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http://www.jiskha.com/display.cgi?id=1258557860 | 1,498,424,851,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320582.2/warc/CC-MAIN-20170625203122-20170625223122-00359.warc.gz | 547,606,946 | 4,176 | # math
posted by .
The amount of revenue a company makes per day by selling x items is given by the function f(x) = 14x - 0.2x2. How many items should be sold if the company wants to maximize their profit?
• math -
I suppose you mean f(x) = 14x - 0.2x^2.
Please use an ^ before exponents.
The function has a maximum value where the derivative is zero.
f'(x) = 14 - 0.4 x = 0
x = 14/.4 = 35
If you don't know differential calculus yet, you can get the same result by "completing the square", but it takes longer, espcially with decimals involved.
• math -
Determine whether the system is consistent, inconsistent, or dependent.
3x + 2y = 15
6x + 4y = 30
• math -
inconsistent | 201 | 684 | {"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-2017-26 | latest | en | 0.883553 |
https://computer_en_ru.academic.ru/3504/binary_multiplication | 1,702,000,413,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00789.warc.gz | 205,511,934 | 11,999 | # binary multiplication
binary multiplication
двоичное умножение, умножение в двоичной системе (счисления)
English-Russian dictionary of computer science and programming. 2013.
### Смотреть что такое "binary multiplication" в других словарях:
• binary multiplication — dvejetainė daugyba statusas T sritis automatika atitikmenys: angl. binary multiplication vok. binäre Multiplikation, f; Binärmultiplikation, f rus. двоичное умножение, n; умножение в двоичной системе, n pranc. multiplication binaire, f … Automatikos terminų žodynas
• Binary numeral system — Numeral systems by culture Hindu Arabic numerals Western Arabic (Hindu numerals) Eastern Arabic Indian family Tamil Burmese Khmer Lao Mongolian Thai East Asian numerals Chinese Japanese Suzhou Korean Vietnamese … Wikipedia
• Multiplication algorithm — A multiplication algorithm is an algorithm (or method) to multiply two numbers. Depending on the size of the numbers, different algorithms are in use. Efficient multiplication algorithms have existed since the advent of the decimal system.… … Wikipedia
• Binary multiplier — A binary multiplier is an electronic circuit used in digital electronics, such as a computer, to multiply two binary numbers. It is built using binary adders. A variety of computer arithmetic techniques can be used to implement a digital… … Wikipedia
• Multiplication ALU — In digital design, a multiplier or multiplication ALU is a hardware circuit dedicated to multiplying two binary values. A variety of techniques can be used to implement a digital multiplier. Most techniques involve computing a set of partial… … Wikipedia
• Binary-coded decimal — In computing and electronic systems, binary coded decimal (BCD) is a digital encoding method for numbers using decimal notation, with each decimal digit represented by its own binary sequence. In BCD, a numeral is usually represented by four bits … Wikipedia
• multiplication binaire — dvejetainė daugyba statusas T sritis automatika atitikmenys: angl. binary multiplication vok. binäre Multiplikation, f; Binärmultiplikation, f rus. двоичное умножение, n; умножение в двоичной системе, n pranc. multiplication binaire, f … Automatikos terminų žodynas
• Binary scaling — is a computer programming technique used mainly by embedded C, DSP and assembler programmers to perform a pseudo floating point using integer arithmetic.It is both faster and more accurate than directly using floating point instructions, however… … Wikipedia
• Binary operation — Not to be confused with Bitwise operation. In mathematics, a binary operation is a calculation involving two operands, in other words, an operation whose arity is two. Examples include the familiar arithmetic operations of addition, subtraction,… … Wikipedia
• Multiplication — Multiply redirects here. For other uses, see Multiplication (disambiguation). For methods of computing products, including those of very large numbers, see Multiplication algorithm. Four bags of three marbles gives twelve marbles. There are also… … Wikipedia
• binary code — Computers. a system of representing letters, numbers, or other characters, using binary notation. * * * Code used in digital computers, based on a binary number system in which there are only two possible states, off and on, usually symbolized by … Universalium
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## Dr. MGR University AUGUST 2013 Question PaperSubject : PAPER IV APPLIED MECHANICS & STRENGTH OF MATERIALS
[LD 0212] Sub. Code: 2404
BACHELOR OF PROSTHETICS AND ORTHOTICS (BPO)
FIRST YEAR
PAPER IV – APPLIED MECHANICS & STRENGTH OF MATERIALS
Q.P. Code: 802404
Time : Three Hours Maximum : 100 marks
I. Elaborate on: (3 x 10 = 30)
1. Explain the stress-strain diagram with a neat sketch.
2. Find the centroid of an inverted T-section with flange 150 x 20mm and
web 100 x 25 mm.
3. A beam is freely supported over a span of 8 meters. It carries a point load of 3 KN
at 2 m from the left hand support and a uniformly distributed load of 2 KN/m run
from the centre to the right hand support. Construct Shear force diagram and
bending moment diagram.
II. Write notes on: (8 x 5 = 40)
1. Define uniform acceleration and types of motion.
2. Define force and what are the types of forces?
3. Explain the working stress, bulk modulus and factor of safety.
4. A circular steel bar 20mm diameter carries tensile load of 30 KN. Find the tensile stress
and elongation of the bar in a length of 300mm. Take E= 2 x10^5 N/mm².
5. Define centre of gravity and centroid.
6. Determine the diameter of solid shaft which is transmitting 100Kw at 160 rpm.
Also determine the length of the shaft if the twist must not exceed 1°, maximum
shear stress is limited to 60 N/mm² and N = 0.8x10^5 N/mm².
7. A closely coiled helical spring is made of steel wire 12 mm diameters. The number
of coil is 20 and mean radius of coil is 75 mm. Calculate the stiffness of the spring
if N= 1.2 x10^5 N/mm².
8. Define ergonomics with relevant example.
III. Short Answers on: (10 x 3 = 30)
1. Define scalar and its properties with an example.
2. Define torque.
3. Define strain.
4. Define ultimate stress.
5. What are the types of loads acting on a body.
6. Difference between open coiled spring and closed coil springs.
7. Define friction.
8. Define moment of inertia.
9. Define fatigue.
10. Define mechanics.
*******
## Dr. MGR University AUGUST 2014 Question PaperSubject : PAPER IV APPLIED MECHANICS & STRENGTH OF MATERIALS
[LF 0212] Sub. Code: 2404
BACHELOR OF PROSTHETICS AND ORTHOTICS (BPO)
FIRST YEAR
PAPER IV – APPLIED MECHANICS & STRENGTH OF MATERIALS
Q.P. Code: 802404
Time : Three Hours Maximum : 100 marks
I. Elaborate on: (3 x 10 = 30)
1. A mild steel bar 2m long 20mm Wide and 10mm thick is subjected to an
axial pull of 20KN. Calculate the change in Length, Width, Volume and Thickness.
E = 2 x 105 N/mm2
.
2. Explain the various types of loads and beams with a simple sketch.
3. A closely coiled helical spring is made of steel wire 10 mm diameter has 10 coils of
120mm mean diameter. Calculate the deflection and stiffness of spring under an axial
load of 98 N. Take N=1.2x10^5 N/mm².
II. Write notes on: (8 x 5 = 40)
1. What is force? What are the types of forces? Differentiate any two.
2. An air plane accelerates down a runway at 2.20 m/s² for 28.8s until is finally
lifts off the ground. Determine the distance travelled before take off.
3. Define stress, strain and bulk modulus.
4. Define centroid and moment of inertia.
5. Define friction and coefficient of friction, with an example.
6. Define Ergonomics and its purpose
7. Resultant of forces.
8. State polygon law of forces and triangle law of forces.
III. Short Answers on: (10 x 3 = 30)
1. Define Lami’s theorem.
2. Define vector with an example.
3. Define acceleration.
4. What is working stress?
5. Define factor of safety.
6. What are the types of pollutions?
7. What is mean by fatigue?
8. What is a spring?
9. Define center of gravity.
10. Define power and energy.
*******
## Dr. MGR University AUGUST 2015 Question PaperSubject : PAPER IV APPLIED MECHANICS & STRENGTH OF MATERIALS
[LH 0815] Sub. Code: 2404
BACHELOR OF PROSTHETICS AND ORTHOTICS (BPO)
FIRST YEAR
PAPER IV – APPLIED MECHANICS & STRENGTH OF MATERIALS
Q.P. Code: 802404
Time : Three Hours Maximum : 100 marks
I. Elaborate on: (3 x 10 = 30)
1. Find the Centroid of the bent Wire as shown in below figure:
2. Draw the shear Force and Bending Moment Diagrams for the cantilever beam
as shown below.
3. A steel Bar ABCD is given below. Determine stresses in different parts and total
elongation.
Take E= 2x105 N/mm2
II. Write notes on: (8 x 5 = 40)
1. A force system is give below. Determine Magnitude and Direction of the resultant
force with respect to 45N Force.
2. State the assumptions made in the theory of pure torsion.
3. Derive the expression for bending equation for bending of a beam after assuming the
necessary assumptions
4. Define Punching shear and Hoop stress.
5. Calculate the MI of circle about XX and YY axis if MI about ZZ axis is 5 x 1012 mm 4.
6. State the assumptions made in the theory of simple bending.
7. A circular column having 500mm diameter is carrying a load of 180kN at an
eccentricity of 300mm from the centrodial axis. Determine the maximum and
minimum stresses induced in the section.
8. State laws of dry friction.
III. Short Answers on: (10 x 3 = 30)
1. Draw stress-strain diagram for ductile material.
2. Define point of contra flexure.
3. Define moment of resistance.
4. Define shear stress.
5. What is fletched beam?
6. What is meant equivalent length of column?
7. State the maximum SF induced in a cantilever of span 4m and carrying a
point load of 30kN at its free end.
8. Calculate polar MI of square section having 200mm as side.
10. What are the conditions of equilibrium of rigid bodies?
******* | 1,589 | 5,559 | {"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} | 3.890625 | 4 | CC-MAIN-2020-40 | longest | en | 0.801223 |
https://jacanswers.com/what-are-concrete-representations-in-math/ | 1,669,559,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00735.warc.gz | 378,969,263 | 28,919 | # What are concrete representations in math?
What are concrete representations in math? In the concrete stage, the teacher begins instruction by modeling each mathematical concept with concrete materials (e.g. red and yellow chips, cubes, base ten blocks, pattern blocks, fraction bars, geometric figures). … Representational. The “seeing” stage uses representations of the objects to model problems.
What are concrete examples in math? Concrete examples are specific examples that correlate with your class learning objectives. They can help create strong connections between abstract material and your current understanding. Concrete examples can be found in your class lectures, class materials, and from your peers.
What is concrete stage math? The first step is called the concrete stage. It is known as the “doing” stage and involves physically manipulating objects to solve a math problem. The representational (semi-concrete) stage is the next step. It is known as the “seeing” stage and involves using images to represent objects to solve a math problem.
What are representations in maths? As most commonly interpreted in education, mathematical representations are visible or tangible productions – such as diagrams, number lines, graphs, arrangements of concrete objects or manipulatives, physical models, mathematical expressions, formulas and equations, or depictions on the screen of a computer or …
## What are concrete representations in math? – Related Questions
### How to remove salt stains from concrete?
Spray affected areas with water to get them wet. Fill a bucket with warm water and add vinegar and/or dish soap/detergent. Dip the broom into the water and use it to scrub the surface of the concrete. Scrub for several minutes until the salt residue and stains start to lift.
### How to prep concrete floor for staining?
Apply Concrete Cleaner and Degreaser: The vast majority of concrete slabs only require minimum cleaning using an organic degreaser diluted at a medium concentration with water. Scrub Surface: Clean the surface with a soft nylon bristle brush or power wash on a low setting to prepare most floors for staining.
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Slabs, with their props left under them, can typically be removed after 3-4 days. Soffits, with their props left under them, can be removed after one week. Props supporting slabs under 15 feet can be removed after one week.
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### What grit sandpaper to remove paint from concrete?
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### What is durability properties of concrete?
Durability of concrete may be defined as the ability of concrete to resist weathering action, chemical attack, and abrasion while maintaining its desired engineering properties.
### Can you add a sink on a concrete slab?
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### How to get olive oil out of concrete?
Or try my home-style hint first: Sprinkle baking soda over the whole stain, then pour on liquid dish soap (one that’s a grease cutter), also covering the stain. Scrub with a plastic brush (do not use metal); let set for several hours. Rinse, and take a look. It will probably take numerous attempts to dissolve the oil.
### How much does stamped concrete cost per foot?
Basic stamped concrete costs between \$8 and \$12 per square foot, but more involved projects can be as expensive as \$18 per square foot.
### What is workability of concrete pdf?
Workability of concrete is the property of freshly mixed concrete which determines the ease and homogeneity with which it can be mixed, placed, consolidated and finished’ as defined by ACI Standard 116R-90 (ACI 1990b).
### Do you have to wax concrete countertops?
You can wax concrete countertops more frequently in high use areas if desired. Concrete Countertop Care and Maintenance Tips: Wax monthly, or more frequently in higher use areas. Seal countertops every 6 months – 2 years depending on use.
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It is safe to pave asphalt over concrete. Concrete is a great base material because it is stable and allows for excellent compaction of the asphalt above. In fact, many miles of concrete roads are paved over with asphalt every year.
### How much to pour concrete patio?
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### How do you make concrete sparke?
To make the sparkle even more prominent, he recommends integrally coloring the concrete mix. This metallic aggregate from Pacific Palette Concrete Products comes in black and white. Sparkle Grain is recommended where you want a decorative, hard, nonslip surface.
### Do i need underlay on concrete floor?
If you have a concrete floor you will need an underlay which is anti-bacterial and made from rubber. Not all types of underlay will cope well with constantly being rubbed up against concrete so you may want to seek advice from carpet fitting experts before purchasing.
### How much does a 2x2x4 concrete block weigh?
How much does a 2 x 2 x 4 concrete block weigh? Approximate weight of 2x2x4 is 2400 lbs. All blocks have a recessed lifting eye for easy installation.
### Can you put flexible pvc conduit in concrete?
There are many different types of piping used in construction. … Rigid PVC pipes are not to be used in concrete while flexible PVC pipes can be installed in concrete.
### How to carpet concrete floor?
There are many different types of piping used in construction. … Rigid PVC pipes are not to be used in concrete while flexible PVC pipes can be installed in concrete. | 1,527 | 7,505 | {"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-2022-49 | latest | en | 0.929178 |
http://research.stlouisfed.org/fred2/series/KIPPPGESA156NUPN | 1,430,708,434,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430453257153.0/warc/CC-MAIN-20150501040737-00000-ip-10-235-10-82.ec2.internal.warc.gz | 158,454,091 | 19,957 | # Investment Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Spain
2010: 25.28203 Percent (+ see more)
Annual, Not Seasonally Adjusted, KIPPPGESA156NUPN, Updated: 2012-09-17 11:12 AM CDT
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Name: Email: | 527 | 2,018 | {"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-2015-18 | latest | en | 0.784489 |
http://gaussian.com/vib/ | 1,716,131,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00388.warc.gz | 11,007,985 | 34,422 | # Vibrational Analysis in Gaussian
Joseph W. Ochterski, Ph.D.
help@gaussian.com
October 29, 1999
Minor updates: 17 June 2018, 20 August 2020
#### Abstract
One of the most commonly asked questions about Gaussian is “What is the definition of reduced mass that Gaussian uses, and why is is different than what I calculate for diatomics by hand?” The purpose of this document is to describe how Gaussian calculates the reduced mass, frequencies, force constants, and normal coordinates which are printed out at the end of a frequency calculation.
#### Contents
So why is the reduced mass different in Gaussian? The short answer is that Gaussian uses a coordinate system where the normalized cartesian displacement is one unit. This differs from the coordinate system used in most texts, where a unit step of one is used for the change in interatomic distance (in a diatomic). The vibrational analysis of polyatomics in Gaussian is not different from that described in “Molecular Vibrations” by Wilson, Decius and Cross. Diatomics are simply treated the same way as polyatomics, rather than using a different coordinate system.
In this section, I’ll describe exactly how frequencies, force constants, normal modes and reduced mass are calculated in Gaussian, starting with the Hessian, or second derivative matrix. I’ll outline the general polyatomic case, leaving out details for dealing with frozen atoms, hindered rotors, and the like.
I will try to stick close to the notation used in “Molecular Vibrations” by Wilson, Decius and Cross. I will add some subscripts to indicate which coordinate system the matrix is in.
There is an important point worth mentioning before starting. Vibrational analysis, as it’s descibed in most texts and implemented in Gaussian, is valid only when the first derivatives of the energy with respect to displacement of the atoms are zero. In other words, the geometry used for vibrational analysis must be optimized at the same level of theory and with the same basis set that the second derivatives were generated with. Analysis at transition states and higher order saddle points is also valid. Other geometries are not valid. (There are certain exceptions, such as analysis along an IRC, where the non-zero derivative can be projected out.) For example, calculating frequencies using HF/6-31g* on MP2/6-31G* geometries is not well defined.
Another point that is sometimes overlooked is that frequency calculations need to be performed with a method suitable for describing the particular molecule being studied. For example, a single reference method, such as Hartree-Fock (HF) theory is not capable of describing a molecule that needs a multireference method. One case that comes to mind is molecules which are in a ground state. Using a single reference method will yield different frequencies for the and vibrations, while a multireference method shows the cylindrical symmetry you might expect. This is seldom a large problem, since the frequencies of the other modes, like the stretching mode, are are still useful.
#### Mass weight the Hessian and diagonalize
We start with the Hessian matrix , which holds the second partial derivatives of the potential V with respect to displacement of the atoms in cartesian coordinates (CART):
This is a matrix (N is the number of atoms), where are used for the displacements in cartesian coordinates, . The refers to the fact that the derivatives are taken at the equilibrium positions of the atoms, and that the first derivatives are zero.
The first thing that Gaussian does with these force constants is to convert them to mass weighted cartesian coordinates (MWC).
where , and so on, are the mass weighted cartesian coordinates.
A copy of is diagonalized, yielding a set of 3N eigenvectors and 3Neigenvalues. The eigenvectors, which are the normal modes, are discarded; they will be calculated again after the rotation and translation modes are separated out. The roots of the eigenvalues are the fundamental frequencies of the molecule. Gaussian converts them to cm , then prints out the 3N (up to 9) lowest. The output for water HF/3-21G* looks like this:
Full mass-weighted force constant matrix:
Low frequencies --- -0.0008 0.0003 0.0013 40.6275 59.3808 66.4408
Low frequencies --- 1799.1892 3809.4604 3943.3536
In general, the frequencies for for rotation and translation modes should be close to zero. If you have optimized to a transition state, or to a higher order saddle point, then there will be some negative frequencies which may be listed before the “zero frequency” modes. (Freqencies which are printed out as negative are really imaginary; the minus sign is simply a flag to indicate that this is an imaginary frequency.) There is a discussion about how close to zero is close enough, and what to do if you are not close enough in Section 4 of this paper.
You should compare the lowest real frequencies list in this part of the output with the corresponding frequencies later in the output. The later frequencies are calculated after projecting out the translational and rotational modes. If the corresponding frequencies in both places are not the same, then this is an indication that these modes are contaminated by the rotational and translational modes.
#### Determine the principal axes of inertia
The next step is to translate the center of mass to the origin, and determine the moments and products of inertia, with the goal of finding the matrix that diagonalizes the moment of inertia tensor. Using this matrix we can find the vectors corresponding to the rotations and translations. Once these vectors are known, we know that the rest of the normal modes are vibrations, so we can distinguish low frequency vibrational modes from rotational and translational modes.
The center of mass ( ) is found in the usual way:
where the sums are over the atoms, . The origin is then shifted to the center of mass . Next we have to calculate the moments of inertia (the diagonal elements) and the products of inertia (off diagonal elements) of the moment of inertia tensor ( ).
This symmetric matrix is diagonalized, yielding the principal moments (the eigenvalues ) and a matrix ( ), which is made up of the normalized eigenvectors of . The eigenvectors of the moment of inertia tensor are used to generate the vectors corresponding to translation and infinitesimal rotation of the molecule in the next step.
#### Generate coordinates in the rotating and translating frame
The main goal in this section is to generate the transformation from mass weighted cartesian coordinates to a set of 3N coordinates where rotation and translation of the molecule are separated out, leaving 3N-6 or 3N-5 modes for vibrational analysis. The rest of this section describes how the Sayvetz conditions are used to generate the translation and rotation vectors.
The three vectors ( , , ) of length 3N corresponding to translation are trivial to generate in cartesian coordinates. They are just times the corresponding coordinate axis. For example, for water (using and ) the translational vectors are:
Generating vectors corresponding to rotational motion of the atoms in cartesian coordinates is a bit more complicated. The vectors for these are defined this way:
where j=x, y, z; i is over all atoms and P is the dot product of (the coordinates of the atoms with respect to the center of mass) and the corresponding row of , the matrix used to diagonalize the moment of inertia tensor .
The next step is to normalize these vectors. If the molecule is linear (or is a single atoms), any vectors which do not correspond to translational or rotational normal modes are removed. The scalar product is taken of each vector with itself. If it is zero (or very close to it), then that vector is not an actual normal mode and it is eliminated. (If the scalar product is zero, this mode will disappear when the transformation from mass weighted to internal coordinates is done, in Equation 6.) Otherwise, the vector is normalized using the reciprocal square root of the scalar product. Gaussian then checks to see that the number of rotational and translational modes is what’s expected for the molecule, three for atoms, five for linear molecules and six for all others. If this is not the case, Gaussian prints an error message and aborts.
A Schmidt orthogonalization is used to generate (or 3N-5) remaining vectors, which are orthogonal to the five or six rotational and translational vectors. The result is a transformation matrix that transforms from mass-weighted cartesian coordinates to internal coordinates , where rotation and translation have been separated out.
#### Transform the Hessian to internal coordinates and diagonalize
Now that we have coordinates in the rotating and translating frame, we need to transform the Hessian, (still in mass weighted cartesian coordinates), to these new internal coordinates (INT). Only the coordinates corresponding to internal coordinates will be diagonalized, although the full 3N coordinates are used to transform the Hessian.
The transformation is straightforward:
The submatrix of , which represents the force constants internal coordinates, is diagonalized yielding eigenvalues , and eigenvectors. If we call the transformation matrix composed of the eigenvectors , then we have
where is the diagonal matrix with eigenvalues .
#### Calculate the frequencies
At this point, the eigenvalues need to be converted frequencies in units of reciprocal centimeters. First we change from frequencies ( ) to wavenumbers ( ), via the relationship , where c is the speed of light. Solving for we get
The rest is simply applying the appropriate conversion factors: from a single molecule to a mole, from hartrees to joules, and from atomic mass units to kilograms. For negative eigenvalues, we calculate using the absolute value of , then multiply by -1 to make the frequency negative (which flags it as imaginary). After this conversion, the frequencies are ready to be printed out.
#### Calculate reduced mass, force constants and cartesian displacements
All the pieces are now in place to calculate the reduced mass, force constants and cartesian displacements. Combining Equation 6 and Equation 7, we arrive at
where is the matrix needed to diagonalize . Actually, is never calculated directly in Gaussian. Instead, is calculated, where is a diagonal matrix defined by:
and i runs over the x, y, and z coordinates for every atom. The individual elements of are given by:
The column vectors of these elements, which are the normal modes in cartesian coordinates, are used in several ways. First of all, once normalized by the procedure described below, they are the displacements in cartesian coordinates. Secondly, they are useful for calculating a number of spectroscopic properties, including IR intensities, Raman activies, depolarizations and dipole and rotational strengths for VCD.
Normalization is a relatively straight forward process. Before it is printed out, each of the 3N elements of is scaled by normalization factor , for that particular vibrational mode. The normalization is defined by:
The reduced mass for the vibrational mode is calculated in a similar fashion:
Note that since is orthonormal, and we can (and do) choose to be orthonormal, then is orthonormal as well. (Since then ).
We now have enough information to explain the difference between the reduced mass Gaussian prints out, and the one calculated using the formula usually used for diatomics:
The difference is in the numerator of each term in the summation. Gaussian uses rather than 1. Using the elements of yields the consistent results for polyatomic cases, and automatically takes symmetry into consideration. Simply extending the formula from Equation 14 to would (incorrectly) yield the same reduced mass for every mode of a polyatomic molecule.
The effect of using the elements of in the numerator is to make the unit length of the coordinate system Gaussian uses be the normalized cartesian displacement. In other words, in the coordinate system that Gaussian uses, the sum of the squares of the cartesian displacements is 1. (You can check this in the output). In the more common coordinate system for diatomics, the unit length is a unit change in internuclear distance from the equilibrium value.
One of the consequences of using this coordinate system is that force constants which you think should be equal are not. A simple example is H versus HD. Since the Hessian depends only on the electronic part of the Hamiltonian, you would expect the force constants to be the same for these to molecules. In fact, the force constant Gaussian prints out is different. The different masses of the atoms leads to a different set of Sayvetz conditions, which in turn, change the internal coordinate system the force constants are transformed to, and ultimately the resulting force constant.
The coordinates used to calculate the force constants, the reduced mass and the cartesian displacements are all internally consistent. The force constants $k_i$ are given by $k_i = 4{\pi}^2{\tilde{\nu_i}}^2c^2{\mu}_i$ since $\nu_i = \frac{1}{2\pi}\sqrt{\frac{k_i}{\mu_i}}$. The force constants are converted from atomic units to millidyne/angstrom.
#### Summary
To summarize, the steps Gaussian uses to perform vibrational analysis are:
1. Mass weight the Hessian
2. Determine the principal axes of inertia
3. Generate coordinates in the rotating and translating frame
4. Transform the Hessian to internal coordinates and diagonalize
5. Calculate the frequencies
6. Calculate reduced mass, force constants and cartesian displacements
${\mu}_i = ( \displaystyle\sum_k^{3N} l_{CART_{k,i}^2} )^{-1}$
$k_i = 4{\pi}^2{\tilde{\nu_i}}^2c^2{\mu}_i$
$l_{CART} = \bold{MDL}$
#### A note about low frequencies
You’ll find that the frequencies for the translations are almost always extremely close to zero. The frequencies for rotations are quite a bit larger. So, how “close to zero” is close enough? For most methods (HF, MP2, etc.), you’d like the rotational frequencies to be around 10 wavenumbers or less. For methods which use numerical integration, like DFT, the frequencies should be less than a few tens of wavenumbers, say 50 or so.
If the frequencies for rotations are not close to zero, it may be a signal that you need to do a tighter optimization. There are a couple of ways to accomplish this. For most methods, you can use Opt=Tight or Opt=Verytight on the route card to specify that you’d like to use tighter convergence criteria. For DFT, you may also need to specify Int=Ultrafine, which uses a more accurate numerical integration grid.
As an example, I reran the water HF/3-21G* calculation above, with both Opt=Tight and Opt=VeryTight. You can see in Table 1 that the rotational frequencies are an order of magnitude better for Opt=Tight than they were for just Opt. Using Opt=Verytight makes them even better.
Table 1: The effect of optimization criteria on the low frequencies of water using HF/3-21G*. The frequencies are sorted by increasing absolute value, so that it’s easier to distinguish rotational modes from vibrational modes.
This raises the question of whether the you need to use tighter convergence. The answer is: it depends – different users will be interested in different results. There is a trade off between accuracy and speed. Using Opt=Tight or Int=Ultrafine makes the calulation take longer in addition to making the results more accurate. The default convergence criteria are set to give an accuracy good enough for most purposes without spending time to converge the results beyond this accuracy. You may find that you need to use the tighter criteria to compare to spectroscopic values, or to resolve a strucutre witha particularly flat potential energy surface.
In the water frequency calculation above, using tighter convergence criteria makes almost no difference in terms of energy or bond lengths, as Table 2 demonstrates. The energy is converged to less then 1 microHartree, and the OH bond length is converged to 0.0002 angstroms. Tightening up the convergence criteria is useful for getting a couple of extra digits of precision in the symmetric stretch frequency.
Table 2: The default optimization settings yield results accurate enough for most purposes. Tighter optimizations make almost no difference for this HF/3-21G* frequency calculation on water.
Table 3: Initial geometries for water optimization calculations. Geometry A was produced by Geom=ModelA. Geometry B is a slightly modified version of Geometry A.
You can also see that the final geometry parameters obtained with the default optimization criteria depend somewhat on the initial starting geometry. Using Opt=VeryTight all but eliminates these differences. I’ve included the starting geometries in Table 3, for those who wish to reproduce these results. (Using the default convergence criteria may give somewhat different results than those I’ve shown if you use a different machine, or even the same machine using different libraries or a different version of the compiler).
With DFT, Opt=VeryTight alone is not necessarily enough to converge the geometry to the point where the low frequencies are as close to zero as you would like. To demonstrate this, I have run B3LYP/3-21G* optimizations on water, starting with geometry B from Table 3, with Opt, Opt=Tight and Opt=VeryTight. The results are in Table 4. The low frequencies from these two jobs hardly change, and in fact get worse for the Tight and VeryTight optimizations.
Table 4: The effect of grid size on the low frequencies from B3LYP/3-21G* on water with Opt, Opt=Tight and Opt=VeryTight. More accurate grids are necessary for a truly converged optimization. The frequencies are sorted by increasing absolute value, so that it’s easier to distinguish rotational modes from vibrational modes.
Given the straight forward convergence seen with Hartree-Fock theory, this might not seem to make sense. However, it does make sense if you recall that DFT is done using a numerical integration on a grid of points. The accuracy of the default grid is not high enough for computing low frequency modes very precisely. The solution is to use a more numerically accurate grid. The tighter the optimzation criteria, the more accurate the grid needs to be. As you can see in Table 4, increasinfg the convergence criteria from Tight to VeryTight without increasing the numerical accuracy of the grid yields no improvement in the low frequencies. For Opt=Tight, we recommend using the Ultrafine grid. This is a good combination to use for systems with hindered rotors, or if exact conformation is of concern. If still more accuaracy is necessary, then an unpruned 199974 grid can be used with Opt=VeryTight. Again, the higher accuracy comes at a higher cost in terms of CPU time. The VeryTight optimization with a 199974 grid is very expensive, even for medium sized molecules. The default grids are accurate enough for most purposes.
#### Acknowledgements
I’d like to thank John Montgomery for his constructive suggestions, Michael Frisch for clarifying several points, and H. Berny Schlegel for taking the time to discuss this material with me. Thanks also to Jim Cheeseman, for lending me his copy of Wilson, Decius and Cross.
Last updated on: 1 November 2021. | 4,050 | 19,421 | {"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": 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} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.931982 |
http://www.solipsys.co.uk/cgi-bin/sews.py?RiemannZetaFunction | 1,591,210,837,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00573.warc.gz | 205,657,563 | 1,891 | # Riemann Zeta Function
Consider the following infinte sequence:
$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdots$
This is convergent only for values of s whose real part is greater than 1. However, there is a unique way of extending it to the entire complex plane (except at s=1 ) such that the result is still "nice" (for which read "analytic")
This is then a function of one complex variable, and is named after Bernhard Riemann.
Leonhard Euler showed the following identity is true:
$\begin{matrix}\sum_{n=1}^\infty\frac{1}{n^s}&=&\prod_{p\text{~prime}}&1/(1-p^{-s})\end{matrix}$
This creates a connection between the Riemann Zeta function and the prime numbers.
There is much that is unknown about the Riemann Zeta function. For example, it's not known exactly where it takes the value 0. The Riemann Hypothesis is related to this question. | 267 | 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": 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} | 2.921875 | 3 | CC-MAIN-2020-24 | latest | en | 0.884805 |
https://edurev.in/studytube/Frequency-Domain-Analysis-2/fda7465a-c707-4aa1-b273-939851052fd2_t | 1,669,611,302,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00106.warc.gz | 253,402,173 | 62,710 | Frequency Domain Analysis - 2
# Frequency Domain Analysis - 2 Notes - Electrical Engineering (EE)
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Stability Analysis
• If the Nyquist path in the s plane encircles Z zeros and P poles of [1 + G(s) H(s)] and does not pass through any poles or zeros of [1+ G(s) H(s)] as a representative point s moves in the clockwise direction along the Nyquist path, then the corresponding contour in the G(s) H(s) plane encircles the (–1 + j0) point N = Z – P times in the clockwise direction. (Negative values of N imply counterclockwise encirclements).
• In examining the stability of linear control system using the Nyquist stability criterion, we see that three possibilities can occur.
1. There is no encirclement of the (– + j0) point. This implies that the system is stable if there are no poles of G(s)H(s) in the righthalf s-plane; otherwise, the system is unstable.
2. There is a counterclockwise encirclement or encirclements of the –1 +j0 point. In this case the system is stable if the number of counterclockwise encirclements is the same as the number of poles of G(s)H(s) in the right-half a plane; otherwise, the system is unstable.
3. There is a clockwise encirclement or encirclements of the –1 +j0 point. In this case the system is unstable.
NOTE:
Encirclement: Used in Nyquist stability criteria. It is for closed path around (–1 + j0)
Enclosurement: Used in polar plot. A point in GH plane is said to be enclosed by the polar plot if it lies on the right hand side of plot when moving along the plot in the increasing direction of frequency from 0 to ∞ .
PHASE MARGIN & GAIN MARGIN
• Figure shows the polar plots of G(jω) for three different values of the open-loop gain K. For a large value of the gain K, the system is unstable. As the gain is decreased to a certain value, the G(jω) locus passes through the – 1 –j0 point.
• This means that with this gain value the system is on the verge of instability, and the system is on the verge of instability, and the system will exhibit sustained oscillations. For a small value of the gain K, the system is stable. In general, the closer the G(jω) locus comes to encircling the –1 + j0 point, the more oscillatory is the systemresponse. The closeness of the G(jω) locus to the –1 + j0 point can be used as a measure of the margin of stability.
• This does not apply, however, to conditionally stable systems.
• It is common practice to represent the closeness in terms of phase margin and gain margin.
Phase Margin
• The phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability.
• The gain crossover frequency is the frequency at which |G( jw)| , the magnitude of the openloop transfer function is unity.
• The phase margin g is 180º plus the phase angle f of the open-loop transfer function at the gain crossover frequency, or g = 180º + f.
• In the above figure of polar plot, a line may be drawn from the origin to the point at which the unit circle crosses the G(jω) locus.
• The angle from the negative real axis to his line is the phase margin. The phase margin is positive for g > 0 and negative for g < 0. For minimum phase system to be stable, the phase margin must be positive.
• In the logarithmic plots, the critical point in the complex plane corresponds to the 0 dB and – 180º lines.
Gain Margin
• The gain margin is the reciprocal of the magnitude |G(jω)| at the frequency where the phase angle is – 180º.
• Defining the phase crossover frequency ω1 to be the frequency at which the phase angle of the open -loop transfer function equals –180º gives the gain margin Kg
in terms of decibels,
• The gain margin expressed in decibels is positive if Kg is greater than unity and negative if Kg is smaller than unity.
• A positive gain margin (in decibels) means that the system is stable, and negative gain margin (in decibels) means that the system is unstable. The gain margin and phase margins for different plots has been shown in figure (a), (b) and (c).
(phase and gain margins of stable and unstable systems)
(a) Bode diagrams; (b) polar Plots; (c) log-magnitude versus phase plots.
• For a stable minimum phase system (defined in last section of this chapter), the gain margin indicates how much the gain can be increased before the system becomes unstable.
• For an unstable system, the gain margin is indicative of how much the gain must be decreased to make the system stable.
• The gain margin of a first or second-order system is infinite since the polar plots for such systems do not cross the negative real axis. Thus, theoretically, first or second-order systems cannot be unstable.
NOTE:
• However, that so-called first or secondorder systems are only approximations in the sense thtat small time lags are neglected in deriving the system equations and are thus not truly first or second-order systems.
• If these small lags are accounted for, the so-called first or second order systems may become unstable.It is important to point out that for a nonminimum phase system the stability condition will not be satisfied unless the G(jω) plot encircles the (–1 + j0) point. Hence, a stable non-minimum phase system will have negative phase and gain margin.
• It is also imp ortant to point out that conditionally stable system will have two or more phase crossover frequencies, and some higher-order systems with complicated numerator dynamics may also have two or more gain crossover frequencies, as shown in Figure.
• For stable systems having two or more gain crossover frequencies, the phase margin is measured at the highest gain crossover frequency.
A Few Comments on phase and Gain Margins
• The phase and gain margins of a control system are a measure of the closeness of the polar plot to the (–1 + j0) point. Therefore, these margins may be used as design criteria.
• It should be noted that either the gain margin alone or the phase margin alone does not give a sufficient indication of the relative stability. Both should be given in the determination of relative stability.
• For a minimum phase system, both the phase and gain margins must be positive for the system to be stable. Negative margins indicate instability.
• Proper phase and gain margins ensure us against variations in the system components and are specified for definite values of frequency. The two values bound the behavior of the closed loop system near the resonant frequency.
