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8,098,752,880,805,181,000 | Multiplying and Dividing Fractions DIGITAL TASK CARDS BUNDLE
Multiplying and Dividing Fractions DIGITAL TASK CARDS BUNDLE
Grade Levels
Common Core Standards
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1. Fractions as Division Digital Task Cards *Make sure to check out the preview for a look at all 20 task cards! ♻♻♻♻GO PAPERLESS!♻♻♻♻ This product includes 2 digital activities: *Digital Task Cards using Google Slides *Automatic Grading Tool using Google Forms Digital Task Cards 21 Task Cards Incl
2. Multiplying Fractions DIGITAL TASK CARDS Updated 10/18/17 to include Google Forms Grading Tool! Thanks for your support and make sure to re-download to gain access to this great new feature! This product includes 2 digital activities: *Digital Task Cards using Google Slides *Automatic Grading Tool
3. Fraction Multiplication as Scaling Digital Task Cards *Make sure to check out the preview for a look at all 20 task cards! ♻♻♻♻GO PAPERLESS!♻♻♻♻ Updated 6/25/18 to include Google Forms Grading Tool! Thanks for your support and make sure to re-download to gain access to this great new feature! Th
4. Multiplying Mixed Numbers DIGITAL TASK CARDS ♻♻♻♻GO PAPERLESS!♻♻♻♻ Updated 6/25/18 to include Google Forms Grading Tool! Thanks for your support and make sure to re-download to gain access to this great new feature! This product includes 2 digital activities: *Digital Task Cards using Google Slid
5. Dividing Unit Fractions DIGITAL TASK CARDS ♻♻♻♻GO PAPERLESS!♻♻♻♻ Updated 6/25/18 to include Google Forms Grading Tool! Thanks for your support and make sure to re-download to gain access to this great new feature! This product includes 2 digital activities: *Digital Task Cards using Google Slides
Bundle Description
Multiplying and Dividing Fractions (5.NF.B Cluster) Digital Task Cards BUNDLE for use with Google Classroom
SAVE 25% PERCENT BY PURCHASING THESE PRODUCTS IN A BUNDLE!
These 5 sets of Google Slides and Google Forms Activities include over 100 total problems covering the 5th Grade Common Core Math Standards in the NBT.B Cluster.
Apply and extend previous understandings of multiplication and division.
CCSS.MATH.CONTENT.5.NF.B.3
Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
CCSS.MATH.CONTENT.5.NF.B.4
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
CCSS.MATH.CONTENT.5.NF.B.5
Interpret multiplication as scaling (resizing), by:
CCSS.MATH.CONTENT.5.NF.B.5.A
Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
CCSS.MATH.CONTENT.5.NF.B.5.B
Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (n × a)/(n × b) to the effect of multiplying a/b by 1.
CCSS.MATH.CONTENT.5.NF.B.6
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
CCSS.MATH.CONTENT.5.NF.B.7
Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
Total Pages
100+
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8,347,598,749,771,175,000 | Weighted Moving Average
Weighted Moving Average (WMA) is one of the configurations of simple moving average which accounts not only for price values but also their weight.
Calculated as per formula:
weighted_moving_average1
where: Pi — price value for the number of i-periods, (today i =1),
Wi — weight value for price for the number of i-periods.
In simpler words, elements with an account of their values are summed and divided for the sum of weights of those elements, thus, generally speaking, arithmetical average of those elements is calculated.
It is accepted that weight changes according to linear function where W1 takes the largest weight and then calculation uses simple arithmetical progression, for instance: 1, 2, 3, 4, 5, 6...; (or any other: 0,5, 0,75, 1, 1,25). Such representation is called Linear Weighted Moving Average, (LWMA). Let's take period equal to 5:
weighted_moving_average2 ,
where: P1 и P2 — are the prices for today and yesterday.
Some configurations may use more complicated formula with non-linear distribution, involving logarithmic, parabolic and other functions, for example, if following is accounted:
- the number of ticks in bar;
- the length of passed distance in candle (High - Low)
- weight average against the distance; - the size of candle body (|Close - Open|).
Price can also be different: Close, Open, High, Low, Median Price, Typical Price.
Application of WMA
Example of Weighted Moving Average
Weighted Moving Average is usually applied in the same cases in which simple moving average is applied for technical analysis purposes. Though under similar entrance and exit market alerts LWMA responds to price change faster because weight is accounted for the latest periods. That allows not to miss lucky moments for entering the market during important economic news, interventions and other significant moves.
For stock market analysis it is recommended to use parameters equal to 7 and 14, for currency market – 5 and 20. As you can see on the image, the larger period is, the smoother moving average is and the bigger fluctuation range is has.
Sine-Weighted Moving Average ( SWMA) uses sine function during its calculation as weight (W). Thanks to SWMA, it is possible to filter noises, determine bottom and top with a higher precision.
Pros and cons of WMA
Due to taking in account weight of elements, WMA is more sensitive towards price change in contrast to simple moving average, which allows getting entrance and exit alerts faster. However, as any other MA, weight also has a certain delay.
It is better to apply it in short- and mid-term strategies, because the latest price changes has the biggest weight. In other words, at high time-frame WMA looks smoother because of low market noise and it does not provide such clear alerts.
WMA is more sensitive towards change of prices
WMA is better to apply in short- and mid-term strategies
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100,637,948,586,101,920 | The 99 Club
The 99 Club is a mental-oral starter at Thirsk Community Primary School which aims to raise standards in maths through encouraging pupils to improve their mental calculations when attempting quick-fire multiplication and division problems.
The idea is that with repeated practice, the scheme should result in increased speed and confidence when tackling mental maths problems without relying on written workings and methods.
All pupils will begin at the 11 Club and work their way up, having three opportunities per week during to answer all calculations at their current level unaided and within the allotted time of five minutes.
If all of the calculations are answered correctly, the child moves up to the next level!
The initial 11 Club involves eleven problems which involve doubling numbers up to ten ie. 5+5, 8+8. The 22 Club then adds eleven further questions involving repeated addition for numbers from one to ten ie. 3+3+3+3, 5+5+5+5+5, while the 33 Club begins to introduce times tables.
Division facts are added by the time a pupil reaches the 77 Club, and in the 88 Club and 99 Club, pupils will be tackling a range of mixed multiplication and division problems.
The full breakdown of The 99 Club levels is as follows:
11 Club - 11 questions involving doubling numbers from one to ten
22 Club - 22 questions involving repeated addition of numbers from one to ten
33 Club - 33 questions introducing the 2x, 3x, 5x and 10x tables
44 Club - 44 questions adding the 1x, 4x and 6x tables
55 Club - 55 questions adding the 7x and 8x tables
66 Club - 66 questions adding the 9x, 11x and 12x tables
77 Club - 77 questions consisting of inverse division facts
88 Club - 88 questions of mixed multiplication and division facts
99 Club - 99 questions of mixed multiplication and division facts
The ultimate challenge is to complete all 99 questions of the 99 Club unaided, with no errors and within five minutes!
The 99 Club is not designed for the Early Years Foundation Stage, but if staff feel that children are ready to attempt it, they will be given the opportunity to enter the 11 Club!
Sheets are available to download below for pupils to practise their level at home. You can support your child by finding out which level they are working at and helping them to practise the relevant times tables.
Some especially skilled mathematicians at Thirsk Community Primary may even make it on to the newly-introduced BronzeSilver and even Gold Clubs which include over one hundred problems to complete in the five minute, including some very tricky maths involving squares, square roots and cubed numbers! Gosh!
Good luck, and keep practising! smile
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9,143,754,141,922,515,000 | viernes, 24 de febrero de 2012
Resumen capitulo 2
Transformaciones geométricas.
Habitualmente un paquete gráfico permite al usuario especificar que parte de una imagen definida se debe de visializar y donde esta parte se debe colocar en el dispositivo del visualización. Se compone por coodenadas.
Transformaciones bidimensionales.
Traslacion:En un objeto para cambiar su posicion a lo largo de la trayectoria de una linea recta de una direccion de coordenadas a otra. Asi que se convierte al agregar distancias de traslacion t y t con (x’ = x + tx’) o (y’ = y + ty), en ocasiones las ecucaciones de transformación matricianal se expresan en terminos de vectores en renglon de coordenadas en vez de vectores de columna.
Los poligonos se trasladas al sumar el vector de traslacion a la posición de coordenadas.
Rotación: Se aplica una rotación bidimensional en un objeto al cambiar su posición a lo largo de la trayectoria de una circunferencia en el plano de xy, para generar una rotación especificamos un angulo y la posición del punto de rotación en torno el cual se girara el objeto
Eslacion: Es una transformación de escalacion la cual altera el tamaño de un objeto. Se puede realizar esta operación para polígonos al multiplicar los valores de coordenadas de cada vértice por los factores de escalacion para producir coordenadas transformadas.
Coordenadas homogéneas y representación matricial
En las aplicaciones de diseño y de creación de imágenes, realizamos traslaciones, rotaciones y escalaciones para ajustar los componentes de la imagen en sus posiciones apropiadas. En este tema consideramos cómo se pueden volver a formular las representaciones de la matriz de modo que se pueden procesar de manera eficiente esas secuencias de transformación.
Composición de transformaciones bidimensionales.
Con las representaciones de matriz del tema anterior, podemos establecer una matriz para cualquier secuencia de transformaciones como una matriz de transformación compuesta al calcular el producto de la matriz de las transformaciones individuales.
P y P’ como vectores de columna de coordenadas homogéneas. Podemos verificar este resultado al calcular el producto de la matriz para las dos agrupaciones asociativas. Asimismo, la matriz de transformación compuesta para esta secuencia de transformaciones es:
Formula equivalente para lograr la rotación
Formula equivalente para lograr la escalacion.
Rotacion del punto pivote general
Se necesitan seguir 3 pequeños pasos a gran escala para desarrollarla.
1. Traslade el objeto de modo que se mueva la posición del punto pivote al origen de las coordenadas.
2. Gire el objeto con respecto del origen de las coordenadas.
3. Traslade el objeto de manera que se regrese el punto pivote a su posición original.
Escalacion del punto fijo general
Son una conjunto de secuencias las cuales se especifican en el siguiente dibujo.
1. Traslade el objeto de modo que el punto fijo coincida con el origen de las coordenadas.
2. Escale el objeto con respecto del origen de las coordenadas
3. Utilice la traslación inversa del paso 1 para regresar el objeto a su posición original.
Transformación ventana-área de vista.
Algunos paquetes gráficos permiten que el programador especifique coordenadas de primitivas de salida en un sistema de coordenadas de mundo de punto flotante, usando las unidades que sean relevantes para el programa de aplicación: angstroms, micras, metros, millas, años luz, etcétera. Se emplea el término de mundo porque el programa de aplicación representa un mundo que se crea o presenta interactivamente para el usuario.
Transformaciones de composición general y de eficiencia computacional.
Una transformación bidimensional general, que representa una combinación de traslaciones rotaciones y escalaciones se puede expresar como una matriz de 3x3.
Así como las transformaciones bidimensionales se pueden representar con matrices de 3x3 usando coordenadas homogéneas, las transformaciones tridimensionales se pueden representar con matrices de 4x4, siempre y cuando usemos representaciones de coordenadas homogéneas de los puntos en el espacio tridimensional.
Así, en lugar de representar un punto como (x,y,z), lo hacemos como (x, y, z, W), donde dos de estos cuádruplos representan el mismo punto si uno es un multiplicador distinto de cero del otro.
Las transformaciones geométricas son transformaciones afines. Esto es, pueden expresarse como una función lineal de posiciones de coordenadas. Traslación, rotación y escalación son transformaciones afines. Transforman líneas paralelas en líneas paralelas y posiciones de coordenadas finitas en posiciones finitas.
Representación matricial de transformación tridimensionales.
Así como las transformaciones bidimensionales se pueden representar con matrices de3 X 3 usando coordenadas homogéneas, las transformaciones tridimensionales se puedenrepresentar con matrices de 4 X 4, siempre y cuando usemos representaciones decoordenadas homogéneas de los puntos en el espacio tridimensional. Así, en lugar derepresentar un punto como (
x, y, z ), lo hacemos como (x, y, z, W ), donde dos de estoscuádruplos representan el mismo punto si uno es un multiplicador distinto de cero del otro:no se permite el cuádruplo (0, 0, 0, 0). Como sucede en el espacio bidimensional, larepresentación estándar de un punto (x, y, z, W ) con W ≠
0 se indica (x/W, y/W, z/W, 1).
La transformación de un punto a esta forma se denomina homogeneización, igual queantes. Además los puntos cuya coordenada W es cero se llaman puntos en el infinito.También existe una interpretación geométrica. Cada punto en el espacio tridimensional serepresenta con una línea que pasa por el origen en el espacio de cuatro dimensiones, y lasrepresentaciones homogeneizadas de estos puntos forman un subespacio tridimensional deun espacio de cuatro dimensiones definido por la ecuación
W = 1.
El sistema de coordenadas tridimensionales que se usará en los siguientes ejemplos esde mano derecha como se ilustra en la figura 2.16- Por convención las rotaciones positivasen el sistema de mano derecha son tales que, al ver hacia un eje positivo desde el origen,una rotación de 90˚ en sentido contrario al giro de las manecillas del reloj transformará uneje positivo en otro.
Composición de transformaciones tridimensionales
En este apartado se analizará la forma de componer matrices de transformación tridimensionales usando un ejemplo. El objetivo es transformar los segmentos de línea dirigida P1 P2 y P1 P3 de su posición inicial en la parte (a) a su posición finalen la parte (b). De esta manera, el punto P1 se trasladará al origen P1 P2 quedará en el ejepositivo y P1 P3 quedará en la mitad del eje positivo del plano (x, y). Las longitudes de laslíneas no se verán afectadas por la transformación.Se presentan dos formas de lograr la transformación deseada. El primer método escomponer las transformaciones primitivas T,Rx, R y y Rz. Este método, aunque es algotedioso, es fácil de ilustrar y su comprensión nos ayudará en nuestro conocimiento de las transformaciones. El segundo método, que utiliza las propiedades de las matrices ortogonales especiales que se analiza en la sección anterior, se explica de manera mas breve pero es más abstracto.
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} | 0c090d63199a0a01e3b08e4a255778a0 |
839,154,417,353,597,400 | [HOME] MAM2000 (Dimension) [Prev][Up][Next]
Dot Product in Three Dimensions
Geometrically, we know that two vectors are perpendicular if the Pythagorean Theorem holds, i.e. the square of the length of (a,b,c) plus the square of the length of (x,y,z) equals the square of the length of (a,b,c) - (x,y,z) = (a-x,b-y,c-z). This means that a2 + b2 + c2 + x2 + y2 + z2 = (a-x)2 + (b-y)2 + (c-z)2 = a2 - 2ax + x2 + b2 - 2by + y2 + c2 - 2cz + z2. From this it follows that 0 = -2ax - 2by - 2cz, so ax + by + cz = 0. Thus two vectors in R3 are perpendicular if and only if their dot product is zero.
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533,278,101,286,126,700 | Find answers, ask questions, and connect with our
community around the world.
Activity Discussion Math Maths
Tagged: ,
• Aashutosh
Member
May 30, 2021 at 10:46 pm
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::
A circle is a set of all those points that lie in a plane that is equidistant from a given point called “center”. It forms a closed two-dimensional figure.
The important basic properties of circles are listed below:
1. The outer line of a circle is equidistant from the center and is called the radius.
2. The diameter of the circle divides the circle into two equal parts.
3. Circles that have equal radii are said to be congruent to each other.
4. Circles that are different in size or having different radii are similar to each other.
5. Diameter of a circle is twice the radius of the circle.
6. The diameter of the circle is the largest chord of that circle.
7. Equal chords and equal circles have the equal circumference
8. The radius drawn perpendicular to the chord bisects the chord
9. A circle can circumscribe a square, trapezium, rectangle, triangle, and kite.
10. A circle can be inscribed in a square, triangle, and kite.
11. The chords that are equidistant from the center are equal in length.
12. The distance from the center of the circle to the longest chord (diameter) is zero
14. The tangents are parallel to each other if they are drawn at the end of the diameter.
15. An isosceles triangle is formed when the radii joining the ends of a chord to the center of a circle.
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2,506,685,838,132,140,500 | Unexpectedly Intriguing!
February 6, 2007
Although we don't go there often, we're not ones to shy away from personal topics here at Political Calculations. We are, after all, the only blog out there that gets into your paycheck, goes into your house to see if you should switch to compact fluorescents and helps you figure out how much diet soda your system can safely handle.
But now we're getting really personal with our latest tool, adapted from math posted by Geek Logik author Garth Sundem at his blog, asking the question "What are the chances your marriage will last?"
While the Geek Logik blog post contains three separate equations for helping decide various marital topics (the other two answer the questions "should we get married" and "how many kids should we have"), we were intrigued by the statistics that underlie the question of marital sustainability. Here's what Garth wrote about the data:
... the first is based on solid statistics -- an 11,000-person study by the CDC that expolored factors that help and hurt a marriage's chances of working (for example, they found that if a woman is married before age 24, her chances of staying married for 15 years decreased by 30%). These statistics were easy to write in math terms, and the equation does fairly accurately predict your chances of being married at time "T". Granted there are other factors that might help or hurt your specific marriage, but the CDC study found that, for most people, these are the biggest factors. Remember that the average for all marriages is only about 50% and if you get a low number, please accept my very best wishes in bucking the odds.
There's not much more than to go straight to the math, captured in our tool below:
Personal Data
Input Data Values
Her Age at Time of Marriage
Current Combined Years of Post-High School Education
Number of Kids from This Marriage
How Religious is the Couple?
(On a scale of 1-10 with 10 being "the Pope")
Combined Number of Divorces of Couple's Parents
Combined Previous Marriages
The Anniversary (Years of Marriage) for Which to Calculate the Probability
Will You Still Be Married?
Calculated Results Values
Probability at Given Year of Anniversary
Having coded the math, let's reassure you that the result isn't processed through any sort of normal probability distribution. It is, at best, an approximation. Just change the default "religiousness" value to 10 (aka "the pope") and you'll get a better than 100% probability level! Aside from these quirks of math however, you'll still be able get a somewhat realistic approximation of the odds that you'll be married for your "Xth" anniversary over a pretty wide range of the distribution curve.
Now that you've seen the generic probability that you'll still be married to your current spouse at the anniversary of your marriage that you entered, you may have more questions than answers. If the probability is really low, that might be a good place to begin a conversation with your spouse. If the probability is high, you may already have a strong foundation for a successful marriage. Just remember that it never hurts to make it stronger.
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4,605,250,094,556,951,600 | Plateforme N°1 de soutien en mathématique Post-bac Prépa
Résolution du problème suivant :
Bases mathématiques-Opérations-entre-ensembles-Partitions-MPSI
Ali Mkhida
Ali Mkhida
Dr. Agrégé en Mathématique & fondateur de Qoosmo.
Chapitre mentionné dans l'article :
Bases mathématiques-Opérations-entre-ensembles-Partitions-MPSI
Partitions
Définition
Bases mathématiques-Opérations-entre-ensembles-Partitions-MPSI
Une famille $\left(F_i\right)_{i \in I}$ de parties de $E$ est une partition de $E$ si
– Aucune partie n’est vide
$$
\forall i \in I, \quad F_i \neq \varnothing .
$$
– Deux parties distinctes sont disjointes
$$
\forall(i, j) \in I^2, \quad i \neq j \Rightarrow F_i \cap F_j=\varnothing .
$$
– La réunion est égale à $E$
$$
\cup_{i \in I} F_i=E .
$$
Exemple
La famille $\left(\left[n, n+1[)_{n \in \mathbb{Z}}\right.\right.$ est une partition de $\mathbb{R}$ : les parties sont non vides, disjointes et de réunion $\mathbb{R}$. Il s’agit de partitionner l’ensemble des réels selon leur partie entière.
Exemple
Les parties $2 \mathbb{Z}$ et $\{2 k+1, k \in \mathbb{Z}\}$ forment une partition de $\mathbb{Z}$ (selon la parité).
En fait, ces exemples relèvent d’un cadre plus général que nous détaillons maintenant.
Définition
Une relation d’équivalence sur un ensemble $E$ est une relation binaire $\sim$ qui vérifie les propriétés suivantes:
– est réflexive, c’est-à-dire, tout élément $x \in E$ est en relation avec lui-même pour :
$$
\forall x \in E, \quad x \sim x .
$$
– est symétrique, c’est-à-dire pour tous les éléments $x$ et $y \in E$ tels que $x$ est en relation avec $y$, on a aussi $y$ en relation avec $x$
$$
\forall x, y \in E, \quad x \sim y \quad \Leftrightarrow \quad y \sim x .
$$
– est transitive, c’est-à-dire pour tous les éléments $x, y$ et $z \in E$ tels que $x$ est en relation avec $y$ et $y$ est en relation avec $z$, on a aussi $x$ en relation avec $z$
$$
\forall x, y, z \in E, \quad x \sim y \text { et } y \sim z \quad \Rightarrow \quad x \sim z .
$$
Deux éléments en relation sont dits équivalents. La classe d’équivalence d’un élément $x$ pour la relation $\sim$ est l’ensemble des éléments de $E$ équivalents à $x$ pour $\sim$, à savoir
$$
\{y \in E, y \sim x\} .
$$
Exemple
Congruence sur les entiers
Soit $p \in \mathbb{N} \backslash\{0\}$. Deux entiers $m$ et $n$ sont congruents modulo $p$ si $p$ divise $m-n$. On note alors $m \equiv n[p]$ ou plus simplement $m=n[p]$ en gardant à l’esprit que ce signe d’égalité n’est pas une véritable égalité.
On vérifie immédiatement que la congruence modulo $p$ est une relation d’équivalence sur $\mathbb{Z}$ et que la classe d’équivalence de $a$ est
$$
\{a+b p, b \in \mathbb{Z}\}
$$
Généralisons l’exemple précédent aux réels (où, rappelons-le, la notion de divisibilité est peu pertinente puisque tout réel non nul divise tous les réels).
Exemple
Deux réels $x$ et $y$ sont congruents modulo $2 \pi$ si la différence $x-y$ est un multiple entier de $2 \pi$. On définit ainsi une relation d’équivalence sur $\mathbb{R}$ (le choix de $2 \pi$ correspond à une utilisation courante pour les fonctions trigonométriques, on peut bien évidemment choisir n’importe quel autre réel non nul).
Proposition
Soit une relation d’équivalence sur un ensemble $E$. Les classes d’équivalence pour $\sim$ forment une partition de $E$.
Démonstration
Il est clair qu’une classe d’équivalence est non vide et que $E$ est inclus dans la réunion des classes d’équivalence (car chaque élément est inclus dans sa propre classe d’équivalence). Montrons que deux classes d’équivalence distinctes sont disjointes. Pour cela, considérons $x$ et $y \in E$ tels que les classes d’équivalence de $x$ et $y$ ne soient pas disjointes et $z \in E$ à la fois équivalent à $x$ et $y$. Alors, par transitivité, la classe de $x$ est la classe de $z$; de même, la classe de $y$ est la classe de $z$. Ainsi, les classes de $x$ et de $y$ sont égales.
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6,445,490,058,772,371,000 | being a math tutor
Right now in my geometry class, I have an A+, but friend is failing that class. I offered to be her math tutor, to possibly meet in the library after school 3 days a week. Does anybody have any ideas on how to make our study sessions more productive (I haven't started yet -- maybe sometime next week)?
1. 👍
2. 👎
3. 👁
1. First, I suggest you talk with her geometry teacher who can probably offer you some good advice about how to help your friend. Both of you should go in together so that the teacher knows that s/he has your friend's consent to share that information with you.
Use as many visuals as possible. I think one of the problems that some students have with geometry is that they have trouble visualizing the shapes.
You could start each session by working together on problems she doesn't understand on that day's assignment. You could work through the first problem together. Then ask her to tackle the second problem while you stop her if she goes astray.
1. 👍
2. 👎
👤
Ms. Sue
2. Thank you very much for the advice! This will help me very much.
:-)
1. 👍
2. 👎
3. Good luck! I'm sure you'll be able to help your friend.
1. 👍
2. 👎
👤
Ms. Sue
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5,496,841,960,242,996,000 | Mandelbrot ... but with no main cardioid!
• 6 Replies
• 572 Views
0 Members and 1 Guest are viewing this topic.
Offline quadralienne
• *
• Fractal Fanatic
• ***
• Posts: 26
• infinite border, finite area
« on: June 20, 2019, 09:34:19 PM »
Once upon a time I stumbled across
https://commons.wikimedia.org/wiki/File:Parameter_plane_and_Mandelbrot_set_for_f(z)_%3D_z%5E4_%2B_m*z.png
and I worked through the C code (languages without native complex numbers make me sad) to convert it to a MathMap expression:
unit filter JustCircles ()
Mmax = 4096;
Escape = 3;
Escape2 = Escape * Escape;
if r < 1.0 || abs(ri:[x-4/3,y]) < 1/3 || abs(ri:[x+4/3*cos(pi/3),abs(y)-4/3*sin(pi/3)]) < 1/3 then
rgba:[0,0,0,1]
else
Mc = ri:[x,y];
ang = atan(-y,-x)/3.0;
rad = (0.0625 * (y*y+x*x)) ^ (1/6);
Mz = ri:[rad * cos(ang), rad * sin(ang)];
Mcount = 0;
while ((Mz[0]*Mz[0]+Mz[1]*Mz[1]) < Escape2) && (Mcount < Mmax) do
Mcount = Mcount + 1;
Mz = Mz ^ 4 + Mz * Mc;
end;
if Mcount < Mmax then
dist = log( Mcount + 1 - (log(log(abs(Mz)))/log(Escape)) ) / log(Mmax);
hue = pmod(arg(Mz)/(2*pi), 1);
sat = (1 - dist) ^ 2;
val = sqrt(dist);
toRGBA(hsva:[ hue, sat, val, 1 ])
else
rgba:[0,0,0,1]
end
end
end
filter JustCircles_p ()
JustCircles(xy:[0.25 + 2.5 * x, 2.5 * y])
end
Linkback: https://fractalforums.org/share-a-fractal/22/mandelbrot-but-with-no-main-cardioid/2890/
Offline Adam Majewski
• *
• Fractal Flamingo
• ****
• Posts: 330
« Reply #1 on: June 21, 2019, 04:19:23 PM »
You are right that c has also complex type and in new programs I use it.
Offline chronologicaldot
• *
• Fractal Friend
• **
• Posts: 10
• Unconventional Formulaic Object
• Personal Website
« Reply #2 on: July 29, 2019, 10:07:43 PM »
That's really cool! It's hard to tell whether it truly follows the Mandelbrot pattern on the tips or if it's now a bunch of successively shrinking circles.
There are no bad fractal parameters. There are simply those that haven't been tweaked enough.
Offline Adam Majewski
• *
• Fractal Flamingo
• ****
• Posts: 330
Offline 3DickUlus
• *
• 3f
• ******
• Posts: 1724
• Digilantism
« Reply #4 on: July 31, 2019, 05:16:00 AM »
https://upload.wikimedia.org/wikipedia/commons/0/03/Parameter_plane_and_Mandelbrot_set_for_f%28z%29_%3D_z%5E4_%2B_m%2Az.png
You can use BBC code for image URLs with weird characters and for wiki images use the image URL not the image page url ;)
Code: [Select]
[img]https://upload.wikimedia.org/wikipedia/commons/0/03/Parameter_plane_and_Mandelbrot_set_for_f%28z%29_%3D_z%5E4_%2B_m%2Az.png[/img]
[url=https://upload.wikimedia.org/wikipedia/commons/0/03/Parameter_plane_and_Mandelbrot_set_for_f%28z%29_%3D_z%5E4_%2B_m%2Az.png]https://upload.wikimedia.org/wikipedia/commons/0/03/Parameter_plane_and_Mandelbrot_set_for_f%28z%29_%3D_z%5E4_%2B_m%2Az.png[/url]
« Last Edit: July 31, 2019, 05:55:29 AM by 3DickUlus, Reason: typo »
Offline quadralienne
• *
• Fractal Fanatic
• ***
• Posts: 26
• infinite border, finite area
« Reply #5 on: August 02, 2019, 10:48:13 PM »
It's not just circles, it's Mandelbrotty all over!
Offline quadralienne
• *
• Fractal Fanatic
• ***
• Posts: 26
• infinite border, finite area
« Reply #6 on: August 02, 2019, 11:21:43 PM »
Here's an 8192x zoom at 1.98835 + 0.183i ... uh, fuzzy because single precision!
xx
Upload limits per day for Main and User Galleries?
Started by Anon on Discuss Fractalforums
13 Replies
799 Views
Last post September 22, 2017, 07:02:49 AM
by claude
question
Upload limits per day for Main and User Galleries (Sticky Topic/Post)
Started by Anon on Discuss Fractalforums
3 Replies
360 Views
Last post February 17, 2018, 08:13:59 PM
by 3DickUlus
xx
Mandelbrot Burning Ship Mandelbrot Mandelbrot hybrid 3
Started by claude on Fractal Image Gallery
0 Replies
377 Views
Last post January 17, 2018, 12:26:38 AM
by claude
xx
Mandelbrot Burning Ship Mandelbrot Mandelbrot hybrid
Started by claude on Fractal Image Gallery
0 Replies
358 Views
Last post January 16, 2018, 11:30:30 PM
by claude
xx
Mandelbrot Burning Ship Mandelbrot Mandelbrot hybrid 2
Started by claude on Fractal Image Gallery
0 Replies
314 Views
Last post January 17, 2018, 12:10:56 AM
by claude
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-6,792,524,150,457,260,000 | Resultado de 3(2x+1)-4=11
Solución simple y rápida para la ecuación 3(2x+1)-4=11. Nuestra respuesta es comprensible y explicada paso a paso.
Si no es lo que está buscando, escriba sus propios datos.
Resultado de 3(2x+1)-4=11:
3(2x+1)-4=11
Movemos todos los personajes a la izquierda:
3(2x+1)-4-(11)=0
Sumamos todos los números y todas las variables.
3(2x+1)-15=0
Multiplicar
6x+3-15=0
Sumamos todos los números y todas las variables.
6x-12=0
Movemos todos los términos que contienen x al lado izquierdo, todos los demás términos al lado derecho
6x=12
x=12/6
x=2
El resultado de la ecuación 3(2x+1)-4=11 para usar en su tarea doméstica.
Ver soluciones similares:
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-970,508,963,681,632,000 | mersenneforum.org
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Old 2009-06-13, 19:11 #1
ATH
Einyen
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Default Mersenne primes have highly composite p-1?
http://www.mersenneforum.org/showpos...6&postcount=85
Quote:
Originally Posted by akruppa View Post
Maybe there's a simple explanation why Mersenne primes M[I]p[/I] tend to have highly composite p-1. Trivially, 2p-1 and 2p-2 have no common factor, and the latter is 2(2p-1-1) which has a lot of algebraic factors if p-1 is highly composite, and so will have a lot of small prime factors, thus slightly reducing the probability that 2p-1 has small prime factors. The effect can't be very strong, though: divisors of 2p-1 are of form 2kp+1 and few factors of 2p-1-1 will be of this form. Still, it's the only thing I can think of how the number of divisors of p-1 might enter the picture.
If this effect is the reason for the smoother-than-expected p-1 in prime M[I]p[/I], then M[I]p[/I] with smooth p-1should simply have a slightly better chance of surviving trial division, but among the trial-divided candidates, the probability that M[I]p[/I] is prime should be independent of the smoothness of p-1 again.
Alex
http://www.mersenneforum.org/showpos...&postcount=345
Quote:
Originally Posted by akruppa View Post
I have no idea how to quantify this. An empirical test is the best I can think of. Only relatively small divisors should be affected, so one might check if those 2[I]p[/I] where p-1 has at least n divisors are more likely to survive trial division to, say, 240.
Alex
I took the list from 20M to 30M from the "Factoring limits" list, which is all those that has no known factor below 2^66 (20M) to 2^68 (30M).
There are 587,252 primes from 20M to 30M and of them 369,166 have known factors while 218,086 are in the factoring limit list and have no known factors. I made a small program to test number of factors in p-1 for all 587,252 primes and see if there was a difference:
Code:
p-1 factors A B
---------------------------------------------
2 factors 15617=4.23% 8278=3.80%
3 factors 52822=14.31% 29439=13.50%
4 factors 81916=22.19% 47299=21.69%
5 factors 80449=21.79% 48040=22.03%
6 factors 59649=16.16% 36367=16.68%
7 factors 37016=10.03% 22935=10.51%
8 factors 20575=5.57% 12596=5.78%
9 factors 10774=2,92% 6684=3.06%
10 factors 5394=1.46% 3265=1.50%
11 factors 2571=0.70% 1686=0.77%
12 factors 1280=0.35% 784=0.36%
13 factors 612=0.17% 369=0.17%
14 factors 265=0.07% 169=0.08%
15 factors 127 92
16 factors 57 39
17 factors 28 22
18 factors 7 8
19 factors 5 5
20 factors 0 6
21 factors 1 1
22 factors 1 1
23 factors 1 1
----------------------------------------------
Total 369166(100%) 218086(100%)
Column A are the exponents where 2^p-1 has factors below 2^66-2^68, and B where 2^p-1 have no factors below 2^66-2^68. Looking at the percentages there is no real difference in number of exponents with higher number of factors of p-1 in column B. Maybe we have to check at a lower factor level like 2^40 like Alex suggests.
Last fiddled with by ATH on 2009-06-13 at 19:19
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Old 2009-06-13, 20:04 #2
akruppa
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I rearranged a little to show the survival rate of exponents p depending on the number of factors of p-1:
Code:
p-1 factors #p #survivors #suvival rate
-----------------------------------------------------
2 factors 23895 8278 0.346
3 factors 82261 29439 0.358
4 factors 129215 47299 0.366
5 factors 128489 48040 0.374
6 factors 96016 36367 0.379
7 factors 59951 22935 0.383
8 factors 33171 12596 0.380
9 factors 17458 6684 0.383
10 factors 8659 3265 0.377
11 factors 4257 1686 0.396
12 factors 2064 784 0.380
13 factors 981 369 0.376
14 factors 434 169 0.389
So it looks like those p with few factors in p-1 do, in fact, have a lower chance of surviving trial division. ATH, I assume the number of factors is the number of prime factors with multiplicity in p-1? It might be interesting to make such a table for the number of proper divisors of p-1 as well.
My hypothesis isn't very convincing, though. By the same argument, 2[I]p[/I]-4 and 2[I]p[/I]-1 have at most the factor 3 in common, so the number of divisors in p-2 (and p-3 and p-4 etc.) should also affect the probability that Mp is prime.
Alex
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Old 2009-06-14, 23:04 #3
ATH
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Quote:
Originally Posted by akruppa View Post
So it looks like those p with few factors in p-1 do, in fact, have a lower chance of surviving trial division. ATH, I assume the number of factors is the number of prime factors with multiplicity in p-1? It might be interesting to make such a table for the number of proper divisors of p-1 as well.
If you mean distinct prime factors, here is the list:
Code:
p-1 factors Total(20M-30M) numbers without factors to 2^66-2^68
--------------------------------------------------------------------
2 factors 49855 17960 = 36.02%
3 factors 173824 63988 = 36.81%
4 factors 218645 81127 = 37.10%
5 factors 118143 44684 = 37.83%
6 factors 25228 9692 = 38.42%
7 factors 1548 629 = 40.63%
8 factors 9 6 (=66.67%)
--------------------------------------------------------------------
Total 587252 218086
There is a clear rising percentage of numbers "surviving" trialfactor to 2^66-2^68, the more distinct prime factors p-1 has.
If you mean all factors (not just prime factors) then the list is extensive, here is whole list (not counting 1 and p-1 as factors of p-1): mersennetest.txt
Here is the list abbriviated by combining the factor-categories with low number of members in them:
Code:
p-1 factors Total(20M-30M) numbers without factors to 2^66-2^68
--------------------------------------------------------------------
2 factors 23895 8278 = 34.64%
4 factors 12264 4577 = 37.32%
6 factors 76473 27329 = 35.74%
7-8 factors 3355 1230 = 36.66%
10 factors 47504 17911 = 37.70%
12-13 factors 985 376 = 38.17%
14 factors 98842 35686 = 36,10%
16 factors 5377 2038 = 37.90%
18 factors 10600 3991 = 37.65%
19-22 factors 68947 25980 = 37.68%
23-26 factors 2651 946 = 35.68%
28 factors 1955 753 = 38.52%
30 factors 66115 24718 = 37.39%
31-34 factors 13651 5150 = 37.73%
36-38 factors 12384 4728 = 38,18%
40-46 factors 49363 18813 = 38.11%
48-54 factors 3795 1435 = 37.81%
58 factors 4090 1563 = 38.22%
61-62 factors 24317 9256 = 38.06%
64-70 factors 12353 4747 = 38.43%
73-78 factors 6840 2622 = 38.33%
79-94 factors 18451 6969 = 37.77%
96-106 factors 1528 577 = 37.76%
108-110 factors 1234 465 = 37.68%
118 factors 2806 1091 = 38.88%
124-126 factors 4652 1819 = 39.10%
128-142 factors 4558 1761 = 38.64%
148-158 factors 1635 634 = 38.78%
160-178 factors 951 361 = 37.96%
180-190 factors 2707 1076 = 39.75%
194-214 factors 659 272 = 41.27%
218-254 factors 1294 508 = 39.26%
258-286 factors 546 230 = 42.12%
292-318 factors 147 58 = 39.46%
322-358 factors 141 57 = 40.43%
376-382 factors 108 43 = 39.81%
394-430 factors 48 28 = 58.33%
446-478 factors 20 7 = 35.00%
502-574 factors 11 3 = 27.27%
--------------------------------------------------------------------
Total 587252 218086 (=37.14%)
The trend is not so clear here, since there is so many categories with more or less members in. But 39+% happens only for >124 factors and 40+% only for >194factors.
Last fiddled with by ATH on 2009-06-14 at 23:11
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Old 2009-06-15, 13:11 #4
ATH
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Combined the categories on the last list even more (the one with all factors of p-1 except 1 and p-1):
Code:
p-1 factors Total(20M-30M) numbers without factors to 2^66-2^68
--------------------------------------------------------------------
2-6 factors 112632 40184 = 35.68%
7-13 factors 51844 19517 = 37.65%
14-18 factors 114819 41715 = 36.33%
19-22 factors 68947 25980 = 37.68%
23-30 factors 70721 26417 = 37.35%
31-46 factors 75398 28691 = 38.05%
48-62 factors 32202 12254 = 38.05%
64-94 factors 37644 14338 = 38.09%
96-142 factors 14778 5713 = 38.66%
148-254 factors 7246 2851 = 39.35%
258-382 factors 942 388 = 41.19%
394-574 factors 79 38 = 48.10%
--------------------------------------------------------------------
Total 587252 218086 (=37.14%)
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-998,444,846,774,835,700 | For the Students of Hindu Vedic Astrology by Dr. A. Shanker
Recent Posts
20130205
Encyclopedia of Vedic Astrology: Tajik Shastra and Annual Horoscopy: The Muntha, Chapter XI, Part - 1
Dr. Shanker Adawal
The Muntha
Part 1
1. The Muntha & its Progression: The Muntha is an important mathematical concept in Varshphal connecting the annual chart with the birth chart. At the time of birth, it is located in lagna/ Ascendant. Each year, the Muntha moves one rashi (sign) in direct motion. When the second year of life begins, the Muntha falls in the 2nd house from lagna. At the 3rd year of life, the Muntha will be in the 3rd house from lagna and so on. Because of its progression by one sign each year, the Muntha is also termed as the progressed ascendant. In the analysis of an annual chart, the Muntha is extremely important. The lord of Muntha is also very significant while analyzing an annual chart and is one of the five contenders for the post of lord of the year Since the Muntha progresses one sign or 30 degrees in a year, it progresses every month by 30/ 12 = 2.5 or 2 degree 30 min. Similarly, daily motion of the Muntha may also be calculated by dividing 2 degree 30 min by 30 (number of days in a month), giving a value of 5 min. These values are important if one intends to go into closer timing of events during a given year.
2. Calculation of the Muntha:
1. Note the number of sign wherein the lagna falls in the birth chart. Say if lagna is in Gemini, the number of sign is 3.
2. Add to it the number of completed years elapsed between the birth and the current year for which the Muntha is to be calculated.
3. Divide the total by 12 (total numbers of signs). The remainder so obtained would be the sign in which the Muntha is located in the annual chart.
4. If the remainder is zero, it should be treated as 12 sign or Pisces.
Based on the above, the Muntha in the example chart No. 21 of previous chapter would be in 3rd house of Aries and in chart No 2, in 2nd house of Sagittarius.
3. The Results of Muntha in different Houses
A well-placed and well-aspected Muntha strengthens the house it is in. The Muntha gives results according to its location in different houses, its association with different planets, and the disposition of the lord of the sign in which the Muntha is located. (This is some what similar to the position of dispositor in Parashari system). The general results are based on the “Tajika Neelkanthi”, the famous treatise by Acharya Neelkanth. The Muntha is considered very auspicious in houses 9, 10 & 11. In houses 1, 2, 3, & 5 it yields good results through personal efforts of the native. In the remaining houses 4, 6, 7, 8, & 12 in the annual chart, it is considered inauspicious. However experience shows that in 7th house, the Muntha does not give bad results except illness to the native. The general results of positioning of the Muntha in various houses are as under:
1. Lagna: It is considered to yield very good results and indicates dominance/ victory over opponents, dignity, favours from rulers, high status or a new job/ source of income, increase in power, comforts & money and good health through own efforts. It may also denote change in place, position or residence, transfer or the birth of a child.
Chart No. 29: PV Narsimha Rao: Born on Tuesday, 28 Jun 1921 at 1-02 PM at Warangal with Virgo Lagna; 70th annual chart for 1990-91
The Muntha is in lagna in own house with Mercury, Lagnesh of birth & annual chart and with benefics. The native became Prime Minister of India on 21 Jun 1991. The sudden & unexpected event was due to mutual aspect with 8-9th lord Saturn. However the Saturn’s aspect on Muntha and Muntha lord also gave illness in first part of the year.
Continue…
Shanker Adawal
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-4,610,528,528,350,969,300 | Slow Sum – Revisited
Good morning. I received a comment from Dmytro regarding the post Slow Sum suggesting the use of a priority queue instead of a stream. I appreciate the comment and suggestion.
Suppose we have a list of N numbers,
and repeat the following operation until we're left with only a single number:
Choose any two numbers and replace them with their sum.
Moreover, we associate a penalty with each operation equal to the value of the new number,
and call the penalty for the entire list as the sum of the penalties of each operation.
For example, given the list [1, 2, 3, 4, 5],
we could choose 2 and 3 for the first operation,
which would transform the list into [1, 5, 4, 5] and incur a penalty of 5.
The goal in this problem is to find the worst possible penalty for a given input.
Input:
An array arr containing N integers, denoting the numbers in the list.
Output format:
An int representing the worst possible total penalty.
Constraints:
o 1 ≤ N ≤ 10^6
o 1 ≤ Ai ≤ 10^7, where *Ai denotes the ith initial element of an array.
o The sum of values of N over all test cases will not exceed 5 * 10^6.
The description of the problem is the same as far as I can tell. I just copied it from the previous post.
1,2,3,4,5
main <<< arr: [1, 2, 3, 4, 5]
main <<< output: 50
4,2,1,3
main <<< arr: [4, 2, 1, 3]
main <<< output: 26
The test codes are the same. This time I did a screen capture using the implementation of the function of interest using a priority queue. The two test cases return the same values.
/**
* Test scaffolding
*
* @throws IOException
*/
public static void main(String[] args) throws IOException {
// **** open buffered reader ****
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
// **** read input line and split values ****
String[] strs = br.readLine().trim().split(",");
// **** close buffered reader ****
br.close();
// **** create and populate array of integers ****
int[] arr = Arrays.stream(strs).mapToInt(Integer::parseInt).toArray();
// ???? ????
System.out.println("main <<< arr: " + Arrays.toString(arr));
// **** call function and display resuly ****
System.out.println("main <<< output: " + getTotalTime(arr));
}
Our test scaffold reads the single input line holding the values for the input array to the function of interest. The contents of the int[] array are displayed to make sure all is well so far.
The function of interest is called and the result is displayed.
/**
* Using a stream.
*
* Execution O(n log(n)) - Space: O(n)
*/
static int getTotalTime0(int[] arr) {
// **** sanity check(s) ****
if (arr.length == 1) return 0;
// ???? ????
System.out.println("<<< arr:: " + Arrays.toString(arr));
// **** sort array in descending order - O(n log(n)) ****
int[] rev = Arrays.stream(arr)
.boxed()
.sorted(Comparator.reverseOrder())
.mapToInt(Integer::intValue)
.toArray();
// ???? ????
System.out.println("<<< rev:: " + Arrays.toString(rev));
// **** initialization ****
int penalty = rev[0] + rev[1];
int penalties = penalty;
// **** loop counting penalties - O(n) ****
for (int i = 2; i < rev.length; i++) {
// **** generate penalty ****
penalty += rev[i];
// **** add penalty ****
penalties += penalty;
}
// **** return penalties ****
return penalties;
}
This is the previous implementation of the function of interest using Arrays.stream to sort the input array `arr` and generate a new int[] `rev` in which the values are sorted in monotonically descending order.
Please take a look at the comments section of the function in which the execution and space orders are listed.
/**
* Using a priority queue instead of a stream.
*
* Execution: O(n) - Space: O(n)
*/
static int getTotalTime(int[] arr) {
// **** sanity check(s) ****
if (arr.length == 1) return 0;
// **** initialization - O(n * log(n)) ****
PriorityQueue<Integer> rev = new PriorityQueue<>(arr.length, (a,b) -> b - a);
for (int i : arr) rev.add(i);
int penalty = rev.poll() + rev.poll();
int penalties = penalty;
// **** loop counting penalties - O(n) ****
while (!rev.isEmpty()) {
// **** generate penalty ****
penalty += rev.poll();
// **** add penalty ****
penalties += penalty;
}
// **** return penalties ****
return penalties;
}
This is the new implementation of the function of interest using a priority queue.
We start by performing a sanity check. There is no reason to proceed if the number of integers in the input int[] `arr` is 1.
We then initialize a priority queue with an initial capacity that matches the size of the input array and a comparator that will allow us to pull items in monotonically decreasing order. We then insert into the priority queue all the elements in the `arr`.
We then declare and initialize the `penalty` and `penalties` variables. We could have used a single variable but I used two to ease debugging by displaying variables during execution. I guess that at this point in the game I could have modified the code to use a single variable, but the code is in a different machine and I already pushed it to GitHub.
A loop is then entered. The loop will consume all the entries in the priority queue while updating the `penalties` value.
When all is said and done, our function returns the value in the `penalties` variable.
Hope you enjoyed solving this problem in a different way as suggested by Dmytro as much as I did. The entire code for this project can be found in my GitHub repository.
Please note that the code here presented might not be the best possible solution. In most cases, after solving the problem, you can take a look at the best accepted solution provided by the different websites (i.e., HackerRank, LeetCode). Since this problem came from a Facebook website, the number of test cases is limited. I believe in this case only two test cases were used to verify the solution.
If you have comments or questions regarding this, or any other post in this blog, please do not hesitate and leave me a note below. I will reply as soon as possible.
Keep on reading and experimenting. It is one of the best ways to learn, become proficient, refresh your knowledge and enhance your developer / engineering toolset.
Thanks for reading this post, feel free to connect with me John Canessa at LinkedIn.
Enjoy;
John
Leave a Reply
Your email address will not be published. Required fields are marked *
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Find the values of k so that lines are perpendicular using symetric equations
• Thread starter soulja101
• Start date
• #1
62
0
Homework Statement
Line 1: x-3/3k+1=Y+6/2=Z+3/2K
Line 2: x+7/3=y+8/-2k=z+9/-3
Homework Equations
Cross product and dot product
The Attempt at a Solution
vector equation for line 1: (x,y,z)=(4,-4,-3)+k(3,0,2)
vector equation for line 2: (x,y,z)=(-4,-8,-12)+k(0,-2,0)
The product of (3,0,2)and (0,-2,0) is zero so they are perpendicular but i don't know how to find k :confused:
Answers and Replies
• #2
HallsofIvy
Science Advisor
Homework Helper
41,795
925
Homework Statement
Line 1: x-3/3k+1=Y+6/2=Z+3/2K
a) Use parentheses: you mean, I think (x-3)/(3k+1)= (y+6)/2= (z+ 3)/2k
b) Don't use y and Y or k and K for the same thing. Those are different symbols and typically mean different values.
Setting each of those equal to t, (x-3)/(3k+1)= t so x- 3= (3k+1)t or x= 3+ (3k+1)t, (y+6)/2= t so y+ 6= 2t or y= -6+ 2t, and (z+3)/2k= t so z+ 3= 2kt or z= -3+ 2kt
Line 2: x+7/3=y+8/-2k=z+9/-3
(x+7)/3= t so x+ 7= 3t or x= -7+ 3t, (y+8)/(-2k)= t so y+8= -2kt or y= -8- 2kt, and (z+9)/(-3)= t so z+ 9= -3t or z= -9- 3t.
Homework Equations
Cross product and dot product
The Attempt at a Solution
vector equation for line 1: (x,y,z)=(4,-4,-3)+k(3,0,2)
No, that is not an equation for line 1. For one thing, k is given in the symmetric equation: one value of k corresponds to one line so without another parameter, this would be just a single point, not a line.
vector equation for line 2: (x,y,z)=(-4,-8,-12)+k(0,-2,0)
Same comment.
The product of (3,0,2)and (0,-2,0) is zero so they are perpendicular but i don't know how to find k :confused:
You don't have equations for the lines. As I said above your vector equation should be
(x, y, z)= (-7, -8, -3)+ t(3k+1, 2, 2k)t. The parameter t determines the point, k is fixed for a line.
For line 2, (x, y, z)= (-7, -8, -9)+ t(3, -2k, -3).
In order that the lines be perpendicular you must have (3k+1, 2, 2k).(3, -2k, -3)= 9k+ 3- 4k- 6k= -k+ 3= 0.
