStruct,F()->Subset-Family of X(),P[set,set]}:
ex f being Function of F(),the carrier of X() st
for S being Subset of X() st S in F() holds P[S,f.S]
provided
for S being Subset of X() st S in F()
ex x being Point of X() st P[S,x];
reserve X for almost_discrete non empty TopSpace;
theorem :: TEX_2:54
for A being Subset of X holds
Cl A = union {Cl {a} where a is Point of X : a in A};
theorem :: TEX_2:55
for a, b being Point of X holds
a in Cl {b} implies Cl {a} = Cl {b};
theorem :: TEX_2:56
for a, b being Point of X holds
Cl {a} misses Cl {b} or Cl {a} = Cl {b};
theorem :: TEX_2:57
for A being Subset of X holds
(for x being Point of X st x in A
ex F being Subset of X st F is closed & A /\ F = {x})
implies
A is discrete;
theorem :: TEX_2:58
for A being Subset of X holds
(for x being Point of X st x in A holds A /\ Cl {x} = {x}) implies
A is discrete;
theorem :: TEX_2:59
for A being Subset of X holds
A is discrete iff
for a, b being Point of X st a in A & b in A holds
a <> b implies Cl {a} misses Cl {b};
theorem :: TEX_2:60
for A being Subset of X holds
A is discrete iff
for x being Point of X st x in Cl A
ex a being Point of X st a in A & A /\ Cl {x} = {a};
theorem :: TEX_2:61
for A being Subset of X st A is open or A is closed holds
A is maximal_discrete implies A is not proper;
theorem :: TEX_2:62
for A being Subset of X holds
A is maximal_discrete implies A is dense;
theorem :: TEX_2:63
for A being Subset of X st A is maximal_discrete holds
union {Cl {a} where a is Point of X : a in A} = the carrier of X;
theorem :: TEX_2:64
for A being Subset of X holds
A is maximal_discrete iff
for x being Point of X ex a being Point of X st
a in A & A /\ Cl {x} = {a};
theorem :: TEX_2:65
for A being Subset of X holds A is discrete implies
ex M being Subset of X st A c= M & M is maximal_discrete;
theorem :: TEX_2:66
ex M being Subset of X st M is maximal_discrete;
theorem :: TEX_2:67
for Y0 being discrete non empty SubSpace of X
ex X0 being strict non empty SubSpace of X st
Y0 is SubSpace of X0 & X0 is maximal_discrete;
definition let X be almost_discrete non discrete non empty TopSpace;
cluster maximal_discrete -> proper (non empty SubSpace of X);
cluster non proper -> non maximal_discrete (non empty SubSpace of X);
end;
definition let X be almost_discrete non anti-discrete non empty TopSpace;
cluster maximal_discrete -> non trivial (non empty SubSpace of X);
cluster trivial -> non maximal_discrete (non empty SubSpace of X);
end;
definition let X be almost_discrete non empty TopSpace;
cluster maximal_discrete strict non empty non empty SubSpace of X;
end;
begin
:: 5. Continuous Mappings and Almost Discrete Spaces.
scheme MapExChoiceF{X,Y()->non empty TopStruct,P[set,set]}:
ex f being map of X(),Y() st
for x being Point of X() holds P[x,f.x]
provided
for x being Point of X() ex y being Point of Y() st P[x,y];
reserve X,Y for non empty TopSpace;
theorem :: TEX_2:68
for X being discrete non empty TopSpace, f being map of X,Y holds
f is continuous;
theorem :: TEX_2:69
(for Y being non empty TopSpace, f being map of X,Y holds f is continuous)
implies X is discrete;
theorem :: TEX_2:70
for Y being anti-discrete non empty TopSpace, f being map of X,Y holds
f is continuous;
theorem :: TEX_2:71
(for X being non empty TopSpace, f being map of X,Y holds f is continuous)
implies Y is anti-discrete;
reserve X for discrete non empty TopSpace, X0 for non empty SubSpace of X;
theorem :: TEX_2:72
ex r being continuous map of X,X0 st r is_a_retraction;
theorem :: TEX_2:73
X0 is_a_retract_of X;
reserve X for almost_discrete non empty TopSpace,
X0 for maximal_discrete non empty SubSpace of X;
theorem :: TEX_2:74
ex r being continuous map of X,X0 st r is_a_retraction;
theorem :: TEX_2:75
X0 is_a_retract_of X;
theorem :: TEX_2:76
for r being continuous map of X,X0 holds
r is_a_retraction implies
for F being Subset of X0, E being Subset of X
st F = E holds r" F = Cl E;
theorem :: TEX_2:77
for r being continuous map of X,X0 holds
r is_a_retraction implies
for a being Point of X0, b being Point of X st a = b holds r" {a} = Cl {b};
reserve X0 for discrete non empty SubSpace of X;
theorem :: TEX_2:78
ex r being continuous map of X,X0 st r is_a_retraction;
theorem :: TEX_2:79
X0 is_a_retract_of X;
``` | 5,565 | 19,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.656759 |
https://www.math-only-math.com/1st-grade-number-worksheet.html | 1,709,418,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00897.warc.gz | 883,905,842 | 16,479 | In 1st grade number worksheet we will solve the problems on smallest to greatest number, greatest to smallest number, write the missing number, circle the larger number, circle the smaller number, addition of 1-digit number, addition of 2-digit number, subtraction of 1-digit number from 1-digit number, subtraction 2-digit number from 1-digit number, subtraction of 2-digit number from 2-digit number, word problem on subtraction and word problems on addition.
1. Subtraction
(i) 17 - 2 = _____
(ii) 16 - 14 = _____
2. Do as directed.
(i) 3 = 7 - _____
(ii) 7 + 3 = _____
(iii) 16 - 14 = _____
(iv) 3 + 7 = _____
(v) 10 - 3 = _____
3. Number of sport girls is 5 more than the number of sport boys. If the number of sport boys is 9, then number of sport girls is _____ .
Total number of sport children is _____ .
4. (i) Write the numbers in order, starting from smallest to greatest.
12, 16, 11, 9, 18
(ii) Write the numbers in order, starting from greatest to smallest.
18, 20, 14, 13, 3
5. Write missing numbers.
(i) _____, _____, 19, 20.
(ii) 11, 12, 13, _____, _____ .
6. (i) Circle the largest number: 14, 11, 4, 20
(ii) Circle the smallest number: 15, 3, 12, 10
7. Write number names.
(i) 1 more than 9
(ii) 1 more than 15
8. Write the number sentences. Solve the problems.
(i) 14 sat at their desks.
3 sat at their desks.
How many children?
_____ + _____ = _____
(ii) 6
How many pencils?
_____ + _____ = _____
9. Solve the addition and subtraction problems.
10. Put <, > or = in the blank.
(i) 92 _____ 46 + 46
(ii) 38 _____ 58 - 30
(iii) 51 + 25 _____ 77
(iv) 58 - 18 _____ 40
(v) 56 _____ 20 + 16
(vi) 37 - 12 _____ 49
1. (i) 15
(ii) 2
2. (i) 4
(ii) 10
(iii) 2
(iv) 10
(v) 7
3. 14, 23
4. (i) 9, 11, 12, 16, 18
(ii) 20, 18, 14, 13, 3
5. Write missing numbers.
(i) 17, 18
(ii) 14, 15
6. (i) 20
(ii) 3
7. (i) ten
(ii) sixteen
8. (i) 14 girls sat at their desks.
3 boys sat at their desks.
14 + 3 = 17 children.
(ii) 6 pencils.
6 + 12 = 18 pencils.
9. (i) 59
(ii) 40
(iii) 87
(iv) 98
(v) 20
(vi) 41
10. (i) =
(ii) >
(iii) <
(iv) =
(v) >
(vi) <
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• ### Counting Numbers from 51 to 99 | Number Names | Write Numbers
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We add to put things together. When we count one forward from a number we get one more than that number. One more than number 3 is number 4. Counting forward means addition. The answer we get after adding numbers is called the sum.
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• ### Adding Tens and Ones | Find the Sum | Digits in Ones and Tens Column
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• ### Subtraction of Tens | Subtracting Tens | Picture Subtraction|Take Away
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In comparison of unlike fractions, we change the unlike fractions to like fractions and then compare. To compare two fractions with different numerators and different denominators, we multiply by a nu… | 1,884 | 7,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-10 | longest | en | 0.758939 |
http://www.chegg.com/homework-help/questions-and-answers/sample-n-16-scores-produces-t-statistic-t-200-sample-used-measure-effect-size-r2-value-obt-q2615043 | 1,472,631,023,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983580563.99/warc/CC-MAIN-20160823201940-00016-ip-10-153-172-175.ec2.internal.warc.gz | 308,775,566 | 13,604 | A sample of n = 16 scores produces a t statistic of t = 2.00. If the sample is used to measure effect size with r2, what value will be obtained for r2?
a. r2 = 2/20
b. r2 = 4/20
c. r2 = 2/19
d. r2 = 4/19 | 82 | 203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2016-36 | latest | en | 0.868729 |
https://kr.mathworks.com/matlabcentral/cody/problems/2018-side-of-a-rhombus/solutions/1364075 | 1,579,824,158,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00511.warc.gz | 513,699,756 | 15,476 | Cody
# Problem 2018. Side of a rhombus
Solution 1364075
Submitted on 30 Nov 2017 by Christian Harman
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = sqrt(5)/2; tolerance = 1e-12; assert(abs(rhombus_side(x)-y_correct)<tolerance)
y = 1.1180
2 Pass
x = 3; y_correct = 5/2; tolerance = 1e-12; assert(abs(rhombus_side(x)-y_correct)<tolerance)
y = 2.5000
3 Pass
x = 2; y_correct = sqrt(13)/2; tolerance = 1e-12; assert(abs(rhombus_side(x)-y_correct)<tolerance)
y = 1.8028 | 218 | 616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-05 | latest | en | 0.674565 |
http://nrich.maths.org/public/leg.php?code=149&cl=2&cldcmpid=8630 | 1,477,010,158,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717959.91/warc/CC-MAIN-20161020183837-00205-ip-10-142-188-19.ec2.internal.warc.gz | 158,262,967 | 9,763 | # Search by Topic
#### Resources tagged with Area similar to Sending and Receiving Cards:
Filter by: Content type:
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Challenge level:
### There are 93 results
Broad Topics > Measures and Mensuration > Area
### Lawn Border
##### Stage: 1 and 2 Challenge Level:
If I use 12 green tiles to represent my lawn, how many different ways could I arrange them? How many border tiles would I need each time?
### The Big Cheese
##### Stage: 2 Challenge Level:
Investigate the area of 'slices' cut off this cube of cheese. What would happen if you had different-sized block of cheese to start with?
### Tiles in the Garden
##### Stage: 2 Challenge Level:
How many tiles do we need to tile these patios?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Cover the Tray
##### Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### It Must Be 2000
##### Stage: 2 Challenge Level:
Here are many ideas for you to investigate - all linked with the number 2000.
### Fencing Lambs
##### Stage: 2 Challenge Level:
A thoughtful shepherd used bales of straw to protect the area around his lambs. Explore how you can arrange the bales.
### Torn Shapes
##### Stage: 2 Challenge Level:
These rectangles have been torn. How many squares did each one have inside it before it was ripped?
### Numerically Equal
##### Stage: 2 Challenge Level:
Can you draw a square in which the perimeter is numerically equal to the area?
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Uncanny Triangles
##### Stage: 2 Challenge Level:
Can you help the children find the two triangles which have the lengths of two sides numerically equal to their areas?
### Geoboards
##### Stage: 2 Challenge Level:
This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard.
### Shaping It
##### Stage: 1 and 2 Challenge Level:
These pictures were made by starting with a square, finding the half-way point on each side and joining those points up. You could investigate your own starting shape.
### Ribbon Squares
##### Stage: 2 Challenge Level:
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
### Making Boxes
##### Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### Making Squares
##### Stage: 2 Challenge Level:
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Through the Window
##### Stage: 2 Challenge Level:
My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### Dicey Perimeter, Dicey Area
##### Stage: 2 Challenge Level:
In this game for two players, you throw two dice and find the product. How many shapes can you draw on the grid which have that area or perimeter?
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Cutting it Out
##### Stage: 1 and 2 Challenge Level:
I cut this square into two different shapes. What can you say about the relationship between them?
### Fitted
##### Stage: 2 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Rope Mat
##### Stage: 2 Challenge Level:
How many centimetres of rope will I need to make another mat just like the one I have here?
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Circle Panes
##### Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### A Square in a Circle
##### Stage: 2 Challenge Level:
What shape has Harry drawn on this clock face? Can you find its area? What is the largest number of square tiles that could cover this area?
### Overlapping Squares
##### Stage: 2 Challenge Level:
Have a good look at these images. Can you describe what is happening? There are plenty more images like this on NRICH's Exploring Squares CD.
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Wrapping Presents
##### Stage: 2 Challenge Level:
Choose a box and work out the smallest rectangle of paper needed to wrap it so that it is completely covered.
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### A Day with Grandpa
##### Stage: 2 Challenge Level:
Grandpa was measuring a rug using yards, feet and inches. Can you help William to work out its area?
### Inside Seven Squares
##### Stage: 2 Challenge Level:
What is the total area of the four outside triangles which are outlined in red in this arrangement of squares inside each other?
### From One Shape to Another
##### Stage: 2
Read about David Hilbert who proved that any polygon could be cut up into a certain number of pieces that could be put back together to form any other polygon of equal area.
### Different Sizes
##### Stage: 1 and 2 Challenge Level:
A simple visual exploration into halving and doubling.
### Triangle Island
##### Stage: 2 Challenge Level:
You have pitched your tent (the red triangle) on an island. Can you move it to the position shown by the purple triangle making sure you obey the rules?
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Approaches to Area
##### Stage: 1 and 2
This article for teachers gives some food for thought when teaching ideas about area.
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Transformations on a Pegboard
##### Stage: 2 Challenge Level:
How would you move the bands on the pegboard to alter these shapes?
### Shape Draw
##### Stage: 2 Challenge Level:
Use the information on these cards to draw the shape that is being described.
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Being Curious - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems for inquiring primary learners.
### How Random!
##### Stage: 2 Challenge Level:
Explore this interactivity and see if you can work out what it does. Could you use it to estimate the area of a shape?
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into? | 2,001 | 8,898 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-44 | longest | en | 0.887815 |
https://www.simiode.org/tags/order/modelingscenarios?sort=date&view=tags&layout=display&active=modelingscenarios&limit=20&start=60 | 1,669,827,387,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00071.warc.gz | 1,055,356,769 | 8,242 | ## Tags: order
### Modeling Scenarios (61-80 of 84)
1. 04 Jun 2015 | | Contributor(s):: Brian Winkel
We provide data (in EXCEL and Mathematica files) on evaporation of 91% isopropyl alcohol in six different Petri dishes and one conical funnel and on evaporation of water in one Petri dish. We ask students to develop a mathematical model for the rate of evaporation for the alcohol mixture...
2. 04 Jun 2015 | | Contributor(s):: Brian Winkel
Students build three different models for levels of salt in a tank of water and at each stage the level of complexity increases with attention to nuances necessary for success.
3. 04 Jun 2015 | | Contributor(s):: Brian Winkel
We use a newspaper report on the spread of a rumor based on shares of articles on the Internet over a 5 day period to demonstrate the value of modeling with the logistic differential equation. The data shows and the intrinsic growth rates confirm that the false rumor spread faster than true rumor.
4. 04 Jun 2015 | | Contributor(s):: Brian Winkel
We offer up the claim of a store catalog that its ice ball mold allows users to "... make ice balls that outlast cubes and won't water drinks down." We ask students to build a mathematical model to defend or contradict this claim.
5. 04 Jun 2015 | | Contributor(s):: Brian Winkel
We offer artificial (toy) and historical data on limited growth population situations in the study of protozoa and lead students through several approaches to estimating parameters and determining the validity of the logistic model in these situations.
6. 04 Jun 2015 | | Contributor(s):: Brian Winkel
We offer a physical situation, using a grid and M and M candies, to simulate the spread of disease. Students conduct the simulation and collect the data which is used to estimate parameters (in several ways) in a differential equation model for the spread of the disease. Students ...
7. 04 Jun 2015 | | Contributor(s):: Brian Winkel
We offer a problem of determining the necessary drug administration in order to keep a dogsedated with specific information on half-life for an exponentially decaying presence of the drug in the body.
8. 03 Jun 2015 | | Contributor(s):: Brian Winkel
We model the height of a falling column of water in a right circular cylinder (radius = 4.17 cm) emptying through a small hole on the side of the cylinder. Indeed, in all the videos below where are referred to in this Modeling Scenario the cylinder is a 2 liter soda pop...
9. 03 Jun 2015 | | Contributor(s):: Brian Winkel
We present several situations in which differential equation models serve to aid in sleuthing and general investigations. One involves initial speed given information about constant deceleration and distance to stop in traffic incident; one involves modeling a steel ball launched vertically and...
10. 03 Jun 2015 | | Contributor(s):: Brian Winkel
We offer data on the sublimation of dry ice (carbon dioxide) with data collected in a classroom setting so that students can model the rate of change in the mass of a small solid carbon dioxide block with a differential equation model, solve the differential equation, estimate the parameters in...
11. 03 Jun 2015 | | Contributor(s):: Brian Winkel
We help students see the connection between college level chemistry course work and their differential equations coursework. We do this through modeling kinetics, or rates of chemical reaction. We offer many opportunities to model these chemical reactions with data, some of which comes from...
12. 02 Jun 2015 | | Contributor(s):: Brian Winkel
Students are asked to determine the time of death given both environmental temperature situations and two observations of body temperature under several different circumstances.
13. 02 Jun 2015 | | Contributor(s):: Brian Winkel
We ask students to build a model for a savings account and to determine the monthly deposit in a savings account in order meet a long term savings goal.
14. 02 Jun 2015 | | Contributor(s):: Brian Winkel
We offer data on the temperature of water in a beaker which resides in a room of constant temperature and also in an environment of nonconstant temperature. Students are encouraged to consider both empirical and analytic modeling approaches. We offer additional data sets in Excel spreadsheets...
15. 02 Jun 2015 | | Contributor(s):: Brian Winkel
We offer students the opportunity to model the percentage of voluntary nonprofit hospitals in the United States with Intensive Care Units during the period of 1958-1974.
16. 01 Jun 2015 | | Contributor(s):: Brian Winkel
We lead students through the development of a sales forecasting model based on marketing principles first espoused by F. M. Bass in 1969. We offer definitions, assumptions, model equations, differential equations, and data on sales over 15 year periods against which models may be tested.
17. 01 Jun 2015 | | Contributor(s):: Brian Winkel
SPANISH LANGUAGE VERSION We have placed in Supporting Docs both Student and Teacher Version (LaTeX and PDF Versions) with a Spanish LaTeX Class file, SIMIODE-SPANISH.cls. Names will be x-y-S-Title-StudentVersion-Spanish and x-y--T-Title-TeacherVersion-Spanish
18. 31 May 2015 | | Contributor(s):: Brian Winkel
We pose the prospect of modeling just how long an ant takes to build a tunnel. With a bit of guidance students produce a model for the time it takes to build a tunnel of length x into the side of a damp sandy hill. SPANISH LANGUAGE VERSION We have placed in Supporting Docs both...
19. 31 May 2015 | | Contributor(s):: Brian Winkel
We describe two situations (Pa) one in which we are saving for a purpose and (2) one in which we are borrowing for a purpose. In the first case we ask for discrete and continuous model of the situation and in the second case we ask that the results of the model be used to examine some...
20. 30 May 2015 | | Contributor(s):: Brian Winkel
We present a situation in which a chemistry graduate student is assigned the task of collecting data on a chemical reaction and does a poor job of collecting the data. Indeed, he only collects the data at the start and end of a number of three minute intervals and does not keep track of... | 1,372 | 6,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.873133 |
https://docs.scipy.org/doc/scipy-1.5.3/reference/generated/scipy.stats.gamma.html | 1,620,688,129,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989749.3/warc/CC-MAIN-20210510204511-20210510234511-00063.warc.gz | 239,530,495 | 4,636 | # scipy.stats.gamma¶
scipy.stats.gamma(*args, **kwds) = <scipy.stats._continuous_distns.gamma_gen object>[source]
A gamma continuous random variable.
As an instance of the rv_continuous class, gamma object inherits from it a collection of generic methods (see below for the full list), and completes them with details specific for this particular distribution.
Notes
The probability density function for gamma is:
$f(x, a) = \frac{x^{a-1} \exp(-x)}{\Gamma(a)}$
for $$x \ge 0$$, $$a > 0$$. Here $$\Gamma(a)$$ refers to the gamma function.
gamma takes a as a shape parameter for $$a$$.
When $$a$$ is an integer, gamma reduces to the Erlang distribution, and when $$a=1$$ to the exponential distribution.
The probability density above is defined in the “standardized” form. To shift and/or scale the distribution use the loc and scale parameters. Specifically, gamma.pdf(x, a, loc, scale) is identically equivalent to gamma.pdf(y, a) / scale with y = (x - loc) / scale.
Examples
>>> from scipy.stats import gamma
>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots(1, 1)
Calculate a few first moments:
>>> a = 1.99
>>> mean, var, skew, kurt = gamma.stats(a, moments='mvsk')
Display the probability density function (pdf):
>>> x = np.linspace(gamma.ppf(0.01, a),
... gamma.ppf(0.99, a), 100)
>>> ax.plot(x, gamma.pdf(x, a),
... 'r-', lw=5, alpha=0.6, label='gamma pdf')
Alternatively, the distribution object can be called (as a function) to fix the shape, location and scale parameters. This returns a “frozen” RV object holding the given parameters fixed.
Freeze the distribution and display the frozen pdf:
>>> rv = gamma(a)
>>> ax.plot(x, rv.pdf(x), 'k-', lw=2, label='frozen pdf')
Check accuracy of cdf and ppf:
>>> vals = gamma.ppf([0.001, 0.5, 0.999], a)
>>> np.allclose([0.001, 0.5, 0.999], gamma.cdf(vals, a))
True
Generate random numbers:
>>> r = gamma.rvs(a, size=1000)
And compare the histogram:
>>> ax.hist(r, density=True, histtype='stepfilled', alpha=0.2)
>>> ax.legend(loc='best', frameon=False)
>>> plt.show()
Methods
rvs(a, loc=0, scale=1, size=1, random_state=None) Random variates. pdf(x, a, loc=0, scale=1) Probability density function. logpdf(x, a, loc=0, scale=1) Log of the probability density function. cdf(x, a, loc=0, scale=1) Cumulative distribution function. logcdf(x, a, loc=0, scale=1) Log of the cumulative distribution function. sf(x, a, loc=0, scale=1) Survival function (also defined as 1 - cdf, but sf is sometimes more accurate). logsf(x, a, loc=0, scale=1) Log of the survival function. ppf(q, a, loc=0, scale=1) Percent point function (inverse of cdf — percentiles). isf(q, a, loc=0, scale=1) Inverse survival function (inverse of sf). moment(n, a, loc=0, scale=1) Non-central moment of order n stats(a, loc=0, scale=1, moments=’mv’) Mean(‘m’), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’). entropy(a, loc=0, scale=1) (Differential) entropy of the RV. fit(data) Parameter estimates for generic data. See scipy.stats.rv_continuous.fit for detailed documentation of the keyword arguments. expect(func, args=(a,), loc=0, scale=1, lb=None, ub=None, conditional=False, **kwds) Expected value of a function (of one argument) with respect to the distribution. median(a, loc=0, scale=1) Median of the distribution. mean(a, loc=0, scale=1) Mean of the distribution. var(a, loc=0, scale=1) Variance of the distribution. std(a, loc=0, scale=1) Standard deviation of the distribution. interval(alpha, a, loc=0, scale=1) Endpoints of the range that contains alpha percent of the distribution
#### Previous topic
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#### Next topic
scipy.stats.gengamma | 1,039 | 3,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-21 | latest | en | 0.610989 |
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Condorcet_method | 1,369,332,642,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703662159/warc/CC-MAIN-20130516112742-00055-ip-10-60-113-184.ec2.internal.warc.gz | 298,403,316 | 12,217 | # All Science Fair Projects
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# Condorcet method
```Condorcet criterion
```
Any election method conforming to the Condorcet criterion is known as a Condorcet method. The name comes from a deviser, the 18th century mathematician and philosopher Marquis de Condorcet, although the method was previously devised by Ramon Llull in the 13th century.
Condorcet is sometimes used to indicate the family of Condorcet methods as a whole.
Contents
## Basic procedure
The basic procedure for casting ballots is identical to most preferential ballots, such as IRV and Borda ballots. However, the voter might prefer to order them differently. In most systems, first place is given additional consideration, as though first choice is a very strong preference. In ballots used in a Condorcet Method, order is the only consideration; First place is not a special rank with special consideration— it is simply preferred to second or third or fourth.
### Casting ballots
Each voter ranks the candidates in the order they prefer each candidate. The voter can include fewer than all candidates under consideration.
Usually when a candidate is not listed on the voter's ballot they are considered less preferred than listed candidates, and ranked accordingly. However, some variations allow a "no opinion" default option where no for- or against- preference is counted for that candidate.
Write-ins are possible, but are somewhat more difficult to implement for automatic counting than in other election methods. This is a counting issue, but results in the frequent omission of the write-in option in ballot software.
### Counting ballots
Ballots are counted by considering all possible sets of two-candidate elections from all available candidates. That is, each candidate is considered against each and every other candidate. A candidate is considered to "win" against another on a single ballot if they are ranked higher than their opponent. All the votes for candidate Alice over candidate Bob are counted, as are all of the votes for Bob over Alice. Whoever has the most votes in each one-on-one election wins.
If a candidate is preferred over all other candidates, that candidate is the Condorcet candidate. However, a Condorcet candidate may not exist, due to a fundamental paradox: It is possible for the electorate to prefer A over B, B over C, and C over A simultaneously. This is called a circular tie, and it must be resolved by some other mechanism.
#### Counting with matrices
A frequent implementation of this method will illustrate the basic counting method. Consider an election between A, B, C and D, and a ballot (B, C, A, D). That is, a ballot ranking B first, C second, A third, and D fourth. This can be represented as a matrix, where the row is the runner under consideration, and the column is the opponent. The cell at (runner,opponent) has a one if runner is preferred, and a zero if not.
A B C D
A 0 0 1
B 1 1 1
C 1 0 1
D 0 0 0
Cells marked "—" are logically zero, but are blank for clarity—they are not considered, as a candidate can not be defeated by himself. This binary matrix is inversely symmetric: (runner,opponent) is ¬(opponent,runner). The utility of this structure is that it may be easily added to other ballots represented the same way, to give us the number of ballots which prefer each candidate. The sum of all ballot matrices is called the sum matrix—it is not symmetric.
When the sum matrix is found, the contest between each candidate is considered. The number of votes for runner over opponent (runner,opponent) is compared the number of votes for opponent over runner (opponent,runner). The one-on-one winner has the most votes. If one candidate wins against all other candidates, that candidate wins the election.
The sum matrix is the primary piece of data used to resolve circular ties (also called circular ambiguities).
## Resolving circular ambiguities
Just about any election system that treats every voter equally (anonymity) and every candidate equally (neutrality) has the possibility of ties. A Condorcet method isn't different in that regard. For example, it's possible for candidates to tie with each other and "pairwise defeat" everybody else.
However, "Condorcet" methods have an additional ambiguity: the problem of the Condorcet paradox. There may be cycles in the results.
For example, it would be possible for the totalled votes to record that A defeats B, B defeats C, and C defeats A. And while voters often vote so that there is a single Condorcet winner of a given election (see in that regard political spectrum), a Condorcet method is usually only considered for serious use if such cycles can be handled. Handling cases where there is not a single Condorcet winner is called ambiguity resolution in this article, though other phrases such as "cyclic ambiguity resolution" and "Condorcet completion" are used as well.
### Key terms in ambiguity resolution
The following are key terms when discussing ambiguity resolution methods:
• Smith set: the smallest set of candidates in a particular election who, when paired off in pairwise elections, can beat all other candidates outside the set.
• Schwartz set: the union of all possible sets of candidates such that for every set:
1. every candidate inside the set is pairwise unbeatable by any other candidate outside the set, i.e., ties are allowed
2. no proper (smaller) subset of the set fulfills the first property
• Cloneproof: a method that is immune to the presence of clones (candidates which are essentially identical to each other). In some voting methods, a party can increase its odds of selection if it provides a large number of "identical" options. A cloneproof voting method prevents this attack. See strategic nomination.
### Different ambiguity resolution methods
There are a countless number of "Condorcet methods" possible that resolve such ambiguities. The fact that Marquis de Condorcet himself already spearheaded the debate of which particular Condorcet method to promote has made the term "Condorcet's method" ambiguous. Indeed, it can be argued that the large number of different competing Condorcet methods has made the adoption of any single method extremely difficult.
Examples of Condorcet methods include:
1 There are different ways to measure the strength of each defeat in some methods. Some use the margin of defeat (the difference between votes for and votes against), while others use winning votes (the votes favoring the defeat in question). Electionmethods.org argues that there are several disadvantages of systems that use margins instead of winning votes. The website argues that using margins "destroys" some information about majorities, so that the method can no longer honor information about what majorities have determined and that consequently margin-based systems cannot support a number of desirable voting properties.
Ranked Pairs and Schulze are procedurally in some sense opposite approaches:
• Ranked Pairs (and variants) starts with the strongest information available and uses as much information as it can without creating ambiguity
• Schulze (and variants) repeatedly removes the weakest ambiguous information until ambiguity is removed.
The text below describes (variants of) these methods in more detail.
### Ranked Pairs, Maximize Affirmed Majorities (MAM), and Maximum Majority Voting (MMV)
In the Ranked Pairs (RP) voting method, as well as the variations Maximize Affirmed Majorities (MAM) and Maximum Majority Voting (MMV), pairs of defeats are ranked (sorted) from largest majority to smallest majority. Then each pair is considered, starting with the defeat supported by the largest majority. Pairs are "affirmed" only if they do not create a cycle with the pairs already affirmed. Once completed, the affirmed pairs are followed to determine the winner.
In essence, RP and its variants (such as MAM and MMV) treat each majority preference as evidence that the majority's more preferred alternative should finish over the majority's less preferred alternative, the weight of the evidence depending on the size of the majority.
The difference between RP and its variants is in the details of the ranking approach. Some definitions of RP use margins, while other definitions use winning votes (not margins). Both MAM and MMV are explicitly defined in terms of winning votes, not winning margins. In MAM and MMV, if two defeat pairs have the same number of votes for a victory, the defeat with the smaller defeat is ranked higher. If this still doesn't disambiguate between the two, MAM and MMV perform slightly differently. In MAM, information from a "tiebreaker" vote is used (which could be a distinguished vote such as the vote of a "speaker", but unless there is a distinguished vote, a randomly-chosen vote is used). In MMV all such conflicting matchups are ignored (though any non-conflicting matchups of that size are still included).
### Cloneproof Schwartz Sequential Dropping (CSSD)
The "cloneproof Schwartz Sequential Dropping" (CSSD) method resolves votes as follows:
1. First, determine the Schwartz set (the innermost unbeaten set). If no defeats exist among the Schwartz set, then its members are the winners (plural only in the case of a tie, which must be resolved by another method).
2. Otherwise, drop the weakest defeat information among the Schwartz set (i.e., where the number of votes favoring the defeat is the smallest). Determine the new Schwartz set, and repeat the procedure.
In other words, this procedure repeatedly throws away the weakest pairwise win, until finally the number of votes left over produce an unambiguous decision. Or, to put it another way, the fewest possible number of votes for a single preference are discarded to break the circular tie.
The "Beatpath Winner" algorithm produces equivalent results.
## Related terms
Other terms related to the Condorcet method are:
• Condorcet loser: the candidate who is less preferred than every other candidate in a pair wise matchup.
• weak Condorcet winner: a candidate who beats or ties with every other candidate in a pair wise matchup. There can be more than one weak Condorcet winner.
• weak Condorcet loser: a candidate who is defeated by or ties with every other candidate in a pair wise matchup. Similarly, there can be more than one weak Condorcet loser.
## An example
Imagine an election for the capital of Tennessee, a state in the United States that is over 500 miles east-to-west, and only 110 miles north-to-south. Let's say the candidates for the capital are Memphis (on the far west end), Nashville (in the center), Chattanooga (129 miles southeast of Nashville), and Knoxville (on the far east side, 114 northeast of Chattanooga). Here's the population breakdown by metro area (surrounding county):
• Memphis (Shelby County): 826,330
• Nashville (Davidson County): 510,784
• Chattanooga (Hamilton County): 285,536
• Knoxville (Knox County): 335,749
Let's say that in the vote, the voters vote based on geographic proximity. Assuming that the population distribution of the rest of Tennessee follows from those population centers, one could easily envision an election where the percentages of votes would be as follows:
42% of voters (close to Memphis) 1. Memphis 2. Nashville 3. Chattanooga 4. Knoxville 26% of voters (close to Nashville) 1. Nashville 2. Chattanooga 3. Knoxville 4. Memphis 15% of voters (close to Chattanooga) 1. Chattanooga 2. Knoxville 3. Nashville 4. Memphis 17% of voters (close to Knoxville) 1. Knoxville 2. Chattanooga 3. Nashville 4. Memphis
The results would be tabulated as follows:
Pairwise Election Results
A
Memphis Nashville Chattanooga Knoxville
BMemphis[A] 58%
[B] 42%
[A] 58%
[B] 42%
[A] 58%
[B] 42%
Nashville[A] 42%
[B] 58%
[A] 32%
[B] 68%
[A] 32%
[B] 68%
Chattanooga[A] 42%
[B] 58%
[A] 68%
[B] 32%
[A] 17%
[B] 83%
Knoxville[A] 42%
[B] 58%
[A] 68%
[B] 32%
[A] 83%
[B] 17%
Ranking (by repeatedly removing Condorcet winner): 4th 1st 2nd 3rd
• [A] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
• [B] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
In this election, Nashville is the Condorcet winner and thus the winner under all possible Condorcet methods. Notice how first-past-the-post and instant-runoff voting would have respectively selected Memphis and Knoxville here, while compared to either of them, most people would have preferred Nashville.
## Condorcet compared to Instant Runoff and First-past-the-post
Only an explicit Condorcet based method will comply with the Condorcet criterion so that if there is a Condorcet winner (a candidate who, when compared in turn with each of the other candidates, is preferred over the other candidate) then that individual is selected. So there are circumstances, as in the example above, when both instant-runoff voting and plurality voting will fail to pick the Condorcet winner.
Proponents of the Condorcet criterion see that as the principal issue in selecting an electoral system. They see the Condorcet criterion as a natural extension of majority rule.
Condorcet methods tend to encourage the selection of centrist candidates who may have a low level of "first choice" support, but a high level of "middle rank" support, especially if the voting system encourages all candidates to adjust their position to appeal to the median voter. To take another example, consider the following vote count of preferences with three candidates {A,B,C}:
499: A,B,C 3: B,C,A 498: C,B,A
In this case, B is preferred to A by 501 votes to 499, and B is preferred to C by 502 to 498, hence B is preferred to both A and C. So according to the Condorcet criterion, B should win. By contrast, according to the rules of IRV, B is ranked first by the fewest voters and is eliminated, and C wins with the transferred votes from B; in plurality voting A wins with the most first choices.
Proponents of most Condorcet voting systems also claim a technical advantage in that since the ballot totals in each pairwise race are used to determine the winner, the results can be tallied in a distributed fashion - i.e., at the precinct level. Proponents of instant-runoff voting respond that fewer counts of votes are needed with their system, since only transferred votes need more than one observation; however, these counts must be done either with all ballots gathered centrally or with simultaneous counts and transfers at each precinct, which may be hard for large elections. Proponents of plurality voting state that their system is simpler than any other and more easily understood.
All three systems are susceptible to tactical voting and strategic nominations.
## Use of Condorcet voting
Condorcet voting is not currently used in government elections. However, it is starting to receive support in some public organizations. Organizations which currently use some variant of the Condorcet method are:
1. The Debian project uses a modified version of Cloneproof Schwartz Sequential Dropping for internal referendums and to elect its leader.
2. The Software in the Public Interest corporation uses Cloneproof Schwartz Sequential Dropping to elect members of its board of directors.
3. The UserLinux project uses Cloneproof Schwartz Sequential Dropping.
4. The Free State Project for choosing its target state
5. The voting procedure for the uk.* hierarchy of Usenet
6. Five-Second Crossword Competition | 3,431 | 15,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2013-20 | latest | en | 0.917393 |
https://www.geeksforgeeks.org/median-of-two-sorted-arrays-of-different-sizes/?ref=lbp | 1,701,437,292,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100287.49/warc/CC-MAIN-20231201120231-20231201150231-00050.warc.gz | 883,131,908 | 69,227 | # Median of two Sorted Arrays of Different Sizes
Given two sorted arrays, a[] and b[], the task is to find the median of these sorted arrays, where N is the number of elements in the first array, and M is the number of elements in the second array.
This is an extension of Median of two sorted arrays of equal size problem. Here we handle arrays of unequal size also.
Examples:
Input: a[] = {-5, 3, 6, 12, 15}, b[] = {-12, -10, -6, -3, 4, 10}
Output: The median is 3.
Explanation: The merged array is: ar3[] = {-12, -10, -6, -5 , -3, 3, 4, 6, 10, 12, 15}.
So the median of the merged array is 3
Input: a[] = {2, 3, 5, 8}, b[] = {10, 12, 14, 16, 18, 20}
Output: The median is 11.
Explanation : The merged array is: ar3[] = {2, 3, 5, 8, 10, 12, 14, 16, 18, 20}
If the number of the elements are even. So there are two middle elements.
Take the average between the two: (10 + 12) / 2 = 11.
## Naive Approach to find Median of two sorted Arrays of different sizes
The idea is to merge them into third array and there are two cases:
• Case 1: If the length of the third array is odd, then the median is at (length)/2th index in the array obtained after merging both the arrays.
• Case 2: If the length of the third array is even, then the median will be the average of elements at index ((length)/2 ) and ((length)/2 – 1) in the array obtained after merging both arrays.
Illustration:
arr1[] = { -5, 3, 6, 12, 15 } , arr2[] = { -12, -10, -6, -3, 4, 10 }
• After merging them in a third array : arr3[] = { -5, 3, 6, 12, 15, -12, -10, -6, -3, 4, 10}
• Sort arr3[ ] = { -12, -10, -6, -5, -3, 3, 4, 6, 10, 12, 15 }
• As the length of arr3 is odd, so the median is 3
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach` `#include ` `using` `namespace` `std;` `int` `Solution(``int` `arr[], ``int` `n)` `{` ` ``// If length of array is even` ` ``if` `(n % 2 == 0) {` ` ``int` `z = n / 2;` ` ``int` `e = arr[z];` ` ``int` `q = arr[z - 1];` ` ``int` `ans = (e + q) / 2;` ` ``return` `ans;` ` ``}` ` ``// If length if array is odd` ` ``else` `{` ` ``int` `z = round(n / 2);` ` ``return` `arr[z];` ` ``}` `}` `// Driver Code` `int` `main()` `{` ` ``int` `arr1[] = { -5, 3, 6, 12, 15 };` ` ``int` `arr2[] = { -12, -10, -6, -3, 4, 10 };` ` ``int` `i = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` ` ``int` `j = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` ` ``int` `arr3[i + j];` ` ``int` `l = i + j;` ` ``// Merge two array into one array` ` ``for` `(``int` `k = 0; k < i; k++) {` ` ``arr3[k] = arr1[k];` ` ``}` ` ``int` `a = 0;` ` ``for` `(``int` `k = i; k < l; k++) {` ` ``arr3[k] = arr2[a++];` ` ``}` ` ``// Sort the merged array` ` ``sort(arr3, arr3 + l);` ` ``// calling the method` ` ``cout << ``"Median = "` `<< Solution(arr3, l);` `}` `// This code is contributed by SoumikMondal`
## Java
`// Java program for the above approach` `import` `java.io.*;` `import` `java.util.Arrays;` `public` `class` `GFG {` ` ``public` `static` `int` `Solution(``int``[] arr)` ` ``{` ` ``int` `n = arr.length;` ` ``// If length of array is even` ` ``if` `(n % ``2` `== ``0``) {` ` ``int` `z = n / ``2``;` ` ``int` `e = arr[z];` ` ``int` `q = arr[z - ``1``];` ` ``int` `ans = (e + q) / ``2``;` ` ``return` `ans;` ` ``}` ` ``// If length if array is odd` ` ``else` `{` ` ``int` `z = Math.round(n / ``2``);` ` ``return` `arr[z];` ` ``}` ` ``}` ` ``// Driver Code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int``[] arr1 = { -``5``, ``3``, ``6``, ``12``, ``15` `};` ` ``int``[] arr2 = { -``12``, -``10``, -``6``, -``3``, ``4``, ``10` `};` ` ``int` `i = arr1.length;` ` ``int` `j = arr2.length;` ` ``int``[] arr3 = ``new` `int``[i + j];` ` ``// Merge two array into one array` ` ``System.arraycopy(arr1, ``0``, arr3, ``0``, i);` ` ``System.arraycopy(arr2, ``0``, arr3, i, j);` ` ``// Sort the merged array` ` ``Arrays.sort(arr3);` ` ``// calling the method` ` ``System.out.print(``"Median = "` `+ Solution(arr3));` ` ``}` `}` `// This code is contributed by Manas Tole`
## Python3
`# Python3 program for the above approach` `def` `Solution(arr):` ` ``n ``=` `len``(arr)` ` ``# If length of array is even` ` ``if` `n ``%` `2` `=``=` `0``:` ` ``z ``=` `n ``/``/` `2` ` ``e ``=` `arr[z]` ` ``q ``=` `arr[z ``-` `1``]` ` ``ans ``=` `(e ``+` `q) ``/` `2` ` ``return` `ans` ` ``# If length of array is odd` ` ``else``:` ` ``z ``=` `n ``/``/` `2` ` ``ans ``=` `arr[z]` ` ``return` `ans` `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` ` ``arr1 ``=` `[``-``5``, ``3``, ``6``, ``12``, ``15``]` ` ``arr2 ``=` `[``-``12``, ``-``10``, ``-``6``, ``-``3``, ``4``, ``10``]` ` ``# Concatenating the two arrays` ` ``arr3 ``=` `arr1 ``+` `arr2` ` ``# Sorting the resultant array` ` ``arr3.sort()` ` ``print``(``"Median = "``, Solution(arr3))` `# This code is contributed by kush11`
## C#
`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `public` `class` `GFG {` ` ``public` `static` `int` `Solution(``int``[] arr)` ` ``{` ` ``int` `n = arr.Length;` ` ``// If length of array is even` ` ``if` `(n % 2 == 0) {` ` ``int` `z = n / 2;` ` ``int` `e = arr[z];` ` ``int` `q = arr[z - 1];` ` ``int` `ans = (e + q) / 2;` ` ``return` `ans;` ` ``}` ` ``// If length if array is odd` ` ``else` `{` ` ``int` `z = n / 2;` ` ``return` `arr[z];` ` ``}` ` ``}` ` ``// Driver Code` ` ``static` `public` `void` `Main()` ` ``{` ` ``// TODO Auto-generated method stub` ` ``int``[] arr1 = { -5, 3, 6, 12, 15 };` ` ``int``[] arr2 = { -12, -10, -6, -3, 4, 10 };` ` ``// Merge two array into one array` ` ``var` `myList = ``new` `List<``int``>();` ` ``myList.AddRange(arr1);` ` ``myList.AddRange(arr2);` ` ``int``[] arr3 = myList.ToArray();` ` ``// Sort the merged array` ` ``Array.Sort(arr3);` ` ``// calling the method` ` ``Console.Write(``"Median = "` `+ Solution(arr3));` ` ``}` `}` `// This code is contributed by Shubhamsingh10`
## Javascript
`// Javascript program for the above approach` `function` `Solution(arr, n)` `{` ` ` ` ``// If length of array is even` ` ``if` `(n % 2 == 0) ` ` ``{` ` ``var` `z = n / 2;` ` ``var` `e = arr[z];` ` ``var` `q = arr[z - 1];` ` ``var` `ans = (e + q) / 2;` ` ``return` `ans;` ` ``}` ` ` ` ``// If length if array is odd` ` ``else` ` ``{` ` ``var` `z = Math.floor(n / 2);` ` ``return` `arr[z];` ` ``}` `}` `// Driver Code ` `// TODO Auto-generated method stub` `var` `arr1 = [ -5, 3, 6, 12, 15 ];` `var` `arr2 = [ -12, -10, -6, -3, 4, 10 ];` `var` `i = arr1.length;` `var` `j = arr2.length;` `var` `l = i+j;` `// Merge two array into one array` `const arr3 = arr1.concat(arr2);` `// Sort the merged array` `arr3.sort(``function``(a, b) {` ` ``return` `a - b;` `});` `// calling the method` `console.log(``"Median = "` `+ Solution(arr3, l));` `// This code is contributed by Shubham Singh`
Output
```Median = 3
```
Time Complexity: O((N + M) * Log (N + M)), Time required to sort the array of size N + M
Auxiliary Space: O(N + M), Creating a new array of size N+M.
## Median of two sorted arrays of different sizes by Merging Arrays efficiently:
The given arrays are sorted, so merge the sorted arrays in an efficient way and keep the count of elements inserted in the output array or printed form. So when the elements in the output array are half the original size of the given array print the element as a median element. There are two cases:
• Case 1: M+N is odd, the median is at (M+N)/2th index in the array obtained after merging both the arrays.
• Case 2: M+N is even, the median will be the average of elements at index ((M+N)/2 – 1) and (M+N)/2 in the array obtained after merging both the arrays
Illustration:
Given two array ar1[ ]= { 900 } and ar2[ ] = { 5, 8, 10, 20 } , n => Size of ar1 = 1 and m => Size of ar2 = 4
• Loop will run from 0 till 2.
• First iteration : { 900 } { 5, 8, 10, 20 } , m1 = 5
• Second iteration : { 900 } { 5, 8, 10, 20 }, m1 = 8
• Third iteration : { 900 } { 5, 8, 10, 20 }, m1 = 10
• As size of ar1 + ar2 = odd , hence we return m1 = 10 as the median
Below is the implementation of the above approach:
## C++
`// A Simple Merge based O(n) solution to find` `// median of two sorted arrays` `#include ` `using` `namespace` `std;` `/* This function returns median of ar1[] and ar2[].` `Assumption in this function:` `Both ar1[] and ar2[] are sorted arrays */` `int` `getMedian(``int` `ar1[], ``int` `ar2[], ``int` `n, ``int` `m)` `{` ` ``int` `i = 0; ``/* Current index of input array ar1[] */` ` ``int` `j = 0; ``/* Current index of input array ar2[] */` ` ``int` `count;` ` ``int` `m1 = -1, m2 = -1;` ` ``/*loop till (m+n)/2*/` ` ``for` `(count = 0; count <= (m + n) / 2; count++) {` ` ``// store (n+m)/2-1 in m2` ` ``m2 = m1;` ` ``if` `(i != n && j != m) {` ` ``m1 = (ar1[i] > ar2[j]) ? ar2[j++] : ar1[i++];` ` ``}` ` ``else` `if` `(i < n) {` ` ``m1 = ar1[i++];` ` ``}` ` ``// for case when j
## Java
`// A Simple Merge based O(n) solution` `// to find median of two sorted arrays` `import` `java.io.*;` `class` `GFG {` ` ``// Function to calculate median` ` ``static` `int` `getMedian(``int` `ar1[], ``int` `ar2[], ``int` `n, ``int` `m)` ` ``{` ` ``// Current index of input array ar1[]` ` ``int` `i = ``0``;` ` ``// Current index of input array ar2[]` ` ``int` `j = ``0``;` ` ``int` `count;` ` ``int` `m1 = -``1``, m2 = -``1``;` ` ``// Since there are (n+m) elements,` ` ``// There are following two cases` ` ``// if n+m is odd then the middle` ` ``// index is median i.e. (m+n)/2` ` ``if` `((m + n) % ``2` `== ``1``) {` ` ``for` `(count = ``0``; count <= (n + m) / ``2``; count++) {` ` ``if` `(i != n && j != m) {` ` ``m1 = (ar1[i] > ar2[j]) ? ar2[j++]` ` ``: ar1[i++];` ` ``}` ` ``else` `if` `(i < n) {` ` ``m1 = ar1[i++];` ` ``}` ` ``// for case when j ar2[j]) ? ar2[j++]` ` ``: ar1[i++];` ` ``}` ` ``else` `if` `(i < n) {` ` ``m1 = ar1[i++];` ` ``}` ` ``// for case when j
## Python3
`# A Simple Merge based O(n) solution to find` `# median of two sorted arrays` `""" This function returns median of ar1[] and ar2[]. ` `Assumption in this function: ` `Both ar1[] and ar2[] are sorted arrays """` `def` `getMedian(ar1, ar2, n, m):` ` ``i ``=` `0` `# Current index of input array ar1[]` ` ``j ``=` `0` `# Current index of input array ar2[]` ` ``m1, m2 ``=` `-``1``, ``-``1` ` ``for` `count ``in` `range``(((n ``+` `m) ``/``/` `2``) ``+` `1``):` ` ``if``(i !``=` `n ``and` `j !``=` `m):` ` ``if` `ar1[i] > ar2[j]:` ` ``m1 ``=` `ar2[j]` ` ``j ``+``=` `1` ` ``else``:` ` ``m1 ``=` `ar1[i]` ` ``i ``+``=` `1` ` ``elif``(i < n):` ` ``m1 ``=` `ar1[i]` ` ``i ``+``=` `1` ` ``# for case when j
## C#
`// A Simple Merge based O(n) solution` `// to find median of two sorted arrays` `using` `System;` `class` `GFG {` ` ``// Function to calculate median` ` ``static` `int` `getMedian(``int``[] ar1, ``int``[] ar2, ``int` `n, ``int` `m)` ` ``{` ` ``// Current index of input array ar1[]` ` ``int` `i = 0;` ` ``// Current index of input array ar2[]` ` ``int` `j = 0;` ` ``int` `count;` ` ``int` `m1 = -1, m2 = -1;` ` ``// Since there are (n+m) elements,` ` ``// There are following two cases` ` ``// if n+m is odd then the middle` ` ``// index is median i.e. (m+n)/2` ` ``if` `((m + n) % 2 == 1) {` ` ``for` `(count = 0; count <= (n + m) / 2; count++) {` ` ``if` `(i != n && j != m) {` ` ``m1 = (ar1[i] > ar2[j]) ? ar2[j++]` ` ``: ar1[i++];` ` ``}` ` ``else` `if` `(i < n) {` ` ``m1 = ar1[i++];` ` ``}` ` ``// for case when j ar2[j]) ? ar2[j++]` ` ``: ar1[i++];` ` ``}` ` ``else` `if` `(i < n) {` ` ``m1 = ar1[i++];` ` ``}` ` ``// for case when j
## Javascript
`// A Simple Merge based O(n) solution to find` `// median of two sorted arrays` `// This function returns median of ar1[] and ar2[].` `// Assumption in this function:` `// Both ar1[] and ar2[] are sorted arrays ` `function` `getMedian(ar1, ar2, n, m)` `{` ` ` ` ``// Current index of input array ar1[] ` ` ``let i = 0; ` ` ` ` ``// Current index of input array ar2[] ` ` ``let j = 0; ` ` ``let count;` ` ``let m1 = -1, m2 = -1;` ` ``// Since there are (n+m) elements,` ` ``// There are following two cases` ` ``// if n+m is odd then the middle` ` ``// index is median i.e. (m+n)/2` ` ``if` `((m + n) % 2 == 1)` ` ``{` ` ``for``(count = 0; ` ` ``count <= (n + m) / 2; ` ` ``count++)` ` ``{` ` ``if` `(i != n && j != m)` ` ``{` ` ``m1 = (ar1[i] > ar2[j]) ? ` ` ``ar2[j++] : ar1[i++];` ` ``}` ` ``else` `if``(i < n)` ` ``{` ` ``m1 = ar1[i++];` ` ``}` ` ` ` ``// For case when j ar2[j]) ? ` ` ``ar2[j++] : ar1[i++];` ` ``}` ` ``else` `if``(i < n)` ` ``{` ` ``m1 = ar1[i++];` ` ``}` ` ` ` ``// For case when j
Output
```10
```
Time Complexity: O(M + N). To merge both arrays O(M+N) time is needed.
Auxiliary Space: O(1). No extra space is required.
## Median of two sorted arrays of different sizes using Binary Search:
The given two arrays are sorted, so we can utilize the ability of Binary Search to divide the array and find the median.
Median means the point at which the whole array is divided into two parts. Hence since the two arrays are not merged so to get the median we require merging which is costly.
Hence instead of merging, we will use a modified binary search algorithm to efficiently find the median.
Below is the implementation:
## C++
`#include ` `using` `namespace` `std;` `// Method to find median` `double` `Median(vector<``int``>& A, vector<``int``>& B)` `{` ` ``int` `n = A.size();` ` ``int` `m = B.size();` ` ``if` `(n > m)` ` ``return` `Median(B, A); ``// Swapping to make A smaller` ` ``int` `start = 0;` ` ``int` `end = n;` ` ``int` `realmidinmergedarray = (n + m + 1) / 2;` ` ``while` `(start <= end) {` ` ``int` `mid = (start + end) / 2;` ` ``int` `leftAsize = mid;` ` ``int` `leftBsize = realmidinmergedarray - mid;` ` ``int` `leftA` ` ``= (leftAsize > 0)` ` ``? A[leftAsize - 1]` ` ``: INT_MIN; ``// checking overflow of indices` ` ``int` `leftB` ` ``= (leftBsize > 0) ? B[leftBsize - 1] : INT_MIN;` ` ``int` `rightA` ` ``= (leftAsize < n) ? A[leftAsize] : INT_MAX;` ` ``int` `rightB` ` ``= (leftBsize < m) ? B[leftBsize] : INT_MAX;` ` ``// if correct partition is done` ` ``if` `(leftA <= rightB and leftB <= rightA) {` ` ``if` `((m + n) % 2 == 0)` ` ``return` `(max(leftA, leftB)` ` ``+ min(rightA, rightB))` ` ``/ 2.0;` ` ``return` `max(leftA, leftB);` ` ``}` ` ``else` `if` `(leftA > rightB) {` ` ``end = mid - 1;` ` ``}` ` ``else` ` ``start = mid + 1;` ` ``}` ` ``return` `0.0;` `}` `// Driver code` `int` `main()` `{` ` ``vector<``int``> arr1 = { -5, 3, 6, 12, 15 };` ` ``vector<``int``> arr2 = { -12, -10, -6, -3, 4, 10 };` ` ``cout << ``"Median of the two arrays are"` `<< endl;` ` ``cout << Median(arr1, arr2);` ` ``return` `0;` `}`
## Java
`public` `class` `GFG {` ` ``// Method to find median` ` ``static` `double` `Median(``int``[] A, ``int``[] B)` ` ``{` ` ``int` `n = A.length;` ` ``int` `m = B.length;` ` ``if` `(n > m)` ` ``return` `Median(B,` ` ``A); ``// Swapping to make A smaller` ` ``int` `start = ``0``;` ` ``int` `end = n;` ` ``int` `realmidinmergedarray = (n + m + ``1``) / ``2``;` ` ``while` `(start <= end) {` ` ``int` `mid = (start + end) / ``2``;` ` ``int` `leftAsize = mid;` ` ``int` `leftBsize = realmidinmergedarray - mid;` ` ``int` `leftA` ` ``= (leftAsize > ``0``)` ` ``? A[leftAsize - ``1``]` ` ``: Integer` ` ``.MIN_VALUE; ``// checking overflow` ` ``// of indices` ` ``int` `leftB = (leftBsize > ``0``) ? B[leftBsize - ``1``]` ` ``: Integer.MIN_VALUE;` ` ``int` `rightA = (leftAsize < n)` ` ``? A[leftAsize]` ` ``: Integer.MAX_VALUE;` ` ``int` `rightB = (leftBsize < m)` ` ``? B[leftBsize]` ` ``: Integer.MAX_VALUE;` ` ``// if correct partition is done` ` ``if` `(leftA <= rightB && leftB <= rightA) {` ` ``if` `((m + n) % ``2` `== ``0``)` ` ``return` `(Math.max(leftA, leftB)` ` ``+ Math.min(rightA, rightB))` ` ``/ ``2.0``;` ` ``return` `Math.max(leftA, leftB);` ` ``}` ` ``else` `if` `(leftA > rightB) {` ` ``end = mid - ``1``;` ` ``}` ` ``else` ` ``start = mid + ``1``;` ` ``}` ` ``return` `0.0``;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int``[] arr1 = { -``5``, ``3``, ``6``, ``12``, ``15` `};` ` ``int``[] arr2 = { -``12``, -``10``, -``6``, -``3``, ``4``, ``10` `};` ` ``System.out.println(``"Median of the two arrays are"``);` ` ``System.out.println(Median(arr1, arr2));` ` ``}` `}` `// This code is contributed by Hritik`
## Python3
`class` `Solution:` ` ``# Method to find median` ` ``def` `Median(``self``, A, B):` ` ``# Assumption both A and B cannot be empty` ` ``n ``=` `len``(A)` ` ``m ``=` `len``(B)` ` ``if` `(n > m):` ` ``return` `self``.Median(B, A) ``# Swapping to make A smaller` ` ``start ``=` `0` ` ``end ``=` `n` ` ``realmidinmergedarray ``=` `(n ``+` `m ``+` `1``) ``/``/` `2` ` ``while` `(start <``=` `end):` ` ``mid ``=` `(start ``+` `end) ``/``/` `2` ` ``leftAsize ``=` `mid` ` ``leftBsize ``=` `realmidinmergedarray ``-` `mid` ` ``# checking overflow of indices` ` ``leftA ``=` `A[leftAsize ``-` `1``] ``if` `(leftAsize > ``0``) ``else` `float``(``'-inf'``)` ` ``leftB ``=` `B[leftBsize ``-` `1``] ``if` `(leftBsize > ``0``) ``else` `float``(``'-inf'``)` ` ``rightA ``=` `A[leftAsize] ``if` `(leftAsize < n) ``else` `float``(``'inf'``)` ` ``rightB ``=` `B[leftBsize] ``if` `(leftBsize < m) ``else` `float``(``'inf'``)` ` ``# if correct partition is done` ` ``if` `leftA <``=` `rightB ``and` `leftB <``=` `rightA:` ` ``if` `((m ``+` `n) ``%` `2` `=``=` `0``):` ` ``return` `(``max``(leftA, leftB) ``+` `min``(rightA, rightB)) ``/` `2.0` ` ``return` `max``(leftA, leftB)` ` ``elif` `(leftA > rightB):` ` ``end ``=` `mid ``-` `1` ` ``else``:` ` ``start ``=` `mid ``+` `1` `# Driver code` `ans ``=` `Solution()` `arr1 ``=` `[``-``5``, ``3``, ``6``, ``12``, ``15``]` `arr2 ``=` `[``-``12``, ``-``10``, ``-``6``, ``-``3``, ``4``, ``10``]` `print``(``"Median of the two arrays is"``)` `print``(ans.Median(arr1, arr2))` `# This code is contributed by Arpan`
## C#
`using` `System;` `public` `class` `GFG {` ` ``// Method to find median` ` ``static` `double` `Median(``int``[] A, ``int``[] B)` ` ``{` ` ``int` `n = A.Length;` ` ``int` `m = B.Length;` ` ``if` `(n > m)` ` ``return` `Median(B,` ` ``A); ``// Swapping to make A smaller` ` ``int` `start = 0;` ` ``int` `end = n;` ` ``int` `realmidinmergedarray = (n + m + 1) / 2;` ` ``while` `(start <= end) {` ` ``int` `mid = (start + end) / 2;` ` ``int` `leftAsize = mid;` ` ``int` `leftBsize = realmidinmergedarray - mid;` ` ``int` `leftA` ` ``= (leftAsize > 0)` ` ``? A[leftAsize - 1]` ` ``: Int32.MinValue; ``// checking overflow` ` ``// of indices` ` ``int` `leftB = (leftBsize > 0) ? B[leftBsize - 1]` ` ``: Int32.MinValue;` ` ``int` `rightA = (leftAsize < n) ? A[leftAsize]` ` ``: Int32.MaxValue;` ` ``int` `rightB = (leftBsize < m) ? B[leftBsize]` ` ``: Int32.MaxValue;` ` ``// if correct partition is done` ` ``if` `(leftA <= rightB && leftB <= rightA) {` ` ``if` `((m + n) % 2 == 0)` ` ``return` `(Math.Max(leftA, leftB)` ` ``+ Math.Min(rightA, rightB))` ` ``/ 2.0;` ` ``return` `Math.Max(leftA, leftB);` ` ``}` ` ``else` `if` `(leftA > rightB) {` ` ``end = mid - 1;` ` ``}` ` ``else` ` ``start = mid + 1;` ` ``}` ` ``return` `0.0;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `Main()` ` ``{` ` ``int``[] arr1 = { -5, 3, 6, 12, 15 };` ` ``int``[] arr2 = { -12, -10, -6, -3, 4, 10 };` ` ``Console.WriteLine(``"Median of the two arrays are"``);` ` ``Console.WriteLine(Median(arr1, arr2));` ` ``}` `}` `// This code is contributed by Shubham Singh`
## Javascript
``
Output
```Median of the two arrays are
3
```
Time Complexity: O(min(log M, log N)): Since binary search is being applied on the smaller of the 2 arrays
Auxiliary Space: O(1)
## Median of two sorted arrays of different sizes using Priority Queue:
In this Approach we have used Priority Queue (min Heap) to find out the median.