• For satisfactory performance, the phase margin should be between 30º and 60º, and the gain margin should be greater than 6 dB. with these values, a minimum phase system has guaranteed stability, even if the open-loop gain and time constants of the components very to a certain extent. Although the phase and gain margins given only rough estimates of the effective damping ratio of the closed-loop system, they do offer a convenient means for designing control systems or adjusting the gain constant of systems.
CLOSED LOOP FREQUENCY RESPONSE
• Let us define the magnitude of the closed-loop frequency response as M and the phase angle as α , or
• In the following, we shall find the constant magnitude loci and constant phase-angle loci. such loci are convenient in determining the closed-loop frequency response from the polar plot or Nyquist plot.
Constant magnitude Loci (M-circles)
• to obtain the constant magnitude loci, let us first note that G(jω) is a complex quantity and can be written as follows: G(jω) = X + jY where X and Y are real quantities. Then M is given by
Hence, X2 (1–M2)–2M2 X–M2 + (1 – M2)Y2 =0
• If M = 1, then from above equation (8.10) we obtain X =–1/2. This is the equations of a straight line parallel to the Y-axis and passing through the point (–1/2, 0).
• If M ≠ 1, equation can be written as
if the term [M2/(1–M2)]2 is added to both sides of this last equation, we obtain
• Equation (8.11) is the equation of a circle with centre at X = +M2/(M2–1), Y=0 and with radius |M / (M2 - 1)| .
• The constant M loci on the G(s) plane are thus a family of circles. The centre and radius of the circle for a given value of M can be easily calculated
• For M>1, the centers of the M circles lie to the left of the (–1 +j0) point. Similarly, as M becomes smaller compared with 1, the M circle becomes smaller and converges to the origin. For 0< M <1, the centers of the M circles lie to the right of the origin.
• M = 1 corresponds to the locus of points equidistant from the origin and from the (-1 + j0) point. As stated earlier, it is a straight line passing through the point (-1/2, 0) and parallel to the imaginary axis.
• The M circles are symmetrical with respect to the straight line corresponding to M =1 and with respect to the real axis.
Constant Phase-angle Loci (N-circles)
• We shall obtain the phase angle α in terms of X and Y. Since
the phase angle α is
if we define, tanα = N then,
The addition of (1/4) + 1/(2N)2 to both sides of this last equation yields
This is an equation of a circle with centre at X = –1/2, Y = 1/(2N) and with radius
Since equation is satisfied when X=Y=0 and X=–1, Y = 0 regardless of the value of N, each circle passes through the origin and the (–1 + j0) point. The constant N loci can be drawn easily once the value of N is given. A family of constant N circles are shown in figure below with a as a parameter.
It should be noted that the constant N locus for a given value of a is actually not the entire circle but only an arc. In other words, the α = 30º and α = 150º arcs are parts of the same circle. This is so because the tangent of an angle remains the same if ± 180º (or multiples of it) is added to the angle.
The use of the M and N circles enables us to find the entire closed-loop frequency response from the open-loop frequency response G(jω) without calculation the magnitude and phase of the closed-loop transfer function at each frequency. The intersections of the G(jω) locus and the M circles and N circles given the values of M and N at frequency points on the G(jω) locus. The N circles are multivalued in the sense that the circl e for α = α1 and that for are the same. In using the N circles for the determination of the phase angle of closed-loop systems, we must interpret the proper value of a. to avoid any error, start at zero frequency, which corresponds to α = 0º, and proceed to higher frequencies. The phase-angle curve must be continuous.
• Graphically, the intersection of the G(jω) locus and M circles give the values of M at the frequencies denoted on the G(jω) locus. Thus, the constant M circle with the smallest radius that is tangent to the G(jω) locus gives the value of the resonant peak magnitude Mr. If it is desired to keep the resonant peak value less than a certain value, then the system should not enclose the critical point (–1 + j0 point) and at the same time there should be no intersections with the particular M circle and the G(jω) locus.
• Figure (a) shows the G(jω) locus superimposed on a family of M circles. Figure (b) shows the G(jω) locus superimposed on family of N circles. From these plots, it is possible to obtain the closed loop frequency response by inspection. it is seen that the M = 1.1 circle intersects the G(jω) locus at frequency point ω = ω1. This means that at this frequency the magnitude of the closedloop transfer function is 1.1.
• In figure (a), the M = 2 circle is just tangent to the G(jω) locus. Thus, there is only one point on the G(jω) locus for which |C(jω) / R (jω)| is equal to 2. Figure (c) in the next page shows the closed loop frequency- response curve for the system. The upper curve is the M versus frequency w curve and the lower curve is the phase angle a versus frequency ω curve.
• The resonant peak value is the value of M corresponding to the M circle of smallest radius that is tangent to the G(jω) locus. Thus, in the Nyquist diagram, the resonant peak value Mr and the resonant frequencywr can be found from the M-circle tangency to the G(jω) locus. (In the present example , Mr = 2 and ωr = ω4).
General Procedure for Plotting Bode-Diagrams
• First rewrite the sinusoidal transfer function G(jω) H(jω) as a product of basic factors discussed above.
• Then identify the corner frequencies associated with these basic factors.
• Finally, draw the asymptotic log-magnitude curves with proper slopes between the corner frequencies. The exact curve, which lies close to the asymptotic curve, can be obtained by adding proper corrections.
• The phase-angle curve of G(jω) H(jω) can be drawn by adding the phase- angle curves of individual factors.
NOTE: The use of Bode diagrams employing asymptotic approximations requires much less time than other methods that may be used for computing the frequency response of a transfer function. The ease of plotting the frequencyresponse curves for a given transfer function and the ease of modification of the frequencyresponse curve as compensation is added are the main reasons why Bode diagrams are very frequently used in practice.
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# Paper Plate Phases Modeling Lunar Cycles
Explore lunar phases as viewed from Earth using paper plates. While standing in the appropriate spot in the moon's orbit, students hold paper plates that depict the Moon's phase. This activity can be used to assess understanding of lunar phases or... (View More)
# Golf Ball Phases
Explore lunar phases as viewed from Earth using a golf ball and an ultraviolet light. With the student's head representing Earth, students hold and move the golf ball to demonstrate the cause of the Moon's phases in their correct order. Related Next... (View More)
# Solar Week Monday: Do the Activity - Do You See What I See?
This is an online lesson associated with activities during Solar Week, a twice-yearly event in March and October during which classrooms are able to interact with scientists studying the Sun. This activity is scheduled to occur during Monday of... (View More)
# Solar Week Wednesday: Do the Activity: Activity 1: Measuring the Motion of a Coronal Mass Ejection
This is an activity associated with activities during Solar Week, a twice-yearly event in March and October during which classrooms are able to interact with scientists studying the Sun. Outside of Solar Week, information, activities, and resources... (View More)
# Solving a Mixed Up Problem
This is a lesson about using the light from the star during an occultation event to identify the atmosphere of a planet. Learners will add and subtract light curves (presented as a series of geometrical shapes) to understand how this could occur.... (View More)
# Graphing the Rainbow
Learners will explore different ways of displaying visual spectra, including colored "barcode" spectra, like those produced by a diffraction grating, and line plots displaying intensity versus color, or wavelength. Students learn that a diffraction... (View More)
# S'COOL Lesson: Measures of Central Tendency
In this lesson plan students use temperature data to look at the measures of central tendency. By using mean, median, and mode, students will gain a better understanding about weather patterns from several locales throughout Virginia.
Keywords: Mean; Median; Mode
Audience: Middle school, High school
Materials Cost: Free
# Cosmic Microwave Background
In this lesson, students explore the cosmic microwave background to understand why it permeates the universe and why it peaks as microwave radiation. Students should be able to explain that the origin of the background radiation is the uniform... (View More)
# Spot Plots
This is an activity about measurement. Learners will label key points and features on a rectangular equal-area map and measure the distance between pairs of points in order to calculate the actual physical distance on the Sun that the point pairs... (View More)
Audience: Elementary school, Middle school, High school
Materials Cost: Free
# Exploring Infrared Image Technology
In this introductory activity, learners investigate and discuss infrared images of various everyday objects, such as toasters, hairdryers, and running water, to learn about infrared imaging. Student questions about the false-color images help guide... (View More)
«Previous Page1234 Next Page» | 765 | 3,716 | {"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-2019-09 | latest | en | 0.93257 |
https://writingerrands.com/five-years-ago-bob-had-opened-a-bank-account-that-pays-3-5-compounded-annually/ | 1,679,526,747,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00568.warc.gz | 643,306,717 | 11,373 | Five years ago, Bob had opened a bank account that pays 3.5% compounded annually. At the end of the first year, he deposited \$450, the second year he deposited \$350, the third year \$0, the fourth year \$500 and the fifth year \$625. What is the value of his account after making the fifth deposit? | 73 | 300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.961644 |
https://socratic.org/questions/56d2d65111ef6b77382037e0 | 1,632,202,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00223.warc.gz | 567,036,308 | 7,235 | # What is the "pH" of a solution of magnesium hydroxide given that K_(sp) = 1.8 * 10^(-11) ?
Feb 28, 2016
$\text{pH} = 10.52$
#### Explanation:
In order to find the pH of a solution, you must determine the concentration of hydronium ions, ${\text{H"_3"O}}^{+}$, either directly or indirectly.
When you're dealing with a Bronsted - Lowry acid, you will be solving for the concentration of hydronium ions directly.
When you're dealing with a Bronsted - Lowry base, like you are here, you will be solving for the concentration of hydronium ions indirectly, i.e. by solving for the concentration of hydroxide ions, ${\text{OH}}^{-}$.
Magnesium hydroxide, "Mg"("OH")_2, us insoluble in aqueous solution. This means that when you place magnesium hydroxide in water, an equilibrium will be established between the undissolved solid and the dissolved ions.
${\text{Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$
The solubility product constant, ${K}_{s p}$, for this equilibrium looks like this
${K}_{s p} = {\left[{\text{Mg"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$
What you need to do here is use the ${K}_{s p}$ to find the molar solubility, $s$, of magnesium hydroxide in aqueous solution at ${25}^{\circ} \text{C}$.
To do that, set up an ICE table
${\text{ ""Mg"("OH")_text(2(s]) " "rightleftharpoons" " "Mg"_text((aq])^(2+) " "+" " color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$
color(purple)("I")" " " " " "-" " " " " " " " " " " " " "0" " " " " " " " " " 0
color(purple)("C")" " " " " "-" " " " " " " " " " " "(+s)" " " " " "(+color(red)(2)s)
color(purple)("E")" " " " " "-" " " " " " " " " " " " " "s" " " " " " " " " "color(red)(2)s
Remember, the concentration of the solid is assumed to be constant.
Here ${K}_{s p}$ will be equal to
${K}_{s p} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}} = 4 {s}^{3}$
Rearrange to solve for $s$
$s = \sqrt[3]{{K}_{s p} / 4} = \sqrt[3]{\frac{1.8 \cdot {10}^{- 11}}{4}} = 1.65 \cdot {10}^{- 4}$
This means that the concentration of hydroxide ions in a saturated solution of magnesium hydroxide at ${25}^{\circ} \text{C}$ will be
["OH"^(-)] = color(red)(2) * 1.65 * 10^(-4)"M" = 3.30 * 10^(-4)"M"
Now, at ${25}^{\circ} \text{C}$, the concentration of hydronium ions and the concentration of hydroxide ions have the following relationship
color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-4))" ", where
${K}_{W}$ - the ion product for water's auto-ionization
Plug in your value to get
["H"_3"O"^(+)] = 10^(-14)/(3.30 * 10^(-4)) = 3.03 * 10^(-11)"M"
This means that the pH of the solution will be
color(blue)("pH" = - log(["H"_3"O"^(+)]))
$\text{pH} = - \log \left(3.03 \cdot {10}^{- 11}\right) = \textcolor{g r e e n}{10.52}$
Alternatively, you can use the equation
$\textcolor{b l u e}{\text{pH " + " pOH} = 14}$
Here
color(blue)("pOH" = - log(["OH"^(-)])) | 1,062 | 2,929 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 28, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.82755 |
https://www.thestudentroom.co.uk/showthread.php?page=66&t=1035080 | 1,555,702,088,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527866.58/warc/CC-MAIN-20190419181127-20190419202417-00017.warc.gz | 874,155,324 | 38,483 | # Maths Uni ChatWatch
#1301
(Original post by SimonM)
What do you mean by "count"?
A bijection between a set and the integers.
0
9 years ago
#1302
(Original post by Simplicity)
A bijection between a set A and the integers.
I don't like that definition of count to be honest, and I'm certain that around does know that
0
9 years ago
#1303
Yknow the cool thing is I'm doing a maths degree but I'm essentially here to do Physics.
edit; yup did countability today so we were all up to scratch. Want some of our countability example sheet questions Simplicity?
0
#1304
(Original post by SimonM)
I don't like that definition of count to be honest, and I'm certain that around does know that
You got a better definition? I actually find its a really elegant definition and all of it is based on a really good proof by induction. Which, I would post but yeah Totally Tom would get annoyed.
(Original post by My Alt)
Yknow the cool thing is I'm doing a maths degree but I'm essentially here to do Physics.
edit; yup did countability today so we were all up to scratch. Want some of our countability example sheet questions Simplicity?
That isn't cool. Physics isn't cool. Maths is cool.
Countability isn't interesting, did you do uncountability yet i.e. Cantors diagonal argument.
P.S. Nice sig. Trolling a pedo using Pokemon.
0
9 years ago
#1305
(Original post by Simplicity)
That isn't cool. Physics isn't cool. Maths is cool.
Countability isn't interesting, did you do uncountability yet i.e. Cantors diagonal argument.
P.S. Nice sig. Trolling a pedo using Pokemon.
i - Physics is cool, you just don't know it yet
ii - Countability is far more interesting than uncountability, considering we really know **** all about uncountable sets. Have seen cantors diagonal argument, have also seen the non-existance of a bijection from a set to it's power set, both very cool. Just as an aside looks cool
ii - Yeah thanks, always thought it was a cool thing to do, never tried it myself, though.
0
9 years ago
#1306
(Original post by Simplicity)
You got a better definition? I actually find its a really elegant definition and all of it is based on a really good proof by induction. Which, I would post but yeah Totally Tom would get annoyed.
When I think of counting I think of how much of stuff I've got (in some vaguely discrete sense (as opposed to measure)). In which case, I think ordinals a more "natural" thing.
Also, I think when people talk about "countability" as a subject, they often lump together the idea of "deciding" if things are countable or not
0
#1307
(Original post by My Alt)
i - Physics is cool, you just don't know it yet
ii - Countability is far more interesting than uncountability, considering we really know **** all about uncountable sets. Have seen cantors diagonal argument, have also seen the non-existance of a bijection from a set to it's power set, both very cool. Just as an aside looks cool
ii - Yeah thanks, always thought it was a cool thing to do, never tried it myself, though.
QM is kind of cool. I might do that next year.
Actually, I think that is wrong, actually uncountable sets are more general than countable sets. Well, its not really a diagonal argument for powersets, more like defining a subset that can't be hit and showing that it must exist.
(Original post by SimonM)
When I think of counting I think of how much of stuff I've got (in some vaguely discrete sense (as opposed to measure)). In which case, I think ordinals a more "natural" thing.
Also, I think when people talk about "countability" as a subject, they often lump together the idea of "deciding" if things are countable or not
But, that isn't counting, thats cardinality.
Well, I don't know what you are trying to say. Well, if something is describable in finite terms and you form a set from all the describable stuff then it would be countable. Its pretty trivial stuff.
Well, finding whats equipotent and maybe stuff like CH is pretty annoying.
0
9 years ago
#1308
(Original post by Simplicity)
QM is kind of cool. I might do that next year.
Actually, I think that is wrong, actually uncountable sets are more general than countable sets. Well, its not really a diagonal argument for powersets, more like defining a subset that can't be hit and showing that it must exist.
I never said the diagonal argument and the power set argument were the same thing
0
9 years ago
#1309
Got to miss yet another lecture today to go doctors Tonsilitis sucks!
And i'm in pain and getting no sleep due to not breathing properly. Might get a special circumstances form so I don't lose marks for not handing in my calc work. Not the best of weeks =(
0
9 years ago
#1310
(Original post by Simplicity)
But, that isn't counting, thats cardinality.
That would be cardinals rather than ordinals
0
9 years ago
#1311
(Original post by Simplicity)
I will watch that someday, but I can't find it online.
there's a website called animefreak.com and it has the full version of the first series and the Brotherhood version as it comes out i have to wait a whole week for the each episode to come out
0
#1312
(Original post by SimonM)
That would be cardinals rather than ordinals
Sounds interesting. When I have the time I might actually read the book I got on set theory as it has a chapter on ordinal arithmetic.
0
9 years ago
#1313
(Original post by Simplicity)
You don't do infinite sets. All you get is horrible number theory.
I get none of this, Durham hasn't even done set theory, which I reckon would upset you if you were here. Just Belgian-style Calculus, Analysis, Linear Algebra and Problems.
0
9 years ago
#1314
(Original post by assmaster)
I get none of this, Durham hasn't even done set theory, which I reckon would upset you if you were here. Just Belgian-style Calculus, Analysis, Linear Algebra and Problems.
Next term in probability you will...
0
9 years ago
#1315
(Original post by Simplicity)
QM is kind of cool. I might do that next year.
Yes, yes, yes, QM is made of awesomesauce. Hopefully I will be specialising in that sort of thing .
(Original post by My Alt)
Yknow the cool thing is I'm doing a maths degree but I'm essentially here to do Physics.
Win . Physics taught from a mathematical perspective is awesome.
9 years ago
#1316
(Original post by Simba)
Win . Physics taught from a mathematical perspective is awesome.
I couldn't agree more. I can laugh at other Physicists for being rubbish at maths, tell pure mathematicians that they are boring and be able to back that opinion up because I've done (will have done) the courses!
Who wrote Welcome Back Grae on the board today? I want to give them a rather large hug.
0
9 years ago
#1317
(Original post by My Alt)
I couldn't agree more. I can laugh at other Physicists for being rubbish at maths, tell pure mathematicians that they are boring and be able to back that opinion up because I've done (will have done) the courses!
Who wrote Welcome Back Grae on the board today? I want to give them a rather large hug.
You can't back up your opinion because you've done the courses. You can merely say "I think it's boring" which tells you nothing about the people. A man should not be judged by the courses he takes at university. (Except Media Studies)
I believe that was Glutamic Acid, writing "WELCOME BACK GREAL"
0
9 years ago
#1318
Yes, it was a slight at pure mathematicians. Glad you bit
That was Glut, cool
edit: yes the "exclaimation mark" was particularly badly done
0
9 years ago
#1319
Me and my friends conversation went like this
him: "who's grael?"
me: "Thats not an l, it's an exclamation mark"
him: "Oh right... who's grae?"
*facepalm*
0
#1320
(Original post by assmaster)
I get none of this, Durham hasn't even done set theory, which I reckon would upset you if you were here. Just Belgian-style Calculus, Analysis, Linear Algebra and Problems.
No wonder Durham has such a high rate of people getting firsts.
(Original post by Hopping Mad Kangaroo)
Next term in probability you will...
That isn't set theory. Yes set theory notation is used and all that stuff, but I doubt you will do stuff like cantors diagnonal argument in probability. Well, unless Durham has a totally different probability course to the one I'm currently suffering from. I can't wait intill next year and few months as then I wouldn't have to do a piece of stats in my life, I hope.
(Original post by Simba)
Yes, yes, yes, QM is made of awesomesauce. Hopefully I will be specialising in that sort of thing.
Well, QM is only interesting to me as I heard that it has some deep connections with number theory particulary the distribution of primes.
(Original post by My Alt)
I couldn't agree more. I can laugh at other Physicists for being rubbish at maths, tell pure mathematicians that they are boring and be able to back that opinion up because I've done (will have done) the courses!
(Original post by SimonM)
You can't back up your opinion because you've done the courses. You can merely say "I think it's boring" which tells you nothing about the people. A man should not be judged by the courses he takes at university. (Except Media Studies)
I don't know about that. If the person in question hasen't taken Maths or maybe Physics then you can conclude that the person in question is an idiot.
(Original post by My Alt)
Yes, it was a slight at pure mathematicians. Glad you bit
Looks like someone can't cut it at real maths.
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47.51% | 2,494 | 10,141 | {"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-2019-18 | latest | en | 0.972885 |
https://maker.pro/analog/tutorial/how-to-make-an-led-simulate-an-incandescent-bulb | 1,709,256,161,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00730.warc.gz | 380,237,964 | 144,877 | Analog
How to Make an LED Simulate an Incandescent Bulb
March 20, 2018 by Robin Mitchell
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Use analog circuitry to give the LED the feel of an incandescent bulb.
Hardware
1 680R Resistor (R10) 1 10K Resistor (R7) 1 47K Resistor (R5) 6 100K Resistor (R2, R3, R4, R6, R8, R9) 1 220K Resistor (R2) 1 100nF Capacitor (C2) 1 10uF Capacitor (C1) 1 LED (D1) 1 2N3904 (Q1) 1 LM358 (U1, U2)
LED lighting has many benefits over traditional incandescent bulbs, including energy efficiency and power output. However, LEDs turn on instantly, and for those who want to retro-fy their LEDs, this circuit will make an LED slowly turn on and slowly turn off, all in analog circuitry!
How It Works
This circuit is composed of three main sub-circuits: a triangular oscillator, a PWM generator, and a capacitor charge/discharge circuit. The first sub-circuit, the triangular oscillator, is composed of op-amps U1A and U1B with U1B being configured as an integrator while U1A is configured as a Schmitt trigger. The output of the integrator is connected to the input of the Schmitt trigger, which results in four conditions:
• If the output voltage of the integrator goes beyond the upper threshold, the Schmitt triggers output switches low
• If the output voltage of the integrator goes beyond the lower threshold, the Schmitt triggers output switches high
• If the output of the Schmitt trigger is high, the output voltage of the integrator increases steadily
• If the output of the Schmitt trigger is low, the output voltage of the integrator decreases steadily
The result of these four conditions is a continuous triangular waveform on the output of U1B. Incidentally, this oscillator also produces a square wave, which can be found on the output of U1A. This circuit is especially useful for those who want a square wave which is in phase with a triangular wave. But why do we have an oscillator for this LED driving circuit? The answer lies in PWM, pulse width modulation.
When incandescent bulbs turn on and off, they do not do so instantly. Instead, they take time to turn on and off, which is something that modern LED lights do not do. Therefore, to simulate this effect with an LED, we will need a circuit to control the brightness efficiently. One of the most common ways to do this with an LED is with the use of a PWM generator. In our circuit, we take the triangular waveform produced by U1A and U1B and feed this into a comparator U2B. The PWM output produced by U2B will be related to the negative input pin such that a higher voltage on this pin will result in a PWM waveform with a lower duty cycle (i.e., more off than on). The smaller the voltage on the negative input pin, the higher the duty cycle will be (i.e., more on than off). Since the LED is connected to a transistor switch (Q1), which is controlled by U2B, a large voltage on the negative pin will make the LED dimmer, and a small voltage will make the LED brighter.
Now that we have an LED whose brightness can be controlled by a voltage, we need to generate a voltage signal that closely resembles old incandescent bulbs. For this circuit, we use a simple RC circuit, which takes approximately 2 seconds to fully charge/discharge. Therefore, when the input to this circuit (ON/OFF) is connected to GND, the LED begins to turn on, and when the input is connected to VCC, the LED gradually turns off.
Construction
This project can be built using many different circuit construction techniques including a breadboard, stripboard, matrix board, and even a PCB. Originally, I designed a PCB for this project, but this board failed for unknown reasons. The goal was to turn this circuit into a small module that could be fitted to convert LED lights, and thus all surface mount parts were used. I successfully made the PCB itself, but when turned on, one of the LM358s produced the all magic smoke! Therefore, a breadboard version is shown here instead.
Author
Robin Mitchell
Graduated from the University Of Warwick in Electronics with a BEng 2:1 and currently runs MitchElectronics. | 967 | 4,063 | {"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-2024-10 | latest | en | 0.892851 |
https://handmadebase.com/how-beautiful-to-hang-photos-on-the-wall-schemes-and-50-photos/ | 1,590,965,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00405.warc.gz | 384,056,999 | 14,112 | How beautiful to hang photos on the wall - schemes and 50 photos
# How beautiful to hang photos on the wall - schemes and 50 photos
Photos help us to keep importantmemories of the events that took place earlier: family holidays, weddings, celebrations, meetings with friends. Previously, they were invested in albums, now they are mostly on electronic media. But why hide them - photos can be hung on the wall - this is a simple and economical decor for the walls! But how nice it would be to see the most importantphotos not once a year, for example, but every day! They may be colored or black and white, glossy or matte, perhaps even with the effect of antiquity, but all are so familiar and pleasing to the eye and heart. In this article we will examine in detail the schemes and ways of how to hang pictures on the wall, and also how to do it correctly. The content of the article:
## Hanging photo
Let us examine in detail all the schemes and methods for arranging photos on the wall. Horizontal line These can be photos of the same size, or from small ones gradually increase (or vice versa). Mark a line and line up the bottom edge of each photo. By the inclined line It looks like the previous method, only here you drawan inclined straight line and the angle of each photograph should touch it approximately. TIP: Take a photo of the same size, or large at the bottom, and reduce to the top. Rectangle or square
• The simplest version of this method is to select all photos of the same size and arrange them at the same distance from each other, then the rectangle will turn out by itself.
• But you can also outline the outline of a rectangle orsquare and select photos so that the edges of the side, top and bottom photos clearly coincide with the lines. Thus, you seem to add a puzzle from photos of different sizes.
• Symmetrical arc This method will look good if the arcwill take place, for example, around a mirror or a clock. Or it can be obtained when attaching a photo to a thread (read on in the methods of attachment). Chaotically ATTENTION: Even hanging photos randomly, keep in mind that they should look harmoniously together. Think about what photos should be next. We form a picture. For example, we put a heart, a flower, a cloudlet out of photos.
## Ways to post photos
We will analyze all the ways how to fix the photo onwall Hang in frames on the wall This is a classic decoration, while the frames can be wooden, metal or colored plastic. Also, each of them can be individually decorated with such various decorative elements as beads, buttons, colored paper, cloth, ribbons, pebbles, rhinestones - everything that comes to mind is important to make the jewelry fit and fit the picture. Arrange within the framework of shelves Everything that concerns the framework from the previous paragraph remains, only here we will not hang them on the walls, but arrange them on shelves, racks, bedside tables and tables. Large photo frame If you have a large and beautiful frame from a mirror or a picture, you can stretch the threads inside and place the photo. Also, this baguette frame can be ordered in any workshop. Fastened onto the wall with double sided tape Simplea way, and less expensive in terms of money than the previous ones, because you don’t need to spend money on a frame, just print out a photo and hang it. Yes, and the time will come faster, holes do not need to drill. Hanging threads on the wall, and attaching a photo to them with clothespins. Two nails were nailed at different ends of the wall, a string was pulled between them, and you hung out the photos, holding clothespins.
• As soon as the photos start to bother, quickly change them to others, here it will not make any problems: you do not need to rearrange the framework, nor unstick the photo from the wall, spoiling the wallpaper.
• You can hang several ropes at different levels, thus decorating the entire wall.
TIP: The rope can be tightened more or less by controlling sagging and placing the photo either on the same straight line or arc. Hang on wooden sticks
• In the photo in the upper corners we make a hole and thread them in a thread, then we take a thin wooden stick in size approximately as the width of the photo.
• And we tie these strings to the edges.
• Now we take another thread, we tie it along the edges of our home-made frame-sticks and hang all our construction on the stud for this thread.
## Where to hang
You can hang a photo best in placesmost attracting attention: for example around the TV, above the sofa, above the bed. Also, photos can fill and beautifully decorate any empty wall. On the whole wall If you have a whole empty wall, then the photos in the best way will fill this space. Pick up photos, scheme and method of hanging and go! Above the sofa Most often, there is an empty space above the sofa, table or nightstand: the shelf is likely to interfere, the choice between the picture or the photos remains.
• If you stopped on photos, then further actions are similar to the previous paragraph, only the places will be a little smaller.
• Although, even on an empty wall, they often hang not from the floor, but just at least a meter from the floor, so everything is the same.
Around the TV Usually the wall on which the TV weighs is empty. But you can beautifully arrange her photo. In the bedroom and above the desk Photos above the desk or at the head of the bed will save you from the problem of buying any decoration. Around the mirror, pictures, windows The mirror will beserve as the center of your exposure, and form a photo collage around you. Not necessarily the mirror should be in the shape of the frame. Perfectly considered round mirror and square frames. On a blank wall in the corridor B 3 or 2room apartments of the old layout is often found a blank wall at the end of a narrow corridor. Usually it is absolutely not functional. But if you hang a photo on it and make the backlight - it will become much more comfortable! In an empty corner, the corners of the walls are often empty, since protruding objects are easy to hook into when moving. But the photos are flat and a collage of them can decorate this part of the apartment in an original way.
• Along the steps or on the wall under the steps (if a private house or a two-story apartment)
• You go up, slowly, up the steps, you look at the photos and you don’t feel any fatigue. Alternatively, you can hang out from bottom to top of the photo as children grow up.
## How to hang
In order to not be disappointed with the result after fixing the photos, you need to initially plan how they will hang, and only then hang them. Let's conditionally divide the planning process into three stages: STEP # 1: Mark up
• First, make a sketch of the hanging scheme on the leaf, then make the appropriate marks on the wall.
• Note that the spotlight will be at eye level, an average of 160 cm,
• Do not reach to the end of the wall, leave centimeters 20.
• STEP №2: Make a template
• Cut out frame sizes from newspapers or leaves and attach them to the wall according to the chosen pattern.
• See how it will actually look, make adjustments if something is not pleasant.
• STEP # 3: Fastened to the wall Now, in place of each paper sketch, you can safely attach selected photos and enjoy the result. Depending on the method of hanging, you may have to drive in nails, in this case, stock up with the necessary tools. You can also attach tags to photos.with inscriptions or just on the wall to write about the image, for example: rest in the mountains, graduation and in the same spirit. Enjoy the memories and do not forget to create new unique life moments! | 1,665 | 7,687 | {"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-2020-24 | latest | en | 0.927585 |
https://www.classace.io/learn/math/1stgrade/equal-parts-halves-thirds-fourths | 1,679,881,714,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946584.94/warc/CC-MAIN-20230326235016-20230327025016-00371.warc.gz | 761,959,104 | 33,137 | Equal Parts: Halves, Thirds, and Fourths
## How to Write Halves, Thirds, and Fourths as Fractions
When you divide a sandwich into 2 equal parts, you get two halves.
Two halves make a whole.
One half is also written as 1/2. 👈 This is called a fraction
So if you eat half of your sandwich, you'll have 1/2 left! 😺
When you divide a pizza into 3 equal parts, you get three thirds.
Three thirds make a whole.
One third is also written as 1/3.
Say you eat a slice. You'd have two out of three slices left.
We can write "two out of three" as the fraction 2/3.
When you divide a sandwich into 4 equal parts, you get four fourths.
Four fourths make a whole.
One fourth is also written as 1/4.
Imagine you eat one fourth, or 1/4, of your sandwich.
You'd have 3/4 left!
Think you got it? Try the practice to see.
Complete the practice to earn 1 Create Credit
10 Create Credits is worth 1 cent in real AI compute time.
1 Create Credit is enough to make 1 image, or get 1 question answered.
Teachers:
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Duplicate | 294 | 1,102 | {"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-2023-14 | longest | en | 0.967232 |
http://mathhelpforum.com/advanced-algebra/16716-eigenvalues-eigen-vectors.html | 1,481,220,520,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00412-ip-10-31-129-80.ec2.internal.warc.gz | 173,217,756 | 12,933 | # Thread: eigenvalues and eigen vectors!
1. ## eigenvalues and eigen vectors!
helloo!
find the eigenvalues and eigen vectors of
A= |1 2|
|5 4|
i dont think its long to work out, but ive just started this topic and cannot work it out
2. Originally Posted by charlotte_usa
helloo!
find the eigenvalues and eigen vectors of
A= |1 2|
|5 4|
i dont think its long to work out, but ive just started this topic and cannot work it out
The eigen values are the roots of the equation (quadratic equation in this case):
$\det(A-\lambda I)=0$
Why is that?
RonL
3. The Eigenvalues are solutions to:
$\left| \begin{array}{cc}k-1&-2\\-5&k-4 \end{array} \right| = 0$
Thus,
$(k-1)(k-4) - 10 = 0$
Now solve.
4. A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.
$Ax=kx$
Where k is the eigenvalue.
They're tied together. Lambda is mostly used, but I will use k.
You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.
$\begin{bmatrix}1&2\\5&4\end{bmatrix}-\overbrace{(-1)}^{\text{eigenvalue}}\begin{bmatrix}1&0\\0&1\end {bmatrix}=\begin{bmatrix}2&2\\5&5\end{bmatrix}$
$\begin{bmatrix}1&2\\5&4\end{bmatrix}-6\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatr ix}-5&2\\5&-2\end{bmatrix}$
$rref\begin{bmatrix}2&2\\5&5\end{bmatrix}=\begin{bm atrix}1&1\\0&0\end{bmatrix}$
$rref\begin{bmatrix}-5&2\\5&-2\end{bmatrix}=\begin{bmatrix}1&\frac{-2}{5}\\0&0\end{bmatrix}$
In the first matrix, we have $x=-y$
Let y=t, then we have x=-t and y=t
In the second one, we have $x=\frac{2}{5}y$
Let y=t, then $x=\frac{2}{5}t$
Therefore, the eigenvectors are
$\begin{bmatrix}-1\\1\end{bmatrix}\cdot{t}$
and
$\begin{bmatrix}\frac{2}{5}\\1\end{bmatrix}\cdot{t}$
5. Originally Posted by galactus
A non zero vector x is called an eigenvector of A if Ax is a scalar multiple of x.
$Ax=kx$
Where k is the eigenvalue.
They're tied together. Lambda is mostly used, but I will use k.
You know your eigenvalues, -1 and 6. They're the roots of your characteristic polynomial.
$\begin{bmatrix}1&2\\5&4\end{bmatrix}-\overbrace{(-1)}^{\text{eigenvalue}}\begin{bmatrix}1&0\\0&1\end {bmatrix}=\begin{bmatrix}2&2\\5&5\end{bmatrix}$
$\begin{bmatrix}1&2\\5&4\end{bmatrix}-6\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatr ix}-5&2\\5&-2\end{bmatrix}$
$rref\begin{bmatrix}2&2\\5&5\end{bmatrix}=\begin{bm atrix}1&1\\0&0\end{bmatrix}$
$rref\begin{bmatrix}-5&2\\5&-2\end{bmatrix}=\begin{bmatrix}1&\frac{-2}{5}\\0&0\end{bmatrix}$
In the first matrix, we have $x=-y$
Let y=t, then we have x=-t and y=t
In the second one, we have $x=\frac{2}{5}y$
Let y=t, then $x=\frac{2}{5}t$
Therefore, the eigenvectors are
$\begin{bmatrix}-1\\1\end{bmatrix}\cdot{t}$
and
$\begin{bmatrix}\frac{2}{5}\\1\end{bmatrix}\cdot{t}$
i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...
6. Originally Posted by college_clare
i understand how to get upto here, but i dont see how you would get the eigenvectors because the teacher didnt explain the steps. i thought that for the 2255 matrix, it would be 2X1+2X2=0 so x1=-X2 so in the eigenvector, the top coordinate would be -1 and the same principle works for the bottom so it would be -1 again so the eigenvector would be (-1/-1) but i think this is wrong...
Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation $Ax = \lambda x$):
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (-1) \left [ \begin{array}{c} a \\ b \end{array} \right ]$
So we have the simultaneous equations
$a + 2b = -a$
and
$5a + 4b = -b$
Solving this for a and b gives a = a and b = -a. So the eigenvector is
$\left [ \begin{array}{c} a \\ -a \end{array} \right ] = a \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ]$
(This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)
We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (6) \left [ \begin{array}{c} a \\ b \end{array} \right ]$
I'll leave it to you to work out the rest.
-Dan
7. Originally Posted by topsquark
Let's try this in a different way. We know the eigenvalues are -1 and 6. So we may write the equation for the -1 eigenvector as (from the eigenvalue equation $Ax = \lambda x$):
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (-1) \left [ \begin{array}{c} a \\ b \end{array} \right ]$
So we have the simultaneous equations
$a + 2b = -a$
and
$5a + 4b = -b$
Solving this for a and b gives a = a and b = -a. So the eigenvector is
$\left [ \begin{array}{c} a \\ -a \end{array} \right ] = a \left [ \begin{array}{c} 1 \\ -1 \end{array} \right ]$
(This is just the negative of galactus' form for the eigenvector. Since the variable a is unspecified we may easily set t = -a and get my form from his.)
We may get an unambiguous eigenvector if we have some condition on the eigenvector, such as orthonormality or something. Since this wasn't mentioned I will leave the eigenvector like this. The 6 eigenvalue is similar:
$\left [ \begin{array}{cc} 1 & 2 \\ 5 & 4 \end{array} \right ] \cdot \left [ \begin{array}{c} a \\ b \end{array} \right ] = (6) \left [ \begin{array}{c} a \\ b \end{array} \right ]$
I'll leave it to you to work out the rest.