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-2,012,722,032,228,191,000 | Wednesday
May 4, 2016
Homework Help: smallest of 3 integers
Posted by Anonymous on Wednesday, June 27, 2012 at 1:15pm.
The sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the product of the integers. What is the smallest of the three integers?
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-237,155,132,277,577,250 | login
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A121523 Number of up steps starting at an even level in all nondecreasing Dyck paths of semilength n. A nondecreasing Dyck path is a Dyck path for which the sequence of the altitudes of the valleys is nondecreasing. 2
1, 3, 10, 33, 103, 315, 941, 2770, 8051, 23171, 66138, 187486, 528365, 1481501, 4135756, 11500721, 31871625, 88054825, 242609585, 666783380, 1828452021, 5003697403, 13667302500, 37267071708, 101455834153, 275797332135 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
a(n) = Sum(k*A121522(n,k), k=1..n). a(n)+A121525(n)=n*fibonacci(2n-1).
LINKS
Table of n, a(n) for n=1..26.
E. Barcucci, A. Del Lungo, S. Fezzi and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
FORMULA
G.f.: z(1-3z+z^2+5z^3-5z^4)/[(1+z)(1-3z+z^2)^2*(1-z-z^2)].
a(n) ~ (5-sqrt(5)) * (3+sqrt(5))^n * n / (5 * 2^(n+2)). - Vaclav Kotesovec, Mar 20 2014
EXAMPLE
a(3)=10 because we have (U)D(U)D(U)D, (U)D(U)UDD, (U)UDD(U)D, (U)UDUDD and (U)U(U)DDD, the up steps starting at even level being shown between parentheses (U=(1,1), D=(1,-1)).
MAPLE
G:=z*(1-3*z+z^2+5*z^3-5*z^4)/(1+z)/(1-3*z+z^2)^2/(1-z-z^2): Gser:=series(G, z=0, 34): seq(coeff(Gser, z, n), n=1..30);
MATHEMATICA
Rest[CoefficientList[Series[x*(1-3*x+x^2+5*x^3-5*x^4)/(1+x)/(1-3*x+x^2)^2 /(1-x-x^2), {x, 0, 20}], x]] (* Vaclav Kotesovec, Mar 20 2014 *)
CROSSREFS
Cf. A001519, A121522, A121525.
Sequence in context: A316411 A292549 A062454 * A115240 A027989 A096483
Adjacent sequences: A121520 A121521 A121522 * A121524 A121525 A121526
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Aug 05 2006
STATUS
approved
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834,479,223,463,730,200 | 18.100B.PracticeFinal
18.100B.PracticeFinal - 18.100B/C Practice Final Exam...
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Unformatted text preview: 18.100B/C Practice Final Exam Monday, December 15, 2008, 1:30–4:30, in Johnson. Closed book, no calculators. YOUR NAME: This is a 180-minute exam. No notes, books, or calculators are permitted. Point values (out of 100) are indicated for each problem. There is a (hard) bonus question, Problem 9, at the end – do not attempt it until you have worked all other problems. (Note, you can achieve the full 100 points without attempting the bonus problem.) Do all the work on these pages. GRADING 1. /10 2. /10 3. /10 4. /15 5. /10 6. /10 7. /15 8. /20 9. /20 TOTAL BONUS /100 1 Problem 1. [10 points] Suppose that x ∈ R satisfies ≤ x ≤ for every > . Show that x = 0 , using only axioms of R as an ordered field. State the axioms you are using. (Note that the Archimedean and least upper bound properties are not ordered field axioms.) 2 Problem 2. [10 points: (a) /5 (b) /5] Let ( a n ) be a sequence of positive real numbers. (a) Suppose that the series ∞ X n =1 a n converges. Prove that ∞ X n =1 √ a n a n +1 also converges.also converges....
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.
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18.100B.PracticeFinal - 18.100B/C Practice Final Exam...
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Unit Converter
Convert 146 Acres to Square Feet
To calculate 146 Acres to the corresponding value in Square Feet, multiply the quantity in Acres by 43560 (conversion factor). In this case we should multiply 146 Acres by 43560 to get the equivalent result in Square Feet:
146 Acres x 43560 = 6359760 Square Feet
146 Acres is equivalent to 6359760 Square Feet.
How to convert from Acres to Square Feet
The conversion factor from Acres to Square Feet is 43560. To find out how many Acres in Square Feet, multiply by the conversion factor or use the Area converter above. One hundred forty-six Acres is equivalent to six million three hundred fifty-nine thousand seven hundred sixty Square Feet.
Definition of Acre
The acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. It is defined as the area of 1 chain by 1 furlong (66 by 660 feet), which is exactly equal to 1⁄640 of a square mile, 43,560 square feet, approximately 4,047 m2, or about 40% of a hectare. The most commonly used acre today is the international acre. In the United States both the international acre and the US survey acre are in use, but differ by only two parts per million, see below. The most common use of the acre is to measure tracts of land. One international acre is defined as exactly 4,046.8564224 square metres.
Definition of Square Foot
The square foot (plural square feet; abbreviated sq ft, sf, ft2) is an imperial unit and U.S. customary unit (non-SI, non-metric) of area, used mainly in the United States and partially in Bangladesh, Canada, Ghana, Hong Kong, India, Malaysia, Nepal, Pakistan, Singapore and the United Kingdom. It is defined as the area of a square with sides of 1 foot. 1 square foot is equivalent to 144 square inches (Sq In), 1/9 square yards (Sq Yd) or 0.09290304 square meters (symbol: m2). 1 acre is equivalent to 43,560 square feet.
Using the Acres to Square Feet converter you can get answers to questions like the following:
• How many Square Feet are in 146 Acres?
• 146 Acres is equal to how many Square Feet?
• How to convert 146 Acres to Square Feet?
• How many is 146 Acres in Square Feet?
• What is 146 Acres in Square Feet?
• How much is 146 Acres in Square Feet?
• How many ft2 are in 146 ac?
• 146 ac is equal to how many ft2?
• How to convert 146 ac to ft2?
• How many is 146 ac in ft2?
• What is 146 ac in ft2?
• How much is 146 ac in ft2? | {
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Acceleration as a function of displacement
1. Nov 13, 2014 #1
In one my classes my lecturer showed us the following derivation of acceleration as a function of displacement
dv/dt = v(dv/dx) = d(.5v^2)/dx
I understand how to get from the first to the second part. But I'm not sure how he got from the second part to the third. Its almost like he integrated v dv on the right hand side?
Any help would be appreciated.
2. jcsd
3. Nov 13, 2014 #2
ShayanJ
User Avatar
Gold Member
[itex]\frac{dv}{dt}=v\frac{dv}{dx}=\frac{1}{2}2v\frac{dv}{dx}=\frac{d}{dx}(\frac{1}{2} v^2)[/itex]
4. Nov 13, 2014 #3
Oh so you're really just reversing the chain rule
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-3,400,955,877,140,714,500 | Number 38500
[ thirty-eight thousand five hundred ]
Properties of number 38500
Cross Sum:
Factorization:
2 * 2 * 5 * 5 * 5 * 7 * 11
Divisors:
1, 2, 4, 5, 7, 10, 11, 14, 20, 22, 25, 28, 35, 44, 50, 55, 70, 77, 100, 110, 125, 140, 154, 175, 220, 250, 275, 308, 350, 385, 500, 550, 700, 770, 875, 1100, 1375, 1540, 1750, 1925, 2750, 3500, 3850, 5500, 7700, 9625, 19250, 38500
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
Base 32:
15j4
sin(38500)
0.21624785196742
cos(38500)
-0.97633849996785
tan(38500)
-0.2214886045921
ln(38500)
10.558413520276
lg(38500)
4.5854607295085
sqrt(38500)
196.21416870349
Square(38500)
Number Look Up
Look Up
38500 which is pronounced (thirty-eight thousand five hundred) is a very unique figure. The cross sum of 38500 is 16. If you factorisate the figure 38500 you will get these result 2 * 2 * 5 * 5 * 5 * 7 * 11. 38500 has 48 divisors ( 1, 2, 4, 5, 7, 10, 11, 14, 20, 22, 25, 28, 35, 44, 50, 55, 70, 77, 100, 110, 125, 140, 154, 175, 220, 250, 275, 308, 350, 385, 500, 550, 700, 770, 875, 1100, 1375, 1540, 1750, 1925, 2750, 3500, 3850, 5500, 7700, 9625, 19250, 38500 ) whith a sum of 104832. The figure 38500 is not a prime number. The figure 38500 is not a fibonacci number. 38500 is not a Bell Number. The number 38500 is not a Catalan Number. The convertion of 38500 to base 2 (Binary) is 1001011001100100. The convertion of 38500 to base 3 (Ternary) is 1221210221. The convertion of 38500 to base 4 (Quaternary) is 21121210. The convertion of 38500 to base 5 (Quintal) is 2213000. The convertion of 38500 to base 8 (Octal) is 113144. The convertion of 38500 to base 16 (Hexadecimal) is 9664. The convertion of 38500 to base 32 is 15j4. The sine of 38500 is 0.21624785196742. The cosine of the number 38500 is -0.97633849996785. The tangent of the number 38500 is -0.2214886045921. The root of 38500 is 196.21416870349.
If you square 38500 you will get the following result 1482250000. The natural logarithm of 38500 is 10.558413520276 and the decimal logarithm is 4.5854607295085. that 38500 is special figure! | {
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3,454,390,731,697,827,000 | NumPyの配列操作をすべて網羅(形状、分割、次元変更、形状変更)
配列の形状変更と操作
NumPy配列の形状を変更したり、複数の配列を結合したり、配列を分割することは、データ処理や分析において非常に重要です。このセクションでは、これらの操作を行う基本的な方法を学びます。
形状変更の基本（reshapeの使用）
reshapeメソッドを使用すると、配列の形状を変更できます。この操作は、元の配列の総要素数が変更後の形状の総要素数と一致する必要があります。
import numpy as np
# 1次元配列の作成
arr = np.arange(1, 10)
print(arr)
# 出力: [1 2 3 4 5 6 7 8 9]
# 1次元配列を3x3の2次元配列に変更
arr_reshaped = arr.reshape((3, 3))
print(arr_reshaped)
# 出力:
# [[1 2 3]
# [4 5 6]
# [7 8 9]]
配列の次元追加と削除
次元の追加(newaxis/expand_dims)
np.newaxisを使用すると、既存の配列に新しい軸（次元）を追加することができます。これは、配列の形状を変更する際に特に便利です。
# 1次元配列
arr = np.arange(4)
print(arr)
# 出力: [0 1 2 3]
# 新しい軸を追加して2次元配列に変更
arr_newaxis = arr[:, np.newaxis]
print(arr_newaxis)
# 出力:
# [[0]
# [1]
# [2]
# [3]]
expand_dimsを使用することで同じことを実現できます。
arr = np.array([0, 1, 2, 3])
expanded_arr = np.expand_dims(arr, axis=0)
print(expanded_arr)
# [[0]
# [1]
# [2]
# [3]]
次元の削除(squeeze)
np.squeeze関数を使用すると、配列から1の次元を削除（圧縮）することができます。これは、不要な次元を削除する際に便利です。
# 不要な次元の削除
arr_squeezed = np.squeeze(arr_newaxis)
print(arr_squeezed)
# 出力: [0 1 2 3]
これらの操作をマスターすることで、NumPy配列をより柔軟に扱うことができるようになります。データの前処理や分析の際に、これらの技術を活用して効率的に作業を進めましょう。
多次元配列を1次元配列へ変形(flatten/ravel)
多次元配列を1次元配列に変形する操作は、データ処理や分析の際によく必要とされます。NumPyでは、このような変形を簡単に行うためにflattenメソッドとravel関数を提供しています。
flattenメソッド
flattenメソッドは、多次元配列を1次元配列に”平滑化”します。このメソッドは元の配列のコピーを作成するため、flattenを使用した後に得られる配列を変更しても、元の配列は影響を受けません。
import numpy as np
# 2次元配列の作成
arr_2d = np.array([[1, 2, 3], [4, 5, 6]])
print("Original 2D array:")
print(arr_2d)
# 出力:
# Original 2D array:
# [[1 2 3]
# [4 5 6]]
# flattenを使用して1次元配列に変形
flattened_arr = arr_2d.flatten()
print("Flattened array:")
print(flattened_arr)
# 出力:
# Flattened array:
# [1 2 3 4 5 6]
ravel関数
ravel関数も配列を1次元に変形しますが、可能な限り元の配列のビューを返します。これはravelflattenよりも効率的に動作する可能性があることを意味しますが、返された配列を変更すると元の配列に影響を与える可能性があります（ビューの場合）。
# ravelを使用して1次元配列に変形
raveled_arr = arr_2d.ravel()
print("Raveled array:")
print(raveled_arr)
# 出力:
# Raveled array:
# [1 2 3 4 5 6]
flattenとravelの主な違いは、flattenが常にデータのコピーを作成するのに対し、ravelは可能な限りビューを返す点です。これにより、ravelはメモリ効率が良く、大きなデータセットでの操作においてパフォーマンスの向上が期待できます。しかし、返された配列を変更する際には、その変更が元の配列に影響を与える可能性があるため注意が必要です詳しくは次の記事を参考にしてください。
配列の結合と分割
配列の結合(concatenate / vstack / hstack)
NumPyでは、np.concatenatenp.vstack（垂直スタック）、np.hstack（水平スタック）などの関数を使用して、複数の配列を結合することができます。
# 配列の結合
arr1 = np.array([[1, 2], [3, 4]])
arr2 = np.array([[5, 6], [7, 8]])
# 垂直方向（行方向）に結合
arr_vstack = np.vstack((arr1, arr2))
print(arr_vstack)
# 出力:
# [[1 2]
# [3 4]
# [5 6]
# [7 8]]
# 水平方向（列方向）に結合
arr_hstack = np.hstack((arr1, arr2))
print(arr_hstack)
# 出力:
# [[1 2 5 6]
# [3 4 7 8]]
配列の分割(split / hsplit / vsplit)
np.splitnp.hsplit（水平分割）、np.vsplit（垂直分割）などの関数を使用して、配列を複数の小さな配列に分割することができます。
# 配列の垂直方向の分割
arr = np.arange(16).reshape((4, 4))
print(arr)
# 出力:
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]
# [12 13 14 15]]
# 2つの配列に垂直方向で分割
upper, lower = np.vsplit(arr, 2)
print(upper)
# 出力:
# [[0 1 2 3]
# [4 5 6 7]]
print(lower)
# 出力:
# [[ 8 9 10 11]
# [12 13 14 15]]
NumPy配列の形状変更チートシート
NumPy配列の形状変更はデータ分析や機械学習のプロジェクトにおいて頻繁に行われる操作です。ここでは、形状変更のための主要な関数とその使用例、および配列操作の際のコツとトラブルシューティングについて解説します。
形状変更のための主要な関数とその使用例
1. reshape: 配列の形状を指定した次元に変更します。
import numpy as np
arr = np.arange(6)
reshaped_arr = arr.reshape((2, 3))
1. flatten / ravel: 配列を1次元に変換します。flattenは新しい配列を返し、ravelは可能な場合はビューを返します。
flattened_arr = reshaped_arr.flatten()
raveled_arr = reshaped_arr.ravel()
1. squeeze: 配列から次元のサイズが1の軸を削除します。
arr = np.array([[[1, 2, 3]]])
squeezed_arr = np.squeeze(arr)
1. expand_dims / np.newaxis: 指定した位置に新しい軸（次元）を追加します。
arr = np.array([1, 2, 3])
expanded_arr = np.expand_dims(arr, axis=0) # np.newaxisを使う場合: arr[np.newaxis, :]
1. concatenate: 既存の配列を軸に沿って結合します。
arr1 = np.array([[1, 2], [3, 4]])
arr2 = np.array([[5, 6]])
concatenated_arr = np.concatenate([arr1, arr2], axis=0)
1. vstack / hstack: 配列を垂直（行方向）または水平（列方向）に積み重ねます。
vstacked_arr = np.vstack([arr1, arr2])
hstacked_arr = np.hstack([arr1, arr2.reshape(2, 1)])
1. split / hsplit / vsplit: 配列を複数の小さな配列に分割します。
arr = np.arange(9).reshape(3, 3)
# 3つの等しいサイズの配列に分割
split_arr = np.split(arr, 3)
配列操作のコツとトラブルシューティング
• 形状の確認: 操作前後でarr.shapeを使用して配列の形状を確認することは、予期しないエラーを避けるために重要です。
• ビューかコピーかを理解する: reshaperavelなどの操作はビューを返すことがありますが、flattenは常にコピーを返します。ビューを変更すると元の配列も変更されるため、意図しない結果を避けるためにこの違いを理解しておくことが重要です。
• エラーメッセージを注意深く読む: 形状変更時には、元の配列のサイズと変更後のサイズが一致しないとエラーが発生します。このような場合、エラーメッセージは問題の原因を特定するのに役立ちます。
• 自動形状推定: 一方の次元を-1に設定すると、NumPyが自動的に適切なサイズを計算します。これは、特定の次元のサイズを計算せずに形状を変更したい場合に便利です。
arr = np.arange(6)
reshaped_arr = arr.reshape(-1, 2) # 2列に自動で形状を変更
これらの関数とコツを駆使することで、NumPy配列の形状変更や操作を効率的に行うことができます。エラーに遭遇した場合は、配列の形状や操作の流れを一つ一つ確認して、問題の原因を特定しましょう。
応用例：NumPyを使ったデータ分析
NumPyはデータ分析において非常に強力なツールです。このセクションでは、実際のデータセットを使って、NumPy配列を活用したデータの前処理と簡単なデータ分析の実行例を紹介します。
実際のデータセットを使った例
ここでは、架空の人口統計データを例に取り上げます。このデータセットには、年齢と年収が含まれているとします。目的は、このデータを使って基本的な統計分析を行うことです。
import numpy as np
# 年齢と年収のデータセット
data = np.array([
[25, 35000],
[30, 40000],
[35, 50000],
[40, 60000],
[45, 80000],
[50, 90000],
[55, 85000],
[60, 75000],
[65, 65000],
[70, 55000]
])
# 列のインデックス
AGE = 0
INCOME = 1
NumPy配列を使ったデータの前処理
データ分析を始める前に、データの前処理を行うことが重要です。前処理には、データのクリーニング、正規化、欠損値の処理などが含まれます。ここでは、簡単な例として、年収データを正規化する手順を示します。
# 年収データの正規化
income_data = data[:, INCOME]
normalized_income = (income_data - income_data.min()) / (income_data.max() - income_data.min())
print("Normalized Income Data:")
print(normalized_income)
簡単なデータ分析の実行例
データの前処理が完了したら、基本的な統計分析を行います。ここでは、年齢と年収の平均値と中央値を計算し、年齢と年収の関係を探るための相関係数を求めます。
# 年齢と年収の平均値
average_age = np.mean(data[:, AGE])
average_income = np.mean(data[:, INCOME])
print(f"Average Age: {average_age}")
print(f"Average Income: {average_income}")
# 年齢と年収の中央値
median_age = np.median(data[:, AGE])
median_income = np.median(data[:, INCOME])
print(f"Median Age: {median_age}")
print(f"Median Income: {median_income}")
# 年齢と年収の相関係数
correlation = np.corrcoef(data[:, AGE], data[:, INCOME])[0, 1]
print(f"Correlation between Age and Income: {correlation}")
この例では、NumPyの基本的な機能を使って、データセットの基本的な統計量を計算し、年齢と年収の間にどのような関係があるかを探りました。NumPyはこれらの基本的な分析だけでなく、より複雑なデータ分析や機械学習アルゴリズムの実装にも使用されます。
データ分析プロジェクトでは、データの理解から始まり、前処理、探索的データ分析（EDA）、モデルの構築と評価、というステップを踏みます。NumPyはこれらのステップの多くで重要な役割を果たし、Pythonでのデータ分析作業を効率的に行うための基盤を提供します。
まとめ
この記事では、NumPyの基本的な使い方から配列操作の応用法、さらには実際のデータセットを使用したデータ分析の例までをカバーしました。NumPyはPythonでの科学計算の基盤となるライブラリであり、データ分析、機械学習、科学研究など幅広い分野で活用されています。
この記事の要点のまとめ
• NumPy配列の基礎: 1次元配列、2次元配列、多次元配列の作成方法と基本的な操作。
• 配列の形状変更と操作: reshapeflattenravelsqueezeexpand_dimsなどの関数を使用した配列の形状変更と次元操作。
• 配列の結合と分割: concatenatevstackhstacksplithsplitvsplitを使用した配列の結合と分割。
• データの前処理と分析: 実際のデータセットを使用した前処理と基本的なデータ分析の例。
コメント
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-910,414,805,128,723,700 | Statistical tests
• For the simple statistical techniques listed below, click to find out what the method is for, download a worked example or download instructions for applying the technique using SPSS, with a test data file to practice on.
• One-way analysis of variance (ANOVA) is a method for testing whether three or more populations have the same mean value. One-way ANOVA is used when quantitative data has been collected from 3 or more independent samples. The analysis involves two variables: the independent variable, which is a factor identifying individuals as belonging to one of three or more groups being compared and the dependent variable, a quantitative variable which is being compared between the three samples.
The null hypothesis states that all the populations have the same mean whereas the alternative hypothesis states that not all means are the same. The alternative hypothesis therefore states that ‘at least two means’ are different. The alternative hypothesis could be true if just one mean is out of line with the rest or if all of them are different to each other. At the conclusion of the test, if the null hypothesis is rejected, then further tests, called ‘post hoc tests’, may be used to determine which means differ significantly from each other.
Two-way or three-way analysis of variance extends this technique to analyses where two or more factors can influence the dependent variable. There are also procedures for applying ANOVA to data from related samples or ‘repeated measures’ data.
Analysis of variance is a PARAMETRIC method. It is based on the assumptions that the data is Normally distributed within each group and that the variance is the same within each group.
Example PDF »
The χ2 test for an association tests for:
A relationship between two categorical variables
or
A difference between groups in how members of the group answer a question which involves choosing a category.
The test is applied to data that can be arranged in a two-way table.
The chi-squared test is a NONPARAMETRIC TEST.
Example PDF » SPSS instructions PDF »
The Friedman test is used to compare data from three or more related samples. It is the non-parametric equivalent of a simple repeated measures analysis of variance (ANOVA). The dependent variable must be either ordinal or a quantitative variable that does not meet the assumptions for a parametric analysis.
The Friedman test is designed to test whether three or more populations have the same median values, using data collected from related samples.
Example PDF »
The Kruskal Wallis test is used to compare three or more independent samples. It is the non-parametric equivalent of a one-way analysis of variance (ANOVA).
It is used when analysing either ordinal data or a quantitative variable that does not meet the assumptions for a parametric test.
The Kruskal Wallis test tests whether three or more populations have the same median values.
Example PDF »
The Mann-Whitney U test is used to compare the medians of two independent samples.
The Mann-Whitney U test is the non-parametric equivalent of the two-sample (independent samples) t-test. The data must record the values of a variable which is either ordinal or a quantitative variable that does not meet the assumptions for a parametric test.
Example PDF » SPSS instructions PDF »
Post hoc tests are sometimes carried out to provide further detail after an analysis of variance. Typically an analysis of variance will have compared three or more groups to see whether their mean response is the same.
If the null hypothesis is rejected then the conclusion is that the mean response is not the same for all groups or treatments. This leads immediately to the question ‘which ones are different?’ and post hoc tests are used to answer this question.
Post hoc tests are designed to provide a way of deciding which pairs of means differ significantly from each other and which do not. There are many variations on post hoc tests, as it seems an area that several statisticians have produced a method for. The outcomes of post hoc tests can be anything ranging from:
• Only the lowest and highest means differ significantly.
• One mean is significantly higher than the rest, but the others do not differ significantly from each other.
• The means of all groups differ significantly from each other.
As it follows an analysis of variance, post hoc testing is a PARAMETRIC method, based on the same assumptions as the analysis of variance.
Example PDF »
Spearman’s correlation coefficient rho (ρ) is used to measure the correlation between two variables when the usual correlation coefficient is not suitable, for one of a number of reasons. Spearman’s rho is a nonparametric measure of correlation, calculated from the ranked data. Spearman’s correlation can be used to measure the association between two variables when:
• Both variables are quantitative but the scatter diagram shows either a non-linear relationship or a small number of points in ‘extreme’ positions that will lead to a value of the correlation coefficient, r, that could be misleading if interpreted in the usual way.
• One variable is quantitative but the other is ordinal
• Both variables are ordinal
The value of Spearman’s rho is between +1 and –1: and the sign and value of ρ are interpreted in the same way as the more conventional correlation coefficient, r. Hence ρ = 0 indicates no association between two variables, ρ = +1 indicates a ‘perfect’ positive association between variables ρ = -.5 indicates a moderate association between variables such that as one increases the other decreases.
Example PDF »
There are three kinds of t-tests:
1. single sample t-test is used to test whether the sample data is from a population whose mean has a certain value.
2. paired t-test is used to compare the means of two related samples.
3. An independent or two-sample t-test is used to compare the means of two independent samples.
The paired-samples and two-sample t-tests are both used to test whether two population means are equal. The paired-samples t-test is used when the data is from related, paired or longitudinal samples, that is, when both sets of measurements have been obtained from the same individuals. The independent-samples t-test is used when the data has been collected from independent samples, that is, when two sets of measurements have been obtained from different individuals.
T- tests are PARAMETRIC tests and are based on assumptions that need to be checked.
Paired example PDF » SPSS instructions PDF »
Two-sample example PDF » SPSS instructions PDF »
The Wilcoxon (Signed Ranks) test is used to compare the medians of two related samples. It is the non-parametric equivalent of the paired-samples t-test.
The data must record the values of a variable which is either ordinal or a quantitative variable that does not meet the assumptions for a parametric test.
Example PDF »
• Other topics | {
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-5,542,587,188,303,438,000 | What is 0.72 percent (calculated percentage %) of number 180?
What is 0.72% of 180?
0.71% of 180 = ? ... 0.73% of 180 = ?
1% of 180 = ?
0.72% × 180 =
(0.72 ÷ 100) × 180 =
(0.72 × 180) ÷ 100 =
129.6 ÷ 100 =
1.296 ≈
1.3;
Answer:
:: written in two ways ::
Not rounded:
0.72% of 180 = 1.296
Rounded to maximum 2 decimals:
0.72% of 180 ≈ 1.3
0.71% of 180 = ? ... 0.73% of 180 = ?
1% of 180 = ?
Signs: % percent, ÷ divide, × multiply, ≈ approximately equal;
Writing numbers: point '.' as a decimal mark;
Proof:
1.296 ÷ 180 =
0.0072 =
0.0072 × 100/100 =
0.0072 × 100% =
(0.0072 × 100)% =
0.72%
0.71% of 180 = ? ... 0.73% of 180 = ?
1% of 180 = ?
% Calculate percentages of given numbers or quantities
Latest calculated numbers percentages
0.72% of 180 = 1.296 Jul 20 20:31 UTC (GMT)
87.4% of 549,217 = 480,015.658 Jul 20 20:31 UTC (GMT)
30% of 2,167 = 650.1 Jul 20 20:31 UTC (GMT)
12% of 2,151 = 258.12 Jul 20 20:31 UTC (GMT)
0% of 1,455.22 = 0 Jul 20 20:31 UTC (GMT)
30% of 216 = 64.8 Jul 20 20:31 UTC (GMT)
16% of 5,150 = 824 Jul 20 20:31 UTC (GMT)
1% of 240 = 2.4 Jul 20 20:31 UTC (GMT)
0% of 1,284.02 = 0 Jul 20 20:31 UTC (GMT)
30% of 2,159 = 647.7 Jul 20 20:31 UTC (GMT)
30% of 215,000 = 64,500 Jul 20 20:31 UTC (GMT)
7.3% of 0.74 = 0.05402 Jul 20 20:31 UTC (GMT)
230% of 3.7 = 8.51 Jul 20 20:31 UTC (GMT)
All users calculated numbers percentages
Percent. Percentages.
Percent
• Percent is one hundredth of a given amount, is an amount calculated with reference to a hundred, hundredth, 1 ÷ 100 = 1/100, ie. a fraction with numerator equal to 1 and denominator to one hundred.
• The word comes from the Latin "per Centum", meaning "by the hundred". The latin word "Centum" means "100", for example a century is 100 years. So, "percent" means "per 100".
• Percent means 1/100, two percent means 2/100, three percent means 3/100, and so on.
• Although a fraction, the percent occurs in writing without the denominator (100), but only with the numerator, followed by the sign %: 1 percent = 1%, two percent = 2%, three percent = 3%, and so on.
Percentage
• Percentage is the "result of multiplying a quantity by a certain percent".
• So 10 percent (10 ÷ 100 = 10/100 = 10%) out of 50 apples is 5 apples (10% of 50 = 10/100 × 50 = 500/100 = 5) - the 5 apples is the percentage.
When do we say percent and when percentage?
• The word percent (or the symbol %) accompanies a specific number: around 60 percent (60%) of the people voted for a change.
• The more general word percentage is used without a number: the percentage of the people that voted for a change was around 60 percent (60%);
• Correct: 60 percent (60%) of the people; Incorrect: 60 percentage of the people;
• Correct: a higher percentage of voters; Incorrect: a higher percent of voters;
• Note: In front of the word "percent" use the number, don't spell it out: "60 percent" is right, "sixty percent" is rather not. Always use figures for ages of people (He's 43 years old), dates (May 26), monetary amounts ($80,000), percentages (60 percent) and ratios (2-to-3).
Percentages converted to decimal numbers:
• 1 percent, 1%, means 1 per 100, 1% = 1/100 = 0.01.
• 50 percent, 50%, means 50 per 100, 50% = 50/100 = 1/2 (half) = 0.5.
• 1 percent of 70 is: 1% of 70 = 1/100 × 70 = 7/10 = 0.7
• 50 percent of 70 is: 50% of 70 = 50/100 × 70 = 1/2 × 70 = 35. | {
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} | 0c090d63199a0a01e3b08e4a255778a0 |
5,825,256,587,747,588,000 | Explicit Differentiation
Calculus Handbook
Feel like "cheating" at Calculus? Check out our Practically Cheating Calculus Handbook, which gives you hundreds of easy-to-follow answers in a convenient e-book.
Derivatives >
What is Explicit Differentiation?
Explicit Differentiation is taking the derivative of an explicit function.
What is an “Explicit Function”?
In an explicit function, one variable is defined completely in terms of the other. This usually means that the independent variable (x) is written explicitly in terms of the dependent variable (y). The general form is: y = f(x). Note that “y” is on one side of the equals sign and “x” is on the other side.
Most of the functions you’re probably familiar with are explicit, like y = x2 or y = 2x + 3. When you first start in calculus, practically all of the functions you work with are going to be in this explicit form, and you’ll use the usual rules for differentiation.
What is “Implicit”?
The opposite of an explicit function is an implicit function, where the variables become a little more muddled. For example, the following equations are implicit:
• x2 + y2 = 1 (x and y are on one side of the equation)
• y*ey = x (two “y”s are on one side of the equation).
Explicit Differentiation vs. Implicit Differentiation
When you have a function that’s in a form like the above examples, it isn’t possible to use the usual rules of differentiation. When that’s the case, you have two choices:
1. Rewrite the equation so that one variable is on each side of the equals sign, then differentiate using the normal rules.
2. Use implicit differentiation.
Sometimes, the choice is fairly clear. For example, if you have the implicit function x + y = 2, you can easily rearrange it, using algebra, to become explicit: y = f(x) = -x + 2. In other cases, it might be easier to just use implicit differentiation.
Example
Let’s say you wanted to differentiate the implicit function x4 + 2y2 = 4.
1. Using explicit differentiation:
Rewrite, using algebra, so that you have one variable on each side of the equals sign:
explicit differentiation
This would give you two derivatives, one for positive values of y, and one for negative values of y.
2. Using implicit differentiation:
Instead of rewriting, you can just go ahead and plug the function in:
implicit differentiation
CITE THIS AS:
Stephanie Glen. "Explicit Differentiation" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/explicit-differentiation/
---------------------------------------------------------------------------
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In a certain school, 40 more than 1/3 of all the students
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In a certain school, 40 more than 1/3 of all the students [#permalink] New post 15 Jul 2004, 09:34
In a certain school, 40 more than 1/3 of all the students are taking a science course and 1/4 of those taking a science course are taking physics. If 1/8 of all the students in the school are taking physics, how many students are in the school?
A) 48
B) 120
C) 240
D) 480
E) 960
Please show the solution :-)
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[#permalink] New post 15 Jul 2004, 10:08
C. plug in the answers, or just do the algebra. Plugging in is fine for this one.
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[#permalink] New post 15 Jul 2004, 10:56
C) 240.
S = 40 + x*(1/3)
P=(1/4)*S = (1/8)X
Solve and get X=240.
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[#permalink] New post 15 Jul 2004, 11:11
Start with Science students
Assume X=all students
Assume Y=Science Students
40+1/3x=y
Now take Physics students
1/8x=1/4y ---> Y=1/2x
Put the two equestions together
40+1/3x=1/2x
80+2/3x=x
80=1/3x
240=x
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[#permalink] New post 17 Jul 2004, 10:08
" 1/4 of those taking a science course are taking physics "
Does it mean that there are some students studying both science and physics? 1/4 { 40 + X*(1/3) } is the number of student who study both science and physics ? Anyone think like me?
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[#permalink] New post 17 Jul 2004, 11:13
HI...
let the total no: of students in the class be X...
science course = X/3 +40 = X +120/3
physics course = X+120/3 *1/4
= X/8 of the total
12x- 8X =8*120
x = 240...
have Fun :)
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[#permalink] 17 Jul 2004, 11:13
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Number properties
Number 3278
three thousand two hundred seventy eight
Properties of the number 3278
Factorization 2 * 11 * 149
Divisors 1, 2, 11, 22, 149, 298, 1639, 3278
Count of divisors 8
Sum of divisors 5400
Previous integer 3277
Next integer 3279
Is prime? NO
Previous prime 3271
Next prime 3299
3278th prime 30319
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 110011001110
Octal 6316
Duodecimal 1a92
Hexadecimal cce
Square 10745284
Square root 57.253820833199
Natural logarithm 8.0949887593038
Decimal logarithm 3.5156089492345
Sine -0.96843212549142
Cosine -0.24927739230857
Tangent 3.8849577032346
Number 3278 is pronounced three thousand two hundred seventy eight. Number 3278 is a composite number. Factors of 3278 are 2 * 11 * 149. Number 3278 has 8 divisors: 1, 2, 11, 22, 149, 298, 1639, 3278. Sum of the divisors is 5400. Number 3278 is not a Fibonacci number. It is not a Bell number. Number 3278 is not a Catalan number. Number 3278 is not a regular number (Hamming number). It is a not factorial of any number. Number 3278 is a deficient number and therefore is not a perfect number. Binary numeral for number 3278 is 110011001110. Octal numeral is 6316. Duodecimal value is 1a92. Hexadecimal representation is cce. Square of the number 3278 is 10745284. Square root of the number 3278 is 57.253820833199. Natural logarithm of 3278 is 8.0949887593038 Decimal logarithm of the number 3278 is 3.5156089492345 Sine of 3278 is -0.96843212549142. Cosine of the number 3278 is -0.24927739230857. Tangent of the number 3278 is 3.8849577032346
Number properties
Examples: 3628800, 9876543211, 12586269025
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9,021,872,415,785,952,000 | Sorting (asymptotic) complexity problem
This is a discussion on Sorting (asymptotic) complexity problem within the C++ Programming forums, part of the General Programming Boards category; Hi all, I have some problems with "asymptotic notations". I have already learned the thery about that but there is ...
1. #1
Beginning game programmer Petike's Avatar
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Question Sorting (asymptotic) complexity problem
Hi all,
I have some problems with "asymptotic notations". I have already learned the thery about that but there is still something unclear for me "in practice".
Let me show you some simple example:
Let's say we have the "Selection" sorting algorithm. The fact is it has the time complexity of "O(n^2) (squared)". But what does it exactly mean?
Does it mean that at worst case this algorithm makes n^2 comparisons or n^2 selections or n^2 assignments or n^2 "something else"?
In other words, how can I simply and logically find out the complexity of some algorithm?
Thanks.
Petike
2. #2
and the hat of int overfl Salem's Avatar
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Just n^2 operations - it could be anything, or any scalar number of anything.
3 compares, a move and an addition would count as one "operation".
quicksort typically has more operations per iteration than say bubble sort. But because it's O(nLogn) rather than O(n^2), it soon starts winning over simpler (but more expensive) approaches.
Last edited by Salem; 01-19-2009 at 12:08 AM. Reason: fix late night typos
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3. #3
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The term O(n^2) means that if you double the number of elements the algorithm works on, it will be four times more work.
This is because these algoorithms usually has a double for-loop (or some such) in the middle of the algorithm, where both loops (essentially) range from 0 to n-1.
But the O(n^2) doesn't necessarily make a good comparison for the actual time it takes for one algorithm to solve a problem compared to another algorithm - they can vary dramatically even tho' they both have O(n^2) - one may be very dumb in the number of copies/compares or other operations it takes to do something. Of course, an algorithm that is O(log(n) * n) will (almost certainly) be dramatically quicker even if it is clumsier than some O(n^2) algorithm, if n is larger than a few dozen.
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4. #4
Beginning game programmer Petike's Avatar
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Thanks to both.
It is now much clearer to me.
Petike
5. #5
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one other thing to note about big Oh notation is that it doesn't really describe the exact amount of time increase new elements will have, instead it describes more the shape of the curve (if that makes any sense).
So for example, you won't see a O (n^3 + 4n^2 + 6n) algorithm, instead, it will be just written as O (n^3). That is one reason (along with others) that two n^3 algorithms with the same initial execution speeds will deviate from each other.
6. #6
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The reason for that is that once you start talking less significant factors in the running time such as an x-squared part of something that is x-cubed, or the constant factor of something that is O(n), you start running into things that are varying solely due to implementation differences, i.e. how well optimised the code is.
You might have two algorithms where you conclude that ones takes n*n + 3n + 9 operations, and another that is n*n + 4n + 5 operations, and decide that this is why the first one is faster. However you then switch compilers and the second one comes out faster for the exact same source code. Due to things like that, all you can really say is that since they are both O(n*n), they take approximately the same running time.
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7. #7
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I agree with what iMalc says. Another way to put it: consider O(n^2) compared with O(n^2 + n) when n is 1000 (1 million and 1 million one thousand) and then change to 10000 - It's now ten million for the first case, and ten million ten thousand for the second case. In the first case, it's 0.1%, in the second case the + n contributes less 0.01% - now, if you measure that precisely, you don't want O(n), you want "CPU clock-cycles" as the measure. And change your data ever so slightly (replace all Jones with Smith on the name list), and you will almost certainly change the overall execution time by more than 0.01%!
Big-O notation is for "the things that change dramatically", not the details. It's a "big picture" tool.
Consider a real-time OS where the task-switch is O(n), where n is number of tasks. Now change it the task-switch code so that it is O(1) - it doesn't take much to convince you that it is better - even if the ACTUAL task-switch code itself is 10% slower [assuming you have more than 1 task, that is].
Another real case is "open file" in FAT vs. NTFS - if you have a directory that is sorted by name, it take O(log(n)) to find the right file. FAT has files stored in the order they are created, so the time to find the right one is O(n) - because you may have to search through every one of them.
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8. #8
Beginning game programmer Petike's Avatar
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Question One another question...
Hi again,
I would have another question.
It should not be too hard to find out the best and the worst case of some algorithm.
But how can I find out the average case of some algorithm?
In other words, does there exist any general rule to finding out the average case of algoritms?
Thanks.
Petike
9. #9
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Just give it several sets of data that isn't the worst and isn't the best - or you can statistically calculate it from the general behaviour of the code - for example if your best case is an already sorted array, and the worst case is one that is sorted in opposite order, then something with a random set of data will be "average", right? So if "every other element needs swapping", you get "half as many operations each outer loop".
But often, worst case is the only one that really matters, because that is what you have to design to cope with, unless you know for some reason that you will never ever have the worst case.
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Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
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-3,751,809,631,433,601,000 | ГДЗ номер 1560 Виленкин 6 класс математика решебник
Виленкин Н.Я. математика 6
Условие задачи №1560
Выполните действие:
Решение задачи №1560
а) 0,38 • 2/19 = 38/100 • 2/19 = 1/25 = 0,04;
б) 3,16 : 4/7 = 3 16/100 : 4/7 = 316/100 • 7/4 = 553/100 = 5,53;
в) 3/8 — 0,48 = 3 • 0,125 — 0,48 = 0,375 — 0,48 = -0,105;
г) 0,169 : 13/14 = 169/1000 • 14/13 = 182/1000 = 0,182;
д) 13,13 : 1 2/11 = 13 13/100 : 1 2/13 = 121/100 = 1,21;
е) 232,3 : 33 2/3 = 232 3/10 : 22 2/3 = 2323/10 • 3/101 = 69/10 = 6,9.
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-478,501,778,131,568,640 | MPE Home Metamath Proof Explorer < Previous Next >
Nearby theorems
Mirrors > Home > MPE Home > Th. List > lspsnel6 Structured version Visualization version GIF version
Theorem lspsnel6 19206
Description: Relationship between a vector and the 1-dim (or 0-dim) subspace it generates. (Contributed by NM, 8-Aug-2014.) (Revised by Mario Carneiro, 8-Jan-2015.)
Hypotheses
Ref Expression
lspsnel5.v 𝑉 = (Base‘𝑊)
lspsnel5.s 𝑆 = (LSubSp‘𝑊)
lspsnel5.n 𝑁 = (LSpan‘𝑊)
lspsnel5.w (𝜑𝑊 ∈ LMod)
lspsnel5.a (𝜑𝑈𝑆)
Assertion
Ref Expression
lspsnel6 (𝜑 → (𝑋𝑈 ↔ (𝑋𝑉 ∧ (𝑁‘{𝑋}) ⊆ 𝑈)))
Proof of Theorem lspsnel6
StepHypRef Expression
1 lspsnel5.a . . . 4 (𝜑𝑈𝑆)
2 lspsnel5.v . . . . 5 𝑉 = (Base‘𝑊)
3 lspsnel5.s . . . . 5 𝑆 = (LSubSp‘𝑊)
42, 3lssel 19147 . . . 4 ((𝑈𝑆𝑋𝑈) → 𝑋𝑉)
51, 4sylan 561 . . 3 ((𝜑𝑋𝑈) → 𝑋𝑉)
6 lspsnel5.w . . . . 5 (𝜑𝑊 ∈ LMod)
76adantr 466 . . . 4 ((𝜑𝑋𝑈) → 𝑊 ∈ LMod)
81adantr 466 . . . 4 ((𝜑𝑋𝑈) → 𝑈𝑆)
9 simpr 471 . . . 4 ((𝜑𝑋𝑈) → 𝑋𝑈)
10 lspsnel5.n . . . . 5 𝑁 = (LSpan‘𝑊)
113, 10lspsnss 19202 . . . 4 ((𝑊 ∈ LMod ∧ 𝑈𝑆𝑋𝑈) → (𝑁‘{𝑋}) ⊆ 𝑈)
127, 8, 9, 11syl3anc 1475 . . 3 ((𝜑𝑋𝑈) → (𝑁‘{𝑋}) ⊆ 𝑈)
135, 12jca 495 . 2 ((𝜑𝑋𝑈) → (𝑋𝑉 ∧ (𝑁‘{𝑋}) ⊆ 𝑈))
142, 10lspsnid 19205 . . . . 5 ((𝑊 ∈ LMod ∧ 𝑋𝑉) → 𝑋 ∈ (𝑁‘{𝑋}))
156, 14sylan 561 . . . 4 ((𝜑𝑋𝑉) → 𝑋 ∈ (𝑁‘{𝑋}))
16 ssel 3744 . . . 4 ((𝑁‘{𝑋}) ⊆ 𝑈 → (𝑋 ∈ (𝑁‘{𝑋}) → 𝑋𝑈))
1715, 16syl5com 31 . . 3 ((𝜑𝑋𝑉) → ((𝑁‘{𝑋}) ⊆ 𝑈𝑋𝑈))
1817impr 442 . 2 ((𝜑 ∧ (𝑋𝑉 ∧ (𝑁‘{𝑋}) ⊆ 𝑈)) → 𝑋𝑈)
1913, 18impbida 794 1 (𝜑 → (𝑋𝑈 ↔ (𝑋𝑉 ∧ (𝑁‘{𝑋}) ⊆ 𝑈)))
Colors of variables: wff setvar class
Syntax hints: wi 4 wb 196 wa 382 = wceq 1630 wcel 2144 wss 3721 {csn 4314 cfv 6031 Basecbs 16063 LModclmod 19072 LSubSpclss 19141 LSpanclspn 19183
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1869 ax-4 1884 ax-5 1990 ax-6 2056 ax-7 2092 ax-8 2146 ax-9 2153 ax-10 2173 ax-11 2189 ax-12 2202 ax-13 2407 ax-ext 2750 ax-rep 4902 ax-sep 4912 ax-nul 4920 ax-pow 4971 ax-pr 5034
This theorem depends on definitions: df-bi 197 df-an 383 df-or 827 df-3an 1072 df-tru 1633 df-ex 1852 df-nf 1857 df-sb 2049 df-eu 2621 df-mo 2622 df-clab 2757 df-cleq 2763 df-clel 2766 df-nfc 2901 df-ne 2943 df-ral 3065 df-rex 3066 df-reu 3067 df-rmo 3068 df-rab 3069 df-v 3351 df-sbc 3586 df-csb 3681 df-dif 3724 df-un 3726 df-in 3728 df-ss 3735 df-nul 4062 df-if 4224 df-pw 4297 df-sn 4315 df-pr 4317 df-op 4321 df-uni 4573 df-int 4610 df-iun 4654 df-br 4785 df-opab 4845 df-mpt 4862 df-id 5157 df-xp 5255 df-rel 5256 df-cnv 5257 df-co 5258 df-dm 5259 df-rn 5260 df-res 5261 df-ima 5262 df-iota 5994 df-fun 6033 df-fn 6034 df-f 6035 df-f1 6036 df-fo 6037 df-f1o 6038 df-fv 6039 df-riota 6753 df-ov 6795 df-0g 16309 df-mgm 17449 df-sgrp 17491 df-mnd 17502 df-grp 17632 df-lmod 19074 df-lss 19142 df-lsp 19184
This theorem is referenced by: lspsnel5 19207 lsmelval2 19297 dihjat1lem 37231
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8,756,164,180,133,542,000 | Back to results
Mathematical problem-solving and chess
Four soccer players on a pitch and one soccer ball
It’s not a mistake. Why is a story about chess illustrated with an image about soccer? Because both involve mathematical problem-solving
Playing a competitive game is all about solving problems – which involves a surprising amount of maths. Lanella Sweet explains why playing chess builds students’ maths capabilities.
The score is 2-2 in the dying seconds of the Girls’ Firsts Soccer match, and Wesley’s star forward Matilda is fouled inside the box and awarded a penalty. She knows she’s more accurate shooting to her right – but, she reasons, the goalkeeper is surely aware of this as well. What are the probabilities – or the payoff – if Matilda shoots right, with the goalkeeper covering that direction? She, and the goalkeeper, have to figure out the optimal tactic.
Matilda and her teammates have a quick discussion, a mixture of competitive passion, problem solving and not a little mathematical thinking. Matilda, who also plays competitive chess, decides to shoot left. Her calculation? Her chance of scoring is greater.
The benefits of chess
There’s plenty of mathematical calculation in most competitive games, not least in chess. While researchers continue to debate the transfer of learning from chess to maths, studies have found that learning chess has a positive impact on students’ ability to tackle novel problem-solving tasks and even maths tasks. Farhad Kazemia and colleagues found that chess significantly improves students’ mathematical problem-solving ability, while Giovanni Sala and colleagues found that even short-term chess practice enhances students’ mathematical abilities.
Students who experience thinking through the many facets of a chess game use a variety of critical thinking tools along the way. Chess requires students to collect data on the different positions on the chess board during a game and analyse the many alternatives for their next move, anticipating the probable consequences of their move, including the probable responses of their opponent – essentially planning ahead. Chess also offers a context in which students can develop their meta-cognitive ability to ‘think about thinking’ and use this to examine options and analyse their consequences.
Wesley’s chess program
Chess taught as a ‘subject’ one period a week ignites student interest and general foundational teaching of the rules and how pieces move on a chess board is beneficial. Teaching the game essentials is a starting point but doesn’t develop students’ complex real problem-solving techniques. As Sala and colleagues observe, understanding the basic rules of chess is not sufficient to develop students’ cognitive skills. It’s the more complex approach to the game that does this, by exposing students’ to opportunities to analyse and abstract in a competitive situation where the better their analysis, the more likely they are to play well. Chess, somewhat like video games, encourages players to learn ‘what works’ and rewards them when they find solutions, as Steven Coshutt observes in ‘Video games, social media and the brave new world of online learning.’ [article yet to be published]
The more students practise chess, collecting and mentally sorting complex data, the more they develop their ability to focus, analyse and choose between multiple options. While the skills and attributes students develop by playing chess transfer across the curriculum, Sala and colleagues note that they transfer particularly strongly to mathematics problem solving.
Attitudes
Students who play chess develop not only important skills that can be applied to other curriculum areas like mathematics and problem solving, but also important attitudes. When students are able to think about thinking they develop positive attitudes about themselves and their abilities – described in education literature as self-efficacy.
This self-efficacy can be seen particularly in the calm approach chess-playing students take to problem solving. This may be in part because they have played in multiple chess competitions and tournaments, but it’s also because they have developed the skill of using the chess clock appropriately to analyse data and consider their options. Unlike many maths ‘games’ that are really just computation sessions that reward – or punish – students on the basis of their speed of recall in identifying correct answers, chess creates genuine, often novel, problems and values the process of thinking, regardless of whether the player wins or loses a particular game. In fact, according to Cathy Seeley in Faster isn’t Smarter, fast number fact recall at the expense of problem solving and conceptual experiences often gives students a distorted idea not just of mathematics but also their ability to do mathematics. In chess, students learn to take ‘enough’ time before they make their next move, ‘enough’ being that amount of time necessary to have considered the data and the options.