The Idea is simple, just push the elements into a single Priority Queue from both arrays. Now we have to find median from priority queue by performing a simple traversal through it upto median.
Illustration:
A[ ] = {-2, 3, 4, 5}, N = 4 & B[ ] = {-4, -1, 7, 8, 9}, M = 5
Step 1: Adding elements to priority queue(pq) from array
• pq.push(-2)
• pq.push(3)
• pq.push(4)
• pq.push(5)
After adding array A elements to priority queue it will look as pq = {-2, 3, 4, 5}
Step 2: Adding elements to priority queue(pq) from array B
• pq.push(-4)
• pq.push(-1)
• pq.push(7)
• pq.push(8)
• pq.push(9)
After adding array B elements to priority queue it will look as pq = {-4, -2, -1, 3, 4, 5, 7, 8, 9}
Step 3: Now we have to find median from Priority Queue based on following conditions:
• if N+M is odd
• then traverse priority queue upto (n+m)/2 by popping element by element
• Then display median as pq.top()
• if N+M is even
• then traverse priority queue upto (n+m)/2 && ((n+m)/2)-1
• Then median = average of both top values of priority queue
In this case the median is 4
Below is the implementation of the above problem:
## C++
`#include ` `using` `namespace` `std;` `// Method to find median` `double` `Median(vector<``int``>& A, vector<``int``>& B)` `{` ` ``int` `i;` ` ``int` `n = A.size();` ` ``int` `m = B.size();` ` ``// initializing Priority Queue (Min Heap)` ` ``priority_queue<``int``, vector<``int``>, greater<``int``> > pq;` ` ``// pushing array A values to priority Queue` ` ``for` `(i = 0; i < n; i++)` ` ``pq.push(A[i]);` ` ``// pushing array B values to priority Queue` ` ``for` `(i = 0; i < m; i++)` ` ``pq.push(B[i]);` ` ``int` `check = n + m;` ` ``double` `count = -1;` ` ``double` `mid1, mid2;` ` ``while` `(!pq.empty()) {` ` ``count++;` ` ``// returning mid value if combined length(n+m) is` ` ``// odd` ` ``if` `(check % 2 != 0 && count == check / 2) {` ` ``double` `ans = pq.top();` ` ``return` `ans;` ` ``}` ` ``// maintaining mid1 value if combined length(n+m) is` ` ``// even where we need to maintain both mid values in` ` ``// case of even combined length` ` ``if` `(check % 2 == 0 && count == (check / 2) - 1)` ` ``mid1 = pq.top();` ` ``// now returning the mid2 value with previous` ` ``// maintained mid1 value by 2` ` ``if` `(check % 2 == 0 && count == check / 2) {` ` ``mid2 = pq.top();` ` ``double` `ans = (mid1 + mid2) / 2;` ` ``return` `ans;` ` ``}` ` ``pq.pop();` ` ``}` ` ``return` `0.00000;` `}` `// Driver code` `int` `main()` `{` ` ``vector<``int``> arr1 = { -2, 3, 4, 5 };` ` ``vector<``int``> arr2 = { -4, -1, 7, 8, 9 };` ` ``cout << ``"Median of the two arrays are"` `<< endl;` ` ``cout << Median(arr1, arr2);` ` ``return` `0;` `}`
## Java
`import` `java.util.PriorityQueue;` `import` `java.util.Scanner;` `import` `java.util.Vector;` `public` `class` `Main {` ` ``// Method to find median` ` ``public` `static` `double` `Median(Vector A,` ` ``Vector B)` ` ``{` ` ``int` `i;` ` ``int` `n = A.size();` ` ``int` `m = B.size();` ` ``// initializing Priority Queue (Min Heap)` ` ``PriorityQueue pq` ` ``= ``new` `PriorityQueue();` ` ``// pushing array A values to priority Queue` ` ``for` `(i = ``0``; i < n; i++) {` ` ``pq.add(A.get(i));` ` ``}` ` ``// pushing array B values to priority Queue` ` ``for` `(i = ``0``; i < m; i++) {` ` ``pq.add(B.get(i));` ` ``}` ` ``int` `check = n + m;` ` ``double` `count = -``1``;` ` ``double` `mid1 = -``1``, mid2 = -``1``;` ` ``while` `(!pq.isEmpty()) {` ` ``count++;` ` ``// returning mid value if combined length(n+m)` ` ``// is odd` ` ``if` `(check % ``2` `!= ``0` `&& count == check / ``2``) {` ` ``double` `ans = pq.peek();` ` ``return` `ans;` ` ``}` ` ``// maintaining mid1 value if combined` ` ``// length(n+m) is even where we need to maintain` ` ``// both mid values in case of even combined` ` ``// length` ` ``if` `(check % ``2` `== ``0` ` ``&& count == (check / ``2``) - ``1``) {` ` ``mid1 = pq.peek();` ` ``}` ` ``// now returning the mid2 value with previous` ` ``// maintained mid1 value by 2` ` ``if` `(check % ``2` `== ``0` `&& count == check / ``2``) {` ` ``mid2 = pq.peek();` ` ``double` `ans = (mid1 + mid2) / ``2``;` ` ``return` `ans;` ` ``}` ` ``pq.poll();` ` ``}` ` ``return` `0.00000``;` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``Vector arr1 = ``new` `Vector();` ` ``arr1.add(-``2``);` ` ``arr1.add(``3``);` ` ``arr1.add(``4``);` ` ``arr1.add(``5``);` ` ``Vector arr2 = ``new` `Vector();` ` ``arr2.add(-``4``);` ` ``arr2.add(-``1``);` ` ``arr2.add(``7``);` ` ``arr2.add(``8``);` ` ``arr2.add(``9``);` ` ``System.out.println(``"Median of the two arrays are"``);` ` ``System.out.println(Median(arr1, arr2));` ` ``}` `}`
## Python3
`# Python code for the above approach` `import` `heapq` `def` `Median(A, B):` ` ``n, m ``=` `len``(A), ``len``(B)` ` ``# initializing Priority Queue (Min Heap)` ` ``pq ``=` `[]` ` ``# pushing array A values to priority Queue` ` ``for` `i ``in` `range``(n):` ` ``heapq.heappush(pq, A[i])` ` ``# pushing array B values to priority Queue` ` ``for` `i ``in` `range``(m):` ` ``heapq.heappush(pq, B[i])` ` ``check ``=` `n ``+` `m` ` ``count ``=` `-``1` ` ``mid1, mid2 ``=` `-``1``, ``-``1` ` ``while` `pq:` ` ``count ``+``=` `1` ` ``# returning mid value if combined length(n+m) is odd` ` ``if` `check ``%` `2` `!``=` `0` `and` `count ``=``=` `check ``/``/` `2``:` ` ``return` `pq[``0``]` ` ``# maintaining mid1 value if combined length(n+m) is even` ` ``# where we need to maintain both mid values in case of` ` ``# even combined length` ` ``if` `check ``%` `2` `=``=` `0` `and` `count ``=``=` `(check ``/``/` `2``) ``-` `1``:` ` ``mid1 ``=` `pq[``0``]` ` ``# now returning the mid2 value with previous maintained` ` ``# mid1 value by 2` ` ``if` `check ``%` `2` `=``=` `0` `and` `count ``=``=` `check ``/``/` `2``:` ` ``mid2 ``=` `heapq.heappop(pq)` ` ``return` `(mid1 ``+` `mid2) ``/` `2` ` ``heapq.heappop(pq)` ` ``return` `0.00000` `# Driver code` `arr1 ``=` `[``-``2``, ``3``, ``4``, ``5``]` `arr2 ``=` `[``-``4``, ``-``1``, ``7``, ``8``, ``9``]` `print``(``"Median of the two arrays are"``)` `print``(Median(arr1, arr2))` `# This code is contributed by karthik`
## C#
`using` `System;` `using` `System.Collections.Generic;` `namespace` `MedianOfTwoSortedArrays {` `class` `Program {` ` ``static` `double` `Median(List<``int``> A, List<``int``> B)` ` ``{` ` ``int` `n = A.Count;` ` ``int` `m = B.Count;` ` ``int` `i;` ` ``// initializing Priority Queue (Min Heap)` ` ``var` `pq = ``new` `SortedSet<``int``>();` ` ``// pushing array A values to priority Queue` ` ``for` `(i = 0; i < n; i++)` ` ``pq.Add(A[i]);` ` ``// pushing array B values to priority Queue` ` ``for` `(i = 0; i < m; i++)` ` ``pq.Add(B[i]);` ` ``int` `check = n + m;` ` ``double` `count = -1;` ` ``double` `mid1, mid2;` ` ``while` `(pq.Count != 0) {` ` ``count++;` ` ``// returning mid value if combined length(n+m)` ` ``// is odd` ` ``if` `(check % 2 != 0 && count == check / 2) {` ` ``double` `ans = pq.Min;` ` ``return` `ans;` ` ``}` ` ``// maintaining mid1 value if combined` ` ``// length(n+m) is even where we need to maintain` ` ``// both mid values in case of even combined` ` ``// length` ` ``if` `(check % 2 == 0` ` ``&& count == (check / 2) - 1) {` ` ``mid1 = pq.Min;` ` ``// now returning the mid2 value with` ` ``// previous maintained mid1 value by 2` ` ``if` `(check % 2 == 0 && count == check / 2) {` ` ``mid2 = pq.Min;` ` ``double` `ans = (mid1 + mid2) / 2;` ` ``return` `ans;` ` ``}` ` ``}` ` ``pq.Remove(pq.Min);` ` ``}` ` ``return` `0.00000;` ` ``}` ` ``// Driver Code` ` ``static` `void` `Main(``string``[] args)` ` ``{` ` ``List<``int``> arr1 = ``new` `List<``int``>{ -2, 3, 4, 5 };` ` ``List<``int``> arr2 = ``new` `List<``int``>{ -4, -1, 7, 8, 9 };` ` ``Console.WriteLine(``"Median of the two arrays are"``);` ` ``Console.WriteLine(Median(arr1, arr2));` ` ``Console.ReadLine();` ` ``}` `}` `}`
## Javascript
`// Method to find median` `function` `Median(A, B) {` ` ``let n = A.length;` ` ``let m = B.length;` ` ``//initializing Priority Queue (Min Heap)` ` ``let pq = ``new` `PriorityQueue();` ` ``//pushing array A values to priority Queue` ` ``for` `(let i = 0; i < n; i++) {` ` ``pq.push(A[i]);` ` ``}` ` ``//pushing array B values to priority Queue` ` ``for` `(let i = 0; i < m; i++) {` ` ``pq.push(B[i]);` ` ``}` ` ``let check = n + m;` ` ``let count = -1;` ` ``let mid1, mid2;` ` ``while` `(!pq.isEmpty()) {` ` ``count++;` ` ``//returning mid value if combined length(n+m) is odd` ` ``if` `(check % 2 !== 0 && count === Math.floor(check / 2)) {` ` ``let ans = pq.top();` ` ``return` `ans;` ` ``}` ` ``//maintaining mid1 value if combined length(n+m) is even` ` ``//where we need to maintain both mid values in case of even combined length` ` ``if` `(check % 2 === 0 && count === (check / 2) - 1) {` ` ``mid1 = pq.top();` ` ``}` ` ``//now returning the mid2 value with previous maintained mid1 value by 2` ` ``if` `(check % 2 === 0 && count === check / 2) {` ` ``mid2 = pq.top();` ` ``let ans = (mid1 + mid2) / 2;` ` ``return` `ans;` ` ``}` ` ``pq.pop();` ` ``}` ` ``return` `0.00000;` `}` `// Priority Queue (Min Heap) implementation` `class PriorityQueue {` ` ``constructor() {` ` ``this``.data = [];` ` ``}` ` ``push(value) {` ` ``this``.data.push(value);` ` ``this``.bubbleUp(``this``.data.length - 1);` ` ``}` ` ``pop() {` ` ``let result = ``this``.data[0];` ` ``let end = ``this``.data.pop();` ` ``if` `(``this``.data.length > 0) {` ` ``this``.data[0] = end;` ` ``this``.bubbleDown(0);` ` ``}` ` ``return` `result;` ` ``}` ` ``top() {` ` ``return` `this``.data[0];` ` ``}` ` ``isEmpty() {` ` ``return` `this``.data.length === 0;` ` ``}` ` ``bubbleUp(index) {` ` ``let parent = Math.floor((index + 1) / 2) - 1;` ` ``if` `(parent >= 0 && ``this``.data[parent] > ``this``.data[index]) {` ` ``[``this``.data[parent], ``this``.data[index]] = [``this``.data[index], ``this``.data[parent]];` ` ``this``.bubbleUp(parent);` ` ``}` ` ``}` ` ``bubbleDown(index) {` ` ``let child1 = (index + 1) * 2 - 1;` ` ``let child2 = (index + 1) * 2;` ` ``let minIndex = index;` ` ``if` `(child1 < ``this``.data.length && ``this``.data[child1] < ``this``.data[minIndex]) {` ` ``minIndex = child1;` ` ``}` ` ``if` `(child2 < ``this``.data.length && ``this``.data[child2] < ``this``.data[minIndex]) {` ` ``minIndex = child2;` ` ``}` ` ``if` `(minIndex !== index) {` ` ``[``this``.data[index], ``this``.data[minIndex]] = [``this``.data[minIndex], ``this``.data[index]];` ` ``this``.bubbleDown(minIndex);` ` ``}` ` ``}` `}` `// Driver code` `let arr1 = [-2, 3, 4, 5];` `let arr2 = [-4, -1, 7, 8, 9];` `console.log(``"Median of the two arrays are"``);` `console.log(Median(arr1, arr2));`
Output
```Median of the two arrays are
4
```
Time Complexity: O(max(N, M)*log(max(N, M))): Since the priority queue is implemented from two arrays
Auxiliary Space: O(N+M): for storing two array values in the priority queue.
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Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks! | 13,277 | 37,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-50 | latest | en | 0.793216 |
https://nigerianscholars.com/tutorials/optics-and-optical-phenomena/refractive-index/ | 1,660,283,122,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00057.warc.gz | 399,820,446 | 19,938 | # Refractive Index
## Refractive Index
The speed of light and therefore the degree of bending of the light depends on the refractive index of material through which the light passes. The refractive index (symbol $$n$$) is the ratio of the speed of light in a vacuum to its speed in the material.
### Definition: Refractive index
The refractive index (symbol $$n$$) of a material is the ratio of the speed of light in a vacuum to its speed in the material and gives an indication of how difficult it is for light to get through the material.
$n = \frac{c}{v}$ where $\begin{array}{rl} n &= \text{refractive index (no unit)} \\ c &= \text{speed of light in a vacuum } (\text{3.00} \times \text{10}^{\text{8}}\text{ m·s^{-1}}) \\ v &= \text{speed of light in a given medium } (\text{m·s^{-1}}) \\ \end{array}$
Using the definition of refractive index $n = \frac{c}{v}$ we can see how the speed of light changes in different media, because the speed of light in a vacuum, $$c$$, is constant.
If the refractive index, $$n$$, increases, the speed of light in the material, $$v$$, must decrease. Therefore light travels slower through materials of high refractive index, $$n$$.
The table below shows refractive indices for various materials. Light travels slower in any material than it does in a vacuum, so all values for $$n$$ are greater than $$\text{1}$$.
Medium Refractive Index Vacuum $$\text{1}$$ Helium $$\text{1.000036}$$ Air* $$\text{1.0002926}$$ Carbon dioxide $$\text{1.00045}$$ Water: Ice $$\text{1.31}$$ Water: Liquid ($$\text{20}$$$$\text{°C}$$) $$\text{1.333}$$ Acetone $$\text{1.36}$$ Ethyl Alcohol (Ethanol) $$\text{1.36}$$ Sugar solution ($$\text{30}\%$$) $$\text{1.38}$$ Fused quartz $$\text{1.46}$$ Glycerine $$\text{1.4729}$$ Sugar solution ($$\text{80}\%$$) $$\text{1.49}$$ Rock salt $$\text{1.516}$$ Crown Glass $$\text{1.52}$$ Sodium chloride $$\text{1.54}$$ Polystyrene $$\text{1.55}$$ to $$\text{1.59}$$ Bromine $$\text{1.661}$$ Sapphire $$\text{1.77}$$ Glass (typical) $$\text{1.5}$$ to $$\text{1.9}$$ Cubic zirconia $$\text{2.15}$$ to $$\text{2.18}$$ Diamond $$\text{2.419}$$ Silicon $$\text{4.01}$$
Table: Refractive indices of some materials. $$n_{\text{air}}$$ is calculated at standard temperature and pressure (STP).
## Example: Refractive Index
### Question
Calculate the speed of light through glycerine which has a refractive index of $$\text{1.4729}$$.
### Step 1: Determine what is given and what is required
We are given the refractive index, $$n$$ of glycerine and we need to calculate the speed of light in glycerine.
### Step 2: Determine how to approach the problem
We can use the definition of refractive index since the speed of light in vacuum is a constant and we know the value of glycerine’s refractive index.
### Step 3: Do the calculation
$n = \frac{c}{v}$Rearrange the equation to solve for $$v$$ and substitute in the known values: \begin{align*} v &= \frac{c}{n} \\ &= \frac{\text{3} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$}}{\text{1.4729}} \\ &= \text{2.04} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$} \end{align*}
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This is a lesson from the tutorial, Optics and Optical Phenomena and you are encouraged to log in or register, so that you can track your progress. | 1,013 | 3,267 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-33 | latest | en | 0.699308 |
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# PRMO 2016 Problem 2 | Number Theory
Try this beautiful interesting problem based on Number Theory from PRMO 2016 Problem 2.
## Number Theory Problem: PRMO 2016 Problem 2
The five digit number $2 a 9 b 1$ is a perfect square. Find the value of $a^{b-1}+b^{a-1}$.
### Key Concepts
Properties of Perfect Squares
Divisibility Rules of different numbers
Finding the square root of a number
## Suggested Book | Source | Answer
Challenge and Thrill of Pre College Mathematics
PRMO 2016
50
## Try with Hints
An odd perfect square is of the form $8k+1$
Hence $8| 2a9b0$
Hence $8 | 2a000 + 9b0$
$8 | 900 +b0$
$8 | b4$
Therefore the possible values of b are $6,2$
So possible numbers are $2a921,2a961$
Now check for the possible values of $a$ and calculate the square root to check if it is a perfect square
The only valid solution is $a=5$
Hence calculate the required expression.
## Subscribe to Cheenta at Youtube
Try this beautiful interesting problem based on Number Theory from PRMO 2016 Problem 2.
## Number Theory Problem: PRMO 2016 Problem 2
The five digit number $2 a 9 b 1$ is a perfect square. Find the value of $a^{b-1}+b^{a-1}$.
### Key Concepts
Properties of Perfect Squares
Divisibility Rules of different numbers
Finding the square root of a number
## Suggested Book | Source | Answer
Challenge and Thrill of Pre College Mathematics
PRMO 2016
50
## Try with Hints
An odd perfect square is of the form $8k+1$
Hence $8| 2a9b0$
Hence $8 | 2a000 + 9b0$
$8 | 900 +b0$
$8 | b4$
Therefore the possible values of b are $6,2$
So possible numbers are $2a921,2a961$
Now check for the possible values of $a$ and calculate the square root to check if it is a perfect square
The only valid solution is $a=5$
Hence calculate the required expression.
## Subscribe to Cheenta at Youtube
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 562 | 1,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-06 | longest | en | 0.701586 |
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Assignment Help Finance Basics
##### Reference no: EM131128822
Use Figure 20.1, which lists prices of various IBM options. Use the data in the figure to calculate the payoff and the profits for investments in each of the following Jul expiration options, assuming that the stock price on the expiration date is \$195
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I got 95% on my college Algebra midterm which boosted my grade back up to an A. I was down to a C and worried when I found your software. I credit your program for most of what I learned. Thanks for the quick reply.
Susan Freeman, OH | 457 | 1,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.96125 |
https://math.answers.com/questions/Is_10_over_50_equal_to_20_over_100 | 1,701,651,266,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00109.warc.gz | 437,794,034 | 45,195 | 0
# Is 10 over 50 equal to 20 over 100?
Updated: 9/23/2023
Wiki User
10y ago
Yes, 10/50 is equivalent to 20/100, because 10/50 * 2/2 = 20/100
Wiki User
7y ago
Wiki User
7y ago
Yes, it is. | 84 | 197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-50 | latest | en | 0.787478 |
https://www.solutionspile.com/ExpertAnswers/below-demonstrates-partial-information-about-these-calculations-threat-category-sle-rate-of-freque-pa832 | 1,713,368,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817158.8/warc/CC-MAIN-20240417142102-20240417172102-00436.warc.gz | 900,964,174 | 6,263 | # (Solved): below demonstrates partial information about these calculations. Threat category SLE Rate of freque ...
below demonstrates partial information about these calculations. Threat category SLE Rate of frequency ARO ALE 1. Internal hardware failure \$5,000 1 per week 52 \$260,000 2. DDoS attack \$75,000 1 per year 1 \$75,000 3. Phishing attack \$500 1 per week 52 \$26,000 4. City-wide power outage \$2,500 1 per quarter 4 \$10,000 5. Employee vandalism \$5,000 1 per 6 months 2 \$10,000 6. Brute-force attack \$500 1 per month 12 \$6000 7. Data manipulation \$5,000 1 per year 1 \$5000 8. Ransomware \$1,500 1 per week 52 \$78,000 9. Eavesdropping \$2,500 1 per quarter 4 \$10,000 10. Tornado \$250,000 1 per 20 years 0.05 \$12,500Using the following formula to perform a cost - benefit analysis ( CBA ) , ?the company is calculating whether investing in this risk control technology ( NGFW ) , ?which costs \$ 6 , 000 ?annually, is cost - effective to mitigate the attack. A positive CBA number indicates a cost - effective investment, and a negative number indicates a poor investment. CBA = ?ALE ( pre - control ) – ?ALE ( post - control ) – ?ACS Where, • ?ALE ( pre - control ) = ?the annualized loss expectancy of the risk before the implementation of the risk control • ?ALE ( post - control ) = ?the ALE examined after the risk control has been in place for a period of time • ?Annual Cost ( ACS ) = ?the annual cost of the risk control Based on the formula, what is the CBA in this scenario? Is it cost - effective for the company to invest in this security technology? Explain your reasoning.
We have an Answer from Expert | 449 | 1,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-18 | latest | en | 0.843434 |
http://personal.cityu.edu.hk/~bsapplec/psychrom.htm | 1,513,004,327,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513611.22/warc/CC-MAIN-20171211144705-20171211164705-00402.warc.gz | 212,205,408 | 3,269 | 1 Psychrometry Psychrometry is the science of studying the thermodynamic properties of moist air and the use of these properties to analyze conditions and processes involving moist air. Moist air is a mixture of dry air and water vapour. In atmospheric air, water vapour content varies from 0 to 3% by mass. 1.1 Thermodynamic Properties of Air 1.1.1 Dry Bulb Temperature 1.1.2 Wet Bulb Temperature 1.1.3 Humidity Ratio/Moisture Content Humidity ratio w (kg/kg) of a given moist air sample is defined as the ratio of the mass of water vapour (mw) to the mass of dry air (ma) contained in the sample. (1) When the dry air and water vapour occupy the same volume and temperature, by applying the characteristic equation of state for perfect gas, Eqn. (1) becomes: (2) where Pw = partial pressure of water vapour in moist air Pat = atmospheric pressure of moist air 1.1.4 Relative Humidity Relative humidity (Æ ) is defined as the ratio of the mole fraction of the water vapour (xw) in a given moist air sample to the mole fraction of water vapour in an air sample of saturated moist air (xws) at the same temperature and pressure. (3) By definition, the mole fraction of the water vapour (xw) is the ratio of the number of mole of water vapour in a given moist air sample to the total number of dry air and water vapour. (4) When the dry air and water vapour occupy the same volume and temperature, by applying the characteristic equation of state for perfect gas, Eqn. (3) becomes: (5) where Pw = partial pressure of water vapour in moist air Pws = partial pressure of water vapour in saturated moist air Relative humidity is usually expressed as percentage (%). 1.1.5 Degree of Saturation/Percentage Saturation Degree of saturation (m ) is the ratio of the humidity ratio of moist air (w) to the humidity ratio of saturated moist air (ws) at the same temperature and pressure. (6) From Eqn. (2), Eqn. (6) becomes (7) The difference between relative humidity Æ and degree of saturation m is usually less than 2%. Percentage saturation is degree of saturation when expressed in percentage. 1.1.6 Specific Volume/Moist Volume Specific volume v (m3/kg) is defined as the total volume V (m3) of the dry air and water vapour mixture per kg of dry air. (8) where ma = mass of dry air, kg 1.1.7 Specific Enthalpy The enthalpy of moist air is defined as the sum of its internal energy and the product of its pressure and volume. Specific enthalpy h (kJ/kg) of moist air is defined as the total enthalpy of the dry air and water vapour mixture per kg of moist air. (9) where ha = specific enthalpy of dry air, kJ/kg hw = specific enthalpy of water vapour, kJ/kg w = moisture content, kg/kg 1.1.8 Dew Point Temperature Dew point temperature is the temperature at which moist air becomes saturated (100% relative humidity) with water vapour when cooled at constant pressure, i.e. temperature at which condensation of moisture begins when the moist air is cooled. 1.2 Psychrometric Charts A psychrometric chart graphically represents the thermodynamic properties of moist air as shown. | 758 | 3,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-51 | latest | en | 0.860926 |
https://www.aavintrichy.com/interesting-facts/how-much-milk-should-4-month-old-baby-drink.html | 1,643,020,896,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00317.warc.gz | 672,447,409 | 12,296 | # How Much Milk Should 4 Month Old Baby Drink?
## How much does a 4-month-old drink per feed?
At 4 months, babies usually take 4 to 6 ounces per feeding. At 6 months, babies may be taking up to 8 ounces every 4 to 5 hours.
## How many ounces should a 4-month-old drink in a day?
Indeed, babies should still drink about 4 -6 ounces per feeding when they’re 4 months old. Once they turn 6 months old, they may take up to 8 ounces every four or five hours, says the AAP.
## How do I calculate how much milk my baby needs?
Take your baby’s weight in ounces and divide that number by 6 (132 / 6 = 22). This figure represents how many ounces of breast milk that your baby should be getting in one day. Based on the example above, the baby should be taking in about 22 ounces of breast milk in a 24-hour period.
## How much should a 4-month-old drink in 24 hours?
At this stage your baby may be drinking up to 6 to 8 ounces of formula per feeding every four to six hours — but limit the total intake to no more than 32 to 36 ounces per 24 – hour period. That’s the upper daily intake for babies 6 months and younger, according to the American Academy of Pediatrics (AAP).
You might be interested: Often asked: How Many Calories In A Litre Of Milk?
## How much should a 4 month old weigh?
Baby weight chart by age
Baby age Female: 50th percentile weight Male: 50th percentile weight
3 months 12 lb 14 oz (5.8 kg) 14 lb 1 oz (6.4 kg)
4 months 14 lb 3 oz (6.4 kg) 15 lb 7 oz (7.0 kg)
5 months 15 lb 3 oz (6.9 kg) 16 lb 9 oz (7.5 kg)
6 months 16 lb 1 oz (7.3 kg) 17 lb 8 oz (7.9 kg)
## When do I stop feeding my baby every 3 hours?
If your baby has surpassed his birth weight and is steadily gaining weight, you can stop feeding every 2 to 3 hours during the night and instead feed on demand. Premature and jaundiced babies may sleep through their hunger., which means you must wake them to feed.
## What are the milestones for a 4 month old?
Movement/Physical Development
• Pushes down on legs when feet are on a hard surface.
• May be able to roll over from tummy to back.
• Can hold a toy and shake it and swing at dangling toys.
• Brings hands to mouth. video icon.
• When lying on stomach, pushes up to elbows.
## How often should a 4 month old feed?
Typically four ounces about four to six times a day. Breastfeeding: How often should a 4 – month – old nurse? Feedings are still typically about every three or four hours, but each breastfed baby may be slightly different.
## How often should a 4 month old eat solids?
Broadly, most babies eat: 4 to 6 months: 3 to 4 tablespoons of cereal once a day, and 1 to 2 tablespoons of a vegetable and fruit 1 or 2 times a day. 7 months: 3 to 4 tablespoons of cereal once a day, 2 to 3 tablespoons of a vegetable and fruit twice a day, and 1 to 2 tablespoons of a meat and protein food once a day.
## How do I know my baby is full when breastfeeding?
Signs of a Full Baby Once your baby is full, she will look like she’s full! She will appear relaxed, content, and possibly sleeping. She will typically have open palms and floppy arms with a loose/soft body, she may have the hiccups or may be alert and content.
## How do I know if baby is getting enough milk?
Signs your baby is getting enough milk Your baby’s cheeks stay rounded, not hollow, during sucking. They seem calm and relaxed during feeds. Your baby comes off the breast on their own at the end of feeds. Their mouth looks moist after feeds.
## How much milk should a baby drink by weight?
Calculate the amount of breastmilk a baby should take in, based on weight. Oz’s per feeding = oz’s per day divided by number of feedings. Example: 9 lb 8 oz baby needs 24.8 oz’s a day. If feeding 8 times a day: 24.8 divided by 8 equals 3.1 oz’s per feeding.
## Can you overfeed a baby formula?
health visitor if: Overfeeding of formula is common as the teat stimulating the roof of the baby’s mouth makes her suck. It is hard for your baby to digest large volumes, so try not to overfeed.
## How do you know when to increase formula?
Most babies are satisfied with 3 to 4 ounces (90–120 mL) per feeding during the first month and increase that amount by 1 ounce (30 mL) per month until they reach a maximum of about 7 to 8 ounces (210–240 mL). If your baby consistently seems to want more or less than this, discuss it with your pediatrician.
You might be interested: Often asked: A Glass Of Milk Contains How Much Protein?
## Can you overfeed newborn?
While it is certainly possible to overfeed a baby, most infant nutrition experts agree that it is fairly uncommon. As we noted earlier, babies are innately capable of self-regulating their intake; they eat when they’re hungry and stop when they’re full. | 1,208 | 4,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-05 | latest | en | 0.923383 |
https://talkchess.com/forum3/viewtopic.php?f=7&t=56168&sid=e7bdb619ba22af115f0fc26fe7d62d45&start=50 | 1,628,090,985,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00377.warc.gz | 559,468,093 | 8,413 | ## Piece weights with regression analysis (in Russian)
Discussion of chess software programming and technical issues.
Moderators: hgm, Dann Corbit, Harvey Williamson
Forum rules
This textbox is used to restore diagrams posted with the [d] tag before the upgrade.
PK
Posts: 876
Joined: Mon Jan 15, 2007 10:23 am
Location: Warsza
Contact:
### Re: Piece weights with regression analysis (in Russian)
Can You give similar output for Stockfish? It's interesting to see if it will differ from human values.
Michel
Posts: 2236
Joined: Sun Sep 28, 2008 11:50 pm
### Re: Piece weights with regression analysis (in Russian)
The logistic regression methods for tuning are really nice and I enjoyed reading Buro's paper. Obviously from a practical viewpoint such tuning methods work quite well.
Question: is there any research into why such statistically correct evaluation functions are expected to work?
The problem I have is that "statistical correctness" does not propagate through the game tree. In other words: if the leaf evaluation is statistically correct then there is a priori no reason why the result of a 1-ply search would be statistically correct.
The propagation through the game tree is via min/max functions and these have bad statistical properties...
Ideas=science. Simplification=engineering.
Without ideas there is nothing to simplify.
Xann
Posts: 127
Joined: Sat Jan 22, 2011 6:14 pm
Location: Lille, France
### Re: Piece weights with regression analysis (in Russian)
Hi Michel,
Michel wrote:The logistic regression methods for tuning are really nice and I enjoyed reading Buro's paper. Obviously from a practical viewpoint such tuning methods work quite well.
Question: is there any research into why such statistically correct evaluation functions are expected to work?
The problem I have is that "statistical correctness" does not propagate through the game tree. In other words: if the leaf evaluation is statistically correct then there is a priori no reason why the result of a 1-ply search would be statistically correct.
The propagation through the game tree is via min/max functions and these have bad statistical properties...
I fully agree with you, and propagating a random eval will quickly convince the sceptic. Machine learning in general expects you to use its results directly. Sometimes a bias such as a cost matrix (http://cse-wiki.unl.edu/wiki/index.php/ ... e_Learning) is introduced.
I haven't seen search/eval interaction modelled anywhere, but don't follow papers closely anymore. It seems very difficult, but could lead to biased error functions that give better results when combined with deep searches.
One element as to why it works at all for me is that regression can compensate its drawbacks by handling a large number of weights, even in the millions.
Fabien.
brtzsnr
Posts: 433
Joined: Fri Jan 16, 2015 3:02 pm
Contact:
### Re: Piece weights with regression analysis (in Russian)
WinPooh wrote:My article on a polular Russian IT portal:
http://habrahabr.ru/post/254753/
It is in Russian, but I believe the text is quite understandable even without translation.
I must be doing something wrong. For 7442 games at 40/2+0.05 (two seconds for 40 moves plus 0.05 increment) between zurichess and itself I get very strange values. Using Texel's Tuning method I get good values.
Code: Select all
``````./pgnlearn.exe ../../pgn/games.pgn
Games: 7442
Created file: .mat
[ 12807 x 5 ]
Iter 0: [ 0 0 0 0 0 ] -> 0.693147
Iter 1000: [ 0.173482 0.210955 0.288834 0.391243 0.17849 ] -> 0.690838
Iter 2000: [ 0.174092 0.212807 0.2914 0.472017 0.187411 ] -> 0.69083
Iter 3000: [ 0.174207 0.213172 0.29191 0.487982 0.188865 ] -> 0.69083
Iter 4000: [ 0.17423 0.213245 0.292011 0.49115 0.18915 ] -> 0.69083
Iter 5000: [ 0.174234 0.213259 0.292031 0.491779 0.189206 ] -> 0.69083
Iter 6000: [ 0.174235 0.213262 0.292035 0.491904 0.189218 ] -> 0.69083
Iter 7000: [ 0.174236 0.213263 0.292036 0.491928 0.18922 ] -> 0.69083
Iter 8000: [ 0.174236 0.213263 0.292036 0.491933 0.18922 ] -> 0.69083
Iter 9000: [ 0.174236 0.213263 0.292036 0.491934 0.18922 ] -> 0.69083
Iter 10000: [ 0.174236 0.213263 0.292036 0.491934 0.18922 ] -> 0.69083
PIECE VALUES:
Pawn: 100
Knight: 122.399
Bishop: 167.61
Rook: 282.339
Queen: 108.6
Press ENTER to finish
``````
WinPooh
Posts: 259
Joined: Fri Mar 17, 2006 7:01 am
Location: Russia
Contact:
### Re: Piece weights with regression analysis (in Russian)
brtzsnr wrote:
WinPooh wrote:My article on a polular Russian IT portal:
http://habrahabr.ru/post/254753/
It is in Russian, but I believe the text is quite understandable even without translation.
I must be doing something wrong. For 7442 games at 40/2+0.05 (two seconds for 40 moves plus 0.05 increment) between zurichess and itself I get very strange values. Using Texel's Tuning method I get good values.
Code: Select all
``````./pgnlearn.exe ../../pgn/games.pgn
Games: 7442
Created file: .mat
[ 12807 x 5 ]
Iter 0: [ 0 0 0 0 0 ] -> 0.693147
Iter 1000: [ 0.173482 0.210955 0.288834 0.391243 0.17849 ] -> 0.690838
Iter 2000: [ 0.174092 0.212807 0.2914 0.472017 0.187411 ] -> 0.69083
Iter 3000: [ 0.174207 0.213172 0.29191 0.487982 0.188865 ] -> 0.69083
Iter 4000: [ 0.17423 0.213245 0.292011 0.49115 0.18915 ] -> 0.69083
Iter 5000: [ 0.174234 0.213259 0.292031 0.491779 0.189206 ] -> 0.69083
Iter 6000: [ 0.174235 0.213262 0.292035 0.491904 0.189218 ] -> 0.69083
Iter 7000: [ 0.174236 0.213263 0.292036 0.491928 0.18922 ] -> 0.69083
Iter 8000: [ 0.174236 0.213263 0.292036 0.491933 0.18922 ] -> 0.69083
Iter 9000: [ 0.174236 0.213263 0.292036 0.491934 0.18922 ] -> 0.69083
Iter 10000: [ 0.174236 0.213263 0.292036 0.491934 0.18922 ] -> 0.69083
PIECE VALUES:
Pawn: 100
Knight: 122.399
Bishop: 167.61
Rook: 282.339
Queen: 108.6
Press ENTER to finish
``````
Probably it is due to my gradient solver - it is very simple and straightforward, looks like some local minimum was caught. Another reason can be PGN file - the parser can't handle comments or folded lines of moves. | 2,117 | 6,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.935165 |
https://macmost.com/calculating-difficult-dates-in-mac-numbers.html | 1,675,676,027,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500334.35/warc/CC-MAIN-20230206082428-20230206112428-00314.warc.gz | 376,189,375 | 12,127 | # Calculating Difficult Dates In Mac Numbers
Learn how to calculate dates like the first or last day of a month, the first or last day-of-week of a month, or the nth day-of-week of a month.
Here are the formulas used in this tutorial:
```=EOMONTH(B2,0)
=EOMONTH(B3,−1)+1
=DATE(YEAR(B4),MONTH(B4),1)
=WEEKDAY(B5)
=EOMONTH(B6,0)−MOD(WEEKDAY(EOMONTH(B6,0))−4,7)
=EOMONTH(B7,0)−MOD(WEEKDAY(EOMONTH(B7,0))−1,7)
=DATE(YEAR(B8),MONTH(B8),1)+MOD(4−WEEKDAY(DATE(YEAR(B8),MONTH(B8),1)),7)
=DATE(YEAR(B9),MONTH(B9),1)+MOD(1−WEEKDAY(DATE(YEAR(B9),MONTH(B9),1)),7)
=DATE(YEAR(B10),MONTH(B10),1)+MOD(4−WEEKDAY(DATE(YEAR(B10),MONTH(B10),1)),7)+14
``` | 262 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-06 | longest | en | 0.528001 |
https://www.dripcyplex.com/precisely-why-a-person-need-to-by-no-means-purchase-a-great-on-line-port-device-method/ | 1,603,730,153,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00071.warc.gz | 699,630,447 | 8,478 | # Precisely why A person Need to By no means Purchase A great On-line Port Device Method!
Taking part in on the internet slot equipment has grow to be progressively well-known, as on the internet casinos have grown in recognition. This development in on-line gaming has witnessed an enhance in the number of gamers hunting for an easy way to hit the million jackpots and turn into one particular of the few high rollers who succeed in on the internet slots. A lot of are tempted to purchase an on the internet slot method which claims to be in a position to make the purchaser regular massive income. The truth of on the web slot device systems nevertheless, is that the promises will not match the hype. Slot equipment continue to be video games of likelihood, and just like roulette and craps, there is no program that can assure you regular jackpots. Don’t buy an on the web slot machine program. Read on and uncover out why!
Simple fact: You Are not able to Utilize a Method to On the internet Slots to Make Standard Revenue
There is no way to make assured revenue from mathematically harmful online games, and on the web slot equipment are such video games. In arithmetic, you know just what will take place. Games of likelihood are the actual reverse. You never know what will happen next. If you did, then of system, it would not be a recreation of possibility. On-line slots are a sport of possibility, so mathematical programs are not able to be utilized. Period of time.
On the web Slots Do Work To A Mathematical Method!
The winning combos produced by on the internet slot devices are produced by a Random Number Generator (RNG). In online slot machines, RNG’s are not actually random, since they are the end result of a mathematical method. If you understood the formulation utilized in any on-line on line casino slot equipment and the value of the very last random quantity produced, you would be in a position to estimate the following random variety that would be created, but of system, you can not. Why? slotxo is the velocity at which the RNG calculates profitable mixtures. The RNG is actually a sequence of codes prepared into the software of the sport chip. It generates quantities and it does it quite speedily. In reality, at least one hundred figures every single second can be created. In an on-line casino slot device, every single a single of individuals quantities corresponds to a consequence on the reels. The effect of this for the participant is a random selection from a discipline of quantities that will establish the outcome of the play.
Why You Can’t Defeat On the web On line casino Slot Machines
On-line slot devices RNG’s generate a random generation of a number from the field of figures in the program, at minimum every 1-hundredth of a 2nd. The RNG is usually generating figures even when it is idle. Even if the programmer of the on-line slot device realized the sequence in which the quantities are currently being produced, by the time he calculates what the following variety is the machine will have moved on, as we all know all computer systems can crunch figures a lot quicker than any man or woman. Even though it is not entirely random by the mother nature of its programming, a programmer even if he understood the sequence would not be in a position preserve up with the machine, so what possibility would a participant have?
Reality is you are unable to use a mathematical technique in on-line slot machines. So a system that tells you it can assure slot machine jackpots persistently is lying. | 719 | 3,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-45 | latest | en | 0.929646 |
https://www.physicsforums.com/threads/neutrons-decay-kintetic-energy.462429/ | 1,590,575,769,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392142.20/warc/CC-MAIN-20200527075559-20200527105559-00534.warc.gz | 863,342,587 | 16,920 | # Neutron's decay, kintetic energy
## Homework Statement
What is the kinetic energy given to the proton in the decay of the neutron when a) the electron has negligibly small kinetic energy; b)antineutrino has small negiligibly energy
## Homework Equations
The decay reaction
$$n\Rightarrow p+e^{-}+\bar{\nu}$$
now I try to work out the Q-value (I assume that neutron is at rest)
$$m_{n}c^2=m_{p}c^2+T_{p}+m_{e}c^2+T_{e}+m_{\bar{nu}}c^2+T_{\bar{\nu}}$$
then movin all the therms with c^2 to the left side I get
$$(m_{n}-m_{p}-m_{e}-m_{\bar{\nu}})c^2=T_{p}+T_{\bar{\nu}}$$
so
$$Q=T_{p}+T_{\bar{\nu}}$$
the Q value can be easily calculated but what with the other kinetic energies on the right side? It this correct?
## The Attempt at a Solution
Related Advanced Physics Homework Help News on Phys.org
lanedance
Homework Helper
sorry what is the T?
is your decay is as follows:
$$n\Rightarrow p+e^{-}+\bar{\nu}$$
when one particle has negligible KE, then you can just consider a momentum energy balance for the other 2 particles
so, say there is excess energy E, and the 2 particles of interest (with masses m1 & m2) will move in oppposite dierctions along a single axis (with speeds v1 & v2), then they must satisfy:
conservation of energy
$$E = m_1 v_1^2 + m_2 v_2^2$$
conservation of monemtum
$$m_1 v_1 + m_2 v_2 = 0$$
T stands for kinetic energy
vela
Staff Emeritus
Homework Helper
For part (a), you're assuming the electron is at rest, so what does that tell you about the momenta of the antineutrino and proton?
Instead of separating out the kinetic energy at the beginning of the problem, I'd try solving for the total energy of the proton and then subtract off the rest energy at the end to get the kinetic energy.
that their momenta are equal?
vela
Staff Emeritus
Homework Helper
Yes, equal in magnitude but opposite in direction. | 551 | 1,853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-24 | longest | en | 0.865419 |
https://www.mycoursehelp.com/QA/a-potato-that-may-be-approximated-as-a-5/115582/1 | 1,642,732,547,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00628.warc.gz | 959,412,879 | 8,455 | ### Create an Account
Home / Questions / A potato that may be approximated as a 57 cm solid sphere with the properties
# A potato that may be approximated as a 57 cm solid sphere with the properties
A potato that may be approximated as a 5.7-cm solid sphere with the properties ρ = 910 kg/m3 ,
cp = 4.25 kJ/kg·K, k = 0.68 W/m·K, and α = 1.76 × 10-7 m2/s. Twelve such potatoes initially at 25°C are to be cooked by placing them in an oven maintained at 250°C with a heat transfer coefficient of 95 W/m2 ·K. The amount of heat transfer to the potatoes by the time the center temperature reaches 100°C is
(a) 56 kJ
(b) 666 kJ
(c) 838 kJ
(d) 940 kJ
(e) 1088 kJ
Jul 23 2020 View more View Less | 221 | 698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-05 | latest | en | 0.888385 |
https://people.ece.cornell.edu/land/courses/ece4760/FinalProjects/s1999/jason/476final.html | 1,722,803,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00761.warc.gz | 352,922,303 | 5,018 | A m -Controller Based Thermostat
A Final Project Report for Professor Land's
Spring 99 EE 476
Link to this document with resources in .zip format
Introduction
The goal of our final project was to design a thermostat using an Atmel AT90S8535 microcontroller. The thermostat was to compute the current temperature once per second and then send an on/off signal to a heating device which would then regulate the temperature to a desired target temperature. In addition, the thermostat was required to keep a time of day clock, as well as record the maximum and minimum temperature values that have occurred.
Setup
We accomplished our goal by using an analog thermistor-based temperature sensitive circuit whose output voltage was inversely proportional to the temperature of the system. In order for the microcontroller to obtain the current temperature, the thermistor voltage was input to the ADC of the 8535 and then converted to a temperature using a simple linear equation involving fixed point mathematics.
In addition, the thermostat also required the use of both an LCD and a keyboard for the purpose of user interaction. Specifically, the LCD performed the following display functions:
1. display the current temperature
2. display the target temperature
3. display the min and max temperatures
4. display the time of day
5. displaying keyboard input for verification
The keyboard was used for:
1. setting the target temperature
2. setting the time of day
3. selecting the view min/max mode
4. clearing the min/max values
As you can see, this project combined many if not all of the concepts taught over the course of the semester. These include timing, use of a keyboard (including debouncing it), use of an LCD, fixed point mathematics, analog circuit design and analog-to-digital conversion.
Thermistor Calibration
The first step in creating a thermistor based temperature controller is the calibration of the thermistor, that is figuring out how the amount of voltage drop across the thermistor corresponds to its temperature. To come up with this data we used the calibration setup circuit shown in Figure 1.
Figure 1. Thermistor Calibration Setup
We then heated the thermistor up from 30 to 90 degrees Celsius (since this setup did not have a safe method for cooling the thermistor to down below ambient temperature) and recorded both the voltage across the thermistor and its temperature. These values are plotted in Figure 2. We accurately knew the temperature of the thermistor because our calibration setup included a thermometer which was thermally coupled to the thermistor. Find our experimental data in Appendex 1.
One complication which arose in this part of the project was that because the thermistor is not a perfect device per se, it requires on the order of ten or so seconds for its resistance to accurately respond to a change in temperature. Because of this effect, there was a thermal lag which existed between the thermistor and the thermometer used to know its temperature. Nonetheless, we found that the way to minimize this source of error was to take calibration data both during a heat up cycle a cool down cycle. This is effective in minimizing this source of error because the thermal lag is effectively canceled out by the fact that during the heat up cycle there is a positive thermal gradient between the thermometer and the thermistor while during the cool down cycle there is a negative thermal gradient.
Therefore we were able to accurately obtain values correlating the voltage drop across the thermistor to the thermistor’s temperature. We then converted these analog voltages (Vread) to 10-bit binary values such as would appear at the output of the ADC using the following equation:
For a reference we use Aref = 4.5V. Appendix 1 shows how these 10-bit ADC values correspond to the temperature of the thermistor (in Celsius).
The Code
The code used in implementing a thermostat included many separate routines from timing to analog-to-digital conversion. A brief description of each follows.
Timing: any useful thermostat is programmable and at the very minimum should accept a target temperature. Optionally, it could apply this target temperature for a specified time interval (for example, as in a toaster), or it could have several target temperatures which change during the course of the day (as in a building thermostat). Since this requires accurate, programmable timing, a useful and practical step is to include a clock display on the LCD. We used Timer1 with a clock division of 64, a Compare/MatchA value of 62,500, and the Clear Timer1 Counter flag set, to yield precise timing intervals of 1 second. After implementing our conversion routines, we are able use this to accurately sample temperatures at one second intervals.
Recording: the thermostat was also desired to keep minimum and maximum temperatures. This was a simple comparison routine which looked at the current min and max temperatures in memory and replaced them if the new temperature exceeded either of the current min or max. We also included a button option which allowed us to reset the min and max temperatures at any time.
Control: In order to turn on or turn off a heater, a control segment was coded which applied either a logical 1 or 0 to PortB0 depending on whether the current temperature was above or below the user input target temperature. Because the microcontroller was not capable of driving the power transistor which was necessary for the application of running a heater, we used an LED to indicate the status of PortB0. If given another fifteen minutes we could have used two transistors in a Darlington configuration to increase the current gain of the transistor switch and in this manner control the heater.
Analog-to-Digital Conversion: Using the one second time base described above, we triggered the ADC (in single step mode). The 8535’s straightforward conversion routine converted the voltage at its input to a 10-bit binary value using equation 1 above. The temperature was then calculated using the obtained ADC value to a value in degrees celsius using equation 2 below. This value was then used in the record and control routines.
Temperature(C) = -0.1143x(10-bit ADC value) + 129 (2)
Display routines: Display routines were also triggered once every second in one of two states: the primary state and the display hi/lo temperature mode. The 16-character LCD display has enough room to output two temperatures and the time, with room to spare as follows:
48->59C P0 16:46 (LCD 1)
Here you see the current temperature heading to the target temperature (in Celsius), an extra indicator (P0) which will be later explained, and then the current time. The colon blinks at a rate of once per second- on for one second, off for the next.
In Hi/Lo Display mode, this LCD is formatted as follows:
22L 48H P0 16 46 (LCD2)
This indicates a low of 22 degrees and a high equal to our current temperature (48 degrees from LCD1 above). The extra indicator ‘P0’ is an optional advanced feature that indicates the current programmed setting which is running.
Two additional states also exist in our basic thermostat design, one each to set the clock and the other to set the target temperature. These routines are straightforward using a two digit entry and echo routine for input. The clock uses this routine twice, once for the hours and once for the minutes (the seconds are reset to zero when the time is set).
The code is designed for an Atmel ATS908535 Microcontroller, an 8-bit 32-register RISC microcontroller with A/D converter. Find source code here.
Conclusion
Our goal of implementing a thermostat using an Atmel AT90S8535 microcontroller was successfully achieved. The implementation included pretty much every aspect of the course described previously as well as including some research outside of the scope of this course. However because we were forced to change projects with only two and a half weeks left until demo (due to the limitations of these microcontrollers- there were not enough ports and memory for our original choice of a project), we were not able to implement all of the options (such as programmable by time of day) that we had hoped to include. Nonetheless the project was a great success and was demoed on 5/3/99 without any hitches.
Stephen Shannon
Jason Pettiss
May 3rd, 1999
(C)1999 | 1,716 | 8,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.898166 |
https://www.sportsbookreview.com/forum/nominated-posts/3692406-march-22-2022-13-666-nominated-post.html | 1,670,573,563,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00577.warc.gz | 1,049,213,728 | 11,654 | ## March 22, 2022 = 13 and 666 : Nominated Post
Originally posted on 03/07/2022:
(3+2+2) = 7 and (2+0+2+2) = 6 so 7 and 6. Take 76ers -7 tonight. | 69 | 147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | longest | en | 0.889888 |
https://math.answers.com/Q/What_is_10_over_3_as_an_improper_fraction | 1,718,671,142,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00596.warc.gz | 332,442,323 | 46,715 | 0
# What is 10 over 3 as an improper fraction?
Updated: 9/21/2023
Wiki User
12y ago
10/3 = 3 and 1/3
Wiki User
12y ago | 51 | 125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.916463 |
https://docs.microsoft.com/zh-cn/dotnet/api/system.tuple-3.item1?view=netframework-4.7.2 | 1,571,706,211,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00215.warc.gz | 459,879,768 | 10,019 | # Tuple<T1,T2,T3>.Item1Tuple<T1,T2,T3>.Item1Tuple<T1,T2,T3>.Item1Tuple<T1,T2,T3>.Item1 Property
## 定义
``````public:
property T1 Item1 { T1 get(); };``````
``public T1 Item1 { get; }``
``member this.Item1 : 'T1``
``Public ReadOnly Property Item1 As T1``
T1 T1 T1 T1
## 示例
``````using System;
public class Example
{
public static void Main()
{
Tuple<string, double, int>[] scores =
{ Tuple.Create("Jack", 78.8, 8),
Tuple.Create("Abbey", 92.1, 9),
Tuple.Create("Dave", 88.3, 9),
Tuple.Create("Sam", 91.7, 8),
Tuple.Create("Ed", 71.2, 5),
Tuple.Create("Penelope", 82.9, 8),
Tuple.Create("Linda", 99.0, 9),
Tuple.Create("Judith", 84.3, 9) };
var result = ComputeStatistics(scores);
Console.WriteLine("Mean score: {0:N2} (SD={1:N2}) (n={2})",
result.Item2, result.Item3, result.Item1);
}
private static Tuple<int, double, double> ComputeStatistics(Tuple<string, double, int>[] scores)
{
int n = 0;
double sum = 0;
// Compute the mean.
foreach (var score in scores)
{
n += score.Item3;
sum += score.Item2 * score.Item3;
}
double mean = sum / n;
// Compute the standard deviation.
double ss = 0;
foreach (var score in scores)
{
ss = Math.Pow(score.Item2 - mean, 2);
}
double sd = Math.Sqrt(ss/scores.Length);
return Tuple.Create(scores.Length, mean, sd);
}
}
// The example displays the following output:
// Mean score: 87.02 (SD=0.96) (n=8)
``````
``````Module Example
Public Sub Main()
Dim scores() =
{ Tuple.Create("Jack", 78.8, 8),
Tuple.Create("Abbey", 92.1, 9),
Tuple.Create("Dave", 88.3, 9),
Tuple.Create("Sam", 91.7, 8),
Tuple.Create("Ed", 71.2, 5),
Tuple.Create("Penelope", 82.9, 8),
Tuple.Create("Linda", 99.0, 9),
Tuple.Create("Judith", 84.3, 9) }
Dim result = ComputeStatistics(scores)
Console.WriteLine("Mean score: {0:N2} (SD={1:N2}) (n={2})",
result.Item2, result.Item3, result.Item1)
End Sub
Private Function ComputeStatistics(scores() As Tuple(Of String, Double, Integer)) _
As Tuple(Of Integer, Double, Double)
Dim n As Integer = 0
Dim sum As Double = 0
' Compute the mean.