-Dan
i work out that 6b=15a so b=5/2a and a=a so then i get a final eigenvector of a (1 and 5/2)
but earlier it was suggested that the answer was 2/5 so have i gone wrong?
8. Originally Posted by charlotte_usa
i work out that 6b=15a so b=5/2a and a=a so then i get a final eigenvector of a (1 and 5/2)
but earlier it was suggested that the answer was 2/5 so have i gone wrong?
You are getting an eigenvector of
$a \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ]$
Again, this is simply a modification of galactus' answer. Let a = (2/5)t. Then
$a \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ] = \frac{2}{5} t \left [ \begin{array}{c} 1 \\ \frac{5}{2} \end{array} \right ] = t \left [ \begin{array}{c} \frac{2}{5} \\ 1 \end{array} \right ]$
which is the same as galactus' eigenvector. So the two are essentially the same.
-Dan | 2,401 | 6,903 | {"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": 37, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2016-50 | longest | en | 0.700818 |
https://www.calculator.net/horsepower-calculator.html?force=90&forceunit=newton&distance=20&distanceunit=mile&time=1&timeunit=hour&calctype=def&x=0&y=0 | 1,656,198,792,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00530.warc.gz | 739,516,895 | 4,965 | # Horsepower Calculator
Horsepower is a measurement of power, or the rate at which work is done. This calculator can be used to compute horsepower according to its definition and to convert between different power units. To find the engine horsepower of a vehicle, please use our Engine Horsepower Calculator.
## Horsepower Calculation Based on Definition
By definition Power = force×
distance time
## Result
The power is 804.672 watts.
It is equivalent to:
1.0791 mechanical horsepowers
1.094 metric horsepowers
1.0786 electrical horsepowers
0.08203 boiler horsepowers
Force: newton kilonewton pound-force kilogram-force Distance: meter kilometer mile yard Time: second minute hour day
## Horsepower Converter
Converts between different units of power.
Amount: Convert From: Convert To: Watt Kilowatt Mechanical Horsepower Metric Horsepower Electrical Horsepower Boiler Horsepower BTU/h Watt Kilowatt Mechanical Horsepower Metric Horsepower Electrical Horsepower Boiler Horsepower BTU/h
### Horsepower
Horsepower is a unit of measurement of power developed by engineer James Watt in the late 18th century. Although its original purpose was to compare the output of steam engines with the power of horses (hence its name), it has since been adopted as a unit of measurement for all sorts of engines used to power things such as vehicles, lawn mowers, boats, chainsaw, and airplanes.
Horsepower is commonly used today to refer to the potential work output of a vehicle's engine, and as such is a widely cited measurement for performance comparisons. Although horsepower is an important metric for doing so, it should not be the only consideration in determining a vehicle's ability as there are other factors such as power-to-weight ratio, torque, drivetrain, and forced induction. For more information or to do calculations on vehicle horsepower, use the Engine Horsepower Calculator.
When measuring horsepower output of anything, to ensure precise comparisons between them, it is important to be consistent in the methodology used to determine horsepower.
### Watt
Horsepower is not recognized in the International System of Units (SI); SI uses a measurement of power (a rate at which energy is generated or used) called a watt, which is named after James Watt. A watt is defined as one joule per second and is generally used to quantify a rate of energy transfer for lower rates of power consumption such as lightbulb or smartphone charging. On the other hand, kilowatts are commonly used for larger scale measures such as appliances and devices like refrigerators or servers. Energy is often presented using units of kilowatt hours (kWh), usually within the context of electric utilities delivering energy to household consumers.
### BTU
The British Thermal Unit (BTU) is another unit of energy, and one unit of BTU is the amount needed to heat one pound of water by one degree Fahrenheit. Coincidentally, one BTU is also roughly the amount of energy released by burning one match. BTU is commonly used to compare the energy inherent in different fuels. 1 BTU is equivalent to 0.293 watt-hour or 1,055 joules.
### Different Horsepowers
When the term horsepower is used, people are probably referring to mechanical horsepower. However, there are other lesser known definitions of the word.
• Mechanical Horsepower, hp(I)—550 foot-pounds per second, approximately 745.7 watts. The most common definition of horsepower, mechanical horsepower, is the horsepower James Watt invented in 1782.
• Metric Horsepower, hp(M)—75 kgf-m per second, approximately 735.499 watts
• Boiler Horsepower, hp(S)—34.5 pounds of water evaporated per hour at 212 degrees Fahrenheit, approximately 9,809.5 watts. This definition is mainly used to denote a boiler's capacity to deliver steam to a steam engine.
• Electrical Horsepower—746 watts. This definition is mainly used for electrical machines. | 818 | 3,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-27 | latest | en | 0.914625 |
https://forums.devx.com/printthread.php?s=1405a9f5f9b1bdd779f21a9b7f203bc1&t=90964&pp=15&page=1 | 1,611,368,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531702.36/warc/CC-MAIN-20210123001629-20210123031629-00657.warc.gz | 336,644,647 | 2,740 | # for, if,
• 03-20-2001, 08:05 PM
CoolMoDee
for, if,
My Problem is this....i run this program, i run it with a rate of 10 dollars
and hour with 22 hours of work for a total of \$220. Well i have an if statment
saying
if(total > 201 && total < 321)
it is supposed to go in to the 5% tax range, but it still goes into the 0%
tax range. I dont know why it does this.. please look @ the code and tell
me what is wrong..... Thanx A Lot CoolMoDee
#include <iostream>
int main()
{
int core, hours;
double rate, rate1, ot, x, total, totaltax;
cin>>rate;
cout<<"How many hours did you work?";
cin>>hours;
cout<<setiosflags(ios::fixed|ios::showpoint)<< setprecision(
2 );
if(hours >40)
{
x = hours - 40;
rate1 = rate * 40;
ot = x * (rate * 1.5);
total = rate1 + ot;
cout<<"You made " <<rate1 <<" at time and " <<ot
<<" at time and a half." <<endl;
cout<<"\n That is a total of " <<total <<" before
taxes\n";
}
if(hours <= 40)
{
rate1 = rate * hours;
cout<<"You made " <<rate1 <<" at time before taxes.";
}
double totaltaxf, totaltaxsevenfive, totaltaxten,
totaltax1threefive, totaltaxonesix;
if(total < 201)
{
cout<<"Taxes are 0% You get to
take home " <<rate1 <<" .\n" <<endl;
goto relog;
}
cout<<setiosflags(ios::fixed|ios::showpoint)<< setprecision(
2 );
if(total > 201 && total < 321)
{
totaltaxf = total - (total * .05);
cout<<"Taxes are 5% " <<totaltaxf <<" .\n"
<<endl;
}
return 0;
}
}
• 03-21-2001, 08:29 AM
Timo Peichl
Re: for, if,
if (hours<=40) you never set total;
CoolMoDee schrieb:
>
> My Problem is this....i run this program, i run it with a rate of 10 dollars
> and hour with 22 hours of work for a total of \$220. Well i have an if statment
> saying
>
> if(total > 201 && total < 321)
>
> it is supposed to go in to the 5% tax range, but it still goes into the 0%
> tax range. I dont know why it does this.. please look @ the code and tell
> me what is wrong..... Thanx A Lot CoolMoDee
>
> #include <iostream>
>
> int main()
> {
> int core, hours;
> double rate, rate1, ot, x, total, totaltax;
> cin>>rate;
> cout<<"How many hours did you work?";
> cin>>hours;
> cout<<setiosflags(ios::fixed|ios::showpoint)<< setprecision(
> 2 );
> if(hours >40)
> {
> x = hours - 40;
> rate1 = rate * 40;
> ot = x * (rate * 1.5);
> total = rate1 + ot;
> cout<<"You made " <<rate1 <<" at time and " <<ot
> <<" at time and a half." <<endl;
> cout<<"\n That is a total of " <<total <<" before
> taxes\n";
> }
> if(hours <= 40)
> {
> rate1 = rate * hours;
> cout<<"You made " <<rate1 <<" at time before taxes.";
>
> }
> double totaltaxf, totaltaxsevenfive, totaltaxten,
> totaltax1threefive, totaltaxonesix;
> if(total < 201)
> {
> cout<<"Taxes are 0% You get to
> take home " <<rate1 <<" .\n" <<endl;
> goto relog;
> }
> cout<<setiosflags(ios::fixed|ios::showpoint)<< setprecision(
> 2 );
> if(total > 201 && total < 321)
> {
> totaltaxf = total - (total * .05);
> cout<<"Taxes are 5% " <<totaltaxf <<" .\n"
> <<endl;
> }
> return 0;
> }
> } | 990 | 2,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-04 | latest | en | 0.78166 |
http://shpsoftware.com/standard-error/interpretation-of-standard-error-in-regression.php | 1,526,954,109,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00347.warc.gz | 276,202,488 | 7,389 | Home > Standard Error > Interpretation Of Standard Error In Regression
# Interpretation Of Standard Error In Regression
## Contents
In essence this is a measure of how badly wrong our estimators are likely to be. And that means that the statistic has little accuracy because it is not a good estimate of the population parameter. That is, should narrow confidence intervals for forecasts be considered as a sign of a "good fit?" The answer, alas, is: No, the best model does not necessarily yield the narrowest When running your regression, you are trying to discover whether the coefficients on your independent variables are really different from 0 (so the independent variables are having a genuine effect on Check This Out
The confidence interval so constructed provides an estimate of the interval in which the population parameter will fall. That's a good one! More commonly, the purpose of the survey is such that standard errors ARE appropriate. Coefficient of determination The great value of the coefficient of determination is that through use of the Pearson R statistic and the standard error of the estimate, the researcher can
## Standard Error Of Estimate Interpretation
Moreover, neither estimate is likely to quite match the true parameter value that we want to know. Given that the population mean may be zero, the researcher might conclude that the 10 patients who developed bedsores are outliers. The ANOVA table is also hidden by default in RegressIt output but can be displayed by clicking the "+" symbol next to its title.) As with the exceedance probabilities for the Suppose the sample size is 1,500 and the significance of the regression is 0.001.
• HyperStat Online.
• Another use of the value, 1.96 ± SEM is to determine whether the population parameter is zero.
• The central limit theorem suggests that this distribution is likely to be normal.
• Jim Name: Nicholas Azzopardi • Wednesday, July 2, 2014 Dear Mr.
• However, S must be <= 2.5 to produce a sufficiently narrow 95% prediction interval.
even if you have ‘population' data you can't assess the influence of wall color unless you take the randomness in student scores into account. In regression with a single independent variable, the coefficient tells you how much the dependent variable is expected to increase (if the coefficient is positive) or decrease (if the coefficient is A model for results comparison on two different biochemistry analyzers in laboratory accredited according to the ISO 15189 Application of biological variation – a review Što treba znati kada izračunavamo koeficijent Standard Error Of Prediction Here is an example of a plot of forecasts with confidence limits for means and forecasts produced by RegressIt for the regression model fitted to the natural log of cases of
Quant Concepts 194.502 προβολές 14:01 Statistics 101: Simple Linear Regression (Part 1), The Very Basics - Διάρκεια: 22:56. Jim Name: Jim Frost • Tuesday, July 8, 2014 Hi Himanshu, Thanks so much for your kind comments! I could not use this graph. http://people.duke.edu/~rnau/regnotes.htm The smaller the standard error, the closer the sample statistic is to the population parameter.
But there is still variability. The Standard Error Of The Estimate Is A Measure Of Quizlet Posted byAndrew on 25 October 2011, 9:50 am David Radwin asks a question which comes up fairly often in one form or another: How should one respond to requests for statistical However, if one or more of the independent variable had relatively extreme values at that point, the outlier may have a large influence on the estimates of the corresponding coefficients: e.g., However, the difference between the t and the standard normal is negligible if the number of degrees of freedom is more than about 30.
## Standard Error Of Regression Formula
Specifically, it is calculated using the following formula: Where Y is a score in the sample and Y’ is a predicted score. http://andrewgelman.com/2011/10/25/how-do-you-interpret-standard-errors-from-a-regression-fit-to-the-entire-population/ For example, you have all 50 states, but you might use the model to understand these states in a different year. Standard Error Of Estimate Interpretation An R of 0.30 means that the independent variable accounts for only 9% of the variance in the dependent variable. Standard Error Of Regression Coefficient A second generalization from the central limit theorem is that as n increases, the variability of sample means decreases (2).
Likewise, the residual SD is a measure of vertical dispersion after having accounted for the predicted values. http://shpsoftware.com/standard-error/interpretation-of-standard-error-in-regression-analysis.php For some statistics, however, the associated effect size statistic is not available. here For quick questions email [email protected] *No appts. So most likely what your professor is doing, is looking to see if the coefficient estimate is at least two standard errors away from 0 (or in other words looking to Linear Regression Standard Error
However, if the sample size is very large, for example, sample sizes greater than 1,000, then virtually any statistical result calculated on that sample will be statistically significant. Standard error: meaning and interpretation. Unlike R-squared, you can use the standard error of the regression to assess the precision of the predictions. this contact form Recall that the regression line is the line that minimizes the sum of squared deviations of prediction (also called the sum of squares error).
Does he have any other options?jrc on Should Jonah Lehrer be a junior Gladwell? Standard Error Of Estimate Calculator Since variances are the squares of standard deviations, this means: (Standard deviation of prediction)^2 = (Standard deviation of mean)^2 + (Standard error of regression)^2 Note that, whereas the standard error of Accessed September 10, 2007. 4.
## Does he have any other options?Chris G on Should Jonah Lehrer be a junior Gladwell?
In fact, if we did this over and over, continuing to sample and estimate forever, we would find that the relative frequency of the different estimate values followed a probability distribution. With a P value of 5% (or .05) there is only a 5% chance that results you are seeing would have come up in a random distribution, so you can say Most of these things can't be measured, and even if they could be, most won't be included in your analysis model. What Is A Good Standard Error share|improve this answer answered Dec 3 '14 at 20:11 whauser 1237 add a comment| up vote 2 down vote If you can divide the coefficient by its standard error in your
KeynesAcademy 136.894 προβολές 13:15 Interpreting Regression Coefficients in Linear Regression - Διάρκεια: 5:41. R-Squared and overall significance of the regression The R-squared of the regression is the fraction of the variation in your dependent variable that is accounted for (or predicted by) your independent When this happens, it is usually desirable to try removing one of them, usually the one whose coefficient has the higher P-value. navigate here For example, the regression model above might yield the additional information that "the 95% confidence interval for next period's sales is \$75.910M to \$90.932M." Does this mean that, based on all
Quant Concepts 4.156 προβολές 6:46 The Most Simple Introduction to Hypothesis Testing! - Statistics help - Διάρκεια: 10:58. Moreover, if I were to go away and repeat my sampling process, then even if I use the same \$x_i\$'s as the first sample, I won't obtain the same \$y_i\$'s - But since it is harder to pick the relationship out from the background noise, I am more likely than before to make big underestimates or big overestimates. It can allow the researcher to construct a confidence interval within which the true population correlation will fall.
The numerator is the sum of squared differences between the actual scores and the predicted scores. So, ditch hypothesis testing. That's what the standard error does for you. Lane DM.
If you are regressing the first difference of Y on the first difference of X, you are directly predicting changes in Y as a linear function of changes in X, without | 1,751 | 8,230 | {"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-2018-22 | longest | en | 0.876512 |
https://ca.answers.yahoo.com/question/index?qid=20201026021104AAAKVyg | 1,606,761,939,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00419.warc.gz | 225,756,105 | 18,795 | # A 220 g block on a 60.0 cm -long string swings in a circle on a horizontal, frictionless table at 85.0 rpm. What is the speed of the block?
Relevance
• oubaas
Lv 7
1 month ago
V = ω*r = 2PI/60*n*r = 0.10472*85*0.60 = 5.34 m/sec
• Ash
Lv 7
1 month ago
v = rω
v = (0.600 m)(85.0 rpm * 2𝜋 rad/rev * 1 min/ 60 s)
v = 5.34 m/s | 145 | 329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-50 | latest | en | 0.622249 |
https://codeforces.com/problemset/problem/959/E | 1,713,127,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00270.warc.gz | 158,111,886 | 14,332 | E. Mahmoud and Ehab and the xor-MST
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of n vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight (where is the bitwise-xor operation). Can you find the weight of the minimum spanning tree of that graph?
The weight of the minimum spanning tree is the sum of the weights on the edges included in it.
Input
The only line contains an integer n (2 ≤ n ≤ 1012), the number of vertices in the graph.
Output
The only line contains an integer x, the weight of the graph's minimum spanning tree.
Example
Input
4
Output
4
Note
In the first sample: The weight of the minimum spanning tree is 1+2+1=4. | 242 | 938 | {"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} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.877772 |
https://www.varsitytutors.com/gre_math-help/data-analysis?page=4 | 1,642,331,521,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00040.warc.gz | 1,146,391,715 | 51,284 | # GRE Math : Data Analysis
## Example Questions
### Example Question #31 : Data Analysis
If you throw a die three times, what is the probability of getting a 1 on at least one of the three throws?
5/6
25/36
1/3
125/216
91/216
91/216
Explanation:
Let's first find the probability of not getting a 1 on any of the three throws. Prob(no 1 on first throw) = 5/6. We need Prob(no 1 on first throw AND no 1 on second throw AND no 1 on third throw). "And" signifies multiplication in probability, so Prob(no 1 on all three throws) = 5/6 * 5/6 * 5/6 = 125/216. Now, to find the probability of getting a 1 on at least one throw, we simply take 1 – Prob(no 1 on all three throws) = 1 – 125/216 = 91/216.
### Example Question #21 : Outcomes
Every day is either rainy or sunny. Mondays are rainy with probability 3/5. Tuesdays are sunny with probability 1/4. Wednesdays are also sunny with probability 2/3. What is the probability that the weather is the same on Monday, Tuesday, and Wednesday?
1/8
13/60
1/15
15/19
3/20
13/60
Explanation:
We want either rainy, rainy, rainy or sunny, sunny, sunny.
P(rain on Mon AND rain on Tues AND rain on Wed) = 3/5 * 3/4 * 1/3 = 9/60
P(sun on Mon AND sun on Tues AND sun on Wed) = 2/5 * 1/4 * 2/3 = 4/60
P(same weather) = P(all 3 rainy days OR all 3 sunny days)
= P(all 3 rainy days) + P(all 3 sunny days)
= 9/60 + 4/60 = 13/60.
### Example Question #21 : Probability
16 dogs and 21 cats are up for adoption at the pound. 12 of these animals are brown and 19 are female. What is the probability of randomly selecting a brown animal?
19/37
Cannot be determined
12/37
16/19
12/19
12/37
Explanation:
The problem gives us extra information. We know 12 animals are brown, and there are 16 + 21 = 37 animals in total. Then the probability of choosing a brown animal is simply 12/37.
### Example Question #22 : Probability
What is the probability of pulling 2 spades out of a standard deck of cards without replacement?
Explanation:
There are 52 cards in a standard deck. 13 of them are spades.
So there is a chance of pulling out the first spade. That leaves 51 cards and 12 spades left. Now there is a chance to get a second spade.
When you combine probabilities, multiply the two individual probabilites together.
### Example Question #21 : Probability
What is the probability of fliiping a coin 4 times and getting 2 heads.
Explanation:
There are 4 flips of the coin. chance that the first is a heads, the second is a heads, the third is a tails, and the fourth is a tails. There are 6 permutations that give a combined 2 heads and 2 tails in whatever order.
### Example Question #21 : Probability
A bag contains 4 blue marbles, 2 black marbles, 1 clear marble, and 3 red marbles. What is the probability of pulling out 3 red marbles without replacement?
Explanation:
There are 10 marbles in the bag. So, the first probability of pulling out the first red marble is ; the next is ; the final is .
Multiply the individual probabilities together to find the total outcome.
### Example Question #31 : Data Analysis
A study was conducted to determine the effectiveness of a vaccine for the common cold (Rhinovirus sp.). 1000 patients were studied. Of those, 500 received the vaccine and 500 did not. The patients were then exposed to the Rhinovirus and the results were tabulated.
Table 1 shows the number of vaccinated and unvaccinated patients in each age group who caught the cold.
In order to determine the effectiveness of the vaccine, scientists want to show the impact of the vaccine on a patient's odds of catching a cold.
A patient who receives the vaccination is __________ to catch Rhinovirus than an unvaccinated patient.
80% more likely
just as likely
20% more likely
20% less likely
80% less likely
80% less likely
Explanation:
According to the table, 50 unvaccinated patients caught Rhinovirus.
18 + 4 + 5 + 4 + 19 = 50
250 unvaccinated patients caught Rhinovirus.
63 + 32 + 29 + 51 + 75 = 250
Next, we need to determine the percentage relationship between the number of unvaccinated versus vaccinated patients that caught the virus.
50/250 = 20%
100% – 20% = 80%
Because only 20% as many vaccinated patients exposed to the cold virus caught the cold, this reflects an 80% reduction in the likelihood of a patient to catch the cold after receiving the vaccine.
### Example Question #22 : Probability
A standard deck of cards contains 52 cards and 13 of each suit. Megan chooses two cards, without replacing. What is the probability that Megan chooses two diamonds?
Explanation:
For the first card, the probability of choosing a diamond is . With no replacing, the probability of choosing another diamond is . Multilying these probabilities yields .
This reduces to .
### Example Question #31 : Probability
You see someone correctly guess a card from a 52 card deck and you want to out-do him in an equally "odds-defying" feat. How many sequential coin flips would you have to predict to do something "less-likely" than guessing the correct card from a 52 card deck?
Explanation:
You are trying to have "worse" odds than a chance. 5 correct coin flip guesses would be a chance, not quite as unlikely.
6 correct guesses however would be a chance, thus "outdoing" the odds of the card guess.
### Example Question #33 : Data Analysis
A fair six-sided die is thrown.
Quantity A: The probability that the die will show either a "6" or an odd number.
Quantity B:
Quantity B is greater.
Quantity A is greater.
The relationship cannot be determined from the information given.
The two quantities are equal. | 1,438 | 5,625 | {"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} | 4.59375 | 5 | CC-MAIN-2022-05 | latest | en | 0.881842 |
https://www.jiskha.com/display.cgi?id=1242922833 | 1,503,571,570,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00453.warc.gz | 929,997,383 | 4,139 | # Financial
posted by .
I need help with defining Step-fixed costs and Shadow cost centers and an example of each one.
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More Similar Questions | 563 | 2,627 | {"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-2017-34 | latest | en | 0.95813 |
https://www.thinkadvisor.com/2011/06/01/chicken-or-egg-risk-tolerance-as-a-driver-of-finan/ | 1,571,797,859,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00268.warc.gz | 1,127,774,870 | 17,502 | This is the premiere blog posting on AdvisorOne by Geoff Davey, a pioneer in using psychometrics to assess clients’ risk profiles. A former advisor in his native Australia, Davey cofounded risk tolerance technology firm FinaMetrica in 1999; advisor users of his Web-based risk-profiling system are worldwide.
Many studies have shown that risk tolerance correlates positively with income and wealth. The correlations are not strong, usually around 0.3, but they seem to be universal.
There is a temptation to think that higher income and/or higher wealth lead to higher risk tolerance. However, there is always a danger in trying to read a cause and effect relationship into a correlation. To know for sure we would need to conduct a longitudinal study measuring risk tolerance, income and wealth as we went along.
Failing that, we can conduct a thought experiment. Suppose that Bill and Bob have different appetites for risk. Presented with a choice between taking a certain \$100 and a 50/50 gamble of winning \$0 or \$X, Bill will take the gamble when X is \$250 but Bob won't take the gamble until it reaches \$300. Looking at any single \$250 gamble choice, Bill has a 50% chance of being no worse off than Bill. However, if Bill and Bob are presented with a series of such choices, the longer the series runs the more certain it is that Bill will finish up better off than Bob. With a series of 10, Bill has an 83% chance of being no worse off than Bob and by the time we get to a series of 100 that chance has increased to 98%. Over 10 choices, Bill will finish with \$1,000 but Bob could expect to have \$1,250, though he may have nothing or \$2500.
Now suppose that Bill and Bob both started with a kitty of \$1,000 and that rather than the choices being framed from a base of \$100, they were framed from a base of 10% of the kitty at the time. For 10 choices, Bob’s kitty grows to \$2,593 but Bill’s grows to an expected average of \$3,260 and 62% of the time will be greater than \$2,590. At worst Bill will have \$1,000 and at best \$9,300.
Overall, by taking more risk Bill can expect to be significantly better off.
So how does this relate to real life? Clearly, life’s choices are rarely as simple as in our example and rather than
a series of identical choices we face a series of mainly different choices where there are usually more than two alternatives—and those alternatives will often include the possibility of losses. Further, the range of outcomes is often not clear and they must be estimated rather than calculated. Finally, we may make cognitive errors in assessing the situation and in identifying and evaluating the alternatives.
As we know from experience, risky choices take many forms and occur in different contexts including employment, borrowing, insurance and investment. For the riskier alternatives to be considered there would be a commensurately greater expected reward, but this will come with the possibility of an unfavorable outcome. The more risk tolerant amongst us will need less of an incentive to take the riskier alternatives. If we continue that pattern over time, all other things being equal, we should finish up better off.
So my hypothesis is that risk tolerance is a driver of financial success rather than the converse. | 705 | 3,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-43 | latest | en | 0.974822 |
https://converter.ninja/volume/metric-teaspoons-to-imperial-quarts/834-brteaspoon-to-imperialquart/ | 1,601,070,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00490.warc.gz | 332,495,018 | 5,322 | # 834 metric teaspoons in imperial quarts
## Conversion
834 metric teaspoons is equivalent to 3.66908706162878 imperial quarts.[1]
## Conversion formula How to convert 834 metric teaspoons to imperial quarts?
We know (by definition) that: $1\mathrm{brteaspoon}\approx 0.00439938496598176\mathrm{imperialquart}$
We can set up a proportion to solve for the number of imperial quarts.
$1 brteaspoon 834 brteaspoon ≈ 0.00439938496598176 imperialquart x imperialquart$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{imperialquart}\approx \frac{834\mathrm{brteaspoon}}{1\mathrm{brteaspoon}}*0.00439938496598176\mathrm{imperialquart}\to x\mathrm{imperialquart}\approx 3.6690870616287876\mathrm{imperialquart}$
Conclusion: $834 brteaspoon ≈ 3.6690870616287876 imperialquart$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 imperial quart is equal to 0.272547362110312 times 834 metric teaspoons.
It can also be expressed as: 834 metric teaspoons is equal to $\frac{1}{\mathrm{0.272547362110312}}$ imperial quarts.
## Approximation
An approximate numerical result would be: eight hundred and thirty-four metric teaspoons is about three point six seven imperial quarts, or alternatively, a imperial quart is about zero point two seven times eight hundred and thirty-four metric teaspoons.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 418 | 1,549 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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-2020-40 | latest | en | 0.702973 |
http://stackoverflow.com/questions/10899881/c-programming-bitwise-operators-and-knowing-when-to-utilize | 1,410,747,325,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657102753.15/warc/CC-MAIN-20140914011142-00011-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 260,953,128 | 17,769 | # C Programming - Bitwise operators and knowing when to utilize
I have read in the K&R II C Programming ANSI C book that the ">>" and "<<" operators control bits, and of course with me being a noob, I don't understand when to use them. I got interested in figuring out how to build packets manually and I came across the following snippet:
``````unsigned short csum(unsigned short *buf, int nwords)
{
unsigned long sum;
for(sum=0; nwords>0; nwords--)
sum += *buf++;
sum = (sum >> 16) + (sum &0xffff);
sum += (sum >> 16);
return (unsigned short)(~sum);
}
``````
I know that this calculates the checksum, but I don't understand what is going on here. XD
Obviously this is out of my skill range, but I figured I can use this snippet as a scapegoat to figure out some unanswered questions. When do you know when to use the bitwise operators to achieve a certain value, why not just add (+) or subtract (-)? Also, why is there a hexadecimal `&0xffff` there next to `sum`, if there are no operators with the two?
P.S. What does `~sum` mean?
-
That's not a question, that's a whole bunch. :)
1. You use bitwise operators when you want to view a number as a collection of bits, rather than an integer. It's much easier to say "I want this bit-pattern shifted two bits to the left" than to create the mathematically equivalent operation. They're conceptually different; if you think of the number as bits, using a bit-operator makes more sense.
2. The `& 0xffff` makes sure the value is 16 bits, by masking off all higher bits. This assumes the system's `unsigned long` is at least 16 bits wide, which is a pretty safe assumption. The `&` (bitwise `AND`) is often used for this purpose. Look at the truth table for logical conjunction and think "false is 0, true is 1" to see how this works.
3. The `&` before the hexadecimal constant is C's bitwise AND operator, which is used to do the masking I describe above. Basically, for single-bit variables `a & b`, the result is `1` if and only if both `a` and `b` are 1. The operator applies this logic to each pair of bits in its input terms.
4. The `~` operator is C's bitwise inversion, it "flips" the bits of its argument. It is commonly used to create masks.
-
Why does the hex have the `&` infront of it? If I were to remove it, what would that mean to the program? – user569322 Jun 5 '12 at 15:17
@Ken The author of the code just omitted the space between the operator `&` and the hex-literal. Whitespace is not always necessary to separate tokens from each other, this is such a case. If you removed it, it wouldn't compile anymore because there's no operator between the two operands. – Daniel Fischer Jun 5 '12 at 15:23
Everything you're talking about has to do with operations on the bit level. For example "var >> num" shifts the var to the right by num (meaning it devides the var by 2^num). Also the ~var inverts the var at bit level (eg. if var = 5 in bit notation= 101 ----> ~var = 010)
-
So, all of the 1's become 0's and vice versa? – user569322 Jun 5 '12 at 15:16
– Florin Stingaciu Jun 5 '12 at 15:17
When do you know when to use the bitwise operators to achieve a certain value
Use bitwise operators when you need to operate on individual bits of an object, and simple integer arithmetic would either not be sufficient, or would less clearly describe the intent of the code.
I know this sounds facile, but it really as simple as that.
why is there a hexadecimal `&0xffff` there next to `sum`
`&` is the bitwise-and operator. In this case, it's used to achieve a bitmask.
What does `~sum` mean?
`~` is the bitwise-inverse operation; it inverts the value of each bit.
I would hope that each of these operators is explained in whatever book you're using to learn C.
-
But why is arithmetic not used in this case? What does it mean to the program? – user569322 Jun 5 '12 at 15:38
@Ken: I don't know how to answer that. The nature of the calculation pretty much requires bitwise operations. – Oliver Charlesworth Jun 5 '12 at 15:54 | 1,029 | 4,008 | {"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-2014-41 | latest | en | 0.942942 |
https://www.hvacrschool.com/mistakes-in-measuring-ohms/ | 1,603,324,675,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878662.15/warc/CC-MAIN-20201021235030-20201022025030-00241.warc.gz | 754,846,473 | 42,457 | ## Mistakes in Measuring Ohms
In HVAC and electrical school, one of the first things you learn about electricity is Ohm’s law:
Volts = Amps x Ohms
Pretty simple Right? and Watt’s law is just as easy:
Watts = Volts x Amps
With this new found knowledge the student walks confidently into the real world with two equations and some elementary Algebra skills expecting to be able to predict Volts, Amps, Watts and Ohms using this secret knowledge.
Then they Ohm out a residential compressor or other single phase motor windings for the first time…
Start to Common = 4.2 Ohms
Run to Common = 2.6 Ohms
Start to Run = 6.9 Ohms
So they measure their voltage, grab a calculator and calculate.
Run Winding 243V ÷ 2.6Ω = 93.4A
Start Winding 243V ÷ 4.2Ω = 57.85A
Well, THAT makes no sense…. so they stop using Ohms law and settle with believing that electricity is magic and ohms law is broken.
In actuality, Ohms law does work it’s just that the loads we measure vary and don’t all behave in the same way.
In the case above, a compressor/fan motor/contactor coil etc… is an inductive load. That means that its job is to convert electrical energy to electromagnetic force. While an inductive load does have SOME resistance when de-energized that can be measured with an Ohmmeter, the majority of the electrical resistance only shows up once current is applied.
In an inductive (magnetic) load, this resistance that shows up once it’s powered on is called inductive reactance and is still measured in Ohms.
In the case of the example above, the run winding is connected directly between L1 & L2 and when the circuit is completed by the contactor in the example above the run winding WILL actually draw really high amps for a split second (first electrical cycle) until the inductive reactance kicks in.
In the start winding the current is limited by a combination of resistance, inductive reactance and the capacitance of the capacitor. Add in the fact that the applied Voltage across the run winding is actually higher than the L1 – L2 voltage due to back EMF (CEMF) and its enough to confuse anyone.
Then you throw in Power Factor to the mix in an inductive load… this means that even when we multiply Volts x Amps in an inductive AC circuit what we see in VA isn’t necessarily what you get in Watts.
It takes a lot of measurements and math to figure this all out and unless you are an engineer it’s much easier to measure what you have on a functioning component rather than attempting to calculate amperage and wattage based on voltage and ohms.
However…
We are taught that resistive loads (loads that create light and heat) are much more simple. There is none of this inductive reactance or power factor nonsense in a resistive load like a light bulb or a heat strip.
But wait… there’s more
So this light bulb is measuring 12.2 Ohms confirmed with two different meters and it is rated at 14.4 volts. So we do the simple math:
Amps = 14.4v ÷ 12.2Ω
Therefore Amps = 1.18
So we put it to the test by feeding this bulb exactly 14.4V from a DC power supply
Annnnddd… Not even close
We expected an amperage of 1.18 and got an amperage of 0.1
SO WHAT THE HECK IS GOING ON?
Incandescent light bulbs are a resistive load but they are also made with a filament of the element tungsten which has PTCR (positive temperature coefficient resistor) properties. This means that the resistance of a tungsten incandescent light bulb goes up by 10 to 15 times from its cold temperature to its hot temperature. In the case of the bulb above we now its cold resistance is 12.2 ohms, but by working ohms law backward we can also tell that its HOT resistance is 142.57 ohms which means that the hot resistance is 11.6 times higher than cold in this particular instance.
Not all resistive loads behave in this way though. Let’s look at a heat strip.
Amps = 240v ÷ 15.5Ω
Amps = 15.48
Watts = 240v x 15.48
Watts = 3715 based on Ohms & Watts law
Don’t worry… This is a 7.2kw (7200 Watt) heat strip divided into two 3600 watt strips and we are only reading one half (3.6kw)
To bench test it further I applied one-tenth of the designed voltage (24 Volts) to see how it would respond.
1.532a @ 24v = 15.32a @ 240v
The reason the math still doesn’t line up perfectly is that even in a heat strip the resistance increases as the temperature increases but in a much smaller fashion than in a light bulb.
When cold the math predicts it is a 3.7kw heater, when warm at 24v applied it predicts 3.67kw and at 240v the resistance increases to its full rating at 3.6kw.
All of this to say that Ohms & Watts law are useful and accurate but they are impacted by real “under load” forces in such a way that it much more realistic to make measurements on a functioning device and work backward than to use ohms to work upwards from an ohm bench test and a voltage.
— Bryan
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### Daily Tech Tip
Get the (near) daily Tech Tip email right in your inbox! | 1,260 | 5,050 | {"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-2020-45 | longest | en | 0.927437 |
www.withtheworks.online | 1,723,277,200,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00801.warc.gz | 835,602,908 | 173,099 | top of page
What is a Delta & Zeta Figs
​
DELTA and ZETA ARE the 4th and 6th letters of the Greek Alphabet.
​
Zeta is also a function, in number theory, an infinite series given by. where z and w are complex numbers and the real part of z is greater than zero.
The ZETA & DELTA Figures are the result of basic concept of four handicapping factors that can affect performance.
​
Bruno DeJulio has spent decades analyzing works and providing information and services to horseplayers at racingwithbruno.com, but dating way back to 1988 when Bruno learned how to make his own figures, building exclusive handicapping information has been his passion.
Bruno has been working on a different kind of fig lately. Early & LateFigures, called ZETA
The DELTA Figures are a combination figure that takes into account standard handicapping tools including of speed, and class — but all of that is controlled for the conditions of the actual race in question. Whereas traditional figures best describe the past performance, DELTA Figures incorporate that information into a forward-looking figure that predicts how an entry will perform based on previous form, under the conditions of a specific race. The lowest Delta Figures are the strongest.
The ZETA figure takes into account the best early pace figure from strictly the last two races on today's surface and coverts, in connection with the surface, the early pace Delta on the main track. The Turf Zeta figure is a last fraction derived from strictly the last two races on the turf and makes the number, as in the final time Delta fig and Zeta Early Pace fig, the lower the better.
​
A Zeta Delta Fig looks like this on the main:
ZETA/DELTA
23 - 45
The first fig is the Zeta fig and on the main track, the lower the better.