Problem-solving approaches
The skills and attitudes of chess players prompted me to observe the problem-solving approaches to a mathematical problem by chess-playing and non-chess-playing students, and I found that Seeley is right: faster isn’t necessarily smarter. I noticed that the chess-playing students were more likely to:
• expect they could understand the problem
• fully interrogate and comprehend the problem to be solved
• sort through data to identify information necessary to solving the problem
• analyse abstract information
• identify whether further information or clarification of supplied information was necessary, and
• undertake these steps before tackling the actual problem.
The non-chess-playing students, in contrast, were more likely to:
• respond quickly, often without fully understanding the problem
• form a superficial understanding of the information necessary to solving the problem
• identify possible solutions before fully understanding the problem to be solved, and
• make inadequate estimations that their solution effectively solved the problem.
It was evident that preparation, analysing, and identifying the key information relevant to the problem were fundamental strategies for the chess-playing problem solvers.
Students who have more than foundational experience in playing chess and extensive experience in managing themselves in chess competitions and tournaments seem to approach problem-solving tasks using skills and attitudes that enable them to fully examine and solve mathematical problems. Not only this, the stronger the chess-playing skills of the students, the better they seem to be at understanding and solving mathematical problems.
And what of the final score in that Girls’ Firsts Soccer match? Wesley won 3-2. Checkmate.
Lanella Sweet is an enrichment teacher at Wesley’s Elsternwick Campus.
Elsternwick Junior School students take a chess class for one period a week for one term. Students can also join Elsternwick’s Chess Club for after-school coaching, as part of the Community College program.
Interested in reading more on maths in sport? Try ‘When to take the two’ on optimal tactics in Rugby League; or ‘What is the absolute limit for human athletes?’ and ‘What it will take to run a sub-two-hour marathon?’ on the maths and science of improvements in athletic performance. | {
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7,464,955,741,400,161,000 | Factorial Simplification
Time Limit: 10000MS
Memory Limit: 65536K
Description
Peter is working on a combinatorial problem. He has carried out quite lengthy derivations and got a resulting formula that is a ratio of two products of factorials like this:
\frac{p_1!p_2!...p_n!}{q_1!q_2!...q_m!}
This does not surprise Peter, since factorials appear quite often in various combinatorial formulae, because n! represents the number of transpositions of n elements — one of the basic combinatorial objects. However, Peter might have made a mistake in his derivations. He knows that the result should be an integer number and he needs to check this first. For an integer result Peter wants to simplify this formula to get a better feeling of its actual combinatorial significance. He wants to represent the same number as a product of factorials like this.
r_1!^{s_1}r_2!^{s_2}...r_k!^{s_k}t
where all ri are distinct integer numbers greater than one in the descending order (ri > ri+1 > 1), si and t are positive integers. Among all the possible representations in this form, Peter is interested in one where r1 is the largest possible number, among those in the one where s1 is the largest possible number; among those in the one where r2 is the largest possible number; among those in the one where s2 is the largest possible number; etc, until the remaining t cannot be further represented in this form. Peter does not care about the actual value of t. He wants to know what is the factorial-product part of his result.
Input
The first line of the input file contains two integer numbers n and m (1 <= n,m <= 1000). The second line of the input file contains n integer numbers pi (1 <= pi <= 10 000) separated by spaces. The third line of the input file contains m integer numbers qi (1 <= qi <= 10 000) separated by spaces.
Output
On the first line of the output write a single integer number k. Write k = -1 if the ratio of the given factorial products is not an integer. Write k = 0 if the ratio is an integer but it cannot be represented in the desired form. Write k > 0 followed by k lines if the ratio can be represented by a factorial product as described in the problem statement. On each of the following k lines write two integers ri and si (for i = 1 ... k) separated by a space.
Sample Input
#1 1 2
6
4 4
#2 1 2
6
3 4
#3 4 2
9 2 2 2
3 4
Sample Output
#1 -1
#2 0
#3 2
7 1
2 2
Hint
Source
Northeastern Europe 2010
提交代码 | {
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7,851,640,181,963,102,000 |
Results 1 to 5 of 5
Thread: Differentiable, continuous function... "show that"
1. #1
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Differentiable, continuous function... "show that"
If f is continuous on [0,1] and differentiable at a point x \in [0,1], show that, for some pair m,n \in \mathbb{N},
\left | \frac{f(t)-f(x)}{t-x}\right | \leq n whenever 0 \leq |t-x| \leq \frac{1}{m}
Since it's differentiable at x I know \lim_{t \to x}\frac{f(t)-f(x)}{t-x} exists...
And it's continuous so I know for that for any \epsilon > 0 there's a \delta > 0 so that |f(t)-f(x)| < \epsilon when |t-x| < \delta
Not sure how to put it all together...
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2. #2
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Try using the mean value theorem instead.
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3. #3
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Quote Originally Posted by HallsofIvy View Post
Try using the mean value theorem instead.
Wouldn't that require the function to be differentiable on (0,1) ? I only have differentiability at a single point in [0,1]...
I don't follow :S
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4. #4
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Oops, right, I didn't see that! Sorry. I guess you were right to use the definition of "differentiable at a point".
Let n be any integer strictly larger than f'(x)+1. Certainly there exist [itex]\delta> 0[/itex] such that \frac{|f(t)- f(x)||}{|t-x|}< f'(x)+ 1< n And, since \frac{1}{\delta} is a real number, there exist [tex]m> \frac{1}{\delta}[/itex]. If |t- x|< \frac{1}{m}< \delta then |\frac{f(t)- f(x)}{x-t}|< n.
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5. #5
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Quote Originally Posted by HallsofIvy View Post
Certainly there exist \delta> 0 such that \frac{|f(t)- f(x)|}{|t-x|}< f'(x)+ 1< n
when |t-x| < delta?
I don't get this bit
Thanks for your help, I'm almost there
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-3,554,082,898,517,134,000 | Kalkulator cotangens, arcus cotangens, cotangens hiperboliczny i area cotangens. - Calcoolator.pl - Kalkulator online.
Kalkulator cotangens, arcus cotangens, cotangens hiperboliczny i area cotangens.
Dzięki kalkulatorowi funkcji trygonometrycznej COTANGENS obliczysz wartości dowolnej funkcji cotagensa. Oprócz wyników w odpowiedzi kalkulator narysuje również wykres wybranej funkcji. Możesz wybrać jedną z gotowych funkcji np. cotagens, cot2 - cotagens kwadrat, arccot - arcus cotagens, coth - cotagens hiperboliczny, arcoth - funkcja odwrotna do coth, możesz też wprowadzić własną funkcję np. cot(x)*cot(x)*cot(x) dla cot3(x), cot(2x), cot(x+3), cot(x^2) itp.
Do obliczeń możesz wykorzystywać:
cyfry 0-9, np. 123,45. Bardzo duże liczby mogą być zapisane jako 2.5E20 dla 2.5*1020, bardzo małe liczby mogą być zapisane w postaci 3E-10 dla 3*10-10. Miejsca dziesiętne mogą zawierać do 12 cyfr bo przecinku;
możesz używać (. ,) kropkę lub przecinek jako separator liczb dziesiętnych np. 1.5 lub 1,5;
nawiasy ( ) [ ] { } < > np. {[(1+x)/(2-x)+1]*3}/(2*x^2) , mogą być używane w dowolnej ilości. Każdy otwarty nawias musi zostać zamknięty. Nie ma znaczenia jaki rodzaj nawiasów wybierzesz;
# jako separator wielu wartości wyjściowych np. cot(pow(x#2));
+ Plus, np. cot(x+1);
- Minus, np. cot(1-x);
* Razy, znak mnożenia może być pominięty gdy znajduje się pomiędzy literą a liczbą np. możesz zapisać cot(2x) zamiast cot(2*x), ale nie wolno zapisać xcot(x) lub cot(ex);
/ : Dzielenie. 1/x lub 1:x;
e liczba Eulera: 2.718281828459;
pi π, Pi: 3.1415926535898;
pi2 π/2, Pi/2: 1.5707963267949;
sq2 pierwiastek kwadratowy z 2: 1.4142135623731;
go relacja złotej proporcji: 1.6180339887499;
d stała Feigenbauma - delta: 4.6692016091030;
^ lub pow Potęga, np. cot(x^2) lub cot(pow(x#2)) dla cot(x2). Pierwiastek może być zapisany jako np. x^(1/2) lub x^0,5 dla pierwiastka kwadratowego z x, dla funkcji wykładniczej jak ta: ex zapis może mieć postać e^x;
Pierwiastek wartości ujemnej może być zapisany jedynie gdy licznik potęgi wynosi 1 a mianownik jest inny (np. x^(1/3) ). Aby obliczyć ujemną wartość x dla np. x^(2/3) , trzeba zapisać tą funkcję w postaci (x^(1/3))^2;
sqr Pierwiastek kwadratowy np. cot(sqr(x)) jest tym samym co cot(x^(1/2));
exp Wykładnik, np. cot(exp(x)) jest tym samym co cot(e^x);
log Logarytm naturalny, np. cot(log(x));
log10 Logarytm dziesiętny, np. cot(log10(x));
logn Logarytm o podstawie n, np. cot(logn(2#x)) dla logarytmu binarnego;
Użyj również:
deg konwertuje liczbę w radianach na odpowiadającą w stopniach, np. cot(deg(pi))
rad Konwertuje stopnie na liczbę w radianach, np. cot(rad(180))
Aby uzyskać wynik wybierz odpowiednią funkcję lub wpisz własną oraz w polu Wartości wprowadź wartość lub wartości oddzielone spacją (np.: 2 5 8 10 ).
Funkcje trygonometryczne - COTANGENS
cot(x) - cotagens
cot2(x) - cotagens kwadrat
arccot(x) - arcus cotagens
coth(x) - cotagens hiperboliczny
arcoth(x) - area cotagens hiperboliczny
Własna funkcja
Wartości
Przykładowe wartości:
Wynik w formacie:
Tylko wynik Tabela Format CSV
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CATENARY Kalkulator - Krzywa łańcuchowa
Dzięki kalkulatorowi funkcji trygonometrycznej CATENARY obliczysz wartości krzywej łańcuchowej catenary. Oprócz wyników w odpowiedzi kalkulator narysuje również wykres wybranej funkcji. Możesz wybrać gotową funkcję np. cat(?#x) gdzie za znak ? podstawisz np. 2 dla catenary dwa co równe będzie 2*cosh(x/2)lub wprowadzić własną funkcję np. cat(2#x^2) dla 2*cosh((x^2)/2), cat(2#x+3) dla 2*cosh((x+3)/2) itp.
Kalkulator spadku napięcia z prądu dla obwodów jednofazowych i trójfazowych
Dzięki kalkulatorowi obliczysz spadki napięć dla obwodów jednofazowych i trójfazowych prądu przemiennego wyliczonych z prądu znamionowego.
Obliczysz również długość przewodu, średnicę przewodu, pole przekroju przewodu, napięcie lub prąd.
COSINUS Kalkulator
Dzięki kalkulatorowi funkcji trygonometrycznej COSINUS obliczysz wartości dowolnej funkcji cosinusa. Oprócz wyników w odpowiedzi kalkulator narysuje również wykres wybranej funkcji. Możesz wybrać jedną z gotowych funkcji np. cosinus, cos2 - cosinus kwadrat, arccos - arcus cosinus, cosh - cosinus hiperboliczny, arcosh - funkcja odwrotna do cosh, możesz też wprowadzić własną funkcję np. cos(x)*cos(x)*cos(x) dla cos3(x), cos(2x), cos(x+3), cos(x^2) itp.
Układ równań 3 niew.
Kalkulator oblicza wartości x, y i z za pomocą metody wyznaczników.
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Dzięki temu programowi do rysowania wykresów funkcji online możesz narysować dowolną funkcję. Na wykresie możliwe jest umieszczenie aż trzech funkcji. Do większość równań i obliczeń zawartych na tej stronie możesz sporządzić wykres przy pomocy tego narzędzia. Narzędzie rysuje:
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- średnie statystyczne (Średnia arytmetyczna, Średnia geometryczna, Średnia harmoniczna, Średnia kwadratowa,) ,
- rozkłady dyskretne (Rozkład dwumianowy, Rozkład Poissona, Rozkład geometryczny, Rozkład logarytmiczny, Rozkład równomierny,...),
- liczby stałe (e, pi, relacja złotej proporcji, stała Feigenbauma, ...),
- krzywe (Krzywa dzwonowa Gaussa, Krzywa trójkątna, Krzywa kwadratowa, Krzywa półelipsy, Krzywa serpentynowa,...),
- podstawowe operacje arytmetyczne,
- wielomiany, itd.
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-6,374,577,950,918,904,000 | Combinatória
Combinatória é a parte da matemática que se preocupa em agrupar e contar coleções de objetos, seguindo certos critérios de contagem.
Tais contagens podem ser feitas de duas maneiras: princípio aditivo e princípio multiplicativo.
Princípio aditivo
O princípio aditivo ou princípio da Inclusão-Exclusão é uma forma de contagem do número de elementos que pertencem à união de dois ou mais conjuntos não necessariamente disjuntos.
Para dois conjuntos A e B, o número de elementos dessa união é dado por:
n(A união B) = n(A) + n(B) – n(A intersecção B)
Para três conjuntos A, B e C tem-se que o número da união é dado por:
n(A união B união C) = n(A) + n(B) + n(C) – n(A intersecção B) – n(A intersecção C) – n(B intersecção C) + n(A intersecção B intersecção C).
Princípio multiplicativo
O princípio multiplicativo ou princípio fundamental da contagem diz que se há x modos de se tomar uma decisão A e, tomada essa decisão há y modos de se tomar uma decisão B, então o número de modos de se pode tomar sucessivamente as decisões A e B é x . y.
Considerando x, y, z modos para, respectivamente, três decisões, tem-se x . y . z.
De uma forma geral, tendo-se x1 maneiras de se tomar a decisão 1, x2 maneiras de se tomar a decisão 2, . . . , xn maneiras de se tomar a decisão n, então se terá ao todo:
x1 . x2 . . . xn possibilidades.
Exemplo:
— Uma pessoa tem em seu guarda-roupa 3 calças, 6 blusas e 2 pares de sapatos, todos diferentes. De quantas maneiras distintas ela poderia se vestir usando uma peça de cada?
Como ela deve usar uma calça, uma blusa e um par de sapato, então o número total é dado por: 3 . 6 . 2 = 36.
Fatorial (!)
O fatorial de um número natural n, representado por n! (lê-se: “n fatorial ou fatorial de n”), é igual ao produto sucessivo desse número pelos seus antecessores até a unidade.
O fatorial de n é dado por:
n! = n.(n – 1).(n – 2).(n – 3). . .3.2.1
Assim, 5! = 5 . 4 . 3 . 2 . 1 = 120.
Por convenção, 0! = 1 e 1! = 1.
É fácil notar que 6! = 6 . 5 . 4 . 3 . 2 . 1 = 6 . 5! = 6 . 5. 4 . 3 . 2 . 1 = 6 . 5 . 4! = 6 . 5 . 4 . 3 . 2 . 1 = 6 . 5 . 4 . 3!
Então, se quando se está multiplicando pelos antecessores do número, parar antes do número 1, completa-se com o símbolo do fatorial.
Daí, o fatorial de n, pode ser escrito, por exemplo, como n! = n.(n – 1).(n – 2).(n – 3)! ou n! = n.(n – 1)!
Isto é muito útil para simplificar algumas expressões.
Exemplos:
6! / 4! = 6.5.4! / 4! = 6 . 5 = 30.
(7! + 6!) / 5! = 7.6.5! - 6! / %! = 5!(7.6 + 6) / 5! = 7 . 6 + 6 = 42 + 6 = 48.
Análise Combinatória
Permutações Simples
Dado um conjunto com n elementos no qual se deseja ordená-los, o número total de agrupamentos que pode ser feito é igual a:
n(n − 1)(n − 2) . . . 1, pois para a primeira posição pode-se ter n maneiras, para a segunda posição pode-se ter n − 1 maneiras; a terceira posição pode-se ter n − 2 maneiras, e assim sucessivamente até a última posição que só terá uma maneira de se escolher.
Portanto, o número de ordens em que se pode colocar n objetos distintos é n(n − 1)(n − 2) . . . 1 = n!
Chama-se permutação simples de n elementos distintos aos agrupamentos dos n elementos de modo que cada agrupamento difere do outro apenas pela ordem de seus elementos.
O número de permutação simples é dado por: Pn = n!
Exemplo:
— Com os dígitos 1, 2, 3 e 5, quantos números de quatro algarismos distintos podem ser formados?
Algarismos distintos quer dizer diferentes; qualquer exemplo que se faça tem exatamente 4 elementos e, por exemplo, 3215 e 3251 diferem apenas pela ordem dos elementos.
Logo, trata-se de uma permutação simples de 4 elementos (os números 1, 2, 3 ou 5) e, portanto:
P4 = 4! = 4 . 3 . 2 . 1 = 24.
Arranjos Simples
Seja um conjunto com n elementos dos quais, se deseja formar agrupamentos de p elementos distintos, com p menor ou igual n, onde cada agrupamento difere do outro pela natureza ou pela ordem de seus elementos.
Neste caso se tem um arranjo simples de n elementos tomados p a p.
O número de arranjos simples é dado por: arranjo = An, p = n! / (n - p)
Exemplo:
— Com os dígitos 1, 2, 3, 4, 5 e 6, quantos números de quatro algarismos distintos podem ser formados?
Qualquer exemplo usará apenas quatro dos seis elementos. Tanto 2345 é diferente de 2354 (pela ordem) como 2345 é diferente de 2346 (pela natureza dos elementos, pois, neste caso, não foram usados os mesmos elementos) e, portanto, trata-se de um arranjo simples (pois cada exemplo usa algarismos distintos) de 6 elementos tomados 4 a 4.
A6, 4 = 6! / (6 - 4)! = 6.5.4.3.2! / 2 = 6 . 5 . 4 . 3 = 360.
Combinação Simples
Seja um conjunto com n elementos dos quais, se deseja formar agrupamentos de p elementos distintos, com p menor ou igual n, onde cada agrupamento difere do outro apenas pela natureza de seus elementos.
Neste caso se tem uma combinação simples de n elementos tomados p a p.
O número dessas combinações simples é dado por: arranjo = Cn,p = Cn, p = n! / p!(n - p)
Observações:
Cn,p = Cn,n-p ex.: C7,4 = C7,7-4 = C7,3
Cn,0 = Cn,n = 1 Cn,1 = n, qualquer que seja o natural n maior ou igual 1.
Exemplo:
— Quantos subconjuntos com 2 elementos possui um conjunto com 5 elementos?
Considerando os elementos a, b, c, d, f, um conjunto com dois seria, por exemplo, {a, b} e ele logicamente é diferente de, por exemplo, de {a, c}, pela natureza de seus elementos, mas os conjuntos {a, b} e {b, a} não são diferentes, mas sim o mesmo conjunto.
Então, trata-se de uma combinação simples de 5 elementos tomados 2 a 2.
C5, 2 = 5!/2!(5-2)! = 5.4.3! / 2!3! = 5.4 / 2! = 20 / 2 = 10.
Permutação Circular ou Cíclica
Dado um conjunto com n elementos onde se deseja ordená-los de maneira que o primeiro e o último se encontrem, isto é, tenham a forma.
A-B-C-D
Neste caso se tem uma permutação circular que é diferente das permutações simples, pelo fato de que rodando os elementos não se tem outro agrupamento, pois, por exemplo, com os elementos A, B, C e D, em círculo, tem-se que ABCD, BCDA, CDAB e DABC são todos iguais e, portanto, correspondem a um único agrupamento.
O número de permutações cíclicas é dado por: PCn = (n – 1)!
Exemplo:
— Quantas rodas de ciranda podem ser formadas com 5 crianças?
Como a primeira e a última estarão se encontrando em circulo tem-se uma permutação circular dos 5 elementos, isto é, PC5 = (5 – 1)! = 4! = 24.
Permutação com Repetição
Supondo que se tem n elementos para permutar, sendo que q1 desses elementos são de um mesmo tipo, q2 de outro tipo, q3 de outro tipo, e assim por diante, onde q1 + q2 + . . . + qp menor ou igualn.
Neste caso se tem um permutação com repetição de n elementos onde se tem q1 iguais a tipo, q2 de outro tipo, . . .
O número de permutações com repetição é dado por: PRn,q1,q2,qp = n! / q1!q2!...qp!
Exemplo:
— Quantos são os anagramas da palavra ARARAS?
Anagrama é a junção de letras tendo significado ou não (se fizer sentido é uma palavra), neste caso há seis elementos que serão permutados, porém 3 são de um mesmo tipo (letra A) e 2 de outro tipo (letra R), portanto tem-se uma permutação com repetição de 6 elementos com 3 e 2 tipos de repetições, logo há: PR6_3,2 = 6! / 3!2! = 6.5.4.3! / 3!2! = 6 . 5 . 4 / 2 = 120 / 2 = 60.
Arranjo com Repetição
Seja um conjunto com n elementos e sendo p um número inteiro positivo menor ou igual a n, chama-se arranjo com repetição dos n elementos tomados p a p, a qualquer sequência de p elementos, onde pode haver repetições de elementos e sendo p o número máximo de repetições.
O número desses arranjos é dado por: ARn,p = np (n elevado a p).
Exemplo:
Com os dígitos 1, 2, 4, 5, 7, e 9; quantos números de 3 algarismos podem ser formados?
Como não foi dito que os três algarismo devem ser distintos, então, por exemplo, 242 pode ser um desses números, e um número difere do outro tanto pela natureza com pela ordem de seus elementos, então se trata de um arranjo com repetição de 6 elementos 3 a 3.
AR6,3 = 63 = 6 . 6 . 6 = 216.
Combinação com Repetição
Seja um conjunto com n elementos dos quais, se deseja formar agrupamentos de p elementos não necessariamente distintos, onde cada agrupamento difere do outro apenas pela natureza de seus elemento, então se tem uma combinação com repetição de n elementos tomados p a p.
O número dessas combinações é dado por: CRnp = CRn,p = CRn,p = Cn + p – 1,p
Exemplo:
— Quantas são as soluções inteiras e não-negativas da equação x + y + z = 4?
De uma forma geral, pode-se considerar a equação como x1 + x2 + . . . + xn = p.
Assim, n é o número de incógnitas e p o resultado da soma, então tem-se:
CR3,4 = C3 + 4 – 1,4 = C6,4 = C6,2 = 6! / 2!(6 - 2)! = 6.5.4! / 2!4! = 6 . 5 / 2 = 30 / 2 = 15.
Se a questão fosse: quantas são as soluções inteiras positivas da equação x + y + z = 4?
Se tomaria x = a + 1, y = b + 1, z = c + 1, pois não poderia ter zero já que teria que ser positiva, daí a equação ficava:
a + 1 + b + 1 + c + 1 = 4 ou a + b + c = 4 – 3 ou a + b + c = 1, assim se teria n = 3 e p = 1, daí:
CR3,1 = C3+1–1,1 = C3,1 = 3, que seriam {(1, 1, 2)}; (1, 2, 1); (2, 1, 1)}.
Observação:
A principal diferença entre arranjo e combinação é que no arranjo, os agrupamentos, por exemplo, ABC e ACB são diferentes, ou seja, a ordem importa e na combinação a ordem nãoimporta.
Assim, se em um grupo de 5 pessoas, 2 forem escolhidas para ir a igreja, falar dos escolhidos tanto faz dizer Ana e Bia como Bia e Ana.
Logo, trata-se de uma combinação.
Porém, se 2 forem escolhidos para presidente e vice, Ana e Bia (Ana é a presidente), já Bia e Ana (Bia é que é a presidente).
Logo, trata-se de um arranjo.
Exercícios Resolvidos
R01 — Quantos são os divisores de 210 . 39? Quantos divisores são pares?
Cada potência de 2 multiplicada por cada potência de 3 representa um divisor, assim 20 . 30= 1 . 1 = 1 é um divisor, 21 . 30 = 2 . 1 = 2, outro, 20 . 31 = 1 . 3 = 3 e, assim por diante, então os divisores dependem das potências, isto é, no caso do 2 os expoentes podem ser E(2) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} e os de 3, E(3) = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
Então, há 11 possibilidades para a base 2 e 10 possibilidades para a base 3, então 11 . 10 = 110 divisores.
Para que o divisor seja para é necessário que o 2 não desapareça, isto é, que o expoente do 2 não seja zero, assim haverá 10 possibilidades para cada e, 10 . 10 = 100 divisores.
R02 — De quantas formas podemos pintar o quadro abaixo com as cores amarelo, preto e vermelho, sem que dois quadrados consecutivos tenham a mesma cor e que nem o primeiro nem o último sejam pintados de amarelo?
seis quadrados
Começa-se pelas condições impostas que é para o primeiro e o sexto há apenas duas opções (vermelho ou preto), a segunda posição pode-se pintar com qualquer uma das três cores exceto a que foi pintada a primeira, para terceira e quarta a mesma coisa, porém na quinta posição não se poder pintar com a mesma cor da quarta nem da sexta, então só há uma opção.
Assim tem-se: 2 . 2 . 2 . 2 . 1 . 2 = 32 modos diferentes.
R03 — Quantos são os anagramas da palavra “PRATO” que começam por consoante?
Cada anagrama corresponde a uma ordem das 5 letras. Para formar um anagrama começando por consoante se deve começar por P, R ou T.
Se começar por P significa que ele não irá trocar de lugar com as demais, P __ __ __ __ e apenas quatro elementos irão permutar.
O mesmo ocorre, com o fato de se começar por R __ __ __ __ ou T __ __ __ __, totalizando “três casos” iguais, logo se tem:
P4 + P4 + P4 = 3 . P4 = 3 . 4! = 3 . 24 = 72.
R04 — De quantas formas pode-se ordenar 8 pessoas de modo que duas determinadas pessoas não fiquem juntas?
O número total que se pode ordenar é dado pela permutação de todos os oito elementos P8, mas isto inclui os casos em que duas determinadas estejam juntas.
Supondo que Ana e Bia, por exemplo, sempre ficassem juntas, elas formariam apenas um elemento ficando as duas mais seis, totalizando então sete: P7, mas as duas também poderiam permutar entre si, neste caso, permutação das duas: P2.
O número que as duas ficariam juntas é P7 . P2.
Então, para que duas não fiquem juntas tem-se o número total menos os casos em que estão juntas. Logo, se tem:
P8 – P2 . P7 = 8! – 2! . 7! = 8 . 7! – 2 . 7! = 7!(8 – 2) = 6 . 7! = 6 . 7 . 6! = 42 . 720 = 30 240.
R05 — De quantas formas podemos acomodar 3 pessoas em 5 cadeiras?
Neste caso, tem-se cinco cadeiras: A, B, C, D, E das quais se usará apenas 3 (onde sentarão as três pessoas).
Claro que a escolha ABC é diferente de ABD, pois são cadeiras diferentes.
ABC e ACB embora sejam as mesmas cadeiras são agrupamentos diferentes, pois as pessoas são distintas e sentando em posições diferentes formam outro agrupamento.
Logo, tem-se um arranjo simples de 5 elementos tomados 3 a 3.
A5,3 = 5! / (5 – 3)! = 5.4.3.2! / 2! = 5 . 4 . 3 = 60.
R06 — Numa reta há 6 pontos e em outra reta paralela a esta, 5 pontos. Quantos triângulos podem ser formados com esses pontos? E quantos quadriláteros?
reta com 6 pontos
reta com 5 pontos
Como as retas são paralelas, pegando dois pontos de uma reta e um ponto da outra reta forma-se um triângulo, e como, por exemplo, os vértices A e B da primeira reta forma com o vértice C da segunda reta um triângulo e o vértice C da segunda reta forma com os vértices A e B da primeira reta o mesmo triângulo. Logo, trata-se de uma combinação simples onde se escolhe 2 pontos na primeira reta e 1 na segunda reta (como é ‘e’ então multiplica-se) ou(como é ‘ou’ soma-se) 1 ponto na primeira reta e 2 pontos na segunda reta.
Daí, C6,2 . C5,1 + C6,1 . C5,2 = 6! / 2!(6 – 2)! . 5 + 6 . 5! / 2!(5 – 2)! = 15 . 5 + 6 . 10 = 75 + 60 = 135.
Para formar quadriláteros, é preciso escolher 2 pontos em cada reta e portanto, C6,2 . C5,2 =6! / 2!(6 – 2)! . 5! / 2!(5 – 2)! = 15 . 10 = 150.
R07 — Quantos são os anagramas da palavra ANAGRAMA, que mantêm juntas as letras ANGM nesta ordem? E que tenham as letras NGRM juntas?
Como as letras ANMG devem ficar juntas elas formam apenas um elemento que com as letras AAAN restantes, formam 5 elementos que serão permutadas, mas há repetição da letra A (3 vezes, já que a letra A que está formando um elemento em ANMG não está sozinha).
Logo, trata-se de uma permutação com repetição de 5 elementos com 3 repetidos.
PR5,3 = 5! / 3! = 5.4.3! / 3! = 5 . 4 = 20.
Considerando que as letras NMGR, podem ser permutadas entre si, tem-se P4 maneiras e, como elas formam um elemento que junto das demais AAAA, formam 5 elementos com repetição das quatro letras A, tem-se PR5,4 maneiras.
Como acontece uma coisa e outra, tem-se, portanto o produto dos dois, isto é:
P4 . PR5,4 = 4! . 5! / 4! = 5! = 120.
R08 — Seja A um conjunto com 4 elementos e outro B, com 3 elementos. Quantas são as funções de A em B? E quantas são sobrejetoras?
Para ser função, nenhum elemento do primeiro conjunto pode sobrar, contudo eles podem ter o mesmo correspondente e portanto, no máximo quatro repetições e daí, AR3,4 = 34 = 3 . 3 . 3 . 3 = 81.
Supondo A = {1, 2, 3, 4} e B = {6, 7, 8} tais conjuntos e como para ser sobrejetora, não pode sobrar elementos no segundo conjunto tem-se que do total de funções deve-se retirar aquelas que não se correspondem com os 3 elementos de B, ou seja, aquelas que se correspondem apenas com 1 ou com 2 elementos.
Para que todos os elementos do conjunto A se correspondam com 2 de B, tem-se C3,2 (para a escolha dos dois dentre os 3) e um dos dois seria repetido no máximo 4 vezes, daí AR2,4, mas nesse caso conta-se duas vezes {(1, 6); (2, 6); (3, 6); (4, 6)}, {(1, 7); (2, 7); (3, 7); (4, 7)} e {(1, 8); (2, 8); (3, 8); (4, 8)}, quando se escolheu dso três elementos de o {6,7}, {7, 8} e {6, 8}. Para que todos os elementos do conjunto A se correspondam com 1 de B, tem-se C3,1(para escolher um dos três) e este seria repetido no máximo 4 vezes, daí AR1,4.
Assim, C3,2 . AR2,4 – C3,1 . AR1,4 = 3 . 24 – 3 . 14 = 3 . 16 – 3 . 1 = 48 – 3 = 45.
Então, o número de funções sobrejetoras é 81 – 45 = 36.
De uma forma geral se n(A) = m e n(B) = n o número de funções sobrejetoras de A em B é dado por:
somatorio 0 a p . Cn,n – p . (n – p)m.
R09 — Num parque há 5 tipos de brinquedos que podem usar o mesmo bilhete de entrada. Com 3 bilhetes de quantas formas se pode brincar neste parque usando todos os bilhetes?
Sendo os brinquedos A, B, C, D, E, se os bilhetes forem utilizados na forma ABC ou BAC dá no mesmo, pois são os mesmos brinquedos. Então, trata-se de uma combinação, mas pode-se brincar no mesmo brinquedo até 3 vezes, logo é com repetição.
Portanto, uma combinação com repetição de 5 tomados 3 a 3.
CR5,3 = C5 + 3 – 1,3 = C7,3 = 7! / 3! . (7–3)! = 7 . 6 . 5 . 4! / 3! . 4! = 7 . 6 . 5 / 3 . 2 . 1 = 35.
R10 — Para jogar dominó em uma mesa retangular é preciso 4 pessoas. Tendo de pessoas, de quantas maneiras pode-se realizar esse jogo?
Para escolher as quatro pessoas que irão jogar tem-se uma combinação de 6 tomados 4 a 4 (pois na escolha tanto faz ABCD com ACBD) e para os quatro que irão jogar tem-se uma permutação circular destes.
Como C6,4 = C6,2, tem-se:
C6,2 . PC4 = 6! / (6 – 2)! . (4 – 1)! = 6 . 5 / 2 . 3! = 15 . 6 = 90.
R11 — Sabendo-se que C8,p+2 / C8,p+1 = 2, determine o valor de p.
Como – (p + 2) = – p – 2 e – (p + 1) = – p – 1, tem-se:
8! / (p+2)! (8 – (p+2))! / 8! / (p+1)! (8 – (p+1))! = 8! / (p+2)! (8 – (p+2))! . (p + 1)! (8 – p – 1)! / 8! = 8! / (p+2)(p+1)!(6 – p)! . (p + 1)! (7 – p)! / 8!(7 – p) / (p + 2) = 2, e dai tem-se:
7 – p = 2.(p + 2) implica 7 – p = 2p + 4 implica 7 – 4 = 2p + p implica 3 = 3p e, portanto, p = 1.
R12 — Quantos coquetéis (mistura de duas ou mais bebidas) podem ser feitos a partir de 7 ingredientes distintos?
Como é para se formar os coquetéis com pelo menos dois ingredientes, tem-se com 2, 3, 4, 5, 6 ou 7 vitaminas diferentes, ou seja,
C7,2 + C7,3 + C7,4 + C7,5 + C7,6 + C7,7 = 7 . 6 / 2! + 7 . 6 . 5 / 3! + 7 . 6 . 5 / 3! + 7 . 6 / 2! + 7 + 1 = 21 + 35 + 35 + 21 + 7 + 1 = 120.
Observações:
C7,4 = C7,3 C7,5 = C7,2 C7,6 = C7,1 = 7 C7,7 = C7,0 = 1
Exercícios Propostos
P01 — Quantos divisores de 210 . 39 formam quadrados perfeitos?
P02 — Em uma sala há 6 lâmpadas com seis interruptores distintos. De quantos modos pode ser iluminada essa sala?
P03 — De quantas maneiras pode-se dispor 4 homens e 4 mulheres em uma fila, sem que dois homens fiquem juntos?
P04 — Lançam-se três dados. Em quantos dos resultados possíveis, a soma dos pontos é 12?
P05 — Quantos inteiros entre 1000 e 10000 inclusive, não são divisíveis por 2 nem por 5?
P06 — De quantas formas pode-se ter o 1o, 2o e 3o lugares de um campeonato com 10 times?
P07 — Com as letras da palavra ADEUS, se pode formar:
a) quantos anagramas?
b) quantos anagramas que começam com a letra D?
c) quantos anagramas que começam com vogal?
d) quantos anagramas que começam com consoante e terminam em vogal?
P08 — De quantos modos pode-se ordenar 2 livros de matemática, 3 de português e 4 de física, de modo que os livros de uma mesma matéria fiquem sempre juntos e, além disso, os de física fiquem sempre na mesma ordem?
P09 — Quantos são os anagramas da palavra INDEPENDENTE:
a) começados por IND?
b) começados por IND e terminados em T?
c) que contenham as letras I e P sempre juntas?
d) que contenham as letras I e P sempre juntas nesta ordem?
e) que contenham as letras I e P sempre juntas e termine em TE?
P10 — Com os dígitos 1, 2, 3, 4 e 5, quantos números de 5 algarismos distintos e maiores que 30 000 se pode formar?
P11 — Quantos números pares de três algarismos distintos pode-se formar com os dígitos 1, 3, 5, 6, 8 e 9?
P12 — Quantos números ímpares, compreendidos entre 300 e 4 000 e com todos os algarismos distintos, pode-se formar com os dígitos 1, 3, 5, 6, 7 e 9?
P13 — Quantas matrizes quadradas de ordem 3 pode-se formar usando os dígitos 1, 2 e 3, cada um uma vez e seis zeros?
P14 — Um trem é constituído de 1 locomotiva e 6 vagões distintos, sendo um deles restaurante. Sabendo que a locomotiva deve ir à frente e que o vagão-restaurante não pode ser colocado imediatamente após a locomotiva, encontre o número de modos diferentes para montar a composição?
P15 — Quantos números de 3 algarismos distintos pode-se formar com os 10 primeiros números naturais?
P16 — De quantas maneiras pode-se escolher 3 representantes de um grupo de 10 pessoas?
P17 — Uma empresa tem 3 diretores e 5 gerentes. Quantas comissões de 5 pessoas podem se formadas contendo no mínimo 1 diretor?
P18 — Uma sociedade tem um conselho administrativo formado por 12 membros, sendo 3/4 de brasileiros e os demais estrangeiros. Quantas comissões de 5 conselheiros podem ser formadas com 3 brasileiros?
P19 — De quantas maneiras distintas um grupo de 10 pessoas pode ser divididos em 3 grupos de 5, 3 e 2 pessoas?
P20 — Com os dígitos 0, 1, 2, 3, 4, 5 e 6 são formados números de 4 algarismos distintos. Quantos são divisíveis por 5?
P21 — Com os dígitos 0, 1, 2, 3, 4, 5 e 6 são formados números de 4 algarismos. Quantos são divisíveis por 5?
P22 — Qual o número de diagonais do decágono?
P23 — Calcular o número de múltiplos de 9 com 4 algarismos distintos que podem ser formados com os dígitos 2, 3, 4, 6 e 9?
P24 — Considerando que a loteria esportiva tenha 13 jogos, quantos são os possíveis resultados?
P25 — Sabendo que as placas de carro são formadas por 3 letras e 4 números, qual o número máximo de carros que podem ser emplacados em uma cidade onde só pode começar por K ou L?
P26 — Colocando em ordem crescente todos os números de 5 algarismos distintos obtidos com 1, 3, 4, 6 e 7, que posição ocupa o número 61 473?
P27 — Quantos são os naturais ímpares com 5 algarismos distintos?
P28 — Quantos são os anagramas da palavra ESTUDAR que começam com vogal? Que começam e terminam em vogal? Que tenham as vogais juntas?
P29 — De quantas maneiras pode-se ordenar 5 livros de Matemática, 3 livros de Química e 2 livros de Física, todos diferentes, de forma que os livros de uma mesma disciplina fiquem juntos?
P30 — De quantas formas pode-se ordenar 6 moças e 4 rapazes de modo que as moças permaneçam juntas?
P31 — De quantas formas pode-se ordenar 8 pessoas de modo que duas determinadas pessoas não fiquem juntas?
P32 — Quantos são os anagramas da palavra MATEMÁTICA de forma que as vogais e as consoantes sempre fiquem alternadas?
P33 — Quantos são os anagramas da palavra ÁLGEBRA que não possuem 2 vogais juntas?
P34 — De quantas formas podemos pintar o quadro abaixo com as cores: verde, amarelo, azul e branco, sem que dois quadrados consecutivos tenham a mesma cor?
seis quadrados
P35 — Quantas são as raízes inteiras não negativas da equação x + y + z = 6?
P36 — Quantas são as raízes inteiras positivas da equação x + y + z = 7?
P37 — Quantos subconjuntos com 3 elementos possui um conjunto com n elementos?
P38 — Quantos são os anagramas da palavra COMBINATÓRIA? Que alternam consoantes e vogais? Que possuem as vogais juntas?
P39 — Quantos segmentos de reta podem ser formados com extremidades em 15 pontos dados?
P40 — Quantas diagonais possui um polígono convexo com 8 lados?
P41 — De quantas maneiras podemos pedir um sorvete de três bolas se dispomos de 5 sabores diferentes?
P42 — Quantos são os anagramas da palavra ESCOLA que terminam em vogal?
P43 — Quantos são os números com 5 algarismos não repetidos formados com 1, 2, 3, 4, 5? E que sejam ímpares? E que sejam maiores que 34 125?
P44 — Quantos são os números com 10 algarismos? E se os algarismos forem distintos?
P45 — De quantas maneiras podemos arrumar 9 pessoas em 3 quartos cada quarto com 3 pessoas?
fonte:hpdemat.apphb.com
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-6,748,734,944,872,839,000 | 60 plus 91 percent
Here we will teach you how to calculate sixty plus ninety-one percent (60 plus 91 percent) using two different methods. We call these methods the number method and the decimal method.
We start by showing you the illustration below so you can see what 60 + 91% looks like, visualize what we are calculating, and see what 60 plus 91 percent means.
60 plus 91 percent
The dark blue in the illustration is 60, the light blue is 91% of 60, and the sum of the dark blue and the light blue is 60 plus 91 percent.
Calculate 60 plus 91 percent using the number method
For many people, this method may be the most obvious method of calculating 60 plus 91%, as it entails calculating 91% of 60 and then adding that result to 60. Here is the formula, the math, and the answer.
((Number × Percent/100)) + Number
((60 × 91/100)) + 60
54.6 + 60
= 114.6
Remember, the answer in green above is the sum of the dark blue plus the light blue in our illustration.
Calculate 60 plus 91 percent using the decimal method
Here you convert 91% to a decimal plus 1 and then multiply it by 60. We think this is the fastest way to calculate 91 percent plus 60. Once again, here is the formula, the math, and the answer:
(1 + (Percent/100)) × Number
(1 + (91/100)) × 60
1.91 × 60
= 114.6
Number Plus Percent
Go here if you need to calculate any other number plus any other percent.
61 plus 91 percent
Here is the next percent tutorial on our list that may be of interest.
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8,090,868,574,895,376,000 | Matematică
dumitruluminit
2024-04-22 01:06:27
Aflati perimetrul unui paralelogram cu laturile: a) 3.5 cm si 4.5 cm b)9 radical din 5 cm si radical din 180 cm
Răspunsuri la întrebare
ecaterina2001
2024-04-22 05:59:02
A) P paralelogram = 2( L + l) = 2 ( 3,5 + 4,5) = 2 · 8 = 16 cm b) P paralelogram = 2 ( L + l) = 2 ( 9√5 + √180) = 2 ( 9√5 + √2²·3²·5) = 2 ( 9√5 + 6√5) = 2 · 15√5 = 30√5 cm Sper ca ti-am fost de ajutor!
Adăugați un răspuns | {
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2,219,359,131,911,872,300 | subject
Mathematics, 16.11.2020 18:30 stevenssimeon30
a quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (-4,6).what is the vertex of the function defined as g(x)=f(-x)+3
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a quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (-4,6)...
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Pandolfini's Puzzler #24 - The 50-50 Problem
Pandolfini's Puzzler #24 - The 50-50 Problem
brucepandolfini
Jan 10, 2014, 12:00 AM 13 Scholastics
Professor: Unless we’re Rip Van Winkle, and can sleep for a very long time, every day we have to make decisions. Some decisions are hard to make, others are easy.
---
Lucian: I guess a hard decision would be where one has many choices.
---
Zephyr: I’ve heard you say that when you have many choices it can be like having no choice at all. You might get lost in all the possibilities.
---
Professor: So does that mean the easiest type of decision to make is when you have but one possibility?
---
Lucian: “Yes,” I think so. But even with a single choice I might not like the consequences.
---
Zephyr: Yeah, the consequences could be very bad. You might be in zugzwang or something.
zugzwang.jpeg
---
Professor: That’s funny. I’ll have to add that to my repertory of jokes.
---
Lucian: Professor, I didn’t know you like to tell jokes.
---
Everybody seemed to smile.
---
Professor: I do, before I tell another joke, I’d like to clarify something. In some cases, if you have a single choice, or a single move, you don't really have any decisions to make. I tell you what. Let’s look at a position with Black to play. Here, there aren’t many moves (or choices). Nor is it a position with just one move. Rather, it’s a position where Black has two moves.
---
---
Professor: Again, it’s Black’s turn, and clearly he has two possibilities. He can move his king to h1, shifting it apparently farther away from the action. Or he can move his king to f1, keeping it apparently closer to the action. There’s a wrong way to go. And there’s a right way to go. The wrong way loses. The right way draws.
---
Zephyr: Wow! Black has a 50 percent chance of drawing.
---
Lucian: Which means, he has a 50 percent chance of losing.
---
Professor: That’s why I’m going to call this the 50-50 problem.
---
Question: How can Black move and draw?
---
Professor: Or, restating the question, should the black king move closer to the action or farther away?
---
Zephyr: From the way you’ve set that problem up, Professor, I feel you’re being a little tricky.
---
Lucian: Maybe it has something to do with what you mean by “closer” and “farther.”
---
Professor: Perhaps you’re right, Lucian. Anyhow, can either of you, or the both of you, tell me which move is right, and which move is actually closer? By the way, you’re going to have to explain why.
---
Answers Below - Try to solve ProfessorPando's Puzzle first!
---
ANSWER #24
---
The right move is 1…Kh1! But before we see why, let’s examine what happens if the king instead moves to f1.
---
---
After 1…Kf1, let’s say White plays 2. Kf3.
---
If Black tries 2…Ke1, play might continue 3. Ke4 Ke2 4. Kd5 Kd3 5. Kc6 Kc4 6. Kb7 Kb5 7. Kxa7 Kc6 8. Kb8, and the pawn will queen.
---
Or if Black tries 2…Kg1, play might continue 3. Ke4 Kf2 4. Kd5 Ke3 5. Kc6 Kd4 6. Kb7 Kc5 7. Kxa7, and the pawn soon queens anyway.
---
Now let’s look at what happens after the right move, 1…Kh1!
---
---
After 1…Kh1, a possible line is 2. Kf3 Kh2 3. Ke4 Kg3 4. Kd5 Kf4 5. Kc6 Ke5 6. Kb7 Kd6 7. Kxa7 Kc7, and the position is drawn, since 8. Ka8 Kb6 9. a7 Kc7 is stalemate!
---
Take note
---
It’s not always obvious which path is closer or farther away on a chessboard. By trying 1…Kf1, it looked on the surface as if Black were moving the king closer to the action. But it was in fact moving the king farther away from something else: the key diagonal of retreat, that is, the b8-h2 diagonal. That diagonal contains the c7-square. Immediately after the a-pawn falls, Black’s king can set up a draw by occupying c7. White’s king is then unable to escape from the corner. In other words, by moving farther away, Black was actually moving closer.
RELATED STUDY MATERIAL
• Here's another draw in king and pawn endgames that you should know;
• Or, if you prefer, a more detailed explanation in video form;
• If you want a real challenge, do Benzoo's king and pawn endgame course - parts one, two and three!
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Insert zeros after each elements in the vector. Number of zeros is specified as the input parameter. For example: x = [1 ...
1년 이상 전
해결됨
First N Perfect Squares
*Description* Return the first N perfect squares *Example* input = 4; output = [ 1 4 9 16 ];
1년 이상 전
해결됨
Wind Chill Computation
On a windy day, a temperature of 15 degrees may feel colder, perhaps 7 degrees. The formula below calculates the "wind chill," i...
1년 이상 전
해결됨
Is it an Armstrong number?
An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. Fo...
1년 이상 전
해결됨
The sum of the numbers in the vector
eg. [1,2,3]---->SUM=6
1년 이상 전
해결됨
Find the longest sequence of 1's in a binary sequence.
Given a string such as s = '011110010000000100010111' find the length of the longest string of consecutive 1's. In this examp...
1년 이상 전
해결됨
Find common elements in matrix rows
Given a matrix, find all elements that exist in every row. For example, given A = 1 2 3 5 9 2 5 9 3 2 5 9 ...
1년 이상 전
해결됨
Who has power to do everything in this world?
There is only one person who is older than this universe. He is Indian version of Chuck Norris.
1년 이상 전
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3,490,922,510,668,079,600 | Dreieck 7 7 9
Spitzwinkligen gleichschenklig dreieck.
Seiten: a = 7 b = 7 c = 9
Fläche: T = 24.12985619132
Umfang: p = 23
Semiperimeter (halb Umfang): s = 11.5
Winkel ∠ A = α = 49.99547991151° = 49°59'41″ = 0.87325738534 rad
Winkel ∠ B = β = 49.99547991151° = 49°59'41″ = 0.87325738534 rad
Winkel ∠ C = γ = 80.01104017697° = 80°37″ = 1.39664449467 rad
Höhe: ha = 6.89438748323
Höhe: hb = 6.89438748323
Höhe: hc = 5.36219026474
Mittlere: ma = 7.26329195232
Mittlere: mb = 7.26329195232
Mittlere: mc = 5.36219026474
Inradius: r = 2.09881358185
Umkreisradius: R = 4.56992735604
Scheitelkoordinaten: A[9; 0] B[0; 0] C[4.5; 5.36219026474]
Schwerpunkt: SC[4.5; 1.78773008825]
Koordinaten des Umkreismittel: U[4.5; 0.7932629087]
Koordinaten des Inkreis: I[4.5; 2.09881358185]
Äußere Winkel des Dreiecks:
∠ A' = α' = 130.00552008849° = 130°19″ = 0.87325738534 rad
∠ B' = β' = 130.00552008849° = 130°19″ = 0.87325738534 rad
∠ C' = γ' = 99.99895982303° = 99°59'23″ = 1.39664449467 rad
Berechnen Sie ein anderes Dreieck
Wie haben wir dieses Dreieck berechnet?
Die Berechnung des Dreiecksfortschritts in zwei Phasen. Die erste Phase ist so, dass wir versuchen, alle drei Seiten des Dreiecks aus den Eingabeparametern zu berechnen. Die erste Phase unterscheidet sich für die verschiedenen eingegebenen Dreiecke. Die zweite Phase ist die Berechnung von Andere Merkmale des Dreiecks wie Winkel, Fläche, Umfang, Höhe, Schwerpunkt, Kreisradien usw. Einige Eingabedaten führen auch zu zwei bis drei korrekten Dreieckslösungen (z. B. wenn das angegebene Dreieck und zwei Seiten angegeben sind) - in der Regel sowohl ein akutes als auch ein stumpfes Dreieck ergeben.