For Each score In scores
n+= score.Item3
sum += score.Item2 * score.Item3
Next
Dim mean As Double = sum / n
' Compute the standard deviation.
Dim ss As Double = 0
For Each score In scores
ss = Math.Pow(score.Item2 - mean, 2)
Next
Dim sd As Double = Math.Sqrt(ss/scores.Length)
Return Tuple.Create(scores.Length, mean, sd)
End Function
End Module
' The example displays the following output:
' Mean score: 87.02 (SD=0.96) (n=8)
``````
## 注解
• 通过对Item1属性`GetType`返回的值调用方法。By calling the `GetType` method on the value that is returned by the Item1 property.
• 通过检索Type Tuple<T1,T2,T3>表示对象的对象, 并从该数组中检索由其Type.GetGenericArguments方法返回的第一个元素。By retrieving the Type object that represents the Tuple<T1,T2,T3> object, and retrieving the first element from the array that is returned by its Type.GetGenericArguments method. | 910 | 2,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-43 | latest | en | 0.377242 |
https://www.kaysonseducation.co.in/questions/p-span-sty_3345 | 1,674,950,044,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00737.warc.gz | 859,271,020 | 13,073 | A metal wire of diameter 1 mm is held on two knife edges separated by a distance of 50 cm. The tension in the wire is 100 N. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats per second. The tension in the wire is then reduced to 81 N. When the two are excited, beats are heard at the same rate. What is the frequency of the fork? : Kaysons Education
# A Metal Wire Of Diameter 1 Mm Is Held On Two Knife Edges Separated By A Distance Of 50 Cm. The Tension In The Wire Is 100 N. The Wire Vibrating With Its Fundamental Frequency And A Vibrating Tuning Fork Together Produce 5 Beats Per Second. The Tension In The Wire Is Then Reduced To 81 N. When The Two Are Excited, Beats Are Heard At The Same Rate. What Is The Frequency Of The Fork?
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## Question
### Solution
Correct option is
95 Hz
Let N be the frequency of the tuning fork. Then, the frequency of the wire, when the tension is 100 N will be (N + 5) and when the tension is 81 N, it is (N – 5); since in each case 5 beats are heard per second. Hence
subtracting (ii) from (i) we have
Using this value of m in (i) or (ii) gives N = 95 Hz.
#### SIMILAR QUESTIONS
Q1
Standing waves are produced by the superposition of two waves
and
Where x and y are expressed in metres and t is in seconds. What is the amplitude of a particle at x = 0.5 m. Given
Q2
The transverse displacement of a string fixed at both ends is given by
Where x and y are in metres and t is in seconds. The length of the string is 1.5 m and its mass is 3.0 × 10–2 kg. What is the tension in the string?
Q3
A wire of density ρ is stretched between two clamps a distance L apart, while being subjected to an extension lY is the Young’s modulus of the material of the wire. The lowest frequency of transverse vibrations of the wire is given by
Q4
A pipe of length 20 cm is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 425 Hz source? The speed of sound = 340 ms –1.
Q5
A pipe of length 20 cm is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1700 Hz source? The speed of sound = 340 ms–1
Q6
A wire has a mass 30 g and linear density 4 × 10 –2 kg m–1. It is stretched between two rigid supports and vibrates in its fundamental mode with a frequency of 50 Hz. What is the speed of transverse waves on the wire?
Q7
A steel rod of length 1.0 m is clamped rigidly at its middle. What is the frequency of the fundamental mode for the longitudinal vibrations of the rod? The speed of sound in steel = 5 × 103 ms –1
Q8
Two tuning forks A and B produce 10 beats per second when sounded together. On slightly loading fork A with a little wax, it was observed that 15 beats are heard per second. If the frequency of fork B is 480 Hz, what is the frequency of fork A before it was loaded?
Q9
Two sitar strings A and B are slightly out of tune and produce beats of frequency 6 Hz. When the tension in string A is slightly decreased, the beat frequency is found to be reduced to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Q10
An ambulance blowing a siren of frequency 700 Hz is travelling slowly towards a vertical reflecting wall with a speed of 2 ms–1. The speed of sound is 350 ms–1. How many beats are heard per second? | 1,024 | 3,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-06 | latest | en | 0.777622 |
https://axiomsofchoice.org/positive_real_step_function?rev=1395396676&do=diff | 1,596,539,836,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00363.warc.gz | 217,795,973 | 4,254 | # Differences
This shows you the differences between two versions of the page.
positive_real_step_function [2013/08/18 19:39]nikolaj created positive_real_step_function [2014/03/21 11:11] (current) 2013/08/18 19:40 nikolaj created2013/08/18 19:39 nikolaj created Next revision Previous revision 2013/08/18 19:40 nikolaj created2013/08/18 19:39 nikolaj created Line 1: Line 1: ===== Positive real step function ===== ===== Positive real step function ===== - ==== Definition ==== + ==== Set ==== - | @#88DDEE: $\langle X,\Sigma\rangle\in\mathrm{MeasurableSpace}(X)$ | + | @#55CCEE: context | @#55CCEE: $\langle X,\Sigma\rangle\in\mathrm{MeasurableSpace}(X)$ | - | @#55EE55: $f\in \mathcal T^+$ | + | @#55EE55: postulate | @#55EE55: $f\in \mathcal T^+$ | - | @#88DDEE: $f\in \mathcal T$ | @#88DDEE: see [[Real step function]] | + | @#55CCEE: context | @#55CCEE: $f\in \mathcal T$ | @#55CCEE: context | @#55CCEE: ...[[Real step function]] | | $x\in X$ | | $x\in X$ | - | @#55EE55: $f(x)\ge 0$ | + | @#55EE55: postulate | @#55EE55: $f(x)\ge 0$ | ==== Discussion ==== ==== Discussion ==== - ==== Context ==== + ==== Parents ==== === Subset of === === Subset of === [[Real step function]] [[Real step function]] | 464 | 1,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-34 | latest | en | 0.420165 |
https://physics.stackexchange.com/revisions/78e3904c-089e-4fa2-9c2c-a2587a2aa607/view-source | 1,571,762,796,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822458.91/warc/CC-MAIN-20191022155241-20191022182741-00389.warc.gz | 656,190,871 | 1,687 | ```It is not uncommon for units of a different physical entity to be used to measure a related physical entity. e.g. distance is generally measured in meters; but it is also measured in light years which is the distance traveled by light in a year. The important thing is that there should be a consistent way to convert one unit to another.
Someone pointed out that Torque is a vector (defined as a cross product) while Work is a scalar (defined as a dot product). However, that can't be "the (only) reason" for different units. Units are defined for "magnitude of a vector", which by itself is a scalar. So, the reason you can't use Joules for torque is because there is no consistent way of converting Newton-meters to Joules and vice versa.
There are 2 types of units viz., the basic/elementary units for mass, distance and time and the compound/derived units such Newton, Joule, etc for physical phenomenon that are derived from the basic units.
So, 1 Newton is the amount of Force required to increase the velocity of 1 Kg of point mass by 1 m/sec in 1 sec, in the direction of the change in velocity. 1 Joule is the amount of work done when a force of 1 Newton moves any point mass by a distance of 1m.
For a unit of Joule to be used for a unit of Torque, you would need a unit of Torque to always perform 1 Joule of work, which is not true.
``` | 318 | 1,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-43 | latest | en | 0.964041 |
https://www.numwords.com/words-to-number/en/5183 | 1,563,257,390,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524503.7/warc/CC-MAIN-20190716055158-20190716081158-00195.warc.gz | 793,113,972 | 1,851 | NumWords.com
# How to write Five thousand one hundred eighty-three in numbers in English?
We can write Five thousand one hundred eighty-three equal to 5183 in numbers in English
< Five thousand one hundred eighty-two :||: Five thousand one hundred eighty-four >
Ten thousand three hundred sixty-six = 10366 = 5183 × 2
Fifteen thousand five hundred forty-nine = 15549 = 5183 × 3
Twenty thousand seven hundred thirty-two = 20732 = 5183 × 4
Twenty-five thousand nine hundred fifteen = 25915 = 5183 × 5
Thirty-one thousand ninety-eight = 31098 = 5183 × 6
Thirty-six thousand two hundred eighty-one = 36281 = 5183 × 7
Forty-one thousand four hundred sixty-four = 41464 = 5183 × 8
Forty-six thousand six hundred forty-seven = 46647 = 5183 × 9
Fifty-one thousand eight hundred thirty = 51830 = 5183 × 10
Fifty-seven thousand thirteen = 57013 = 5183 × 11
Sixty-two thousand one hundred ninety-six = 62196 = 5183 × 12
Sixty-seven thousand three hundred seventy-nine = 67379 = 5183 × 13
Seventy-two thousand five hundred sixty-two = 72562 = 5183 × 14
Seventy-seven thousand seven hundred forty-five = 77745 = 5183 × 15
Eighty-two thousand nine hundred twenty-eight = 82928 = 5183 × 16
Eighty-eight thousand one hundred eleven = 88111 = 5183 × 17
Ninety-three thousand two hundred ninety-four = 93294 = 5183 × 18
Ninety-eight thousand four hundred seventy-seven = 98477 = 5183 × 19
One hundred three thousand six hundred sixty = 103660 = 5183 × 20
Sitemap | 426 | 1,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-30 | latest | en | 0.788537 |
https://jp.mathworks.com/matlabcentral/profile/authors/3871842-unai-san-miguel | 1,590,486,313,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390755.1/warc/CC-MAIN-20200526081547-20200526111547-00422.warc.gz | 361,851,116 | 23,070 | Community Profile
# Unai San Miguel
### Nordex-ACCIONA
26 2016 年以降の合計貢献数
#### Unai San Miguel's バッジ
Code to plot von mises and tresca yield conditions
This plot can be generated with some analytic work, by getting the ellipse in parametric equations, generating points along tha...
5ヶ月 前 | 0
Scattered interpolation with weight factors
You could add some random points near the point you rely more on. Close enough to effect the interpolation and separated enough ...
6ヶ月 前 | 0
curve fitting toolbox: how to fit a multivariate b spline to data (regression)
I think it is the same as for the d=1 case: you have to define a convenient knot sequence, and in turn a spline vector space, an...
11ヶ月 前 | 0
With the Nurbs2IGES toolbox found in MATLAB File Exchange https://www.mathworks.com/matlabcentral/fileexchange/12087-nurbs2iges ...
11ヶ月 前 | 0
| 採用済み
N-Dimensional smoothing spline weighting
You can use a cell of two set of weights, one in each direction of the input csaps({1:M, 1:N}, I, [], {wi, wj}); with wi an ...
1年以上 前 | 0
How can I represent and visualize my four dimensional system in a meaningful way?
3 dimensions are really hard to plot, it is almost always better to switch to 2D, and better contours than filled contours. If y...
1年以上 前 | 0
how can i work with the coefs of cscvn?
There is no need to use the coefficients of the function, because the Curve Fitting Toolbox provides with functions to derive an...
2年弱 前 | 0
Error finding and graphing max values of data (Error using horzcat Dimensions of matrices being concatenated are not consistent. )
Your |TEK00055.csv| data has three data points with maximum y-values: >> xmax xmax = 6.8720 6.8730 6.9890...
2年弱 前 | 0
| 採用済み
multidimensional numerical integration without large matrices
I don't like |meshgrid| even for 2 variables, so I would use |ndgrid| instead [X1, X2, X3, X4] = ndgrid(x1, x2, x3, x4); ...
'stacking' 2D graphs on top of each other.
There are |plot3|, |contour3| versions of |plot|, |contour| for which you would have to add a third coordinate for each 2D plot,...
| 採用済み
How to plot sin(theta)*cos(phi) in spherical coordinates
|sin(theta) * cos(phi)| is the result of the matrix multiplication of the two terms. Maybe you wanted |sin(theta) .* cos(phi)|? ...
2年以上 前 | 0
difficulty in printing for 'cell' inputs.
If you are OK with the contents and format of |c| maybe you can substitute the |fprintf| line by: cellfun(@(x) fprintf(fid2...
2年以上 前 | 0
How do i find the intersection points between a surface and a patch?
It is hard to find the intersection of a surface and a plane (in general), as it is explained <http://web.mit.edu/hyperbook/Patr...
2年以上 前 | 0
| 採用済み
Angle between two vectors
You have two vectors , determine the angle between these two vectors For example: u = [0 0 1]; v = [1 0 0]; The a...
2年以上 前
Calling multiple function files to plot in one figure
Although I don't know exactly what kind of plots you want to do, or what are the variables involved, sometimes it helps to use t...
2年以上 前 | 0
Constructing an Offset-Curve
The subject is not trivial, a nice description can be found in <http://web.mit.edu/hyperbook/Patrikalakis-Maekawa-Cho/node210.ht...
| 採用済み
How to display the contour line values from a contour plot generated using the curve fitting tools?
If, in the auto-generated code from the gui you specify |h| (the handler to the plots) as an aditional output variable you can a...
| 採用済み
Revolution 2D spline to 3D surface
It depends on what you mean by _analytical expression_. For example, it is not possible to express a circle and thus a revolutio...
fnval for tensor splines
|fnval(pp, [x(:)'; y(:)'])|? From the <https://es.mathworks.com/help/curvefit/fnval.html fnval> doc page (...) For an m...
Fit polynomial to data with prescribed second derivative
The <http://es.mathworks.com/help/curvefit/csape.html help page for csape> says that you would have to use |csape(x, [-0.001, y,...
Closed spline fit of airfoil coordinates
You are working with curves, not functions. Your |airfoil| is a (planar) curve, or in the Splines toolbox wording a 2-valued...
jupyter notebook vs live scripts
What are the differences between Live Scripts and Jupyter notebooks?
### 3
Storing the output of spap2 into an array then looping and graphing that output
The output of |spap2| is a struct, something like: >>sp sp = form: 'B-' knots: [1x25 double] ... ...
3年以上 前 | 0
statistics (PCA) on piecewise B spline for Shape Model problem using spline toolbox
|fncmb| takes first both spline functions to the same vector space (same breaks and order in this case). If the breaks are not t...
3年以上 前 | 0
B-Spline least squares (spap2)
Hi, |spap2| returns the B-form of the spline, and if you don't supply a knot vector it will use C2 continuity (k-2 = 4-2 in y...
3年以上 前 | 0
What is the best way to remove a knot from a (univariate) spline?
In my problem, I have a planar curve, i.e., a 2-valued univariate spline function |s|. I insert |k-1| times a value in the knot ... | 1,390 | 5,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.8348 |
https://www.physicsforums.com/threads/how-to-check-if-function-is-differentiable-at-a-point.734863/ | 1,508,684,438,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825264.94/warc/CC-MAIN-20171022132026-20171022152026-00573.warc.gz | 996,008,785 | 13,939 | # How to check if function is differentiable at a point
1. Jan 25, 2014
### Fabio010
The question is to check where the following complex function is differentiable.
$$w=z \left| z\right|$$
$$w=\sqrt{x^2+y^2} (x+i y)$$
$$u = x\sqrt{x^2+y^2}$$
$$v = y\sqrt{x^2+y^2}$$
Using the Cauchy Riemann equations
$$\frac{\partial }{\partial x}u=\frac{\partial }{\partial y}v$$
$$\frac{\partial }{\partial y}u=-\frac{\partial }{\partial x}v$$
my results:
$$\frac{x^2}{\sqrt{x^2+y^2}}=\frac{y^2}{\sqrt{x^2+y^2}}$$
$$\frac{x y}{\sqrt{x^2+y^2}}=0$$
solutions says that it's differentiable at (0,0). But doesn't it blow at (0,0)?
Last edited: Jan 25, 2014
2. Jan 26, 2014
### Staff: Mentor
What do you mean with "blow"?
The limit of those expressions for x,y -> 0 is well-defined and zero. | 294 | 786 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-43 | longest | en | 0.74273 |
https://ch.mathworks.com/matlabcentral/cody/problems/74-balanced-number/solutions/170744 | 1,579,981,790,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00516.warc.gz | 373,134,591 | 15,696 | Cody
# Problem 74. Balanced number
Solution 170744
Submitted on 30 Nov 2012 by Francis Esmonde-White
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
%% n = 13722; assert(isequal(isBalanced(n),true))
2 Pass
%% n = 23567414; assert(isequal(isBalanced(n),true))
3 Pass
%% n = 20567410; assert(isequal(isBalanced(n),false))
4 Pass
%% n = 1; assert(isequal(isBalanced(n),true))
5 Pass
%% n = 11111111; assert(isequal(isBalanced(n),true))
6 Pass
%% n = 12345678; assert(isequal(isBalanced(n),false))
7 Pass
%% n = 12333; assert(isequal(isBalanced(n),false))
8 Pass
%% n = 9898; assert(isequal(isBalanced(n),true))
9 Pass
%% n = 469200; assert(isequal(isBalanced(n),false))
10 Pass
%% n = 57666; assert(isequal(isBalanced(n),true)) | 290 | 874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-05 | latest | en | 0.410112 |
http://www.johnrzaleski.com/ | 1,500,791,386,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424287.86/warc/CC-MAIN-20170723062646-20170723082646-00458.warc.gz | 451,454,872 | 35,995 | ### Rowing Data Analytics: Reducing and Studying the Rowing Workout
In my last post (“Rowing Data…”) I discussed the steps associated with downloading the Garmin Vivoactive HR data from Garmin Connect to an Excel spreadsheet. In this post, I’m going to take the reader through the analysis of the data as a tutorial and guide for assessing certain elements of these data.
Raw data in Excel format are shown in Figure 1. I am going to focus on distance (column M), speed (column N), and heart rate (column O).
I normally like to study discrete, time-based data by translating the time component from the Zulu time (column L) into a relative time from the start of the workout. Furthermore, I like to translate these into units of seconds as the base unit.
To do so, we can take advantage of some powerful capabilities contained within formulas inside of Microsoft Excel. For example, the start time listed in column L begins with the entry:
2017-07-08T14:09:31.000Z
The next entry is:
2017-07-08T14:09:34.000Z
These are “Zulu” time or absolute time references. We wish all future times to be keyed or made in reference to the first time. In order to do so, we need to translate this entry into a time in seconds. We can do so by parsing each element of the entry. These entries are listed sequentially in column L2 and L3, respectively.
Each element is translated into seconds by parsing the hours, minutes and seconds using the following formula:
=MID(\$A\$2,12,2)*60*60+MID(\$A\$2,15,2)*60+MID(\$A\$2,18,2)
The first component extracts the time in hours and translates into seconds. The second component extracts the “minutes” and translates into seconds. The third component extracts the “seconds” element by itself. The total time is the superposition of all three individual components.
Thus, what I normally do is to copy the contents of the initial spreadsheet into a new sheet adjacent to the original and then begin working on the data. Presently, I am in the process of developing an application that will perform this function automatically. Yet, here I am “walking the track” associated with analyzing the data in order to chronicle the mathematics surrounding the process.
The hour, minute and second can be extracted as separate columns. Let us copy the contents of column L in the original spreadsheet into a new sheet within the existing workbook and place the time in column A of that new sheet. Thus, the entries in this sheet would appear as follows:
ns1:Time Absolute Time (seconds) Relative Time (seconds) 2017-07-08T14:09:31.000Z 50971 0 2017-07-08T14:09:34.000Z 50974 3 2017-07-08T14:09:35.000Z 50975 4
The Absolute time in the middle column is the time in seconds represented by the left-hand column relative to Midnight Zulu time. The right-hand column is the time relative to the first cell entry in the middle column. Thus, zero corresponds to 50971-50971. The entry for three seconds corresponds to the difference between 50974 (second entry) and 50971 (first entry), and so on.
I also created some columns to validate parameter entries. For instance, the reported total distance and speed (in units of meters and meters per second, respectively), in column M and N and the heart rate, in column O, are referred to next. I created a new column O in the new spreadsheet to provide a derived estimate of total distance, which I computed as the integral of speed over time. The incremental distance, dS, is equal to the speed at that time, dV, multiplied by the time differential between the current time and the previous time stamp, dt. Then, the total distance is the integral, or the summation of this incremental distance and all prior distances. I reflect this as column G in the new worksheet, shown in Figure 2.
What follows now are plots of the raw and derived data. First, the heart rate measurement over time is shown in Figure 3. Note that the resting rate is shown at first. Once the workout intensifies, heart rate increases and remains relatively high throughout the duration of the workout.
The total distance covered over time is shown in Figure 4. This tends to imply a relatively constant speed during the workout due to the linear behavior over the 8700+ meters.
The reported speed, as measured via GPS, shows variability but is typically centered about 1.85 meters per second. The speed over time is shown in Figure 5.
The GPS coordinates are also available through the Excel data. I have subtracted out the starting location in order to provide a relative longitude-latitude plot of the workout, shown in Figure 6.
In my next post I will focus on the athletic aspects of the workout related to training.
## An Introduction to Garmin Connect
For those who use the Garmin Connect Dashboard (https://connect.garmin.com), to synchronize their Garmin fitness devices, there is a fairly straightforward method for downloading higher-frequency data from the workout relative to heart rate, distance and GPS location that can be directly imported into Microsoft Excel for further analysis.
From inside of Garmin Connect (Figure 1), select a specific activity. In this example, I am picking my latest rowing workout, shown by the red arrow.
Once you have selected the specific workout, click on it and this will take you to the details of that workout. Once there, navigate over to the gear on the right-hand side, as shown by the red arrow.
The drop-down box from the arrow shows a number of export options. Select “Export to TCX”, shown by red arrow in Figure 3.
Then, change the suffix from .TCX to .XML, as shown in Figure 6. Accept the change when prompted.
Now, open Microsoft Excel. Select the Data tab, as shown in Figure 7.
On the left-hand side, select Get External Data “From Other Sources”, and scroll down to “From XML Data Import”, as shown in Figure 8.
A dialog box will open. Navigate to your newly-created XML file. Select it, and click the series of “OK” buttons in the dialogs that come up, including the one placing the location of the start in cell \$A\$1. Once completed, the contents of the file will be imported into your spreadsheet. Heart rate data will be contained in column O, as shown in Figure 9. Distance & speed are contained in columns M & N, respectively. GPS latitude & longitude are contained in columns P & Q, respectively. Average speed is contained in column R.
In my next post on the subject I will describe how to manipulate these data for further analysis.
## Sculling and Rowing
I am a rower and sculler. I first cut my teeth in the sport over 30 years ago while at college rowing on the Charles River. I had been looking for the longest time for a device that I could use to track my heart, stroke rate, and also support GPS mapping of my workout while on the water. There are professional devices that track stroke rate and the like, such as Speed Coach GPSStroke Coach and Coxmate GPS. These are all excellent pieces of equipment, by the way. But, I am not in varsity rowing any more and I was looking for a piece of equipment that could support my rowing “habit” both for indoor and outdoor rowing (aside: I also possess a Concept 2 ergometer, which I love) while also serving the utilitarian purpose of being a good watch that can track heart rate full time.
When I row, however, I am really interested in being able to map the analytics to the motion. The Vivoactive HR enables me to do this as well as to post-process the data. I am into data. As a Chief Analytics Officer in the healthcare field for a medical device and real-time patient surveillance company, it is important to me to be able to access and understand the information collected during an activity. The connectivity and access to data provided by the Vivoactive HR are phenomenal.
The figure above details an example analytics screen, which shows the map of the workout, heart rate, stroke rate, distance traveled at each measurement point, and allows tracking the entire workout with a cross-hair that is dynamic and interactive on the web screen. The unit supports many other types of workouts, including running, biking, pool, golf, walking, indoor rowing on ergometer, SUP rowing, XC skiing, indoor walking, indoor biking, and indoor running, and tracks sleep. The unit can be submerged in water and the battery life is amazing. I normally live with the unit on my wrist, and after 3 days of continuous use, battery is down to, perhaps 80%. I will take it off for an hour or so to charge, and it is good-to-go. I highly recommend this unit for the avid professional or veteran rower (like myself).
## Update June 29th, 2017: Comparison among NK, Coxmate, Minimax
Robin Caroe of RowPerfect kindly left me a comment to this post last evening and provided an updated article on comparison among the NK, Coxmate GPS and Catapult Minimax which contains quite valuable data on performance related to these products. I have provided the hyperlink to the article above. Technological differences in sampling rate (e.g.: 5 Hz for NK versus 10 Hz for Coxmate) are important for accuracy. I must say that I was very close to purchasing the Coxmate GPS prior to investigating the Garmin. Upon reading the brochure for the Minimax S4, I am intrigued. The Minimax offers an update rate on the GPS that provides for precision in terms of location. In the Rowperfect article, of the key measures of performance identified, (1) heart rate & heart rate variability; (2) force and length of stroke; and, (3) GPS update rate are important measures for the elite athlete. In the case of the Minimax, GPS update on the order of 100 times per second (10 milliseconds) can reveal boat pitch, roll & yaw. Highly impressive. I would agree, though, that this level of accuracy and precision would be important for the competitive athlete. Yet, in my case (non-competitive, casual athlete), I still love my Garmin. I am able to see and track my position very accurately, monitor my stroke and heart rate, and in terms of heart rate variability, I can write an algorithm in R or Matlab to monitor that measure fairly directly.
### Filtering of Arterial Blood Pressure Signal Artifact using the Extended Kalman Filter
In an earlier post, I had discussed some mathematical techniques for mitigating alarm fatigue.
Expanding on the mathematical techniques employed, another reason for filtering of data includes the smoothing of artifact or spikes that are due to signal errors or other issues associated with signal acquisition.
Figure 1 depicts several seconds of raw arterial blood pressure (ABP) data obtained from a patient within the MIMIC II physiologic waveform database. [1,2]
This figure shows a raw signal with a tracking signal based on the extended Kalman filter (EKF) overlaid. In this case, the signal error and the process noise are very small (signal noise 0.1 mmHg, process noise 0.5 mmHg). With these settings, the filter tracks the actual signal very closely, and makes it appear as if there is not difference between signal measurement and track.
The full analysis is available at the following link in PDF form:
ABP Tracking via EKF
[1] M. Saeed, M. Villarroel, A.T. Reisner, G. Clifford, L. Lehman, G.B. Moody, T. Heldt, T.H. Kyaw, B.E. Moody, R.G. Mark.Multiparameter intelligent monitoring in intensive care II (MIMIC-II): A public-access ICU database. Critical Care Medicine 39(5):952-960 (2011 May); doi: 10.1097/CCM.0b013e31820a92c6.
[2] Goldberger AL, Amaral LAN, Glass L, Hausdorff JM, Ivanov PCh, Mark RG, Mietus JE, Moody GB, Peng C-K, Stanley HE. PhysioBank, PhysioToolkit, and PhysioNet: Components of a New Research Resource for Complex Physiologic Signals.Circulation 101(23):e215-e220 [Circulation Electronic Pages; http://circ.ahajournals.org/cgi/content/full/101/23/e215]; 2000 (June 13).
### Alarm Fatigue
“Hospital staff are exposed to an average of 350 alarms per bed per day, based on a sample from an intensive care unit at the Johns Hopkins Hospital in Baltimore.”[1]
From the same survey, almost 9 in 10 hospitals indicated they would increase their use of patient monitoring, particularly of Capnography and pulse oximetry, if false alarms could be reduced. [2]
“Of those hospitals surveyed that monitor some or all patients with pulse oximetry or Capnography, more than 65 percent have experienced positive results in terms of either a reduction in overall adverse events or in reduction of costs.”[3]
### Attenuating Alarm Signals
The problem with attenuating alarm data is achieving the balance between communicating the essential, patient-safety specific information that will provide proper notification to clinical staff while minimizing the excess, spurious and non-emergent events that are not indicative of a threat to patient safety. In the absence of contextual information, the option is usually to err on the side of excess because the risk of missing an emergent alarm or notification carries with it the potential for high cost (e.g.: patient harm or death).
### Analysis
The purpose of this study is to look at the and some of the Mathematical Techniques for Mitigating Alarm Fatigue: techniques and options available for evaluating real-time data. The objective is to suggest a dialog for further research and investigation into the use of such techniques as appropriate. Clearly, patient safety, regulatory, staff fatigue and other factors must be taken into account in terms of aligning on a best approach or practice (if one can even be identified). These aspects of alarm fatigue are intentionally omitted from the discussion at this point (to be taken up at another time) so that a pure study of the physics of the parameter data and techniques for analyzing can be explored.
### References
[1] Ilene MacDonald, “Hospitals rank alarm fatigue as top patient safety concern”, Fierce Healthcare. January 22, 2014.
[2] Wong, Michael; Mabuyi, Anuj; Gonzalez, Beverly; “First National Survey of Patient-Controlled Analgesia Practices.” March-April 2013, A Promise to Amanda Foundation and the Physician-Patient Alliance for Health & Safety.
[3] Ibid.
# Designers Choice: A Small Sloop
Originally designed by the naval architects Sparkman & Stephens, and built circa Edison New Jersey in the late ”70s through mid-’80s, the Designers Choice (“DC”) is a fibreglass-hulled sailboat with length overall (LOA) of 14′ 10.5″, length at the waterline of 12′ 9″ and beam of 6′ 1″. She weighs in at 315 lbs.
The draft of the DC varies from 5″ (centerboard up) to 3′ 0″ (centerboard down). Aft freeboard is 1′ 2″.
The mast is tall and the sail area of the mainsail is 82 sq ft; that of the jib is 28 sq ft. Crew capacity is 900 lbs. In my experience, 3 adults and 3 children can be comfortable on board.
# DC Standard Features
She features:
• Black anodized aluminum spars.
• Grooved mast with loose footed mainsail fitted with luff slugs.
• Stainless steel chain plates, headstay & shrouds.
• Deluxe heave duty fittings.
• Four-part mainsheet with quick release cam cleat on centerboard trunk.
• All hardware mounted with through-bolts or drilled and tapped into aluminum backing plates.
• Controllable outhaul, boom vang and Cunningham.
• Kick-up rudder with foam-filled floating black anodized aluminum tiller and universal hiking stick.
• 1.25″ vinyl rub-rail.
• Non-leaking centerboard pin above the waterline and cockpit sole for easy access.
• Hand laid-up heavy duty mat and roving hull construction
• White gelcoat finish.
• Molded-in skid-resistant side seats and cockpit sole.
• Large covered stowage locker under afterdeck.
• Durable dacron mainsail and jib.
• Jib window and jiffy reefing are standard.
# My DC
I have owned my DC since 2003. I have had 3 sailboats in my life and this is a decent little craft. She was built in 1979, making her 38 years. Several photographs are included below. A copy of the original Howmar Designers Choice is provided for download, as well.
### Opioid-induced respiratory depression and monitoring patients diagnosed with obstructive sleep apnea
HIMSS Future Care posting:
Managing patients on the general care floor (GCF) who are either at risk or “diagnosed with obstructive sleep apnea (OSA), and those in particular who meet the requirements of the STOP-BANG criteria for OSA, can be quite challenging. The ECRI Institute, a federally-certified patient safety and research organization, has identified in its 2017 list of Top 10 Health Technology Hazards “Undetected Opioid-Induced Respiratory Depression” as Number 4 [1]. Opioids used for treatment of acute postoperative pain is rather commonplace, and patients at-risk for OSA, if left unattended, can experience anoxic brain injury or death.
### Medical Device Plug and Play
Originally posted at MedicalConnectivity.com, the discussion surrounding plug-and-play medical devices focused on the ability to have true interoperability from a semantic and physical perspective. This post was originally written in 2009 surrounding the need for better medical device plug and play interoperability and integration, in much the same way a USB-enabled accessory purchased for a standard computer is recognized by the drivers once plugged into the computer.
## Will the future of medicine rely on nanotechnology for treatment of disease?
The average size of the avian influenza virus is on the order of 100 nanometers, or 0.1 microns, which is of the order of nanotechnology. That a virus so small can wreak such havoc on the human body is a testament to the complex mechanisms associated with these infections. The ability to ward off such infections is also a testament to the awesome nature of the human immune system. By comparison, the width of a typical human hair is on the order of 100,000 nanometers (estimates put the range at 50,000 – 150,000, depending on the specific type of hair).
Now, consider the field of nanotechnology which focuses on the manufacture and fielding of mechanical and electronic devices of microscopic size, typically on the order of 100 nanometers or smaller. The National Cancer Institute (NCI) provides a fairly detailed overview of the use of nanotechnology in cancer treatment, and the NCI Alliance for Nanotechnology in Cancer is an initiative that provides a focal point for public and private investigation for the application of nanotechnology to the treatment of cancer. Researchers and companies have been investigating the manufacture of devices of this order of magnitude and smaller for application in the treatment of disease. A major focus for nanotechnology in healthcare is, not surprisingly, the treatment of cancer. Specific methods and modes of delivery vary. Examples include outfitting little “robots” with markers that will burrow into and attach themselves to cancerous cells for the purpose of enabling treatment and destruction of malignant cells. A major benefit of this approach versus traditional methods of radiation and chemotherapy is that the malignancies can be targeted directly without attacking or otherwise molesting healthy cells. This is a major advancement, since many of the current therapies that attack cells indiscriminately will kill both healthy as well as malignant cell material. When battling this terrible disease the last thing needed is to destroy those healthy cells upon which the individual depends for sustenance and survival. Thus, nanotechnology provides a mechanism for delivering targeted, customized, tailored therapy.
## What about nanotechnology for diagnosis?
While we are on the cutting edge of the application of these technologies, the vision is real, and it is extremely promising. Treatment is only one aspect of nanotechnology use. Diagnosis is another area, in which nanoparticles can be used to assist in imaging of potential malignancies. While almost a cliché, the aging of the baby-boomer population will drive a number of these new technologies, applications, and initiatives. It is almost a tautology that early diagnosis of disease translates into a higher likelihood of survival. Technologies that support early diagnosis are, therefore, of great value and will enable better, more efficient, and more accurate treatment of disease going forward. As a member of this generation (albeit, at the tail end), I am very encouraged and supportive of this research. I recall some 17 years ago when my mother passed away from breast cancer that the use of exotic technologies such as nanotechnology was barely an inkling. Indeed, the three oft-used mechanisms for treating cancer have remained surgery, irradiation, or poisoning (chemotherapy). It has only been within the past 10 years or so in which alternative therapies have been devised and discovered that are not simply variants of these three. Research into the targeted treatment of cancer by destroying the genetic material within malignant cells so that they cannot reproduce or cannot receive nourishment is an astonishing advancement and offers great future promise—a testament to human ingenuity, talent, innovation, and creativity. As in vitro and in vivo medicine evolve, such future-looking technologies will be essential in terms of early diagnoses and intervention.
## Can medical device integration facilitate diagnosis and treatment?
One cannot control what one cannot measure. In vivo measurements are necessary to determine whether any treatment paradigm is working: comparison pre- and post-treatment to determine the correlation and association of a treatment modality to establish intended effect. In later posts, I discuss the use of data taken from medical devices at the point of care to facilitate clinical decision making. These data, whether obtained from the patient externally or internally form the basis for identifying the state of the patient and trends towards improvement or decompensation.
## Medicine & Healthcare Information Technology In the Future
When it comes to medicine, healthcare information technology has become an integral part of care. Estimates by the U.S. Census Bureau expect the population of Americans aged 65 and older to increase by more than a factor of two between 2010 and 2050 [1]. This is not good news for the state of medicine in the U.S., particularly when it comes to providing quality care for these patients.
Estimates of healthcare expenditure increases between 2007 and 2017 show an increase to nearly 20% of GDP in this period [2]. These estimates were made prior to the recent financial crisis that began during the Fall of 2008. Further compounding this increasing demand and the concomitant increase in costs is the availability of allied healthcare professionals. Some studies [3] identify the likely decrease in the number of physicians entering any number of key specialty areas, including cardiology (20% decrease by 2020), geriatrics (35% of current demand met today), rheumatology (38 day average wait for a new appointment), and primary care (on the verge of collapse). Those of us who are baby boomers are on the leading edge of this demand and, in order to mitigate and minimize the cost impacts on our children, it is our challenge and responsibility to innovate and meet these challenges without passing along unnecessary burdens to our children and grandchildren.
## Aging & Future Medicine
For most of us, aging means more frequent and severe afflictions, from chronic ailments to more acute care needs and longer hospital stays. Medicine and medical care becomes an extremely important part of life as we age. Taking care of our health by improving diet, exercising, and maintaining an otherwise active lifestyle is essential to ensure a high quality life. Even with increased vigilance chronic ailments can affect us later in life, brought on both by our genetics and consequentially due to the lifestyles we’ve led in our youths. Ailments such as dementia, coronary artery disease, Alzheimer’s, myocardial infarction, congestive heart failure, macular degeneration, osteoporosis, hypertension, chronic obstructive pulmonary disease, diabetes, and others take their toll. Managing chronic diseases is costly from a logistical perspective in terms of time and money. However, even more to the point, effective and quality oversight of patients with chronic ailments requires regular review, screening, and monitoring of patients. This is further complicated by the need to serve patients who lack the means or are physically incapable of leaving their homes for extended periods. Telehealth and remote monitoring are a means by which a case manager—an individual assigned to oversee the care of chronically ill patients within a home-health setting—can review patient information on a regular basis (for example, daily) and support both the patient and the primary care provider. Furthermore, Intensive care units and emergency departments are becoming more crowded. Healthcare information technology is helping to ameliorate the challenges through improved workflow and more reliable & dense information. Individuals with insurance are going to EDs because they cannot find satisfaction in terms of prompt scheduling with their gatekeepers (family practitioners). The quantity of individuals with chronic ailments is on the rise (stroke, CHF, diabetes, COPD, etc.) This is in part due to the fact that people are living longer. At the same time the Medicare and SS systems will not be able to sustain the growth in population over age 65. This means that working individuals will increasingly bear the financial burden for us “boomers.” As a result of increased longevity and the fiscal challenges, the retirement age will increase.
## Where does healthcare & medicine go? To the home!
So, what do we do? Well, several things: first, technology in the form of remote data collection, reporting devices and software will become more prevalent: glucometers, BP cuffs, spirometers and associated software will be more readily available for direct communication with personalized electronic health records. If the purpose of a typical visit is to take BP and diabetic assessments, this can be handled most by collecting data at the point of care (home) and transmitting to the physician’s office for assessment. Such also applies to nursing and assisted living facilities. Next, the technical infrastructure required to transmit and store these data will be required. Paying for this infrastructure could come from a number of sources. One possibility: most everyone nowadays has access to cable television. Cable companies could offer devices that integrate with existing modems to collect and transmit data to the FP, together with complementary emails to next of kin (e.g. “Your mother’s BP as of 8:10 this morning was 145/89”). Other technologies that can be used to evaluate and monitor chronic ailments such as macular degeneration can further reduce costs by providing video cameras at point of care whereby opthalmologists can review retinal changes without requiring an elderly individual to be transported at expense and time to a hospital or office. In addition, support for remote consults via VoIP and video can be supported over the same network. This empowers the remote provider with the ability to interact with the patient All of these technologies are in use in remote pockets around the world today. But, they will become more prevalent. These technology implementations will reduce costs and provide for more personalized care in comfortable settings (homes). Of course, nothing takes the place of the tactile hands-on. But, for routine visits the above will be invaluable. In terms of the software technologies, personalized medicine will become the norm (eventually). Telehealth will be key. But, also, support for automated workflow in the acute care environment will need to be augmented. This means fully integrating all data into the enterprise HIS.
The U.S. Department of Health and Human Services through its Office of the National Coordinator for Health Information Technology, published operational scenarios focused on providing key information to assist in harmonizing standards on the implementation, certification, and policy implications for robust remote patient monitoring [4]. Included in this assessment are requirements on interacting with personalized health records and enterprise health information systems. The approaches to advancing remote monitoring include both seamless communication from medical devices at the point of care (i.e., in a patient’s home setting) and with a case manager and primary care provider both through electronic transfer, storage, and display of health information and remote video and audio interaction with patients in the same home health setting.
Healthcare Information Technology is not the silver bullet, but those described above are key enablers for remote health monitoring. Of course, the use of technology carries with it the implication that sufficient underlying infrastructure exists. This is not always the case in remote areas of the country. Satellite, cable, and fiber optic technologies are fairly extensive within the continental United States, but pockets and regions exist in which this is not the case. Therefore, a combined effort to extend the communications infrastructure must continue together with a unified effort to standardize and train and “in-service” individual care providers on these technologies must occur. One of the best mechanisms for enabling this is through the local hospitals and their satellite clinics.
So, how long do we have? Well, the sooner the better. Successful telehealth and remote monitoring programs exist throughout the United States and worldwide today. We should ensure that our elected representatives direct healthcare expenditures towards several specific areas to promote growth and alignment to meet the objectives of remote monitoring. These include continuing alignment on electronic personalized health records, expansion of our underlying communications infrastructure, and promoting common standards of communication among these records so that, regardless of location, a patient can communicate his or her information to any physician and allied health professional within the country. In summary: healthcare information technology employing common storage, homogeneous communication, standardized formats is necessary to provide the type of support necessary for the future of medicine.
References
[1] Source: Population Division, U.S. Census Bureau, August 14th, 2008; Table 12: “Projections of the population by Age and Sex for the United States: 2010 to 2050 (NP2008-T12)”
[2] Cinda Becker, “Slow: Budget Danger Ahead,” Modern Healthcare, March 3rd 2008. | 6,402 | 30,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-30 | longest | en | 0.894827 |
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# Class 10 Maths MCQs for Constructions
Welcome to your Class 10 Maths MCQs for Constructions
1.
Constructions
. To divide a line segment AB in the ratio 3:4, first, a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is:
2.
Constructions
To divide a line segment AB of length 7.6cm in the ratio 5:8, a ray AX is drawn first such that ∠BAX forms an acute angle and then points A1, A2, A3, ….are located at equal distances on the ray AX and the point B is joined to:
3.
Constructions
To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is:
4.
Constructions
To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1, A2, A3, ....are located at equal distances on the ray AX and the point B is joined to
5.
Constructions
To construct a triangle similar to a given ΔPQR with its sides 5/8 of the similar sides of ΔPQR, draw a ray QX such that ∠QRX is an acute angle and X lies on the opposite side of P with respect to QR. Then locate points Q1, Q2, Q3, … on QX at equal distances, and the next step is to join:
6.
Constructions
To construct a triangle similar to a given ΔPQR with its sides, 9/5 of the corresponding sides of ΔPQR draw a ray QX such that ∠QRX is an acute angle and X is on the opposite side of P with respect to QR. The minimum number of points to be located at equal distances on ray QX is:
7.
Constructions
To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A1, A2, A3, ... and B1, B2, B3, ... are located at equal distances on ray AX and BY, respectively. Then the points joined are
8.
Constructions
To construct a triangle similar to a given ΔABC with its sides 3/7 of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1, B2, B3, ... on BX at equal distances and next step is to join:
9.
Constructions
To construct a pair of tangents to a circle at an angle of 60° to each other, it is needed to draw tangents at endpoints of those two radii of the circle, the angle between them should be
10.
Constructions
To divide a line segment PQ in the ratio m:n, where m and n are two positive integers, draw a ray PX so that ∠PQX is an acute angle and then mark points on ray PX at equal distances such that the minimum number of these points is:
11.
Constructions
To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is:
12.
Constructions
To draw a pair of tangents to a circle which are inclined to each other at an angle of 45°, it is required to draw tangents at the endpoints of those two radii of the circle, the angle between which is:
13.
Constructions
To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:
14.
Constructions
To divide a line segment AB in the ratio p : q (p, q are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is
15.
Constructions
8. A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of ___________ from the centre. | 1,016 | 3,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.938768 |
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test/384085: Interest Formulas If you invest \$P, and you earn interest only on the amount you invested (the \$P), then ________________ interest is earned. If you invest \$P and you earn interest not only on the \$P, but also on the interest gained, then ______________ interest is earned. Let P = principal, r = annual nominal interest rate (in decimal form), t = time of the investment (in years), A = amount in the account after t years. THEN… Simple interest: A = _________________________________________ Compound interest: A = _____________________________________ Ideally, what do you want to happen to the value of n in the compound interest formula if the formula represented the amount of money you have in the bank?1 solutions Answer 271873 by Theo(3464) on 2010-12-10 08:05:20 (Show Source): You can put this solution on YOUR website!If you invest \$P, and you earn interest only on the amount you invested (the \$P), then SIMPLE interest is earned. If you invest \$P and you earn interest not only on the \$P, but also on the interest gained, then COMPOUND interest is earned. Let P = principal, r = annual nominal interest rate (in decimal form), t = time of the investment (in years), A = amount in the account after t years. THEN… SIMPLE INTEREST A = P + (P * r * t) Example: P = 10,000 r = .10 t = 5 years A = 10,000 + (10,000 * .10 * 5) = 10,000 + 5,000 = 15,000 COMPOUND INTEREST A = P * (1+r)^t Example: P = 10,000 r = .10 t = 5 years A = 10,000 * (1.10)^5 = 10,000 * 1.61051 = 16,1051 You make more money with compound interest than you do with simple interest, assuming the same investment in the same time frame with the same rate of return. LAST QUESTION Ideally, what do you want to happen to the value of n in the compound interest formula if the formula represented the amount of money you have in the bank? I don't see "n" anywhere in your question. If "n" is the number of compounding periods per year, than, ideally, you would want n to be as large as possible. If n = 1, then the number of compounding periods per year is 1. If n = 12, then the number of compounding periods per year is 12. When n is greater than 1, you multiply the number of years by n and you divide the nominal interest rate by n. The nominal interest rate is the annual interest assuming only 1 compounding period per year. In the compounding example above, n was equal to 1 (not shown in the formula. with n in the formula, the formula becomes: A = P * (1 + (r/n))^(t*n) When n was equal to 1, you got: A = P * (1 + r) ^ n That's what we solved earlier. When n is 12, the formula becomes: A = P * (1 + (r/12))^(t*12) Same compounding example but with number of compounding periods per year equal to 12. P = 10,000 r = .1 t = 5 r/12 = .1/12 = .008333333333 t*12 = 5*12 = 60 formula becomes: A = 10,000 * (1.0083333333) ^ 60 = 16453.08934 with monthly compounding, the future value is 16,453.08934 with yearly compounding, the future value is 16,16,105.1 Your savings are greater when the number of compounding periods per year are greater. The most compounding periods per year you can get is with continuous compounding. That formula is A = P * e^rt) e is the well known scientific constant of 2.718281828..... Here you use the annual interest rate and the number of years. You get: A = 10,000 * e^(.1*5) = 10,000 * e^.5 = 16,487.21271 That's the theoretical maximum number of compounding periods per year. If you compounded daily, you would get close to that. Assume daily compounding with 365 days per year. P = 10,000 i = .1 / 365 = .000273973 t = 5 * 365 = 1825 A = P * (1.000273973)^1825 = 16486.08362 That's pretty close to the theoretical maximum, but not exactly. Bottom Line is the more the compounding periods per year, the greater your savings will be. Without knowing what you meant by "n", this is the best I can come up with based on what I know about compounding. I can't think of what else "n" can be, given the formulas as presented.
Polynomials-and-rational-expressions/384076: I don't understand this part. Where do you use the common factor in an expression consisting of various terms? The actual question is What is a common factor? Where do you use the common factor in an expression consisting of various terms? What is the Greatest Common Factor? 1 solutions Answer 271867 by Theo(3464) on 2010-12-10 07:39:09 (Show Source): You can put this solution on YOUR website!A common factor is a factor that can be divided evenly into each of the terms of an expression. Take the expression 9 + 27 + 54. 3 can be divided evenly into 9 and into 27 and into 54, so 3 is a common factor. 9 can be divided evenly into 9 and into 27 and into 54, so 9 is also a common factor. The greatest common factor is 9 because that is the largest factor that can be divided evenly into each of the terms of the expression. You use the common factor in an expression consisting of various terms in order to simplify the expression so that it can be solved easier. Consider the expression (3x^3 - 17x^2 + 18x - 12) / (x^2 - 5x + 6) Consider that you know that the numerator in this expression is equal to (x^2 - 5x + 6) * (3x-2) The expression becomes ((3x-2) * (x^2 - 5x + 6)) / (x^2 - 5x + 6). the common factor of (x^2 - 5x + 6) cancels out and you are left with a result of (3x-2). Knowing what the common factor is made the division much simpler than not knowing what the common factor is.
Miscellaneous_Word_Problems/384063: I greatly NEED some help URGENTLY with this problem. i tried to do it forever but i dont get it so would somebody PLEASE HELP ME! I REALLY NEED A QUICK ANSWER. THANK YOU! Lesson 4—4 Two loan balances can be approximated by the equations given in the table. The time x is in years. i will separate things in table with a (/) Loan Amount / Interest rate / Years / Equation \$200,000 / 7% / 20 yrs / y= -9,734.7x + 218,761 \$250,000 / 6.25% / 15 yrs / y= -16,474x + 266,478 1)Use Matrices and the inverse matrix to find when the loan balances will be approximately the same for the two loans before the final payment is made. 2)What will the approximate balance of each loan be at that time? 3)Why might your estimate of the time when the loan balances will be the same be somewhat inaccurate? THANK YOU SO MUCH IN ADVANCE! 1 solutions Answer 271855 by Theo(3464) on 2010-12-10 04:48:56 (Show Source): You can put this solution on YOUR website!The 2 equations that approximate the solution to this are: y = -9734.7 * x + 218,761 y = -16474 * x + 266,478 If you graph both these equations, you will see when they cross. The graph of both of these equations is shown below: You can see from the graph that the approximate intersection point is somewhere around 7 years. You can calculate that by making the equations equal to each other. When you do that, you are saying that the value of y in the first equation is equal to the value of y in the second equation. you get: -9734.7x + 218761 = -16474 * x + 266,478 Solve this equation algebraically to get an answer of: x = 7.080408945 y = 149835.343 The approximation given by the graph is pretty close. x is the number of years when the balances will be approximately equal. y i the outstanding balance in each account at that time. The estimate will be somewhat inaccurate because the declining balance equation is not a straight line. What the equations show are probability the best fit of a straight line to the actual balances as they drop in time. The actual point where the balances are equal can be calculated through the use of Present Value of A Loan Amount Formulas, or through the use of a mechanized program that will do the calculations for each year of each loan. I used Excel to do that for you, after having calculated what the monthly payments would be using a financial calculator, and the results are shown below: ``` Remaining Balance year Loan1 Loan2 0 \$200,000.00 \$250,000.00 1 \$199,616.07 \$249,158.53 2 \$199,229.90 \$248,312.67 3 \$198,841.47 \$247,462.41 4 \$198,450.79 \$246,607.72 5 \$198,057.82 \$245,748.58 6 \$197,662.56 \$244,884.96 7 \$197,264.99 \$244,016.84 8 \$196,865.10 \$243,144.21 9 \$196,462.89 \$242,267.03 10 \$196,058.32 \$241,385.28 11 \$195,651.40 \$240,498.93 12 \$195,242.10 \$239,607.98 13 \$194,830.41 \$238,712.38 14 \$194,416.33 \$237,812.11 15 \$193,999.82 \$236,907.16 16 \$193,580.89 \$235,997.50 17 \$193,159.52 \$235,083.09 18 \$192,735.68 \$234,163.93 19 \$192,309.38 \$233,239.97 20 \$191,880.58 \$232,311.21 21 \$191,449.29 \$231,377.60 22 \$191,015.48 \$230,439.14 23 \$190,579.14 \$229,495.79 24 \$190,140.25 \$228,547.52 25 \$189,698.80 \$227,594.31 26 \$189,254.78 \$226,636.14 27 \$188,808.17 \$225,672.98 28 \$188,358.95 \$224,704.81 29 \$187,907.12 \$223,731.59 30 \$187,452.64 \$222,753.30 31 \$186,995.52 \$221,769.91 32 \$186,535.73 \$220,781.41 33 \$186,073.26 \$219,787.75 34 \$185,608.09 \$218,788.92 35 \$185,140.20 \$217,784.89 36 \$184,669.59 \$216,775.63 37 \$184,196.23 \$215,761.11 38 \$183,720.11 \$214,741.31 39 \$183,241.21 \$213,716.20 40 \$182,759.52 \$212,685.75 41 \$182,275.02 \$211,649.93 42 \$181,787.69 \$210,608.72 43 \$181,297.52 \$209,562.08 44 \$180,804.50 \$208,509.99 45 \$180,308.59 \$207,452.42 46 \$179,809.79 \$206,389.35 47 \$179,308.09 \$205,320.74 48 \$178,803.45 \$204,246.56 49 \$178,295.87 \$203,166.78 50 \$177,785.34 \$202,081.39 51 \$177,271.82 \$200,990.34 52 \$176,755.31 \$199,893.60 53 \$176,235.78 \$198,791.16 54 \$175,713.23 \$197,682.97 55 \$175,187.62 \$196,569.02 56 \$174,658.95 \$195,449.25 57 \$174,127.20 \$194,323.66 58 \$173,592.34 \$193,192.21 59 \$173,054.37 \$192,054.86 60 \$172,513.25 \$190,911.59 61 \$171,968.98 \$189,762.36 62 \$171,421.53 \$188,607.15 63 \$170,870.90 \$187,445.92 64 \$170,317.05 \$186,278.65 65 \$169,759.96 \$185,105.29 66 \$169,199.63 \$183,925.82 67 \$168,636.03 \$182,740.21 68 \$168,069.14 \$181,548.43 69 \$167,498.95 \$180,350.44 70 \$166,925.43 \$179,146.20 71 \$166,348.56 \$177,935.70 72 \$165,768.33 \$176,718.89 73 \$165,184.72 \$175,495.74 74 \$164,597.70 \$174,266.23 75 \$164,007.25 \$173,030.31 76 \$163,413.36 \$171,787.95 77 \$162,816.01 \$170,539.12 78 \$162,215.17 \$169,283.79 79 \$161,610.83 \$168,021.92 80 \$161,002.96 \$166,753.48 81 \$160,391.55 \$165,478.43 82 \$159,776.57 \$164,196.74 83 \$159,158.00 \$162,908.37 84 \$158,535.82 \$161,613.29 85 \$157,910.02 \$160,311.47 86 \$157,280.56 \$159,002.87 87 \$156,647.43 \$157,687.45 88 \$156,010.61 \$156,365.18 * 89 \$155,370.07 \$155,036.03 * 90 \$154,725.80 \$153,699.95 91 \$154,077.77 \$152,356.92 92 \$153,425.96 \$151,006.88 93 \$152,770.35 \$149,649.82 94 \$152,110.91 \$148,285.69 95 \$151,447.63 \$146,914.45 96 \$150,780.47 \$145,536.08 97 \$150,109.43 \$144,150.52 98 \$149,434.47 \$142,757.75 99 \$148,755.57 \$141,357.72 100 \$148,072.71 \$139,950.40 101 \$147,385.87 \$138,535.75 102 \$146,695.03 \$137,113.73 103 \$146,000.15 \$135,684.31 104 \$145,301.22 \$134,247.44 105 \$144,598.21 \$132,803.09 106 \$143,891.10 \$131,351.22 107 \$143,179.87 \$129,891.78 108 \$142,464.49 \$128,424.74 109 \$141,744.93 \$126,950.06 110 \$141,021.18 \$125,467.71 111 \$140,293.21 \$123,977.63 112 \$139,560.99 \$122,479.79 113 \$138,824.49 \$120,974.14 114 \$138,083.71 \$119,460.66 115 \$137,338.60 \$117,939.29 116 \$136,589.14 \$116,410.00 117 \$135,835.31 \$114,872.75 118 \$135,077.09 \$113,327.49 119 \$134,314.44 \$111,774.18 120 \$133,547.34 \$110,212.78 121 \$132,775.77 \$108,643.25 122 \$131,999.70 \$107,065.54 123 \$131,219.10 \$105,479.61 124 \$130,433.94 \$103,885.43 125 \$129,644.21 \$102,282.94 126 \$128,849.87 \$100,672.11 127 \$128,050.90 \$99,052.89 128 \$127,247.26 \$97,425.23 129 \$126,438.94 \$95,789.10 130 \$125,625.90 \$94,144.44 131 \$124,808.12 \$92,491.22 132 \$123,985.57 \$90,829.39 133 \$123,158.22 \$89,158.90 134 \$122,326.05 \$87,479.71 135 \$121,489.02 \$85,791.78 136 \$120,647.11 \$84,095.05 137 \$119,800.29 \$82,389.49 138 \$118,948.52 \$80,675.05 139 \$118,091.79 \$78,951.67 140 \$117,230.06 \$77,219.32 141 \$116,363.31 \$75,477.95 142 \$115,491.49 \$73,727.50 143 \$114,614.60 \$71,967.94 144 \$113,732.58 \$70,199.22 145 \$112,845.43 \$68,421.28 146 \$111,953.09 \$66,634.09 147 \$111,055.56 \$64,837.58 148 \$110,152.78 \$63,031.72 149 \$109,244.74 \$61,216.45 150 \$108,331.40 \$59,391.73 151 \$107,412.74 \$57,557.51 152 \$106,488.72 \$55,713.73 153 \$105,559.30 \$53,860.35 154 \$104,624.47 \$51,997.31 155 \$103,684.18 \$50,124.58 156 \$102,738.41 \$48,242.08 157 \$101,787.11 \$46,349.79 158 \$100,830.27 \$44,447.64 159 \$99,867.85 \$42,535.58 160 \$98,899.82 \$40,613.56 161 \$97,926.14 \$38,681.53 162 \$96,946.77 \$36,739.44 163 \$95,961.70 \$34,787.23 164 \$94,970.88 \$32,824.86 165 \$93,974.28 \$30,852.27 166 \$92,971.86 \$28,869.40 167 \$91,963.60 \$26,876.20 168 \$90,949.46 \$24,872.63 169 \$89,929.40 \$22,858.61 170 \$88,903.39 \$20,834.11 171 \$87,871.39 \$18,799.06 172 \$86,833.38 \$16,753.42 173 \$85,789.31 \$14,697.12 174 \$84,739.15 \$12,630.11 175 \$83,682.86 \$10,552.33 176 \$82,620.41 \$8,463.74 177 \$81,551.77 \$6,364.26 178 \$80,476.89 \$4,253.85 179 \$79,395.74 \$2,132.45 180 \$78,308.28 (\$0.00) 181 \$77,214.48 182 \$76,114.30 183 \$75,007.71 184 \$73,894.65 185 \$72,775.11 186 \$71,649.03 187 \$70,516.39 188 \$69,377.13 189 \$68,231.24 190 \$67,078.65 191 \$65,919.35 192 \$64,753.28 193 \$63,580.41 194 \$62,400.70 195 \$61,214.10 196 \$60,020.59 197 \$58,820.11 198 \$57,612.63 199 \$56,398.10 200 \$55,176.50 201 \$53,947.76 202 \$52,711.86 203 \$51,468.75 204 \$50,218.38 205 \$48,960.73 206 \$47,695.73 207 \$46,423.36 208 \$45,143.56 209 \$43,856.30 210 \$42,561.53 211 \$41,259.21 212 \$39,949.29 213 \$38,631.73 214 \$37,306.49 215 \$35,973.51 216 \$34,632.76 217 \$33,284.18 218 \$31,927.74 219 \$30,563.39 220 \$29,191.08 221 \$27,810.76 222 \$26,422.39 223 \$25,025.93 224 \$23,621.31 225 \$22,208.51 226 \$20,787.46 227 \$19,358.12 228 \$17,920.45 229 \$16,474.38 230 \$15,019.89 231 \$13,556.90 232 \$12,085.39 233 \$10,605.29 234 \$9,116.56 235 \$7,619.14 236 \$6,112.98 237 \$4,598.05 238 \$3,074.27 239 \$1,541.61 240 \$0.00 ``` The asterisks (*) show you that the balances become equal somewhere between the 88th and 89th month. This would be somewhere between 7.33333333 years and 7.41666666667 years. The actual graph of the declining balance is not a straight line. It drops slowly in the early years and drops a lot faster in the later years. You can link to the following webiter and scroll down to the bottom and a graph of a remaining balance on a loan will show up. You can see that it is not a straight line. http://www.tvmcalcs.com/calculators/apps/excel_loan_amortization
Polynomials-and-rational-expressions/384066: SIMPLIFY the expression: (1/x^2+6x+8)−(1/x^2−2x−8) and give your answer in the form of f(x)/g(x) My solution was (1/(x-4)(x+4)), but it was incorrect. 1 solutions Answer 271843 by Theo(3464) on 2010-12-10 03:51:49 (Show Source): You can put this solution on YOUR website!The solution that I came up with is: I checked the solution out by assuming x was equal to 2 and solving both the original expression and the final expression. They came out to the same answer leading to the conclusion that I simplified correctly. My manual calculations are in the picture shown below: first I factored. then I found common denominator. then I combined numerator under same denominator. then I performed calculations on numerator. result was final answer.