The lower the faster early the horse is, and in this case got a 23 Zeta Pace figure and a 45 total Delta.
A Zeta Delta Fig looks like this on the turf:
ZETA/DELTA
33 - 45
The first Fig on the turf is the late strength of the horses finish. Thus, the 33 is the closing fig. turf racing is all about finishing and the late closing kick is the key. The 45 is the overall Delta
​
DELTA Fig Key
DELTA followed by ‘d’ is computed off dirt race when no turf race available
DELTA followed by ’t’ is computed off turf race when no dirt race available
DELTA in parenthesis, '( )' is computed off race more than 90 days
These DELTA and ZETA figures are incorporated into all of Bruno’s products he provides at Racingwithbrunocom, WiththeWorks.online, Brunowiththeworks.com, Equibase and Brisnet.com.
​
The DELTA and ZETA Figures are a new generation of figures, which filtered through a lens of todays conditions becomes a strong, additional tool players may use in their race analysis, of which, we bet on ourselves.
f.
bottom of page | 667 | 2,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.938233 |
https://www.physicsforums.com/threads/non-fractional-negative-exponents.113577/ | 1,600,960,427,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00699.warc.gz | 983,201,986 | 16,533 | # (non)fractional Negative Exponents
We got thirty questions dealing with zero and negative exponents, but I'm having trouble with the negative exponent questions.
An example of this is:
a^3b-2/3^-1a^4b^-3
(-3x^-2y^3/3^-3x^4y^8)^-2
Sorry if these two examples look messy I don't know how to write it out like people write equations on this site.
Related Precalculus Mathematics Homework Help News on Phys.org
Hurkyl
Staff Emeritus
Gold Member
You could still use parentheses correctly. And spaces are terrific for readability. E.G. I think you meant to write:
a^3 b^-2 / (3^-1 a^4 b^-3)
for the first one.
Thanks for the tip Hurkyl. Do you have any ideas on how to solve them?
You should know (and remember) these relations when dealing with exponents:
$$x^{a}x^{b}=x^{a+b}$$ and $$\frac{x^{a}}{x^{b}}=x^{a-b}$$.
And, of course, $$x^{a}y^{b}\neq xy^{a+b}$$ and the same for the fractions.
Hint: If you click on the LaTeX images above you can see the code and learn how to use it in this forum.
Thanks for the tip Assyrian.
But if you have a solution to the above problems it would definitely give me a kick start to my homework.
Sorry, but solutions are not just given away. You have to show your work. What have you done so far on these problems?
Well I think I've actually figured out the first one, I'm working on the second one rigt now but i might have it in a minute. here's my work for the first one by the wya, tell me if I got it right:
$$\frac{a^{3}b^{-2}} {3^{-1}a^{4}b^{-3}}$$
$$=3a^{-1}b$$
$$=\frac{3b} {a}$$
if you cant see th LaTeX, I probobly put it in wrong, here's my work in plain text:
(a^3 b^-2) / (3^-1 a^4 b^-3)
= 3a^-1 b
=(3b) / a
That looks correct. And yes, I can see the LaTeX.
Thanks assyrian :)
Well I've looked at a few math help sites and a few tutorials, got some help from a teacher today, and didn't understand the lesson in calss yesterday, but I finally understand it :p
talk about dedication to the "arts" :)
Can you explain something to me though?
I have a question that's written like this $$x^{2}y^{-3}$$ but I have no clue as to what they're asking of me.
I don't know what the question is either. Do they want you to simplify $$x^{2}y^{-3}$$ or what?
I have no clue and they don't tell you what you're supposed to do on the worksheet, which really sucks.
Well I thank you for your assistance Assyrian, but I'm going to go watch Harry Potter and the Goblet of Fire now while I redue all of my homework.
Thanks again, bye.
Interesting....I'm soon to go and watch the same movie. | 729 | 2,552 | {"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} | 3.765625 | 4 | CC-MAIN-2020-40 | latest | en | 0.919786 |
https://astronomy.stackexchange.com/questions/48322/how-close-are-we-to-observing-all-of-the-sky-all-of-the-time | 1,718,766,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00712.warc.gz | 100,298,295 | 40,276 | # How close are we to observing all of the sky all of the time?
I was musing on how amateur astronomy is still a heavy contributor to the overall field, and how on any given night there will be plenty of small telescopes pointing into the void and capturing the sort of chance stuff the large telescopes never see because they are observing something specific elsewhere. And this made me wonder, with the combined set of professional and amateur telescopes, do we come close to seeing all of the sky all of the time?
Another way of phrasing the spirit of the question is: If some transient event occurs over the course of seconds, and is not visible to the naked eye, approximately how likely would we (meaning humans) be to notice it?
• I'd be surprised if the answer isn't "we're already doing it". The real question is how much detail we have of the sky all of the time, i.e. how luminous can a transient be and still be undetected. Commented Jan 27, 2022 at 9:56
• One might also ask what is "all the time"? How long can the time be between subsequent images to still count as "all the time"? A normal camera has a dead time between two subsequent images and you need two synced cameras with an offset of half their acquisition time to really have a continuous data feed. Commented Jan 27, 2022 at 10:10
• Surveyed depth and number of filters/passbands measured matter for the answer. NEO surveys e.g. ATLAS or facilities like ZTF already cover the visible sky each night but only to a relatively shallow depth or in a single filter (which normally prevents classification of static transients) Commented Jan 28, 2022 at 1:09
• all of the sky all of the time isn't possible, as far as I understand: we can never see what's on the other side of the sun from us until we move around in our orbit Commented Jan 28, 2022 at 9:49
• On the ameteur astronomy angle of the question - which the answers don't really tackle - the largest near earth asteroid discovered so far this year, 2022 CO6, has just been discovered by amateurs. That's right - the largest one, magnitude 25.6. That's larger than the Chelyabinsk asteroid which caused damage to1000s of buildings and associated injuries. In addition - they discovered it a 5 days before close approach - the most warning time any survey has given for a close approach this year so far. Well done MAP (Alain M aury, Georges A ttard and Daniel P arrott). Très bien! Commented Feb 15, 2022 at 8:42
To answer the first phrasing of the question: "Not yet". Approximate answer to the second phrasing below, after discussion. Adding to GrapefruitIsAwesome's answer, there are a couple of problems in addition to Daylight and Cloud Cover:
• Moonlight: Once every 28 days, the moon is full and makes the whole sky glow (in the same way that the sun does during the day, but to a much lesser extent). As this makes it impossible to see faint objects, all surveys shut down for 3-7 days for this period (depending on the survey and what they consider to be faint), leaving the sky more or less unmonitored. Let's say this means we only see around 85% of what we would otherwise see.
• Opposition Effect: For non luminous objects, the amount of light received from them by an Earth based optical telescope is heavily influenced by their solar elongation (angle between the Sun, the object and the observer). I.e. even if it is night and we are looking at the right part of the sky as an asteroid whizzes by, we get to see very little of the sunlight reflecting off it. This means there is in fact a very limited part of the sky where we manage to observe non-luminous objects. Because of this, over half of the asteroid discoveries we make are in a narrow cone facing directly away from the sun, which covers less than 4% of the sky. See Wikipedia. Let's say this means we only see around 8% of non luminous objects we would otherwise see.
In fact these issues are related. The moon is so bright when it is full because it is in opposition, and significantly dimmer the rest of the time. However, the Sun is much brighter than the moon, and never dims, so Daylight is definitely a bigger problem. It's easy to forget how big an issue this is.
Because we see the day/night cycle from the spinning earth it's easy to think "I only need to wait a few hours to see that bit of the sky". Unfortunately of course it's you that's moved. The daylight stays in roughly the same direction in the sky. To see that part of the universe you need to wait months, not hours for it to be night... and the bit of space between Earth and the sun, you never get to see with an ordinary telescope*. Let's say this means we only see around 50% of what we would otherwise see.
So in an attempt to give an approximate answer to "how likely would we be to notice it":
• For luminous objects: P(not full moon) 0.85 x P(not in daylight) 0.5 = ~43%
• For non luminous objects: P(not full moon) 0.85 x P(illuminated enough) 0.08 = ~4%
Note these approximate numbers ignore cloud cover, airglow and various other lesser factors, and additionally limited by the sensitivity of current telescopes.
* For this reason there's a space telescope being built (NEOSM) that will orbit at Sun-Earth L1, the point at which Earth and the Sun's gravity cancel out. It will look back towards Earth with the Sun behind it, so that we can at least see into the part of space between it's orbit and the Earth. For reference it's about 1% closer to the sun than we are, which may not sound like much, but that's still about 4 times further than the moon. That's the LaGrange point on the opposite side of Earth to the one that Webb is currently orbiting, L2. Both are around 1.5 million km from Earth
• worth noting that you can't solve this from Earth of course. Space telescopes have a much better view. Being in space means blue sky isn't in the way, but they still can't look too close to the sun or the instruments get fried. To see the whole sky the whole time, you really need telescopes orbiting the Sun... however the data backhaul then becomes a problem - the further the distance from Earth, the lower the bandwidth available to send back what they've seen. Tricky! Commented Jan 29, 2022 at 13:13
If we limit ourselves to defining "all of the time" to at least once per day while it is night to avoid the issue raised by Aaron F of only observing optically when the sun isn't in the sky (avoids day time and polar regions with 24 hours sunlight) there are couple that I was able to quickly identify:
1. Zwicky Transient Facility (ZTF) (Thank you astrosnapper)
The Zwicky Transient Facility (ZTF) is a public-private partnership aimed at a systematic study of the optical night sky. Using an extremely wide-field of view camera, ZTF scans the entire Northern sky every two days.
1. Asteroid Terrestrial-impact Last Alert System (ATLAS) (again astrosnapper)
ATLAS is an asteroid impact early warning system developed by the University of Hawaii and funded by NASA. It consists of four telescopes (Hawaii ×2, Chile, South Africa), which automatically scan the whole sky several times every night looking for moving objects.
1. The All-Sky Automated Survey for Supernovae (ASAS-SN)
We are changing that with our "All-Sky Automated Survey for Supernovae" (ASAS-SN) project, which is now automatically surveying the entire visible sky every night down to about 18th magnitude, more than 50,000 times deeper than human eye.
There are probably more, so I think we do cover all of the visible sky possible several times per day, depending on our definitions.
A couple of thoughts:
Daylight
From a given site you are limited to surveying optical only in darkness so you won't have coverage during the daytime. This is only worsened in polar regions during local summer (with a corresponding benefit in winter).
Cloud Cover
You can't survey when it is cloudy, but if you have multiple geographic sites one location may be clear while another is clouded out. This redundancy also helps with issue like failure and maintenance time.
• See also Kiso observatory. Scans the visible sky in 2 hours: en.wikipedia.org/wiki/… Commented Jan 29, 2022 at 11:50 | 1,896 | 8,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.957017 |
https://math.stackexchange.com/questions/1476978/can-a-set-be-neither-open-nor-closed/1477023 | 1,701,724,152,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00766.warc.gz | 430,214,159 | 38,866 | # Can a set be neither open nor closed?
Can a set be neither open nor closed? An example would do. I cant think of any.
• $[0,1)$ in the usual topology on $\Bbb R$. Oct 12, 2015 at 19:55
• @Brian M. Scott why not make that an answer? Oct 12, 2015 at 19:56
• Now I want to encourage the author to think of sets that are both open and closed at the same time. Oct 12, 2015 at 20:03
• @IvanNeretin I know those. $\Bbb R$ for example Oct 12, 2015 at 20:08
• Fun fact: a topology in which every set is either open, closed, or both is called a door space. en.wikipedia.org/wiki/Door_space Oct 12, 2015 at 20:21
One very familiar example is the set $[0,1)$ in the usual topology of $\Bbb R$: it’s not open, because it does not contain any nbhd of $0$, and it’s not closed, because $1$ is in its closure.
• Does that mean singletons are neither open nor closed, too? I apologize for following up an old question. Jun 6, 2020 at 9:25
• @Invisible I believe (in the usual topology of $\Bbb R$), singleton sets $\{r\}$ for some $r \in \Bbb R$ are equivalent to closed intervals $[r, r]$, which are closed but not open. Jul 14, 2020 at 5:06
• In particular $\{r\}$ is not open because it does not contain any nbhd of $r$, and it is closed because its complement $(-\infty,r) \cup (r,\infty)$ is open because it's a union of open sets. Jul 14, 2020 at 5:12
Another example: the rationals $\mathbb{Q}$ as a subset of $\mathbb{R}$ with the usual topology. It's not open, because every interval contains irrationals. It's not closed, because every irrational is a limit of rationals.
As the other answers have already pointed out, it is possible and in fact quite common for a topology to have subsets which are neither open nor closed. More interesting is the question of when it is not the case. A door topology is a topology satisfying exactly this condition: every subset is either open or closed (just like a door).
Conversely, we can ask whether subsets can be both open and closed, and this is the more well-known property of connectedness: a connected space is one where the only closed-and-open sets (clopen sets) are $\emptyset$ and $X$, which are always clopen in any topology. Thus in a connected door topology, you have $A$ is open iff $A$ is not closed, except for $A=X$ or $A=\emptyset$, where $A$ is both open and closed.
The most common door topology one comes across is the discrete topology, where every subset is both open and closed. A nontrivial example of a connected door topology is given by the collection of open sets $\scr U\cup\{\emptyset\}$ given any ultrafilter $\scr U$.
Any non-empty, proper subset of a topological space endowed with the indiscrete topology.
• Clarification: "trivial" here meaning indiscrete, not discrete - arguably both are trivial, although the former is somehow more trivial. Oct 12, 2015 at 21:24 | 775 | 2,847 | {"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} | 3.4375 | 3 | CC-MAIN-2023-50 | latest | en | 0.949796 |
http://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.1125.1/holomorphic/4.4.1125.1-145.1-d | 1,571,406,534,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00132.warc.gz | 266,193,429 | 5,468 | # Properties
Base field $$\Q(\zeta_{15})^+$$ Weight [2, 2, 2, 2] Level norm 145 Level $[145, 145, w^{2} - 6]$ Label 4.4.1125.1-145.1-d Dimension 1 CM no Base change no
# Related objects
## Base field $$\Q(\zeta_{15})^+$$
Generator $$w$$, with minimal polynomial $$x^{4} - x^{3} - 4x^{2} + 4x + 1$$; narrow class number $$2$$ and class number $$1$$.
## Form
Weight [2, 2, 2, 2] Level $[145, 145, w^{2} - 6]$ Label 4.4.1125.1-145.1-d Dimension 1 Is CM no Is base change no Parent newspace dimension 6
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q$.
Norm Prime Eigenvalue
5 $[5, 5, -w^{2} + 1]$ $\phantom{-}1$
9 $[9, 3, w^{3} + w^{2} - 4w - 3]$ $-4$
16 $[16, 2, 2]$ $\phantom{-}3$
29 $[29, 29, -w^{3} - w^{2} + 2w + 3]$ $\phantom{-}1$
29 $[29, 29, -w^{2} + w + 3]$ $\phantom{-}2$
29 $[29, 29, w^{3} - w^{2} - 4w + 2]$ $\phantom{-}2$
29 $[29, 29, 2w^{3} + w^{2} - 7w]$ $\phantom{-}2$
31 $[31, 31, -2w + 1]$ $-10$
31 $[31, 31, 2w^{2} - 5]$ $\phantom{-}4$
31 $[31, 31, 2w^{3} + 2w^{2} - 6w - 3]$ $\phantom{-}4$
31 $[31, 31, 2w^{3} - 8w + 1]$ $\phantom{-}4$
59 $[59, 59, w^{3} + w^{2} - 2w - 5]$ $-10$
59 $[59, 59, -w^{3} + 2w^{2} + 4w - 5]$ $\phantom{-}4$
59 $[59, 59, -3w^{3} + 10w - 4]$ $\phantom{-}4$
59 $[59, 59, -2w^{3} - w^{2} + 7w - 2]$ $\phantom{-}4$
61 $[61, 61, 4w^{3} + w^{2} - 13w - 1]$ $-8$
61 $[61, 61, 2w^{3} - w^{2} - 5w + 2]$ $\phantom{-}6$
61 $[61, 61, -3w^{3} - w^{2} + 8w]$ $\phantom{-}6$
61 $[61, 61, 3w^{3} - w^{2} - 10w + 5]$ $\phantom{-}6$
89 $[89, 89, w^{3} + w^{2} - w - 4]$ $\phantom{-}6$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
5 $[5, 5, -w^{2} + 1]$ $-1$
29 $[29, 29, -w^{3} - w^{2} + 2w + 3]$ $-1$ | 894 | 1,699 | {"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} | 2.8125 | 3 | CC-MAIN-2019-43 | latest | en | 0.274601 |
http://megasoft-rapid.com/West-Virginia/error-propagation-when-dividing-by-a-constant.html | 1,547,694,014,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00585.warc.gz | 148,328,537 | 7,846 | Moonlight Computing opens in the evening to provide Computer & iPhone repairs when all other shops are closed.
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# error propagation when dividing by a constant Liberty, West Virginia
These rules only apply when combining independent errors, that is, individual measurements whose errors have size and sign independent of each other. Bitte versuche es später erneut. Wird geladen... The indeterminate error equation may be obtained directly from the determinate error equation by simply choosing the "worst case," i.e., by taking the absolute value of every term.
This also holds for negative powers, i.e. Please try the request again. It can show which error sources dominate, and which are negligible, thereby saving time you might otherwise spend fussing with unimportant considerations. Multiplying by a Constant What would be your guess: can an American Corvette get away if chased by an Italian police Lamborghini?
The top speed of the Corvette
We'd have achieved the elusive "true" value! 3.11 EXERCISES (3.13) Derive an expression for the fractional and absolute error in an average of n measurements of a quantity Q when General function of multivariables For a function q which depends on variables x, y, and z, the uncertainty can be found by the square root of the squared sums of the If we knew the errors were indeterminate in nature, we'd add the fractional errors of numerator and denominator to get the worst case. So our answer for the maximum speed of the Corvette in km/h is: 299 km/h ± 3 km/h.
More precise values of g are available, tabulated for any location on earth. It should be derived (in algebraic form) even before the experiment is begun, as a guide to experimental strategy. The student may have no idea why the results were not as good as they ought to have been. If you are converting between unit systems, then you are probably multiplying your value by a constant.
Suppose n measurements are made of a quantity, Q. Under what conditions does this generate very large errors in the results? (3.4) Show by use of the rules that the maximum error in the average of several quantities is the What should we do with the error? The data quantities are written to show the errors explicitly: [3-1] A + ΔA and B + ΔB We allow the possibility that ΔA and ΔB may be either
If this error equation is derived from the indeterminate error rules, the error measures Δx, Δy, etc. The number "2" in the equation is not a measured quantity, so it is treated as error-free, or exact. The system returned: (22) Invalid argument The remote host or network may be down. A one half degree error in an angle of 90° would give an error of only 0.00004 in the sine. 3.8 INDEPENDENT INDETERMINATE ERRORS Experimental investigations usually require measurement of a
Die Bewertungsfunktion ist nach Ausleihen des Videos verfügbar. The coefficients will turn out to be positive also, so terms cannot offset each other. Your email Submit RELATED ARTICLES Simple Error Propagation Formulas for Simple Expressions Key Concepts in Human Biology and Physiology Chronic Pain and Individual Differences in Pain Perception Pain-Free and Hating It: All rights reserved.
the relative error in the square root of Q is one half the relative error in Q. The answer to this fairly common question depends on how the individual measurements are combined in the result. With errors explicitly included: R + ΔR = (A + ΔA)(B + ΔB) = AB + (ΔA)B + A(ΔB) + (ΔA)(ΔB) [3-3] or : ΔR = (ΔA)B + A(ΔB) + (ΔA)(ΔB) the relative determinate error in the square root of Q is one half the relative determinate error in Q. 3.3 PROPAGATION OF INDETERMINATE ERRORS.
It will be interesting to see how this additional uncertainty will affect the result! Then the error in any result R, calculated by any combination of mathematical operations from data values x, y, z, etc. When two quantities are multiplied, their relative determinate errors add. Constants If an expression contains a constant, B, such that q =Bx, then: You can see the the constant B only enters the equation in that it is used to determine
Then our data table is: Q ± fQ 1 1 Q ± fQ 2 2 .... Example: If an object is realeased from rest and is in free fall, and if you measure the velocity of this object at some point to be v = - 3.8+-0.3 A simple modification of these rules gives more realistic predictions of size of the errors in results. Indeterminate errors show up as a scatter in the independent measurements, particularly in the time measurement.
Schließen Ja, ich möchte sie behalten Rückgängig machen Schließen Dieses Video ist nicht verfügbar. It is the relative size of the terms of this equation which determines the relative importance of the error sources. When two numbers of different precision are combined (added or subtracted), the precision of the result is determined mainly by the less precise number (the one with the larger SE). It can suggest how the effects of error sources may be minimized by appropriate choice of the sizes of variables.
Please note that the rule is the same for addition and subtraction of quantities. Let Δx represent the error in x, Δy the error in y, etc. Do this for the indeterminate error rule and the determinate error rule. What is the error in the sine of this angle?
In summary, maximum indeterminate errors propagate according to the following rules: Addition and subtraction rule. This is easy: just multiply the error in X with the absolute value of the constant, and this will give you the error in R: If you compare this to the How precise is this half-life value? Multiplying this result by R gives 11.56 as the absolute error in R, so we write the result as R = 462 ± 12.
So, rounding this uncertainty up to 1.8 cm/s, the final answer should be 37.9 + 1.8 cm/s.As expected, adding the uncertainty to the length of the track gave a larger uncertainty You will sometimes encounter calculations with trig functions, logarithms, square roots, and other operations, for which these rules are not sufficient. Adding these gives the fractional error in R: 0.025. Example: We have measured a displacement of x = 5.1+-0.4 m during a time of t = 0.4+-0.1 s.
For example, to convert a length from meters to centimeters, you multiply by exactly 100, so a length of an exercise track that's measured as 150 ± 1 meters can also | 1,511 | 6,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-04 | latest | en | 0.900666 |
https://crypto.stackexchange.com/questions/30221/is-there-any-reason-not-to-use-single-key-em-with-aes-and-a-constant-key | 1,713,187,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00700.warc.gz | 170,499,616 | 41,315 | Is there any reason not to use Single-Key EM with AES and a constant key?
I've read recently the paper "Minimalism in Cryptography: The Even-Mansour Scheme Revisited" by Dunkelman, Keller and Shamir where it is claimed that the following construction (Single-Key EM) is secure:
$$C=K\oplus E(P\oplus K)$$
with $E$ being a publicly known fixed permutation.
Is there any reason not to use the above scheme (as block cipher) with AES-128 and key $0^{128}$ as $E$?
As for the reasons why somebody may want to use it:
• It can provide smaller implementations, enabling the hard-coding of the round constants
• It may be faster for the same reason
• It is more key-agile, allowing for much faster key switches (e.g. in a hashing / KFB scenario)
• I guess that you might rather want to refer to the algorithm as Single-Key EM in the title as well. XTS seems to have some additional properties that are not included in this scheme (and is treated as a mode of encryption rather than a block cipher). Nov 1, 2015 at 16:27
• @MaartenBodewes, fixed. Note that XTS $\neq$ XEX $\neq$ Single-Key EM as XTS uses two keys, XEX is XTS with one key and Single-Key EM is the here described scheme. Nov 1, 2015 at 17:09
This is considered in §6 of Bogdanov et al., who go on to devise an alternative 2-round AES-based Even-Mansour cipher—$\text{AES}^2$. The problem is, essentially, that 1-round Even-Mansour is only secure up to $2^{n/2}$ blockcipher queries, for an $n$-bit block. Specifically, a collision between $\text{SEM}_K(P) \oplus P$ and $E(P) \oplus P$ immediately reveals $K$ (cf. §4.2 of Dunkelman-Keller-Shamir).
In general, you want an Even-Mansour cipher targeting $n$-bit security to have block size $2n$. This is the case with, e.g., Salsa20 and ChaCha's core functions, which are Even-Mansour-like, target 256-bit security, and work on 512-bit blocks. Several permutation-based CAESAR candidates, such as Prøst or Minalpher, work similarly. | 534 | 1,946 | {"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} | 2.5625 | 3 | CC-MAIN-2024-18 | longest | en | 0.939165 |
https://www.physics.wisc.edu/~craigm/idl/mpfittut.html | 1,432,597,480,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928729.99/warc/CC-MAIN-20150521113208-00319-ip-10-180-206-219.ec2.internal.warc.gz | 909,327,489 | 7,141 | Markwardt IDL Page
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Tutorial: 1D Curve Fitting in IDL using MPFITFUN and MPFITEXPR
I have found curve-fitting in IDL to be somewhat of a frustrating process. There are a number of hoops you have to jump through that just make data analysis a pain. Furthermore, the IDL supplied curve fitting routine, called CURVEFIT, is not as robust as I would like. I have found that I can crash the entire IDL session with some fairly simple data and models. I have translated MINPACK-1, a very nice curve fitting package into IDL. The fitting programs are called MPFIT, MPFITFUN, and MPFITEXPR (and can be downloaded here). Here I present a short tutorial on how to use MPFITFUN and MPFITEXPR.
MPFITEXPR is generally the easiest to use interactively at the command line, while MPFITFUN is most commonly used in programs.
The first step in any analysis process is to collect your data. You are in charge of that, since only you know the specific details of how your experiment is run. In general, you will have three sets of numbers:
• The "X" values - these are the independent variables of the experiment.
• The "Y" values - these are the "measured" dependent variables.
• The "Error" values - this is typically the 1-sigma uncertainty in your measurement.
Of course you will have your own data, but I will provide some sample data (6 kb), in the form of an IDL "SAVE" file. It contains three variables, which you can imagine might represent the rate measured by a detector: t, r, and rerr, corresponding to a time, rate, and error in the rate. The error is simply the Poisson statistical error. My example below will use these variables.
Here is a plot of the data with errors:
### What to do if you don't have error bars
A proper experimenter should always assign error bars to their data. After all, a data point with larger errors should be weighted less in the fit, compared to a point with small error bars. However, I can foresee that under circumstances which only you can judge, the error bars may not be relevant. In that case, you may set the "Error" term to unity and proceed with the fit. Be aware that your data may not be properly weighted, and the error estimates produced by MPFITFUN/EXPR will not be correct. Bevington (Ch. 6.4) has an approach that allows you to assign error bars once you know the best-fit sum-of-squares. This number is returned through the BESTNORM keyword.
## Choosing a Model
By fitting a curve to your data, you are assuming that a particular model best represents the data. This again is up to you because of course, only you can assign an interpretation to your own data. Interpretations aside however, we can try to see how well a particular model fits by, well, just fitting it!
In this example, it is pretty clear that there is a fairly constant level of about 1000, with a "hump" near 2.5. I speculate that a constant plus Gaussian will fit that curve quite nicely. [ I in fact generated the data using that model. ]
How do you construct a model that IDL will understand? When you use MPFITEXPR, you only need to supply an IDL expression which computes the model. Here's how I would do the constant plus Gaussian model:
```IDL> expr = 'P[0] + GAUSS1(X, P[1:3])'
```
The variable expr now contains an IDL expression which takes a constant value "P[0]" and adds a Gaussian "GAUSS1(X, P[1:3])". The GAUSS1 function is a one dimensional Gaussian curve, whose source code can be downloaded.
There are a few important things to notice here. First, the name of the independent variable is always "X", no matter what it is called in your session. When MPFITEXPR executes your expression, it substitutes the correct independent variable for "X" in the expression. Second, all of the parameters are stored in a single array variable called "P". Again, you are free to name the parameter array anything you like in your own session, but in the expression it must appear as "P".
When you use MPFITFUN instead, you need to construct an IDL function which does the same thing as the expression above. You should deposit the following function definition into a text file called mygauss.pro:
```FUNCTION MYGAUSS, X, P
RETURN, P[0] + GAUSS1(X, P[1:3])
END
```
and compile with:
```IDL> .comp mygauss
```
You will need to decide for yourself how to arrange your parameter values. In my example, I decided that parameter 0, the first parameter, would be the constant value, while parameters 1 through 3 would be the parameters of the Gaussian (the three parameters to GAUSS1 are, in order: mean, sigma, and area under curve). If two parts of the expression require the same parameter value, then just type it in that way! This is a very elegant way to share parameter values between several different model components.
## Choosing Starting Values
You need to at least give MPFITFUN/EXPR a starting point in the parameter space. A rough guess is fine for most problems. I can enter my guess into the IDL session like this:
```IDL> start = [950.D, 2.5, 1., 1000.]
```
Those four numbers mean that the constant value will start at 950, and the Gaussian will start with a mean of 2.5, a sigma of 1, and an area of 1000. Since the data is double precision, I force the starting values to be double as well (or else MPFIT will complain). It is the fitting program's job to iterate until it finds the best solution it can.
Choosing the starting values can be somewhat of an art. For some particularly nasty problems with deep local minima, the proper choice of the starting parameters may mean the difference between converging to the global minimum or a local one. Again, only you can make this judgment.
## Fitting the Curve
Finally we can fit the curve using MPFITEXPR or MPFITFUN on the command line:
```IDL> result = MPFITEXPR(expr, t, r, rerr, start) or
IDL> result = MPFITFUN('MYGAUSS', t, r, rerr, start)
```
This will tell MPFITEXPR or MPFITFUN to fit the time/rate/error data using the model specified in the expression expr and starting at start. The routine will print diagnostic messages showing its progress, and finally it should converge to an answer. When it is done, we can print the results:
```IDL> print, result
997.61864 2.1550703 1.4488421 3040.2411
```
which means that the best-fit constant level is 997, the mean of the "hump" is 2.15 with a width of 1.45, and the area under the hump is 3040. That's all there is too it!
### Verifying the Fit
As a final step in the fitting process, we can make a plot of the data and overlay a fitted model:
```IDL> ploterr, t, r, rerr
IDL> oplot, t, result(0)+gauss1(t, result(1:3)), color=50, thick=5
```
In the oplot command above, I substituted the proper names for the independent variable and the parameter array. The color and thick keywords make the fitted curve stand out a little better. The results are excellent:
### Fixing Parameters
Now let's say that I have learned that the constant level should be fixed at 1000 exactly. I need to redo the analysis, and "freeze" the constant to 1000. One way to do that would be to rewrite the expression, and hard-code the value of 1000. Another more natural way to achieve the same thing is to "fix" the value to 1000 within the logic of MPFIT itself. All of the MPFIT functions understand a special keyword called PARINFO which allows you to do this.
You pass an array of structures through the PARINFO keyword, one structure for each parameter. The structure describes which parameters should be fixed, and also whether any constraints should be imposed on the parameter (such as lower or upper bounds). The structures must have a few required fields. You can do this by replicating a single one like this:
```IDL> pi = replicate({fixed:0, limited:[0,0], limits:[0.D,0.D]},4)
```
A total of four structures are made because there are four parameters. Once we have the blank template, then we can fill in any values we desire. For example, we want to fix the first parameter, the constant:
```IDL> pi(0).fixed = 1
IDL> start(0) = 1000
```
I have reset the starting value to 1000 (the desired value), and "fixed" that parameter by setting it to one. If fixed is zero for a particular parameter, then it is allowed to vary. Now we run the fit again, but pass pi to the fitter using the PARINFO keyword:
```IDL> result = MPFITEXPR(expr, t, r, rerr, start, PARINFO=pi)
IDL> result = MPFITFUN('MYGAUSS', t, r, rerr, start, PARINFO=pi)
```
You interpret the results the same way as before. It should be clear that the first parameter remained fixed at 1000 rather than varying to 997.
### Specifying Constraining Bounds
All of the fitting procedures here also allow you to impose lower and upper bounding constraints on any combination of the parameters you choose. This might be important, say, if you need to require a certain parameter to be positive, or be constrained between two fixed values. The technique again uses the PARINFO keyword. You see above that in addition to the fixed entry, there are some others, including limited and limits. They work in a similar fashion to fixed.
For example, let us say we know a priori that the Gaussian mean must be above a value of 2.3. I need to fill that information into the PARINFO structure like this:
```IDL> pi(1).limited(0) = 1
IDL> pi(1).limits(0) = 2.3
```
Here, for parameter number 1, I have set limited(0) equal to 1. The limited entry has two values corresponding to the lower and upper boundaries, respectively. If limited(0) is set to 1, then the lower boundary is activated. The boundary itself is found in limits(0), where I entered the value of 2.3. The same logic applies to the upper limits (which for each parameter are specified in limited(1) and limits(1)). You can have any combination of lower and upper limits for each parameter. Just make sure that you set both the limited and limits entries: one enables the bound, and the other gives the actual boundary value.
## Concluding Remarks
Well, those are the basics of fitting with MPFITEXPR and MPFITFUN. You will need some practice before you can feel comfortable, which is true for anything new. I have documented the usage of each function in the header of the program file. If you need to find more information about the techniques I used above, you may find it there. If you are concerned about error analysis, then you will want to check the parameters called PERROR, COVAR and BESTNORM, which return the 1-sigma parameter errors, covariance matrix and the best-fit chi squared value.
### References
Bevington, P. R. and Robinson, D. K. 1992, Data Reduction and Error Analysis for the Physical Sciences, 2nd Ed., McGraw-Hill, Inc.
### Files
Oct 02 2012 140 kb mpfit.pro REQUIRED Oct 02 2012 36 kb mpfitfun.pro Recommended Dec 17 2007 31 kb mpfitexpr.pro Aug 03 1998 6 kb fakedata.sav Sample Data Oct 13 2001 2 kb gauss1.pro Gaussian Model | 2,657 | 10,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2015-22 | longest | en | 0.93272 |
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# Pre-Cal 20S January 22, 2009
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Arithmetic series.
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### Pre-Cal 20S January 22, 2009
1. 1. The Story of Young Gauss ... really this time ... Photo Source: Karl Gauss (1777–1855)
2. 2. Series: The sum of numbers in a sequence to a particular term in a sequence. Example: denotes the sum of the first 5 terms. denotes the sum of the first n terms. Artithmetic Series: The sum of numbers in an arithmetic sequence given by is the sum to the nth term n is the quot;rankquot; of the nth term a is the first term in the sequence d is the common difference
3. 3. Find the sum of the first 6 terms of each series: 5 + 8 + 11 + ...
4. 4. Find the sum of the first 6 terms of each series: -6 - 8 - 10 - ...
5. 5. Find the sum of the first 60 terms of each series: 5 + 8 + 11 + ... -6 - 8 - 10 - ...
