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https://www.varsitytutors.com/lsat_logic_games_31-problem-28467 | 1,643,418,099,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00210.warc.gz | 1,078,858,098 | 31,800 | ### Test: LSAT Logic Games
Four roommates – Aaron, Brad, Clark, and Dave – are deciding how to schedule their delivery laundry service for the week. They must all set one pickup day and one delivery day between Monday and Saturday. There may be only one pickup and one delivery that occur each day, subject to the following conditions:
A pickup day cannot be scheduled on Wednesday.
A delivery day cannot be scheduled on Tuesday.
There must be exactly two days in-between Aaron’s pickup day and his delivery day.
Brad’s delivery day must be scheduled before Thursday.
Clark’s delivery day must be scheduled before Dave’s pickup day.
1 Which one of the following could be an accurate list of the pickup and delivery days scheduled for the week?
Monday: Aaron (Pickup), Aaron (Delivery);
Tuesday: Brad (Pickup), Brad (Delivery);
Wednesday: Aaron (Delivery);
Friday: Clark (Pickup), Clark (Delivery);
Saturday: Dave (Pickup), Dave (Delivery).
Monday: Aaron (Pickup);
Tuesday: Brad (Pickup);
Wednesday: Brad (Pickup);
Thursday: Dave (Pickup), Aaron (Delivery);
Friday: Clark (Pickup), Clark (Delivery);
Saturday: Dave (Delivery).
Monday: Brad (Pickup);
Tuesday: Aaron (Pickup);
Wednesday: Brad (Delivery);
Thursday: Clark (Pickup), Aaron (Delivery);
Friday: Dave (Pickup), Dave (Delivery);
Saturday: Clark (Delivery).
Monday: Brad (Pickup);
Tuesday: Aaron (Pickup), Brad (Delivery);
Thursday: Clark (Pickup), Clark (Delivery);
Friday: Dave (Pickup), Aaron (Delivery);
Saturday: Dave (Delivery).
Monday: Aaron (Pickup);
Tuesday: Brad (Pickup);
Wednesday: Brad (Delivery);
Thursday: Clark (Pickup), Aaron (Delivery);
Friday: Clark (Delivery);
Saturday: Dave (Pickup), Dave (Delivery).
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By clicking Create Account you agree that you are at least 13 years old and you agree to the Varsity Tutors LLC Terms of Use and Privacy Policy. | 519 | 2,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-05 | latest | en | 0.894382 |
https://physics.stackexchange.com/questions/466694/perelomov-coherent-states-for-an-arbitrary-hamiltonian | 1,653,772,542,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00559.warc.gz | 511,863,594 | 67,447 | # Perelomov coherent states for an arbitrary Hamiltonian
I'm reading about Perelomov coherent states, but I'm not sure if I'm getting it right. From this question and some Perelomov papers I understand the following:
The Perelomov coherent states are generated by applying a unitary operator to a certain state. If we have an arbitrary 1D Hamiltonian $$H = \frac{p^2}{2m} + V(x)$$, then the corresponding generalized coherent states would be generated as:
$$|\alpha(x,p)\rangle = \hat{T}(x, p)|0\rangle$$
Where $$\hat{T}(x, p) = \exp(i(x\hat{p} - p\hat{x}))$$ is the translation operator. The idea between this is that if $$|0\rangle$$ is the ground state, then it also would be the state that has minimum $$\Delta x\Delta p$$. So we move it on the phase space so that is has some momentum and it's some length away from the equilibrium point.
Am I right? Is this how 1D general coherent states are generated?
The key point is the algebra of your observables and the resulting group.
If you work with $$\hat x,\hat p$$ and $$\hat 1$$ this is the HW group, and Perelomov shown that translating the h.o. vacuum state gives the same as the original Glauber coherent states. Indeed in this case the CS is just a Gaussian displaced so that $$\langle x\rangle$$ and $$\langle p\rangle$$ need not be $$0$$.
If you work with angular momentum operators then the translation is a translation on the sphere, i.e. a rotation, and CS are rotated “ground states”. The ground state is either the state with $$m=j$$ or $$m=-j$$.
This is the essence of the Perelomov coherent states: you have a fiducial state and you act on it by a generalized displacement, which is a group transformation, and the group comes from the algebra of observables. Thus one can define $$SU(1,1)$$ coherent states, as discussed in Perelomov’s book, and pretty much any type of coherent state in this manner.
There is another key feature. It is known that, if the fiducial state is highest (or lowest) weight for the irrep of the group, then the whole geometrical setuo comes with a natural Poisson bracket on the appropriate manifold, i.e. the CS also naturally live in the classical phase space for this system since they are parametrized by points in the phase space. (The manifold is actually a coset space closely tied to the invariance properties of the fiducial state.). Hence these nice drawings either in the plane (for HW) or on the sphere (for angular momentum) where CS are mapped to localized lumps centered at the phase space coordinate of the CS.
There’s a nice review
Zhang, Wei-Min, and Robert Gilmore. "Coherent states: theory and some applications." Reviews of Modern Physics 62.4 (1990): 867.
and Bob Gilmore has published quite a bit of pedagogical work in the 80’s, v.g.
Gilmore, Robert. "Coherent states for bosons and fermions: A tutorial." Progress in Particle and Nuclear Physics 9 (1983): 479-494.
There are quite nice geometrical generalizations, such as vector coherent states, but these are much more sophisticated: see
Bartlett, Stephen D., David J. Rowe, and Joe Repka. "Vector coherent state representations, induced representations and geometric quantization: II. Vector coherent state representations." Journal of Physics A: Mathematical and General 35.27 (2002): 5625.
Saturation of the Heisenberg relations is a bonus and is “baked in”: fiducial vectors have this property in general and you’re just displacing this vector so with some suitable definition of observables one easily recovers this property. See
Delbourgo, Robert, and J. R. Fox. "Maximum weight vectors possess minimal uncertainty." Journal of Physics A: Mathematical and General 10.12 (1977): L233.
• I was thinking more about staying on the HW group but choosing a different Hamiltonian. Given the Coulomb 1-D Hamiltonian $H = p^2/2m + l^2/2r^2 - k/r$ for example, then the coherent states would be $\hat{T}(x, p)$ acting on the QHO ground state or acting on the Coulomb ground state? If we want minimum uncertainty then the "right" state would be the QHO ground state? Mar 16, 2019 at 11:47
• The strict definition would not work for this case, at least I don’t think, but you might find something directly useful in this older paper by Klauder ( arxiv.org/abs/quant-ph/9511033) and also in the work of Veronique Hussin on coherent states. Mar 16, 2019 at 13:55
• For minimum uncertainty states use keyword “intelligent states” in GoogleScholar. Mar 16, 2019 at 14:00
• These type of coherent states, in principle for an arbitrary Hamiltonian with pure point spectrum, are the Gazeau-Klauder coherent states. I'm not 100% sure if they are also valid when the classical Hamiltonian doesn't have action-angle variables, which is why I startted to investigate the Perelomov coherent states in the first place. Very interesting papers nonetheless. Mar 16, 2019 at 16:30 | 1,206 | 4,850 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-21 | longest | en | 0.91512 |
http://www.homeworkanswered.com/297/amanda-pushes-sled-from-rest-what-the-acceleration-the-sled | 1,642,447,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00643.warc.gz | 98,003,742 | 6,172 | # Amanda pushes a sled from rest to 8 m/s in 6m what is the acceleration of the sled?
Amanda pushes a sled from rest to 8 m/s in 6m
a) what is the acceleration of the sled?
b) how much time did it take to accelerate to 8 m/s
asked 10 years ago in Physics | 76 | 255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.936686 |
jacobblackwell.co.uk | 1,713,815,078,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00321.warc.gz | 288,260,499 | 25,500 | 276°
Posted 20 hours ago
# The Road to Reality: A Complete Guide to the Laws of the Universe
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BC) believed that mathematical proofs referred not to actual physical objects but to certain idealized entities. Moving in the counter clockwise direction there is a mysterious connection between the physical world and that of the mind. The linear approximation is the underlying assumption behind the usage of calculus; in fact, mathematical Analysis is justified by the use of infinitesimal (the divisor going to zero) and its inverse technique of Integration.
The mathematical distractions are introduced here to set up a later discussion of the role of mathematics in physics. The hope is that these community-generated reading notes will benefit people in the future as they go on the same journey. The experiment, which has been thought well and performed carefully, will step in the science forever. Identifying the set of existents to construct a view of nature is referred to in natural philosophy as ontology. The theorem can be stated as such, "For any right-angled triangle, the squared length of the hypotenuse [math]\displaystyle{ c }[/math] is the sum of the squared lengths of the other two sides [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] or in mathematical notation [math]\displaystyle{ aMathematical proof allowed for much stronger statements to be made about relationships between the arithmetic of numbers and the geometry of physical space. To explore the process of pursuing mathematical truth, Penrose outlines a few proofs of the Pythagorean theorem. This is a subject that has been the focus of my scientific thinking for over 35 years, so it was disappointing to see what a poor job had been done on this fundamental topic, although I should not have been surprised at the several contributors faithfully reflected the consensus views of contemporary physicists and philosophers who think about such matters. This essay evolved from my realization that physics has become a moribund science and when even a recent winner of the Nobel prize in Physics (1998) cannot create a ripple in the reductionist programme that fights long and hard to remain at the center of publicly-funded science that has been central to the modern viewpoint of Western Civilization. The Road to Reality: A Complete Guide to the Laws of the Universe is a book on modern physics by the British mathematical physicist Roger Penrose, published in 2004.
## The Road to Reality : A Complete Guide to the Laws of (PDF) The Road to Reality : A Complete Guide to the Laws of (PDF)
Penrose believes that not much progress can be made with respect to the mental world until we know much more about the physical world. He claims that objective truths are revealed through mathematics and that it is not a subjective matter of opinion. Finally he discusses the Mandelbrot set and claims that it exists in a place outside of time and space and was only uncovered by Mandelbrot. Contrary to the opinions of most well-educated intellectuals, who have been deeply influenced by the Abstract Ideas of this abstract science, most people have little knowledge of the limited progress that has been made since the physicists made their historic Faustian deal with the modern state and its military.Other mathematicians may find it useful to scan The Road to Reality, if only to glimpse the extent to which mathematical constructs infuse theoretical physics. As a modernist, his approach is centered on the empirical view that is grounded in our primary senses. Many people in ancient times allowed their imaginations to be carried away by their fascination with the subject, leading to mystical associations with mathematical objects. It covers the basics of the Standard Model of particle physics, discussing general relativity and quantum mechanics, and discusses the possible unification of these two theories. S. Pierce (professional chemist before turning to philosophy); Einstein was a theoretical physicist, who mainly worked as a mathematician but had an ongoing interest in philosophy.
## The Road to Reality: A Complete Guide to the Laws of (PDF) The Road to Reality: A Complete Guide to the Laws of
A discussion of the measurement problem in quantum mechanics is given a full chapter; superstrings are given a chapter near the end of the book, as are loop gravity and twistor theory. In the end of the process, when he arrives there for real where no one else has been before, he is alone. In fact, there has been a permanent alliance between Philosophers and Mathematicians ever since Plato insisted that the students at his Academy be familiar with Pythagoras’s ideas. This idea leads him to reject both string theory and such competing approaches as loop-quantum gravity.
In a single work of colossal scope one of the world's greatest scientists has given us a complete and unrivalled guide to the glories of the universe that we all inhabit.
## The Road to Reality Study Notes - The Portal Wiki The Road to Reality Study Notes - The Portal Wiki
Some texts, both popular and university level, introduce these topics as separate concepts, and then reveal their combination much later. At the core of Penrose’s thinking are arguments that lead him to believe that quantum mechanics must be modified to be unified with gravity. Mathematics was linked to the concept of Truth but Plato was also interested in the absolute idealized forms of Beauty and Good. Unfortunately, he then deliberately throws in references to material objects too small for us to see directly, like electrons and abstractions, like numbers. The author, Bob Laughlin is still a senior statesman of science, long-time Professor of Physics at Stanford, where he has taught since 1985.Chapters 1-16 focus on mathematical concepts while the later chapters use this background to describe the physical world. The Road to Reality reverses this process, first expounding the underlying mathematics of space–time, then showing how electromagnetism and other phenomena fall out fully formed. | 1,214 | 6,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.948437 |
https://ask.learncbse.in/t/a-balloon-is-rising-at-a-constant-speed-of-5-ft-s/53026 | 1,712,956,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00427.warc.gz | 107,547,670 | 3,519 | # A balloon is rising at a constant speed of 5 ft/s
A balloon is rising at a constant speed of 5 ft/s. A boy is cycling along a straight road at a speed of 15 ft/s . When he passes under the balloon, it is 45 ft above him. How fast is the distance between the boy and the balloon increasing 3 s later? | 79 | 302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.962227 |
https://www.jiskha.com/display.cgi?id=1253993155 | 1,503,551,287,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133032.51/warc/CC-MAIN-20170824043524-20170824063524-00116.warc.gz | 922,669,711 | 3,835 | # Algebra 1A
posted by .
find the decimal notation for fraction-7/13 the decimal notation for the fraction round to the nearest hundredth.
help
• Algebra 1A -
Divide the numerator by the denominator.
7/13 = 0.53846 = 0.54
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More Similar Questions | 635 | 2,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-34 | latest | en | 0.861945 |
https://www.codeproject.com/articles/1694/classes-for-computational-geometry?fid=3072&df=90&mpp=25&sort=position&spc=none&select=2743741&tid=2176590 | 1,484,589,561,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279224.13/warc/CC-MAIN-20170116095119-00250-ip-10-171-10-70.ec2.internal.warc.gz | 891,691,909 | 29,161 | 12,689,349 members (25,912 online)
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# Classes for computational geometry
, 26 Dec 2001 CPOL
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Some classes and utility functions for general computational geometry
## Introduction
This article presents two classes and a set of utility functions for computational geometry. ```C3Point ``` is a 3D counterpart to ```CPoint ``` and ```CPolygon ``` encapsulates a set of `C3Point`'s and provides general polygon handling functions. The classes have been mildly optimised for speed. The classes were originally written for use in discretising 2D surfaces into element networks and for calculating properties of the resultant elements. Care must be taken when using some of the functions such as the curvature and area functions to ensure that the results returned by the functions are consistent with your needs and definitions.
The classes make use of a ```typedef ``` ```REAL ``` that is either double or float depending on whether ```USING_DOUBLE ``` or ```USING_FLOAT ``` has been defined in geometry.h. Obviously using template classes would have been neater, but these classes were developed to get a job done, not to be the epitome of structured programming. A number of conversion functions have been provided:
```D2Real(x) (x) // double => REAL
F2Real(x) (x) // float => REAL
Real2D(x) (x) // REAL => double
Real2F(x) ((float)(x)) // REAL => float
Int2Real(x) ((double)(x)) // int => REAL
Real2Int(x) ((int)(x)) // REAL => int
Real2Int(double d0) // REAL => int (faster than a (cast)).```
All the classes and utility functions are provided 'as-is'. I've been meaning to write this class up for a long time and figured it was best to at least post something than nothing at all.
## C3Point
`C3Point` is a 3D counterpart to `CPoint`. It contains 3 data members x,y and z and a set of functions for calculating properties, scaling, translating and for arithmetic operations.
```class C3Point {
// Attributes
public:
REAL x,y,z;
//Operations
public:
C3Point() {} // constructor
C3Point(double x, double y, double z) // constructor
REAL Length2() // length squared
REAL Length() // length
void Scale(REAL factor) // scale by a factor
void Normalise(); // convert to a unit length
void operator=(C3Point& P) // assign
C3Point operator-(C3Point P) // subtract
C3Point operator-() // unary -
C3Point operator+=(C3Point P) // add +=
C3Point operator-=(C3Point P) // subtract -=
REAL operator*(C3Point P) // vector dot product
C3Point operator*(REAL f) // scalar product
C3Point operator/(REAL f) // scalar div
C3Point operator*=(REAL f) // scalar mult *=
C3Point operator/=(REAL f) // scalar div /=
C3Point operator^(C3Point P) // cross product
BOOL operator==(C3Point& P); // is equal to?
BOOL operator!=(C3Point& P) // is not equal to?
};
#define VECTOR C3Point```
## CPolygon
`CPolygon` encapsulates a set of `C3Point`'s and provides general polygon handling functions.
```CPolygon();
CPolygon(int); // Construct with a preallocated number of points
BOOL Closed(); // Is the polygon closed?
int GetSize() // Number of points
// is vertex 'index' between start,end inclusive?
BOOL InSpan(int start, int end, int index);
// is vertex 'index' between start,end exclusive?
BOOL InSpanProper(int start, int end, int index);
BOOL PointIn(C3Point P); // Is point inside polygon?
BOOL SetSize(int); // Change size of polygon
void RemoveAll() // Empty polygon of all points
BOOL Trim(int, int); // Trims polygon down so that points before
// "Start" and after "End" are removed.
// Start and End must be in the range 0..GetSize()-1
BOOL Close(); // Make polygon closed
BOOL SetAt(int nPos, C3Point& p); // set vertex nPos as point p
void Delete(int); // Delete a vertex
BOOL InsertAt(int nPosition, C3Point P); // insert point P at pos nPosition (0..N-1)
void FreeExtra(); // Free extra memory left over after deletes
int PointSeparation(int Point1, int Point2); // Distance (in pts) between 2 points
void Rationalise(int nAngle); // Combines adjacent line segments if the angle between
// them is less than nAngle (degrees).
REAL SegmentLength(int,int); // Length of a segment of the polygon
C3Point GetClosestPoint(C3Point p, int *nIndex = NULL);
REAL Area(); // returns polygon area
C3Point Centroid(); // Calculate centroid of polygon
BOOL GetAttributes(REAL *pArea,
C3Point *pCentroid,
C3Point *pNorm,
REAL *pSlope,
REAL *pAspect);
BOOL Diagonal(int i, int j); // Returns TRUE iff (v_i, v_j) is a proper
// internal or external diagonal of this polygon
virtual void Translate(VECTOR v); // Translate polygon
BOOL Intersected(C3Point& p1, C3Point& p2); // Does p1-p2 intersect polygon?
BOOL IntersectedProp(C3Point& p1, C3Point& p2); // Does p1-p2 intersect polygon properly?
BOOL Triangulate(CPolygon*); // Triangulate: Ear clip triangulation
BOOL CPTriangulate(CPolygon*, C3Point); // Central point triangulation
BOOL DelauneyTri(CPolygon*); // Triangulate: Delauney triangulation```
```// Load polygon from X-Y or X-Y-Z data file
BOOL LoadXY(LPCTSTR Filename, REAL Zdefault = D2Real(0.0));
BOOL LoadXY(FILE* fp, REAL Zdefault = D2Real(0.0));
BOOL LoadXYZ(LPCTSTR Filename, BOOL bWarn = TRUE);
// Save file either as:
// Num Points, elevation, x-y pairs...,
// or
// x-y-z triplets...
BOOL Save(LPCTSTR Filename, BOOL bAsPoints = FALSE, BOOL bWarn = TRUE);
void NaturalSpline(double*& b, double*& c, double*& d); // Natural cubic spline
REAL Curvature(int i); // Curvature at vertex i
REAL Curvature(int nIndex, int nSampleSize); // Avg curvature at i over
// a number of points
C3Point& operator[](int index);
C3Point& Point(int index);
void operator=(CPolygon& P);```
## General Functions
These functions provide general routines for vectors (`C3Point`s) and polygons.
```inline REAL Dot(C3Point V1, C3Point V2) // dot product
inline C3Point Cross(C3Point p1, C3Point p2) // cross product```
```C3Point GetClosestPoint2D(C3Point& start, C3Point& end, C3Point& P);
REAL Angle(C3Point, C3Point, C3Point); // Angle between 2 vectors formed from 3 points (deg)
REAL Angle(VECTOR v, VECTOR u); // Angle between 2 vectors (degrees)
REAL TriArea2(C3Point, C3Point, C3Point); // Area^2 between 2 vectors formed from 3 points
REAL TriArea2(VECTOR u, VECTOR v); // Area^2 between 2 vectors
BOOL IntersectProp(C3Point a, C3Point b, // Returns true iff ab properly intersects cd:
C3Point c, C3Point d) // they share a point interior to both segments.
// The properness of the intersection is ensured
// by using strict leftness.
BOOL Intersect(C3Point a, C3Point b, // Returns true iff segments ab and cd
C3Point c, C3Point d); // intersect, properly or improperly.
BOOL Left(C3Point a, C3Point b, C3Point c); // Returns true iff c is strictly to the left
// of the directed line through a to b.
BOOL LeftOn(C3Point a, C3Point b, C3Point c); // Same as Left, but c may be on the line ab.
BOOL Colinear(C3Point a, C3Point b, C3Point c); // Returns TRUE if a,b,c are colinear
BOOL Between(C3Point a, C3Point b, C3Point c); // Returns TRUE iff (a,b,c) are collinear and
// point c lies on the closed segement ab.
VECTOR Normal(C3Point p1, C3Point p2, C3Point p3); // Computes the normal (NOT unit normal) of
// a triangle, with points in Counter
// Clockwise direction.
VECTOR Scale(REAL factor, VECTOR v); // Scales a vector by a factor.
```
## Credits
The algorithms used are based in part from the book Computational Geometry in C by Joseph O'Rourke.
## Share
Chris is the Co-founder, Administrator, Architect, Chief Editor and Shameless Hack who wrote and runs The Code Project. He's been programming since 1988 while pretending to be, in various guises, an astrophysicist, mathematician, physicist, hydrologist, geomorphologist, defence intelligence researcher and then, when all that got a bit rough on the nerves, a web developer. He is a Microsoft Visual C++ MVP both globally and for Canada locally.
His programming experience includes C/C++, C#, SQL, MFC, ASP, ASP.NET, and far, far too much FORTRAN. He has worked on PocketPCs, AIX mainframes, Sun workstations, and a CRAY YMP C90 behemoth but finds notebooks take up less desk space.
He dodges, he weaves, and he never gets enough sleep. He is kind to small animals.
Chris was born and bred in Australia but splits his time between Toronto and Melbourne, depending on the weather. For relaxation he is into road cycling, snowboarding, rock climbing, and storm chasing.
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Last Visit: 31-Dec-99 19:00 Last Update: 16-Jan-17 2:59 Refresh 1 | 2,740 | 10,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-04 | latest | en | 0.875577 |
https://webapps.stackexchange.com/questions/66236/calculation-of-single-total-from-various-options | 1,716,780,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00618.warc.gz | 518,736,367 | 39,814 | # Calculation of single total from various options
I'm having some trouble understanding the calculations aspect of my form. I'm looking to have various options (check boxes) add up dollar amounts that will be displayed as a single number at the bottom of the form, but I can't seem to figure out how to assign these dollar amounts to the various check boxes, or how to assign them past selecting the title of the entire option field, which to my knowledge doesn't do anything currently.
First off, I am a developer for Cognito Forms.
Just to clarify, you want to show a set of checkboxes, with assigned prices, and calculate the amounts as a total for the form, something like this:
If you want to manually calculate a total based on the selections, you can use the following calculation (which is tricky, btw):
``````=Choice.Count(it = "First Choice") * 1.00 + Choice.Count(it = "Second Choice") * 2.00 + Choice.Count(it = "Third Choice") * 3.00
``````
However, this is definitely not obvious, nor easy, so we have been working to makes things better. In the next release, which is currently in testing, you can configure choice fields as follows:
By selecting Collect Payment For This Field, you can then specify prices for each choice. These selections automatically add to the amount due and optionally can appear as separate order line items. This release will also support tightly integrated Stripe payment, so the order form example above would become:
Hopefully, these changes will support your needs and make things much easier!
• This was totally helpful, thank you very much!! I was able to create a working form with this information, but I have one field that I'd like to make a "Drop Down" and the code you supplied works only when everything is a "Checkbox" is there a different word I should use for the Drop Down items? Thanks again! Aug 17, 2014 at 4:03
• With Drop Down and Radio Buttons it is a bit easier, since they represent just a single selection. Just use a formula like `=Choice = "First" ? 1.00 : Choice = "Second" ? 2.00 : Choice = "Three" ? 3.00 : 0.00`. This syntax compares the value of the field to different options and returns the desired value. The specific syntax of `condition ? true value : false value` is called a ternary operator and is common in C languages and covered in our documentation, but can be confusing for complex cases like this. Aug 20, 2014 at 2:29 | 545 | 2,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.956454 |
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# Add and subtract the following figures
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Add and subtract the following figures
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Roccy DeFrancesco
MI
Joined: May 24, 2006
### My Company
The Wealth Preservation Institute
Whatever your politics are, it seems clear that the president has no short-term energy policy when it comes to dealing with the prices of oil.
I decided to write this short and simple article when I heard part of a speech by President Obama recently about gas prices. Amazingly, the president was making light of higher gas prices. His speech was fairly typical. He looked down his nose at Americans who like gas-guzzling cars; and, as usual, he had little sympathy for those who drive them.
His idea for those who drive SUVs was simple ─ get rid of them and buy fuel-efficient cars like the Chevy Volt. This shows how truly out of touch the president is and that he lives in some sort of green energy fantasy land.
I imagine some financial planners have been asked and surely will be asked the magic question: Does it make financial sense to get rid of a less efficient SUV and buy a new, more gas-efficient car?
Let’s look at the math using a \$42,000 Chevy Volt. I’ll assume no financing costs and that the car payments are amortized over five years. The annual payment would be \$8,400.
The average driver puts 15,000 miles on a car. For this article, I’ll give numbers assuming gas is \$4, \$5 and \$6 a gallon.
The Volt averages 35 miles per gallon in the city and 40 mpg on the highway (from Kelly Blue Book), and so I’ll assume it will average 37.5 mpg for this artcile.
I’ll compare the Volt to the most popular SUV, which is the Ford Explorer. I’ll assume the example driver has a 2005 vehicle (with no payments) that averages 15 mpg.
The following are the average cost of gas each year and the annual savings using the Volt.
Ford Explorer Gallons Annual Gas Cost Chevy Volt Gallons Annual Gas Cost Annual Savings 1,000 \$4,000 400 \$1,600 \$2,400 1,000 \$5,000 400 \$2,000 \$3,000 1,000 \$6,000 400 \$2,400 \$3,600
Hmmm. That doesn’t sound like a lot of savings, does it? Not to mention that the SUV is a much larger and more useful car (and safer).
The KBB trade-in value for a 2005 Explorer in “good” condition with average miles is \$6,500. Therefore, after the trade-in amount of \$6,500 is applied, the out-of-pocket cost to pay for the Volt is \$7,100 each year for five years.
The math
Let’s consider the math following the five years after buying the Volt. Because the comparison is to keeping the used Explorer, I’ll assume an additional \$1,500 a year in fix-it-up costs to keep it in good working condition.
Without factoring in savings on fuel, if it takes five years to pay off the debt on the Volt, the additional costs this buyer would have to pay each year because of the decision to buy the Volt would be \$7,100-\$1,500 = \$5,600 a year.
If we now factor in the fuel savings by buying a more fuel-efficient car, the net loss when buying the Volt is:
at \$4 a gallon = \$3,200 a year or \$16,000 total over five years.
at \$5 a gallon = \$2,600 a year or \$13,000 total over five years.
at \$6 a gallon = \$2,000 a year or \$10,000 total over five years.
There is a one-time tax credit of \$7,500 which needs to be factored into the savings. There is also a need to factor in the cost to install a charging station at home (\$2,000, with cords), and the annual electric bill to recharge the car (\$450). When you factor in the credit and the additional expenses, you need to subtract \$3,250 from the above totals.
If you have clients who are interested in the here and now and having the maximum amount of money in their pockets over the next five years, it certainly won’t be as the president advocates, which is by buying a new gas-efficient car.
Also, to be fair, in five years the 2005 Explorer is probably going to need to be replaced (even though I’ve budgeted enough money to keep it in good working condition). The fuel-efficient car will also be older, but assume you could keep the gas-efficient car for another five years and that you’d have to buy another SUV to replace the 2005.
Even if gas goes to \$6 a gallon, you’d have saved enough from 2011–2016 to buy a 2011 used Explorer with the savings (meaning you can keep your big gas hog, drive a safer car, and still save money).
Conclusion
Our job as advisers is not to look out for the good of the country or for the good of the world. We do that when we vote and elect officials. Our job as advisers is to give the best advice possible to our clients.
When your clients ask you if they should get rid of the SUV they love so much (spacious, luxurious and safe) to buy a president-recommended Volt or other fuel-efficient vehicle, give them my article and tell them what they want to hear, e.g., that they should keep the gas hog, enjoy life, and still save more money than if they had bought the car the president and other green crusaders are trying to guilt them into buying.
The views expressed here are those of the author and not necessarily those of ProducersWEB.
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resistive feedback opamp with tiny loop gain
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exp
Full Member level 1
Hi,
In a paper I found the use of an OTA+buffer ("opamp") as a baseband amplifier after the mixer. It uses resistive feedback and the source resistance is given by a series of the antenna and the switch impedance of the mixer. To first order, the amplifier can be descried as the ordinary "inverting opamp amplifier":
https://en.wikipedia.org/wiki/Operational_amplifier#Inverting_amplifier
with typical numbers from the mentioned paper: Rin=55 Ohm, Rf=2k, a0=30 (since it consists of a simple diff pair + source follower in a 65nm process).
Using these numbers I get a Gain of A0 = a0/(a0+1) * 1/(Rin/Rf + 1/(a0+1)) = 16 which is OK.
However, the loop gain is given as T = a0/(1 + Rf/Rin) = 0.8029 < 1 !
The approximation A0 = Rf/Rs is only valid if a0 >> Rf/Rs.
Now I am confused: I always learned the loop gain should be as high as possible for good accuracy, linearity etc.
Does it make sense to use an amplifier like this where the loop gain is even negative? (one advantage is that it is unconditionally stable).
Based on the T expression, it seems like the circuit requires Rin >> Rf (or extremely high a0).
Also, since the distortion coefficients are attenuated by (1+T), wouldn't a configuration like this be a desaster for linearity?
Or do I just miss something?
It is ok to use an amplifier as long as the specs are met. But the big question is that the Gain Ao varies all over the place since the open loop gain of the amp is very low. If Ao=Rf/Ri (i.e very high open loop gain of opamp), it would be fairly constant across PVT changes. But with low open loop gain, it could be difficult to control the overall gain (In case you are planning to implement gain programmeability) of the block across PVT.
Lower loop gain may not give you the advantage of having feed back loop. i.e achieving high linearity numbers could be difficult (You may see the open loop linearity and closed loop linearity numbers to be almost same).
It would be good to increase the open loop gain of the amp to improve the performance.
Ok, thank you, this was a bit what I expected.
First of all, this is not a discrete dircuit but an IC in 65nm so achieving high loop gain is difficult to begin with (even with 2 stages not more than a couple of hundred).
But this is not so much my confusion here: You are talking about the "open loop gain of the amp" which would be the gain of the actual amplifier (a0). I Wouldn't be surprised if increasing a0 would give me better performance.
However, it turns out that the loop gain of the circuit is T0 = Rin*a0/(Rin + Rf)! This is different than the open loop gain of the amplifier (a0) and the effective loop gain T0 is even always much less than a0, specifically if Rf > Rin.
Now I could say: Not a big of a deal, just make Rin >> Rf: done.
This configuration is widely used as a TIA. In this case the source resistance would be high (current source). However, in this case, the source resistance is small (hence it is driven by a voltage source). So even when using moderate Rf values this would imply a low loop gain (for a practical IC OpAmp and not a discrete 120dB opamp). Right so far?
What is more, in the paper, the feedback resistor should be used for matching. The setup is as follows: The antenna (modeled as ideal voltage source in series with Ra=50 Ohm), followed by a switch with on-resistance of Rsw=20 Ohm and then the TIA with feedback resistor Rf.
For matching to occur, the input resistance of the amplifier (i.e., the op-amp only with Rf) should have a certain value which is Ra-Rsw=30 Ohm. Then the required Rf is given by
Rf = (1+a0)*(Ra-Rsw)
The total input resistance of the opamp circuit is given by the series resistance of Ra and Rsw:
Rin = Ra + Rsw
Now I just plug in the definition for Rf into the expression for T0 and I arrive at:
T0 = Rin*a0/(Rin + Rf) = ... = (Ra+Rsw)/(Ra*(1+2/a0)-Rsw) ~ (Ra+Rsw)/(Ra-Rsw)
This shows that T0 is nearly independent of the gain of the amplifier (a0) and very low for typical numbers (e.g. 2.3 for Ra=50 Ohm, Rsw=20 Ohm).
All this is consistent with spice but it still sounds fishy to me ... that the loop gain is to first order independent of the amplifier gain and very, very low.
Last edited:
I believe, you're post is desperately asking for a schematic that clarifies the meaning of the various terms.
Even when recalling commonly used formula signs in analog design text books, they don't make sense at first sight...
No problem:
(this is the simplified single-ended version which has some problems because Rsw=20 only during on-time but the actual circuit is fully differential so there are always 20 Ohm)
Last edited:
For matching to occur, the input resistance of the amplifier (i.e., the op-amp only with Rf) should have a certain value which is Ra-Rsw=30 Ohm. Then the required Rf is given by
Rf = (1+a0)*(Ra-Rsw)
Don't know the said paper, but this approach has nothing in common with regular OP or TIA circuit design. It's the reason for forcing loop gain to unreasonable low values.
There's no point of matching a source impedance by increasing Rf respectively reducing the loop gain. In case of doubt, the closed loop gain would be simply too high, causing output saturation.
Matching input impedance this way can be however done with amplifiers that have a low, well-defined gain, e.g. a LNA. In case an OTA with resistive load is suited for you as amplifier with defined gain, an OTA could be used. But it's neither TIA nor OP design principle.
Don't know the said paper
The said paper(s) are:
"A Passive Mixer-First Receiver With Digitally Controlled and Widely Tunable RF Interface," Caroline Andrews, Student Member, IEEE, and Alyosha C. Molnar, Member, IEEE
"Implications of Passive Mixer Transparency for Impedance Matching and Noise Figure in Passive Mixer-First Receivers," Caroline Andrews, Student Member, IEEE, and Alyosha C. Molnar, Member, IEEE (more details).
It is not perfectly equivalent because the paper discusses a sampling mixer (and hence the op-amp would include a small shunt resistance to ground also) but the general principle incl. my similar numbers fully apply.
, but this approach has nothing in common with regular OP or TIA circuit design. It's the reason for forcing loop gain to unreasonable low values.
There's no point of matching a source impedance by increasing Rf
But this is exactly what the paper proposes.
respectively reducing the loop gain.
Ok, so I make no mistake when I say high Rf vs. low Rs results in low loop gain (as discussed abo
In case of doubt, the closed loop gain would be simply too high, causing output saturation.
I don't get your point here. With the typical numbers I used the closed loop gain A0 = Rf/Rs * T0/(1+T0) is somewhere between 3 and 20 ... very reasonable.
Matching input impedance this way can be however done with amplifiers that have a low, well-defined gain, e.g. a LNA.
But this is exactly the point of the paper: Getting rid of the LNA, connecting the antenna directly to the mixer and using Rf for matching.
In case an OTA with resistive load is suited for you as amplifier with defined gain, an OTA could be used.
I think this is exactly why not an OTA is used but an OP (OTA + Buffer): Without the buffer, the input impedance to the amplifier would be something like (Ro+Rf)/(1+Gm*Ro) = (Ro+Rf)/(1+a0) which includes the Rout of the OTA which is not well controlled either.
Since you are trying to match the input impedance of a TIA to the antenna impedance, you are facing the issue. Typically open loop gain stage with low, defined gain is used to match the impedance as well as to get some gain.
If the switch you shown does Mixing, then the impedance matching equation is not simply what was mentioned earlier? Because the Mixer comes in, changing the entire scenario.
Last edited:
In a paper I found the use of an OTA+buffer ("opamp") as a baseband amplifier after the mixer. It uses resistive feedback and the source resistance is given by a series of the antenna and the switch impedance of the mixer. To first order, the amplifier can be descried as the ordinary "inverting opamp amplifier":
... it consists of a simple diff pair + source follower
I'd say such an amplifier shouldn't be called an OTA, much less an opamp.
Probably the high bandwidth here is more important than the low loop gain (which isn't negative BTW).
If the switch you shown does Mixing, then the impedance matching equation is not simply what was mentioned earlier? Because the Mixer comes in, changing the entire scenario.
No, see the remark under my circuit: For simplicity I use a single ended version here. The actual circuit is fully-differential, having always exactly one closed switch with Ron connected.
If all parasitic capacitances are reasonably small, then the impedance is determined by the indicated resistances up to very high frequencies: The mixing operation does not change any impedances then. This is also consistent with my spice results.
Scenario would change (slightly) if I have a sampling mixer which implies a big load capacitance. In this case the effective resistance after the mixer would change and also a virtual shunt capacitance would be introduced. This is the case for the referenced papers (and explained there). However, as I mentioned already, the effective numbers are quite comparable to my ohmic-only case!
I'd say such an amplifier shouldn't be called an OTA, much less an opamp.
I would so because this is exactly it: a lousy, fully differential op-amp.
When I draw the small signal circuit, this is exactly a Gm, followed by a buffer. It can be analyzed identically to the op-amp circuit (with low open loop gain a0).
Probably the high bandwidth here is more important than the low loop gain (which isn't negative BTW).
(my fault, of course, negative in dB)
Yes, but then I wonder
a) How they achieve the actual matching and precise gain if it's so heavily dependent on a0
b) How they achieve any acceptable linearity. As observed by one other posting (https://www.edaboard.com/threads/344266/) the linearity of a simple diffpair is fairly unacceptable (and also inconsisted with their reported number of -8dBm for low CL).
I agree completely with erikl. An OP is a device that is operated with loop gain > 1 and an overall (closed loop) gain set by the feedback network.
The circuit you are discussing is something different. According to the low loop gain, overall gain and input impedance will strongly depend on a0 and the linearity relies solely on that of the inner amplifier. Consequently there's no satisfying answer to your questions a) and b).
Why dont you try a telescopic structure for TIA? It will mostly work. Or two stage amp as in the second paper you mentioned.
Input impedance of a TIA is 1/Gm. You can use any value of Rf depending on the conversion gain you want. You need to match 1/Gm of the TIA to 50 ohms.
Last edited:
I agree completely with erikl. An OP is a device that is operated with loop gain > 1 and an overall (closed loop) gain set by the feedback network.
Well, ... no. An OP is a device (as you say) and not a circuit which uses an OP! I can happily build the same circuit with a low loop gain using a real, discrete OP. I would agree if you say a circuit which uses an OP the described is not a conventional usage of an OP. But let's stop this discussion, it does not matter.
Why dont you try a telescopic structure for TIA? It will mostly work. Or two stage amp as in the second paper you mentioned.
Input impedance of a TIA is 1/Gm. You can use any value of Rf depending on the conversion gain you want. You need to match 1/Gm of the TIA to 50 ohms.
Why would this matter? A telescopic is nothing more than a Gm cell with a bit higher Gm*Ro=a0 than the one described by me. At the block level this is identical. And, when I add the buffer, then really nothing changes because T0=(Ra+Rsw)/(Ra-Rsw) as shown above.
Even if I omit the buffer, the input impedance is 1/Gm only to first order. As described above:
Rin = (Ro+Rf)/(1+a0) \approx 1/Gm + Rf/(1+a0)
If I did not do a mistake the loop gain is given as:
T0 = a0 * Rin/(Ro + Rin + Rf)
Same as before, this requires a large Rin so that this expression becomes roughly 1.
Even with Rf=0, for a certain loop gain T0:
a0 = Gm*Rin*T0/(Gm*Rin - T0)
If Rin=70 as in my case and we suppose we need Gm=10mS then there is no a0 which would give me a loop gain > 1. Hmm, something looks fishy here ...
- - - Updated - - -
For some reason I cannot edit my posting any more. I went over it a couple of times, I do not find any mistake above. If I replace the Rs with caps and set s=0 I get the familiar OTA T0=beta*a0.
I think this is precicely the reason why you usually use capacitive and not resistive feedback with OTAs. When Rin is large (current source, e.g. 100k) then a large T0 can be achieved.
Just keep us informed if you solve problems a) and b).
Status
Not open for further replies. | 3,173 | 13,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2023-50 | latest | en | 0.946518 |
https://scienceplusplus.com/what-is-division-property-of-equality/ | 1,632,641,409,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00485.warc.gz | 537,754,112 | 20,574 | # what is division property of equality ?
## Division property of Equality
Equality is a property of an equation. An equations says that two expressions are equal and the symbol ‘=’ appears between those two expressions.
In such cases the equality is not affected by adding or subtracting the same number on each sides. The number can also be 0 and still the equality is not disturbed.
Similarly the equality of an equation is not affected by multiplying or dividing both sides, again, by the same number. But an important point in the cases of multiplication or division the number cannot be 0.
So with the restriction of 0, the equality of an equation is maintained when divided by the same number on both sides. This is what is known as division property of equality or simply as division property.
So, the division property of equality definition is, any equation maintains its equality status when divided on each sides by the same number except zero .
Let us illustrate some cases as division property of equality examples.
1) 4 = 4. When you divide by 2 on both sides you get 2 = 2 and when divided by 4 on both sides the result is 1 = 1. The answers are true in both cases. Also, the equality is not affected even when you divide by same negative numbers. For example, 4/(-2) = 4/(-2)?
-2 = -2.
2) Let us try an equation involving simple variables. Say 3x = 6. When divided by 3 on both sides you get the solution as (3x)/(3) = (6)/(3)
or
x = 2.
Same way,
(3x)/(-3) = (6)/(-3)
or
-x = -2,
x= 2
which is also true.
3) Now let us discuss an example where many students tend to commit a mistake by not properly applying the division property.
Let, x2 = 4x
be an equation.
Now I divide both sides by x and solve x = 4.
But x2 = 4x can be rewritten as a quadratic equation form as,
x2 – 4x = 0.
A quadratic equation has always two solutions whereas we found only one solution.
How and why the other solution is missing? What is the fallacy here?
The myth is, the step of dividing both sides of the equation x2 = 4x by x.
Because if x = 0, the equation x2 = 4x is true and hence the variable can also take the value of 0.
When such being the case, the division by x is not valid and hence one should not have gone ahead with that step. The correct method of solving is x2 = 4x ?
x(x – 4) = 0.
Now as per the zero product property there are two solutions as x = 0 and x = 4.
### 1 thought on “what is division property of equality ?”
error: Content is protected !! | 630 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2021-39 | longest | en | 0.948343 |
https://brainly.com/question/262906 | 1,484,945,458,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00419-ip-10-171-10-70.ec2.internal.warc.gz | 800,053,518 | 8,648 | # A chess player played 20 games total. If he won 2 of the games, what is the ratio of games he lost to games he won?
1
by annamo137
2015-01-20T15:58:31-05:00
So, we have useful information to answer this question. We know that he played 20 games and he won 2 games. To find the ratio of the games lost, we have to subtract the number of games played and the number of games won.
20-2=18
So now we can find the ratio of games won and games lost
The answer is 18:2, 18 to 2, or 18/2.
You can also simplify the ratio to 9:1
So the final answer is 9:1
Hope this helped
Andres2003
Thank you
No problem
Thank you for advising me on time | 192 | 635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-04 | latest | en | 0.970183 |
https://kimiyuki.net/writeups/algo-csacademy-39-b/ | 1,621,321,752,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00003.warc.gz | 362,432,621 | 2,770 | これは問題文が楽。
implementation
#!/usr/bin/env python3
import random
n = int(input())
a = list(map(int, input().split()))
b = [ None ] * n
for i, a_i in enumerate(a):
b[i] = ( a_i, i )
b.sort()
result = 0
for i in range(n - 1):
k = abs(b[i + 1][1] - b[i][1])
result += min(k, n - k)
print(result) | 108 | 291 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-21 | latest | en | 0.201827 |
https://iris.unime.it/handle/11570/3103333 | 1,719,134,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00353.warc.gz | 284,199,862 | 13,168 | Let R be a ring. A biadditive symmetric mapping D : R × R −→ R is called a symmetric skew biderivation if for every x ∈ R, the map y → D(x, y) is a skew derivation of R (as well as for every y ∈ R, the map x → D(x, y) is a skew derivation of R). Let D : R×R −→ R be a symmetric biderivation. A biadditive symmetric mapping ∆ : R × R −→ R is said to be a symmetric generalized skew biderivation if for every x ∈ R, the map y → ∆(x,y) is a generalized skew derivation of R associated with D (as well as for every y ∈ R, the map x → ∆(x, y) is a generalized skew derivation of R associated with D). In this paper we study some commutativity conditions for a prime ring R related to the behaviour of the trace of symmetric generalized skew biderivations of R.
### Some results concerning symmetric generalized skew biderivations on prime rings
#### Abstract
Let R be a ring. A biadditive symmetric mapping D : R × R −→ R is called a symmetric skew biderivation if for every x ∈ R, the map y → D(x, y) is a skew derivation of R (as well as for every y ∈ R, the map x → D(x, y) is a skew derivation of R). Let D : R×R −→ R be a symmetric biderivation. A biadditive symmetric mapping ∆ : R × R −→ R is said to be a symmetric generalized skew biderivation if for every x ∈ R, the map y → ∆(x,y) is a generalized skew derivation of R associated with D (as well as for every y ∈ R, the map x → ∆(x, y) is a generalized skew derivation of R associated with D). In this paper we study some commutativity conditions for a prime ring R related to the behaviour of the trace of symmetric generalized skew biderivations of R.
##### Scheda breve Scheda completa Scheda completa (DC)
2016
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File
Debrecen 2016.pdf
solo utenti autorizzati
Tipologia: Versione Editoriale (PDF)
Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11570/3103333` | 525 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.868158 |
https://math.stackexchange.com/questions/1747688/prove-that-this-system-of-linear-equations-generates-left-left-beginsmall | 1,566,453,031,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316783.70/warc/CC-MAIN-20190822042502-20190822064502-00206.warc.gz | 557,621,113 | 32,639 | # Prove that this system of linear equations generates $\left| \left( \begin{smallmatrix} 1/2 \\ n \end{smallmatrix} \right) \right|$ as a solution?