Pythagorean-theorem/384062: if the legs of a right angle are 9 and 17 what is the hypotenuse?1 solutions Answer 271840 by Theo(3464) on 2010-12-10 03:27:45 (Show Source): You can put this solution on YOUR website!c = hypotenuse a = the shorter leg b = the longer leg formula is: c^2 = a^2 + b^2 becomes: c^2 = 9^2 + 17^2 = 370 c = sqrt(370) = 19.23538406
logarithm/383497: A savings bond will pay \$5,000 at maturity 15 years from now. How much should you be willing to pay for the note now if money is worth 4.11% compounded semiannually?1 solutions Answer 271535 by Theo(3464) on 2010-12-09 06:25:19 (Show Source): You can put this solution on YOUR website!will pay \$5000 in 15 years. interest rate per year is 4.11% compounded semiannual. formula to use is f = p * (1+i)^n your time periods need to be semi-years. because of that, you need to divide your interest rate by 2 and you need to multiply your years by 2. in your formula: f = 5000 p = what you want to find i = .0411 / 2 = .02055 n = 15 * 2 = 30 your formula becomes: 5000 = p * (1.02055)^30 your answer should be that you are willing to pay \$2,716.072716 for the savings bond. Let's see if that works. 1.02055^30 = 1.840893276 your formula becomes 5000 = p * 1.840893276 divide both sides of this equation by 1.840893276 and you get: p = 5000 / 1.840893276 = \$2,716.072716 We're good.
Miscellaneous_Word_Problems/383501: I hope someone can help me. I am lost. I am to reduce the rational expression to the lowest terms. And assume that the variables represent only numbers which the denominators are nonzero. I do not undertand please to step by step. a^2-b^2/a-b Thank you so much!1 solutions Answer 271534 by Theo(3464) on 2010-12-09 06:18:45 (Show Source): You can put this solution on YOUR website!problem is: (a^2 - b^2)/(a-b) (a^2 - b^2) can be factored to be (a-b) * (a+b) your expression becomes: ((a-b) * (a+b)) / (a-b) The (a-b) in the numerator and denominator cancel out and you are left with: 1 * (a+b) / 1 = (a+b) you had to know that (a^2 - b^2) is equivalent to (a+b) * (a-b). when you take (a+b) and multiply it by (a-b), you get: a^2 + ab - ab - b^2 which becomes a^2 - b^2 because the +ab and -ab cancel out.
Miscellaneous_Word_Problems/383500: could someone help me work this out step by step so I can understand it. I am lost.I am suppose to build up the rational expression into an equivalent rational expression with the indicated denominator. Please help. Thank you -7yt/3x=?/18xyt 1 solutions Answer 271533 by Theo(3464) on 2010-12-09 06:12:48 (Show Source): You can put this solution on YOUR website!-7yt/3x=?/18xyt it helps to separate what is being multiplied by each other by *. it also helps to surround operations that are going to be performed together by parentheses. your formula becomes: (-7*y*t) / (3*x) = ? / (18*x*y*t) multiply both sides of this equation by (18*x*y*t) you get: (18*x*y*t) * ((-7*y*t) / (3*x)) = ? since a * (b/c) = (a*b) / c, then your expression above is equivalent to: ((18*x*y*t) * (-7*y*t)) / (3*x) = ? (18*x*y*t) * (-7*y*t) is equal to (-126 * x * y^2 * t^2) this is because 18 * -7 = -126 and y * y = y^2 and t * t = t^2. (-126 * x * y^2 * t^3) / (3 * x) is equal to (-42 * y^2 * t^2) this is because -126/3 = -42 and x/x = 1 which is not shown but implied. your answer should be ? = (-42 * y^2 * t^2) you can confirm it by assigning values to each of the variables at random and seeing if the final expression is true after simplification. Assume x = 2, y = 3, t = 4 your original expression of (-7*y*t) / (3*x) = ? / (18*x*y*t) becomes (-7*y*t) / (3*x) = (-42 * y^2 * t^2) / (18*x*y*t) which becomes (-7*3*4) / (3*2) = (-42 * 3^2 * 4^2) / (18*2*3*4) this resolves to -84/6 = -6048 / 432 which resolves finally to -14 = -14 confirming that we did the calculations correctly.
Age_Word_Problems/383490: Please help me solve this word problem: Formulas: y=be^rt; A=P(1+(r/n)^nt George has \$65 to invest. a. The bank gives him 8.2% continuous interest, how long will it take for George to accrue \$100? b. At 7.6% interest compound monthly, how much money will George have in 12 years?1 solutions Answer 271527 by Theo(3464) on 2010-12-09 05:45:30 (Show Source): You can put this solution on YOUR website!y = b * e^rt sounds like this is the continuous compounding formula. george has \$65 to invest. at 8.2% continuous interest, how long will it take for george to accrue \$100? y = \$100 b = \$65 r = .082 t = time in years formula is \$100 = \$65 * e^(.082*t) divide both sides of this formula by 65 to get: 100/65 = e^(.082*t) take log of both sides to get: log(100/65) = log(e^(.082*t)) by laws of logarithms, log(x^y) = y*log(x). formula becomes log(100/65) = .082*t * log(e) divide both sides of equation by log(e) to get: log(100/65) / log(e) = .082*t divide both sides of this equation by .082 to get: t = (log(100/65) / log(e)) / .082 log(100/65) = .187086643 log(e) = log(2.718281828) = .434294482 log(100/65) / log(e) = .430782916 .430782916 / .082 = 5.253450196 you get t = (log(100/65) / log(e)) / .082 = 5.253450196. If we did this right, then t = 5.253450196 years your original formula is: 100 = 65 * e^(.082*t) this becomes: 100 = 65 * e^(.082*5.253450196) e represents the scientific constant of 2.718281828 100 = 65 * e^(.082*5.253450196) becomes: 100 = 65 * 2.718182818^(.082*5.253450196). Use your calculator to see that 100 = 100, making t = 5.253450196 correct. -------------------------------- second problem. At 7.6% interest compound monthly, how much money will George have in 12 years? Formula they show is: A=P(1+(r/n)^nt A = future value you want to find. P = \$65.00 r = .076 / 12 = .006333333 t = 12 * 12 = 144 the n they are showing is equal to 12. to compound monthly you take the number of years and multiply by 12 and you take the annual interest rate and divide it by 12. formula becomes: A = 65 * (1 + (.076/12))^(12*12) this becomes: A = 65 * (1.00633333)^144 = 161.3395912. In 12 years, george will have that much.
Inequalities/383493: PLEASE HELP! I have solved this one but something looks wrong. Determine whether these numbers are solutions of the inequality : x-1 (greater then or equal to) 7; -5,1,4,20 I substituted -5,1,4,20 for x in the inequality and solved accordingly. I found there was NO solution for any of the numbers. Am I wrong and thank you.1 solutions Answer 271526 by Theo(3464) on 2010-12-09 05:29:30 (Show Source): You can put this solution on YOUR website!if x is -5, 1, or 4, then x-1 >= 7 is false. if x is 20, then x-1 >= 7 is true, because 19 >= 7 -5-1 = -6 1-1 = 0 4-1 = 3 20-1 = 19
Linear-equations/380971: Good Sunday Morning! I am trying to find the slope-intercept form of the equation of the line through the point (-1, 2), parallel to the line -4x- 9y = -7. Thanks for the help! Denise 1 solutions Answer 270346 by Theo(3464) on 2010-12-05 09:36:43 (Show Source): You can put this solution on YOUR website!If the line is parallel then it has the same slope. The line it is parallel to is the line given by the equation -4x - 9y = -7 convert that line to the slope intercept form by solving for y. you will get y = (-4/9)*x + (7/9) The slope of that line is equal to (-4/9) That will be the slope of your new line. The general equation for the slope intercept form is y = mx + b where m is the slope and b is the y intercept. Since you have the slope, the general form becomes y = (-4/9)*x + b To find the y intercept, you substitute the point values that the line passes through. The point values that you have that the line passes through are (-1,2). That means that x = -1 and y = 2. Substitute in the equation of y = (-4/9)*x + b to get: 2 = (-4/9)*(-1) * b Simplify to get: 2 = (4/9) * b Solve for b to get: b = 2 - (4/9) = (18/9) - (4/9) = (14/9) The equation of your parallel line should be: y = (-4/9)*x + (14/9) graph both equations and you will see that they are parallel. To prove that the slope intercept form of your original equation is good, you need to substitute the (x,y) values derived from the slope intercept form of the equation and see if the original equation is true. Your original equation is -4x-9y=-7. The slope intercept form of this equation is y = (-4/9)*x + (7/9) If we let x = 18, then y = (-4/9)*18 + (7/9) = -7.22222222 If we let x = 18 and y = -7.22222222, then -4x - 9y = -7 becomes -72 + 65 = -7 the translation from the standard form of the equation (-4x-9y=-7) to the slope intercept form of the equation ( y = (-4/9)*x + (7/9)) is good. The line parallel to it is therefore good because we calculated that correctly based on the slope intercept form of the original equation.
Evaluation_Word_Problems/380967: A student receives his grade report from a local community college, but the GPA is smudged. He took the following classes: a 2 hour credit art, a 3 hour credit history, a 4 hour credit science course, a 3 hour credit mathematics course, and a 1 hour science lab. He received a B in the art class, an A in the history class, a C in the science class, a B in the math class, and an A in the science lab. What was his GPA if the letter grades are based on a 4 point scale? (A = 4, B = 3, C = 2, D = 1, F = 0)1 solutions Answer 270345 by Theo(3464) on 2010-12-05 09:08:17 (Show Source): You can put this solution on YOUR website!A = 4, B = 3, C = 2, D = 1, F = 0 The following chart shows you how the calculation is set up. ```Class Hours Grade Grade Points Grade Point Hours Art 2 B 3 6 History 3 A 4 12 Science 4 C 2 8 Mathematics 3 B 3 9 Science Lab 1 A 4 4 ``` A student receives his grade report from a local community college, but the GPA is smudged. He took the following classes: a 2 hour credit art, a 3 hour credit history, a 4 hour credit science course, a 3 hour credit mathematics course, and a 1 hour science lab. He received a B in the art class, an A in the history class, a C in the science class, a B in the math class, and an A in the science lab. What was his GPA if the letter grades are based on a 4 point scale? (A = 4, B = 3, C = 2, D = 1, F = 0) The total grade point hours are equal to 39 The total hours are equal to 13 The grade point average = 39/13 = 3 This equates to an overall average of B. You multiply the grade points by the hours to get the grade point hours. The total grade point hours divided by the total hours equals the grade point average.
Triangles/380867: hey i would like to know how to find the area of a geometric figure 10 yd 13yd1 solutions Answer 270342 by Theo(3464) on 2010-12-05 08:38:19 (Show Source): You can put this solution on YOUR website!I presume the figure is a rectangle? It has length of 10 yards and a width of 13 yards? The area would be length * width = 10 * 13 = 130 square yards. You posted this under triangles? If this figure is a triangle, then we need to make some assumptions. The area of a triangle is 1/2 * b * h where b is the base and h is the height. If you assume that the base is 10 yards and the height is 13 yards, then the area would be 1/2 * 10 * 13 = 65 square yards. If you assume that the base is 13 yards and the height is 10 yards, you would get the same area equal to 1/2 * 13 * 10 = 65 square yards.
Polynomials-and-rational-expressions/380855: Worker efficiency. In a study of worker efficiency at Wong Laboratories it was found that the number of components assembled per hour by the average worker t hours after starting work could be modeled by the formula N(t) = 3t^3 + 23t^2 + 8t a) Rewrite the formula by factoring the right-hand side completely. b) Use the factored version of the formula to find N(3). c) What is the time at which the workers are most efficient. d) What is the maximum number of components assembled per hour during an 8-hour shift.1 solutions Answer 270341 by Theo(3464) on 2010-12-05 08:31:26 (Show Source): You can put this solution on YOUR website!the formula you gave is N(t) = 3t^3 + 23t^2 + 8t I'll make x = t so this can be graphed and referenced easily. the equation becomes n(x) = 3x^3 + 23x^2 + 8x the graph of this formula is shown below: the formula as given doesn't make any sense. there is no maximum when x is greater than 0. x cannot be less than 0 so any values on the graph where x is less than 0 are to be ignored. your graph would make more sense if the equation was: N(t) = -3t^3 + 23t^2 + 8t which I would translate to: n(x) = -3x^3 + 23x^2 + 8x. that graph is shown below: now you have a maximum when x is greater than 0 which is more realistic. I'll solve for n(x) = -3x^3 + 23x^2 + 8x. x can be factored out to get: n(x) = x * (-3x^2 + 23x + 8) (-3x^2 + 23x + 8) factors out to be (3x+1) * (-x+8) the completely factored equation becomes: n(x) = x * (3x+1) * (-x+8) when x = 0, n(x) = 0 as it should because no time worked = no components produced. when x = 3, n(x) = (3) * (10) * (5) = 150 using the factored equation. when x = 3, n(x) = -3(3^3) + 23*(3^2) + 8*3 = -81 + 207 + 24 = 150 you get the same answer with the factored equation and the non factored equation which confirms that the factorization is good. I believe we need to use calculus to find the maximum / minimum point of this cubic equation. I couldn't find any formula not using calculus on the web, so we'll use calculus. The maximum point occurs when the derivative of the equation equals 0. the equation is -3x^3 + 23x^2 + 8x The derivative of this equation is -9x^2 + 46x + 8 This derivative is 0 when x = 5.279477927 That could not be factored easily, so I used the quadratic formula of: x = (-b +/- sqrt(b^2-4ac))/(2a) When x = 5.279477927, n(x) = 241.8493507 I'll add a line at n(x) = 241.8493507 to the graph of the equation and it should confirm that is the maximum point. the graph looks like this: The graph confirms the maximum point. Since x represents t which represents the number of hours after starting work, then the maximum amount of output per hour occurs 5.279477927 hours after work starts. The maximum amount of output per hour during the 8 hour shift becomes 241.8493507 units per hour which occurs at that time. Best I can do in the short period of time allotted to the solving of this problem. I'm not a calculus guru, but I know a little bit about it, and the little bit I know was helpful in this case. If you were not to use calculus, then I don't really know how to find the maximum point of the cubic equation other than by iteration.
Linear_Equations_And_Systems_Word_Problems/380854: When soft drinks sold for \$0.80 per can at football games, approximately 6500 cans were sold. When the price was raised to \$1.00 a can, the demand dropped to 4000. Assume that the relationship between the price p and the demand y is linear. Write a linear function giving the demand y as a function of p.1 solutions Answer 270337 by Theo(3464) on 2010-12-05 07:12:17 (Show Source): You can put this solution on YOUR website!let y = number of cans sold. let x = price per can. when x = .8, y = 6500 when x = 1, y = 4000 you have 2 points from which you can generate a linear equation. the point slope form of the linear equation is y = mx + b. m is the slope and b is the y intercept. the general form of the slope of the line is (y2-y1)/(x2-x1) you have 2 points. they are: (.8,6500) and (1,4000) x1 = .8 y1 = 6500 x2 = 1 y2 = 4000 your slope is (y2-y1)/(x2-x1) which becomes (4000-6500) / (1-.8). this becomes -2500 / .2 which becomes -12500. that's your slope and the slope intercept form of your equation becomes: y = -12500*x + b to find the y intercept, substitute one of the points for y and x. we'll use (.8,6500) the equation becomes: 6500 = -12500*(.8) + b that becomes 6500 = -10000 + b solve for b to get b = 16500 your equation becomes: y = -12500*x + 16500 when x = .8, y becomes 6500 when x = 1, y becomes 4000 The slope is -12500 the y intercept is 16500 the equation is y = -12500*x + 16500 graph this equation and it looks like this: when x = 0, y = 16500 when x = .8, y = 6500 when x = 1, y = 4000 It's hard to see from the graph, but if you plot real careful the intersection of x = 1 and y = 4000, you will see that they intersect on the line of y = -12500*x + 16500. The more exact way is to simply solve the equation of y = -12500*x + 16500 when x = .8 you get y = -12500*(.8) + 16500 which becomes y = -10000 + 16500 which becomes y = 6500.
Money_Word_Problems/380958: A retailer purchased 70 reams of typewriter paper for \$9,730. At the end of the week, he found that there are still 13 reams left, the cost of which was \$1,807. Find the cost of goods sold.1 solutions Answer 270336 by Theo(3464) on 2010-12-05 06:44:11 (Show Source): You can put this solution on YOUR website!he bought 70 for 9730 at a cost of 9730/70 = 139 apiece. he had 13 left at a cost of 1807. cost of goods sold is 9730 - 1807 = 7923 number of goods sold is 70 - 13 = 57 57 * 139 = 7923 number confirm each other. 7923 is your answer.
Angles/380897: if m<1 is 75 what is the measure of its complementary angle1 solutions Answer 270334 by Theo(3464) on 2010-12-05 06:41:36 (Show Source): You can put this solution on YOUR website!the measure of an angle plus its complement is equal to 90 degrees. if the angle is 75 degrees, then its complement must be equal to 15 degrees because 75 + 15 = 90.
Exponential-and-logarithmic-functions/380911: The function f(x)=9x+1 is one-to-one. Find the inverse1 solutions Answer 270333 by Theo(3464) on 2010-12-05 06:38:53 (Show Source): You can put this solution on YOUR website!f(x) = 9x + 1 let y = f(x) you get y = 9x + 1 let y = x and x = y you get x = 9y + 1 solve for y to get y = (x-1)/9 that's your inverse function. if it is a true inverse function, it will be a reflection of the original function about the line y = x. to show that, then graph the functions y = x, y = 9x+1, y = (x-1)/9. that graph is shown below: you can pretty much eyeball it and see that the graph of y = (x-1)/9 is a reflection of y = 9x+1 about the line y = x. a more exact interpretation is that f(x,y) in the normal equation should be equal to f(y,x) in the inverse equation. how you find that out is as follows: let x = 5 in the normal equation. then you get y = 9x + 1 which becomes 46 in your normal equation, when x = 5, then y = 46 in your inverse equation, when x = 46, you should get y = 5. the y in the normal equation becomes the x in the inverse equation. the x in the normal equation becomes the y in the inverse equation. when x = 46, y = (x-1)/9 becomes y = (46-1)/9 becomes y = 45/9 becomes 5. f(x,y) = f(5,46) in your normal equation. f(y,x) = f(46,5) in your inverse equation. this proves they are inverse equations. the inverse equation undoes what the normal equation does. normal equation takes 5 and makes it 46. inverse equation takes 46 and makes it 5.
Exponential-and-logarithmic-functions/380916: write the expression (log base of b)(2y+5)-4(log base of b)(y+3) as a single logarithm1 solutions Answer 270329 by Theo(3464) on 2010-12-05 06:28:20 (Show Source): You can put this solution on YOUR website!that expression would be shown as: log(b,(2y+5)) - 4*log(b,(y+3)) in general x * log(y) = log(y^x) your expression becomes: log(b,(2y+5)) - log(b,((y+3)^4)) in general log(x) - log(y) = log(x/y) your expression becomes: log(b,((2y+5)/(y+3)^4)) to show you how this works, we will let b = 10 because your calculator can do logs to the base of 10 (usually called the LOG function). we will let y = 5 (chosen at random small enough to calculate easily). your original expression becomes: log(2y+5) - 4*log(y+3) the base of 10 is implied. substituting 5 for y, we get: log(2*5+5) - 4*log(5+3) which becomes: log(15) - 4*log(8) which becomes -2.436268689 looking at our final expression of: log(b,(2y+5)/(y+3)^4), we get: log((2y+5)/(y+3)^4). the base of 10 is implied. substituting 5 for y, we get: log(15/8^4) which becomes log(15/4096). using our calculator, we get log(15/4096) = log(.003662109) which equals -2.436268689 we get the same answer either way, so the translation is good, and the answer to your question is: log(b,2y+5) - 4*log(b,y+3) = log(b,(2y+5)/(y+3)^4) which looks like:
Exponential-and-logarithmic-functions/380915: Write the expression log2(6x/y) as a sum and/ or difference of logarithms1 solutions Answer 270327 by Theo(3464) on 2010-12-05 06:10:44 (Show Source): You can put this solution on YOUR website!log(2,6x/y) is equal to the log of 6x/y to the base of 2. it means the same as what you wrote as log2(6x/y). the notation, however, is more in line with the way the algebra.com formula generators work. using the algebra.com formula generator, log(2,6x/y) will show up as All you do is put 3 { in front of it and 3 } behind it. in general, log(x/y) = log(x) - log(y) your equation becomes: log(2,6x/y) = log(2,6x) - log(2,y) in general, log (x*y) = log(x) + log(y) your equation becomes: log(2,6x/y) = log(2,6) + log(2,x) - log(y) the concept is the same regardless of the base. if the base were 10, then it would be shown as: log(10,6x/y) = log(10,6) + log(10,x) - log(10,y) to show you how it works, we'll use log to the base 10 because your calculator can handle that. also, log(10,x) is normally shown as log(x). the base of 10 is implied. let's take log(6*15/30) this should be translated to log(6) + log(15) - log(30) which is the same treatment we provided above. using our calculator, we get log(6*15/30) = log(3) = .477121255 using our calculator again, we get log(6) + log(15) - log(30) = : .77815125 + 1.176091259 - 1.477121255 = .477121255 we get the same answer, as we should. same concepts works with any base, so the answer to your question is: log(2,6x/y) = log(2,6) + log(2,x) - log(2,y).
Exponential-and-logarithmic-functions/380907: For f(x)=8x+6 and g(x)= x squared find (g of f)(x)1 solutions Answer 270326 by Theo(3464) on 2010-12-05 05:57:20 (Show Source): You can put this solution on YOUR website!you are given: f(x) = 8x+6 g(x) = x^2 you want to find: g(f(x)) start with g(x) = x^2 replace x with f(x) to get g(f(x)) = (f(x))^2 replace f(x) with 8x+6 to get g(8x+6) = (8x+6)^2 solve by squaring 8x+6 to get g(f(x)) = 64x^2 + 96x + 36
Probability-and-statistics/380950: Investing is a game of chance. Suppose there is a 39% chance that a risky stock investment will end up in a total loss of your investment. Because the rewards are so high, you decide to invest in five independent risky stocks. Find the probability that at least one of your five investments becomes a total loss. I'm so confused! Thank you so much for your help! 1 solutions Answer 270309 by Theo(3464) on 2010-12-05 04:39:25 (Show Source): You can put this solution on YOUR website!probability of total loss of one risky investment = .39 you take 5 risky investments, each with a probability of total loss of .39. since the probability of total loss of each investment is .39, then the probability that the investment will not be a total loss is 1 - .39 = .61 the probability that none of the 5 investments will be a total loss is .61^5 = .08445963. the probability that at least one of these investments will be a total loss is 1 - .61^5. that probability would be .91554037
Equations/380936: How do I covert 1256kg to grams? I came up with 1256000 Is this correct? Thank you!1 solutions Answer 270306 by Theo(3464) on 2010-12-05 04:16:53 (Show Source): You can put this solution on YOUR website!to find the conversion factors, it's convenient to go to www.google.com and put in the conversion you are looking for. In your case, you want to convert from kilograms to grams. The google search string entered would be: kilograms to grams. the answer returned would be: 1 kilogram = 1000 grams If you want to convert from kilograms to grams, you would have to multiply the number of kilograms by 1000. If you start with 1256 kg, then your answer would bge 1,256,000 grams. You did good. | 13,668 | 40,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2013-20 | latest | en | 0.405419 |
https://1library.net/document/y81wdm0z-vanishing-power-values-commutators-derivations-prime-rings.html?utm_source=related_list | 1,606,643,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00058.warc.gz | 173,162,429 | 26,695 | # Vanishing Power Values of Commutators with Derivations on Prime Rings
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## Full text
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Volume 2009, Article ID 582181,8pages doi:10.1155/2009/582181
## Derivations on Prime Rings
### Basudeb Dhara
Department of Mathematics, Belda College, Belda, Paschim Medinipur 721424, India
Correspondence should be addressed to Basudeb Dhara,basu dhara@yahoo.com
Received 30 August 2009; Accepted 14 December 2009
Recommended by Howard Bell
LetRbe a prime ring of charR /2,da nonzero derivation ofRandρa nonzero right ideal ofR
such thatdx, xn,y, dym t
0 for allx, yρ, wheren≥0,m≥0,t≥1 are fixed integers. If ρ, ρρ /0, thendρρ0.
Copyrightq2009 Basudeb Dhara. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
### 1. Introduction
Throughout this paper, unless specifically stated,Ralways denotes a prime ring with center ZRand extended centroidC,Qthe Martindale quotients ring. Letnbe a positive integer. For given a, bR, let a, b0 a and let a, b1 be the usual commutator abba, and inductively forn >1,a, bn a, bn−1, b. Bydwe mean a nonzero derivation inR.
A well-known result proven by Posner1states that ifdx, x, y 0 for allx, yR, then R is commutative. In2, Lanski generalized this result of Posner to the Lie ideal. Lanski proved that ifUis a noncommutative Lie ideal ofRsuch thatdx, x, y 0 for allxU, yR, then eitherRis commutative or charR 2 andRsatisfiesS4, the standard identity in four variables. Bell and Martindale III3studied this identity for a semiprime ringR. They proved that ifRis a semiprime ring anddx, x, y 0 for allxin a non-zero left ideal ofRandyR, thenRcontains a non-zero central ideal. Clearly, this result says that ifRis a prime ring, thenRmust be commutative.
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In continuation of these previous results, it is natural to consider the situation when dx, xn,y, dymt0 for allx, yρ,n, m≥0, t≥1 are fixed integers. We have studied this identity in the present paper.
It is well known that any derivation of a prime ringRcan be uniquely extended to a derivation ofQ, and so any derivation ofRcan be defined on the whole ofQ. MoreoverQis a prime ring as well asRand the extended centroidCofRcoincides with the center ofQ. We refer to6,7for more details.
Denote byQCC{X, Y}the free product of theC-algebraQandC{X, Y}, the free
C-algebra in noncommuting indeterminatesX, Y.
### Prime Ring
We need the following lemma.
Lemma 2.1. Letρbe a non-zero right ideal ofRandda derivation ofR. Then the following conditions are equivalent:id is an inner derivation induced by somebQsuch thatbρ 0;iidρρ0
(for its proof refer to [8, Lemma]).
We mention an important result which will be used quite frequently as follows.
Theorem 2.2 see Kharchenko 9. Let R be a prime ring, d a derivation on R and I a non-zero ideal of R. If I satisfies the differential identity fr1, r2, . . . , rn, dr1, dr2, . . . , drn
0 for anyr1, r2, . . . , rnI,then eitheriIsatisfies the generalized polynomial identity
fr1, r2, . . . , rn, x1, x2, . . . , xn 0, 2.1
oriidisQ-inner, that is, for someqQ, dx q, xandI satisfies the generalized polynomial identity
fr1, r2, . . . , rn,
q, r1
,q, r2
, . . . ,q, rn
0. 2.2
Theorem 2.3. Let R be a prime ring of char R /2 and d a derivation of R such that
dx, xn,y, dymt 0 for all x, yR, where n ≥ 0, m ≥ 0, t ≥ 1 are fixed integers.
ThenRis commutative ord0.
Proof. LetRbe noncommutative. Ifdis notQ-inner, then by Kharchenko’s Theorem9
gx, y, u, v u, xn,y, vmt0, 2.3
for allx, y, u, vR. This is a polynomial identity and hence there exists a fieldF such that RMkFwithk >1,andRandMkFsatisfy the same polynomial identity10, Lemma
1. But by choosingue12, xe11, ve11andye21, we get
0u, xn,y, vmt −1tne11 −te22
, 2.4
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Now, letdbeQ-inner derivation, saydadafor someaQ, that is,dx a, x for allxR, then we have
a, xn1,y,a, ymt0, 2.5
for allx, yR. Sinced /0,a /Cand henceRsatisfies a nontrivial generalized polynomial identityGPI. By11, it follows thatRCis a primitive ring withHSocRC/0,andeHe is finite dimensional overCfor any minimal idempotenteRC. Moreover we may assume thatHis noncommutative; otherwise,Rmust be commutative which is a contradiction.
Notice thatHsatisfiesa, xn1,y,a, ymt 0see10, Proof of Theorem 1. For any idempotenteHandxH,we have
0 a, en1,ex1−e,a, ex1−emt. 2.6
Right multiplying bye, we get
0 a, en1,ex1−e,a, ex1−emte
a, en1,ex1−e,a, ex1−emt−1
· {a, en1ex1−e,a, ex1−emeex1−e,a, ex1−ema, en1e}
a, en1,ex1−e,a, ex1−emt−1
·
⎧ ⎨
a, en1 ⎛ ⎝m
j0 −1j
m
j
a, ex1−ejex1−ea, ex1−emj ⎞ ⎠e
⎛ ⎝m
j0 −1j
m
j
a, ex1−ejex1−ea, ex1−emj
a, en1e ⎫ ⎬ ⎭
a, en1,ex1−e,a, ex1−emt−1
·
⎧ ⎨ ⎩0−
⎛ ⎝m
j0 −1j
m
j
ex1−eajex1−eaex1−emj ⎞ ⎠ae
⎫ ⎬ ⎭
a, en1,ex1−e,a, ex1−emt−1 ⎛ ⎝m
j0
m
j
ex1−eam1 ⎞ ⎠e
−2ma, en1,ex1−e,a, ex1−emt−1ex1−eam1e
t2mtex1eam1te.
2.7
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Soacommutes with all idempotents inH. SinceHis a simple ring, either His generated by its idempotents orH does not contain any nontrivial idempotents. The first case gives aCcontradictingd /0. In the last case,His a finite dimensional division algebra overC. This implies thatH RC Q andaH. By 10, Lemma 2, there exists a field F such thatHMkFandMkFsatisfiesa, xn1,y,a, ym
t. Then by the same argument as
earlier,acommutes with all idempotents inMkF, again giving the contradictionaC, that
is,d0. This completes the proof of the theorem.
Theorem 2.4. LetRbe a prime ring of charR /2,da non-zero derivation ofRandρa non-zero right ideal ofRsuch thatdx, xn,y, dymt0 for allx, yρ, wheren≥0, m≥0, t≥1 are fixed
integers. Ifρ, ρρ /0, thendρρ0.
We begin the proof by proving the following lemma.
Lemma 2.5. Ifdρρ /0 anddx, xn,y, dymt0 for allx, yρ, m, n≥0, t≥1 are fixed
integers, thenRsatisfies nontrivial generalized polynomial identity (GPI).
Proof. Suppose on the contrary thatRdoes not satisfy any nontrivial GPI. We may assume thatRis noncommutative; otherwise,Rsatisfies trivially a nontrivial GPI. We consider two cases.
Case 1. Suppose that disQ-inner derivation induced by an elementaQ. Then for any xρ,
a, xXn1,xY,a, xYmt 2.8
is a GPI forR, so it is the zero element inQCC{X, Y}. Expanding this, we get
a, xXn1m
j0 −1j
m
j
a, xYjxYa, xYmj
m
j0 −1j
m
j
a, xYjxYa, xYmja, xXn1
AX, Y 0,
2.9
whereAX, Y a, xXn1,xY,a, xYmt−1. Ifax andx are linearlyC-independent for somexρ,then
axXn1m
j0 −1j
m
j
a, xYjxYa, xYmj
m
j0 −1j
m
j
axYjxYa, xYmja, xXn1
AX, Y 0.
(5)
Again, sinceaxandxare linearlyC-independent, above relation implies that
xYa, xYma, xXn1AX, Y 0, 2.11
and so
xYaxYmaxXn1AX, Y 0. 2.12
Repeating the same process yields
xYaxYmaxXn1t0 2.13
inQCC{X, Y}. This implies thatax 0, a contradiction. Thus for anyxρ,axandxare
C-dependent. Thenaαρ0 for someαC. Replacingawithaα, we may assume that 0. Then by Lemma2.1,dρρ0, contradiction.
Case 2. Suppose thatdis notQ-inner derivation. If for allxρ,dxxC, thendx, x 0 which implies thatR is commutative see13. Therefore there exists xρ such that dx/xC, that is,xanddxare linearlyC-independent.
By our assumption, we have thatRsatisfies
dxX, xXn,xY, dxYmt0. 2.14
By Kharchenko’s Theorem9,
dxXxr1, xXn,xY, dxYxr2mt0, 2.15
for allX, Y, r1, r2∈R. In particular forr1r2 0,
dxX, xXn,xY, dxYmt0, 2.16
which is a nontrivial GPI for R, because x and dx are linearly C-independent, a contradiction.
We are now ready to prove our main theorem.
Proof of Theorem2.4. Suppose thatdρρ /0,then we derive a contradiction. By Lemma2.5,R is a prime GPI ring, so is alsoQby14. SinceQis centrally closed overC, it follows from 11thatQis a primitive ring withHSocQ/0.
By our assumption and by7, we may assume that
(6)
is satisfied byρQand hence byρH. Letee2ρHandyH. Then replacingxwitheand ywithey1−ein2.17, then right multiplying it bye,we obtain that
0de, en,ey1−e, dey1−emte
de, en,ey1−e, dey1−emt−1
·
⎧ ⎨
de, en m
j0 −1j
m
j
dey1−ejey1−edey1−emje
m
j0 −1j
m
j
dey1−ejey1−edey1−emjde, ene ⎫ ⎬ ⎭.
2.18
Now we have the fact that for any idempotente,dy1−eey1−ede,edee0 and so
0de, en,ey1−e, dey1−emt−1
·
⎧ ⎨ ⎩0−
m
j0 −1j
m
j
ey1−edejy1−edey1−emjdee ⎫ ⎬ ⎭.
2.19
Now since for any idempotenteand for anyyR,1−edey 1−edey, above relation gives
0de, en,ey1−e, dey1−emt−1
·
⎧ ⎨ ⎩−e
m
j0
m
j
y1−edejy1−edey1−emjdee ⎫ ⎬ ⎭
de, en,ey1−e, dey1−emt−1 ⎧ ⎨ ⎩−e
m
j0
m
j
y1−edem1e ⎫ ⎬ ⎭
de, en,ey1−e, dey1−emt−1−2mey1−edem1e
−2mey1−edem1te.
2.20
(7)
ThenJ J/JlHJ,a primeC-algebra with the derivationdsuch thatdx dx, for all
xJ. By assumption, we have that
dx, x
n,
y, dy
m
t
0, 2.21
for all x, yJ. By Theorem 2.3, we have either d 0 or ρH is commutative. Therefore we have that either dρHρH 0 or ρH, ρHρH 0. Now dρHρH 0 implies that 0 dρρHρH dρρHρH and so dρρ 0. ρH, ρHρH 0 implies that 0 ρρH, ρHρH ρ, ρHρHρH and soρ, ρHρ 0,then 0 ρ, ρρHρ ρ, ρρHρ implying thatρ, ρρ 0. Thus in all the cases we have contradiction. This completes the proof of the theorem.
### Semiprime Ring
In this section we extend Theorem2.3to the semiprime case. LetRbe a semiprime ring and Ube its right Utumi quotient ring. It is well known that any derivation of a semiprime ring R can be uniquely extended to a derivation of its right Utumi quotient ringUand so any derivation ofRcan be defined on the whole ofU7, Lemma 2.
By the standard theory of orthogonal completions for semiprime rings, we have the following lemma.
Lemma 3.1see16, Lemma 1 and Theorem 1or7, pages 31-32. LetRbe a 2-torsion free semiprime ring and P a maximal ideal of C. Then P U is a prime ideal of U invariant under all derivations ofU. Moreover,{P U|P is a maximal ideal ofCwithU/P U 2-torsion free}0.
Theorem 3.2. LetRbe a 2-torsion free semiprime ring and da non-zero derivation ofRsuch that
dx, xn,y, dymt0 for allx, yR,n, m≥0, t≥1 fixed are integers. ThendmapsRinto its center.
Proof. Since any derivationdcan be uniquely extended to a derivation inU,and Rand U satisfy the same differential identities7, Theorem 3, we have
dx, xn,y, dymt0, 3.1
for allx, yU. LetPbe any maximal ideal ofCsuch thatU/P U is 2-torsion free. Then by Lemma3.1,P Uis a prime ideal ofUinvariant underd. SetU U/P U.Then derivationd canonically induces a derivationdonUdefined bydx dxfor allxU. Therefore,
dx, x
n,
y, dy
m
t
0, 3.2
(8)
{P U | P is a maximal ideal of CwithU/P U 2-torsion free} 0. ThusdUU, U 0. Without loss of generality, we havedRR, R 0. This implies that
0dR2R, R dRRR, R RdRR, R dRRR, R. 3.3
ThereforeR, dRRR, dR 0. By semiprimeness of R, we have R, dR 0, that is, dRZR. This completes the proof of the theorem.
### References
1 E. C. Posner, “Derivations in prime rings,” Proceedings of the American Mathematical Society, vol. 8, pp. 1093–1100, 1957.
2 C. Lanski, “Differential identities, Lie ideals, and Posner’s theorems,” Pacific Journal of Mathematics, vol. 134, no. 2, pp. 275–297, 1988.
3 H. E. Bell and W. S. Martindale III, “Centralizing mappings of semiprime rings,” Canadian
Mathematical Bulletin, vol. 30, no. 1, pp. 92–101, 1987.
4 I. N. Herstein, “A note on derivations,” Canadian Mathematical Bulletin, vol. 21, no. 3, pp. 369–370, 1978.
5 V. De Filippis, “On derivations and commutativity in prime rings,” International Journal of Mathematics
and Mathematical Sciences, no. 69–72, pp. 3859–3865, 2004.
6 K. I. Beidar, W. S. Martindale III, and A. V. Mikhalev, Rings with Generalized Identities, vol. 196 of
Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 1996.
7 T. K. Lee, “Semiprime rings with differential identities,” Bulletin of the Institute of Mathematics Academia
Sinica, vol. 20, no. 1, pp. 27–38, 1992.
8 M. Breˇsar, “One-sided ideals and derivations of prime rings,” Proceedings of the American Mathematical
Society, vol. 122, no. 4, pp. 979–983, 1994.
9 V. K. Kharchenko, “Differential identities of prime rings,” Algebra i Logika, vol. 17, no. 2, pp. 155–168, 1978.
10 C. Lanski, “An Engel condition with derivation,” Proceedings of the American Mathematical Society, vol. 118, no. 3, pp. 731–734, 1993.
11 W. S. Martindale III, “Prime rings satisfying a generalized polynomial identity,” Journal of Algebra, vol. 12, pp. 576–584, 1969.
12 I. N. Herstein, Topics in Ring Theory, The University of Chicago Press, , Chicago, Ill, USA, 1969. 13 H. E. Bell and Q. Deng, “On derivations and commutativity in semiprime rings,” Communications in
Algebra, vol. 23, no. 10, pp. 3705–3713, 1995.
14 C.-L. Chuang, “GPI’s having coefficients in Utumi quotient rings,” Proceedings of the American
Mathematical Society, vol. 103, no. 3, pp. 723–728, 1988.
15 C. Faith and Y. Utumi, “On a new proof of Litoff’s theorem,” Acta Mathematica Academiae Scientiarum
Hungaricae, vol. 14, pp. 369–371, 1963.
Updating...
Updating... | 4,773 | 12,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-50 | latest | en | 0.773021 |
https://books.google.com.jm/books?qtid=217e78c1&lr=&id=i-Y2AAAAMAAJ&sa=N&start=130 | 1,674,864,184,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00707.warc.gz | 161,812,117 | 5,983 | Books Books
When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Elements of Geometry and Conic Sections - Page 10
by Elias Loomis - 1857 - 226 pages
## Mathematical Cranks
Underwood Dudley - Mathematics - 1992 - 388 pages
...prefer to leave the term undefined — but others were just fine: When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to...
## The Poincaré Half-plane: A Gateway to Modern Geometry
Saul Stahl - Mathematics - 1993 - 320 pages
...conclusions, and that Euclid felt the need to acknowledge their efforts. 10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to...
## Mathematics: The Science of Patterns: The Search for Order in Life, Mind and ...
Keith Devlin - Mathematics - 1996 - 228 pages
...is a line that lies evenly with the points on itself. Definition 10 When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the equal angles is right and the straight line standing on the other is called a perpendicular to...
## Foundations and Fundamental Concepts of Mathematics
Howard Whitley Eves - Mathematics - 1997 - 370 pages
...are straight lines, the angle is called a rectilinear angle. 10. When a straight line erected on a straight line makes the adjacent angles equal to one another, each of the equal angles is called a right angle, and the straight line standing on the other is called a perpendicular...
## Geometry: Plane and Fancy
David A. Singer - Mathematics - 1998 - 176 pages
...review what we know about right angles. Euclid's tenth definition says: When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to...
## Space from Zeno to Einstein: Classic Readings with a Contemporary Commentary
Nick Huggett - Philosophy - 1999 - 292 pages
...containing the angle are straight, the angle is called rectilineal. 10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to...
## Mathematical Expeditions: Chronicles by the Explorers
Reinhard Laubenbacher, David Pengelley - Mathematics - 2000 - 292 pages
...parallel postulate itself. Euclid, from Elements BOOK I. DEFINITION 10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to...
## A Course in Modern Geometries
Judith Cederberg - Mathematics - 2004 - 472 pages
...containing the angle are straight, the angle is called rectilineal. 10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and Reprinted with permission of Cambridge University Press from The Thirteen... | 759 | 3,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-06 | latest | en | 0.869753 |
https://www.businessinsider.in/a-professor-tested-the-prisoners-dilemma-on-his-students-by-bribing-them-with-extra-credit-points/articleshow/48120046.cms | 1,716,521,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00600.warc.gz | 600,806,166 | 21,608 | A professor tested the 'Prisoner's Dilemma' on his students by bribing them with extra credit points
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3. A professor tested the 'Prisoner's Dilemma' on his students by bribing them with extra credit points
A professor tested the 'Prisoner's Dilemma' on his students by bribing them with extra credit points
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A professor at the University of Maryland threw his class a curveball when he allowed them to each choose how many extra credit points they wanted to add to their final paper grade.
Sounds generous right? Well, not really.
Here's what Professor Dylan Selterman positioned for his students:
Select whether you want 2 points or 6 points added onto your final paper grade. But there's a small catch: if more than 10% of the class selects 6 points, then no one gets any points.
Shahin Rafikian, a junior at the University, snapped a picture of the question and tweeted it. Now his tweet has over 7,000 retweets, mostly because people are agonizing over making the right choice.
Here's why they're having a hard time!
Selterman is demonstrating the "Prisoner's Dilemma and the Tragedy of the Commons," Buzzfeed News reports. It's concept in game theory in which a signaling problem inevitably leads to an unfavorable outcome.
Here's how it works: Two conspiring prisoners are arrested and questioned in separate rooms, with no way to communicate. If they both confess, they will be sentenced to two years in prison. If neither confess, they will both receive only one year in prison. The latter is obviously the best option, explains the Concise Encyclopedia of Economics.
The catch: If one confesses and the other doesn't, the prisoner who confesses will go free and the silent prisoner will serve a 3 year sentence. Thus presents the problem - the temptation of going free. Ideally, neither prisoner would confess, but by presenting the temptation, it's now more likely both will confess and both will end up serve a longer sentence.
In the case of Selterman's class, the students are the prisoners. All would benefit from two additional points on their final papers, but the majority will give in to the temptation of six additional points. As a result, no one will get extra credit.
He's been presenting the dilemma to students since 2008, but Selterman says only one class has successfully received an extra two points each.
It was not this year's class. | 534 | 2,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-22 | latest | en | 0.955339 |
http://mathhelpforum.com/calculus/140342-factorial-function-decreasing-print.html | 1,529,666,374,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864391.61/warc/CC-MAIN-20180622104200-20180622124200-00554.warc.gz | 199,564,588 | 3,010 | # Is the factorial function decreasing?
• Apr 20th 2010, 01:28 PM
DBA
Is the factorial function decreasing?
I need to know if the factorial function is decreasing.
This is the second condition for the Alternating Series Test.
The Alternating Series:
$\displaystyle \Sigma^{\infty}_{k = 1} (-1)^{k-1}\frac{k!}{(2k-1)!}$
I need to show that
$\displaystyle f(x) = \frac{x!}{(2x-1)!}$
decreasing.
Normally, I just take the first derivative but since I have a factorial I don't know.
Can someone let me know how I have to do that?
Thanks.
• Apr 20th 2010, 01:35 PM
TKHunny
Factorials have a symbiotic relationship with the Ratio Test.
Divide any term by the term before it and see how it goes.
$\displaystyle \frac{\frac{(k+1)!}{(2(k+1)-1)!}}{\frac{k!}{(2k-1)!}}$
Simplify that. >1, increasing, <1, decreasing
• Apr 20th 2010, 01:56 PM
DBA
Does it mean that I do not use the Alternating Series Test at all?
Do I just use the Ratio Test and take the limit from the Ratio or do I use the ratio just to determine if it is decreasing?
And how do I find the ratio?
$\displaystyle \frac{(k+1)!}{(2(k+1)-1)!} * \frac{(2k-1)!}{k!} = \frac{(2k-1)!}{(2(k+1)-1)!} * (k+1)$
I do not know how to simplify the last term.
Can I write
$\displaystyle (2(k+1)-1)! = (2k+2-1)! = (2k+1)!$ | 424 | 1,279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-26 | latest | en | 0.855217 |
https://www.mrexcel.com/board/threads/clarify-named-range-offset-match.23316/ | 1,721,835,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00592.warc.gz | 781,594,345 | 21,071 | # clarify named range (=OFFSET/=MATCH)
#### buz
##### Board Regular
Drange w/ refers to is =OFFSET(Hrs!\$A\$2,0,0,End-1,4)
End w/ refers to is =MATCH(9.9999999E+307,Hrs!\$D:\$D)
In the Drange I see the sheet/row/col but loose it with End-1,4). What does End-1,4 refer to? Ibelieve the 'D' in Drange refers to col D on sheet hrs?????
In the =MATCH the sheet/range is understood, but what does the 9.9999999+307 refer to?
The point is I would like to name some more ranges using the OFFSET/MATCH routine. I can imagine for the Drange using perhaps another col heading on the other sheet like 'Erange refers to =OFFSET(SHEET!\$A\$2,0,0,End-1,4). Then it's a matter of the End part - how End was choosen for the =MATCH and what another good =MATCH name would be.
You see, it would help if I understood how things inter-relate.
Looks like a strange question to me - but hey - I gotta ask.
tfyh
### Excel Facts
Can Excel fill bagel flavors?
You can teach Excel a new custom list. Type the list in cells, File, Options, Advanced, Edit Custom Lists, Import, OK
The End name returns the last row with a number in column D of sheet Hrs.
Then, the DRange name, "stands" in Hrs!A2, extends down End-1 rows (Height), and to the left 4 columns (Width).
The D in DRange is just the name, you can change that to whatever you want.
In the =MATCH the sheet/range is understood, but what does the 9.9999999+307 refer to?
9.9999999...+307 is for the largest number that Excel can handle ... in the match function it is intended to pick up the row number of the last entry in column D
Regards!
Yogi
So to name another range on another sheet i would choose names other than Drange and End?
On 2002-10-04 00:40, buz wrote:
So to name another range on another sheet i would choose names other than Drange and End?
Yes. Suppose that we have 3 worksheets in Book1 and the following current data areas are frequently changing--that is, expanding or crimping:
Sheet1: A1:E300
Sheet2: F60:H2000
Sheet3: B2:J659
You can define appropriate names by means of dynamic formulas (usually with OFFSET) for each of the above ranges...
The intermediate steps can be put in a separate worksheet which I often name Admin.
In Admin you compute with respect to each range the first and last cells for each range. We use here MATCH, not COUNT or COUNTA. Two rules:
(1) If a range has a numeric column (or row), which is ideal, we use:
=MATCH(BigNum,x!ColRef)
where BigNum is 9.99999999999999E+307 and ColRef (e.g., C:C) is of numeric type and x is the sheetname. Thus we pick out a numeric column from the range of interest to compute the current maximum extent of the range.
(2) If a range is all text, we use:
=MATCH(REPT("z",90),x!ColRef)
where REPT("z",90) plays the same role as BigNum does.
(3) If a range is mixed, we use:
=MAX(MATCH(BigNum,x!ColRef),MATCH(REPT("z",90),x!ColRef))
Since 9.99999999999999E+307 is an important number and hard to type when needed, we define a name for it: BigNum.
See:
http://www.mrexcel.com/board/viewtopic.php?topic=16207&forum=2&7
http://www.mrexcel.com/board/viewtopic.php?topic=24122&forum=2
Then, the DRange name, "stands" in Hrs!A2, extends down End-1 rows (Height), and to the left 4 columns (Width).
Your description of the width??????? is it to the 'left' four columns or to the right?
Seems like it may be just a typo - but guesing if I changed it from
=OFFSET(Hrs!\$A\$2,0,0,End-1,4)
to
=OFFSET(Hrs!\$A\$2,0,0,End-1,6)
then it would be picking up 6 columns?
On 2002-10-08 01:13, buz wrote:
Then, the DRange name, "stands" in Hrs!A2, extends down End-1 rows (Height), and to the left 4 columns (Width).
Your description of the width??????? is it to the 'left' four columns or to the right?
Seems like it may be just a typo - but guesing if I changed it from
=OFFSET(Hrs!\$A\$2,0,0,End-1,4)
to
=OFFSET(Hrs!\$A\$2,0,0,End-1,6)
then it would be picking up 6 columns?
To the right...
The last argument, here 6, indicates width, put otherwise, the number of columns the deined range consists of.
Thus the defined range runs here from column A to F inclusive.
This message was edited by Aladin Akyurek on 2002-10-08 01:59
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# Does wind speed and gusts affect descent rate?
## Recommended Posts
Quagmirian
***I vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped
Ahh, the problem here is that a canopy's airpseed stays the same regardless of windspeed. The bullet analogy is more suited to a skydiver exiting a moving aeroplane compared to a balloon or a helicopter. In this case, the airspeeds are different, so descent rate is affected.
true
i was only answering format's thought experiment
otoh... by analogy with this experiment, even a round parachute (with zero forward drive) will change its decent rate when going through shear layers, and it will float down always slightly slower regardless of whether u go from faster layer to slower or vice versa
##### Share on other sites
sky12345
i vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped
Non-relevant. Bullet is moving relative to the air mass. Aerodynamic forces on the bullet have to be calculated as a whole, and can't purely be calculated purely on horizontal and vertical movement separately without reference to the other. The simple way to put it, is that there are significant lift and drag forces on a high speed object (relative to the air mass, not the earth or solar system) which are different from one dropped from a standing start, and thus affect the time-of-fall.
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sky12345
***If I am to make a maple seed out of aluminium or other material and/or size to match density to water in proportion, as it does have proportion in air...
and it emulates behavior in water exactly like real one in the air...
You are positive that it will hit bottom of, say 2 feet deep pool at the same time as in another pool, same depth, with artificial steady water current (of any speed)?
Give me a "yes" and I quit and thank you for your patience.
i vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped
Not the same thing at all. And easily proved using regular Newtonian physics.
...
The only sure way to survive a canopy collision is not to have one.
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pchapman
***
i vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped
Non-relevant. Bullet is moving relative to the air mass. Aerodynamic forces on the bullet have to be calculated as a whole, and can't purely be calculated purely on horizontal and vertical movement separately without reference to the other. The simple way to put it, is that there are significant lift and drag forces on a high speed object (relative to the air mass, not the earth or solar system) which are different from one dropped from a standing start, and thus affect the time-of-fall.
Because drag goes as v^2 at the Reynolds numbers involved, the problem becomes non linear and the horizontal and vertical components of drag are no longer independent. Easily proved, and I think I posted the proof some years ago in the "freefall drift" discussion on here.
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The only sure way to survive a canopy collision is not to have one.
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sky12345
******I vagely remember the episode of mythbusters where they proved that a bullet shot horizontally falls slower than just dropped
Ahh, the problem here is that a canopy's airpseed stays the same regardless of windspeed. The bullet analogy is more suited to a skydiver exiting a moving aeroplane compared to a balloon or a helicopter. In this case, the airspeeds are different, so descent rate is affected.
true
i was only answering format's thought experiment
otoh... by analogy with this experiment, even a round parachute (with zero forward drive) will change its decent rate when going through shear layers, and it will float down always slightly slower regardless of whether u go from faster layer to slower or vice versa
But format's thought experiment explicitly excluded shears.
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The only sure way to survive a canopy collision is not to have one.
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JeffCa
***
And these effects have been exhaustively discussed a decade ago. Quit being lazy, look it up and stop wasting everyone's time.
"No class, we won't be doing the tickertape cart experiment this year, we'll just be looking up what should happen. It would be wasting our time to do or discuss anything that has already been done. Read the lab report from last year's students. No, no, don't try to think about what might happen if a wave passes through a double-slit, that would be lazy and it has already been done. Look it up."
1. If he really wanted to work it out for himself instead of just arguing about it, he could have done that. However, he simply argues with everyone who tells him what's what.
2. He is NOT my student. My students are expected to show initiative in finding stuff for themselves.
3. I responded to the question posed in the thread title by telling the questioner where to find the answer.
...
The only sure way to survive a canopy collision is not to have one.
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sky12345
>Since you can't make a 2' deep pool with steady current of a viscous fluid that has the same speed all the way to the bottom
sure u can
just move the entire pool (with still water in it)
boom! for the object dropped from above in it its tye perfectly steady current of viscous fluid
or just drop the object in nonmoving still pool with same horizontal speed
its relativity
e = mc2
maybe u should take physics 101 mr "physics professor"
Yep. Unfortunately it doesn't model the atmosphere.
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The only sure way to survive a canopy collision is not to have one.