Jetzt wissen wir, dass die Längen aller drei Seiten des Dreiecks das Dreieck eindeutig bestimmen. Als nächstes berechnen wir ein anderes seine Eigenschaften - dasselbe Verfahren wie Berechnung des Dreiecks von den bekannten drei Seiten SSS .
a=7 b=7 c=9
1. Der Dreiecksumfang ist die Summe der Längen seiner drei Seiten
p=a+b+c=7+7+9=23
2. Semiperimeter des Dreiecks
Der Halbmesser des Dreiecks ist die Hälfte seines Umfangs. Das Semiperimeter erscheint häufig in Formeln für Dreiecke, denen ein eigener Name gegeben wird. Durch die Dreiecksungleichung ist die längste Seitenlänge eines Dreiecks kleiner als das Semiperimeter.
s=2p=223=11.5
3. Das Dreiecksgebiet mit Herons Formel
Die Formel von Heron gibt die Fläche eines Dreiecks an, wenn die Länge aller drei Seiten bekannt ist. Es ist nicht erforderlich, zuerst Winkel oder andere Abstände im Dreieck zu berechnen. Die Formel von Heron funktioniert in allen Fällen und Arten von Dreiecken gleich gut.
T=s(sa)(sb)(sc) T=11.5(11.57)(11.57)(11.59) T=582.19=24.13
4. Berechnen Sie die Höhe des Dreiecks aus seinem Inhalt.
Es gibt viele Möglichkeiten, die Höhe des Dreiecks zu ermitteln. Der einfachste Weg ist von der Fläche und Grundlänge. Die Fläche eines Dreiecks ist die Hälfte des Produkts aus der Länge der Basis und der Höhe. Jede Seite des Dreiecks kann eine Basis sein; Es gibt drei Basen und drei Höhen. Die Dreieckshöhe ist das senkrechte Liniensegment von einem Scheitelpunkt zu einer Linie, die die Basis enthält.
T=2aha ha=a2 T=72 24.13=6.89 hb=b2 T=72 24.13=6.89 hc=c2 T=92 24.13=5.36
5. Berechnung der inneren Winkel des Dreiecks mit einem Kosinusgesetz
Das Gesetz des Cosinus ist nützlich, um die Winkel eines Dreiecks zu finden, wenn wir alle drei Seiten kennen. Die Kosinusregel, auch als Kosinusgesetz bekannt, bezieht alle drei Seiten eines Dreiecks mit einem Winkel eines Dreiecks ein. Das Gesetz des Kosinus ist die Extrapolation des Satzes von Pythagoras für jedes Dreieck. Der Satz von Pythagoras funktioniert nur in einem rechtwinkligen Dreieck. Der Satz des Pythagoras ist ein Sonderfall des Kosinussatzes und kann daraus abgeleitet werden, weil der Kosinus von 90 ° 0 ist. Es ist am besten, den Winkel gegenüber der längsten Seite zuerst zu finden. Mit dem Cosinusgesetz gibt es auch kein Problem (wie mit dem Sinusgesetz) mit stumpfen Winkeln, da die Cosinusfunktion für stumpfe Winkel negativ, für rechte Null und für spitze Winkel positiv ist. Wir verwenden auch den inversen Kosinus, der als Arkuskosinus bezeichnet wird, um den Winkel aus dem Kosinuswert zu bestimmen.
a2=b2+c22bccosα α=arccos(2bcb2+c2a2)=arccos(2 7 972+9272)=49°5941" b2=a2+c22accosβ β=arccos(2aca2+c2b2)=arccos(2 7 972+9272)=49°5941" γ=180°αβ=180°49°5941"49°5941"=80°37"
6. Inradius
Ein Kreis eines Dreiecks ist ein Kreis, der jede Seite berührt. Ein Incircle-Center heißt Incenter und hat einen Radius mit dem Namen inradius. Alle Dreiecke haben einen Mittelpunkt, der immer innerhalb des Dreiecks liegt. Der Mittelpunkt ist der Schnittpunkt der drei Winkelhalbierenden. Das Produkt aus Inradius und Semiperimeter (halber Umfang) eines Dreiecks ist seine Fläche.
T=rs r=sT=11.524.13=2.1
7. Umkreisradius
Der Umkreis eines Dreiecks ist ein Kreis, der durch alle Eckpunkte des Dreiecks verläuft, und der Umkreis eines Dreiecks ist der Radius des Umkreises des Dreiecks. Der Mittelpunkt (Mittelpunkt des Kreises) ist der Punkt, an dem sich die senkrechten Winkelhalbierenden eines Dreiecks schneiden.
R=4 rsabc=4 2.098 11.57 7 9=4.57
8. Berechnung des Medians
Ein Median eines Dreiecks ist ein Liniensegment, das einen Scheitelpunkt mit dem Mittelpunkt der gegenüberliegenden Seite verbindet. Jedes Dreieck hat drei Mediane, die sich alle im Schwerpunkt des Dreiecks schneiden. Der Schwerpunkt unterteilt jeden Median im Verhältnis 2: 1 in Teile, wobei der Schwerpunkt doppelt so nahe am Mittelpunkt einer Seite liegt wie am gegenüberliegenden Scheitelpunkt. Wir verwenden den Satz von Apollonius, um die Länge eines Medians aus den Längen seiner Seite zu berechnen.
ma=22b2+2c2a2=22 72+2 9272=7.263 mb=22c2+2a2b2=22 92+2 7272=7.263 mc=22a2+2b2c2=22 72+2 7292=5.362
Berechnen Sie ein anderes Dreieck
Schauen Sie sich auch die Sammlung von mathematischen Beispielen und Problemen unseres Freundes an:
Weitere Informationen zu Dreiecken oder weitere Details finden Sie unter Dreieck über Dreiecke lösen | {
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Math Help - Parallel Proof
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Parallel Proof
Prove that nonvertical lines are parallel if and only if their slopes are equal.
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2. #2
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Lines can be written in the form \displaystyle y = mx + c, where \displaystyle m is the gradient and \displaystyle c is the \displaystyle y intercept.
So we could write the equation of two lines as
\displaystyle y=m_1x + c_1 and \displaystyle y=m_2x + c_2.
You should know that parallel lines never cross. These two lines will cross when their equations are equal. So
\displaystyle m_1x + c_1 = m_2x + c_2
\displaystyle m_1x - m_2x = c_2 - c_1
\displaystyle x(m_1-m_2) = c_2 - c_1
\displaystyle x = \frac{c_2 - c_1}{m_1-m_2}.
Obviously there is no solution when the denominator is \displaystyle 0, in other words, where \displaystyle m_1 = m_2.
So the only time when there will not be a point of intersection is when the two gradients are equal.
In other words, the lines are parallel when their gradients are equal.
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If a square mirror has a 30-inch diagonal, what is the area of the mir
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Bunuel wrote:
If a square mirror has a 30-inch diagonal, what is the area of the mirror, in inches?
A. 225
B. 450
C. 600
D. 750
E. 900
Kudos for correct solution.
Let's name square sides of s, using pythagoras 30^2=s^2+s^2, this ends to s=\sqrt{450}, which is \sqrt{2*3^2*5^2}, hence square sides are 15\sqrt{2}, 15\sqrt{2}*15\sqrt{2}=450, hence, i think the correct answer is B.
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If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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Bunuel wrote:
If a square mirror has a 30-inch diagonal, what is the area of the mirror, in inches?
A. 225
B. 450
C. 600
D. 750
E. 900
Kudos for correct solution.
Diagonal of a mirror = \(\sqrt{2}\)*side
Therefore
\(\sqrt{2}\)*side = 30
Side = 30/\(\sqrt{2}\)
Area = 30/\(\sqrt{2}\) * 30/\(\sqrt{2}\) = 900/2 = 450
Option B
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Re: If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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(30/sqrt 2)^2 = 450
Correct answer - B
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Re: If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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New post 10 Jun 2016, 13:15
Bunuel wrote:
AbdurRakib wrote:
If a square mirror has a 30-inch diagonal, what is the area of the mirror, in inches?
A. 225
B. 450
C. 600
D. 750
E. 900
Merging topics. Please search before posting.
Thanks But,
Actuality I searched on the Google but found only this link if-a-square-mirror-has-a-20-inch-diagonal-what-is-the-99359.html
I'll search it next time on GMATCLUB to avoid mistake
Thanks again
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If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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New post 13 Jun 2016, 23:18
Bunuel wrote:
If a square mirror has a 30-inch diagonal, what is the area of the mirror, in inches?
A. 225
B. 450
C. 600
D. 750
E. 900
Kudos for correct solution.
If a square has a side a, then the length of the diagonal is \(a* \sqrt{2}\)
Area of square would be \(a^2\)
\(a\sqrt{2}\) = 30;
squaring both sides
2\(a^2\) = 30*30 = 900
\(a^2\) = 450 --> This is the area.
Answer: B.
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If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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New post 06 Nov 2017, 11:23
Bunuel wrote:
If a square mirror has a 30-inch diagonal, what is the area of the mirror, in inches?
A. 225
B. 450
C. 600
D. 750
E. 900
Kudos for correct solution.
Given a square's diagonal, we need side lengths to calculate area.
The relationship between a square's side, s, and its diagonal, d,* is given by
\(s\sqrt{2} = d\)
\(s = \frac{d}{\sqrt{2}}\)
The side of square (d = 30), therefore, is \(\frac{30}{\sqrt{2}}\).
Leave it; no need to rationalize the denominator because it needs to be squared. Square the side length to find area:
\((\frac{30}{\sqrt{2}}\) * \(\frac{30}{\sqrt{2}})\) = \(\frac{30*30}{2}\) = \(\frac{900}{2}=450\)
Answer B
*Although \(d = s\sqrt{2}\) probably should be in memory, it is easily derived. Two sides, \(s\), of a square, form a right isosceles triangle. Pythagorean theorem hence yields:
\(s^2 + s^2 = d^2\)
\(2s^2 = d^2\)
\((\sqrt{2})(\sqrt{s^2})=\sqrt{d^2}\)
\((\sqrt{2})s
= d\), or \(s\sqrt{2}= d\)
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Re: If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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New post 08 Nov 2017, 16:30
Bunuel wrote:
If a square mirror has a 30-inch diagonal, what is the area of the mirror, in inches?
A. 225
B. 450
C. 600
D. 750
E. 900
Since the diagonal of a square is 30:
diagonal = side√2
30 = side√2
Squaring the entire equation, we have:
30^2 = side^2 x 2
900/2 = side^2
450 = side^2 = area
Answer: B
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Re: If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink]
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New post 09 Nov 2017, 08:09
Area = (diag 1)* (diag 2)/2
so, Area of the mirror = 30*30/2= 450 sq inch
Re: If a square mirror has a 30-inch diagonal, what is the area of the mir [#permalink] 09 Nov 2017, 08:09
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3,457,230,484,010,531,300 | Roman numerals 2+
Add up the number writtens in Roman numerals. Write the results as a roman numbers.
Result
DCCLXXXVIII + DCCLXVII = (Correct answer is: MDLV) Wrong answer
CCXXII + CMLXVIII = (Correct answer is: MCXC) Wrong answer
DXXXIV + DCCCLXV = (Correct answer is: MCCCXCIX) Wrong answer
CCCLXXXIV + CMXCVII = (Correct answer is: MCCCLXXXI) Wrong answer
DCCXLVII + CCLXII = (Correct answer is: MIX) Wrong answer
DVIII + CMXXVIII = (Correct answer is: MCDXXXVI) Wrong answer
DCCCLXXX + DCCLXXXVII = (Correct answer is: MDCLXVII) Wrong answer
CDXXVIII + DCXCII = (Correct answer is: MCXX) Wrong answer
CCCLXXV + DCCXCI = (Correct answer is: MCLXVI) Wrong answer
CDXXXVIII + CXXXVIII = (Correct answer is: DLXXVI) Wrong answer
Solution:
DCCLXXXVIII + DCCLXVII = 788 + 767 = 1555 = MDLV
CCXXII + CMLXVIII = 222 + 968 = 1190 = MCXC
DXXXIV + DCCCLXV = 534 + 865 = 1399 = MCCCXCIX
CCCLXXXIV + CMXCVII = 384 + 997 = 1381 = MCCCLXXXI
DCCXLVII + CCLXII = 747 + 262 = 1009 = MIX
DVIII + CMXXVIII = 508 + 928 = 1436 = MCDXXXVI
DCCCLXXX + DCCLXXXVII = 880 + 787 = 1667 = MDCLXVII
CDXXVIII + DCXCII = 428 + 692 = 1120 = MCXX
CCCLXXV + DCCXCI = 375 + 791 = 1166 = MCLXVI
CDXXXVIII + CXXXVIII = 438 + 138 = 576 = DLXXVI
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-6,866,651,650,328,858,000 | notes 14 - Examples Submitted by gfj100 on Wed, 11/11/2009...
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Examples Submitted by gfj100 on Wed, 11/11/2009 - 10:07 Example 1 The probability of a student getting an A in this course is 0.25 (Not True!) and the probability of getting a B is 0.30 (again Not True!). What is the probability of getting an A or a B? According to Rule 4.a , P(A or B) = P(A) + P(B) = 0.25 + 0.30 = 0.55. In this example events A and B are mutually exclusive since a student cannot get both an A and a B for a course, but instead can only get one grade. Take a look at the Venn diagram example below: Example 2 The probability of a student in this course getting an A is 0.25; the probability of being female is 0.60. What is the probability of getting an A or being female? . ...... Can't say? Well, we could add probabilities in example 1 since the two events were mutually exclusive, but in this example getting an A and being female may both occur. So we just cannot simply add the probability of one event to the probability of the other event - if we did we would be counting twice the
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notes 14 - Examples Submitted by gfj100 on Wed, 11/11/2009...
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Ask a homework question - tutors are online | {
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-4,825,471,495,086,472,000 | Consecutive Sampling- Definition, Example, & Advantages
Consecutive sampling: Definition
Consecutive sampling is defined as a non-probability sampling technique where samples are picked at the ease of a researcher more like convenience sampling, only with a slight variation. Here, the researcher selects a sample or group of people, conducts research over a period, collects results, and then moves on to another sample.
This sampling technique gives the researcher a chance to work with multiple samples to fine-tune his/her research work to collect vital research insights.
Select your respondents
In most of the sampling techniques in research, a researcher will finally infer the study by concluding that the experiment and the data analysis will either accept the null hypothesis or disapprove it and accept the alternative explanation.
A null hypothesis means a statistical theory in which no significant difference exists between the set of variables involved in the research or experiment. In the mathematical terms, the original or default statement is often represented by H0. If the null hypothesis is accepted, then a researcher will not make any changes in opinions or actions. The null hypothesis is indirect or implicit.
An alternative hypothesis is the opposite of the null hypothesis. In this statistical hypothesis, there is a relationship between the two variables involved in the study or research. An alternative explanation is accepted when a null hypothesis is rejected. An alternative hypothesis the testing is direct and explicit. H1 denotes an alternative theory.
However, in consecutive sampling, there is a third option available. Here, a researcher can accept the null hypothesis, if not the null hypothesis, then its alternative hypothesis. If neither of them is applicable, then a researcher can select another pool of samples and conduct the research or the experiment once again before finally making a research decision.
Consecutive sampling example
Here is an easy to understand example of consecutive sampling
• One of the most common examples of a consecutive sample is when companies/ brands stop people in a mall or crowded areas and hand them promotional leaflets to purchase a luxury car.
• In this example, the people walking in the mall are the samples, and let us consider them as representative of a population.
• Now, the researcher hands these people an advertisement or a promotional leaflet. A few of them agree to stay back and respond to the questions asked by the promotion executive (we can consider him/her as a researcher).
• The responses are collected and analyzed, but there is no conclusive result that people would want to buy that car based on the features described in the leaflet.
• The promotion executive now asks questions to another group of people who analyze the details of the car and its features and show a keen interest in buying the luxury car. Thus, this group of people has provided conclusive results for purchasing the vehicle.
However, there is a downside to this sampling method. You cannot consider the sample to be representative of the entire population. In this example, not all people who have taken this leaflet were interested in buying the car.
Here is where sampling bias comes into the picture. So to overcome this bias, consecutive sampling should be used in tandem with probability sampling.
Select your respondents
Advantages of consecutive sampling
Here are the four advantages of consecutive sampling
• In a consecutive sampling technique, the researcher has many options when it comes to sampling size and sampling schedule. The sample size can vary from a few to a few hundred, that the kind of range of sample size we are talking about here.
• In this sampling technique, sampling schedule is completely dependent on the nature of the research, a researcher is conducting. If a researcher is unable to obtain conclusive results with one sample, he/she can depend on the second sample and so on for drawing conclusive results.
• In consecutive sampling, a researcher can fine-tune his/her researcher. Due to its repetitive nature, minor changes and adjustments can be made right at the beginning of the research to avoid considering research bias.
• Very little effort is needed from the researcher’s end to carry out the research. This technique is not time-consuming and doesn’t require an extensive workforce. | {
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-7,108,469,360,486,555,000 | Search by Topic
Resources tagged with Games similar to Up and Across:
Filter by: Content type:
Stage:
Challenge level: Challenge Level:1 Challenge Level:2 Challenge Level:3
There are 94 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Games
problem icon
Diamond Collector
Stage: 3 Challenge Level: Challenge Level:1
Collect as many diamonds as you can by drawing three straight lines.
problem icon
Games Related to Nim
Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
problem icon
Fifteen
Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
problem icon
Diamond Mine
Stage: 3 Challenge Level: Challenge Level:1
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
problem icon
Learning Mathematics Through Games Series: 2.types of Games
Stage: 1, 2 and 3
This article, the second in the series, looks at some different types of games and the sort of mathematical thinking they can develop.
problem icon
Quadrilaterals Game
Stage: 3 Challenge Level: Challenge Level:1
A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .
problem icon
Got It
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
problem icon
Nim-interactive
Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
problem icon
Multiplication Tables - Matching Cards
Stage: 1, 2 and 3 Challenge Level: Challenge Level:1
Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score.
problem icon
Diagonal Dodge
Stage: 2 and 3 Challenge Level: Challenge Level:1
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
problem icon
Jam
Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
To avoid losing think of another very well known game where the patterns of play are similar.
problem icon
Khun Phaen Escapes to Freedom
Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
problem icon
Learning Mathematics Through Games Series: 1. Why Games?
Stage: 1, 2 and 3
This article supplies teachers with information that may be useful in better understanding the nature of games and their role in teaching and learning mathematics.
problem icon
Learning Mathematics Through Games Series: 4. from Strategy Games
Stage: 1, 2 and 3
Basic strategy games are particularly suitable as starting points for investigations. Players instinctively try to discover a winning strategy, and usually the best way to do this is to analyse. . . .
problem icon
Snail Trails
Stage: 3 Challenge Level: Challenge Level:1
This is a game for two players. You will need some small-square grid paper, a die and two felt-tip pens or highlighters. Players take turns to roll the die, then move that number of squares in. . . .
problem icon
Winning Lines
Stage: 2, 3 and 4 Challenge Level: Challenge Level:1
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
problem icon
Sliding Puzzle
Stage: 1, 2, 3 and 4 Challenge Level: Challenge Level:1
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
problem icon
Patience
Stage: 3 Challenge Level: Challenge Level:1
A simple game of patience which often comes out. Can you explain why?
problem icon
Squayles
Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2
A game for 2 players. Given an arrangement of matchsticks, players take it is turns to remove a matchstick, along with all of the matchsticks that touch it.
problem icon
Conway's Chequerboard Army
Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
problem icon
Making Maths: Snake Pits
Stage: 1, 2 and 3 Challenge Level: Challenge Level:1
A game to make and play based on the number line.
problem icon
Property Chart
Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2
A game in which players take it in turns to try to draw quadrilaterals (or triangles) with particular properties. Is it possible to fill the game grid?
problem icon
Online
Stage: 3 Challenge Level: Challenge Level:1
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
problem icon
Jam
Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
A game for 2 players
problem icon
Pentanim
Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
problem icon
Nim
Stage: 4 Challenge Level: Challenge Level:1
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
problem icon
Got it for Two
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
Got It game for an adult and child. How can you play so that you know you will always win?
problem icon
Estimating Angles
Stage: 2 and 3 Challenge Level: Challenge Level:1
How good are you at estimating angles?
problem icon
First Connect Three for Two
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
problem icon
Nim-like Games
Stage: 2, 3 and 4 Challenge Level: Challenge Level:1
A collection of games on the NIM theme
problem icon
One, Three, Five, Seven
Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
problem icon
Lambs and Tigers
Stage: 3 Challenge Level: Challenge Level:1
Investigations based on an Indian game.
problem icon
First Connect Three
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
problem icon
Shapely Pairs
Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
A game in which players take it in turns to turn up two cards. If they can draw a triangle which satisfies both properties they win the pair of cards. And a few challenging questions to follow...
problem icon
Spiralling Decimals for Two
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
Spiralling Decimals game for an adult and child. Can you get three decimals next to each other on the spiral before your partner?
problem icon
Some Games That May Be Nice or Nasty for Two
Stage: 2 and 3 Challenge Level: Challenge Level:1
Some Games That May Be Nice or Nasty for an adult and child. Use your knowledge of place value to beat your oponent.
problem icon
Flip Flop - Matching Cards
Stage: 1, 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
A game for 1 person to play on screen. Practise your number bonds whilst improving your memory
problem icon
The Triangle Game
Stage: 3 and 4 Challenge Level: Challenge Level:1
Can you discover whether this is a fair game?
problem icon
Spiralling Decimals
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
Take turns to place a decimal number on the spiral. Can you get three consecutive numbers?
problem icon
Matching Fractions Decimals Percentages
Stage: 2 and 3 Challenge Level: Challenge Level:1
An activity based on the game 'Pelmanism'. Set your own level of challenge and beat your own previous best score.
problem icon
Dicey Operations
Stage: 2 and 3 Challenge Level: Challenge Level:1
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
problem icon
Some Games That May Be Nice or Nasty
Stage: 2 and 3 Challenge Level: Challenge Level:1
There are nasty versions of this dice game but we'll start with the nice ones...
problem icon
Tangram Pictures
Stage: 1, 2 and 3 Challenge Level: Challenge Level:1
Use the tangram pieces to make our pictures, or to design some of your own!
problem icon
Factors and Multiples for Two
Stage: 2 and 3 Challenge Level: Challenge Level:1
Factors and Multiples game for an adult and child. How can you make sure you win this game?
problem icon
Dicey Operations for Two
Stage: 2 and 3 Challenge Level: Challenge Level:1
Dicey Operations for an adult and child. Can you get close to 1000 than your partner?
problem icon
Intersection Sudoku 1
Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2
A Sudoku with a twist.
problem icon
The Remainders Game
Stage: 2 and 3 Challenge Level: Challenge Level:2 Challenge Level:2
A game that tests your understanding of remainders.
problem icon
Square it for Two
Stage: 1, 2, 3 and 4 Challenge Level: Challenge Level:1
Square It game for an adult and child. Can you come up with a way of always winning this game?
problem icon
Square It
Stage: 1, 2, 3 and 4 Challenge Level: Challenge Level:1
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
problem icon
Ratio Sudoku 2
Stage: 3 and 4 Challenge Level: Challenge Level:1
A Sudoku with clues as ratios. | {
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data was given in columns in the order {y, x, z}, with y dependent on x, z
points = {{144, 18, 52}, {142, 24, 40}, {124, 12, 40}, {64, 30,
48}, {96, 30, 32}, {74, 26, 56}, {136, 26, 24}, {54, 22,
64}, {92, 22, 16}, {96, 14, 64}, {92, 10, 56}, {82, 10, 24}, {76,
6, 48}, {68, 6, 32}};
equation2 = A + Bx + Cz + D*x^2 + Ez^2 + Fx*z;
nlm = NonlinearModelFit[points, equation2, {A, B, C, D, E, F}, {x, z}, Method -> NMinimize]
I get errors, and nothing that gives the values of the coefficients A,B,C,D,E,F...
What am I doing wrong?
share|improve this question
Fit[RotateRight[#, 1] & /@ points, {1, x, z, x x , z z, x z}, {x, z}] – Dr. belisarius Jun 2 '14 at 3:36
Belisarius: thank you very much, I cut and pasted what you wrote and shift+entered and got a result that seems to work. I'm still trying to figure out how all of your syntax works...but I'll take it for now! You're a boss! – Matthew Lee Jun 2 '14 at 4:43
points = {{144, 18, 52}, {142, 24, 40}, {124, 12, 40}, {64, 30, 48}, {96, 30,
32}, {74, 26, 56}, {136, 26, 24}, {54, 22, 64}, {92, 22, 16}, {96, 14,
64}, {92, 10, 56}, {82, 10, 24}, {76, 6, 48}, {68, 6, 32}};
Previous answer from belisarius:
fit = Fit[RotateRight /@ points, {1, x, z, x x, z z, x z}, {x, z}]
(* -9.44205 + 0.824875*x + 0.000781796*x^2 + 0.175003*z - 0.0102413*x*z + 0.00150438*z^2 *)
Alternate
fit == ((lmf =
LinearModelFit[RotateRight /@ points, {x, z, x^2, z^2, x*z}, {x, z}]) // Normal)
(* True *)
Note that built-in symbols (e.g., E) cannot be used as user-defined symbols. As a general rule, all user-defined symbols should start with a lower case letter to avoid naming conflicts with built-in symbols.
equation2 = a + b*x + c*z + d*x^2 + e*z^2 + f*x*z;
(nlm = NonlinearModelFit[RotateRight /@ points,
equation2, {a, b, c, d, e, f}, {x, z}, Method -> NMinimize]) // Normal // Quiet
(* -9.49492 + 0.825614*x + 0.000777544*x^2 + 0.175796*z - 0.0102451*x*z + 0.00150132*z^2 *)
nlm["BestFitParameters"]
(* {a -> -9.49492, b -> 0.825614, c -> 0.175796, d -> 0.000777544, e -> 0.00150132, f -> -0.0102451} *)
However, since you said that y is the dependent variable, RotateLeft should be used above rather than RotateRight
RotateRight[{y, z, x}]
(* {x,y,z} *)
RotateLeft[{y, z, x}]
(* {z,x,y} *)
share|improve this answer
Thank you for your input. This method also works, I shouldn't have used A,B,C,D,E,F in the example, my code had greek letters, but when I copy and pasted the syntax was very busy, so I substituted capital letters to make it look cleaner in the forum. Your method looks great and I can see how it works. Thank you again for your time, good sir! – Matthew Lee Jun 2 '14 at 5:02
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6,917,876,730,641,124,000 | ch 10 summarizing the data
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Ch. 10: Summarizing the Data. Criteria for Good Visual Displays. Clarity Data is represented in a way closely integrated with their numerical meaning. Precision Data is not exaggerated. Efficiency Data is presented in a reasonably compact space. Frequency Distribution Example.
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criteria for good visual displays
Criteria for Good Visual Displays
• Clarity
• Data is represented in a way closely integrated with their numerical meaning.
• Precision
• Data is not exaggerated.
• Efficiency
• Data is presented in a reasonably compact space.
measures of central tendency determining the median
Measures of Central Tendency: Determining The Median
• Arrange scores in order
• Determine the position of the midmost score: (N+1)*.50
• Count up (or down) the number of scores to reach the midmost position
• The median is the score in this (N+1)*.50 position
measures of central tendency the arithmetic mean
Measures of Central Tendency: The Arithmetic Mean
• The balancing point in the distribution
• Sum of the scores divided by the number of scores, or
measures of central tendency the mode
Measures of Central Tendency: The Mode
• The most frequently occurring score
• Problem: May not be one unique mode
symmetry and asymmetry
Symmetry and Asymmetry
• Symmetrical (b)
• Asymmetrical or Skewed
• Positively Skewed (a)
• Negatively Skewed (c)
comparing the measures of central tendency
Comparing the Measures of Central Tendency
• If symmetrical: M = Mdn = Mo
• If negatively skewed: M < Mdn Mo
• If positively skewed: M > Mdn Mo
measures of spread types of ranges
Measures of Spread:Types of Ranges
• Crude Range: High score minus Low score
• Extended Range: (High score plus ½ unit) minus (Low score plus ½ unit)
• Interquartile Range: Range of midmost 50% of scores
descriptive vs inferential formulas
Descriptive vs. Inferential Formulas
• Use descriptive formula when:
• One is describing a complete population of scores or events
• Symbolized with Greek letters
• Use inferential formula when:
• Want to generalize from a sample of known scores to a population of unknown scores
• Symbolized with Roman letters
variance descriptive vs inferential formulas
Descriptive Formula
Inferential Formula
Called the “unbiased estimator of the population value”
Variance: Descriptive vs. Inferential Formulas
the normal distribution
The Normal Distribution
Standard Normal Distribution: Mean is set equal to 0, Standard deviation is set equal to 1
standard scores or z scores
Standard Scores or z-scores
• Raw score is transformed to a standard score corresponding to a location on the abscissa (x-axis) of a standard normal curve
• Allows for comparison of scores from different data sets.
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-4,784,814,129,572,596,000 | Médianes dans un triangle
On veut montrer que les trois médianes dans un triangle sont concourantes, et préciser la position de leur point d’intersection.
Soit ABC un triangle.
Tracer les médianes [CI] et [AJ] de ce triangle. Elles se coupent en G.
Tracer le point D, symétrique de B par rapport à G.
1. Montrer que (CI) et (DA) sont parallèles.
Dest le symétrique de B par rapport à G donc G est le milieu de [BD].
Dans le triangle ADB:
I est le milieu de [AB] et G est le milieu de [BD], donc (AD)//(GI) d'après le premier théorème des milieux.
donc (CI)//(AD).
2. Montrer que (AJ) et (CD) sont parallèles.
Dans le triangle CDB:
J est le milieu de [BC] et G est le milieu de [BD], donc (GJ)//(CD) d'après le premier théorème des milieux.
donc (AJ)//(AD).
3. Montrer que GADC est un parallélogramme.
(CI)//(AD) et (AJ)//(AD) donc GADC est un parallélogramme par définition.
4. (BG) coupe (AC) au point O.
Montrer que O est le milieu de [AC].
GADC est un parallélogramme, or les diagonales d'un parallélogramme se coupent en leur milieu, donc O est le milieu de [AC].
Que représente (BO) pour ABC ?
O est le milieu de [AC], donc (BO) est la médiane issue de B de ABC.
5. Conclusion : que peut on dire des médianes des trois côtés du triangle ?
Les trois médianes (AJ), (CI) et (BO) se coupent au point G. Elles sont concourantes.
6. a ) Monter que BG = 2GO
GADC est un parallélogramme, donc O est le milieu de [GD], donc GO = OD.
G est le milieu de [BD] donc BG = GD.
donc BG = 2GO.
b ) Donc GO = BO
Conclusion : les trois médianes d'un triangle sont concourantes.
Leur point d'intersection est le centre de gravité du triangle.
Il est situé au tiers de chaque médiane.
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} | 0c090d63199a0a01e3b08e4a255778a0 |
6,953,658,497,810,924,000 | Friday
December 13, 2013
Homework Help: math
Posted by Chris on Monday, October 3, 2011 at 11:48pm.
construct-angle-f-so-that-m-angle-f-equals-2-m-angle-c
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} | 0c090d63199a0a01e3b08e4a255778a0 |
4,756,376,511,029,416,000 | 二分查找法基本原理和实践
概述
前面算法系列文章有写过分治算法基本原理和实践，对于分治算法主要是理解递归的过程。二分法是分治算法的一种，相比分治算法会简单很多，因为少了递归的存在。
在计算机科学中，二分查找算法（英语：binary search algorithm），也称折半搜索算法（英语：half-interval search algorithm）、对数搜索算法（英语：logarithmic search algorithm）[2]，是一种在有序数组中查找某一特定元素的搜索算法。搜索过程从数组的中间元素开始，如果中间元素正好是要查找的元素，则搜索过程结束；如果某一特定元素大于或者小于中间元素，则在数组大于或小于中间元素的那一半中查找，而且跟开始一样从中间元素开始比较。如果在某一步骤数组为空，则代表找不到。这种搜索算法每一次比较都使搜索范围缩小一半。
二分查找算法在情况下的复杂度是对数时间。二分查找算法使用常数空间，无论对任何大小的输入数据，算法使用的空间都是一样的。除非输入数据数量很少，否则二分查找算法比线性搜索更快，但数组必须事先被排序。尽管特定的、为了快速搜索而设计的数据结构更有效（比如哈希表），二分查找算法应用面更广。
二分查找算法有许多中变种。比如分散层叠可以提升在多个数组中对同一个数值的搜索。分散层叠有效的解决了计算几何学和其他领域的许多搜索问题。指数搜索将二分查找算法拓宽到无边界的列表。二叉搜索树和B树数据结构就是基于二分查找算法的。
入门 demo
对二分法的概念了解后，下面来看一道示例：
153. 寻找旋转排序数组中的最小值
已知一个长度为 n 的数组，预先按照升序排列，经由 1 到 n 次 旋转 后，得到输入数组。例如，原数组 nums = [0,1,2,4,5,6,7] 在变化后可能得到：
若旋转 4 次，则可以得到 [4,5,6,7,0,1,2]
若旋转 7 次，则可以得到 [0,1,2,4,5,6,7]
注意，数组 [a[0], a[1], a[2], ..., a[n-1]] 旋转一次 的结果为数组 [a[n-1], a[0], a[1], a[2], ..., a[n-2]] 。
给你一个元素值 互不相同 的数组 nums ，它原来是一个升序排列的数组，并按上述情形进行了多次旋转。请你找出并返回数组中的最小元素 。
示例 1：
输入：nums = [3,4,5,1,2]
输出：1
解释：原数组为 [1,2,3,4,5] ，旋转 3 次得到输入数组。
示例 2：
输入：nums = [4,5,6,7,0,1,2]
输出：0
解释：原数组为 [0,1,2,4,5,6,7] ，旋转 4 次得到输入数组。
示例 3：
输入：nums = [11,13,15,17]
输出：11
解释：原数组为 [11,13,15,17] ，旋转 4 次得到输入数组。
提示：
n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
nums 中的所有整数 互不相同
nums 原来是一个升序排序的数组，并进行了 1 至 n 次旋转
下面来看一下我写的一个失败版的答案，此时的我还没入门二分法：
class Solution {
public int findMin(int[] nums) {
int left = 0;
int right = nums.length-1;
while (left<=right) {
int middle = left + (right -left)/2;
if (nums[middle] > nums[left]) {
left = middle + 1;
}else {
right = right-1;
}
}
return nums[left];
}
}
输入：[4,5,6,7,8,9,10,0,1,2,3]
输出：10
结果：0
可以看到结果是不对，那这里的问题是什么呢？都说失败是成功之母，我们只有分析清楚为啥我们的解法会存在问题，才能更好地明白二分法的精髓。
先从答案分析，这里输出 10，为啥会是 10。
看上面这张图，代码逻辑写的是 middle > left，那么 left = middle +1； 这个逻辑这么写是没有问题的。
接着看，当不满足 middle > left，说明 middle 处于最小值的右半部分，这时候我们让 right--。那如果 right 就是最小值呢，这时候就会错过最小值。
还有如果 middle 是最大值呢？那么 left= middle +1 就是最小值，此时你再去计算 middle ，就直接把最小值错过了。比如输入数组：[5,6,7,8,9,0,1,2,3,4]；
还要考虑一种特殊情况，如果此时只有两个元素了，有两种情况 [1，2]，[2，1] ，这时候如果按照 right--，就会直接取到第一个元素。所以在 middle 和 left 相等的时候也要在做额外的判断。
完整版通过代码如下：
class Solution {
public int findMin(int[] nums) {
int left = 0;
int right = nums.length-1;
while (left<right) {
int middle = left + (right -left)/2;
if (nums[middle] > nums[left] && nums[middle] > nums[right]) {
left = middle +1;
// 说明最小值就在最右边，此时处于只有两个元素的时候 }
else if(middle == left && nums[left] > nums[right]) { left = right; } else { right = right-1; } } return nums[left]; } }
当你看到这段代码后，你懵逼了，这还是二分法嘛，分析下来这么复杂。
那我们来看下官方给的代码：
class Solution {
public int findMin(int[] nums) {
int low = 0;
int high = nums.length - 1;
while (low < high) {
int pivot = low + (high - low) / 2;
// 最小值一定是在和 high 在一个区间内的，所以这里要判断 pivot 和 high 的大小关系，不能去判断和 low 的关系
if (nums[pivot] < nums[high]) { high = pivot; } else { low = pivot + 1; } } return nums[low]; } }
是不是觉得官方代码简洁易懂。
那为啥这两个解法的代码会差这么多，答案在于 middle 到底是应该和 left 比较，还是应该和 right 比较。
这也说明了方向的选择的重要性。可是我们应该怎么选择呢。这个主要是在分析问题的时候要想清楚。个人觉得也可以这么理解：
本题是找最小值的。从最小值到最右端，这其实就是单调递增的，因此我们只要关注右半部分，抛弃左半部分就好。
那么本题错误原因就是跟左边进行比较，你再怎么找，最后得出值都不在这一部分上，就导致你得添加很多额外的逻辑来确保可以找到值。
PS：对于二分法要时刻关注只有两个元素的情况。这时候 middle = left。这时候注意 left 和 right 之间的关系。
通过这道题目相信大家已经对二分法有一定的认识了。
二分法思想
二分查找的思想就一个：逐渐缩小搜索区间。 如下图所示，它像极了「双指针」算法，left 和 right 向中间走，直到它们重合在一起：
根据看到的中间位置的元素的值 nums[mid] 可以把待搜索区间分为三个部分：
• 情况 1：如果 nums[mid] = target，这时候我们直接返回即可。
• 情况 2： target 在 mid 左半部分 [left..mid - 1]，此时分别设置 right = mid - 1 ；
• 情况 3： target 在 mid 右半部分 [mid+1..right]，此时分别设置 left = mid + 1。
这样就可以获得二分法基本模板：
class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1; // 确保 left 和 right 都在数组可取范围内
while (left <= right) { // < 还是 <= 按照自己的习惯即可
int mid = left + (right -left)/2;
if (nums[mid] == target) { // 找到结果就返回
return mid;
}else if(nums[mid] > target) {
right = mid-1;
} else {
left = mid +1;
}
}
// 退出循环就说明没找到
return -1; } }
虽然我们看到的写法有很多，但思想就这一个；为什么总是有朋友觉得二分难？因为有很多二分的写法，虽然都对，但是对于新手朋友们来说有一定干扰，因为不同的写法其实对应着不同的前提和应用场景，比起套用模板，审题、练习和思考更重要。「二分查找」就几行代码，完全不需要记忆，也不应该用记忆的方式解题.
下面解释一些细节：
1、模板的结束条件是 left <= right ，也就是结果一定是在 while 里面找到的。否则就是没找到。
2、有些学习资料上说 while (left < right) 表示区间是 [left..rihgt) ，为什么你这里是 [left..rihgt]？
区间的右端点到底是开还是闭，完全由编写代码的人决定，不需要固定。主要还是看你 left 和 right 的取值。 如果 right = nums.length ; 那么其实 right 这个位置是取不到的，也就是开区间了。所以开闭就是看点位能不能取到。
3、有些学习资料给出了三种模板，例如「力扣」推出的 LeetBook 之 「二分查找专题」，应该如何看待它们？
回答：三种模板其实区别仅在于退出循环的时候，区间 [left..right] 里有几个数。
• while (left <= right) ：退出循环的时候，right 在左，left 在右，区间为空区间，所以要讨论返回 left 和 right；
• while (left < right) ：退出循环的时候，left 与 right 重合，区间里只有一个元素，这一点是我们很喜欢的；
• while (left + 1 < right) ：退出循环的时候，left 在左，right 在右，区间里有 2 个元素，需要编写专门的逻辑。这种写法在设置 left 和 right 的时候不需要加 1 和减 1。
看似简化了思考的难度，但实际上屏蔽了我们应该且完全可以分析清楚的细节。退出循环以后一定要讨论返回哪一个，也增加了编码的难度。
我个人的经验是：
• while (left <= right) 用在要找的数的性质简单的时候，把区间分成三个部分，在循环体内就可以返回；
• while (left < right) 用在要找的数的性质复杂的时候，把区间分成两个部分，在退出循环以后才可以返回；
• 完全不用 while (left + 1 < right) ，理由是不会使得问题变得更简单，反而很累赘。
很多题目在二分法的基础上有变化，我们要学会灵活变化。还要理解题意。
示例：
给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。
请必须使用时间复杂度为 O(log n) 的算法。
示例 1:
输入: nums = [1,3,5,6], target = 5
输出: 2
示例 2:
输入: nums = [1,3,5,6], target = 2
输出: 1
示例 3:
输入: nums = [1,3,5,6], target = 7
输出: 4
示例 4:
输入: nums = [1,3,5,6], target = 0
输出: 0
示例 5:
输入: nums = [1], target = 0
输出: 0
提示:
• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums 为无重复元素的升序排列数组
• -104 <= target <= 104
class Solution {
public int searchInsert(int[] nums, int target) {
int left =0;
int right = nums.length -1;
while (left<=right) {
int mid = left + (right-left)/2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid]>target) {
right = mid-1;
}else {
left = mid+1;
}
}
// 没找到，那么 left 就是它所处的位置
return left;
}
}
注意一点：二分法只是用于有序数组，如果是无序的，此时是无法确定边界的，这时候我们就需要自己创造条件，找到数组的有序部分。
比如下面两道，大家可以自己找二分法题目去练习。
33. 搜索旋转排序数组
81. 搜索旋转排序数组 II
关于二分法的理论就讲到这里了，剩下的就是靠大家多多练习了。
算法系列文章：
滑动窗口算法基本原理与实践
广度优先搜索原理与实践
深度优先搜索原理与实践
双指针算法基本原理和实践
分治算法基本原理和实践
动态规划算法原理与实践
算法笔记
参考文章
https://leetcode-cn.com/problems/search-insert-position/solution/te-bie-hao-yong-de-er-fen-cha-fa-fa-mo-ban-python-/
posted @ 2021-07-25 23:20 huansky 阅读(2958) 评论(0编辑 收藏 举报 | {
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Decimals: Multiplying by 100
In this worksheet, students multiply numbers by 100 by shifting digits two places to the left in a grid.
'Decimals: Multiplying by 100' worksheet
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Decimals
Difficulty level:
Worksheet Overview
When we multiply numbers by 100, we do not simply add 00 to the end.
When the numbers are decimals, this rule does not work.
54.6 × 100 does not equal 54.600
When we multiply numbers by 100, each digit must move two places to the left.
When we look at the columns in a Th, H, T, U grid we notice that Thousands are 100 times bigger than Tens, Hundreds are 100 times bigger than units, Tens are 100 times bigger than tenths etc...
*Note that the U for Units is another term for Ones.
e.g. Work out:
54.6 × 10 = 5460
Here is 54.6 in a Th, H, T, U grid.
Place value grid
Answer:
Each digit moves two places to the left.
Remember to fill in that 0 in the units column.
So 54.6 × 100 = 5460
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1,907,933,579,612,044,500 |
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Math Forum - Pre-Calculus
Pre-Calculus Help Forum: Pre-calculus does not involve calculus, but explores topics that will be applied in calculus
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Math Help - If two random variable have teh joint density
1. #1
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If two random variable have teh joint density
If two random variable have the joint density given by
f(x,y) = \begin{cases} 2e^{-2y} e^{-x} & 0<x< \infty , o<y<\infty\\<br />
0 & otherwise \end{cases}
Are X and Y independent ?
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2. #2
Moo
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Hello,
If you can write the joint pdf as the product of a function which only depends on x, and a function which only depends on y, then they're independent.
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3. #3
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You can see that they are indep by inspection and that the mariginal densities are
f_X(x)=e^{-x}I(x>0)
and
f_Y(y)=2e^{-2y}I(y>0)
an exponential and a gamma.
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4. #4
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Quote Originally Posted by Moo View Post
Hello,
If you can write the joint pdf as the product of a function which only depends on x, and a function which only depends on y, then they're independent.
i didnt understand .......could u please elaborate
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5. #5
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Quote Originally Posted by wolfyparadise View Post
i didnt understand .......could u please elaborate
You have been told how to do this. You are expected to have sufficient understanding of things so that explanations such as those given make sense.
Go back to your textbook or class notes and review this material. Also, click on 5.2 here: Probability density function - Wikipedia, the free encyclopedia
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6. #6
MHF Contributor matheagle's Avatar
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Quote Originally Posted by wolfyparadise View Post
i didnt understand .......could u please elaborate
no mite
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Greatest Common Factor
Suppose that in some distant part of the universe that a star with four orbiting planets.One planet makes a trip around the star in 6 years, the second planet 9 years, the third takes 15 years, and the fourth takes 28 ...
Greatest Common Factor
There are 40 girls and 32 boys who want to participate in the relay race. If each team must have the same number
of girls and boys, what is the greatest number of teams that can ...
Greatest Common Factor
The greatest common factor of a number and 48 is 24. The sum of the numbers digits is 15. find the number
Greatest Common Factor
volunteers at a bake sale want to sell slices of banana nut bread and raisin bread packaged together. They have 63 slices of banana nut bread and 45 slices of raisin bread, and plan to use all the bread. What is the greatest number of packages they can put ...
GCF
The Coverall Paint Company wants to sell paint in can sizes that most people would use up completely. The company has found out that most people paint wall surfaces of 300m2, 450m2, or 600m2...
Greatest Common Factor
There are 15 boys and 12 girls in a classroom. For a class trip, they need to be split up into the largest possible even ...
Greatest Common Factor
Two even numbers have a greatest common factor of 150...
GCF
find the gcf of 78 and42
Greatest Common Factor
The length of a rectangle is 7 ft greater than the width. The area is 60ft^2...
Greatest Common Factor
sarim has $1 in coins. one-fifth of the coins are dimes, two-fifteenths are nickels, and two-thirds are ...
GCF
sides of square are lengthened by 8cm, the area becomes 256cm^2. Find the length of a side of the original square
GCF
7TH & 8TH GRADE CLASS IS PLANNING A DINNER. THERE ARE 168 STUDENTS IN THE 7TH GRADE & 112 STUDENTS IN THE 8TH GRADE. EACH GRADE SITS IN A DIFFERENT ROOM. HOW MANY CHAIRS ARE NEEDED IN EACH ROOM IF THE TOTAL NUMBER OF TABLES PER ROOM CANNOT EXCEED 15 AND THE TOTAL NUMBER OF CHAIRS PER TABLE CANNOT EXCEED 15. EACH TABLE IN EACH ROOM SHOLD HAVE THE SAME NUMBER OF CHAIRS
Greatest Common Factor
The width of a rectangle is 6 inches less than twice its ...
Greatest Common Factor
the gcf of two numbers is 4 one of the numbers is 36 find the other number that is between.
GCF
Brandon and carlos are members of the high school track team. They run different events; both events are multiple laps. At a recent track meets, carlos\' 3rd place time for his event was 8 minutes and 32 seconds. Coincidentally, carlos\' lap time was exactly twice as long as Brandon\'s. Find their lap times - they both greater than 30 seconds. (Hint: convert time to second first)
Greatest Common Factor
Twelve more than the square of a number is seven times the ...
Greatest Common Factor
Ellen pours one-half of tea from a pot. Juanita pours one-third of the remaining tea and leaves four-cups of tea in the ...
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-8,087,511,004,069,725,000 | Teacher Notes
Knowing how do you decompose an improper fraction into a sum of a whole number and a number less than 1 is the first step of understanding a mixed number. You should start with using visual models to build understanding. When adding and subtracting mixed numbers we are joining or separating parts within the same whole.
Student Knowledge Goals
I know that mixed numbers can be written as fractions.
I can use the properties of operations to solve addition and subtraction problems involving mixed numbers with like denominators.
I can use what I know about whole number addition and subtraction to solve problems with mixed numbers.
I can use what I know about adding and subtracting fractions to solve problems with mixed numbers.
I know that mixed numbers can be combined or separated (composed and decomposed).
I can use a variety of strategies for adding and subtracting mixed numbers.
I understand that mixed numbers can be combined or separated (composed and decomposed).
Vocabulary
improper fraction
mixed number
Lessons
__Engage NY Module 5 F-29__ – Estimate sums and differences using benchmark numbers.
__Engage NY Module 5 F-30__ – Add a mixed number and a fraction.
__Engage NY Module 5 F-31__ – Add mixed numbers.
__Engage NY Module 5 F-32__ – Subtract a fraction from a mixed number.
__Engage NY Module 5 F-33__ – Subtract a mixed number from a mixed number.
__Engage NY Module 5 F-34__ – Subtract mixed numbers.
Student Video Lessons
__Learn Zillion__ – Add and subtract mixed numbers with like denominators
__Virtual Nerd__ – Understand a fraction a/b with a > 1 as a sum of fractions 1/b
__Study Jams__ – Adding and Subtract mixed numbers
Online Problems and Assessments
__Khan Academy__ – Questions and Video Lessons
Add and subtract mixed numbers with like denominators
Online Games
__Adding Subtracting Mixed Numbers__
Printables
__Fraction Bars__
__Benchmark Fraction Strips__
__Adding Subtracting Mixed Numbers__
Assessment Task 1
Assessment Task 2
Assessment Task 3
Assessment Task 4
Assessment Task 5
Assessment Task 6
Assessment Task 7
Assessment Task 8 | {
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-8,300,796,214,402,981,000 | fbpx
Extending Decimal Place Value Understanding Through Problem Solving
Models and manipulatives build understanding of decimal place value
Written by Donna Boucher
Donna has been a teacher, math instructional coach, interventionist, and curriculum coordinator. A frequent speaker at state and national conferences, she shares her love for math with a worldwide audience through her website, Math Coach’s Corner. Donna is also the co-author of Guided Math Workshop.
I was planning with my 5th-grade team today, and they’re getting ready to move into decimal place value. The 5th-grade standard is to extend decimal place value from the hundredths place to the thousandths place. We decided to start with a very quick Ticket In the week before the unit actually starts to get a feel for what the kiddos retained from 4th grade.
On Day 1, we’re going to focus on reading decimal numbers, starting out with decimals to the hundredths (4th-grade skill) and extending to the thousandths (5th-grade skill).
Write the following on the board/document camera: (2 x 100) + (3 x 10) + (6 x 1) + (1 x 0.1) + (5 x 0.01). Talk with your partner about how you would write this number in standard form. Accept responses and have a class discussion about which the class feels is the correct way to write the number. Be sure that they come to the conclusion that it should be written 236.15. Who thinks they can read the number? Accept responses. You may have students that try to read it two hundred thirty-six point fifteen because they often hear adults use that shortcut. Tell kiddos that mathematicians are more precise and would read it two hundred thirty-six and fifteen hundredths. We say the word and when we reach the decimal point, and we always include the decimal place value name.