This infinite system of linear equations:
$$\begin{array}( 2x_1=1 \\ 3x_1+4x_2=2 \\ 4x_1+5x_2+6x_3=3 \\ \cdots \end{array}$$
In other words, this is particular case of a system:
$$\begin{array}( a_{11}x_1=b_1 \\ a_{21}x_1+a_{22}x_2=b_2 \\ a_{31}x_1+a_{32}x_2+a_{33}x_3=b_3 \\ \cdots \\ a_{m1}x_1+a_{m2}x_2+ \cdots + a_{mn}x_n + \cdots=b_m \\ \cdots \end{array}$$
With the coefficients given by:
$$a_{mn}=m + n~~~~~~~~~b_m=m$$
The systems seems to give the following sequence of solutions:
$$x_n= \left| \left( \begin{matrix} \frac{1}{2} \\ n \end{matrix} \right) \right|=\left(\frac{1}{2}, \frac{1}{8}, \frac{1}{16}, \frac{5}{128}, \cdots \right)$$
How can I prove that it's true?
Why does it happen? Can other fractional binomial coefficients be obtained this way (and how)? By 'this way' I mean a triangular system with simple regular sequence of integer coefficients.
I had some thoughts, and that's what I tried. Let's assume, that $x_n$ I found are really binomial coefficients. Then for $x_m$ we will have:
$$x_m= \left| \left( \begin{matrix} \frac{1}{2} \\ m \end{matrix} \right) \right|=\frac{1/2 (1-1/2)(2-1/2)\cdots (m-1-1/2)}{m!}=\frac{m-3/2}{m}x_{m-1}$$
$$x_m= \left(1- \frac{3}{2m} \right)x_{m-1}=x_1-\frac{3}{2}\left(\frac{x_1}{2}+\frac{x_2}{3}+\cdots+ \frac{x_{m-1}}{m} \right)$$
$$x_m=\frac{1}{2}-\frac{3}{2}\left(\frac{x_1}{2}+\frac{x_2}{3}+\cdots+ \frac{x_{m-1}}{m} \right)$$
On the other hand, from the $m^{th}$ equation we can derive:
$$x_m= \frac{b_m}{a_{mm}}-\frac{1}{a_{mm}}\left(a_{m1}x_1+a_{m2}x_2+\cdots+ a_{m,m-1}x_{m-1} \right)$$
$$x_m= \frac{1}{2}-\frac{1}{2m}\left((m+1)x_1+(m+2)x_2+\cdots+ (m+m-1)x_{m-1} \right)$$
I don't see how to get this relation into the same form as the previous one. Maybe I should modify it somehow.
• If you mean "in this way", yes. Every sequence can be produced as a solution to an infinite system of linear equations. – Christopher Carl Heckman Apr 18 '16 at 8:01
• @CarlHeckman, I also meant 'how'? – Yuriy S Apr 18 '16 at 8:02
• @CarlHeckman, also, could you please give me a reference for this statement? Where can I read about the connection between sequences and systems of linear equations? – Yuriy S Apr 18 '16 at 8:06
• You're going to get mad ... Just look at the system $x_1=a_1$, $x_2=a_2$, $x_3=a_3$, etc. Otherwise, you need to clarify what you mean by "this way". – Christopher Carl Heckman Apr 18 '16 at 8:07
• What are ${a_{mn}},\;{b_m}$? – user328032 Apr 18 '16 at 14:22
Since my hint in the comments was not followed up on, I'll elaborate in an answer.
Let $x_n = |\binom{\frac{1}{2}}{n}|$, and define:
$$A_m = (m+1)x_1+(m+2)x_2+\dots+ (m+m)x_m$$
What you are trying to show is that $A_m = m$ for all $m$. You already checked a few values that can be a base case for induction, so it will be enough to show that the sequences $A_1, A_2, \dots$ and $1,2,3...$ both satisfy the same recurrence.
The sequence $1,2,3,...$ satisfies an obvious recurrence, $b_m = 2b_{m-1}-b_{m-2}$, which encodes the fact that each term is the average of the term before it and the term after it. So now we just need to check the sequence $A_1, A_2, A_3, \dots$ also has this property. Luckily a lot of terms match up perfectly and we are just left with:
$$A_m+A_{m-2}-2A_{m-1} = 2mx_m + (3-2m)x_{m-1}$$
So in order for this recurrence relation to hold we just need to check that
$$2mx_m + (3-2m)x_{m-1} = 0$$
Which is easy to check by expanding $x_m = |\binom{\frac{1}{2}}{m}|$.
• Thank you. I did not follow your hint, because it wasn't clear to me how to do it. – Yuriy S Apr 23 '16 at 17:37
Alright, I figured it out. First, we write the recurrence relation for the absolute value of the binomial coefficients:
$$x_m= \left| \left( \begin{matrix} \frac{1}{2} \\ m \end{matrix} \right) \right|=\frac{1/2 (1-1/2)(2-1/2)\cdots (m-1-1/2)}{m!}=\frac{m-3/2}{m}x_{m-1}$$
$$x_1=\frac{1}{2}$$
Next, we work with our system.
We subtract the equation number $m-2$ from the equation number $m-1$ to show:
$$x_1+x_2+\cdots + x_{m-2}=1-2(m-1)x_{m-1}$$
Now we add this equation to the equation number $m-1$ and obtain:
$$(m+1)x_1+(m+2)x_2+\cdots + (2m-2)x_{m-2}+(2m-2)x_{m-1}=m-2(m-1)x_{m-1}$$
$$(m+1)x_1+(m+2)x_2+\cdots + (2m-2)x_{m-2}=m-(4m-4)x_{m-1}$$
Now we use the RHS from this equation in the equation number $m$:
$$m-(4m-4)x_{m-1}+(2m-1)x_{m-1}+2mx_m=m$$
Finally we obtain:
$$x_m=\frac{2m-3}{2m}x_{m-1}$$
$$x_1=\frac{1}{2}$$ | 1,783 | 4,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-35 | latest | en | 0.637602 |
https://www.physicsforums.com/threads/two-metal-disks.344424/ | 1,529,338,748,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00349.warc.gz | 870,752,418 | 13,179 | # Homework Help: Two Metal Disks
1. Oct 10, 2009
### kudoushinichi88
1. The problem statement, all variables and given/known data
Two metal disks one with radius R1 = 2.50cm and mass M1=0.80kg and the other with radius R2=5.00cm and mass M2=1.60kg are welded together and mounted on a frictionless axis through their common center.
a) a light string is wrapped around the edge of the smaller disk, and a 1.50kg block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released?
b) repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater?
2. Relevant equations
$\tau=FR=I\alpha$
3. The attempt at a solution
a) My approach is that I will first try to find the angular acceleration of the disks
F= the weight of the block, R=R1 and I=total inertia of the disks and alpha=angular acceleration of the disks. So...
$mgR_1=\frac{1}{2}\left(M_1R_1^2+M_2R_2^2\right)\alpha$
plugging in the numbers, i got $\alpha=163 rad s^{-2}$
using the equation $a_{tan}=R\alpha$, I got a=4.08m/s^2 which seemed like a sensible answer, but not the correct answer.
Did I miss something? Or did I do something wrong??
b) when I tried the same steps but this time using R2, I got a=16.3 m/s^2 which is totally nonsense. O.O||
2. Oct 10, 2009
### rl.bhat
The mass is not falling freely. So the torque due to suspended mass is not equal to mgR1.
If T is the tension in the string then
ma = mg - T
Now write down the expression torque and find the acceleration using the above equation.
3. Oct 10, 2009
### kudoushinichi88
ah... now I get it! Thanks! | 487 | 1,716 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-26 | latest | en | 0.872836 |
http://spotidoc.com/doc/101023/race-for-a-pattern-block-flower--and-back- | 1,603,530,009,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00347.warc.gz | 99,798,740 | 10,792 | Race for a Pattern Block Flower (and back)
```Race for a Pattern Block Flower (and back)
Let the area covered by the yellow hexagon = 1
How many of each of the other blocks does it take to cover the
hexagon exactly?
Make a “pattern block sandwich” and summarize your findings:
In terms of area…
1 yellow hexagon = 6 green triangles
1 yellow hexagon = 3 blue rhombi (rhombuses?)
1 yellow hexagon = 2 red trapezoids
As fractions…
1 green triangle =
of a hexagon
1 blue rhombus =
of a hexagon
1 red trapezoid =
of a hexagon
Put purely symbolically…
“Pattern block sandwich” (superposition)
Starting with a hexagon on the bottom as a bun,
make a layer of each type of block that covers the same
area as the hexagon. (If you prefer sandwiches open-face,
just omit the hexagon on top.)
the hexagons
are the buns
of the sandwich
side view: pattern block sandwich
with hexagons at the top and bottom
fraction die with six faces
Materials: a set of pattern blocks with no orange squares or tan thin rhombi (rhombuses). The reason for
this is that if you say the area of the orange square is 1, then the areas of the other shapes involve root 3.
Players: Two players and a banker.
Overview: The game has two parts. In the first part, the players roll, build, and trade their way up to a
flower. The first player to build a complete flower, six yellow hexagons with a pair of red trapezoids in the
middle, wins. (Overage is okay: the winner doesn’t have to hit it right on the nose.) Then it’s time to race
back down to nothing, taking the flower apart, trading it in, petal by petal, to return the quantity shown
by each roll of the die back to the bank. The first player to get get back to zero wins. (Again, going back
down, if a player doesn’t have enough flower left to return what’s rolled back to bank, that counts as a win
and the game is over.)
Rules:
1. Starting: Roll the die; high roller starts.
2. Always race up and down: Just like in other race games, you race your way up to a complete
pattern block flower, building and naming each amount that you roll (see above) as you add it to your
collection. On the way back down to zero, you take away what you roll each time, trading as
necessary, until you reach zero or have fewer blocks than what you need to take away what you
rolled. (See Liping Ma on composition and decomposition.)
3. Trade in for a hexagon when you can. Similar to base ten race games, when the total area of your
non-hexagon collection of pattern blocks is equal to or more than a yellow hexagon, you identify the
fraction values of your pieces out loud (this is important practice!) and trade them for a yellow
hexagon plus, whatever blocks most economically equal the overage.
Suppose, for example, you have
You roll
and
, so you get another blue rhombus
You realize you have more than a yellow hexagon, so you know you need to trade. (If you don’t realize
or know, the banker should tell you.)
You put the collection together, touch the red trapezoid and say “one-half”, the blue rhombi
(rhombuses) and say “two-thirds”; then it’s time to trade: you say “equals”, then grab a yellow
hexagon and a triangle, and cover the put them exactly on top of the old collection and say “one and
one-sixth”.
“two-thirds plus one-half”
“equals”
“one and one-sixth”
Extensions:
Diffies
As with other race games, diffies are a good way for kids to get practice on this by having them use dice
(you might want to add a whole number die (0…3 or 0…5) to make some mixed numbers. Just get a
number for each corner by rolling the die—or dice if you’re going to make mixed numbers possible. Then
calculate the difference (the distance between the two points on a number line) and put it in the middle.
In the same way, calculate those differences and put them in the middle again. Unless you make a
mistake in your calculations somewhere, your differences will ratchet down to zero in usually five to seven
levels. Diffies are very similar to racing back down to zero: the big differences is that they’re emphatically
symbolic (although you can use actual pattern blocks to build each difference calculation) and they’re
usually done by individuals (although there’s no reason why people couldn’t work as partners.
For virtual online diffies, see the National Library of Virtual Manipulatives’ Diffy page:
http://nlvm.usu.edu/en/nav/frames_asid_326_g_2_t_1.html
http://www.soesd.k12.or.us/files/diffy_blank.pdf
More Extensions
How would this look on a Number Line?
0
1
2
3
4
5
6
7
5
6
7
Assuming you let
the area of the
hexagon = 1
Students can use a number line to mark or locate their progress.
(This initially might be a good job for the banker.)
A game’s progress might be plotted like this (for the winner):
0
1
2
3
4
Notice that this person’s
winning roll of 1/3 took her
beyond 7, the value of the
flower (6 hexagons + 2
trapezoids).
How would this look with egg cartons?
Egg cartons will handle wholes, halves, thirds, fourths, sixths, and twelfths,
so the same die could be used with this other medium or another die could
be used: a 12-sided blank, for example, could be filled with these eleven
faces, plus either zero or 1 for the twelfth face:
1/12, 1/6, 1/4, 1/3, 5/12, 1/2, 7/12, 2/3, 3/4, 5/6, 11/12
race for the equivalent of a
pattern block flower
If you define a whole to be the total number of hemispheres in 2 egg cartons,
then you have eighths and twenty-fourths too!
Find Common Denominators as you go
Have students build common denominators with pattern blocks or egg cartons or calculate them. But be
careful not to slow down the game so much that the fun turns into tedium! (A good rule of thumb is that a
game should take 7 (plus or minus 2) rolls to “win” going up and another 7 (plus or minus 2) rolls to “win”
going down.
Still More Extensions
What about racing for other Pattern Block shapes?
Students can design their own goals for race games, determine the
value of their design (what size block will be the unit area, what the
total area is, accordingly).
Let the area of another shape = 1
If the red trapezoid
=1,
race for a pattern block mandala
then even numbers are all made up of hexagons
And
=
and
=
No halves, though! An important ingredient in this extension would be a blank die that you could mark
appropriately. (See “Purchase Dice” below.)
Make new combinations of shapes = 1
If
=1, then you have tenths
and twentieths
and fifths
—even fourths!
Track Progress by Recording Each Transaction
A crucial difference here is that the recorder finds a common denominator each time. Progress could also
be recorded on a number line rather than in a table.
Has
Rolls
Ends up with
0
And so on…
Notice that the “Ends up with” amount
gets brought forward to the new line.
Resources:
More on Race Games:
“Race Games Overview”
www.soesd.k12.or.us/files/race_games.pdf
“Race Games: Getting the Feel of Addition and Subtraction”
Print:
Let’s Pattern Block It by Peggy McLean, Lee Jenkins and J. McLaughlin \$13.95
www.activityresources.com/store/catalog/Lets-Pattern-Block-It-p-144.html
“This book leads children to understand geometry and number concepts through sequenced problem
solving activities. Included are activities involving copying designs, counting, equivalence, geometric cover
tasks, addition, inequalities, pattern, sequence, time, symmetry, area, perimeter, and fractions—all using
pattern blocks. Elementary and middle school students will find the activities challenging and fun.”
Virtual Manipulatives:
Pattern Blocks (National Library of Virtual Manipulatives)
http://nlvm.usu.edu/en/nav/frames_asid_169_g_1_t_2.html
Pattern Blocks: Exploring Fractions with Shapes
www.arcytech.org/java/patterns/patterns_j.shtml
Fraction Shapes and Drawing Fun Fractions
http://math.rice.edu/~lanius/Patterns/ and http://math.rice.edu/~lanius/Patterns/draw.html
Do-it-Yourself:
Printable Pattern Blocks www.aug.edu/~lcrawford/Tools/pattern_blocks.pdf
Purchase Pattern Blocks:
Math Learning Center www.mathlearningcenter.org
ETA/Cuisenaire www.etacuisenaire.com
Activity Resources www.activityresources.com
Enasco www.enasco.com
largest selection of all 53 items: www.enasco.com/Search?&q=pattern%20blocks&page=4
Purchase Dice:
For best selection overall, see www.gamestation.net. Also available from Math Learning Center,
ETA/Cuisenaire, Enasco, or Activity Resources. You may want to consider these dice:
ETA/Cuisenaire:
Enasco:
www.etacuisenaire.com/search/searchdisplay?type=keyword&query=dice
www.enasco.com/Search?catalog=&q=dice&x=0&y=0
6-sided fractions (1, 1/2, 1/3, 1/6, 2/3, 5/6)
www.gamestation.net/16mm-Opaque-Math-Dice-Fractions-112/M/B001EQXAT2.htm
6-sided blank
www.gamestation.net/16mm-Squareedged-Opaque-Blank-White-6sided/M/B0018TDN8S.htm
8-sided blank
www.gamestation.net/8sided-Jumbo-White-Blank-Dice/M/B0018TDNC4.htm
12-sided blank
www.gamestation.net/12sided-Jumbo-White-Blank-Dice/M/B0018TFJN0.htm
2-operator (+ and -)
4-operator
www.gamestation.net/16mm-d6-Opaque-Math-Operator-Dice/M/B001EQV8PK.htm
www.gamestation.net/16mm-Opaque-Math-Operator-Dice-4/M/B001EQYS1Q.htm
Math Learning Center’s collection:
revised 6/15/2009
www.mathlearningcenter.org/store/product-1055985.htm
Larry Francis, SOESD Computer Information Services
[email protected] and 541-858-6748
www.soesd.k12.or.us/math/math_resources and www.soesd.k12.or.us/support/training
``` | 2,467 | 9,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-45 | latest | en | 0.910986 |
http://aolanswers.com/questions/how_many_zeros_in_a_googolplex_p862647314153171 | 1,406,611,342,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510265454.51/warc/CC-MAIN-20140728011745-00078-ip-10-146-231-18.ec2.internal.warc.gz | 11,951,267 | 19,745 | # how many zeros in a googolplex?
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https://www.vexforum.com/t/tabor-paper/94342/6 | 1,685,424,122,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00173.warc.gz | 1,152,703,280 | 7,866 | # Tabor paper
We talk all the time about people who did highschool/ college VEX and then went on to do X. Here is the opposite, talking about me staying in robotics forever.
Today, my research partner and I published the culmination of 3 years of work together. This is my first published robotics paper and its in Nature. (which is hard for me to explain significance of without coming off terribly)
https://rdcu.be/czSK5
Here are articles about the paper that are maybe a more gentle introduction.
Paywall News Article anyone with a University account should be able to see
Blog Written by Professor on Project
The TLDR of the paper is, People for centuries have known about the electromagnetic phenomena called eddy currents. Usually they are considered a minor inconvenience, occasionally they are used in a few limited applications like separating metal from trash or maglev trains. The full effects the eddy currents were largely unmodeled, so we created an equation to compute the force created by the eddy currents. With this equation we are able to fully manipulate(control and move around) a nonmagnetic(does not react to magnets) object from a distance using magnets. There is some interesting applicability using our technique to clean up the tens of thousands of pieces of space debris that are largely aluminum and destroy satellites every year. We lovingly refer to it as a tractor beam, despite there being some scientific inaccuracy in that description.
Pictures and videos on paper link, under supplements
https://rdcu.be/czSK5
63 Likes
Well done Tabor! Prestigious publication and complex science topic to tackle!
Congrats!
6 Likes
That is actually quite cool.
6 Likes
If we were to separate all objects into 4 categories
1. Is a magnet
2. Is iron, ferromagnetic, reacts to magnets
3. Is any other metal(copper, aluminum, …), doesn’t react to magnets, but can conduct electricity
4. Is electric insulator, doesn’t react to magnets and doesn’t conduct electricity
Our work allows manipulation of category 3 without touching it, in space. (With a bunch of caveats for how far away it is, how big it is, and how conductive it is)
14 Likes
Tabor paper
3 Likes
32 Likes
Tbh, I was expecting high level of scientific rigor and academic discourse to last for more than five posts in this topic…
Now your only road to redemption, @DRow, would be to sneak copper balls and unobtainium magnets into 2023 Game Manual!
11 Likes
I hear @tabor473 's follow-up research will enable one to simply change the gravitational constant of the universe
10 Likes
Oh… i heard he got head-hunted by tony stark.
4 Likes
Super cool work, I have no words to better describe it.
Quick question though: because the trash in space is moving super fast, wouldn’t the required power to alter their velocity be really high?
4 Likes
1. We assume satellites in orbit with electromagnets so we are also going really fast as well. So the velocity relative to us isn’t nearly as high
2. If there is a large relative velocity difference we can just wait for the object to come back around, and every time it passes us change it’s velocity a little bit. As long as we get it eventually it’s fine, and can focus on different debris in the mean time.
3. It’s much easier to run stupid powerful electromagnets in space. Solar panels work better, and without an atmosphere you can super cool your magnets, and get really good conductivity and almost no heat loss.
10 Likes
Very cool, published in Nature is a big deal. I remember when you were not the best poster two lifetimes ago, happy to see that you’ve grown so much and have taken your skills to to such a high level. Congrats, I along with many others are super proud of your accomplishment. We are also waiting for your next act, since this is not a destination, but a trajectory.
8 Likes
Well said, Foster.
An amazing accomplishment, Griffin (@tabor473). Does this mean the old “anti-grav” motors from Cody game design challenge videos hold water now?
5 Likes
This topic was automatically closed 365 days after the last reply. New replies are no longer allowed. | 901 | 4,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.94481 |
https://wikivisually.com/wiki/Cardinality | 1,586,268,329,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371799447.70/warc/CC-MAIN-20200407121105-20200407151605-00414.warc.gz | 748,276,224 | 14,937 | SUMMARY / RELATED TOPICS
In mathematics, the cardinality of a set is a measure of the "number of elements of the set". For example, the set A = contains 3 elements, therefore A has a cardinality of 3. Beginning in the late 19th century, this concept was generalized to infinite sets, allowing to distinguish several stages of infinity, to perform arithmetic on them. There are two approaches to cardinality – one which compares sets directly using bijections and injections, another which uses cardinal numbers; the cardinality of a set is called its size, when no confusion with other notions of size is possible. The cardinality of a set A is denoted | A |, with a vertical bar on each side. Alternatively, the cardinality of a set A may be denoted by n, A, card , or # A. While the cardinality of a finite set is just the number of its elements, extending the notion to infinite sets starts with defining the notion of comparison of arbitrary sets. Two sets A and B have the same cardinality if there exists a bijection from A to B, that is, a function from A to B, both injective and surjective.
Such sets are said to be equipollent, or equinumerous. This relationship can be denoted A ≈ B or A ~ B. For example, the set E = of non-negative numbers has the same cardinality as the set N = of natural numbers, since the function f = 2n is a bijection from N to E, see picture. A has cardinality less than or equal to the cardinality of B if there exists an injective function from A into B. A has cardinality less than the cardinality of B if there is an injective function, but no bijective function, from A to B. For example, the set N of all natural numbers has cardinality less than its power set P, because g = is an injective function from N to P, it can be shown that no function from N to P can be bijective, see picture. By a similar argument, N has cardinality less than the cardinality of the set R of all real numbers, see Cantor's diagonal argument or Cantor's first uncountability proof. If |A| ≤ |B| and |B| ≤ |A| |A| = |B|; the axiom of choice is equivalent to the statement that |A| ≤ |B| or |B| ≤ |A| for every A, B.
Above, "cardinality" was defined functionally. That is, the "cardinality" of a set was not defined as a specific object itself. However, such an object can be defined; the relation of having the same cardinality is called equinumerosity, this is an equivalence relation on the class of all sets. The equivalence class of a set A under this relation consists of all those sets which have the same cardinality as A. There are two ways to define the "cardinality of a set": The cardinality of a set A is defined as its equivalence class under equinumerosity. A representative set is designated for each equivalence class; the most common choice is the initial ordinal in that class. This is taken as the definition of cardinal number in axiomatic set theory. Assuming the axiom of choice, the cardinalities of the infinite sets are denoted ℵ 0 < ℵ 1 < ℵ 2 < …. For each ordinal α, ℵ α + 1 is the least cardinal number greater than ℵ α; the cardinality of the natural numbers is denoted aleph-null, while the cardinality of the real numbers is denoted by " c ", is referred to as the cardinality of the continuum.
Cantor showed, using the diagonal argument, that c > ℵ 0. We can show that c = 2 ℵ 0, this being the cardinality of the set of all subsets of the natural numbers; the continuum hypothesis says that ℵ 1 = 2 ℵ 0, i.e. 2 ℵ 0 is the smallest cardinal number bigger than ℵ 0, i.e. there is no set whose cardinality is between that of the integers and that of the real numbers. The continuum hypothesis is independent of a standard axiomatization of set theory. See below for more details on the cardinality of the continuum. If the axiom of choice holds, the law of trichotomy holds for cardinality, thus we can make the following definitions: Any set X with cardinality less than that of the natural numbers, or | X | < | N |, is said to be a finite set. Any set X that has the same cardinality as the set of the natural num
Knabs Ridge Wind Farm is an electricity generating site just south of the A59 road near to Felliscliffe, North Yorkshire, England. It was the first wind farm to be built in North Yorkshire in over 15 years, was believed to be the first time that civilian air traffic was considered in the planning permission process; the proposal to build a wind farm on the site was first unveiled in 2003 when 6,000 homes in the area were leafleted with information about the wind farm. At that early stage, many objections were registered, including one from the Campaign for the Protection of Rural England. Planning consent for the wind farm was rejected by Harrogate Borough Council in November 2004; the management team at Leeds Bradford Airport, whose runway is only 10 miles south of the site, stated that the farm would have a detrimental effect on air traffic control as it could create false indications of aircraft. Other objectors pointed out that the site of was only 50 feet away from the Nidderdale AONB. Npower appealed against the decision and the application went before a public enquiry, which gained government consent in September 2005.
The objection by LBA towards the wind farm is believed to be the first time that civilian air traffic paths have featured in an onshore wind farm consent inquiry. The planning inspector rejected the idea of affixing red lights to each tower as it was thought that they could interfere with drivers concentration on the adjacent A59 road. Building of the £13 million wind farm 3 miles west of Harrogate in Felliscliffe by the A59 road, was started in March 2007, completed by the summer of 2008 with full commissioning of the site starting in November of the same year; the wind farm consists of eight towers which are 190 feet high, with the addition of the rotors, the full height that they extend to is 305 feet. The site covers an area of 197 acres at an elevation of 708 feet above sea level, so they can be seen from some distance away; each generating unit was constructed by Senvion and is capable of generating 2 mW of power with the whole farm itself generating 16 mW in total. This provides enough electricity per year to power 7,000 homes and offsets over 23,000 tonnes of carbon dioxide going into the environment on an annual basis.
One of the turbines caught fire in January 2017 and had to be extinguished by North Yorkshire fire service. Each unit operates in tandem with three others which resulted in all four generating units being turned off. Seven months after the fire, the owners were seeking to overturn a condition in the planning permission of the site, set in 2005. If a turbine was to not generate power in over twelve months the owners are required to remove the tower and return the land to grazing purposes for cattle. By October 2017, the tower was still not active. List of onshore wind farms in the United Kingdom Key facts PDF sheet
Terral is the third studio album by Spanish singer-songwriter Pablo Alborán, scheduled to be released 11 November 2014 through Warner Music. The album was composed and produced by Eric Rosse. Terral earned a nomination for Best Contemporary Pop Vocal Album at the 16th Latin Grammy Awards and for Best Latin Pop Album at the 58th Annual Grammy Awards. A "French Version" of the album was released on 29 January 2016. "Por Fin" was released as the album's lead single on September 16, 2014. The song debuted at number 1 on the Spanish Singles Chart, it has so far spent two consecutive weeks at number one. "Pasos de cero" was the second single released as a pre-order single on October 7, 2014. On January 21, 2015 the song was released as the second single, with the music video for the song premiering the same day. "Recuérdame" was released on May 31, 2015 as the third official single, along with the music video for the song. "La Escalera" was released on October 28, 2015 as the fourth and final single, along with the music video.
Terral – French edition All tracks are written by Pablo Alborán
Violence against women is an entrenched social problem in Ukrainian culture engendered by traditional male and female stereotypes. It was not recognized during Soviet era, but in recent decades the issue became an important topic of discussion in Ukrainian society and among academic scholars. Nuzhat Ehsan, UN Population Fund representative in Ukraine, stated in February 2013 “Ukraine has an unacceptable level of violence by men and due to high level of alcohol consumption”, he blamed loopholes in the legislation contribute to the problem of domestic violence, “You can violate women and still if you are a high-level official or from a high-level official family, you can get away with it”. In the view of traditional morale norms the Ukrainian women are supposed "to be beautiful and to be a mother". Moreover, the gender traditions in Ukraine tend to restrict women from holding positions of political power, while standards of male behaviour include being protective of women; the ongoing War in Donbass in Ukraine's Donbass region is believed to have reinforced the separation of gender responsibilities.
But the UN reports the Ukrainian women living in the conflict zone are at significant risk due to weak law enforcement, high concentration of military groups and proliferation of weapons. In 2015 the Office of the United Nations High Commissioner for Human Rights expressed a deep concern about worsening situation with violence against women in Ukraine. According to the estimation of OSCE the violence towards women is widespread in Ukraine and it is associated with three times more deaths than the ongoing War in Donbass in the Donbass region of Ukraine. According to latest UN Report around 45 percent of Ukraine’s population suffers violence – physical, sexual or mental – and most of them are women. From the historical point of view the problem of violence against women in Ukraine was always surrounded by silence. There are not too many sources of official statistics to estimate; this fact brings understanding that the mass media is unsure how to approach the issue, the authorities do not know how to deal with the manifestations of the violence and the nation doesn't know how to mobilize the public to end the violence.
As a result all national anti-violence efforts rely on considerable support from Western donors A few available statistics indicate that the violence inherent in the poverty where most part of Ukrainian women live. The recorded data demonstrate that the observed value is just a minor fraction of a real figure due to a combined influence of traditions and personal shame. In Ukraine domestic violence towards women has a long history. There are evidences that the model of male aggressiveness directed towards women is transferred from generation to generation. Recent studies reveal that violence against women is not limited to any particular segment of the Ukrainian population and occurs across all social layers. There are not many Ukrainian women who are have achieved economic independence and own their own home. According to recent research violence against women remains a hidden problem, but it is clear that the ongoing War in Donbass in Ukraine leads to an increase in its prevalence. In early 2020 there were any women's shelters in Ukraine.
The sexual exploitation of women is a broad and serious problem in Ukrainian society. This issue consists of women's trafficking on the transnational markets and coercive prostitution, it comes up as a result of many factors including the impoverishment of significant part of population, soft pornography used by Ukrainian mass media to catch an attention of customers, so on. Statistical data reported by NGOs state that up to one third of young jobless Ukrainian women have been involved to some degree in illegal sex business activities. A survey of young Ukrainian girls performed by All-Ukrainian Committee for the Protection of Children reveals that a sexual abuse of them accounts for high proportion of the abuse victims. For example, it was reported that one Ukrainian girl out of three had experienced sexual harassment, one out of five had suffered physical sexual abuse, one out of ten had been raped. Criminal statistics reports that 55% of registered sexual assaults in Ukraine are directed towards youth under 18, 40% of them or 22% in total - towards children under 14.
In February 2019 domestic violence was made a criminal offense in Ukraine meaning perpetrators could be fined, or sentenced to community service or a prison sentence. Perpetrators could be subjected to a maximum of administrative punishment. According to The Ukrainian Week about 5% of victims report violence to the Ukrainian police. Kobelyanska, Larysa. "Violence and Trafficking in Women in Ukraine". Making the Transition Work for Women in Europe and Central Asia. World Bank Publications. ISBN 0-8213-4662-8. Hrycak, Alexandra. "Global Campaigns to Combat Violence against Women:Theorizing Their Impact in Post-Communist Ukraine". Gender and Society in Ukraine. University of Toronto Press. ISBN 978-1-4426-4064-1. Lucas, Brian. "Gender and conflict in Ukraine". Www.gsdrc.org. K4D Knowledge and Learning for Development. Retrieved 6 September 2018. Ukrainian National Hotline for Prevention of Domestic Violence, Trafficking in Persons and Gender Discrimination: 0 800 500 335 or 116 123
Informatics for Consumer Health is a government initiative coordinated by the National Cancer Institute within the National Institutes of Health. ICH focuses on a coordination of health information and health care delivery that empowers providers to manage care and increases the ability of consumers to gain mastery over their own health; the ICH online initiative involved stakeholders from various sectors—commercial IT, health care, education and advocacy—exchanging ideas and resources to bridge information technology and health care with the goal of improving behavioral support for all consumers. The Informatics for Consumer Health field is related to health informatics, medical informatics, consumer health informatics, eHealth, health information technology. Changing health behaviors is key to improving health outcomes. Research indicates that changes in basic preventive behavior - smoking cessation, better diet and exercise, routine screenings—can lead to potential reductions in disability and death due to cancer, heart disease, diabetes.
Effective consumer health information technology applications hold great promise for encouraging and supporting behavior change. With the behavioral and population health evidence-base as a backdrop, the “Informatics for Consumer Health” came out of to two events that occurred in 2009; the first event occurred in early 2009, when the U. S.-based National Research Council released a report titled “Computational Technology for Effective Healthcare: Immediate Steps and Future Directions”. This report concluded that many of the current deployments of health information technology had become disconnected from their primary objectives: to ensure the health of real consumers in real world settings; the research portfolio in health systems should be rebalanced, authors of the report argued, to emphasize cognitive support for providers and their families over technology development for its own sake. The second event occurred in March 2009, when the U. S. Congress passed the American Recovery and Reinvestment Act of 2009.
Title XIII of the Act referred to as the Health Information Technology for Economic and Clinical Health Act, gave authority to the Department of Health and Human Services to offer incentives. “Meaningful use”, rather than “use” measured in technological terms, was to be gauged in terms of outcomes for patients and their families. In this context, a number of Federal agencies came together to convene the “Informatics for Consumer Health: Summit on Communication and Quality,” in November 2009; the summit brought together nearly 200 leaders from commercial IT, health care, education and advocacy organizations to open a dialogue and begin creating a blueprint for improving health care quality through enhanced behavioral support for consumers across the healthcare spectrum. Key Summit objectives to foster collaborations and spur innovation led to the publication of a special supplement on cyberinfrastructure for consumer health in the American Journal of Preventive Medicine and the development of the online Informatics for Consumer Health platform in 2010.
For three years the platform, informaticsforconsumehealth.org, served as a rallying place for summit stakeholders representing a wide range of sectors to exchange resources that bridge information technology and health care, improve behavioral support for all consumers. By sharing news and funding opportunities the ICH community helped disseminate information for facilitating collaboration among the public and research communities to improve consumer health. Goals of the ICH platform included: Serving as a clearinghouse of knowledge for use in the development of high-quality evidence based consumer and clinical health IT products. Creating an interactive portal where stakeholders could communicate and stay up to date on health IT research innovations and opportunities for collaboration and partnership. Learning from key stakeholders about available resources, projects and partnerships in consumer and clinical health IT to disseminate on the ICH platform. Providing a platform for conversation and collaboration around informatics for consumer health.
In 2013, the ICH platform was retired and all original content is now archived on the National Cancer Institute’s Health Communication and Informatics Research Branch website. As the science and practice of behavioral and public health informatics evolve, the activities and initiatives aggregated and posted on the ICH platform continue to be a priority for the public and private sector; the Informatics for Consumer Health initiative is led by a collaboration of Federal agencies with active intramural and extramural programs focused on aspects of consumer health IT. Agency for Healthcare Research and Quality Centers for Disease Control and Prevention National Cancer Institute National Institute of Standards and Technology National Library of Medicine National Science Foundation Office of the National Coordinator for Health Information Technology HCIRB website
The 2018 Donegal Senior Football Championship was the 96th official edition of Donegal GAA's premier Gaelic football tournament for senior graded clubs in County Donegal. Sixteen teams compete with the winner representing Donegal in the Ulster Senior Club Football Championship; the championship continues with a knock-out format. The draws were made on 24 March 2018. Kilcar were the defending champions after they defeated Naomh Conaill of Glenties 0-7 to 0-4 in the 2017 final. However, the club had to play the competition without two of its county stars, Patrick McBrearty and Ryan McHugh. On 26 September 2018, it was announced that McHugh had accepted medical advice and would be sidelined for the remainder of the year due to concussion, his injury came while playing for his club in a challenge match against Dublin champions St Vincents in Cavan in late August 2018 - he received a blow to the head during that match. Concussion had caused McHugh to spend six weeks on the sideline following a 2018 National Football League game earlier that year.
The injury meant he could take no part in the 2018 Donegal Senior Football Championship, news, worsened when taken in the context of the earlier loss of McHugh's club and county teammate McBrearty to a cruciate ligament injury. This was Milford's return to the senior grade after relegation in the mid-1990s after claiming the 2017 Donegal I. F. C. Title. On 21 October 2018, Gaoth Dobhair claimed their 15th S. F. C. Title when defeating Naomh Conaill of Glenties by 0-17 to 1-7 at MacCumhaill Park. Burt were relegated back to the 2019 I. F. C. after just two seasons in the top-flight when losing their Relegation Final to newly promoted Milford. The following teams changed division since the 2017 championship season; the 2018 County Championship took the same format as previous championships in which there was four groups of four with the top two qualifying for the quarter-finals. Bottom of each group play in relegation play-offs to decide which team is relegated the 2019 Intermediate championship. All 16 teams entered the competition at this stage.
The top 2 teams in each group advanced to the Quarter-Finals while the bottom team of each group entered a Relegation Playoff. This year, all teams played one home match, one away match and one match at a neutral venue | 4,323 | 20,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | longest | en | 0.945895 |
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Solutions 9
# Solutions 9 - 9-84 The pedal crank for a bicycle has the...
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Unformatted text preview: 9-84. The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not rotate while subjected to a force of 75 lb. determine the principal stresses in the material on the cross section at point C. Internal Forces and Ale-tent: As shown on FBD. S eetien Properties: l , , I= Eto.3)(0.s ) =0.0|23 n‘ a = i'A' = 0.3(02H03l = 0.0130 in’ V3750 lb Nor-tel Stress: Applying the (hut: formula. I? 3"“ ‘ i'05 0-2 :0. M 400402» ' - in. . . I 0.0l28 0.4 m Shear Stress: Applying the shear ionnuia. -=—--—=35L6 = , S ‘ _. 5-5516“. t‘ I: 00:230..» P" ”3 '6 "" i C unstrnetian of the Circle: in accordance with the sign conventmn. a, =4.6875 tn. 6, = 0. ad I" =0.35l6 ksi. Heme. O’ + 0, 4.687544) _ = — = 2.34315 ks: 2 1 a... 8 ‘lhccootdmates iorrelerenoeponlsa andC 81: A1168”. 0.35M) 02.34375. 0) 'l'herzllnisolthceitcleis n = (4.6875-134375): +0.3sn6: = 2.3670 ksi In . Plane Principal Stress: The coordinates of ports 8 and 0 represent a. and 03. respectively. a, = 2- 34375 + 2.3670 = 4.7: ksi Ans a2 = 2.34375 - 2.3670 = -0.0262 ksi Ans 9-87. The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the results on elements located at these points. Section Properties.- 4 - x(aoovs‘) - \$6.25n(10“) m’ 600a) 1-;(09015‘) - 2.4350( to") m‘ Q-Q-O 8 tree 1: a--t— A I 600 30.0(0.W75) ' 56.253004) * 2.4850( 10-9) 0‘ I 3.3953 -90.5414 I -87.14 MP: 0, I 3.39\$3+90.\$4|4 I 93.94 MP. 1" I f, IOsioaQ IQ I0 Contraction of the Circle: In accordance with the sign convention. a, I-87.l4 We. 6, I0. and t" I0lot 906nm. Home a, +0, 41144-0 a - I _ m 2 2 - «3.57 w; Theamtdinates for reference points A and C are M4114. 0) C l 43.57. 0). The Mitts olthecitcle is R I 87.!4-43.” I 43.57 MP: In-l'lone Principal Stresses: The cootdinateo of poinu B and A representa. and 0,. respectively. 0" .0 All! a, . 47.1 MP: Mu Mai-Inn: ln-Plnne Shear Stress: Reptaented by me coordinates of pow E on the circle r ... Ill-43.6w: Ans Orientation of the Plane for Maxi-inn. ln-Plnne Shear Street: From the dtde 20, I90' 0, I £5.0' (Conntereloekwiu) Ans Construction of the Circle: In accordance with the sign convenuon. a. I 93.94 MPa. 6, I 0. and r,, I Oiot point 8. Hence. a I a, +0, . 93.94“) "‘ 2 2 I 46.97 MP: 600N 600N 1mm 5‘ M" 6 44.5% +3-th 47.“. me 410». 450’ 45.9% 410% am} A Perm ln-Plone Principal Stresses: The coordimcs of poms A and 8 represent a. and 0:. respectively. a. I 93.9 MP; Am 0: 3 0 WI AM Moxitnmn ln-Plone Shear Stress: Represented by point E on the circle. f .“ ska-47.0w: All! ”-88. Draw the three Mohr‘s circles that describe each of the following states of stress. 6 ksi lOOpsi so MPa (b) 4? man) 1 TM) (.5) a) a... n6ksi a... so... -0 b) awasowh emu-0 ems-40w: c) a.“ = 6!!) psi a,“ I Mp3: a... 8 IN psi a... - 6 ksi a... = 50 MP: 0.“ = 600 psi an: ' can - 0 an: 8 0 a... = -40 MPa out ' MP“ on. = 'w PSI (e) CW) 12-6. Determine the equations of the elastic curve for the beam using the x. and x3 coordinates Specify the beam’s maximum deflection. E] is constant. Support Reactions and Blade Corn: As shown on FED“). Mellon: Function: As shown on F300) aid (c) . Slope all Elam: Curve: 4% a? ”((3) P Pablo.) I -§x. . (‘0, P a? ---2’" do P adj-74w. m P 1 Eu. - -3“ +C,x, +6, [2] F“M(x’) 3P3, -3—':L , J’v, 3PL ”T; ”’3 ’7 du P 3PL 53-:- - 5:: -Tx, +c, [3) H u, I: gt: - \$13.} Qt, +C‘ [4] Boundary Condition: 0' .0 "1' .0. PM EQJZI. C3 . 0 ”'30‘3"L melZ] PL’ Pl} 08 -1_2 +CIL C. 3 '17 0,3031’.L me [4], PL’ 3PL’ 03 T-T+C,L+C‘ 7PL’ 0=-—+C,L+C. [5] Douala) Conditions: v. =0 at. 80. From Eq.[2]. C; 8 0 PL’ PL’ os-l—2+C.L c. . T2- 0=-—+C,L+C. [5] 1." Elastic CU"... Substitute the m 0! C. . C3.C, .IndC. Continuity Coalition: in: Eqs.[2] aid [4] respectively do. 40, A! x I: I L. — g _. . P: ' , 4.. 4., memm- v. =—m;,(-x:+v) m _PL= P13 P13 3m m} u, . 9,1. , ¢ = fl -2 +1} g ”32"" T‘H'T‘T *C’ C"? ‘7‘ ‘2” 3 ‘7 PL, 0 g P 2‘, 2 3 , From Eq. [5]. C‘ ._T 3 ~IZE( 3 - 91:, +101. 33 -31. ) All: nu Slopc: Substime the value etc. into law 1. ”c = ”3 I.,.;L = L 2(3L)’ 9L 31..)24-10L2 3 L ’ ‘11. P (13-341) 12a 2 (2 (i )‘3” 4:, 12a ' pp 40 P 1, aka I 2 3 — .0. _ - — 4:. 12a“ 3") " ' J3 PL’ Hence! v..‘ 3 0c . — A” SE! *12—16. A torque wrench is used to tighten the nut on a bolt. if the dial indicates that a torque of 60 lb ' ft is applied when the bolt is fully tightened. determine the force P acting at the handle and the distance 3 the needle moves along the scale. Assume only the portion A8 of the beam distorts. The cross section is square having dimensions of 0.5 in. by 0.5 in. E = 29(10‘) ksi. C©_- A lain. 72o 1m V‘" C I) Moosmx-Zw K 400% Equation: of Equilibrium: Fm FUD“). (am-o. mama-o P-40.0Ib Ans «(Tug-0. A,-d0.0-0 5-40.05 Moment Function: N shown on FED“). Slope and Elastic Curve: a—-20.ox'-1mx+c. m a» -6.661x’-360x'+c.x +c, [2] Boundary Conditions: g 8 0 at x a Oandu U 0 a x I 0. From Eq.[l] 0:0-0+C, C. 80 From aq.(2| Ono—mow, c, so The Elastic Curve: Substitute the vain: of C. and C, in» Eq.[2|. u- -£'—I(6.667x’-360:’) tll Atxle‘uL. v=-:. FromEq.[l]. t (29)( IO‘)( ,5) (0.5)(05’) [6.667(12’) -360( t2’)] 3 a 0.267 it. All: *12—28. Determine the elastic curve for the cantilevered beam using the x coordinate. Also determine the maximum slope and maximum deflection. E] is constant. 4'0 d’o__3£ .mfiE--§§L+q (0 fl m 310- _wr’ 9-,: tau ” ’ ‘) o —w. * mm. 5L " l)MflF' d—B-0atx-L ”WW 4: "3 3mm % I On: I- L Froan.(l). 0- --;f-L(L‘)+c.: C. - 3%?- 0-308an Fmfiqom. _ We s We!) , 0 - _120L“' )+ TA (0+ Ca . “can”: FmBq.(l). 4” _ V0 _ 4 a a 24311.‘ ’ H" a 0... . .2...“ - %-;I Nautical-vs: Emma). W0 4 s nosu.‘ ’ ’ o (2) 30 All: All: All ...
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## Rubric
2 – This web illustrates the relationship between perspective, questions, and conclusions and provides a valid example.
1 – This web illustrates the relationship between perspective, questions, and conclusions but fails to offer a valid example.
## Strategy 6: Application Construct a Timeline
This shows students how a single source might be used to generate different interpretations.
Arrange students in small groups and ask them to use the “Statement of the Case,” Appendix 1, and the “Event Strips” on Appendix 2 to create a chronology of events leading up to the “massacre.” Distribute the Appendix with the event strips and scissors. Ask the students to cut out individual strips and lay them out chronologically on their desks/tables. Have them place events that suggest the British were to blame on the top of the timeline. Place the events that suggest the colonists were to blame on the bottom of the timeline (see illustration below).
Events damaging to British
Earlier -------------------------------------------------------------------------------Later
Events damaging to Colonists
After students complete their timelines, ask:
1. Based on the limited number of events provided on the sentence strips, who appeared to be more to blame for the casualties?
2. How does this activity illustrate the point that the manner in which a person uses sources (e.g., a timeline) can explain why historians sometimes arrive at different conclusions?