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>Not the same thing at all. And easily proved using regular Newtonian physics.
how a bullet fired horizontally and a bullet dropped in the moving stream not the same thing? the result is the same.. it moves vertically slightly slower than bullet dropped in still medium
>horizontal and vertical components of drag are no longer independent
what do u mean independent? they r just projections of a vector, how they can be independent?
>But format's thought experiment explicitly excluded shears.
can u read b4 replying? when i finished with formats experiment, i introduced my own one, with shears.... reading comp probs?
>Yep. Unfortunately it doesn't model the atmosphere.
again can u read? i was talking about producing uniform stream in formats exp. it has nothing to do with the atmosphere, its a thought experiment!!
i had F- in physics but even an idiot like me knows it better than "physics professor"
hereby i demote u of your title and advise u to go to a middle school and start it from the ground up
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k_marr08
I know when facing into the wind, ground speed decreases when wind speed increases, but does descent rate decrease as well? Does the added wind speed provide more lift?
In a steady state flight, when nothing changes, there should be no affect of wind speed on descent rate. This is only true if nothing changes at all. At the same time, air density and temperature changes with altitude and these will act as changing inputs.
It helps to think about the parachute and skydiver as two bodies. These bodies have different momentum (different mass) and different drag. When both bodies get the same additional input, their relative speed changes differently and this causes the canopy to surge forward or the skydiver to swing forward. This is similar to what is happening when we apply inputs to the canopy. These changes will affect angle of attack, which will affect lift coefficient and therefore the descent rate. I understand that this formula is for a fixed wing aircraft, but the principle still applies.
k_marr08
And, a related question, if one were to downwind a landing, would their descent rate increase, or would only their ground speed increase?
See above.
k_marr08
If the answer is that the descent rate is NOT effected by wind speed, then why do gusts "feel" like they affect descent rate? Is there some additional element of gusts that actually do affect descent rate?
Gusts affect landings because they cause change in angle of attack more than they affect lift. Think of two bodies with different masses getting the same impulse applied to them.
Regards, Alexander. http://staticlineinteractive.com
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sjc
In a steady state flight, when nothing changes, there should be no affect of wind speed on descent rate.
So, how come I've timed ~10% difference (with a stopwatch) in flight_time with your Canopy Glide Simulator between 'No wind' and '13mph wind'?
(steady 'no input'; 'half input'; full input' flight)
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>how a bullet fired horizontally and a bullet dropped in the moving stream not the
>same thing? the result is the same.. it moves vertically slightly slower than bullet
>dropped in still medium
Just as your parachute would fly differently if it were flying at 700 mph vs. 0 mph airspeed.
However, if you drop a bullet in zero speed air (standing on the ground) vs. drop a bullet in 500mph air (standing inside a 747 in flight) they will fall the same way in the same time.
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we're talking about different things.. i didnt mean the quantitative difference between an object moving at 700mph and 20mph.. i meant qualitatative diff
ok,,, one last try, more succinctly... two da vinci parachutes (essentially a round parachute with fully formed shape) r released from a balloon and from an airplane, from the same altitude simultaneously. which one will land first?
my answer: the one dropped from the balloon
this is the same answer, qualitatively, as with mythbusters bullet
once u agree with that theres no reason not to agree that a round descends slightly slower through shear layers than still air
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format
So, how come I've timed ~10% difference (with a stopwatch) in flight_time with your Canopy Glide Simulator between 'No wind' and '13mph wind'?
(steady 'no input'; 'half input'; full input' flight)
The short answer is: the model is not perfect. The longer answer is, that there are multiple variables at play and the model must accommodate the steady state as well as the response to the inputs. This requires a compromise which results in parameters that produce overall behavior. By definition it is a mathematical approximation of the reality. As the model becomes better we'll see better results.
In general, the goal of the simulator is to support the discussions such as this one. I was happy to see that it was brought up in the conversation.
Regards, Alexander. http://staticlineinteractive.com
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sjc
***So, how come I've timed ~10% difference (with a stopwatch) in flight_time with your Canopy Glide Simulator between 'No wind' and '13mph wind'?
(steady 'no input'; 'half input'; full input' flight)
The short answer is: the model is not perfect. The longer answer is, that there are multiple variables at play and the model must accommodate the steady state as well as the response to the inputs. This requires a compromise which results in parameters that produce overall behavior. By definition it is a mathematical approximation of the reality. As the model becomes better we'll see better results.
thats very poor excuse
the descent rates in no wind and in steady winds MUST match 100% ***or else*** u should ditch your 'model' completely
u may not get simulation of control inputs or turns 100% right,,, nobody can,,, but not to match the descent rates is like writing a calculator program that when entered 2*2 gives 4.4. "yeah i know it must be 4 but my model is not perfect"
"new version of my calculator will be better. 2*2 will be 4.2"
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sky12345
thats very poor excuse
the descent rates in no wind and in steady winds MUST match 100% ***or else*** u should ditch your 'model' completely
u may not get simulation of control inputs or turns 100% right,,, nobody can,,, but not to match the descent rates is like writing a calculator program that when entered 2*2 gives 4.4. "yeah i know it must be 4 but my model is not perfect"
"new version of my calculator will be better. 2*2 will be 4.2"
I understand your feelings, though I disagree with the logic. If you would like to discuss this further we can do it in some other venue (at least not this thread).
Regards, Alexander. http://staticlineinteractive.com
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sky12345
>Not the same thing at all. And easily proved using regular Newtonian physics.
how a bullet fired horizontally and a bullet dropped in the moving stream not the same thing?
[...]
>horizontal and vertical components of drag are no longer independent
what do u mean independent? they r just projections of a vector, how they can be independent?
Ok, although I've been snickering at some of the stuff in this thread, I'll give you a serious answer about how it works to help you and others learn this.
Let's say a heavy round ball is dropped vertically, vs. dropped from an airplane moving fast horizontally.
Drag tends to change with the square of the speed. I'll skip all the real world numbers and we'll just say that something falling with speed 1 has a drag of 1 squared, which =1. Speed 2 gives drag 4, in some system of units.
Lets say we compare the drag when both balls are falling speed 3 vertically, but one ball is still moving forwards fast, lets say speed 10.
The drag on the ball dropping vertically is 3 squared =9.
For the ball moving forward, we can't just say that the vertical speed is 3 so the drag that affects how fast it is accelerating downwards is 3.
Instead we have to look at the whole velocity vector. Speed 3 down and 10 forward = 10.44 speed by Pythagoras, at an angle 16.70 degrees below the horizon. Total drag is 10.44 squared = 109.
The vertical component of that is 109*sin(16.70 deg) = 109*.287 = 31.28.
Voila. The ball falling straight down will accelerate straight down based on a drag of 9, while the ball arcing down with a lot of horizontal velocity, will accelerate downwards (at that particular moment in its trajectory) as befits a drag of over 31. So it will be accelerating downwards a lot slower.
There's a huge difference in the drag values, even when one is looking only at the vertical direction, despite both balls moving in the vertical direction the same speed.
At slow overall speeds you might be able to make approximate calculations for the vertical speed only, just by looking at the vertical speed of the object, treating the horizontal dimension as independent. After all, we might use a skydiving table that says that after 5 seconds of fall, a jumper will have dropped roughly 366 ft. In reality, that will differ a little depending on the forward speed of the airplane (even for jumpers all on their belly and 'falling straight').
If you want the calculations to be correct, you have to do the trigonometry and not just treat horizontal and vertical movement separately.
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Another perspective: imagine a no-wind day and say you're trying to land on a moving vehicle. It's obvious that your descent rate will be the same no matter whether you're landing in the direction of this vehicle moving or in the opposite direction (let's assume that the vehicle is huge and you can easily land on it in those conditions).
Now the only thing left to understand is that the ground is your vehicle moving in the opposite direction to the wind, because you know, all movement is relative and air moving wrt earth is the same as earth moving wrt to the air.
Does this make sense?
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***THANK U***
this is a perfect explanation
even an idiot like me understood
if only all physics professors were like u!
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sjc
***
thats very poor excuse
the descent rates in no wind and in steady winds MUST match 100% ***or else*** u should ditch your 'model' completely
u may not get simulation of control inputs or turns 100% right,,, nobody can,,, but not to match the descent rates is like writing a calculator program that when entered 2*2 gives 4.4. "yeah i know it must be 4 but my model is not perfect"
"new version of my calculator will be better. 2*2 will be 4.2"
I understand your feelings, though I disagree with the logic. If you would like to discuss this further we can do it in some other venue (at least not this thread).
my logic is that its not ok to sell bullshit
your prog fails an elementary bs check
u disagree? its ok 4u to sell bs?!
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sky12345
>Not the same thing at all. And easily proved using regular Newtonian physics.
how a bullet fired horizontally and a bullet dropped in the moving stream not the same thing? the result is the same.. it moves vertically slightly slower than bullet dropped in still medium
>horizontal and vertical components of drag are no longer independent
what do u mean independent? they r just projections of a vector, how they can be independent?
IF drag were a linear function of velocity then the horizontal component of drag would be independent of the vertical component of velocity and the vertical component of drag would be independent of the horizontal component of velocity.
However, drag is NOT a linear function of velocity, so the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.
Quote
>
Why am I not surprised.
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The only sure way to survive a canopy collision is not to have one.
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Quote
However, drag is NOT a linear function of velocity, so the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.
Soo, this is actually the dominating reason why trackers stay longer in the air, right? Not "lift"? Also, this must work for larger canopies (most canopies?)
And in general, what reading do you suggest on canopy aerodynamics?
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unkulunkulu
Quote
However, drag is NOT a linear function of velocity,[...]
Soo, this is actually the dominating reason why trackers stay longer in the air, right?
No. Trackers or inclined barn doors or airplane wings all still work due to a combination of lift (defined as the force generated perpendicular to the line of flight) and drag (defined as the force generated parallel to the line of flight).
This principle about 'not being able to treat vertical motion while ignoring horizontal motion' is about aerodynamic forces in general.
It applies equally well to a round cannonball that is generating no lift at all. If it is zooming forward while dropping a little, with all its speed it is generating a lot of drag, which does indeed slow down its acceleration vertically.
(Which if you don't understand the whole picture, is like some kind of magic anti-gravity lift. But it isn't lift since there's no force perpendicular to the line of flight.)
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unkulunkulu
Quote
However, drag is NOT a linear function of velocity, so the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.
Soo, this is actually the dominating reason why trackers stay longer in the air, right? Not "lift"? Also, this must work for larger canopies (most canopies?)
No, but since Mr. Chapman has already answered correctly I'll refrain from duplication.
...
The only sure way to survive a canopy collision is not to have one.
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>the horizontal component of velocity does have an effect on the vertical component of drag and the vertical component of velocity does have an effect on the horizontal component of drag.
its like saying that if we have a triangle with sides a b c and we blow it up 2x the increase of length of a has an effect on length of c!
thats bs,, no causality whatsoever,, the reason c became 2x is not because a or b became 2x, its because we blew the whole triangle up 2x!
same here,, as pchapman perfectly explained, greater velocity = greater drag (in quadratic proportion), and its vertical component is greater than when the object is dropped with no horizontal speed
total velocity is the primary thing, the cause of the effect,, not some mysterious crossbreeding btw horiz and vert components lol
components r arbitrary imaginary things
velocity is a real one
u saying that horiz component of speed has "an effect" on vert comp of drag is like saying that the growing child became taller because the height from crotch to floor increased! lol
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Your F- in physics is showing, Mr. Anonymous Troll.
This has all been discussed in DZ.COM a decade ago, as I have repeatedly stated:
www.dropzone.com/cgi-bin/forum/gforum.cgi?post=1158261#1158261
(In fact Newton was aware of it in 1676; he gives an example in Principia).
...
The only sure way to survive a canopy collision is not to have one.
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# Understanding air compressor measurements: work, power, and flow
After learning about the basics of physics, you might want to know more about understanding air compressor measurements regarding matter.
This information is very useful when determining the appropriate size and power you need for a particular application. In this article, we will explain the basics of measuring work, power and volume rate of flow.
## How is mechanical work measured
Mechanical work may be defined as the product of a force and distance over which the force operates on an object. Like heat, work involves energy transferred from one body to another. The difference is that it involves force rather than temperature. An example of this is when gas becomes compressed in a cylinder with a moving piston.
Compression occurs as a result of force moving the piston. Energy, therefore, transfers from the piston to the gas. This energy transfer is work in the thermodynamic sense of the word. The result of work can have many forms, such as changes in the potential, kinetic, or thermal energy.
Mechanical work associated with changes in the volume of a gas mixture is one of the most important processes in engineering thermodynamics. The SI unit for work is the Joule: 1 J = 1 Nm = 1 Ws.
## Measuring power
Power is work performed per unit of time. It is a measure of how quickly work becomes completed. The SI unit for power is the Watt: 1 W = 1 J/s. For example, power or energy flow to a compressor's drive shaft is numerically similar to system heat emissions, plus heat applied to compressed gas.
## Measuring flow rate
The volumetric flow rate of a system is a measure of the volume of fluid flowing per unit of time. It may be calculated as the product of the cross-sectional area of the flow and the average flow velocity. The SI unit for volume rate of flow is m3/s.
However, the unit liter/second (l/s) is also frequently used when referring to the volume rate of flow (also called the capacity) of a compressor. It is either stated as Normal liter/second (Nl/s) or as free air delivery (l/s). With Nl/s, the air flow rate is recalculated to "the normal state." That is, conventionally chosen as 1.013 bar(a) and 0 °C. The Normal unit Nl/s is primarily used when specifying a mass flow.
For free air delivery (FAD) the compressor's output flow rate is recalculated to a free air volume rate at the standard inlet condition (inlet pressure 1 bar(a) and inlet temperature 20 °C). The relation between the two volume rates of flow is (note that the simplified formula above does not account for humidity).
## Free air delivery
The following example illustrates free air delivery (FAD). What does FAD = 39l/s for a compressor working at 13 bar mean? How long does it take to fill a 390L tank at a pressure of 13 bar? To calculate this, we need to go back to the inlet conditions. Which is 1 bar.
When we start with an empty vessel, after 1 second there are 39 liters in the vessel at 1 bar. Then, after 10 seconds the pressure inside the vessel is 1 bar. Following this, the pressure is 2 bar after 20 seconds. Therefore, after 130 seconds it becomes filled at 13 bar.
Next, the difference between reference conditions and normal conditions. Reference conditions use 1bar, 20 °C, 0% Relative Humidity (RH).
Normal conditions involve 1atm = 1,01325bar, 0 °C, 0% RH. The next definition is SER or Specific Energy Requirement. This means the amount of energy that is required to deliver 1 liter FAD at a certain pressure.
## Go with the flow for air compressor measurements
Specifying your compressed air system by flow and pressure – not kW or horsepower – is the best way to match its performance to your needs. Compressor sizing should match your business requirements more precisely than just going by kW rating.
## Purchasing the right size equipment
There's a lot of technical terms covered in this article pertaining to mechanical work, power, and flow. Understanding this information is important for investing in the right equipment for your application. If you purchase equipment that's either too large or too small, there's the risk of inefficiency.
What's important to consider is how much force you'll need to move an object to complete a given job in a given time frame. As mentioned above, this is expressed in flow and pressure. In addition to liters per second (l/s), flow is represented in cubic feet per minute (cfm) or cubic meters per hour (m3/h). These measurements all pertain to speed.
Pressure is both shown as bar, mentioned above, or pounds per square inch (psi). If you need to move heavy objects, you'll need more pressure. You'll also want to determine whether you need all day air delivery and if there are different requirements for your applications. This context is useful when it comes to determining size and choosing between fixed speed and variable speed drive (VSD) machines.
## Fixed speed vs. variable speed drive (VSD)
When researching air compressors, you'll come across fixed speed and VSD equipment. These terms refer to how the engine operates. As suggested, fixed speed machines only operate at one speed, while VSD compressors change speed depending on demand. Each has their own advantages, depending on your workflow and needs.
A fixed speed machine is generally cheaper to purchase, while a VSD machine offers efficiency advantages. The latter provides operational energy cost savings. If you're undecided on what makes the most sense for your needs, feel free to get in touch. Our team is happy to help assess what makes the most sense.
We understand that there's not a one size fits all solution for every customer, and offer tailor made solutions.
After learning about the basics of physics here, you might want to know more concerning the physical units used to measure different aspects of matter. This can be very helpful when dealing with compressed air. In this article, we will explain the basics of measuring work, power and volume rate of flow.
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# Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
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Chapter 8 8-1 (a) Thread depth = 2 . 5 mm Ans. Width = 2 . 5 mm Ans. d m = 25 1 . 25 1 . 25 = 22 . 5 mm d r = 25 5 = 20 mm l = p = 5 mm Ans. (b) Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. d m = 22 . 5 mm d r = 20 mm l = p = 5 mm Ans. 8-2 From Table 8-1, d r = d 1 . 226 869 p d m = d 0 . 649 519 p ¯ d = d 1 . 226 869 p + d 0 . 649 519 p 2 = d 0 . 938 194 p A t = π ¯ d 2 4 = π 4 ( d 0 . 938 194 p ) 2 Ans . 8-3 From Eq. ( c ) of Sec. 8-2, P = F tan λ + f 1 f tan λ T = Pd m 2 = Fd m 2 tan λ + f 1 f tan λ e = T 0 T = Fl / (2 π ) Fd m / 2 1 f tan λ tan λ + f = tan λ 1 f tan λ tan λ + f Ans . Using f = 0 . 08, form a table and plot the efficiency curve. λ , deg. e 0 0 10 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 1 0 50 , deg. e 5 mm 5 mm 2.5 2.5 2.5 mm 25 mm 5 mm
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Chapter 8 205 8-4 Given F = 6 kN, l = 5 mm, and d m = 22 . 5 mm, the torque required to raise the load is found using Eqs. (8-1) and (8-6) T R = 6(22 . 5) 2 5 + π (0 . 08)(22 . 5) π (22 . 5) 0 . 08(5) + 6(0 . 05)(40) 2 = 10 . 23 + 6 = 16 . 23 N · m Ans . The torque required to lower the load, from Eqs. (8-2) and (8-6) is T L = 6(22 . 5) 2 π (0 . 08)22 . 5 5 π (22 . 5) + 0 . 08(5) + 6(0 . 05)(40) 2 = 0 . 622 + 6 = 6 . 622 N · m Ans . Since T L is positive, the thread is self-locking. The efficiency is Eq. (8-4): e = 6(5) 2 π (16 . 23) = 0 . 294 Ans . 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom seg- ment of the screws must be in compression. Where as tension specimens and their grips must be in tension. Both screws must be of the same-hand threads. 8-6 Screws rotate at an angular rate of n = 1720 75 = 22 . 9 rev/min (a) The lead is 0.5 in, so the linear speed of the press head is V = 22 . 9(0 . 5) = 11 . 5 in/min Ans . (b) F = 2500 lbf/screw d m = 3 0 . 25 = 2 . 75 in sec α = 1 / cos(29 / 2) = 1 . 033 Eq. (8-5): T R = 2500(2 . 75) 2 0 . 5 + π (0 . 05)(2 . 75)(1 . 033) π (2 . 75) 0 . 5(0 . 05)(1 . 033) = 377 . 6 lbf · in Eq. (8-6): T c = 2500(0 . 06)(5 / 2) = 375 lbf · in T total = 377 . 6 + 375 = 753 lbf · in/screw T motor = 753(2) 75(0 . 95) = 21 . 1 lbf · in H = Tn 63 025 = 21 . 1(1720) 63 025 = 0 . 58 hp Ans .
206 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 8-7 The force F is perpendicular to the paper. L = 3 1 8 1 4 7 32 = 2 . 406 in T = 2 . 406 F M = L 7 32 F = 2 . 406 7 32 F = 2 . 188 F S y = 41 kpsi σ = S y = 32 M π d 3 = 32(2 . 188) F π (0 . 1875) 3 = 41 000 F = 12 . 13 lbf T = 2 . 406(12 . 13) = 29 . 2 lbf · in Ans . (b) Eq. (8-5), 2 α = 60 , l = 1 / 14 = 0 . 0714 in, f = 0 . 075, sec α = 1 . 155, p = 1 / 14 in d m = 7 16 0 . 649 519 1 14 = 0 . 3911 in T R = F clamp (0 . 3911) 2 Num Den Num = 0 . 0714 + π (0 . 075)(0 . 3911)(1 . 155) Den = π (0 . 3911) 0 . 075(0 . 0714)(1 . 155) T = 0 . 028 45 F clamp F clamp = T 0 . 028 45 = 29 . 2 0 . 028 45 = 1030 lbf Ans . (c) The column has one end fixed and the other end pivoted. Base decision on the mean diameter column. Input: C = 1.2, D = 0.391 in, S y = 41 kpsi, E = 30(10 6 ) psi, L = 4 . 1875 in, k = D / 4 = 0 . 097 75 in, L / k = 42 . 8. For this J. B. Johnson column, the critical load represents the limiting clamping force for bucking. Thus, F clamp = P cr = 4663 lbf . (d) This is a subject for class discussion. 8-8 T = 6(2 . 75) = 16 . 5 lbf · in d m = 5 8 1 12 = 0 . 5417 in l = 1 6 = 0 . 1667 in, α = 29 2 = 14 . 5 , sec 14 . 5 = 1 . 033 1 4 " 3 16 D . " 7 16 " 2.406" 3"
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Chapter 8 207 Eq. (8-5): T = 0 . 5417( F / 2) 0 . 1667 + π (0 . 15)(0 . 5417)(1 . 033) π (0 . 5417) 0 . 15(0 . 1667)(1 . 033) = 0 . 0696 F Eq. (8-6): T c = 0 . 15(7 / 16)( F / 2) = 0 . 032 81 F T total = (0 . 0696 + 0 . 0328) F = 0 . 1024 F F = 16 . 5 0 . 1024 = 161 lbf Ans .
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https://pdfcookie.com/documents/materials-variance-formulas-68v4y53on7vg | 1,680,376,746,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00132.warc.gz | 512,993,628 | 5,888 | # Materials Variance Formulas
• November 2019
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Materials Variance Formulas – Total, Price, and Quantity
AQ = Actual Quantity AP = Actual Price SQ = Standard Quantity SP = Standard Price F = Favorable is when the actual is lower than the standard. U = Unfavorable is when the actual is higher than the standard. Standard Quantity (SQ) Formula Actual Units Produced X Units of material required per unit Step 1 (7,250 X 4) Step 2 = 29,000 Note: Actual Quantity, Actual Price, and Standard Price will all be given to you in the problem.
Step 1 Step 2 Step 3
Total Materials Variance Formula (AQ X AP) - (SQ X SP) (30,000 X \$5.05) - (29,000 X \$5.50) \$151,500 - \$159,500 = \$8,000 F
Step 1 Step 2 Step 3
Materials Price Variance Formula (AQ X AP) - (AQ X SP) (30,000 X \$5.05) - (30,000 X \$5.50) \$151,500 - \$165,000 = \$13,500 F
Step 1 Step 2 Step 3
Materials Quantity Variance Formula (AQ X SP) - (SQ X SP) (30,000 X \$5.50) - (29,000 X \$5.50) \$165,000 - \$159,500 = \$5,500 U
MJC 12/2012
Page 1
Materials Variance Formulas – Total, Price, and Quantity How to Determine Amounts for Variance Formulas Exercise Example: The standard cost of Product- A manufactured by Morrison Pet Supply Company includes four units of direct material at \$5.50 per unit. During August, 30,000 units of direct materials are purchased at a cost of \$5.05 per unit. And 30,000 units of direct materials are used to produce 7,250 units of Product -A. AQ = 30,000 units of direct materials used to produce 7,250 units of Product-A AP = \$5.05 Actual price paid per unit of direct materials SQ = (7,250 units of Product-A produced X 4 units Estimated direct material to produce one unit of Product-A) which equals 29,000 total units of direct material. SP = \$5.50 Estimated price to be paid per unit of direct material.
MJC 12/2012
Page 2
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# Number 13569967
thirteen million five hundred sixty nine thousand nine hundred sixty seven
### Properties of the number 13569967
Factorization 13569967 Divisors 1, 13569967 Count of divisors 2 Sum of divisors 13569968 Previous integer 13569966 Next integer 13569968 Is prime? YES (883872nd prime) Previous prime 13569949 Next prime 13569991 13569967th prime 247887821 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 110011110000111110101111 Octal 63607657 Duodecimal 4664ba7 Hexadecimal cf0faf Square 184144004381089 Square root 3683.7436121424 Natural logarithm 16.423369599973 Decimal logarithm 7.132578791526 Sine 0.88916251498811 Cosine -0.45759154487384 Tangent -1.9431358051715
Number 13569967 is pronounced thirteen million five hundred sixty nine thousand nine hundred sixty seven. Number 13569967 is a prime number. The prime number before 13569967 is 13569949. The prime number after 13569967 is 13569991. Number 13569967 has 2 divisors: 1, 13569967. Sum of the divisors is 13569968. Number 13569967 is not a Fibonacci number. It is not a Bell number. Number 13569967 is not a Catalan number. Number 13569967 is not a regular number (Hamming number). It is a not factorial of any number. Number 13569967 is a deficient number and therefore is not a perfect number. Binary numeral for number 13569967 is 110011110000111110101111. Octal numeral is 63607657. Duodecimal value is 4664ba7. Hexadecimal representation is cf0faf. Square of the number 13569967 is 184144004381089. Square root of the number 13569967 is 3683.7436121424. Natural logarithm of 13569967 is 16.423369599973 Decimal logarithm of the number 13569967 is 7.132578791526 Sine of 13569967 is 0.88916251498811. Cosine of the number 13569967 is -0.45759154487384. Tangent of the number 13569967 is -1.9431358051715 | 586 | 1,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.674668 |
https://www.bbsmax.com/A/qVdenpx1zP/ | 1,606,854,849,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00534.warc.gz | 543,186,167 | 14,244 | ## 第1题
``````["1", "2", "3"].map(parseInt)
``````
``````parseInt('1', 0);
parseInt('2', 1);
parseInt('3', 2);
``````
## 第2题
``````[typeof null, null instanceof Object]
``````
typeof 返回一个表示类型的字符串.
instanceof 运算符用来检测 constructor.prototype 是否存在于参数 object 的原型链上.
typeof 的结果请看下表:
``````type result
Undefined "undefined"
Null "object"
Boolean "boolean"
Number "number"
String "string"
Symbol "symbol"
Host object Implementation-dependent
Function "function"
Object "object"
``````
## 第3题
``````[ [3,2,1].reduce(Math.pow), [].reduce(Math.pow) ]
``````
`arr.reduce(callback[, initialValue])`
reduce接受两个参数, 一个回调, 一个初始值.
## 第4题
``````var val = 'smtg';
console.log('Value is ' + (val === 'smtg') ? 'Something' : 'Nothing');
``````
## 第5题
``````var name = 'World!';
(function () {
if (typeof name === 'undefined') {
var name = 'Jack';
console.log('Goodbye ' + name);
} else {
console.log('Hello ' + name);
}
})();
``````
``````var name = 'World!';
(function () {
var name;
if (typeof name === 'undefined') {
name = 'Jack';
console.log('Goodbye ' + name);
} else {
console.log('Hello ' + name);
}
})();
``````
## 第6题
``````var END = Math.pow(2, 53);
var START = END - 100;
var count = 0;
for (var i = START; i <= END; i++) {
count++;
}
console.log(count);
``````
## 第7题
``````var ary = [0,1,2];
ary[10] = 10;
ary.filter(function(x) { return x === undefined;});
``````
``````if (!Array.prototype.filter) {
Array.prototype.filter = function(fun/*, thisArg*/) {
'use strict';
if (this === void 0 || this === null) {
throw new TypeError();
}
var t = Object(this);
var len = t.length >>> 0;
if (typeof fun !== 'function') {
throw new TypeError();
}
var res = [];
var thisArg = arguments.length >= 2 ? arguments[1] : void 0;
for (var i = 0; i < len; i++) {
if (i in t) { // 注意这里!!!
var val = t[i];
if (fun.call(thisArg, val, i, t)) {
res.push(val);
}
}
}
return res;
};
}
``````
``````0 in ary; => true
3 in ary; => false
10 in ary; => true
``````
## 第8题
``````var two = 0.2
var one = 0.1
var eight = 0.8
var six = 0.6
[two - one == one, eight - six == two]
``````
IEEE 754标准中的浮点数并不能精确地表达小数
## 第9题
``````function showCase(value) {
switch(value) {
case 'A':
console.log('Case A');
break;
case 'B':
console.log('Case B');
break;
case undefined:
console.log('undefined');
break;
default:
console.log('Do not know!');
}
}
showCase(new String('A'));
``````
switch 是严格比较, String 实例和 字符串不一样.
``````var s_prim = 'foo';
var s_obj = new String(s_prim);
console.log(typeof s_prim); // "string"
console.log(typeof s_obj); // "object"
console.log(s_prim === s_obj); // false
``````
## 第10题
``````function showCase2(value) {
switch(value) {
case 'A':
console.log('Case A');
break;
case 'B':
console.log('Case B');
break;
case undefined:
console.log('undefined');
break;
default:
console.log('Do not know!');
}
}
showCase2(String('A'));
``````
`String(x) does not create an object but does return a string, i.e. typeof String(1) === "string"`
## 第11题
``````function isOdd(num) {
return num % 2 == 1;
}
function isEven(num) {
return num % 2 == 0;
}
function isSane(num) {
return isEven(num) || isOdd(num);
}
var values = [7, 4, '13', -9, Infinity];
values.map(isSane);
``````
``````7 % 2 => 1
4 % 2 => 0
'13' % 2 => 1
-9 % % 2 => -1
Infinity % 2 => NaN
``````
## 第12题
``````parseInt(3, 8)
parseInt(3, 2)
parseInt(3, 0)
``````
## 第13题
``````Array.isArray( Array.prototype )
``````
## 第14题
``````var a = [0];
if ([0]) {
console.log(a == true);
} else {
console.log("wut");
}
``````
## 第15题
``````[]==[]
``````
`==` 是万恶之源, 看上图
## 第16题
``````'5' + 3
'5' - 3
``````
`+` 用来表示两个数的和或者字符串拼接, `-`表示两数之差.
``````> '5' + 3
'53'
> 5 + '3'
'53'
> 5 - '3'
2
> '5' - 3
2
> '5' - '3'
2
``````
## 第17题
``````1 + - + + + - + 1
``````
``````1 + (a) => 2
a = - (b) => 1
b = + (c) => -1
c = + (d) => -1
d = + (e) => -1
e = + (f) => -1
f = - (g) => -1
g = + 1 => 1
``````
## 第18题
``````var ary = Array(3);
ary[0]=2
ary.map(function(elem) { return '1'; });
``````
``````Array.prototype.map = function(callback, thisArg) {
var T, A, k;
if (this == null) {
throw new TypeError(' this is null or not defined');
}
var O = Object(this);
var len = O.length >>> 0;
if (typeof callback !== 'function') {
throw new TypeError(callback + ' is not a function');
}
if (arguments.length > 1) {
T = thisArg;
}
A = new Array(len);
k = 0;
while (k < len) {
var kValue, mappedValue;
if (k in O) {
kValue = O[k];
mappedValue = callback.call(T, kValue, k, O);
A[k] = mappedValue;
}
k++;
}
return A;
};
``````
## 第19题
``````function sidEffecting(ary) {
ary[0] = ary[2];
}
function bar(a,b,c) {
c = 10
sidEffecting(arguments);
return a + b + c;
}
bar(1,1,1)
``````
``````function sidEffecting(ary) {
ary[0] = ary[2];
}
function bar(a,b,c=3) {
c = 10
sidEffecting(arguments);
return a + b + c;
}
bar(1,1,1)
``````
## 第20题
``````
var a = 111111111111111110000,
b = 1111;
a + b;
``````
## 第21题
``````var x = [].reverse;
x();
``````
`The reverse method transposes the elements of the calling array object in place, mutating the array, and returning a reference to the array.`
## 第22题
``````Number.MIN_VALUE > 0
``````
`true`
## 第23题
``````[1 < 2 < 3, 3 < 2 < 1]
``````
`````` 1 < 2 => true;
true < 3 => 1 < 3 => true;
3 < 2 => false;
false < 1 => 0 < 1 => true;
``````
## 第24题
``````// the most classic wtf
2 == [[[2]]]
``````
`both objects get converted to strings and in both cases the resulting string is "2"` 我不能信服...
## 第25题
``````3.toString()
3..toString()
3...toString()
``````
``````var a = 3;
a.toString()
``````
## 第26题
``````
(function(){
var x = y = 1;
})();
console.log(y);
console.log(x);
``````
y 被赋值到全局. x 是局部变量. 所以打印 x 的时候会报 `ReferenceError`
## 第27题
``````var a = /123/,
b = /123/;
a == b
a === b
``````
## 第28题
``````var a = [1, 2, 3],
b = [1, 2, 3],
c = [1, 2, 4]
a == b
a === b
a > c
a < c
``````
## 第29题
``````var a = {}, b = Object.prototype;
[a.prototype === b, Object.getPrototypeOf(a) === b]
``````
## 第30题
``````function f() {}
var a = f.prototype, b = Object.getPrototypeOf(f);
a === b
``````
f.prototype is the object that will become the parent of any objects created with new f while Object.getPrototypeOf returns the parent in the inheritance hierarchy.
f.prototype 是使用使用 new 创建的 f 实例的原型. 而 Object.getPrototypeOf 是 f 函数的原型.
``````
a === Object.getPrototypeOf(new f()) // true
b === Function.prototype // true
``````
## 31
``````function foo() { }
var oldName = foo.name;
foo.name = "bar";
[oldName, foo.name]
``````
## 第32题
``````"1 2 3".replace(/\d/g, parseInt)
``````
`str.replace(regexp|substr, newSubStr|function)`
• match 首先是匹配的字符串
• p1, p2 .... 然后是正则的分组
• offset match 匹配的index
• string 整个字符串
``````parseInt('1', 0)
parseInt('2', 2)
parseInt('3', 4)
``````
## 第33题
``````function f() {}
var parent = Object.getPrototypeOf(f);
f.name // ?
parent.name // ?
typeof eval(f.name) // ?
typeof eval(parent.name) // ?
``````
## 第34题
``````var lowerCaseOnly = /^[a-z]+\$/;
[lowerCaseOnly.test(null), lowerCaseOnly.test()]
``````
## 第35题
``````[,,,].join(", ")
``````
`[,,,] => [undefined × 3]`
## 第36题
``````var a = {class: "Animal", name: 'Fido'};
a.class
``````
## 第37题
``````var a = new Date("epoch")
``````
## 第38题
``````var a = Function.length,
b = new Function().length
a === b
``````
## 第39题
``````var a = Date(0);
var b = new Date(0);
var c = new Date();
[a === b, b === c, a === c]
``````
• 如果不传参数等价于当前时间.
• 如果是函数调用 返回一个字符串.
## 第40题
``````var min = Math.min(), max = Math.max()
min < max
``````
## 第41题
``````function captureOne(re, str) {
var match = re.exec(str);
return match && match[1];
}
var numRe = /num=(\d+)/ig,
wordRe = /word=(\w+)/i,
a1 = captureOne(numRe, "num=1"),
a2 = captureOne(wordRe, "word=1"),
a3 = captureOne(numRe, "NUM=2"),
a4 = captureOne(wordRe, "WORD=2");
[a1 === a2, a3 === a4]
``````
``````var myRe = /ab*/g;
var str = 'abbcdefabh';
var myArray;
while ((myArray = myRe.exec(str)) !== null) {
var msg = 'Found ' + myArray[0] + '. ';
msg += 'Next match starts at ' + myRe.lastIndex;
console.log(msg);
}
// Found abb. Next match starts at 3
// Found ab. Next match starts at 9
``````
## 第42题
``````var a = new Date("2014-03-19"),
b = new Date(2014, 03, 19);
[a.getDay() === b.getDay(), a.getMonth() === b.getMonth()]
``````
JavaScript inherits 40 years old design from C: days are 1-indexed in C's struct tm, but months are 0 indexed. In addition to that, getDay returns the 0-indexed day of the week, to get the 1-indexed day of the month you have to use getDate, which doesn't return a Date object.
``````a.getDay()
3
b.getDay()
6
a.getMonth()
2
b.getMonth()
3
``````
## 第43题
``````if ('http://giftwrapped.com/picture.jpg'.match('.gif')) {
'a gif file'
} else {
'not a gif file'
}
``````
String.prototype.match 接受一个正则, 如果不是, 按照 `new RegExp(obj)` 转化. 所以 `.` 并不会转义
## 第44题
``````function foo(a) {
var a;
return a;
}
function bar(a) {
var a = 'bye';
return a;
}
[foo('hello'), bar('hello')]
``````
# 总结
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10. Linux Doxygen的安装和使用 | 4,584 | 11,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-50 | latest | en | 0.172749 |
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## A solenoid 15.0 cm long has 600 turns and carries a current of 5.0 A. if 2.0 cm segment of a current balance arm is balanced inside the solenoid when the magnetic force on the segment 1x10-5 N. the current in it is O a. 100 mA O b. 2 mA Oc. 300 mA Od. 20 mA
#### STEP 1
1. The solenoid is assumed to be ideal, meaning it has a uniform magnetic field inside. 2. The length of the solenoid is $l = 15.0 \, \text{cm} = 0.15 \, \text{m}$. 3. The number of turns in the solenoid is $N = 600$. 4. The current through the solenoid is $I_s = 5.0 \, \text{A}$. 5. The segment of the current balance arm inside the solenoid is $L = 2.0 \, \text{cm} = 0.02 \, \text{m}$. 6. The magnetic force on the segment is $F = 1 \times 10^{-5} \, \text{N}$. 7. The magnetic permeability of free space is $\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}$.
#### STEP 2
To find the current in the segment of the current balance arm, we will follow these steps: 1. Calculate the magnetic field inside the solenoid using the formula $B = \mu_0 \frac{N}{l} I_s$. 2. Use the formula for the magnetic force on a current-carrying conductor $F = B I L$ to solve for the current $I$ in the segment.
#### STEP 3
Calculate the magnetic field inside the solenoid using the formula: $B = \mu_0 \frac{N}{l} I_s$
#### STEP 4
Substitute the given values into the formula: $B = (4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}) \frac{600}{0.15 \, \text{m}} (5.0 \, \text{A})$
#### STEP 5
Simplify the expression inside the parentheses: $B = (4\pi \times 10^{-7}) \frac{600}{0.15} (5.0)$
#### STEP 6
Calculate the fraction: $\frac{600}{0.15} = 4000$
#### STEP 7
Substitute back into the equation: $B = (4\pi \times 10^{-7}) (4000) (5.0)$
#### STEP 8
Simplify further: $B = 4\pi \times 10^{-7} \times 20000$
#### STEP 9
Calculate the product: $B = 8\pi \times 10^{-3} \, \text{T}$
#### STEP 10
Calculate the value of $8\pi$: $8\pi \approx 25.1327$
#### STEP 11
Substitute back into the equation: $B \approx 25.1327 \times 10^{-3} \, \text{T}$
#### STEP 12
Convert to standard form: $B \approx 2.51327 \times 10^{-2} \, \text{T}$
#### STEP 13
Use the formula for the magnetic force on a current-carrying conductor: $F = B I L$
#### STEP 14
Rearrange the formula to solve for the current $I$: $I = \frac{F}{B L}$
#### STEP 15
Substitute the given values into the formula: $I = \frac{1 \times 10^{-5} \, \text{N}}{2.51327 \times 10^{-2} \, \text{T} \times 0.02 \, \text{m}}$
#### STEP 16
Calculate the product in the denominator: $2.51327 \times 10^{-2} \times 0.02 = 5.02654 \times 10^{-4}$
#### STEP 17
Substitute back into the equation: $I = \frac{1 \times 10^{-5}}{5.02654 \times 10^{-4}}$
#### STEP 18
Calculate the division: $I \approx 1.989 \times 10^{-2} \, \text{A}$
#### STEP 19
Convert to mA: $I \approx 19.89 \, \text{mA}$
##### SOLUTION
Round to two significant figures: $I \approx 20 \, \text{mA}$ The current in the segment of the current balance arm is $20 \, \text{mA}$. | 1,075 | 3,010 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.659918 |
https://ramseyelbasheer.io/2022/06/16/18-non-cliche-datasets-for-beginner-data-scientists-to-build-a-strong-portfolio/ | 1,656,142,361,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00372.warc.gz | 529,205,748 | 20,119 | # 18 Non-Cliché Datasets for Beginner Data Scientists to Build a Strong Portfolio
Original Source Here
# 18 Non-Cliché Datasets for Beginner Data Scientists to Build a Strong Portfolio
## Unique datasets ranging from microbiology to sports
You might have already developed a slightly sickened feeling when you constantly saw people using the same datasets over and over (and over) again.
You just can’t help it — everyone wants the easy thing. Beginners use datasets like Titanic, Iris, and Ames Housing Dataset because they are stupidly straightforward; most course creators and bloggers use them because they are just a single Google search away (or even bookmarked).
In 115+ articles I have written, I honestly can’t remember using any of those clichés (if my memory is letting me down and did use them one or more times, I apologize!). This is mainly thanks to the many hours I have spent searching for a good dataset and delivering content in novel ways to my precious audience.
Today, I have decided to share a list of curated example datasets I used in my posts and as part of my learning. Enjoy!
# Regression datasets
## 1️. Diamond prices and carat regression
My favorite from this list is the diamonds dataset. It is ideal in length for practice (+50k samples) and has multiple targets you can predict as a regression or a multi-class classification task:
🎯 Targets: ‘carat’ or ‘price’
📦Dimensions: (53940, 10)
⚙Missing values: No
📚Starter notebook
## 2️. Age of Abalone shells
This is a unique dataset from the field of zoology. The task is to predict the age of Abalone shells (a type of mollusk) using several physical measurements. Traditionally, their age is found by cutting through their cone, staining them, and counting the number of rings inside the shell under a microscope.
For zoologists, this might be fun, but for data scientists, not so much:
🎯 Target: ‘Rings’
📦Dimensions: (4177, 9)
⚙Missing values: No
📚Starter notebook
## 3️. King county house sales
This is the dataset for those who are still interested in real estate and house prices regression:
🎯 Target: ‘price’
📦Dimensions: (21613, 17)
⚙Missing values: Yes
📚Starter notebook
## 4️. Cancer death rate
This dataset challenges you to find cancer mortality rate per capita (100,000) using several demographic variables:
🎯 Target: ‘TARGET_deathRate’
📦Dimensions: (3047, 33)
⚙Missing values: Yes
## 5️. Life expectancy
How long will a person live? This is one of the hardest questions unanswered in science. Several studies have been undertaken to understand human life and longevity, and this dataset provided by WHO (World Health Organization) is one of them:
🎯 Target: ‘Life expectancy.’
📦Dimensions: (2938, 21)
⚙Missing values: Yes
📚Starter notebook
## 6️. Car prices
The title says it all — predict car prices using variables like mileage, fuel type, transmission, and several domain-specific features. This is also an excellent dataset for pumping out your feature engineering muscles:
🎯 Target: ‘selling_price’
📦Dimensions: (8128, 12)
⚙Missing values: Yes
📚Starter notebook
# Binary classification
## 7️. NBA rookie stats
The first binary classification dataset in the list requires you to predict if a rookie basketball player will last more than 5 years in the league:
🎯 Target: ‘TARGET_5Yrs’
📦Dimensions: (8128, 12)
⚙Missing values: Yes
📚Starter notebook
## 8️. Stroke prediction
Another medical dataset asks you to predict whether a patient will have a stroke or not based on their history with interesting features:
🎯 Target: ‘stroke’
📦Dimensions: (5110, 11)
⚙Missing values: Yes
📚Starter notebook
## 9️. Water potability
Safe drinking water is the most basic human right and a major influencer on health. Using this dataset, you should classify water bodies into potable (drinkable) and not potable using several chemical properties:
🎯 Target: ‘Potability’
📦Dimensions: (3276, 10)
⚙Missing values: Yes
📚Starter notebook
## 10. Smart grid stability
This is an augmented version of the “Electrical Grid Stability Simulated Dataset” created by Vadim Arzamasov. It is donated to UCI and made available on Kaggle. You will be predicting the stability of 4-node smart grid systems (whatever they mean):
🎯 Target: ‘stabf’
📦Dimensions: (60000, 13)
⚙Missing values: No
📚Starter notebook
## 1️1. IBM HR analytics & employee attrition
This fictional dataset created by IBM datasets tasks you to uncover which factors lead to employee attrition (whether they will leave their role):
🎯 Target: ‘Attrition’
📦Dimensions: (1470, 35)
⚙Missing values: No
📚Starter notebook
## 1️2. Can I eat this mushroom?
Another one-of-a-kind dataset is classifying mushrooms into edible and poisonous. It also presents a unique challenge — all features are categorical:
🎯 Target: ‘class’
📦Dimensions: (8124, 23)
⚙Missing values: Yes
📚Starter notebook
## 1️3️. Banknote authentication
Even though this dataset has very few features, I wanted to include it because the task is really interesting — using physical attributes of banknotes, you should classify them into forged or original:
🎯 Target: ‘class’
📦Dimensions: (1372, 5)
⚙Missing values: No
📚Starter notebook
Predict whether a person will end up earning more than 50k using factors like age, education, background, gender, marital status, etc.:
🎯 Target: ‘income’
📦Dimensions: (48842, 15)
⚙Missing values: Yes
📚Starter notebook
# Multi-class classification datasets
## 1️5️. Yeast classification
This dataset will give you a small taste from the world of microbiology. You are tasked to classify a fungus called yeast into species:
🎯 Target: ‘class_protein_localization’
📦Dimensions: (1484, 9)
⚙Missing values: No
## 1️6️. Kaggle TPS May 2021
Kaggle hosts monthly competitions called the “Tabular Playground Series” with beginner-to-medium difficult tasks. The most important point is that a new synthetic dataset of considerable size is created each month using the CTGAN framework. This one is from the May edition.
🎯 Target: ‘target’
📦Dimensions: (100000, 52)
⚙Missing values: No
📚Starter notebook
## 1️7️. Kaggle TPS June 2021
A similar dataset with more features and samples:
🎯 Target: ‘target’
📦Dimensions: (200000, 77)
⚙Missing values: No
📚Starter notebook
## 1️8️. Diamonds, again
Just mentioning the diamonds dataset again because it has three categorical features, which can be multi-class targets on their own:
🎯 Targets: ‘cut’, ‘color’, ‘clarity’ | 1,670 | 6,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-27 | latest | en | 0.910764 |
http://www.coderanch.com/t/481330/Oracle-OAS/answer-correct | 1,467,093,775,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396459.32/warc/CC-MAIN-20160624154956-00148-ip-10-164-35-72.ec2.internal.warc.gz | 452,515,050 | 10,689 | This week's book giveaway is in the Cloud/Virtualizaton forum.We're giving away four copies of Mesos in Action and have Roger Ignazio on-line!See this thread for details.
Win a copy of Mesos in Action this week in the Cloud/Virtualizaton forum!
Sona Patel
Ranch Hand
Posts: 75
Hi...
I am reading about using 'order by' clause. I came across following question -
The results from an SQL query are shown here.
Which of the following SQL statements could not have produced this output?
A. select deptno,dname,loc from dept order by 2 asc, 1 asc, 3 desc;
B. select deptno,dname,loc from dept order by 3 asc;
C. select deptno,dname,loc from dept order by 2 asc;
D. select deptno,dname,loc from dept order by 2 asc, 3 asc, 1 desc;
I think correct ans is (B.). But book says it is (C.). What do you think?
Jan Cumps
Bartender
Posts: 2588
11
Hi Sona,
If you put an exam question on the Ranch, you have to tell where you got the question from.
Sona Patel
Ranch Hand
Posts: 75
Its from OCP introduction to oracle 9i by Jason Couchman...
Fatih Keles
Ranch Hand
Posts: 182
Hi Sona,
I agree with you, and also my local 11g r1 database does.
Regards,
Fatih.
Sona Patel
Ranch Hand
Posts: 75
Hey Fatih...
Thanks...
How ignorant of me...i forgot i can try it out. thanks for reminding me
Regards,
Eric Pascarello
author
Rancher
Posts: 15385
6
swapnil kachave
Greenhorn
Posts: 27
order by 2 asc means it doing order by dname in ascending order.
if you check the alphabetic order of colom 2 you will find the answer c is right.
if you are saying the answer is (b) then check the order by statement which say order by LOC in ascending order. which is not in alphabeticall order so that answer (c) is right.
Sona Patel
Ranch Hand
Posts: 75
Hi Swapnil...
Option (A),(C),(D) will produce the output as shown in the question.
Option (B) will not produce the same output right ??
Question is :
Which of the following SQL statements could not have produced this output?
So the ans is (B) could not have produced the same out put.
swapnil kachave
Greenhorn
Posts: 27
Sona Patel wrote:Hi Swapnil...
Option (A),(C),(D) will produce the output as shown in the question.
Option (B) will not produce the same output right ??
Question is :
Which of the following SQL statements could not have produced this output?
So the ans is (B) could not have produced the same out put.
Yes..
If you go for b then check the order by clouse where you specify the different column no and output not contain that column in sorted oreder
Vinod Tiwari
Ranch Hand
Posts: 466
1 | 694 | 2,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2016-26 | latest | en | 0.901633 |
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Population of Republic of Bondia = Total population of six islands
Total population of six islands = 8,628,709
10% of 8,628,709 = 862,870
Islands which have less than 10% of population of Bondia = Brosnan and Dalton
Out of Brosnan and Dalton, we need to know which island’s area is more than 20% of the area of Bondia.
Total area of Bondia = 26315
20% of 26315 = (10% of 26315) x 2 = 2631.5 x 2 ≈ 5200 km^2
Island with area more than 5200 = Dalton
Brosnan has an area of 5079 which is less than 5200 so that cannot be the answer. | 175 | 536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-33 | latest | en | 0.919322 |
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# Critical Speed Calculation Ball Mill
How to calculate ball mill efficiency.Mill speed - critical speed - paul o abbeo matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar rolling mill, its rotational speed is important to proper and efficient mill operationcalculating how fast a jar needs to.
• ### How To Calculate Critical Speed Of Ball Mill
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How can i determine the best rpm for dry ball milling critical speed in rpm 42.3sqrtd - d with d the diameter of the mill in meters and d the diameter of the largest grinding ball you will use for experiment also.
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The critical speed of a sag mill is the speed in rpms at which centrifugal force causes the material being ground to be held against the inside of the shell.This speed is only dependent on the.
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Optimization of mill performance by using online ball and pulp measurements j o u r n a l p a p e r the journal of the southern african institute of mining and metallurgy volume 110 non-refereed paper march 2010 135 table i influence of speed and liner design on load dynamics mill speed soft design aggressive design.
• ### Formula For Ball Mill Critical Speed Ecole Creperie84
Critical speed the critical speed for a grinding mill is defined as the.The basic formula for this is the bond formula.Hence a rod mill generally precedes a ball mill in a grinding circuit especially where a fine size product is required.
• ### Metric Critical Speed Calculator Nook Industries
Metric critical speed calculator.Critical speed enter any two of the following and click on the calculate button.
• ### Critical Speeds Of Rotating Shafts With Single Loads
Critical speeds of rotating shafts with single loads when calculating critical speeds, the weight or mass of the rotating cylinder or shaft is assumed to be zero or add 12 to 23 of the rotating shaft to the load mass.Keep in mind that a shaft with more than one load or distributed loads may have an infinite number of critical speeds.
• ### Ball Mill Speed Calculation Formula For Wet Grinding
Ball mill speed calculation formula for wet grinding.20199192rotation speed calculation of ball mill critical speed when the ball mill cylinder is rotated, there is no relative slip between the grinding medium and the cylinder wall, and it just starts to run in a state of rotation with the cylinder of the mill this instantaneous speed of the mill is as follows n0 mill working speed, rmin. | 995 | 4,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-34 | latest | en | 0.779172 |
http://exxamm.com/QuestionSolution5/Aptitude/What+would+be+the+volume+that+Mr+Harshwardhan+should+train+where+he+would+be+indifferent+between+setting+up+a+comedy+l/2247645583 | 1,553,494,161,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203755.18/warc/CC-MAIN-20190325051359-20190325073359-00540.warc.gz | 76,527,589 | 10,134 | What would be the volume that Mr. Harshwardhan should train, where he would be indifferent between setting up a comedy l
Question Asked by a Student from EXXAMM.com Team
Q 2247645583. Directions: Study the information carefully to answer the following questions.
Mr. Harshwardhan is facing a decision problem. He has excellent training products but is not sure about the demand for his abilities. He wants to set up a training centre to provide training programmes of Dance, Action and comedy level. His financial advisor Mr. Anil kapoor told him that if he wants to set up a comedy level training centre, the total cost would be on two counts - the first would be a fixed would cost which is Rs 2 lakh per annum. Besides, it would also entail a variable cost of training per candidate. This would be Rs 1000 per candidate. He further estimated that if a training centre is set up for conducting Action and comedy level training programmes, the total fixed cost would be Rs 3.2 lakh per annum and the cost of training per candidate will be Rs 750. Mr. Anil kapoor motivates Mr. Harshwardhan to set up a combined training centre for Dance, Action and comedy, the fixed cost of which is Rs 5 lakh per annum and the cost of providing training per candidate is Rs 500. You as a expert from FTII, have some decision making abilities. Please help Mr. Harshwardhan by answering the following questions:"
What would be the volume that Mr. Harshwardhan should train, where he would be indifferent between setting up a comedy level and comedy and action level training centre?
IBPS-PO 2017 Mock Mains
A
495
B
490
C
480
D
475
E
None of these
HINT
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)
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https://cr4.globalspec.com/thread/76801/Neutral?sort=linear&order=asc | 1,701,471,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00547.warc.gz | 232,440,719 | 24,528 | Previous in Forum: Arc Suppression Next in Forum: Broyce ELR "Test Button" No Trip
Associate
Join Date: Oct 2009
Location: kuruman NC. SOUTH AFRICA
Posts: 51
# Neutral
03/23/2012 4:00 PM
Hi Guys
there is alot of confusion regarding the term flowding neutral.
can somebody please explain the flowding neutral in detail please.
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Dan Segami
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### Comments rated to be Good Answers:
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2
Guru
Join Date: Apr 2010
Location: Mid Western USA - The Corn Belt
Posts: 1439
#1
### Re: Neutral
03/23/2012 4:22 PM
Hi Danny,
I assume you are referring to a "floating" neutral.
If that is the case have a look at some of these.
Regards - KJK
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Good Answer (Score 2)
Power-User
Join Date: Nov 2011
Posts: 171
#2
### Re: Neutral
03/23/2012 4:31 PM
If your neutral is disconnected from ground or not connected to the ground, it is called that your neutral is a floating neutral.
Analysis of a floating vs. grounded output
Power-User
Join Date: Apr 2010
Location: Dominican Republic
Posts: 189
#3
### Re: Neutral
03/24/2012 8:06 AM
A floating neutral can be a serious problem. Suppose you have a breaker panel with 2 Line inputs and a neutral. The voltage between each line is 240 and the voltage between each line and the neutral is 120. You have single breakers feeding loads that require 120Volts. These 120Volt loads have one line fed by the breaker and a neutral. The double breakers have 2 lines and no neutral for 240Volt loads. Now suppose the Neutral gets loose or oxidized or somehow disconnected in the panel or maybe even out where the power comes from. The 240Volt loads will be unaffected however the 120V loads can be in serious trouble. With this Floating neutral condition you will discover that one of the two lines will go from 120Volts up to 180 or 190 and the other line will go down to 50 or 60 volts. Half of your 120Volt equipment will go up in smoke and the other half will not function due to a low voltage condition.
So, be careful with floating neutrals.
John
2
Commentator
Join Date: Jul 2010
Location: North West Province, South Africa
Posts: 67
#4
### Re: Neutral
03/24/2012 4:15 PM
Hi Danny,
You are 100% right - in South Africa there seems to be a lot of confusion regarding the cause and effects of a floating neutral.
As you are aware, in SA our supply line voltage is phase to phase 400v and phase to neutral 230v (may vary slightly from area to area). The supply comes from the secondary of the stepdown transformer where the output is connected in Star where the Star point becomes the Neutral and this is earthed at this point.
Now in order to explain the phenomenon of a 'floating neutral' we should look at a building which is supplied by from a three phase + neutral + earth supply connection - everything would operate correctly if the neutral remained connected.
Should the neutral be disconnected (in SA usually stolen) the building wiring system would now be effectively connected in Star (all the neutral wires from the various 3 phase circuits taken to a common neutral bar). So long as the three phases are balanced (equally loaded) the neutral would be stable but as soon as an unbalanced load is connected the unbalance affects the voltage at the Star point (the neutral bar) and this is your 'floating neutral' which would continually vary according to the unbalance across the three phases.
The practical effect of this is that instead of the phase to neutral voltage being 230vac you are likely to measure between 280 & 320vac which is why so much damage is done to appliances etc. connected to a system where the neutral has been disconnected.
In other countries where I have worked a Protective Multiple Earthing (PME) system is used where the neutral line is earthed at multiple points e.g. at every distribution board and this helps to overcome this problem.
In SA the cable theft problem (in a lot of cases just the neutral & earth wires) has become a major problem not only because of the loss of cable but also the resultant damage to the end user's appliances.
Hope this helps you!