Write the decimal place value names above the number and ask if students notice anything about the place value names. If they don’t notice the symmetry between the whole and decimal place value names, draw the arrows as shown below to illustrate the relationship.
Change the number to 1,236.158 by adding a 1 in the thousands place and an 8 in the thousandths place. Turn and talk to your partner. How should we label the place value position names of the new digits? You’re hoping that they will conclude that since the 1 is in the thousands place, then the 8 must be the thousandths place. Remember, you’re wanting them to see the relationship between the place value positions.
Continue practicing reading decimal numbers to the thousandths. Be sure to write the numbers with the place value names labeled to support the learning. Try letting the student who reads the number correctly be the one to write the next number.
Give a quick Ticket Out to determine how well students understood the concept.
Day 2 moves us into comparing decimals. Again, we’ll frame it in a problem-solving context.
Give each pair of students the two place value mats shown below, manipulative money (3 dollar bills, 10 dimes, and 10 pennies), and base-10 blocks (3 flats, 10 rods, and 10units).
Models and manipulatives build understanding of decimal place value
Huh. I wonder which is bigger, 0.9 or 0.09? Let’s read these decimals to make sure we’re clear on them. Nine-tenths and nine-hundredths. Great! I’ve given you some materials. You may use these materials to prove your answer. You also need to be able to explain in words and pictures how you decided which is the bigger decimal.
Give students some time to work in pairs on the problem, and then come back together to discuss. See if students can make any generalizations.
Give students additional numbers to build and compare. Here are some good ones to try: 0.17 and 0.4; 1.08 and 1.18; 2.2 and 2.24; 1.5 and 1.67
Be sure to bring the class back together to discuss their results and strategies for comparing. If you feel they have really grasped this lesson, you might try…
Huh. I wonder which is bigger, 0.102 or 0.14? 🙂
Click here to grab your mats! Looking for additional decimal resources? Check out my Common Core: Decimals & Fractions Using Models and Manipulatives unit. It’s a best-seller!
12 Comments
1. MS. OCD
Awesome! Thank you!
Reply
• Donna Boucher
My pleasure!
Reply
2. TheElementary MathManiac
Great ideas! I love connecting this to money.
Reply
• Donna Boucher
It’s something that really clicks with the kiddos! They ALL know money! Ha ha.
Reply
3. Deb K.
Donna, these are really great ideas that I’ll save for next year. I love your tickets in and out too! Is there any way you could share “blank” ones that we could use for other concepts? We’re working on %, decimals, fractions now and I’d love to use your tickets! As always, you have great ideas for teaching math concepts.
Debbie
Reply
• Donna Boucher
Hey, Debbie! That’s actually why I uploaded them as Word documents, not PDF files. You can download them and change them however you want!
Reply
4. SiouxGeonz
I just learned about a neat way to show place value with decimals (and do a little diagnosis of understanding of fractions), tho’ it’s a candy lesson and edible lessons have their issues… I s’pose it could be done with strips of paper, though.
Start with 100 Twizzlers and show that place… then ten… then 1…
Then … into Decimal land! what would 1/10 of the twizzler be? Break it off (but you can’t eat it yet…) That’s going ot be the tenths place.
Then… take that little piece~ can you find 1/10 of it?
Nifty room for discussion…
Reply
• Donna Boucher
100 Twizzlers?! Yum! 🙂
Reply
5. Tami O'Keefe
Do you have any mats that are not decimal related., I managed to find some in my closet at school but your decimal ones are way cuter.
Reply
• Donna Boucher
Oddly enough, Tami, I actually didn’t have a cute one for whole numbers. It was easy enough for me to modify the decimal one, though. Grab it here.
Reply
6. Raul
You’re amazing Donna, thank you
Reply
• Savannah
Hello!
Thank you so much for the detailed information you share on your blog. Plus the resources! I was wondering if you have a post on how you would suggest introducing decimals to students in fourth? I’ve searched your posts but didn’t quite find one on this topic. Your insights are greatly appreciated!
Reply
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2,558,079,489,214,104,000 | How Many 1/3 Cup Servings are in 4 Cups?
If you’re wondering how many 1/3 cup servings are in 4 cups, you’ve come to the right place. This seemingly simple question is actually a bit more complicated than it appears on the surface. Fortunately, we’re here to break down the math and help you figure out the answer.
How Much is 1/3 in a Cup?
Before we can answer the main question, let’s start with a basic definition. When we talk about 1/3 cup, we’re referring to a measurement of volume that equals exactly one-third of a cup. This can be written as a fraction (1/3) or as a decimal (0.3333).
How Many 1/3 Cups are in 3/4 Cup?
To figure out how many 1/3 cup servings are in 4 cups, let’s start by looking at a slightly simpler question: how many 1/3 cups are in 3/4 cup?
To solve this problem, we need to divide 3/4 by 1/3. Let’s break it down step by step:
• First, we’ll convert both fractions to decimals. 3/4 is equal to 0.75, and 1/3 is equal to 0.3333.
• Next, we’ll divide 0.75 by 0.3333. The answer is approximately 2.25.
• So, there are 2 and a quarter (or 2 1/4) servings of 1/3 cup in 3/4 cup.
How Much is 3/4 Cup Serving?
Now that we know how many 1/3 cup servings are in 3/4 cup, let’s take a quick look at what a standard 3/4 cup serving looks like. This measurement is equal to 12 tablespoons or 6 fluid ounces.
What is 1/3 of 4 Cups?
Now, let’s move on to the main question: how many 1/3 cup servings are in 4 cups?
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To solve this problem, we can use the same process that we used for the previous question. Here are the steps:
• First, we’ll convert 4 cups to 12/3 cups (since there are 3 cups in a quart). This gives us 12 servings of 1/3 cup.
• Next, we’ll multiply 12 by the number of 1/3 cup servings in 3/4 cup (which we found to be 2.25 in the previous section). The answer is approximately 27.
• So, there are approximately 27 servings of 1/3 cup in 4 cups.
How Many 1/4 Make 1/3?
Another way to think about this question is to consider how many 1/4 cup servings are in a 1/3 cup serving. To determine this, we can divide 1/3 by 1/4:
• First, we’ll invert the second fraction to make it a divisor. 1/4 becomes 4/1.
• Next, we’ll multiply 1/3 by 4/1. The answer is 4/12 or 1/3.
• So, there are exactly 1 and one-third (or 1 1/3) servings of 1/4 cup in 1/3 cup.
How Many 1/3 Cups are in 3/4 Cup?
We already covered this question in a previous section, but it’s worth reiterating. There are 2 and a quarter (or 2 1/4) servings of 1/3 cup in 3/4 cup.
1/3 Times 4 in Fraction
When we multiply 1/3 by 4, we can simplify the expression by re-writing 4 as 12/3 (since 4 cups is the same as 12/3 cups):
1/3 x 12/3 = (1 x 12) / (3 x 3) = 12/9 = 4/3
So, 1/3 times 4 is equal to 4/3, which is also equal to 1 and one-third (or 1 1/3).
How Many 1/3 Cups are in 1 Cup?
To determine how many 1/3 cup servings are in 1 cup, we can use the same process that we used for the main question. Here are the steps:
• First, we’ll convert 1 cup to 3/3 cups (since there are 3 cups in a quart). This gives us 3 servings of 1/3 cup.
• Next, we’ll multiply 3 by the number of 1/3 cup servings in 3/4 cup (which we found to be 2.25 in a previous section). The answer is approximately 6.75.
• So, there are approximately 6 and three-quarter (or 6 3/4) servings of 1/3 cup in 1 cup.
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How Many 1/3 are There in 4?
If we’re asking how many 1/3’s are in the number 4 (rather than the measurement of 4 cups), the answer is 12. This is because 4 can be expressed as 12/3, which is the same as 12 servings of 1/3 cup.
How Much Does 1/3 of a Cup Add to 1/4 of a Cup?
To determine how much 1/3 cup adds to 1/4 cup, we need to add the two fractions together. Here are the steps:
• First, we’ll find a common denominator for the two fractions. The lowest common multiple of 3 and 4 is 12.
• Next, we’ll convert 1/3 and 1/4 to fractions with a denominator of 12. 1/3 becomes 4/12, and 1/4 becomes 3/12.
• Finally, we’ll add the two fractions together. 4/12 + 3/12 = 7/12.
• So, 1/3 cup adds 7/12 of a cup to 1/4 cup.
1/3 Cup 4 Times
If you need 1/3 cup four times, you’ll need a total of 1 and one-third cups (or 1 1/3 cups). This is because 1/3 + 1/3 + 1/3 + 1/3 = 4/3, which is equal to 1 and one-third.
Can I Use 1 Cup for a 3/4 Cup?
If a recipe calls for 3/4 cup and you only have 1 cup, you can use it as a substitute. Just keep in mind that you’ll end up with slightly more than 3/4 cup (since 1 cup is greater than 3/4 cup). To be precise, you’ll have 1 and one-third cups (or 1 1/3 cups).
Conclusion
In conclusion, there are approximately 27 servings of 1/3 cup in 4 cups. We hope that this guide has helped you understand how to calculate the number of 1/3 cup servings in various measurements, as well as how to add and subtract fractions. If you have any other questions about cooking or baking, feel free to reach out to us for more information. Happy cooking!
Related Post: The Ribbon vs Bow Debate: Which One is the Best For You? | {
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-7,053,725,251,735,142,000 | 2
$\begingroup$
Let the set $S_{n}$ = {$(x,y):x,y \in \mathbb{O}$} such that $x+y=n$ where $\mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples: $S_{10} =\{(3,7),(5,5),(7,3)\}$ and $f(10) = 3$.
$S_{14} = \{(3,11), (5,9), (7,7), (9,5),(11,3)\}$ and $f(14) = 5$
$S_{32} = \{(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)\}$
and $f(32) = 14$
It turns out that $f(n) = ((n/2) - 2)$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x \neq 3$ and $y\neq 3$.
Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3. They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).
There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n \to\infty$?
My guess is that as $n \to \infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.
$\endgroup$
0
$\begingroup$
The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $\left\lfloor \frac{n}{6}\right\rfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $\left\lfloor \frac{n}{6}\right\rfloor -1$ values of $q$ where $q$ is divisible by $3$.
So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:
$$g(n)=2\left\lfloor \frac{n}{6}\right\rfloor -2$$
The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:
$$g(n)=2\left\lfloor \frac{f\left(n\right)+2}{3}\right\rfloor -2$$
Your guess about the limit for $\frac{g(n)}{f\left(n\right)}$ is also right:
$$\frac{g(n)}{f\left(n\right)}=\frac{2\left\lfloor \frac{n}{6}\right\rfloor -2}{\frac{n}{2}-2}=\frac{\frac{n}{3}-2}{\frac{n}{2}-2}=\frac{2n-12}{3n-12}$$
$$\lim_{n\rightarrow\infty}\frac{2n-12}{3n-12}=\frac{2\infty-12}{3\infty-12}=\frac{2\infty}{3\infty}=\frac{2}{3}\left(\frac{\infty}{\infty}\right)=\frac{2}{3}$$
$\endgroup$
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4,443,198,002,984,576,000 |
Results 1 to 11 of 11
Math Help - solve for variable when vector a and b have a 45 degree angle between them
1. #1
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solve for variable when vector a and b have a 45 degree angle between them
vector a = (2,5) and vector b = (-1,t)
solve for t when the vectors have a 45 degree angle between them
any help would be appreciated.
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2. #2
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If I remember correctly you can use the dot product
a.b = |a||b|cos \theta
-2+5t = \sqrt{29} \cdot \sqrt{1+t^2} \cdot \dfrac{\sqrt{2}}{2}
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3. #3
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hm, yeah thats what I figured. I can't figure out how to solve from there on.
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4. #4
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Where are you having trouble? It just ends up becoming a quadratic after you square both sides.
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5. #5
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-2+5t = \dfrac{\sqrt{58(1+t^2)}}{2}
10t-4 = \sqrt{58+58t^2}
100t^2-80t+16 = 58+58t^2
42t^2-80t-42 = 0
21t^2-40t-21 = 0
This one does factor but it might be easier to use the quadratic formula
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6. #6
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I tried that but Ididn't get the correct answer. the answer is t=7/3
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7. #7
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Yes, that is the positive answer you get when you solve the quadratic.
21t^2 - 40 t - 21 = 0
t = \displaystyle\frac{40 \pm \sqrt{(-40)^ 2- 4(21)(-21)}}{2\times 21}
= \displaystyle\frac{40 \pm 58}{42}
Last edited by Quacky; January 11th 2011 at 09:39 AM. Reason: Clarified formula usage
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8. #8
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ah okay, i'll give it another go
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9. #9
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I've editted the post. Where was your mistake? I assume it was probably just a sign error or something.
Last edited by Quacky; January 11th 2011 at 09:38 AM. Reason: Politeness.
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10. #10
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Quote Originally Posted by colorado View Post
I tried that but Ididn't get the correct answer. the answer is t=7/3
Please show your working rather than just giving an answer. Then it is easier to diagnose what help you actually need.
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11. #11
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noted
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3,146,686,093,161,063,000 | 2
$\begingroup$
In the formulation of the Tragedy of the Commons below, what does $n$ exactly represent? Is a threshold? The text is from plato.stanford.edu website on Kuhn's Prisoner's dilemma.
Also, it says that $\bf D$ strongly dominates $\bf C$ for all players, and so rational players would choose $\bf D$ and achieve $0$, while preferring that everyone would choose $\bf C$ and obtain $C+B$.
I don't understand the last part. The reason $\bf D$ dominates $\bf C$ is because either the player gets $B$ or $0$ which are better from $C+B$ or $C$. Why would rational players prefer everybody to cooperate?
enter image description here
$\endgroup$
3
$\begingroup$
what does 𝑛 exactly represent? Is a threshold?
Yes, $n$ is the threshold above which the benefit $B$ realizes. Think of this as a situation where a public good with benefit $B$ is provided only if more than half of the population votes for it. $C$ is the cost of voting, and $n=\lceil\frac12N\rceil$ is the smallest integer greater than half of the population size ($N$).
Why would rational players prefer everybody to cooperate?
While a rational player would choose $\mathbf{D}$ regardless of what others would choose, he/she would still prefer the outcome with benefit $B$ to the outcome with $0$ benefit. Hence, he/she would prefer everyone else chooses $\mathbf{C}$, i.e. bearing the cost of providing benefit $B$, while he/she enjoys $B$ without incurring any cost by choosing $\mathbf{D}$.
Additionally, suppose there are $N$ players, and that $n<N-1$. Then with all the other $N-1$ players playing $\mathbf{C}$, it doesn't really matter if the remaining player chooses $\mathbf{C}$ or $\mathbf{D}$. So arguably it's less morally objectionable for the last player to choose $\mathbf{D}$.
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Topic: norm
Replies: 8 Last Post: Mar 10, 2013 1:45 PM
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fom
Posts: 1,968
Registered: 12/4/12
Re: norm
Posted: Mar 10, 2013 7:30 AM
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On 3/10/2013 6:26 AM, fom wrote:
> On 3/9/2013 3:58 PM, David C. Ullrich wrote:
>> On Sat, 09 Mar 2013 14:47:45 -0500, quasi <[email protected]> wrote:
>>
>>> novis wrote:
>>>> quasi wrote:
>>>>> novis wrote:
>>>>>
>>>>>> Suppose A is a p x q columnwise orthonormal matrix and suppose
>>>>>> x is any vector in R^p. Then what is the relation between
>>>>>> ||x|| and ||Ax|| ?
>>>>>
>>>>> A is a p x q matrix, so regarded as a function,
>>>>>
>>>>> A maps R^q to R^p.
>>>>>
>>>>> Thus,
>>>>>
>>>>> x is in R^q
>>>>>
>>>>> not in R^p as you specified, and
>>>>>
>>>>> Ax is in R^p
>>>>>
>>>>> Also, since A is columnwise orthonormal, it follows that
>>>>> p >= q.
>>>>>
>>>>> As far as norm comparison, since A is orthonormal,
>>>>>
>>>>> |Ax| = |x|
>>>>>
>>>>> where the norms are the usual Euclidean norms in R^p and R^q,
>>>>> respectively.
>>>>
>>>> Well I was talking about A transpose x or ||A'x||. Can you please
>>>> show how ||x||=||A'x||?
>>>
>>> OK, I missed your use of the symbol ' denoting transpose.
>>>
>>> So A' is a map from R^p to R^q.
>>>
>>> As before, since A is columnwise orthonormal, rank(A) = q,
>>> hence p >= q.
>>>
>>> For x in R^p, A'x is in R^q, and yes, it's true that
>>>
>>> |A'x| = |x|.
>>
>> I don't think so...
>
> Why not?
>
answered by quasi
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-9,158,422,996,429,729,000 | What is 34% of 158?
Solution and how to solve it
34% of 158 is 53.72
Calculating the percent of a number is simple, but can be a bit tricky if you aren't careful. Luckily, it only requires a few basic operations to get to the solution: multiplication and division. If you haven't learned what percentage is yet, or would like a little refresher, feel free to check out our introduction to percentage page. To solve percentage problems, it may be useful to use a calculator. But, they can also be solved by hand or in your head (if you practice enough). So, how did we get to the solution that 34 percent of 158 = 53.72?
• Step 1: As we know, a whole of something is equal to 100%. In this case, we want to find what 34% of 158 is. We know that 100% of 158 is, well, just 158.
• Step 2: If 100% of 158 is 158, then we can get 1% of 158 by dividing it by 100. Let's do 158 / 100. This is equal to 1.58. Now we know that 1% is 1.58.
• Step 3: Now that we know what 1% of 158 is, we just need to multiply it by 34 to get our solution! 1.58 times 34 = 53.72. That's all there is to it!
When is this useful?
Percentage is one of the most commonly used math concepts in day-to-day life. You can use it to calculate a gratuity on a restaurant bill, or to grade your score on an exam. It is useful to know your percentages well!
Other Percent of 158 Calculations Other 34% Calculations
What is 24% of 158? What is 34% of 148?
What is 25% of 158? What is 34% of 149?
What is 26% of 158? What is 34% of 150?
What is 27% of 158? What is 34% of 151?
What is 28% of 158? What is 34% of 152?
What is 29% of 158? What is 34% of 153?
What is 30% of 158? What is 34% of 154?
What is 31% of 158? What is 34% of 155?
What is 32% of 158? What is 34% of 156?
What is 33% of 158? What is 34% of 157?
What is 35% of 158? What is 34% of 159?
What is 36% of 158? What is 34% of 160?
What is 37% of 158? What is 34% of 161?
What is 38% of 158? What is 34% of 162?
What is 39% of 158? What is 34% of 163?
What is 40% of 158? What is 34% of 164?
What is 41% of 158? What is 34% of 165?
What is 42% of 158? What is 34% of 166?
What is 43% of 158? What is 34% of 167?
What is 44% of 158? What is 34% of 168?
Understanding Percentage Increase and Decrease
Another important application of percentage calculations is understanding how to calculate percentage increases and decreases. When you want to find out how much something has increased or decreased in percentage terms, you can use the following formula:
Percentage Change = (New Value - Old Value) / Old Value × 100%
Percentage change is used in various fields such as finance, economics, and science to measure the growth or decline of a specific value. It helps us to better understand and compare the changes in values over time.
Real-World Applications of Percentage Calculations
Here are some common real-world applications of percentage calculations:
• Discounts: When shopping, you may encounter discounts offered by stores. You can calculate the final price of an item after applying the percentage discount.
• Interest Rates: Banks and financial institutions use percentage calculations to determine the interest rate on loans or savings accounts. Knowing how to calculate interest can help you make informed decisions about your finances.
• Tax Rates: Tax rates are often expressed as a percentage. Being able to calculate the amount of tax you need to pay based on a given percentage can help you better manage your personal or business finances.
• Data Analysis: In data analysis, you may need to calculate the percentage change between two values or the percentage of a specific value in a dataset. This can provide valuable insights and make data-driven decisions.
Tips for Mastering Percentage Calculations
Here are some tips to help you become proficient in percentage calculations:
• Practice: The more you practice percentage problems, the better you'll get at them. Try solving different types of percentage problems to improve your skills.
• Understand the Concept: Make sure you have a clear understanding of the concept of percentages and how they work. This will make it easier to apply percentage calculations to real-world problems.
• Use Tools: There are many tools available, such as calculators and online resources, that can help you solve percentage problems. Make use of these tools to double-check your answers or to practice solving problems.
Comparing Percentages
Understanding how to compare percentages is important when making decisions or evaluating options. Let's use the example of 34% of 158 to demonstrate how to compare percentages.
• Greater Than: Is 34% of 158 greater than some other percentage of 158? To compare, calculate the other percentage and see which value is larger.
• Less Than: Is 34% of 158 less than another percentage of 158? Perform the same comparison as above, but check if the value is smaller instead.
• Equal To: Is 34% of 158 equal to another percentage of 158? Calculate both percentages and compare the values to determine if they are equal.
Percentage Increase and Decrease
Percentage increase and decrease are essential concepts for understanding how values change over time. Here's how to calculate the percentage increase or decrease using the example of 34% of 158:
• Percentage Increase: To calculate the percentage increase from an original value to a new value, divide the difference between the new and original values by the original value, and then multiply by 100. For example, if the original value was 158 and the new value is the result of a 34% increase, the calculation would be (((158 * (34/100)) + 158) - 158) / 158 * 100.
• Percentage Decrease: To calculate the percentage decrease, follow the same process as percentage increase, but subtract the percentage instead of adding it. The calculation would be (158 - (158 - (158 * (34/100)))) / 158 * 100.
Converting Percentages to Fractions and Decimals
Percentages can be converted to fractions and decimals for various mathematical operations or to express values in different forms. Here's how to convert 34% to a fraction and a decimal:
• Percentage to Fraction: To convert 34% to a fraction, simply write 34 as the numerator and 100 as the denominator. Then, simplify the fraction if possible.
• Percentage to Decimal: To convert 34% to a decimal, divide 34 by 100. This will give you the decimal equivalent of 34%. | {
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900,899,714,781,177,000 | Last updated on November 4th, 2023 at 06:43 am
PART 66 module 1 sample | maths exam | easa part 66 mathematics Exam Overall rating: ★★★★☆ 4.4 based on 507 reviews.
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Sample Questions.
The following Quiz contain 35 questions during 30 min.
Easa Part 66 Module 1 Questions
Category A – Mathematics Exams ( 16 questions 20 min),
Category B1 – Mathematics Exams ( 32 questions 40 min),
Category B2 – Mathematics Exams ( 32 questions 40 min),
Category B3 – Mathematics Exams ( 28 questions 35 min) ,
EASA PART 66 Academy provide all necessary martials for preparation of module 01 both easa part 66 mathematics questions and easa part 66 mathematics pdf books. This module is mandatory for A1 Airplane Turbine, B1.1 Airplane Turbine, B1.2 Airplane Piston, B1.3 Helicopter Turbine and for B2 Avionics.
Mathematics exam questions contain trees Sub-Modules:
Sub-Module 01 –Easa part 66 mathematics questions Arithmetic: contain also Arithmetical terms and signs, methods of multiplication and division, fractions and decimals, factors and multiples, weights, measures and conversion (ratio and proportion, averages and percentages, areas and volumes, squares, cubes, square and cube roots.
Sub-Module 02 – Mathematics exam questions Algebra Evaluating simple algebra expressions, addition, subtraction, multiplication and division, use of brackets, simple algebraic fractions; Linear equations and their solutions; Indices and powers, negative and fractional indices; Binary and other applicable numbering systems; Simultaneous equations and second-degree equations with one unknown; Logarithms.
For Sub-Module-03, EASA PART 66 Academy provide best explanation for EASA PART 66 mathematics questions Geometry with Simple geometrical constructions, Graphical representation; nature and uses of graphs, graphs of equations/functions and Simple trigonometry; trigonometrical relationships, rectangular and polar coordinates. All of this content available as easa part-66 pdf download, based on easa part 66 syllabus and regulations. Arithmetic in Aviation Maintenance is the branch of mathematics dealing with the properties and manipulation of numbers.
Performing arithmetic calculations with success requires an understanding of the correct methods and procedures. Arithmetic may be thought of as a set of tools. The aviation maintenance professional will need these tools to successfully complete the maintenance, repair, installation, or certification of aircraft equipment. Thus, easa part 66 mathematics questions Arithmetic is the basis for all aspects of mathematics. Math is used in measuring and calculating service ability of close tolerance engine components, when calculating the weight and balance for the installation of new avionics and more.
A sound knowledge and manipulation of mathematic principles is used on a regular basis during nearly all aspects of aircraft maintenance. The level to which an aviation maintenance student is required to have knowledge of arithmetic is listed on mathematics exam questions page, according to the certification being sought. A description of the applicable knowledge levels is presented and will be included at the beginning of each sub-module in the module.
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3,442,515,760,866,663,000 | "Airplanes in the Sky"
by Benjamin Kuipers
Many years ago, I was teaching second-semester calculus in the Fall term. This class tends to have a bimodal distribution of students, some stronger in STEM fields, and some stronger in the humanities and social sciences. These students taught me an important lesson.
The Problem
At one point, I was teaching a topic called "related rates", and I put the following problem on the board.
One Response
One group of students saw a right triangle, 400 miles on one side and 300 miles on the other. They also saw the Pythagorean Theorem, that lets them calculate that the hypotenuse is 500 miles (the first answer). And they knew (or learned) that the trick behind related rates problem is to differentiate the Pythagorean equation to get another equation involving rates (velocities), that they could solve for the second answer.
Pythagorean Theorem
This is the answer I was looking for, that the Calculus class is trying to teach.
A Different Response
The second group of students had a very different response to this problem. They didn't see the right triangle and the Pythagorean Theorem. They saw airplanes in the sky. And this led to a whole new set of questions.
The Two Cultures
A number of years ago (1959), British scientist and novelist C. P. Snow gave an important lecture that was published as a short book called The Two Cultures and the Scientific Revolution. (Very much worth reading!)
At the time, the problem was that the British intelligentsia only considered the humanities worthy of attention, and scorned the value of scientific knowledge. More than six decades later, the pendulum may have swung toward the other extreme.
Given a description of a situation . . .
Both of these approaches can provide powerful, useful conclusions. They complement each other, and can be used together. Often, useful abstractions can only be found after the original description of the situation has been expanded to include new factors. (Read "The Blind Men and the Elephant".)
Being able to use both approaches is important for everyone, but is especially important for research in artificial intelligence.
Written 3 July 2016 (after several decades of using this example when teaching AI).
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4,488,274,747,582,913,500 | Feed on
Posts
Comments
Math Help
Suppose a football team faces a third down and 5 yards to go situation.
Suppose that the team REALLY needs to make a first down (and that they always punt on 4th down no matter what).
Suppose their coaches love statistics. In deciding whether to execute a running play or passing play they look back at their own data to find that:
(1) On average when they choose to pass the football, they gain 7 yards per play.
(2) On average when they choose to run the football, they gain 3 yards per play.
If they are to be solely statistically driven does it follow that they should select to throw the football?
Of course that answer is no! But here’s my point for today, or better yet a question.
Suppose you have a child that reads very well and generally excels in the language aspects of learning. This child however has an extremely difficult time doing math. Since she is “worse” (DO take note of the scare quotes all of you Ricardians out there) at math does it follow that as good parents we ought to convince her to spend more time working on math?
Suppose you answer, as I would based on my football example, that you should NOT necessarily do this. Is there a difference between football and childrens’ education that suggests otherwise?
6 Responses to “Math Help”
1. sherlock says:
Always punting on 4th down and loving statistics seems to be a bit of a contradiction. 😉
Sorry I have nothing to add to the real meaning of the post.
2. Harry says:
Wintercow gives us a puzzle.
In children’s education there are at least two coaches — the parent and the teacher. Also, here I assume we are talking about arithmetic , but maybe Wintercow has more advanced math in mind.
Like the statistically-driven football coach, the elementary school teacher may be driven by pedagogical theory — not necessarily the latest — it may go back to Dewey.
So, I am going to fudge and ask,”Work harder on what?” If it means doing more exercise sheets with two pairs of six bees equals how many bees, and the kid counts the bees, then that may be time wasted, even if that is what professional educators say is the best way to learn the times tables.
Division is a different matter. In “science time” you could divide one (dead) bee in half with a sharp scalpel and teach dissection technique and insect anatomy at the same time.
Returning to the football analogy, if it is third and five to go, Krugman would recommend the Hail Mary stimulus pass, which never worked.
3. sherlock says:
The “safer” method is to just always punt on 4th down. The coach would receive no real criticism from the media/public as that is the way it has typically always been done. The coach can pass the blame on to his players or “relying on his defense”. A smarter coach realizes that on third down you really have two opportunites to obtain a 1st down and that you can use both downs to achieve the necessary yardage (maybe two running plays?). However, making the decision to go for it on 4th down (when it does increase your chances of winning the game) puts the coach’s neck on the line. If the 4th down fails, all the blame is put on the coach for making that decision. So even a smarter coach who realizes that it isn’t the best option to have one standard to go by in all situations may still make the decison to not go for it because he doesn’t want the blame.
Replace “coach” with “teacher” and you have yourself an analogy. A teacher may not want to try something new or develop methods based off an individual’s skill and ability, even if does have a better chance of succeding. They don’t want their head on the line so they “stick to the book” and hand the kid her mulitplcation tables.
4. drobviousso says:
Well, lets see. In the football example, the average gain is useless. What you want to use to decide is your historical % of gaining 5 yards on 3rd or 4th and 5 in “competitive football” (usually defined as first and third quarters with the score within 14 points and outside the red zone). Then, you pick randomly from a mixed bag weighted by the two historical percentages.
This, of course, assumes that the defense you are going up against is equally as good at stopping the run as they are at stopping the pass, and that all run plays are created the same and that all pass plays are created the same. You should probably add in play-action passes, screen passes, and draws. If you want to add in fake punts, you’ll need to get rid of all that 3rd down data and use exclusively 4th down data, which will probably reduce your sample size down to useless. Weather has been shown to increase variance but not favor one type of play, so you can ignore that, but you should probably include home field (dis)advantage for your stadium.
This all assumes you are working to maximize a successful play metric, and not a net expected points added metric or net expected win added metric, in which case this is a trick question. You just look up the proper mixed strategy for your down, distance, LoS, time, and score differential and pick based on *that* mixed strategy, no matter what the game situation is.
Now, how do we apply that to child rearing? I would say the corollary is to get Selfish Reasons to Have More Kids and ask your child if they want help with homework, or if they want you to read a book to them, or do they want to go collect bugs with you.
All of that is the long way to answer the question ” Is there a difference between football and childrens’ education that suggests otherwise?” Yes, we have better data on football, football isn’t a zero sum game, and you don’t need to account for the football’s agency.
5. Harry says:
Attempts with irony aside, one’s kid’s education is not a game, and is not confined to what they learn in school.
From the day one realizes he or she is a child, on through the moment one is a parent and until your child grows up, one has an immensely complicated problem with child rearing. You read books, you get advice from parents, you try to apply what happened to you, since that was perfect, or you do the opposite, which might be logical.
If you participated in The Right Stuff, a dating service that facilitates Ivy League graduates to meet and, er, breed, then from day one you buy educational toys for the nursery and get your kid doing word problems before he can walk, or, if the other side of the brain is working, writing musical scores, just like Mozart.
Or, if you are a little less Type A, and have not gotten bred to a high performance herd sire, you might be the Chinese Mother from Hell, which may very well result in your children living
Or, if you are a Hells Angel, you have a few children, get thrown in jail, and let the mothers or the state worry about your kids. If they are lucky, they might get adopted by a good family.
Then there are the rest of us, who do the best we can as often as we are willing and able.
So one’s daughter is strong in verbal skills, but finds other things difficult, in particular, math. Conventional wisdom (like the stat football coach) tells us girls are normally right-sided brain people, more verbal, than left-sided boys given to logic — or is it the other way around?
So this idea becomes embedded in our thinking, and may prejudice how we handle the situation.
Who knows? Not I. Do I push my kid, or do I let things happen? How do I know what I did made a difference? The difference?
TUW is, if anything epistemology Center. And that does not mean we don’t know anything.
6. chuck martel says:
Comparing a single football predicament with the on-going development of a child is hardly valid from the outset. However, everyone has different talents and individuals who excel at a certain activity not only possess talents in that line, if they are successful they are usually enthusiastic about it as well. Budding musicians don’t need to be forced to practice their scales. Potential NHL stars show up at the rink no matter what. That being said, it’s important to stress to the young that there are things that they may not particularly enjoy doing and that they probably will never become exceptionally adept at. Nevertheless, they must be encouraged as much as possible to acquire at least a rudimentary skill in things that are a part of normal life. Practical applications of mathematics, for instance, are necessary and should be mastered by everyone. The ability to express coherent thoughts “on paper” is also important. It’s very satisfying for a person that knows they’re not a math expert to be able to solve a practical math problem, more satisfying than it might be for a person with a natural talent for working with numbers.
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Math Help - Trig integral with x^2
1. #1
Junior Member
Joined
Feb 2009
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Trig integral with x^2
I want to evaluate the following 2 integrals: \frac{2}{a} \int_{\frac{-a}{4}}^{\frac{3a}{4}} x \sin^2 \left[\frac{n\pi}{a} \left(x + \frac{a}{4}\right) \right] \, dx = \frac{a}{4}
I think this is correct.
But x^2 is more difficult: \frac{2}{a} \int_{\frac{-a}{4}}^{\frac{3a}{4}} x^2 \sin^2 \left[ \frac{n\pi}{a} \left(x + \frac{a}{4}\right) \right] \, dx
Using integration by parts I let:
u=x^2 \Rightarrow du = 2x dx
dv = sin^2 \left[ \frac{n\pi}{a} \left(x + \frac{a}{4} \right) \right] \, dx
\Rightarrow v = \frac{x}{2} - \frac{a}{2n\pi} \sin \left[\frac{n\pi}{a} \left(x + \frac{a}{4} \right) \right] \cos \left[ \frac{n\pi}{a} \left(x + \frac{a}{4} \right) \right] = \frac{x}{2} - \frac{a}{4n\pi} \sin \left[ \frac{2n\pi}{a} \left(x + \frac{a}{4}\right) \right]
(double angle formula used)
< x^2 > = uv - \int v du = \frac{2}{a} \left[x^2 \, \frac{x}{2} - \frac{a}{4n\pi} \sin \left[ \frac{2n\pi}{a} \left( x + \frac{a}{4} \right) \right] \right]_{\frac{-a}{4}}^{\frac{3a}{4}} \left. - \int_{\frac{-a}{4}}^{\frac{3a}{4}} 2x \left( \frac{x}{2}-\frac{a}{4n\pi} \sin \left[ \frac{2n\pi}{a} \left(x+\frac{a}{4} \right) \right] \right)dx\right ]
=\frac{2}{a}\left [ \frac{27a^3}{128}-\frac{-a^3}{128}-\int_{\frac{-a}{4}}^{\frac{3a}{4}}x^3dx+\frac{a}{4n\pi}\int_{\f rac{-a}{4}}^{\frac{3a}{4}}2xsin[\frac{2n\pi}{a}(x+\frac{a}{4})])dx\right ]
=\frac{2}{a}\left [ \frac{7a^3}{32}-\frac{5a^4}{64}+\frac{a}{2n\pi}\left ( \frac{a^2}{4n^2\pi^2}sin[\frac{2n\pi}{a}(x+\frac{a}{4})])-\frac{a}{2n\pi}xcos[\frac{2n\pi}{a}(x+\frac{a}{4})])\right )_{\frac{-a}{4}}^{\frac{3a}{4}}\right ]
Sine term vanishes at both limits while cosine term as 1 at both limits:
=\frac{2}{a}\left [ \frac{7a^3}{32}-\frac{5a^4}{64}+\frac{a}{2n\pi}\left ( \frac{-3a^2}{8n\pi}+\frac{a^2}{8n\pi} \right )\right ]=\frac{7a^2}{16}-\frac{a^2}{4n^2\pi^2}-\frac{5a^3}{32}=\frac{a^2(7n^2\pi^2-4)}{16n^2\pi^2}-\frac{5a^3}{32}
This is obviously wrong since the units don't add up. How can there be a^2 and a^3 at the same time? They both have the dimension of meters. My answer should be in meters squared. I spend hours checking and still can't find my mistake.
I plug the equation into a graphic calculator, I think the answer should be: \frac{a^2(7n^2\pi^2-24)}{64n^2\pi^2}
Last edited by mr fantastic; September 19th 2009 at 06:34 PM. Reason: Improved readability (added spacing, larger brackets) and fixed some latex
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2,061,060,651,376,892,700 | Take the 2-minute tour ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.
From , the proofs by the probabilistic method are often said to be non-constructive.
However, a proof by probabilistic method indeed designs a randomized algorithm and uses it for proving existence. Quoted from p103 of Randomized Algorithms By Rajeev Motwani, Prabhakar Raghavan:
We could view the proof by the probabilistic method as a randomized algorithm. This would then require a further analysis bounding the probability that the algorithm fails to find a good partition on a given execution. The main difference between a thought experiment in the probabilistic method and a randomized algorithm is the end that each yields. When we use the probabilistic method, we are only concerned with showing that a combinatorial object exists; thus, we are content with showing that a favorable event occurs with non-zero probability. With a randomized algorithm, on the other hand, efficiency is an important consideration - we cannot tolerate a miniscule success probability.
So I wonder if randomized algorithms are viewed as not constructive, although they do output a solution at the end of each run, which may or may not be an ideal solution.
How is an algorithm or proof being "constructive" defined?
Thanks!
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2
Since there isn't any agreed-on definition of "constructive" as a technical term, and there isn't any central authority to give the definition of "constructive", and since different people will have different definitions (possibly depending on which subfield of computer science or mathematics they come from), I really don't think there can be a definitive answer to this question. – Peter Shor Oct 24 '12 at 12:33
I just ask about its most common meaning for proofs and algorithms. I think randomized algorithms are constructive, but proving by the probabilistic method isn't although it has a randomized algorithm inside, and therefore confused. – Tim Oct 24 '12 at 12:39
According to wikipedia, which doesn't mention time complexity, almost all proofs using the probabilistic algorithm would be constructive, since they give (very inefficient) algorithms. It depends on context. – Peter Shor Oct 24 '12 at 12:42
@PeterShor: isn't "constructive" approximately as well-defined a term as "logic" itself is? Without clarification, I would have assumed that a constructive result was one which involved ZF set theory and used constructive logic. – Niel de Beaudrap Oct 24 '12 at 13:02
I never heard "constructive" used to describe algorithms, only proofs. – Raphael Oct 25 '12 at 8:21
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2 Answers
up vote 8 down vote accepted
The probabilistic method is typically used to show that the probability of some random object having a certain property is non-zero, but doesn't exhibit any examples. It does guarantee that a "repeat-until-success" algorithm will eventually terminate, but does not give an upper bound on the runtime. So unless the probability of a property holding is substantial, an existence proof by the probabilistic method makes a very poor algorithm.
In point of fact, probabilistic algorithms aren't actually constructive existence proofs, so much as they are algorithms to produce constructive existence proofs. The output is an object of the sort which it was meant to prove the existence of; but the fact that it will eventually yield one ("there will exist an iteration in which it yields an example — except with probability zero...") is not enough to be constructive; it will only be satisfactory to someone who already accepts that non-zero-probability-without-construction suffices for existence. Conversely, if you do have a good bound on the run-time, then there's in principle no excuse not to run it in order to actually produce an example. A good probabilistic algorithm still isn't a constructive proof, but a good plan to obtain a constructive proof.
Note that this idea, that a randomized algorithm is a proof strategy (as opposed to a proof in itself) to demonstrate an existential quantification, is not unlike the idea that induction is a good proof strategy to show a universal quantification (over the natural numbers). This analogy may seem compelling, as induction is essentially the heart of recursion as a computational technique. (For any positive integer $n$, if you want to decide whether $n^2$ is a sum of the consecutive odd numbers preceding $2n+1$, you can reduce this to investigating whether $(n-1)^2$ is a sum of the consecutive odd numbers preceding $2n-1$, and so forth.) Induction is essentially an algorithmic proof-strategy which we have elevated to a theorem, allowing us to have the knowledge without explicitly computing it each time. However, induction is accepted constructively because it is already an axiom(-scheme) of Peano arithmetic, and one which is independent of the other axioms. By contrast, there is no rule of inference or axiom which allows the probabilistic method to prove existence constructively, or to constructively prove that probabilistic algorithms produce existence proofs, or anything along these lines. You simply cannot prove that there are examples of a class of object from the fact that there is a probabilistic algorithm to construct it, unless you already accept that proposition, either as an axiom, or from other premises.
Of course, one might adopt a philosophical position intermediate to constructivism and the classical approach to existence, and say that what one wants is not constructions per se but construction-schema which are allowed to fail with any probability less than one; that would make any probabilistic construction "schematic", if not completely constructive. Where one wishes to draw the line, to say that they find an existence proof "satisfactory", ultimately depends on how much intuition (in a non-philosophical sense) they wish to gain from proofs.
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Uniform proof complexity is a field devoted (among else) to the study of constructive notions of proofs, and their relation to complexity classes. For each of the popular (uniform) circuit complexity classes, one can define a theory in which everything which is provable has a "backing" in an algorithm in this complexity class. Randomized algorithms are accommodated through versions of the pigeonhole principle (oddly enough).
Unfortunately I'm not an expert so I can't say much more, other than point you to the book by Cook and Nguyen (same Cook as in Cook's theorem) and to the work of Emil Jeřábek, especially his thesis on randomized computation.
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4,807,359,190,144,383,000 | Everyday maths 1
Everyday maths 1
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Free course
Everyday maths 1
1.1 Changing units
Sometimes you will need to change between millimetres and centimetres, or centimetres and metres. For example, you might need to do this if you were fitting a kitchen or measuring a piece of furniture.
The diagram below shows you how to convert between metric units if you’re calculating any of the following:
• length, which is a measurement of how long something is
• mass (sometimes referred to as weight), which is a measurement of how heavy something is
• volume (sometimes referred to as capacity), which is a measurement of how much space something takes up.
Described image
Figure 8 A conversion chart for length, mass and volume
Starting with the smallest, metric measures of length are in millimetres, centimetres and metres. These three measurements are all related:
10 millimetres (or mm, for short) = 1 centimetre (cm)
100 cm = 1 metre (m).
Please take a look at the example below on how to carry out simple metric conversions.
Hint: 10 millimetres = 1 centimetre, 100 centimetres = 1 metre
Example: Making Christmas cards
You are making Christmas cards for a craft stall. You want to add a bow, which takes 10 cm of ribbon, to each card. You plan to make 50 cards. How many metres of ribbon do you need?
Method
First you need to work out how many centimetres of ribbon you need:
• 10 × 50 = 500 cm
Notice that the question asks how many metres of ribbon you need, rather than centimetres. So you need to divide 500 cm by 100 to find out the answer in metres:
• 500 ÷ 100 = 5 m
Do you remember the metric conversion diagram at the start of this session?
Described image
Figure 9 A conversion chart for length
Now try the following activity. Remember to check your answers once you have completed the questions.
Activity 2: Measuring lengths
1. You are fitting kitchen cabinets. The gap for the last cabinet is 80 cm. The sizes of the cabinets are shown in millimetres. Which size should you look for?
2. Thirty children in a class each need 20 cm of string for a project. How many metres of string will they use all together?
3. You want to buy 30 cm of fabric. The fabric is sold by the metre. What should you ask for?
Answer
You will have found it useful to refer to the metric conversion diagram for this activity.
1. To convert from centimetres to millimetres, you need to multiply the figure in centimetres by 10. The size is 80 cm, so the answer is:
• 80 × 10 = 800 mm
2. Thirty children each need 20 cm of string. To find the total in centimetres you would do the following:
• 30 × 20 = 600 cm
However, the question asked for how much string is needed in metres, not centimetres. To convert from centimetres to metres, you need to divide the figure in centimetres by 100. So if you need 600 cm, the answer is:
• 600 ÷ 100 = 6 m
1. To convert from centimetres to metres, you need to divide the figure in centimetres by 100. The length of fabric you need size is 30 cm, so the answer is:
• 30 ÷ 100 = 0.3 m
Now have a go at the following quickfire activity, using the conversion chart above if needed.
Activity 3: Converting lengths
What are these lengths in another unit of measurement? Select the correct answers from the list of options below.
1. 20 mm = ? cm
a.
200 cm
b.
2 cm
c.
0.2 cm
The correct answer is b.
1. 450 mm = ? m
a.
45 m
b.
4.5 m
c.
0.45 m
The correct answer is c.
1. 0.5 cm = ? mm
a.
5 mm
b.
15 mm
c.
50 mm
The correct answer is a.
1. 400 cm = ? m
a.
0.4 m
b.
4 m
c.
40 m
The correct answer is b.
Summary
In this section you have looked at measuring and calculating length. You have used different metric measurements, such as kilometres, metres and centimetres. You can now:
• measure and understand the sizes of objects
• understand different units of measurement.
FSM_1
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-8,637,921,480,189,881,000 | Fonctions statistiques (référence)
Remarque : Nous faisons de notre mieux pour vous fournir le contenu d’aide le plus récent aussi rapidement que possible dans votre langue. Cette page a été traduite automatiquement et peut donc contenir des erreurs grammaticales ou des imprécisions. Notre objectif est de faire en sorte que ce contenu vous soit utile. Pouvez-vous nous indiquer en bas de page si ces informations vous ont aidé ? Voici l’article en anglais à des fins de référence aisée.
Pour obtenir des informations détaillées sur une fonction, cliquez sur son nom dans la première colonne.
Remarque : Les marques de version indiquent la version d’Excel dans laquelle la fonction a été publiée pour la première fois. Ces fonctions ne sont pas disponibles dans les versions antérieures. Par exemple, le marqueur de version 2013 indique que cette fonction est disponible dans Excel 2013 et versions ultérieures.
Fonction
Description
ECART.MOYEN
Renvoie la moyenne des écarts absolus observés dans la moyenne des points de données.
MOYENNE
Renvoie la moyenne de ses arguments.
AVERAGEA
Renvoie la moyenne de ses arguments, nombres, texte et valeurs logiques inclus.
MOYENNE.SI
Renvoie la moyenne (arithmétique) de toutes les cellules d’une plage qui répondent à des critères donnés.
MOYENNE.SI.ENS
Renvoie la moyenne (arithmétique) de toutes les cellules qui répondent à plusieurs critères.
LOI.BETA.N
Excel 2010
Renvoie la fonction de distribution cumulée.
BETA.INVERSE.N
Excel 2010
Renvoie l’inverse de la fonction de distribution cumulée pour une distribution bêta spécifiée.
LOI.BINOMIALE.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire discrète suivant la loi binomiale.
LOI.BINOMIALE.SERIE
Excel 2013
Renvoie la probabilité d’un résultat d’essai à l’aide d’une distribution binomiale.
LOI.BINOMIALE.INVERSE
Excel 2010
Renvoie la plus petite valeur pour laquelle la distribution binomiale cumulée est inférieure ou égale à une valeur de critère.
LOI.KHIDEUX.N
Excel 2010
Renvoie la fonction de densité de distribution de la probabilité suivant une loi bêta cumulée.
LOI.KHIDEUX.DROITE
Excel 2010
Renvoie la probabilité unilatérale de la distribution khi-deux.
LOI.KHIDEUX.INVERSE
Excel 2010
Renvoie la fonction de densité de distribution de la probabilité suivant une loi bêta cumulée.
LOI.KHIDEUX.INVERSE.DROITE
Excel 2010
Renvoie l’inverse de la probabilité unilatérale de la distribution khi-deux.
CHISQ.TEST
Excel 2010
Renvoie le test d’indépendance.
INTERVALLE.CONFIANCE.NORMAL
Excel 2010
Renvoie l’intervalle de confiance pour une moyenne de population.
INTERVALLE.CONFIANCE.STUDENT
Excel 2010
Renvoie l’intervalle de confiance pour la moyenne d’une population, à l’aide d’une distribution t de Student.
COEFFICIENT.CORRELATION
Renvoie le coefficient de corrélation entre deux séries de données.
NB
Détermine les nombres compris dans la liste des arguments.
NBVAL
Détermine le nombre de valeurs comprises dans la liste des arguments.
NB.VIDE
Compte le nombre de cellules vides dans une plage.
NB.SI
Compte le nombre de cellules qui répondent à un critère donné dans une plage.
NB.SI.ENS
Compte le nombre de cellules à l’intérieur d’une plage qui répondent à plusieurs critères.
COVARIANCE.PEARSON
Excel 2010
Renvoie la covariance, moyenne des produits des écarts pour chaque série d’observations.
COVARIANCE.STANDARD
Excel 2010
Renvoie la covariance d’échantillon, moyenne des produits des écarts pour chaque paire de points de deux jeux de données.
SOMME.CARRES.ECARTS
Renvoie la somme des carrés des écarts.
LOI.EXPONENTIELLE.N
Excel 2010
Renvoie la distribution exponentielle.
LOI.F.N
Excel 2010
Renvoie la distribution de probabilité F.
LOI.F.DROITE
Excel 2010
Renvoie la distribution de probabilité F.
INVERSE.LOI.F.N
Excel 2010
Renvoie l’inverse de la distribution de probabilité F.
INVERSE.LOI.F.DROITE
Excel 2010
Renvoie l’inverse de la distribution de probabilité F.
F.TEST
Excel 2010
Renvoie le résultat d’un test F.
FISHER
Renvoie la transformation de Fisher.
FISHER.INVERSE
Renvoie l’inverse de la transformation de Fisher.
PREVISION
Calcule une valeur par rapport à une tendance linéaire.
Remarque : Dans Excel 2016, cette fonction a été remplacée par PREVISION.LINEAIRE dans le cadre des nouvelles fonctions de prévision, mais elle reste disponible à des fins de compatibilité avec les versions antérieures.
PREVISION.ETS
Excel 2016
Renvoie une valeur future en fonction des valeurs existantes (historiques) à l’aide de la version AAA de l’algorithme de lissage exponentiel (Exponential Smoothing, ETS).