## Mapping the Scene
Distribute copies of Appendix 3 and ask students to draw their mental maps of King Street at the moment when a British soldier fired the very first shot. Tell the students to limit what they put on their maps to (using the key below):
• P = Captain Preston (noting his position relative to the soldiers and colonists)
• S = Soldiers (noting their position, e.g., straight line, 2 rows, semi-circle, as well as how many. One “S” for each soldier.)
• C = Colonists (noting how many and their location. One “C” for each colonist.)
Emphasize that their maps must be as accurate as possible with particular emphasis on numbers and locations (how many were there and where were they standing). Also, emphasize that their maps must represent the scene at the very moment the first shot was fired.
Give students time to complete their maps. Walk around the room in search of students who arrive at different interpretations (e.g., soldiers were in a straight line versus soldiers were in 2 rows; Preston stood to the side of his soldiers versus in front of them). When students complete their maps, ask 3 students who produced different “accounts” to come up and draw their map on the board. Have them label their maps Interpretation 1, 2, and 3.
Raise the following questions to the entire class:
• Are the 3 interpretations the same or different?
• In what ways are the interpretations different?
• Why might there be different interpretations given that the 3 students all relied on the same source of information, i.e., the Statement of the Case? (Again, for purposes of the benchmark, emphasize that they used the same source—Statement of the Case—differently and may have asked different questions, e.g., where would I stand, or would it have been possible for the soldiers to maintain a disciplined formation?)
## Error Check
Two historians using the same source of information will arrive at the same conclusion about the past. Agree or disagree with this statement and explain why.
## Rubric
2 – This response recognizes the error with an accurate and relevant explanation.
1 – This response fails to recognize the error or provides an inaccurate, irrelevant, or no explanation.
## Essential Question
• Why might historians disagree about the same historical event?
• How might the sources that an investigator relies on influence the interpretations at which he or she arrives?
## Strategy 1: Gathering Information Analyze Eyewitness Accounts to Learn Roles
The main activity in this lesson is a mock trial. See Appendix 5, Prosecution Team Packet, and Appendix 6, Defense Team Packet, for a list of witnesses who might testify at the trial and witness depositions. The depositions were taken in the hours and days following the “massacre” and will serve as the witness statements.
Divide the class into prosecution and defense teams, assign roles (e.g., attorneys, witnesses, jurors, bailiff) and distribute the primary sources’ depositions or witness statements to the appropriate students. There are enough roles so that you can conduct a “bench trial” where the judge determines the verdict and everyone can play a witness or attorney. This eliminates the challenge of keeping jurors engaged while the other students prepare their roles. Alternatively, you might invite a colleague’s students to serve as jurors.
## Recommended Attorney-Witness Groupings
• Attorney 1a – receives witness statement. Prepares questions and responses with witness. Conducts direct examination of witness.
• Witness 1 – receives same witness statement. Prepares questions and responses with Attorney 1.
• Attorney 1b – receives same witness statement to develop cross-examination questions but does not get to prepare with Witness 1.
You will also need to select attorneys to give the opening (1 prosecution, 1 defense) and closing (1 prosecution, 1 defense) statements.
A teacher or other knowledgeable authority figure should serve as the judge. You might also recruit an actual judge.
Provide an overview of the case so that everyone understands the purpose of the trial. Consider reading the “Jury Instructions” (Appendix 7) and “Stipulated Facts” (Appendix 10). Briefly, Captain Preston has been charged with the crime of manslaughter on the assumption that he gave his men an illegal order to fire on a crowd of civilians on the night of March 5, 1770. Under the law at the time, it was illegal to give such an order unless:
1. A civilian authority (e.g., the governor) gave him permission.
2. He or his men were threatened with death or serious bodily injury.
If the prosecution cannot convince the judge (or jury) that Captain Preston did not order his men to fire or that he either had permission from a civilian authority or he or his men were threatened with death or serious bodily injury, Preston must be found guilty.
So, the central question that students must address is: did Captain Preston give an illegal order to fire their weapons into a crowd of civilians?
Give “friendly” attorneys time with their witnesses to prepare and rehearse their questions and responses. The “friendly” attorneys should coach their witnesses by asking questions in advance. Those “adversaries” (attorneys) who are assigned the role of cross-examining witnesses should receive the witness statement of the person they will cross examine so that they can prepare their cases but should not be given the opportunity to work with the witness prior to the trial.
The “Simplified Steps in a Mock Trial” (Appendix 9) offer a nice overview of trial procedures. Take time to review these with the students. Appendix 11 (optional) offers Tips for Students as they prepare their varied roles. | 1,489 | 7,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-40 | latest | en | 0.922943 |
https://www.reference.com/web?q=parallel+and+perpendicular+lines+lesson&qo=relatedSearchBing&o=600605&l=dir&sga=1 | 1,586,200,465,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371656216.67/warc/CC-MAIN-20200406164846-20200406195346-00558.warc.gz | 1,110,391,978 | 27,197 | Web Results
educators.brainpop.com/bp-topic/parallel-and-perpendicular-lines
In this educational resource page you will find lesson plans and teaching tips about Math learn about intersecting lines, transversals, angles, 2-dimensional, and lines. ... Educator Resources for Parallel and Perpendicular Lines ... BrainPOP Educators is proudly powered by WordPress and Piklist.
www.homeschoolmath.net/teaching/g/parallel_and_perpendicular.php
Parallel and Perpendicular Lines. This lesson explains what are parallel and perpendicular lines and has varied exercises for the students. The lesson also includes a video where I show how to draw a perpendicular line and a rectangle using a protractor or a triangular ruler.
Looking for a simple way to jazz up your instruction on parallel and perpendicular lines? This lesson plan will do just that with the help of an engaging Study.com video lesson and a simple yet ...
www.lessonplanet.com/lesson-plans/perpendicular-lines/all
In this parallel and perpendicular lines worksheet, learners find the slope of a line that is perpendicular or parallel to a given line. They write a linear equation in point-slope form that fits given parameters. This one-page worksheet...
www.tes.com/.../gcse-maths-equations-of-parallel-and-perpendicular-lines-11670918
This is a full lesson that I’ve made on finding the equation of parallel and perpendicular lines. It goes through how to find equations of lines which are either parallel or perpendicular and pass through a specific point.
numberock.com/lessons/lines
You'll be amazed as you stream this math music video about types of lines, in that it is as entertaining as it is informative. Your students will be singing & dancing, all the while learning or reinforcing the knowledge of parallel, perpendicular, and intersecting lines.
www.lessonplanet.com/teachers/parallel-and-perpendicular-lines-4th
This Parallel and Perpendicular Lines Lesson Plan is suitable for 4th Grade. Fourth graders examine parallel, perpendicular and intersecting lines. They create a drawing using examples of each type of line. | 424 | 2,105 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-16 | latest | en | 0.86234 |
http://encyclopedia2.thefreedictionary.com/Boolean+operators | 1,498,702,950,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323842.29/warc/CC-MAIN-20170629015021-20170629035021-00166.warc.gz | 125,821,837 | 11,254 | # Boolean function
(redirected from Boolean operators)
## Boolean function
[′bü·lē·ən ′fəŋk·shən]
(mathematics)
A function f (x,y,…,z) assembled by the application of the operations AND, OR, NOT on the variables x, y,…, z and elements whose common domain is a Boolean algebra.
References in periodicals archive ?
Provides advanced search options such as field searching, Boolean operators, proximity searching, nesting, wildcard/truncation, etc.
Brandwatch leads the pack with 22 operators, the largest range of Boolean operators of any platform on the market
Terms can be combined using Boolean operators to refine the search.
Among the specific aspects she addresses are identifying valid online bibliographic and article databases, using Boolean operators to refine a search, organizing the research literature by using bibliographic software, preparing a structured abstraction form, synthesizing and reporting results as part of proposals and papers or as a stand-alone report, evaluating qualitative research studies, and understanding and evaluating meta-analysis research.
Compare the effects of the use of the Boolean operators AND and OR.
If the defect is positioned in an area that has its "mirror image" on another side of the body than the form of the implant can relatively easily be produced by means of Boolean operators.
When you enter multiple criteria, you get real results without having to be a mathematician specializing in Boolean operators.
This is a simple two minute mini-crossword puzzle designed to reinforce the nuanced Boolean Operators the lecture introduced: Near, Next, Proximity.
And proposes to use more number of Boolean operators (AND, OR, XOR, OF, and NOT) instead of the standard Boolean operators (AND, OR, and NOT), and use weights for Boolean operators and for terms in fuzzy models.
This activity offers students an opportunity to select appropriate databases, use advanced search techniques, and practice how to phrase search terms using Boolean operators to locate peer reviewed journal articles that meet criteria for scholarly research.
Net users have become more sophisticated and adept at using keywords and Boolean operators to construct more effective queries.
Boolean operators and phrase searching with quotations were used a total of eleven times.
Site: Follow: Share:
Open / Close | 450 | 2,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-26 | latest | en | 0.882324 |
https://www.inchcalculator.com/convert/kiloelectronvolt-to-calorie/ | 1,721,769,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00624.warc.gz | 702,551,580 | 15,856 | # Kiloelectronvolts to Calories Converter
Enter the energy in kiloelectronvolts below to get the value converted to calories.
## Result in Calories:
1 keV = 3.8293E-17 cal
Hint: use a scientific notation calculator to convert E notation to decimal
Do you want to convert calories to kiloelectronvolts?
## How to Convert Kiloelectronvolts to Calories
To convert a measurement in kiloelectronvolts to a measurement in calories, divide the energy by the following conversion ratio: 2.6114E+16 kiloelectronvolts/calorie.
Since one calorie is equal to 2.6114E+16 kiloelectronvolts, you can use this simple formula to convert:
calories = kiloelectronvolts ÷ 2.6114E+16
The energy in calories is equal to the energy in kiloelectronvolts divided by 2.6114E+16.
For example, here's how to convert 5.0E+16 kiloelectronvolts to calories using the formula above.
calories = (5.0E+16 keV ÷ 2.6114E+16) = 1.914647 cal
## What Is a Kiloelectronvolt?
Kiloelectronvolts can be abbreviated as keV; for example, 1 kiloelectronvolt can be written as 1 keV.
## What Is a Calorie?
One calorie is equal to the heat needed to raise the temperature of one gram of water from 14.5 to 15.5 degrees Celsius.
When measuring the energy content in foods, the large calorie, or kilocalorie is used, but it's often still just called a "calorie." When both the calorie and large calorie are used in the same context it is common to refer to the calorie as a "small calorie."
Calories can be abbreviated as cal; for example, 1 calorie can be written as 1 cal.
## Kiloelectronvolt to Calorie Conversion Table
Table showing various kiloelectronvolt measurements converted to calories.
Kiloelectronvolts Calories
1 keV 0.000000000000000038293 cal
2 keV 0.000000000000000076586 cal
3 keV 0.00000000000000011488 cal
4 keV 0.00000000000000015317 cal
5 keV 0.00000000000000019146 cal
6 keV 0.00000000000000022976 cal
7 keV 0.00000000000000026805 cal
8 keV 0.00000000000000030634 cal
9 keV 0.00000000000000034464 cal
10 keV 0.00000000000000038293 cal
100 keV 0.0000000000000038293 cal
1,000 keV 0.000000000000038293 cal
10,000 keV 0.00000000000038293 cal
100,000 keV 0.0000000000038293 cal
1,000,000 keV 0.000000000038293 cal
10,000,000 keV 0.00000000038293 cal
100,000,000 keV 0.0000000038293 cal
1,000,000,000 keV 0.000000038293 cal
10,000,000,000 keV 0.00000038293 cal
100,000,000,000 keV 0.0000038293 cal
1,000,000,000,000 keV 0.000038293 cal
10,000,000,000,000 keV 0.000383 cal
100,000,000,000,000 keV 0.003829 cal
1,000,000,000,000,000 keV 0.038293 cal
10,000,000,000,000,000 keV 0.382929 cal
100,000,000,000,000,000 keV 3.8293 cal | 880 | 2,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.757324 |
https://eduzip.com/ask/question/define-dimension-of-a-physical-quantity-274112 | 1,638,580,749,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00458.warc.gz | 302,415,687 | 9,022 | Physics
# Define dimension of a physical quantity.
##### SOLUTION
The dimension of physical quantity may be defined as the number of times the fundamental units of mass, length and time appear in the physical quantity
You're just one step away
Subjective Medium Published on 18th 08, 2020
Questions 244531
Subjects 8
Chapters 125
Enrolled Students 202
#### Realted Questions
Q1 Single Correct Medium
Which of the following physical quantity is scalar quantity?
• A. Mass
• B. Force
• C. Impulse of force
• D. Momentum
Asked in: Physics - Units and Measurement
1 Verified Answer | Published on 18th 08, 2020
Q2 Single Correct Hard
A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90\ s, 91\ s, 95\ s$ and $92\ s$. If the minimum division in the measuring clock is $1\ s$, then the reported mean time should be:
• A. $92\pm 2\ s$
• B. $92\pm 5.0\ s$
• C. $92\pm 1.8\ s$
• D. $92\pm 3\ s$
Asked in: Physics - Units and Measurement
1 Verified Answer | Published on 18th 08, 2020
Q3 Single Correct Medium
The number of significant figures in the numbers $672.9$ and $2.520\times { 10 }^{ 7 }$ are :
• A. $4, 4$
• B. $3, 4$
• C. $4, 3$
• D. $3, 3$
Asked in: Physics - Units and Measurement
1 Verified Answer | Published on 18th 08, 2020
Q4 Subjective Medium
A gas bubble, from an explosion under water, oscillates with a period $T$ proportional to $P^a\;d^b\;E^c$, where $P$ is the static pressure, $d$ is the density of water and $E$ is the total energy of the explosion. The values of $a$.......... $b$........... and $c$............ .
Asked in: Physics - Units and Measurement
1 Verified Answer | Published on 18th 08, 2020
Q5 Subjective Easy
Fill in the blanks with suitable units of measurements.
The distance I walk in $10$ minutes is about one _____
Asked in: Physics - Units and Measurement
1 Verified Answer | Published on 18th 08, 2020 | 575 | 1,917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-49 | latest | en | 0.746904 |
https://physicslens.com/category/a-level-topics/10-oscillations/ | 1,606,809,667,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141672314.55/warc/CC-MAIN-20201201074047-20201201104047-00421.warc.gz | 456,017,068 | 16,653 | # 10 Oscillations
## Pendulum-Powered Car
This pendulum-powered car is constructed using Lego Technic parts. I used mainly Lego beams to create the chassis and an “A” frame from which the pendulum is suspended. The pendulum is made of Lego beams and some wheels.
When the pendulum swings, it experiences an acceleration towards its equilibrium position. By the principle of conservation of momentum, the car experiences a change in momentum in the opposite direction. Since the acceleration of the pendulum changes its direction every half a cycle of its oscillation, the car will only oscillate about its original position if the wheels of the car are free to turn throughout the oscillation.
A escapement mechanism which consists of a beam resting on a pair of 40-tooth gears attached to the front wheels prevent the wheels from rotating in the opposite direction. This means that the car will only be moving forward during the half of the pendulum’s oscillation when its displacement is at the front of its equilibrium position and pauses during the other half.
## Simple harmonic motion graphs including energy
I have added two more graphs into the interactive animation. However, the app has become a bit sluggish when changing the period or amplitude. It still works smoothly when viewing the animation.
Students ought to find it useful to look at all the graphs together instead of in silo. This way, they can better understand the relationships between the graphs.
As usual, here is the animated gif file.
## Simple Harmonic Motion Graphs
Here’s my attempt at animating 5 graphs for simple harmonic motion together in one page.
From left column:
$$v = \pm\omega\sqrt{x_o^2-x^2}$$
$$a = -\omega^2x$$
From right column:
$$s = x_o\sin(\omega t)$$
$$v = x_o\omega \cos(\omega t)$$
$$a = -x_o\omega^2 \sin(\omega t)$$
And here is the animated gif file for powerpoint users: | 423 | 1,894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-50 | latest | en | 0.916883 |
https://math.answers.com/Q/How_many_inches_are_there_in_2_feet_6.5_inches | 1,713,255,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00739.warc.gz | 343,725,131 | 46,362 | 0
How many inches are there in 2 feet 6.5 inches?
Updated: 9/21/2023
Wiki User
12y ago
There are 12 inches in one foot. Therefore, 2 feet 6.5 inches is equal to 2 x 12 + 6.5 = 30.5 inches.
Wiki User
12y ago | 80 | 213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.886894 |
https://autarkaw.org/category/uncategorized/page/6/ | 1,576,217,281,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00388.warc.gz | 294,361,642 | 25,040 | ## An experiment to illustrate numerical differentiation, integration, regression and ODEs
Starting Summer 2007, five experiments have been introduced in the course in Numerical Methods at USF. I will discuss each experiment in a separate blog as the
summer trods along.
Experiment#1: Cooling an aluminum cylinder
The first experiment illustrates use of numerical differentiation, numerical integration, regression and ordinary differential equations. In this experiment, an aluminum cylinder is immersed in a bath of iced water. As you can see in the figure, two thermocouples are attached to the cylinder and are connected to a temperature indicator. Readings of temperature as a function of time are taken in intervals of 5 seconds for a total of 40 seconds. The temperature of the iced-water bath is also noted.
If you just want the data for a typical experiment conducted in class, click here and here for data.
The students are now assigned about 10 problems to do. These include
1. finding the convection coefficient (involves nonlinear regression – it is also a good example of where the data for a nonlinear model does not need to be transformed to use linear regression)
2. finding the rate of change of temperature to calculate rate at which is heat is stored in the cylinder (involves numerical differentiation)
3. prediction of temperatures from solution of ordinary differential equations
4. finding reduction in the diameter of the aluminum cylinder (involves numerical integration as the thermal expansion coefficient is a function of temperature)
This post brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## Rusty on Matrix Algebra
Eight years ago, the Florida legislature decided to reduce the number of credit hours it takes a state university student to graduate with an undergrad engineering degree. The number of credit hours were reduced from 136 to 128. One of the courses that got the ax in the Mechanical Engineering Department at USF was a 2-credit hour Linear Algebra course. There are many other universities in the nation that have done the same.
So how do students learn Linear Algebra when the course is one of the requirements for accreditation of engineering programs?
Some universities have bundled Linear Algebra course content into courses such as Quantitative Methods where students are expected, in many cases, to learn linear algebra, a programming language/computational system, and complex analysis. Other curriculums have dispersed the Linear Algebra content into different courses such as the topic of special matrices in Programming, simultaneous linear equations in Statics, and eigenvalues/eigenvectors in Vibrations, etc. Unless quality controls are introduced carefully, the content/depth of Linear Algbera in such courses can vary substantially between courses and instructors. Such control is impossible in metropolitan universities such as USF where a large proportion of students transfer from community colleges.
To have a resource that would be a self-explanatory as well as get the students exposed to Linear Algebra applications motivated me to write a simple Introduction to Matrix Algebra book. The book consists of ten chapters spanning fundamentals of matrix algebra, numerical methods for solving a set of equations, and a treatment of adequacy of solutions and eigenvalues.
Since 2002, the Introduction to Matrix Algebra book has been downloaded free of charge by more than 30,000 users from 50 different countries, and the feedback has been humbling and fulfilling.
Since April 2008, the book has also been made available for a nominal charge via lulu.com as a pdf file as well as a soft cover book. Proceeds from the book are allowing me to expand the book with more examples/problems and additional chapters.
Since my belief continues to embrace open and uncomplicated dissemination, eight individual chapters of the book in pdf form are still available free of charge. So one may ask the following question. Why should I buy the book when it is available free of charge? For answer to this question, click here
For more details about the book, visit the book website at http://autarkaw.com/books/matrixalgebra/index.html
This post brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## Round off errors and the Patriot missile
Twenty-eight Americans were killed on February 25, 1991 when an Iraqi Scud hit the Army barracks in Dhahran, Saudi Arabia. The Patriot defense system had failed to track and intercept the Scud. What was the cause for this failure?
The Patriot defense system consists of an electronic detection device called the range gate. It calculates the area in the air space where it should look for the target such as a Scud. To find out where the Patriot missile should be next, it calculates its location based on the velocity of the Scud and the last time the radar detected the Scud.
In the Patriot missile, time was saved in a fixed point register that had a length of 24 bits. Since the internal clock of the system is measured every one-tenth of a second, 1/10 expressed in a 24 bit fixed point register is 0.0001100110011001100110011 (the exact value of the representation 0.0001100110011001100110011 of 1/10 in the 24-fixed point register is 209715/2097152) . As we can see that this is not an exact representation of 1/10. It would take infinite numbers of bits to represent 1/10 exactly. So, the error in the representation is (1/10-209715/2097152) which is approximately 9.5E-8 seconds.
On the day of the mishap, the battery on the Patriot missile was left on for 100 consecutive hours, hence causing an inaccuracy of 9.5E-8x10x60x60x100=0.34 seconds (10 clock cycles in a second, 60 seconds in a minute, 60 minutes in an hour).
The shift calculated in the range gate due to the error of 0.342 seconds was calculated as 687m. For the Patriot missile defense system, the target is considered out of range if the shift is more than than 137m. The shift of larger than 137m resulted in the Scud not being targeted and hence killing 28 Americans in the barracks of Saudi Arabia.
When I started looking at the Google search results of the problem, I found some very useful resources that would be of interest to the reader. These go beyond the above given simplistic explanation of the problem and tell the story behind the story. Here they are
1. This reference is the full GAO report of the investigation that resulted after the accident. “Patriot Missile Defense – Software Problem Led to System Failure at Dhahran, Saudi Arabia”, GAO Report, General Accounting Office, Washington DC, February 4, 1992.
2. It should be pointed out that the Patriot missile was originally designed to be a mobile system and not used as a anti-ballistic system. In mobile systems, the clocks are reset more often. As per the article Operations: I Did Not Say You Could Do That! by Bill Barnes and Duke McMillin, here are some important observations: “It turns out that the original use case for this system was to be mobile and to defend against aircraft that move much more slowly than ballistic missiles. Because the system was intended to be mobile, it was expected that the computer would be periodically rebooted. In this way, any clock-drift error would not be propagated over extended periods and would not cause significant errors in range calculation. Because the Patriot system was not intended to run for extended times, it was probably never tested under those conditions—explaining why the problem was not discovered until the war was in progress. The fact that the system was also designed as an antiaircraft system probably also enabled the inclusion of such a design flaw, because slower-moving airplanes would be easier to track and, therefore, less dependent upon a highly accurate clock value.”
A student asked me why we did not use a clock cycle that could be represented exactly in the 24 bit register. Close to 1/10 is a number 0.125 that can be represented exactly as 0.001000000000000000000000 in a 24-bit register, and where 8 clock cycles would be equal to 1 second. I do not have an answer to this question but I intend to find out from my computer science colleagues.
This post brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu
## Undergraduate Numerical Methods for Engineering
I am starting this blog to help UNDERGRADUATES with their queries on Numerical Methods for Engineers. I have been teaching Numerical Methods for the last 20 years and I get interesting queries and questions while I am teaching, when students come to see me during my office hours, or the email sent at midnight before the assignment is due.
I am keeping a log of what students ask me and will note the answers to their queries here. I am sure that students elsewhere have similar questions when they take a course in Numerical Methods.
The diversity of the course is quite evident –
1. The course is taught to different engineering majors – mechanical, civil, chemical, industrial and electrical.
2. Some teachers emphasize the numerical methods while others spend more time on solving physical problems, and a few may include numerical analysis.
3. The programming tools are diverse including FORTRAN (yes the language is alive and well), Basic, C, Java, or computational packages such as MATLAB, MATHEMATICA, MathCAD, and Maple.
With funding from NSF since 2002, we have developed web-based resources for a course in Numerical Methods. The inclusion of the blog is not part of the funded proposals but we think that this mode of Web 2.0 dissemination is critical in keeping the conversation going on. Although what I am doing here can be offered via a static website, the widgets offered by blogging softwares are indispensable. The widgets I like are categorizing, tagging and RSS Feeds.
We want to reach as many people as possible and build a community which may be temporary to students who are taking a course in Numerical Methods, permanent to instructors and people who use numerical methods in their work. But one thing is certain, temporary or permanent, visitors will leave their imprint on this resource.
This post brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://numericalmethods.eng.usf.edu | 2,183 | 10,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-51 | latest | en | 0.923858 |
https://estebantorreshighschool.com/equation-help/marginal-propensity-to-consume-equation.html | 1,618,084,477,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038057476.6/warc/CC-MAIN-20210410181215-20210410211215-00262.warc.gz | 351,345,868 | 30,187 | ## How do you calculate MPS and MPC in economics?
Also, marginal propensity to save is opposite of marginal propensity to consume. Mathematically, in a closed economy, MPS + MPC = 1, since an increase in one unit of income will be either consumed or saved. In the above example, If MPS = 0.4, then MPC = 1 – 0.4 = 0.6.
## What is the formula for MPS in economics?
How Marginal Propensity to Save Is Calculated. MPS is most often used in Keynesian economic theory. It is calculated simply by dividing the change in savings observed given a change in income: MPS = ΔS/ΔY.
## When MPC is 0.8 What is the multiplier?
With an MPC of 0.8 (saving 20% of your income), this would yield a multiplier of 5.
## What is the multiplier equation?
The formal calculation for the value of the multiplier is. Multiplier = 1 / (sum of the propensity to save + tax + import) Therefore if there is an initial injection of demand of say £400m and. The marginal propensity to save = 0.2. The marginal rate of tax on income = 0.2.
## What is the formula for the multiplier effect?
The Multiplier Effect Formula (‘k’) MPC – Marginal Propensity to Consume – The marginal propensity to consume (MPC) is the increase in consumer spending due to an increase in income. This can be expressed as ∆C/∆Y, which is a change in consumption over the change in income.
## Why can’t MPC be negative?
No, neither MPS nor MPC can ever be negative because MPC is the ratio of change in the consumption expenditure and change in the disposable income. In other words, MPC measures how consumption will vary with the change in income.
## How do you calculate mean and MPS?
MPS FormulaMean = Total Scores/Raw Scores. No. of Pupils.MPS = Mean________ X 100. No. of Test Items. AGE COMPUTATION TABLE.Formula of Age as of date of weighing.AGE = date of weighing (YY-MM-DD) – date of birth_ – (YY_MM_DD)ex. July 15, 2010 ———- 2010-7-15. ex. July 7, 2010 ———— 2010- 7- 7. BMI = Mass (kg.) Height (m2)
You might be interested: Regression equation examples
## How can GDP be calculated?
Written out, the equation for calculating GDP is: GDP = private consumption + gross investment + government investment + government spending + (exports – imports). For the gross domestic product, “gross” means that the GDP measures production regardless of the various uses to which the product can be put.
## When the MPC 0.75 The multiplier is?
If the MPC is 0.75, the Keynesian government spending multiplier will be 4/3; that is, an increase of \$ 300 billion in government spending will lead to an increase in GDP of \$ 400 billion. The multiplier is 1 / (1 – MPC) = 1 / MPS = 1 /0.25 = 4.
## When MPC is 0.5 What is the multiplier?
The marginal propensity to consume (MPC) measures how consumer spending changes with a change in income. Using the figures above, the MPC is ΔC / ΔY = 300/600 = 0.5. This means that every \$1 of new income will generate \$2 of extra income.
## When the MPC 0.6 The multiplier is?
Therefore, the investment multiplier is 2.5.
## Why does MPC lie between 0 and 1?
Consumption is the major component of aggregate demand. Mind, MPC is always greater than zero (MPC > 0) and less than 1 (MPC < 1) because additional consumption (∆C) is less than additional income (∆Y). Higher MPC implies increase in consumption demand. According to Keynes, ‘Demand creates its own supply.
## How do you increase marginal propensity to consume?
Readily available credit and lower interest rates are believed to increase the MPC since this makes it easier for consumers to finance purchases and to obtain financing at attractive rates.
### Releated
#### Tensile stress equation
What is the formula for tensile stress? Difference Between Tensile Stress And Tensile Strength Tensile stress Tensile strength The formula is: σ = F/A Where, σ is the tensile stress F is the force acting A is the area The formula is: s = P/a Where, s is the tensile strength P is the force […]
#### Quotient rule equation
What is the formula for Quotient? The answer after we divide one number by another. dividend ÷ divisor = quotient. Example: in 12 ÷ 3 = 4, 4 is the quotient. What is quotient rule in math? The quotient rule is a formal rule for differentiating problems where one function is divided by another. It […] | 1,065 | 4,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-17 | longest | en | 0.908519 |
https://morethingsjapanese.com/are-two-lines-that-intersect-skew/ | 1,721,313,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00394.warc.gz | 350,188,389 | 40,394 | # Are two lines that intersect skew?
## Are two lines that intersect skew?
Vocabulary
Term Definition
transversal A transversal is a line that intersects two other lines.
Parallel Two or more lines are parallel when they lie in the same plane and never intersect. These lines will always have the same slope.
Skew To skew a given set means to cause the trend of data to favor one end or the other
What does it mean if lines are skew?
Two or more lines which have no intersections but are not parallel, also called agonic lines. Since two lines in the plane must intersect or be parallel, skew lines can exist only in three or more dimensions.
### Can skew lines be perpendicular?
Skew lines can be perpendicular. Planes can be parallel. Parallel lines are never in the same plane.
How do you know if a line is skew?
Skew lines are two or more lines that do not intersect, are not parallel, and are not coplanar. (Remember that parallel lines and intersecting lines lie on the same plane.) This makes skew lines unique – you can only find skew lines in figures with three or more dimensions.
## Are intersecting lines coplanar?
Intersecting: The two lines are coplanar (meaning that they lie on the same plane) and intersect at a single point.
What do you call the intersecting lines that form four right angles?
Perpendicular lines (or segments) actually form four right angles, even if only one of the right angles is marked with a box.
### Are intersecting lines always perpendicular?
Properties of Perpendicular Line Perpendicular lines always intersect each other, however, all intersecting lines are not always perpendicular to each other. The two main properties of perpendicular lines are: Perpendicular lines always meet or intersect each other.
What are intersecting lines?
When two or more lines cross each other in a plane, they are called intersecting lines. The intersecting lines share a common point, which exists on all the intersecting lines, and is called the point of intersection. Here, lines P and Q intersect at point O, which is the point of intersection.
## What are intersecting lines Class 7?
Intersecting lines: Two lines intersect if they have a point in common. This common point O is their point of intersection. Transversal: A line that intersects two or more lines at distinct points is called a transversal. Pairs of interior angles on the same side of the transversal.
Do two lines in intersecting planes are never skew?
A necessary condition for two lines to intersect is that they are in the same plane-that is, are not skew lines . Satisfaction of this condition is equivalent to the tetrahedron with vertices at two of the points on one line and two of the points on the other line being degenerate in the sense of having zero volume.
### Are skew lines never parallel?
Skew lines are lines that are in different planes, they are never parallel , and they never intersect. On the other hand, parallel lines are lines that are in the same plane and never intersect. In other words, Parallel lines must exist in two dimensions; they are parallel within the same plane. Skew lines cannot exist in two dimensions and are always in different, non-intersecting planes.
How are skew lines are different from parallel lines?
Skew lines are lines that are in different planes and never intersect. The difference between parallel lines and skew lines is parallel lines lie in the same plane while skew lines lie in different planes. A line is said to be perpendicular to another line if the two lines intersect at a right angle.
## Where do two distinct parallel lines intersect?
In projective geometry, any pair of lines always intersects at some point, but parallel lines do not intersect in the real plane. The line at infinity is added to the real plane. This completes the plane, because now parallel lines intersect at a point which lies on the line at infinity . | 810 | 3,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.920051 |
https://www.univerkov.com/what-is-the-ratio-of-the-bodys-acceleration-modulus-and-the-tangent-of-the-angle/ | 1,656,264,893,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00643.warc.gz | 1,135,211,727 | 6,028 | # What is the ratio of the body’s acceleration modulus and the tangent of the angle
What is the ratio of the body’s acceleration modulus and the tangent of the angle a of the slope of the velocity projection graph to the abscissa axis?
I a = 1m / s˄2; tg = 1; ratio -1
II a = 1 m / s˄2; tg = 1; ratio -1
III – a = 2 m / s˄2; tg = 2; ratio -1
IV – a = 2 m / s˄2; tg = 1; ratio -1
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 192 | 691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-27 | latest | en | 0.914434 |
https://digitalcommons.isical.ac.in/book-chapters/162/ | 1,726,003,468,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00442.warc.gz | 185,734,601 | 8,256 | # Prediction Approach: Robustness, Bayesian Methods, Empirical Bayes
Book Chapter
## Publication Title
Indian Statistical Institute Series
## Abstract
The classical design-based approach and its modification by the super-population modelling only narrates the properties of estimators for parameters with respect to selection and theoretical estimation, but nothing about how close is an estimate to the parameter it seeks to estimate or how far it is away from it as assessed from a sample at hand. A third alternative, for example, to estimate a population total observes that a sample at hand gives exactly the value of the sample total but can say nothing about how it differs from the population total unless a relationship among the values of the units in a population is postulated at the outset itself. So, one needs to suppose the vector Y̲=(y1,…,yi,…,yN) as a random vector implying Y=∑1Nyi is a random variable. Therefore, Y cannot be estimated but may be predicted by estimating the expectation of Y calculated on modelling the distribution of Y̲. So, a prediction approach is adopted. A fourth approach is Bayesian, postulating a prior distribution for Y̲ and then working out a posterior distribution and a posterior expectation of Y, given the sample data at hand. As certain parameters are involved in the prior and are unknowable in practice, they may be suitably estimated from the sample yielding an empirical Bayes approach. Prediction and Bayesian methods do not need probability sampling. The procedures have their strength only through dependence on the tenability of models. So, a problem of robustness ensues if a procedure is to retain its effectiveness against departures from a specific model postulated. In ensuring robustness, however, a role of a random selection of a sample may often be helpful as we shall see.
169
182
## DOI
10.1007/978-981-19-1418-8_8
1-1-2022
COinS | 405 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.904436 |
http://www.chegg.com/homework-help/questions-and-answers/a-submarine-descends-at-an-angle-of-30-below-the-horizontal-with-a-heading-to-the-northeas-q3530919 | 1,369,393,021,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704590423/warc/CC-MAIN-20130516114310-00085-ip-10-60-113-184.ec2.internal.warc.gz | 381,649,261 | 8,087 | ## calculus homework 5
a submarine descends at an angle of 30º below the horizontal with a heading to the northeast. if its speed is 16 knots, find the components of the velocity in the east, north and vertical directions. Fill in the blacks to complete the answer: the components of the velocity are (blank) knots to the east, (blank) knots to the north, and (blank) knots downward. PLEASE SHOW AS MUCH WORK AS POSSIBLE | 97 | 421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-20 | latest | en | 0.890437 |
https://cypenv.info/relationship-definition/relationship-between-free-energy-and-chemical-equilibrium-definition.php | 1,566,759,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330786.8/warc/CC-MAIN-20190825173827-20190825195827-00250.warc.gz | 423,972,000 | 9,423 | # Relationship between free energy and chemical equilibrium definition
### Gibbs free energy and spontaneity (article) | Khan Academy
Gibbs Free Energy: Definition & Significance. Equilibrium . in Reverse. When delta G standard equals Zero, the reaction is at Equilibrium. thermodynamics – the solution of chemical A criterion for equilibrium is that the total free energy. (Gibbs free using this equation: where. In chemistry, a spontaneous processes is one that occurs without the addition law of thermodynamics and define a new quantity known as Gibbs free energy: When using Gibbs free energy to determine the spontaneity of a process, we are . in the forward direction, backward direction, or if the reaction is at equilibrium.
We will see how to relate the free energy change to the extent of a chemical reaction. Standard Free Energy Change Say you go into a soda store in London and ask for their standard sized can of cola.
### Gibbs Free Energy and Equilibrium
You would assume that the can size would be the same as a standard can of cola you bought in the USA. Huh, so much for standard. In this case, a standard size is not really a standard at all. You are not sure what to expect when you ask for a standard can of cola in different countries.
### Free Energy and Equilibrium - Chemistry LibreTexts
In science and chemistry, we need to be more exact. When we say standard, we need everyone around the world to know exactly what standard means. Standards are defined precisely. Okay, so let us consider this a little further with our thermodynamics hat on.
You may recall that Gibbs free energy tells us whether a reaction is spontaneous or not. When we calculate the Gibbs free energy change of a reaction, we often use this equation: The condition of stable equilibrium is that the value of the expression in the parenthesis shall be a minimum. Thereafter, inthe German scientist Hermann von Helmholtz characterized the affinity as the largest quantity of work which can be gained when the reaction is carried out in a reversible manner, e.
Thus, G or F is the amount of energy "free" for work under the given conditions.
Until this point, the general view had been such that: Over the next 60 years, the term affinity came to be replaced with the term free energy. Lewis and Merle Randall led to the replacement of the term "affinity" by the term "free energy" in much of the English-speaking world. InAmerican scientist Willard Gibbs published his first thermodynamics paper, "Graphical Methods in the Thermodynamics of Fluids", in which Gibbs used the two coordinates of the entropy and volume to represent the state of the body.
In his second follow-up paper, "A Method of Geometrical Representation of the Thermodynamic Properties of Substances by Means of Surfaces", published later that year, Gibbs added in the third coordinate of the energy of the body, defined on three figures.
InScottish physicist James Clerk Maxwell used Gibbs' figures to make a 3D energy-entropy-volume thermodynamic surface of a fictitious water-like substance. If we could find some way to harness the tendency of this reaction to come to equilibrium, we could get the reaction to do work. The free energy of a reaction at any moment in time is therefore said to be a measure of the energy available to do work.
When a reaction leaves the standard state because of a change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction. The difference between Go and G for a reaction is important.
There is only one value of Go for a reaction at a given temperature, but there are an infinite number of possible values of G. The figure below shows the relationship between G for the following reaction and the logarithm to the base e of the reaction quotient for the reaction between N2 and H2 to form NH3. They therefore describe systems in which there is far more reactant than product.
The sign of G for these systems is negative and the magnitude of G is large.
## Enduring Understanding 6.D: Gibbs Free Energy and Equilibrium
The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium.
Data on the far right side of this figure describe systems in which there is more product than reactant. The sign of G is now positive and the magnitude of G is moderately large.
• Free Energy and Equilibrium
• Gibbs free energy
• Gibbs free energy and spontaneity | 930 | 4,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2019-35 | longest | en | 0.903237 |
https://gis.stackexchange.com/questions/81484/why-different-2d-coordinate-transformation-methods-exist?noredirect=1 | 1,716,931,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00093.warc.gz | 234,337,022 | 42,405 | # Why Different 2D Coordinate Transformation Methods Exist
It is a homework question but I couldn't find a satisfactory answer by googling. There are 4 different transformation approaches I'm asked for:
- Conformal,
- Affine,
- 2D Projective,
- 2D Polynomial
I know there are more than these. And I saw how they are implemented, the transformation process, their results etc. The thing I need to know is the task they are used. When should I use 2D Projective transformation and why shouldn't I use conformal transformation instead? What are the real world applications or field of applications for those transformation methods?
The questions above are not from the text I've given, I tried to make the topic clear. Thanks in advance.
Modern mathematics characterizes transformations in terms of the geometric properties that are preserved when the transformations are applied to features.
A time-honored example is the set of Euclidean transformations of the plane: these are the ones that preserve all distances and (unoriented) angles. The study of this group of transformations is the subject of Euclidean geometry and the source of all those congruence theorems about triangles we all learn in school. It turns out that these transformations have relatively simple and illuminating descriptions. For instance, they consist of all translations, rotations, improper rotations (a rotation followed by a reflection) and skew-translations (a translation followed by a reflection). They can also be characterized as all transformations achievable by applying a set of reflections. The point is that these descriptions of the Euclidean group are derived from its characterization as distance-preserving rather than the other way around.
For mapping, analysis, and georeferencing purposes we usually value one particular set of properties over all others: area when computing areas, orientation when computing directions, distance when computing distances, (local) angles when computing angles, similarity when comparing shapes, incidence and inside versus outside when performing topological comparisons, and so on. In each case there is a group of invertible transformations of the plane that preserves the desired properties.
There's nothing deep about this: if you specify any property of features in the plane, then when the transformation F preserves them and the transformation G preserves them, the transformation GF (F followed by G) must preserve them, too. There is always a transformation that will preserve any property: the identity transformation (everything stays put). So you form the collection of all such transformations. If you insist that they be invertible--that is, they can be undone--then this collection forms a mathematical object called a "group." The deep part is that any group of transformations can be conceived of as defining a "property" of features in the plane. A modern mathematician would say, for instance, that "shape" is whatever is preserved by the group of Euclidean transformations and isotheties (see below). That's the definition of "shape"!
Although the foregoing answers the question--you choose among transformations that preserve properties which matter for your analysis--it ducks the more practical issue of how to find these groups of transformations and how to compute with them. Here is a partial guide to this menagerie organized by the property they are intended to preserve. I describe each group, how to compute with it, and how it is usually used in practice.
• Size and orientation. This is the group of translations. A translation is given by a vector (e, f); it acts on an arbitrary Cartesian point (x, y) by sending it to (x + e, y + f). The translations shift features around without rotating them or changing their sizes. When coupled with isotheties (see below), this is the group that pans and zooms maps. It often is used to provide "world coordinates" for images (where rotation is not needed).
• Size (distance). This is the group of Euclidean transformations, or isometries. It includes the translations as well as all reflections and rotations. A rotation through angle q sends (x, y) to (x * cos(q) - y * sin(q), x * sin(q) + y * cos(q)). This is the group to use when comparing shapes and studying the usual Euclidean properties of shape, including length and curvature.
• Shape. In addition to the Euclidean transformations, this group includes all isotheties: these amount to a uniform rescaling with respect to a central point. For instance, an isothety with respect to the origin (0, 0) sends (x, y) to (a*x, a*y) for some non-zero number a. This group is used when lengths and areas only need to be compared rather than measured on an absolute scale.
• Incidence. These are the projective transformations. They correspond to what a camera records when it views features on a planar surface in three dimensions: all lines are rendered as lines, but they can be foreshortened and rotated. All the foregoing are projective transformations, but there exist some additional ones. The most general is given by a collection of numbers (a, b, c, d, e, f, g, h) and sends (x, y) to the value (u, v) with u = (a*x + b*y + c)/(g*x + h*y + 1) and v = (d*x + e*y + f)/(g*x + h*y + 1). There is a mild restriction on the possible values of (a, b, ..., h). Projective transformations evidently are useful for re-rendering maps made from cameras (such as aerial or satellite photos) to present them from alternative points of view. The most general "world file" formats used to provide coordinates for images or raster datasets specify projective transformations and usually use the numbers (a, b, ..., h) to do so.
• Area. Some projective transformations preserve areas: the special affine transformations. These are the ones that can be written in terms of four numbers a, b, c, d for which a*d - b*c is either +1 or -1. They send (x, y) to (a*x + b*y), (c*x + d*y). These are useful for converting among images of area-preserving projections. For instance, any two cylindrical equal-area projections of an ellipsoid differ by a special affine transformation.
• Angle. These are the conformal transformations. Unlike the preceding ones, they cannot be parameterized with a finite set of real numbers. Conformal transformations are best understood as transformations of the complex numbers, the analytic ones. They are precisely the ones that have a complex derivative. All have formulas in terms of power series and integrals. They often are obtained as solutions of differential equations. Because preserving angles tends to render shapes fairly accurately (at least small shapes), conformal transformations are sometimes referred to (incorrectly) as "shape-preserving." There are lots more of them than there are Euclidean transformations, though, so they provide more flexibility in transforming maps than just rotations and translations. The (small) price paid is that larger shapes are distorted.
• Circular arcs (including arcs of "infinite radius" circles--straight lines). These are the Mobius transformations. In terms of the complex number z = x + y*i (with i^2 = -1), they can all be written in terms of four complex numbers a, b, c, and d with a*d - b*c = 1; z is sent to the complex number (a*z + b)/(c*z + d). Mobius transformations are useful when working with configurations of circular and linear features. (For instance, images of lines of latitude and longitude in a Stereographic projection of the sphere are all portions of circles and lines in the plane.) Being complex analytic (except where c*z + d = 0), they are conformal almost everywhere.
• Incidence, nearness, and inside/outside. These are the continuous transformations. Many cannot even be written in terms of formulas: their defining quality is that around any point (x, y) there will always be infinitely many points (x', y') close to it which do not become greatly separated when they are transformed. Continuous (invertible) transformations provide an enormously rich and flexible way to turn almost any map into anything else while still maintaining important relationships among features, such as incidence, intersection, disjointness, and containment. This provides opportunities to simplify such analyses and make them more efficient.
This list is by no means exhaustive, but it does cover most of the groups of transformations commonly used in two dimensions.
Polynomial transformations do not fit in this list for several reasons. First, most polynomial functions are not invertible, so polynomials tend to be useful only within bounded regions of the plane. Second, polynomials do not preserve any practically useful property. (There is a deeper sense in which they preserve something, but that is a formal algebraic structure which rarely has a direct application in GIS or geography.) Polynomial functions are used solely because they are relatively simple to specify and to calculate, since they can be computed using a small number of multiplications and additions. There are enough of them that they can be used as a relatively flexible way to approximate any continuous function (q.v.), thereby overcoming the difficulty that exact formulas do not exist for most continuous functions.
Typically, polynomial transformations are not used by themselves, but rather as building blocks of various kinds of splines. A spline is locally polynomial, but its coefficients are allowed to change in a controlled fashion throughout the plane. Splines are used in "warping" one image to correspond to another, for instance. In this fashion any map can be made to correspond as closely as you would like to any other map of the same features, no matter how arbitrary the distortion.
There are different 2D adjustment methods because there are different sources of deformation.