6. 6. Find the sum of the arithmetic series ... 5 + 9 + 13 + ... + 201 -8 - 5 - 2 + ... + 139
7. 7. Find the number of terms in each of the following arithmetic sequences. -2, -8, -14, ..., -206 x + 2, x + 9, x + 16, ... , x + 303
8. 8. Complete each arithmetic sequence by finding the missing arithmetic means. 1, ____, 25 14, ____, ____, 32
9. 9. Complete each arithmetic sequence by finding the missing arithmetic means. -1.5, ____, ____, ____, 4.5 -3, ____, ____, -60
10. 10. What is the domain and range of each relation? (f) y 4 2 x -3
11. 11. Identify all the intercepts for each relation? (f) y 4 2 x -3
12. 12. Complete each arithmetic sequence by finding the missing arithmetic means. 2, ____, ____, ____, ____, 107 m + 40, ____, ____, ____, m + 4 | 582 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-30 | latest | en | 0.829843 |
http://blog.zorilestore.ro/fb6x9bu/if-and-only-if-statements-examples-efa916 | 1,623,952,503,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487630518.38/warc/CC-MAIN-20210617162149-20210617192149-00392.warc.gz | 8,893,429 | 15,702 | if and only if statements examples
WELDER B½ G½ In case, if none of the conditions return true, then the code inside of El… The screenshot below shows the IF AND function in Excel: If you'd like to return some value in case the logical test evaluates to FALSE, supply that value in the value_if_false argument. I want the fifth row to automatically add the" x" to be the total of the "X" only if it equals to four or three x's in a column. =IF(AND(B7=0,B8>20000),Yes,IF(AND(B7>0,B8>200000),Yes,No)) In our example, it would be conceivable to give an even higher discount to goods that already have a reduced price if there are only a few of them left. 0-6=1 I'm unsure why the following formula is not working. An if statement can be followed by an optional else if...else statement, which is very useful to test various conditions using single if...else if statement. An "if and only if" statement is also called a necessary and sufficient condition. It is often used to conjoin two statements which are logically equivalent. When posting a question, please be very clear and concise. N Y Y GG Only one branch will trigger an action. The formula is missing "" back to back. Two and two makes 5. THE OUTPUT WILL BE 3323 6 66 86 64 20 89 68 42 16 31 Rate; IF(OR(J4"MOLCO",Y4"SHORE",W4=3,"P") Hello! 3323 6 66 86 64 20 89 68 42 16 31 If I have a triangle, then my polygon has only three sides. Proof: Part 1: P )Q. More formula examples can be found in Excel nested IF AND statements. To form a conditional statement, we could say “if P then Q.”. Amount (B2) should be greater than or equal to $100, Delivery date (D2) is within the current month. Let’s take example 5 and see how we can reverse the logic and use a ‘less than or equal to’ operator to construct the formula so that it still results in a ‘5% Discount’ for all customers whose total spend exceeds$50 and ‘No Discount’ for all the other customers. SP15. The goal is to use the correct formula depending on what I have in D3 (Call or Put). Example. In Excel 2003 and lower, no more than 30 arguments are allowed, with a total length not exceeding 1,024 characters. But there are many cases when we want to test multiple outcomes with IF statement. Y N N AA If column E = Eagle and column D = Platinum 5 or Platinum 7 and column P = 0-75 then column Q = .75% Example 6: Using ‘less than or equal to’ operator within the IF statement. Hello! If Statements. 5. is it because it two different types of functions? If column E = Eagle and column D = Select 6 and column P = 76-80 then column Q = 1.1% But how would i write 10 cells. please suggest how to solve it. In this case, the DAYS function is part of the test: The current data is compared with the specified date (in F2) and the difference in the number of days is displayed.IF checks whether the count is higher than 30. The statement or statements in the then-statement and the else-statement can be of any kind, including another if statement nested inside the original if statement. Biconditional. 6. In cell G, you can choose the conduit type via a drop down list. When we need to execute a block of statements only when a given condition is true then we use if statement. I was think the If/And/or formula would work but can't seem to get the correct combination of things. D9=SP3, In that scenario above im choosing anything but TSA in A9 ..and again, if D7,D8 and D9 contain any one or more of the following, in any combination: SP3, SP6, SP7, SP10, SP11, SP15 then this time give me the value in L21....(N/A could be a choice too but since I have that SP3 in there, then L21 value should still show up in L28), The combo to leave L28 at zero would be anything but TSA in A9 and any one or more of the following combinations in D7, D8 and D9: SP2 , LPWC, A9 = anything but TSA if expression, statements, end evaluates an expression, and executes a group of statements when the expression is true.An expression is true when its result is nonempty and contains only nonzero elements (logical or real numeric). ABCD int number = 5; if (number > 0) { cout << "The number is positive." i want count sunday of till date in excel, =COUNTIFS(L9:AO9,"sun")it give me 4 sunday, whole month of june-2020(date 01/06/2020 to 30/06/2020) Today, we are going to look at how you can use IF together with the AND function to evaluate two or more conditions at the same time. JAI HOMOEO STORE LUCKNOW 09ASEPS8853K1ZW 01/07/2019 T0000481 2037.00 Local 34011110 152.55 152.55 9.00 13.73 9.00 13.73 0.00 =IF(B27=7,B27=11,B27=14,B27<=16),4) iff if and only if Biconditional A way of writing two conditionals at once: both a conditional and its converse.. For example, the statement "A triangle is equilateral iff its angles all measure 60°" means both "If a triangle is equilateral then its angles all measure 60°" and "If all the angles of a triangle measure 60° then the triangle is equilateral". Another way to think of this sort of statement is as an equivalence between the statements A and B: whenever A … then just simply copy and paste it down the Column D, the moving ribbon on the bottom of the blog page is crap, cant read because of that stupid thing. Give an example of the source data and the expected result. Proof. If column E = Eagle and column D = Platinum 5 or Platinum 7 and column P = 76-80 then column Q = .50% More formula examples can be found in Excel nested IF AND statements. Even statements that do not at first look like they have this form conceal an implication at their heart. IF Brazil + Tennis then Column Y Example 1: Basic ‘If’ Test the company named Cyberspace) with an amount exceeding a certain number, say $100. so, for my homework im supposed to do this: I ALSO WANT TO USE THE VLOOKUP OR ANY OTHER FORMULA A THAT WILL DO THE SEARCH AND A MATCH. An if statement is one of a few ways to do something when a condition is set. Compose your response just once, save it as a template and reuse whenever you want. The following are examples of this kind of statement: Three other statements are related to any conditional statement. I WANT TO USE THE IF AND FORMULA TO CHECK IF I GET 2 NUMBER THAT ARE THE SAME THEN In English: If you are thirsty, take a drink. JAI HOMOEO STORE LUCKNOW 09ASEPS8853K1ZW 01/07/2019 T0000478 2904.00 Local 30039014 111.62 111.62 6.00 6.70 6.00 6.70 0.00 No No No No No Yes Yes No No No Yes Yes Yes No IN D2 put the following formula: =IF(AND(B2="yes",C2="yes"),"yes","no") The IF function accepts 3 bits of information: =IF(logical_test, [value_if_true], [value_if_false]) logical_test: This is the condition for the function to check. Trying to build an if and statement If in a column there are multiple Letters but each letter = a word. I recommend here and here. I'm stuck trying to get the following results basded on all possible combinations (456!) NATIONAL HOMEO STORE RAIBAREILY 09ANTPS5913L1Z2 01/07/2019 T0000477 8056.00 Local 29362940 218.88 218.88 9.00 19.70 9.00 19.70 0.00 0.00 989.00 =IF(AND(L11>0, L11=18.5, L11=25, L11=30, L11<100, "Obese")))))))). Here, if statement may be a single statement or a compound statement enclosed in curly braces (that is, a block). WELDER B½ I want to pick out the rates from the table based on Name and Effective Date. As an example, let's make a formula that checks if B2 is "delivered" and C2 is not empty, and depending on the results, does one of the following: =IF(AND(B2="delivered", C2<>""), "Closed", ""). NATIONAL HOMEO STORE RAIBAREILY 09ANTPS5913L1Z2 01/07/2019 T0000477 8056.00 Local 30039015 144.65 144.65 2.50 3.62 2.50 3.62 0.00 However, the second if statement in the example above, has an else built onto it. =IF($A3"",IF($E3"","Completed","In Progress"),IF($A3"",IF($E3="",IF(AND($I3"","Pending Approval","In Progress"),"Completed")),"")), =IF(AND(A3"",E3="",I3=""),"IN PROGRESS",IF(AND(A3"",E3"",I3=""),"COMPLETED",IF(AND(A3"",E3="",I3""),"PENDING APPROVAL",IF(AND(A3="",E3=""),"")))), i have to use three conditions in a cell. Unable to open Outlook window" error, Outlook Quick Parts and AutoText: how to create, edit and use, Merge data from duplicate rows based on a unique column, How to compare data in two Google sheets or columns. However, I want to add an extra piece =if(D5="Next Question", " ", " "). The IF function is the main logical function in Excel and is, therefore, the one to understand first. ; If the test expression is evaluated to false, statements inside the body of if are not executed. If A3 and E3 shows âblankâ show D3 as âBlankâ. In Excel, there are many ways to use If statements. How can I do this with excel formula? 5. The statements get executed only when the given condition is true . To demonstrate the approach, we will be calculating a bonus of 5% for "Closed" sales with the amount greater than or equal to $100. If I have a pet goat, then my homework will be eaten. =if(and(A1="Y",B1="N",C1="N"),"AA",if(and(A1="N",B1="N",C1="Y"),"BB",if(and(A1="Y",B1="N",C1="Y"),"CC",if(and(A1="N",B1="Y",C1="N"),"DD",if(and(A1="Y",B1="Y",C1="N"),"EE",if(and(A1="Y",B1="Y",C1="Y"),"FF",if(and(A1="N",B1="Y",C1="Y"),"GG",if(and(A1="N",B1="N",C1="N"),"HH","")))))))), hello sir, For use in Office 2010 uses cookies to provide you with several examples on how to get in case. ( value_if_false = 0 ) { cout < < endl ; } else { cout 0 ) then 103 (. This process is referred to as decision making in ' C. 3 variables for that cell and I come! % '' small '' value in B6 is red '' if and only if statements examples C has any date in it non-blank! If-Then statement: three other statements are true, its converse Summary has 4 drop downs, which will! After an if statement to produce a value for three different cells or error occurred 2 if... An example of the data values are identical BILL WWISE and PERCETAGEWISE Total KA formula JANNA HAI help! Decides which statement to produce a value for three different cells that true., contrapositive, and only if '' theorem, we defined multiple conditions in a value..., if formula ( D2 ) is greater than or equal to zero if n = and. Month by year the count of installations by classification back with findings but are. Your proof looking at the structure of the if and only if '' theorem, we prove! As nested if and only if all rich people are happy, then all of most. 744.46 7 pet goat, then the converse, inverse, and then see some examples of use! Statements gets executed only when the result you want to add Y & Z be! Not get it to work are evaluated in order, and tomorrow would still be Monday greater! A can be found in Excel if tutorials and still discover new every. Going to describe all if the polygon is a mammal '' and X is a,... Without errors or delays called the converse, contrapositive, and still new... With multiple possible locations use some help on this B ca n't seem to figure how... P then Q. ” 15-Jan-18 would return 8 % '' ( *. Some examples of propositions are-7 + 4 = 10 ; Apples are black Total if and only if statements examples... 2: Q ) P. Therefore, P, Q give you any advice, sorry a drink that! 30039014 111.62 111.62 6.00 6.70 6.00 6.70 6.00 6.70 0.00 2 let 's flag dates 10-Sep-2018... 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Math and science::Topology
# Subbasis
Let $$X$$ be a set. A subbasis for a topology on $$X$$ is a collection of subsets of $$X$$ whose union equals $$X$$.
Let $$\mathcal{S}$$ be a subbasis for a topology on $$X$$. The set of all unions of finite intersections of elements of $$\mathcal{S}$$ is a topolgy on $$X$$. This topology is said to be generated by the subbasis $$\mathcal{S}$$.
It can be proved that the generated set forms a valid topology.
### Basis vs subbasis
A subbasis is an even more lightweight description of a topology compared to a basis. A subbasis can generate a basis for the same topology: the collection of all finite intersections of subbasis elements forms a basis. | 180 | 710 | {"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} | 2.96875 | 3 | CC-MAIN-2023-14 | latest | en | 0.800613 |
https://www.geeksforgeeks.org/find-perimeter-shapes-formed-1s-binary-matrix/amp/?ref=rp | 1,601,252,438,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00247.warc.gz | 823,420,817 | 22,672 | Find perimeter of shapes formed with 1s in binary matrix
Given a matrix of N rows and M columns, consist of 0’s and 1’s. The task is to find the perimeter of subfigure consisting only 1’s in the matrix. Perimeter of single 1 is 4 as it can be covered from all 4 side. Perimeter of double 11 is 6.
```
| 1 | | 1 1 |
```
Examples:
```Input : mat[][] =
{
1, 0,
1, 1,
}
Output : 8
Cell (1,0) and (1,1) making a L shape whose perimeter is 8.
Input : mat[][] =
{
0, 1, 0, 0, 0,
1, 1, 1, 0, 0,
1, 0, 0, 0, 0
}
Output : 12
```
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
The idea is to traverse the matrix, find all ones and find their contribution in perimeter. The maximum contribution of a 1 is four if it is surrounded by all 0s. The contribution reduces by one with 1 around it.
Algorithm for solving this problem:
1. Traverse the whole matrix and find the cell having value equal to 1.
2. Calculate the number of closed side for that cell and add, 4 – number of closed side to the total perimeter.
Below is the implementation of this approach:
`// C++ program to find perimeter of area coverede by ` `// 1 in 2D matrix consisits of 0's and 1's. ` `#include ` `using` `namespace` `std; ` `#define R 3 ` `#define C 5 ` ` ` `// Find the number of covered side for mat[i][j]. ` `int` `numofneighbour(``int` `mat[][C], ``int` `i, ``int` `j) ` `{ ` ` ``int` `count = 0; ` ` ` ` ``// UP ` ` ``if` `(i > 0 && mat[i - 1][j]) ` ` ``count++; ` ` ` ` ``// LEFT ` ` ``if` `(j > 0 && mat[i][j - 1]) ` ` ``count++; ` ` ` ` ``// DOWN ` ` ``if` `(i < R-1 && mat[i + 1][j]) ` ` ``count++; ` ` ` ` ``// RIGHT ` ` ``if` `(j < C-1 && mat[i][j + 1]) ` ` ``count++; ` ` ` ` ``return` `count; ` `} ` ` ` `// Returns sum of perimeter of shapes formed with 1s ` `int` `findperimeter(``int` `mat[R][C]) ` `{ ` ` ``int` `perimeter = 0; ` ` ` ` ``// Traversing the matrix and finding ones to ` ` ``// calculate their contribution. ` ` ``for` `(``int` `i = 0; i < R; i++) ` ` ``for` `(``int` `j = 0; j < C; j++) ` ` ``if` `(mat[i][j]) ` ` ``perimeter += (4 - numofneighbour(mat, i ,j)); ` ` ` ` ``return` `perimeter; ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[R][C] = ` ` ``{ ` ` ``0, 1, 0, 0, 0, ` ` ``1, 1, 1, 0, 0, ` ` ``1, 0, 0, 0, 0, ` ` ``}; ` ` ` ` ``cout << findperimeter(mat) << endl; ` ` ` ` ``return` `0; ` `} `
`// Java program to find perimeter of area ` `// coverede by 1 in 2D matrix consisits ` `// of 0's and 1's ` `class` `GFG { ` ` ` ` ``static` `final` `int` `R = ``3``; ` ` ``static` `final` `int` `C = ``5``; ` ` ` ` ``// Find the number of covered side ` ` ``// for mat[i][j]. ` ` ``static` `int` `numofneighbour(``int` `mat[][], ` ` ``int` `i, ``int` `j) ` ` ``{ ` ` ` ` ``int` `count = ``0``; ` ` ` ` ``// UP ` ` ``if` `(i > ``0` `&& mat[i - ``1``][j] == ``1``) ` ` ``count++; ` ` ` ` ``// LEFT ` ` ``if` `(j > ``0` `&& mat[i][j - ``1``] == ``1``) ` ` ``count++; ` ` ` ` ``// DOWN ` ` ``if` `(i < R - ``1` `&& mat[i + ``1``][j] == ``1``) ` ` ``count++; ` ` ` ` ``// RIGHT ` ` ``if` `(j < C - ``1` `&& mat[i][j + ``1``] == ``1``) ` ` ``count++; ` ` ` ` ``return` `count; ` ` ``} ` ` ` ` ``// Returns sum of perimeter of shapes ` ` ``// formed with 1s ` ` ``static` `int` `findperimeter(``int` `mat[][]) ` ` ``{ ` ` ` ` ``int` `perimeter = ``0``; ` ` ` ` ``// Traversing the matrix and ` ` ``// finding ones to calculate ` ` ``// their contribution. ` ` ``for` `(``int` `i = ``0``; i < R; i++) ` ` ``for` `(``int` `j = ``0``; j < C; j++) ` ` ``if` `(mat[i][j] == ``1``) ` ` ``perimeter += (``4` `- ` ` ``numofneighbour(mat, i, j)); ` ` ` ` ``return` `perimeter; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = {{``0``, ``1``, ``0``, ``0``, ``0``}, ` ` ``{``1``, ``1``, ``1``, ``0``, ``0``}, ` ` ``{``1``, ``0``, ``0``, ``0``, ``0``}}; ` ` ` ` ``System.out.println(findperimeter(mat)); ` ` ``} ` `} ` ` ` `// This code is contributed by Anant Agarwal. `
`# Python 3 program to find perimeter of area ` `# covered by 1 in 2D matrix consisits of 0's and 1's. ` ` ` `R ``=` `3` `C ``=` `5` ` ` `# Find the number of covered side for mat[i][j]. ` `def` `numofneighbour(mat, i, j): ` ` ` ` ``count ``=` `0``; ` ` ` ` ``# UP ` ` ``if` `(i > ``0` `and` `mat[i ``-` `1``][j]): ` ` ``count``+``=` `1``; ` ` ` ` ``# LEFT ` ` ``if` `(j > ``0` `and` `mat[i][j ``-` `1``]): ` ` ``count``+``=` `1``; ` ` ` ` ``# DOWN ` ` ``if` `(i < R``-``1` `and` `mat[i ``+` `1``][j]): ` ` ``count``+``=` `1` ` ` ` ``# RIGHT ` ` ``if` `(j < C``-``1` `and` `mat[i][j ``+` `1``]): ` ` ``count``+``=` `1``; ` ` ` ` ``return` `count; ` ` ` `# Returns sum of perimeter of shapes formed with 1s ` `def` `findperimeter(mat): ` ` ` ` ``perimeter ``=` `0``; ` ` ` ` ``# Traversing the matrix and finding ones to ` ` ``# calculate their contribution. ` ` ``for` `i ``in` `range``(``0``, R): ` ` ``for` `j ``in` `range``(``0``, C): ` ` ``if` `(mat[i][j]): ` ` ``perimeter ``+``=` `(``4` `-` `numofneighbour(mat, i, j)); ` ` ` ` ``return` `perimeter; ` ` ` `# Driver Code ` `mat ``=` `[ [``0``, ``1``, ``0``, ``0``, ``0``], ` ` ``[``1``, ``1``, ``1``, ``0``, ``0``], ` ` ``[``1``, ``0``, ``0``, ``0``, ``0``] ] ` ` ` `print``(findperimeter(mat), end``=``"\n"``); ` ` ` `# This code is contributed by Akanksha Rai `
`using` `System; ` ` ` `// C# program to find perimeter of area ` `// coverede by 1 in 2D matrix consisits ` `// of 0's and 1's ` `public` `class` `GFG ` `{ ` ` ` ` ``public` `const` `int` `R = 3; ` ` ``public` `const` `int` `C = 5; ` ` ` ` ``// Find the number of covered side ` ` ``// for mat[i][j]. ` ` ``public` `static` `int` `numofneighbour(``int``[][] mat, ``int` `i, ``int` `j) ` ` ``{ ` ` ` ` ``int` `count = 0; ` ` ` ` ``// UP ` ` ``if` `(i > 0 && mat[i - 1][j] == 1) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// LEFT ` ` ``if` `(j > 0 && mat[i][j - 1] == 1) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// DOWN ` ` ``if` `(i < R - 1 && mat[i + 1][j] == 1) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``// RIGHT ` ` ``if` `(j < C - 1 && mat[i][j + 1] == 1) ` ` ``{ ` ` ``count++; ` ` ``} ` ` ` ` ``return` `count; ` ` ``} ` ` ` ` ``// Returns sum of perimeter of shapes ` ` ``// formed with 1s ` ` ``public` `static` `int` `findperimeter(``int``[][] mat) ` ` ``{ ` ` ` ` ``int` `perimeter = 0; ` ` ` ` ``// Traversing the matrix and ` ` ``// finding ones to calculate ` ` ``// their contribution. ` ` ``for` `(``int` `i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(``int` `j = 0; j < C; j++) ` ` ``{ ` ` ``if` `(mat[i][j] == 1) ` ` ``{ ` ` ``perimeter += (4 - numofneighbour(mat, i, j)); ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``return` `perimeter; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main(``string``[] args) ` ` ``{ ` ` ``int``[][] mat = ``new` `int``[][] ` ` ``{ ` ` ``new` `int``[] {0, 1, 0, 0, 0}, ` ` ``new` `int``[] {1, 1, 1, 0, 0}, ` ` ``new` `int``[] {1, 0, 0, 0, 0} ` ` ``}; ` ` ` ` ``Console.WriteLine(findperimeter(mat)); ` ` ``} ` `} ` ` ` `// This code is contributed by Shrikant13 `
` 0 && (``\$mat``[``\$i` `- 1][``\$j``])) ` ` ``\$count``++; ` ` ` ` ``// LEFT ` ` ``if` `(``\$j` `> 0 && (``\$mat``[``\$i``][``\$j` `- 1])) ` ` ``\$count``++; ` ` ` ` ``// DOWN ` ` ``if` `((``\$i` `< ``\$R``-1 )&& (``\$mat``[``\$i` `+ 1][``\$j``])) ` ` ``\$count``++; ` ` ` ` ``// RIGHT ` ` ``if` `((``\$j` `< ``\$C``-1) && (``\$mat``[``\$i``][``\$j` `+ 1])) ` ` ``\$count``++; ` ` ` ` ``return` `\$count``; ` `} ` ` ` `// Returns sum of perimeter of shapes ` `// formed with 1s ` `function` `findperimeter(``\$mat``) ` `{ ` ` ``global` `\$R``; ` ` ``global` `\$C``; ` ` ``\$perimeter` `= 0; ` ` ` ` ``// Traversing the matrix and finding ones ` ` ``// to calculate their contribution. ` ` ``for` `(``\$i` `= 0; ``\$i` `< ``\$R``; ``\$i``++) ` ` ``for` `( ``\$j` `= 0; ``\$j` `< ``\$C``; ``\$j``++) ` ` ``if` `(``\$mat``[``\$i``][``\$j``]) ` ` ``\$perimeter` `+= (4 - ` ` ``numofneighbour(``\$mat``, ``\$i``, ``\$j``)); ` ` ` ` ``return` `\$perimeter``; ` `} ` ` ` `// Driver Code ` `\$mat` `= ``array``(``array``(0, 1, 0, 0, 0), ` ` ``array``(1, 1, 1, 0, 0), ` ` ``array``(1, 0, 0, 0, 0)); ` ` ` `echo` `findperimeter(``\$mat``), ``"\n"``; ` ` ` `// This code is contributed by Sach_Code ` `?> `
Output:
```12
```
Time Complexity : O(RC).
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## Example Questions
← Previous 1 3
### Example Question #1 : Reaction Calculations And Limiting Reagent
What products will be formed by the reaction between copper (II) fluoride and sodium sulfate?
CuSO4 and NaF
CuS and NaF
Cu2Na and F2SO4
Cu2SO4 and NaF
CuSO4 and NaF
Explanation:
Recognize that this is a double replacement reaction in which ion pairs will switch. After reaction, copper (II) will now be paired with sulfate, and sodium with flouride according to the (unbalanced) equation below. Remember that sulfate has a charge of -2, and flouride of -1.
### Example Question #2 : Reaction Calculations And Limiting Reagent
Which represents the correct balanced equation for the reaction between silver (I) nitrate and magnesium hydroxide?
Explanation:
Silver (I) nitrate is AgNO3. Recognizing this allows us to eliminate two answer choices, which incorrectly substitute nitrogen (N) for nitrate (NO3) or balance the molecular charges incorrectly. Of the two remaining choices, only one is balanced correctly.
### Example Question #3 : Reaction Calculations And Limiting Reagent
Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:
Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:
How many molecules of carbon monoxide are produced per liter of hydrogen, when using the natural gas method of production at STP?
Explanation:
Since the reaction takes place at STP, we can convert liters of hydrogen to moles using a constant.
Using this value, we can find the moles of carbon monoxide produced and convert to molecules by using Avogadro's number.
### Example Question #4 : Reaction Calculations And Limiting Reagent
Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:
Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:
Suppose equal masses of iron and methane were available, along with excess water. What is the difference in the amount of hydrogen molecules that can be produced?
Iron will produce about 1.3 times more hydrogen molecules
Iron will produce about eight times more hydrogen molecules
Methane will produce about eight times more hydrogen molecules
Both methods will produce equal amounts of hydrogen molecules
Methane will produce about eight times more hydrogen molecules
Explanation:
To determine which method will produce more hydrogen molecules, convert equal masses of each reactant into molecules of hydrogen gas. The conversion requires us to convert the initial compound to moles, multiply by the molar ratio from the reaction to find moles of hydrogen, and multiply by Avogadro's number to convert to molecules.
Dividing these solutions, we see that the methane method produces roughly eight times the amount of hydrogen molecules.
### Example Question #5 : Reaction Calculations And Limiting Reagent
Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:
Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:
Suppose of iron react with of steam to create iron oxide and hydrogen. What is the limiting reagent and how much of the excess reactant remains?
Iron is the limiting reagent and of steam will remain
Steam is the limiting reagent and of iron will remain
Iron is a catalyst, and will not be consumed
Iron is the limiting reagent and of steam will remain
Iron is the limiting reagent and of steam will remain
Explanation:
In order to identify an excess reactant, we use stoichiometry to convert atoms of iron to mass of steam.
We have of steam available, but only require to fully react the given iron. Iron is thus the limiting reagent, since it will be fully consumed first.
There will be of excess steam.
### Example Question #6 : Reaction Calculations And Limiting Reagent
How many milliliters of 0.5M NaOH are required to react completely with 10g of BaCO3?
400mL
100mL
200mL
50mL
200mL
Explanation:
First, write a balanced equation for the reaction.
Next, convert 10g BaCO3 to milliliters of NaOH.
### Example Question #7 : Reaction Calculations And Limiting Reagent
Which of the following produces a solution of potassium chloride?
Potassium chloride is not soluble in water
Put of solute into a container and bring the volume to one liter by adding water while stirring
Put of solute into a container and bring the volume to by adding water while stirring
Add of solute to of water
Add of solute to one liter of water
Put of solute into a container and bring the volume to by adding water while stirring
Explanation:
We will be looking for a solution that results in one mole of potassium chloride per ten liters of water.
We will need to find the molecular weight of potassium chloride.
In order to get the desired concentration, we will need to add one-tenth of this amount to one liter of water.
Our ratio, then is:
The only answer to follow this ratio is of potassium chloride in .
### Example Question #8 : Reaction Calculations And Limiting Reagent
If you begin with of aluminum chloride and unlimited silver nitrate, how many grams of silver chloride can be produced?
Explanation:
This is a stoichiometry question requiring us to convert between grams, moles, reactants, and products.
Use the periodic table to find the molar masses of the two compounds in question.
We can use the reaction formula to find the ratio of aluminum chloride to silver chloride. In this case, the ratio is 1:3.
Now we can set up a calculation to convert grams of aluminum chloride to grams of silver chloride, making sure that all units cancel appropriately.
### Example Question #1 : Reaction Calculations And Limiting Reagent
Hydrogen can be produced on a large scale by a method called the Bosch process, by which steam is passed over heated iron shavings. The reaction is shown below:
Alternatively, hydrogen can be produced by reacting steam with natural gas, according to the following equation:
Which hydrogen production method would be more efficient in areas where water is scarce?
Both methods would be equally efficient in the described environment
The natural gas method is more efficient because for every one mole of water consumed, three moles of hydrogen are produced
The iron method is more efficient because it produces four moles of hydrogen gas
The iron method is more efficient because it utilizes a common metal
The natural gas method is more efficient because for every one mole of water consumed, three moles of hydrogen are produced
Explanation:
The question asks us to consider water scarcity. Assuming our goal is to utilize minimal water to produce maximal hydrogen, the natural gas method is most efficient as is produces more hydrogen per mole of water consumed.
One mole of water produces three moles of hydrogen.
Four moles of water produce four moles of hydrogen.
### Example Question #1 : Reaction Calculations And Limiting Reagent
5.6g of manganese reacts with 650mL of 6.0M hydrochloric acid to form manganese (V) chloride and hydrogen gas. Along with the products, a large amount of heat is evolved.
What is the limiting reagent, and how much of the excess reagent will remain after the reaction?
There is no limiting reagent
HCl is limiting; 0.3g excess Mn
Mn is limiting; 565mL excess HCl
HCl is limiting; 0.15g excess Mn
Mn is limiting; 370mL excess HCl | 1,684 | 7,809 | {"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} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.902549 |
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Polya's Four-Step Problem-Solving Process 1. Understand the problem2. Devise a plan3. Carry out the plan4. Looking back 1. Understanding the problem Can you state the problem in your own words 2. Devise a plan Look for a pattern 3. Carrying out the plan Check each step of the plan as you proceed 4. Looking back check the results in the original problem Conjecture a statement throught to be true, but not proven Counterexample example that contradicts the conjecture, shows the conjecture false Arithmetic Sequence an= a1+ d(n-1) Geometric Sequence an = a1* r(n-1) Recursive Sequence Ex: a1=2, a2=3, an=3an-2-an-1, for natural #n>2 must have all 3 parts or will be wrong In logic, a statement is a sentence that is either T or F The negation of a statement is a statement w the opposite true value of the given statement Be careful w quantifiers: Universal: all, every, & no refers to each & every element in a set Existential: some, there exists at least one refers to one or more or passible all elements in a set Truth tables p^q (p and q) - if both are T then its TpVq (p or q) - if both are F then its F Truth Tables Conditional Statements:p --> q (if p then q)Converse: q --> p (if q then p)Inverse:~p --> ~q (if not p then not q)Contrapositive:~q --> ~p (if not q them not p)Biconditional:p <--> q (p iff q) *If 1st is T & 2nd is F then its F* Place Value assigns a value of a digit depending upon its placement in a numeral Definition of an if a is any # and n e N, then an= a*a*...*a Ex: 23= 2*2*2=8 Mayan Numeration System a0=1 a1=20a2=20*18=360a3=202*18=7200...etc Dozen: Base 12 gross = dozen dozen0, 1, 2, 3, 4, 5, 6, 7, 8, 9, T, E Sets P & Q are in one-to-one correspondence if elements of P and Q can be paired so that for each element of P there is exactly one element of Q, & for each element of Q there is exactly one element of P Fundamental Counting Principle If event M can occur in m ways, and after it has occurred, event N can occur in n ways, then event M followed by event N can occure in mn ways Two sets A & B are equivalent A~B iff there exists a 1-1 correspondence btwn the two sets. The cardinal # of a set A, n(A): indicates the # of elelments in set A A set is finite if its cardinal number is a whole # The complement of a set A, written Ac: is the set of all elements in the universal set U that are not in A The empty set is a subset of everyset. Why? for any set A, either {}c A, or {} c A. Suppose{}c A, then there is some element in the empty set that is not in A, but because {} has no elements, it cannot have an element that is not in A.therefore {}c A Inequalities are an application of set concepts "Less Than" using sets: If A and B are finite sets then n(A) is less than n(B), written n(A)n(B) or a>b, which is n(B) b, a-b is a unique c eW such that a=b+c The Number Line Model - adding & subtracting Start at zero facing the (+) directionAdd means stay facing same directionSubtact means turn around(+) # means go forward(-) # means go backwards Expanded Algorrithm: 125 345+ 79 19 add ones 130 add tens+400 add hundreds 549 Left to Right Algorithm 458+8321200 (400+800)80 (50+30)+ 10 (8+2)1200+ 90 (80+10)1290 Authordt1158 ID66397 Card SetMath 365 DescriptionMath 365 Exam Cards Updated2011-02-15T03:13:36Z Show Answers | 958 | 3,282 | {"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-2024-33 | latest | en | 0.817081 |
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### Wave Speed
Wave speed is given in meters per second (the number of waves that pass per second). Wavelength is measured in meters and frequency is measured …
### Force on a Spring
When a force acts on a spring it is stretched or compressed, its length will changes by an amount e from its original length. …
### Force, Mass and Acceleration
An object of constant mass accelerates in proportion to the force applied. Demo In this tutorial you will learn how to calculate the force applied to an object if you …
### Elastic Potential Energy
Stretching or squashing an object can transfer energy into its elastic potential energy store. What is Elastic potential energy? Stretching or squashing an object can transfer …
### Forces and Work Done
The unit for work done is the joule (J), or Newton meter (N-m). One joule is equal to the amount of work that is done when 1 N of force moves an …
### Momentum
All moving objects have momentum. Forces can cause changes in momentum. The total momentum in a collision or explosion is conserved and stays the …
### Wave Period
The period of a wave is the time it takes to complete one cycle. The unit for a wave period is seconds, and it is inversely proportional …
### The Motor Effect
A current-carrying wire or coil can exert a force on a permanent magnet. The force increases if the strength of the magnetic field and/or current increases. This is …
### Acceleration
Acceleration is how quickly the velocity is changing whether it be speed, direction or both. Acceleration is a measure of how quickly the velocity …
### Speed
Speed is how fast an object is moving in no given direction. Velocity is how fast an object is moving in a given direction. …
### Weight on the Moon
Weight is the force acting on an object due to gravity. The weight of an object is defined as the force of gravity on the object …
### Weight on Earth
Weight is the force acting on an object due to gravity. The weight of an object can be defined as the force of gravity on the …
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Chapter 13. Simple Linear Regression Analysis. Simple Linear Regression. 13.1The Simple Linear Regression Model and the Least Square Point Estimates 13.2Model Assumptions and the Standard Error 13.3Testing the Significance of Slope and y-Intercept 13.4Confidence and Prediction Intervals.
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Chapter 13
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## Chapter 13
Simple Linear Regression Analysis
### Simple Linear Regression
13.1The Simple Linear Regression Model and the Least Square Point Estimates
13.2Model Assumptions and the Standard Error
13.3Testing the Significance of Slope and y-Intercept
13.4Confidence and Prediction Intervals
### Simple Linear Regression Continued
13.5Simple Coefficients of Determination and Correlation
13.6Testing the Significance of the Population Correlation Coefficient (Optional)
13.7An F Test for the Model
13.8Residual Analysis (Optional)
13.9Some Shortcut Formulas (Optional)
### The Simple Linear Regression Model and the Least Squares Point Estimates
• The dependent (or response) variable is the variable we wish to understand or predict
• The independent (or predictor) variable is the variable we will use to understand or predict the dependent variable
• Regression analysis is a statistical technique that uses observed data to relate the dependent variable to one or more independent variables
### Objective of Regression Analysis
The objective of regression analysis is to build a regression model (or predictive equation) that can be used to describe, predict and control the dependent variable on the basis of the independent variable
### Example 13.1: Fuel ConsumptionCase #4
• The values of β0 and β1 determine the value of the mean weekly fuel consumption μy|x
• Because we do not know the true values of β0 and β1, we cannot actually calculate the mean weekly fuel consumptions
• We will learn how to estimate β0 and β1 in the next section
• For now, when we say that μy|x is related to x by a straight line, we mean the different mean weekly fuel consumptions and average hourly temperatures lie in a straight line
### Form of The Simple Linear Regression Model
• y = β0 + β1x + ε
• y = β0 + β1x + ε is the mean value of the dependent variable y when the value of the independent variable is x
• β0 is the y-intercept; the mean of y when x is 0
• β1 is the slope; the change in the mean of y per unit change in x
• εis an error term that describes the effect on y of all factors other than x
### Regression Terms
• β0 and β1 are called regression parameters
• β0 is the y-intercept and β1 is the slope
• We do not know the true values of these parameters
• So, we must use sample data to estimate them
• b0 is the estimate of β0 and b1 is the estimate of β1
### The Least Squares Estimates, andPoint Estimation and Prediction
• The true values of β0 and β1 are unknown
• Therefore, we must use observed data to compute statistics that estimate these parameters
• Will compute b0 to estimate β0 and b1 to estimate β1
### The Least Squares Point Estimates
• Estimation/prediction equationŷ = b0 + b1x
• Least squares point estimate of the slope β1
### The Least Squares Point EstimatesContinued
• Least squares point estimate of the y-intercept 0
### Example 13.3: Fuel Consumption Case #2
• From last slide,
• Σyi = 81.7
• Σxi = 351.8
• Σx2i = 16,874.76
• Σxiyi = 3,413.11
• Once we have these values, we no longer need the raw data
• Calculation of b0 and b1 uses these totals
### Example 13.3: Fuel Consumption Case #5
• Prediction (x = 40)
• ŷ = b0 + b1x = 15.84 + (-0.1279)(28)
• ŷ = 12.2588 MMcf of Gas
### Model Assumptions
• Mean of ZeroAt any given value of x, the population of potential error term values has a mean equal to zero
• Constant Variance AssumptionAt any given value of x, the population of potential error term values has a variance that does not depend on the value of x
• Normality AssumptionAt any given value of x, the population of potential error term values has a normal distribution
• Independence AssumptionAny one value of the error term ε is statistically independent of any other value of ε
### Mean Square Error
• This is the point estimate of the residual variance σ2
• SSE is from last slide
### Standard Error
• This is the point estimate of the residual standard deviation σ
• MSE is from last slide
### Testing the Significance of the Slope
• A regression model is not likely to be useful unless there is a significant relationship between x and y
• To test significance, we use the null hypothesis:H0: β1 = 0
• Versus the alternative hypothesis:Ha: β1 ≠ 0
### Testing the Significance of the Slope #2
If the regression assumptions hold, we can reject H0: 1 = 0 at the level of significance (probability of Type I error equal to ) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than
### Testing the Significance of the Slope #4
• Test Statistics
• 100(1-α)% Confidence Interval for β1[b1± t /2 Sb1]
• t, t/2 and p-values are based on n–2 degrees of freedom
### Example 13.6: Fuel ConsumptionCase
• The p-value for testing H0 versus Ha is twice the area to the right of |t|=7.33 with n-2=6 degrees of freedom
• In this case, the p-value is 0.0003
• We can reject H0 in favor of Ha at level of significance 0.05, 0.01, or 0.001
• We therefore have strong evidence that x is significantly related to y and that the regression model is significant
### A Confidence Interval for the Slope
• If the regression assumptions hold, a 100(1-) percent confidence interval for the true slope B1 is
• b1± t/2sb
• Here t is based on n - 2 degrees of freedom
### Example 13.7: Fuel ConsumptionCase
• An earlier printout tells us:
• b1 = -0.12792
• sb1 = 0.01746
• We have n-2=6 degrees of freedom
• That gives us a t-value of 2.447 for a 95 percent confidence interval
• [b1± t0.025 · sb1] = [-0.12792 ± 0.01746] = [-0.1706, -0.0852]
### Testing the Significance of the y-Intercept
If the regression assumptions hold, we can reject H0: 0 = 0 at the level of significance (probability of Type I error equal to ) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than
### Confidence and Prediction Intervals
• The point on the regression line corresponding to a particular value of x0 of the independent variable x is ŷ = b0 + b1x0
• It is unlikely that this value will equal the mean value of y when x equals x0
• Therefore, we need to place bounds on how far the predicted value might be from the actual value
• We can do this by calculating a confidence interval mean for the value of y and a prediction interval for an individual value of y
### Distance Value
• Both the confidence interval for the mean value of y and the prediction interval for an individual value of y employ a quantity called the distance value
• The distance value for a particular value x0 of x is
• The distance value is a measure of the distance between the value x0 of x and x
• Notice that the further x0 is from x, the larger the distance value
### A Confidence Interval for a Mean Value of y
• Assume that the regression assumption holds
• The formula for a 100(1-a) confidence interval for the mean value of y is as follows:
• This is based on n-2 degrees of freedom
### Example 13.9: Fuel ConsumptionCase
• From before:
• n = 8
• x0 = 40
• x = 43.98
• SSxx = 1,404.355
• The distance value is
### Example 13.9: Fuel ConsumptionCase Continued
• From before
• x0 = 40 is 10.72 MMcf
• t = 2.447
• s = 0.6542
• Distance value is 0.1363
• The confidence interval is
### A Prediction Interval for an IndividualValue of y
• Assume that the regression assumption holds
• The formula for a 100(1-) prediction interval for an individual value of y is as follows:
• This is based on n-2 degrees of freedom
### Example 13.9: Fuel ConsumptionCase
• From before
• x0 = 40 is 10.72 MMcf
• t = 2.447
• s = 0.6542
• Distance value is 0.1363
• The prediction interval is
### Which to Use?
• The prediction interval is useful if it is important to predict an individual value of the dependent variable
• A confidence interval is useful if it is important to estimate the mean value
• The prediction interval will always be wider than the confidence interval
### The Simple Coefficient of Determination and Correlation
• How useful is a particular regression model?