PREVISION.ETS.CONFINT
Excel 2016
Renvoie un intervalle de confiance pour la prévision à la date cible spécifiée.
PREVISION.ETS.CARACTERESAISONNIER
Excel 2016
Renvoie la longueur du modèle de répétition détecté par Excel pour la série chronologique spécifiée.
PREVISION.ETS.STAT
Excel 2016
Renvoie une valeur statistique suite à la prévision de la série chronologique.
PREVISION.LINÉAIRE
Excel 2016
Renvoie une valeur future en fonction des valeurs existantes.
FREQUENCE
Calcule la fréquence d’apparition des valeurs dans une plage de valeurs, puis renvoie des nombres sous forme de matrice verticale.
GAMMA
Excel 2013
Renvoie la valeur de fonction Gamma.
LOI.GAMMA.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire suivant une loi Gamma.
LOI.GAMMA.INVERSE.N
Excel 2010
Renvoie, pour une probabilité donnée, la valeur d’une variable aléatoire suivant une loi Gamma.
LNGAMMA
Renvoie le logarithme népérien de la fonction Gamma, Γ(x).
LNGAMMA.PRECIS
Excel 2010
Renvoie le logarithme népérien de la fonction Gamma, Γ(x).
GAUSS
Excel 2013
Renvoie 0,5 de moins que la distribution cumulée normale standard.
MOYENNE.GEOMETRIQUE
Renvoie la moyenne géométrique.
CROISSANCE
Calcule des valeurs par rapport à une tendance exponentielle.
MOYENNE.HARMONIQUE
Renvoie la moyenne harmonique.
LOI.HYPERGEOMETRIQUE.N
Renvoie la probabilité d’une variable aléatoire discrète suivant une loi hypergéométrique.
ORDONNEE.ORIGINE
Renvoie l’ordonnée à l’origine d’une droite de régression linéaire.
KURTOSIS
Renvoie le kurtosis d’une série de données.
GRANDE.VALEUR
Renvoie la k-ième plus grande valeur d’un jeu de données.
DROITEREG
Renvoie les paramètres d’une tendance linéaire.
LOGREG
Renvoie les paramètres d’une tendance exponentielle.
LOI.LOGNORMALE.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire continue suivant une loi lognormale.
LOI.LOGNORMALE.INVERSE.N
Excel 2010
Renvoie l’inverse de la fonction de distribution suivant une loi lognormale cumulée.
MAX
Renvoie la valeur maximale contenue dans une liste d’arguments.
MAXA
Renvoie la valeur maximale d’une liste d’arguments, nombres, texte et valeurs logiques inclus.
MAX.SI
Excel 2016
Renvoie la valeur maximale parmi les cellules spécifiées par un ensemble de conditions ou critères.
MEDIANE
Renvoie la valeur médiane des nombres donnés.
MIN
Renvoie la valeur minimale contenue dans une liste d’arguments.
MIN.SI
Excel 2016
Renvoie la valeur minimale entre les cellules spécifiées par un ensemble de conditions ou de critères.
MINA
Renvoie la plus petite valeur d’une liste d’arguments, nombres, texte et valeurs logiques inclus.
MODE.MULTIPLE
Excel 2010
Renvoie une matrice verticale des valeurs les plus fréquentes ou répétitives dans une matrice ou une plage de données.
MODE.SIMPLE
Excel 2010
Renvoie la valeur la plus courante d’une série de données.
LOI.BINOMIALE.NEG.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire discrète suivant une loi binomiale négative.
LOI.NORMALE.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire continue suivant une loi normale.
LOI.NORMALE.INVERSE.N
Excel 2010
Renvoie, pour une probabilité donnée, la valeur d’une variable aléatoire suivant une loi normale standard.
LOI.NORMALE.STANDARD.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire continue suivant une loi normale standard.
LOI.NORMALE.STANDARD.INVERSE.N
Excel 2010
Renvoie l’inverse de la distribution cumulée normale standard.
PEARSON
Renvoie le coefficient de corrélation d’échantillonnage de Pearson.
CENTILE.EXCLURE
Excel 2010
Renvoie le k-ième centile des valeurs d’une plage, où k se trouve dans la plage comprise entre 0 et 1 exclus.
CENTILE.INCLURE
Excel 2010
Renvoie le k-ième centile des valeurs d’une plage.
RANG.POURCENTAGE.EXCLURE
Excel 2010
Renvoie le rang d’une valeur d’un jeu de données sous forme de pourcentage (0..1, exclues).
RANG.POURCENTAGE.INCLURE
Excel 2010
Renvoie le rang en pourcentage d’une valeur d’une série de données.
PERMUTATION
Renvoie le nombre de permutations pour un nombre donné d’objets.
PERMUTATIONA
Excel 2013
Renvoie le nombre de permutations pour un nombre d’objets donné (avec répétitions) pouvant être sélectionnés à partir du nombre total d’objets.
PHI
Excel 2013
Renvoie la valeur de la fonction de densité pour une distribution normale standard.
LOI.POISSON.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire suivant une loi de Poisson.
PROBABILITE
Renvoie la probabilité que des valeurs d’une plage soient comprises entre deux limites.
QUARTILE.EXCLURE
Excel 2010
Renvoie le quartile d’un jeu de données en fonction des valeurs du centile comprises entre 0..1, exclues.
QUARTILE.INCLURE
Excel 2010
Renvoie le quartile d’une série de données.
MOYENNE.RANG
Excel 2010
Renvoie le rang d’un nombre contenu dans une liste.
EQUATION.RANG
Excel 2010
Renvoie le rang d’un nombre contenu dans une liste.
COEFFICIENT.DETERMINATION
Renvoie la valeur du coefficient de détermination R^2 d’une régression linéaire.
COEFFICIENT.ASYMETRIE
Renvoie l’asymétrie d’une distribution.
COEFFICIENT.ASYMETRIE.P
Excel 2013
Renvoie l’asymétrie d’une distribution : la caractérisation du degré d’asymétrie d’une distribution par rapport à sa moyenne.
PENTE
Renvoie la pente d’une droite de régression linéaire.
PETITE.VALEUR
Renvoie la k-ième plus petite valeur d’une série de données.
CENTREE.REDUITE
Renvoie une valeur centrée réduite.
ECARTYPE.PEARSON
Excel 2010
Calcule l’écart type d’une population à partir de la population entière.
ECARTYPE.STANDARD
Excel 2010
Évalue l’écart type d’une population en se basant sur un échantillon de cette population.
STDEVA
Évalue l’écart type d’une population en se basant sur un échantillon de cette population, nombres, texte et valeurs logiques inclus.
STDEVPA
Calcule l’écart type d’une population à partir de l’ensemble de la population, nombres, texte et valeurs logiques inclus.
ERREUR.TYPE.XY
Renvoie l’erreur type de la valeur y prévue pour chaque x de la régression.
LOI.STUDENT.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire suivant la loi de t de Student.
LOI.STUDENT.BILATERALE
Excel 2010
Renvoie la probabilité d’une variable aléatoire suivant la loi de t de Student.
LOI.STUDENT.DROITE
Excel 2010
Renvoie la probabilité d’une variable aléatoire suivant une loi T de Student.
LOI.STUDENT.INVERSE.N
Excel 2010
Renvoie la valeur d’une variable aléatoire suivant la loi T de Student, en fonction de la probabilité et du nombre de degrés de liberté.
LOI.STUDENT.INVERSE.BILATERALE
Excel 2010
Renvoie, pour une probabilité donnée, la valeur d’une variable aléatoire suivant une loi T de Student.
T.TEST
Excel 2010
Renvoie la probabilité associée à un test T de Student.
TENDANCE
Renvoie des valeurs par rapport à une tendance linéaire.
MOYENNE.REDUITE
Renvoie la moyenne de l’intérieur d’un jeu de données.
VAR.P.N
Excel 2010
Calcule la variance sur la base de l’ensemble de la population.
VAR.S
Excel 2010
Calcule la variance sur la base d’un échantillon.
VARA
Estime la variance d’une population en se basant sur un échantillon de cette population, nombres, texte et valeurs logiques incluses.
VARPA
Calcule la variance d’une population en se basant sur la population entière, nombres, texte et valeurs logiques inclus.
LOI.WEIBULL.N
Excel 2010
Renvoie la probabilité d’une variable aléatoire suivant une loi de Weibull.
Z.TEST
Excel 2010
Renvoie la valeur de probabilité unilatérale d’un test z.
Important : Les résultats calculés des formules et certaines fonctions de feuille de calcul Excel peuvent différer légèrement entre un PC Windows avec une architecture x86 ou x86-64 et un PC Windows RT avec une architecture ARM. En savoir plus sur les différences.
Rubriques connexes
Fonctions Excel (par catégorie)
fonctions Excel (par ordre alphabétique)
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-2,860,235,991,154,670,600 | If $p(x)=5 x-4 x^{2}+3$ then $p(-1)=?$
Question.
If $p(x)=5 x-4 x^{2}+3$ then $p(-1)=?$
(a) 2
(b) –2
(c) 6
(d) –6
Solution:
$p(x)=5 x-4 x^{2}+3$
Putting $x=-1$ in $p(x)$, we get
$p(-1)=5 \times(-1)-4 \times(-1)^{2}+3=-5-4+3=-6$
Hence, the correct answer is option (d).
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6,955,419,653,584,025,000 | What Is an Erlang Calculator?
An Erlang calculator is one of the most useful tools in the call centre toolkit.
An Erlang Calculator is a mathematical calculation that allows you to calculate the number of staff that you need for a given number of calls, to meet a given service level.
It is based on the Erlang C formula (a derivative of the Poisson distribution) that was designed by the Danish Mathematician A.K. Erlang around 100 years ago. The formula is quite involved, but is relatively easy to follow if you studied maths to a reasonable level at school.
You simply enter in the number of phone calls that you receive in a period of time (say per half hour), along with the average duration of the calls and also the service level that you are looking for.
Typical Inputs
• Number of phone calls
• Time period (e.g. per half hour)
• Average Call Duration (Average Handling Time)
• Service Level (Percentage of calls answered within a period of time, e.g. 80% of calls in 20 seconds)
• Some Erlang calculators also include a shrinkage input.
Typical Outputs
• Number of agents (advisors) needed to meet the service level target
Formats of Erlang C Calculators
There are two main formats for Erlang calculators.
1. Excel-based Erlang Calculators
This Excel-based worksheet uses macros, automated input sequences or add-ins, to perform the calculations due to the complexity of the mathematics.
Whilst there are some examples where people have been able to perform the function without using macros (using the Poisson function), these tend to need multiple rows or columns to obtain the required results.
For more information on the Poisson Distribution, read our article: How Is Average Handling Time Distributed? It is not how you think!
Pros
• Easy to build a mini workforce management system.
• Flexible – easy to perform what-if type functions and compare whole days or weeks of calls.
Cons
• Requires macros to be enabled.
• Many do not take account of shrinkage.
• Typically only work up to around 200 agents (700 agents tends to be the absolute limit for double precision floating point numbers).
• Many handmade spreadsheets often contain between 20% and 40% errors – see this article for more details: http://panko.shidler.hawaii.edu/SSR/Mypapers/whatknow.htm
Follow the link for our: Free Erlang C Calculator Excel – Including Shrinkage
Worked Example
Work out the total number of agents (FTE) required for a call volume (including shrinkage)
Calls Reporting Period (mins) Average Call Duration (secs) Required Service Level (%) Target answer time (secs) Shrinkage (%) Agents (FTE)
1000 60 180 80% 20 35% 86
This uses the Excel Formula:
=AgentsFTE(Calls, Reporting_period, Average_Call_duration, Service_level_percent, Service_level_time, Shrinkage)
For example =AgentsFTE(B30,C30,D30,E30,F30,G30)
This formula is more accurate as it includes shrinkage (holidays, training, meetings, etc.) and therefore gives a more realistic staffing requirement.
For a full explanation of shrinkage, read our article: How to Calculate Shrinkage
2. Online Erlang Calculators
A new generation of Erlang calculators has emerged that are available online.
Pros
• These are freely available online and are easy to test.
• Great for what-if type calculations.
• Some can perform calculations in excess of 700 agents.
• Do not require any software download or potentially harmful macros to be installed on your PC.
Cons
• Most calculators do not take account of shrinkage.
• Limited quality checking.
• Errors over 700 agents. Many online Erlang calculators produce wrong results for large number of agents (see below for how to check this).
Follow the link for our: Call Centre Erlang Staffing Calculator – including Shrinkage
Results
The number of agents needed is 86 agents including 35% shrinkage (56 agents before shrinkage)
This would give a Service Level of 82.3% answered in 20 seconds
The Average Speed of Answer (ASA) would be 12.2 seconds
How to spot if your Erlang calculator is giving the wrong results
STEP 1: Enter the following details into an Erlang calculator
14,200 calls per hour, 180-second call duration, 80% of calls handled in 20 seconds (with no shrinkage if the calculator provides it).
STEP 2: Check that the number of agents equals 721
The correct answer should be 721 agents. Many calculators will confidently predict 711 advisors (often with a Service Level Prediction of nan [not a number]).
STEP 3: Put in 100 fewer calls
So, enter: 14,100 calls per hour, 180-second call duration, 80% of calls handled in 20 seconds (with no shrinkage if the calculator provides it).
Every calculator will then give the right answer of 716 advisors.
This tells you that despite putting in a lower call volume, and keeping all of the other variables the same, the number of advisors has actually increased from 711 to 716, so something must be wrong.
Find some other examples that will enable you to check your Erlang Calculator below.
Calls Per Hour Other Inputs Correct result Faulty Erlang Calculators Result
1410 Call duration 180 secs, 80% of calls handled in 20 seconds (no shrinkage) 716 Advisors 716 Advisors
1420 Call duration 180 secs, 80% of calls handled in 20 seconds (no shrinkage) 721 Advisors 711 Advisors
1500 Call duration 180 secs, 80% of calls handled in 20 seconds (no shrinkage) 761 Advisors 751 Advisors
2000 Call duration 180 secs, 80% of calls handled in 20 seconds (no shrinkage) 1011 Advisors 1001 Advisors
This process is vital, to ensure that your staffing calculations are as accurate as possible – with overstaffing causing great financial problems and understaffing risking both customer and advisor satisfaction.
With this in mind, we hope that this has been a great introduction to the Erlang Calculator and that we have inspired you to read more of our content – including the three articles below – so you can do the best possible job in staffing your contact centre.
Follow the link for our: Free Erlang Calculator
To find out more about applying Erlang mathematics to the contact centre, read our articles:
Published On: 29th Mar 2017 - Last modified: 25th Sep 2019
Read more about - Call Centre Planning, , , , , ,
4 Comments
1. Hi Call Centre Helper,
Does the Erlang Calculator “Amount of Calls” refer to Inbound and Outbound calls. Would you consider doing to seperate calculations on Inbound and Outbound calls?
Thank you.
Adri Visser 31 Mar at 7:29 am
2. The Erlang calculator only applies to inbound calls as these are the types of calls that have queuing applied to them.
Jonty Pearce 3 Apr at 4:58 pm
3. Best results if you use this tool if you calculate per interval.
Please correct my statement if this is incorrect.
i tried to compare the calculation of a day overview and per interval on a day overview.
placing same values such as the below
Day Overview
Incoming Calls – 2120
In a period of – 780 minutes (8 hours hoop)
AHT – 629 seconds
Required Service Level – 80%
Target Answer Time – 600 seconds
Max Occupancy – 90%
Shrinkage – 30%
Downloaded the free erlang calculator file on this same site which can calculate per interval view(Day Calculator), placing same data, differs only in time period:
AHT – 629
Reporting Period – 15 minutes
Required Service Level – 80%
Target Answer Time – 600
Shinkage – 30%
The Day overview stated i need 47 agents to be able to achieve the following
Service level at 99.6% answered in 600 seconds
ASA at 44.4 seconds
but the file where i calculated the per interval stated i needed more than 44 agents on 41 intervals if i divided the overall forecast base on 6 weeks trend such as below:
6:00 4
6:15 2
6:30 5
6:45 6
7:00 24
7:15 35
7:30 46
7:45 44
8:00 47
8:15 61
8:30 58
8:45 59
9:00 60
9:15 60
9:30 54
9:45 57
10:00 53
10:15 64
10:30 61
10:45 64
11:00 63
11:15 61
11:30 68
11:45 52
12:00 64
12:15 63
12:30 61
12:45 66
13:00 62
13:15 67
13:30 59
13:45 65
14:00 58
14:15 56
14:30 60
14:45 56
15:00 48
15:15 42
15:30 33
15:45 31
16:00 19
16:15 16
16:30 11
16:45 14
17:00 12
17:15 10
17:30 10
17:45 9
18:00 7
18:15 8
18:30 3
18:45 2
would the result differ base on the time period of the calls that will be coming in or is it because the calculator have a much deeper view on the calls per interval giving a higher staffing requirements than the results in the day view?
i guess the calculation on the day view considered same volume
CherryBlossom 14 Apr at 9:11 am
4. Hi, is there available an erlang calculator tool that considers the period in months. My necessity is to calculate how many agents I need per month.
Thanks in advance!
Rita
rita 15 Jun at 12:02 pm
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-8,850,209,930,613,932,000 | Graphing Piecewise Functions
Graphing Piecewise Functions
• Currently 4.0/5 Stars.
2878 views, 1 rating
Part of video series Piecewise Functions
Taught by mrbrianmclogan
I show how to solve math problems online during live instruction in class. This is my way of providing free tutoring for the students in my class and for students anywhere in the world. Every video is a short clip that shows exactly how to solve math problems step by step. The problems are done in real time and in front of a regular classroom. These videos are intended to help you learn how to solve math problems, review how to solve a math problems, study for a test, or finish your homework. I post all of my videos on YouTube, but if you are looking for other ways to interact with me and my videos you can follow me on the following pages through My Blog, Twitter, or Facebook.
How to graph a piecewise function
• How do you graph a piecewise function?
• What is a piecewise function?
• How do you graph f(x) if f(x) = -1 when x <= 0, f(x) = 2x when 0 < x <= 3, and f(x) = 6 when x > 3?
• How do you know where points are open or closed circles in a piecewise function?
This lesson shows how to graph a piecewise function in 3 pieces. Each of the 3 pieces is graphed for only a limited domain. All steps involved in graphing each of the 3 pieces are explained and shown. This is a very good introduction to piecewise functions.
• Currently 4.0/5 Stars.
Reviewed by MathVids Staff on February 05, 2012.
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-2,842,387,341,923,926,000 |
+0
+1
220
3
avatar+10
A solid consists of a right circular cone of height 12 cm surmounted on a right circular cylinder of the same height and radius 5 cm. Determine the total area and volume of a solid.
Aug 11, 2018
#2
avatar+98129
+2
The cone will have 1/3 of the volume of the cylinder
The volume of the cylinder is pi * radius^2 * height = pi * (5)^2 * 12 = pi * 25 * 12 = 300 pi cm^3
So the volume of the cone is 300pi / 3 = 100 pi cm^3
So....the volume of the solid is [ 300 + 100] pi cm^3 = 400pi cm^3
The lateral area of the cylinder is 2pi * radius * height = 2pi * 5 * 12 = 120 pi cm^2
The area of the bottom of the cylinder is pi * radius^2 = pi * 5^2 = 25 pi cm^2
The slant height of the cone is √ [ height^2 + radius^2 ]= √ [12^2 + 5^2] =√169 = 13 cm
So....the lateral area of the cone is pi * radius * slant height = pi * 5 * 13 = 65 pi cm^2
So...the total area of the solid is [ 120 + 25 + 65] pi = 210 pi cm^2
cool cool cool
Aug 11, 2018
15 Online Users
avatar
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-1,427,992,153,934,917,600 | Liberal Arts Blog — Moon Math — Synchronous Rotation, The Tides, The Phases
John Muresianu
4 min readJun 10, 2024
--
Liberal Arts Blog — Monday is the Joy of Math, Statistics, Shapes, and Numbers Day
Today’s Topic: Moon Math — synchronous rotation, the tides, the phases
How precise and how detailed should an 8th grader’s understanding of tides and the moon phases be? How about a 12th grader? How about a 5th grader? a 3rd grader? How about a Harvard graduate?
Have you ever wrestled with this challenge and come up short? Did you ever claw your way out of a dark tunnel of confusion to see the light of understanding? Did a great teacher ever lead you to the light? How?
Could AI help here? Could AI put all this into a sonnet form and set it to a catchy folk tune?
Experts — please chime in. Correct, elaborate, elucidate.
SYNCHRONOUS ROTATION: the moon spins on its axis exactly one time for every rotation of the earth — why? so what? what does a man on the moon see?
1. “It is the natural consequence of tidal friction.”
2. “The Moon has tidal bulges similar to those on Earth. It is thought that the Moon once rotated much faster than it does today.”
3. “The friction created by the stretching and squeezing of the Moon caused the Moon’s rate of rotation to slow down until its rotational period was the same as its orbital period.”
NB: “Most of the far side of the Moon was not seen until 1959, when photographs of most of the far side were transmitted from the Soviet spacecraft Luna 3.”
“When Earth is observed from the Moon. Earth does not appear to move across the sky. It remains in the same place while showing nearly all its surface as it rotates on its axis.”
WHY ARE THERE TWO HIGH TIDES NOT ONE HIGH TIDE?’
1. The difference in distance between the side of the earth closest to the moon and the side farthest away means that the gravitational force is less on the side furthest from the moon.
2. Remember that in Newton’s gravitational equation the force of gravity is inversely proportional to the square of the distance.
3. The solid mass of the earth is more strongly attached to itself than is the ocean which lags creating the high tide on the far side of the earth.
WHEN IS THE WAXING HALF MOON VISIBLE? THE WANING HALF MOON?
1. The waxing half moon rises at about noon, and sets at midnight.
2. The waning half moon rises at midnight, and sets at noon.
3. If you see a beautiful half moon in the evening, it’s a waxing half moon and it should look like a backward C (ie. a D).
Moon Phases — NASA Science
https://lasp.colorado.edu/outerplanets/math/synchronous_rotation.pdf
Ocean’s Tides Explained
How the tides REALLY work
Libration — Wikipedia
QUOTE OF THE MONTH — Have you made your own Bible yet?
“Make your own Bible. Select and collect all the words and sentences that in all your readings have been to you like the blast of a trumpet.”
- Ralph Waldo Emerson
My spin — then periodically review, re-rank, and exchange your list with those you love. I call this the “Orion Exchange” because seven is about as many as any human can digest at a time. Game?
ATTACHMENTS BELOW:
#1 A graphic guide to justice (9 metaphors on one page).
#2 “39 Songs, Prayers, and Poems: the Keys to the Hearts of Seven Billion People” — Adams House Senior Common Room Presentation, (11/17/20)
#3 Israel-Palestine Handout
NB: Palestine Orion (Decision) — let’s exchange Orions, let’s find Rumi’s field (“Beyond all ideas of right and wrong, there is a field. Meet me there” Rumi, 13 century Persian Sufi mystic)
Last four years of posts organized thematically:
Updated PDFs — Google Drive
YOUR TURN
Please share the coolest thing you learned this week related to math, statistics, or numbers in general.
Or, even better, the coolest or most important thing you learned in your life related to math.
This is your chance to make someone else’s day. And to consolidate in your memory something you might otherwise forget. Or to think more deeply than otherwise about something dear to your heart. Continuity is key to depth of thought.
--
--
John Muresianu
Passionate about education, thinking citizenship, art, and passing bits on of wisdom of a long lifetime. | {
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-8,683,132,147,684,106,000 | Drexel dragonThe Math ForumDonate to the Math Forum
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Topic: The Monoid Category in DC Proof
Replies: 7 Last Post: Nov 11, 2012 10:37 PM
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Dan Christensen
Posts: 3,395
Registered: 7/9/08
The Monoid Category in DC Proof
Posted: Nov 9, 2012 10:43 AM
Click to see the message monospaced in plain text Plain Text Click to reply to this topic Reply
What is a category?
Use the imagery of a directed graph, we can define a category as a
logical structure that satisfies the following axioms (in my DC Proof
notation):
1. ALL(f):[f @ arrows => source(f) @ nodes]
where:
@ = epsilon (set/class membership)
arrows is a class
nodes is a class
source(f) is the source node for arrow f
2. ALL(f):[f @ arrows => target(f) @ nodes]
where:
target(f) is the target node of arrow f
3. ALL(f):ALL(g):[f @ arrows & g @ arrows
=> [target(f)=source(g) => comp(f,g) @ arrows]]
where:
comp(f,g) is the composition of arrow f followed by and arrow g
Note the order of arguments -- a minor departure from the usual
presentation. In a graphical sense, the comp(f,g) operator is similar
to the addition of vectors with the head of arrow f being place at the
tail of arrow g.
4. ALL(f):ALL(g):[f @ arrows & g @ arrows
=> [target(f)=source(g)
=> source(comp(f,g))=source(f) & target(comp(f,g))=target(g)]]
The source and target of the composition of two arrows must be
compatible with the the targets and sources of those two arrows.
5. ALL(f):ALL(g):ALL(h):[f @ arrows & g @ arrows & h @ arrows
=> [target(f)=source(g) & target(g)=source(h)
=> comp(comp(f,g),h)=comp(f,comp(g,h))]]
Composition of arrows is associative.
6, ALL(a):ALL(b):[a @ nodes & b @ nodes
=> ALL(f):[f @ arrows(a,b) <=> f @ arrows & source(f)=a &
target(f)=b]]
where:
arrows(a,b) is the class of arrows with source a and target b
Although this construction is not referred to by other axioms here,
traditionally, it is part of the definition of a category. It may be
referred to in other definitions and theorems of category theory.
7. ALL(a):[a @ nodes => id(a) @ arrows]
where:
id(a) is the identity arrow for node a
8. ALL(f):[f @ arrows => ALL(a):[a @ nodes => [source(f)=a =>
comp(id(a),f)=f]]]
9. ALL(f):[f @ arrows => ALL(a):[a @ nodes => [target(f)=a =>
comp(f,id(a))=f]]]
What is a monoid?
The monoid (m,+) is an algebraic structure on a set m that satisfies
the following axioms:
1. 0 @ m
2. ALL(a):ALL(b):[a @ m & b @ m => a+b @ m]
where:
+ is a binary operator on m
3. ALL(a):[a @ m => a+0=a & 0+a=a]
0 is the identity element in m with respect to +
4. ALL(a):ALL(b):ALL(c):[a @ m & b @ m & c @ m => a+b+c=a+(b+c)]
+ is associative
What is the monoid category?
The monoid category m is a category based on the monoid (m,+).
Before constructing the monoid category, it should be noted that on
every set m, we can construct a unique identity function i such that:
ALL(a):[a @ m => i(a)=a]
Now reconstruct the monoid in terms of nodes, arrows, comp, etc.
The only element of the set of nodes is the underlying set m:
1. ALL(a):[a @ nodes <=> a=m]
The set of arrows is the set of unary functions on the set m such
that:
2. ALL(f):[f @ arrows <=> EXIST(a):[a @ m & ALL(b):[b @ m => f(b)=b
+a]]]
Since m is a set, we can construct the set of arrows using the axioms
of set theory.
The source of all arrows is just m itself:
3. ALL(f):[f @ arrows => source(f)=m]
Likewise, for the target of all arrows:
4. ALL(f):[f @ arrows => target(f)=m]
Since there is only one node m, we can define the set of arrows with
source a and target b as follows:
5. ALL(f):[f @ arrows(m,m) <=> f @ arrows]
We define the composition of these arrows is defined as follows:
6. ALL(f):ALL(g):[f @ arrows & g @ arrows => comp(f,g) @ arrows]
7. ALL(f):ALL(g):ALL(h):[f @ arrows & g @ arrows & h @ arrows =>
[comp(f,g)=h <=> ALL(a):[a @ m => h(a)=g(f(a))]]]
We define the id function on the only node m at follows:
8. id(m)=i
where:
i is the identity function on m as defined in above.
It can be formally shown (in 549 lines in DC Proof) that this
reconstruction of the monoid does indeed meet all the requirements of
category.
Comments?
Dan
Download my DC Proof 2.0 software at http://www.dcproof.com
Point your RSS reader here for a feed of the latest messages in this topic.
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Chapter 5: Analytic Trigonometry. Section 5.1a: Fundamental Identities HW: p. 451-452 1-7 odd, 27-49 odd. Is this statement true?. This identity is a true sentence, but only w ith the qualification that x must be in the d omain of both expressions .
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Chapter 5 analytic trigonometry
Chapter 5: Analytic Trigonometry
Section 5.1a: Fundamental Identities
HW: p. 451-452 1-7 odd, 27-49 odd
Is this statement true?
This identity is a true sentence, but only
with the qualification that x must be in the
domain of both expressions.
If either side of the equality is undefined (i.e., at x = –1), then
the entire expression is meaningless!!!
The statement is a trigonometric identity
because it is true for all values of the variable for which both
sides of the equation are defined.
The set of all such values is called the domain of validity of
the identity.
Basic Trigonometric Identities
Reciprocal Identities
Quotient Identities
is in the domain of validity of exactly three of the basic
identities. Which three?
Basic Trigonometric Identities
Reciprocal Identities
Quotient Identities
For exactly two of the basic identities, one side of the equation
is defined at and the other side is not. Which two?
Basic Trigonometric Identities
Reciprocal Identities
Quotient Identities
For exactly three of the basic identities, both sides of the
equation are undefined at . Which three?
Pythagorean Identities
Recall our unit circle:
P
What are the coordinates of P?
sint
(1,0)
cost
So by the Pythagorean Theorem:
Divide by :
Pythagorean Identities
Recall our unit circle:
P
What are the coordinates of P?
sint
(1,0)
cost
So by the Pythagorean Theorem:
Divide by :
Pythagorean Identities
Given and , find and .
We only take the positive answer…why?
Cofunction Identities
Can you explain why each of these is true???
Odd-Even Identities
If , find .
Sine is odd
Cofunction Identity
Simplifying Trigonometric Expressions
Simplify the given expression.
How can we support this answer graphically???
Simplifying Trigonometric Expressions
Simplify the given expression.
Graphical support?
Simplifying Trigonometric Expressions
Simplify the given expressions to either a constant or a basic
trigonometric function. Support your result graphically.
Simplifying Trigonometric Expressions
Simplify the given expressions to either a constant or a basic
trigonometric function. Support your result graphically.
Simplifying Trigonometric Expressions
Use the basic identities to change the given expressions to ones
involving only sines and cosines. Then simplify to a basic
trigonometric function.
Simplifying Trigonometric Expressions
Use the basic identities to change the given expressions to ones
involving only sines and cosines. Then simplify to a basic
trigonometric function.
Simplifying Trigonometric Expressions
Use the basic identities to change the given expressions to ones
involving only sines and cosines. Then simplify to a basic
trigonometric function.
Let’s start with a practice problem…
Simplify the expression
How about some
graphical support?
Quick check of your algebra skills!!! of a
Factor the following expression (without any guessing!!!)
What two numbers have a product of –180 and a sum of 8?
Rewrite middle term:
Group terms and factor:
Divide out common factor:
Write each expression in factored form as an algebraic of a
expression of a single trigonometric function.
e.g.,
Let
Substitute:
Factor:
“Re”substitute for your answer:
Write each expression in factored form as an algebraic of a
expression of a single trigonometric function.
e.g.,
Write each expression in factored form as an algebraic of a
expression of a single trigonometric function.
e.g.,
Let
Write each expression in factored form as an algebraic of a
expression of a single trigonometric function.
e.g.,
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5,732,281,672,126,141,000 | Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):
how do you solve 3^x = 20 on a graphing calculator?
OpenStudy (anonymous):
well answer is \[x=\frac{\ln(20)}{\ln(3)}\] for sure
OpenStudy (anonymous):
by change of base formula if you want to solve \[b^x=A\] for x the answer is \[x=\frac{\ln(A)}{\ln(b)}\]
OpenStudy (anonymous):
thx!
OpenStudy (anonymous):
in english, the log of the total divided by the log of the base. you can use \[x=\frac{\log(A)}{\log(b)}\] to they both work
OpenStudy (anonymous):
*too
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Join our real-time social learning platform and learn together with your friends!
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Join our real-time social learning platform and learn together with your friends! | {
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1,827,105,135,793,723,400 | Lesson plan
Finding Perimeter
In this lesson, students will find the perimeter of figures using cheese crackers. Next, they'll design a floor plan for their cheese cracker dream house, to help them practice and retain the formula for finding perimeter.
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Students will be able to find the perimeter of a given geometric figure.
(10 minutes)
1. Play the video Perimeter, by Math Antics, for the class to introduce the concept of perimeter.
2. Once the video is complete, ask for a volunteer to tell you the definition of perimeter.
3. Write the definition on the board.
4. Draw a few figures on the board, along with the lengths of their sides.
5. Challenge students to find the perimeter of the geometrical figures you’ve drawn.
6. Each time a student answers, correctly or incorrectly, explain the reasoning behind each figure’s perimeter.
(20 minutes)
• Start by showing your students how to use unit squares to count the perimeter units of a figure. For example, find an object in the classroom, such as a book, to measure.
• Using the cheese crackers (or the unit of your choice), model how students would find the perimeter of the book by placing crackers side-by-side, all around the book.
• Write the unit measurements on the board, and then ask the class to find the perimeter.
• For example, if the object you measured is 3 crackers on one side, by 7 crackers on another side, the perimeter (total) would be 3 + 3 + 7 + 7, or 20 crackers.
(20 minutes)
• Once you’ve finished modeling finding the perimeter, tell students to make a 3x5 array (rectangle) with the cheese crackers.
• As the class creates the rectangle, ask students to find the perimeter when they’re done, and record the perimeter in their math journals.
• Once the class is done, demonstrate how to draw a 3x5 rectangle, and how to find the perimeter, using a projector or document camera.
• Have students create a 2x10 rectangle this time, and repeat the process. Walk around the class to support students, as they need it.
• Pass out copies of the Find the Perimeter worksheet (see attached).
• Model two of the problems for the class on the board, with a document camera or with a projector.
(20 minutes)
• Have students finish the rest of the Find the Perimeter worksheet independently. If students get stuck, have them raise their hand for your assistance.
• As students complete the worksheet, check their answers.
• Students who successfully complete the worksheet should move on to their application project, creating a Dream House.
• Tell the class that each person will create a floor plan for their dream house, using cheese crackers to design and measure their layout.
• Once they’ve created a floor plan, have students record it in their math journal. The math journal should include a drawing of the floor plan, and the perimeter.
• Enrichment: Students who need more of a challenge can build multiple floor plans (for a multi-level dream house) during independent working time. You can also have these students find the area in addition to the perimeter.
• Support: Arrange students who are struggling into a small group to work with you. Complete the Find the Perimeter worksheet as a group, using cheese crackers as manipulatives.
(5 minutes)
• Review your students’ worksheets and dream house floor plans to assess their understanding of perimeter.
(10 minutes)
• Students will complete “Perimeter Four Square Review” as their journal reflection.
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8,233,918,745,728,852,000 | Curl (mathematics)
From HandWiki
Short description: Circulation density in a vector field
Depiction of a two-dimensional vector field with a uniform curl.
In vector calculus, the curl is a vector operator that describes the infinitesimal circulation of a vector field in three-dimensional Euclidean space. The curl at a point in the field is represented by a vector whose length and direction denote the magnitude and axis of the maximum circulation.[1] The curl of a field is formally defined as the circulation density at each point of the field.
A vector field whose curl is zero is called irrotational. The curl is a form of differentiation for vector fields. The corresponding form of the fundamental theorem of calculus is Stokes' theorem, which relates the surface integral of the curl of a vector field to the line integral of the vector field around the boundary curve.
Curl F is a notation common today to the United States and Americas. In many European countries, particularly in classic scientific literature, the alternative notation rot F is traditionally used, which is spelled as "rotor", and comes from the "rate of rotation", which it represents. To avoid confusion, modern authors tend to use the cross product notation with the del (nabla) operator [math]\displaystyle{ \nabla \times \mathbf{F} }[/math] [2] which also reveals the relation between curl (rotor), divergence, and gradient operators.
Unlike the gradient and divergence, curl as formulated in vector calculus does not generalize simply to other dimensions; some generalizations are possible, but only in three dimensions is the geometrically defined curl of a vector field again a vector field. This deficiency is a direct consequence of the limitations of vector calculus; on the other hand, when expressed as an antisymmetric tensor field via the wedge operator of geometric calculus, the curl generalizes to all dimensions. The unfortunate circumstance is similar to that attending the 3-dimensional cross product, and indeed the connection is reflected in the notation [math]\displaystyle{ \nabla \times }[/math] for the curl.
The name "curl" was first suggested by James Clerk Maxwell in 1871[3] but the concept was apparently first used in the construction of an optical field theory by James MacCullagh in 1839.[4][5]
Definition
The components of F at position r, normal and tangent to a closed curve C in a plane, enclosing a planar vector area [math]\displaystyle{ \mathbf{A} = A\mathbf{\hat{n}} }[/math].
Right-hand rule
Convention for vector orientation of the line integral
The thumb points in the direction of [math]\displaystyle{ \mathbf{\hat{n}} }[/math] and the fingers curl along the orientation of C
The curl of a vector field F, denoted by curl F, or [math]\displaystyle{ \nabla \times \mathbf{F} }[/math], or rot F, is an operator that maps Ck functions in R3 to Ck−1 functions in R3, and in particular, it maps continuously differentiable functions R3R3 to continuous functions R3R3. It can be defined in several ways, to be mentioned below:
One way to define the curl of a vector field at a point is implicitly through its projections onto various axes passing through the point: if [math]\displaystyle{ \mathbf{\hat{u}} }[/math] is any unit vector, the projection of the curl of F onto [math]\displaystyle{ \mathbf{\hat{u}} }[/math] may be defined to be the limiting value of a closed line integral in a plane orthogonal to [math]\displaystyle{ \mathbf{\hat{u}} }[/math] divided by the area enclosed, as the path of integration is contracted indefinitely around the point.
More specifically, the curl is defined at a point p as[6][7]
[math]\displaystyle{ (\nabla \times \mathbf{F})(p)\cdot \mathbf{\hat{u}} \ \overset{\underset{\mathrm{def}}{}}{{}={}} \lim_{A \to 0}\frac{1}{|A|}\oint_C \mathbf{F} \cdot \mathrm{d}\mathbf{r} }[/math]
where the line integral is calculated along the boundary C of the area A in question, |A| being the magnitude of the area. This equation defines the projection of the curl of F onto [math]\displaystyle{ \mathbf{\hat{u}} }[/math]. The infinitesimal surfaces bounded by C have [math]\displaystyle{ \mathbf{\hat{u}} }[/math] as their normal. C is oriented via the right-hand rule.
The above formula means that the projection of the curl of a vector field along a certain axis is the infinitesimal area density of the circulation of the field projected onto a plane perpendicular to that axis. This formula does not a priori define a legitimate vector field, for the individual circulation densities with respect to various axes a priori need not relate to each other in the same way as the components of a vector do; that they do indeed relate to each other in this precise manner must be proven separately.
To this definition fits naturally the Kelvin–Stokes theorem, as a global formula corresponding to the definition. It equates the surface integral of the curl of a vector field to the above line integral taken around the boundary of the surface.
Another way one can define the curl vector of a function F at a point is explicitly as the limiting value of a vector-valued surface integral around a shell enclosing p divided by the volume enclosed, as the shell is contracted indefinitely around p.
More specifically, the curl may be defined by the vector formula
[math]\displaystyle{ (\nabla \times \mathbf{F})(p) \overset{\underset{\mathrm{def}}{}}{{}={}} \lim_{V \to 0}\frac{1}{|V|}\oint_S \mathbf{\hat{n}} \times \mathbf{F} \ \mathrm{d}S }[/math]
where the surface integral is calculated along the boundary S of the volume V, |V| being the magnitude of the volume, and [math]\displaystyle{ \mathbf{\hat{n}} }[/math] pointing outward from the surface S perpendicularly at every point in S.
In this formula, the cross product in the integrand measures the tangential component of F at each point on the surface S, together with the orientation of these tangential components with respect to the surface S. Thus, the surface integral measures the overall extent to which F circulates around S, together with the net orientation of this circulation in space. The curl of a vector field at a point is then the infinitesimal volume density of the net vector circulation (i.e., both magnitude and spatial orientation) of the field around the point.
To this definition fits naturally another global formula (similar to the Kelvin-Stokes theorem) which equates the volume integral of the curl of a vector field to the above surface integral taken over the boundary of the volume.
Whereas the above two definitions of the curl are coordinate free, there is another "easy to memorize" definition of the curl in curvilinear orthogonal coordinates, e.g. in Cartesian coordinates, spherical, cylindrical, or even elliptical or parabolic coordinates: [math]\displaystyle{ \begin{align} & (\operatorname{curl}\mathbf F)_1=\frac{1}{h_2h_3}\left (\frac{\partial (h_3F_3)}{\partial u_2}-\frac{\partial (h_2F_2)}{\partial u_3}\right ), \\[5pt] & (\operatorname{curl}\mathbf F)_2=\frac{1}{h_3h_1}\left (\frac{\partial (h_1F_1)}{\partial u_3}-\frac{\partial (h_3F_3)}{\partial u_1}\right ), \\[5pt] & (\operatorname{curl}\mathbf F)_3=\frac{1}{h_1h_2}\left (\frac{\partial (h_2F_2)}{\partial u_1}-\frac{\partial (h_1F_1)}{\partial u_2}\right ). \end{align} }[/math]
The equation for each component (curl F)k can be obtained by exchanging each occurrence of a subscript 1, 2, 3 in cyclic permutation: 1 → 2, 2 → 3, and 3 → 1 (where the subscripts represent the relevant indices).
If (x1, x2, x3) are the Cartesian coordinates and (u1, u2, u3) are the orthogonal coordinates, then [math]\displaystyle{ h_i = \sqrt{\left (\frac{\partial x_1}{\partial u_i} \right )^2 + \left (\frac{\partial x_2}{\partial u_i} \right )^2 + \left (\frac{\partial x_3}{\partial u_i} \right )^2} }[/math] is the length of the coordinate vector corresponding to ui. The remaining two components of curl result from cyclic permutation of indices: 3,1,2 → 1,2,3 → 2,3,1.
Intuitive interpretation
Suppose the vector field describes the velocity field of a fluid flow (such as a large tank of liquid or gas) and a small ball is located within the fluid or gas (the centre of the ball being fixed at a certain point). If the ball has a rough surface, the fluid flowing past it will make it rotate. The rotation axis (oriented according to the right hand rule) points in the direction of the curl of the field at the centre of the ball, and the angular speed of the rotation is half the magnitude of the curl at this point.[8]
The curl of the vector at any point is given by the rotation of an infinitesimal area in the xy-plane (for z-axis component of the curl), zx-plane (for y-axis component of the curl) and yz-plane (for x-axis component of the curl vector). This can be clearly seen in the examples below.
Usage
In practice, the two coordinate-free definitions described above are rarely used because in virtually all cases, the curl operator can be applied using some set of curvilinear coordinates, for which simpler representations have been derived.
The notation ∇ × F has its origins in the similarities to the 3-dimensional cross product, and it is useful as a mnemonic in Cartesian coordinates if is taken as a vector differential operator del. Such notation involving operators is common in physics and algebra.
Expanded in 3-dimensional Cartesian coordinates (see Del in cylindrical and spherical coordinates for spherical and cylindrical coordinate representations),∇ × F is, for F composed of [Fx, Fy, Fz] (where the subscripts indicate the components of the vector, not partial derivatives):
[math]\displaystyle{ \nabla \times \mathbf{F} = \begin{vmatrix} \boldsymbol{\hat\imath} & \boldsymbol{\hat\jmath} & \boldsymbol{\hat k} \\[5pt] {\dfrac{\partial}{\partial x}} & {\dfrac{\partial}{\partial y}} & {\dfrac{\partial}{\partial z}} \\[10pt] F_x & F_y & F_z \end{vmatrix} }[/math]
where i, j, and k are the unit vectors for the x-, y-, and z-axes, respectively. This expands as follows:[9]:43
[math]\displaystyle{ \nabla \times \mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} = \begin{bmatrix}\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\end{bmatrix} }[/math]
Although expressed in terms of coordinates, the result is invariant under proper rotations of the coordinate axes but the result inverts under reflection.
In a general coordinate system, the curl is given by[1]
[math]\displaystyle{ (\nabla \times \mathbf{F} )^k = \frac{1}{\sqrt{g}} \varepsilon^{k\ell m} \nabla_\ell F_m }[/math]
where ε denotes the Levi-Civita tensor, the covariant derivative, [math]\displaystyle{ g }[/math] is the determinant of the metric tensor and the Einstein summation convention implies that repeated indices are summed over. Due to the symmetry of the Christoffel symbols participating in the covariant derivative, this expression reduces to the partial derivative:
[math]\displaystyle{ (\nabla \times \mathbf{F} ) = \frac{1}{\sqrt{g}} \mathbf{R}_k\varepsilon^{k\ell m} \partial_\ell F_m }[/math]
where Rk are the local basis vectors. Equivalently, using the exterior derivative, the curl can be expressed as:
[math]\displaystyle{ \nabla \times \mathbf{F} = \left( \star \big( {\mathrm d} \mathbf{F}^\flat \big) \right)^\sharp }[/math]
Here and are the musical isomorphisms, and is the Hodge star operator. This formula shows how to calculate the curl of F in any coordinate system, and how to extend the curl to any oriented three-dimensional Riemannian manifold. Since this depends on a choice of orientation, curl is a chiral operation. In other words, if the orientation is reversed, then the direction of the curl is also reversed.
Examples
Vector field F(x,y)=[y,-x] (left) and its curl (right).
Example 1
The vector field
[math]\displaystyle{ \mathbf{F}(x,y,z)=y\boldsymbol{\hat{\imath}}-x\boldsymbol{\hat{\jmath}} }[/math]
can be decomposed as
[math]\displaystyle{ F_x =y, F_y = -x, F_z =0. }[/math]
Upon visual inspection, the field can be described as "rotating". If the vectors of the field were to represent a linear force acting on objects present at that point, and an object were to be placed inside the field, the object would start to rotate clockwise around itself. This is true regardless of where the object is placed.
Calculating the curl:
[math]\displaystyle{ \nabla \times \mathbf{F} =0\boldsymbol{\hat{\imath}}+0\boldsymbol{\hat{\jmath}}+ \left({\frac{\partial}{\partial x}}(-x) -{\frac{\partial}{\partial y}} y\right)\boldsymbol{\hat{k}}=-2\boldsymbol{\hat{k}} }[/math]
The resulting vector field describing the curl would at all points be pointing in the negative z direction. The results of this equation align with what could have been predicted using the right-hand rule using a right-handed coordinate system. Being a uniform vector field, the object described before would have the same rotational intensity regardless of where it was placed.
Vector field F(x,y)=[0,−x2] (left) and its curl (right).
Example 2
For the vector field
[math]\displaystyle{ \mathbf{F}(x,y,z)=-x^2\boldsymbol{\hat{\jmath}} }[/math]
the curl is not as obvious from the graph. However, taking the object in the previous example, and placing it anywhere on the line x = 3, the force exerted on the right side would be slightly greater than the force exerted on the left, causing it to rotate clockwise. Using the right-hand rule, it can be predicted that the resulting curl would be straight in the negative z direction. Inversely, if placed on x = −3, the object would rotate counterclockwise and the right-hand rule would result in a positive z direction.
Calculating the curl:
[math]\displaystyle{ {\nabla} \times \mathbf{F} = 0 \boldsymbol{\hat{\imath}} + 0\boldsymbol{\hat{\jmath}} + {\frac{\partial}{\partial x}}\left(-x^2\right) \boldsymbol{\hat{k}} = -2x\boldsymbol{\hat{k}}. }[/math]
The curl points in the negative z direction when x is positive and vice versa. In this field, the intensity of rotation would be greater as the object moves away from the plane x = 0.
Descriptive examples
• In a vector field describing the linear velocities of each part of a rotating disk, the curl has the same value at all points, and this value turns out to be exactly two times the vectorial angular velocity of the disk (oriented as usual by the right-hand rule). More generally, for any flowing mass, the linear velocity vector field at each point of the mass flow has a curl (the vorticity of the flow at that point) equal to exactly two times the local vectorial angular velocity of the mass about the point.
• For any solid object subject to an external physical force (such as gravity or the electromagnetic force), one may consider the vector field representing the infinitesimal force-per-unit-volume contributions acting at each of the points of the object. This force field may create a net torque on the object about its center of mass, and this torque turns out to be directly proportional and vectorially parallel to the (vector-valued) integral of the curl of the force field over the whole volume.
• Of the four Maxwell's equations, two—Faraday's law and Ampère's law—can be compactly expressed using curl. Faraday's law states that the curl of an electric field is equal to the opposite of the time rate of change of the magnetic field, while Ampère's law relates the curl of the magnetic field to the current and the time rate of change of the electric field.
Identities
Main page: Vector calculus identities
In general curvilinear coordinates (not only in Cartesian coordinates), the curl of a cross product of vector fields v and F can be shown to be
[math]\displaystyle{ \nabla \times \left( \mathbf{v \times F} \right) = \Big( \left( \mathbf{ \nabla \cdot F } \right) + \mathbf{F \cdot \nabla} \Big) \mathbf{v}- \Big( \left( \mathbf{ \nabla \cdot v } \right) + \mathbf{v \cdot \nabla} \Big) \mathbf{F} \ . }[/math]
Interchanging the vector field v and operator, we arrive at the cross product of a vector field with curl of a vector field:
[math]\displaystyle{ \mathbf{v \ \times } \left( \mathbf{ \nabla \times F} \right) =\nabla_\mathbf{F} \left( \mathbf{v \cdot F } \right) - \left( \mathbf{v \cdot \nabla } \right) \mathbf{ F} \ , }[/math]
where F is the Feynman subscript notation, which considers only the variation due to the vector field F (i.e., in this case, v is treated as being constant in space).
Another example is the curl of a curl of a vector field. It can be shown that in general coordinates
[math]\displaystyle{ \nabla \times \left( \mathbf{\nabla \times F} \right) = \mathbf{\nabla}(\mathbf{\nabla \cdot F}) - \nabla^2 \mathbf{F} \ , }[/math]
and this identity defines the vector Laplacian of F, symbolized as 2F.