Regards, Keith
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Interested in this topic? By joining CR4 you can "subscribe" to | 1,061 | 4,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.8982 |
https://www.dummies.com/article/business-careers-money/business/accounting/general-accounting/how-to-prepare-a-cost-volume-profit-analysis-168044/ | 1,696,381,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00312.warc.gz | 818,243,615 | 18,529 | ##### Managerial Accounting For Dummies
Contribution margin indicates how sales affects profitability. Cost-volume-profit analysis helps you understand different ways to meet your net income goals. When running a business, a decision-maker or managerial accountant needs to consider how four different factors affect net income:
• Sales price
• Sales volume
• Variable cost
• Fixed cost
The graphs provide a helpful way to visualize the relationship among cost, volume, and profit. However, when solving problems, you’ll find that plugging numbers into formulas is much quicker and easier.
## Draft a cost-volume-profit graph
Pemulis Basketballs sells basketballs for \$15 each. The variable cost per unit of the basketballs is \$6. Pemulis had total fixed costs of \$300 per year.
Fixed costs are represented by a horizontal line because no matter the sales volume, fixed costs stay the same. Total variable costs are a diagonal line, starting at the origin (the point in the lower-left corner of the graph where there are zero sales).
Total costs (the sum of total variable costs and total fixed costs) are a diagonal line starting at the \$300 mark because when the company makes and sells zero units, total costs equal the fixed costs of \$300. Total costs then increase with volume. Finally, total sales forms a diagonal line starting at the origin and increasing with sales volume.
The figure shows when the company will earn net income or incur a loss. When the sales curve exceeds total costs, the company earns net income (represented by the shaded right side of the X). However, if total sales is too low to exceed total costs, then the company incurs a net loss (the shaded left side of the X).
The higher the sales volume — that is, the more sales volume moves to the right of the graph — the higher the company’s net income.
Dropping numbers into the chart shows exactly how much income can be earned at different sales levels. Assuming Pemulis has a sales price of \$15 per unit, a variable cost per unit of \$6, and total fixed costs of \$300, what happens if Pemulis sells 60 basketballs?
Total sales come to \$900 (60 units x \$15). Total variable costs multiply to \$360 (60 units x \$6). Add these total variable costs to total fixed costs of \$300 to get total costs of \$660.
Total sales (\$900) sits on the Total sales line. Total costs (\$660) sits on the Total cost line. The difference between these amounts (\$240) represents the net income from selling 60 units.
## Try out the total contribution margin formula
The following formula, based on total contribution margin, follows the same structure as the contribution margin income statement.
Net income = Total contribution margin – Fixed costs
Assume that Pemulis Basketballs sells 60 units for \$15 each for total sales of \$900. The variable cost of each unit is \$6 (so total variable costs come to \$6 x 60, or \$360), and total fixed costs are \$300. Using the contribution margin approach, you can find the net income in two easy steps.
1. Calculate total contribution margin.
Use the formula to compute total contribution margin, subtracting total variable costs from total sales:
Total contribution margin = Total sales – Total variable costs = (60 x \$15) – (60 x \$6) = \$540
This total contribution margin figure indicates that selling 60 units increases net income by \$540.
2. To calculate net income, subtract the fixed costs from the total contribution margin.
Just plug in the numbers from Step 1:
Net income = Total contribution margin – Fixed costs = \$540 – \$300 = \$240
Subtracting fixed costs of \$300 from total contribution margin of \$540 gives you net income of \$240.
## The contribution margin per unit formula
If you know the contribution margin per unit, the following approach lets you use that information to compute net income. Here’s the basic formula equating net income with contribution margin per unit:
Net income = (Sales volume x Contribution margin per unit) –Fixed costs
Say Pemulis Basketballs now wants to use this formula. It can simply plug in the numbers — 60 units sold for \$15 each, variable cost of \$6 per unit, fixed costs of \$300 — and solve. First compute the contribution margin per unit:
Contribution margin per unit = Sales price per unit – Variable costs per unit = \$15 – \$6 = \$9
Next, plug contribution margin per unit into the net income formula to figure out net income:
Net income = (Sales volume x Contribution margin per unit) –Fixed costs = (60 x \$9) – \$300 = \$240
## The contribution margin ratio formula
If you want to estimate net income but don’t know total contribution margin and can’t find out the contribution margin per unit, you can use the contribution margin ratio to compute net income.
You can compute contribution margin ratio by dividing total contribution margin by total sales. So if your contribution margin is 540 and your sales is 900, your contribution margin ratio is 60 percent:
This means that 60 cents of every sales dollar directly increases net income. After you know the contribution margin ratio, you’re ready for the net income formula:
Net income = (Sales x Contribution margin ratio) – Fixed costs
To calculate net income for the earlier example company, plug the contribution margin ratio of 60 percent into the formula:
Net income = (Sales x Contribution margin ratio) –Fixed costs = (900 x 60%) – \$300 = \$240
Mark P. Holtzman, PhD, CPA, is Chair of the Department of Accounting and Taxation at Seton Hall University. He has taught accounting at the college level for 17 years and runs the Accountinator website at www.accountinator.com, which gives practical accounting advice to entrepreneurs. | 1,208 | 5,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-40 | longest | en | 0.907934 |
https://www.physicsforums.com/threads/differentiating-mult-variable-equation.11115/ | 1,477,186,421,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00103-ip-10-171-6-4.ec2.internal.warc.gz | 962,343,275 | 13,985 | # Differentiating mult-variable equation
1. Dec 16, 2003
### dbrag
Any ideas for solving this, im having trouble using implicit differentiation along with using log differentiation, thanx!:
x^3 + x tan^-1 y = e^y
2. Dec 16, 2003
### himanshu121
Differentiate w.r.t x
$$\frac{df(y)}{dx}=\frac{df(y)}{dy}\frac{dy}{dx}$$
3. Dec 17, 2003
### HallsofIvy
Staff Emeritus
In other words, the derivative of y with respect to x is:
3x2+ tan-1y+ xy'/(1+ y2)= eyy'. Now solve for y'.
I see no reason to use "logarithmic differentiation". | 181 | 538 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-44 | longest | en | 0.751822 |
https://financial-dictionary.thefreedictionary.com/front-end+ratio | 1,660,361,640,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571869.23/warc/CC-MAIN-20220813021048-20220813051048-00362.warc.gz | 241,016,319 | 12,120 | front-end ratio
Front-End Ratio
A ratio of an individual's monthly mortgage expenses to his/her monthly income. The expenses used in this calculation are usually the principal, interest, taxes, and insurance that an individual owes on a monthly basis. Mortgage lenders often use front-end ratios to determine whether an individual has sufficient income in order to qualify for a mortgage. Generally speaking, lenders look for a front-end ratio of less than 0.30 - 0.33. Persons with ratios in excess of that have more difficulty securing mortgages. See also: Back-end ratio.
front-end ratio
A mortgage qualification calculation prepared by taking the proposed monthly mortgage payments, plus real estates taxes and insurance, and dividing that number by the borrower's gross monthly income without reduction for taxes.
Example: Steve makes \$4,000 per month. The mortgage for a home he would like to buy would result in payments of \$1,100 per month. His front-end ratio is \$1,100 \$4,000 27.5%. This is an acceptable ratio for lenders, who would prefer to keep it at 29 percent or lower.
The Complete Real Estate Encyclopedia by Denise L. Evans, JD & O. William Evans, JD. Copyright © 2007 by The McGraw-Hill Companies, Inc.
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References in periodicals archive ?
The front-end ratio calculates the yearly gross income devoted towards making the monthly payment, which consist of three components which are principal, interest and insurance.
If an employee earning those wages applied for a mortgage, his or her front-end ratio would only allow for \$630 in housing costs, which would include taxes, homeowner's insurance and home association dues, in addition to the monthly mortgage payment.
In line with other studies (Quercia, McCarthy, and Wachter 2003), we measure mortgage consumption in two ways: (1) as the dollar amount of the monthly mortgage payment, and (2) as the ratio of the monthly mortgage payment to monthly household income, referred to here as the front-end ratio. The monthly mortgage payment is derived from administrative data at the time of origination, and includes principal, interest, taxes, insurance, and private mortgage insurance.
As shown in Table 2, the average mortgage payment for borrowers in our sample is \$815, based on an average purchase price of \$102,007, with a resulting average front-end ratio of 22.6% (ranging from 7.7% to 51.6%).
(10) Most real estate lenders cap the front-end ratio at 28 percent and
Administration (FHA) will in some circumstances allow a front-end ratio
While both of these studies analyzed the correlation between the back-end ratio and default risk, an analysis of 179 FHA loans originated in Utah between 2000 and 2001 estimated the impact of a borrower's front-end ratio as well.
If the lender says that the front-end ratio cannot exceed 32%, this means that his PITI divided by his gross monthly income must equal 32% or less.
There are two mortgage qualification ratios widely used--the front-end ratio and the back-end ratio.
The front-end ratio (or housing-cost-to-income ratio) is monthly housing expenses (principal, interest, taxes, and insurance, or PITI) divided by gross monthly income.
(The front-end ratio is the monthly housing debt (PITI) divided by the borrower's monthly gross income; the back-end ratio is a borrower's total monthly debt divided by monthly gross income.) Rather, we've found the nonhousing ratio - the difference between the front- and back-end ratio - to be much more reliable because it accurately measures the borrower's cash-flow situation.
Hans noted that front-end ratios were lifted by a move in the sales mix toward higher-margin categories, especially digital photo processing.
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Lies all around (Posted on 2011-06-04)
When Mr. Mathwiz visited the IoK&L* he was surrounded by 20 or 21 male inhabitants, who formed a perfect circle and then each said a single word (either "Knight" or "Liar") about the man on his right.
The wiz then evaluated correctly the exact numbers of liars and the truth-tellers.
What were those numbers?
*IoK&L, is an imaginary island, inhabitated by Liars and by Truth-Tellers a.k.a. Knights.
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Knights and liars | Comment 1 of 10
If all of the inhabitants were of the opposite type, then they would have said the same thing. Therefore, whatever they said, there are two solutions with opposite types for each person. Make the number of people x. If one solution has k knights, then there are x-k liars, so the other solution has x-k knights and k liars. For Mr. Mathwiz to know the number of knights and liars, x-k and k must be the same number. Therefore, x=2k, so there are an even number of inhabitants. That means that there are 20 inhabitants, so x=20 and k=10. Therefore, there are 10 knights and 10 liars.
Edited on June 4, 2011, 2:20 pm
Posted by Math Man on 2011-06-04 14:18:18
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https://www.physicsforums.com/threads/kinematics-question-rocket-launch.708866/ | 1,709,612,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948217723.97/warc/CC-MAIN-20240305024700-20240305054700-00104.warc.gz | 929,420,846 | 13,914 | # Kinematics question - Rocket launch
In summary, the conversation discusses a homework question about the vertical acceleration and height of a rocket launched from the surface of the Earth. The first 10 seconds of its motion is given by an equation and the task is to find the height at t=10s and the speed when it reaches 200m above the surface. The student attempts to solve the problem but struggles with finding the correct answer. It is suggested to use calculus instead of kinematic equations to solve for the changing acceleration.
Kinematics question -- Rocket launch
## Homework Statement
A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by ay=(2.70m/s3)t, where the +y-direction is upward.
A.What is the height of the rocket above the surface of the Earth at t = 10s?
B.What is the speed of the rocket when it is 200m above the surface of the earth?
## Homework Equations
y=y(not)+V(not y)*t-(1/2)*g*t^2
v=V(not y)+a*t
## The Attempt at a Solution
I am stuck on part A and need some help. I used the first equation to find an answer that was wrong. Here is my work:
ay=2.7*(10) =27m/s^3
I then divided that by 9.8 m/s^2 and then multiplied it by 100.
y=0+0-(1/2)*(27/9.8)*(10)^2
y=-860 m
I don't believe the actual answer is going to be negative if the up direction is considered (+). I tried putting (+) 860 in and it said it was wrong. If someone could point me in the right direction for Part A and B that would be much appreciated.
You're given the vertical acceleration of the rocket so it must already take into account all the forces acting, including that due to gravity. Since acceleration is varying over time (it's not constant) you must abandon all your kinematic equations and go back to basics. In other words, a little calculus is called for.
## 1. What is kinematics?
Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.
## 2. What is a rocket launch?
A rocket launch is the process of sending a spacecraft or missile into space by means of a rocket.
## 3. How is the velocity of a rocket calculated during launch?
The velocity of a rocket during launch is calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity (usually 0), a is the acceleration, and t is the time.
## 4. What factors affect the trajectory of a rocket during launch?
The trajectory of a rocket during launch is affected by factors such as the thrust of the rocket, the mass of the rocket, air resistance, and gravity.
## 5. Can kinematics equations be used to predict the motion of a rocket during launch?
Yes, kinematics equations can be used to predict the motion of a rocket during launch as long as the forces acting on the rocket are known and are constant throughout the launch. | 684 | 2,925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-10 | latest | en | 0.946859 |
http://de.metamath.org/mpeuni/cdleme31sdnN.html | 1,659,969,196,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00679.warc.gz | 12,654,364 | 4,284 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cdleme31sdnN Structured version Visualization version GIF version
Theorem cdleme31sdnN 34693
Description: Part of proof of Lemma E in [Crawley] p. 113. (Contributed by NM, 31-Mar-2013.) (New usage is discouraged.)
Hypotheses
Ref Expression
cdleme31sdn.c 𝐶 = ((𝑠 𝑈) (𝑄 ((𝑃 𝑠) 𝑊)))
cdleme31sdn.d 𝐷 = ((𝑡 𝑈) (𝑄 ((𝑃 𝑡) 𝑊)))
cdleme31sdn.n 𝑁 = if(𝑠 (𝑃 𝑄), 𝐼, 𝐶)
Assertion
Ref Expression
cdleme31sdnN 𝑁 = if(𝑠 (𝑃 𝑄), 𝐼, 𝑠 / 𝑡𝐷)
Distinct variable groups: 𝑡, 𝑡, 𝑡,𝑃 𝑡,𝑄 𝑡,𝑈 𝑡,𝑊 𝑡,𝑠
Allowed substitution hints: 𝐶(𝑡,𝑠) 𝐷(𝑡,𝑠) 𝑃(𝑠) 𝑄(𝑠) 𝑈(𝑠) 𝐼(𝑡,𝑠) (𝑠) (𝑡,𝑠) (𝑠) 𝑁(𝑡,𝑠) 𝑊(𝑠)
Proof of Theorem cdleme31sdnN
StepHypRef Expression
1 cdleme31sdn.n . 2 𝑁 = if(𝑠 (𝑃 𝑄), 𝐼, 𝐶)
2 biid 250 . . 3 (𝑠 (𝑃 𝑄) ↔ 𝑠 (𝑃 𝑄))
3 vex 3176 . . . 4 𝑠 ∈ V
4 cdleme31sdn.d . . . . 5 𝐷 = ((𝑡 𝑈) (𝑄 ((𝑃 𝑡) 𝑊)))
5 cdleme31sdn.c . . . . 5 𝐶 = ((𝑠 𝑈) (𝑄 ((𝑃 𝑠) 𝑊)))
64, 5cdleme31sc 34690 . . . 4 (𝑠 ∈ V → 𝑠 / 𝑡𝐷 = 𝐶)
73, 6ax-mp 5 . . 3 𝑠 / 𝑡𝐷 = 𝐶
82, 7ifbieq2i 4060 . 2 if(𝑠 (𝑃 𝑄), 𝐼, 𝑠 / 𝑡𝐷) = if(𝑠 (𝑃 𝑄), 𝐼, 𝐶)
91, 8eqtr4i 2635 1 𝑁 = if(𝑠 (𝑃 𝑄), 𝐼, 𝑠 / 𝑡𝐷)
Colors of variables: wff setvar class Syntax hints: = wceq 1475 ∈ wcel 1977 Vcvv 3173 ⦋csb 3499 ifcif 4036 class class class wbr 4583 (class class class)co 6549 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-rex 2902 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-nul 3875 df-if 4037 df-sn 4126 df-pr 4128 df-op 4132 df-uni 4373 df-br 4584 df-iota 5768 df-fv 5812 df-ov 6552 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 1,239 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.133338 |
https://www.midasbridge.com/en/blog/casestudy/traffic-load-consideration-to-different-types-of-bridges | 1,695,623,313,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00823.warc.gz | 994,139,692 | 37,584 | BLOG CASE STUDY
## 1. Introduction to Traffic Load Consideration
Loads on bridges can be permanent and transient. Live Loads on bridges are typically dynamic in nature. These could be due to traffic, rolling stocks, locomotive, etc. The vertical and horizontal components could be analyzed as static analysis with appropriate dynamic amplification factors to account for the dynamic effects based on the type of bridges adhering to code provisions. Live loads applied on the structures, unlike permanent loads, do not have a fixed position on the bridge structures and can be placed at critical positions as the traffic moves over a span.
Hence, we would require the Moving Load Analysis, typically a Static Load Analysis finding the worst load effects out of various possible positions along the bridge based on the Influence Line analysis or influence surface analysis.
## 2. Significance of Influence Line Analysis
• Definition: The influence line for a response function is given by the deflected shape of the released structure due to a unit displacement (or rotation) at the location and the response function's direction.
1. Response Function = support reaction, axial force, shear force, or Influence Line Definitions bending moment.
2. Influence Line/Influence surface = graph of a response function of a structure as a function of the position of a downward unit load moving across the structure.
For example, in fig.1, for the two-span continuous bridge, to obtain the maximum sagging at point “A” span, which is at the mid-span, would be loaded to create a positive influence.
Similarly, if the second span is loaded, this would have a reliving effect for the maximum sagging moment at A.
Fig 2.1 Calculation of maximum/minimum moments due to a concentrated vehicle load (P)
(source Analysis ref. manual)
Fig 2.2 Calculation of maximum/minimum moments due to a distributed lane load (W)
(source Analysis ref. manual)
The effect of a moving load on the magnitude of certain functions of a structure, such as support reactions, deflection, and shear force and moment, at a section of the structure, varies with the moving load position.
In the program, a set of concentrated loads would be applied considering all the loading conditions along the moving direction of the traffic to find the maximum and minimum values.
Hence, Influence line analysis is important in determining the critical position of moving loads.
## 3. Moving Load analysis for various types of bridges
To find the most critical design parameters (member forces, displacements, and support reactions) in the analysis of a bridge structure, all the conditions of vehicle loads must be considered.
Especially, when a number of design vehicle load groups and traffic lanes are involved, all the conditions that may affect the design parameters must be analyzed:
b) if only the critical-case design vehicle load group is to be applied among the load groups;
d) what load reduction factor is to be applied if a number of traffic lanes are loaded.
Defining Moving Load definition as per the various standards is 4 step process in midas Civil.
Once the moving load analysis is completed the results will be generated using the influence line results.
Fig 3.1 Process flow for defining the moving load definition in midas Civil
I. Selecting the Moving Load Code: Moving Load can be selected through different standards.
II. Defining Traffic Line/Surface Lanes: Specify the centerline of the vehicle considering the
eccentric positions of the vehicles.
The distribution of the vehicular load is by two methods:
Lane element: Traffic Load would be applied directly to the reference elements (Vertical Concentrated load and moment). This method is applied in the case of line beam models.
Cross Beams: Traffic load will be applied at the actual location and will get distributed to longitudinal girders through Cross-Beams. Generally, this method is recommended for the Grillage Models.
Fig.3.2 Load distribution through Lane element and Cross Beams (source: online manual)
Another way of defining traffic lanes would be through the Moving Load Optimization Option.
Moving Load Optimization will automatically consider the placements of the lanes in the transverse and longitudinal direction along the entire carriageway considering the positions on the Left, Right, and Center.
Defining the individual traffic line lanes can be avoided through the moving load optimization option. Hence, the lanes are loaded in such a manner that it gives the most critical moving load envelops.
Fig.3.3 Vehicular position for Moving Load Optimization (source: online manual)
The standard load models according to different standards can be defined directly added.
However, in case any User-defined loads need to be added then that case such loads can be defined through three options based on the moving traffic load code as below
1. Truck/Lane
• Moving Load cases option allows combining the different classes of vehicles in one moving load case considering the minimum and maximum lanes to be loaded.
• Hence, the repetitive task of creating the individual load cases and then making a combination of all the vehicles can be avoided.
Independent
- Vehicles applied independently
• In the case of complex models consisting of numerous elements, running moving load analysis can be time-consuming. However, with the moving load analysis control selecting specific girder group elements can be chosen to reduce the analysis time.
• Hence, with the Moving load analysis control option variety of parameters such as selection filters for specific result group output can be controlled and hence the run-time of the model.
• It is always recommended to cross verify the program results with simple manual calculations
Fig.3.4 Moving Load Case definition in midas Civil
## 4. Case-Studies
### i. Line Beam Models - Case Study 1 (Rail Seat Loads)
Purpose: 2D Line beam model was created to obtain the maximum reactions at the sleeper or the rail pad under the rail.
Configuration
• Span: Multi-span continuous
• Alignment: Straight
• Section: UIC 60 kg/m rail
• Carriageway: Single Track
• Application: Traffic Line Lane
User-defined rail load can be directly specified using the User-defined vehicular load option and dynamic amplification factor for the same can be accounted for by just defining the scale factor.
However, scale factors can be used as Impact factor/CDA, Load combination factor, etc.
Dynamic amplification can however be applied at the time of creating the load combination in an excel sheet. However, applying it when creating the moving load cases would be easy to get the final results.
The results of the moving load cases can be converted to static load cases.
From the envelope of all load positions, the load position most critical for a particular node can be extracted into a static load case.
### i. Line Beam Models - Case Study 2 (Pedestrian Bridge)
Purpose: To directly incorporate the pedestrian load using the code reference and also to model the traffic live load on curvature.
The moving load on the curvature can be directly assigned just by specifying the elements numbers which are in a sequence that generally do not lie on the straight line.
Pedestrian load intensity is dependent on the loaded area which is predefined in midas Civil.
If for the continuous structures, different load intensities need not be assigned in case only a few or all the spans are loaded since midas Civil automatically calculates these intensities for the different loading patterns.
• Configuration
Span: 4 spans continuous
Alignment: Curved
Section: Steel box girder
Carriageway: 2 x 2.5m wide shared path
The location of the concentrated loads for the maximum sagging moments, shear, or the reactions can be obtained along with the influence line diagram
### i. Line Beam Models - Case Study 3 (Road Bridge)
Purpose: To carry out the structural assessment of the bridge for the T44/L44 vehicles.
In this case, the moving load optimization was used in the 2D line beam model in order to avoid the manual definition of the individual lanes.
However, this method can only be used when the individual lanes to be defined have the same properties.
Configuration
• Span: 2 spans continuous
• Alignment: Straight
• Section: Box girder
• Traffic lanes: 3 lanes
### ii. Grillage Model - Case Study 1 (Straight Bridge)
Purpose: To find the critical response due to moving loads in Grillage Models.
When the combining vehicles have similar vehicular configuration, traffic lane width, and wheel spacing, then the Moving load optimization can be used.
If different vehicles are required to be combined, different traffic lanes need to be created.
Configuration
• Span: 1 span simply supported
• Alignment: Straight
• Section: Super-T + Deck
• Carriageway: 6m, 3 lanes
• Application: Traffic lane for grillage, Combining vehicles
### ii. Grillage Model - Case Study 2 (Skew grillage)
Purpose: To find the critical response due to moving loads in Grillage Models.
Incorrect skew will leave some portion of the deck outside the traffic lane extents and hence traffic loads will not be run in this portion of the deck.
When the multiple lanes having the same vehicular configuration needs to be combined in the direction of the left skew for the right skew can be done by directly selecting the load effect as “Independent
Once the analysis is completed, the concurrent forces due to moving loads can be obtained.
Configuration
Span: 6 spans simply supported
Alignment: Skewed (30 degree)
Section: Plank + deck
Carriageway: 9m, 2 lanes
Application: Skew deck
### ii. Grillage Model - Case Study 3 (Rail+ Road Bridge)
Purpose: Due to different vehicular loads, moving load optimization cannot be used. Hence, in order that transverse lane optimization is required to be defined.
Since the available carriageway width is more than that of the design width, transverse lane optimization can be used.
However, when defining the Traffic lane for the railway loading transverse lane definition is not to be selected since the track position is fixed.
Configuration
• Span: 5 spans simply supported
• Alignment: Straight
• Section: Multi-girder bridge
• Moving load: 350LA + SM1600
• Carriageway: 1 Track + 1 Lane
### iii. Planer Models- Case Study 1 (Box Culvert)
Purpose: To perform the load rating on the bridge structure
### Configuration
• Span: Single span
• Alignment: Straight
• Section: Solid slab roof and walls.
• Moving load: T44, M1600, HLP400
• Carriageway: 12m wide (multi-lane)
• Application: Traffic surface lanes, optimization, permit vehicles.
The permit truck can be defined using the User-defined vehicular loads by inputting the multiple axles and spacing’s.
Eccentricity: Input the distance of the Permit Vehicle to the Ref. Lane. By looking towards the direction the vehicle travels, the right side is considered positive and the left side is considered negative.
fig. Influence Surface-Mxx
## SUMMARY
• Moving load analysis based on Influence line/surface
• International codes, predefined and user-defined vehicles
• Traffic line lanes, transverse floating of vehicles within a lane
• Moving load optimization – auto lane definition
• Moving load case – several scenarios can be combined in one load case.
• Moving load analysis control – filter the results, reduce run time
• Variety of result outputs
• Diagrams/contours for force, displacement, stresses, reaction, etc.
• Result tables, concurrent forces/reaction
• Moving load tracer - visualize vehicle positions, write load to file.
Q1. How do you determine eccentricity in traffic lanes?
A) Eccentricity is the distance from the centerline of the traffic lane to the reference elements used to define traffic lanes.
Q2. How can you define braking loads?
A) When BS or Euro code is selected, braking loads can be obtained when creating a static load case from the moving load tracer tab. See below: Although this functionality is currently not available for all codes, you may have to load it manually.
Q3. In using Traffic Load Optimization for transverse vehicle position within the lane, does MIDAS individually float the vehicle positions in adjacent lanes to generate the most onerous results? Eg. Lane 1: vehicle left-hand side of the lane, Lane 2: vehicle right-hand side of the lane, etc.). Note on Slide 55, all loads were either LHS or RHS. For a central girder, will it consider some lanes LHS, others RHS?
A) Yes, Midas will float the vehicles individually in each lane to produce the worst effects.
Q4. Is it possible to input numerous vehicles on the carriageway simultaneously based on actual scenarios?
A) Yes, different vehicle types can be combined together in a carriageway. Although in a multi-lane loaded scenario, within an individual traffic lane only one vehicle type will be loaded.
Q5. Is there any way to tabulate results for several beams after I use Moving Load Tracer?
A) If you wish to create a static load case for several beams, this can be done by Batch conversion (available in the moving load tracer tab). After obtaining static load cases, you need to analyze the model again. The results can now be obtained for each static load case in a tabular or graphical form.
Q6. Is there a way to define vehicular moving load to a curve & ramping up at the same time?
A) Yes, this can be defined in the same way as you would do for a flat bridge. Although additional horizontal load effects won’t be considered by the software automatically.
Q7. How about the moving load on the curved bridge with superelevation?
A) Moving load can be run on a curved bridge with superelevation. Although an only vertical components of wheel loads will be considered. For rail traffic loads, an eccentricity can be input to account for unequal wheel load distribution. See below:
Watch the Full Webinar Video Here: | 2,896 | 13,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2023-40 | latest | en | 0.901134 |
https://www.get-digital-help.com/nested-if-functions/?replytocom=367476 | 1,652,704,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00727.warc.gz | 929,489,995 | 22,814 | Author: Oscar Cronquist Article last updated on October 13, 2021
Nested IF statements in a formula are multiple combined IF functions so more conditions and outcomes become possible. They all are more or less complicated to read and thankfully there are better alternatives.
You are allowed to use up to 254 nested IF statements but before you do that make sure you read this article, I promise you it will save you a lot of time.
This article will provide step-by-step tutorials for your specific scenario. If you can't find what you are looking for here please comment and I will add the missing part.
## 1. How to simplify nested IF functions based on a numerical range
You don't need to use multiple IF statements if you want to check if a cell value is in a given numerical range, it is enough to simply use two logical expressions in the first argument.
Nested IF function formula in cell C3:
=IF(B3>=0, IF(B3<=10, TRUE, FALSE), FALSE)
The formula above demonstrates a nested IF function. The following formula simplifies the formula above, it returns TRUE if the value in cell B3 is equal to or greater than 0 and is equal to or smaller than 10.
Formula in cell C7:
=IF((B3>=0)*(B3<=10), TRUE, FALSE)
Each logical expression is encapsulated with parentheses, this makes sure that the comparisons are made before multiplying the expressions. In other words, the parentheses determine the order of calculation.
The asterisk between the logical expressions means that both logical expressions must return TRUE for the logical test argument to return TRUE.
### 1.1 Explaining formula in cell C7
The formula in cell C7 is easier to understand and troubleshoot than the formula in cell C7.
#### Step 1 - First logical expression
The larger than character and equal sign are logical operators, they let you compare values and returns boolean values True or False.
B3>=0
becomes
5>=0
and returns boolean value True.
#### Step 2 - Second logical expression
B3<=10
becomes
5<=10
and returns boolean value True.
#### Step 3 - Multiply logical expressions - AND logic
We must check that both conditions are met and to do that we use the asterisk to multiply the expressions.
Use parentheses to control the order of calculations, we need to compare values before we multiply.
(B3>=0)*(B3<=10)
becomes
True * True
and returns 1. 1 is the numerical equivalent to True and False is 0 (zero).
AND logic returns True only if both conditions are met.
TRUE * TRUE = TRUE (1)
TRUE * FALSE= FALSE(0)
FALSE* TRUE = FALSE(0)
FALSE* FALSE= FALSE(0)
#### Step 4 - Evaluate IF function
The IF function returns one value if the logical test evaluates to TRUE and another value if the logical test returns FALSE.
IF(logical_test, [value_if_true], [value_if_false])
IF((B3>=0)*(B3<=10), TRUE, FALSE)
becomes
IF(1, TRUE, FALSE)
and returns TRUE.
## 2. How to simplify nested IF statements based on numerical ranges
The picture above shows you nested IF statements that allow you to return different outcomes depending on the value in column B.
Nested If functions formula in cell C3:
=IF(B3<10, IF(B3>=0, "Group 1", ""), IF(B3<20, IF(B3>=10, "Group 2", ""), IF(B3<30, IF(B3>=20, "Group 3", ""), "")))
A value equal to or greater than 0 and smaller than 10 is in "Group 1".
A value equal to or greater than 10 and smaller than 20 returns "Group 2".
A value equal to or greater than 20 and smaller than 30 returns "Group 3".
The simplified formula in cell C11:
=VLOOKUP(B11, \$E\$11:\$F\$14, 2, TRUE)
This formula is considerably smaller and doesn't grow bigger if you need more criteria.
### 2.1 Explaining formula in cell C11
#### Step 1 - Build the table
The table above defines the numerical ranges and what value to return. Note, that there are no gaps between the ranges.
#### Step 2 - VLOOKUP function
The VLOOKUP function lets you search the leftmost column for a value and return another value on the same row in a column you specify.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
lookup_value - A value.
table_array -
The range you want to use, remember that the VLOOKUP function always looks in the leftmost column in your specified range.
col_index_num -
The column number which contains the return value.
[range_lookup] -
True or False. True - approximate match, the leftmost column must be in ascending order. False - Exact match.
#### Step 3 - Populate arguments
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
lookup_value - B11
table_array - \$E\$11:\$F\$14
col_index_num -
2
[range_lookup] -
True.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
becomes
VLOOKUP(B11, \$E\$11:\$F\$14, 2, TRUE)
#### Step 4 - Evaluate VLOOKUP function
VLOOKUP(B11, \$E\$11:\$F\$14, 2, TRUE)
becomes
VLOOKUP(5, {0, "Group 1"; 10, "Group 2"; 20, "Group 3"; 30, 0}, 2, TRUE)
and returns "Group 1" in cell C11.
The last argument TRUE lets you perform an approximate match meaning it matches an item equal to or next smaller item if no exact match is found. This is why it is so important to have the table sorted in ascending order.
Value 5 matches no value in the table, however, it is between 0 (zero) and 10. The next smaller value is 0 (zero), the corresponding value in column F on the same row is "Group 1".
### 2.2 How to add criteria
Imagine that you have 5 different groups, you now need to add six more nested IF statements. The formula grows considerably, however, the VLOOKUP function is really useful in this case.
The VLOOKUP lets you easily group numbers using a simple function instead of constructing a mega formula that is hard to follow and troubleshoot.
Need more groups? No, problem. Add groups to the first table and then adjust the cell reference in the second argument in the VLOOKUP function.
=VLOOKUP(B10, \$B\$3:\$C\$7, 2, TRUE)
Note, the first table must have the first column sorted from small to large. Make sure you use TRUE in the fourth VLOOKUP argument. This means that it only needs an approximate match.
## 3. How to simplify nested IF functions based on conditions
These nested IF statements in cell C3 check if a value is equal to a condition and returns another value if True.
Nested IF function formula in cell C3:
=IF(B3="V", "Level 1", IF(B3="D", "Level 2", IF(B3="S", "Level 3", IF(B3="T", "Level 4"))))
You can use the VLOOKUP function in this case as well as an alternative to nested IF functions. You need to find an exact match in this case so change the fourth VLOOKUP argument to FALSE.
The simplified formula in cell C8:
=VLOOKUP(B8, \$E\$8:\$F\$11, 2, FALSE)
The VLOOKUP function looks for the value in the first column (E8:F11) and returns the corresponding value in the second column. That is why I use 2 in the third VLOOKUP argument.
### 3.1 Explaining formula in cell C11
#### Step 1 - Build table
The table above defines the conditions and what value to return.
#### Step 2 - VLOOKUP function
The VLOOKUP function lets you search the leftmost column for a value and return another value on the same row in a column you specify.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
lookup_value - A value.
table_array -
The range you want to use, remember that the VLOOKUP function always looks in the leftmost column in your specified range.
col_index_num -
The column number which contains the return value.
[range_lookup] -
True or False. True - approximate match, the leftmost column must be in ascending order. False - Exact match.
#### Step 3 - Populate arguments
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
lookup_value - B8
table_array - \$E\$8:\$F\$11
col_index_num -
2
[range_lookup] -
FALSE.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
becomes
VLOOKUP(B8, \$E\$8:\$F\$11, 2, TRUE)
#### Step 4 - Evaluate VLOOKUP function
VLOOKUP(B8, \$E\$8:\$F\$11, 2, TRUE)
becomes
VLOOKUP("V", {"V", "Level 1";"D", "Level 2";"S", "Level 3";"T", "Level 4"}, 2, TRUE)
and returns "Level 1" in cell C8.
Use VLOOKUP to calculate discounts, commissions, tariffs, charges, shipping costs, packaging expenses or bonuses
Have you ever tried to build a formula to calculate discounts based on price? The VLOOKUP function is much easier […]
## 4. How to simplify nested IF functions based on date ranges
The above image demonstrates how to use multiple date ranges with a short and simple VLOOKUP function. A date range consists of two dates, since these date ranges are contiguous the end date also represents the start date for the next range.
Formula in cell C3:
=VLOOKUP(B3, \$E\$3:\$F\$6, 2, TRUE)
The first cell range is between 1/1/2017 and 2/15/2017. The second cell range is between 3/1/2017 and 6/1/2017.
### 4.1 Explaining formula in cell C11
Excel dates are really not much different from numerical values, in fact, they are numerical values formatted as dates.
1 is 1/1/1900 and 1/1/2000 is 36526. We can use the same technique described in section 2 to extract the correct quarter.
#### Step 1 - Build a table
The table above defines the date ranges and what value to return. Note, that there are no gaps between the date ranges.
#### Step 2 - VLOOKUP function
The VLOOKUP function lets you search the leftmost column for a value and return another value on the same row in a column you specify.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
lookup_value - A value.
table_array -
The range you want to use, remember that the VLOOKUP function always looks in the leftmost column in your specified range.
col_index_num -
The column number which contains the return value.
[range_lookup] -
True or False. True - approximate match, the leftmost column must be in ascending order. False - Exact match.
#### Step 3 - Populate arguments
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
lookup_value - B3
table_array - \$E\$3:\$F\$6
col_index_num -
2
[range_lookup] -
True.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
becomes
VLOOKUP(B3, \$E\$3:\$F\$6, 2, TRUE)
#### Step 4 - Evaluate VLOOKUP function
VLOOKUP(B3, \$E\$3:\$F\$6, 2, TRUE)
becomes
=VLOOKUP(42795, {42736, "Quarter 1"; 42826, "Quarter 2"; 42917, "Quarter 3"; 43009, "Quarter 4"}, 2, TRUE)
and returns "Group 1" in cell C3. | 2,689 | 10,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.872356 |
http://www.cut-the-knot.org/Curriculum/Geometry/AsymmetricPropeller.shtml | 1,722,671,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00018.warc.gz | 34,738,295 | 4,958 | # Asymmetric PropellerWhat Is It? A Mathematical Droodle
Created with GeoGebra
Explanation
### Asymmetric Propeller
This is a generalization of a theorem about three equilateral triangles.
Vertices of three similar triangles are attached to the corresponsing vertices of the fourth triangle similar to the first three. All four have the same orientation. Three pairs of the unattached vertices of the outside triangles are joined and the midpoints of those segments are connected. The triangle thus obtained is similar to the four given triangles.
Here's a proof with complex numbers.
In the above diagram, I placed the origin O at one of the vertices and identified the straight line segments as complex numbers directed (unconventionally) from a (small yellow) square to a (small yellow) circle. X, Y, Z are the vertices of the "midpoint" triangle. The four given triangles are similar to a generic triangle with sides 1, k, and (1-k), where k is a complex number. The factors a, b, c, and d define the rotation and size of the triangles. We have
1. X = ((ka) + (a - c + kd))/2,
2. Y = (a - (1-k)c - kb)/2,
3. Z = ((a - c + d) + (a - (1-k)c + (1-k)b))/2,
from where we can compute the sides
1. YZ = Z - Y = (a + b - c + d)/2,
2. YX = X - Y = k(a + b - c + d)/2,
3. XZ = Z - X = (1-k)(a + b - c + d)/2.
So that triangle XYZ is indeed similar to the generic triangle (1, k, (1-k)).
### References
1. American Mathematical Monthly, v 75, n 7, 1968, pp 732-739
2. L. Bankoff, P. Erdös and M. Klamkin, The Asymmetric Propeller, Mathematics Magazine, v. 46, n. 5, 1973, pp 270-272
3. M. Gardner, The Asymmetric Propeller, The College Mathematics Journal, v. 30, n. 1, 1999, (18-22) | 492 | 1,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.864109 |
http://oeis.org/A108348 | 1,585,600,639,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00359.warc.gz | 134,717,803 | 4,645 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A108348 Numbers of the form p^k + p^(k-1) + ... + p + 1 (where p is a prime and k>=0) in ascending order. 3
1, 3, 4, 6, 7, 8, 12, 13, 14, 15, 18, 20, 24, 30, 31, 32, 38, 40, 42, 44, 48, 54, 57, 60, 62, 63, 68, 72, 74, 80, 84, 90, 98, 102, 104, 108, 110, 114, 121, 127, 128, 132, 133, 138, 140, 150, 152, 156, 158, 164, 168, 174, 180, 182, 183, 192, 194, 198, 200 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS A proper subset of A002191 (e.g. 28 is in A002191, but not in this sequence). a(15)=31 admits two representations: 31=2^4+2^3+2^2+2+1=5^2+5+1. Are there other numbers with two or more representation? I have checked all the sums of primes up to prime number 56873 to a sum total >= 10^100 and have not come across another number that has multiple representations. - Patrick Schutte (patrick(AT)onyxsa.co.za), Mar 28 2007 LINKS M. F. Hasler, Table of n, a(n) for n = 1..1000 EXAMPLE a(2)=3=2+1 since a(1)=1 and 2 is not expressible in the required form. PROG (PARI) A108348(n)={ local(m=1, a=[m]); while( #a 1 and a <> b then Print("a ", a, " ap ", ap, " at ", at, " "); Print("b ", b, " bp ", bp, " bt ", bt, " "); Print("---------------- "); fi; bp := bp + 1; until bt > at; b := NextPrime(b); until b >=a; ap := ap + 1; until at > 10^100; a := NextPrime(a); until a >FNum; end; # Patrick Schutte (patrick(AT)onyxsa.co.za), Mar 28 2007 (Haskell) a108348 n = a108348_list !! (n-1) a108348_list = 1 : f [2..] where f (x:xs) = g a000040_list where g (p:ps) = h 0 \$ map ((`div` (p - 1)) . subtract 1) \$ iterate (* p) (p ^ 2) where h i (pp:pps) | pp > x = if i == 0 then f xs else g ps | pp < x = h 1 pps | otherwise = x : f xs -- Reinhard Zumkeller, Nov 26 2013 CROSSREFS Cf. A002191. Cf. A000040, A090503. Sequence in context: A007609 A285703 A002191 * A085149 A239458 A007370 Adjacent sequences: A108345 A108346 A108347 * A108349 A108350 A108351 KEYWORD nonn AUTHOR Franz Vrabec, Jul 01 2005 STATUS approved
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Last modified March 30 16:16 EDT 2020. Contains 333127 sequences. (Running on oeis4.) | 939 | 2,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-16 | latest | en | 0.70378 |
http://research.stlouisfed.org/fred2/series/M16072USM351SNBR?rid=257 | 1,430,661,536,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430448950892.36/warc/CC-MAIN-20150501025550-00000-ip-10-235-10-82.ec2.internal.warc.gz | 178,409,459 | 19,541 | # Composite Index of Twelve Leading Indicators, Original Trend, Short List for United States
1969-12: 0 Index 1963=100 (+ see more)
Monthly, Seasonally Adjusted, M16072USM351SNBR, Updated: 2012-08-20 9:31 AM CDT
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Source: NBER
This NBER data series m16072 appears on the NBER website in Chapter 16 at http://www.nber.org/databases/macrohistory/contents/chapter16.html.
NBER Indicator: m16072
Release: NBER Macrohistory Database
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(a) Composite Index of Twelve Leading Indicators, Original Trend, Short List for United States, Index 1963=100, Seasonally Adjusted (M16072USM351SNBR)
Source: NBER
This NBER data series m16072 appears on the NBER website in Chapter 16 at http://www.nber.org/databases/macrohistory/contents/chapter16.html.
NBER Indicator: m16072
Composite Index of Twelve Leading Indicators, Original Trend, Short List for United States
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Name: Email: | 609 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2015-18 | latest | en | 0.772307 |
http://www.utteraccess.com/forum/lofiversion/index.php/t1691398.html | 1,556,266,713,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578762045.99/warc/CC-MAIN-20190426073513-20190426095513-00476.warc.gz | 318,242,064 | 3,019 | Full Version: Formula for Specific Items Within A Drop Down
Karol
I have a form with a drop down. Lets say Id 1 is Flavor 1, Id 2 is flavor 2. I want to be able to give a sum total for each flavor (which would be the description). So the example would be flavor 1 with an ID of 1 has 50 hits for the day so it will total 50. However, Flavor 2 with an id of 2 only has 25 hits so its total for the day is 25. I do have a table (tblprovide) that the drop down pulls from. So my question is within the form text box how would I create a formula so it would calculate Id 1 within a text box with the label of flavor 1 and a formula for flavor 2 within another text box with the label of flavor 2?
xample: Flavor 1 50
Flavor 2 25
niesz
Define "hits" ... and how it is derived.
My guess is the row source for the combobox would be based off of an aggregate query.
merlicky
I guess I’m a little confused by your request. My interpretation is that you want to count how many times each flavor was selected from a ComboBox and display the result in another TextBox. If I understand correctly, there are a couple things you would need to do:
. You need to store how many times each flavor was selected. To do this you could use a “Hits” field in your table for each flavor. You could use the ComboBox’s AfterUpdate Event to write code to increase the value in the “Hits” field.
2. You need to have the ControlSource for your TextBox based on a Query that combines the “Flavor”, “ID”, and “Hits” fields into one value.
Karol
OK. I do want to count how many times each flavor was selected from the combobox. I would have a query that would add 1 for each record. So, If I have another combo box on the the form by month. The user select the month of July the total records for flavor 1 would show a total of 50.
So, what would the formula look like?
merlicky
There is no magic formula in this instance. It sound like you will need to create a table to store the hits and the date of the hits for each flavor. Then, it entirely depends on how you have your table/s, query/s, and form/s set up and what your intention is.
Karol
Now, I am confused. There has to some sort of formula to count the number of records for flavor 1. Somehow I am thinking that within the control box for flavor 1 it would have =[Hits].Query!Flavor1
Would that work?
merlicky
You are not going to store each individual hit as a seperate record, the Hits field will keep the total number of hits for each flavor.
Yre not going to store each individual hit as a seperate record, the Hits field will keep the total number of hits for each flavor.
Your table would need fields like these:
| Flavor | Hits | HitDate |
The SQL you would need is something like: "SELECT Hits FROM MyTable WHERE HitDate=MyInputDate"
or
"SELECT Sum(Hits) AS TotalHits FROM MyTable WHERE Month(HitDate)=Month(MyInputDate) AND Year(HitDate)=Year(MyInputDate)"
But again, where you put this SQL is going to depend on how you have your database set up. What is the RecordSource for your form? Are the results and the ComboBox where you keep track of hits on the same form? ect...
mike60smart
Hi
Can you post your Db so we can take a look?
Mike
Karol
Here is a copy of a test db.
jmcwk
See attached and open the Query
Karol
OK. I see what you did with the query. Now how do I show this by description on the form? I want to show one text box with the sum total for flavor 1 and another text box with the total for flavor 2?
jmcwk
Is this just for informational purposes ? By Total do you mean the Total excluding the Date Grouping ? Flavor 1 would have 50 and flavor2 would have 24
Is far as displaying I would use a Listbox and then set your Criteria in the Query against the FlavorID See Attached, open your form and select April from the Combo
mike60smart
Hi
The only way you can show the two flavor's is by means of a Pivot Table
See attached
Open frmTest
Mike | 970 | 3,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-18 | latest | en | 0.957574 |
https://www.daniweb.com/programming/software-development/threads/450120/double-array-problem | 1,537,481,025,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156622.36/warc/CC-MAIN-20180920214659-20180920235059-00139.warc.gz | 710,394,380 | 13,041 | Okay, here is my problem. The instructions are I would have to store the minimum points and grades in a five row two column array. The procedure should display the grade corresponding to the number of points entered by the user.
``````Minimum Points Maximum Points Grades
0 299 F
300 349 D
350 399 C
400 449 B
450 500 A
``````
This is the code I have, and the book is somewhat helpful..my problem is when I enter the points, nothing is being displayed..what I am I doing wrong? Also, we have not discussed the Exit format yet..Any help and specific explaination is greatly appreciated.
``````Dim strGrades(,) As String = {{"F", "0"},
{"D", "300"},
{"C", "350"},
{"B", "400"},
{"A", "450"}}
Dim intSearchPoints As Integer
Dim intRow As Integer
Dim intNumRows As Integer
Dim intFound As Integer
Integer.TryParse(txtPoints.Text, intSearchPoints)
intRow = 0
intFound = "600"
Do Until intFound = "500" OrElse intRow = intNumRows
If strGrades(intRow, 0) = intSearchPoints Then
intFound = "500"
Else
intRow = intRow + 1
End If
Loop
If intFound = "500" Then
Else
End If
``````
2
Contributors
2
Replies
12
Views
5 Years
Discussion Span
Last Post by siaosituimoloaublood
A couple of things:
use the boolean result of TryParse to test if there is a number being entered,
``````Dim ValidTest as Boolean = Integer.TryParse(txtPoints.Text, intSearchPoints)
If Not ValidTest Then
MsgBox "Please Enter a Valid Number"
Exit Sub
End If
``````
use a for loop whenever you know the exact number of items to check, then simply check if the value at the loop variable index is greater than intSearchPoints, if so the grade is the value at the loop variable minus 1 index, something like this should work:
``````lblGrade.Text = "Invalid Points(0-500 only)"
For I = 1 to strGrades.Count -1
Exit Sub
End If
Next
If intSearchPoints <= 500 Then
End If
``````
Edited by tinstaafl
I will test that out and post the final result..thanks again
Here is what I came up with..The maximum value doesnt display the grade, but it was satisfactory effort..according to my instructor :), but I really do appreciate the help
``````Dim strGrades(,) As String = {{"F", "0"},
{"D", "300"},
{"C", "350"},
{"B", "400"},
{"A", "450"}}
'declaring variables
Dim intPoints As Integer
Dim intRows As Integer
Dim intNumRows As Integer
Dim intFound As Boolean
'converting text variable to integer
Integer.TryParse(txtPoints.Text, intPoints)
'search for the point in the frist colum of the array
'continue searching until the point is or each
'row has been searched
intRows = 0
intFound = False
Do Until intFound = True OrElse intRows = intNumRows
If strGrades(intRows, 1) = intPoints Then
intFound = True
Else
intRows = intRows + 1
End If
Loop
'determine whether the point was found
'displays outcome
If intFound = True Then | 781 | 3,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-39 | longest | en | 0.785022 |
https://chem.libretexts.org/Courses/Heartland_Community_College/CHEM_120%3A_Fundamentals_of_Chemistry/02%3A_Atoms_and_Elements/2.04%3A_Neutrons%3A__Elemental_Isotopes_and_Mass_Number_Calculations | 1,702,137,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00224.warc.gz | 200,485,912 | 32,763 | # 2.4: Neutrons: Isotopes and Mass Number Calculations
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Learning Objectives
• Describe the location, charge, and relative mass of the neutron.
• Define isotope and mass number.
• Determine the number of protons, neutrons, and electrons in a specific isotope of an element.
• Represent a single isotope of an element using the three main elemental symbolisms.
The final subatomic particle was not discovered until 1932. In order to account for the neutral charge of an atom as a whole, the number of positively-charged protons and negatively-charged electrons found within an atom must be equal. Therefore, any remaining subatomic particles must be uncharged, so as to not upset this established charge balance. Indeed, neutrons, which were named as a result of their neutral charge, do not possess any electrical properties. Consequently, these subatomic particles, which are symbolized using the notation "n0," were incredibly difficult to detect. Neutrons are also located in the nucleus of an atom, and the mass of a neutron was found to be just slightly greater than the mass of a proton.
Each subatomic particle exists to serve a specific purpose. As stated in the previous section, the number of valence electrons present in an atom dictates the reactivity of that element. The number of protons found within an atom defines the identity of that atom, and all of an atom's protons collectively attract the surrounding electrons, keeping the latter bound to the atom. Recall, however, that all protons, which each bear a +1 charge, are densely-packed into the central region of an atom. Therefore, each positively-charged proton must be strongly repelled by every other proton in the nucleus, and, furthermore, the combined strength of these repulsive forces is substantial enough to splinter the nucleus. However, neutrons effectively act as "nuclear glue" and allow the protons to exist in close physical proximity to one another. In other words, neutrons are the subatomic particle responsible for maintaining the structural integrity of the nucleus.
Finally, recall that every atom of a certain element must have a defined number of protons and electrons. Every atom of carbon, C, that exists in the known universe is defined to contain 6 protons, because its atomic number is 6, and must also contain 6 electrons, in order for the atom to maintain an overall net neutral charge. However, the number of neutrons within an atom of an element is not defined by the atomic number of that element. In fact, the number of neutrons present in an element can vary from atom to atom. The "glue" analogy found within the previous paragraph can be extended to explain this phenomenon. While a minimum amount of glue is required to adhere one object to another, a small amount of excess glue will not prevent those objects from sticking together, but a large excess of glue could prove to be problematic. Likewise, each element must contain a minimum number of neutrons to hold the nucleus together, but could contain a small number of additional neutrons without sacrificing the structural integrity of the nucleus. However, a nucleus that contains too many neutrons will become unstable and undergo radioactive decay, which will be discussed in Chapter 9 of this text.
### Mass Number
The mass number of an atom is equal to the total number of protons and neutrons contained in its nucleus. This definition can be represented in an equation, as shown below.
Mass Number = # of Protons + # of Neutrons
The true mass of an atom is an incredibly small quantity. To simplify the numerical values being used, the mass of a single proton is assigned a value of 1 atomic mass unit, or amu. As the mass of a neutron is approximately the same as the mass of a proton, each neutron that is present is also given a value of 1 amu. Since the mass of an electron is 1/2,000th of the mass of a proton, any contribution that electrons make to the overall mass of an atom is negligible. Therefore, the number of electrons present in an atom are ignored when calculating the mass number of that atom.
Example $$\PageIndex{1}$$
Use a periodic table to calculate the mass number of a hydrogen atom that contains 2 neutrons.
Solution
The mass number of an atom is calculated by adding together the number of protons and neutrons that are found within that atom. The number of neutrons is given, but the number of protons must be determined from the atomic number for the element. In this case, hydrogen (H) has an atomic number of 1 and, therefore, every atom of hydrogen will contain 1 proton. The equation shown above can then be applied, as follows.
Mass Number = # of Protons + # of Neutrons
Mass Number = 1 + 2
Therefore, this particular atom of hydrogen will have a mass number of 3.
Note that the mass number calculated in Example $$\PageIndex{1}$$ does not match the number underneath the elemental symbol and name for hydrogen on the periodic table. This discrepancy can be explained by a subtle, but incredibly important, piece of information: The calculation performed in Example $$\PageIndex{1}$$ was done for a single atom of hydrogen. However, the periodic table is intended to represent all of the atoms of hydrogen in the known universe. Since every existing atom of hydrogen must contain 1 proton, the atomic number that is written above hydrogen's elemental symbol truly does represent every atom of hydrogen.
However, recall that the number of neutrons contained in an element can vary from atom to atom. Changing the number of neutrons present in an atom will, in turn, cause these individual atoms of hydrogen to have different calculated mass numbers. These individual "versions" of an element are called isotopes, which are defined as atoms of an element that have the same atomic numbers and, therefore, contain the same number of protons, but different mass numbers, and, therefore, contain differing numbers of neutrons. Three isotopes of hydrogen are modeled in Figure $$\PageIndex{1}$$. Most hydrogen atoms have one proton, one electron, and do not contain any neutrons, but less common isotopes of hydrogen can contain either one or two neutrons. Hydrogen is unique, in that its isotopes are given special names, which are also shown below in Figure $$\PageIndex{1}$$.
For spatial reasons, listing the mass numbers for all of an element's isotopes within a single box on the periodic table is impractical. Instead, a weighted average, called an atomic mass average, is calculated. A weighted average takes into account not only the mass number of each isotope, but also how prevalent, or common, that isotope is in nature, relative to each of that element's other isotopes. Therefore, an atomic mass average is a quantity that truly represents all isotopes of a given element, making it appropriate for inclusion on the periodic table.
Example $$\PageIndex{2}$$
Use a periodic table to determine the following information for an atom that has an atomic number of 74 and a mass number of 186.
1. Elemental symbol
2. Elemental name
3. Number of protons contained in the atom
4. Number of electrons contained in the atom
5. Number of neutrons contained in the atom
Solutions
1. The atomic number of an element is found above the elemental symbol within a box on the periodic table. The element with an atomic number of 74 is symbolized as W.
2. The element with an atomic number of 74 is named tungsten.
3. The number of protons present in an atom is defined by the element's atomic number. Therefore, every atom of tungsten contains 74 protons.
4. Since an atom must have an overall neutral charge, the number of protons and electrons found within an atom of an element must be equal. Therefore, every atom of tungsten also contains 74 electrons.
5. The mass number of an atom is calculated by adding together the number of protons and neutrons that are found within that atom, as shown below.
Mass Number = # of Protons + # of Neutrons
Since the mass number was provided, this equation can be rearranged to determine the number of neutrons contained in this specific isotope of tungsten.
186 = 74 + # of Neutrons
186 - 74 = # of Neutrons
Therefore, this particular atom of tungsten contains 112 neutrons.
### Elemental Symbolisms
In total, 252 stable isotopes have been isolated for 80 different elements. Factoring in the number of unstable isotopes that have been observed causes the total number of known elemental isotopes to increase substantially. While each of hydrogen's three most common isotopes has a unique name, it would ultimately be highly impractical to establish different names for every isotope of every element that has been shown to exist. Therefore, scientists utilize three different elemental symbolisms to refer to specific elemental isotopes. The first two symbolisms are very similar, in that each includes the elemental name, or elemental symbol, of an element, followed by a dash and a numerical value, which corresponds to the mass number of a particular isotope of that element. In the third type of elemental symbolism, which is called a nuclear symbol, the mass number of the isotope is positioned as a superscript before an elemental symbol, and the atomic number of the element is written directly underneath the mass number. It is important to note the difference between an isotope and an elemental symbolism. Figure $$\PageIndex{2}$$ models these three different elemental symbolisms, which all represent the same isotope, since each has an identical mass number.
Example $$\PageIndex{3}$$
Write the nuclear symbol of the isotope that is described in Example $$\PageIndex{2}$$.
Solutions
In a nuclear symbol, the mass number of the isotope is positioned as a superscript before an elemental symbol, and the atomic number of the element is written directly underneath the mass number.
The isotope in Example $$\PageIndex{2}$$ has an atomic number of 74, a mass number of 186, and is symbolized as W. When this information is incorporated into this notation, the nuclear symbol shown below results.
$$\ce{^{186}_{74}W}$$
Example $$\PageIndex{4}$$
Determine how many protons, electrons, and neutrons are present in an atom of each of the following isotopes.
1. $$^{40}_{19}\ce{K}$$
2. Zinc-65
Solutions
1. This isotope is symbolized using a nuclear symbol. In this notation, the atomic number of the isotope is written as a subscript. Since the atomic number indicates both the number of protons and the number of electrons present in an atom, this isotope contains 19 protons and 19 electrons. The number of neutrons in the isotope can be calculated from its mass number, which is written as a superscript in a nuclear symbol.