1) deformations due to acquisition of an image by a remote sensor (central projection or push-broom)
2) deformations due to the representation of the Earth surface on a 2D plane
3) local errors
My rule is to use the simplest method possible (the one that needs the lesser number of GCP's) when it fits to the deformation type. For instance, if you only need to rescale ,translate or rotate a map, your best choice is an affine transform. On the other hand, if you move from one known coordinate system to another known coordinate system, you should use the existing transforms to do it (and you don't need GCP's). For raw images, you should rather use orthorectification models. And if you don't know what type of errors you have in front of you, 2D polynomial can be usefull (first or second order is usually enough). | 2,209 | 10,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.925799 |
https://math.answers.com/math-and-arithmetic/Use_euclid_division_algorithm_to_find_HCF_of_135_and_225 | 1,722,668,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00861.warc.gz | 311,569,463 | 47,830 | 0
# Use euclid division algorithm to find HCF of 135 and 225?
Updated: 9/15/2023
Wiki User
10y ago
225=135*1+110
135=110*1+25
110=25*4+10
25=10*2+5
10=5*2+0
so, the HCF of 135 and225 is 5.
Wiki User
10y ago
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Q: Use euclid division algorithm to find HCF of 135 and 225?
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135.4286
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16 + ? = 135 In order to find the unknown in this example, subtract 16 from 135. 135 - 16 = 119 Check: 16 + 119 = 135!!!
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### What is 45 of 300?
To find 45 percent of a number, multiply the number by 0.45. In this instance, 0.45 x 300 = 135. Therefore, 45 percent of 300 is equal to 135. | 504 | 1,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-33 | latest | en | 0.713755 |
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Prisoner's Visitors Problem (Posted on 2009-01-21)
There are 60 records given to you which correspond to a prisoner who is imprisoned for 60 days. He has 6 relatives one of whom visits him daily and the others visit him every ith day from the day of his imprisonment (i=2,3,4,5,6 for these 5 relatives).
Every record is sealed with the day number on it which indicates the number of days he is jailed when the record is filed with the names of visitors on that particular day. You have to make a new record which should be filled with the following details:
visitor name - number of visits after 60 days
Assume that no other relatives visited him at all, names of these 6 relatives are different and you don't know their names. Find the minimum number of records that need to be checked to make the new record correctly. Find the number of ways you can choose the minimum number of records and you can still make the new record correctly.
See The Solution Submitted by Praneeth Rating: 4.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
program code | Comment 4 of 7 |
here is the Qbasic program I used to find all the solutions
DATA 3,1,2,1,2,1
DATA 3,2,1,1,2,1
DATA 3,2,2,1,1,1
DATA 3,2,2,1,2,1
filen% = 1
OPEN "./perp.txt" FOR OUTPUT AS filen%
DIM mlts(1 TO 6, 1 TO 3)
DIM fls(1 TO 3)
DIM cnts(1 TO 6)
DIM ans(1 TO 4, 1 TO 6)
FOR i = 1 TO 4
FOR j = 1 TO 6
NEXT j
NEXT i
sols = 0
FOR f1 = 2 TO 58
FOR f2 = f1 + 1 TO 60
IF f2 <> f1 THEN
FOR f3 = f2 + 1 TO 60
IF f3 <> f2 AND f3 <> f1 THEN
fls(1) = f1
fls(2) = f2
fls(3) = f3
FOR i = 1 TO 6
tot = 0
FOR j = 1 TO 3
num = fls(j)
IF num MOD i = 0 THEN
mlts(i, j) = 1
tot = tot + 1
ELSE
mlts(i, j) = 0
END IF
NEXT j
cnts(i) = tot
NEXT i
pass = 1
FOR k = 1 TO 6
IF ans(1, k) <> cnts(k) THEN
pass = 0
END IF
NEXT k
IF pass = 0 THEN
pass = 1
FOR k = 1 TO 6
IF ans(2, k) <> cnts(k) THEN
pass = 0
END IF
NEXT k
END IF
IF pass = 0 THEN
pass = 1
FOR k = 1 TO 6
IF ans(3, k) <> cnts(k) THEN
pass = 0
END IF
NEXT k
END IF
IF pass = 1 THEN
sols = sols + 1
PRINT #filen%, f1, f2, f3
END IF
END IF
NEXT f3
END IF
NEXT f2
NEXT f1
PRINT #filen%, "found "; sols; " solutions"
CLOSE filen%
Posted by Daniel on 2009-01-21 22:22:03
Search: Search body:
Forums (2) | 849 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-24 | latest | en | 0.872671 |
http://smallbusiness.chron.com/calculate-business-turnover-12433.html | 1,529,852,601,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866965.84/warc/CC-MAIN-20180624141349-20180624161349-00516.warc.gz | 291,884,022 | 11,702 | # How to Calculate Business Turnover
by William Adkins; Updated May 16, 2018
Unlike items such as buildings, fixtures and manufacturing equipment, some assets a business owns must be continuously disposed of and replaced. This ongoing cycle is called turnover. The term is principally used in reference to accounts receivable or inventory. Tracking the rate of turnover of such assets is important, because they often involve large amounts of money. Slow turnover increases costs and the amount of working capital required, which in turn reduces the profitability of the firm.
### Overview of Business Turnover
Businesses often extend credit to customers. The amount a customer owes is listed on the company's books under accounts receivable. This is a sound business practice, because extending credit gives customers flexibility when making purchases and helps drive sales. However, if customers do not pay in a timely manner, the business can find itself with a lot of money tied up and unavailable. The accounts receivable turnover ratio measures the average time it takes for a business to receive payment from customers.
The situation with inventory turnover is similar to accounts receivable turnover. A firm needs to keep sufficient stock on hand to meet customer demand. If the inventory levels are too low, the business will often run short, resulting in dissatisfied customers. At the same time, management wants to keep inventory levels in check to avoid needlessly tying up working capital. Inventory turnover ratio measures how often a company replaces its inventory during an accounting period. For investors, business turnover ratios are a useful tool for assessing how efficiently a company manages its operations and utilizes working capital.
### Calculation of the Accounts Receivable Turnover Ratio
The account receivable turnover ratio measures how quickly the business collects the average balance of credit extended to its customers. To figure the average accounts receivable balance, add the total amount customers owe at the start of the accounting period to the ending balance and divide by 2. Divide total sales on credit for the period by this average balance. Exclude cash receipts.
Suppose the average balance of outstanding credit is \$75,000 and your business has \$900,000 in sales on credit for a year. Dividing \$900,000 by \$75,000 gives you an accounts receivable turnover ratio of 12, which means it takes about one month to collect payment.
### Calculation of the Inventory Turnover Ratio
To calculate the inventory turnover ratio, divide the cost of goods sold (COGS) by the average dollar value of the inventory during the accounting period. The average value of inventory is figured by adding the value of the inventory at the beginning of the accounting period to the ending value and dividing by 2. Suppose COGS equals \$2 million and the average inventory is \$400,000. Dividing \$2 million by \$400,000 equals an inventory ratio of 5. In other words, your business sells its on-hand inventory about five times each year. The higher the inventory turnover ratio, the faster the turnover. The goal is to maximize turnover as much as possible without running short on needed items for customers.
#### Tip
• If you are having difficulty computing turnover or keeping efficient financial records, you may want to consult with an accountant.
#### About the Author
Based in Atlanta, Georgia, W D Adkins has been writing professionally since 2008. He writes about business, personal finance and careers. Adkins holds master's degrees in history and sociology from Georgia State University. He became a member of the Society of Professional Journalists in 2009.
#### Photo Credits
• Creatas/Creatas/Getty Images | 722 | 3,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-26 | latest | en | 0.946429 |
https://conversion.org/length/shaku-japan/nautical-league | 1,726,888,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00619.warc.gz | 154,026,296 | 7,290 | # shaku (Japan) to nautical league conversion
Conversion number between shaku (Japan) and nautical league [NL; nl] is 5.4541091258154 × 10-5. This means, that shaku (Japan) is smaller unit than nautical league.
### Contents [show][hide]
Switch to reverse conversion:
from nautical league to shaku (Japan) conversion
### Enter the number in shaku (Japan):
Decimal Fraction Exponential Expression
shaku (Japan)
eg.: 10.12345 or 1.123e5
Result in nautical league
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 shaku (Japan) = (exactly) ((10/33)) / (5556) = 5.4541091258154 × 10-5 nautical league
• 1 nautical league = (exactly) (5556) / ((10/33)) = 18334.8 shaku (Japan)
• ? shaku (Japan) × ((10/33) ("m"/"shaku (Japan)")) / (5556 ("m"/"nautical league")) = ? nautical league
### High precision conversion
If conversion between shaku (Japan) to metre and metre to nautical league is exactly definied, high precision conversion from shaku (Japan) to nautical league is enabled.
Decimal places: (0-800)
shaku (Japan)
Result in nautical league:
?
### shaku (Japan) to nautical league conversion chart
Start value: [shaku (Japan)] Step size [shaku (Japan)] How many lines? (max 100)
visual:
shaku (Japan)nautical league
00
100.00054541091258154
200.0010908218251631
300.0016362327377446
400.0021816436503262
500.0027270545629077
600.0032724654754892
700.0038178763880708
800.0043632873006523
900.0049086982132338
1000.0054541091258154
1100.0059995200383969
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## Multiple conversion
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## Details about shaku (Japan) and nautical league units:
Convert Shaku (Japan) to other unit:
### shaku (Japan)
Definition of shaku (Japan) unit: ≡ 10/33 m .
Convert Nautical league to other unit:
### nautical league
Definition of nautical league unit: ≡ 3 nmi.
← Back to Length units | 612 | 2,002 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.6633 |
https://www.physicsforums.com/threads/inverse-function-thm.141348/ | 1,580,070,479,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00473.warc.gz | 1,024,171,916 | 18,699 | # Inverse Function Thm
## Main Question or Discussion Point
Im not sure whether this is a "Homework Question", but it is a question regarding the proof of the Inverse Function Theorem. It starts like this:
Let k be the linear transformation Df(a). Then k is non-singular, since det(f '(a)) != 0. Now D((k^-1(f(a))) = D(k^-1)(f(a)) (Df(a)) = k^-1 (Df(a)) is the identity linear transformation.
Heres what i dont understand:
If the theorem is true for k^-1 (f) then it is clearly true for f. Therefore we may assume at the outset the k is the identity.
Can anyone explain this?
mathwonk
Homework Helper
you are trying to prove a certain function is a local homeomorphism. if it is them composing it with an invertible linear map will not chNGE THIS, AND ALSO IOF IT IS NOT COMPOSING WITH an invertible linear map will not changw that.
so we may compose it with an invertible linear map before starting the proof.
i.e. if we wnT to prove f is invertible, and if L is kown to be invertible, then if we prove fT is invertible, we may conclude that also fTT^I(-1) = f is invertible.
the purpose of this reduction is to be able to simplify the derivative.
Ok so f:Rn->Rn. and by the fact that k: Rn->Rn is a homeomorphism on
an open set:
If k^-1 (f) is a homeomorphism on an open set then f is a
homeomorphism on an open set. Thus it suffices to prove that k^-1 (f)
is a homeomorphism on an open set. (An open set containing the point a
where d is continuously differentiable).
But why can you assume that k is the identity map?
matt grime
Homework Helper
Becuase you've just shown that the result is true for arbitrary k (satisfying the hypotheses) if and only if it is true for the identity.
This is perfectly normal. Any result in linear algebra about a vector v can often be translated to showing it for the zero vector only.
The analytic version is to simply rescale so that Df, which just a matrix of derivatives, is the identity.
ok so if the theorem is true for k = I then it is true for arbitrary k and from what I said before we can conclude that it is true for f?
matt grime
Homework Helper
It's just a change of basis argument - draw a picture in 2-d for the y=f(x) case to see what it's saying: if the slope is non-zero at a point we may assume that it is 1. I.e. if f'(0)=2, say, then f(x)/2 is a function whose derivative is 1 at x=0. (the general case is more complicated, it is not just dividing by a number, but the principle is the same).
You can also assume that a=(0,0,..,0) as well, by similar arguments.
Last edited:
Well actually if its true for k = I then k^-1 f reduces to f and therefore its true for f. It seems a little too simplified of an assumption. Is my logic correct?
ok the one variable case makes sense. If its true for f(x)/2 then it is true for f(x).
ok then i think my post number 7 is flawed because when were assuming that k = I were changing the nature of the function (like from f(x) to f(x)/2) so its actually what I said in post num 5 thats true right?
matt grime
Homework Helper
ak416 said:
Well actually if its true for k = I then k^-1 f reduces to f and therefore its true for f. It seems a little too simplified of an assumption. Is my logic correct?
No, you're missing the point. The assumption is not that 'because k=I, we then have that k^-1f=f' at all. I mean, it's true, but not relevant.
The assumption is that we may assume f satisfies Df(a)=I, because if it didn't the function k^-1f would satisfy Df(a)=I, and if k^-1f is invertible, so is f.
We may always assume in these cases that a=(0,0,..,0), and Df(a)=I if it helps, and other things too just by a change of coordinates.
Last edited:
Ok im still not sure. I understand that its true for arbitrary k if and only if it is true for k=I. But we are not sure that k = I. And by assuming k = I arent you changing the function?
Im thinking we are supposed to somehow use the fact that D(k^-1 f)(a) = k^-1 Df(a) is the identity linear transformation.
matt grime said:
The assumption is that we may assume f satisfies Df(a)=I, because if it didn't the function k^-1f would satisfy Df(a)=I, and if k^-1f is invertible, so is f.
Ok so if Df(a) is not I, then you move on to k^-1 f which satisfies D(k^-1 f) = I. And if you can prove it for D(k^-1 f) then you proved it for f. But what makes you assume k = I?
mathwonk
Homework Helper
i thionk i understand your question. you are puzzled because they are changing notation. i.e. if the derivative k of f is not I then consider the derivative of k^-1f which is I. then call that new derivative k gain, to save letters. now the derivative of k^-1f , which is I, is still being called "k", although that is confusing.
get it?
i.e. instead of saying "we can assume k = I" they should more accurately have said "thus we only have to prove the result for functions whose derivative is I. so if k is the derivative, we may assume k =I".
matt grime
Homework Helper
I am really baffled by these questions. We are allowed to assume that k=I since we have shown that we may replace f by a function g(x)=Df(a)^-1f(x) that has Dg(a)=I, and that the inverse function theorem will be true for f if and only if it is true for g. Thus replacing f with g we can assume Dg=I from the beginning. What part of that don't you understand?
matt grime
Homework Helper
ak416 said:
Ok im still not sure. I understand that its true for arbitrary k if and only if it is true for k=I. But we are not sure that k = I. And by assuming k = I arent you changing the function?
Yes, we are changing the function. But it doesn't matter the result is true for the original function if and only if it is true for the one we replace it by.
ya so we now prove it for the function g = Df(a)^-1 f which has Dg(a) = I. Ok that makes sense. But in the next part of the proof it says:
Whenever f(a+h) = f(a) we have
|f(a+h)-f(a)-k(h)|/|h| = |h|/|h| = 1
So is he still talking about the original f and the original k, or is he talking about the g and k = Dg(a). This is what confused me, but from what you guys are saying im assuming that hes talking about the g. | 1,653 | 6,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-05 | latest | en | 0.907699 |
https://blog.finxter.com/5-best-ways-to-check-list-partition-into-pairs-where-sum-is-multiple-of-k-in-python/ | 1,721,800,708,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00127.warc.gz | 109,375,508 | 21,385 | # 5 Best Ways to Check List Partition into Pairs Where Sum is Multiple of K in Python
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π‘ Problem Formulation: Given a list of integers, the challenge is devising a program in Python that can determine whether it’s possible to partition the list into pairs such that the sum of the elements in each pair is a multiple of a specified integer `k`. For example, with the input list `[3, 1, 2, 6]` and `k=5`, a desired output would be `True` since pairs (3, 2) and (1, 6) sum to 5 and 7 respectively, both multiples of 5.
## Method 1: Using a Frequency Map and Matching
This method involves creating a frequency map to count occurrences of each residue class modulo `k`. Pairs are formed by matching each element’s complement in the frequency map. It is suitable for small to medium-sized lists, as the complexity increases with the list size and diversity of remainders.
Here’s an example:
```from collections import Counter
def can_partition_to_k_multiple_pairs(lst, k):
freq_map = Counter(x % k for x in lst)
for remainder in freq_map:
complement = (k - remainder) % k
if freq_map[remainder] != freq_map[complement]:
return False
return True
result = can_partition_to_k_multiple_pairs([3, 1, 2, 6], 5)
print(result)```
Output:`True`
This Python function `can_partition_to_k_multiple_pairs` generates a frequency map that counts the occurrences of each number’s remainder when divided by `k`. It then checks if each number’s necessary complement (which, when added to it, would be a multiple of `k`) has the same frequency; if not, partitioning isn’t possible.
## Method 2: Sorting and Two-Pointer Technique
By sorting the list and using a two-pointer technique, this method efficiently finds complements that add up to a multiple of `k`. Suitable for large lists, it offers good performance but requires the list elements to be mutable and sortable.
Here’s an example:
```def can_partition_to_k_multiple_pairs(lst, k):
lst.sort()
left, right = 0, len(lst) - 1
while left < right:
if (lst[left] + lst[right]) % k != 0:
return False
left += 1
right -= 1
return True
result = can_partition_to_k_multiple_pairs([3, 1, 2, 6, 4, 5], 5)
print(result)```
Output: `True`
After sorting the list, this function uses two pointers starting at opposite ends. If the sum of the elements at the pointer indices is not a multiple of `k`, the list cannot be partitioned as required. Pointers are moved inward on a successful check until they meet.
## Method 3: Greedy Pair Selection
The greedy method iteratively selects pairs that add up to a multiple of `k`. It’s a straightforward approach that is best suited when each element in the list has a clear and unique pairing.
Here’s an example:
```def can_partition_to_k_multiple_pairs(lst, k):
while lst:
num = lst.pop()
has_pair = False
for val in lst:
if (num + val) % k == 0:
lst.remove(val)
has_pair = True
break
if not has_pair:
return False
return True
result = can_partition_to_k_multiple_pairs([3, 1, 2, 6, 5], 5)
print(result)```
Output: `True`
This function removes an element from the list and then searches for a pair that, when added to the removed element, results in a multiple of `k`. It continues this process until it finds pairs for all elements or determines that it’s not possible.
## Method 4: Dynamic Programming for Partitioning
Dynamic programming can solve this problem by keeping track of possible sums with subsets of the list. It is a complex yet powerful solution best applied when the elements in the list and the value of `k` are reasonably small.
Here’s an example:
```# This method is more abstract and generally not preferred for this specific problem,
# but serves as an example of how a DP approach might be conceptualized.
```
Due to the complexity and the need for a tailored DP solution, a specific code example is not provided. This method typically involves creating a DP table that records whether a certain sum is achievable with a subset of the numbers and then checking if this translates to a valid partition.
## Bonus One-Liner Method 5: Using List Comprehensions and All
This one-liner is a compact yet less readable version using list comprehensions and the built-in `all()` function. It essentially compresses the logic of Method 1 into a single line of code.
Here’s an example:
```can_partition_to_k_multiple_pairs = lambda lst, k: all(lst.count(x) == lst.count(k - x % k) for x in set(lst))
result = can_partition_to_k_multiple_pairs([3, 1, 2, 6], 5)
print(result)```
Output: `True`
This one-liner uses a lambda function which, through a list comprehension, checks that for each unique value `x`, there is an equal count of its complement in the list. It leverages Python’s expressive capabilities but may suffer in terms of readability and performance.
## Summary/Discussion
• Method 1: Frequency Map and Matching. Utilizes a counter for efficient matching of pairs. Does not sort the data. May not perform well for lists with a wide range of values.
• Method 2: Sorting and Two-Pointer Technique. Offers a performance boost for large lists that are mutable and sortable. May not be applicable for lists with complex or non-comparable data types.
• Method 3: Greedy Pair Selection. Straightforward and easy to implement. Performance may degrade with large lists or when no early pairings are found.
• Method 4: Dynamic Programming for Partitioning. Powerful and generalizes well. Often overkill for this type of problem and requires a deep understanding of DP to implement correctly.
• Bonus One-Liner Method 5: Using List Comprehensions and All. Concise and Pythonic. However, not recommended for complex or large datasets due to potential readability and performance issues. | 1,336 | 5,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.795819 |
http://mathhelpforum.com/algebra/15304-compound-interest-problem-print.html | 1,513,493,766,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948593526.80/warc/CC-MAIN-20171217054825-20171217080825-00145.warc.gz | 180,917,539 | 3,388 | # compound interest problem
• May 23rd 2007, 01:12 PM
imppy725
compound interest problem
A syntesizer is advertised at $500 down and$100 per month for 18 months. If interest is charged at 18% per annum compounded monthly, what is the cash proce of the synthesizer?
hi guys, i need to know what does $500 down mean and how to calculate it. I know it has something to do with the Annuity formula: A = R[(1+i)^n -1] / i thanks in advance! Michael • May 23rd 2007, 01:18 PM janvdl Quote: Originally Posted by imppy725 A syntesizer is advertised at$500 down and $100 per month for 18 months. If interest is charged at 18% per annum compounded monthly, what is the cash proce of the synthesizer? hi guys, i need to know what does$500 down mean and how to calculate it.
I know it has something to do with the Annuity formula:
A = R[(1+i)^n -1] / i
Michael
I think 500 is discount.
$1800 = (P - 500)(1 + \frac{18}{1200} )^ \frac{3}{2}$
There now it looks better. :) Solve for P :)
I get P = 21623 :D
• May 23rd 2007, 01:28 PM
imppy725
Quote:
Originally Posted by janvdl
I think 500 is discount.
$A = \frac{(P - 500)(1 + \frac{18}{1200} )^ \frac{3}{2}}{12} + 1800$
Something doesnt look right here. Why are both A and P unknown?
is ur formula the same as mines? if yes, can you plug in numbers u sing my variables, because im getting kind of confused. thanks, and can you tell me how I should be dealing with the discount?
• May 23rd 2007, 01:29 PM
janvdl
Quote:
Originally Posted by imppy725
is ur formula the same as mines? if yes, can you plug in numbers u sing my variables, because im getting kind of confused. thanks, and can you tell me how I should be dealing with the discount?
You'll see i've edited my post :)
• May 23rd 2007, 01:31 PM
imppy725
Quote:
Originally Posted by janvdl
You'll see i've edited my post :)
will you walk me through it please? because i sort of want it in my math language, you know...the stuff i've learnt
• May 23rd 2007, 01:37 PM
janvdl
Quote:
Originally Posted by imppy725
will you walk me through it please?
$1800 = (P - 500)(1 + \frac{18}{1200} )^ \frac{3}{2}$
In the end you pay 100 for 18 months = 1800. That is A's value.
P is the original price but you got 500 discount. Therefore: (P - 500)
The interest is over 18 months = 1.5 years = $\frac{3}{2}$ years.
But interest is compounded monthly so its 18% x $\frac{1}{12}$
Therefore:
(P - 500) = 1760
P = 2260
Hope it helps. :) | 750 | 2,435 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-51 | longest | en | 0.928352 |
http://articles.sun-sentinel.com/1998-01-27/lifestyle/9801230245_1_second-thoughts-nine-seconds-heart-attack | 1,521,620,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00412.warc.gz | 26,321,646 | 9,617 | # Second Thoughts
## Every Few Seconds, Some Body Comes Up With Another ``every . . . Seconds'' Statistic.
January 27, 1998|By MICHAEL PRECKER The Dallas Morning News
Once every 86,400 seconds, this newspaper is delivered to your door with another day of news, sports, funnies and all the rest.
But a lot is happening in the meantime. An American is born every nine seconds. Another dies every 13 seconds. Yet another has his or her car stolen every 23 seconds.
You get the idea.
It's just a mathematical equation _ the total number of times something happens divided by the number of seconds in that time period.
But the resulting statistic (every 25 seconds, an American suffers a heart attack) usually seems far more dramatic than the original (1.25 million Americans suffer heart attacks each year).
So let's synchronize our watches, tick off the seconds and see what's happening every:
0.1 second _ Somebody buys a roll of Life Savers.
0.25 second _ A can of Spam is consumed in the United States.
0.5 second _ Somebody buys a Barbie doll.
1 second _ The U.S. government spends about \$53,300.
1.5 seconds _ Somebody dials 911 from a cellular phone.
2 seconds _ A small comet may be plunging into the Earth's atmosphere.
3 seconds _ Somebody in the world buys a bottle of Chanel No. 5.
4 seconds _ A larceny or theft is committed in the United States.
5 seconds _ Somebody in the world buys a Star Trek book.
8 seconds _ Another U.S. Baby Boomer turns 50.
9 seconds _ An American is born.
12 seconds _ An American needs blood.
13 seconds _ An American dies.
13 seconds _ A couple get married in the United States.
19 seconds _ A violent crime is committed in the United States.
23 seconds _ A car is stolen in the United States.
25 seconds _ An American has a heart attack.
27 seconds _ A couple get divorced in the United States.
|
|
|
Please note the green-lined linked article text has been applied commercially without any involvement from our newsroom editors, reporters or any other editorial staff. | 458 | 2,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-13 | longest | en | 0.915121 |
https://devlibrary.in/nios-class-12-geography-chapter-3/ | 1,686,246,255,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655092.36/warc/CC-MAIN-20230608172023-20230608202023-00168.warc.gz | 235,342,647 | 59,326 | # NIOS Class 12 Geography Chapter 3 Dynamic Surface of The Earth
NIOS Class 12 Geography Chapter 3 Dynamic Surface of The Earth, Solutions to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Geography Chapter 3 Dynamic Surface of The Earth and select need one. NIOS Class 12 Geography Chapter 3 Dynamic Surface of The Earth Question Answers Download PDF. NIOS Study Material of Class 12 Geography Notes Paper 316.
## NIOS Class 12 Geography Chapter 3 Dynamic Surface of The Earth
Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Geography Chapter 3 Dynamic Surface of The Earth, NIOS Senior Secondary Course Geography Solutions for All Chapter, You can practice these here.
### Dynamic Surface of The Earth
Chapter: 3
GEOGRAPHY
INTEXT QUESTION 3.1
Q. Fill in the blanks:
1. Isostasy means_______.
Ans: The state of being balance.
2. Airy considered the density of different columns to be ________.
Ans: Same.
3. Pratt considered landblocks of various height to be different in terms of their________.
Ans: Density.
4. According to Airy there is ________ root below the mountain and _______ beneath the plain.
Ans: Deeper, lower.
5. Pratt postulated the concept of _______ root formation but a________ of compensation.
Ans: No, level.
6. Endogenetic forces often________ the crustal balance.
Ans: Disturb.
7. Regular earthquakes and volcanic eruptions along a particular belt does not signify________ but a sort of continuous ________.
Ans. Any balance, adjustment is needed.
INTEXT QUESTIONS 3.2
Q.1. Fill in the blanks:
(a) Alfred Wegener termed the super continent as_______.
Ans: Pangaea.
(b) Primordial ocean was known as _______.
Ans: Panthalasa.
(c) Pangaea was broken into two_______ in the north and _______ in the south.
Ans: Lauresia (Angaraland), Gondwanaland.
(d) North and south America drifted towards ________.
Ans: west.
(e) Tethys sea emerged between _______ and ______ by filling up of the water of _______.
Ans: Angaraland, Gondwanaland, Panthalasa.
Q.2. Name three evidences of continental drift put forwarded by Wegener.
Ans. (a) Jig saw fit.
(b) geological similarities.
(c) coal and vegetation evidences.
Q.3. Name two evidences of continental drift, put not mentioned by Wegener.
Ans. (a) Evidences from paleomagnetism.
INTEXT QUESTIONS 3.3
Q.1. Fill in the blanks:
(a) The uppermost outer_____ layer of the earth is called_____.
Ans: solid, plate.
(b) Crust and upper part of mantle upto an average depth of ______ is ______.
Ans: 100km, solid.
(c) Lithosphere includes ______ and _____.
Ans: upper solid mantle, crust.
(d) Tectonics is sort of ______ of lithosphere plates.
Ans: movement.
(e) The concept of convectional current was first explained by_____ in _____.
Ans: Arthur Holmes, 1928-29.
(f) Convectional currents are classified into ______ and _____; they _____ and _____ respectively.
Ans: rising, falling, diverge, converge.
(g) Plate boundaries are associated with ______, ______ and ______.
Ans. newly formed mountain systems, oceanic ridge, trenches.
Q.3. Name some important minor plates.
Ans. (a) Arabian plate.
(b) Philippines plate.
(c) Cocos plate.
(d) Nazca plate.
(e) Caribbean plate.
(f) Scotia plate.
Q.4. Enumerate different types of place boundaries.
Ans. (a) Divergent boundaries.
(b) Convergent boundaries.
(c) Fracture or transform fault boundaries.
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https://www.sciencefacts.net/illuminance.html | 1,708,515,168,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00334.warc.gz | 1,023,807,083 | 39,002 | Home / Physics / Illuminance
# Illuminance
Illuminance represents the quantity of light falling onto a given surface area. Conceptually, it signifies the brightness that an observer would perceive on a surface under the influence of incident light. It is not merely about the inherent properties of the surface itself but rather the amount of light it receives, a crucial distinction that underlines its significance in various fields.
## Formula
The formula for illuminance serves as the quantitative backbone for understanding how light interacts with surfaces. Mathematically, illuminance (E) is expressed as the ratio of luminous flux (Φ) incident onto a surface to the surface area (A) receiving that flux. This core formula can be articulated as follows:
$E = \frac{\Phi}{A}$
Here, luminous flux (Φ) is measured in lumens (lm), and the surface area (A) is measured in square meters (m²). This formula elucidates how the amount of light energy falling onto a surface is distributed over the given area, quantifying the intensity of light that the surface receives.
## Unit
Lux (lx) is the commonly used unit across diverse fields. Lux represents the light intensity that falls on a surface per unit area. One lux equals one lumen per square meter, i.e., 1 lx = 1 lm/m². In the CGS system, illuminance is measured in the unit known as phot, which is equivalent to 10,000 lux.
Another unit popular in the United States is foot-candles (fc). The foot-candle is defined as one lumen per square foot.
1 fc = 1 lm/ft2 = 10.76 lux
## Examples
Practical examples showcasing various illuminance values illustrate the importance of understanding these units. For instance, indoor office spaces commonly aim for illuminance levels between 300 to 500 lux to ensure comfortable working conditions without causing glare or visual discomfort. On the other hand, public outdoor areas require higher illuminance, often ranging from 10 to 20 lux for pedestrian pathways and up to hundreds of lux for sports stadiums or high-traffic zones.
## Applications
Illuminance is essential in many different areas. It is a measurement that helps design spaces well, helps scientists with their work, and makes things look better in various industries. Knowing about illuminance is vital when designing how to light up a place. It helps to decide where and how bright the lights should be to make sure everything is just right for what people are doing, and it also helps save energy and make things look nice. Whether it is offices needing good light or art shows needing a distinct vibe, knowing about illuminance allows designers to create places that fit what they are meant for.
It is not just about spaces, in any case. Illuminance is a big deal in photography, movies, and screens. Knowing illuminance is critical for photographers and filmmakers to make great pictures or scenes. It helps them set up the camera right and ensure everything looks fantastic. In screens and projectors, illuminance decides how bright and clear things appear, making sure pictures or videos look just how they should. Controlling illuminance well makes a huge difference in how things look and feel in these creative and techie areas, making the experience better for everyone.
## Luminance vs. Illuminance
Luminance and illuminance, often intertwined but inherently distinct, encapsulate crucial aspects of light measurement. Luminance refers to the intensity of light emitted or reflected from a surface, gauging how bright an object appears. It is inherently tied to the surface and typically measured in candelas per square meter (cd/m²). It focuses on the inherent properties of the surface and how it emits or reflects light, irrespective of the environment.
In contrast, illuminance centers on the amount of light falling onto a surface, emphasizing the intensity of light incident on a specific area. Illuminance represents the light flux per unit area, revealing how much light the surface receives rather than how the surface emits or reflects light. It pertains more to the environmental conditions and how light interacts with the surface, influencing visibility and perceived brightness.
Article was last reviewed on Wednesday, January 31, 2024 | 860 | 4,240 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-10 | latest | en | 0.916294 |
https://www.teacherspayteachers.com/Product/Place-Value-I-Have-Who-Has-Billions-Place-2058909 | 1,547,709,576,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658844.27/warc/CC-MAIN-20190117062012-20190117084012-00070.warc.gz | 950,078,294 | 20,827 | # Place Value I Have, Who Has - Billion's Place
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Place Value I Have, Who Has - Billion's Place
Your students will love this Place Value I Have, Who Has Game! Each student will have a place value card at their desk. Then the student with the star on their card begins the Place Value I Have, Who Has Game. He/she will read their place value card, and the other students look at their place value cards to see if they have the answer to the question. They read the answer on their card and read the next question on their place value card. Then students will analyze their cards to see if they have the answer. The game continues in that manner.
This Place Value I Have, Who Has game has a total of 32 various place value cards. This place value game covers standard form, written form, expanded form, greater than/less than, and other place value clues (My number has a 2 in the billion's place).
You may also use these Place Value I Have Who Has game at centers or math stations. Students will just start with the place value card with star and place all the place value cards in order. Students could do this individually or with a partner/group.
This fun place value game will make learning come alive in your classroom!
You can find my other I games here!
I Have, Who Has:
Addition I Have, Who Has – Two digits and Two digits
Addition I Have, Who Has – Three digits and Two digits
Addition I Have Who Has-Three digits and Three digits
Multiplication I Have, Who Has - Two digits by One digit
Multiplication I Have, Who Has-Two digits by Two digits
Multiplication I Have Who Has-Three digit by One digit
Multiplication I Have Who Has-Three digits by Two digits
Division I Have Who Has Two digits by One digit
Division I Have Who Has Three digits by Two digits
Equivalent Fractions I Have Who Has
Telling Time Quarter Hour I Have Who Has
Multiplication Patterns I Have Who Has
Fraction Comparison I Have, Who Has
Subtraction I Have, Who Has Three Digits less Two
Place Value I Have, Who Has - Hundred's Place
Place Value I Have, Who Has - Thousand's Place
Place Value I Have, Who Has - Hundred Thousand's Place
Place Value I Have, Who Has - Million's Place
Place Value I Have, Who Has - Hundred Million's Place
Place Value I Have, Who Has - Billion's Place
I Have, Who Has, Add and Subtract Fractions
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I Have, Who Has, Capacity Customary Conversion I Have, Who Has, Length Customary Conversion I Have, Who Has, Capacity Metric Conversion I Have, Who Has, Length Metric Conversion I Have, Who Has, Converting Improper Fractions and Mixed Numbers I Have, Who Has, Weight Conversion
I Have, Who Has, Division 2 digits by 1 digit with Remainder
Bingo:
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Multiplication Bingo Two Digits by One Digit
Division Bingo Three by One Digit
Subtraction Bingo Two Decimals
Addition Bingo Two Digits and Two Digits
Addition Bingo Three Digits and Three Digits
Subtraction Bingo Three Digits from Three Digits
QR Codes:
Subtraction Two Digits One Decimal Task Cards With QR Codes
Multiplication Two by One Digit, with QR Codes
Subtraction Two Digits Task Cards with QR Codes
Other:
Goemetry Match Game
Fraction Match Game
Scavenger Hunt Subtraction Three Digits
Scavenger Hunt Multiplication Two Digits
Scavenger Hunt Four Digits
Scavenger Hunt Division Three by two Digits
Scavenger Hunt Find the Area
Scavenger Hunt Lines and Angles
Scavenger Hunt Division Three by two Digits
Scavenger Hunt Fraction Comparison with Uncommon Denominators
Tic-Tac-Toe Subtraction Two Digits
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 892 | 3,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-04 | latest | en | 0.881152 |
https://community.boredofstudies.org/search/289226/ | 1,569,240,061,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576355.92/warc/CC-MAIN-20190923105314-20190923131314-00221.warc.gz | 423,174,704 | 10,078 | # Search results
1. ### Graphing Q
https://imgur.com/a/tLEFY the magflux graph is positive SINE what is the second graph according to this question
Anyone experienced with the experiments done with cooled supeconductors, can you explain the experiment with a little detail where they place a magnet over an already cooled superconductor? I have an explanation myself but theres alot of confliction between a couple people i know.
3. ### Are type 1 and type 2 superconductors ...
Are type 1 and type 2 superconductors specifically in our syllabus?
4. ### Someone help with this Q!
If i place a magnet over an already cooling superconductor. The magnet repels by the production of eddy currents, but what keeps it in the position?
5. ### Help with this Q!
If i gave you a magnetic flux vs time graph as a positive cosine non phase shifted function, that is its initially at a maximum. What is the induced emf graph?
6. ### Chances of Plank Einstein Debate?
Hello! Could anyone give their thoughts on the possibility of the Plank Einstein debate on science's involvement in political affairs question coming up? Its such a burden to remember, even if its just key points.. In my opinion it has no physics involvement at all
7. ### How do you do this q?
https://imgur.com/a/TZscN ^ Thank you :)
8. ### Relationships ..?
Compose a short story that shows how relationships lead to provocative discoveries. ^ Can such relationship be a relationship with land? Like for example someone who has sentimental value with a certain place, and the discovery of it being destroyed or something? Or are they just talking...
9. ### Help with this question please
I know that at maximum speed the back emf that a motor has is approximately equal to the supply emf, (which ultimately causes its speed to cap). But im confused with the words "it draws a current of 4A". Its probably not a complicated issue at all, just a little confused, any help would be...
10. ### 2017 Trials anyone?
Anyone got any of their schools 2017 ext1 maths trials? (Not cssa or anything like that) | 462 | 2,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-39 | latest | en | 0.94082 |
https://whatdoesmean.net/what-is-equivalent-fraction/ | 1,722,870,944,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00686.warc.gz | 487,111,830 | 44,475 | # What is equivalent fraction?
## What Does equivalent fraction Mean
In mathematics , the expression that refers to a division is called a fraction . The fraction 1/3, for example, implies that the number 1 is divided into 3 (or, in other words, 1 divided by 3). Two or more equivalent elements , as long as they are similar or equal .
To construct a mathematical fraction we need to have two components : a numerator and a denominator . In the previous paragraph, example 1/3 is mentioned , which we must read "one third"; in this case we have a numerator of value 1 and a denominator that is worth 3 . The meaning of such a pair is that we are in front of the third part of an integer, a quantity that to reach the other must be multiplied by three.
It is worth mentioning that the numerators and denominators must always be whole numbers with the exception of zero, that is, elements of the set that has the natural numbers from minus infinity to most infinity . Without delving into overly technical issues, it is enough to observe the concept of fraction to understand this rule: since in itself it expresses a ratio, and that the process of dividing its numerator by its denominator many times gives us a result with a comma, it would not be logical build it with decimal numbers .
To read a fraction it is necessary to know a special type of word : the numeral . When we write a number we have two options: use the appropriate figures according to the system used or write their names in words, and for this there are numerals.
The numerals are proper names to designate the numbers; in other words, they are nouns that serve to refer to them through the written or spoken language. There is more than one type of numeral, and the use of one or the other depends on the mathematical concept that we want to express in words. For example, cardinal numerals (also known by the name of common numerals ) are what we use every day to mention numbers when we need to count objects: one, two, three, and so on.
In the case of fractions, both the equivalents and any other, the cardinal numerals are used to refer to their numerator. On the other hand, there are fractional numerals , which are also known as partitive numerals , which serve to express the division of a whole into several parts: half, third, fourth, and so on. The denominator of a fraction is read using these terms.
The equivalent fractions , thus, are those which, although written differently, represent a same amount . 5/10 , 15/30 and 20/40 , to name a few, are equivalent fractions. Let's see a check that is obtained by dividing their numerators by their denominators:
5/10 = 0.5
15/30 = 0.5
20/40 = 0.5
It can be stated that these fractions ( 5/10 , 15/30 and 20/40 ) are equivalent fractions since all three indicate the same amount: 0.5 .
An easy way to find out if two or more fractions are equivalent is to multiply the numerator and denominator of each one by the same number. This process is known by the name of amplification .
Returning to the previous example, we can try the number 3 :
(5 x 3) / (10 x 3) = 15/30 = 0.5
(15 x 3) / (30 x 3) = 45/90 = 0.5
(20 x 3) / (40 x 3) = 60/120 = 0.5
The simplification is a similar process, but based on the division of the numerator and denominator by the same number. It is important to note that in order to complete this operation, the two terms must be divisible by the number in question. If the result is the same, then we are dealing with equivalent fractions. We can do the test with the previous examples and number 5 :
(5/5) / (10/5) = 1/2 = 0.5
(15/5) / (30/5) = 3/6 = 0.5
(20/5) / (40/5) = 4/8 = 0.5
La utilidad de las fracciones equivalentes reside en la posibilidad de hallar una versión más pequeña de otra, que nos vuelva menos complicado un determinado cálculo, por ejemplo. Por otro lado, reconocer dos o más fracciones equivalentes en una operación puede simplificarla si nos permite eliminarlas o asociarlas.
Go up | 980 | 3,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-33 | latest | en | 0.954065 |
https://encyclopedia2.thefreedictionary.com/particular+solutions | 1,566,628,942,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00537.warc.gz | 455,380,691 | 12,587 | Particular Solution
(redirected from particular solutions)
particular solution
[pər¦tik·yə·lər sə′lü·shən]
(mathematics)
A solution to an ordinary differential equation obtained by assigning numerical values to the parameters in the general solution. Also known as particular integral.
Particular Solution
A particular solution of a differential equation F (x, y, y′, . . ., y(n)) = 0 is a solution y = Φ(x) obtained from the general solution y = Φ(x, C1, . . ., Cn) of the equation for some specific choice of the arbitrary constants C1, . . ., Cn. For example, the general solution of the equation y + y = 0 is yC1 cos x + C2 sin x; if we set C1 = 2 and C2 = –1, we obtain the particular solution y = 2 cos x – sin x.
References in periodicals archive ?
On the other hand, the question of the stability of particular solutions of the Einstein equations in the wake of the groundbreaking proof of the stability of the Minkowski space-time due to D.
While there may be some getting used to the process of using these particular solutions or even some functionality shortcomings, it seems likes physicians are moving forward with the solutions they have on hand.
Therefore the obtained particular solutions for the bimorph converter were compared with the literature-derived solution.
The intent is to look at problems and alternatives, not to advocate particular solutions.
As a theological ideal, Christianity may be regarded as an absolute; but, as a human endeavor, it demands particular solutions to these various conflicts.
Cirnu, Determination of particular solutions of non-homogeneous linear differential equations by discrete deconvolution, UPB Scientific Bulletin, A, 69 (2007) 2, 3-16
As agreed between the parties, Accelerance will act as Oxagile representative in the USA matching the clients' demands for particular solutions with Oxagile programming talent and business expertise, putting special emphasis on such areas as VOD and IP video technology, system integration services, SaaS development and business software solutions.
With common agreement, effective and particular solutions can be found upon which we can all agree on," Christofias said.
0] are some particular solutions of (2) and T is arbitrary element of the ring [R.
Unlike many NGOs, we have a broad perspective that must go beyond the interests of narrow, particular solutions.
Overall, the deliberations revealed three key findings: (1) the words "achievement gap" hold almost no meaning for the people with the most at stake: the students, parents, and other residents of communities where the achievement gap is most pronounced; (2) while educational experts see the achievement gap as a national problem, citizens see it as a local problem with particular solutions that reflect specific local factors; and (3) forum participants across the nation felt that responsibility for helping minority and low-income students succeed rested not just with educators and schools--the traditional focus of action on education matters--but also with parents and other adults, with local institutions other than schools, and with broad community involvement and individual commitment.
The capability to deliver these particular solutions and implement them quickly was an overwhelmingly positive factor in our success.
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Open / Close | 686 | 3,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-35 | latest | en | 0.889489 |
https://forums.bmwmoa.org/showthread.php?40954-test&p=515305&viewfull=1 | 1,516,533,927,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00499.warc.gz | 677,898,697 | 14,767 | 1. A? No, B. No, false, true, no false. Oh wait it's an essay question?
2. I had my test yesterday, and it involved 9ft. of camera lead and two orifices..
I passed the test; let me give you more details....
never mind...
3. .... thus far, all of this has been a testimony to the amount of extra time we have on our hands - Bob
4. Originally Posted by tourunigo
.... thus far, all of this has been a testimony to the amount of extra time we have on our hands - Bob
5. The answer is 42!
anyone....?
6. 42, yep, I get it.......
(ZZ9 plural Z Alpha)
7. Originally Posted by Gilly
42, yep, I get it.......
(ZZ9 plural Z Alpha)
I'm going to be held back a year!
8. ## From my 9th grade Algebra teacher...
a is most nearly always equal to 2, while
b is most nearly always equal to 3.
x? Well that's a whole 'nother story...
I'm going with a=2.
9. Originally Posted by Braddog
a is most nearly always equal to 2, while
b is most nearly always equal to 3.
x? Well that's a whole 'nother story...
I'm going with a=2.
Please, leave my X out of this.
10. Originally Posted by 86755
The answer is 42!
anyone....?
Too many Pan Galactic Gargle Blasters last night.
This is a test of Roy's Emergency Broadcast system. If you are reading this, Thank a teacher.
11. I've been accused of being testy, and this D@*# test does nothing to disprove this dirty rotten rumor...
12. ## test
Getting some good milage out of one word...
13. Originally Posted by thompsonr
Getting some good milage out of one word...
Wasn't that the test??
14. ## hmmm...
.... I think that I am seeing a pattern in all of this behavior
15. I have heard of this: Male pattern blandness.
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http://www.jiskha.com/display.cgi?id=1351476963 | 1,462,562,564,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461862047707.47/warc/CC-MAIN-20160428164727-00139-ip-10-239-7-51.ec2.internal.warc.gz | 602,968,628 | 3,758 | Friday
May 6, 2016
# Homework Help: Math
Posted by Steven on Sunday, October 28, 2012 at 10:16pm.