The curl of the gradient of any scalar field φ is always the zero vector field
[math]\displaystyle{ \nabla \times ( \nabla \varphi ) = \boldsymbol{0} }[/math]
which follows from the antisymmetry in the definition of the curl, and the symmetry of second derivatives.
The divergence of the curl of any vector field is equal to zero:
[math]\displaystyle{ \nabla\cdot(\nabla\times\mathbf{F})=0. }[/math]
If φ is a scalar valued function and F is a vector field, then
[math]\displaystyle{ \nabla \times ( \varphi \mathbf{F}) = \nabla \varphi \times \mathbf{F} + \varphi \nabla \times \mathbf{F} }[/math]
Generalizations
The vector calculus operations of grad, curl, and div are most easily generalized in the context of differential forms, which involves a number of steps. In short, they correspond to the derivatives of 0-forms, 1-forms, and 2-forms, respectively. The geometric interpretation of curl as rotation corresponds to identifying bivectors (2-vectors) in 3 dimensions with the special orthogonal Lie algebra [math]\displaystyle{ \mathfrak{so} }[/math](3) of infinitesimal rotations (in coordinates, skew-symmetric 3 × 3 matrices), while representing rotations by vectors corresponds to identifying 1-vectors (equivalently, 2-vectors) and [math]\displaystyle{ \mathfrak{so} }[/math](3), these all being 3-dimensional spaces.
Differential forms
Main page: Differential form
In 3 dimensions, a differential 0-form is simply a function f(x, y, z); a differential 1-form is the following expression, where the coefficients are functions:
[math]\displaystyle{ a_1\,dx + a_2\,dy + a_3\,dz; }[/math]
a differential 2-form is the formal sum, again with function coefficients:
[math]\displaystyle{ a_{12}\,dx\wedge dy + a_{13}\,dx\wedge dz + a_{23}\,dy\wedge dz; }[/math]
and a differential 3-form is defined by a single term with one function as coefficient:
[math]\displaystyle{ a_{123}\,dx\wedge dy\wedge dz. }[/math]
(Here the a-coefficients are real functions of three variables; the "wedge products", e.g. dxdy, can be interpreted as some kind of oriented area elements, dxdy = −dydx, etc.)
The exterior derivative of a k-form in R3 is defined as the (k + 1)-form from above—and in Rn if, e.g.,
[math]\displaystyle{ \omega^{(k)}=\sum_{\scriptstyle{i_1\lt i_2\lt \cdots\lt i_k} \atop \forall \scriptstyle{i_\nu\in 1,\ldots,n}} a_{i_1,\ldots,i_k}\,dx_{i_1}\wedge \cdots\wedge dx_{i_k}, }[/math]
then the exterior derivative d leads to
[math]\displaystyle{ d\omega^{(k)}=\sum_{\scriptstyle{j=1} \atop \scriptstyle{i_1\lt \cdots\lt i_k}}^n\frac{\partial a_{i_1,\ldots,i_k}}{\partial x_j}\,dx_j \wedge dx_{i_1}\wedge \cdots \wedge dx_{i_k}. }[/math]
The exterior derivative of a 1-form is therefore a 2-form, and that of a 2-form is a 3-form. On the other hand, because of the interchangeability of mixed derivatives, e.g. because of
[math]\displaystyle{ \frac{\partial^2}{\partial x\,\partial y}=\frac{\partial^2}{\partial y\,\partial x}, }[/math]
the twofold application of the exterior derivative leads to 0.
Thus, denoting the space of k-forms by Ωk(R3) and the exterior derivative by d one gets a sequence:
[math]\displaystyle{ 0 \, \overset{d}{\longrightarrow} \; \Omega^0\left(\mathbb{R}^3\right) \, \overset{d}{\longrightarrow} \; \Omega^1\left(\mathbb{R}^3\right) \, \overset{d}{\longrightarrow} \; \Omega^2\left(\mathbb{R}^3\right) \, \overset{d}{\longrightarrow} \; \Omega^3\left(\mathbb{R}^3\right) \, \overset{d}{\longrightarrow} \, 0. }[/math]
Here Ωk(Rn) is the space of sections of the exterior algebra Λk(Rn) vector bundle over Rn, whose dimension is the binomial coefficient (nk); note that Ωk(R3) = 0 for k > 3 or k < 0. Writing only dimensions, one obtains a row of Pascal's triangle:
0 → 1 → 3 → 3 → 1 → 0;
the 1-dimensional fibers correspond to scalar fields, and the 3-dimensional fibers to vector fields, as described below. Modulo suitable identifications, the three nontrivial occurrences of the exterior derivative correspond to grad, curl, and div.
Differential forms and the differential can be defined on any Euclidean space, or indeed any manifold, without any notion of a Riemannian metric. On a Riemannian manifold, or more generally pseudo-Riemannian manifold, k-forms can be identified with k-vector fields (k-forms are k-covector fields, and a pseudo-Riemannian metric gives an isomorphism between vectors and covectors), and on an oriented vector space with a nondegenerate form (an isomorphism between vectors and covectors), there is an isomorphism between k-vectors and (nk)-vectors; in particular on (the tangent space of) an oriented pseudo-Riemannian manifold. Thus on an oriented pseudo-Riemannian manifold, one can interchange k-forms, k-vector fields, (nk)-forms, and (nk)-vector fields; this is known as Hodge duality. Concretely, on R3 this is given by:
• 1-forms and 1-vector fields: the 1-form ax dx + ay dy + az dz corresponds to the vector field (ax, ay, az).
• 1-forms and 2-forms: one replaces dx by the dual quantity dydz (i.e., omit dx), and likewise, taking care of orientation: dy corresponds to dzdx = −dxdz, and dz corresponds to dxdy. Thus the form ax dx + ay dy + az dz corresponds to the "dual form" az dxdy + ay dzdx + ax dydz.
Thus, identifying 0-forms and 3-forms with scalar fields, and 1-forms and 2-forms with vector fields:
• grad takes a scalar field (0-form) to a vector field (1-form);
• curl takes a vector field (1-form) to a pseudovector field (2-form);
• div takes a pseudovector field (2-form) to a pseudoscalar field (3-form)
On the other hand, the fact that d2 = 0 corresponds to the identities
[math]\displaystyle{ \nabla\times(\nabla f) = \mathbf 0 }[/math]
for any scalar field f, and
[math]\displaystyle{ \nabla \cdot (\nabla \times\mathbf v)=0 }[/math]
for any vector field v.
Grad and div generalize to all oriented pseudo-Riemannian manifolds, with the same geometric interpretation, because the spaces of 0-forms and n-forms at each point are always 1-dimensional and can be identified with scalar fields, while the spaces of 1-forms and (n − 1)-forms are always fiberwise n-dimensional and can be identified with vector fields.
Curl does not generalize in this way to 4 or more dimensions (or down to 2 or fewer dimensions); in 4 dimensions the dimensions are
0 → 1 → 4 → 6 → 4 → 1 → 0;
so the curl of a 1-vector field (fiberwise 4-dimensional) is a 2-vector field, which at each point belongs to 6-dimensional vector space, and so one has
[math]\displaystyle{ \omega^{(2)}=\sum_{i\lt k=1,2,3,4}a_{i,k}\,dx_i\wedge dx_k, }[/math]
which yields a sum of six independent terms, and cannot be identified with a 1-vector field. Nor can one meaningfully go from a 1-vector field to a 2-vector field to a 3-vector field (4 → 6 → 4), as taking the differential twice yields zero (d2 = 0). Thus there is no curl function from vector fields to vector fields in other dimensions arising in this way.
However, one can define a curl of a vector field as a 2-vector field in general, as described below.
Curl geometrically
2-vectors correspond to the exterior power Λ2V; in the presence of an inner product, in coordinates these are the skew-symmetric matrices, which are geometrically considered as the special orthogonal Lie algebra [math]\displaystyle{ \mathfrak{so} }[/math](V) of infinitesimal rotations. This has (n2) = 1/2n(n − 1) dimensions, and allows one to interpret the differential of a 1-vector field as its infinitesimal rotations. Only in 3 dimensions (or trivially in 0 dimensions) we have n = 1/2n(n − 1), which is the most elegant and common case. In 2 dimensions the curl of a vector field is not a vector field but a function, as 2-dimensional rotations are given by an angle (a scalar – an orientation is required to choose whether one counts clockwise or counterclockwise rotations as positive); this is not the div, but is rather perpendicular to it. In 3 dimensions the curl of a vector field is a vector field as is familiar (in 1 and 0 dimensions the curl of a vector field is 0, because there are no non-trivial 2-vectors), while in 4 dimensions the curl of a vector field is, geometrically, at each point an element of the 6-dimensional Lie algebra [math]\displaystyle{ \mathfrak{so}(4) }[/math].
The curl of a 3-dimensional vector field which only depends on 2 coordinates (say x and y) is simply a vertical vector field (in the z direction) whose magnitude is the curl of the 2-dimensional vector field, as in the examples on this page.
Considering curl as a 2-vector field (an antisymmetric 2-tensor) has been used to generalize vector calculus and associated physics to higher dimensions.[10]
Inverse
Main page: Helmholtz decomposition
In the case where the divergence of a vector field V is zero, a vector field W exists such that V = curl(W). This is why the magnetic field, characterized by zero divergence, can be expressed as the curl of a magnetic vector potential.
If W is a vector field with curl(W) = V, then adding any gradient vector field grad(f) to W will result in another vector field W + grad(f) such that curl(W + grad(f)) = V as well. This can be summarized by saying that the inverse curl of a three-dimensional vector field can be obtained up to an unknown irrotational field with the Biot–Savart law.
See also
References
1. 1.0 1.1 Weisstein, Eric W.. "Curl". http://mathworld.wolfram.com/Curl.html.
2. ISO/IEC 80000-2 standard Norm ISO/IEC 80000-2, item 2-17.16
3. Proceedings of the London Mathematical Society, March 9th, 1871
4. Collected works of James MacCullagh
5. Earliest Known Uses of Some of the Words of Mathematics tripod.com
6. Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISBN:978-0-521-86153-3
7. Vector Analysis (2nd Edition), M.R. Spiegel, S. Lipschutz, D. Spellman, Schaum's Outlines, McGraw Hill (USA), 2009, ISBN:978-0-07-161545-7
8. Gibbs, Josiah Willard; Wilson, Edwin Bidwell (1901), Vector analysis, Yale bicentennial publications, C. Scribner's Sons, http://hdl.handle.net/2027/mdp.39015000962285?urlappend=%3Bseq=179
9. Arfken, George Brown (2005). Mathematical methods for physicists. Weber, Hans-Jurgen (6th ed.). Boston: Elsevier. ISBN 978-0-08-047069-6. OCLC 127114279. https://www.worldcat.org/oclc/127114279.
10. McDavid, A. W.; McMullen, C. D. (2006-10-30). "Generalizing Cross Products and Maxwell's Equations to Universal Extra Dimensions". arXiv:hep-ph/0609260.
Further reading
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-8,352,815,575,793,834,000 | Sunday
March 29, 2015
Homework Help: Algebra 2
Posted by HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!! on Wednesday, February 25, 2009 at 8:08pm.
how do not understand how to do this
Log X + Lob (X-3) = 1
I know I do this
Log X(X-3)=1
then I do this
Log X^(2) - 3X = 1
then I do this
2 Log (X-3X) = 1
then
2 Log (-2X) = 1
then
(2 Log (-2X) = 1)(1/2)
Log (-2X) = 1/2
then
(Log (-2X) = 1/2)(-1/2)
Log X = (-1/4)
then
by def
10^ (-1/4) = X
I get something under one and calculator tells me I can't do that I think I'm doing something wrong what do I do?
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-6,734,665,798,861,087,000 |
Results 1 to 3 of 3
Like Tree2Thanks
• 1 Post By HallsofIvy
• 1 Post By Prove It
Thread: Adding Rational Expressions
1. #1
Newbie
Joined
Nov 2012
From
United States
Posts
17
Adding Rational Expressions
Problem: \frac{x - 8}{x^2 - 4} + \frac{3}{x^2-2x}
Answer: \frac{x-3}{x(x+2)}, x\not=2
What I ended up with: \frac{x^2-5x+6}{x(x^2 - 4)}
Any help on how to do problems like this would be great as well. I feel like I was doing everything correctly, but now that I'm at the end I don't see any way to make my answer look like the one in the book.
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2. #2
MHF Contributor
Joined
Apr 2005
Posts
14,730
Thanks
1034
Re: Adding Rational Expressions
Very close. All you need to do is observe that x^2- 5x+ 6= (x- 2)(x- 3).
Even if you are not good at factoring in general, you should recognize that x^2- 4= (x- 2)(x+ 2), which you apparently did to get the "least common denominator" and then check for common factors in the numerator and denominator. And you can do that by evaluating 0^2- 5(0)+ 6= 6 so x- 0= x is not a factor. (-2)^2- 5(-2)+ 3= 17 so x+ 2 is not a factor. But 2^2- 5(2)+6= 0 so x- 2 is a factor of x^2- 5x+ 6.
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Re: Adding Rational Expressions
Quote Originally Posted by fogownz View Post
Problem: \frac{x - 8}{x^2 - 4} + \frac{3}{x^2-2x}
Answer: \frac{x-3}{x(x+2)}, x\not=2
What I ended up with: \frac{x^2-5x+6}{x(x^2 - 4)}
Any help on how to do problems like this would be great as well. I feel like I was doing everything correctly, but now that I'm at the end I don't see any way to make my answer look like the one in the book.
Notice that \displaystyle \begin{align*} x^2 - 4 = (x - 2)(x + 2) \end{align*} and \displaystyle \begin{align*} x^2 - 2x = x(x - 2) \end{align*}. So the lowest common denominator is \displaystyle \begin{align*} x(x-2)(x+2) \end{align*}. You should get into the habit of keeping things factorised until the end, as it makes cancelling a lot easier.
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8,850,922,001,561,759,000 | Converting 1.5 Miles to Kilometers: The Ultimate Guide for Easy Calculation and Conversion
Understanding the Conversion: 1.5 Miles to Kilometers Explained
Understanding the Conversion: 1.5 Miles to Kilometers Explained
The conversion from miles to kilometers is an important concept for anyone who frequently deals with measurements. It allows for easy comparison and understanding of distances in different units. In this article, we will explain the conversion from 1.5 miles to kilometers in detail, providing you with the necessary knowledge to make accurate calculations.
Miles to Kilometers Conversion Formula
To convert miles to kilometers, you can use a simple formula. Multiply the number of miles by 1.60934 to get the equivalent distance in kilometers. In the case of converting 1.5 miles to kilometers, you would perform the following calculation: 1.5 miles * 1.60934 = 2.414 kilometers.
It’s important to note that the conversion factor of 1.60934 is an approximate value. The exact conversion factor for miles to kilometers is defined as 1.609344, but for simplicity, it is often rounded to 1.60934 in calculations. Keep this in mind when performing conversions to ensure accuracy.
Understanding the miles to kilometers conversion is crucial in many scenarios, such as when planning a trip or understanding distances in different countries. By knowing how to convert 1.5 miles to kilometers, you can quickly and easily make accurate measurements and comparisons. Remember to use the conversion formula of 1.60934 for the best results.
Why Knowing How to Convert 1.5 Miles to Km is Essential
The Importance of Understanding Distance Conversions
As travel becomes more accessible and international borders become easier to cross, it is crucial to have a good grasp of distance conversions. One commonly encountered conversion is from miles to kilometers. Knowing how to convert 1.5 miles to km can be especially essential, as it can help you plan your trips, determine commuting distances, or simply understand distances mentioned in conversations or online resources.
The Simple Formula: Miles to Kilometers Conversion
To convert miles to kilometers, you can use a simple formula: multiply the number of miles by 1.60934. With this formula, converting 1.5 miles to km is a breeze. Simply multiply 1.5 by 1.60934, resulting in 2.414 kilometers.
Practical Applications and Benefits
Understanding how to convert 1.5 miles to km can have a range of practical applications. For travelers, this conversion skill can help estimate journey lengths, optimize travel itineraries, and plan for transportation and rental costs. For fitness enthusiasts, converting distances can assist in tracking progress, comparing workouts, or setting personal goals. Additionally, having a good understanding of distance conversions is crucial for professionals in shipping and logistics, as incorrect measurements can lead to inefficient transportation planning, added costs, and potential customer dissatisfaction.
By mastering the conversion from miles to kilometers, you are arming yourself with a valuable skill that can enhance your travel experiences, improve fitness tracking, and even benefit professional endeavors. So, the next time you come across the measurement of 1.5 miles, you’ll be prepared to swiftly convert it to kilometers, expanding your understanding of distance and making informed decisions.
Mastering the Calculation: Converting 1.5 Miles to Kilometers
Why Convert Miles to Kilometers?
Converting miles to kilometers is essential for individuals who travel internationally or work in a global environment. Kilometers are the standard unit of distance used in most countries outside of the United States, making it important to be able to convert distances accurately. One common conversion is converting 1.5 miles to kilometers. Learning how to master this calculation is not only useful for everyday life but can also come in handy for various professional fields such as logistics, transportation, and sports.
The Conversion Formula
To convert miles to kilometers, you can use a simple multiplication formula. The conversion factor is 0.62137119, which represents the number of kilometers in one mile. So, to convert 1.5 miles to kilometers, you would multiply 1.5 by the conversion factor. This can be expressed as:
1.5 miles * 0.62137119 = 1.60934 kilometers
It is essential to note that this is an approximate conversion, as the conversion factor has decimals. However, for most practical purposes, this level of precision is sufficient.
Benefits of Understanding the Calculation
Mastering the calculation to convert 1.5 miles to kilometers has several advantages. Firstly, it allows for better understanding and appreciation of international distances. Additionally, it helps in planning and estimating travel times more accurately. It also provides a common foundation for communication with individuals who are accustomed to using kilometers. Understanding this calculation can boost your confidence when dealing with measurements on both local and global scales, making it an essential skill for anyone working or traveling internationally.
The Advantages of Converting 1.5 Miles to Kilometers
Accurate Measurement Conversion
You may also be interested in: Converting 98.3°F to Celsius: The Quick and Easy Guide for Accurate Temperature Conversions
Converting 1.5 miles to kilometers opens up a wide range of advantages, especially when it comes to accurate measurement conversion. While miles are commonly used in the United States, kilometers are the standard unit of measurement in most other parts of the world. By converting miles to kilometers, you can communicate and understand distances more effectively, whether you are traveling abroad or working on international projects.
Improved Navigation
Converting 1.5 miles to kilometers can also greatly improve navigation, particularly when using digital navigation tools and maps. Most GPS systems and online mapping services use kilometers as the default unit of measurement. By converting miles to kilometers, you can easily input and follow directions, making navigation simpler and more accurate. This is especially helpful for travelers exploring foreign cities or planning road trips in countries using the metric system.
Enhanced Fitness Tracking
If you are an avid runner, cyclist, or fitness enthusiast, converting 1.5 miles to kilometers can be beneficial for tracking your progress and setting goals. Many fitness tracking devices and apps offer metric measurements, and converting miles to kilometers allows you to take full advantage of these features. By converting distances, you can accurately monitor your pace, track improvements, and compare your performance with others globally.
Overall, the advantages of converting 1.5 miles to kilometers are plentiful. From accurate measurement conversion to improved navigation and enhanced fitness tracking, understanding and utilizing kilometers opens doors to wider opportunities both locally and internationally. Whether you’re planning a trip abroad, using digital mapping tools, or pursuing fitness goals, converting miles to kilometers can provide a valuable edge in various aspects of life.
You may also be interested in: Converting 53°F to Celsius: The Easy and Accurate Method Explained
Quick and Easy Methods for Converting 1.5 Miles to Km
1. Multiply by 1.60934
Converting miles to kilometers is a straightforward process. One of the quickest and easiest methods is to multiply the distance in miles by 1.60934. In the case of converting 1.5 miles to kilometers, simply multiply 1.5 by 1.60934, which equals 2.414 km. This method is widely used and can be easily done with a calculator or even mentally.
You may also be interested in: Quick and Easy Conversion: Mastering the 99 F to C Equation for Temperature Conversions
2. Use an Online Converter
Another quick and easy method for converting 1.5 miles to kilometers is to utilize an online conversion tool. There are numerous websites and apps available that offer simple and accurate mile-to-kilometer conversions. To convert 1.5 miles to kilometers using an online converter, input the value in miles and the tool will provide the equivalent value in kilometers instantly. This method is convenient when you need to convert various distances quickly without the need for manual calculations.
3. Memorize Popular Conversion Factors
For those who frequently need to convert miles to kilometers, it can be helpful to memorize some common conversion factors. One popular conversion factor is that 1 mile is roughly equivalent to 1.6 kilometers. By keeping this conversion factor in mind, you can easily estimate the equivalent distance in kilometers. In the case of converting 1.5 miles to kilometers, you can quickly estimate that it is approximately 2.4 kilometers using mental arithmetic.
Converting 1.5 miles to kilometers doesn’t have to be a complicated task. By using the methods mentioned above, you can quickly and easily obtain the equivalent value in kilometers. Whether you choose to multiply by the conversion factor, use an online converter, or memorize popular conversion factors, these methods will save you time and effort when converting distances between miles and kilometers.
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-7,070,700,196,449,720,000 |
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lisa is starting a movie collecion. blu-rays cost $20 each and DVD's cost $14 each. If she buys 12 movies for $186, how many movie did she buy?
Guest May 18, 2017
Sort:
2+0 Answers
#1
avatar+84168
+1
Call the number of $20 movies, x
Then the number of $14 movies = 12 - x
And we have
20x + 14 (12 - x) = 186 simplify
20x + 168 - 14x = 186
6x + 168 = 186 subtract 168 from both sides
6x = 18 divde both sides by 6
x = 3 = number of $20 movies
And 12 - 3 = 9 = the number of $14 movies
cool cool cool
CPhill May 18, 2017
#2
avatar
+1
Let 'b' represent the number of Blue Rays
Since she bought 12 movies, we know that the number of DVDs is '12 - b'
So b + (12 - b) = 12
She spent $186, and the price of Blue Rays are $20 while DVDs are $14
So 20b + 14(12 - b) = 186
To find b, lets make the coefficient of b equal 20 in both equations
(b + (12 - b) = 12) * 20
20b + 20(12 - b) = 240
Now lets subtract the second equation from the first one.
20b + 20(12 - b) = 240
- 20b + 14(12 - b) = 186
-------------------------------
6(12 - b) = 54
6(12 - b) = 54
72 - 6b = 54
72 - 54 = 6b
(18 = 6b) / 6
3 = b
We now know that there are 3 Blue Rays
And since the number of DVDs is 12 - b, we know that there are 9 DVDs
So therefore, Lisa bought 3 Blue Rays and 9 DVDs for $186
Guest May 18, 2017
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-3,546,048,444,222,436,000 | Example 22 - Let f be one-one onto f(1) = a, f(2) = b - Finding Inverse
Slide42.JPG
1. Chapter 1 Class 12 Relation and Functions
2. Serial order wise
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Transcript
Example 22 Let f : {1, 2, 3} {a, b, c} be one-one and onto function given by f (1) = a, f(2) = b and f (3) = c. Show that there exists a function g : {a, b, c} {1, 2, 3} such that gof= IX and fog = IY, where, X = {1, 2, 3} and Y = {a, b, c}. Finding gof So, gof = { (1, 1) , (2, 2), (3, 3) } = IX = Identity function on set X = {1, 2, 3} Finding fog fog = { (a, a) , (b, b), (c, c) } = IY = Identity function on set Y = {a, b, c}
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6,495,967,693,854,550,000 | dharr
Dr. David Harrington
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I am a professor of chemistry at the University of Victoria, BC, Canada, where my research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.
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These are replies submitted by dharr
@dharr For the EqN for all parameters 1, the three equations have a common factor, that can be divided out so that you could work on solving a simpler system. If this still works with parameters in, that could afford a considerable simplification.
@MaPal93 In general, I expect the solvability to depend on the parameters. This seems like a very complicated problem with too many parameters - how about solving a simpler version of the problem with a smaller set of parameters first? But if I were doing this problem I would step back and analyze the matrices. Are lambda's eigenvalues? if so a lot can be said about their possible values depending on the structure of the matrix of which they are eigenvalues, e.g., we know many classes of matrices that are guaranteed to have one or more real eigenvalues.
@dharr For solution 1 you can find real solutions as follows (but real and positive is harder).
restart
EqN in startup code.
Since most lambda__1 and lambda++2 work, you can look at the solution as a quadratic in lambda_3
sol1 := [[((2*`λ__1`+1)*`λ__2`+`λ__1`)*`λ__3`^2+((2*`λ__1`+1)*`λ__2`^2+(2*`λ__1`^2-6*`λ__1`+2)*`λ__2`+`λ__1`^2+2*`λ__1`)*`λ__3`+4*`λ__2`^2*`λ__1`+(4*`λ__1`^2+8*`λ__1`)*`λ__2`], {(2*`λ__1`+1)*`λ__2`+`λ__1` <> 0}]
[[((2*lambda__1+1)*lambda__2+lambda__1)*lambda__3^2+((2*lambda__1+1)*lambda__2^2+(2*lambda__1^2-6*lambda__1+2)*lambda__2+lambda__1^2+2*lambda__1)*lambda__3+4*lambda__1*lambda__2^2+(4*lambda__1^2+8*lambda__1)*lambda__2], {(2*lambda__1+1)*lambda__2+lambda__1 <> 0}]
p := collect(sol1[1][], `λ__3`)
((2*lambda__1+1)*lambda__2+lambda__1)*lambda__3^2+((2*lambda__1+1)*lambda__2^2+(2*lambda__1^2-6*lambda__1+2)*lambda__2+lambda__1^2+2*lambda__1)*lambda__3+4*lambda__1*lambda__2^2+(4*lambda__1^2+8*lambda__1)*lambda__2
d := discrim(p, `λ__3`)
4*lambda__1^4*lambda__2^2+8*lambda__1^3*lambda__2^3+4*lambda__1^2*lambda__2^4+4*lambda__1^4*lambda__2-52*lambda__1^3*lambda__2^2-52*lambda__1^2*lambda__2^3+4*lambda__1*lambda__2^4+lambda__1^4-20*lambda__1^3*lambda__2-42*lambda__1^2*lambda__2^2-20*lambda__1*lambda__2^3+lambda__2^4+4*lambda__1^3-52*lambda__1^2*lambda__2-52*lambda__1*lambda__2^2+4*lambda__2^3+4*lambda__1^2+8*lambda__1*lambda__2+4*lambda__2^2
Real solutions when discriminant is zero
ans := solve(d, `λ__2`)
-(lambda__1+2)/(2*lambda__1+1), -lambda__1, (-lambda__1^2+(lambda__1^4-32*lambda__1^3+222*lambda__1^2-32*lambda__1+1)^(1/2)+15*lambda__1-1)/(2*lambda__1+1), -(lambda__1^2+(lambda__1^4-32*lambda__1^3+222*lambda__1^2-32*lambda__1+1)^(1/2)-15*lambda__1+1)/(2*lambda__1+1)
example
sol := {`λ__1` = 1}; sol := `union`(sol, {`λ__2` = (eval(ans, sol))[3]}); sol := `union`(sol, {`λ__3` = solve(eval(p, sol), `λ__3`)[2]}); simplify(eval(EqN, sol))
{lambda__1 = 1}
{lambda__1 = 1, lambda__2 = 13/3+(1/3)*160^(1/2)}
{lambda__1 = 1, lambda__2 = 13/3+(1/3)*160^(1/2), lambda__3 = -2*(13+4*10^(1/2))/(2*10^(1/2)+7)}
{0}
Download EqNsol.mw
@MaPal93 Some answers:
1) what is exactly Eqn? could you load my solution instead of hard coding expressions?
I definitely did not reenter this by hand! This was extracted from your worksheet by opening a new worksheet with the same Maple engine - try EqN; in your own worksheet to check.
2) is this system indeterminate (infinitely many solutions)?
Yes.
3) would it be preferred if I chose backsubstitute = true?
Probably the answer is too complicated to analyse, and maybe too slow to achieve - but you could try.
4) what happened to my parameters P := indets(MyEqs, name) minus MyVars ? I was hoping to obtain solutions as functions of those.
I just looked at what you were passing to the solver, i.e., EqN. Looks like in the line EqN := ((`~`[((numer@evala)@(:-Norm))@numer])@eval)(MyEqs, `=`~(P, 1)) you set all the parameters to 1.
5) is this solving approach (a) reliable and (b) preferable to engine=groebner or standard solve()?
I don't see why the answer wouldn't be reliable. It worked for the case I tried at random. I don't know much about different engines; solve probably passes to one or other of these.
6) how to enforce RealDomain?
If you want only real solutions, try RealTriangularize in the RegularChains package, which was a recent suggestion somewhere here on Mapleprimes. But I really don't know much about solving polynomial equations beyond the bivariate case.
@MaPal93 Here's my understanding. You chose backsubstitute=false, so you can carry out the backsubstitution yourself, e.g.
restart
EqN := 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-(2*`λ__1`^2*`λ__2`*`λ__3`+2*`λ__1`*`λ__2`^2*`λ__3`+2*`λ__1`*`λ__2`*`λ__3`^2+4*`λ__1`^2*`λ__2`+`λ__1`^2*`λ__3`+4*`λ__1`*`λ__2`^2-6*`λ__1`*`λ__2`*`λ__3`+`λ__1`*`λ__3`^2+`λ__2`^2*`λ__3`+`λ__2`*`λ__3`^2+8*`λ__1`*`λ__2`+2*`λ__1`*`λ__3`+2*`λ__2`*`λ__3`)*(4*`λ__1`^5*`λ__2`^2*`λ__3`^2+8*`λ__1`^4*`λ__2`^3*`λ__3`^2+8*`λ__1`^4*`λ__2`^2*`λ__3`^3+4*`λ__1`^3*`λ__2`^4*`λ__3`^2+8*`λ__1`^3*`λ__2`^3*`λ__3`^3+4*`λ__1`^3*`λ__2`^2*`λ__3`^4+16*`λ__1`^5*`λ__2`^2*`λ__3`+4*`λ__1`^5*`λ__2`*`λ__3`^2+32*`λ__1`^4*`λ__2`^3*`λ__3`-4*`λ__1`^4*`λ__2`^2*`λ__3`^2+8*`λ__1`^4*`λ__2`*`λ__3`^3+16*`λ__1`^3*`λ__2`^4*`λ__3`-4*`λ__1`^3*`λ__2`^3*`λ__3`^2-16*`λ__1`^3*`λ__2`^2*`λ__3`^3+4*`λ__1`^3*`λ__2`*`λ__3`^4+4*`λ__1`^2*`λ__2`^4*`λ__3`^2+8*`λ__1`^2*`λ__2`^3*`λ__3`^3+4*`λ__1`^2*`λ__2`^2*`λ__3`^4+16*`λ__1`^5*`λ__2`^2+8*`λ__1`^5*`λ__2`*`λ__3`+`λ__1`^5*`λ__3`^2+32*`λ__1`^4*`λ__2`^3-8*`λ__1`^4*`λ__2`^2*`λ__3`+4*`λ__1`^4*`λ__2`*`λ__3`^2+2*`λ__1`^4*`λ__3`^3+16*`λ__1`^3*`λ__2`^4-8*`λ__1`^3*`λ__2`^3*`λ__3`+126*`λ__1`^3*`λ__2`^2*`λ__3`^2-2*`λ__1`^3*`λ__2`*`λ__3`^3+`λ__1`^3*`λ__3`^4+8*`λ__1`^2*`λ__2`^4*`λ__3`-28*`λ__1`^2*`λ__2`^3*`λ__3`^2-34*`λ__1`^2*`λ__2`^2*`λ__3`^3+2*`λ__1`^2*`λ__2`*`λ__3`^4+9*`λ__1`*`λ__2`^4*`λ__3`^2+18*`λ__1`*`λ__2`^3*`λ__3`^3+9*`λ__1`*`λ__2`^2*`λ__3`^4+64*`λ__1`^4*`λ__2`^2+32*`λ__1`^4*`λ__2`*`λ__3`+4*`λ__1`^4*`λ__3`^2+64*`λ__1`^3*`λ__2`^3-88*`λ__1`^3*`λ__2`^2*`λ__3`-10*`λ__1`^3*`λ__2`*`λ__3`^2+4*`λ__1`^3*`λ__3`^3+16*`λ__1`^2*`λ__2`^3*`λ__3`+2*`λ__1`^2*`λ__2`^2*`λ__3`^2+4*`λ__1`^2*`λ__2`*`λ__3`^3+8*`λ__1`*`λ__2`^4*`λ__3`+18*`λ__1`*`λ__2`^3*`λ__3`^2+12*`λ__1`*`λ__2`^2*`λ__3`^3+2*`λ__1`*`λ__2`*`λ__3`^4+2*`λ__2`^4*`λ__3`^2+4*`λ__2`^3*`λ__3`^3+2*`λ__2`^2*`λ__3`^4+64*`λ__1`^3*`λ__2`^2+32*`λ__1`^3*`λ__2`*`λ__3`+4*`λ__1`^3*`λ__3`^2+32*`λ__1`*`λ__2`^3*`λ__3`+4*`λ__1`*`λ__2`^2*`λ__3`^2+8*`λ__1`*`λ__2`*`λ__3`^3+8*`λ__2`^3*`λ__3`^2+8*`λ__2`^2*`λ__3`^3+32*`λ__1`*`λ__2`^2*`λ__3`+8*`λ__1`*`λ__2`*`λ__3`^2+8*`λ__2`^2*`λ__3`^2), -(2*`λ__1`^2*`λ__2`*`λ__3`+2*`λ__1`*`λ__2`^2*`λ__3`+2*`λ__1`*`λ__2`*`λ__3`^2+4*`λ__1`^2*`λ__2`+`λ__1`^2*`λ__3`+4*`λ__1`*`λ__2`^2-6*`λ__1`*`λ__2`*`λ__3`+`λ__1`*`λ__3`^2+`λ__2`^2*`λ__3`+`λ__2`*`λ__3`^2+8*`λ__1`*`λ__2`+2*`λ__1`*`λ__3`+2*`λ__2`*`λ__3`)*(4*`λ__1`^4*`λ__2`^2*`λ__3`^3+8*`λ__1`^3*`λ__2`^3*`λ__3`^3+8*`λ__1`^3*`λ__2`^2*`λ__3`^4+4*`λ__1`^2*`λ__2`^4*`λ__3`^3+8*`λ__1`^2*`λ__2`^3*`λ__3`^4+4*`λ__1`^2*`λ__2`^2*`λ__3`^5+16*`λ__1`^4*`λ__2`^2*`λ__3`^2+4*`λ__1`^4*`λ__2`*`λ__3`^3+32*`λ__1`^3*`λ__2`^3*`λ__3`^2-4*`λ__1`^3*`λ__2`^2*`λ__3`^3+8*`λ__1`^3*`λ__2`*`λ__3`^4+16*`λ__1`^2*`λ__2`^4*`λ__3`^2-4*`λ__1`^2*`λ__2`^3*`λ__3`^3-16*`λ__1`^2*`λ__2`^2*`λ__3`^4+4*`λ__1`^2*`λ__2`*`λ__3`^5+4*`λ__1`*`λ__2`^4*`λ__3`^3+8*`λ__1`*`λ__2`^3*`λ__3`^4+4*`λ__1`*`λ__2`^2*`λ__3`^5+48*`λ__1`^4*`λ__2`^2*`λ__3`+8*`λ__1`^4*`λ__2`*`λ__3`^2+`λ__1`^4*`λ__3`^3+96*`λ__1`^3*`λ__2`^3*`λ__3`-40*`λ__1`^3*`λ__2`^2*`λ__3`^2+4*`λ__1`^3*`λ__2`*`λ__3`^3+2*`λ__1`^3*`λ__3`^4+48*`λ__1`^2*`λ__2`^4*`λ__3`-40*`λ__1`^2*`λ__2`^3*`λ__3`^2+102*`λ__1`^2*`λ__2`^2*`λ__3`^3-2*`λ__1`^2*`λ__2`*`λ__3`^4+`λ__1`^2*`λ__3`^5+8*`λ__1`*`λ__2`^4*`λ__3`^2+4*`λ__1`*`λ__2`^3*`λ__3`^3-2*`λ__1`*`λ__2`^2*`λ__3`^4+2*`λ__1`*`λ__2`*`λ__3`^5+`λ__2`^4*`λ__3`^3+2*`λ__2`^3*`λ__3`^4+`λ__2`^2*`λ__3`^5+32*`λ__1`^4*`λ__2`^2+8*`λ__1`^4*`λ__2`*`λ__3`+64*`λ__1`^3*`λ__2`^3+88*`λ__1`^3*`λ__2`^2*`λ__3`+32*`λ__1`^3*`λ__2`*`λ__3`^2+4*`λ__1`^3*`λ__3`^3+32*`λ__1`^2*`λ__2`^4+88*`λ__1`^2*`λ__2`^3*`λ__3`-32*`λ__1`^2*`λ__2`^2*`λ__3`^2-12*`λ__1`^2*`λ__2`*`λ__3`^3+4*`λ__1`^2*`λ__3`^4+8*`λ__1`*`λ__2`^4*`λ__3`+32*`λ__1`*`λ__2`^3*`λ__3`^2-12*`λ__1`*`λ__2`^2*`λ__3`^3+8*`λ__1`*`λ__2`*`λ__3`^4+4*`λ__2`^3*`λ__3`^3+4*`λ__2`^2*`λ__3`^4+128*`λ__1`^3*`λ__2`^2+32*`λ__1`^3*`λ__2`*`λ__3`+128*`λ__1`^2*`λ__2`^3+32*`λ__1`^2*`λ__2`*`λ__3`^2+4*`λ__1`^2*`λ__3`^3+32*`λ__1`*`λ__2`^3*`λ__3`+32*`λ__1`*`λ__2`^2*`λ__3`^2+8*`λ__1`*`λ__2`*`λ__3`^3+4*`λ__2`^2*`λ__3`^3+128*`λ__1`^2*`λ__2`^2+32*`λ__1`^2*`λ__2`*`λ__3`+32*`λ__1`*`λ__2`^2*`λ__3`)}
Check out first solution, solve order lambda__3 > lambda__2 > lambda__1
sol1 := [[((2*`λ__1`+1)*`λ__2`+`λ__1`)*`λ__3`^2+((2*`λ__1`+1)*`λ__2`^2+(2*`λ__1`^2-6*`λ__1`+2)*`λ__2`+`λ__1`^2+2*`λ__1`)*`λ__3`+4*`λ__2`^2*`λ__1`+(4*`λ__1`^2+8*`λ__1`)*`λ__2`], {(2*`λ__1`+1)*`λ__2`+`λ__1` <> 0}]
[[((2*lambda__1+1)*lambda__2+lambda__1)*lambda__3^2+((2*lambda__1+1)*lambda__2^2+(2*lambda__1^2-6*lambda__1+2)*lambda__2+lambda__1^2+2*lambda__1)*lambda__3+4*lambda__1*lambda__2^2+(4*lambda__1^2+8*lambda__1)*lambda__2], {(2*lambda__1+1)*lambda__2+lambda__1 <> 0}]
Choose any lambda__1
sol := {`λ__1` = 4}
{lambda__1 = 4}
Lambda_2 is now found from {(2*`λ__1`+1)*`λ__2`+`λ__1` <> 0}
{(2*lambda__1+1)*lambda__2+lambda__1 <> 0}
solve(eval({(2*`λ__1`+1)*`λ__2`+`λ__1` = 0}, sol), `λ__2`)
{lambda__2 = -4/9}
So any value except the above, e.g.,
sol := `union`(sol, {`λ__2` = 2})
{lambda__1 = 4, lambda__2 = 2}
So lambda__3 now found from
sol1[1][]; eval(%, sol); solve(%, `λ__3`); sol := `union`(sol, {`λ__3` = %[1]})
((2*lambda__1+1)*lambda__2+lambda__1)*lambda__3^2+((2*lambda__1+1)*lambda__2^2+(2*lambda__1^2-6*lambda__1+2)*lambda__2+lambda__1^2+2*lambda__1)*lambda__3+4*lambda__1*lambda__2^2+(4*lambda__1^2+8*lambda__1)*lambda__2
22*lambda__3^2+80*lambda__3+256
-20/11+((12/11)*I)*7^(1/2), -20/11-((12/11)*I)*7^(1/2)
{lambda__1 = 4, lambda__2 = 2, lambda__3 = -20/11+((12/11)*I)*7^(1/2)}
Check
simplify(eval(EqN, sol))
{0}
NULL
Download EqNsol.mw
@occipitalbaker Thanks.
I did something similar before here but for the case where the walk had to start and end at a particular vertex. (Perhaps that is the same problem?) There is a significant difference between just finding the length and finding the walk itself. Are you OK with just the length?
@Axel Vogt Vote up. The second one can be done without a change of variable.
restart;
i1:=(sin(x + sqrt(x)))/(1 + x);
i2:=(x*BesselJ(0, x^2))/(1 + x);
sin(x+x^(1/2))/(1+x)
x*BesselJ(0, x^2)/(1+x)
int(i2, x = 0 .. infinity);
int2:=evalf(%);
(1/4)*MeijerG([[1/4, 3/4, 1], []], [[1, 3/4, 1/2, 1/2, 1/4], []], 1/4)/Pi^3
.3233985132
Download int.mw
@rockyicer You're welcome. If you want to convert the .lib file to .mla, you can use the LibraryTools:-ConvertVersion command.
Yes, development is always going on - new integration methods were added in the latest (2023) version, see
https://www.maplesoft.com/support/help/Maple/view.aspx?path=updates%2fMaple2023%2fAdvancedMath
@mmcdara If you try Eigenvalues(SQRT),Eigenvalues(Q_L); you will see that two of the eigenvalues are negative of what they should be, i.e., the method has effectively chosen the wrong sqrt. I'm guessing that can be solved by a different starting guess, but not sure how exactly that could be done. Note that for the Eigenvalue method, you can work completely in reals (and more efficiently) by telling it C is symmetric:
LEC := proc(C)
local L, E;
L, E := LinearAlgebra:-Eigenvectors(Matrix(evalf(C),shape=symmetric)):
E . DiagonalMatrix(sqrt~(L)) . E^(-1)
end proc:
But if you use a numerical version of my routine above (with shape = symmetric) then you can avoid the numerical inverse (just a transpose now) and it might be more efficient (but the normalization will slow it down)
Msqrtf:=proc(A::Matrix(square,datatype=float))
local evs,evals,evecs,U,i,j,k,g,Lambda;
uses LinearAlgebra;
evs:=Eigenvectors(Matrix(A,shape=symmetric),output='list');
evals:=table();evecs:=table();
j:=0;
for i in evs do
if i[2] = 1 then
j:=j+1;
evals[j]:=i[1];
evecs[j]:=Normalize(i[3][],'Euclidean');
else
g:=GramSchmidt(i[3],conjugate=false,normalized=true);
for k to i[2] do
j:=j+1;
evals[j]:=i[1];
evecs[j]:=g[k];
end do;
end if;
end do;
Lambda:=DiagonalMatrix(map(sqrt,convert(evals,list)));
U:=convert(convert(evecs,list),Matrix);
U.Lambda.U^%T;
end proc:
@sand15 Thanks - I remembered that and had a quick look for it but didn't find it. The emphasis there was on the numerical case, but here on symbolic. For symbolic use, the method here requires that the exact eigenvalues can be found. Here even though the matrices are large, the eigenvalues and eigenvectors are easily found (for M1..M3, the eigenvalues are actually integers). The method here avoids finding a matrix inverse, or more accurately the inverse is just the hermitian transpose so is easy to compute.
My recollection is that Maple's method can suceed in more cases, at the expense of being slow. The solution is presumably for MatrixPower and/or MatrixFunction to do more preselection of different algorithms depending on the matrix type.
convert(expr,elementary) is intended to convert hypergeoms etc to elementary functions, but doesn't seem to work on these cases.
@lcz Isomorphic triangulations don't necessarily correspond to isomorphic binary trees, so I think a different approach is needed.
@OtherAlloyMonkey I left all the equations in terms of beta, and made all the constants exact. Then it is easy to see that the 5th and 6th rows of the matrix are identical, and therefore the determinant must be zero.
Model_Derivation_3.mw
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-3,248,574,030,282,502,700 | Econ1-Fall2010-PS4 - Econ 1 PS#4 Budget Constraints\/Indifference Curves Fall 2010 1 Every morning Professor Gordon buys a bagel and a cup of coffee from
Econ1-Fall2010-PS4 - Econ 1 PS#4 Budget...
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Econ 1 PS #4: Budget Constraints/Indifference Curves Fall 2010 1.Every morning Professor Gordon buys a bagel and a cup of coffee from Peabody’s. At Peabody’s the price of bagels is $2.00 and the price of a cup of coffee is $1.00. a.Assuming that Professor Gordon has $20 to spend on coffee and bagels, sketch her budget constraint, putting coffee on the x-axis and bagels on the y-axis. Clearly label a) the x-intercept, b) the y-intercept, c) and the slope of the budget constraint. b.Now, suppose that Peabody’s wants to encourage customer loyalty. To do so they start the following program: every time someone purchases 10 cups of coffee, they get the 11thcup for free. Sketch Professor Gordon’s new budget constraint under the assumption that she still has $20 to spend on coffee. As above, put coffee on the x-axis and bagels on the y-axis. Clearly label a) the x-intercept, b) the y-intercept, c) and the slope of the budget constraint. In addition, if there are any changes in the slope of the budget constraint, clearly identify the coordinates (i.e. the points) where
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1,196,958,292,784,113,700 | 6
$\begingroup$
Right now I’m teaching precalculus in high school and I want to propose a project to my students about polynomial functions. They already know enough about quadratic functions and we study variation models recently. We start working with polynomials this week and I really want to do a 2 weeks project for them with this topic but I’m falling short in ideas.
I already have a bacteria growth project prepared for them in exponentials functions but I want also to make something that meaningful to them in this topic too.
Any ideas?
Thank you in advance.
$\endgroup$
1
• $\begingroup$ Any suggestions? $\endgroup$
– Grouper
Nov 4, 2017 at 17:23
6 Answers 6
3
$\begingroup$
Given a circle, cut it with 1 straight line. You obtain 2 pieces.
Now cut it with 2 straight lines. If you pick the second line correctly, you obtain 4 pieces.
What is the largest number of pieces you can obtain with $n$ lines?
The answer is a cubic (OEIS A000125, MathWorld:Cake Number), although you might not manage to prove that to the students' satisfaction. It can be used to explore identifying probable polynomials using iterated finite differences, and extrapolation of polynomials using finite differences. I remember that aspect of it making for the most interesting project we did in secondary school maths.
$\endgroup$
3
$\begingroup$
You can identify polyhedra with polynomials. Let $a_0$ be the number of vertices, $a_1$ the number of edges and $a_2$ the number of faces, then the polynomial for a 3D polyhedron is $p_3(x)=-x^3+a_2x^2-a_1x+a_0$, for a 2D polyhedron (an n-gon) it is $p_2(x)=x^2-a_1x+a_0=x^2-nx+n$, a 1D polyhedron (a line) $p_1(x)=-x+a_0=-x+2$
If you want to get a prism with an n-gon base, multiply the base with the line segment: $$p_\text{prism}(x)=(x^2-nx+n)(-x+2)$$
If you want a cube, take the third power of a line: $$p_\text{cube}(x)=(-x+2)^3$$
You can also explore higher dimensional polyhedra (polytopes) and read off the number of $k$-dimentional cells from its polynomial. Say, how many edges does a fourdimensional cube have? No problem, just compute $(-x+2)^4$ and read off the first power coefficient. Is there some way to compute a pyramid with a given basis like it worked for a prism? Explore examples!
$\endgroup$
1
$\begingroup$
Perhaps curve fitting of experimental data with more and more terms in a polynomial?
first term: average, constant
second term: slope, line
third term: curvature, quadratic
fourth term: rate of change of curvature, cubic
(and maybe up to 4th order)
Can show that (depending on the data) that most of the explanation can come from very simple models. Also gives a little intuition about value of a constant (average) versus line (increasing/decreasing quantity), etc.
This might be more than what they need or are ready for, but it is an idea. Definitely don't do something to be cool that is too much for them. (I wonder if any projects are needed. But if you do some, make sure they work for the students.)
P.s. You don't need any fancy stats modeling programs, you can do it in excel.
http://www.statisticshowto.com/excel-multiple-regression/
$\endgroup$
2
• $\begingroup$ You are right, I don´t want to do something that is too hard on them but I like your idea. Thx. $\endgroup$
– Grouper
Nov 6, 2017 at 1:02
• $\begingroup$ So I am really interested in this idea and want to maybe do some data fitting exercises without making it too complex. In searching based on this thread I found this which may work. Creating a Polynomial function to fit a table $\endgroup$ Nov 7, 2017 at 0:33
1
$\begingroup$
I think doing something with polygonal numbers might be fun and instructive. You could ask them to write down the sequence of triangular numbers $1, 3, 6, 10, 15, \dots$ and have them see that the first difference increaseas by one and the second difference is constant. This suggests that the sequence is given by a quadratic. They have three points and so the could solve for the coefficients of the quadratic. The square numbers are easy. The pentagonal numbers can be handled in the same way as the triangular numbers. The might even by able to figure out the formula for the $k$-gonal numbers.
They could then maybe explore a famous theorem due to Fermat that says every positive integer $n$ can be written as the sum of $k$ $k$-gonal numbers (if we allow 0 to be a $k$-gonal number for all $k$).
Among other things, it will reinforce the following:
• figuring out patters
• solving systems of equations
• multiplication as area
• there is a unique degree $n$ polynomial through $n+1$ points (as long as the points have different $x$-coordinates
• math has these things called theorems that can be pretty cool
$\endgroup$
1
• 1
$\begingroup$ Building on this, what about characterizing an $n^{th}$ degree polynomial as one whose sequence of successive differences becomes constant after $n$ steps? Students do this a lot by intuition, but they usually must be led to the general observation. That is, the successive differences of a linear function are constant, the differences of the differences of a quadratic are constant, the diff's of the diff's of the diff's of a cubic are constant, etc. Since this is a precalculus course, this would be a nice precursor to the derivative concept. $\endgroup$
– Nick C
Nov 10, 2017 at 21:31
0
$\begingroup$
This may not be easy to make clear without computation, but the intersection of two ellipses leads to a 4th-degree polynomial.
enter image description here
(Image from Elliet Noma.)