Mass Number = # of Protons + # of Neutrons
40 = 19 + # of Neutrons
40 - 19 = # of Neutrons
Therefore, there are 21 neutrons in this isotope of potassium (K).
2. This isotope is represented using the second symbolism shown in Figure $$\PageIndex{2}$$. When using this notation, the name of the element must be used to find its atomic number. Since zinc (Zn) has an atomic number of 30, this isotope contains 30 protons and 30 electrons. The number of neutrons in the isotope can again be calculated from its mass number, which is the numerical value written after the dash in both representations shown in Figure $$\PageIndex{2}$$.
Mass Number = # of Protons + # of Neutrons
65 = 30 + # of Neutrons
65 - 30 = # of Neutrons
Therefore, there are 35 neutrons in this isotope of zinc (Zn).
Exercise $$\PageIndex{1}$$
Determine how many protons, electrons, and neutrons are present in an atom of each of the following isotopes.
1. $$^{60}_{27}\ce{Co}$$
2. Uranium-238
3. Na-25
This isotope is represented using a nuclear symbol. In this notation, the atomic number of the isotope is written as a subscript. Since the atomic number indicates both the number of protons and the number of electrons present in an atom, this isotope contains 27 protons and 27 electrons. The number of neutrons in the isotope can be calculated from its mass number, which is written as a superscript in a nuclear symbol.
Mass Number = # of Protons + # of Neutrons
60 = 27 + # of Neutrons
60 - 27 = # of Neutrons
Therefore, there are 33 neutrons in this isotope of cobalt (Co).
This isotope is represented using the second symbolism shown in Figure $$\PageIndex{2}$$. When using this notation, the name of the element must be used to find its atomic number. Since uranium (U) has an atomic number of 92, this isotope contains 92 protons and 92 electrons. The number of neutrons in the isotope can again be calculated from its mass number, which is the numerical value written after the dash in both representations shown in Figure $$\PageIndex{2}$$.
Mass Number = # of Protons + # of Neutrons
238 = 92 + # of Neutrons
238 - 92 = # of Neutrons
Therefore, there are 146 neutrons in this isotope of uranium (U).
This isotope is represented using the first symbolism shown in Figure $$\PageIndex{2}$$. When using this notation, the symbol of the element must be used to find its atomic number. Since sodium (Na) has an atomic number of 11, this isotope contains 11 protons and 11 electrons. The number of neutrons in the isotope can again be calculated from its mass number, which is the numerical value written after the dash in both representations shown in Figure $$\PageIndex{2}$$. | 3,503 | 14,456 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | longest | en | 0.601864 |
http://mathhelpforum.com/calculus/138371-question-about-points-inflection-derivatives.html | 1,508,379,403,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823214.37/warc/CC-MAIN-20171019012514-20171019032341-00056.warc.gz | 204,670,506 | 11,345 | 1. ## question about points of inflection derivatives (urgent)
y= 1/3 x^3 + x^2-15x +3
y'= x^2 +2x-15 =0
(x+5) (x-3)
x=-5 x=3
y'' = 2x +2 =0
check for max or min
2(-5)+2 =-8< 0 local max
2(3) +2= 8 >o local min
check for poi
y'''= 2 not= 0 so we have a poi check if its stationary
2x+2=0 2X=-2 x=- 2/2 x=-1
non stationary point of infecton
so if the was x=0 we have a stationary point of inflection?
because i the book is they have this
y=(x+4)^3
y'=3(x+4)^2 =o x=-4
y''= 6(x+4)=0 x=-4
y'''=6 is not = to 0 we have a poi
the order of the derivative is an odd number so at x=-4 we have a poi since y=0 when x=-1 the point is stationary
bit confused because i thought that when the 2nd derviative result is calculated and x=0 then it is stationary when its x not = 0 nonstationary.
2. really need help on this
3. Originally Posted by matlondon
y= 1/3 x^3 + x^2-15x +3
y'= x^2 +2x-15 =0
(x+5) (x-3)
x=-5 x=3
y'' = 2x +2 =0
check for max or min
2(-5)+2 =-8< 0 local max
2(3) +2= 8 >o local min
check for poi
y'''= 2 not= 0 so we have a poi check if its stationary
2x+2=0 2X=-2 x=- 2/2 x=-1
non stationary point of infecton
so if the was x=0 we have a stationary point of inflection?
because i the book is they have this
y=(x+4)^3
y'=3(x+4)^2 =o x=-4
y''= 6(x+4)=0 x=-4
y'''=6 is not = to 0 we have a poi
the order of the derivative is an odd number so at x=-4 we have a poi since y=0 when x=-1 the point is stationary
bit confused because i thought that when the 2nd derviative result is calculated and x=0 then it is stationary when its x not = 0 nonstationary.
Hi matlondon,
$f(x)=(x+4)^3$
$f'(x)=3(x+4)^2=0\ for\ x=-4$
$f''(x)=6(x+4)=0\ for\ x=-4$
$f''(x)$ is neither positive or negative at x=-4, so it is a saddle point, a stationary point of inflexion.
That's as far as you need to go to know the difference.
$f(x)=\frac{x^3}{3}+x^2-15x+3$
$f'(x)=x^2+2x-15=(x+5)(x-3)$
This is zero at $x=-5,\ x=3$
$f''(x)=2x+2=max\ at\ x=-5,\ min\ at\ x=3$
$f''(x)=0\ for\ x=-1$
There is a non-stationary point of inflexion at $x=-1$
4. f ' ' ' = 2 not equal to 0 so we have a poi
not we plug in the x-1 into the first derivative to see if it stationary or non.
f'(x)=-1^2+2(-1)-15= -18 this is non stationary
so if the same the equation and the answer was zero then it would a a stationary poi?
5. ## stationary point
Stationary points only when the first derivative =0
6. Originally Posted by matlondon
f ' ' ' = 2 not equal to 0 so we have a poi
not we plug in the x-1 into the first derivative to see if it stationary or non.
f'(x)=-1^2+2(-1)-15= -18 this is non stationary
so if the same the equation and the answer was zero then it would a a stationary poi?
Hi matlondon,
no need to check the third derivative,
that's a different ball game.
If a point of inflexion is stationary, then the first and second derivatives are zero for the same x.
Calculate the first derivative.
This is zero at a maximum or minimum or saddle point.
The saddle point is another term for a stationary point of inflexion.
To know whether the point is a max, min or stationary point of inflexion,
use the value of x that causes the first derivative to be zero.
Plug this x into the 2nd derivative equation.
If the answer is negative, the point is a local maximum.
If the answer is positive, then it's a local minimum.
if the answer is zero, it's a stationary point of inflexion.
This applies to functions of a single variable, such as x.
It's more complex for functions of more than one variable. | 1,183 | 3,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-43 | longest | en | 0.857083 |
https://comicvine.gamespot.com/profile/vuviper/blog/estimating-reaction-time-from-travel-velocities/64526/ | 1,596,762,939,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737050.56/warc/CC-MAIN-20200807000315-20200807030315-00232.warc.gz | 268,393,215 | 30,208 | # Estimating Reaction Time from Travel Velocities
For many superspeeders we are never given a quantifiable measurement of their reaction time, but I think we can overcome this by looking at their speed feats.
(Defn: Reaction Time:=The time required to perceive a stimulus and respond)
(Assumption 1: Minimum reaction time for even the simplest tasks is .1 secondshttp://www.humanbenchmark.com/tests/reactiontime/allresults.php)
(Assumption 2: Current records for travel speed represent a reasonable model for the upper limit of the speed acheivable with a reaction time of .1 seconds)
Space Travel:
The flight speed record for a manned rocket in spave is Mach 30 (~10,000 m/s) (http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19690029435_1969029435.pdf)
(Assumption 3: The maximum speed obtained in an environment is inversely proportional to the reaction time)
We can then construct a coefficient relating space travel speed(V) with reaction time(t), V*t= 1,000 m/s^2.
From this you could estimate that someone capable of traveling at light speed(2.998x10^8 m/s) from space would need a reaction time of 3.34 microseconds.
Atmospheric Travel:
Obviously their are more obstacles one must react to during atmospheric travel than in space travel, therefore a new coefficient must be created.
The flight airspeed record is about 980 m/s (http://en.wikipedia.org/wiki/Flight_airspeed_record#Timeline)
Our new coefficient is then V*t=98 m/s^2
Lets look at a light speed flyer again, to flight at light speed safely in the atmosphere, the necessary reaction time is .33 microseconds
Landspeed Travel:
Travel on land is the most dangerous as land is more densely crowded, to safely manuever one requires an even greater reaction time.
The land speed record is about 340 m/s, V*T=34 m/s^2 (http://en.wikipedia.org/wiki/Land_speed_record)
City Travel:
Some places require better manueverability than others. The Nuerburgring is considered one of the most demanding race tracks and will be used for this first order approximation 40 m/s (http://en.wikipedia.org/wiki/List_of_Nordschleife_lap_times_%28racing%29#1984_.28and_later.29_Combined_variants_.28VLN_.26_24h.29)
V*T=4 m/S^2
High maneuverability travel:
For Speedsters traveling in doors, in between people, while performing tasks, etc we will use another coefficient and estimate it using the human running speed record 10.4 m/s (Usain Bolt)
V*T=1 m/s^2
Closing statements:
Yes the speed records used are not all analogous to the situations that the heroes find themselves in, but they are a good enough reference for a first approximation. Remember, this only tells you that the speedster has a reaction time that is at least that fast. Also, this is just an estimate to allow us some idea of reaction times in ABSENSE of any other information, in comic statements or feats always trump, comics don't make sense so they don't have to follow these rules. | 690 | 2,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-34 | latest | en | 0.896063 |
https://www.codeproject.com/Articles/43607/Solving-ordinary-differential-equations-in-C?msg=4335113 | 1,508,635,853,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00444.warc.gz | 867,874,149 | 46,522 | 13,198,465 members (65,869 online)
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# Solving ordinary differential equations in C++
, 19 Oct 2011
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This article explains a framework for solving ordinary differential equations, which is based on template metaprogramming.
Attention: A new version of odeint exists, which is decribed here.
## Introduction
This article introduces the C++ framework odeint for solving ordinary differential equations (ODEs), which is based on template meta-programming. I think this framework has some nice advantages over existing code on ODEs, and it uses templates in a very elegant way. Furthermore, odeint has a STL-like syntax, and is independent of a specific container. The whole project lives in the boost sandbox; the code presented here is more or less a snapshot of the current development.
## Background
I will only very briefly describe ordinary differential equations. If there is some interest in a more detailed explanation of ODEs, I can extend this part in future versions of the article.
### Ordinary Differential Equations
An ordinary differential equation describes the evolution of some quantity x in terms of its derivative. It often takes the form:
`d x(t) / d t = f( x(t) , t )`
The function f defines the ODE, and x and f can be vectors. Associated with every ODE is an initial value problem (IVP) that is the ODE, and an initial value x(t0)=x0. The solution of the initial value problem is the temporal evolution of x(t), with the additional condition that x(t0)=x0, and it can be shown that every IVP has a unique solution. Of course, ordinary differential equations are not restricted to temporal problems, hence the variable t can be replaced by another quantity, like a spatial coordinate. ODEs and their relative PDEs (partial differential equation) are very important in nearly all scientific disciplines. All major equations in physics fall in this class, like Newton's law for classical physics, the Maxwell's equations for electromagnetism, the Schrödinger equation and its relativistic generalizations for the quantum world, or Einstein's equation for general relativity. Very often, the spatial axes of a PDE are discretized, and an ODE on a lattice results.
In the following figure, an example of an ODE from chaos theory is shown: the famous Lorenz attractor. It was the first example of a deterministic chaotic system, and triggered a huge amount of scientific work. The Lorenz attractor is described by a set of coupled ordinary differential equations:
```d x1 / dt = sigma * ( x2 - x1 )
d x2 / dt = R * x1 - x2 - x1 * x3
d x3 / dt = x1 * x2 - b * x3```
with:
```sigma = 10
R = 28
b = 8 / 3```
This equation has no analytical solution, such that it can only be solved numerically.
### Numerical Solutions of ODEs
To solve ODEs numerically, various methods exist; all of them discretize the time. The easiest method is surely the explicit Euler scheme, which writes the derivative as the difference quotient:
`d x(t) / d t = x(t+dt) - x(t) / dt`
Then, some basic algebraic manipulations yield:
`x(t+dt) = x(t) + dt*f (x(t) , t )`
such that x(t+dt) can be obtained from the knowledge of x at the time t. Note again, that x does not have to be a scalar, but can be a vector. The rest is very simple. One chooses an initial condition x(t=0)=x0, and iteratively applies the Euler step to x(t) to obtain the evolution of x(t). Of course, more advanced solvers exist, and the most commonly used solver is probably the Runge-Kutta method of fourth order.
In the following source code, the function f(x(t)) defining the ODE is synonymously called system or dynamical system.
## The Main Ideas
Now, we come to the design aspects of the library. One clearly needs two main building blocks:
1. A stepper function, which calculates the state x(t+d t) from the knowledge of x(t).
2. An integrate function, which performs the iteration, hence applies the stepper function many times.
The whole library lives in the namespace `boost::odeint::numeric`.
### Stepper Methods
We try to avoid `abstract `classes, since this costs some extra performance. Instead, we use generic programming and concepts which describe how a type or class has to behave, and whose methods have to be available. For the stepper function, we need the following ingredients:
```template<
class Container ,
class Time = double ,
class Traits = container_traits< Container >
>
class ode_step
{
// provide basic typedefs
//
public:
typedef unsigned short order_type;
typedef Time time_type;
typedef Traits traits_type;
typedef typename traits_type::container_type container_type;
typedef typename traits_type::value_type value_type;
// public interface
public:
// return the order of the stepper
order_type order_step() const;
// standard constructor
ode_step( void );
// adjust the size of the internal arrays
void adjust_size( const container_type &x );
// performs one step
template< class DynamicalSystem >
void do_step( DynamicalSystem &system ,
container_type &x ,
const container_type &dxdt ,
time_type t ,
time_type dt );
// performs one step
template< class DynamicalSystem >
void do_step( DynamicalSystem &system ,
container_type &x ,
time_type t ,
time_type dt );
};```
The container type defines how the state is stored. It has to fulfill some basic requirements like the value type and it should be default constructible. The `container_traits` type abstracts the resize functionality of the sequence concept. Furthermore, it knows how to obtain the appropriate `begin()` and `end()` iterators. Its usage will be shown below. The `adust_size` method is responsible for adjusting the size of any internal containers, which store the results of intermediate calculations.
The `do_step` functions calculates x(t+dt) from the knowledge of x(t), and they transform x(t) in-place, meaning that x(t) is replaced by x(t+dt) within `do_step`. The argument `system` defines the ODE. This can be a simple function with three arguments: `system( state , dxdt , t )`, or a class with the appropriate brackets `operator()(state,dxdt,t)`. Such a class is also known as a functor. Two versions of `do_step` exist: one which calculates x(t+dt) with the explicit knowledge of f(x(t)), and one which calculates this value internally. In principle, the second version only calls the first version. Both functions exist, because in many situations, the user needs the knowledge of f(x(t)) and will call the system method with x(t). To avoid a double call of `system`, the `do_step` version with the explicit declaration of f(x(t)) is introduced.
Note, that the above class definition is not found in the code; it is only used here to show the stepper functionality. All specific classes possessing these `typedef`s and methods are said to be stepper classes; we also say that all classes with these definitions fulfill the stepper concept.
Now, we look at a specific method, the Euler solver:
```template<
class Container ,
class Time = double ,
class Traits = container_traits< Container >
>
class stepper_euler
{
// provide basic typedefs
public:
typedef unsigned short order_type;
typedef Time time_type;
typedef Traits traits_type;
typedef typename traits_type::container_type container_type;
typedef typename traits_type::value_type value_type;
// private members
private:
container_type m_dxdt;
// public interface
public:
// return the order of the stepper
order_type order_step() const { return 1; }
// standard constructor, m_dxdt is not adjusted
stepper_euler( void )
{
}
stepper_euler( const container_type &x )
{
}
// adjust the size of m_dxdt
void adjust_size( const container_type &x )
{
}
// performs one step with the knowledge of dxdt(t)
template< class DynamicalSystem >
void do_step( DynamicalSystem &system ,
container_type &x ,
const container_type &dxdt ,
time_type t ,
time_type dt )
{
detail::it_algebra::increment( traits_type::begin(x) ,
traits_type::end(x) ,
traits_type::begin(dxdt) ,
dt );
}
// performs one step
template< class DynamicalSystem >
void do_step( DynamicalSystem &system ,
container_type &x ,
time_type t ,
time_type dt )
{
system( x , m_dxdt , t );
do_step( system , x , m_dxdt , t , dt );
}
};```
This definition is straightforward, only some notes:
Adjusting the size is done in the `container_traits`-type. The standard traits type is defined as:
```template< class Container >
struct container_traits
{
typedef Container container_type;
typedef typename container_type::value_type value_type;
typedef typename container_type::iterator iterator;
typedef typename container_type::const_iterator const_iterator;
static void resize( const container_type &x , container_type &dxdt )
{ dxdt.resize( x.size() ); }
static bool same_size( const container_type &x1 , const container_type &x2 )
{ return (x1.size() == x2.size()); }
static void adjust_size( const container_type &x1 , container_type &x2 )
{ if( !same_size( x1 , x2 ) ) resize( x1 , x2 ); }
static iterator begin( container_type &x ) { return x.begin(); }
static const_iterator begin( const container_type &x ) { return x.begin(); }
static iterator end( container_type &x ) { return x.end(); }
static const_iterator end( const container_type &x ) { return x.end(); }
};```
which will work with all (STL) containers, fulfilling the sequence and the container concept. But for other containers, these requirements are not sufficient. For example `std::tr1::array` does not possess a `resize` method. The container traits are also responsible to for the `begin()` and `end()` iterators. For SGI containers this is trivial, but for matrix types like Blitz++ or MTL matrices the `begin()` and `end()` are not defined, but can be obtained in different ways.
All iterator operations have been put in their own functions. This has the advantage that the code is small and clear, since no iterators need to be defined in the stepper functions. Furthermore, it is be possible to specialize and optimize these iterator functions for specific iterators or containers.
The adjust size functionality is necessary if the system size changes during the evolution; think, for example, of an ODE defined on a network with a non-constant number of nodes.
This is not the end of the story of the stepper functions. odeint also provides an advanced stepper concept which calculates the numerical errors made during each step. With this error calculation, it is possible to adapt the step-size dt during each step in the iteration. The concepts behind the adaptive step size integration will not be described in this article.
The following code demonstrates the usage of the stepper function:
```#include <iostream>
#include <boost/numeric/odeint.hpp>
#define tab "\t"
using namespace std;
using namespace boost::numeric::odeint;
typedef vector< double > state_type;
const double sigma = 10.0;
const double R = 28.0;
const double b = 8.0 / 3.0;
void lorenz( state_type &x , state_type &dxdt , double t )
{
dxdt[0] = sigma * ( x[1] - x[0] );
dxdt[1] = R * x[0] - x[1] - x[0] * x[2];
dxdt[2] = x[0]*x[1] - b * x[2];
}
int main( int argc , char **argv )
{
const double dt = 0.01;
state_type x(3);
x[0] = 1.0 ;
x[1] = 0.0 ;
x[2] = 0.0;
stepper_euler< state_type > stepper;
double t = 0.0;
for( size_t oi=0 ; oi<10000 ; ++oi,t+=dt )
{
stepper.do_step( lorenz , x , t , dt );
cout << x[0] << tab << x[1] << tab << x[2] << endl;
}
}```
In some cases, it is desirable to define the ODE as a class. In our example, `lorenz` could also be defined as a class, having the appropriate bracket operator:
```class lorenz_class
{
public:
void operator()( state_type &x , state_type &dxdt , double t )
{
dxdt[0] = sigma * ( x[1] - x[0] );
dxdt[1] = R * x[0] - x[1] - x[0] * x[2];
dxdt[2] = x[0]*x[1] - b * x[2];
}
}```
which is then called via:
```lorenz_class lorenzo;
stepper.do_step( lorenzo , x , t , dt );```
With classes more complicated dynamical systems (ODEs) can be created.
### Integration Function
The next main idea is to provide one or more functions to automatically iterate the state of the ODE, with the possibility to observe the state in each step. This is done with the `integrate_const` function:
```template<
class Stepper ,
class DynamicalSystem ,
class Observer
>
size_t integrate_const(
Stepper &stepper ,
DynamicalSystem &system ,
typename Stepper::container_type &state ,
typename Stepper::time_type start_time ,
typename Stepper::time_type end_time ,
typename Stepper::time_type dt ,
Observer &observer
)
{
size_t iteration = 0;
while( start_time < end_time )
{
observer( start_time , state , system );
stepper.do_step( system , state , start_time , dt );
start_time += dt;
++iteration;
}
observer( start_time , state , system );
return iteration;
}```
which solves the ODE with constant steps `dt` in the time interval `start_time`, `end_time`. The initial state has also to be provided. Furthermore, the concept of an observer is introduced, although this is not exactly an Observer pattern. Again, the observer can have any type, it only has to possesses the bracket operator, `operator()( time , state , system )`. For example, lambda expressions from boost can be used:
```integrate_const( stepper , lorenz , x , 0.0 , 10.0 , dt ,
cout << _1 << tab << _2[0] << "\n" );```
or:
```vector< double > lx;
back_insert_iterator< vector<double> > iter( lx );
integrate_const( stepper , lorenz , x , 0.0 , 10.0 , dt , var(*iter++) = _2[0] );```
## Installation
The whole library is header only, meaning that no special installation process has to be carried out. On unixoid systems, do the following:
1. Unpack the sources : `unzip odeint.zip`
2. Change to the odeint directory : `cd odeint`
3. Set the boost environment variable : `export \$BOOST_ROOT=path_to_boost`
4. Compile the lorenz example : `make`
Now, an executable `lorenz` should be present. The boost sources are needed for the lambda expressions in the example. You can download them from boost.org and the environment variable should point to the root directory of boost, in which you will find the subdirectories bin.v2, boost, doc, lib, more, ...
For MSVC odeint will also work, but at the moment I don't have a step to step guide on how to compile the example.
## Summary and Outlook
I have shown you some design aspects, and the usage of odeint - a C++ framework for solving ordinary differential equations. It is very easy to use and very flexible. The performance is also very good. We have made some test run, where odeint is compared with the gsl. As a test system, the Lorenz system was used. It turned out that odeint is about two times faster than the gsl routines.
The source code provided with this article is a snapshot of the current development. Some methods for adaptive step size control exist and some methods to solve Hamiltonian system. In the following table, an overview over existing methods is given. S means that the stepper fulfills the simple stepper concept described above, whereas E means that the stepper fulfills the Error stepper concept, which is not introduced here but needed for adaptive step size control.
Method Class name Order Error (order) Stepper concept Euler `stepper_euler` 1 No S Runge Kutta 4 `stepper_rk4` 4 No S Runge Kutta Cash Karp `stepper_rk5_ck` 5 Yes 4 E Runge Kutta Fehlber `stepper_rk78_fehlberg` 7 Yes 8 S,E Midpoint `stepper_midpoint` Var. No S Symplectic Euler `hamiltonian_stepper_euler` 1 No S Symplectic Runge-Kutta `hamiltonian_stepper_rk` 6 No S
The whole library is not complete, and will be extended in the near future. We have a small Todo list, which should be completed before the final release:
• Adaptors for Blitz++, MTL, `boost::ublas `and `boost::multi_array`
• Documentation
• Solvers for stiff systems
• Abstract the iterator algebra
If you like to contribute to this library, or have some interesting points, criticisms, or suggestions, you are cordially invited.
## History
• 9.11.2009 - Initial document
• 16.3.2010 - Update, introducing the container traits, replacing the examples
• 23.3.2010 - Updated source code and article
## Share
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## You may also be interested in...
First Prev Next
solve state-space model? Mohammed Alloboany30-Dec-15 0:47 Mohammed Alloboany 30-Dec-15 0:47
My vote of 5 Edward Keningham11-Aug-12 3:22 Edward Keningham 11-Aug-12 3:22
My vote of 1 RedDK22-Feb-12 6:30 RedDK 22-Feb-12 6:30
look here: odeint v2 - Solving ordinary differential equations in C++[^]. But is also boost and Linux^^. Seriously, there are no problems with windows.
Excellent stuff Espen Harlinn11-Feb-12 5:07 Espen Harlinn 11-Feb-12 5:07
speed question Antediluvian999923-Jan-12 5:27 Antediluvian9999 23-Jan-12 5:27
Re: speed question Antediluvian999923-Jan-12 5:28 Antediluvian9999 23-Jan-12 5:28
Re: speed question Antediluvian999923-Jan-12 5:47 Antediluvian9999 23-Jan-12 5:47
Re: speed question Antediluvian999923-Jan-12 6:02 Antediluvian9999 23-Jan-12 6:02
Re: speed question Antediluvian999915-Feb-12 14:58 Antediluvian9999 15-Feb-12 14:58
Re: speed question Antediluvian999916-Feb-12 23:46 Antediluvian9999 16-Feb-12 23:46
My vote of 1 RedDK20-Oct-11 5:55 RedDK 20-Oct-11 5:55
Re: My vote of 1 AndyUk0624-Oct-11 4:33 AndyUk06 24-Oct-11 4:33
Re: My vote of 1 RedDK24-Oct-11 6:04 RedDK 24-Oct-11 6:04
Adaptive Example safariman11-May-10 7:55 safariman 11-May-10 7:55
Re: Adaptive Example safariman13-May-10 6:24 safariman 13-May-10 6:24
Re: Adaptive Example safariman13-May-10 12:04 safariman 13-May-10 12:04
Re: Adaptive Example safariman14-May-10 8:36 safariman 14-May-10 8:36
Re: Adaptive Example - BUG FIX gintack28-Jun-11 0:48 gintack 28-Jun-11 0:48
Re: Adaptive Example safariman13-May-10 12:19 safariman 13-May-10 12:19
Compiling in Windows pokemaniac1029-Apr-10 1:40 pokemaniac10 29-Apr-10 1:40
Use with C# KenJohnson27-Mar-10 8:21 KenJohnson 27-Mar-10 8:21
Re: Use with C# KenJohnson28-Mar-10 12:19 KenJohnson 28-Mar-10 12:19
Just a remark [modified] M.Kor24-Mar-10 5:24 M.Kor 24-Mar-10 5:24
How to compile it? caviiar30-Nov-09 12:38 caviiar 30-Nov-09 12:38
Re: How to compile it? caviiar1-Dec-09 10:04 caviiar 1-Dec-09 10:04
Re: How to compile it? caviiar3-Dec-09 10:13 caviiar 3-Dec-09 10:13
Re: How to compile it? caviiar3-Dec-09 13:35 caviiar 3-Dec-09 13:35
I suggest you remind yourself about numerical analysis. MicroImaging12-Nov-09 15:45 MicroImaging 12-Nov-09 15:45 | 4,819 | 18,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | latest | en | 0.908386 |
https://www.sophia.org/tutorials/et1410-module-3-basic-op-amp-circuits-and-active-filters-lab-31 | 1,481,028,103,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541905.26/warc/CC-MAIN-20161202170901-00181-ip-10-31-129-80.ec2.internal.warc.gz | 993,228,465 | 15,458 | ### Free Educational Resources
• > ET1410 MODULE 3 BASIC OP-AMP CIRCUITS AND ACTIV...
+
# ET1410 MODULE 3 BASIC OP-AMP CIRCUITS AND ACTIVE FILTERS LAB 3.1
(0)
Author: Carol 96
##### Description:
http://theperfecthomework.com/et1410-module-3-basic-op-amp-circuits-and-active-filters-lab-3-1/
What is the purpose of a comparator?
What type of circuit uses hysteresis to avoid rapid switching due to noise?
Describe the transfer curve of a basic comparator circuit.
Transcribe the comparator waveforms from Plot 22-1 to the Plot below.
Describe how the threshold voltage changes the comparator output.
Transcribe the comparator transfer curve from Plot 22-2 to the Plot below.
Describe how the threshold voltage changes the transfer curve for a comparator.
Assume the circuit in Figure 22-2 had VREF set to zero volts. How would you expect the output to be affected by varying the dc offset control on the generator?
Transcribe the comparator transfer curve from Plot 22-3 to the Plot below.
Would a sinusoidal input to the comparators produce the same transfer curve as a triangle waveform? Explain.
Transcribe the Schmitt trigger waveform from Plot 22-4 to the Plot below.
Transcribe the Schmitt trigger transfer curve from Plot 22-5 to the Plot below.
Summarize the important differences between a comparator and a Schmitt trigger.
Assume the input signal in Figure 22-4 could have as much as 100 mVpp noise. In order to avoid multiple tripping due to noise, you need to set the trip points at least 100 mV apart. What is the minimum value of resistance that the potentiometer can be set? Assume the output saturates at 13 V
How is the output of a summing amplifier calculated?
List four common applications of summing amplifiers.
What is the major advantage of using summing amplifiers to connect different sources?
What is the purpose of the 7493 IC used in the first summing circuit?
Transcribe the Weighted DAC Summing Amplifier waveform from Plot 23-1 to the Plot below.
The step generator in Figure 23-4 forms negative falling steps starting at zero volts and going to a negative voltage (approximately –4.4 V). Explain why.
Assume that all three inputs to the summing amplifier (QA, QB, and QC) in Figure 23-4 are 4.5V. Compute the output voltage from the summing amplifier.
(more)
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Tutorial | 670 | 2,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2016-50 | longest | en | 0.841513 |
https://brainmass.com/math/basic-calculus/calculus-differential-equations-vectors-circle-equati-119514 | 1,477,275,899,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719463.40/warc/CC-MAIN-20161020183839-00301-ip-10-171-6-4.ec2.internal.warc.gz | 835,778,920 | 17,825 | Share
Explore BrainMass
# Calculus - Differential Equations - Vectors - Circle Equati
Question (1)
First find a general solution of the differential equation dy/dx=3y. Then find a particular solution that satisfies the initial condition that y(1) = 4
Question (2)
Solve the initial value problem dy/dx=y^3 , y(0) = 1
Question (3)
Find the centre and radius of the circle described in the equation
2x2 + 2y2 - 6x + 2y = 3
Question (4)
Given a = 2 i + 3 j , b = 3 i + 5 j , and c = 8 i + 11 j express c in the form ra + sb where r and s are scalars
Question (5)
Given a = < 4 , - 3 , - 1 > and b = < 1 , 4 , 6 > find a × b
For detailed description of the questions in mathematical font and format, please see the attached question file.
Thank you
#### Solution Preview
Solution (1)
The general solution is y = c e3x
The Particular solution is y = 0.199148 e3x
NOTE: For detailed step by step solutions of all these question, please see the attached solution file.
Solution (2)
First we find the general solution ...
#### Solution Summary
Solution to the posted problems is given with step by step instructions so that the student could easily understand and use these solutions as model solutions to solve other similar problems.
\$2.19 | 339 | 1,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-44 | longest | en | 0.846692 |
https://codetuto.com/2017/02/direction-vectors-godot-engine/ | 1,653,510,874,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00401.warc.gz | 233,854,348 | 12,081 | # Direction vectors in Godot Engine
There is a previous article where we discussed movement and rotation basics of Godot Engine. In this article, we are going to look at the rotation of game objects using direction vectors in Godot Engine in detail.
We know that rotations are expressed in Euler angles. For 2D games, this is enough because there are only two axes in the world and the rotation will be along the z-axis. In 3D, setting them in Euler will be disastrous if there are multiple axes are involved because after some rotations, Gimbal Lock comes into place and our rotations will be out of control. So for 3D rotations, we use Quaternion instead of Euler. If you don’t know what is a Quaternion, then I say you don’t need to. We only need to know how to convert between Euler and Quaternion and how to specify the rotation as a Quaternion etc.
3D rotation of a game object is said as orientation. Because most of the times a game object rotation will depend on external factors such as movement direction, attack direction, enemy positions etc. So we calculate the Vector3 direction at which the game object should orientate.
### Local directional vectors
In Unity3D, there are two vector properties in the Transform component. But in Godot, there are no named directional vectors. But the same can be accessed from the transform basis. To know more about how transform and matrix are useful in Godot engine, you can visit the official tutorial page here.
Also Read: Create a Catch The Egg game in Godot engine - 4
The directional vectors can be accessed from the transform basis like this.
Forward: -get_transform().basis.z
Left: get_transform().basis.x
Up: get_transform().basis.y
### Moving an object in a direction
First, we need to find the motion vector by multiplying the unit direction vector with a speed value. We can then add the motion vector to its position to get the object moving in that direction.
```var speed = 5
var dir = get_transform().basis.x
var motion = dir * speed
var pos = get_translation()
set_translation(pos + motion)```
Another way using the transform origin.
```var speed = 5
var t = get_transform()
var dir = t.basis.x
var motion = dir * speed
t.origin += motion
set_transform(t)```
Another way using translate
```var speed = 5
var dir = get_transform().basis.x
var motion = dir * speed
translate(motion)```
### Find the direction vector to another object
`var dir = (target_obj.get_transform().origin - get_transform().origin).normalized()`
To get the direction towards another object, we should subtract its position from the other object’s position. This will get the direction vector with a magnitude which is the distance between the two objects. We normalize the vector to get the direction unit vector.
### Look at a direction
Godot has a built-in method to look at objects.
`look_at(target_obj.get_transform().origin, Vector3(0,1,0))`
The look_at method accepts the target position to rotate to. The second argument is the up vector and usually, it is the world up vector which is Vector3(0,1,0). But how can we rotate an object to a direction by passing a direction vector? It is simple, we add the direction to its position so we get a position apart from its origin in that direction.
`look_at(get_transform().origin + dir, Vector3(0,1,0))`
Thanks for reading, I’ll try to include some advanced tutorials that use these type of calculations.
Also Read: How to automate Godot Android build process in 5 easy steps
[Total: 30 Average: 3.3/5] | 793 | 3,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-21 | latest | en | 0.889655 |
http://mathhelpforum.com/discrete-math/206092-i-would-really-appreciate-if-someone-could-please-help-me-question.html | 1,480,953,293,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541697.15/warc/CC-MAIN-20161202170901-00379-ip-10-31-129-80.ec2.internal.warc.gz | 171,972,699 | 9,045 | # Thread: I would really appreciate it if someone could please help me with this question.
1. ## I would really appreciate it if someone could please help me with this question.
Define a sequence a0 a1 a2. . . recursively by setting a0 = 1 and an+1= 3 – 1/an for all n= 0, 1, 2, 3, . . . use induction to prove that 0 < an < an + 1 < 3 for all n = 0, 1, 2, 3, . . .
Attached.
3. ## Re: I would really appreciate it if someone could please help me with this question.
hey thanks alot. but i am still having a bit of difficulties with the -a^2(subscript n). i understand that the [3a(subscript n) -1]/a(subscript n) came from the a(subscript n+1) but i don't understand how -a(subscript n) gives u that -a^2(subscript n)
Attached | 221 | 733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-50 | longest | en | 0.94659 |
https://shikshahouse.com/mcq/10-wave-optics/ | 1,618,543,016,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088471.40/warc/CC-MAIN-20210416012946-20210416042946-00014.warc.gz | 549,076,673 | 32,698 | # Class 12th Physics – 10 Wave Optics MCQs
#### Light exhibits polarization because ____.
Correct! Wrong!
Polarization is a phenomenon exhibited by all transverse waves. Since light waves are transverse in nature, they also exhibit polarization.
#### When waves from two coherent sources, each producing waves of intensity Io, interfere constructively at 'P' and destructively at 'Q', then the intensity of waves at 'P' is ____ and at Q is ____.
Correct! Wrong!
When two waves interfere constructively, the intensity becomes four times the intensity of each wave. When they interfere destructively, the intensity becomes zero.
#### In which of the following Polaroids are not used.
Correct! Wrong!
Polaroids are used to control the intensity of light. In case of telescopes, no such Polaroid is required.
#### As a wave propagates from one refractive medium to another, there will be no change in ____ of the wave.
Correct! Wrong!
As a wave propagates from one refractive medium to another, there will be no change in frequency of the wave.
#### Interference of waves can be explained with ____.
Correct! Wrong!
Interference of waves can be explained with the principle of superposition.
#### If v1 and v2 are the speeds of a wave in two different media of refractive indices n1 and n2 respectively and n1 < n2, then ____
Correct! Wrong!
If v1 and v2 are the speeds of a wave in two different media of refractive indices n1 and n2 respectively and n1 < n2, then v1 >v2
#### If the angles made by incident wavefront and reflecting wavefront with the reflecting surface are "θ1" and "θ2", then ____.
Correct! Wrong!
The angle, the incident wavefront makes with the reflecting surface is the angle of incidence.
The angle, the reflecting wavefront makes with the reflecting surface is angle of reflection
According to laws of reflection θ1 = θ2.
#### The shape of the wave fronts formed on the surface of water is ____
Correct! Wrong!
The shape of the wave fronts formed on the surface of water is circular.
#### Speed of light in a medium of refractive index n1 is 2.5 × 108 m/s and in another medium of refractive index n2 is 2 × 108 m/s, then ____
Correct! Wrong!
Since speed of light is less in a medium of high refractive index, n2 > n1
#### According to Huygens, the amplitude of secondary wavelets is ___ in the forward direction and ____ in the backward direction.
Correct! Wrong!
According to Huygens, the amplitude of secondary wavelets is maximum in the forward direction and zero in the backward direction.
#### When unpolarised light incident at Brewster's angle, the reflected light is polarized with its ____ to the plane of incidence.
Correct! Wrong!
When unpolarised light incident at Brewster's angle, the reflected light is completely polarized with its electric vector perpendicular to the plane of incidence.
#### In air sound travels with a speed of 340 m/s. The diameter of the spherical sound wave front generated by a point source of sound after 1/10 th of a second is ____
Correct! Wrong!
The speed of the wave front in a medium is equal to the speed of the wave in the same medium. That means, the wave front travels at a speed of 340 m/s. In 1/10th of a second, the radius of the spherical wave front becomes "340 × 1/10 = 34 m. The diameter of the wave front after 1/10th of a second must be 68 m.
#### ' P' is a point in space and S1 and S2 are two coherent sources generating waves of wavelength 1 cm. If the path difference between S1P and S2P is 5 cm, then waves interfere ____ at P.
Correct! Wrong!
Path difference
S1P − S2P = 5 cm
Since wavelength λ = 1 cm
S1P − S2P = 5λ
For path difference equal to integer multiples of wavelength, waves interfere constructively.
#### Two sources are said to be coherent if ____.
Correct! Wrong!
Two sources are said to be coherent if frequency of waves produced is same for both and the phase difference between the displacements produced by each of the waves does not change.
#### If 'θ' is the angle an incident wave front makes with the refracting surface, then the angle of incidence is ____
Correct! Wrong!
If 'θ' is the angle an incident wavefront makes with the refracting surface, then the angle of incidence is 'θ'.
#### When unpolarized light incident on the boundary between two transparent media and if the reflected wave is perpendicular to the refracted wave, then ___
Correct! Wrong!
When the reflected and refracted waves are perpendicular to each other, the reflected wave is completely polarized. This is the condition of Brewster's law.
#### If λ1 is the wavelength of light in medium one and λ2 is the wavelength of light in medium two, as the wave travels from medium one to medium two, sini/sinr= ____
Correct! Wrong!
If λ1 is the wavelength of light in medium one and λ2 is the wavelength of light in medium two, as the wave travels from medium one to medium two, sini/sinr=λ21
#### When light incident on a transparent interface at polarizing angle the intensity of reflected light is I1 and the intensity of refracted light is I2, then ____.
Correct! Wrong!
At polarizing angle, the reflected light is completely polarized and contain light with 'E' vector perpendicular to the plane of incidence. But the refracted light contains light that is still not completely polarized. Hence the intensity of refracted light is greater than the intensity reflected light.
#### The angular width of central maxima in Fraunhoffer diffraction when light of wavelength 5500A is used and the slit width is 11 × 10−5 cm
Correct! Wrong!
θ=2λ/a
=2×5500×10−10/11×10−5×10−2
#### When a slit is illuminated with a light of wavelength 6000A, the first minima is obtained at θ = 20° the slit width is ____.
Correct! Wrong!
The angular position of the first minima is given by
θ=λ/a
⇒a=λ/θ
=(6000×10−10/20×2π)×360
=6000×1010×360×7/20×2×22
= 17.2 × 10−4 mm θ=λ/a
⇒a=λ/θ
=(6000×10−10/20×2π)×360
=6000×1010×360×7/20×2×22
= 17.2 × 10−4 mm
#### When a wave is passing from a denser medium to rarer medium an angle of incidence of 0000, then after refraction, ____
Correct! Wrong!
When a wave is passing from a denser medium to rarer medium an angle of incidence of 00, then after refraction, waves travel without bending.
#### Huygens principle is used to ____
Correct! Wrong!
Huygens principle is used to determine the shape and position of the wave front at a later time.
#### The refractive index of a material is equal to the tangent of the polarizing angle. This is known as ____.
Correct! Wrong!
According to Brewster's law, the refractive index of a material is equal to the tangent of the polarizing angle.
#### a) Primary waves can travel in all directions in a given medium. b) Secondary waves can travel only in the forward direction but not in backward direction
Correct! Wrong!
Primary waves can travel in all directions in a given medium. Secondary waves can travel only in the forward direction but not in backward direction.
#### Huygens' Principle is applicable ____
Correct! Wrong!
Huygens' principle is useful for all types of waves.
#### If at a point "P" interfering waves from two coherent sources are differing in phase by 7π4, then at "P" ____.
Correct! Wrong!
The phase difference in the given case = 7π4
The general expression for constructive interference is phase difference = 2nπ
The general expression for destructive interference is phase difference = (2n + 1)π
Since the given phase difference is not in any of these forms, there will be no interference at "P".
#### a) Each point of the primary wave acts as a source of a secondary disturbance b) Common tangent to all secondary wavelets gives the new position of the wave
Correct! Wrong!
According to Huygens principle, each point of the primary wave acts as a source of a secondary disturbance and the common tangent to all secondary wavelets gives the new position of the wave.
#### In a medium, the speed of the wave front is ____ the speed of the wave in that medium.
Correct! Wrong!
In a medium, the speed of the wave front is equal to the speed of the wave in that medium.
#### A plane wave front, after travelling for a long time, will behave like a ____
Correct! Wrong!
A plane wave front, after travelling for a long time, will behave like a plane wave front only
#### Wave front is an imaginary surface representing locus of points oscillating ____.
Correct! Wrong!
Wave front is an imaginary surface representing locus of points oscillating in phase
Class 12th Physics - 10 Wave Optics MCQs
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Topic: pigeonhole lemma
David Renshaw (Sep 09 2020 at 23:38):
Can anyone help me to simplify my proof of this pigeonhole lemma?
import data.finset.basic
import data.nat.basic
import tactic
lemma pidgeonhole :
∀ s: finset ℕ, ∀ f : ℕ → {x : ℕ // x ∈ s}, ∃ a b : ℕ, (a ≠ b ∧ f a = f b) :=
begin
apply @finset.induction ℕ (λ s: finset ℕ, ∀ f : ℕ → {x : ℕ // x ∈ s}, ∃ a b : ℕ, (a ≠ b ∧ f a = f b)),
{
intro,
exfalso,
exact (finset.not_mem_empty (f 0).val) (f 0).property,
},
intros n s hn hip f,
cases classical.em (∃k:ℕ, (∀ m:ℕ, k ≤ m → (f m).val ≠ n)) with hbounded hinf,
{
obtain ⟨k0, hk0⟩ := hbounded,
let fk: ℕ → {x // x ∈ s} := (λn:ℕ, ⟨(f (n + k0)).val, _⟩),
{
obtain ⟨a, ha⟩ := hip fk,
obtain ⟨b, hb⟩ := ha,
use (a + k0),
use (b + k0),
cases hb,
split,
{
intro heq,
apply hb_left,
},
simp at hb_right,
exact subtype.ext hb_right,
},
have hk0a := (hk0 (n + k0)) (nat.le_add_left _ _),
exact finset.mem_of_mem_insert_of_ne (f (n + k0)).property hk0a,
},
{
have hall : ∀k:ℕ, ¬(∀ m:ℕ, k ≤ m → (f m).val ≠ n) := not_exists.mp hinf,
let hk0 := hall 0,
have : (∃ m:ℕ, ¬(0 ≤ m → (f m).val ≠ n)) := not_forall.mp hk0,
obtain ⟨m0, hm0⟩ := this,
have hnn : 0 ≤ m0 ∧ ¬ (f m0).val ≠ n := not_imp.mp hm0,
have hfm0 : (f m0).val = n := not_not.mp hnn.2,
let hk1 := hall (m0 + 1),
have : (∃ m:ℕ, ¬(m0 + 1 ≤ m → (f m).val ≠ n)) := not_forall.mp hk1,
obtain ⟨m1, hm1⟩ := this,
have hnn1 : m0 + 1 ≤ m1 ∧ ¬ (f m1).val ≠ n := not_imp.mp hm1,
have hfm1 : (f m1).val = n := not_not.mp hnn1.2,
use m0,
use m1,
split,
{
cases hnn1,
linarith,
},
have : (f m0).val = (f m1).val := by rw [hfm0, hfm1],
exact subtype.eq this,
}
end
Scott Morrison (Sep 09 2020 at 23:44):
Very mathlib style would be to deduce this from src#infinite_pigeonhole. :-) I appreciate that's not what you're asking for!
Scott Morrison (Sep 09 2020 at 23:47):
By "simplify" do you mean "but still using induction on finset"? Because my first suggest would be to not do that. :-)
Scott Morrison (Sep 09 2020 at 23:49):
Given the result you're after, you could just intro s, say its card is n, and restrict f to range (n+1), and use a finite pigeonhole principle.
Alexandre Rademaker (Sep 09 2020 at 23:50):
That on the same formulation of the lemma ? Or are you proposing another formulation ?
Scott Morrison (Sep 09 2020 at 23:51):
I was suggesting to either use from a mathlib, or prove from scratch if that's the exercise, the usual finite pigeonhole principle for maps between finsets. Then reduce what @David Renshaw wants to prove to that.
Scott Morrison (Sep 09 2020 at 23:53):
There is src#finset.card_image_of_injective. Someone should PR the easy consequence that actually looks like the finite pigeonhole principle, and make sure to add a doc-string with the word "pigeonhole"!
David Renshaw (Sep 10 2020 at 00:02):
thanks! I tried searching mathlib for "pidgeonhole" but that was before I figured out that I was spelling it wrong. :face_palm:
Kyle Miller (Sep 10 2020 at 00:08):
I simplified your proof using some tricks I know. I didn't know about obtain!
import data.finset.basic
import data.nat.basic
import tactic
lemma pidgeonhole (s : finset ℕ) (f : ℕ → {x : ℕ // x ∈ s}) :
∃ a b : ℕ, (a ≠ b ∧ f a = f b) :=
begin
refine finset.induction (λ f, _) (λ n s hn hip f, _) s f,
{ exfalso,
exact (finset.not_mem_empty (f 0).val) (f 0).property, },
by_cases h : ∃ k : ℕ, ∀ m : ℕ, k ≤ m → (f m).val ≠ n,
{ obtain ⟨k0, hk0⟩ := h,
let fk: ℕ → {x // x ∈ s} := λ n : ℕ, ⟨(f (n + k0)).val, begin
have hk0a := hk0 (n + k0) (nat.le_add_left _ _),
exact finset.mem_of_mem_insert_of_ne (f (n + k0)).property hk0a,
end⟩,
obtain ⟨a, b, hne, hfk⟩ := hip fk,
use [a + k0, b + k0],
split,
{ revert hne,
contrapose, push_neg,
{ simp only [subtype.mk_eq_mk, subtype.val_eq_coe] at hfk,
exact subtype.ext hfk, } },
{ push_neg at h,
obtain ⟨m0, _, hk0⟩ := h 0,
obtain ⟨m1, hne, hk1⟩ := h (m0 + 1),
use [m0, m1],
split,
{ linarith [hne], },
{ apply subtype.eq,
rw [hk0, hk1], }, },
end
Scott Morrison (Sep 10 2020 at 00:15):
obtain is pretty awesome. :-)
David Renshaw (Sep 10 2020 at 00:15):
how would you use an existential without obtain?
Scott Morrison (Sep 10 2020 at 00:16):
some combination of have and Exists.some, I guess?
Kyle Miller (Sep 10 2020 at 00:16):
I always would do rcases hip fk with ⟨a, b, hne, hfk⟩ rather than obtain ⟨a, b, hne, hfk⟩ := hip fk.
Kyle Miller (Sep 10 2020 at 00:18):
It's pretty much the same, but I like how with obtain the expression is at the end, since it opens up having a larger expression being decomposed while still having somewhat readable code.
Bryan Gin-ge Chen (Sep 10 2020 at 00:20):
Before I learned to use rcases, I would write horrible terms starting with exists.elim. I still need to get in the habit of using obtain though.
Kyle Miller (Sep 10 2020 at 00:36):
@David Renshaw Found a way to simplify it even more:
lemma pigeonhole (s : finset ℕ) (f : ℕ → (↑s : set ℕ)) :
∃ a b : ℕ, a ≠ b ∧ f a = f b :=
begin
classical,
by_contra hc,
push_neg at hc,
have hinj : function.injective f,
{ intros a b,
contrapose,
exact hc a b, },
apply not_injective_infinite_fintype f hinj,
end
(The (↑s : set ℕ) is another way to write {x : ℕ // x ∈ s})
cool!
Scott Morrison (Sep 10 2020 at 00:39):
At this point, you're essentially relying on an existing proof of the pigeonhole principle, and this proof is just providing the glue.
David Renshaw (Sep 10 2020 at 00:41):
That's good enough for me! Working through the induction was a fun exercise, but I'm mostly concerned with learning how to do things idiomatically.
David Renshaw (Sep 10 2020 at 00:45):
ah... push_neg is quite convenient
Kyle Miller (Sep 10 2020 at 00:48):
I wouldn't really call these the pigeonhole principle, but maybe I've been too influenced by a Dijkstra essay. His formulation was, essentially, that if you have a function $f : A \to B$ with $B$ a finite set, then the maximum cardinality of a preimage is at least $\lvert A\rvert / \lvert B\rvert$. So, for $f : \mathbb{N} \to B$, I'd like $\exists b \in B, \lvert f^{-1}(b)\rvert = \infty$.
David Renshaw (Sep 10 2020 at 00:51):
You're saying that there's a stronger lemma that better deserves the name pigeonhole? That sounds like a reasonable stance to me.
Kyle Miller (Sep 10 2020 at 01:01):
You can say that there's a hole that contains at least the average number of pigeons per hole. I'm not sure the best way to formalize that in Lean, though.
Following @Scott Morrison's suggestion, it seems like it might be worth creating an analogue of not_injective_infinite_fintype for fintype.card_le_of_injective and adding documentation comments that these are the classic pigeonhole principles for finitely many and infinitely many pigeons. All I really did with my last simplification was using push_neg to recover the hidden function.injective.
Bryan Gin-ge Chen (Sep 10 2020 at 01:06):
(Issue #2772 is also about the pigeonhole principle.)
Kyle Miller (Sep 10 2020 at 01:10):
(Ah, found it: #2272)
Julian Berman (Sep 10 2020 at 01:39):
Kyle Miller said:
(The (↑s : set ℕ) is another way to write {x : ℕ // x ∈ s})
what do these notations mean?
Julian Berman (Sep 10 2020 at 01:40):
(or how would I look up // other than grepping around mathlib which takes me to some pretty low-level files)
David Renshaw (Sep 10 2020 at 01:42):
One thing you can do is set_option pp.notation false
David Renshaw (Sep 10 2020 at 01:45):
I find that when I jump-to-definition on the opening { of {x : ℕ // x ∈ s}, I get properly taken the definition of subtype. It does not work on the // or the closing }.
(in emacs)
Julian Berman (Sep 10 2020 at 01:46):
aha cool, that helps, thanks (pp.notation works here), lemme try jump to definition
Julian Berman (Sep 10 2020 at 01:47):
@David Renshaw awesome, that works too as long as yeah I do it on the { and not // which is what I tried before. Appreciated!
Kyle Miller (Sep 10 2020 at 04:01):
Bryan Gin-ge Chen said:
(Issue #2272 is also about the pigeonhole principle.)
Ok, here they are! #4096
Kevin Buzzard (Sep 10 2020 at 06:11):
Adrian Mathias told me that the pigeonhole principle was that if there were two pigeons in one hole, then one hole contained two pigeons. He then remarked that this idea could be generalized
Last updated: May 14 2021 at 00:42 UTC | 2,704 | 8,305 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | longest | en | 0.688864 |
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# Functional Programming in Swift
This post written by Guanshan Liu
Guanshan is an iOS Developer currently working as an iOS Engineer at Alibaba Inc. Before working at Alibaba, Guanshan worked with 2K Games on Civilization Revolution 1 and 2 for iOS. He has a Masters in Software Engineering from the University of York, UK as well as a Bachelors of Engineering in Information Security from Nanjing University of Aeronautics and Astronautics. Find him on twitter, @guanshanliu.
# Thoughts on Functional Programming in Swift
Like most of you, I have to use Objective-C at my day job. I could only craft my Swift skills at night. Swift is not a purely functional language. It can be use imperatively because all frameworks from Apple are written in Objective-C at the time of writing. However, it is also functional, learning from modern languages like Haskell, F#, etc. At the beginning when reading Swift blogs and tutorials, many people talked about terms like functors and monads, those sounded alien-like to me. I started to learn Haskell to understand what they were talking about. I’m not an expert, and still learning functional programming and Swift, but I wanted to share what I’ve learned so far. Hopefully it will save you some troubles to get into the functional programming world.
# Key Concepts
## Higher-order functions
One key concept of functional programming is higher-order functions. According to Wikipedia, a higher-order function:
• takes one or more functions as an input
• outputs a function
In Swift, functions are first-class citizens, like structs and classes, as we can see in the following example:
``let addFive = { \$0 + 5 }addFive(7) // Output: 12``
We define a function as an inline closure, and then assign it to an inline closure. Swift provides shorthand argument to it. Instead of by name, we can refer to the arguments by number, \$0, \$1 and so on.
``func addThreeAfter(f: Int -> Int) -> Int -> Int { return { f(\$0) + 3 }}let addEight = addThreeAfter(addFive)addEight(7) // Output: 15``
The argument type `Int -> Int` means it is a function that takes an Int as an argument, and returns an Int. If a function requires more than one arguments, for example, it has two Int argument, we can define the type as `(Int, Int) -> Int`.
The return type of `addThreeAfter` is also a function. It equivalents to `func addThreeAfter(f: Int -> Int) -> (Int -> Int)`.
``[addFive, addEight].map { \$0(7) } // Output: [12 15]``
`map` is a special function for container type, such as Array, Optional. It unwraps values from the container type, then applies a transform for each, and wraps them again. We use `map` here to pass 7 as an argument to each function in the array, and wraps the results into a new array.
In summary, we can assign functions to variables, store them in data structures, pass them as arguments to other functions, and even return them as the values from other functions.
## Pure functions
A function is called a pure function if it has no observable side effects. But what the heck is side effect?
A function or expression is said to have a side effect if, in addition to returning a value, it also modifies some state or has an observable interaction with calling functions or the outside world.
— from Wikipedia
The two functions `addFive` and `addEight` are pure functions. They don’t modify the input value, or change any global state. In the example below, `addFive2` modifies the input, so it is not a pure function.
``func addFive2(inout a: Int) -> Int { a += 5 return a}var a = 7addFive2(&a)a // Output: 12``
Functions that access or modify a database, the local file system, or printing strings to the screen are also considered impure. Because they modify the state of the database records, file system, and display buffers respectively. Sometimes side effects are desirable. Without side effects, we could not interact with the program.
Haskell introduces types like IO to separate pure and impure layers. But in Swift, we don’t need to worry too much about this separation. We could still use Cocoa APIs as usual. But, I strongly encourage the use of pure functions whenever possible. With pure functions, the code is more predictable. A pure function will always return the same value if given the same input. It’s easy to test, too. We don’t need to mock a data structure to satisfy its internal state requirements to test a function.
# Imperative & Functional Programming
All above are very theoretical. You may want to know how functional programming with help us solve problems in a better, clearer, or less error-prone way. What are the benefits we could gain from it?
First, you need to understand that we could do almost anything in imperative programming, and functional programming does not extend the possibilities of what we could do.
Second, if you come from the imperative programming world, some functional code is harder to understand at first, especially those with custom operators. It is not because functional programming is hard and obscure, but it’s because our mindsets are trained from years of imperative programming practices.
Take a look at this example of imperative code, which we can rewrite as functional code. These two do exactly the same things.
``// Imperativevar source = [1, 3, 5, 7, 9]var result = [Int]()for i in source { let timesTwo = i * 2 if timesTwo > 5 && timesTwo < 15 { result.append(timesTwo) }}result // Output: [6, 10, 14]``
``// Functionallet result2 = source.map { \$0 * 2 } .filter { \$0 > 5 && \$0 < 15 }result2 // Output: [6, 10, 14]``
It is arguable which one is clearer. But from this simple example you can see the main difference between imperative and functional programming. In imperative programming, we instruct the computer how to do something:
1. Iterate through `source`
2. Get the result from the element, and multiply by 2
3. Compare it with 5 and 15
4. If it is bigger than 5 and less than 15, put it into `result`
However, in functional programming, we describe what to do:
1. Transform each element in `source` to itself multiplied by 2
2. Only select the ones with value bigger than 5 and less than 15
I’m not going to persuade you functional programming is better. It’s your personal preference. After all, good code is all about writing code that:
1. Works as intended
2. Is clear to you and your team
# An Example: Reverse Polish Notation Calculator
I like Swift and functional programming, because it enables me to solve a problem in a different perspective. There is usually more than one way to solve a problem. Exploring a better solution helps us grow to become good developers.