There are 285 7 th graders there are 19 more girls than boys what are the totals for each
• Math - Ms. Sue, Sunday, October 28, 2012 at 10:38pm
Let b = boys.
b + b + 19 = 285
2b = 266
b = 133 | 109 | 296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2016-18 | longest | en | 0.965223 |
https://www.abebooks.com/9781598632903/Beginning-Math-Concepts-Game-Developers-1598632906/plp | 1,521,769,331,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648113.87/warc/CC-MAIN-20180323004957-20180323024957-00578.warc.gz | 773,051,731 | 16,252 | ## Beginning Math Concepts for Game Developers (Applied Mathematics)
### Ph.D. John P Flynt; Boris Meltreger
3.5 avg rating
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Successful game programmers understand that in order to take their skills beyond the basics, they must have an understanding of central math topics; however, finding a guide that explains how these topics relate directly to games is not always easy. Beginning Math Concepts for Game Developers is the solution! It includes several hands-on activities in which basic math equations are used for the creation of graphs and, ultimately, animations. By the time you finish, you will have developed a complete application from the ground up that allows you to endlessly explore game development scenarios for 2D games. If you have a basic understanding of programming essentials and a desire to hone your math skills, then get ready to take a unique journey that examines what is possible when you combine game development with basic math concepts.
"synopsis" may belong to another edition of this title.
John P. Flynt, Ph.D., works in the software development industry, has taught at colleges and universities, and has authored courses and curricula for several college-level game development programs. His academic background includes work in information technology, the social sciences, and the humanities. Among his works are In the Mind of a Game, Simulation and Event Modeling for Game Developers (with co-author Ben Vinson), and Software Engineering for Game Developers. John lives in the foothills near Boulder, Colorado.
Review:
1. GETTING STARTED WITH C# AND THE MATH LIBRARY A. C# as a Game Development Language B. Setting up a Project C. Inspecting the Math Library D. Guess a Number 2. FUNCTIONS AND METHODS A. Understanding Functions as Patterns B. Creating Lab for Exploring Functions a.Generating Date for a Table b.Developing Classes c.Equations and Methods 3. CONCEPTS BEHIND FUNCTIONS A. Number Domains B. Restricted Values C. Handling Exceptions D. Making use of a List to Store Function Output E. Fields and Properties F. Division By Zero and Other Mysteries 4. EXTENDING THE LAB WITH A COORDINATE PLANE A. How to Graph B. Putting the List Values to Work C. Using a Flag D. Closing E. The True Value Game 5. LOCAL AND WORLD SPACE IN CARTESIAN TERMS A. Spawning a Cartesian Plane. B. Learning How to Do Things Twice C. Understanding Grids D. Axes and How to Make Them E. Making Your Point F. Finding the Curve 6. CHANGING THINGS: LINES, SLOPES, AND METHODS A. What Counts As a Valid Function B. Constant Functions C. How to Make a Profit D. Linear Things E. Making Things Visible F. Method Overloading H. The Factory Game 7. QUADRATICS AND OTHER FUNCTIONS A. Parabolas B. Minimum and Maximum C. Absolute Values D. Discontinuous Functions E. Stair Steps and Other Antics F. The Table Game 8. LIMITS AND METHODS FOR THEM A. Talk of Limits B. Bicycle Tires C. Different forms of Limits D. Continuity E. Infinity F. Creating Graphics That Merge 9. ANIMATING THE WORLD A. Threads and Timers B. Eliminating Flicker C. Derived Classes D. Working with Arrays of Continuous Values E. Queues and Coordinates F. Event Generation G. Event Detection 10. IN TO THE GAME A. Derivation and Acceleration B. Controlling Flight C. Multiplying Complexity D. A Target Game E. Extended Intelligence in Games
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Book Description Course Technology Ptr, 2006. Paperback. Condition: Brand New. paperback/cd-rom edition. 319 pages. 9.00x7.25x0.75 inches. In Stock. Seller Inventory # __1598632906 | 1,567 | 6,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-13 | latest | en | 0.910054 |
https://zbmath.org/?q=an:0881.60034 | 1,611,492,697,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00101.warc.gz | 1,054,093,767 | 11,641 | # zbMATH — the first resource for mathematics
Estimates for the quantiles of smooth conditional distributions and the multidimensional invariance principle. (English. Russian original) Zbl 0881.60034
Sib. Math. J. 37, No. 4, 706-729 (1996); translation from Sib. Mat. Zh. 37, No. 4, 807-831 (1996).
This is the first half of the proof of the theorem given by the author in 1995 in his Bielefeld University preprint devoted to the rates in invariance principles for independent centered random vectors (r.v.’s) $$\xi_1,\ldots,\xi_N\in\mathbb{R}^d$$. Their laws $${\mathcal L}(\xi_k)$$ belong to a certain class $${\mathcal A}_d(\tau)$$ with $$\tau\geq 1$$ and $$\text{cov} \xi_k=I$$, $$k=1,\ldots,n$$. The claim is that for $$\alpha>0$$ one can construct on a probability space $$(\Omega,{\mathcal F},\mathbb{P})$$ independent r.v.’s $$X_1,\ldots,X_n$$ and independent standard Gaussian r.v.’s $$Y_1,\ldots,Y_n$$, s.t. $${\mathcal L}(X_k)={\mathcal L}(\xi_k)$$, $$k=1,\ldots,n$$, and $E\exp\Biggl(\frac{c_1(\alpha)\Delta(X,Y)}{\tau d^3\log^+d}\Biggr)\leq\exp\Bigl(c_2(\alpha)d^{9/4+\alpha}(\log^+(n/\tau^2))\Bigr)$ where $$c_1(\alpha),c_2(\alpha)>0$$ depend only on $$\alpha$$ and $$\Delta(X,Y)=\max_{1\leq k\leq n}\Bigl|\sum_{j=1}^k X_j- \sum_{j=1}^k Y_j\Bigr|$$. This theorem is a refinement of a result by U. Einmahl [J. Multivariate Anal. 28, No. 1, 20-68 (1989; Zbl 0676.60038)]. The proof is based on the study of quantiles of smooth conditional distributions for vectors whose laws belong to $${\mathcal A}_d(\tau)$$.
##### MSC:
60F17 Functional limit theorems; invariance principles 62A01 Foundations and philosophical topics in statistics
##### Keywords:
multidimensional invariance principle
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##### References:
[1] A. Yu. Zaîtsev, ”Estimation of the Lévy-Prokhorov distance in the central limit theorem for random vectors with finite exponential moments,” Teor. Veroyatnost. i Primenen.,31, No. 2, 203–220 (1986). [2] A. A. Borovkov, ”On the rate of convergence in the invariance principle,” Teor. Veroyatnost. i Primenen.,18, No. 2, 217–234 (1973). · Zbl 0323.60031 [3] M. Csorgo and P. Révész, Strong Approximations in Probability and Statistics, Academic Press, New York (1981). [4] S. Csorgo and P. Hall, ”The Komlós-Major-Tusnády approximations and their applications,” Austral. J. Statist.,26, No. 2, 189–218 (1984). · Zbl 0557.60028 · doi:10.1111/j.1467-842X.1984.tb01233.x [5] Yu. V. Prokhorov, ”Convergence of random processes and the limit theorems of probability theory,” Teor. Veroyatnost. i Primenen.,1, No. 2, 157–214 (1956). [6] A. V. Skorokhod, Studies in the Theory of Random Processes [in Russian], Kievsk. Univ., Kiev (1961). · Zbl 0215.53501 [7] J. Komlós, P. Major, and G. Tusnády, ”An approximation of partial sums of independent RV’s and the sample DF. I,” Z. Wahrscheinlichkeitstheor. Verw. Geb., No. 32, 111–131 (1975). · Zbl 0308.60029 · doi:10.1007/BF00533093 [8] J. Komlós, P. Major, and G. Tusnády, ”An approximation of partial sums of independent RV’s and the sample DF. II,” Z. Wahrscheinlichkeitstheor. Verw. Geb., No. 34, 34–58 (1976). · Zbl 0307.60045 [9] A. I. Sakhanenko, ”On the rate of convergence in the invariance principle for nonidentically distributed variables with exponential moments,” in: Trudy Inst. Mat. (Novosibirsk), Novosibirsk,3, 4–49 (1984). · Zbl 0541.60024 [10] E. Berger, Fast Sichere Approximation von Partialsummen Unabhängiger und Stationärer Ergodischer Folgen von Zufallsvectoren, Dissertation, Universität Göttingen (1982). [11] I. Berkes and W. Philipp, ”Approximation theorems for independent and weakly dependent random vectors,” Ann. Probab.,7, 29–54 (1979). · Zbl 0392.60024 · doi:10.1214/aop/1176995146 [12] U. Einmahl, ”A useful estimate in the multidimensional invariance principle,” Probab. Theory Related Fields,76, No. 1, 81–101 (1987). · Zbl 0608.60029 · doi:10.1007/BF00390277 [13] U. Einmahl, ”Strong invariance principles for partial sums of independent random vectors,” Ann. Probab.,15, 1419–1440 (1987). · Zbl 0637.60041 · doi:10.1214/aop/1176991985 [14] W. Philipp, ”Almost sure invariance principles for sums of B-valued random variables,” Lecture Notes in Math.,709, 171–193 (1979). · Zbl 0418.60013 · doi:10.1007/BFb0071957 [15] U. Einmahl, ”Extensions of results of Komlós, Major and Tusnády to the multivariate case,” J. Multivariate Anal.,28, 20–68 (1989). · Zbl 0676.60038 · doi:10.1016/0047-259X(89)90097-3 [16] A. Yu. Zaîtsev, ”A multidimensional version of the Hungurian construction,” in: Abstracts: Second All-Russian School-Colloquium on Stochastic Methods (Îoshkar-Ola, 1995), TVP, Moscow, 1995, pp. 54–55. [17] A. Yu. Zaîtsev, Multidimensional Version of the Results of Komlós, Major and Tusnády for Vectors with Finite Exponential Moments [Preprint], Univ. Bielefeld 95-055, FRG, Bielefeld (1995). [18] V. V. Yurinsky, ”On approximation of convolutions by normal laws,” Teor. Veroyatnost. i Primenen.,22, No. 4, 653–667 (1977). · Zbl 0399.60022 [19] A. Yu. Zaîtsev, ”On the Gaussian approximation of convolutions under multidimensional analogues of S. N. Bernstein inequality conditions,” Probab. Theory Related Fields,74, No. 4, 535–566 (1987). · Zbl 0612.60031 · doi:10.1007/BF00363515 [20] V. A. Statulevičius, ”On large deviations,” Z. Wahrscheinlichkeitstheor. Verw. Geb., No. 6, 133–144 (1966). · Zbl 0158.36207 [21] L. Saulis and V. Statulyavichus, Limit Theorem on Large Deviations [in Russian], Mokslas, Vil’nyus (1991). [22] A. Yu. Zaîtsev, ”An improvement of the U. Einmahl estimate in the multidimensional invariance principle,” in: I. A. Ibragimov and A. Yu. Zaîtsev (eds.), Probability Theory and Mathematical Statistics, Proceedings of the Euler Institute Seminars Dedicated to the Memory of Kolmogorov, Gordon and Breach, Amsterdam, 1996, pp. 109–116. · Zbl 0873.60020
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2,070 | 6,176 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.622513 |
https://percent.info/minus/18/how-to-calculate-920-minus-18-percent.html | 1,725,711,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00856.warc.gz | 433,567,089 | 2,735 | 920 minus 18 percent
This is where you will learn how to calculate nine hundred twenty minus eighteen percent (920 minus 18 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 920 minus 18 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 920 of something.
920
(100%)
18 percent means 18 per hundred, so for each hundred in 920, you want to subtract 18. Thus, you divide 920 by 100 and then multiply the quotient by 18 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(920 ÷ 100) × 18
= 165.6
We made a pink square that we put on top of the image shown above to illustrate how much 18 percent is of the total 920:
The dark blue not covered up by the pink is 920 minus 18 percent. Thus, we simply subtract the 165.6 from 920 to get the answer:
920 - 165.6
= 754.4
The explanation and illustrations above are the educational way of calculating 920 minus 18 percent. You can also, of course, use formulas to calculate 920 minus 18%.
Below we show you two formulas that you can use to calculate 920 minus 18 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
920 - ((920 × 18/100))
920 - 165.6
= 754.4
Formula 2
Number × (1 - (Percent/100))
920 × (1 - (18/100))
920 × 0.82
= 754.4
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
921 minus 18 percent
Here is the next percent tutorial on our list that may be of interest. | 419 | 1,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-38 | latest | en | 0.894931 |
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1 Chapter 1 Notes on the SHARP EL-738 calculator General The SHARP EL-738 calculator is recommended for this module. The advantage of this calculator is that it can do basic calculations, financial calculations and statistical calculations. Most of the keys can perform two functions. To perform a function written on the key, you simply press the key. To perform a function written on the surface just above the key, first press the orange 2ndF key to activate it to perform the function when pressed. 7
2 CHAPTER 1. NOTES ON THE SHARP EL-738 CALCULATOR 8
3 Contents 1 Notes on the SHARP EL-738 calculator 7 2 Normal Calculation Mode Switch on your calculator The SET UP menu Selecting a MODE Normal mode Calculator keys Financial Calculator Mode Normal mode Interest rates Simple interest Simple discount Compound interest Annuities Present value Future value Amortisation Statistical mode Mean Standard deviation Linear line
4 10 CONTENTS
5 Chapter 2 Normal Calculation Mode Display screen Key operation keys Cursor keys { Power ON/OFF and clear key Mode key Numerical and arithmetic keys 11
6 CHAPTER 2. NORMAL CALCULATION MODE 2.1 Switch on your calculator Before using your calculator for the first time, reset (initialise) it. Press the RESET switch located on the back of the calculator with the tip of a ball-point pen. After resetting the calculator, the initial display of the NORMAL mode appears. 6 ) *, - / NOTE: Pressing 2ndF M-CLR 1 = will also erase all stored data in the memory and restore the calculator s default setting. Note that I will not write the numbers 0,..., 9 in blocks. All the other functions will be writteninblocks. 2.2 The SET UP menu Press the SET UP key to display the SET UP menu. 6 ) *, - /, 5 2, 4 /, appears on the screen. Press the arrow three times and, ) 6 -! 6 ) *, - / will appear on the screen. A menu item can be selected by using the keys (the selected number will blink). Press the = key. 12
7 2.3. SELECTING A MODE To set the number of decimals, press SET UP 00., 1/ ' 6 ) *, - / appears on the screen. Press 2 to select two decimals 6 ) *, - / If you want four decimals press SET UP 0, 0, 4. (However, we will use two decimals. Press SET UP 0, 0, 2.) NOTE: The calculator uses a decimal point (0.00) where we use the decimal comma (0,00). 2.3 Selecting a MODE Press MODE. The menu display appears 6 ) *, - / 4 ) 5 6 ) 6 13
8 CHAPTER 2. NORMAL CALCULATION MODE Press 0. 6 ) *, - / appears on the screen 2.4 Normal mode The NORMAL mode allows you to perform financial, arithmetic or scientific calculations. 2.5 Calculator keys The keys are classified according to the work they do. The following keys are worth mentioning: ON: ON/C Last key, first row. To switch on the calculator. The ON/C key also clears the screen. To preserve the batteries, the calculator turns itself off after about 10 minutes. OFF:.. + The orange function on the red ON/C key. Press 2ndF ON/C to switch your calculator off. NUMERIC KEYS: 1, 2, , 0 These keys are used to enter numbers. MULTIPLICATION X Second last key, third last row. DIVISION Second last key, sixth row. EQUAL = Last key, last row. 14
9 2.5. CALCULATOR KEYS CLEAR ON/C Last key, first row. BRACKETS ( ) Last key, third last row and second last row. Use the ( and ) keys to place parentheses around parts of expressions. The closing parenthesis ) may be omitted. NEGATIVE +/- Second key, last row. This key is used to enter a negative number or change the sign of a number, while the key is used for the operation of subtraction. Note the different ways in which subtraction, with the long dash, and the sign of the number, with a small dash, are displayed. For example: 3 2 and 3+(-2). Example: Add 8 to 5 Press 8 + +/- 5 = The answer is 3. Subtract 5from8. Press 8 5 = The answer is 13. Subtract 5from 8. Press +/ /- 5 = The answer is 3. Add 5to 8 Press +/ /- 8 = The answer is 13. DELETE: DEL If you made a mistake, press DEL (last key, sixth row) to erase the number and then enter the correct number to continue. If you want to change a number or sign after you have pressed = use the cursor to move to the place where you want to change it. Enter the new number or sign, then press DEL and continue. 15
10 CHAPTER 2. NORMAL CALCULATION MODE INSERT: INS Use the cursor to move to the place where you want to insert a number. Press 2ndF INS (sixth row, last key) and enter the number. The cursor will flicker after the inserted number. TO THE POWER key y x. The 2ndF third key, fourth row Example: Calculate 2 3. Enter the base number first press 2. Then press 2ndF y x 3 = The answer is 8,00. If the power consists of more than one term, use brackets for the power. Example: Calculate (3 2 ) 4 Press ( 3 2ndF y x 2 ) 2ndF y x 4 = The answer is 6 561,00. Example: Calculate 5 2/3 Press 5 2ndF y x ( 2 3 ) = The answer is 2,92. SQUARE: (x 2 ) Use the power key. Example: Calculate 4 2. Press 4 2ndF y x 2 = The answer is 16. Example: Calculate 10 1 Press 10 2ndF y x +/- 1 = The answer is 0,10. Calculate Press 1 5 2ndF y x 2 = The answer is 0,04. 16
11 2.5. CALCULATOR KEYS SQUARE ROOT: x Use the x key. 2ndF third key, forth row. Example: Calculate means Press 2 2ndF x 64 = The answer is 8. Example: Calculate Press 3 2ndF x 64 = The answer is 4. Example: Calculate Press 4 2ndF x 3 2ndF y x 3 = The answer is 2,28. NUMERIC FRACTIONS Example: Calculate Press = The answer is Calculate Press = The answer is 1,77. LOGARITHM to the base e: ln Example: Calculate ln 3. Press 2ndF ln, (second key, second last row) 3 = The answer is 1,10. Example: Calculate ln ( ) Press 2ndF ln ( = The answer is 0,17. 17
12 CHAPTER 2. NORMAL CALCULATION MODE THE EXPONENTIAL FUNCTION: e x The inverse of ln. Example: Calculate e 1,10. Press 2ndF e x (2nd key, last row) 1.10 = The answer is 3. MEMORY: M+ The calculator has 11 temporary memories (A-H and X-Z), one independent memory (M) and one last answer memory (ANS). Temporary memory To store a value, press STO and the variable in which you want to store it. Example: Store 17 in A. Press 17 STO (fifth row, second key) A (first key, fourth row). % ) 6 ) *, - / % appears on the screen. If you want to recall the value stored in A, press RCL (first key, fifth row) A. ) 6 ) *, - / % appears on the screen. Store a value in the independent memory M+. Example: Store 19, 21 and 25 in M+. Press 19 M+ (third key, fifth row). 21 M+ 25 M+ To find the answer, press RCL M+. The answer is
13 2.5. CALCULATOR KEYS To clear the register. press 2ndF M-CLR. 6 ) *, - / appears on the screen. Press 0. 6 ) *, - / ; appears on the screen. Press 0. 6 ) *, - / appears on the screen. ERROR If ERROR 1 appears on the screen after you did a calculation press the key and the cursor will flicker where you made the mistake, press DEL and continue by pressing =. PERCENTAGE (Fourth key, sixth row) Example: Calculate 25% of R Press ndF %. The answer is 450,00. PERMUTATION Determine 5! 19
14 CHAPTER 2. NORMAL CALCULATION MODE Example: factorial n! Press 5 2ndF n! (first key, fourth row) = The answer is 120. Example: Permutations n P r Determine 10 P 5. Press 10 2ndF n P r 5 = The answer is Example: Combination n P r Determine 4 C 2. Press 4 2ndF n C r 2 = The answer is 6. 20
15 Chapter 3 Financial Calculator Mode Number of periods Interest rate per year Future value Present value Payment 21
16 CHAPTER 3. FINANCIAL CALCULATOR MODE 3.1 Normal mode TheNORMALmodeallowsustousethefinancial keys. The financial keys N, 1/Y, PV, PMT, FV can only be used when the exponent in the applicable formula consists of a single number (not a product or sum of numbers). Before using the financial keys, first clear the register by pressing 2ndF M-CLR 0, Interest rates Simple interest I = Prt Determine the amount of interest received if R1 200 is invested for 4 years at 14% simple interest per year. I = Prt = % 4 = ,14 4 = 672,00 The interest received is R672,00. We cannot use the financial keys because there is no exponent in the formula. Key in as = The answer is 672,00. S = P (1 + rt) Determine the accumulated amount for if R2 400 is invested for 42 months at a 9% simple interest rate per year. S = ( ) 1+9% = ( ) 1+0, = 3 156,00. The accumulated amount is R3 156,00. Key in as ( = The answer is 3 156,00. 22
17 3.2. INTEREST RATES Determine the simple interest rate if R3 600 accumulates to R5 760 in five years time. The simple interest rate is 12%. Key in as ( ) 5 = The answer is 0,12, that is, 12%. S = P (1 + rt) = (1 + r 5) 1+5r = r = r = ( = 0, Simple discount P = S (1 dt) Determine the present value of a promissory note that is worth R months later, and the applicable discount rate is 10,24% per year. The present value is R2 180,00. Key in as ( = The answer is 2 180,00. P = S (1 dt) P = ( ) 1 0, = 2 180,00 Determine the time under consideration (in months) if a simple interest rate of 11,76% is equivalent to a 10,25% simple discount rate. 23
18 CHAPTER 3. FINANCIAL CALCULATOR MODE By manipulating S = P (1 + rt) andp = S (1 dt) we get r = d 1 dt and t = ( 1 d ) d. r Substituting the values, we get ( ) t = 1 0,1025 0,1025 0,1176 = 1,25. The time under consideration is 1,25 years, that is, 15 months. Key in as ( ) = The answer is 1,25, that is, 15 months Compound interest S = P ( 1+ j ) tm m or S = P (1 + r) t m We use our financial keys to do the calculations because there is only one exponent in the formula: S = P (1 + r) t NB: The interest rate must be entered into the calculator as a percentage and NOT as a decimal because the calculator has been preprogrammed to automatically divide the interest rate by a hundred. Remember that it is convention to enter either the present value or future value as a negative amount. Calculate the future value if R5 000 is invested for five years at 15% per year compounded monthly. 24
19 3.2. INTEREST RATES S = P (1 + r) t = ( 1+ 0,15 12 = ,91 ) 5 12 The future value is R10 535,91. Key in as 2ndF CA (to clear the register). First enter the number of compounding periods. 2ndF P/Y (second key, third row) 12 ENT (sixth key, fifth row) ON/C +/ PV 15 I/Y 5 2ndF P/Y (first key, third row) N To check if you have entered the correct values press RCL and the financial key that you want to check. If the value is incorrect, enter the new value, press the financial key and continue. COMP FV The answer is ,91. Determine the time under consideration if R5 000 is invested at 15% per year, compounded half yearly, and the accumulated amount is R S = P (1 + r) t = ( 1+ 0,15 2 t = 4,79 The time under consideration is 4,79 years. Key in as 2ndF CA 2ndF P/Y 2 ENT ON/C ) t 2 +/ FV PV 15 l/y 25
20 CHAPTER 3. FINANCIAL CALCULATOR MODE COMP N N = appears on the screen. Because the number of compounding periods is half yearly, divided the answer by two. Press 2 = appears on the screen. 3.3 Annuities Present value P = Ra [ n i ] = R (1+i) n 1 i(1+i) n Calculate the present value of R1 600 quarterly payments for five years at an interest rate of 20% per year, compounded quarterly. The present value is R19 939,54. Key in as 2ndF CA 2ndF P/Y 4 ENT ON/C P = 1 600a 5 4 0,20 4 = ,54. +/ PMT 5 2ndF P/Y N 20 I/Y COMP PV The answer is , Future value S = Rs n i = R [ (1+i) n 1 26 i ].
21 3.3. ANNUITIES Determine the future value of R400 monthly payments made for five years at 16% interest per year, compounded monthly. The future value is R36 414,21. Key in as 2ndF CA 2ndF P/Y 12 ENT ON/C S = 400s ,16 12 = ,21 +/ 400 PMT 5 2ndF P/Y N 16 I/Y COMP FV The answer is , Amortisation Draw up an amortisation schedule for a loan of R5 000 which is repaid in annual payments over five years at an interest rate of 15% per year. Key in as 2ndF CA 2ndF P/Y 1 ENT ON/C P = Ra n i = Ra 5 0,15 R = 1 491,58 +/ PV 5 2ndF P/Y N 15 I/Y COMP PMT 27
22 CHAPTER 3. FINANCIAL CALCULATOR MODE appears on the screen. Press AMRT (fourth row, first key) 1 (Down arrow) 1 ENT Press BALANCE = appears on the screen. Press PRINCIPAL = appears on the screen. Press INTEREST = appears on the screen. Press [ 2 ENT ]twice - 6 ) ) *, - / appears on the screen. Press BALANCE = Press PRINCIPAL = Press INTEREST = Press [ 3 ENT ]twice - 6 ) 4 6 2! 6 ) *, - / appears on the screen. Press BALANCE = Press PRINCIPAL = Press INTEREST = Press [ 4 ENT ]twice Press BALANCE = Press Press PRINCIPAL = INTEREST = Press [ 5 ENT ]twice 28
23 3.3. ANNUITIES Press BALANCE = 0.02 Press Press PRINCIPAL = INTEREST =
24 CHAPTER 3. FINANCIAL CALCULATOR MODE 30
25 Chapter 4 Statistical mode Display screen Key operation keys Cursor keys { Power ON/OFF and clear key Mode key Statistical keys 31
26 CHAPTER 4. STATISTICAL MODE Given a data set, the calculator s STAT function can be used to calculate certain statistical values such as the average (mean), standard deviation and the equation of a linear line. Change to the stat mode. Press MODE 1, 0 5 J= J 6 ) *, - / 5 6 ) 6 appears on the screen. 4.1 Mean Determine the mean of the following values: 25; 30; 26; 15; 40; 35 Key the data in 25 DATA The calculator displays, ) 6 ) ) *, - / 5 6 ) 6 This means that it accepted the first data point. Keep on entering the data until the last one. 6 should be displayed. 30 DATA 26 DATA 15 DATA 40 DATA 35 DATA Calculate the mean. 32
27 4.2. STANDARD DEVIATION Press ON/C 5 J= J 6 ) *, - / 5 6 ) 6 appears on the screen. Press RCL (first key, fifth row) x (first key, seventh row) N 6 ) *, - / 5 6 ) 6 & # appears on the screen. The mean is 28, Standard deviation Determine the standard deviation of the above data. Without re-entering the data, press ON/C RCL sx (second key, seventh row). The standard deviation is 8, Linear line Please note: The calculator takes the a-value as the y-intercept and the b-value as the slope. Determine the equation of a linear line y = bx + a Press MODE 1 1 appears on the screen. 33
28 CHAPTER 4. STATISTICAL MODE 5 J= J 6 ) *, - / 5 6 ) 6 Determine the equation for the straight line passing through the points (1 ; 3) and (3 ;7). Key in as 1 x,y - (fifth key, fifth row) 3 DATA 3 x,y 7 DATA RCL a (fourth key, sixth row) a = 1 appears on the screen. RCL b (fifth key, sixth row) b = 2 appears on the screen. The equation for the straight line is y =2x +1 34
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E fx-92b Collège 2D+ User s Guide CASIO Worldwide Education Website http://edu.casio.com CASIO EDUCATIONAL FORUM http://edu.casio.com/forum/ Contents Important Information... 2 Sample Operations... 2 Initializing | 10,268 | 41,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-49 | latest | en | 0.793184 |
https://math.stackexchange.com/questions/933212/good-book-for-self-studying-binary-relations | 1,653,694,766,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00354.warc.gz | 434,404,677 | 67,279 | # Good book for self-studying Binary Relations
I am studying economics and I frequently encounter Binary Relations. But without any good knowledge of it, I get confused.
Here is some background, if it's helpful:
I know calculus(single and multi-variable). I have taken semester-long rigorous(definition-theorem-proof style) courses in optimization theory, linear algebra, probability theory and statistics. But, I am not very good at writing proofs myself.
I will be obliged if I will get some good text teaching me binary relations as google does not help me.
Edit:
As required in comments, I add pages from a microeconomics text. Here is preface requiring a course in abstract algebra that focuses mainly on binary relations. here is first chapter of that book which uses binary relations. I hope these links will be helpful. I feel handicapped while doing exercises with binary relations. So please suggest me what shall I do.
• I'm not sure there's a lot to be said about binary relations, per se: without more structure they are probably not very interesting. Maybe you could give some more details about the kind of things you want to know. Sep 16, 2014 at 3:48
• @NateEldredge I do not know this will help or not: We usually use binary relations to describe preference ordering. Sep 16, 2014 at 3:50
• The general subject is called Graph Theory, a huge field. But without knowing what it is you have difficulty with, I doubt one can give useful advice. Sep 16, 2014 at 3:50
• To be frank, what you need to add is not your personal background, but an example of the type of binary relations which you encounter during your study of economics. It suffice if you have an excerpt from a book or paper which is causing you difficulties, or perhaps a reference to where one can find such an example. From your previous comment it may be that some books on order theory may be useful, but it could also be that more generally looking at graph theory (in particular those of directed graphs) is what you need. We can't tell without more info. Sep 17, 2014 at 12:27
• The preface does not require an abstract algebra course that focuses mainly on binary relations but the binary relations material contained in the first weeks of an undergraduate abstract algebra course. Sep 21, 2014 at 4:30
Based on the text you have provided, I'm not sure there is a book that does what you want. In particular, except the formal definition of a relation (which you don't really need), no particularly advanced knowledge is assumed. By that I mean:
• either the definitions are provided by the text, or
• they are easily found on Wikipedia and understanding them does not require a lot of mental aerobics.
If you insist on a book, you should be able to find this material in the intro chapter of any "Abstract Algebra" textbook, or in some section of most textbooks on "Discrete Mathematics".
Strictly speaking, it looks like your subject of interest is elementary order theory, but a book on order theory is probably more expensive than the other options and you will almost surely only use the first couple pages of it anyway.
• Please let me know that if this is this book an order thery book? And thank you so much for so good answer. Sep 21, 2014 at 10:34
• It is a book about lattice theory, which is a subfield of order theory. It is certainly broad enough to contain what you are looking for. But as I said in my answer, I suspect you will really only be interested in pages 1 and 2 (and possibly 3, which has some good examples). Sep 22, 2014 at 1:24
This might be more advanced then you want, but a possibility is the book "Theory of Relations" by Roland Fraisse. As he says (roughly) the theory of relations isn't really the same as graph theory because in graph theory, you care more about which vertices are connected. In more abstract relation theory, the situation is more symmetric, with the two options (the relation holding or not) are on more equal footing.
• Looks like an interesting book, +1, though I agree that it's probably not what's needed here. Sep 21, 2014 at 4:43 | 918 | 4,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.963158 |
http://neoprogrammics.com/perpetual_calendar_algorithms/Computing_The_Time_Elements_hhmmss.php | 1,498,188,544,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319992.22/warc/CC-MAIN-20170623031127-20170623051127-00487.warc.gz | 300,102,069 | 2,290 | ### Computing The Time Elements (hh, mm, ss)
There are slightly different procedures for computing the time elements (`hh,mm,ss`) depending on the starting argument and the type of time we are computing. e.g. is it an actual time of day or a fraction of a day converted into equivalent time elements.
All times used here will be expressed in the universal standard `00`h to `24`h format.
There are three basic sources of fractions that will be converted into time elements.
• If the starting argument is a general Julian Day number ((`JD`), to compute the time elements from its decimal part, Algorithm 2a applies.
• If the starting argument is a time interval expressed as a fraction of a day (`DayFrac`), then Algorithm 2b applies.
• If the starting argument is a time interval expressed in decimal (`hours`), then Algorithm 2c applies.
Algorithm 2a:
Given any general JD number, compute the time elements (`hh,mm,ss`) from its decimal part. These time elements are will all equate to integer values with the seconds value (`ss`) rounded-off to the nearest second.
``` uu = JD + 0.5
hours = 24*(uu - floor(uu))
hh = floor(hours)
minutes = 60*(hours - hh)
mm = floor(minutes)
seconds = 60*(minutes - mm)
ss = floor(seconds + 0.5)
```
RANDOM EXAMPLE 1:
Given:
`JD = 2448103.872299210834`
To find the time of day elements (`hh,mm,ss`) from the decimal part of any general JD number:
``` JD = 2448103.872299210834
uu = JD + 0.5 = 2448104.372299210834
hours = 24*(uu - floor(uu)) = 8.93518105522
hh = floor(hours) = 08
minutes = 60*(hours - hh) = 56.1108633131
mm = floor(minutes) = 56
seconds = 60*(minutes - mm) = 06.652
ss = floor(seconds + 0.5) = 07
```
So the time of day elements, in this case, work out to:
`hh:mm:ss = 08:56:07 = 08h 56m 07s`
resolved to the nearest second.
Algorithm 2b:
Given a (±`DayFrac`) value, compute the corresponding time elements (`TimeSignVal, hh, mm, ss`)
``` if DayFrac < 0 then TimeSignVal = -1 else TimeSignVal = 1
hours = TimeSignVal * 24 * DayFrac
hh = floor(hours)
minutes = 60*(hours - hh)
mm = floor(minutes)
seconds = 60*(minutes - mm)
ss = floor(seconds + 0.5)
```
or
``` TimeSignVal = (DayFrac < 0)? -1:1
hours = TimeSignVal * 24 * DayFrac
hh = floor(hours)
minutes = 60*(hours - hh)
mm = floor(minutes)
seconds = 60*(minutes - mm)
ss = floor(seconds + 0.5)
```
RANDOM EXAMPLE 2:
Given:
`DayFrac = 0.4039646880429179`
To find the time elements (`TimeSignVal, hh,mm,ss`) corresponding to day fraction (`±DayFrac`):
``` TimeSignVal = (DayFrac < 0)? -1:1 = 1
hours = TimeSignVal * 24 * DayFrac = 9.69515251303
hh = floor(hours) = 09
minutes = 60*(hours - hh) = 41.7091507818
mm = floor(minutes) = 41
seconds = 60*(minutes - mm) = 42.549
ss = floor(seconds + 0.5) = 43
```
So the time elements corresponding to the fraction of a day, in this case, work out to:
`hh:mm:ss = 09:41:43 = 09h 41m 43s`
resolved to the nearest second.
Algorithm 2c:
Given an (`±hours`) value, compute the corresponding time elements (`TimeSignVal, hh,mm,ss`).
``` if hours < 0 then TimeSignVal = -1 else TimeSignVal = 1
uu = TimeSignVal * hours
hh = floor(uu)
ww = 60*(uu - hh)
mm = floor(ww)
ss = 60*(ww - mm)
```
or
``` TimeSignVal = (hours < 0)? -1:1
uu = TimeSignVal * hours
hh = floor(uu)
ww = 60*(uu - hh)
mm = floor(ww)
ss = 60*(ww - mm)
```
RANDOM EXAMPLE 3:
Given:
`hours = 1.40803817175960`
To find the time elements (`hh,mm,ss`) corresponding to the decimal (`hours`) value:
``` TimeSign = (1.40803817175960 < 0)? -1:1 = 1
hours = 1.40803817175960
hh = floor(hours) = 01
ww = 60*(hours - hh) = 24.4822903056
mm = floor(ww) = 24
ss = 60*(ww - mm) = 28.937
```
So the time elements, corresponding to the hours value, in this case, work out to:
`hh:mm:ss = 01:24:28.937 = 01h 24m 28.937s`
resolved to the nearest millisecond. | 1,228 | 3,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-26 | latest | en | 0.708588 |
https://www.peel520.net/what-is-a-good-beginning-salary/ | 1,718,876,954,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00055.warc.gz | 810,446,801 | 8,576 | # What is a good beginning salary?
07/02/2021 Off By admin
## What is a good beginning salary?
PayScale estimates the typical graduate with zero to five years experience makes \$48,400. The National Association of Colleges and Employers (NACE) calculates that the preliminary average starting salary for graduates from the class of 2018 is about \$Feb 2019
## How much is 50000 a year an hour?
For example, if you make \$50,000 per year, work 40 hours per week and take two weeks of paid vacation, your hourly wage is \$25 (\$50,000 divided by 2,000 working hours).
How much rent can you afford on 50k?
A simple rule of thumb is you shouldn’t spend more than 1/3 of your after tax salary on rent. As an example, your annual salary is 50K that leaves you with \$4,166/month. After taxes, you should have around \$3,270. One third of 3270 is about \$980, and that’s what your monthly rent should be on 50K a year.
How much per year is 18 dollars an hour?
In this case, you can quickly compute the annual salary by multiplying the hourly wage by 2000. Your hourly pay of 18 dollars is then equivalent to an average annual income of \$36,000 per year.
### How much is 20.00 an hour annually?
Assuming 40 hours a week, that equals 2,080 hours in a year. Your hourly wage of \$20.00 would end up being about \$41,600 per year in salary.
### Is \$20 hr a good salary?
\$20 an hour is \$41,600 annually if you can get full time hours. If your benefits come with insurance and a retirement account, and if your cost of living means you can find somewhere to live for under \$900 a month, it should be fine for someone who wants to pay the bills, save a little, and live simply but comfortably.
Is \$24 an hour a good salary?
Assuming all things equal, \$24 per hour would be slightly above the median household income in the US. It is also worthwhile to look at sites like Glassdoor to see what others make in your field. You can filter by company, location, job, etc.
Is 20 dollars an hour good?
How Much Is 20 Dollars An Hour Per Year? If you are working a full-time job, you will be working 40 hours per week on average. 40 hours multiplied by 52 weeks is 2,080 working hours in a year. \$20 per hour multiplied by 2,080 working hours per year is an annual income of \$41,600 per year.
## What jobs make \$50 an hour?
10 jobs that pay \$50 an hourComputer and information research scientist. Computer hardware engineer. Judge and hearing officer. Mathematician. Nuclear engineer. Physicist and astronomer. Political scientist. Sales manager. | 623 | 2,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.948781 |
https://www.weegy.com/?ConversationId=86C15164 | 1,596,605,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735909.19/warc/CC-MAIN-20200805035535-20200805065535-00048.warc.gz | 821,958,359 | 9,146 | 3(4y- 1) = 16y
Question
Asked 10/9/2012 12:16:59 PM
Updated 4/13/2014 9:16:55 AM
This conversation has been flagged as incorrect.
Edited by andrewpallarca [4/13/2014 9:16:53 AM]
s
Original conversation
User: 3(4y- 1) = 16y
Weegy: 3(4y)-3(1)=16y ... Distribute 12y - 3 = 16y ... Multiply
Question
Asked 10/9/2012 12:16:59 PM
Updated 4/13/2014 9:16:55 AM
This conversation has been flagged as incorrect.
Edited by andrewpallarca [4/13/2014 9:16:53 AM]
Rating
3
3(4y- 1) = 16y;
12y - 3 = 16y;
12y - 16y = 3;
-4y = 3;
y = -3/4
Added 4/13/2014 9:16:55 AM
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Questions asked by the same visitor
solve for y 12y - 3 = 16y
Weegy: 12y - 3 = 16y; 12y - 16y = 3; -4y = 3; y = -3/4 (More)
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Updated 4/13/2014 9:16:36 AM
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 701 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-34 | latest | en | 0.833545 |
http://esolangs.org/wiki/Backslash_Calculus | 1,679,740,563,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00462.warc.gz | 16,065,088 | 6,395 | Backslash Calculus
Backslash Calculus is a programming language invented by User:Tailcalled after reading this reddit comment.
Syntax
Backslash Calculus is based on lambda calculus, but it only uses two characters: backslash and space. This is the basic syntax:
``` E = \ \ E - abstraction
| \ \\ E E - application
| \\ - innermost variable
| \\\ - the variable one level more out
| ... - ...
```
Computational Class
Backslash Calculus is clearly Turing-complete, even if one ignores its simple translation to lambda calculus. Here are the SKI combinators:
``` S = \ \ \ \ \ \ \ \\ \ \\ \\\\ \\ \ \\ \\\ \\
K = \ \ \ \ \\\
I = \ \ \\
```
Alternative alphabets
Tailcalled imagined a few alternative alphabets for Backslash Calculus.
One can replace backslash with ' and space with ". The SKI combinators will look like this with that alphabet:
``` S = '"'"'"'"'"'"'"''"'"''"''''"''"'"''"'''"''
K = '"'"'"'"'''
I = '"'"''
```
Brainfuck
One can change the syntax a bit to make Backslash Calculus look more like brainfuck:
``` E0 = + E1 -
| [ E0 E0 ]
| .
| ,.
| ,,. <- unary-ish
| ...
E1 = > E0 <
| [ E1 E1 ]
| .
| ,.
| ,,. <- unary-ish
| ...
```
The SKI combinators will look like this:
``` S = +>+[[,,..][,..]]-<-
K = +>[,.]<-
I = +.-
``` | 373 | 1,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-14 | latest | en | 0.766388 |
http://jlebl.wordpress.com/2009/09/11/r-is-uncountable/ | 1,419,041,599,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802769321.94/warc/CC-MAIN-20141217075249-00107-ip-10-231-17-201.ec2.internal.warc.gz | 146,321,245 | 16,508 | # The Spectre of Math
## September 11, 2009
### R is uncountable
Filed under: Mathematics,Teaching — jlebl @ 4:34 pm
Today I did the “$\mathbb{R}$ is uncountable” proof in class. I got a few “understanding nods”, and only one person seemed to be falling asleep (visibly). One person stormed out the moment I finished the proof, but I think it was because I went about 30 seconds over, rather than being angry at Cantor. All in all, I think that was success. Last time I did this a few years ago to a similar audience I got many angry stares. I think they were also not small Kroneckers, but more angry as in “I have to learn this nonsense?” I always find that this theorem has a quality about it that makes people either like it a lot or hate it a lot. I think much more so than anything else they will see in this class. I already had a question in office hours whose answer depended on the axiom of choice. I guess that is more subtle than uncountability of $\mathbb{R}$, but also I don’t talk about AC in class.
I did something like Cantor’s original proof of 1874. I looked at Wikipedia (which has a simplified version of Cantor’s proof) and I looked at baby Rudin, which has essentially the same proof in completely different language and in greater generality. Though I did it without using contradiction, since proofs by contradiction are evil.
Proof: Given a countable set $\{ x_1, x_2, \ldots \} \subset \mathbb{R}$, you construct a sequence of closed intervals $[a_j,b_j]$ such that $[a_{j+1},b_{j+1}] \subset (a_j,b_j)$, and such that $x_j \notin (a_{j+1},b_{j+1})$. The intersection of all the intervals is nonempty (compactness) and can’t contain any element of the countable set. QED
That’s more analysis-like than the diagonal proof, which is more computer-science-like.
## 7 Comments »
1. Wow, I definitely wouldn’t be angry at being taught that…
I read Amir Aczel’s “pop-math” book on Cantor (“The mystery of the Aleph”), and despite some mumbo-jumbo in the book (Kabbala? Wtf?), I found it fascinating. I hope to someday understand that stuff properly (I mean, above the “pop-math” level..).
Comment by Mauro — September 11, 2009 @ 5:40 pm
2. … since proofs by contradiction are evil.
This leaves me puzzled since I love proofs by contradiction.
I guess it is because my wife is right and I am evil myself :-)
And, btw, should not your proof start by saying
“Let us assume that R is countable and write it as {x1,x2,…}”
If not, I perceive a missing logical link.
If yes, it is a proof by contradiction.
Maybe you are evil, too, but you are hiding the fact artfully.
All the best, Stan
PS: nice blog; ahoj
Comment by stan — September 30, 2009 @ 9:50 am
• that’s why you’re evil :) The way I said it is without contradiction. I proved that, “given a countable subset X of R, then R-X is nonempty.” Therefore R is uncountable. The reason why such a proof is less evil is that the set X is something that is just fine and exists. If you start with the contradiction statement R=X, then you are arguing about something that doesn’t exist. It’s a subtle difference.
I do contradiction proofs as well, but I find it often clearer to avoid them if possible. I like to work with things that are consistent and exist. Working with nonexistent objects makes it easier to do something wrong.
But, yes, I am evil.
Comment by jlebl — September 30, 2009 @ 5:04 pm
3. I’m not going to call this a proof, but this is quite convincing that R is not countable:
Suppose that R is countable, i.e. that R = { x(k) | k = 0, 1, 2, … }. Now cover x(k) with an interval of length 2^(-k), e.g. [x(k)-2^(-k-1), x(k)+2^(-k-1)]. Then the union of these intervals should of course cover all of R. But the total length of the intervals is just 2^(-0) + 2^(-1) + 2^(-2) + … = 2, and thus cannot cover all of R, the length of which is infinite. Hence, the assumption that R is countable is false.
Comment by Per Persson — August 6, 2010 @ 7:17 pm
• That’s actually close to a proof I guess. What’s left is to prove that a countable collection of intervals that covers R cannot have finite measure (length). That actually doesn’t seem too difficult.
Comment by jlebl — August 7, 2010 @ 2:34 am
4. Yes, we need continuity from below: If E(k) is an increasing (i.e. E(k)⊆E(k+1)) sequence of measurable sets, then m(∪E(k)) = lim m(E(k)). Here E(k) will be the union of the k first intervals.
Another remark is that we can get the measure of the union of the intervals as small as we want by taking the lengths be epsilon/2^(k+1) with epsilon small.
Comment by Per Persson — August 7, 2010 @ 12:14 pm
• even weaker thing is needed, all you need is an inequality for the measures. So subaditivity is all you need. The standard outer measure construction will suffice (that is, you need to prove that the standard outer measure defined starting with say half-open intervals is subadditive). No need to worry about measurable sets.
Comment by jlebl — August 7, 2010 @ 2:30 pm
The Rubric Theme. Create a free website or blog at WordPress.com. | 1,353 | 5,048 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2014-52 | latest | en | 0.970358 |
https://physics.stackexchange.com/questions/478729/how-to-show-the-metric-is-unchanged-under-an-orthogonal-transform | 1,597,303,936,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00136.warc.gz | 436,656,061 | 32,857 | # How to show the metric is unchanged under an orthogonal transform?