• One measure of usefulness is the simple coefficient of determination
• It is represented by the symbol r2
### Calculating The Simple Coefficient ofDetermination
• Total variation is given by the formula (yi-ȳ)2
• Explained variation is given by the formula (ŷi-ȳ)2
• Unexplained variation is given by the formula (yi-ŷ)2
• Total variation is the sum of explained and unexplained variation
• r2 is the ratio of explained variation to total variation
### What Does r2 Mean?
The coefficient of determination, r2, is the proportion of the total variation in the n observed values of the dependent variable that is explained by the simple linear regression model
### The Simple Correlation Coefficient
• The simple correlation coefficient measures the strength of the linear relationship between y and x and is denoted by r
• Where b1 is the slope of the least squares line
### Example 13.13: Fuel Consumption Case
The value of the simple correlation coefficient (r) is not the slope of the least square line
That value is estimated by b1
High correlation does not imply that a cause-and-effect relationship exists
It simply implies that x and y tend to move together in a linear fashion
Scientific theory is required to show a cause-and-effect relationship
### Testing the Significance of thePopulation Correlation Coefficient
• The simple correlation coefficient (r) measures the linear relationship between the observed values of x and y from the sample
• The population correlation coefficient (ρ) measures the linear relationship between all possible combinations of observed values of x and y
• r is an estimate of ρ
### Testing ρ
• We can test to see if the correlation is significant using the hypothesesH0: ρ = 0Ha: ρ ≠ 0
• The statistic is
• This test will give the same results as the test for significance on the slope coefficient b1
### An F Test for Model
• For simple regression, this is another way to test the null hypothesisH0: β1 = 0
• This is the only test we will use for multiple regression
• The F test tests the significance of the overall regression relationship between x and y
### Mechanics of the F Test
• To test H0: β1= 0 versus Ha: β1 0 at the a level of significance
• Test statistics based on F
• Reject H0 if F(model) > Fa or p-value < a
• Fa is based on 1 numerator and n-2 denominator degrees of freedom
### Example 13.15: Fuel Consumption Case
F-test at = 0.05 level of significance
Test Statistic
Reject H0 at level of significance, since
Fais based on 1 numerator and 6 denominator degrees of freedom
### Residual Analysis #1
• Checks of regression assumptions are performed by analyzing the regression residuals
• Residuals (e) are defined as the difference between the observed value of y and the predicted value of y, e = y - ŷ
• Note that e is the point estimate of ε
• If the regression assumptions are valid, the population of potential error terms will be normally distributed with a mean of zero and a variance σ2
• Furthermore, the different error terms will be statistically independent
### Residual Analysis #2
• The residuals should look like they have been randomly and independently selected from normally distributed populations having mean zero and variance σ2
• With any real data, assumptions will not hold exactly
• Mild departures do not affect our ability to make statistical inferences
• In checking assumptions, we are looking for pronounced departures from the assumptions
• So, only require residuals to approximately fit the description above
### Residual Plots
• Residuals versus independent variable
• Residuals versus predicted y’s
• Residuals in time order (if the response is a time series)
### Constant Variance Assumptions
• To check the validity of the constant variance assumption, examine residual plots against
• The x values
• The predicted y values
• Time (when data is time series)
• A pattern that fans out says the variance is increasing rather than staying constant
• A pattern that funnels in says the variance is decreasing rather than staying constant
• A pattern that is evenly spread within a band says the assumption has been met
### Assumption of Correct FunctionalForm
• If the relationship between x and y is something other than a linear one, the residual plot will often suggest a form more appropriate for the model
• For example, if there is a curved relationship between x and y, a plot of residuals will often show a curved relationship
### Normality Assumption
• If the normality assumption holds, a histogram or stem-and-leaf display of residuals should look bell-shaped and symmetric
• Another way to check is a normal plot of residuals
• Order residuals from smallest to largest
• Plot e(i) on vertical axis against z(i)
• Z(i) is the point on the horizontal axis under the z curve so the area under this curve to the left is (3i-1)/(3n+1)
• If the normality assumption holds, the plot should have a straight-line appearance
### Independence Assumption
• Independence assumption is most likely to be violated when the data are time-series data
• If the data is not time series, then it can be reordered without affecting the data
• Changing the order would change the interdependence of the data
• For time-series data, the time-ordered error terms can be autocorrelated
• Positive autocorrelation is when a positive error term in time period i tends to be followed by another positive value in i+k
• Negative autocorrelation is when a positive error term in time period i tends to be followed by a negative value in i+k
• Either one will cause a cyclical error term over time
where | 3,591 | 14,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-17 | longest | en | 0.775963 |
https://ventolaphotography.com/can-we-use-merge-sort-on-linked-list/ | 1,708,660,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00415.warc.gz | 619,979,341 | 15,143 | Table of Contents
## Can we use merge sort on linked list?
Merge sort is one of the most famous divide-and-conquer sorting algorithms. This algorithm can be used to sort values in any traversable data structure (i.e., a linked list).
## How do you sort a linked list by sorting merge?
Algorithm for Sorting a Linked List Using Merge Sort
1. Step 1: Dividing the Lists Into Two Smaller Sublists. We will keep on recursively dividing the lists into two smaller sublists until the size of each sublist becomes 1.
2. Step 2: Merging the Sublists to Produce a Sorted List.
How do you sort a singly linked list?
Algorithm
1. Define a node current which will point to head.
2. Define another node index which will point to node next to current.
3. Compare data of current and index node.
4. Current will point to current.
5. Continue this process until the entire list is sorted.
Which sorting algorithm is best for sorting a linked list?
Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.
### Why we use merge sort in linked list?
Why is Merge Sort preferred for Linked Lists?
• In the case of linked lists, the nodes may not be present at adjacent memory locations, therefore Merge Sort is used.
• Unlike arrays, in linked lists, we can insert items in the middle in O(1) extra space and O(1) time if we are given a reference/pointer to the previous node.
### How do you sort two linked lists?
Write a SortedMerge() function that takes two lists, each of which is sorted in increasing order, and merges the two together into one list which is in increasing order. SortedMerge() should return the new list.
How do you reverse a singly linked list using recursion?
The recursive approach to reverse a linked list is simple, just we have to divide the linked lists in two parts and i.e first node and the rest of the linked list, and then call the recursion for the other part by maintaining the connection.
Is merge sort or quick sort suitable for sorting linked list justify?
Merge sort is faster in these situations because it reads the items sequentially, typically making log2(N) passes over the data. There is much less I/O involved, and much less time spent following links in a linked list. Quicksort is fast when the data fits into memory and can be addressed directly.
## What is singly linked list?
A singly linked list is a type of linked list that is unidirectional, that is, it can be traversed in only one direction from head to the last node (tail). Each element in a linked list is called a node. A single node contains data and a pointer to the next node which helps in maintaining the structure of the list.
## What is merge sort algorithm?
Merge sort is a sorting algorithm based on the Divide and conquer strategy. It works by recursively dividing the array into two equal halves, then sort them and combine them. It takes a time of (n logn) in the worst case.
Can quick sort applied on singly linked list?
QuickSort can be implemented both on Arrays as well as on Linkedlists.
Why merge sort is suitable for linked list?
Why is Merge Sort preferred for Linked Lists? In the case of linked lists, the nodes may not be present at adjacent memory locations, therefore Merge Sort is used. | 724 | 3,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-10 | latest | en | 0.884372 |
http://www.internetdict.com/answers/what-is-the-length-of-a-decimeter.html | 1,516,157,946,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886794.24/warc/CC-MAIN-20180117023532-20180117043532-00159.warc.gz | 475,423,234 | 7,632 | # What is the Length of a Decimeter?
A decimeter, abbreviated dm, is equal to 10 centimeters or 100 millimeters, or 1/10 of a meter. It is a unit of measurement of the metric system. | 51 | 183 | {"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-2018-05 | latest | en | 0.887794 |
https://www.physicsforums.com/threads/npn-transistor.208004/ | 1,529,943,710,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868135.87/warc/CC-MAIN-20180625150537-20180625170537-00605.warc.gz | 873,917,235 | 14,557 | Homework Help: NPN Transistor
1. Jan 9, 2008
beesher
1. The problem statement, all variables and given/known data
an npn transistor is used to drive a constant current I through three ‘white’ LEDs, connected in series. If the required current value is I = 25 mA, and the current-gain of the transistor is β = 300, find the value of the resistor R, giving your answer to 3 significant figures. You may assume for the operating conditions specified that the base-emitter voltage of the transistor is 600mV, and that its saturation voltage is 150mV.
I should point out that the base of the transistor is connected to a voltage divider, with a resistor of 10k above a resistor R.
2. Relevant equations
Base Current = $$\frac{collector current}{gain}$$
3. The attempt at a solution
Hi guys looking for a bit of help on a past paper for my instrumentation class.
In this problem, I've tried to find the base current by dividing 25 by 300, and I'm a bit stumped as to the next step. Do i simply need to to divide the base current by the saturation voltage, or is that completely wrong?
2. Jan 9, 2008
blochwave
It's been a while since working with transistor circuits, and I think I completely forget what saturation currents and voltages were, whoo
But anyways there should be some voltage going into that voltage divider, so you can figure out Vin(voltage at base) and the base current via that, and base current times gain is collector current. Since the base current depends on R, you can figure out R. There may be more nuances in there but I think that's the idea
3. Jan 9, 2008
beesher
Circuit..
Here's a circuit diagram if that can give anyone a better idea of the circuit in question.
File size:
84.9 KB
Views:
154
4. Jan 9, 2008
Staff: Mentor
You set the 25mA by setting the emitter voltage, because that determines the current through the emitter resistor. Ve is Vb - 600mV, so that tells you what you need for Vb. But in setting Vb, you need to account for some base current that bypasses the base resistor R.
BTW, the base resistor R and the emitter resistor "68R" cannot possibly be referring to the same "R". Does the diagram mean that the emitter resistor is 68 Ohms?
Also, keep in mind that the current through the emitter resistor is really the LED current plus the base current. Granted, 1/300 is a small number, but to be accurate (and receive full points for the answer), you need to keep track of that current.
That should be enough info to get you going again. Post up your calcs when you're done, and we'll take a look.
5. Jan 9, 2008
beesher
Cheers
Ok, thanks very much I'll get it finished tomorrow. Oh and yes the 68R is simply referring to Ohms.
Edit; Still can't figure it out :(
Last edited: Jan 10, 2008 | 687 | 2,757 | {"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} | 3.5625 | 4 | CC-MAIN-2018-26 | latest | en | 0.933607 |
https://groups.google.com/g/alt.fan.douglas-adams/c/595nPukE-Jo/m/koaAJ3tPBtEJ?hl=en | 1,686,105,653,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00226.warc.gz | 323,564,333 | 134,530 | # Why 42 ?
16049 views
### Mark J Cherkas
Nov 1, 1993, 7:50:35 PM11/1/93
to
I am new to this group so bear with this beginners question:
Why is the answer 42 ?
Has Douglas Adams ever explained this ?
### Kampen van F
Nov 2, 1993, 6:50:41 AM11/2/93
to
NO, HELP, NOT AGAIN THIS NEVER ENDING DISCUSSION!!!
PLEASE NO! (Spare you precious bandwith...)
In binary it appears to be 101010 (VERY CONSPICUOUS)
Als there is some fooling arount with 2b OR (NOT 2b) (possibly with
some additional parantheses, I did not bother to check this).
(To be or not to be),
Sigh, it's very very very complicated.
YOU DON"T want to KNOW
Greetings
Florentijn van Kampen
--
+--------------------------------------------------------------------+
| Florentijn van Kampen | "O dear", says God, "I hadn't |
| fvka...@cs.vu.nl | thought of that", and promptly |
| (...) The World is Music (...)| vanishes in a puff of logic" |
### Robert Svebeck
Nov 2, 1993, 3:52:26 AM11/2/93
to
This is one-of-the-many explenations:
Douglas Adams The Hitchhikers Guide to the Galaxy
||||||| ||||| ||| ||||||||||| ||||| || ||| ||||||
7 +5 +3 +11 +5 +2 +3 +6 = 42 !
______..________________________
| (~ ) / /
| ___)( / Robert Svebeck /
|__(_____)__/_________________/
Nov 3, 1993, 2:51:46 AM11/3/93
to
In Article <2b4asr\$b...@syzygy.socs.uts.edu.au>, mjch...@socs.uts.EDU.AU
The answer to this is very simple. It was a joke. It had to be a number, an
ordinary, smallish number, and I chose that one. Binary representations,
base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
stared into the garden and thought '42 will do' I typed it out. End of story.
Best,
Currently in Santa Fe, NM | ada...@nic.cerf.net (current)
### Jack Death
Nov 2, 1993, 11:53:42 AM11/2/93
to
>
>The answer to this is very simple. It was a joke. It had to be a number, an
>ordinary, smallish number, and I chose that one. Binary representations,
>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>stared into the garden and thought '42 will do' I typed it out. End of story.
>
>Best,
>
>Currently in Santa Fe, NM | ada...@nic.cerf.net (current)
Hmmmm... I kinda figured it was something like that, but then
again if the REAL answer to life the universe and everything WERE
42, it would be a common number to _guess_ or _Pick_, wouldn't it?
I think that whether you, Mr. Adams, picked it intentionaly or not,
you hit the right answer anyways. (Besides isn't it Fun to see just
how many poeple are looking for this number to show up everywhere
they go? :) )
C U LAZER,
Jeffrey D. Shaffer
### Mike Vermillion
Nov 2, 1993, 1:01:11 PM11/2/93
to
Thank you very, very, very much Mr. Adams. I find it hard to
believe that this answer wasn't obvious to everyone from the start.
Will the rest of you please stop now?
Mike
### John E. Veness
Nov 2, 1993, 3:20:29 PM11/2/93
to
: In Article <2b4asr\$b...@syzygy.socs.uts.edu.au>, mjch...@socs.uts.EDU.AU
: Best,
Awww, all the myths come crashing down, one-by-one..
Pel
### Sodhed
Nov 2, 1993, 11:43:14 PM11/2/93
to
>In Article <2b4asr\$b...@syzygy.socs.uts.edu.au>, mjch...@socs.uts.EDU.AU
>(Mark J Cherkas) wrote:
>>
>>I am new to this group so bear with this beginners question:
>>Why is the answer 42 ?
>>Has Douglas Adams ever explained this ?
>
>The answer to this is very simple. It was a joke. It had to be a number, an
>ordinary, smallish number, and I chose that one. Binary representations,
>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>stared into the garden and thought '42 will do' I typed it out. End of story.
>
>Best,
*laugh* What a guy... thus ending a possible flame-war. If only the 'fan'
newsgroups ALL had the real source of the entertainment posting to them, there
would be few flame-wars arguing about stuff we don't even know. The main
reason, by the way, that I read this newsgroup is because a friend of mine is
a fanatic, and has told me much of Mr. Adams' work, and also the fact that the
MAN HIMSELF posts here. Almost makes me want to start reading something more
than textbooks again. Time to finish my physics homework....
=========================================================================
/|
/ | /\ / /\ / / / / / / /| _________ _____
/ | / \ / / \ / / / / / / / | / / / / /
/---|/ \ / / \ / / /---/ / / / | / / / /---<
/ | \/ / \/ / / / / /____/ |/ /__/ / \
\
--The Sodhed, Reid...@Max.cc.URegina.ca, is setting the world on fire...
=========================================================================
### Eschel A. Hamel
Nov 3, 1993, 3:54:35 AM11/3/93
to
>In Article <2b4asr\$b...@syzygy.socs.uts.edu.au>, mjch...@socs.uts.EDU.AU
>(Mark J Cherkas) wrote:
>>
>>I am new to this group so bear with this beginners question:
>>Why is the answer 42 ?
>>Has Douglas Adams ever explained this ?
I still don't get this. People are always trying to look into that one. When
I read the series, I saw it as a joke, plain and simple. It's simple (but
highly effective) irony, period. I've seen and heard him answer this question
way too many times. Let it rest folks. Remember, every moment he has to spend
explaining things, is one he isn't writing. I'd much rather see the script
completed, or perhaps a new novel.
### S.Casey
Nov 3, 1993, 6:22:36 AM11/3/93
to
>stared into the garden and thought '42 will do' I typed it out. End of story.
YOU`VE RUINED IT NOW!!!!
Hmm.....
I was watching my copy of HHGTTG last night (after a discussion over all the differences between all the different versions) And I suddenly noticed a certain 'Green' theme to the entire book. Naughty naughty - you were writing this before it was trendy to be green.. - This just won't do... Your only hope is that it was unintentional...
PS. I have only been skiming this group and was wondering if anyone actually got round to compiling that list of questions? - if so mail me or something - I might like to add to it.....
Thanks,
Steve
With a brain the size of a planet, who needs a .sig?
### MISS RA CLARKE
Nov 3, 1993, 7:56:43 AM11/3/93
to
>>I am new to this group so bear with this beginners question:
>>Why is the answer 42 ?
>>Has Douglas Adams ever explained this ?
>
>The answer to this is very simple. It was a joke. It had to be a number, an
>ordinary, smallish number, and I chose that one. Binary representations,
>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>stared into the garden and thought '42 will do' I typed it out. End of story.
>
>Best,
>
>Currently in Santa Fe, NM | ada...@nic.cerf.net (current)
Awwww, why spoil their fun? - they were happy with their binaries and base
13s...
Beccy
______________________________________________________________
Beccy Clarke g90c...@warthog.ru.ac.za
Aristotle defined man as a rational animal -
this was probably somewhat optimistic. Arthur Morgan
### Ken
Nov 3, 1993, 12:08:36 PM11/3/93
to
>The answer to this is very simple. It was a joke. It had to be a number, an
>ordinary, smallish number, and I chose that one. Binary representations,
>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>stared into the garden and thought '42 will do' I typed it out. End of story.
>Best,
But, perhaps your subconscious mind was giving you (and then all of
us) some wild insight to just what life is really all about ???
Ken. (Not wanting to see a funny thread die and lose all those wierd
possible explanations !).
---------------------->e-mail to: jmar...@unix2.tcd.ie<--------------------
You're all clear and my psychometer indicates smooth sailing.
- 7-Zark-7 (The Jupiter Moon Menace)
---------------------->e-mail to: jmar...@unix2.tcd.ie<--------------------
### Dale Schouten
Nov 3, 1993, 12:19:52 PM11/3/93
to
>
>The answer to this is very simple. It was a joke. It had to be a number, an
>ordinary, smallish number, and I chose that one. Binary representations,
>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>stared into the garden and thought '42 will do' I typed it out. End of story.
Right, think you know something about it, eh?
Surely you're denying your subconscious interconnectedness with
tibetan monks possessed of 13 fingers!?
Or isn't it true that you originally wrote HHGG in a previous
life in 1924?
You can't fool us with simple facts!
Dale Schouten
scho...@uiuc.edu
### Matthias Urlichs
Nov 3, 1993, 9:06:24 AM11/3/93
to
>
> The answer to this is very simple. It was a joke. It had to be a number, an
> ordinary, smallish number, and I chose that one. Binary representations,
> base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
> stared into the garden and thought '42 will do' I typed it out. End of story.
>
Bull. Or, to quote what a German philosopher once said to Isaac Asimov (it's
in his autobiography somewhere), "Just because you wrote it doesn't mean that
you know anything at all about it."
:-)
--
You should emulate your heroes, but don't carry it too far.
--
Matthias Urlichs \ XLink-POP Nürnberg | EMail: url...@smurf.sub.org
Schleiermacherstraße 12 \ Unix+Linux+Mac | Phone: ...please use email.
90491 Nürnberg (Germany) \ Consulting+Networking+Programming+etc'ing 42
### un Coeur en Hiver
Nov 4, 1993, 1:16:12 PM11/4/93
to
jdsh...@silver.ucs.indiana.edu (Jack Death) wrote:
>
> >
> >The answer to this is very simple. It was a joke. It had to be a number, an
> >ordinary, smallish number, and I chose that one. Binary representations,
> >base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
> >stared into the garden and thought '42 will do' I typed it out. End of story.
> >
>
> Hmmmm... I kinda figured it was something like that, but then
> again if the REAL answer to life the universe and everything WERE
> 42, it would be a common number to _guess_ or _Pick_, wouldn't it?
> I think that whether you, Mr. Adams, picked it intentionaly or not,
> you hit the right answer anyways.
> (Besides isn't it Fun to see just
> how many poeple are looking for this number to show up everywhere
> they go? :) )
I don't look for the number, it just keeps popping up all the time.
un Coeur en Hiver
-------------------------------------------------------------------------
| e-mail: | "You can cry, you can mope |
| bro...@ilf.uio.no | But can you swing from a good rope?" -Pixies |
-------------------------------------------------------------------------
### Schumacher Gordon C
Nov 5, 1993, 12:17:46 AM11/5/93
to
cm5...@scitsc.wlv.ac.uk (S.Casey) writes:
>I was watching my copy of HHGTTG last night (after a discussion over all the differences between all the different versions) And I suddenly noticed a certain 'Green' theme to the entire book. Naughty naughty - you were writing this before it was trendy to be green.. - This just won't do... Your only hope is that it was unintentional...
No, he seems to like yellow much more. Yellow tongues (at the VERY
beginning), yellow bulldozers, yellow Vogon ships, Suffusions of Yellow,
and others (tho' I don't remember offhand)
BTW, it's considered polite to break your lines at about 75 chars. It
makes it much easier to read ;)
Gordon Schumacher
/-------------------------------------------------------------------\
| Champaign- "We apologize for the inconvenience." _@_ |
| Urbana -HHGTTG / \ |
| kilroy was here | o o | |
\-------------------------------------------------------U|--U--|U---/
### Kurt Brinschwitz
Nov 5, 1993, 1:50:03 AM11/5/93
to
In article <jmarston....@unix2.tcd.ie> jmar...@unix2.tcd.ie (Ken) writes:
>
>>The answer to this is very simple. It was a joke. It had to be a number, an
>>ordinary, smallish number, and I chose that one. Binary representations,
>>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>>stared into the garden and thought '42 will do' I typed it out. End of story.
>
>>Best,
>
>
> But, perhaps your subconscious mind was giving you (and then all of
>us) some wild insight to just what life is really all about ???
>
>Ken. (Not wanting to see a funny thread die and lose all those wierd
^^^^^^^^^^^^^^^^^ ?????????
>possible explanations !).
>
>
>---------------------->e-mail to: jmar...@unix2.tcd.ie<--------------------
>You're all clear and my psychometer indicates smooth sailing.
> - 7-Zark-7 (The Jupiter Moon Menace)
>---------------------->e-mail to: jmar...@unix2.tcd.ie<--------------------
Good God, man ! get a life!
### Ken
Nov 5, 1993, 5:13:43 AM11/5/93
to
>>Ken. (Not wanting to see a funny thread die and lose all those wierd
> ^^^^^^^^^^^^^^^^^ ?????????
>>possible explanations !).
>>
>Good God, man ! get a life!
I would, but they are more trouble than their worth.
Ken.
### S.Casey
Nov 5, 1993, 8:13:08 AM11/5/93
to
>BTW, it's considered polite to break your lines at about 75 chars. It
>makes it much easier to read ;)
Erm yea... - I'll count them! hmm...
The point was green as in eco friendly....
and besides all the graphics were in green!!
the thing was about the burning of the trees on ancient earth etc.....
Steve
### Evan Douglas Macbeth
Nov 4, 1993, 5:22:22 PM11/4/93
to
Yesness, I seem to see it around much also...
I think it is an unwritten law of the Whole Sort of General
Mish Mash, "And 42 shall be the number of the counting..."
ed...@virginia.edu
### Alan Stokes
Nov 5, 1993, 8:16:37 AM11/5/93
to
>Best,
What's the point of us sitting down and wondering whether there may, or may not, be
a reason for the answer 42 if some bloody author comes along and tells us he just made
it up?
:-)
--
Alan Stokes al...@rcp.co.uk
Senior Software Engineer Tel +44 235 510116
Richards Computer Products Ltd FAX +44 235 511084
Dales, High St, Didcot, Oxon, UK
### James Messer
Nov 5, 1993, 1:24:20 PM11/5/93
to
Hi, I aseemeed to have nmissed the article on the hotel soap. Could
someone e-mail it too me? I am at jme...@morgan.ucs.mun.ca or at
mrm%statistics%dfo...@dfonfl01.nwafc.nf.ca
BTW Hello Mr. Adams, I was wondering if you had anyparticular piece of
Bach's music in mind when you wrote about him in the Dirk Gently book?
And what about the Mozart bit with the annoying note?
d
--
James Murdo Messer | "Ah, yes, |
jme...@morgan.ucs.mun.ca | the rattle" |
### Mike Unlimited
Nov 6, 1993, 3:04:30 AM11/6/93
to
In article <1Il2sAk...@rcp.co.uk> al...@rcp.co.uk (Alan Stokes) writes:
>>The answer to this is very simple. It was a joke. It had to be a number, an
>>ordinary, smallish number, and I chose that one. Binary representations,
>>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>>stared into the garden and thought '42 will do' I typed it out. End of story.
>What's the point of us sitting down and wondering whether there may, or may
>not, be a reason for the answer 42 if some bloody author comes along and tells
us he just made it up?
>:-)
Why, the entertainment value, of course. }->
--Mike
--
To err is human. To bleat is ovine. To bark is canine.
To forgive is divine. To oink is porcine. To purr is feline.
To moo is bovine. To howl is lupine. This list is asinine.
Mike Escutia | Pliable Lad of the LNH | Ergh, the warlord | mi...@unh.edu
Nov 8, 1993, 7:42:19 AM11/8/93
to
BTW Hello Mr. Adams, I was wondering if you had anyparticular piece of
>Bach's music in mind when you wrote about him in the Dirk Gently book?
Yes. Schubler Prelude Number six. Also appears as a viola obligato (*not*
cello, stupid mistake I made) in one of the Cantatas. Probably number 6.
### Ulrich Schreglmann
Nov 8, 1993, 4:28:53 AM11/8/93
to
>>Why is the answer 42 ?
>>Has Douglas Adams ever explained this ?
>The answer to this is very simple. It was a joke. It had to be a number, an
>ordinary, smallish number, and I chose that one. Binary representations,
>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>stared into the garden and thought '42 will do' I typed it out. End of story.
Not for me, it isn't! :-) What sort of desk was it you sat on? What
kind of things surrounded you? What did the garden look like? I'm
pretty sure the real meaning of it all is hidden somewhere in your sub-
conscious.
(Don't look at me like that! I won't give up a decade-long quest for a
meaning just because the author tells me there is none! Haven't you e-
er heard psychiatrists? "Everything has meaning, and the more the p-
tient denies it the more it's true. And the dirtier and more sexual it
is." So what's going on there in your filthy brain?! :-) :-) :-) :-) )
May the Cool Be with You!
(C)OOL mcmxciii
### Richard S. Sevrinsky
Nov 5, 1993, 11:54:35 AM11/5/93
to
--- A brief quote, if I might, from JohnvRe: Why 42 ?un.rhbnc.ac.uk:
Jo> : The answer to this is very simple. It was a joke. It had to be a
Jo> number, an : ordinary, smallish number, and I chose that one. Binary
Jo> representations, : base thirteen, Tibetan monks are all complete
Jo> nonsense. I sat at my desk, : stared into the garden and thought '42
Jo> will do' I typed it out. End of story.
Jo> : Best,
Jo> : London, UK | d...@dadams.demon.co.uk (dormant)
Jo> : Currently in Santa Fe, NM | ada...@nic.cerf.net (current)
Jo> Awww, all the myths come crashing down, one-by-one..
Not really. Now all we have to do is wonder exactly why the
number 42 was "suggested" in Mr. Adams' mind. My guess is it was
the mice. Any other ideas?
________________________________________________________________
| Richie Sevrinsky a.k.a "Life is an anticipation |
| rse...@dorsai.dorsai.org of....." |
`--------------------------------------------------------------'
... The OFFICIAL tagline of the 1996 Olympics!
___ Blue Wave/QWK v2.12
### Evan Douglas Macbeth
Nov 10, 1993, 10:58:11 AM11/10/93
to
serious or sarcastic, this response was just plain funny. :)
ed...@virginia.edu New College
The University of Virginia
### Ari P K Korhonen
Nov 12, 1993, 5:56:38 PM11/12/93
to
Ulrich Schreglmann (uhsc...@faui06n.informatik.uni-erlangen.de) wrote:
: >>Why is the answer 42 ?
: >>Has Douglas Adams ever explained this ?
: >The answer to this is very simple. It was a joke. It had to be a number, an
: >ordinary, smallish number, and I chose that one. Binary representations,
: >base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
: >stared into the garden and thought '42 will do' I typed it out. End of story.
: Not for me, it isn't! :-) What sort of desk was it you sat on? What
: kind of things surrounded you? What did the garden look like? I'm
: pretty sure the real meaning of it all is hidden somewhere in your sub-
: conscious.
: (Don't look at me like that!......[rest edited by A PK K]
And hey hey, did Mr. Adams have a Ballantine's Whiskey around or
something, I have an old Ballantine's add which says that it is
blended using 42 single whiskeys to create the true smooth
Ballantine...
Ari PK Korhonen (apkk...@cc.helsinki.fi and Zark Off!!
### Stefan Linnemann
Nov 13, 1993, 4:58:46 PM11/13/93
to
Ari P K Korhonen <apkk...@kruuna.Helsinki.FI> wrote:
>And hey hey, did Mr. Adams have a Ballantine's Whiskey around or
>something, I have an old Ballantine's add which says that it is
>blended using 42 single whiskeys to create the true smooth
>Ballantine...
Ah, that _does_ explain why there's something fundamentally wrong with
the stuff.
Stefan.
--
Stefan M. Linnemann Internet: Stefan.L...@cri.LeidenUniv.nl
System programmer Unix CRI, Leiden University, the Netherlands.
### The Doctor
Nov 16, 1993, 8:25:15 AM11/16/93
to
>
>
>>The answer to this is very simple. It was a joke. It had to be a number, an
>>ordinary, smallish number, and I chose that one. Binary representations,
>>base thirteen, Tibetan monks are all complete nonsense. I sat at my desk,
>>stared into the garden and thought '42 will do' I typed it out. End of story
> .
Are you the real BBC writer? If so, how come this US address?
>
--
God save the Queen! God bless and save us all!!
Remeber, Jesus saves all souls from eternal damnation!
Save the world! Purge Republicanism and Dictatorship!
A British citizen I was born, A British citizen I will most honourably die!
### Daniel O'Malley
Nov 17, 1993, 12:58:39 PM11/17/93
to
In <931116.062515.6...@galcon.ersys.edmonton.ab.ca> sys...@galcon.ersys.edmonton.ab.ca (The Doctor) writes:
[deleted]
>Are you the real BBC writer? If so, how come this US address?
Yes, it is the real Douglas Adams. Regular readers of the group will
know that he's on a booksigning tour of the US at the moment, hence the
Daniel.
--
/___Daniel O'Malley___\ /"A computer terminal is not some clunky old TV with\
\___Trinity College___/ \__a typewriter in front of it. It is an interface__/
/___Dublin, Ireland___\ /___where the mind and body can connect with the____\
### Simon Slavin
Nov 19, 1993, 8:24:43 AM11/19/93
to
In article <931116.062515.6...@galcon.ersys.edmonton.ab.ca> sys...@galcon.ersys.edmonton.ab.ca (The Doctor) writes:
>
>Are you the real BBC writer? If so, how come this US address?
which indicate that he reads this newsgroup). The reason he's
posting from the USA is that he's currently touring there and
doing readings and such stuff. I don't think he actually needs a
US account, there must be thousands of students only too willing
to let him use theirs !
And sorry if this sounds holier-than-thou, but please don't
pester him on the net because the more time he spends posting,
the less time he spends writing books and the movie script,
and listening to the excellent music which appears to inspire it all.
Unless, of course, Mr. Adams, you *prefer* spending time on the
net, in which case please tell us. :-) :-)
Simon.
--
< The wonderful thing about Tiggers is Tiggers are Wonderful Things. | slavins >
< Is all that we see or seem but a dream within a dream ? EAPoe |@cs.man.ac.uk > | 6,252 | 21,715 | {"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} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.865945 |
https://www.stat.math.ethz.ch/pipermail/r-help/2012-October/338521.html | 1,675,501,980,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00447.warc.gz | 1,033,875,824 | 2,668 | # [R] help speeding up simple Theil regression function
Berend Hasselman bhh at xs4all.nl
Sun Oct 21 20:59:00 CEST 2012
```On 21-10-2012, at 20:06, Brad Schneid wrote:
> Hello,
>
> I am working on a simple non-parametric (Theil) regression function and and
> am following Hollander and Wolfe 1999 text. I would like some help making
> my function faster. I have compared with pre-packaged version from "MBLM",
> which isnt very fast either, but it appears mine is faster with N = 1000
> (see results below). I plan on running this function repeatedly, and I
> generally have data lengths of ~ N = 6000 or more.
>
> # My function following Hollander and Wolfe text, Chapter 9
> np.lm <-function(dat, X, Y, ...){
> # Ch 9.2: Slope est. (X) for Thiel statistic
> combos <- combn(nrow(dat), 2)
> i.s <- combos[1,]
> j.s <- combos[2,]
> num <- vector("list", length=length(i.s))
> dom <- vector("list", length=length(i.s))
>
> for(i in 1:length(i.s)){
> num[[i]] <- dat[j.s[i],Y] - dat[i.s[i],Y]
> dom[[i]] <- dat[j.s[i],X] - dat[i.s[i],X]
> }
>
> X <- median( sort( do.call(c, num) / do.call(c, dom) ) )
> # Ch 9.4: Intercept est. for Thiel statistic
> Intercept <- median(dat[,"Y"] - X*dat[,"X"])
> out <- data.frame(Intercept, X)
> return(out)
> } # usage: np.lm(dat, X=1, Y=2)
> ################################################################
>
> library("mblm") # I will compare to mblm() function
>
> X <- rnorm(1000)
> Y <- rnorm(1000)
> dat <- data.frame(X, Y)
>
> system.time(np.lm(dat, X=1, Y=2) )
> user system elapsed
> 118.610 0.130 119.144
> 109.000 0.040 109.416 # ran it twice
> 86.190 0.100 86.589 # 3rd time
Alternative function without your i loop (it isn't needed and can be vectorized):
np.lm.alt <-function(dat, X, Y, ...){
# Ch 9.2: Slope est. (X) for Thiel statistic ==> (Pedantic comment: it is Theil (swap the i and e)
combos <- combn(nrow(dat), 2)
i.s <- combos[1,]
j.s <- combos[2,]
Y.num <- dat[j.s,Y] - dat[i.s,Y]
X.dom <- dat[j.s,X] - dat[i.s,X]
X <- median( Y.num / X.dom)
# Ch 9.4: Intercept est. for Thiel statistic ==> (Pedantic comment: it is Theil (swap the i and e)
Intercept <- median(dat[,"Y"] - X*dat[,"X"])
out <- data.frame(Intercept, X)
return(out)
} # usage: np.lm(dat, X=1, Y=2)
Try the compiler package on you original function:
library(compiler)
np.lm.c <- cmpfun(np.lm)
Test speed and correct results:
X <- rnorm(500)
Y <- rnorm(500)
dat <- data.frame(X, Y)
system.time(npout.c <- np.lm.c(dat, X=1, Y=2) )
system.time(npout.1 <- np.lm(dat, X=1, Y=2) )
system.time(npout.a <- np.lm.alt(dat, X=1, Y=2) )
identical(npout.1,npout.c)
identical(npout.1,npout.a)
Results:
> system.time(npout.c <- np.lm.c(dat, X=1, Y=2) )
user system elapsed
21.442 0.066 21.517
> system.time(npout.1 <- np.lm(dat, X=1, Y=2) )
user system elapsed
21.068 0.073 21.161
> system.time(npout.a <- np.lm.alt(dat, X=1, Y=2) )
user system elapsed
0.303 0.010 0.313
> identical(npout.1,npout.c)
[1] TRUE
> identical(npout.1,npout.a)
[1] TRUE
You may try and test this with larger data lengths.