See Dave Eberly's "Intersection of Ellipses" (PDF download) for the (formidable) details.
$\endgroup$
0
$\begingroup$
Since a precalculus is, by its nature, designed to prepare students for calculus, I recommend gently introducing students to the concepts of calculus using the tools they already have. In this case, you can use quadratic functions to introduce students to the concept of a tangent line and a limit, without really using that terminology.
See the first project ("How fast does the pumpkin fly?") in the link below.
https://www.dropbox.com/s/60qdz1npspwu7o0/Precalculus%20Projects%20-%20Brendan%20W%20Sullivan.pdf?dl=0
To celebrate autumn, I bought a a small trebuchet with which to launch pumpkins and investigate their flight paths. I am able to control the initial launch height and velocity of the pumpkin, but I need your help to determine where it will hit the ground and how fast it will be traveling at that moment.
I put my trebuchet atop a hill so that the pumpkin is launched from exactly 80 feet above ground level. The launch imparts a vertical velocity of 64 feet per second and a horizontal velocity of 40 feet per second. Because I know the strength of gravity on Earth, the height h of the pumpkin, t seconds after launch, is given by this function: $h(t) = -16t^2 + 64t + 80$.
1. Use factoring or the quadratic formula to determine when the pumpkin returns to ground level. For the purposes of the rest of the project, let's say your answer is $T$ seconds after the launch.
2. Use the given horizontal velocity to identify where the pumpkin will hit the ground. (You may assume there are no effects on the flight path like friction or wind, so the horizontal velocity is constant.)
3. Your next goal is to find the speed of the pumpkin as it hits the ground. Suppose $d > 0$ is a small positive value, and consider the height of the pumpkin at time $t = T - d$ and at time $t = T$. Explain why the ratio $\frac{h(T)-h(T-d)}{d}$ represents the average velocity of the pumpkin over the time interval $T-d\leq t\leq T$. (Possible hint: Think about the units of measurement of the top and bottom of that fraction.)
4. Use a calculator to estimate that ratio for the values $d = 1, d = 0.1, d = 0.01$, and $d = 0.001$. Make a prediction for what you think the vertical velocity of the pumpkin is as it hits the ground.
5. Now, instead of using specifi c numbers, use the given function $h(t)$ to simplify that expression, $\frac{h(T)-h(T-d)}{d}$, as much as possible so that it no longer involves division. Then, substitute $d = 0$ to fi nd the instantaneous vertical velocity of the pumpkin at that moment. Compare to your prediction.
6. In fact, the overall speed of the pumpkin as it hits the ground is $S = \sqrt{V^2+H^2}$, where $V$ is its vertical velocity and $H$ is its horizontal velocity. Find $S$.
7. Find the equation of the line that passes through $(T, 0)$ and has slope $V$. Graph this line together with the function $h(t)$ itself on the domain $0\leq t\leq T$ to verify that you found the tangent line to the graph of $h$ by fi nding that instantaneous velocity.
Extra Credit Follow a procedure similar to the one used above to find the instantaneous vertical velocity at the moment where $t = 2$. Graph $h$ and the tangent line together. Explain why your result makes sense physically.
(Note: $t=2$ is where the maximum height is reached so velocity is 0.)
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9,102,277,893,746,212,000 | IntMath Newsletter: Subitizing, Jacob, Fractals and Donald Duck
[12 Feb 2013]
12 Feb 2013
In this Newsletter:
1. Gong Xi Fa Cai
2. Counting game and Subitizing
3. Jacob’s Staff
4. Creating a Google-like classroom climate
5. Math puzzles
6. Friday math movie
7. Final thought – Perfect days
1. Gong Xi Fa Cai!
subitizing
Happy Year of the Snake to all those enjoying Chinese New Year festivities this week!
Chinese calendars are quite interesting. A “standard” year has around 354 days, and there are leap years containing a leap month, with around 384 days. The calendar is always adjusted so the winter solstice (the shortest day of the year) falls in the 11th month. Months always start on the new Moon (when it is completely in shadow).
Chinese New Year falls on the second new Moon after the December solstice (unless a leap month messes things up). The earliest possible date is Jan 21st, and the latest is Feb 21st (but that won’t happen until 2319).
Read more in (the extensive) The Mathematics of the Chinese Calendar
2. Counting game and subitizing
subitizing
Can you count faster than a chimp? Our "number sense" is important for understanding math, and it begins with our ability to count.
First, try the counting game. All you have to do is count the dots as quickly as you can.
Counting Challenge
Next up is a short article on why I created the game, and some early observations:
Counting Game and Subitizing
Then, for some interesting background, see this article:
Number Sense
3. Jacob’s Staff
Jacob's Staff
Scientists used Jacob’s Staff during the Renaissance to find heights and distances, using trigonometry. It was a forerunner to the sextant.
Math teachers could design some interesting activities using this, so students would learn some math, and some math history.
See: Jacob’s Staff
Here’s a similar activity based on historic math: Noon Day Project. This activity re-creates Eratosthenes’ famous experiment 2000 years ago where he produced a good approximation of the size of the Earth based on measuring shadow lengths at noon in different parts of Egypt.
4. Creating a Google-like classroom climate
Google is well-known for being one of the best companies in the world to work for. Around 7,000 people per day apply to work there. The company offers excellent benefits and a relaxed, almost playful atmosphere that keeps creative juices flowing.
They put a lot of effort into finding the right people, and because Google doesn’t want to lose anyone, they create conditions that encourage success and fulfillment.
The article, Inside Google’s Culture of Success and Employee Happiness, got me thinking about why so many students drop out of school (often because of acute math anxiety, or because they don’t have a feeling of belonging, or a host of other reasons).
Good classroom "climate" is very important for effective learning, especially in math!
Here are some of the things Google does that could apply in class:
Hire (and reward) good teachers! I enjoy living in Asia where education is generally held in high esteem, and teachers are, for the most part, regarded as valuable members of society. Good teachers almost always leads to better learning outcomes. Keep the good ones by giving them purpose, recognition, and reward.
Collect data on (and share) what works: I don’t mean what works for standardized tests, here. I mean quality class activities, different approaches to explaining things, or different resources. What works? What needs changing? Teachers aren’t given enough time to think about what they’re doing, nor enough opportunities to learn from each other. (Google collects huge amounts of data to improve its processes, and creates many situations where workers mingle and share ideas.)
Warm greetings make a huge difference: Google found their best managers were warm and accepting of input from their team, leading to a "15% increase in productivity". Sadly, I see a lot of classrooms where there are no greetings at the beginning of the day, and no friendly talk with students about mutual interests. It’s often no more than "OK, today we’re going to do the quadratic equation…". A sense of belonging is important for learning and for sticking around.
Too much bureaucracy: Google has a "flat" structure where middle managers have quite a bit of responsibility, compared to many other companies. That’s not always the situaion with teachers.
Too many distractions: Many classrooms have too many interruptions. Are those announcements about sport buses – or uniforms – really that crucial? There are already too many things trying to grab everyone’s attention. Focus is good.
In summary, we need to create the best working & learning environment we can, whether it’s in school, or in a company.
5. Math puzzles
The puzzle in the last IntMath Newsletter involved finding the time taken to fill a bath.
Correct answers with reasoning were given by Christopher Buchanan, Christian Mills, Nicos Mavrommatis, S. Nickerson, Thomas A Buckley, Bonnie, Hamid Zandi, Lachezar Borisov, Dineth, Ramesh Babu and Mawanda Ismail.
It was interesting to see the variety of thought processes used to solve the puzzle. Some of you went straight for the algebra, and that’s fine. My approach with this kind of puzzle is to assume a reasonable amount of water for the bath (say 100 liters) and then work the rest of it from there. It makes it easier (for me) to check if my "rate in" and "rate out" values make sense.
Please note: As usual in math, the best answers are those that include reasons!
New puzzle: Peter is an enthusiastic student and runs to school each day, but usually gets tired and walks the rest of the way. His running speed is twice his walking speed. He notices on Monday that he walks twice the time he runs and it took 20 minutes to get to school. On Tuesday, he runs twice the time he walks. How long does it take to get to school on Tuesday?
You can leave your responses, with reasoning, here.
6. Friday math movies
fractal
(a) Fractals – Hunting the Hidden Dimension
See this PBS documentary about the development of fractals and how they are used in computer graphics. (It’s 55 minutes, but good value.)
Friday math movie: Fractals – Hunting the Hidden Dimension
Donald in Mathmagic Land
(b) Donald in Mathmagic Land
This is a classic late-1950s cartoon where Donald learns where math comes from.
Friday math movie: Donald in Mathmagic Land
7. Final thought: A perfect day
I enjoy helping people, and I hope you do, too. I think this sums things up nicely:
You have not lived a perfect day, unless you’ve done something for someone who will never be able to repay you. [Ruth Smeltzer]
Until next time, enjoy whatever you learn.
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13 Comments on “IntMath Newsletter: Subitizing, Jacob, Fractals and Donald Duck”
1. ibrahim ghobrial says:
monday :d (runing)=2(t),d(walking)=1(2t)total distance is 4t assuming walkin speed one and running speed 2 tuesday:d(runing)=2(2x) , d (walking)=1(x) so total time on monday is 3t=20 total distance =4t=5x which gives x=16/3 total time on tuseday is 3x=16 answer is 16
2. liuting says:
In this Newsletter:4. Creating a Google-like classroom climate .–This part describes some phenomenon of the math class .I think these descriptions are very real.The classrooms lack of inspiration and exchange.Math teacher don’t think and find more effective method.
3. Peter Hunter says:
Let Twa=time spent walking on day a(Monday)
Let Tra=time spent running on day a(Monday)=Twa/2
Let x=walking speed (constant on all days). Running speed is defined as 2x.
Let D=the distance to school (a constant)
Computing the time spent walking on Monday is:
20 = Twa + Twa/2 yielding
Twa=40/3 and consequently Tra=20/3.
The distance (speed*time) is therefore:
D= x*(40/3) + 2x*(20/3)
Solving for walking speed, we get:
x=3*D/80.
Now, let Twb=time spent walking on day b(Tuesday) and let Trb=time spent running on Tuesday, or 2*Twb because he ran twice the time he walked the second day.
The distance equation for the second day is therefore
D = x*Twb + (2x)*2*Twb
But since x (walking speed) is constant both days, and was computed to be 3*D/80, we have:
D = (3*D/80)*Twb + 2*(3*D/80)*2*Twb
The D’s cancel and we are left with
Twb=80/15 or 5.333 minutes.
Since we are told he ran twice as much as he walked on Tuesday, Trb=10.667 minutes
Therefore, it took him 5.333 + 10.667 = 16 minutes to get to school on Tuesday.
4. Thomas A Buckley says:
Answer, it takes Peter 16 mins to get to school on Tuesday.
Let Distance to school = D, walking speed = s,
For Tuesday, walk time = tw, run time = tr
From distance = speed x time, for Monday
D = walk distance + run distance
D = s (2×20/3) + 2s (20/3) = s(80/3) . . (1)
For Tue, same distance
D = walk distance + run distance
D = tw ( s ) + tr ( 2s)
D = tw ( s ) + 2tw ( 2s) . . Run time is twice walk time.
D = 5tw (s) . . . (2)
As both (2) and (1) = D
5tw (s) = s (80/3) . . Cancel s
tw = 16/3 Mins. . . And Run time is twice walk time
tr = 2 x (16/3) Mins
Total Tuesday time Walk + Run times
= 16/3 + 2 x (16/3) Mins
= 16 mins = Answer
5. Guido says:
The solution is based on the equation S= VT {1}, where V is the velocity , T is the time and S is the distance. Further the distance on monday is equal to the distance on tuesday. On monday we can write : Trun + Twalk = 20 {2}and Twalk = 2Trun. Substituting in {2} : 3Trun = 20 —–> Trun = 20/3 and consequently Twalk = 40/2. Considering {1} for both monday and tuesday : Vwalk.Twalk(m)+Vrun.Trun(m)=Vwalk.Twalk(t)+Vrun.Trun(t)
Substituting Vrun = 2Vwalk and Twalk=40/2 into the equation:
40/2Vwalk(m)+2Vwalk.20/3=Vwalk.Twalk(t)+2Vwalk.Trun
80/3Vwalk = Vwalk{Twalk(t)+2Trun(t)} On tuesday:Trun=2Twalk
80/3=Twalk(t)+4Twalk
Twalk=16/3
Twalk + Trun = 16/3 +32/3 = 48/3 = 16 min
6. Christopher Buchanan says:
Constants: Run speed = 2x
Walk speed = x
Total distance = z
Run time = t
Monday: Walk time = 2t
t + 2t = 20
t = 20/3
z = 2x*t + x*2t
=(80/3)*x
Tuesday: Walk time = t/2
(80/3)*x = 2x*t + x*t/2
(80/3)*x = (5x/2)*t
t = 32/3
Total Time Tuesday = t + (t/2)
= 16 mins
7. Mawanda Ismail says:
let the walking speed be s and t be the time used on Tuesday
then the running speed is 2s
thus distance (d) = (20/3xs + 2×20/3x2s) ………(i)
or (d) = (t/3x2s + 2xt/3xs)………(ii)
equating (20/3xs + 2×20/3x2s)= (t/3x2s + 2xt/3xs)
20/3 +40/3 = 2t/3 +2t/3
60/3 = 4t/3
60 = 4t
t = 60/4
t = 15
It takes 15 minutes to get to school on Tuesday
8. Nicos Mavrommatis says:
Let Vr:running velocity, Vw:walking velocity, Tmw:walking time on Monday, Tmr:running time on Monday, Ttw:walking time on Tuesday, Ttr:running time on Tuesday, Smw & Smr:walking & running distance on Monday, Stw & Str:walking & running distance on Tuesday, So:distance Home-School, Tm:total time on Monday(=20min), Tt:total time on Tuesday.
Generally: (constant)Velocity (V)= Distance (S)/Time (T) or V=S/T
It is given that:Tm=20min, Tmr=2Tmw (1), Ttr=2Ttw (2),Vr=2Vw (3)
We’re looking for Tt=?
Tm=20′=Tmr+Tmw=Tmr+2Tmr=3Tmr due to (1)
Thus, Tmr=20/3 min and Tmw=40/3 min (4).
So=Smr+Smw=Vr*Tmr+Vw*Tmw=2Vw*(20/3)+Vw*(40/3) due to (3)&(4)
So=80Vw/3 (5)
On Tuesday: So=Str+Stw=Vr*Ttr+Vw*Ttw=2Vw*2Ttw+Vw*Ttw=5Vw*Ttw due to (2)&(3)
Thus, So=80Vw/3=5Vw*Ttw => Ttw=16/3 min. Due to (2):Ttr=32/3 min
The total time Tt on Tuesday is Tt=Ttr+Ttw=(32+16)/3=48/3=16 min.
It takes 16 minutes to get to school on Tuesday.
9. Sieu Do says:
Please correct. You should say Lunar New Year not Chinese New Year. Depending on the longitude of the capitals as well as subtle differences in the way they calculate. In the past, Chinese and Vietnamese may celebrate Lunar New Year on different date.
This is from Computing the Vietnamese lunar calendar
Comparison with Chinese calendar
1985 is one of the few years where Vietnamese and Chinese calendars differ significantly: the Vietnamese New Year was 1 month earlier than the Chinese one. The reason can be detected from the above table. The Winter Solstice 1984 falls on 21/12/1984 Hanoi time, but on 22/12/1984 Beijing time, the same day as the New Moon. The month 11 of the Chinese year must contain the Winter Solstice, so it is not the month from 23/11/1984 to 21/12/1984 like in the Vietnamese calendar, but the one starting 22/12/1984. Consequently, the subsequent months (12, 1,…) also start about one month later than the corresponding months of the Vietnamese calendar. While New Year in Vietnam falls on 21/01/1985, it is on 20/02/1985 in China. The two calendars agree again after a leap month is inserted to the Vietnamese calendar (month from 21/03/1985 to 19/04/1985, as seen above). Also, in year 1984 the Chinese lunar month from 23/11/1984 to 21/12/1984 is the first lunar month after Winter Solstice 1983 that does not contain a Major Term and is therefore a leap month.
In the 21th century there are 3 years where the Lunar New Year begins at different dates in Vietnam and in China. In 2007 the Vietnamese New Year is on 17/02/2007, the Chinese one on 18/02/2007. In 2030 the dates are 02/02/2030 and 03/02/2030, and in 2053 they are 18/02/2053 and 19/02/2053.
10. Murray says:
@Sieu Do: Thanks for the clarification.
11. hamid zandi says:
r=runs w=walk x=path v=velocity X=total path
t=time
we know vr=2*vw
in monday:
tw=2*tr,tw+tr=20min => tr=20/3; tw=40/3
X=xr+xw => X=xr+xw => X=vr*tr+vw*tw => X=vr*20/3+2*vr*40/3
=> X= vr*100/3
in tuesday:
tr=2*tw => X=vr*tr+vw*tw => X=tr*vr*5/4
finnallyn:tr*vr*5/4=vr*100/3=> tr=400/15,tw=200/15=>time=40min
12. Kind says:
I think it took 10 minutes to get to school on tuesday because from the first statement running speed is twice walking speed impling that walking time=twice running time (since speed is inversely proportional to time), on Monday running time=2 times walking time. From the two statements, it implies that whenever his walking time is twice his running time, it takes 20 mins to get to school, therefore, it will take him 10 mins to get to school on tuesday since his running time was twice his walking time and as such less time will be taken to get to school.
13. Miguel Angel Velo says:
Peter is an enthusiastic student and runs to school each day, but usually gets tired and walks the rest of the way. His running speed is twice his walking speed. He notices on Monday that he walks twice the time he runs and it took 20 minutes to get to school. On Tuesday, he runs twice the time he walks. How long does it take to get to school on Tuesday? If v1=walking speed, v2=running speed=2*v1
On Monday: t1=walking time, t2=running time
On Tuesday: t3=walking time 2*t3=running time
Thus: 2*t1*v1 + t2*2v1=e (distance of school)
2v1(t1+t2)=e : t1+t2=20 : 40*v1=e
On Tuesday: t3*v1+2*t3*2v1=e : v1(t3+4t3)=40*v1
For transitive character: 40= t3+4*t3 : 5*t3=40
On tuesday it take 5 minutes.
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4,270,401,671,312,914,000 | 1524318000000
1,524,318,000,000 (one trillion five hundred twenty-four billion three hundred eighteen million) is an even thirteen-digits composite number following 1524317999999 and preceding 1524318000001. In scientific notation, it is written as 1.524318 × 1012. The sum of its digits is 24. It has a total of 15 prime factors and 224 positive divisors. There are 406,483,200,000 positive integers (up to 1524318000000) that are relatively prime to 1524318000000.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 24
• Digital Root 6
Name
Short name 1 trillion 524 billion 318 million
Full name one trillion five hundred twenty-four billion three hundred eighteen million
Notation
Scientific notation 1.524318 × 1012
Engineering notation 1.524318 × 1012
Prime Factorization of 1524318000000
Prime Factorization 27 × 3 × 56 × 254053
Composite number
ω(n) Distinct Factors 4
Total number of distinct prime factors
Ω(n) Total Factors 15
Total number of prime factors
rad(n) Radical 7621590
Product of the distinct prime numbers
λ(n) Liouville Lambda -1
Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n)
μ(n) Mobius Mu 0
Returns:
• 1, if n has an even number of prime factors (and is square free)
• −1, if n has an odd number of prime factors (and is square free)
• 0, if n has a squared prime factor
Λ(n) Mangoldt function 0
Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,524,318,000,000 is 27 × 3 × 56 × 254053. Since it has a total of 15 prime factors, 1,524,318,000,000 is a composite number.
Divisors of 1524318000000
224 divisors
Even divisors 196
Odd divisors 28
4k+1 divisors 14
4k+3 divisors 14
τ(n) Total Divisors 224
Total number of the positive divisors of n
σ(n) Sum of Divisors 5061167247480
Sum of all the positive divisors of n
s(n) Aliquot Sum 3536849247480
Sum of the proper positive divisors of n
A(n) Arithmetic Mean 22594496640.535
Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n))
G(n) Geometric Mean 1234632.7389147
Returns the nth root of the product of n divisors
H(n) Harmonic Mean 67.464127404604
Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,524,318,000,000 can be divided by 224 positive divisors (out of which 196 are even, and 28 are odd). The sum of these divisors (counting 1,524,318,000,000) is 5,061,167,247,480, the average is 225,944,966,40.,535.
Other Arithmetic Functions (n = 1524318000000)
1 φ(n) n
φ(n) Euler Totient 406483200000
Total number of positive integers not greater than n that are coprime to n
λ(n) Carmichael Lambda 12702600000
Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n
π(n) Prime Pi ≈ 56406532674
Total number of primes less than or equal to n
r2(n) Sum of 2 squares 0
The number of ways n can be represented as the sum of 2 squares
There are 406,483,200,000 positive integers (less than 1,524,318,000,000) that are coprime with 1,524,318,000,000. And there are approximately 56,406,532,674 prime numbers less than or equal to 1,524,318,000,000.
Divisibility of 1524318000000
m 2 3 4 5 6 7 8 9
n mod m 0 0 0 0 0 5 0 6
The number 1,524,318,000,000 is divisible by 2, 3, 4, 5, 6 and 8.
Classification of 1524318000000
By Arithmetic functions
• Abundant
Expressible via specific sums
• Polite
• Practical
Other numbers
• Frugal
Base conversion (1524318000000)
Base System Value
2 Binary 10110001011101000011011101101001110000000
3 Ternary 12101201112100020112102020
4 Quaternary 112023220123231032000
5 Quinary 144433300402000000
6 Senary 3124132332521440
8 Octal 26135033551600
10 Decimal 1524318000000
12 Duodecimal 20750b020280
16 Hexadecimal 162e86ed380
20 Vigesimal 2jah97a000
36 Base36 jg9fkwao
Basic calculations (n = 1524318000000)
Multiplication
n×i
n×2 3048636000000
n×3 4572954000000
n×4 6097272000000
n×5 7621590000000
Division
ni
n2 762159000000.000
n3 508106000000.000
n4 381079500000.000
n5 304863600000.000
Exponentiation
ni
n2 2323545365124000000000000
n3 3541822023875085432000000000000000000
n4 5398863063789222475535376000000000000000000000000
n5 8229584147669060025463133273568000000000000000000000000000000
Nth Root
i√n
2√n 1234632.7389147
3√n 11508.671468538
4√n 1111.1402876841
5√n 273.28457255061
1524318000000 as geometric shapes
Circle
Radius = n
Diameter 3048636000000
Circumference 9577572461069.4
Area 7.2996330493562E+24
Sphere
Radius = n
Volume 1.4835949400705E+37
Surface area 2.9198532197425E+25
Circumference 9577572461069.4
Square
Length = n
Perimeter 6097272000000
Area 2.323545365124E+24
Diagonal 2155711188969.4
Cube
Length = n
Surface area 1.3941272190744E+25
Volume 3.5418220238751E+36
Space diagonal 2640196222891.8
Equilateral Triangle
Length = n
Perimeter 4572954000000
Area 1.0061246565215E+24
Altitude 1320098111445.9
Triangular Pyramid
Length = n
Surface area 4.0244986260859E+24
Volume 4.1740772847299E+35
Height 1244600435246.6
Cryptographic Hash Functions
md5 8f5d0e467573244107facb344956b020
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Using the Candidates Test to Determine Absolute Extrema (Topic 5.5)
Unit 5 - Day 4
Learning Objectives
• Justify conclusions about the behavior of function based on its derivative.
Success Criteria
• I can consider both relative extrema and endpoints to determine absolute extrema.
• I can justify an absolute maximum or minimum value.
Quick Lesson Plan
Activity: Are You a Stock Market Master?
pdf.png
docx.png
Lesson Handout
Answer Key
pdf.png
Overview
Students will formalize the circumstances for and the process required when investigating function values at endpoints. Today’s lesson is straightforward and intuitive for most students. After identifying critical values along the x-axis and finding the associated relative extreme values, students investigate the idea of analyzing function behaviors at endpoints to determine absolute (global) extreme values.
Teaching Tips
Study the scoring guidelines for free response question 2018 AB 5c to better teach students the best way to structure their solutions and justifications. Clarify for students that the Candidates Test is used for locating absolute (global) extreme values on a closed interval. (The first derivative test is reserved for relative extrema.)
Exam Insights
An excellent example of the Candidates Test is found in 2018 AB 5c. It is interesting to note that students are able, even at this early point in their studies, to complete the entire AP question!
Student Misconceptions
The need for correct and precise language is evident in these last lessons of Unit 5. Students who write “x = 3 is a local min because the slope changes there,” or similar vague statements, miss the opportunity to earn full points on our assignments and assessments. A careful statement is required. “The function f(x) has a local minimum of 6 at x = 3 where f’(x) changes sign from negative to positive” accurately describes the function value in addition to the x-value where that value occurs. Terms such as “the slope” or “the function” are no longer allowed: students must specifically name all functions and their derivatives.
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-7,411,212,780,514,488,000 | Michael Shulman
regular 2-category
Definition
A 2-category KK is called regular if 1. It is finitely complete, 1. esos are stable under pullback, and 1. Every 2-congruence which is a kernel can be completed to an exact 2-fork.
In particular, the last condition implies that every 2-congruence which is a kernel has a quotient.
Examples
• Cat is regular.
• A 1-category is regular as a 2-category iff it is regular as a 1-category, since the esos in a 1-category are precisely the strong epics.
• Every finitely complete (0,1)-category (that is, every meet-semilattice) is regular.
Factorizations
In StreetCBS the last condition is replaced by
• Every morphism ff factors as fmef\cong m e where mm is ff and ee is eso.
We now show that this follows from our definition. First we need:
Lemma
(Street’s Lemma) Let KK be a finitely complete 2-category where esos are stable under pullback, let e:ABe:A\to B be eso, and let n:BCn:B\to C be a map. 1. If the induced morphism ker(e)ker(ne)ker(e) \to ker(n e) is ff, then nn is faithful. 1. If ker(e)ker(ne)ker(e) \to ker(n e) is an equivalence, then nn is ff.
Proof
First note that ker(e)ker(ne)ker(e)\to ker(n e) being ff means that if a 1,a 2:YAa_1,a_2: Y \rightrightarrows A and δ 1,δ 2:ea 1ea 2\delta_1,\delta_2 : e a_1 \;\rightrightarrows\; e a_2 are such that nδ 1=nδ 2n \delta_1 = n \delta_2, then δ 1=δ 2\delta_1=\delta_2. Likewise, ker(e)ker(ne)ker(e)\to ker(n e) being an equivalence means that given any α:nea 1nea 2\alpha: n e a_1 \to n e a_2, there exists a unique δ:ea 1ea 2\delta: e a_1 \to e a_2 such that nδ=αn \delta = \alpha.
We first show that nn is faithful under the first hypothesis. Suppose we have b 1,b 2:XBb_1,b_2:X \rightrightarrows B and β 1,β 2:b 1b 2\beta_1,\beta_2:b_1\to b_2 with nβ 1=nβ 2n \beta_1 = n \beta_2. Take the pullback
Y r X (a 1,a 2) (b 1,b 2) A×A e×e B×B\array{&Y & \overset{r}{\to} & X \\ (a_1,a_2) & \downarrow && \downarrow & (b_1,b_2)\\ & A\times A & \overset{e\times e}{\to} & B\times B}
Then we have two 2-cells
β 1r,β 2r:b 1rb 2r\beta_1 r, \beta_2 r: b_1 r \;\rightrightarrows\; b_2 r
such that the composites
nea 1nb 1rnβ 1r=nβ 2rnb 2rnea 2n e a_1 \cong n b_1 r \overset{n \beta_1 r = n \beta_2 r}{\to} n b_2 r \cong n e a_2
are equal. By the hypothesis, nβ 1r=nβ 2rn \beta_1 r = n \beta_2 r implies β 1r=β 2r\beta_1 r = \beta_2 r. But rr is eso, since it is a pullback of the eso e×ee\times e, so this implies β 1=β 2\beta_1=\beta_2. Thus, nn is faithful.
Now suppose the (stronger) second hypothesis, and form the pair of pullbacks:
(ne/ne) g n/n C 2 A×A e×e B×B n×n C×C\array{(n e / n e) & \overset{g}{\to} & n / n & \to & C^{\mathbf{2}}\\ \downarrow && \downarrow && \downarrow \\ A\times A & \overset{e\times e}{\to} & B\times B & \overset{n\times n}{\to} & C\times C}
Then gg, being a pullback of e×ee\times e, is eso. We also have a commutative square
(e/e) (ne/ne) g B 2 (n/n).\array{(e/e) & \to & (n e / n e)\\ \downarrow && \downarrow g \\ B^{\mathbf{2}} & \to & (n/n).}
By assumption, (e/e)(ne/ne)(e/e) \to (n e / n e) is an equivalence. Since we have shown that nn is faithful, the bottom map B 2(n/n)B^{\mathbf{2}} \to (n/n) is ff, so since the eso gg factors through it, it must be an equivalence as well. But this says precisely that nn is ff.
Theorem
A 2-category is regular if and only if 1. it has finite limits, 1. esos are stable under pullback, 1. every morphism ff factors as fmef\cong m e where mm is ff and ee is eso, and 1. every eso is the quotient of its kernel.
Proof
First suppose KK is regular; we must show the last two conditions. Let f:ABf:A\to B be any morphism. By assumption, the kernel ker(f)ker(f) can be completed to an exact 2-fork ker(f)AeCker(f) \rightrightarrows A \overset{e}{\to} C. Since ee is the quotient of the 2-congruence ker(f)ker(f), it is eso, and since ff comes with an action by ker(f)ker(f), we have an induced map m:CBm:C\to B with fmef\cong m e. But since the 2-fork is exact, we also have ker(f)ker(e)ker(f)\simeq ker(e), so by Street’s Lemma, mm is ff.
Now suppose that in the previous paragraph ff were already eso. Then since it factors through the ff mm, mm must be an equivalence; thus ff is equivalent to ee and hence is a quotient of its kernel.
Now suppose KK satisfies the conditions in the lemma. Let f:ABf:A\to B be any morphism; we must show that ker(f)ker(f) can be completed to an exact 2-fork. Factor f=mef = m e where mm is ff and ee is eso. Since mm is ff, we have ker(f)ker(e)ker(f)\simeq ker(e). But every eso is the quotient of its kernel, so the fork ker(f)AeCker(f) \rightrightarrows A \overset{e} \to C is exact.
In StreetCBS it is claimed that the final condition in Theorem 1 follows from the other three, but there is a flaw in the proof.
Subobjects
In a regular 2-category KK, we call a ff m:AXm:A\to X with codomain XX a subobject of XX. We write Sub(X)Sub(X) for the preorder of subobjects of XX, as a full sub-2-category of the slice 2-category K/XK/X. Since KK is finitely complete and pullbacks preserve ffs, we have pullback functors f *:Sub(Y)Sub(X)f^*:Sub(Y)\to Sub(X) for any f:XYf:X\to Y.
If gmeg \cong m e where mm is ff and ee is eso, we call mm the image of gg. Taking images defines a left adjoint f:Sub(X)Sub(Y)\exists_f:Sub(X)\to Sub(Y) to f *f^* in any regular 2-category, and the Beck-Chevalley condition is satisfied for any pullback square, because esos are stable under pullback.
Preservation
It is easy to check that if KK is regular, so are:
The slice 2-category K/XK/X does not, in general, inherit regularity, but we have:
Theorem
If KK is regular, so are the fibrational slices Opf(X)Opf(X) and Fib(X)Fib(X).
Revised on December 20, 2009 06:52:38 by Mike Shulman (173.8.161.189) | {
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1,634,361,632,397,595,600 | Top 10 3rd Grade IAR Math Practice Questions
Top 10 3rd Grade IAR Math Practice Questions
If your 3rd-grade student has reviewed the 3rd Grade IAR Math materials and has a relative mastery of all math topics, we suggest solving the Top 10 3rd Grade IAR Math Practice Questions. In fact, using the Top 10 3rd Grade IAR Math Practice Questions, the most important mathematical concepts of the 3rd Grade IAR exam are reviewed and the strengths and weaknesses of your 3rd-grade student are identified. Top 10 3rd Grade IAR Math Practice Questions have been updated in accordance with the 2022 3rd Grade IAR Math exam guidelines. The full answer and explanation are also provided at the end of the post.
Now it’s time to use the Top 10 3rd Grade IAR Math Practice Questions to assess your 3rd-grade student’s readiness for the 3rd Grade IAR Math exam!
The Absolute Best Book to Ace the 3rd Grade IAR Math Test
3rd Grade IAR Math Practice Questions
1- There are \(6\) numbers in the box below. Which of the following list shows only odd numbers from the numbers in the box?
A. \(15, 29, 42\)
B. \(15, 29, 83\)
C. \(15, 30, 42\)
D. \(42, 18, 30\)
2- Mia’s goal is to save \($140\) to purchase her favorite bike.
In January, she saved \($36\).
In February, she saved \($28\).
How much money does Mia need to save in March to be able to purchase her favorite bike?
A. \($28\)
B. \($30\)
C. \($52\)
D. \($76\)
3- Michelle has \(72\) old books. She plans to send all of them to the library in their area. If she puts the books in boxes that can hold \(9\) books, which of the following equations can be used to find the number of boxes she will use?
A. \(72 + 9 =\) ______
B. \(72 × 9 =\) ______
C. \(72 – 9 =\) ______
D. \(72 ÷ 9 =\) ______
4- Eva had \(983\) cards. Then, she gave \(349\) of the cards to her friend Alice. After that, Eva lost \(260\) cards.
Which equation can be used to find the number of cards Eve has now?
A. \(983 – 349 + 260 =\) _____
B. \(983 – 349 –260 =\) _____
C. \(983 +349 +260 =\) _____
D. \(983 +349 –260 =\) _____
5- The length of the following rectangle is \(7\) centimeters and its width is \(3\) centimeters. What is the area of the rectangle?
A. \(10\space cm^2\)
B. \(20\space cm^2\)
C. \(14\space cm^2\)
D. \(21\space cm^2\)
6- Which number is made up of \(4\) hundred, \(8\) tens, and \(4\) ones?
A. \(4,084\)
B. \(484\)
C. \(448\)
D. \(844\)
7- Look at the spinner above. On which color is the spinner most likely to land?
A. RED
B. GREEN
C. YELLOW
D. NONE
8- A group of third-grade students recorded the following distances that they jumped.
A. \(23\)
B. \(24\)
C. \(32\)
D. \(34\)
9- Emma flew \(2,448\) miles from Los Angeles to New York City. What is the number of miles Emma flew rounded to the nearest thousand?
A. \(2,000\)
B. \(2,400\)
C. \(2,500\)
D. \(3,000\)
10- A number sentence such as \(31 + Z = 98\)can be called an equation. If this equation is true, then which of the following equations is not true?
A. \(98 – 31 = Z\)
B. \(98 – Z= 31\)
C. \(Z– 31 = 98\)
D. \(Z+31 = 98\)
Best 3rd Grade IAR Math Prep Resource for 2022
$15.99
Answers:
1- B
An easy way to tell whether a large number is odd or even is to look at its final digit. If the number ends with an odd digit \((1, 3, 5, 7, or\space 9)\), then it’s odd. On the other hand, if the number ends with an even digit or \(0 (0, 2, 4, 6\space or\space 8)\), then it is even.
\(15, 29,\) and \(83\) ends with odd digits, so they are odd numbers.
2- D
She saves \($36\) and \($28\) so she has \($64\).
\($140 – $64 = $76\) she needs to save.
3- D
Michelle divides \(72\) books into \(9\) boxes therefore \(72 ÷ 9\) formula is correct.
4- B
Eva gave \(349\) of her \(983\) cards to her friend so her cards are \(983-349\), then she lost \(260\) cards so now she has \(983-349-260=374\) cards.
5- D
Use are formula of rectangle:
\(Area= length × width\)
Area \(=3cm×7cm =21cm^2\)
6- B
Add \(4\) hundred and \(8\) tens and \(4\) ones
So we have :\(400 + 80 + 4 = 484\)
7- A
The chances of landing on red are \(3\) in \(6\)
The chances of landing on yellow are \(1\) in \(6\)
The chances of landing on the green are \(2\) in \(6\)
The chances of landing on red are more than in other colors.
8- C
\(32\) inches distance most occur in this table.
9- A
10- C
All these equations are true:
\(98 – 31 = Z\)
\(98 – Z = 31\)
\(Z + 31 = 98\)
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167,478,258,530,369,800 | Exercices sur les suites numériques, Exercices de Mathématiques. Université Claude Bernard (Lyon I)
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Exercices sur les suites numériques, Exercices de Mathématiques. Université Claude Bernard (Lyon I)
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Exercices de mathématique sur les suites numériques. Les principaux thèmes abordés sont les suivants: exercices, les modèles de population.
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polycomplet.dvi
3. Suites numériques
Ex. 1 Montrer que la suite définie par u0 = 1 et un+1 = √ 3 + 2un est
convergente.
Ex. 2 Que signifie pour la suite (un) : ∀A ∈ R, ∃N ∈ N, ∀n ∈ N (n ≥ N ⇒ un < A) ?
Ex. 3 On suppose un → l. Est-il vrai que l ≥ 0 lorsque i) un ≥ 0 pour 0 ≤ n ≤ 1000 ? ii) un2 ≥ 0 pour tout n ∈ N ?
Ex. 4 Etudier les limites des suites définies par
un = 32n
(6n)2 , vn = sin(n !)− n2, wn =
n3 − n2 + 2 2n3 + 1− n−2 ·
Ex. 5 On considère les suites (un)n∈N et (vn)n∈N, définies par un = 3n − n! 2n − n3
pour tout n ≥ 0, v0 = 0 et vn+1 = (vn+12 ) 1 3 . Etudier la convergence de cha-
cune de ces suites en précisant la limite lorsqu’elle existe.
Ex. 6 On pose pour tout n ≥ 1
un =
n ∑
k=0
1
k! et vn = un +
1
n · n! ·
Montrer que les suites (un)n≥1 et (vn)n≥1 sont adjacentes. Que peut-on dire de la suite (un) ?
Ex. 7 Soient an = n
n+1 et bn =
n+1 n+2
. Les suites (an) et (bn) sont-elles
adjacentes ?
Ex. 8 Soit (un)n≥0 la suite définie par u0 = 2, u1 = 1 et la relation de récurrence un = −32un−1 + un−2 pour n ≥ 2. Donner l’expression de un pour tout n ≥ 0, et la limite de un lorsque n → +∞.
5
Ex. 9 ∗ Modèles de population. On s’intéresse à des modèles discrets s’appliquant à des populations présentant des phénomènes de synchronisation (par exemple, les insectes se reproduisent à une même période de l’année). On choisit une unité de temps et on note Pn le nombre d’individus au bout de n unités de temps, P0 (> 0) désignant la population initiale. La quantité Tn = Pn+1 − Pn
Pn est le taux d’accroissement de la population entre les instants n
et n+ 1. 1. On suppose que Tn = T pour tout n (loi de Malthus), où T ∈] − 1,+∞[ est une constante. Ecrire l’équation liant Pn+1 et Pn et reconnâıtre le type de la suite (Pn). Etudier sa limite. 2. On suppose que pour tout n
Tn = k(1− Pn P ∗
) (loi de Verhulst),
où k > 0 et P ∗ > 0 sont des constantes. a) On pose
xn = k
k + 1
Pn P ∗
·
Montrer que xn+1 = (k + 1)xn(1− xn), ∀n ≥ 0.
A partir de maintenant on suppose que k = 1/2 et que 0 < P0 < 3P ∗.
b) Montrer que xn → 13 . (On distinguera les cas suivants (i) 0 < x0 < 1/3, (ii) x0 = 1/3, (iii) 1/3 < x0 < 2/3 et (iv) 2/3 ≤ x0 < 1 et on établira dans les cas (i) et (iii) la monotonie de la suite (xn) à partir d’un certain rang.) c) Conclure.
Ex. 10 ∗ Développement décimal illimité Le but de cet exercice est de définir le développement décimal illimité de tout nombre réel l ∈ [0, 10]. 1. Soit (un)n≥0 une suite réelle vérifiant un ∈ {0, 1, 2, ..., 9} pour tout n ≥ 0. On définit deux suites (sn)n≥0 et (s
′ n)n≥0 par
sn = n ∑
k=0
uk · 10−k = u0, u1u2 · · ·un, s′n = sn + 10−n ∀n ≥ 0.
Montrer que les suites (sn) et (s ′ n) sont adjacentes. En déduire qu’elles con-
vergent vers une nombre l ∈ [0, 10]. On écrira l = u0, u1u2 · · · et on dira
6
que u0, u1u2 · · · est un développement décimal illimité du réel l. Que vaut l lorsque un = 9 pour tout n ? 2. Inversement, on se donne un réel quelconque l ∈ [0, 10[ et on cherche à construire un développement décimal illimité de l. On pose u0 = [l] (partie entière de l). Si u0, ..., un sont construits, on pose un+1 = [10
n+1(l− sn)], où sn = u0, u1 · · ·un. Montrer par récurrence sur n que
{
0 ≤ un ≤ 9 sn ≤ l < sn + 10−n
∀n ≥ 0.
En déduire que l = u0, u1u2 · · · 3. Cette question vise à montrer qu’un nombre réel l ∈]0, 10] admet un unique dévelop- pement décimal illimité si et seulement si l n’est pas un nombre décimal. Soit l ∈]0, 10] un réel possédant deux développements décimaux illimités distincts :
l = u0, u1u2 · · · = v0, v1v2 · · · Soit N le premier entier pour lequel un 6= vn. On peut supposer par exemple que uN < vN . a) Montrer que vN = uN + 1. b) Montrer que un = 9 et vn = 0 pour tout n ≥ N + 1. En déduire que l est un nombre décimal. c) Inversement, montrer que tout nombre décimal l ∈]0, 10] admet deux développements décimaux illimités. 4. On dit qu’un développement décimal illimité est périodique (à partir d’un certain rang) s’il existe deux entiers N (le rang) et T (la période) tels que un+T = un pour tout n ≥ N . Le but de cette question est de montrer que les rationnels sont les seuls réels à posséder un développement décimal illimité périodique. a) Déterminer le développement décimal illimité de l = 131/7. (Indication : faire la division de l’école primaire). b) Supposons que le nombre l ∈ [0, 10] soit rationnel : l = p/q, avec p, q ∈ N. Montrer que le développement décimal illimité de l s’obtient en divisant indéfiniment p par q, et que ce développement décimal illimité est périodique. c) Réciproquement, soit l ∈ [0, 10] un réel admettant un développement décimal illimité périodique. Montrer que l est rationnel.
7
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Differential Equation Model
1. Apr 3, 2012 #1
1. The problem statement, all variables and given/known data
Here's a model for the balance owed on a loan with the following conditions:
* Interest accumulated on the loan at a rate of 5.24% per year
* The amount owed at the beginning of the loan was $20,000.
* No payments were made on the loan for the first two years
* After two years, the loan was paid off at $3,000 per year.
The model is given by the equations:
[itex]\frac{dL}{dt} = 0.0524 \ L, \ \ \ 0<t<2[/itex]
[itex]\frac{dL}{dt} = 0.0524 \ L \ - \ 3 , \ \ t>2[/itex]
(L is in thousands)
Using a direction field work out how large repayments should be if the loan is to be paid back in exactly 12 years (i.e., with exactly 10 years of repayments after the first two years with no repayments).
3. The attempt at a solution
Here is the direction field I made for this model
http://img210.imageshack.us/img210/1231/dfield.jpg [Broken]
And the solution for the initial value L(0)=20 is
http://img703.imageshack.us/img703/1116/dfieldi.jpg [Broken]
The root of this solution occurs at t=11.39, so 11 years is how long it will take to pay back the loan.
But how can we work out how large repayments should be if the loan is to be paid back in 12 years? :confused:
Any help is greatly appreciated.
Last edited by a moderator: May 5, 2017
2. jcsd
3. Apr 3, 2012 #2
Ray Vickson
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Change the DE to dL/dt = 0.0524L - R for t > 2, and solve it with R as a symbolic parameter. Then find R that makes L(12) = 0.
RGV
4. Apr 3, 2012 #3
So, [itex]L(12)= 0.0524 \times (12) - R = 0.6288 - R = 0 \implies R = 0.6288[/itex]
Is this what you meant?
5. Apr 3, 2012 #4
Ray Vickson
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No, it is not what I said and not what I meant. Go back and read what I wrote. You need to solve the DE for t > 2.
RGV
6. Apr 3, 2012 #5
I didn't quite understand what you meant by treating R as a "symbolic parameter". I already found the value of R that gives 0. So we can write the equation as
dL/dt = 0.0524 L - 0.6288
I'm not sure how this helps us. Did you mean I have to first solve the DE using separation of variables while ignoring R? I appreciate it if you could maybe explain that a bit more clearly.
7. Apr 3, 2012 #6
epenguin
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It looks like you are using a phase plane plotter as well as the direction field. You can use it or the direction field to plot backwards as well as forwards in time! (And it looks like that is what your plotter is doing since your starting point is in the middle!)
8. Apr 3, 2012 #7
Yes I've used the Matlab add-on called "dfield", my plot shows the initial condition y(0)=20. But how do can we use this plot to figure out how large repayments should be if the loan is to be paid back in exactly 12 years?
9. Apr 3, 2012 #8
Ray Vickson
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I don't have access to Matlab, so I cannot offer technical advice. But if I were doing the question I would avoid the use of a phase plot and would, instead, just write down the formula for the DE solution; it would be a formula that has both R and t in it: L = f(t,R). Then I would set L=0 at t=12 and solve the equation to find R.
I guess if you want to use phase plot methods you could start by guessing a value of R, then make the phase plot for that R value and trace out the solution. If L=0 occurs before t=12 our guessed value if R is too large, so we should decrease it a bit and start over. If L=0 occurs after t=12, our guessed R is too small, so we should increase it a bit and start over. This procedure would be horrible and would take forever, and that is why I would not use it unless I had hours of spare time and nothing better to do.
RGV
Last edited: Apr 3, 2012
10. Apr 3, 2012 #9
epenguin
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Your plot is actually predicting backwards from that t(0) - predicting your what your debt would have been over the previous 30+ years if your first eq. had been applicable over that time!
Now you know where your starting t is for the 'retrodiction'. t = 12. It should be obvious that at your present R = 3 the slopefield will not take you back from (12, 0) to (0, 20) i.e. L = 20 but to something higher. If not obvious make a few trials and you'll soon see. So as you cannot get back to 20 at the right time with that R try different R till it comes right!
You can as RV said calculate it mathematically, which is not very hard depending on what math you know. But using the plotter is not even math, it's common sense.
11. Apr 3, 2012 #10
I agree, that would take a lot of time. So could you explain to me how to find the right value mathematically without using the phase plots? (I didn't quite get your first post)
12. Apr 3, 2012 #11
epenguin
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How did you get the continuous curve that is in your pic?
13. Apr 3, 2012 #12
I just went to the dfield set-up window and typed in the initial condition x=20 when t=0, and it gave me the curve.
14. Apr 3, 2012 #13
epenguin
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It sounds like you have got a programme or calculator that (like most such) doesn't do only d-fields but plots phase planes or paths as well (fancy name for 'solves 1st order d.e.'s in 2 variables').
You have all you need.
(I have to go now.)
Last edited: Apr 4, 2012
15. Apr 3, 2012 #14
Ray Vickson
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Since you titled your thread "differential equation model" I assumed you knew something about differential equations. Is that assumption wrong?
I really need more information from you in order to _guide_ you while avoiding doing your homework for you. What is the course? What is your background? How much calculus have you had? etc.
RGV
16. Apr 3, 2012 #15
I have done basic calculus and some linear algebra. But this is my first course in differential equations. So far I've learned how to solve DEs using the method of separation of variables and integrating factor.
17. Apr 3, 2012 #16
Ray Vickson
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Well, that is plenty enough to solve the simple DE dL/dt = c*L - R with constants c and R. Just apply what you have been taught.
It may be that you are comfortable with solving the DE dL/dt = c*L but uncomfortable with the DE that has the extra '-R' on the right. Is that the case? If so, look instead at the DE for M = L - p, where p is some constant. You ought to be able to figure out what p value to choose in order to get rid of the constants on the right and have just dM/dt = c*M.
Alternatively, you can use an integrating factor.
RGV
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-4,200,850,689,112,684,500 | find derivatives of the following function using differentiation rules HELP! f(x)= x^4 + 3x^2+4x-31 f(x)=2(x+3)e^x f(x)= (2x-1)e^x-3
Asked on by palexiadis
1 Answer | Add Yours
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neela | High School Teacher | (Level 3) Valedictorian
Posted on
To find the derivatives of the following:
i)
f(x)= x^4 + 3x^2+4x-31
Differentiating both sides, we get:
f'(x) = (x^4+3x^2+4x-31)'
f'(x) = (x^4)' +(3x^2)'+(4x)'- (31)', as (f(x)+g(x))' = f'(x)+g'(x).
f'(x) = 4x^3+3*2x+4-0, as d/dx(k x^n ) = knx^(n-1).
f'(x) = 4x^3+6x+4.
ii)
f(x)=2(x+3)e^x.
Differentiating both sides, we get:
f'(x) ={2(x+3)e^x}'
f'(x) = {2(x+3)}' e^x +(2(x+3)(e^x)'
f'(x) = 2e^x+2(x+1)e^x , as (e^x)' = e^x.
f'(x) = 2(x+4)e^x.
iii0
f(x)= (2x-1)e^x-3 . We assume -3 is separate not in power of e.)
Differentiating both sides, we get:
f'(x) = (2x-1)'e^x+(2x-1)(e^x)' -(3)'
f'(x) = 2e^x +(2x-1)e^x-0
f'(x) = 2e^x+2xe^x -e^x.
f'(x) =(2x+1)e^x.
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