Say we have the transform such that $$x^i \rightarrow (x')^{i'}=M^{i'}_i(x)$$ where $$M$$ is an orthogonal rotation matrix
I've been asked to show that a general metric $$g_{ij}(x)$$ invariant under the transformation using the fact that $$M^{i'}_iM^{j'}_j \delta_{i'j'} = \delta_{ij}$$[1]
Here is my thinking,
The line element $$ds^2 = g_{ij}(x)dx^i dx^j$$ becomes $$\bar{g}_{i'j'}(x')d(M^{i'}_i x^i)d(M^{j'}_j x^j)$$, but when I use the fact that $$M$$ is constant so they can be taken out of the derivatives and use relation [1] I get
$$ds^2 = \bar{g}_{i'j'}(x')dx^i dx^j \delta_{ij}$$
which doesn't make sense as the i' and j' indices are not contracted
• I just realised that I implicitly assumed the some properties of the metric in my previous answer. Are you sure the metric is general? Are you sure you are not only concerned with its spatial elements? I cannot see why this should be true for a general, curved geometry. – Ollie113 May 8 '19 at 13:56
• Perhaps consider a LIF and prove it there? – Ollie113 May 8 '19 at 14:04
• @Ollie113 It also specifies that the matrix is a rotation matrix, so I think it means on the spatial coordinates. I'll correct the question with that. – Nonsematter May 8 '19 at 15:24
• I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. – user4552 May 8 '19 at 15:29 | 440 | 1,433 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-34 | latest | en | 0.911371 |
https://brainmass.com/business/finance/capital-budgeting-15031 | 1,656,364,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00339.warc.gz | 194,069,117 | 75,154 | Explore BrainMass
Capital Budgeting
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A ski resort plans to eventually add five new chairlifts. One lift costs \$2 million, preparing slope costs another \$1.3 million. The lift allows 300 additional skiers ,but there are only 40 days a year when the extra capacity will be needed. (The resort sell alls 300 lift tickets on those 40 days.) Running the new lift will cost \$500 a day for the entire 200 days the lodge is open. Assume that the lift tickets at Deer Valley cost \$55 a day and the added cash expenses for each skier-day are \$5. The new lift has an economic life of 20 years. The before-tax required rate of return for the resort. Compute the before-tax NPV of the new lift and advise the managers of th resort the lift will be a profitable investment. | 218 | 959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | longest | en | 0.910992 |
https://www.stat.math.ethz.ch/pipermail/r-help/2005-September/079762.html | 1,653,198,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00767.warc.gz | 1,171,020,542 | 2,370 | # [R] scatterplot3d + density() + polygon(color)
Uwe Ligges ligges at statistik.uni-dortmund.de
Wed Sep 28 08:58:16 CEST 2005
klebyn wrote:
> Hi R Users,
>
> How to use the function polygon()
> together with the package scatterplot3d?
>
> I am trying to color below of the curves
> defined for the function density().
>
> I tried to use the site: R GRAPH GALLERY
> as tutorial.
>
> [figure]:
>
> [code]:
>
> to my case but I do not obtain success.
>
> Somebody could give a tip to me, please?
>
> I am thankful anticipatedly.
>
> Cleber Borges
>
>
>
> #My code test
> ##############
> library(scatterplot3d)
> x=c(0.4, -1.2, .8, -.7, 0)
>
> d1 = density(x[1],bw=1.2, from=-3.0, to=3.0 )
> d2 = density(x[2],bw=0.8, from=-3.0, to=3.0 )
> d3 = density(x[3],bw=0.6, from=-2.5, to=2.5 )
> d4 = density(x[4],bw=0.5, from=-2.0, to=2.0 )
> d5 = density(x[5],bw=0.3, from=-1.5, to=1.5 )
>
> sx = c(d1\$x,d2\$x,d3\$x,d4\$x,d5\$x)
> sy = c(d1\$y,d2\$y,d3\$y,d4\$y,d5\$y)
> sz = c(rep(0.1,512),rep(0.2,512),rep(0.3,512),rep(0.4,512),rep(0.5,512))
>
> scatterplot3d(x=sx,y=sz,z=sy,type='l')
> ##############
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
Example:
library(scatterplot3d)
x <- c(0.4, -1.2, .8, -.7, 0)
d <- vector(length = length(x), mode = "list")
d[[1]] <- density(x[1], bw = 1.2, from = -3.0, to = 3.0)
d[[2]] <- density(x[2], bw = 0.8, from = -3.0, to = 3.0)
d[[3]] <- density(x[3], bw = 0.6, from = -2.5, to = 2.5)
d[[4]] <- density(x[4], bw = 0.5, from = -2.0, to = 2.0)
d[[5]] <- density(x[5], bw = 0.3, from = -1.5, to = 1.5)
x <- lapply(d, "[[", "x")
y <- lapply(d, "[[", "y")
z <- lapply(seq(0.1, 0.5, 0.1), rep, each = 512)
sx <- unlist(x)
sy <- unlist(y)
sz <- unlist(z)
s3d <- scatterplot3d(x = sx, y = sz, z = sy, type = "n")
for(i in rev(seq(along=d))){
s3d_coords <- s3d\$xyz.convert(x[[i]], z[[i]], y[[i]])
polygon(s3d_coords, col = i, border = "black")
}
Uwe Ligges | 842 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.53477 |
https://www.easycalculation.com/engineering/civil/flexible-pavement-structural-number.php | 1,568,542,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571027.62/warc/CC-MAIN-20190915093509-20190915115509-00018.warc.gz | 835,694,356 | 6,975 | English
# Flexible Pavement Structural Number Calculator
Calculate the structural number of a flexible pavement with the values of thickness and layer co-efficients of asphalt, base, sub base and additional layers.
## Structural Number of Flexible Pavement Calculation
Calculate the structural number of a flexible pavement with the values of thickness and layer co-efficients of asphalt, base, sub base and additional layers.
#### Formula:
L=a1ta + b1tb + c1tsb +d1tad Where, L=Structural Number of Flexible pavement, a1=Layer coefficient for asphalt , ta=Asphalt layer thickness, b1=Layer coefficient of base, tb=Base layer thickness , c1=Layer coefficient of sub-base, tsb=Sub-base layer thickness, d1=Layer coefficient of additional layer, tad=Thickness of additional layer
Flexible pavement structural number calculation is made easier here. | 182 | 853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | longest | en | 0.700979 |
https://www.answers.com/Q/How_many_days_does_president_work_in_a_year | 1,717,071,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00833.warc.gz | 563,440,914 | 50,246 | 0
# How many days does president work in a year?
Updated: 9/21/2023
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208 days
260 days.
### If you work 6 days a week how many days do you work in a year?
6 days per week x 52 weeks per year = 312 days per year
264
365 stink'n days
### How many working days in a month?
I'm assuming the question is "How many work days in a month?" Depends on the month and if there are holidays that month or now. February can be 20 work days but if President's Day is on a week day it's 19. August a month of 31 days and no holidays this year has 21 work days.
### How many days does the president get off work?
The last one got 365 days a year off Hopefully the new one can improve on that stat US labor laws do not apply to elected persons; theyb receive an ANNUAL pay regardless of days worked. No President has ever taken a year off since Woodrow Wilson.
### How many work days are there in a year including weekends and holidays?
365 days and a 1/4 and 366 days in a leap year.
365
### How many days do egyptologist work for?
The average egyptologists work 10 hours a day and 300 days per year.
### How many days in year did people work in the 1600s?
Since 90% of the population was in farming they worked 365 days a year.
241 | 361 | 1,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.957788 |
http://mathlair.allfunandgames.ca/normal.php | 1,582,843,789,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00419.warc.gz | 79,739,564 | 1,729 | # Normal Distribution
Math Lair Home > Topics > Normal Distribution
The normal distribution, also called the Gaussian distribution or, informally, as a "bell curve," is a probability distribution that plays a very significant role in the field of statistics. It is used frequently used in the social sciences and the natural sciences.
• The central limit theorem states that the mean of a large number of random variables from the same distribution is distributed approximately normally, regardless of the distribution of the random variable. To take a simple example, if you were to flip a coin a large number of times, the total number of heads would be distributed normally.
• The normal distribution can be applied to describe many phenomena in the natural and human worlds; for example, height and IQ in humans are distributed normally.
• The normal distribution is used to describe the distribution of observational errors (this was its original use). | 177 | 960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-10 | longest | en | 0.94325 |
https://lib.rs/crates/libecdsautil | 1,670,277,338,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00491.warc.gz | 401,566,069 | 4,792 | ## libecdsautil
Safe bindings to libecdsautil for ec25519 signing
### 5 releases
Uses new Rust 2021
0.2.1 Aug 29, 2022 Aug 23, 2022 Aug 19, 2022 Aug 15, 2022 Aug 15, 2022
#518 in Cryptography
Used in gluon-mesh-vpn-key-translate
19KB
301 lines
# libecdsautil
This will contain safe wrappers for libecdsautil-sys, yet currently only provides a representation of fastd public keys based on curve25519-dalek, as well as dalek-ff-group.
## theory
private fastd: 282615cc09656f1f3dbe5fa24b640bfd48d8302f982477d38335c2ffab84c17e
derived public fastd: faebc966b4b255d9383f44fb2abc1b8b4d596ced9951a421af4df97f62aa1a7b
Public fastd keys (u8[32]~256bit) represent a point on the legacy curve. Points on Edwards25519 (as used in Ed25519) should share the same y-coordinate; the x-coordinate can be calculated using a multiplication as in ecc_25519_store_xy_legacy and ecc_25519_load_xy_legacy.
In Ed25519 public keys are stored in "Edwards y"-format, meaning the first 255 bits represent the y coordinate, while the high byte of the last byte gives the sign of x.
In libuecc public keys are stored in what could be called CompressedEdwardsX, as the x coordinate is packed alongside the least significant bit (lsb) of the y-coordinate.
Decompressed points are represented as a four tuple {X,Y,Z,T} with:
$x= {X \over Z}$
$y={Y \over Z}$
$x*y={T \over Z}$
Each of the tuples segments is u32[32]~1024bit
-> But the last 24 are actually zeroes, while it's not within a calculation
Steps to verify:
• use libuecc to decompress a fastd public key
• load decompressed key as EdwardsPoint using rusts curve25519_dalek-crate
• implement compressX for EdwardsPoint which should emit the fastd public key again
• implement decompress for CompressedEdwardsX as opposing to he former
• then fastd keys could be loaded as CompressedEdwardsX and converted to Daleks representation and later fed into the verify-functions of libecdsautil
Left to do for compressed_points.rs
• implement EDWARDS_D
• implement sqrt_ratio_i()
• implement is_negative / is_odd
• implement conditional_negate
• upstream EDWARDS_D
• upstream sqrt_ratio_i() https://github.com/serai-dex/serai/pull/87
• upstream is_negative / is_odd
• upstream conditional_negate
• wait for next upstream release and strip downstream implementations
~2.3–4.5MB
~101K SLoC | 670 | 2,338 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-49 | latest | en | 0.824014 |
https://fr.slideserve.com/dolph/egr-1101-unit-12-lecture-1 | 1,632,326,609,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00546.warc.gz | 315,199,500 | 20,075 | EGR 1101: Unit 12 Lecture #1
# EGR 1101: Unit 12 Lecture #1
Télécharger la présentation
## EGR 1101: Unit 12 Lecture #1
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##### Presentation Transcript
1. EGR 1101: Unit 12 Lecture #1 Differential Equations (Sections 10.1 to 10.4 of Rattan/Klingbeil text)
2. Linear ODE with Constant Coefficients • Given independent variable t and dependent variable y(t), a linear ordinary differential equation with constant coefficients is an equation of the formwhere A0, A1, …, An, are constants.
3. Some Examples • Examples of linear ordinary differential equation with constant coefficients:
4. Forcing Function • In the equation the function f(t) is called the forcing function. • It can be a constant (including 0) or a function of t, but it cannot be a function of y.
5. Solving Linear ODEs with Constant Coefficients • Solving one of these equations means finding a function y(t) that satisfies the equation. • You already know how to solve some of these equations, such as • But many equations are more complicated and cannot be solved just by integrating.
6. A Procedure for Solving Linear ODEs with Constant Coefficients • We’ll use a four-step procedure for solving this type of equation: • Find the transient solution. • Find the steady-state solution. • Find the total solution by adding the results of Steps 1 and 2. • Apply initial conditions (if given) to evaluate unknown constants that arose in the previous steps. • See pages 371-372 in Rattan/Klingbeil textbook.
7. Forcing Function = 0? • If the forcing function (the right-hand side of your differential equation) is equal to 0, then the steady-state solution is also 0. • In such cases, you get to skip straight from Step 1 to Step 3!
8. Some Equations that Our Procedure Can’t Handle • Nonlinear differential equations • Partial differential equations • Diff eqs whose coefficients depend on y or t
9. MATLAB Commands • Without initial conditions: >>dsolve('2*Dy + y = 8') • With initial conditions: >>dsolve('2*Dy + y = 8', 'y(0)=5')
10. MATLAB Commands • Without initial conditions: >>dsolve('D2y+5*Dy+6*y=3*t') • With initial conditions: >>dsolve('D2y+5*Dy+6*y=3*t', 'y(0)=0', 'Dy(0)=0')
11. Today’s Examples • Leaking bucket with constant inflow rate and bucket initially empty • Leaking bucket with zero inflow and bucket initially filled to a given level
12. EGR 1101: Unit 12 Lecture #2 First-Order Differential Equations in Electrical Systems (Section 10.4 of Rattan/Klingbeil text)
13. Review: Procedure • Steps in solving a linear ordinary differential equation with constant coefficients: • Find the transient solution. • Find the steady-state solution. • Find the total solution by adding the results of Steps 1 and 2. • Apply initial conditions (if given) to evaluate unknown constants that arose in the previous steps.
14. Forcing Function = 0? • Recall that if the forcing function (the right-hand side of your differential equation) is equal to 0, then the steady-state solution is also 0. • In such cases, you get to skip straight from Step 1 to Step 3.
15. Today’s Examples • Series RC circuit with constant source voltage • First-order low-pass filter
16. Exponentially Saturating Function • A function of the form where K and are constants, is called an exponentially saturating function. • At t = 0, f(t) = 0. • As t , f(t)K.
17. Exponentially Saturating Function: Time Constant • In , the quantity is called the time constant. • The time constant is a measure of how quickly or slowly the function rises. • The greater is, the more slowly the function approaches its limiting value K.
18. Time Constant Rules of Thumb • For , • When t = , f(t) 0.632 K. (After one time constant, the function has risen to about 63.2% of its limiting value.) • When t = 5 , f(t) 0.993 K. (After five time constants, the function has risen to about 99.3% of its limiting value.) • See next slide for graph.
19. Exponentially Saturating Function: Graph
20. Exponentially Decaying Function • A function of the form where K and are constants, is called an exponentially decaying function. • At t = 0, f(t) = K. • As t , f(t) 0.
21. Exponentially Decaying Function: Time Constant • In , the quantity is called the time constant. • The time constant is a measure of how quickly or slowly the function falls. • The greater is, the more slowly the function approaches 0.
22. Time Constant Rules of Thumb • For , • When t = , f(t) 0.368 K. (After one time constant, the function has fallen to about 36.8% of its initial value.) • When t = 5 , f(t) 0.007 K. (After five time constants, the function has fallen toabout 0.7% of its initial value.) • See next slide for graph.
23. Exponentially Decaying Function: Graph
24. Low-Pass and High-Pass Filters • A low-pass filter is a circuit that passes low-frequency signals and blocks high-frequency signals. • A high-pass filter is a circuit that does just the opposite: it blocks low-frequency signals and passes high-frequency signals. | 1,372 | 5,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-39 | latest | en | 0.816471 |
https://www.odz-m.de/13099784/3-unit-sand-rate.html | 1,603,410,827,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880401.35/warc/CC-MAIN-20201022225046-20201023015046-00291.warc.gz | 851,951,810 | 7,425 | Welcome to visit us!
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# 3 Unit Sand Rate
the rate of flow on a filter depends on the type of filter.A rapid sand filter will have a flow of 2-3 gpm/square foot of filter area.The high rate filter may have 4-6 gpm/square foot applied to the surface.A constant rate flow valve is almost fully closed when a filter is clean so that the desired water level on top of the filter is maintained.
• ### Sand Is Being Dropped At A Rate Of 10 Cubic Feet Per
Given the related rates you have provided, we can calculate $v$ in terms of the radius, $r$, and in terms of time, $t$.$v = 10t$ $v = \pi r^2h$ but because we know that $h = 2r$, w.
• ### Numerical Modeling Of Onset And Rate Of Sand
Unit.Vosges sandstone.Oil-bearing sand.Youngs modulus (e) mpa.Injection rate, and sand mass produced during the experiment.The experiment shows that onset of sand production occurred at a pressure of approximately 5250 psi and a flow rate of approximately 20.7 in 3 /min.Such data are highly useful to understand the sand production.
• ### Rate Analysis For Cement Mortarcalculate Quantity
when, quantity of cement, sand, labour and tools are multiplied by its rates, and a contractors profit of 10-20% is assumed, the total sum of the amount is taken as the cost of 1m 3 of cement mortar, which is shown in the file below.Download: rate analysis for cement mortar.Read more:.
• ### Part 3 Soils
Slowing to a fairly constant rate after several hours.Therefore, a soil listed in the 0.3 intake family, for instance, does not necessarily take water at 0.3 inch per hour--especially at initial intake.For a light irrigation application of 1 inch, the intake rate might be 2 or 3 times the index rate.Each of the soil intake family groups relate.
• ### Sand Calculator How Much Sand Do You Need In Tons
Sand calculator online - estimate the sand required for your construction or landscaping project in weight (pounds, kilograms, tons, tonnes) and volume (cubic ft, cubic yards, cubic meters).If you are wondering 'how much sand do i need', our free sand calculator is here to do the math for you.Information about sand density, common sand types, sand grain sizes, how much a cubic yard of sand.
• ### Beach Sand 1 Cubic Meter Volume To Kilograms
How heavy is beach sand? convert how many kilograms ( kg - kilo ) of beach sand are in 1 cubic meter ( 1 m3 ).The beach sand calculator for exchange of conversion factor 1 cubic meter m3 equals = 1,529.20 kilograms kg - kilo exactly.To convert beach sand measuring unit properties can be useful in repairing beach sand or in productions where beach sand gets applied.
• ### Sand Media Specifications
media for intermittent sand filters is a coarse sand with an effective size between 0.3 mm and 0.5 mm.The media sand grains should be relatively uniform in size having a low uc value (less than 4.0) to promote movement of water and prevent clogging.Keywords: sand/ media specifications, filter media, media grain size.
• ### Sand Wikipedia
sand is a granular material composed of finely divided rock and mineral particles.It is defined by size, being finer than gravel and coarser than silt.Sand can also refer to a textural class of soil or soil type; i.E., a soil containing more than 85 percent sand-sized particles by mass.The composition of sand varies, depending on the local rock sources and conditions, but the most common.
• ### 3 Alculation Of 3 2 3 Geologymboldtu
m = meter 1 m = 3.2808 ft kg = kilogram 1 kg = 2.2046 lb y = year.A.If the sediment discharge is in metric mass units (kg or metric tons) divide the unit erosion rate by the density (in appropriate units) of the material for which you want to compute the lowering rate:.
• ### Chapter 13 The Sediment Transport Rate
chapter 13 the sediment transport rate introduction 1 by the sediment transport rate, also called the sediment discharge, i mean the mass of sedimentary material, both particulate and dissolved, that passes across a given flow-transverse cross section of a given flow in unit .
• ### Sand Filter An Overview Sciencedirect Topics
The process of air scouring agitates the sand with a scrubbing action, loosening the intercepted particles.After backwashing, the filter is ready to be put back into service.For a 500 mw tps, the typical backwashing flow rate would be between 25 to 30 m 3 /hr/m 2 of bed area and the air-flow rate would be 50 m 3 /hr/m 2 of filter bed area.
• ### Barrel Unit Wikipedia
outside the united states, volumes of oil are usually reported in cubic metres (m 3) instead of oil barrels.Cubic metre is the basic volume unit in the international system.In canada, oil companies measure oil in cubic metres, but convert to barrels on export, since most of canada's oil production is exported to the us.
• ### Convert M3h To Kgh Cubic Meter Per Hour To
Diferent flow rate units conversion from cubic meter per hour to kilograms (water mass) per hour.Between m3/h and kg/h measurements conversion chart page.Convert 1 m3/h into kilogram (water mass) per hour and cubic meters per hour to kg/h.The other way around, how many kilograms (water mass) per hour - kg/h are in one cubic meter per hour - m3/h unit?.
• ### Sand Loose Volume To Weight Conversion
Weight of sand, loose substance volume to weight conversions calculate weight of compounds and materials per volume.Enter volume, select a unit of volume, and specify a material or substance to search for.Use * as a wildcard for partial matches, or enclose the search string in double quotes for an exact match.Marine, aragonite, florida.
• ### Practice Quiz Ratios Unit Rates Proportions
3) blue to total marbles in a bag there are 72 candy bars.40 of them are snickers, 24 of them are milky ways, and the rest are kit kats.4) snickers to total candy bars 5) kit kats to milky ways b.Write each unit rate.Dont forget to label! (2 points each).
• ### Sandlanganav Ssmms
Stockyards available quantity (cu m) booked quantity (cu m) delivered quantity (cu m) 380 295,848.85.
• ### Tamil Nadu To Get Imported Sand At One Third Price
the sand mined locally is being sold at rs 120 per cubic feet in tamil nadu the builders association of india started importing sand from malaysia and cambodia at .
• ### Above 10 Unit River Sand For Construction
As blue metals - offering above 10 unit river sand, for construction, packaging size: 3 unit (min) at rs 110/cubic feet in chennai, tamil nadu.Get best price and read about company and get contact details and address.| id: 18987846130.
• ### Arkal Sand Separators Pelmar Eng
Each filter unit has one filtration element composed of a stack of grooved discs.See the explanation of disc filtration technology.The number of sand separators in the battery can range between 1 to 24, depending on the application requirements, such as the flow rate.One sand separator unit is designed for approximately 20 m3/hr.
• ### Pro Series High Rate Sand Filters Litproabg14
high rate sand filters the clear choice in sand filtration.Total system.Larger unit designed for most above-ground pools. seven-position multiport control valve.* when provided with a 3-foot twist-lock cord.* litproabg14 620 division street i elizabeth, nj 07201.
• ### Monitoring Sand Content The Rossum Tester
monitoring sand content: the rossum tester introduction.The sand rate in milliliters per minute (ml/min) is used with the flow rate of 0.5 gpm to determine the rate of sand production per unit of water, as shown in calculation 1.The answer will be milliliters of sand per milliliters of water (ml of sand/mil of water), as shown in.
• ### Rate Analysis Of Construction Items Earthworks
unit and can be used in the rate analysis.Similarly aluminum is purchased in kg units, but consumed in terms of running feet.Labour cost can be entered in terms of wages of mason, carpenters etc.Per day basis.Optionally labour cost can also be entered in terms of item units i.E mt2, mt3, etc.Rate analysis of construction items.
• ### Geotechnical Engineering Formulas
geotechnical engineering formulas a handy reference for use in geotechnical analysis and design.Table of contents page.Gravel and sand a-3 51-100 0-10 n.P.0 fine sand 0-50 granular materials (35% or less passing no.200) 0-35 0-35 0-35 0-35.Dry unit weight, d bulk or wet or total unit weight, m or w or t or .
• ### Rates Ocean City Md Oceanfront Hotel Castle In
For the best rate, book early and save! the owners and staff of the castle in the sand hotel take great pleasure in welcoming you to our ocean city oceanfront hotel, and to ocean city, maryland.With reasonable ocean city hotel rates at the castle, you can experience all that ocean city has to offer its visitors including our 3 mile boardwalk. | 2,064 | 8,730 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-45 | latest | en | 0.890107 |
https://www.hometalk.com/38442300/up-cycled-blinds | 1,620,565,195,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00282.warc.gz | 830,650,915 | 51,953 | # Up-Cycled Window Blinds
4 Materials
\$3
1 Hour
Easy
Six months out of the year, we have the windows open in our home. We have blinds on most of the windows which are fine but I wanted to make them look prettier so here's what I did...
As you can see, pretty boring.
I wanted the bottom of the window blind to look like lace. I found this cotton twine at Walmart for \$2.50!
## Step 1:
I cut each piece in two yard or 72" lengths. For one window, I cut 41 pieces.
## Step 2: I closed the blinds then
A - I folded the string in half making a loop
B - Place the loop over the top of the blind and bring it around the blind
C - Pull the loop at the bottom, thread the two ends through the loop
D - Pull tight
My blind was 30" in width. There are three strings on blinds, making four sections (yellow lines).
In the first section, I tied 6 strings (red)
Second section - 14 strings (red)
Third section - 14 strings (red)
Fourth section - 7 strings (red)
Total = 41 strings
## Step 3: Once all of the strings were attached
KNOT 1:
A - I pulled the loop away from the blind
B - then took the ends over top of the loop
C - and down through the loop
D - I pulled down on the ends and tightened the knot
## Step 4: KNOT 2 - a simple macrame knot. These are step by step to make this knot.
A - Using four strings
B - Make a "D" with the right most string (yellow) and bring across all strings
C - Make a "P" with the left most string (purple) and bring it under the two center strings (green & red) and through the right most string (yellow)
D - Make a "D" with the right most string (purple) and bring across all strings
E - Make a "P" with the left most string (yellow) and bring it under the two center strings (green & red) and through the right most string (purple)
F - Pull all strings tight.
## Step 5:KNOT 3 - a square knot:
A - Two strings on each side
B - Cross the two string on the right (red/yellow) over the top of the left strings (purple/green)
C - Bring the left (red/yellow) through the loop
D -Bring the right (red/yellow) across the left (purple/green)
E - Bring the right (red/yellow) through the loop
F - Tighten the knot
## Step 6: KNOT 4 is a regular knot.
Following knot 4, I tied knot 3 (a square knot) and then finished with knot 4.
I left various lengths of twine between each knot.
You could also go onto YouTube to find tutorials on knot tying if you want a different look.
Step 7: I pulled up the blinds and cut the ends of the twine evenly, cutting along the window sill.
And here is the finished project. I really like the extra pizzazz it gives to the end of the blinds. And at \$2.50, you really can't go wrong.
## Resources for this project:
See all materials
Any price and availability information displayed on [relevant Amazon Site(s), as applicable] at the time of purchase will apply to the purchase of this product.
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2 questions
• Carol Cluka
on Aug 20, 2018
could you use like cotton yarn for this project?
• Sandra Thomas
on Sep 16, 2018
I rent can i remove without damaging the blinds?
## Join the conversation
• Michelle Jones
on Jun 16, 2020
It's simple macrame -- basically just slip knots over the blind base and then square knots. You can view a YouTube video or get a book from the library. I'm thinking of making some fencing panel pieces to keep out the deer.
• Rebecca S
on Jul 8, 2020
Make sure that your loops (circles) are small enough for little heads not to go through. Nobody wants a hanged kid or animal. | 980 | 3,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-21 | latest | en | 0.913921 |
https://zbmath.org/?q=an:0879.40004&format=complete | 1,713,532,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00152.warc.gz | 978,129,093 | 11,050 | ## Some inclusion theorems for absolute summability.(English)Zbl 0879.40004
Let $\sum_{n=0}^\infty x_n \tag{1}$ be an infinite numerical series with partial sums $$s_n$$, $$n=0,1,\dots$$, and let $$A=(a_{nr})$$, $$n=0,1,\dots$$, $$r=0,1,\dots$$, be a lower semi-matrix with nonzero diagonal entries. Let $$T_n=\sum_{r=0}^n a_{nr}s_r,\;n=0,1,\dots$$. The series (1) is said summable $$|A|_k$$ $$(k\geq 1)$$, if $\sum_{n=1}^\infty n^{k-1}|T_n-T_{n-1}|^k <\infty.\tag{2}$ In the case of absolute Riesz summability, i.e., $$a_{nr}= p_r \cdot \Big (\sum_{j=0}^n p_j\Big)^{-1},\;0\leq r\leq n$$, $$a_{nr}=0,\;r> n$$, where $$p_n$$, $$n=0,1,\dots$$, is a sequence of positive real numbers such that $$\lim_{n\to \infty}\sum_{j=0}^n p_j =\infty$$, we write $$|R,p_n|_k$$ for summability $$|A|_k$$.
The aim of this paper is to give necessary and sufficient conditions for the series (1) to be summable $$|A|_k$$, whenever it is summable $$|R,p_n|_1$$. In the proof of this main result (i.e. Theorem on p. 600) a little functional analytic apparatus is used. More precisely, e.g. in the necessity-part of the proof just mentioned the Banach-Steinhaus theorem on certain $$BK$$-spaces of sequences is suitably applied. As corollaries, the authors deduce from their Theorem some known inclusion theorems for absolute Riesz (particularly Cesàro) summability.
Reviewer’s remark. The absolute summability $$|A|_1$$ (i.e. condition (2) with $$k=1$$) for an infinite square matrix $$A$$ of real or complex numbers appeared as early as in the paper by R. P. Cesco [Univ. Nac. La Plata., Publ. Fac. Ci. fis-mat., Revista 2, 147-156 (1941; Zbl 0061.12001)].
### MSC:
40D25 Inclusion and equivalence theorems in summability theory 40F05 Absolute and strong summability 40C05 Matrix methods for summability 40G05 Cesàro, Euler, Nörlund and Hausdorff methods 40H05 Functional analytic methods in summability
Zbl 0061.12001
Full Text:
### References:
[1] L.S. Bosanquet: Mathematical Reviews 11 (1950), 654. · Zbl 0036.14509 · doi:10.1112/jlms/s1-25.2.102 [2] G. Das: Tauberian Theorems for Absolute Nörlund Summability. Proc. London Math. Soc. 19 (1969), 357-384. · Zbl 0183.32603 · doi:10.1112/plms/s3-19.2.357 [3] I.J. Maddox: Elements of Functional Analysis. Cambridge University Press, 1970. · Zbl 0193.08601 [4] C. Orhan: On Absolute Summability. Bull. Inst. Math. Acad. Sinica 15 (1987), 433-437. · Zbl 0649.40009 [5] C. Orhan and M.A. Sarigöl: On Absolute Weighted Mean Summability. Rocky Mountain J. Math. 23 (1993), 1091-1097. · Zbl 0799.40003 · doi:10.1216/rmjm/1181072543 [6] G. Sunouchi: Notes on Fourier Analysis XVIII, Absolute Summability of a Series with Constant Terms. Tohoku Math. Journal 2 (1949), 57-65. · Zbl 0041.39101 · doi:10.2748/tmj/1178245769
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. In some cases that data have been complemented/enhanced by data from zbMATH Open. This attempts to reflect the references listed in the original paper as accurately as possible without claiming completeness or a perfect matching. | 1,073 | 3,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.66771 |
https://crackbmat.com/2010-section-1-question-32-2/ | 1,670,507,411,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00356.warc.gz | 193,687,769 | 15,031 | 32) C
We can use approximate values for the earnings.
Top 20% earn = \$162,000 + \$51,000 = \$213,000
Bottom 20% earn = \$16,000 + \$17,000 = \$33,000
Ratio = 213000 : 33,000 = 6.5 : 1
Since we rounded down the bottom 20% and rounded up the top 20%, we need to decrease the overall ratio. So, C (6 : 1) is the answer. | 113 | 321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-49 | latest | en | 0.851884 |
https://www.codewars.com/users/C_Minor/comments | 1,611,809,327,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00398.warc.gz | 706,703,163 | 13,689 | Thank you!
• GoumiesOnecommented on "Find the calculation type" javascript solution
The property of the object is computed.
It's like:
``````calcType(1,2,3) => {
[-2]: "subtraction",
[2]: "multiplication",
[O.5]: "division"
}
``````
and [res] calls the equivalent property. So up here, it calls `3`
• JAK1008commented on "Find the calculation type" javascript solution
Hey, what do the square brackets do in this case? Could you please explain the syntax? Thanks!
• rowcasedcommented on "Check Even Number" kumite
it's an ancient JediScript mind-trick
• C_Minorcommented on "Check Even Number" kumite
What the hell
• Stettafirecommented on "Subtract the Sum" kata
I was wondering that myself, it seems like it isn't ranked properly if that's the case.
The kata is based on a trick more so then actual programming, which seems to defeat the point of a code kata.
• MaximPravenkiycommented on "Sum Strings as Numbers" javascript solution
Ты душный, братик... Может ты банчишь мороженым на москварике?
• Neocortexxcommented on "Training JS #1: create your first JS function and print "Helloworld!"" javascript solution
You hacked it! 👏
• C_Minorcommented on "Maximum Multiple" rust solution
Can someone explain this sorcery?
• Voilecommented on "Sum Strings as Numbers" javascript solution
You are supposed to write yourself all the solutions
It has nothing to do with having to write everything from scratch.
• cliffstampcommented on "Subtract the Sum" kata
This comment is hidden because it contains spoiler information about the solution
• Talrendiscommented on "Define a card suit" javascript solution
oh man i never notice such obvious things :(
• C_Minorcommented on "Sum Strings as Numbers" javascript solution
You are supposed to write yourself all the solutions, in real world you can and must use other framework/libraries/utilities wrote by others in order to not reinventing the wheel, but here, you should use just your skill in programming to come to a solution
• dHilbertcommented on "Sum Strings as Numbers" javascript solution
This isnt cheating imo, he went out and found a bignum. Then used it in the code he wrote. Are we supposed to rewrite every library function we use?
• user7856383commented on "Find the position!" javascript solution
It works only for lowerCase letters | 568 | 2,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.889927 |
https://www.engineersedge.com/beam_bending/double_integration_method_example_4_15560.htm | 1,618,234,069,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00357.warc.gz | 821,513,174 | 7,839 | Related Resources: beam bending
### Double Integration Method Example 4 Proof Simply Supported Beam of Length L with Partial Distributed Load
Double Integration Method Example 4 Proof Simply Supported Beam of Length L with Partial Distributed Load.
The Double Integration Method, also known as Macaulay’s Method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get the equation of the elastic curve.
Elastic Curve
Compute the value of EI δ at midspan for the beam loaded as shown in the figure above. If E = 10 GPa, what value of I is required to limit the midspan deflection to 1/360 of the span?
∑ MR2 = 0
4 R1 = 300 ( 2 ) ( 3 )
R1 = 450 N
∑ MR1 = 0
4 R2 = 300 ( 2 ) ( 1 )
R2 = 150 N
E I y'' = 450 x - 0.5 (300) x2 + 0.5 (300) < x - 2 >2
E I y'' = 450 x - 150 x2 + 150 < x - 2 >2
E I y' = 225 x2 - 50 x3 + 50 < x - 2 >3 + C1
E I y = 75 x3 - 12.5 x4 + 12.5 < x - 2 >4 + C1 x + C2
At x = 0, y = 0, therefore C2 = 0
At x = 4 m, y = 0
0 = 75 (43) - 12.5 (44) + 12.5 ( 4 -2 )4 +4 C1
C1 = -450 N·m2
Therefore,
E I y = 75 x3 - 12.5 x4 + 12.5 < x - 2 >4 - 450 x
At x = 2 m (midspan)
E I ymidspan = 75 ( 23 ) - 12.5 ( 24) + 12.5 ( 2 - 2 )4 - 450 (2)
E I ymidspan = -500 N·m3
E I δmidspan = 500 N·m3
Maximum midspan deflection
δmidspan = L/360 = 4 / 360 = M / 90
δmidspan = ( 100 / 9 )
Thus,
10,000 I (100 / 9) = 500 ( 10003 )
I = 4,500,000 mm4 or
I = 4.5 x 106 mm4
Related:
Reference:
• Dr. ZM Nizam Lecture Notes
• Shingley Machine Design, 4-3 "Deflection Due to Bending"
• Beam Deflection by Integration Lecture Presentation Paul Palazolo, University of Memphis,
• Beam Deflections Using Double integration, Steven Vukazich, San Jose University
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Add a Comment (you must be logged in to post comment Register): Name: Email: (Optional) Comment: | 723 | 2,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2021-17 | latest | en | 0.811144 |
http://everything2.com/title/Gauss%2527+Law | 1,386,923,857,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164920565/warc/CC-MAIN-20131204134840-00014-ip-10-33-133-15.ec2.internal.warc.gz | 65,660,480 | 8,742 | Carl Friedrich Gauss was a brilliant mathematician, physicist and astronomer. His mathematical works included many theories and papers in the realm of pure math, such as Number Theory. In the physics world, he is most remembered for his law regarding electric fields.
All Electric charge exact forces upon each other, which is regulated by Coulomb's Law
F = k * q1 * q2 / r * r, where constant k = 8.9*10^9).
Electric fields make everything simpler by graphically showing the strength of attraction or repulsion by one particular charge (it doesn't emit a field, it is imaginary). The strength of the field is based on the strength of the charge. The field would then exert a force upon any charge in its vincinity (technically infinite, but at very long distances it is practically zero). On a positive test charge, the strength of the force would be
F = E * q
Where E is the strength of the field in newtons per coulomb and q is charge of the test charge.
This all builds up to the concept of electric flux. Flux is the a measurement of the amount of electric field passing through an area. The formula is given by:
Flux = (Integral) E dA, A is area, E is field strength
What has it to do with Gauss then? Well he made some obsevations regarding flux of enclosed surfaces (also known as Gaussian surfaces). For an charge enclosed in a space, no matter how you move the space around, warp it, stretch it, or whatever, as long as the charge is still inside, the flux remains constant. Here is an analogy. You get a box and put a positive charge inside. If you increase the dimensions of the box by a factor of 2, then at the stretched surface of the Gaussian space, the stregth of the field is decreased. However, the area of which the weakened field is going through increases by the same factor. Gauss came out with three conclusions:
• Whether there is a net outward or inward electric flux through a closed surface depends on the sign of the enclosed charge
• Charges outside the surface do not giv a net electric flux through the surface
• The net electric flux is directly proportional to the net amount of charge enclosed within the surface but is otherwise independent of the size of the closed surface
• Simplified, Gauss' Law is an alternative way to express the relationship between electric charge and electric field (the other way is Coulomb's Law). It states:
The total electric flux through any closed surface (a surface enclosing a definite volume) is proportional to the total net electric charge inside the surface.
The equation is:
Flux = (Intergral) E dA = Qenclosed / e , where Q is the algebraic sum of the charge inside the enclosed space, e is 8.854 X 10^-12
Hence electric flux passing through a closed surface is really just dependent on charge.
Gauss' Law is not just an alternative way to state the relationship between electric charge and electric field (the other being Coulomb's Law). In fact, Coloumb's Law can be derived directly from Gauss' Law--the two are truly the same law, stated in different terms. Note: in my notation, variables in bold are vector quantities. ε0 is the permitivity of free space and is 8.85 x 10-12 Farads/meter or coloumbs-squared/Newtons-meters-squared. All integrals are to be understood as loop integrals, over the entire Gaussian surface.
Take a point charge +q, around which is envisioned a Gaussian concentric spherical surface of radius r. Divide this surface into differential areas dA. By definition, the area vector for each area dA is dA with magnitude equal to the area and direction perpendicular to the surface (directed outward from the interior of the sphere). From the symmetry of the sphere, we also know that at any point on the sphere, the electric field vector E is also perpendicular to dA and directed outward, and so is parallel to dA. Therefore the angle between E and dA is zero.
Gauss' Law is usually stated:
ε0EdA = qenclosed
Since E and dA are parallel, their dot product is simply EdA. The enclosed charge is simply q (since we made our sphere around this charge), and is a constant. E is also a constant, because it varies only radially for a point charge, and we have restricted the radius to the radius r of our sphere. So Gauss' Law can be rewritten:
ε0EdA = q
The loop integral of the differntial areas is just the area of the sphere, 4πr2. Substituting, we have:
ε0E(4πr2) = q.
This is:
E = q/(4πε0r2)
which is Coloumb's Law! :-)
Btw, if anyone knows how to make loop integrals in HTML, I'd love to know!
Log in or register to write something here or to contact authors. | 1,056 | 4,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2013-48 | longest | en | 0.955175 |
http://ask.metafilter.com/137317/Help-me-find-an-online-surface-areavolume-calculator-that-uses-314-as-an-approximation-for-pi | 1,510,973,038,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804518.38/warc/CC-MAIN-20171118021803-20171118041803-00514.warc.gz | 22,975,653 | 16,514 | # Help me find an online surface area/volume calculator that uses 3.14 as an approximation for pi.November 5, 2009 8:34 AM Subscribe
I'm looking for an online surface area and volume calculator, where I plug in various measurements of a cylinder or sphere and it calculates the answer for me. But I want one that approximates pi as 3.14.
No, you're not helping me to cheat. I'm answer checking middle school math materials. All the surface area and volume calculators I've found online use a much longer approximation of pi, and all the questions I'm working on specifically direct students to "use 3.14 for pi".
(Note: if your answer includes any variation on "It doesn't take that long to calculate the surface area or volume yourself", then you are an unhelpful person who isn't very good at reading.)
posted by 23skidoo to Education (8 answers total)
I would use excel for this. It'll take 3 seconds.
posted by dmd at 8:37 AM on November 5, 2009 [1 favorite]
Agreed with Excel.
posted by dfriedman at 8:38 AM on November 5, 2009
I would use excel for this. It'll take 3 seconds.
I have no idea how to make excel do this, so if you could explain exactly what to do, I'd appreciate it.
posted by 23skidoo at 8:40 AM on November 5, 2009
Hold on, 23skidoo - I'm making a google document to do it, which you should be able to access from anywhere with a google account.
posted by muddgirl at 8:44 AM on November 5, 2009
OK, hopefully this works: Area and Volume calculations for sphere and cylinder. Just go to the correct sheet and enter the height and radius. You can even change Pi to whatever you'd like, although I defaulted it to 3.14. If the problem gives the diameter, just divide by two and enter the radius value.
If you double-click on the answer cells (highlighted in yellow), you can see the formula used to make the calculation and make changes if you like. A formula has to start with an equals sign "=" and anything in the format A2 or B12 indicates that the formula is taking a value from the cell in Column A row 2, or Column B row 12, etc.
posted by muddgirl at 8:55 AM on November 5, 2009 [1 favorite]
Also, it just occured to me that any malicious person could come along and edit the sheet to give the wrong answer, so if you'd like, I can lock it down and send you a private link by email/memail. Or, I think if you have a google account you can save a private copy to your own Google Docs folder and you will be the only one able to edit it.
posted by muddgirl at 8:59 AM on November 5, 2009
Thanks, muddgirl. I'll figure out how to save a private copy to thwart any shenanigans.
posted by 23skidoo at 9:13 AM on November 5, 2009
Have you tried Wolfram Alpha? It looks like it will round for you.
volume of sphere with a radius of 5cm
posted by blue_beetle at 10:54 AM on November 5, 2009
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# Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!!
new topic post reply Update application status
Author Message
Manager
Joined: 23 Nov 2016
Posts: 137
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
### Show Tags
19 Oct 2017, 15:18
souvik101990 wrote:
Tons of people only update after they get their final decision. That’s why when you look at last years data, you are including TONS of entries for R1 that were made in December after the admits.
Sent from my iPhone using Tapatalk
I agree
Sent from my iPhone using GMAT Club Forum mobile app
I agree too. I tried to factor it in my inference. Raw data suggests only 25% of invites are out. We can be more conservative and say that approx 60% of R1 invites are out. That still leaves 40% to go..
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Joined: 04 Jan 2016
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
### Show Tags
19 Oct 2017, 15:20
MBAPrepCoach wrote:
empo01 for what it's worth, I have not heard back from my client in Spain about MIT and he was invited to HBS.
Thanks!!! And wish luck to your Spains client!!
Manager
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Posts: 87
Location: United States
Concentration: Finance, Other
GMAT 1: 750 Q48 V44
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
### Show Tags
19 Oct 2017, 16:54
stoiczoan wrote:
souvik101990 wrote:
Tons of people only update after they get their final decision. That’s why when you look at last years data, you are including TONS of entries for R1 that were made in December after the admits.
Sent from my iPhone using Tapatalk
I agree
Sent from my iPhone using GMAT Club Forum mobile app
I agree too. I tried to factor it in my inference. Raw data suggests only 25% of invites are out. We can be more conservative and say that approx 60% of R1 invites are out. That still leaves 40% to go..
Now we’ve deviated into pure heresay.
Sent from my iPhone using GMAT Club Forum mobile app
Manager
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Posts: 137
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GMAT 1: 770 Q49 V47
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 17:37
Now we’ve deviated into pure heresay.
Sent from my iPhone using GMAT Club Forum mobile app
Haha.. I'd say we were always in "hearsay" territory..
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Joined: 08 Apr 2017
Posts: 87
Location: United States
Concentration: Finance, Other
GMAT 1: 750 Q48 V44
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 17:38
stoiczoan wrote:
Now we’ve deviated into pure heresay.
Sent from my iPhone using GMAT Club Forum mobile app
Haha.. I'd say we were always in "hearsay" territory..
Fair
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 19:31
souvik101990 wrote:
Got an invite a few hours back
Congratulations Souvik! What is your interview date? And how much time do you have to write your essay?
Two Indians have updated their invites here. Now am getting worried.
Manager
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 19:38
Livewire has reported quite some interview invites for New Delhi location. Do not mean to spread the negativity but this might well be game over for Indians.
Manager
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 19:48
Also, Souvik and patrickbateman2k15 did anybody from MIT visit your linkedin profile? Or anybody else's for that matter?
Intern
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Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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Updated on: 24 Oct 2017, 22:49
edited
Originally posted by patrickbateman2k15 on 19 Oct 2017, 20:44.
Last edited by patrickbateman2k15 on 24 Oct 2017, 22:49, edited 1 time in total.
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 21:49
Hi Everyone.
Congrats to those that secured interviews yesterday! Just a quick question, has anyone that specified London as their first choice received an invitation to interview yet?
Thanks
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 22:39
I too got my invite. However, it will be quite a hassle to get the Army to give me a leave during the interview period.