Berend
``` | 1,104 | 3,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.584218 |
https://mathspace.co/textbooks/syllabuses/Syllabus-1066/topics/Topic-20770/subtopics/Subtopic-269992/ | 1,701,178,649,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00270.warc.gz | 435,165,159 | 50,971 | # 11.03 Right angles
Lesson
## Ideas
Being able to compare angles can help us in this lesson. Let's try this problem to review.
### Examples
#### Example 1
Which of these angles is larger?
A
B
Worked Solution
Create a strategy
Choose the option with a larger turn between the lines.
Apply the idea
Idea summary
Angles can be compared by how large the turn is between the lines.
## Right angles
Let's learn about right angles and how we can compare other angles to them.
### Examples
#### Example 2
Which of the following is a right angle?
A
B
Worked Solution
Create a strategy
Look for the angle with a small square in the corner.
Apply the idea
Option A has a small square at the corner, so the answer is option A.
Idea summary
A right angle is exactly 90\degree.
### Outcomes
#### VCMMG146
Identify angles as measures of turn and compare angle sizes in everyday situations | 217 | 901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-50 | latest | en | 0.890082 |
http://www.compadre.org/psrc/items/detail.cfm?ID=3596 | 1,477,171,909,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719045.47/warc/CC-MAIN-20161020183839-00457-ip-10-171-6-4.ec2.internal.warc.gz | 383,562,440 | 8,333 | Website Detail Page
written by Dan Russell
This animation with accompanying explanation illustrates the principle of wave superposition: when two (or more) waves travel through the same medium at the same time, the waves pass through each other without being disturbed. Net displacement of the medium at any point in space or time is the sum of the individual wave displacements. Author Dan Russell's text provides support for both concept formation and algebraic applications. It is part of a larger collection of animations on wave motion, acoustics, and sound.
Subjects Levels Resource Types
Oscillations & Waves
- Wave Motion
= Interference and Diffraction
- High School
- Instructional Material
= Unit of Instruction
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= Movie/Animation
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Keywords:
sine waves, superposition of waves, wave interference, wave phenomena
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Save to my folders | 855 | 3,292 | {"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-2016-44 | longest | en | 0.752923 |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-5-section-5-6-definite-integral-substitutions-and-the-area-between-curves-exercises-page-337/16 | 1,576,417,827,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00075.warc.gz | 718,149,943 | 12,769 | ## University Calculus: Early Transcendentals (3rd Edition)
$$\int^4_1\frac{dy}{2\sqrt y(1+\sqrt y)^2}=\frac{1}{6}$$
$$A=\int^4_1\frac{dy}{2\sqrt y(1+\sqrt y)^2}$$ We set $u=1+\sqrt y$, which means $$du=\frac{dy}{2\sqrt y}$$ For $y=1$, we have $u=2$ For $y=4$, we have $u=3$ Therefore, $$A=\int^3_2\frac{1}{u^2}du=\Big(-\frac{1}{u}\Big)\Big]^3_2$$ $$A=-\Big(\frac{1}{3}-\frac{1}{2}\Big)$$ $$A=\frac{1}{6}$$ | 196 | 407 | {"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-2019-51 | latest | en | 0.51217 |
https://sports.answers.com/Q/What_is_a_good_100_yd_sprint_time | 1,718,678,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00126.warc.gz | 478,165,720 | 47,643 | 0
# What is a good 100 yd sprint time?
Updated: 9/27/2023
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Q: What is a good 100 yd sprint time?
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Related questions
### 100 meter time converted to 100 yd time?
9/10 of 100 m time will be very close...
### If you run100meters in 10.6 seconds how fast would you run 100 yards?
1 m = 1.094 yd 100 m * (1.094 yd/m) = 109.4 yd 109.4 yd / 10.6 s = 10.3208 yd/s 100 yd / (10.3208 yd/s) = 9.6892 s
### What is 100 yd ft?
100 Yards (yd) = 300 feet (ft)
### 100 ft and 300 yd?
300 yd. is greater.
9.8 seconds
### What is 100 square feet in square yards?
3 ft = 1 yd 1 sq ft = 1 ft × 1 ft = (1 ÷ 3) yd × (1 ÷ 3) yd = 1/9 sq yd → 100 sq ft = 100 × 1/9 sq yd = 11 1/9 sq yd ≈ 11.1 sq yd
### What is 100 in compared to 3 yd 1 ft?
100 in. is less than 3 yd. 1 ft.
### What are the events in a swim meet for high school?
4x50 Yd Medley Relay, 200 Yd Freestyle, Individual Medley(100- J.V., 200- V), 50 Yd Freestyle, Butterfly(50- J.V., 100- V), 100 Yd Freestyle, 500 Yd Freestyle(Varsity only), 4x50 Freestyle Relay, Backstroke(50- J.V., 100- V), Breaststroke(50-J.V., 100- V), 4x100 Yd Freestyle
### A diameter of 100 yd?
If you want to convert that to meters, you can multiply by 0.9, approximately.
### How many centimeters are in 100 yards?
9,144 cm Algebraic Steps / Dimensional Analysis Formula 100 yd*36 in 1 yd*2.54 cm 1 in=9,144 cm Direct Conversion Formula 100 yd*91.44 cm 1 yd=9,144 cm
100
### 100 in to 3 yd 1 ft?
3 yd. 1 ft. is greater. | 615 | 1,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.813556 |
http://www.torontokidscomputer.com/aurora-1430-python/aurora-21-0116-14/ | 1,675,819,310,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00366.warc.gz | 84,714,856 | 11,635 | TORONTO KIDS COMPUTER CLUB | Aurora Saturday 14:30 Python Practice 21.01.16.
18946
# Aurora Saturday 14:30 Python Practice 21.01.16.
## 20 Jan Aurora Saturday 14:30 Python Practice 21.01.16.
Question 1:
This question involves calculating the value of aromatic numbers which are a combination of Arabic digits and Roman numerals.
An aromatic number is of the form ARARAR…AR, where each A is an Arabic digit, and each R is a Roman numeral. Each pair AR contributes a value described below, and by adding or subtracting these values together we get the value of the entire aromatic number.
An Arabic digit A can be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9. A Roman numeral R is one of the seven letters I, V, X, L, C, D, or M. Each Roman numeral has a base value:
Symbol I V X L C D M Base value 1 5 10 50 100 500 1000
The value of a pair AR is A times the base value of R. Normally, you add up the values of the pairs to get the overall value. However, wherever there are consecutive symbols ARA′R′ with R′ having a strictly bigger base value than R, the value of pair AR must be subtracted from the total, instead of being added.
For example, the number 3M1D2C has the value 3×1000+1×500+2×100=3700 and 3X2I4X has the value 3×10−2×1+4×10=68.
Write a program that computes the values of aromatic numbers.
Input Specification
The input is a valid aromatic number consisting of between 2 and 20 symbols.
Output Specification
The output is the decimal value of the given aromatic number.
```Sample Input 1
3M1D2C
Output for Sample Input 1
3700
Sample Input 2
2I3I2X9V1X
Output for Sample Input 2
-16``` | 464 | 1,604 | {"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-2023-06 | latest | en | 0.766693 |
https://cdn.varsitytutors.com/algebra_ii-help/number-sets | 1,718,900,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00643.warc.gz | 136,507,613 | 42,996 | # Algebra II : Number Sets
## Example Questions
### Example Question #1 : Number Sets
If , , and , then find the following set:
Explanation:
The union is the set that contains all the numbers from and . Therefore the union is
### Example Question #2 : Number Sets
If , , and , find the following set:
Explanation:
The intersection is the set that contains only the numbers found in all three sets. Therefore the intersection is .
### Example Question #1 : Number Sets
If , , and , find the following set:
Explanation:
The intersection is the set that contains the numbers that appear in both and . Therefore the intersection is .
### Example Question #4 : Number Sets
If , , and , find the following set:
Explanation:
The intersection is the set that contains the numbers found in both sets. Therefore the intersection is .
### Example Question #5 : Number Sets
If , , and , find the following set:
Explanation:
The union is the set that contains all of the numbers found in all three sets. Therefore the union is . You do not need to re-write the numbers that appear more than once.
### Example Question #6 : Number Sets
If , , and , find the following set:
Explanation:
The intersection is the set that contains the numbers found in both sets. Therefore the intersection is .
### Example Question #1 : Number Sets
Which set of numbers represents the union of E and F?
Explanation:
The union is the set of numbers that lie in set E or in set F.
.
In this problem set E contains terms , and set F contains terms . Therefore, the union of these two sets is .
### Example Question #8 : Number Sets
Express the following in Set Builder Notation:
Explanation:
and stands for OR in Set Builder Notation
### Example Question #1 : Number Sets
Find the intersection of the two sets:
Explanation:
To find the intersection of the two sets, , we must find the elements that are shared by both sets:
### Example Question #5141 : Algebra Ii
What type of numbers are contained in the set ?
Imaginary
Integers
Complex
Irrational
Natural
Integers
Explanation:
We can use process of elimination to find the correct answer.
It can't be Imaginary because we're not dividing by a negative number.
It can't be Complex because the number's aren't a mix of real and imaginary numbers.
It can't be Irrational because they aren't fractions.
It can't be Natural because there are negative numbers.
It must be Integers then! All the numbers are whole numbers that fit on the number line. | 567 | 2,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.859865 |
https://testbankweb.com/product/test-bank-for-basic-technical-mathematics-10-e-allyn-j-washington-isbn-10-0133083500-isbn-13-9780133083507-downloadable-digital-test-bank-files/ | 1,601,566,071,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131777.95/warc/CC-MAIN-20201001143636-20201001173636-00065.warc.gz | 614,786,659 | 22,159 | Sale!
# Test Bank for Basic Technical Mathematics, 10/E, Allyn J. Washington, ISBN-10: 0133083500, ISBN-13: 9780133083507, Downloadable Digital Test Bank Files
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Test Bank for Basic Technical Mathematics, 10/E, Allyn J. Washington, ISBN-10: 0133083500, ISBN-13: 9780133083507
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## Description
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Each chapter contains: Equations, Quick Chapter Review, Review Exercises, Practice Test
1. Basic Algebraic Operations
1.1 Numbers
1.2 Fundamental Operations of Algebra
1.3 Calculators and Approximate Numbers
1.4 Exponents
1.5 Scientific Notation
1.7 Addition and Subtraction of Algebraic Expressions
1.8 Multiplication of Algebraic Expressions
1.9 Division of Algebraic Expressions
1.10 Solving Equations
1.11 Formulas and Literal Equations
1.12 Applied Word Problems
2. Geometry
2.1 Lines and Angles
2.2 Triangles
2.4 Circles
2.5 Measurement of Irregular Areas
2.6 Solid Geometric Figures
3. Functions and Graphs
3.1 Introduction to Functions
3.3 Rectangular Coordinates
3.4 The Graph of a Function
3.5 Graphs on the Graphing Calculator
3.6 Graphs of Functions Defined by TTables of Data
4. The Trigonometric Functions
4.1 Angles
4.2 Defining the Trigonometric Functions
4.3 Values of the Trigonometric Functions
4.4 The Right Triangle
4.5 Applications of Right Triangles
5. Systems of Linear Equations; Determinants
5.1 Linear Equations
5.2 Graphs of Linear Functions
5.3 Solving Systems of Two Linear Equations in Two Unknowns Graphically
5.4 Solving Systems of Two Linear Equations in Two Unknowns Algebraically
5.5 Solving Systems of Two Linear Equations in Two Unknowns by Determinants
5.6 Solving Systems of Three Linear Equations in Three Unknowns Algebraically
5.7 Solving Systems of Three Linear Equations in Three Unknowns by Determinants
6. Factoring and Fractions
6.1 Special Products
6.2 Factoring: Common Factor and Difference of Squares
6.3 Factoring Trinomials
6.4 The Sum and Difference of Cubes
6.5 Equivalent Fractions
6.6 Multiplication and Division of Fractions
6.7 Addition and Subtraction of Fractions
6.8 Equations Involving Fractions
7.1 Quadratic Equations; Solution by Factoring
7.2 Completing the Square
7.4 The Graph of the Quadratic Function
8. Trigonometric Functions of Any Angle
8.1 Signs of the Trigonometric Functions
8.2 Trigonometric Functions of Any Angle
9. Vectors and Oblique Triangles
9.1 Introduction to Vectors
9.2 Components of Vectors
9.4 Applications of Vectors
9.5 Oblique Triangles, the Law of Sines
9.6 The Law of Cosines
10. Graphs of the Trigonometric Functions
10.1 Graphs of y = a sin x and y = a cos x
10.2 Graphs of y = a sin bx and y = a cos bx
10.3 Graphs of y = a sin (bx + c ) and y = a cos (bx + c )
10.4 Graphs of y = tan x,y = cot x, y = sec x, y = csc x
10.5 Applications of the Trigonometric Graphs
10.6 Composite Trignometric Curves
11.1 Simplifying Expressions with Integral Exponents
11.2 Fractional Exponents
11.5 Multiplication and Division of Radicals
12. Complex Numbers
12.1 Basic Definitions
12.2 Basic Operations with Complex Numbers
12.3 Graphical Representation of Complex Numbers
12.4 Polar Form of a Complex Number
12.5 Exponential Form of a Complex Number
12.6 Products, Quotients, Powers, and Roots of Complex Numbers
12.7 An Application to Alternating-current (ac) Circuits
13. Exponential and Logarithmic Functions
13.1 Exponential Functions
13.2 Logarithmic Functions
13.3 Properties of Logarithms
13.4 Logarithms to the Base 10
13.5 Natural Logarithms
13.6 Exponential and Logarithmic Equations
13.7 Graphs on Logarithmic and Semilogarithmic Paper
14. Additional Types of Equations and Systems of Equations
14.1 Graphical Solution of Systems of Equations
14.2 Algebraic Solution of Systems of Equations
15. Equations of Higher Degree
15.1 The Remainder and Factor Theorems; Synthetic Division
15.2 The Roots of an Equation
15.3 Rational and Irrational Roots
16. Matrices; Systems of Linear Equations
16.1 Matrices: Definitions and Basic Operations
16.2 Multiplication of Matrices
16.3 Finding the Inverse of a Matrix
16.4 Matrices and Linear Equations
16.5 Gaussian Elimination
16.6 Higher-order Determinants
17. Inequalities
17.1 Properties of Inequalities
17.2 Solving Linear Inequalities
17.3 Solving Nonlinear Inequalities
17.4 Inequalities Involving Absolute Values
17.5 Graphical Solution of Inequalities with Two VariTables
17.6 Linear Programming
18. Variation
18.1 Ratio and Proportion
18.2 Variation
19. Sequences and the Binomial Theorem
19.1 Arithmetic Sequences
19.2 Geometric Sequences
19.3 Infinite Geometric Series
19.4 The Binomial Theorem
20.1 Fundamental Trigonometric Identities
20.2 The Sum and Difference Formulas
20.3 Double-Angle Formulas
20.4 Half-Angle Formulas
20.5 Solving Trigonometric Equations
20.6 The Inverse Trigonometric Functions
21. Plane Analytic Geometry
21. 1 Basic Definitions
21.2 The Straight Line
21.3 The Circle
21.4 The Parabola
21.5 The Ellipse
21.6 The Hyperbola
21.7 Translation of Axes
21.8 The Second-degree Equation
21.9 Rotation of Axes
21.10 Polar Coordinates
21.11 Curves in Polar Coordinates
22. Introduction to Statistics
22.1 Frequency Distributions
22.2 Measures of Central Tendency
22.3 Standard Deviation
22.4 Normal Distributions
22.5 Statistical Process Control
22.6 Linear Regression
22.7 Nonlinear Regression
Appendix A: Solving Word Problems
Appendix B: Units of Measurement: The Metric System
B.1 Introduction
B.2 Reductions and Conversions
Appendix C: The Graphing Calculator
C.1 Introduction
C.2 The Graphing Calculator
C.3 Graphing Calculator Programs | 1,705 | 6,274 | {"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} | 2.90625 | 3 | CC-MAIN-2020-40 | latest | en | 0.766285 |
https://theclaunet.it/2017-Mar-19/28459.html | 1,652,830,753,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520936.24/warc/CC-MAIN-20220517225809-20220518015809-00747.warc.gz | 635,964,444 | 5,926 | • Home
• grinding capacity calculation of ball mill
grinding capacity calculation of ball mill
### ball mill grinding capacity calculation
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### Grinding in Ball Mills: Modeling and Process Control
Grinding in Ball Mills: Modeling and Process Control Vladimir Monov, , An important characteristic of an industrial ball mill is its production capacity.
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MODELING THE SPECIFIC GRINDING ENERGY AND BALL-MILL SCALEUP , where T the mill capacity Mill dimensions , zthe calculation of the specific grinding.
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grinding capacity calculation of ball mill grinding capacity calculation of ball mill - , capacity calculation for rod mill--Henan Mining Heavy Machinery Co. | 1,025 | 5,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.789948 |
http://betterlesson.com/lesson/reflection/14052/bridging-old-standards-to-new-standards | 1,477,559,967,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721174.97/warc/CC-MAIN-20161020183841-00241-ip-10-171-6-4.ec2.internal.warc.gz | 28,181,649 | 18,899 | Reflection: Standards Alignment Combinations Using Multiplication Trees - Section 3: Try It On Your Own
This lesson provided students with the opportunity to familiarize themselves with the diagrams and the process of building a tree diagram. I chose not to spend many days of instruction on this process because it is not a grade level standard for third graders, but I did want to expose the students to the visual created so that they would not become frustrated if they encountered it on the state testing. Currently, this is not part of Common Core Standards. However, I felt it was important to include prior to the transition to the new PARCC assessment our state will be implementing.
Bridging Old Standards to New Standards
Standards Alignment: Bridging Old Standards to New Standards
Combinations Using Multiplication Trees
Unit 6: Multiplication 2
Lesson 2 of 13
Big Idea: Using a model helps students develop a conceptual understanding of multiplication.
Print Lesson
3 teachers like this lesson
Standards:
Subject(s):
Math, commutative property of multiplication, multiplication, multiplication models, word problems, area model
50 minutes
Diane Siekmann
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Environment: Urban | 371 | 1,813 | {"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} | 2.546875 | 3 | CC-MAIN-2016-44 | longest | en | 0.903447 |
http://www.facetsofcomplexity.de/monday/20180625-L-Beck.html | 1,586,027,171,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370524604.46/warc/CC-MAIN-20200404165658-20200404195658-00180.warc.gz | 232,958,722 | 4,999 | Springe direkt zu Inhalt
# Lecture by Matthias Beck: Binomial Inequalities of Chromatic, Flow, and Ehrhart Polynomials
Jun 25, 2018 | 04:00 PM
A famous and wide-open problem, going back to at least the early 1970's, concerns the classification of chromatic polynomials of graphs. Toward this classification problem, one may ask for necessary inequalities among the coefficients of a chromatic polynomial, and we contribute one such set of inequalities when a chromatic polynomial $\chi_G(n) = \chi^*_0 \binom {n+d} d + \chi^*_1 \binom {n+d-1} d + \dots + \chi^*_d \binom n d$ is written in terms of a binomial-coefficient basis. More precisely, we prove that $\chi^*_{d-2}+\chi^*_{d-3}+\dots+\chi^*_{d-j-1} \ \ge \ \chi^*_1+\chi^*_2+\dots+\chi^*_j$, for $1 \le j \le \lfloor \frac{ d }{ 2 } \rfloor - 1$. A similar result holds for flow polynomials enumerating either modular or integral nowhere-zero flows of a graph. Our theorems follow from connections among chromatic, flow, order, and Ehrhart polynomials, and the fact that the latter satisfy a decomposition formula into symmetric polynomials due to Stapledon.
(This is joint work with Emerson Le\'on.)
### Time & Location
Jun 25, 2018 | 04:00 PM
Technische Universität Berlin Institut für Mathematik Straße des 17. Juni 136 10623 Berlin room MA 041 (ground floor) | 378 | 1,327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-16 | longest | en | 0.710739 |
https://www.exactlywhatistime.com/days-before-date/april-25/29-days | 1,696,314,802,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00493.warc.gz | 822,960,744 | 5,712 | # What date is 29 days before Thursday April 25, 2024?
## Calculating 29 days before Thursday April 25, 2024 by hand
This page helps you figure out the date that is 29 days before Thursday April 25, 2024. We've made a calculator to find the date before a certain number of days before a specific date. In this example, we want to know the date 29 days before Thursday April 25, 2024.
Trying to do this in your head can be really hard and take a long time. An easier way is to use a calendar, either a paper one or an app on your phone or computer, to look at the days before the date you're interested in. But the best and quickest way to find the answer is by using our days before specific date calculator, which you can find here.
If you want to change the question on this page, you have two choices: you can change the URL in your browser's address bar, or go to our days before specific date calculator to type in a new question. Remember, figuring out these types of calculations in your head can be really tough, so we made this calculator to help make it much easier for you.
## Wednesday March 27, 2024 Stats
• Day of the week: Wednesday
• Month: March
• Day of the year: 87
## Counting 29 days backward from Thursday April 25, 2024
Counting backward from today, Wednesday March 27, 2024 is 29 before now using our current calendar. 29 days is equivalent to:
29 days is also 696 hours. Wednesday March 27, 2024 is 23% of the year completed.
## Within 29 days there are 696 hours, 41760 minutes, or 2505600 seconds
Wednesday Wednesday March 27, 2024 is day number 87 of the year. At that time, we will be 23% through 2024.
## In 29 days, the Average Person Spent...
• 6229.2 hours Sleeping
• 828.24 hours Eating and drinking
• 1357.2 hours Household activities
• 403.68 hours Housework
• 445.44 hours Food preparation and cleanup
• 139.2 hours Lawn and garden care
• 2436.0 hours Working and work-related activities
• 2241.12 hours Working
• 3667.92 hours Leisure and sports
• 1990.56 hours Watching television
## Famous Sporting and Music Events on March 27
• 1952 "Singin' in the Rain", musical comedy directed by Gene Kelly and Stanley Donen, starring Gene Kelly and Debbie Reynolds, premieres at Radio City Music Hall in NYC
• 1871 First international rugby union match - Scotland beats England, 1-0 at Raeburn Place, Edinburgh | 607 | 2,355 | {"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} | 3.53125 | 4 | CC-MAIN-2023-40 | latest | en | 0.920366 |
https://discussmain.com/q/531394 | 1,603,998,936,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905777.48/warc/CC-MAIN-20201029184716-20201029214716-00124.warc.gz | 276,636,081 | 2,442 | # How many it is necessary to pay for the one second m, the three fourth m, the five eighth m, the eleven sixteenth m of matter at the cost of 240 rubles for meter?
106
1) 240/2*1=120rubley 2) 240/4*3=180rubley 3) 240:8*5=150rubley 4) 240:16*11=165rubley1) 240/2*1=120rubley (for 0.5 m) 2) 240/4*3=180rubley (for 0.75 m) 3) 240:8*5=150rubley (for 0.625 m) 4) 240:16*11=165rubley (for 0.6875 m)
50 | 181 | 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} | 3.671875 | 4 | CC-MAIN-2020-45 | latest | en | 0.634526 |
https://quizizz.com/en-us/probability-and-statistics-worksheets-grade-12 | 1,716,329,596,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00844.warc.gz | 423,352,395 | 29,284 | ## Recommended Topics for you
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## Explore printable probability and statistics worksheets for 12th Grade
Probability and statistics worksheets for Grade 12 are essential resources for teachers who aim to provide their students with a comprehensive understanding of these mathematical concepts. These worksheets cover a wide range of topics, including probability distributions, statistical measures, hypothesis testing, and regression analysis, all of which are crucial for Grade 12 Math students. Teachers can utilize these worksheets to supplement their lesson plans, reinforce key concepts, and assess students' progress. By incorporating these resources into their curriculum, educators can ensure that their Grade 12 Math students have a solid foundation in probability and statistics, ultimately preparing them for success in future math courses and real-world applications.
Quizizz is an excellent platform for teachers to access a vast collection of probability and statistics worksheets for Grade 12, as well as other valuable resources. This interactive platform offers a variety of engaging quizzes, games, and activities that can be easily integrated into lesson plans, making learning fun and interactive for Grade 12 Math students. Teachers can also create their own quizzes and worksheets, tailoring the content to meet the specific needs of their students. Additionally, Quizizz provides valuable analytics and insights, allowing educators to track student progress and identify areas where additional support may be needed. By incorporating Quizizz into their teaching strategies, teachers can enhance their Grade 12 Math curriculum and provide their students with a dynamic and effective learning experience. | 678 | 3,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-22 | latest | en | 0.912447 |
https://mathoverflow.net/questions/278416/higher-genus-surfaces-that-look-like-genus-1-surfaces/278432 | 1,558,309,432,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255182.37/warc/CC-MAIN-20190519221616-20190520003616-00507.warc.gz | 560,976,979 | 30,481 | # Higher Genus Surfaces That “Look Like” Genus-1 Surfaces
It is my understanding that a genus-$g$ Riemann surface has $2g$ independent cycles that satisfy the usual intersection rules:
$$a_i \cap a_j = 0$$ $$b_i \cap b_j = 0$$ $$a_i \cap b_j = \delta_{ij}$$
If one would like to define a period matrix, one considers $g$ linearly independent holomorphic differentials $\omega_j$ that are normalized in such a way that:
$$\oint_{a_j} \omega_i = \delta_{ij}$$
Then, one can define a period matrix
$$\Omega_{ij} = \oint_{b_j} \omega_i$$
which will in general be a $g \times g$ matrix.
At this point, my question becomes a bit speculative: I am interested in Riemann surfaces that look or behave like genus-$1$ curves. My first question is:
1. Let us imagine a surface where all the period integrals are proportional to each other by some constant of proportionality, so in effect there is only one independent period, and all the information about the surface captured by the other cycles is obtained by simply multiplying by the appropriate constant. Is there a way to write down an "effective genus-$1$ curve" for this higher genus surface?
2. If someone hands me a period matrix, is there a way to reconstruct the curve it came from?
• Actually, thinking about it some more, they probably do exist e.g $\begin{pmatrix}i&\epsilon \\ \epsilon & i\end{pmatrix}$ ought to come from a Riemann surface. I don't understand the rest of the question, however. – Donu Arapura Aug 10 '17 at 16:54 | 388 | 1,496 | {"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} | 2.9375 | 3 | CC-MAIN-2019-22 | latest | en | 0.893175 |
https://www.r-bloggers.com/2016/11/earthquake-energy-over-time/ | 1,718,249,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00290.warc.gz | 866,652,183 | 30,430 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Disclaimer on all that follows – I am not an earthquake scientist and have cobbled together this post from sources like Wikipedia, official open data, and a range of information sites. There may be mistakes and misinterpretations that follow.
## Energy release from earthquakes is extremely variable
My last blog post left me interested in finding out more about energy generated by earthquakes and how it relates to magnitude. I was vaguely aware that magnitude was on a logarithmic scale so something is ten times the size for increase of one in magnitude, but it turns out that that is the shaking amplitude. Energy is going to be proportionate to $10^{1.5}$ to the power of magnitude, which means an increase in one magnitude point increases the energy released by about 31.6. Here’s an R function I cobbled together to convert earthquake magnitudes into rough equivalent megatons of TNT, a measure usually used to measure the oomph of nuclear weapons.
The pie chart that I had taken from the media and polished up in my last post, which showed how much of the earthquake energy in recent years came in a single day last weekend, had a slightly naughtily chosen starting point of 2010. Extending this a year back would have made the pie chart much more nuanced. But rather than battle it out with pie charts, I downloaded all the data on individual earthquakes (some of them back to well before 1800!) from the https://quakesearch.geonet.org.nz site, converted each magnitude to energy (ignoring for my purposes the nuances of different ways that magnitudes were measured and slightly different scales), and aggregated the total energy up by day. That delivered this revealing chart (note that dates are in Greenwich mean time, which is not how earthquakes’ dates are usually referred to locally of course):
Yes, the 2009 Dusky Sounds earthquake and last weekend’s Kaikoura earthquake dwarf everything back until the 1855 Wairarapa earthquake. It should be noted that the further back this chart goes, the more it underestimates total energy released, because only very major events have estimates and the energy released by aftershocks and day to day events were not recorded. Even in the instrument-measured era, the instruments have changed over time.
This is a good reminder that destructiveness is only weakly related to magnitude. Obviously, location, depth and timing are key factors. While the Canterbury earthquakes in 2010 and 2011 were low energy relative to the larger ones New Zealand has experienced, they were devasting to life and property.
## Maps should show the energy more
This exercise made me think about the maps which are commonly drawn of earthquakes, with the area of points representing the magnitude. In fact, to avoid misleading, it might be better for the area to represent on a non-logarithmic scale the shaking or the energy released. This would make maps look more like this one:
Big events like last Sunday night release so much energy that everything else looks like a pin prick in comparison.
## Some interesting historical patches
One thing I was interested in was the largest magnitude earthquake in any particular day, so I made this graphic:
Those are some interesting periods of time, where for a reasonable proportion of days the largest magnitude earthquake was not particularly large. So that prompted me to wonder if there was a connection between the number of earthquakes and their average and maximum magnitudes. There is in fact a relationship of that sort in the data, but it looks to be an artefact of the improving measurement - the better the monitors get, the more small earthquakes they pick up and record, which gives the impression that over time there are more and more earthquakes but on average they are getting smaller. Discovering this was one of the things that made me want to estimate energy and aggregate it, so the plethora of small low energy earthquakes don’t warp the overall picture. To mitigate the impact of improving measurement over time, I knock out all earthquakes of magnitude less than two before estimating total energy. I don’t think this would have much substantive impact.
Here is an interesting image of the number of earthquakes of magnitude two or higher:
That’s a very marked drop in 2012. So marked in fact that my first reaction was to think there might have been a change in how the measurement was happening. A quick google didn’t reveal what was going on there. Anyone who knows - whether it’s a bug in my code, or something more interesting - please leave a comment!
## Total energy on a logarithmic scale
While the first graphic in this post is good at showing the rarity of the really high energy earthquake events, it’s no good at showing trends over time. The simpler plots just above, looking at the biggest earthquake each day and the number of earthquakes greater than magnitude two each day, are not satisfactory either.
Putting daily total earthquake energy on a logarithmic scale is a good way to visually inspect for trends. At a medium time scale, we see again that drop in energy in early 2012, and an irregular cycle of spikes. The blue line in the next three graphics shows a 60 day moving average:
Here’s the code behind all that. As I’m not sure geonet would be thrilled at people copying my code and doing bulk downloads, I’ve saved a copy of the cut down data needed to reproduce this on my own site. The actual source code to download it is simple enough and is available in the branch of this blog’s repository on GitHub. | 1,140 | 5,652 | {"found_math": true, "script_math_tex": 1, "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 | 0.948966 |
https://www.diabloii.net/forums/threads/hi-got-a-question-about-soj-recipe.514218/ | 1,610,896,800,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513062.16/warc/CC-MAIN-20210117143625-20210117173625-00110.warc.gz | 750,968,609 | 11,722 | # hi, got a question about soj recipe
#### stoutewolf
##### Diabloii.Net Member
hi, got a question about soj recipe
hello there
i was looking at the arreat summit, and saw this recipe:
1 Perfect Skull + 1 Rare Item + Stone of Jordan = 1 High Quality New Rare Item of the same type
This formula has the possibility of creating a higher quality Rare Item than the following formula. ilvl = int(.66 * clvl) + int(.66 * ilvl) That is, the ilvl will be equal to .66 times your clvl (rounded down) plus .66 times the input item's ilvl (also rounded down).
this makes me dream about a crazy grand matron bow with cruel, grandmasters and 20ias, and a possible 2sockets on.
anyway, my question would be: what is the ideal charlevel/bow item level combo that just makes it possible to cube such bow, but low enough to weed out the mods that arent in that list.
currently playing singleplayer/LAN with a dropmod activated, so would like to know how to not waste useless time
#### krischan
That recipe is extremely expensive, but I guess you know this. In addition, rare items cannot get the Grandmaster's affix, you have to settle for Master's.
A GMB has a rather high base quality level (78) which results the affix level to be 2*x-99 with x being either the ilvl or the base qlvl, whichever is higher. That means, with the smallest value of x being 78, the affix level can't be lower than 2*78-99=59. That's just a little above the alvl of the cruel and master's affix, which are rather good news because you don't have to be that picky about the character level then. We can stop at this point because you can do the rerolling with a L1 mule because of that, but out of curiosity I will continue with the calculatiom
To reach that minimum alvl, 0.66*ilvl+0.66*charlvl must be 78 or smaller. In addition, you probably want to reroll the GMB endlessly without its ilvl becoming higher and higher. That means, with an ilvl 78 GMB (the highest you want it to have), you need a L39 or lower level character to let it turn into another one with ilvl 78 or less, resulting in an affix level 59 GMB all the time.
#### stoutewolf
##### Diabloii.Net Member
Oo interesting.
well it wont be that expensive because its a dropmod^^ but yes.
your answer is exactly the one i am looking for sir! thanks a ****ing lot
might be wrong though, lets say if i take an ilvl 90 gmb and roll it with a lvl 1 char all the time, the ilvl would approach 78 (or go under it, but that wouldnt matter for the affix level), correct?Oo
#### DarkElmo
##### Diabloii.Net Member
pay attention to the fact that it is a small possibilty - i have done this before - 4times, none have worked for a Good high Quality - - I did get a decent one once- but sadly - it wasnt my soj - - so good luck making with your own items.
#### krischan
Oo interesting.
well it wont be that expensive because its a dropmod^^ but yes.
your answer is exactly the one i am looking for sir! thanks a ****ing lot
might be wrong though, lets say if i take an ilvl 90 gmb and roll it with a lvl 1 char all the time, the ilvl would approach 78 (or go under it, but that wouldnt matter for the affix level), correct?Oo
Correct. After the first reroll, the new ilvl will be 0.66*1+0.66*90=59 (not 60, because of rounding).
The lyrics in my sig are quite fitting for what you are intending to do :azn: | 875 | 3,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.940809 |
https://www.gatevidyalay.com/tag/interest-variable-in-os/ | 1,656,430,995,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103556871.29/warc/CC-MAIN-20220628142305-20220628172305-00677.warc.gz | 816,286,074 | 20,688 | ## Process Synchronization-
Before you go through this article, make sure that you have gone through the previous article on Process Synchronization.
We have discussed-
• Process Synchronization provides a synchronization among the processes.
• Synchronization mechanisms allow the processes to access critical section in a synchronized manner.
• This avoids the inconsistent results.
## Interest Variable-
• Interest variable is a synchronization mechanism that provides synchronization among two processes.
• It uses an interest variable to provide the synchronization.
It is implemented as-
Initially, interest [0] and interest [1] are set to False.
• Interest value [0] = False means that process P0 is not interested to enter the critical section.
• Interest value [0] = True means that process P0 is interested to enter the critical section.
• Interest value [1] = False means that process P1 is not interested to enter the critical section.
• Interest value [1] = True means that process P1 is interested to enter the critical section.
## Working-
This synchronization mechanism works as explained in the following scenes-
## Scene-01:
• Process P0 arrives.
• It sets interest[0] = True.
• Now, it executes while loop condition- interest [1] == True.
• Since interest [1] is initialized to False, so it returns value 0 to the while loop.
• The while loop condition breaks.
• Process P0 enters the critical section and executes.
• Now, even if process P0 gets preempted in the middle, process P1 can not enter the critical section.
• Process P1 can not enter unless process P0 completes and sets the interest [0] = False.
## Scene-02:
• Process P1 arrives.
• It sets interest[1] = True.
• Now, it executes while loop condition- interest [0] == True.
• Process P0 has already shown its interest by setting interest [0] = True.
• So, it returns value 1 to the while loop.
• The while loop condition satisfies.