Let me show you a functional example. It is a calculator for algebraic expressions of reverse polish notation, or RPN in short. (It is a Swift implementation of the Haskell example in Learn You a Haskell for Great Good.)
A RPN expression of (3 + 5) 2 is 3 5 + 2 . You may think of this as a stack of numbers. We go through the RPN expression from left to right. When encountering a number, we push it onto the stack. When we encounter an operator, we pop two numbers from the stack, use the operator with those two numbers, and then push the result back onto the stack. When reaching the end of the expression, the only one number left on the stack is the result (assuming the RPN expression is valid). For more explanation about RPN, please check on Wikipedia.
We want a function that returns the result for an RPN expression.
``func solveRPN(expression: String) -> Double { // Process the expression and return the result}``
Given a RPN expression String “3 5 + 2 *”, first we need to transform it into an array of elements that we can process. There are two kind of elements, operand and operator. The Enum data type in Swift comes in handy for defining the element. We name it `RPNElement`.
``enum RPNElement { case Operand(Double) case Operator(String, (Double, Double) -> Double)}``
Next, we split the expression into an array of Strings, then `map` it into an RPNElement array.
``extension String { func toRPNElement() -> RPNElement { switch self { case "*": return .Operator(self, { \$0 * \$1 }) case "+": return .Operator(self, { \$0 + \$1 }) case "-": return .Operator(self, { \$0 - \$1 }) default: return .Operand(Double(self.toInt()!)) } }}func stringToRPNElement(s: String) -> RPNElement { return s.toRPNElement()}func solveRPN(expression: String) -> Double { let elements = expression.componentsSeparatedByString(" ").map(stringToRPNElement) // Further process }``
Next, we will go through the array and process it according to how RPN works, as I described earlier. We `reduce` the array into a Double array. Assuming the expression is valid, the Double array should only contain one element. It will be the result we want.
``func solveRPN(expression: String) -> Double { let elements = expression.componentsSeparatedByString(" ").map(stringToRPNElement) let results = elements.reduce([]) { (acc: [Double], e: RPNElement) -> [Double] in switch e { case .Operand(let operand): return [operand] + acc case .Operator(let op, let f): let r = f(acc[0], acc[1]) return [r] + acc[2..<acc.count] } } return results.first ?? 0}solveRPN("3 5 + 2 *") // Output: 16``
# Where to Go From Here?
If you are interested in learning more about functional programming, I highly recommend the following two books:
Even though the second book is written for Haskell, but the concepts also apply to Optional in Swift as well. Besides, it explains Functor, Applicative, Monad in details.
Your support on Patreon allows me to make better tutorials more often.
This entry was posted in Guest Post, iPhone Development, Programming, Swift and tagged , , by Guanshan Liu. Bookmark the permalink.
Guanshan is an iOS Developer currently working as an iOS Engineer at Alibaba Inc. Before working at Alibaba, Guanshan worked with 2K Games on Civilization Revolution 1 and 2 for iOS. He has a Masters in Software Engineering from the University of York, UK as well as a Bachelors of Engineering in Information Security from Nanjing University of Aeronautics and Astronautics. Find him on twitter, @guanshanliu.
## 3 thoughts on “Functional Programming in Swift”
1. Look at the line case Operator(String, (Double, Double) -> Double)
I’m not sure why the string value of the operator is stored. It is not referred to at any point. It gets assigned to a variable in this statement, but is not used:
case .Operator(let op, let f):
let r = f(acc[0], acc[1])
return [r] + acc[2..<acc.count]
}
I tested it in a playground and it worked fine, then removed the String from the enum and modified all the places that then put xcode in a tizz, and it worked again.
Good tutorial, btw.
2. Earlier in the tutorial you have:
A RPN expression of (3 + 5) 2 is 3 5 + 2
The “3 5 + 2” is a typo and should be replaced by “3 5 + 2 *” what you have later in the tutorial
3. This implementation appears to be incorrect.
I believe this line
let r = f(acc[0], acc[1])
should be
let r = f(acc[1], acc[0])
because currently “3 5 -” evaluates to 2 where it should be -2. or is my understanding of RPN wrong? | 2,745 | 11,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.935126 |
http://www.finderchem.com/what-is-more-decimeter-or-centimeter.html | 1,500,671,198,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423809.62/warc/CC-MAIN-20170721202430-20170721222430-00383.warc.gz | 432,657,078 | 7,280 | # What is more decimeter or centimeter?
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Centimeter: more ... A centimeter (UK: centimetre) is a measure of length. There are 100 centimeters in a meter. 2.54 cm = 1 inch. The abbreviation is cm.
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Related Questions
Recent Questions | 1,388 | 5,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-30 | latest | en | 0.625721 |
http://mizar.uwb.edu.pl/version/current/html/proofs/pnproc_1/18 | 1,571,625,924,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00362.warc.gz | 129,828,974 | 1,552 | let P be set ; :: thesis: for m1, m2, m3 being marking of P holds (m1 + m2) + m3 = m1 + (m2 + m3)
let m1, m2, m3 be marking of P; :: thesis: (m1 + m2) + m3 = m1 + (m2 + m3)
let p be object ; :: according to PNPROC_1:def 1 :: thesis: ( p in P implies p multitude_of = p multitude_of )
assume A1: p in P ; :: thesis:
then A2: ((m1 + m2) + m3) . p = ((m1 + m2) . p) + (m3 . p) by Def4
.= ((m1 . p) + (m2 . p)) + (m3 . p) by ;
(m1 + (m2 + m3)) . p = (m1 . p) + ((m2 + m3) . p) by
.= (m1 . p) + ((m2 . p) + (m3 . p)) by
.= ((m1 . p) + (m2 . p)) + (m3 . p) ;
hence p multitude_of = p multitude_of by A2; :: thesis: verum | 281 | 614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-43 | latest | en | 0.500705 |
https://kumamotojet.com/mw/index.php?title=Blockbusters:_a_TV_Quiz_Show_Game_in_the_Classroom | 1,603,652,588,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00122.warc.gz | 389,968,747 | 6,365 | # Blockbusters: a TV Quiz Show Game in the Classroom
Textbook and specific lesson: None
Goal: work as a team to solve word clues/puzzles
Preparation: a list of words File:Blockbusterswordlist.docx
Class Time: Varies, Low academic = 45-50 mins. High academic = 25-30 mins
Draw a 5 x 5 grid on the board and fill it with all the letters of the alphabet, except X, in random order, as in the diagram here. File:Blockbustersgameboardex.docx
Divide students into 2 teams and write an A and a horizontal arrow at the left side of the grid to indicate that Team A are going across, a B and a downwards-facing arrow at the top to indicate that Team B are going down.
Team A have to get a horizontal line across e.g. U-J-A-T-E to win. Team B have to get a vertical line down to win. They will of course get in each other’s way as they do this so it doesn’t have to be a straight line; it just has to be connected e.g. Team A could go U-J-I-P-H-Q-G. (The connected squares must be above, below or to the side, not in a diagonal relationship to each other)
Team A picks a letter. They would usually start in the furthest left column but they can begin anywhere they like. The teacher just has to think of a word beginning with that letter and a short and simple way to explain it:
For A you might think of ‘artist’ and say ‘for example, Picasso’.
For F you might think of ‘forty’ and say ’20 + 20’.
For K you might think of ‘king’ and say ‘the queen’s husband’.
In the grid above, imagine Team A has picked ‘F’ and the teacher has said ‘it’s a sport’. Either team can answer. It’s simply a matter of the first person to shout it out. If team A says ‘football’ first, they win and get to pick the next letter- logically ‘B’ for them. If Team B wins they get to pick the next letter – logically ‘V’ for them as they work their way down. However, as before, with each new question, either team can answer and change the direction of the game.
If students don’t get an answer immediately keep giving further clues or even start spelling it out on the board. If it’s too hard for them, then change to another easier word starting with that letter. If it’s too easy and students are flying through the game too quickly throw in a long, hard word or give cryptic explanations for simple words. One of my favorites for ‘V’ is ‘Everyone has one, though you can’t see it’ (Answer: Voice)
As you go along, the teacher must fill in the squares to show each team’s progression, e.g. with an ‘X’ over the letter. If possible use different coloured markers; if not indicate the difference in teams by a different pattern in each square e.g. Xs vs. Os. Continue like this until one team has made a fully connected line, however convoluted, all the way across or down.
Though of course the scope is endless and it’s easy just to think of words as you go along, you want to avoid words that are hard to explain like ‘make’ or ‘truth’ or ‘since’ or ‘only’. Simple action verbs and concrete nouns are best. Here are some suggestions | 732 | 3,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-45 | latest | en | 0.954165 |
http://www.onbarcode.com/tech/748/17/ | 1,656,471,406,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00543.warc.gz | 106,955,556 | 8,657 | # how to print barcode in crystal report using vb.net ALGORITHMS in Java Generation USS Code 128 in Java ALGORITHMS
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The original assumption, that a TM does exist that can determine whether any particular program will run to completion when presented with any arbitrary input data, must be incorrect. That assumption led to the contradictory state illustrated by NotH. Therefore, computer scientists conclude that there can be no one algorithm that can determine whether any particular program will run to completion, or fail to run to completion, for every possible set of inputs. It would be very nice to have a program to which we could submit new code for a quick determination as to whether it would run to completion given any particular set of inputs. Alas, Turing proved that this cannot be. One can and should write test programs, but one will never succeed in writing one program which can test every program. The halting problem is one of the provably unsolvable problems in computing (Turing, Alan, On computable Numbers with an Application to the Entscheidungsproblem , Proceedings of the London Mathematical Society, 2:230 265, 1936). No one algorithm will ever be written to prove the correct or incorrect execution of every possible program when presented with any particular set of inputs. While no such algorithm can be successful, knowing that allows computer scientists to focus on problems for which there are solutions. SUMMARY An algorithm is a specific procedure for accomplishing some job. Much of computer science has to do with finding or creating better algorithms for solving computational problems. We usually describe computational algorithms using pseudocode, and we characterize the performance of algorithms using the term order of growth or theta. The order of growth of an algorithm tells us, in a simplified way, how the running time of the algorithm will vary with problems of different sizes. We provided examples of algorithms whose orders of growth were (lg n), n, n(lg n), n2, 2n and n!. Algorithm development should be considered an important part of computing technology. In fact, a better algorithm for an important task may be much more impactful than any foreseeable near-term improvement in computing hardware speed. The Turing machine is a formal mathematical model of computation, and the Church Turing thesis maintains that any algorithmic procedure to manipulate symbols can be conducted by some Turing machine. We gave example Turing machines to perform the simple binary operations of complementing and incrementing a binary number. Some problems in computing are provably unsolvable. For instance, Turing proved that it is impossible to write one computer program that can inspect any other program and verify that the program in question will, or will not, run to completion, given any specific set of inputs. While the Holy Grail of an algorithm to prove the correctness of programs has been proven to be only a phantom in the dreams of computer scientists, computer scientists at least know that is so, and can work instead on practical test plans for real programs. REVIEW QUESTIONS 2.1 Write pseudocode for an algorithm for finding the square root of a number. 2.2 Write pseudocode for finding the mean of a set of numbers. 2.3 Count the primitive operations in your algorithm to find the mean. What is the order of growth of your mean algorithm 2.4 Write pseudocode for finding the median of a set of numbers. 2.5 What is the order of growth of your algorithm to find the median 2.6 Suppose that your algorithm to find the mean is (n), and that your algorithm to find the median is (n lg n), what will be the execution speed ratio between your algorithm for the mean and your algorithm for the median when the number of values is 1,000,000 2.7 A sort routine which is easy to program is the bubble sort. The program simply scans all of the elements to be sorted repeatedly. On each pass, the program compares each element with the one next to it, and reorders the two, if they are in inverse order. For instance, to sort the following list: 67314
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Copyright © OnBarcode.com . All rights reserved. | 1,417 | 6,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-27 | latest | en | 0.890977 |
http://evolutionoftruth.com/goldensection/history.htm | 1,579,692,421,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606975.49/warc/CC-MAIN-20200122101729-20200122130729-00210.warc.gz | 54,713,417 | 4,586 | ### The Phi Nest™ now has its own domain with new content at http://goldennumber.net Please update your bookmarks.
The Phi-Nest™ on the Golden Number Home Phi for Neo'phi'tes Fibonacci Series Divine Proportion/ Golden Section/ Golden Mean Credit Cards Energy Geometry History Life Human Hand Human Face Human Body Heartbeat Animals 1 Animals 2 Plants DNA Mathematics Means Music Number 89 Penrose Tiling Quasi-crystals Population Growth Powers of Phi Solar System Spirals Stock Markets The Bible Theology Feedback Meet the Phi Guy Do It Yourself!!! Search the Site Links Books & More
# History
While the proportion known as the Golden Mean has always existed in mathematics and in the physical universe, it is unknown exactly when it was first discovered and applied by mankind. It is reasonable to assume that it has perhaps been discovered and rediscovered throughout history, which explains why it goes under several names.
### Uses in architecture date to the ancient Egyptians and Greeks
Euclid (365BC - 300BC), in "Elements," referred to dividing a line at the 0.6180399... point as dividing a line in the extreme and mean ratio. This later gave rise to the use of the term mean in the golden mean.
The Egytians used both pi and phi in the design of the Great Pyramids. The Greeks, who called it the Golden Section, based the entire design of the Parthenon on this proportion. It was also used by Phidias, a famous Greek Sculptor. (1)
### The Fibonacci Series was discovered around 1200 AD
Leonardo Fibonacci, an Italian born in 1175 AD (2) discovered the unusual properties of the numerical series that now bears his name, but it's not certain that he even realized its connection to phi and the Golden Mean. His most notable contribution to mathematics was a work known as Liber Abaci, which became a pivotal influence in adoption by the Europeans of the Arabic decimal system of counting over Roman numerals. (3)
### It was first called the "Divine Proportion" in the 1500's
Da Vinci provided illustrations for a dissertation published by Luca Pacioli in 1509 entitled "De Divina Proportione" (1), perhaps the earliest reference in literature to another of its names, the "Divine Proportion." This book contains drawings made by Leonardo da Vinci of the five Platonic solids. It was probably da Vinci who first called it the "sectio aurea," which is Latin for golden section.
The Renaissance artists used the Golden Mean extensively in their paintings and sculptures to achieve balance and beauty. Leonardo Da Vinci, for instance, used it to define all the fundamental proportions of his painting of "The Last Supper," from the dimensions of the table at which Christ and the disciples sat to the proportions of the walls and windows in the background.
Johannes Kepler (1571-1630), discoverer of the elliptical nature of the orbits of the planets around the sun, also made mention of the "Divine Proportion," saying this about it:
"Geometry has two great treasures: one is the theorem of Pythagoras; the other, the division of a line into extreme and mean ratio. The first we may compare to a measure of gold; the second we may name a precious jewel."
### The term "Phi" was not used until the 1900's
It wasn't until the 1900's that American mathematician Mark Barr used the Greek letter phi to designate this proportion. Phi is the first letter of Phidias (1), who used the golden ratio in his sculptures, as well as the Greek equivalent to the letter "F," the first letter of Fibonacci. By this time this ubiquitous proportion was known as the golden mean, golden section and golden ratio as well as the Divine proportion.
### Recent appearances of Phi in math and physics
Phi continues to open new doors in our understanding of life and the universe. It appeared in Roger Penrose's discovery in the 1970's of "Penrose Tiles," which first allowed surfaces to be tiled in five-fold symmetry. It appeared again in the 1980's in quasi-crystals, a newly discovered form of matter.
### Phi as a door to understanding life
The description of this proportion as Golden and Divine is fitting perhaps because it is seen by many to open the door to a deeper understanding of beauty and spirituality in life. That's an incredible role for a single number to play, but then again this one number has played an incredible role in human history.
(1) Page 25
(2) Page 157
(3) Page 158
Phi - The Golden Number
A source to some of Net's "phi-nest" information on the
Golden Section / Mean / Proportion / Ratio / Number,
Divine Proportion, Fibonacci Series and Phi (1.6180339887...) | 1,022 | 4,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-05 | longest | en | 0.938965 |
https://sailboat.guide/wildfire | 1,685,706,337,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00704.warc.gz | 537,392,958 | 13,141 | # Wildfire
1968
Designer
Ian Proctor
Builders
Gmach & Co. Ltd. (J. L. Gmach & Co.)
O'Day Corp.
Association
Classic & Vintage Racing Dinghy Association
# Built
?
Hull
Monohull
Keel
Lifting
Rudder
?
Construction
FG
### Dimensions
Length Overall
16 9 / 5.1 m
Waterline Length
14 11 / 4.6 m
Beam
6 3 / 1.9 m
Draft
0 11 / 0.3 m 3 10 / 1.2 m
Displacement
620 lb / 281 kg
Ballast
250 lb / 113 kg
### Rig and Sails
Type
Sloop
Reported Sail Area
165′² / 15.3 m²
Total Sail Area
?
Sail Area
?
P
?
E
?
Air Draft
?
Sail Area
?
I
?
J
?
Forestay Length
?
Make
?
Model
?
HP
?
Fuel Type
?
Fuel Capacity
?
### Accomodations
Water Capacity
?
Holding Tank Capacity
?
?
Cabins
?
Hull Speed
8.1 kn
Classic: 5.19 kn
### Hull Speed
The theoretical maximum speed that a displacement hull can move efficiently through the water is determined by it's waterline length and displacement. It may be unable to reach this speed if the boat is underpowered or heavily loaded, though it may exceed this speed given enough power. Read more.
Formula
Classic hull speed formula:
Hull Speed = 1.34 x √LWL
A more accurate formula devised by Dave Gerr in The Propeller Handbook replaces the Speed/Length ratio constant of 1.34 with a calculation based on the Displacement/Length ratio.
Max Speed/Length ratio = 8.26 ÷ Displacement/Length ratio.311
Hull Speed = Max Speed/Length ratio x √LWL
8.12 knots
Classic formula: 5.19 knots
Sail Area/Displacement
36.3
>20: high performance
### Sail Area / Displacement Ratio
A measure of the power of the sails relative to the weight of the boat. The higher the number, the higher the performance, but the harder the boat will be to handle. This ratio is a "non-dimensional" value that facilitates comparisons between boats of different types and sizes. Read more.
Formula
SA/D = SA ÷ (D ÷ 64)2/3
• SA: Sail area in square feet, derived by adding the mainsail area to 100% of the foretriangle area (the lateral area above the deck between the mast and the forestay).
• D: Displacement in pounds.
36.33
<16: under powered
16-20: good performance
>20: high performance
Ballast/Displacement
40.2
>40: stiffer, more powerful
### Ballast / Displacement Ratio
A measure of the stability of a boat's hull that suggests how well a monohull will stand up to its sails. The ballast displacement ratio indicates how much of the weight of a boat is placed for maximum stability against capsizing and is an indicator of stiffness and resistance to capsize.
Formula
Ballast / Displacement * 100
40.21
<40: less stiff, less powerful
>40: stiffer, more powerful
Displacement/Length
82.1
<100: Ultralight
### Displacement / Length Ratio
A measure of the weight of the boat relative to it's length at the waterline. The higher a boat’s D/L ratio, the more easily it will carry a load and the more comfortable its motion will be. The lower a boat's ratio is, the less power it takes to drive the boat to its nominal hull speed or beyond. Read more.
Formula
D/L = (D ÷ 2240) ÷ (0.01 x LWL)³
• D: Displacement of the boat in pounds.
• LWL: Waterline length in feet
82.05
<100: ultralight
100-200: light
200-300: moderate
300-400: heavy
>400: very heavy
Comfort Ratio
5.3
<20: lightweight racing boat
### Comfort Ratio
This ratio assess how quickly and abruptly a boat’s hull reacts to waves in a significant seaway, these being the elements of a boat’s motion most likely to cause seasickness. Read more.
Formula
Comfort ratio = D ÷ (.65 x (.7 LWL + .3 LOA) x Beam1.33)
• D: Displacement of the boat in pounds
• LWL: Waterline length in feet
• LOA: Length overall in feet
• Beam: Width of boat at the widest point in feet
5.32
<20: lightweight racing boat
20-30: coastal cruiser
30-40: moderate bluewater cruising boat
40-50: heavy bluewater boat
>50: extremely heavy bluewater boat
Capsize Screening
2.9
>2.0: better suited for coastal cruising
### Capsize Screening Formula
This formula attempts to indicate whether a given boat might be too wide and light to readily right itself after being overturned in extreme conditions. Read more.
Formula
CSV = Beam ÷ ³√(D / 64)
• Beam: Width of boat at the widest point in feet
• D: Displacement of the boat in pounds
2.94
<2: better suited for ocean passages
>2: better suited for coastal cruising
### Notes
This boat features a hollow daggerboard secured in place by a hatch. Ballast is lowered into the keel using small bags of lead shot. It’s unknown how similar to the orginal version the later O’Day model is. Thanks to the folks at CVRDA.com for providing information on the WILDFIRE as well as some other English boats of this era. (Another boat that has a similar feature is the Milne designed VULCAN.)
### For Sale
Have a sailboat to sell? | 1,304 | 4,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.789973 |
http://www.topperlearning.com/forums/ask-experts-19/a-storage-battery-of-emf-8v-intrnal-resistance-05-ohm-is-physics-electric-charges-and-fields-55848/reply | 1,488,006,965,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171670.52/warc/CC-MAIN-20170219104611-00138-ip-10-171-10-108.ec2.internal.warc.gz | 651,927,271 | 37,844 |
Question
Fri May 25, 2012 By: Namami Gaur
# A storage battery of emf 8V & intrnal resistance 0.5 ohm is being charged by a 120V d.c..Supply using a series resistor of 15.5 ohm.What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit.
Fri May 25, 2012
Resistance of the circuit = 15.5 + 0.5 = 16 ohms
Current = 8/16 = 0.5 ampere
the terminal voltage of the battery will be the same as the voltage across the 15.5 Ohms load
Using ohms law again
V = I x R
V = 15.5 x 0.5 = 7.75 V
This is the terminal voltage.
Related Questions
Fri September 16, 2016
# please upload the imp problems of all the chapters in physics
Thu September 15, 2016
| 226 | 725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-09 | longest | en | 0.937426 |
https://tolstoy.newcastle.edu.au/R/e2/help/07/05/16397.html | 1,596,563,791,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735881.90/warc/CC-MAIN-20200804161521-20200804191521-00443.warc.gz | 537,551,704 | 4,894 | # Re: [R] Weighted least squares
From: John Fox <jfox_at_mcmaster.ca>
Date: Wed, 09 May 2007 07:16:37 -0400
> -----Original Message-----
> Sent: Wednesday, May 09, 2007 2:21 AM
> To: John Fox
> Cc: R-help_at_stat.math.ethz.ch
> Subject: Re: [R] Weighted least squares
>
> Thanks John,
>
> That's just the explanation I was looking for. I had hoped
> that there would be a built in way of dealing with them with
> R, but obviously not.
>
> Given that explanation, it stills seems to me that the way R
> calculates n is suboptimal, as demonstrated by my second example:
>
> summary(lm(y ~ x, data=df, weights=rep(c(0,2), each=50)))
> summary(lm(y ~ x, data=df, weights=rep(c(0.01,2), each=50)))
>
> the weights are only very slightly different but the
> estimates of residual standard error are quite different (20
> vs 14 in my run)
>
Observations with 0 weight are literally excluded, while those with very small weight (relative to others) don't contribute much to the fit. Consequently you get very similar coefficients but different numbers of observations.
I hope this helps,
John
>
> On 5/8/07, John Fox <jfox_at_mcmaster.ca> wrote:
> >
> > I think that the problem is that the term "weights" has different
> > meanings, which, although they are related, are not quite the same.
> >
> > The weights used by lm() are (inverse-)"variance weights,"
> reflecting
> > the variances of the errors, with observations that have
> low-variance
> > errors therefore being accorded greater weight in the
> resulting WLS regression.
> > What you have are sometimes called "case weights," and I'm
> unaware of
> > a general way of handling them in R, although you could
> regenerate the
> > unaggregated data. As you discovered, you get the same coefficients
> > with case weights as with variance weights, but different
> standard errors.
> > Finally, there are "sampling weights," which are inversely
> > proportional to the probability of selection; these are
> accommodated by the survey package.
> >
> > To complicate matters, this terminology isn't entirely standard.
> >
> > I hope this helps,
> > John
> >
> > --------------------------------
> > John Fox, Professor
> > Department of Sociology
> > McMaster University
> > Hamilton, Ontario
> > 905-525-9140x23604
> > http://socserv.mcmaster.ca/jfox
> > --------------------------------
> >
> > > -----Original Message-----
> > > From: r-help-bounces_at_stat.math.ethz.ch
> > > [mailto:r-help-bounces_at_stat.math.ethz.ch] On Behalf Of hadley
> > > wickham
> > > Sent: Tuesday, May 08, 2007 5:09 AM
> > > To: R Help
> > > Subject: [R] Weighted least squares
> > >
> > > Dear all,
> > >
> > > I'm struggling with weighted least squares, where
> something that I
> > > had assumed to be true appears not to be the case.
> > > Take the following data set as an example:
> > >
> > > df <- data.frame(x = runif(100, 0, 100)) df\$y <- df\$x + 1 +
> > > rnorm(100, sd=15)
> > >
> > > I had expected that:
> > >
> > > summary(lm(y ~ x, data=df, weights=rep(2, 100)))
> summary(lm(y ~ x,
> > > data=rbind(df,df)))
> > >
> > > would be equivalent, but they are not. I suspect the
> difference is
> > > how the degrees of freedom is calculated - I had expected
> it to be
> > > sum(weights), but seems to be sum(weights > 0). This seems
> > > unintuitive to me:
> > >
> > > summary(lm(y ~ x, data=df, weights=rep(c(0,2), each=50)))
> > > summary(lm(y ~ x, data=df, weights=rep(c(0.01,2), each=50)))
> > >
> > > What am I missing? And what is the usual way to do a linear
> > > regression when you have aggregated data?
> > >
> > > Thanks,
> > >
> > >
> > > ______________________________________________
> > > R-help_at_stat.math.ethz.ch mailing list
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
> > >
> >
> >
> >
>
R-help_at_stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 09 May 2007 - 11:23:41 GMT
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Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help. Please read the posting guide before posting to the list. | 1,275 | 4,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-34 | latest | en | 0.953593 |
https://www.reference.com/web?q=2.25+as+mixed+number&qo=contentPageRelatedSearch&o=600605&l=dir | 1,555,886,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532929.54/warc/CC-MAIN-20190421215917-20190422001917-00294.warc.gz | 810,453,528 | 14,372 | Web Results
A mixed number, like 3 and 1/2, is a combination of a whole number and a fraction and indicates a number that lies between consecutive whole numbers. Whole numbers are integers, counting numbers ...
www.fractioncalculator.pro/.../_2.25_as-a-mixed-number
2.25 as a mixed number. Welcome! Here is the answer to the question: 2.25 as a mixed number or what is 2.25 as a fraction. Use the decimal to fraction converter/calculator below to write any decimal number as a fraction.
www.mathway.com/popular-problems/Pre-Algebra/103229
Pre-Algebra. Convert to a Fraction 2.25. ... Reduce the fractional part of the mixed number. Convert the mixed number into an improper fraction first by multiplying the denominator by the whole number part and add the numerator to get the new numerator.
A ratio is a comparison of parts in a whole. Both parts of a ratio will work out less than 1, so would never be a mixed number. eg ratio of oil to petrol of 1:25, total number of parts = 26, so ...
www.reference.com/math/2-25-fraction-2975341e0fed6e2b
The decimal 2.25 is equal to the fraction 9/4. The decimal first needs to be converted to the basic fraction 2 1/4 before being converted to an improper fraction. The top of a fraction is called the numerator while the bottom is known as the denominator. The whole number standing before a fraction is referred to as an integer.
brainly.com/question/3681750
Just put the decimal over 100. 2.25=2 25/100 Reduce 25/100 to 1/4. You end up getting 2 1/4. Change 2 1/4 into a mixed number. Final answer: 9/4
sciencing.com/write-56-mixed-number-decimal-8477306.html
A mixed number is any whole number with a fraction portion. If the upper number – the numerator – were bigger than the lower number – the denominator – also known as an improper fraction, you would divide the denominator into the numerator and calculate how many times it goes in, resulting in a whole number.
coolconversion.com/.../Convert-decimal_2.25_to-fraction
Convert 2.25 to Fraction. Here you can find a decimal to fraction chart and also write any decimal number as a fraction. ... As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 13/ 10 is also equal to 1 3 / 10 when expressed as a mixed number. Conversion table ... | 581 | 2,327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-18 | latest | en | 0.883473 |
http://www.apug.org/forum/index.php?threads/ci-of-blank-film.8238/ | 1,469,823,772,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831771.10/warc/CC-MAIN-20160723071031-00056-ip-10-185-27-174.ec2.internal.warc.gz | 291,082,219 | 27,532 | # CI of Blank Film
Discussion in 'B&W: Film, Paper, Chemistry' started by Kirk Keyes, Apr 8, 2005.
1. ### Kirk KeyesMember
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Sorry to all concerned if this thread brings up sad memories of another recent thread, but there was a big conceptual issue in that thread that went unresolved. As many of you know, I am big on finding analogies to help better our (my?) understanding of complicated concepts, and I thought of one that may have helped in that other thread. For some reason, that thread has been closed, so I am posting it here.
The opinion was stated several times in the other thread that a sheet of blank film cannot have a CI. In defending this opinion, it was said that the film must have at least two areas of exposure in order for the film to have recieved a development of a certain CI. I disagree with the opinion - CI is not dependant on exposure, and therefore a blank sheet can and will have whatever CI (or gradient or gamma or...) that it was processed to.
Here's my proof:
So let's take a sheet of film that has been exposed by contact printing it with a step tablet. Sufficient exposure was given to create a suitable range of film densities that will allow us to measure and calculate a CI. Let's say it has a CI of 0.60, just to give it a value.
I hope everyone will agree that this sheet, has a CI of 0.60 - we determined this by making measurements of all the wedge steps that were exposed onto the precessed film.
OK - and I think we will all agree that this sheet - the entire sheet of film - has a CI of 0.60. So now let's get our scissors out, and start cutting this sheet of film into smaller pieces. Let's cut it in half. We all agree that both halfs of our film have a CI of 0.60 still - the act of cutting the film will not change the contrast index of that peice of film.
So let's make a few more cuts. Say we separate out a few of the steps, one has a density of say 1.50, and another one a density of 0.74, and a third has a density of say 0.31. All of these steps still have a CI of 0.60. Cutting the film does not change the CI of our film.
OK, so now, let's cut out that step that has not recieved sufficient exposure to have gained any density at all - it is at the base+fog level of density. It is a "blank" piece of film. I hope that no one will argue that this film has a CI of 0.00, because it does not - it cannot. It has the same CI as the rest of that sheet of film. And therefore a peice of film does not have to have at least two different exposures to have a CI. It does to actually measure the CI, and they need to be the right exposures, but a films CI is purely dependant on the processing conditions.
So lets take this info and extend it to a second sheet of the same film as the one we just exposed with the step wedge and processed. That first precessed sheet of film had a CI of 0.60, the entire sheet did. And in the same processing batch, we had also run this second sheet. If the first sheet that we processed had a CI of 0.60, then we can be pretty certain that the second sheet does as well. (At least to the extent of our ability to measure CI when we consider sheet to sheet variability and the experimental error of measurement.)
Now if we had made an identical exposure of a step wedge on this second sheet, we could actually verify that it did receive the proper development to achieve a CI of 0.60 by measureing it and doing the calculations. And it should be very close, if not identical (remember the experimental error of the density readings as well as sheet to sheet variations.) If we had not used a step wedge, but had given enough of a fogging exposure to create a density on the sheet of 1.00, that's fine - as this film still has a CI of 0.60. If we had given a much larger exposure, then the density may well be 1.8, that's fine as well, as the film still has a CI of 0.60. If we had given no exposure to this second sheet, all's still fine - it also is processed to a CI of 0.60. CI is dependant on development, and not exposure.
Think about all the people that do roll film zone testing. They are using a series of exposures, each on an individual frame. They process the roll, measure densities, and calculate a CI. I hope no one will argue that each frame on that roll has been processed to a CI of 0.00 - although that is actually what has been argued here. While you can't calculate the CI of that roll using just one frame, each frame has been processed to some CI and it can be measured.
Let's stick some more sheets or rolls of the same film in the development batch and process them simultaneously, say in a drum or tank. Each roll in that run will have the same CI. If we did not make a proper set of exposures on some of those films, we cannot determine the actual CI the run was processed to. But, thanks to the science of process control, we don't need to actually have prepared a set of exposures to make the measurement of CI on each and every run. (This of course assumes that you do have control of your processing!!)
So I hope you all can see, it is very easy to give no exposure to a peice of film and process it so some CI value.
And I think the issue that was being pointed out about Davis' phrase "SBR", is that it should probably called "Subject Illumination Range - SIR", as an incident meter can only measure the light that falls on it - the amount of illumination, not "brightness".
Kirk - www.keyesphoto.com
2. ### Donald MillerMember
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How many angels can dance on the head of a pin? Does the pin exist if no one recognizes it for a pin? Does the pin exist outside of objective awareness? Is an angel and angel if someone sees it and mistakes it for a pixie? What if an angel is not seen, is it still dancing on the head of a pin?
3. ### OrnelloInactive
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Attached from Kodak. "It is a measure of the degree of development." Kodak replaced gamma with CI a number of years ago. Gamma used only the straight-line portion of the curve, but good negatives make use of the toe, below the straight-line portion. CI includes the toe, whereas gamma did not.
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4. ### sankingMember
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Kirk,
I left out the content of your message because it belongs in a category I label as "masturbation intellectuel, inutile et sans plaisir" as we might say in French. You fellows who continue to annoy the rest of us with this nonesense would most likely be a much happier lot if you lost the mental and just used your hands.
With all due respect,
Sandy
5. ### mikeprySubscriber
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Sandy, I didn't know you had it in you! Well said, chap!
6. ### OleModeratorStaff MemberModerator
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The CI is the slope of a line. Said line has a slope in every single point of the line, so even the "0" point has a slope - and an unexposed film can be developed to a given CI - just like one exposed to a uniform "zone V".
What's so difficult to understand about that, and why do so many otherwise intelligent people feel so upset by it?
7. ### JorgeInactive
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yeah but it requires at least 2 point to first determine that slope. If this is not true, then here is a set of coordinates on an x,y frame. 2, 5 please tell me the slope......
I now understand that saying that if you cannot dazzle them with your brilliance.....(fill in the blank)
Processing two things that appear to be the same but are not does not yield the same results. if I put water in one cup, and I put water and a bag of tea in another cup and then put both of them in a microwave oven to heat, I did not make two cups of tea, I made one cup of tea and one cup of warm water. The glaring error in this page long example is that after all the cutting, in essence the blank part of the film has a significant difference from the other parts..it did not receive any exposure, a essential part in determining CI. I hate to admit it but King is right, this is mental masturbation to the highest degree, but I think Don Miller said it best.....if you want to know, read his response on the SBR thread.
8. ### OleModeratorStaff MemberModerator
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Jorge, the correct CI can only be determined by developing a standard test negative. When using any other negative you are assuming that they will be developed to the same CI by the same processing. It doesn't matter what scene, SBR or whatever you have captured on that negative - or even whether it has been exposed at all.
It takes 2 points to determine a slope, but any one exposure will correspond to one density.
9. ### sankingMember
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Ole,
Definitions.
Gradient -- degree of inclination, inclined surface, rising or descending by regular degrees of inclination, and other similar.
Slope -- inclination from the horizontal, deviation from the horizontal, and other similar
Definitions of gamma, CI, and C-Bar always include the use of one of these two words, either slope or gradient. At least all that I have seen, and I have looked at about ten sources this evening, include one of these terms.
Fact. A sheet of unexposed film that has been developed for n period of time will graph on a horizontal line, i.e. at right angle to the vertical or parallel with the horizon. Get the point? No slope, no gradient!!! And no CI.
This discussion shows considerable ignorance of language on the part of some people who in other respects appear to be perfectly intelligent.
Sandy King
10. ### OrnelloInactive
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Kodak said: "It is a measure of the degree of development." Where is your question? The H&D curves in the scan above show gamma values based on development time. In essence, the gamma (or CI/G-bar) values represent those times for reference purposes.
11. ### OleModeratorStaff MemberModerator
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Sandy,
A sheet of unexposed film has received only one exposure - zero. It will not plot on a horizontal line. In fact it will not plot on a line at all, but in a point.
My native language may not be yours, but I'm fluent in several sciences.
12. ### Kirk KeyesMember
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Sandy, sorry - you got that fact wrong. Since you are looking up definitions, please check the mathematical definition of "line". I think you will find, in cartesian space (2 dimensional space, as we use for graphing our film plots) one data value can at most be described as a point on a graph. It is not a line. It requires 2 points, to create a line. You have no "rise over run" with one point and you cannot calculate a slope with only one point.
Your single sheet with only one exposure on it developed for n period of time will graph as a single point. Not a line, horizontal or otherwise.
Kirk
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Huh?
15. ### Kirk KeyesMember
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Sorry Ornello - what did you not understand there? Maybe you should see Ole's answer, he was a little less wordy, but the same answer.
16. ### smieglitzSubscriber
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Let's do another "thought experiment". I have several sheets of film that have received some uniform exposure. I develop these to some CI. I hand one of these sheets to any one of you so that you can record the density value of that sheet. Based on that information alone, you cannot tell me the CI to which that film has been developed. If I give you another sheet having received a different exposure and a different development (and consequently a different CI), you cannot tell me what the CI is for that film. You can only tell me the density. That density BTW may end up being identical to the first.
I can take any of those sheets with different densities, including a blank film into the darkroom and produce any tone up to and including the dMax of the paper I choose to print upon. In fact, I don't even need to use a negative to get that result.
None of the information you can relate to me about these films has any utility in isolation.
My point is that all of the information derived from a single measurement is useless in practice. And that is what the original poster was interested in-how to understand and use the CI measurement practically.
So, OK. I'll concede the point about a blank film theoretically having a possible CI other than 0. Point taken. Point useless.
Perhaps what we really need here instead of a CI measurement is a BS scale.
Bottom line: I'll not be letting anyone else develop my film, thank you very much.
17. ### sankingMember
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Thanks. You just proved my point. No slope, no gradient. Nothing but a point.
So what is your point? Are you claiming that a point is a CI?
Sandy
18. ### Kirk KeyesMember
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Donald, good question. But you will have to prove to me that angels even exist before I can try to answer this one.
Sandy, sorry you feel that way. I'm not sure what you have against masterbation, intellectual or otherwise, but are you saying that my analysis of the issue is incorrect? If so, I look forward to hearing why.
And for non-sense, you surely must be addressing the people that suggest that a blank sheet of film cannot be developed to a particular CI. I sure hope so.
I think that Jeff got it - "When we think of the useful data provided by the CI number, it becomes counterintuitive that a blank piece of film can be ascribed a CI or make use of the number, the last point being the crucial one. In isolation, we cannot determine, or make use of the CI of a blank piece of film, even though in theory it must have one, if it has been developed."
Sandy, Mike, Donald, if you would care to correct my analysis, please feel free to do so. As I've said many times on this forum, I enjoy learning new things, and if I'm going around with incorrect ideas on these sorts of subject, I hope someone will help me be better informed. Otherwise, please continue with the masturbation jokes.
Kirk
19. ### OleModeratorStaff MemberModerator
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No Sandy, I disproved your point. Read the whole thread over again, particularly those entries you didn't write yourself.
Or go and find a textbook in basic calculus.
Or just look here: http://en.wikipedia.org/wiki/Calculus
20. ### Kirk KeyesMember
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Ole - calculus is not even needed here. No rates of change with only one point. What is needed is plain old simple geometry. Or actually just algebra.
21. ### sankingMember
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Stephen,
No, you have not proven me wrong on anything, though I at least have been gracious enough to recognize that I might be wrong on some things. You, on the other hand, appear to be so high on your own arrogance that I suspect you have to get up on a ladder to scratch your ass when it itches.
A little click of ass kissers who have joined in a mutual admiration society? Jorge and I, for example?
So Stephen, just a thought. Any chance that you might be wrong on this blank film thing? And let's always return to that, shall we, as the thread that brought us together?
Sandy King
22. ### OrnelloInactive
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Kodak defines 'CI' in such a way that it can indeed be applied to blank film, as it represents a degree of development.
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23. ### Kirk KeyesMember
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Sandy - I started this thread. Perhaps instead of asking Steve if he is wrong on this issue, perhaps you could tell me where I went wrong.
While I wait for your analysis, I think I'll go surfing for some good porn to masturbate to. (Sorry about the bad grammar.)
Kirk
24. ### JorgeInactive
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AH, but there is the rub Ole, this thing that you are assuming. The reality of it is very simple, I will send you a blank negative and you will tell me to what CI I developed it, you can send me a negative you have made and I will tell you what CI you developed it. Lets see who gets closer......
I also disagree that the "correct" CI has to be determined by a standard test negative. While I am sure Il douche bag will disagree, we do our "own" CI every time we expose a piece of film to a step wedge with known density gradient.
There comes a time when talking about theoretical ideas has to yield to the reality....Arguing that a piece of blank film has a CI just because it was "assumed" it was developed as the test strip done before IMO is ludicrous.
In the end some might enjoy this kind of pseudo intellectual discussion...I dont, I think I have explained as well as I can what I mean, I leave you all to hash this out.
Here is hoping we get the "Ignore this thread" button soon....
25. ### Kirk KeyesMember
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Hi Jorge!
I didn't mean to imply that a lot of usefull information could be derived from a blank piece of film developed to a particular CI. And you are right, you cannot determine the CI of a processing run merely from a blank sheet of film. I beleive I said that in my orginal post. I would be a fool to argue otherwise.
Steve did point out that one can determine the base+fog level from that blank sheet of film. That's about all.
But the sheet is still associated to a particular CI.
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http://oeis.org/A141183 | 1,586,074,059,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370529375.49/warc/CC-MAIN-20200405053120-20200405083120-00343.warc.gz | 145,068,518 | 4,394 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A141183 Primes of the form -x^2+6*x*y+2*y^2 (as well as of the form 7*x^2+10*x*y+2*y^2). 8
2, 7, 11, 19, 43, 79, 83, 107, 127, 131, 139, 151, 167, 211, 227, 239, 263, 271, 283, 307, 347, 359, 431, 439, 479, 491, 503, 523, 547, 563, 571, 607, 659, 739, 743, 787, 811, 827, 887, 919, 967, 1019, 1031, 1051, 1063, 1091, 1151, 1163, 1187, 1223, 1231, 1283, 1319, 1327 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Discriminant = 44. Class = 2. Binary quadratic forms a*x^2+b*x*y+c*y^2 have discriminant d=b^2-4ac and gcd(a,b,c)=1. Also primes of form 11*u^2-v^2. The transformation {u,v}={-3*x-y,10*x+3*y} yields the form in the title. - Juan Arias-de-Reyna Mar 20 2011. Also primes p equal -1 mod 4 and = 1, 3, 4, 5, or 9 mod 11. - Juan Arias-de-Reyna Mar 20 2011. REFERENCES Z. I. Borevich and I. R. Shafarevich, Number Theory, Academic Press, NY, 1966. D. B. Zagier, Zetafunktionen und quadratische Koerper LINKS Juan Arias-de-Reyna, Table of n, a(n) for n = 1..10000 N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references) EXAMPLE a(4)=19 because we can write 19= -1^2+6*1*2+2*2^2 (or 19=7*1^2+10*1*1+2*1^2). MATHEMATICA Select[Prime[Range[250]], # == 2 || # == 11 || MatchQ[Mod[#, 44], Alternatives[7, 19, 35, 39, 43]]&] (* Jean-François Alcover, Oct 28 2016 *) CROSSREFS Cf. A038872 (d=5). A038873 (d=8). A068228, A141123 (d=12). A038883 (d=13). A038889 (d=17). A141111, A141112 (d=65). A141182 (d=44). For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link. Sequence in context: A106905 A097159 A139603 * A308724 A103182 A160698 Adjacent sequences: A141180 A141181 A141182 * A141184 A141185 A141186 KEYWORD nonn AUTHOR Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales (sergarmor(AT)yahoo.es), Jun 13 2008 STATUS approved
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Last modified April 5 03:58 EDT 2020. Contains 333238 sequences. (Running on oeis4.) | 933 | 2,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-16 | latest | en | 0.52445 |
http://www.codebymath.com/index.php/welcome/lesson/arduino-etch-a-sketch | 1,524,468,848,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945855.61/warc/CC-MAIN-20180423070455-20180423090455-00605.warc.gz | 386,334,319 | 5,000 | # Lesson goal: Make an electronic etch-a-sketch
Previous: Linearly scale a potentiometer | Home | Next: Read a digital input
In a previous lesson, you learned how to read (and scale) a voltage, from a potentiometer. In this lesson we'll read and scale two potentiometers to make an electronic etch-a-sketch.
# Now you try. Scale the voltages from each potentiometer so you can draw in the entire plot area! Then draw something cool!
Wire up this circuit, using two potentiometers. Run the above code, and turn the potentiometers along their full range of motion. You should be able to "draw" with the potentiometers in a small square near the center of the screen.
Reformulate the x = v1 and y = v2 lines to scale v1 and v2 to fill the whole screen. For each, use a linear expression of the form $y=mx+b$, or y=m*v1+b and y=m*v2+b.
Instead of plotting points, you can also try plotting small circles via circle(x,y,2). Dismiss.
Show a friend, family member, or teacher what you've done! | 252 | 993 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-17 | latest | en | 0.872328 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-2nd-edition/chapter-3-derivatives-3-2-working-with-derivatives-3-2-exercises-page-143/29 | 1,531,696,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589022.38/warc/CC-MAIN-20180715222830-20180716002830-00333.warc.gz | 925,595,151 | 13,450 | ## Calculus: Early Transcendentals (2nd Edition)
$a.$ At $t=0$; $b.$ Positive; $c.$ Decreasing; $d.$ The graph is on the figure below.
$a.$ The rate of change of the charge $Q'$ is the greatest at the point where the tangent to the graph of $Q(t)$ has the steepest upward slope which is for $x=0$; $b.$ It is positive because the charge is only increasing. $c.$ $Q'$ is decreasing because we see that as time goes by (as $t$ becomes bigger) the slope of the tangent to the graph is smaller and smaller. $d.$ The general shape of the graph of $Q'$ can be deduced from the shape of the graph of $Q$. We see that it starts from some positive value and keeps decreasing and asymptotically approaches zero because the graph of $Q$ becomes increasingly horizontal as $t$ grows and thus the slope of its tangent tends to zero. The graph is on the figure below. | 220 | 854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-30 | longest | en | 0.949832 |
https://math.stackexchange.com/questions/2215649/number-of-digits-of-n-factorial-in-base-b | 1,569,141,382,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00293.warc.gz | 568,158,644 | 27,715 | # Number of digits of N factorial in base B [duplicate]
Given the decimal integers N and B. How can the number of digits in N! in base B be calculated without calculating the value of N!?
Example: If N = 5 and B = 2, then N! = 5! = 120. Representing the value in base B, we get 1111000 which has 7 digits.
But I'm trying to get the result without calculating the value of N!.
## marked as duplicate by Chris Culter, Community♦Apr 3 '17 at 8:40
• Hint: $\lceil\log_bN!\rceil$. – Yves Daoust Apr 3 '17 at 6:50
• A base-$b$ number of $d$ digits lies between $b^{d-1}$ and $b^d-1$. – Yves Daoust Apr 3 '17 at 7:09
The number of digits is $\lfloor\log_b N!\rfloor+1$. This can be rewritten as $$\Big\lfloor\sum_{r=1}^N\log_br\Big\rfloor+1.$$ That sum should be possible to calculate on computer even for relatively large $N$. However, you can also approximate it using Stirling's formula:$$\ln N!\approx(N+1/2)\ln N-N+\ln\sqrt{2\pi}.$$ To convert this to what you want, use the fact that $\log_bx=\frac{\ln x}{\ln b}$.
This approximation is pretty good. For $\log_25!$ it gives $6.883$, whereas the actual value is $6.907$. | 363 | 1,123 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-39 | latest | en | 0.783567 |
surlysubgroup.com | 1,685,408,975,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00767.warc.gz | 598,411,039 | 31,535 | On Marginal and Effective Tax Rates
By Sam Brunson
A friend sent me a cartoon that’s been going around his Facebook feed. The cartoon’s message amounts, roughly, to: undocumented immigrants are far better-off than legal residents because taxes.
The various assertions the cartoon makes are an amazing collection of racist and wrong. In fact, there’s too much racist, and too much wrong, to address. But it’s worth pointing out that, although the cartoon uses numbers, its numbers are wildly, wildly wrong.
If you want to hate-read the whole thing, you can see it here, but for my purposes, we’re going to look at these two panels:
Okay, I said I was only going to look at numbers, but it’s worth noting the racism that underlies the whole thing: our harmed legal resident is a white guy, while our rapacious illegal interloper is a person of color. (Side note: why am I even bothering analyzing something so transparently pointless and wrong? If nothing else, to provide numerical ammo when your racist uncle—or aunt, for that matter—brings it up at dinner.)
So why don’t the numbers work? Because John’s not paying taxes at anywhere near a 30% effective or marginal rate. (Marginal rate is the rate of tax you pay on your last dollar of income. Effective rate is the percentage of your income you pay in taxes.)
How much does John pay in taxes? Let’s assume he doesn’t itemize and just has his wage income. So we start at \$52,000. According to the cartoon, he’s married with two kids, and I’ll assume he files a joint return. That means he gets a standard deduction—in 2017, that’s \$12,700—and he gets four personal exemptions, at \$4,050 a pop.
Subtract that from his income, and he has taxable income of \$23,100. That puts him in the 15-percent marginal rate. Of course, that doesn’t mean he pays taxes at a 15% rate. He pays 10% on his first \$18,650, and 15% on the rest. Ultimately, he owes federal income taxes of \$2,532.50.
He also owes payroll taxes. (Note that I’m going to grossly simplify here: payroll taxes today mean that he gets social security payments later. But let’s ignore that.) As long as he’s an employee (and the cartoon talks about withholding on his wages, so he’s an employee), he pays 6.2% of his wages, for \$3,224.
After taxes, then, John has \$46,243.50 left. (I’m leaving out state income taxes, mostly because I don’t want to figure out Arizona tax law. But he’s not going to owe \$10,000 in Arizona—at most he’ll owe 2.59% on a portion of his income.)
What about Juan? According to the cartoon, he doesn’t pay federal taxes.[fn1] Broadly speaking, that’s probably not true.[fn2] But let’s pretend it is. How much would Juan owe in federal taxes? We have the same standard deduction and personal exemptions. On \$31,200 of income, Juan would have \$2,300 of taxable income, and would owe federal income taxes of \$230. He would also owe \$1,934.40 in payroll taxes.
Which is to say, even if Juan were not paying federal taxes (and note that, if he’s not paying his share of payroll taxes, that’s because his employer has chosen to break the law and not withhold), that’s a loss of just over \$2,000.
Ignoring for a moment the racism, and pretending that the cartoonist is sincere, the cartoonist’s errors underscore two issues to me. First, people have no idea the difference between effective and marginal rates, and will use whichever is most convenient for their point.
And second, people vastly overestimate the amount they pay in taxes.
[fn1] FWIW, later the cartoon complains about various state benefits received by undocumented immigrants; those immigrants, though, do pay sales and property taxes, which largely fund the services the cartoon complains about.
[fn2] Also, if the undocumented immigrants’ employers are paying them under the table, why isn’t the cartoonist complaining about the employers? Oh yeah, racism. Right. | 903 | 3,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-23 | latest | en | 0.964709 |
http://en.m.wikipedia.org/wiki/Principal_branch | 1,369,283,220,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702810651/warc/CC-MAIN-20130516111330-00043-ip-10-60-113-184.ec2.internal.warc.gz | 92,125,867 | 6,064 | # Principal branch
In mathematics, a principal branch is a function which selects one branch, or "slice", of a multi-valued function. Most often, this applies to functions defined on the complex plane: see branch cut.
One way to view a principal branch is to look specifically at the exponential function, and the logarithm, as it is defined in complex analysis.
The exponential function is single-valued, where $e^z$ is defined as:
$e^z=e^a \cos b +i e^a \sin b$
where $z = a + bi$ .
However, the periodic nature of the trigonometric functions involved makes it clear that the logarithm is not so uniquely determined. One way to see this is to look at the following:
$\operatorname{Re}(\log z)=\log \sqrt{a^2 + b^2}$
and
$\operatorname{Im}(\log\ z) = \arctan(b/a) + 2\pi k$
where k is any integer.
Any number log(z) defined by such criteria has the property that elog(z) = z.
In this manner log function is a multi-valued function (often referred to as a "multifunction" in the context of complex analysis). A branch cut, usually along the negative real axis, can limit the imaginary part so it lies between −π and π. These are the chosen principal values.
This is the principal branch of the log function. Often it is defined using a capital letter, Log(z).
A more familiar principal branch function, limited to real numbers, is that of a positive real number raised to the power of 1/2.
For example, take the relation y = x1/2, where x is any positive real number.
This relation can be satisfied by any value of y equal to a square root of x (either positive or negative). When y is taken to be the positive square root, we write $y = \sqrt x$.
In this instance, the positive square root function is taken as the principal branch of the multi-valued relation x1/2.
Principal branches are also used in the definition of many inverse trigonometric functions. | 457 | 1,877 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2013-20 | latest | en | 0.896334 |
https://www.mrexcel.com/board/threads/displaying-multiple-values-without-using-macros-or-rows-function.377554/ | 1,679,914,396,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00575.warc.gz | 1,004,655,998 | 20,061 | # Displaying Multiple Values without using Macros or ROWS function
#### ltzhao
##### New Member
Hello Gurus out there,
I am working with a software itself doesn't support Macros and ROWS function.
My Goal:
is to display a set of data based on filtered information.
My Data base:
Company Department Name
A X John
A Y Joe
A X Jane
B Y Bob
C Z Kate
A X Kerri
Based on user selection of Company and Department, I want to be able to display the relavent names.
If user chose Company A, and Department X, I want to be able to display
John
Jane
Kerri
I've used the ROWS, Index, Small combination that works perfectly (Please see sample below). However, since the this software doesn't support the ROWS function, and doesn't support Macros, I am stuck. Can anyone provide any alternatives to this? Your help is highly appreciated.
Sample Formula
IF(\$B\$2="Company",IF(ROWS(C\$15:C15)<\$B\$14,INDEX('Department!\$F\$3:\$F\$691,SMALL(IF('Department!\$B\$3:\$B\$691=\$B\$11,ROW(Department!\$F\$3:\$F\$691)-ROW('WW DELMIA HC ROSTER 2009'!\$F\$3)+1),ROWS(C\$15:C15))),""),IF(ROWS(C\$15:C15)<\$B\$14,INDEX('Department!\$F\$3:\$F\$691,SMALL(IF('Department!\$A\$3:\$A\$691=\$B\$11,ROW('Department!\$F\$3:\$F\$691)-ROW('Department!\$F\$3)+1),ROWS(C\$15:C15))),""))
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#### jim may
##### Well-known Member
Why not use an auto-filter?
#### ltzhao
##### New Member
I am not using the auto filter because I have to create this table in another file that's being used by another program which doesn't support ROWS functions and does NOT support Macros. And it has limitations on filtered rows to 500 rows, which is only a fraction of the database I have.
I have tried 3 different ways to solve the problem so far (ROWS, Macros, and Filters within the software itself), none has worked. I am looking for another way around this.
#### mikerickson
##### MrExcel MVP
That other app doesn't sound like Excel.
Where does Excel come in?
Is Excel the source of the data and giving it to the other app?
If so, use Excel's filters to filter the data and then export a .txt to the other app.
#### ltzhao
##### New Member
The other application is Xcelsius, which uses excel as a basis to do reporting dashboards.
It supports most excel functions, but not some.
Is there another way to get around this?
#### Peter_SSs
##### MrExcel MVP, Moderator
OK, we cannot use ROWS(), but can we use ROW()?.
D2 and I2 copied down.
Excel Workbook
ABCDEFGHI
1CompanyDepartmentName3CompanyDepartmentNames
2AXJohn1AXJohn
3AYJoe1Jane
4AXJane2Kerri
5BYBob2
6CZKate2
7AXKerri3
8
List
#### ltzhao
##### New Member
Thank you Jim, Mike and Peter.
Your help is very much appreciated. Unfortunately, the software doesn't support ROW() function either. However, using Peter's suggestion, and some manual manipulation, I got it to work at least. It's not pretty, and it will require manual management in the future, but at least it got around the problem.
Thanks once again for all of your help. It's highly appreciated.
#### Peter_SSs
##### MrExcel MVP, Moderator
I've not used Xcelsius but Googling tells me that AND, COUNTA, IF, INDEX, MATCH and MAX are available. If so, try this with D2 and I2 copied down.
Excel Workbook
ABCDEFGHI
1CompanyDepartmentName3CompanyDepartmentNames
2AXJohn1AXJohn
3AYJoe1Jane
4AXJane2Kerri
5BYBob2
6CZKate2
7AXKerri3
8
List
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Go back | 1,256 | 4,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.862994 |
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