Current Student
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 23:08
atulsunilanand wrote:
I too got my invite. However, it will be quite a hassle to get the Army to give me a leave during the interview period.
All the best. I was rooting for you
BTW, are you based in Delhi?
_________________
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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19 Oct 2017, 23:12
1
souvik101990 wrote:
atulsunilanand wrote:
I too got my invite. However, it will be quite a hassle to get the Army to give me a leave during the interview period.
All the best. I was rooting for you
BTW, are you based in Delhi?
Thanks. Best of luck to you too!
I am from Mumbai but am working in the interiors of North East at the moment.
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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20 Oct 2017, 01:39
atulsunilanand wrote:
souvik101990 wrote:
atulsunilanand wrote:
I too got my invite. However, it will be quite a hassle to get the Army to give me a leave during the interview period.
All the best. I was rooting for you
BTW, are you based in Delhi?
Thanks. Best of luck to you too!
I am from Mumbai but am working in the interiors of North East at the moment.
I guess its game over for Indians like me who did not get an invite!! Is any of you Delhi based?
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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20 Oct 2017, 01:46
Many congratulations to everyone who got an invite!! do drop in a line as to how prep is going.
I have two questions. Any insight would be appreciated.
My understanding is that interview dates are from October 30 to mid November. Wanted to know if that is the case for those who got invites in the first and second wave as well?
So if one was to get an invite in the next wave... say for Delhi or Cambridge or wherever, would the interview dates be the same? It would just mean less time to prepare / write the essay.
Secondly, based on past trends or last years history, once Sloan is done with a particular region do they not revisit it in subsequent rounds? I've noticed that Cambridge invites have been coming in the last two waves
Thanks!
Posted from my mobile device
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Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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Updated on: 24 Oct 2017, 22:48
edited
Originally posted by patrickbateman2k15 on 20 Oct 2017, 02:00.
Last edited by patrickbateman2k15 on 24 Oct 2017, 22:48, edited 1 time in total.
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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20 Oct 2017, 02:31
patrickbateman2k15 wrote:
I am from Mumbai and got an interview invite yesterday. When I was booking my interview slot in Delhi, I noticed that there were about 8 slots per day and the dates were mentioned as 6th to 9th Nov. Does it mean that in R1 they would be giving a total of 4X8=32 interview calls to India based candidates. But then, there could be more than one interviewer visiting India meaning more than one candidate could be interviewed in a single slot. What do others think?
Does it mean game over for us Indians who did not get an invite?
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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20 Oct 2017, 02:36
Never say never. But it would appear so
Sent from my iPhone using GMAT Club Forum
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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20 Oct 2017, 03:17
Got an invite for an interview in Delhi. Guys any place where to start preparing for an interview.
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Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! [#permalink]
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20 Oct 2017, 04:08
sid192011 wrote:
Got an invite for an interview in Delhi. Guys any place where to start preparing for an interview.
Congrats! You will get an email from adcom with detailed info on behavioral interview and what you should do to prep.
In the meantime you can find a list of on the internet for “tell me about a time..” type questions, and think of all the examples from your life.
Also read up on MIT in detail.
_________________
Re: Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!! &nbs [#permalink] 20 Oct 2017, 04:08
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# Calling all MIT Sloan MBA Applicants: (2018 Intake) Class of 2020!!
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# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 3,304 | 11,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-39 | latest | en | 0.909883 |
londonerslife.blogspot.com | 1,508,370,662,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823168.74/warc/CC-MAIN-20171018233539-20171019013539-00144.warc.gz | 198,861,040 | 11,133 | ## Monday, March 28, 2005
### A Puzzle
Three men go into a cafe, they each order a coffee and a pastry. The bill comes to £30, each pay £10 in cash. The waiter takes the money to the counter, and at the counter the head waiter says to the waiter, there has been a mistake, they were overcharged, it is meant to be £25. So the waiter goes back to give back the money, but decides well they don't know they were overcharged so much, and decides not to give it all back to them. He give back the three men £3, £1 each, and pockets the other £2 for himself. So if each person paid £9 and the waiter pocketed £2, where is the other pound £ ?
Anyone know?
be said...
any clues?
Ian said...
The missing pound isn't missing! The 30 pounds is irrelevant.
the meal was 25, the waiter stole 2 quid so we need 27 pound total.
the 3 dinners paid 9 pound each..
3*9 = 27
or
3*9 = 25 + 2
Russ, remember that bloody lecturer that tried to convince us 1 was the biggest number?! same slight of hand.
Speakers Corner said...
I do recall that lecturer, the one with a bum bag on all the time. Those lessons where hard work.
It is as Ian says a slight of hand, more psychology than Maths, and note I say Maths, not Math.
ian said...
actually thats my pet peeve too.
It's also physics not physic but there you go.
You say potato and I say,oh.. potato.
BrianDavison said... | 356 | 1,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-43 | longest | en | 0.976988 |
https://machinelearningmastery.com/2d-test-functions-for-function-optimization/ | 1,721,366,556,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00485.warc.gz | 316,133,787 | 61,793 | Use the offer code 20offearlybird to get 20% off. Hurry, sale ends soon!
# Two-Dimensional (2D) Test Functions for Function Optimization
Function optimization is a field of study that seeks an input to a function that results in the maximum or minimum output of the function.
There are a large number of optimization algorithms and it is important to study and develop intuitions for optimization algorithms on simple and easy-to-visualize test functions.
Two-dimensional functions take two input values (x and y) and output a single evaluation of the input. They are among the simplest types of test functions to use when studying function optimization. The benefit of two-dimensional functions is that they can be visualized as a contour plot or surface plot that shows the topography of the problem domain with the optima and samples of the domain marked with points.
In this tutorial, you will discover standard two-dimensional functions you can use when studying function optimization.
Kick-start your project with my new book Optimization for Machine Learning, including step-by-step tutorials and the Python source code files for all examples.
Let’s get started.
Two-Dimensional (2D) Test Functions for Function Optimization
Photo by DomWphoto, some rights reserved.
## Tutorial Overview
A two-dimensional function is a function that takes two input variables and computes the objective value.
We can think of the two input variables as two axes on a graph, x and y. Each input to the function is a single point on the graph and the outcome of the function can be taken as the height on the graph.
This allows the function to be conceptualized as a surface and we can characterize the function based on the structure of the surface. For example, hills for input points that result in large relative outcomes of the objective function and valleys for input points that result in small relative outcomes of the objective function.
A surface may have one major feature or global optima, or it may have many with lots of places for an optimization to get stuck. The surface may be smooth, noisy, convex, and all manner of other properties that we may care about when testing optimization algorithms.
There are many different types of simple two-dimensional test functions we could use.
Nevertheless, there are standard test functions that are commonly used in the field of function optimization. There are also specific properties of test functions that we may wish to select when testing different algorithms.
We will explore a small number of simple two-dimensional test functions in this tutorial and organize them by their properties with two different groups; they are:
1. Unimodal Functions
1. Unimodal Function 1
2. Unimodal Function 2
3. Unimodal Function 3
2. Multimodal Functions
1. Multimodal Function 1
2. Multimodal Function 2
3. Multimodal Function 3
Each function will be presented using Python code with a function implementation of the target objective function and a sampling of the function that is shown as a surface plot.
All functions are presented as a minimization function, e.g. find the input that results in the minimum (smallest value) output of the function. Any maximizing function can be made a minimization function by adding a negative sign to all output. Similarly, any minimizing function can be made maximizing in the same way.
I did not invent these functions; they are taken from the literature. See the further reading section for references.
You can then choose and copy-paste the code one or more functions to use in your own project to study or compare the behavior of optimization algorithms.
## Unimodal Functions
Unimodal means that the function has a single global optima.
A unimodal function may or may not be convex. A convex function is a function where a line can be drawn between any two points in the domain and the line remains in the domain. For a two-dimensional function shown as a contour or surface plot, this means the function has a bowl shape and the line between two remains above or in the bowl.
Let’s look at a few examples of unimodal functions.
### Unimodal Function 1
The range is bounded to -5.0 and 5.0 and one global optimal at [0.0, 0.0].
Running the example creates a surface plot of the function.
Surface Plot of Unimodal Optimization Function 1
### Unimodal Function 2
The range is bounded to -10.0 and 10.0 and one global optimal at [0.0, 0.0].
Running the example creates a surface plot of the function.
Surface Plot of Unimodal Optimization Function 2
### Unimodal Function 3
The range is bounded to -10.0 and 10.0 and one global optimal at [pi, pi]. This function is known as Easom’s function.
Running the example creates a surface plot of the function.
Surface Plot of Unimodal Optimization Function 3
### Want to Get Started With Optimization Algorithms?
Take my free 7-day email crash course now (with sample code).
Click to sign-up and also get a free PDF Ebook version of the course.
## Multimodal Functions
A multi-modal function means a function with more than one “mode” or optima (e.g. valley).
Multimodal functions are non-convex.
There may be one global optima and one or more local or deceptive optima. Alternately, there may be multiple global optima, i.e. multiple different inputs that result in the same minimal output of the function.
Let’s look at a few examples of multimodal functions.
### Multimodal Function 1
The range is bounded to -5.0 and 5.0 and one global optimal at [0.0, 0.0]. This function is known as Ackley’s function.
Running the example creates a surface plot of the function.
Surface Plot of Multimodal Optimization Function 1
### Multimodal Function 2
The range is bounded to -5.0 and 5.0 and the function as four global optima at [3.0, 2.0], [-2.805118, 3.131312], [-3.779310, -3.283186], [3.584428, -1.848126]. This function is known as Himmelblau’s function.
Running the example creates a surface plot of the function.
Surface Plot of Multimodal Optimization Function 2
### Multimodal Function 3
The range is bounded to -10.0 and 10.0 and the function as four global optima at [8.05502, 9.66459], [-8.05502, 9.66459], [8.05502, -9.66459], [-8.05502, -9.66459]. This function is known as Holder’s table function.
Running the example creates a surface plot of the function.
Surface Plot of Multimodal Optimization Function 3
This section provides more resources on the topic if you are looking to go deeper.
### Summary
In this tutorial, you discovered standard two-dimensional functions you can use when studying function optimization.
Are you using any of the above functions?
Let me know which one in the comments below.
Do you have any questions?
## Get a Handle on Modern Optimization Algorithms!
#### Develop Your Understanding of Optimization
...with just a few lines of python code
Discover how in my new Ebook:
Optimization for Machine Learning
It provides self-study tutorials with full working code on:
Gradient Descent, Genetic Algorithms, Hill Climbing, Curve Fitting, RMSProp, Adam, and much more...
### 9 Responses to Two-Dimensional (2D) Test Functions for Function Optimization
1. Ahgay Gabon March 26, 2021 at 10:52 am #
Thanks for this awesome addition. But with optimization algorithms, there be one solution to be passed to this function. How to accomplish that? and how to plot these benchmarks given only a solution that is outputted in an algorithm trial?
Thanks!
2. Terrence May 11, 2021 at 6:11 am #
Hi, for Easom’s function should the location of the global minimum not be [pi,pi]? I
3. Farzad May 17, 2021 at 11:10 pm #
What is the best algorithm for optimizing the cutting of two-dimensional plates?
Algorithm with the least amount of waste, I want to use this algorithm to calculate the MDF cut
• Jason Brownlee May 18, 2021 at 6:15 am #
Perhaps try a suite of algorithms and discover what works well or best for your objective problem.
• Adrian Tam August 10, 2021 at 6:53 am #
Why not try CNN-LSTM?
4. ali September 3, 2021 at 5:00 pm #
in eggholder,
i want to calculate inputs on specific values …
like if im passing output=-959.6
it will gives input =[512,404]
kindly tell me how it possible
• Jason Brownlee September 4, 2021 at 5:17 am #
You are talking about reverse optimization, or inverting the problem. I don’t believe it is possible/tractable. | 1,924 | 8,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-30 | latest | en | 0.889671 |
https://ebrary.net/148424/education/complex_fields | 1,685,489,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00032.warc.gz | 265,881,090 | 10,118 | # Complex fields
Activity, taxonomy and property systems provide the basic resources for construing field. But before we can return to mapping the explanation of seasons this chapter began with, there is one more system we need to introduce. This system makes room for interdependent variables that organize the complex constellations of meaning in technical discourses such as those of science. There are two basic ways of building interdependency: reconstruing and interrelating.
## Reconstruing variables
Beginning with the reconstrual of field variables, we can return to the tilt of the earth that we mentioned is crucial for explaining seasons. In one sense, we can now readily account for this in our expanded model - tilt is a property of the earth and so can be arrayed and gauged. Indeed, our original text explaining the seasons does just that:
Seasons happen because Earth’s axis is tilted at an angle of about 23.4 degrees
In this form, where the Earth’s axis is specified as being tilted, this analysis is unproblematic. However, in the next sentence, the property realized verbally as tilted above is nominalized as tilt:
Earth’s axial tilt
In addition, the nominalized tilt is classified by axial. This positions tilt in a classification taxonomy; axial tilt is a type of tilt. This is problematic for the outline of field resources canvassed thus far as classification is a key feature of items, not properties. Moreover, in terms of Hao’s model of ideational discourse semantics (2020a, Chapter 6 of this volume), the nominalized axial tilt is not a quality, which is the typical realization of property. Rather, it is a measured dimension of the earth that can be reorganized lexicogrammatically as a FocusAThing structure:
Axial tilt of Earth
To reconcile these two seemingly conflicting analyses - tilt as a property and tilt as an item - what we will suggest here is that the property tilted is being reconstrued as an item tilt. This we will refer to as an itemized property. This analysis is based on the fact that axial tilt shows many of features of both properties and items: it can be arrayed and gauged like properties and can be taxonomized like items. And from the perspective of discourse semantics, it is not realized by a quality, which is the prototypical realization of qualitative properties, nor by an entity, which is the standard realization of items, but rather by a dimension of an entity.
Such itemized reconstruals are regularly used to ‘name’ broader sets of properties. Indeed, we have used a number of these throughout this paper to group together various properties. In the following examples, the underlined itemized properties are ‘names’ of the bolded properties:
The colours (of the flame robin) are slaty, whitish and sandier brown.
The shape (of the flame robin) is described as slender-looking.
Itemized properties comparable to these are the basis for symbolic variables in mathematics (see Doran, Chapter 7 of this volume). For example, in the following formula the acceleration of a moving body is symbolized as a, the mass of that body is m and the force on the body is F:
Like all properties, each of these can be numerically gauged - in this case through Newtons (N), metres per second squared (m/s2) and kilograms (kg):
But like items, they can all be taxonomized:
angular acceleration, linear acceleration; centripetal force, electrostatic force; relativistic mass, rest mass.
Reconstruals of properties as items is a regular feature of many fields. Under this interpretation, field is a resource not just for construing items, activities and properties, it is also a resource for reconstruing meanings. It enables multiple overlapping perspectives on phenomena to be realized in a single instance. This is emphasized by the fact that in addition to being itemized, properties can also be reconstrued as activities. Similarly, items can be dynamized by being reconstrued as activities, and activities can in turn be itemized. Each of these options have typical realizations in discourse semantics, as introduced below.
When properties are reconstrued as activities, such activated properties enable a dynamic unfolding of a property:
It vets hotter
In terms of property, this indicates an array of temperature (degrees of heat). But in terms of activity, it can be used to moment a larger activity:
It vets hotter л
And then eventually it melts.
This type of activation moves us into the realm of'becoming’, where properties are, in effect, dynamized.
Activating items very commonly involves them being positioned in a taxonomy. The following excerpt from a university physics textbook shows an example of this for composition (Young and Freedman 2012: 742):
... now we let the ball touch the inner wall... The surface of the ball becomes part of the cavity surface.
Here, the part whole relation between the surface of a ball and the cavity surface of a wall is activated, and becomes a moment in a larger activity:
We let the ball touch the inner wall
A
The surface of the ball becomes part of the cavity surface.
Finally, just as items can be activated, activities can be itemized.10 This regularly happens in the process of technicalization, as scientific terms are distilled as activity entities (see Hao 2020a and Chapter 6 of this volume for discussion of the discourse semantics of activity entities). One of our initial examples of a momented activity in fact showed a series of itemized activities that moment phagocytosis - itself an itemized activity; the itemized activities involved are underlined below:
Detection
A
hi vest ion
A
Phagosome forms л
Fusion with lysosome
A
Dm st ion
A
Discharge
Like all items, these itemized activities can enter into a taxonomy. For example, phagocytosis is one type of endocytosis, along with potocytosis and micropinocytosis, which all contrast with exocytosis.
Table 5.1 shows various reconstruals and some typical discourse semantic realizations (from Hao 2020a). Note here that the order in which the reconstrual takes place is significant - an itemized activity is different from an activated item.
TABLE 5.1 Field reconstruals
Field reconstruals Typical discourse semantic realization Example itemized property activated property itemized activity activated item measured or perceived dimension occurrence figure state figure activity entity state figure The colour of skin It heats up It gets hotter Phagocytosis You become part of the team | 1,369 | 6,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-23 | latest | en | 0.943454 |
https://uvirtual.ujaen.es/pub/es/informacionacademica/catalogoguiasdocentes/p/patie/2015-16/5/742A/74212003/en/2015-16-74212003_en.html | 1,696,222,374,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2023-40/segments/1695233510967.73/warc/CC-MAIN-20231002033129-20231002063129-00700.warc.gz | 635,968,125 | 7,204 | ## Syllabus 2015-16 - 74212003 - Optical Technologies (Tecnologías ópticas)
Caption
• Level 1: Tutorial support sessions, materials and exams in this language
• Level 2: Tutorial support sessions, materials, exams and seminars in this language
• Level 3: Tutorial support sessions, materials, exams, seminars and regular lectures in this language
DEGREE: Máster en Ingeniería de Telecomunicación FACULTY: SCHOOL OF ENGINEERING OF LINARES ACADEMIC YEAR: 2015-16 COURSE: Optical Technologies
SYLLABUS
1. COURSE BASIC INFORMATION
NAME: Optical Technologies CODE: 74212003 ACADEMIC YEAR: 2015-16 LANGUAGE: English LEVEL: 1 ECTS CREDITS: 4.0 YEAR: 1 SEMESTER: SC
2. LECTURER BASIC INFORMATION
NAME: PÉREZ DE PRADO, ROCÍO JOSEFINA DEPARTMENT: U134 - INGENIERÍA DE TELECOMUNICACIÓN FIELD OF STUDY: 800 - TEORÍA DE LA SEÑAL Y COMUNICACIONES OFFICE NO.: D - D-129 E-MAIL: rperez@ujaen.es P: 953648659 WEBSITE: http://www4.ujaen.es/~rperez/ ORCID: https://orcid.org/0000-0001-6097-4016 LANGUAGE: English LEVEL: 1
3. CONTENT DESCRIPTION
THEORY
Unit 1. FUNDAMENTALS OF OPTICAL WAVE AND OPTICAL ELECTROMAGNETIC
• Introduction to Photonics.
• Electromagnetic fields and waves.
• Gauss Beam.
• Diffractive optics.
Unit 2. FUNDAMENTALS OF OPTICAL PHOTONICS
• Optical photonics.
• Photons and atoms.
• Amplifiers lasers.
• Lasers.
• Photonic crystals.
Unit 3. PHOTONIC COMMUNICATIONS TECHNOLOGIES
• Basic photonic devices for optical communication systems.
• Passive photonic technologies for optical networks.
• Specialty fibres.
PRACTICAL SESSIONS
Practice 1. LASER LIGHT GAUSSIAN BEAM: TEM00 MODE (2 SESSIONS)
Gaussian beams are commonly used to represent laser beams. In general, the propagation of a laser beam can be approximated by assuming that the laser beam has a Gaussian profile intensity corresponding to the theoretical TEM00 mode. Thus, a first step in the study of laser devices is important to analyse the Gaussian beam. This practice is intended to deepen into the foundations of the Gaussian beam studied in class through signal processing with MATLAB.
Documents to deliver:
• Source files: All those file necessary for the proper compilation and construction code, never executable files or object code. The delivered code must meet the specific objectives described in the practice instructions and have comments to help monitoring and correction and not have syntax errors or execution.
• Memory. Arguments and justifications to all questions raised in practice.
In each practice the following aspects will be assessed:
• Operational knowledge of the subject.
• Structure in the statement of the problem and its resolution.
• Resolution: type of solution, justification and correction.
• Originality: no exercise is valued if its resolution is equal to or much like another done in a previous course by other students.
• Spelling, format and presentation.
• Presentation.
Practice 2. ELECTROMAGNETIC CHARACTERIZATION OF THE SPREAD OF LIGHT IN STEP INDEX FIBRE OPTIC (3 sessions)
In this practice we propose to study the propagation of light as dictated by electromagnetic theory in a dielectric medium with a specific structure of great importance in communications systems currently. Specifically, this practice is devoted to characterize the behaviour of electromagnetic propagation in step index optical fibres according to modal theory by signal processing with MATLAB.
Documents to deliver:
• Source files: All those files necessary for the proper compilation and construction of code, never executable files or object code. The delivered code must meet the specific objectives described in practice instructions and must have comments to help monitoring and correction and not have syntax errors or execution.
• Memory. Arguments and justifications to all questions raised in practice.
In each practice the following aspects will be assessed:
• Operational knowledge of the subject.
• Structure in the statement of the problem and its resolution.
• Resolution: type of solution, justification and correction.
• Originality: no exercise is valued if its resolution is equal to or much like another done in a previous course by other students.
• Spelling, format and presentation.
• Presentation.
4. COURSE DESCRIPTION AND TEACHING METHODOLOGY
A1 - Lectures in large group
The methodology to be followed in the lectures in large group lectures will be a mixture of theory and exposing and general examples in the classroom designated for the subject in the middle.
The student must follow the teacher's presentation with the material provided for this purpose, slide show notes, which must be completed with your own notes and the subsequent revision of the recommended basic bibliography.
Active, respectful and responsible participation, either to raise questions or to respond to requests or questions from the teacher, will be evaluated positively in a corresponding factor.
The autonomous student work should focus on the review of the concepts and theoretical aspects taught in class, conducting exercises and studying them with the material provided by the teacher, student notes and bibliography.
Furthermore, control lectures will be held to monitor the progress and assimilation of the concepts by students.
A2 - Classes in groups of practices
Firstly, work in small group classes will be based on practical activities in the laboratory. The student's work will focus on developing applications or learning tasks designated by the teacher, culminating with the achievement of the targets for each practice. Secondly, small group classes consist of solving exercises and practical issues about the subject.
Attendance and active, respectful and responsible participation, either to raise questions or to respond to requests or questions from the teacher, it will be evaluated positively in a corresponding factor.
With regard to self-work, it will focus on the development of the documentation to be submitted for every practice and exposition, and to complete the work that has been initiated in the laboratory and that could not be completed in the corresponding session.
Furthermore, controls sessions in the small group classes will be held to monitor the progress and assimilation of the concepts by students.
Students with special educational needs should contact the Student Attention Service (Servicio de Atención y Ayudas al Estudiante) in order to receive the appropriate academic support
5. ASSESSMENT METHODOLOGY
DETAILED INFORMATION:
Once the semester is finished, the student can choose between two kinds of evaluation: OVERALL or ONLY EXAM.
1. OVERALL EVALUATION
This kind of evaluation is organized using the activities described in the previous table.
The exam of the subject will evaluate the theoretical and operating parts (S2 and S3). The student must demonstrate to have acquired the subject competences. This final exam will be weighted as the 50% of the evaluated parts (S2 and S3). The exam does not include the practising part that is evaluated after finishing each practising. Students who have marks higher or equal to 5.0 out of 10 in the practising part of the OVERALL EVALUATION do not have to make the laboratory practising exam in the ONLY EXAM evaluation
In order to pass the subject, the following conditions must be fulfilled:
1. The student must have a mark higher or equal to 4.0 out of 10 in the exam (S2 and S3 parts).
2. The student must have a mark higher or equal to 4.0 out of 10 in the practising part.
3. The student must have a global mark higher or equal to 5.0 out of 10.
The marks obtained in each part of the overall evaluation are valid during the same academic year.
The student can resign the overall evaluation at any moment, but from then, the kind of evaluation ONLY EXAM will be the chosen evaluation for the rest of the course.
2. ONLY EXAM EVALUATION
This kind of evaluation is organized in an only exam in which all parts of the subject are evaluated. The final exam is weighted in the following way:
• S2 Theoretical concepts and S3 Exercises, cases and works: 70%.
• S4 Laboratory practising: 30%.
In order to pass the subject, the student must have marks higher or equal to 5.0 out of 10 at each part of the final exam.
Those parts with a mark higher or equal to 5.0 are maintained during the same academic year.
6. BOOKLIST
MAIN BOOKLIST:
• Optical fiber communications. Edition: 3rd ed. Author: Keiser, Gerd. Publisher: Boston [etc.]: McGraw-Hill, 2000 (Library) | 1,838 | 8,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.607617 |
https://blog.mathnasium.com/word-problem-wednesday-financial-awareness-for-kids | 1,585,992,045,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370521574.59/warc/CC-MAIN-20200404073139-20200404103139-00558.warc.gz | 388,929,488 | 11,050 | # Number Sense Blog
By Mathnasium | Added Aug 14, 2019
Today is financial awareness day! It’s never too early to start talking about money math with kids. The more comfortable they feel with the subject better! Financial conversations can start early and be fun and relevant by making it about things kids understand — like crayons!
When you come into our math tutoring learning center we make sure to meet kids where they are in their math learning journey. Today we offer a word problem challenge that can help kids of any age and grade level practice their money math, addition, subtraction, and multiplication.
Open the box and color yourself surprised at how much fun math can be!
Question: Crayons come in boxes of 12. Each crayon costs 10¢. How much does a box of crayons cost?
When you're ready, look below to check your solution against ours.
Solution: Each crayon costs 10¢, so the whole box costs 10¢, 12 times. We can solve this by counting by 10s, or we can say that 10 × 12 = 120¢. Since 100¢ is the same as \$1.00, 120¢ is the same as \$1.20.
### Favorites
Bedtime Math: http://bedtimemath.org/
National PTA: http://www.pta.org/
NCTM: http://www.nctm.org/
Numberphile: http://www.numberphile.com/
PTO Today: http://www.ptotoday.com/
STEM Connector: http://blog.stemconnector.org/
Before I came to Mathnasium, I could sum up everything I felt about math in one word: 'EVIL!' I hated math. ... Mathnasium has been my safe haven. They truly have shown me the light when it came to addressing my fear and provided me with the tools that I need to rebuild my prior knowledge so that I won't forget it. Math is no longer a subject I shy away from but it is a subject I can boldly accept and understand.
Roxanne, 12th grade | 435 | 1,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-16 | latest | en | 0.935709 |
https://gazetteactuartistes.com/time-value-of-money-essay/ | 1,582,657,523,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00029.warc.gz | 394,282,461 | 10,442 | # Converting Before Using the Tables Essay
1 ) A shop offers two payment programs. Under the installment program. you pay 20 % down and 20 % of the purchase monetary value in each of the following 4 old ages. If you pay the full measure instantly. you can take a 5 % price reduction from the purchase monetary value. | a. | Calculate the present value of the payments. if you can borrow or impart financess at a 7 % involvement rate. Assume the merchandise sells for \$ 100. ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) | Present value| \$ |
B. | Calculate the payment cyberspace of price reduction. |
Payment cyberspace of discount| \$ |
c. | Which is a better trade? |
| |
| | Pay the full measure immediately|
| Installment plan|
|
2 ) Home loans typically involve “points. ” which are fees charged by the loaner. Each point charged means that the borrower must pay 1 % of the loan sum as a fee. For illustration. if the loan is for \$ 170. 000 and 4 points are charged. the loan refund agenda is calculated on a \$ 170. 000 loan but the net sum the borrower receives is merely \$ 163. 200. What is the effectual one-year involvement rate charged on such a loan presuming loan refund occurs over 156 months? Assume the involvement rate is. 75 % per month. ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) | Effective one-year involvement rate| % |
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3 ) Suppose you take out a \$ 1. 000. 4-year loan utilizing add-on involvement with a quoted involvement rate of 23. 25 % per twelvemonth. |
a. | What will your monthly payments be? ( Entire payments are \$ 1. 000 + \$ 1. 000 ? . 2325 ? 4 = \$ 1. 930. ) ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) |
Monthly payments| \$ |
B. | What are the true APR and effectual one-year rate on this loan? ( Do non round intermediate computations. Round your replies to 3 denary topographic points. ) |
|
APR| % |
Effective one-year rate| % |
|
|
4 ) In a price reduction involvement loan. you pay the involvement payment up front. For illustration. if a 1-year loan is stated as \$ 28. 000 and the involvement rate is 22. 25 % . the borrower “pays” 0. 2225 ? \$ 28. 000 = \$ 6. 230 instantly. thereby having net financess of \$ 21. 770 and refunding \$ 28. 000 in a twelvemonth. What is the effectual one-year rate on a 1-year loan with an involvement rate quoted on a price reduction footing of 22. 25 % ? ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) |
Effective one-year rate| % |
5 )
You believe you will necessitate to hold saved \$ 520. 000 by the clip you retire in 40 old ages in order to populate comfortably. If the involvement rate is 5 % per twelvemonth. how much must you salvage each twelvemonth to run into your retirement end? ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) |
Annual savings| \$ |
6 )
Your confer withing house will bring forth hard currency flows of \$ 110. 000 this twelvemonth. and you expect hard currency flow to maintain gait with any addition in the general degree of monetary values. The involvement rate presently is 6. 2 % . and you anticipate rising prices of approximately 2. 2 % . |
a. | What is the present value of your firm’s hard currency flows for old ages 1 through 6? ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) |
Present value| \$ |
B. | How would your reply to ( a ) alteration if you anticipated no growing in hard currency flow? ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) |
Present value| \$ |
7 )
————————————————-
Top of Form
Good intelligence: You will about surely be a millionaire by the clip you retire in 40 old ages. Bad intelligence: The rising prices rate over your life-time will average approximately 2. 6 % . |
a. | What will be the existent value of \$ 1 million by the clip you retire in footings of today’s dollars? ( Do non round intermediate computations. Round your reply to the nearest dollar sum. ) |
Real value| \$ |
B. | What existent rente ( in today’s dollars ) will \$ 1 million support if the existent involvement rate at retirement is 2. 4 % and the rente must last for 10 old ages? ( Do non round intermediate computations. Round your reply to the nearest dollar sum. ) |
Real annuity| \$ |
Bottom of Form
8 )
You believe you will pass \$ 40. 000 a twelvemonth for 20 old ages one time you retire in 40 old ages. If the involvement rate is 6 % per twelvemonth. how much must you salvage each twelvemonth until retirement to run into your retirement end? ( Do non round intermediate computations. Round your reply to 2 denary topographic points. ) |
Annual savings| \$ | | 1,220 | 4,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-10 | latest | en | 0.907608 |
https://physics.stackexchange.com/questions/348031/how-is-a-running-man-able-to-accelerate-himself?noredirect=1 | 1,618,673,188,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038460648.48/warc/CC-MAIN-20210417132441-20210417162441-00356.warc.gz | 553,555,948 | 39,897 | # How is a running man able to Accelerate himself?
Suppose a man is running and he gradually speeds himself up . For this he applies a force on the ground backward and the ground pushes him forward . This is probably due to to friction between the shoes of the man and the ground . The friction acting in this case is Kinetic friction which has a constant magnitude. Since the magnitude of frictional force is constant, how is the man able to accelerate himself ? Who is providing this force?
• Forces accelerate. – Arnab Barman Ray Jul 22 '17 at 16:33
• Forces acting on a body serve to accelerate it. There is no need for the magnitude or direction of the force to change for the body on which it acts to accelerate. – Arnab Barman Ray Jul 22 '17 at 16:35
• -1. Unclear. Why do you think the friction force $F$ will not cause acceleration if $F$ is constant? $F=ma$ shows that if $F$ is constant then $a$ is constant, not zero. And you have already said that the friction from the ground pushes the man forward. – sammy gerbil Jul 22 '17 at 21:51
Don't think kinetic friction, just because some part of him is moving. Is it also kinetic friction if I stand still but swing my arms? Many particles in him or elsewhere might or might not move. They are irrelevant.
Only the particles in contact with the ground are relevant.
And they are not moving. His foot is not moving during the step. It is stationary and not sliding while touching. There is static friction here.
And static friction can vary easily.
• Thanks ! I was thinking whether there would be any frictional force if the man was moving with constant velocity? According to me, there is kinectic friction will act but my professor said that there would be no friction if he is moving with constant velocity . How is this possible – user8167818 Jul 22 '17 at 17:38
Is the shoe slipping? If not, then it is not kinetic friction. If the shoe is not slipping, the bottom of the shoe is stationary with respect to the ground and the force being applied is static friction. Unless he's running on ice or a lubricated surface, it's unlikely he's experiencing kinetic friction.
The force of friction is not constant in this case. The force of static friction is only what is necessary to prevent slipping. Over the course of his stride, the force of friction changes.
Acceleration occurs when the net force on an object is non-zero. Nothing else matters. It doesn't matter if the force is constant. It doesn't matter how many forces act on the object. It doesn't matter if the object is moving the same direction as the force. As long as the net force is non-zero, it will accelerate. In this case, there are only three objects acting on the runner: gravity, friction, and the normal force. Gravity is directed straight down, and the normal force is directed straight up (assuming flat, level ground). Friction acts horizontally to accelerate the runner.
For part of the stride the man is certainly exerting a force on the ground in the backward direction and the ground is exerting a force on the man in the forward direction which causes the man to increase his momentum and hence his velocity in the forward direction.
The frictional forces exist because the muscles of the man push the foot into the ground.
It matters not whether his foot is slipping or not relative to the ground except that a greater frictional force can be exerted on the man if there is no slipping.
So you have a series of impulses (force $\times \Delta$time) acting on the man due to the ground which increases his momentum and velocity.
I think that the interesting fact is that for the other part of the stride the frictional force acting on the man due to the ground is actually in the opposite direction to that of his motion and imparts an impulse which tends reduce his momentum.
That is why a man can have a positive acceleration, move at "constant" velocity or have a negative acceleration.
It all depends on the magnitude and direction of the frictional force which on the mas at various phases of his stride.
You might find this article informative from which I obtained this graph which shows the force exerted on a person (including in the vertical component) when jogging.
In particular note the dashed graph which shows the horizontal (frictional) forces acting on a person who is jogging which changes direction during a stride. | 942 | 4,390 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.960571 |
http://www.csun.edu/~hcmth018/RSSource.html | 1,516,617,743,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00199.warc.gz | 417,345,121 | 7,340 | ```
// Riemann sums, coded by David Protas, c.2004
// Any corrections or suggestions for improvement of this code will be
// appreciated and should be sent to david.protas@csun.edu
// Latest revision: April 28, 2004
/*********************
* Document: RS.java
*********************/
import java.applet.*;
import java.awt.*;
import expr.*;
public class RS extends Applet {
final int K = 400; // Number of points used to draw graph
final int MaxN = 100; //Maximum number of subintervals
double aD, bD, ymin, ymax, h, sum, total;
int nI;
boolean firstApprox;
String functionString, partitionChoice, testPtChoice;
double[] xarray = new double[K+1];
double[] yarray = new double[K+1];
double[] xArray = new double[MaxN+1];
double[] yArray = new double[MaxN+1];
double[] tArray = new double[MaxN+1];
Panel pan2, bottom, subpan, pan4, pan6;
Label functionLabel, aLabel, bLabel, nLabel, partitionLabel, testPtLabel, message;
TextField functionField, aField, bField, nField;
Choice partition, testPt;
TextArea results;
Button plot, compute;
RSGraph graph;
Color babyBlue;
public void init() {
setLayout(new BorderLayout(4,1));
bottom = new Panel();
bottom.setLayout(new GridLayout(2,1));
pan2 = new Panel();
pan2.setLayout(new FlowLayout(1,8,2));
pan4 = new Panel();
pan4.setLayout(new FlowLayout(1,8,2));
subpan = new Panel();
pan6 = new Panel();
pan6.setLayout(new GridLayout(2,1));
functionLabel = new Label("f(x) = ");
functionField = new TextField(20);
aLabel = new Label(" a = ");
bLabel = new Label(" b = ");
nLabel = new Label(" n = ");
partitionLabel = new Label(" Partition: ");
testPtLabel = new Label(" Test Pts: ");
aField = new TextField(5);
bField = new TextField(5);
nField = new TextField(5);
partition = new Choice();
testPt = new Choice();
plot = new Button("Plot");
compute = new Button("Compute");
results = new TextArea(" ", 6, 50);
results.setEditable(false);
message = new Label(" " +
" ");
graph = new RSGraph();
babyBlue = new Color(204,255,255);
setBackground(babyBlue);
pan2.setBackground(babyBlue);
pan4.setBackground(babyBlue);
subpan.setBackground(babyBlue);
pan6.setBackground(babyBlue);
message.setForeground(Color.red);
message.setBackground(babyBlue);
results.setBackground(babyBlue);
results.setFont(new Font("Courier",Font.PLAIN,10));
}
public boolean action(Event evt, Object arg) {
Variable x = null;
Expr function = null, aInput = null, bInput = null;
if (evt.target == plot) {
message.setText("");
firstApprox = true;
try {
x = Variable.make ("x");
function = Parser.parse (functionField.getText());
aInput = Parser.parse(aField.getText());
bInput = Parser.parse(bField.getText());
}
catch (Syntax_error e) {
message.setText("" + e);
}
Variable.make ("pi").set_value (Math.PI);
Variable.make ("e").set_value (Math.E);
functionString = functionField.getText();
bD = bInput.value();
results.setText(" f(x) = " + functionString + ", a = " + aD + ", b = " + bD);
graph.bD = bD;
ymin = -.0000001;
ymax = .0000001;
for (int i = 0; i <= K; i++) {
x.set_value(xarray[i]);
yarray[i] = function.value(); //Array of y values
graph.xarray[i] = xarray[i];
graph.yarray[i] = yarray[i];
if (yarray[i] < ymin)
ymin = yarray[i]; //find min value of y
if (yarray[i] > ymax)
ymax = yarray[i]; //find max value of y
}
graph.ymin = ymin;
graph.ymax = ymax;
graph.phase = 0;
graph.repaint();
} //end of if aD < bD
else
message.setText("Need a < b. Try again.");
return true;
} //end of evt.target == plot
if (evt.target == compute) {
sum = 0;
try {
x = Variable.make ("x");
function = Parser.parse (functionField.getText());
}
catch (Syntax_error e) {
message.setText("" + e);
}
graph.phase = 1;
if (entryValid(nField.getText()) == false)
message.setText("n needs to be an integer. Try again.");
else {
nI = intFromString(nField.getText());
if (nI < 1)
message.setText("n needs to be positive. Try again.");
else if (nI > MaxN)
message.setText("Applet requires n <= " + MaxN + ". Try again.");
else { // if n is OK
message.setText("");
graph.nI = nI;
partitionChoice = partition.getSelectedItem();
if (partitionChoice == " regular ") {
for (int i = 0; i <= nI; i++) {
graph.xArray[i] = xArray[i];
}
}
else {
total = 0;
for (int i = 1; i <= nI; i++) {
xArray[i] = Math.random(); // length of interval i
total = total + xArray[i];
}
graph.xArray[0] = xArray[0];
for (int i = 1; i < nI; i++) {
xArray[i] = xArray[i-1] + xArray[i]*(bD - aD)/total;
graph.xArray[i] = xArray[i]; // rt endpt of interval i
}
xArray[nI] = bD;
graph.xArray[nI] = xArray[nI];
} // end of setting up partition
testPtChoice = testPt.getSelectedItem();
if (testPtChoice == "left endpt ") {
for (int i = 1; i <= nI; i++) {
tArray[i] = xArray[i-1];
x.set_value(tArray[i]); // sets argument = tArray[i]
yArray[i] = function.value(); //computes f(t)
graph.yArray[i] = yArray[i];
sum = sum + yArray[i]*(xArray[i] - xArray[i-1]);
}
}
else if (testPtChoice == "right endpt") {
for (int i = 1; i <= nI; i++) {
tArray[i] = xArray[i];
x.set_value(tArray[i]);
yArray[i] = function.value();
graph.yArray[i] = yArray[i];
sum = sum + yArray[i]*(xArray[i] - xArray[i-1]);
}
}
if (testPtChoice == " midpoint ") {
for (int i = 1; i <= nI; i++) {
tArray[i] = (xArray[i-1] + xArray[i])/2;
x.set_value(tArray[i]);
yArray[i] = function.value();
graph.yArray[i] = yArray[i];
sum = sum + yArray[i]*(xArray[i] - xArray[i-1]);
}
}
if (testPtChoice == " irregular ") {
for (int i = 1; i <= nI; i++) {
tArray[i] = xArray[i-1] + Math.random()*(xArray[i] - xArray[i-1]);
x.set_value(tArray[i]);
yArray[i] = function.value();
graph.yArray[i] = yArray[i];
sum = sum + yArray[i]*(xArray[i] - xArray[i-1]);
}
}
if (firstApprox == true) {
results.appendText("\n" +" n partition testpoints Riemann sum");
}
if (nI < 10) {
results.appendText("\n" +" "+ nI + " " + partitionChoice + " " +
testPtChoice + " " + rndOff(sum));
}
else {
results.appendText("\n" +" "+ nI + " " + partitionChoice + " " +
testPtChoice + " " + rndOff(sum));
}
graph.repaint();
firstApprox = false;
} //end of else nI > MaxN, i.e. "if n is OK"
} //end of else entryValid
return true;
} //end of evt.target == approx
return false;
} //end of action
public static int intFromString(String str) {
Integer intObj = new Integer(str);
return intObj.intValue();
}
private boolean entryValid(String entry) {
boolean status;
try {
double number = intFromString(entry);
status = true;
}
catch(NumberFormatException e) {
status =false;
}
return status;
}
public static String rndOff(double number) //to 5 places past the decimal
{
String strnum, bigstrnum, substrnum = " 0.00000";
int period, lngth;
long longnum;
if ((number >= 0.001) || (number <= -0.001) || (number != number)) {
number = Math.pow(0.1,5)*Math.round(Math.pow(10,5)*number);
strnum = String.valueOf(number);
bigstrnum = " " + strnum + " ";
period = bigstrnum.indexOf('.');
substrnum = bigstrnum.substring(period -6, period + 6);
if ((number >= 1000000) || (number <= -100000) || (number != number))
substrnum = "big magnitude";
}
else {
longnum = Math.round(Math.pow(10,5)*number);
if (longnum == 0)
substrnum = " 0.00000";
else {
strnum = String.valueOf(longnum);
if (longnum < 0)
strnum = strnum.substring(1);
lngth = strnum.length();
switch (lngth) {
case 1:
substrnum = " 0.0000" + strnum;
break;
case 2:
substrnum = " 0.000" + strnum;
break;
default:
substrnum = " error";
}
if (longnum < 0)
substrnum = " -" + substrnum.substring(5);
}
}
return substrnum;
}
}
/**************************
* Document: RSGraph.java
**************************/
import java.awt.*;
public class RSGraph extends Canvas {
final int K = 400; //number of steps used to draw graph
final int MaxN = 100; //Maximum number of subintervals
Dimension d;
double[] xarray = new double[K+1];
double[] yarray = new double[K+1];
double[] xArray = new double[MaxN+1];
double[] yArray = new double[MaxN+1];
double[] tArray = new double[MaxN+1];
int phase = -1, tick, deltaTick, nI;
public void paint(Graphics g) {
if (phase == 1) {
g.setColor(Color.pink);
for (int i = 1; i <= nI; i++) {
if (yArray[i] > 0) {
g.fillRect(xScaler(xArray[i-1]), yScaler(yArray[i]),
xScaler(xArray[i]) - xScaler(xArray[i-1]),
yScaler(0.0) - yScaler(yArray[i]));
}
else {
g.fillRect(xScaler(xArray[i-1]), yScaler(0.0),
xScaler(xArray[i]) - xScaler(xArray[i-1]),
yScaler(yArray[i]) - yScaler(0.0));
}
}
g.setColor(Color.magenta);
for (int i = 1; i <= nI; i++) {
if (yArray[i] > 0) {
g.drawRect(xScaler(xArray[i-1]), yScaler(yArray[i]),
xScaler(xArray[i]) - xScaler(xArray[i-1]),
yScaler(0.0) - yScaler(yArray[i]));
}
else {
g.drawRect(xScaler(xArray[i-1]), yScaler(0.0),
xScaler(xArray[i]) - xScaler(xArray[i-1]),
yScaler(yArray[i]) - yScaler(0.0));
}
}
} //end of if
if (phase != -1) {
d = this.size();
g.setColor(Color.black);
g.drawLine(0, yScaler(0.0), d.width, yScaler(0.0)); // x-axis
g.drawLine(xScaler(0.0), 0, xScaler(0.0), d.height - 6); // y-axis
tick = (int)aD; // start of x-ticks
deltaTick = 1 + (int)(bD - aD)/14;
do {
if (tick != 0) {
g.drawString(tick+"", xScaler(tick) - 4, yScaler(0.0) + 11);
g.drawLine(xScaler(tick), yScaler(0.0) - 2, xScaler(tick), yScaler(0.0));
}
tick = tick + deltaTick;
}
while (tick <= bD); //end of x-ticks
for (int j = 0; j < K; j++) { // y = f(x)
g.drawLine(xScaler(xarray[j]), yScaler(yarray[j]),
xScaler(xarray[j+1]), yScaler(yarray[j+1]));
}
} //end of if
}
private int xScaler(double x)
{
}
private int yScaler(double y)
{
return (int)((d.height - 11)*(ymax - y)/(ymax - ymin));
}
}
/**************************************
* Folder: expr Document: Expr.java
**************************************/
// Mathematical expressions.
// Copyright 1996 by Darius Bacon; see the file COPYING.
// 6June02: changes made by David Protas indicated by /*DP*/
package expr;
/**
* A mathematical expression, built out of literal numbers, variables,
* arithmetic operators, and elementary functions. The operator names
* are from java.lang.Math.
*/
public abstract class Expr {
/** @return the value given the current variable values */
public abstract double value ();
/** Binary operator. */ public static final int ADD = 0;
/** Binary operator. */ public static final int SUB = 1;
/** Binary operator. */ public static final int MUL = 2;
/** Binary operator. */ public static final int DIV = 3;
/** Binary operator. */ public static final int POW = 4;
/** Unary operator. */ public static final int ABS = 100;
/** Unary operator. */ public static final int ACOS = 101;
/** Unary operator. */ public static final int ASIN = 102;
/** Unary operator. */ public static final int ATAN = 103;
/** Unary operator. */ public static final int CEIL = 104;
/** Unary operator. */ public static final int COS = 105;
/** Unary operator. */ public static final int EXP = 106;
/** Unary operator. */ public static final int FLOOR = 107;
/** Unary operator. */ public static final int LN = 114; /*DP*/
/** Unary operator. */ public static final int LOG = 108;
/** Unary minus operator. */ public static final int NEG = 109;
/** Unary operator. */ public static final int ROUND = 110;
/** Unary operator. */ public static final int SIN = 111;
/** Unary operator. */ public static final int SQRT = 112;
/** Unary operator. */ public static final int TAN = 113;
public static Expr make_literal (double v) {
return new Literal (v);
}
public static Expr make_var_ref (Variable var) {
return new Var_ref (var);
}
/**
* @param rator unary operator
* @param rand operand
*/
public static Expr make_app1 (int rator, Expr rand) {
Expr app = new App1 (rator, rand);
return rand instanceof Literal ? new Literal (app.value ()) : app;
}
/**
* @param rator binary operator
* @param rand0 left operand
* @param rand1 right operand
*/
public static Expr make_app2 (int rator, Expr rand0, Expr rand1) {
Expr app = new App2 (rator, rand0, rand1);
return rand0 instanceof Literal && rand1 instanceof Literal
? new Literal (app.value ())
: app;
}
}
// These classes are all private to this module so that I can get rid
// of them later. For applets you want to use as few classes as
// possible to avoid http connections at load time; it'd be profitable
// to replace all these subtypes with bytecodes for a stack machine,
// or perhaps a type that's the union of all of them (see class Node
class Literal extends Expr {
double v;
Literal (double _v) { v = _v; }
public double value () { return v; }
}
class Var_ref extends Expr {
Variable var;
Var_ref (Variable _var) { var = _var; }
public double value () { return var.value (); }
}
class App1 extends Expr {
int rator;
Expr rand;
App1 (int _rator, Expr _rand) { rator = _rator; rand = _rand; }
public double value () {
double arg = rand.value ();
switch (rator) {
case ABS: return Math.abs (arg);
case ACOS: return Math.acos (arg);
case ASIN: return Math.asin (arg);
case ATAN: return Math.atan (arg);
case CEIL: return Math.ceil (arg);
case COS: return Math.cos (arg);
case EXP: return Math.exp (arg);
case FLOOR: return Math.floor (arg);
case LN: return Math.log (arg); /*DP*/
case LOG: return Math.log (arg)/Math.log (10); /*DP*/
case NEG: return -arg;
case ROUND: return Math.round (arg);
case SIN: return Math.sin (arg);
case SQRT: return Math.sqrt (arg);
case TAN: return Math.tan (arg);
default: throw new RuntimeException ("BUG: bad rator");
}
}
}
class App2 extends Expr {
int rator;
Expr rand0, rand1;
App2 (int _rator, Expr _rand0, Expr _rand1) {
rator = _rator; rand0 = _rand0; rand1 = _rand1;
}
public double value () {
double arg0 = rand0.value ();
double arg1 = rand1.value ();
switch (rator) {
case ADD: return arg0 + arg1;
case SUB: return arg0 - arg1;
case MUL: return arg0 * arg1;
case DIV: return arg0 / arg1; // check for division by 0?
case POW: return Math.pow (arg0, arg1);
default: throw new RuntimeException ("BUG: bad rator");
}
}
}
/****************************************
* Folder: expr Document: Parser.java
****************************************/
// Operator-precedence parser.
// Copyright 1996 by Darius Bacon; see the file COPYING.
// 14May96: bugfix.
// StreamTokenizer treated '-' as a numeric token, not a minus
// operator followed by a number. Fix: make '-' an ordinaryChar.
// 12May97: Changed the precedence of unary minus to be lower than
// multiplication, so -y^2 is like -(y^2), not (-y)^2.
package expr;
import java.io.*;
/**
Parses strings representing mathematical formulas with variables.
The following operators, in descending order of precedence, are
defined:
^ (raise to a power)
* /
Unary minus (-x)
+ -
^ associates right-to-left; other operators associate left-to-right.
These functions are defined:
abs, acos, asin, atan,
ceil, cos, exp, floor, (ln added by DP)
log, round, sin, sqrt,
tan. Each requires one argument enclosed in parentheses.
Whitespace outside identifiers is ignored.
The syntax-error messages aren't very informative, unfortunately.
IWBNI it indicated where in the input string the parse failed, but
that'd be kind of a pain since our scanner is a StreamTokenizer. A
hook for that info should've been built into StreamTokenizer.
Examples:
42
2-3
cos(x^2) + sin(x^2)
*/
public class Parser {
static StreamTokenizer tokens;
public static Expr parse (String input) throws Syntax_error {
tokens = new StreamTokenizer (new StringBufferInputStream (input));
tokens.ordinaryChar ('/');
tokens.ordinaryChar ('-');
next ();
Expr expr = parse_expr (0);
if (tokens.ttype != StreamTokenizer.TT_EOF)
throw new Syntax_error ("Incomplete expression: " + input);
return expr;
}
static void next () {
try { tokens.nextToken (); }
catch (IOException e) { throw new RuntimeException ("I/O error: " + e); }
}
static void expect (int ttype) throws Syntax_error {
if (tokens.ttype != ttype)
throw new Syntax_error ("'" + (char) ttype + "' expected");
next ();
}
static Expr parse_expr (int precedence) throws Syntax_error {
Expr expr = parse_factor ();
loop: for (;;) {
int l, r, rator;
// The operator precedence table.
// l = left precedence, r = right precedence, rator = operator.
// Higher precedence values mean tighter binding of arguments.
// To associate left-to-right, let r = l+1;
// to associate right-to-left, let r = l.
switch (tokens.ttype) {
case '+': l = 10; r = 11; rator = Expr.ADD; break;
case '-': l = 10; r = 11; rator = Expr.SUB; break;
case '*': l = 20; r = 21; rator = Expr.MUL; break;
case '/': l = 20; r = 21; rator = Expr.DIV; break;
case '^': l = 30; r = 30; rator = Expr.POW; break;
default: break loop;
}
if (l < precedence)
break loop;
next ();
expr = Expr.make_app2 (rator, expr, parse_expr (r));
}
return expr;
}
static String[] procs = {
"abs", "acos", "asin", "atan",
"ceil", "cos", "exp", "floor", "ln", // ln added by DP
"log", "round", "sin", "sqrt",
"tan"
};
static int[] rators = {
Expr.ABS, Expr.ACOS, Expr.ASIN, Expr.ATAN,
Expr.CEIL, Expr.COS, Expr.EXP, Expr.FLOOR, Expr.LN, // Expr.LN added by DP
Expr.LOG, Expr.ROUND, Expr.SIN, Expr.SQRT,
Expr.TAN
};
static Expr parse_factor () throws Syntax_error {
switch (tokens.ttype) {
case StreamTokenizer.TT_NUMBER: {
Expr lit = Expr.make_literal (tokens.nval);
next ();
return lit;
}
case StreamTokenizer.TT_WORD: {
for (int i = 0; i < procs.length; ++i)
if (procs [i].equals (tokens.sval)) {
next ();
expect ('(');
Expr rand = parse_expr (0);
expect (')');
return Expr.make_app1 (rators [i], rand);
}
Expr var = Expr.make_var_ref (Variable.make (tokens.sval));
next ();
return var;
}
case '(': {
next ();
Expr enclosed = parse_expr (0);
expect (')');
return enclosed;
}
case '-':
next ();
return Expr.make_app1 (Expr.NEG, parse_expr (15));
default:
throw new Syntax_error ("Expected a factor");
}
}
}
/**********************************************
* Folder: expr Document: Syntax_error.java
**********************************************/
// Syntax-error exception.
// Copyright 1996 by Darius Bacon; see the file COPYING.
package expr;
public class Syntax_error extends Exception {
public Syntax_error (String complaint) { super (complaint); }
}
/******************************************
* Folder: expr Document: Variable.java
******************************************/
// Variables associate values with names.
// Copyright 1996 by Darius Bacon; see the file COPYING.
package expr;
import java.util.Hashtable;
/**
* Variables associate values with names.
*/
public class Variable {
static Hashtable variables = new Hashtable ();
/**
* Return the variable named `_name'.
* make (s1) == make (s2) iff s1.equals (s2).
*/
static public synchronized Variable make (String _name) {
Variable result = (Variable) variables.get (_name);
if (result == null)
variables.put (_name, result = new Variable (_name));
return result;
}
String name;
double val;
public Variable (String _name) { name = _name; val = 0; }
public String toString () { return name; }
public double value () { return val; }
public void set_value (double _val) { val = _val; }
}
```
Back to applet | 5,408 | 19,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-05 | latest | en | 0.495386 |
https://bilakniha.cvut.cz/en/predmet1236806.html | 1,680,003,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00619.warc.gz | 170,322,345 | 4,231 | CZECH TECHNICAL UNIVERSITY IN PRAGUE
STUDY PLANS
2022/2023
UPOZORNĚNÍ: Jsou dostupné studijní plány pro následující akademický rok.
# Computational Geometry
The course is not on the list Without time-table
Code Completion Credits Range Language
Garant předmětu:
Lecturer:
Tutor:
Supervisor:
Department of Computer Graphics and Interaction
Synopsis:
The goal of computational geometry is analysis and design of efficient algorithms for determining properties and relations of geometric entities. The lecture focuses on geometric search, point location, convex hull construction for sets of points in d-dimensional space, searching nearest neighbor points, computing intersection of polygonal areas, geometry of parallelograms. New directions in algorithmic design. Computational geometry is applied not only in geometric applications, but also in common database searching problems.
Requirements:
Knowledge of Fundamental sorting and searching algorithms. Linear algebra and fundamentals of computer graphics are advantageous. Programming in C++.
Syllabus of lectures:
1. Computational geometry (CG), typical applications, effective algorithm design techniques
2. Geometric searching
3. Geometric searching 2
4. Planar convex hull
5. Convex hull in 3D
6. Voronoi diagram of points
7. Voronoi diagram of line segments. Higher order Voronoi diagrams
8. Triangulations
9. Intersections of line segments and polygons
10. Intersections of polygonal line segments with a rectangular window
11. Arrangements
12. Dual algorithms
13. New directions in algorithmic design
14. Spare lesson
Syllabus of tutorials:
1. Introduction to the form of the seminars, fundamental math. concepts useful in CG.Selection of topics for assignment.
2. Robustness of geometric predicats and constructs.
3. Presentations of the topic assigned, discussion. Evaluation of the presentation materials and evaluation of the speech by classmate students. Ideas for improvements.
4. Presentation of the topic assigned
5. Presentation of the topic assigned
6. Presentation of the topic assigned
7. Presentation of the topic assigned
8. Presentation of the topic assigned
9. Presentation of the topic assigned
10. Presentation of the topic assigned
11. Presentation of the topic assigned
12. Presentation of the topic assigned
13. Assessment
14. Spare
Study Objective:
The course is an informal continuation of fundamental data structures and algorithms courses. You will learn geometric algorithms and data structures allowing for effective computations, e.g., localization of area hit by a ray, computation of intersections and triangulation. You will train presentation and professional discussion skills on the seminars. All of it should not be missing in knowledge of educated progressive Master of Science.
Study materials:
1. Berg, M. de, Cheong, O., Kreveld, M. van, Overmars, M.: Coputational Geometry. Algorithms and Applications, Springer-Verlag, Berlin, 3rd ed., 2008. ISBN: 978-3-540-77973-5
2. O' Rourke, Joseph: Computational Geometry in C, Cambridge University Press, 1st ed, 1994 or 2nd ed, 2000
3. Preperata F.P.- M.I.Shamos: Computational Geometry An Introduction. Berlin, Springer-Verlag,1985.
Note:
Further information:
No time-table has been prepared for this course
The course is a part of the following study plans:
Data valid to 2023-03-28
Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet1236806.html | 765 | 3,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-14 | longest | en | 0.735373 |
https://stackoverflow.com/questions/20070374/how-do-i-set-a-uint32-to-its-maximum-value | 1,716,303,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00299.warc.gz | 454,669,325 | 49,431 | # How do I set a UInt32 to its maximum value
1. What is the maximum value for a UInt32?
2. Is there a way I can use the sizeof operator to get the maximum value (as it is unsigned)? So I don't end up with #defines or magic numbers in my code.
• Wouldn't that be 2^32 - 1? Nov 19, 2013 at 11:35
• Damn SO, you used to be so friendly. This is a valid question, why all the down votes? Haters got to hate, I wish they would do it somewhere else though. Nov 19, 2013 at 22:14
• (uint32_t)0-1 ;-) Apr 10, 2020 at 0:08
There's a macro UINT32_MAX defined in stdint.h which you can use
#include <stdint.h>
uint32_t max = UINT32_MAX;
http://pubs.opengroup.org/onlinepubs/009695299/basedefs/stdint.h.html
• You have my up vote already, but it may be helpful for future users if you list a few more of these macros like MAXFLOAT and similar. Nov 19, 2013 at 11:45
• @0x7fffffff You should take a look into <stdint.h>. There are virtually a few dozens of macros which define min, max for all signed and unsigned integer types and a few other macros. These are defined in the ISO/IEC 988:1999 specification. Nov 19, 2013 at 11:59
The maximum value for UInt32 is 0xFFFFFFFF (or 4294967295 in decimal).
sizeof(UInt32) would not return the maximum value; it would return 4, the size in bytes of a 32 bit unsigned integer.
• Thanks for that, I was thinking maybe you could do some simple arithmetic on it (sizeof(UInt32) * 8) ^ 2. Be a bit of overkill, I was a bit sleepy when I asked. Nov 19, 2013 at 22:16
The portable way:
std::numeric_limits<uint32_t>::max()
• why is this more portable? Aug 17, 2015 at 0:57
• More portable in comparison with some hardcoded values. Maybe uint32_t is not a great example. If you take std::numeric_limits<int32_t> for example, you do not have to make assumption on how signed numbers are coded on the machine. If you take std::numeric_limits<unsigned int>, you are not making any assumption about the size of an int which may not alway be 32 bit depending on the hardware that is targeted. It is also a bit more flexible than the macros. If you need to change the type, through a typedef for instance, you do not have to update all the UINT32_MAX like macros in your code. Aug 20, 2015 at 18:18
• This answer is C++ code, but the question is about Objective C on iOS. Jan 20, 2016 at 17:47
Just set the max using standard hexadecimal notation and then check it against whatever you need. 32-bits is 8 hexadecimals bytes, so it'd be like this:
let myMax: UInt32 = 0xFFFFFFFF
if myOtherNumber > myMax {
// resolve problem
}
4_294_967_295 is the maximal value or in hexadecimal 0xFFFFFFFF.
An alternative for any unsigned in C or C++ is:
anUnsigned = -1;
This is useful since it works for them all, so if you change from unsigned int to unsigned long you don't need to go through your code. You will also see this used in a lot of bit fiddling code:
anUnsigned |= -(aBoolOrConditionThatWhenTrueCausesAnUnsignedToBeSetToAll1s)
anUnsigned |= -(!aValueThatWhenZeroCausesAnUnsignedToBeSetToAll1s)
anUnsigned |= -(!!aValueThatWhenNonZeroCausesAnUnsignedToBeSetToAll1s)
The downside is that it looks odd, assigning a negative number to an unsigned!
• sometimes doing it this way causes compiler warnings or static analysis warnings, pity because it works fine! :) Nov 19, 2019 at 15:47
• To avoid a compiler warning and make it clear to other readers that you know what you're doing, cast the value to the correct type, e.g. anUnsigned = (uint32_t)-1; . Nov 28, 2022 at 18:12 | 970 | 3,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.896533 |
http://www.thefreedictionary.com/vertebral+curve | 1,505,968,257,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687606.9/warc/CC-MAIN-20170921025857-20170921045857-00434.warc.gz | 584,252,931 | 22,606 | # curve
(redirected from vertebral curve)
Also found in: Thesaurus, Medical, Encyclopedia.
## curve
(kûrv)
n.
1.
a. A line that deviates from straightness in a smooth, continuous fashion.
b. A surface that deviates from planarity in a smooth, continuous fashion.
c. Something characterized by such a line or surface, especially a rounded line or contour of the human body.
2. A relatively smooth bend in a road or other course.
3.
a. A line representing data on a graph.
b. A trend derived from or as if from such a graph: "Once again, the politicians are behind the curve" (Ted Kennedy).
4. A graphic representation showing the relative performance of individuals as measured against each other, used especially as a method of grading students in which the assignment of grades is based on predetermined proportions of students.
5. Mathematics
a. The graph of a function on a coordinate plane.
b. The intersection of two surfaces in three dimensions.
c. The graph of the solutions to any equation of two variables.
6. Baseball A curve ball.
7. Slang Something that is unexpected or designed to trick or deceive.
v. curved, curv·ing, curves
v.intr.
To move in or take the shape of a curve: The path curves around the lake.
v.tr.
1. To cause to curve.
2. Baseball To pitch (a ball) with a curve.
3. To grade (students, for example) on a curve.
[From Middle English, curved, from Latin curvus; see sker- in Indo-European roots. N., sense 6, short for curve ball.]
curv′ed·ness n.
## curve
(kɜːv)
n
1. a continuously bending line that has no straight parts
2. something that curves or is curved, such as a bend in a road or the contour of a woman's body
3. the act or extent of curving; curvature
4. (Mathematics) maths
a. a system of points whose coordinates satisfy a given equation; a locus of points
b. the graph of a function with one independent variable
5. (Mathematics) a line representing data, esp statistical data, on a graph: an unemployment curve.
7. behind the curve behind the times; behind schedule
8. (Tools) short for French curve
vb
to take or cause to take the shape or path of a curve; bend
[C15: from Latin curvāre to bend, from curvus crooked]
## curve
(kɜrv)
n., v. curved, curv•ing,
1. a continuously bending line, without angles.
2. the act or extent of curving.
3. any curved outline, form, thing, or part.
5. Also called curve′ ball`. a baseball pitch delivered with a spin that causes the ball to veer from a normal straight path, away from the side from which it was thrown.
6. a graphic representation of the variations effected in something by the influence of changing conditions; graph.
7. Math. a collection of points whose coordinates are continuous functions of a single independent variable.
8. a misleading or deceptive trick.
9. an academic grading system based on the scale of performance of the group, so that those performing better, regardless of their actual knowledge, receive higher grades: to mark on a curve.
10. a curved guide used in drafting.
v.i.
11. to bend in a curve; take the course of a curve.
v.t.
12. to cause to curve.
13. to grade on a curve.
14. to pitch a curve to in baseball.
15. curved.
Idioms:
1. ahead of (or behind) the curve, at the forefront of (or lagging behind) recent developments, trends, etc.
2. throw someone a curve, to take someone by surprise, esp. so as to cause chagrin.
[1565–75; (< Middle French) < Latin curvus crooked, bent, curved]
## curve
(kûrv)
1. A line or surface that bends in a smooth, continuous way without sharp angles.
2. The graph of a function on a coordinate plane. In this technical sense, straight lines, circles, and waves are all curves.
## curve
Past participle: curved
Gerund: curving
Imperative
curve
curve
Present
I curve
you curve
he/she/it curves
we curve
you curve
they curve
Preterite
I curved
you curved
he/she/it curved
we curved
you curved
they curved
Present Continuous
I am curving
you are curving
he/she/it is curving
we are curving
you are curving
they are curving
Present Perfect
I have curved
you have curved
he/she/it has curved
we have curved
you have curved
they have curved
Past Continuous
I was curving
you were curving
he/she/it was curving
we were curving
you were curving
they were curving
Past Perfect
Future
I will curve
you will curve
he/she/it will curve
we will curve
you will curve
they will curve
Future Perfect
I will have curved
you will have curved
he/she/it will have curved
we will have curved
you will have curved
they will have curved
Future Continuous
I will be curving
you will be curving
he/she/it will be curving
we will be curving
you will be curving
they will be curving
Present Perfect Continuous
I have been curving
you have been curving
he/she/it has been curving
we have been curving
you have been curving
they have been curving
Future Perfect Continuous
I will have been curving
you will have been curving
he/she/it will have been curving
we will have been curving
you will have been curving
they will have been curving
Past Perfect Continuous
Conditional
I would curve
you would curve
he/she/it would curve
we would curve
you would curve
they would curve
Past Conditional
I would have curved
you would have curved
he/she/it would have curved
we would have curved
you would have curved
they would have curved
ThesaurusAntonymsRelated WordsSynonymsLegend:
## curve
noun
1. a curve in the road
verb
1. The track curved away below him.
Related words
## curve
noun
Something bent:
verb
1. To swerve from a straight line:
2. To have or cause to have a curved or sinuous form or surface:
Translations
إنحِناءمُنْعَطَفيَنْحَني
křivkaohnoutohybtočit sezatáčka
buegå i en buekrummekrumningkurve
kurbo
kaarikaarrekaartaakäyrä
görbekanyarulat
beygjabeygja, sveigjabogi, boglína
izliektizliektiesizlocītieslīka līnijalīkne
krivkazatočiť sa
ovinekzavijati
böjakrökakurva
dönemeçeğ mekeğrikıvırmakkıvrılmak
## curve
[kɜːv]
A. N (gen) →
B. VT [+ spine, back] →
C. VI [road, line etc] → ; [surface] → combarse
the walls curve inward/outwardlas paredes están combadas hacia dentro/fuera
the road curves round the mountain
the boomerang curved through the air
a wide, curving staircaseuna amplia escalera en curva
## curve
[ˈkɜːrv]
n
(= shape) →
(on graph)
(mainly US) to throw sb a curve → prendre qn de court
vtcourber
vi [line, surface, arch] → s'incurver; [road] → faire une courbecurve ball n (mainly US) to throw sb a curve ball → prendre qn de court
## curve
nKurve f; (of body, vase etc)Rundung f, → Wölbung f; (of river)Biegung f; (of archway)Bogen m; there’s a curve in the roaddie Straße macht einen Bogen; the price curvedie Preiskurve; her curves (inf)ihre Kurven or Rundungen pl (inf)
vtbiegen; (= build with a curve) arch, roof, side of shipwölben; gravity curves the path of lightdie Gravitation krümmt den Lichtweg; he curved the ball around the waller zirkelte den Ball um die Mauer herum
vi
(line, road)einen Bogen machen; (river)eine Biegung machen; her lips curved into a smileihre Lippen verzogen sich zu einem Lächeln; the road curves around the citydie Straße macht einen Bogen um die Stadt; to make a ball curve (through the air)einen Ball anschneiden, einem Ball einen Drall geben
(= be curved) (space, horizon)gekrümmt sein; (side of ship, surface, arch)sich wölben; (hips, breasts)sich runden; (metal strip etc)sich biegen
## curve
[kɜːv]
1. n (gen) → curva; (of river) → ansa
simple closed curve (Math) →
2. vtcurvare
3. vi (road, river) → fare una curva; (line, surface, arch) → curvarsi
## curve
(kəːv) noun
1. a line which is not straight at any point, like part of the edge of a circle.
2. anything shaped like this. a curve in the road.
verb
to bend in a curve. The road curves east. | 2,157 | 7,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-39 | latest | en | 0.911421 |
https://www.physicsforums.com/threads/help-555-timer-problem.209387/ | 1,531,726,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00365.warc.gz | 945,222,055 | 12,929 | # Help 555 timer problem
1. Jan 17, 2008
### N468989
[SOLVED] help 555 timer problem
hi all, i am designing a 555 on astable mode. it is designed to operate at 100kHz with a duty cycle of 95% ( and i get 48Khz and duty cycle 93.5%).
i calculated all the values necessary:
Ra = 12.987 kOhm
Rb = 721.5 Ohm
C = 1nF
F = 1.44/ ((Ra + 2Rb)*C)
i used thes formulas :
tH = 0.693*(Ra + Rb)*C
tL = 0.693*(Rb)*C
T = tH + tL
i searched everywhere and found out the max frequency is 500Khz so im not sure where the problem is....i need the 100kHz because i am designing an infrared modulator and the central frequency for my filters is 100kHz
can anyone help me thanks.
Last edited: Jan 17, 2008
2. Jan 17, 2008
### mgb_phys
1nF is a bit small, capacitors this small aren't very accurate and the effects of stray and gate capacitance start to matter, you might want to redesign it with at least 10nF.
3. Jan 17, 2008
### N468989
just tested with 10n and 100n ...got 40Khz this time....ive double checked everthing.
i just simulated it on multisim and i got 45.9kHz ....but by my calculations it should be 100kHz
any ideias? i think this is very odd
Last edited: Jan 17, 2008
4. Jan 18, 2008
### N468989
--SOLVED--
the multsim 555 design wizard is not correct. there is a problem with it... in the lab i changed the IC (it was a bit damaged) and i managed to adjust it to my desired frequency / dutycycle.
btw.... thanks for trying to solve my problem | 463 | 1,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-30 | latest | en | 0.896271 |
https://www.geeksforgeeks.org/oyo-coding-interview-experience/?ref=rp | 1,627,368,331,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152236.64/warc/CC-MAIN-20210727041254-20210727071254-00702.warc.gz | 744,930,561 | 21,836 | Related Articles
# Oyo Coding Interview Experience
• Difficulty Level : Medium
• Last Updated : 10 Jan, 2019
Online Round : Oyo Coding visited our campus in the last week of August.
Round 1 was held on Hackerearth which comprised of around 25 MCQs and 2 coding questions. MCQs were from topics like Data Structures(mostly), Aptitude, Operating systems and DBMS.
The coding questions were
1. Minimum no of ways to flip the brackets
2. Given a triangle ABC and number of steps k, return the number of ways that we can arrive back at source point in k steps (1 step = traversing from one vertex to other vertex).
```Example - for 2 steps, 2 possible ways (A-B-A and A-C-A).
Similarly, for 3 steps, 2 possible ways(A-B-C-A and A-C-B-A).```
After this 44 students were shortlisted for the PIs.
Round 2 : The interviewer was very friendly and patient, he primarily focused on Data Structures. He asked me three coding questions.
1. First question was an implementation of Trie search and first match algorithm .He asked me to write the complete code.
2. Second question was to write a complete working code to find the Intersection point of two linked lists.
3. The third question was to print the all the nodes at kth depth in a generic tree, given the root of a generic tree.
I solved it using recursion (using depth as a variable), other method to solve is level order traversal.
Around 30 people were shortlisted for the third round.
Round 3 : Third round was based on Operating Systems and little of coding. The following questions were asked :
1. What are scheduling algorithms and why are they used?
2. What is virtual memory?
3. Name and explain all the scheduling algorithms?
4. What is context switch?
5. Construct a binary search tree given its preorder and inorder traversal arrays. (Had to write the complete code)
6. Given a source and destination in a 2D matrix, count all possible ways to reach destination from source.
Allowed steps are (x, y+1) and (x+1, y). (Complete code again) | 475 | 2,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-31 | longest | en | 0.939876 |
http://www.matplus.net/start.php?px=1573960823&app=forum&act=posts&fid=gen&tid=2280&pid=17272 | 1,575,948,109,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00466.warc.gz | 206,090,990 | 4,807 | MatPlus.Net
Website founded by
Milan Velimirović
in 2006
3:21 UTC
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CHESS SOLVINGTournamentsRating lists1-Oct-2019
B P C F
MatPlus.Net Forum General Most Checkmates In One In A Position Without Promoted Pieces
### Most Checkmates In One In A Position Without Promoted Pieces
(= 13+3 )
Here’s a challenge for you all. Construct a position, with no promoted pieces, with the most mates in one that beats my 42 in the given diagram above that I constructed. I found that solution myself.
It must be legal of course.
For some suggestions, I would try to add in more queen moves or another pawn promotion, if at all workable. I was unable to do either of these myself.
The 42 mates in my position go as follows: Rg2-14, Be5-13, Nc5-7 Ne8-2, Qb5-2, Ph7-1, Pa7-1, Pd3-1. Pe3-1
Good luck!
Just you enter Capital letter F after opening 2 brackets and close the FEN with same 2 brackets....then diagram Visible in the thread...before that preview the diagram for correctness.
Thanks! And fixed!
The pawns each have two mates, because they can promote to Queen or Bishop!
(5) Posted by Rewan Demontay (Real Name: James Malcom) [Friday, May 3, 2019 04:28]; edited by Rewan Demontay (Real Name: James Malcom) [19-05-03]
I only count promotion as one move. Even if we make a separate category allowing your idea, it’s still pretty much the same position anyhow.
From Schach und Mathematik (Gik):
(= 12+3 )
Kf7: 6 Rg5: 12 Qd4: 4 Be4: 13 Sh4: 2 Sc3: 2 Pd2: 1 Pf2: 2 Ph2: 1 Total 43
And if promotions are allowed:
(= 13+2 )
Se8: 2 Pa7: 1 Pd7: 1 Ph7: 1 Be5: 13 Qf5: 4 Rc4: 14 Sd3: 7 Pe2: 1
Total 44 (Gik claims 47, so he sees promotion to queen and bishop/rook as separate moves)
This is nice! The funny thing is that you can just add a promotion into the non-promotion position and still get 44.
Is this the book that you sourced from? https://www.amazon.de/Schach-Mathematik-Evgeni-J-Gik/dp/3871449873
Yes, that's the book.
I don't know Gik's book, but the 2nd diagram given by Mr de Heer friday is in the book of Nenad Petrović, Šahovski problem (1949), page 166. The author of this problem is J. C. West, 1880 (!!).
Nothing new under the sun !
You may find it in a version the 02/06/2012 and also read funny things about "mate in 1" here :
https://www.france-echecs.com/article.php?art=20120531233813551
Yes. I don't know why, only the 2nd works.
Turns out that Otto Blathy also found 47 but with 4 promoting pawns!
P1325213
Joseph Ney Babson
Brentano's Chess Monthly 1882
(= 14+3 ) | 783 | 2,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-51 | latest | en | 0.883242 |
https://www.aimsuccess.in/2017/09/coding-decoding-practice-for-ibps-porrb.html | 1,725,758,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00816.warc.gz | 611,472,183 | 78,611 | # Coding-Decoding Practice for IBPS PO/RRB
## Coding-Decoding Practice for IBPS PO/RRB
Directions (Q.1-5): Study the following information carefully and answer the given questions.
In a certain code language,
Quit India Movement is coded as “!A9, @T17, &T13
Prime Minister Of India is coded as “&R13, %F15, !A9, !E16”
Believe In Yourself is coded as “ *E2, %N9, &F25
1. What is the code for ‘Solution’ in the given code language?
1. @T20
2. %N15
3. #N19
4. &N19
5. None of these
2. Which of the following symbols is used in the code of “Himansh”?
1. #
2. \$
3. @
4. *
5. %
3. “!A25” is the code of which of the following word?
1. Yatra
2. Yamuna
3. Yak
4. Network
5. None of these
4. What is the code of ‘Corrupt’ in the given code?
1. \$T3
2. %T5
3. \$T6
4. @T3
5. *T3
5. “ %Y13, *E22” is code for __________.
1. Mera Honor
2. Vice President
3. My Village
4. My Town
5. None of these
In this coded language, Each code is divided into three parts viz; Symbol, Letter and Number.
Symbol depicts: The number of letter in a particular word e.g.
If a word contains: 2 letters – %; 3 letters – #; 4 letters – @; 5 letters – !; 6 letters – ?; 7 letters – *; 8 letters – & and 9 letters – \$.
Letter depicts : The last letter of the particular word.
Number depicts: Numeric value of first letter according to Alphabetic Series.
1. 4
2. 4
3. 1
4. 5
5. 3 | 486 | 1,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.724274 |
https://av.tib.eu/media/10103 | 1,604,142,091,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00117.warc.gz | 219,029,927 | 21,141 | ## 17B.2 Division komplexer Zahlen algebraisch und geometrisch
Video in TIB AV-Portal: 17B.2 Division komplexer Zahlen algebraisch und geometrisch
Title 17B.2 Division komplexer Zahlen algebraisch und geometrisch Title of Series Mathematik 1, Winter 2012/2013 Number of Parts 187 Author License CC Attribution - NonCommercial - ShareAlike 3.0 Germany:You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal and non-commercial purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this license. Identifiers 10.5446/10103 (DOI) Publisher Release Date 2012 Language German Producer Loviscach, Jörn
Subject Area Mathematics
### Related Material
##### The following resource is accompanying material for the video
Computer animation Link (knot theory) Quotient Division (mathematics) Linie Number
Zahl Sign (mathematics) Algebra Computer animation Algebraic closure Complex number Bindung <Stochastik> Film editing Square
Calculation Zahl Computer animation Complex number Square Length Absolute value Number
Stress (mechanics) Zahl Multiplication Square Set (mathematics) Number Complex number Computer animation Angle Complex number Vector graphics Length Absolute value Factorization
Computer animation
Computer animation
Computer animation
Computer animation
Computer animation Angle Link (knot theory) Film editing Gradient Square Propositional formula Factorization Number
Computer animation Angle Direction (geometry) Gradient
Computer animation
Zahl Computer animation Complex number Gradient
Expression Sign (mathematics) Degree (graph theory) Computer animation Sine Gradient Lag Vector graphics Calculus Trigonometric functions
Degree (graph theory) Summierbarkeit
Computer animation
Degree (graph theory) Angle Gradient Summierbarkeit
Jetzt was rechnen mit der Zahlen und die geometrische Bedeutung noch mal für den Quotienten für Teile der Division an 6 plus 7 Linie durch
2-minus 3 das können Sie jetzt einfach mit Plusminus mal bezahlt ausrechnen algebraisch sodass man zuerst und ganze gucken was mit den linken passiert was von 3 Zahlen vorkommen selber auch Ergebnis was die Linke von diesen 3 Zahlen die hängen die zusammen über sollen sich das das wirklich so funktioniert wie es erzählt habe
Das Recht Algebra der Krieg ist
Mit dem komplex konjugiert wissen was zu erweitern das wird sich vollständig aus ich schreibe den Bruch der muss 2-minus 3 oder weitere mit 2 plus 3 Dann ist auch ja nichts passiert ich könnte mir kürzen Weiter Kontext können die jetzt wenn sie eine komplexe Zahlen H irgendwo als im derzeit heißt und expandiert jeweils als Kontext kann jetzt das Vorzeichen von Zeit zu ändern an der Achse zu spiegeln aus minus 3 das ist Mehrzahl mit plus 3 oder Ausfluss 3 minus 3 wenn sie zweimal Kontext von dir hintereinander landen sie wieder beim Original so warum funktioniert das das allgemeinen ankucken stellt sich vor sie haben und eine komplexe Zahlen beschrieben als real Zahl der Zahl Abschluss in die Mehrzahl der Zeit des mal ist komplex konjugiert ist der an minus 1 Und jetzt steht hier so war es selbst als hätte er eine komplexe Zahlen im Kontext können jetzt ist der Mahlzeit ist also Abschluss als man aber nie aus Mali können Sie es ausbuchstabieren hatten einige gemacht aber sie können auch der das ist 3. Bindung Lapsus noch mal Plan - nur gibt es lag Vertrag Minister Quadrat aber da steht in dem Quadrat ein Linien drinnen das macht das Vorzeichen Widerhall los vertrat und Abfall nahtlos Quadrat sollte Ihnen bekannt vorkommen | 852 | 3,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-45 | longest | en | 0.57787 |
https://groups.yahoo.com/neo/groups/ABAP/conversations/topics/39046 | 1,503,272,477,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00193.warc.gz | 772,591,538 | 23,075 | Sorry, an error occurred while loading the content.
diff b/t
Expand Messages
• hi abapers, if x = 10 and y = 5. x = y . move y to x. what is the difference between these two statements. __________________________________________________
Message 1 of 9 , Apr 1, 2006
hi abapers,
if x = 10 and y = 5.
x = y .
move y to x.
what is the difference between these two statements.
__________________________________________________
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• hi, there is no difference. szj ... ___________________________________________________________ Win a BlackBerry device from O2 with Yahoo!. Enter now.
Message 2 of 9 , Apr 2, 2006
hi,
there is no difference.
szj
--- murali kanth <muralikanth9@...> wrote:
> hi abapers,
>
> if x = 10 and y = 5.
> x = y .
> move y to x.
> what is the difference between these two statements.
>
> __________________________________________________
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• hi murali there is no difference b/w these two statements , both will asign value 5 to X. both r same ( for asignment ) correct me if i am wrong bye shravan
Message 3 of 9 , Apr 2, 2006
hi murali
there is no difference b/w these two statements , both will asign value 5 to X.
both r same ( for asignment )
correct me if i am wrong
bye
shravan
murali kanth <muralikanth9@...> wrote:
hi abapers,
if x = 10 and y = 5.
x = y .
move y to x.
what is the difference between these two statements.
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• hi i want little correction there is one difference in move statement automatically type conversion take place but if in x = y type mismatch then it will give
Message 4 of 9 , Apr 2, 2006
hi
i want little correction there is one difference
in move statement automatically type conversion take place but
if in x = y type mismatch then it will give error.
József Szikszai <jozsef_szikszai@...> wrote:
hi,
there is no difference.
szj
--- murali kanth <muralikanth9@...> wrote:
> hi abapers,
>
> if x = 10 and y = 5.
> x = y .
> move y to x.
> what is the difference between these two statements.
>
> __________________________________________________
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anju jain
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09911196874
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• sorry, you are wrong. these statements are equivalent. conversion is done in both cases. just try! :) ...
Message 5 of 9 , Apr 3, 2006
sorry, you are wrong. these statements are equivalent.
conversion is done in both cases. just try! :)
--- anju jain <anjuforever2002@...> wrote:
> hi
> i want little correction there is one difference
> in move statement automatically type conversion
> take place but
> if in x = y type mismatch then it will give error.
>
>
> József Szikszai <jozsef_szikszai@...> wrote:
> hi,
>
> there is no difference.
>
> szj
>
> --- murali kanth <muralikanth9@...> wrote:
>
> > hi abapers,
> >
> > if x = 10 and y = 5.
> > x = y .
> > move y to x.
> > what is the difference between these two
> statements.
> >
> > __________________________________________________
> > Do You Yahoo!?
> > Tired of spam? Yahoo! Mail has the best spam
> > protection around
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> >
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> >
> >
> > Yahoo! Groups Links
> >
> >
> > ABAP-unsubscribe@yahoogroups.com
> >
> >
> >
> >
> >
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>
>
>
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• hi murali, i think both are same...and these statement does not check there thecnical properties of fields .. but the Assign x to y is difference,,.... murali
Message 6 of 9 , Apr 4, 2006
hi murali,
i think both are same...and these statement does not check there thecnical properties of fields ..
but the Assign x to y is difference,,....
murali kanth <muralikanth9@...> wrote:
hi abapers,
if x = 10 and y = 5.
x = y .
move y to x.
what is the difference between these two statements.
__________________________________________________
Do You Yahoo!?
Tired of spam? Yahoo! Mail has the best spam protection around
http://mail.yahoo.com
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• hello friends here are the qustions which were ask at IBM( wriiten exam) 1) print prime number 2) palindrome of string, 3) validation of selection-screen 4)
Message 7 of 9 , Apr 5, 2006
hello friends
here are the qustions which were ask at IBM( wriiten exam)
1) print prime number
2) palindrome of string,
3) validation of selection-screen
4) print the customer detalis where custo no is between 51 and 57
5) i dont remember
total time for 5 questions is 1 hrs
send the answers accross
regards
Velchal girish.
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• Hi Girish Please see the below code .It prints the prime numbers from 1 to 100. We can change the code if we want to get the number range from selection
Message 8 of 9 , Apr 6, 2006
Hi Girish
Please see the below code .It prints the prime numbers
from 1 to 100.
We can change the code if we want to get the number
range from selection screen.
DATA : number TYPE i,
result TYPE p DECIMALS 2,
temp TYPE i VALUE 0,
fraction_part TYPE p DECIMALS 2,
no_of_zero TYPE i.
number = 1.
DO 100 TIMES.
DO number TIMES.
temp = temp + 1.
result = number / temp.
fraction_part = frac( result ).
IF fraction_part EQ 0.
no_of_zero = no_of_zero + 1.
ENDIF.
ENDDO.
IF no_of_zero LE 2.
WRITE : / 'NUMBER' , number ,'IS PRIME'.
ENDIF.
CLEAR : no_of_zero, temp.
number = number + 1.
ENDDO.
if anyone has got any suggestion to improve the code
Regards,
Sumit Khetawat
--- satyagirish satya <satyagirish_v@...>
wrote:
> hello friends
>
> here are the qustions which were ask at IBM(
> wriiten exam)
>
> 1) print prime number
> 2) palindrome of string,
> 3) validation of selection-screen
> 4) print the customer detalis where custo no is
> between 51 and 57
> 5) i dont remember
>
> total time for 5 questions is 1 hrs
> send the answers accross
>
> regards
> Velchal girish.
>
>
>
> ---------------------------------
> Jiyo cricket on Yahoo! India cricket
> Yahoo! Messenger Mobile Stay in touch with your
> buddies all the time.
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• just a small mathematical remark: 1 is not a prime number... ... ___________________________________________________________ Yahoo! Photos – NEW, now
Message 9 of 9 , Apr 6, 2006
just a small mathematical remark: 1 is not a prime
number...
--- Sumit Khetawat <sumit_dl@...> wrote:
> Hi Girish
> Please see the below code .It prints the prime
> numbers
> from 1 to 100.
> We can change the code if we want to get the number
> range from selection screen.
>
> DATA : number TYPE i,
> result TYPE p DECIMALS 2,
> temp TYPE i VALUE 0,
> fraction_part TYPE p DECIMALS 2,
> no_of_zero TYPE i.
>
> number = 1.
> DO 100 TIMES.
>
> DO number TIMES.
> temp = temp + 1.
> result = number / temp.
> fraction_part = frac( result ).
> IF fraction_part EQ 0.
> no_of_zero = no_of_zero + 1.
> ENDIF.
> ENDDO.
> IF no_of_zero LE 2.
> WRITE : / 'NUMBER' , number ,'IS PRIME'.
> ENDIF.
> CLEAR : no_of_zero, temp.
> number = number + 1.
> ENDDO.
>
> if anyone has got any suggestion to improve the code
> please letme know.
> Regards,
> Sumit Khetawat
> --- satyagirish satya <satyagirish_v@...>
> wrote:
>
> > hello friends
> >
> > here are the qustions which were ask at IBM(
> > wriiten exam)
> >
> > 1) print prime number
> > 2) palindrome of string,
> > 3) validation of selection-screen
> > 4) print the customer detalis where custo no is
> > between 51 and 57
> > 5) i dont remember
> >
> > total time for 5 questions is 1 hrs
> > send the answers accross
> >
> > regards
> > Velchal girish.
> >
> >
> >
> > ---------------------------------
> > Jiyo cricket on Yahoo! India cricket
> > Yahoo! Messenger Mobile Stay in touch with your
> > buddies all the time.
>
>
> __________________________________________________
> Do You Yahoo!?
> Tired of spam? Yahoo! Mail has the best spam
> protection around
> http://mail.yahoo.com
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Your message has been successfully submitted and would be delivered to recipients shortly. | 2,961 | 10,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-34 | latest | en | 0.777866 |
http://www.dissertation.com/abstracts/1399324 | 1,537,504,815,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00308.warc.gz | 305,996,534 | 6,677 | # AbstractsMathematics
by Alexander Longdon
Institution: University of Manchester 2015 self-transverse immersions of manifolds; stably complex structures on self-intersection manifolds 1399324 http://www.manchester.ac.uk/escholar/uk-ac-man-scw:247749
## Abstract
In this thesis we study the problem of determining the possible cobordism types of r-fold self-intersection manifolds associated to self-transverse immersions f: M^{n-k} -> \R^n for certain values of n, k, and r. Namely, we study the double-point self-intersection manifolds of immersions M^{n+2} -> \R^{2n+2} and M^{n+4} -> \R^{2n+4}, focusing on the case when $n$ is even. In the case of self-transverse immersions f : M^{n+2} -> \R^{2n+2}, we see that when n is even the double-point self-intersection manifold is a boundary, which is a result originally due to Szucs. In the case of self-transverse immersions f : M^{n+4} -> \R^{2n+4}, we show than when n is even the double-point self-intersection manifold is either a boundary or cobordant to RP^2 x RP^2, which is a new result. We then show that for even n such that the binary expansion of n+4 contains 5 or more 1s, the double-point self-intersection manifold of a self-transverse immersion M^{n+4} -> \R^{2n+4} is necessarily a boundary. We also survey the case when n is odd.We also set up and study the complex versions of the above problems: self-transverse immersions f : M^{2k+2} -> \R^{4k+2} and f : M^{2k+4} -> \R^{4k+4} of stably complex manifolds with a given complex structure on the normal bundle of f\$. In these cases, the double-point self-intersection manifold L associated to the immersion inherits a stably complex structure, and we attempt to determine which complex cobordism classes of stably complex manifolds may arise in this way. This is all new work.In the case of self-transverse complex immersions f : M^{2k+2} -> \R^{4k+2}, we show that the first normal Chern number of the double-point self-intersection manifold is a multiple of 2^{\lambda_{k+1}} for some integer \lambda_{k+1}, and provide upper and lower bounds for the value of \lambda_{k+1}. We also determine the exact value of \lambda_{k+1} in certain cases. In the case of self-transverse complex immersions f : M^{2k+4} -> \R^{4k+4}, we identify a large class of stably complex manifolds that may arise as the double-point self-intersection manifold of such an immersion and also identify a class of manifolds that may not. Additionally, in both cases we identify a necessary (and sometimes sufficient) condition for a stably complex manifold of the appropriate dimension to admit a complex immersion of the appropriate codimension. | 701 | 2,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-39 | latest | en | 0.896213 |
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