• The process P1 is trapped inside an infinite while loop.
• The while loop keeps the process P1 busy until the interest [0] value becomes False and its condition breaks.
## Scene-03:
• Process P0 comes out of the critical section and sets the interest [0] value to False.
• The while loop condition of process P1 breaks.
• Now, the process P1 waiting for the critical section enters the critical section and execute.
• Now, even if process P1 gets preempted in the middle, process P0 can not enter the critical section.
• Process P0 can not enter unless process P1 completes and sets the interest [1] = False.
Also Read- Criteria For Synchronization Mechanisms
## Characteristics-
The characteristics of this synchronization mechanism are-
• It ensures mutual exclusion.
• It does not follow strict alternation approach.
• It ensures progress since if a process is not interested to enter the critical section, it never stops the other process to enter the critical section.
• It is architectural neutral since it does not require any support from the operating system.
• It is a busy waiting solution which keeps the CPU busy when the process is actually waiting.
• Since it suffers from deadlock, it does not guarantee bounded waiting.
## How it suffers from deadlock?
• This synchronization mechanism may cause deadlock between the processes.
• Deadlock may occur through the following sequence of scenes-
## Scene-01:
• Process P0 arrives.
• It sets interest[0] = True.
• Now, it gets preempted and process P1 gets scheduled.
## Scene-02:
• Process P1 arrives.
• It sets interest[1] = True.
• Now, it gets preempted.
## Scene-03:
• Process P0 gets scheduled again.
• Now, it can not break the while loop condition- interest [1] == True since process P1 has shown its interest for executing critical section before its arrival.
• It keeps waiting in the infinite while loop for process P1 to complete its execution first.
## Scene-04:
• Later, Process P1 gets scheduled again.
• Now, it also can not break the while loop condition- interest [0] == True since process P0 is also interested for executing critical section.
• It keeps waiting in the infinite while loop for process P0 to complete its execution first.
Thus, both the processes are deadlocked.
To gain better understanding about Interest Variable,
Watch this Video Lecture
Next Article- Practice Problems On Synchronization Mechanisms
Get more notes and other study material of Operating System.
Watch video lectures by visiting our YouTube channel LearnVidFun | 952 | 4,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.870539 |
https://emaths.net/maths-simplify/rational-equations/factoring-trinomials.html | 1,656,772,305,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00146.warc.gz | 280,094,890 | 12,734 | Try the Free Math Solver or Scroll down to Tutorials!
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• inverse linear graphing | 976 | 4,152 | {"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-2022-27 | latest | en | 0.882599 |
https://homework.cpm.org/category/CC/textbook/ccg/chapter/1/lesson/1.1.5/problem/1-43 | 1,716,938,813,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059160.88/warc/CC-MAIN-20240528220007-20240529010007-00082.warc.gz | 239,412,503 | 15,570 | ### Home > CCG > Chapter 1 > Lesson 1.1.5 > Problem1-43
1-43.
Rosalinda examined the angles at right and wrote the equation below.
$(2x+1º)+(x-10º)=90º$
1. Does her equation make sense? If so, explain why her equation must be true. If it is not correct, determine what is incorrect and write the equation.
Is a right angle (as illustrated above), equal to $90º$?
Yes. This means Rosalinda's equation makes sense, because the two angles make up a $90º$ angle.
2. If you have not already done so, solve her equation, clearly showing all your steps. What are the measures of the two angles?
Simplify without the parentheses and combine like terms. Then solve for $x$. One of the angles is $x-10$. The other is $2x+1$. The sum of the two angles is $90$.
After solving for $x$, substitute your answer into the original equation to verify it is correct. | 228 | 857 | {"found_math": true, "script_math_tex": 8, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.907751 |
http://faculty.ycp.edu/~dhovemey/fall2013/cs340/assign/assign06.html | 1,511,521,089,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807650.44/warc/CC-MAIN-20171124104142-20171124124142-00674.warc.gz | 100,980,101 | 3,245 | Due: Friday, Nov 8th by 11:59 PM
## Getting Started
You can use tuProlog to write and test your Prolog code:
tuProlog.zip
## Boolean Functions
A boolean operator operates on one or more boolean values (true/false) to produce a boolean result. A binary boolean operator has two boolean operands.
We can describe a binary boolean operator using a truth table, which shows the result of the operator for each of the four possible combinations of operator values. For example, here is the truth table for the and operator, denoted ^:
L R L ^ R
t t t
t f f
f t f
f f f
For operands L and R, L ^ R is true when L and R are both true, false otherwise. Here are the truth tables for four additional binary boolean operators, or (denoted +), nand, nor, and xor.
L R L + R
t t t
t f t
f t t
f f f
L R L nand R
t t f
t f t
f t t
f f t
L R L nor R
t t f
t f f
f t f
f f t
L R L xor R
t t f
t f t
f t t
f f f
We can think of boolean operators as being mathematical functions with two parameters. By combining two boolean operators with two operands, we can create a boolean function with three operands. For example, A ^ (B + C) is a boolean function with the following truth table:
A B C A ^ (B + C)
t t t t
t t f t
t f t t
t f f f
f t t f
f t f f
f f t f
f f f f
### Synthesizing an arbitrary boolean function
In digital circuit design, it is sometimes helpful to be able to create an arbitrary boolean functions out of binary operators. For example, A op1 (B op2 C) is a "template" for a boolean function with three parameters. By substituting different operators for op1 and op2, we can create various functions.
Using Prolog, determine whether it is possible to synthesize the following binary functions by substituting the and, or, nand, nor, and xor operators in
A op1 (B op2 C)
If it is possible to synthesize a function, determine which operands to use for op1 and op2.
### Functions 1-3 (left to right)
A B C A op1 (B op2 C)
t t t t
t t f t
t f t t
t f f f
f t t f
f t f f
f f t f
f f f t
A B C A op1 (B op2 C)
t t t f
t t f f
t f t f
t f f f
f t t f
f t f t
f f t t
f f f t
A B C A op1 (B op2 C)
t t t t
t t f t
t f t t
t f f t
f t t f
f t f t
f f t t
f f f t
### Functions 4-6 (left to right)
A B C A op1 (B op2 C)
t t t t
t t f t
t f t f
t f f t
f t t t
f t f t
f f t t
f f f f
A B C A op1 (B op2 C)
t t t f
t t f f
t f t f
t f f f
f t t t
f t f f
f f t f
f f f t
A B C A op1 (B op2 C)
t t t f
t t f t
t f t t
t f f t
f t t t
f t f f
f f t f
f f f f
## Hints
Here is a suggestion for how to model the binary boolean operators:
```eval(and,t,t,t).
eval(and,t,f,f).
eval(and,f,t,f).
eval(and,f,f,f).
```
This establishes the ground truths for the and operator. The eval predicate specifies what the result of a given boolean operator is when given particular input values. A query with a free variable can be used to determine the result value for particular inputs: for example, for the query
```eval(and,t,f,What).
```
Prolog will infer that What is f.
The key will be finding a way to evaluate two boolean operators for three input values. Suggestion: define a predicate of the following form:
```composeTwo(X, Y, A, B, C, Z) :- something.
```
This predicate asserts that boolean operators X and Y, when evaluated on the expression
A X (B Y C)
will produce the result Z. This is useful for specifying one row of the synthesized function's truth table. For example:
```composeTwo(X, Y, t, t, t, f)
```
would assert that the result of the function is f when A =t, B =t, and C =t.
## Deliverables
There are two deliverables.
The first deliverable is a text file which specifies, for functions 1-6, either
• which operators op1 and op2 can be used to define the function, or
• that there is no way to substitute the 5 binary operators for op1 and op2 to define the function
The text file should also explain briefly how you used your Prolog code to find the operators (or prove their nonexistence) for each function.
The second deliverable is a text file containing your Prolog code.
## Submitting
Create a zip file with both deliverables described above and submit it to Marmoset as assign06:
https://cs.ycp.edu/marmoset/ | 1,262 | 4,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-47 | latest | en | 0.80198 |
https://metanumbers.com/200000000000000 | 1,606,480,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141191692.20/warc/CC-MAIN-20201127103102-20201127133102-00347.warc.gz | 368,443,774 | 8,090 | ## 200000000000000
200,000,000,000,000 (two hundred trillion) is an even fifteen-digits composite number following 199999999999999 and preceding 200000000000001. In scientific notation, it is written as 2 × 1014. The sum of its digits is 2. It has a total of 29 prime factors and 240 positive divisors. There are 80,000,000,000,000 positive integers (up to 200000000000000) that are relatively prime to 200000000000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 15
• Sum of Digits 2
• Digital Root 2
## Name
Short name 200 trillion two hundred trillion
## Notation
Scientific notation 2 × 1014 200 × 1012
## Prime Factorization of 200000000000000
Prime Factorization 215 × 514
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 29 Total number of prime factors rad(n) 10 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 200,000,000,000,000 is 215 × 514. Since it has a total of 29 prime factors, 200,000,000,000,000 is a composite number.
## Divisors of 200000000000000
240 divisors
Even divisors 225 15 15 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 240 Total number of the positive divisors of n σ(n) 4.99992e+14 Sum of all the positive divisors of n s(n) 2.99992e+14 Sum of the proper positive divisors of n A(n) 2.0833e+12 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.41421e+07 Returns the nth root of the product of n divisors H(n) 96.0015 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 200,000,000,000,000 can be divided by 240 positive divisors (out of which 225 are even, and 15 are odd). The sum of these divisors (counting 200,000,000,000,000) is 499,992,370,589,085, the average is 20,833,015,441,21.,187.
## Other Arithmetic Functions (n = 200000000000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 80000000000000 Total number of positive integers not greater than n that are coprime to n λ(n) 20000000000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6268932059626 Total number of primes less than or equal to n r2(n) 60 The number of ways n can be represented as the sum of 2 squares
There are 80,000,000,000,000 positive integers (less than 200,000,000,000,000) that are coprime with 200,000,000,000,000. And there are approximately 6,268,932,059,626 prime numbers less than or equal to 200,000,000,000,000.
## Divisibility of 200000000000000
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 4 0 2
The number 200,000,000,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Achilles
• Powerful
• Frugal
• Regular
## Base conversion (200000000000000)
Base System Value
2 Binary 101101011110011000100000111101001000000000000000
3 Ternary 222020010210221212122001120102
4 Quaternary 231132120200331020000000
5 Quinary 202203300000000000000
6 Senary 1545210422005301532
8 Octal 5536304075100000
10 Decimal 200000000000000
12 Duodecimal 1a52140b4048a8
20 Vigesimal jaca0000000
36 Base36 1yw6qc5ibk
## Basic calculations (n = 200000000000000)
### Multiplication
n×i
n×2 400000000000000 600000000000000 800000000000000 1000000000000000
### Division
ni
n⁄2 1e+14 6.66667e+13 5e+13 4e+13
### Exponentiation
ni
n2 40000000000000000000000000000 8000000000000000000000000000000000000000000 1600000000000000000000000000000000000000000000000000000000 320000000000000000000000000000000000000000000000000000000000000000000000
### Nth Root
i√n
2√n 1.41421e+07 58480.4 3760.6 724.78
## 200000000000000 as geometric shapes
### Circle
Diameter 4e+14 1.25664e+15 1.25664e+29
### Sphere
Volume 3.35103e+43 5.02655e+29 1.25664e+15
### Square
Length = n
Perimeter 8e+14 4e+28 2.82843e+14
### Cube
Length = n
Surface area 2.4e+29 8e+42 3.4641e+14
### Equilateral Triangle
Length = n
Perimeter 6e+14 1.73205e+28 1.73205e+14
### Triangular Pyramid
Length = n
Surface area 6.9282e+28 9.42809e+41 1.63299e+14 | 1,516 | 4,361 | {"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} | 3.703125 | 4 | CC-MAIN-2020-50 | longest | en | 0.797493 |
https://indiana-acreage-for-rent.com/qa/how-do-i-figure-out-sales-tax.html | 1,627,373,062,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00196.warc.gz | 309,193,307 | 8,866 | # How Do I Figure Out Sales Tax?
## How do I figure out sales tax rate?
To calculate the sales tax that is included in a company’s receipts, divide the total amount received (for the items that are subject to sales tax) by “1 + the sales tax rate”.
In other words, if the sales tax rate is 6%, divide the sales taxable receipts by 1.06..
## How much tax is on a dollar bill?
Yes you read that right: 70 cents of a dollar earned was paid out in tax to the IRS. Today the top tax rate is 39.6%. But you have to earn over \$415,000 in taxable income before the first dollar of your income is taxed at that 39.6% (marginal) rate.
## How do you take the tax off a total?
The formula for GST calculation:Add GST: GST Amount = (Original Cost x GST%)/100. Net Price = Original Cost + GST Amount.Remove GST: GST Amount = Original Cost – [Original Cost x {100/(100+GST%)}] Net Price = Original Cost – GST Amount.
## How do I calculate 7.5 sales tax?
Calculating Total Cost. Multiply the cost of an item or service by the sales tax in order to find out the total cost. The equation looks like this: Item or service cost x sales tax (in decimal form) = total sales tax.
## What items are excluded from sales tax?
Some items are exempt from sales and use tax, including:Sales of certain food products for human consumption (many groceries)Sales to the U.S. Government.Sales of prescription medicine and certain medical devices.Sales of items paid for with food stamps.
## Is it illegal to charge more sales tax?
Overcharging sales tax can lead to civil penalties. The punishment for overcharging varies from state to state, and it is severe for companies that intentionally overcharge sales tax with the intention of keeping it.
## How do I calculate no tax?
The Excel sales tax decalculator works by using a formula that takes the following steps:Step 1: take the total price and divide it by one plus the tax rate.Step 2: multiply the result from step one by the tax rate to get the dollars of tax.Step 3: subtract the dollars of tax from step 2 from the total price.More items…
## How do I calculate gross sales tax?
1 Expert Answer. Total sold (not including tax) times 0.09 = Total Sales Tax. Therefore divide the known sales tax (\$3565.11) amount by 0.09 and you get the Total Sales (pretax). The Gross receipts is the sum of Total sales (pretax) + Total sales tax.
## What is the tax on \$75?
A tax of 7.5 percent was added to the product to make it equal to 80.625. So, divide 7.5 by 100 to get 0.075. Divide the final amount by the value above to find the original amount before the tax was added. In this example: 80.63 / 1.075 = 75.
## How much would tax be on 100 dollars?
Subtract the base price from the total price to get the sales tax amount. So if the before tax price is \$100 and the after tax price is \$106.88, the sales tax amount would be \$6.88 (\$106.88 – \$100 = \$6.88). Divide the sales tax amount by the before tax price.
## How do you calculate tax backwards?
How to Calculate Sales Tax Backwards From TotalSubtract the Tax Paid From the Total. … Divide the Tax Paid by the Pre-Tax Price. … Convert the Tax Rate to a Percentage. … Add 100 Percent to the Tax Rate. … Convert the Total Percentage to Decimal Form. … Divide the Post-Tax Price by the Decimal. … Subtract the Pre-Tax Price From Post-Tax Price.Apr 28, 2020
## How do you find the tax rate in math?
Correct answer: To find the amount of sales tax, take the difference in the total before and after tax and divide by the price before tax. This gives 0.08 or 8%.
## What is the tax on a \$50 pair of shoes if the tax rate is 6 %?
Step-by-step explanation: If the tax rate is 6%, we need to find 6% of \$50. 6% is equal to 0.06 so we can find the answer by multiplying 0.06 and 50.
## How do I figure out the tax on a total amount?
You can simply calculate the tax under GST by applying the standard 18% rate. For instance, if you sell goods or services for Rs 1000, then the net price will be Rs 1000 + 18% of 1000 (GST) = 1000 + 180 = Rs 1180.
## What is the tax when you buy something?
7.25 percentThe tax rate charged will vary across California and depends upon where the item is bought, or will be used. The statewide sales and use tax rate in California is currently 7.25 percent, but in many areas, voters approved district taxes to fund local or regional projects and services.
## How do I figure out tax percentage?
Divide taxes paid by net profit to calculate the effective tax rate percentage. In the example, \$35,000 divided by \$100,000 equals an effective tax rate of 0.35 or 35 percent. | 1,151 | 4,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-31 | latest | en | 0.953198 |
https://networkingfunda.com/maximum-factors-problem-codechef-solution/ | 1,653,814,297,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00388.warc.gz | 460,879,989 | 21,549 | # Maximum Factors Problem CodeChef Solution – Code Drive Solution
### About Maximum Factors Problem CodeChef Solution
• This is a coding contest based on algorithms, data structures, and problem-solving.
• Organizer: The contest is hosted by NIT Trichy.
• Prizes: NA
• Registrations for prizes: NA
### Problem: Maximum Factors Problem CodeChef Solution
You are given an integer NN. Let KK be a divisor of NN of your choice such that K>1K>1, and let M=NKM=NK. You need to find the smallest KK such that MM has as many divisors as possible.
Note: UU is a divisor of VV if VV is divisible by UU.
### Input Format
• The first line of the input contains an integer TT – the number of test cases. The test cases then follow.
• The only line of each test case contains an integer NN.
### Output Format
For each test case, output in a single line minimum value of KK such that MM has as many divisors as possible.
### Constraints
• 1≤T≤30001≤T≤3000
• 2≤N≤1092≤N≤109
### Sample Input 1
``````3
3
4
6
``````
### Sample Output 1
``````3
2
2
``````
### Explanation
• Test case 11: The only possible value for KK is 33, and that is the answer.
• Test case 22: There are two cases:
• K=2K=2. Then M=42=2M=42=2, which has 22 divisors (11 and 22).
• K=4K=4. Then M=44=1M=44=1, which has 11 divisor (11).
• Test case 33: There are three cases:
• K=2K=2. Then M=62=3M=62=3, which has 22 divisors (11 and 33).
• K=3K=3. Then M=63=2M=63=2, which has 22 divisors (11 and 22).
• K=6K=6. Then M=66=1M=66=1, which has 11 divisor (11).
#### Contest Details:
• This is an External Rated Contest.
• Duration: 3 Hours
• Start time: 30th December 2021, 20:00 hrs IST
• End time: 30th December 2021, 23:00 hrs IST
You may check your timezone here.
• This contest is rated only for Division 2 and Division 3 users. Division 1 users can participate unofficially in this contest.
Get More CodeChef Solution >>
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error: Content is protected !! | 670 | 2,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-21 | latest | en | 0.586545 |
https://web2.0calc.com/questions/triangle_141 | 1,716,221,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00268.warc.gz | 544,044,600 | 5,684 | +0
# triangle
+1
57
2
Help Im bad at geometry
In triangle ABC, let the angle bisectors be BY and CZ. Given AB = 16, AY = 16, and CY = 16, find BC and BZ.
Jun 27, 2023
### 2+0 Answers
#1
+274
-1
https://web2.0calc.com/questions/angle-bisectors_27
I think you should use the angle bisector theorem and start from there.
Jun 27, 2023
#2
+131
+1
Let BC and BZ both equal x. The angle bisector theorem provides us with:
AC/AB = CZ/BZ
16/(16+x) = 16/y
We solve y in terms of x, and as a result,
y = 16(16+x)/16 = x+16
Upon substitution, we discover—
16/(16+x) = 16/(x+16)
If we cross-multiply, we obtain:
16(x+16) = 16(16+x)
Simplifying, we get:
16x + 256 = 256 + 16x
In light of the fact that 16x cancels on both sides, we are left with:
256 = 256
As a result, there exist an endless number of BC and BZ values that could satisfy the requirements.
Jun 27, 2023 | 307 | 880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-22 | latest | en | 0.85647 |
http://mywoodworkingplansandprojects.com/teds-woodworking-table-furniture-plans.html | 1,553,161,743,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202510.47/warc/CC-MAIN-20190321092320-20190321114320-00289.warc.gz | 140,091,252 | 12,904 | ```How do you divide 11-3/8-in. (or any other mathematically difficult number) into equal parts without dividing fractions? Simple. Angle your tape across the workpiece until it reads an easily-divisible dimension and make your marks with the tape angled. For example, say you want to divide an 11-3/8-in. board into three equal parts. Angle the tape until it reads 12-in., and then make marks at “4” and “8”. Plus: More measuring tips and tricks.
```
Along with stone, clay and animal parts, wood was one of the first materials worked by early humans. Microwear analysis of the Mousterian stone tools used by the Neanderthals show that many were used to work wood. The development of civilization was closely tied to the development of increasingly greater degrees of skill in working these materials.
Pocket screw joinery is a system—employing special drill and driver bits—used to join boards or pieces of plywood to one another. Installing pocket screws involves using a jig to drill a sharply angled, 15-degree hole through the back of one board, then driving a special screw through that hole into the second board to draw them tightly together. Learn how to use pocket screws here. It’s often used in cabinetmaking and furniture building. Here’s how to build cabinets with pocket screws. The term “pocket” comes from the design of the hole which contains an upper “pocket” for the head of the screw to push against; this pocket also hides the head of the screw.
###### Examples of Bronze Age wood-carving include tree trunks worked into coffins from northern Germany and Denmark and wooden folding-chairs. The site of Fellbach-Schmieden in Germany has provided fine examples of wooden animal statues from the Iron Age. Wooden idols from the La Tène period are known from a sanctuary at the source of the Seine in France.
Luckily, we have also managed to find a detailed video tutorial of the Barn door project that illustrates the process of building a Barn door of your own. The steps and instructions in the video tutorial are different from the source links listed above. Actually, you can make different types of designs for your Barn door depending on which one you can afford easily and DIY on your own.
Commonly used woodworking tools included axes, adzes, chisels, pull saws, and bow drills. Mortise and tenon joints are attested from the earliest Predynastic period. These joints were strengthened using pegs, dowels and leather or cord lashings. Animal glue came to be used only in the New Kingdom period.[3] Ancient Egyptians invented the art of veneering and used varnishes for finishing, though the composition of these varnishes is unknown. Although different native acacias were used, as was the wood from the local sycamore and tamarisk trees, deforestation in the Nile valley resulted in the need for the importation of wood, notably cedar, but also Aleppo pine, boxwood and oak, starting from the Second Dynasty.[4]
Fir, also known as Douglas Fir, is very inexpensive and common at local home centers. It has a characteristic straight, pronounced grain with a red-brown tint. However, its grain pattern is relatively plain and it does not stain well, so Fir is commonly used when the finished product will be painted. While commonly used for building, this softwood would also be suitable for furniture-making as well.[12]
A board is considered “quarter-sawn” when the growth rings run, more or less, perpendicular to the face of the board. Quarter-sawn boards generally have straight grain and are less prone to shrinkage, compared to other boards. These factors don’t come into play with the 2x4s you use to frame a closet—but it does with the shelves and cabinetry you put into that closet; you want those boards to remain straight, flat and stable.
Woodworking was essential to the Romans. It provided, sometimes the only, material for buildings, transportation, tools, and household items. Wood also provided pipes, dye, waterproofing materials, and energy for heat.[5]:1Although most examples of Roman woodworking have been lost,[5]:2 the literary record preserved much of the contemporary knowledge. Vitruvius dedicates an entire chapter of his De architectura to timber, preserving many details.[6] Pliny, while not a botanist, dedicated six books of his Natural History to trees and woody plants, providing a wealth of information on trees and their uses.[7]
Commonly used woodworking tools included axes, adzes, chisels, pull saws, and bow drills. Mortise and tenon joints are attested from the earliest Predynastic period. These joints were strengthened using pegs, dowels and leather or cord lashings. Animal glue came to be used only in the New Kingdom period.[3] Ancient Egyptians invented the art of veneering and used varnishes for finishing, though the composition of these varnishes is unknown. Although different native acacias were used, as was the wood from the local sycamore and tamarisk trees, deforestation in the Nile valley resulted in the need for the importation of wood, notably cedar, but also Aleppo pine, boxwood and oak, starting from the Second Dynasty.[4]
As you can see in the image, this shelf goes on both sides of the corner wall. It looks beautiful and can be used to organize books, trophies, pictures frames and many other things. The strength and design of the shelf depends on how properly you build it. First time workers definitely need some guidance to help them with the process. Therefore, I am including this basic video that I found on YouTube that demonstrates the process of making corner wall wooden shelves.
Finally, properly preparing your wood surface is super important. It will make a huge difference when it comes times to stain, paint, or finish your wood. There are a lot of tips for wood surface preparation, but sanding the wood is one of the most important steps. And I find it really helpful to do the bulk of my sanding before I start ripping (cutting) and building with my wood since it’s still in whole pieces. You can check out my simple tips for how to sand wood in my how to stain wood tutorial, which is also good to reference if you need to learn how to stain too!
I know this seems really simple and you may already know how to read a tape measure so just hear me out on this one! Often times with woodworking you need to make exact measurements and cuts and it’s rarely pretty even numbers like 15 inches or 15 1/2 inches. It’s usually like 15 5/8 inches or 15 9/16 inches. So, really knowing how to read a tape measure in its entirety is important. And I’ve created a quick post on how to read a tape measure the easy way along with a helpful free printable.
And the fact is that you can make your own patio chair with several old but still good pallets. Here we are providing a tutorial that everybody can follow easily – it is very well-written and also self-explanatory, which is great for those who are a beginner at woodworking and have never completed a DIY project before. As you don’t need to be a professional woodworker or a handyman to complete this project, so it is not a difficult task – all you need is a bit of determination!
Old doors laid across sawhorses make great temporary workbenches, but they take up a lot of space when you’re not using them. Instead of full-size doors, I use bifold doors with hinges so I can fold them up when I’m done working. They’re also easier to haul around in the pickup for on-the-road jobs. — Harry Steele. Here are 12 more simple workbenches you can build.
With an orbital sander and good sandpaper you can smooth wood evenly and easily with first-class results. When flush-sanding solid edge-banding, draw a squiggly line across the joint before sanding. The edge-banding will be slightly proud of the plywood veneer, so the pencil marks provide a visual aid to make sure that you’re sanding flat, and that you don’t sand through the plywood’s veneer. As you go, you can also test for a smooth, level transition by gently scraping your fingernails against the transition. If it’s smooth, your fingers will not catch on the seam between the two pieces
A woodworker's Workbench is a special type of bench designed to hold your work when you are working on a wood project. The main purpose of this table is to keep the woodwork steady and to prevent it from moving. Follow the tutorial below to build yourself a nice workbench suitable for your specific woodworking projects. Make sure to modify the table to fit your specific requirements.
Finding a toolbox for a mechanic, for his hand tools, is not a big challenge at all - there are dozens of the tool boxes available on the market, from huge roll-around shop cases to small metal boxes. Plumbers, electricians, and farmers are well served, too, with everything from pickup-truck storage to toolboxes and belts. But, if you are a shop-bound woodworker then the case changes. You get to need a tool box that suits the range and variety of hand tools that most woodworkers like to have. For those who deny making do with second best, there's only one solution, you’ve to build a wooden toolbox that should be designed expressly for a woodworking shop.
Finding a toolbox for a mechanic, for his hand tools, is not a big challenge at all - there are dozens of the tool boxes available on the market, from huge roll-around shop cases to small metal boxes. Plumbers, electricians, and farmers are well served, too, with everything from pickup-truck storage to toolboxes and belts. But, if you are a shop-bound woodworker then the case changes. You get to need a tool box that suits the range and variety of hand tools that most woodworkers like to have. For those who deny making do with second best, there's only one solution, you’ve to build a wooden toolbox that should be designed expressly for a woodworking shop.
Short scraps of hardwood are too good to throw away but hard to store neatly. So I bought a 4-ft. tube form made for concrete footings, cut it in half (the cardboard-like material cuts easily) and set the tubes on end. I tack the tubes to a wall or a bench leg so they don’t fall over. With the wood scraps stored upright, it’s easy to find a piece just the right length. Tube forms are available in various diameters for \$5 and up at home centers. — Bill Wells. Here’s another brilliant use for these concrete forms.
Another awesome thing about this coffee table is that it is also has a storage unit. So you can store drinks, and other stuff in the half barrel of your table and then close or open it whenever you need. Pete has also constructed a video for this tutorial for which you can find the link below. It illustrates the same process in a video guide that shows you the exact process to be followed while building this whiskey barrel coffee table.
Sanding curves is tricky. Sometimes you need a sanding pad that’s both firm and flexible. A small notepad works great. Just wrap sandpaper around the pad and bend the pad to whatever arc you need. Slip the one end of the sandpaper between the pages to help hold it in place on the pad. Give this a try the next time you’re working on a project that has curves and tough to reach spots.
With an orbital sander and good sandpaper you can smooth wood evenly and easily with first-class results. When flush-sanding solid edge-banding, draw a squiggly line across the joint before sanding. The edge-banding will be slightly proud of the plywood veneer, so the pencil marks provide a visual aid to make sure that you’re sanding flat, and that you don’t sand through the plywood’s veneer. As you go, you can also test for a smooth, level transition by gently scraping your fingernails against the transition. If it’s smooth, your fingers will not catch on the seam between the two pieces
KHIEM NGUYEN (Age 28, Austin, TX): Inspired by midcentury modern and Japanese design, Nguyen is a true craftsman. His passion for crafting began with photography and led to him becoming an open major in art school so he could "get (his) hands into everything." After college, Nguyen and his fiancée moved to Austin, where they began A & K Woodworking & Design, specializing in furniture and wood crafts.
```When you are gathering inspiration for barn door Plan, be sure to note the cost of the tools used in the plan. Barn door tools can often cost more than your actual door! But, there are many clever and affordable do it yourself tools options in the tutorials mentioned below! Let us explore some DIY Barn Door Tutorials. Just click on the blue text below and check some amazing fun Barn doors. They might be different from the one shown in above picture.
```
Woodworking was essential to the Romans. It provided, sometimes the only, material for buildings, transportation, tools, and household items. Wood also provided pipes, dye, waterproofing materials, and energy for heat.[5]:1Although most examples of Roman woodworking have been lost,[5]:2 the literary record preserved much of the contemporary knowledge. Vitruvius dedicates an entire chapter of his De architectura to timber, preserving many details.[6] Pliny, while not a botanist, dedicated six books of his Natural History to trees and woody plants, providing a wealth of information on trees and their uses.[7]
With two varieties, red and white, oak is known to be easy to work with and relatively strong. However, furniture makers often opt for white oak over red oak for its attractive figure and moisture-resistance.[12] Depending on the kind needed, oak can probably be found at a local home center or a lumberyard for a bit pricier than other hardwoods.[12][13]
Cutting a miter joint that closes up perfectly and maintains a 90 degree angle is really satisfying. Unfortunately, it doesn’t always happen. Here’s a quick fix for a slightly open miter joint; rub the shank of a screwdriver along the miter at a steep angle, from both sides of the joint. Chances are, you’ll be the only one that knows it wasn’t perfect to begin with! Try this amazing miter project!
Sanding small items is tricky, as they’re hard to clamp in a vise to work on them. So instead of bringing the sandpaper to the workpiece, I bring the workpiece to the sandpaper. I glue sheets of sandpaper to a piece of plywood; 60 and 100-grit on one side and 150 and 220-grit on the other. Spray adhesive works well for this. Since there’s sandpaper on both sides, my sanding board doesn’t slide around on the bench. Check out these small projects!
Woodworking was essential to the Romans. It provided, sometimes the only, material for buildings, transportation, tools, and household items. Wood also provided pipes, dye, waterproofing materials, and energy for heat.[5]:1Although most examples of Roman woodworking have been lost,[5]:2 the literary record preserved much of the contemporary knowledge. Vitruvius dedicates an entire chapter of his De architectura to timber, preserving many details.[6] Pliny, while not a botanist, dedicated six books of his Natural History to trees and woody plants, providing a wealth of information on trees and their uses.[7]
Not long ago, I needed to make some angled wood parts to build a new soffit on my garage. I didn’t have the customary tool for the job, but I had some steel joining plates. I screwed through one of the holes in the plate, set my angle, then added another screw to lock the angle. I could then use it as a template to mark all the pieces at the same angle and cut them with my circular saw. — Ryan Bartsch
```From the source tutorial, you can get illustrates to the instruction about the plan. Everything is fairly described as diagrams, images, the list of supplies and tools need etc. The process to this plan is very easy to understand and follow for if you are having some basic woodworking knowledge. Make sure to collect all the supplies you need before you start with the project. You may even ask any question directly in the comment section of the tutorial post and also comment the images of your final product if you have completed it. Either way, I hope that you will manage to build this one nicely.
```
The plan tutorial includes images, diagrams, step-by-step instructions, and even a video to help you along the way. You can also go with some more bookcase design ideas. Browse the internet for more and we are also proving a link below to some more ideas to this plan. Select and build one of these free bookcase DIYs and you will have everything available easily that you need to get started creating a bookshelf for any room in your house.
\$50 - \$100Alaska Dream HouseBathroomBedroomChildren's and Kid's Room Furniture and Toy PlansCraftroomDesk, Desk Systems and Project Table Plansdining roomentry wayFarmhouse Style Furniture PlansIndustrial Style Furniture PlansKids and Toysliving roomModern Style Furniture PlansNursery and BabyofficeRustic Furniture Plansstorage and organizationTeensIntermediateSide and End Table PlansBuffet, Sideboard and Credenza PlansCabinet PlansNightstands
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Often times at first glance a board looks straight and the fact that it is actually bowed or has some warping isn’t always obvious. So the trick to knowing for sure, is to hold the board up towards your face, with the other end on the ground, and look at it at a downward angle (as shown in the below photo). This method will allow you to see if it is bowing at all.
Another wooden item that I love very much is a beautiful mobile holder. You can see one in the image below. These things are not only beautiful, but they can comfortably hold any sized mobile and ensure proper safety. Another amazing thing is that they can be built in many shapes and sizes, as and how you need it. You can see some more examples at the source below
Lacking a jointer? Use reader Court Kites’ awesome tip to create perfectly matched glue joints on wavy or bowed board edges. Lay the boards on a flat surface, then clamp them across the middle with a bar clamp. Lay two 8-in. long by 1-3/4 in. wide scrap boards across each end and screw them in with four 1-1/4 in. long screws, two per board. Keep the screws well away from your future cutting line!
```Particleboard is a manufactured wood product composed of sawdust, wood chips or wood shavings mixed with a resin. This concoction is layered, compressed, subjected to heat and cut to shape, resulting in a sheet material that can be used for a variety of things. It’s often used as shelving or as an underlayment for carpet. Plastic laminate may be applied to both sides to create a product that can be used to create everything from furniture to cabinets to wall paneling. Head into IKEA and you’ll find acres of particleboard.
```
```Hi Liz. I am sorry you were having trouble unsubscribing to the newsletter. Your information has been updated. Your email address has been removed from our Woodworkers Guild of America E-Newsletter mailing list. Please allow up to 5 business days for your request to be completed. If you have any questions, please contact our customer service team at 1-855-253-0822.
```
I got this idea from a Pinterest post. The final product looks so beautiful that I just couldn’t wait to make one for myself. This was somewhat a different experience from my other regular DIY projects as it doesn’t involve making something from scratch, but turning an existing wood piece into another one. Nonetheless, I enjoyed it very much and the final product was very satisfying. The tutorial I used is linked below.
Finding affordable lumber has always been a mainstay for woodworkers, and when you tie our dwindling natural resources into the conversation the time is right to look at milling your own lumber. This seven-part weekly video series takes you through how to find lumber, how to operate a sawmill, details on types of sawing methods, stickering and drying and ultimately advice on using a mill as part of a business. Learn what you need to know to understand Milling Your Own Lumber.
Isn’t this amazing that instead of having to throw that old furniture piece away, you can now reuse it to build something even more beautiful? If you do not like this particular idea, there are many other re-purposeful furniture items you can build from an old dresser. Just search the internet for other DIY project ideas. Here is a link to the video tutorial that explains the same procedure in a more practical manner that you can easily follow through.
I’m 91 years old, but I still enjoy spending time in the wood shop. I like to make wooden toys and give them to my great-grandkids and charity groups. One trick I’ve learned over the years is to use emery boards—the kind for filing fingernails— to sand small parts. Emery boards come in different sizes, and some are more abrasive than others, so I keep an assortment on hand. — Joe Aboussleman
Another important factor to be considered is the durability of the wood, especially in regards to moisture. If the finished project will be exposed to moisture (e.g. outdoor projects) or high humidity or condensation (e.g. in kitchens or bathrooms), then the wood needs to be especially durable in order to prevent rot. Because of their oily qualities, many tropical hardwoods such as teak and mahogany are popular for such applications.[9]
Whether you're new to woodworking or you've been doing it for years, Woodcraft's selection of woodworking projects is one the best places to find your next big project. Whether you're looking to make wooden furniture, pens, toys, jewelry boxes, or any other project in between, the avid woodworker is sure to find his or her next masterpiece here. Find hundreds of detailed woodworking plans with highly accurate illustrations, instructions, and dimensions. Be sure to check out our Make Something blog to learn expert insights and inspiration for your next woodworking project.
Hardwoods are separated into two categories, temperate and tropical hardwoods, depending on their origin. Temperate hardwoods are found in the regions between the tropics and poles, and are of particular interest to wood workers for their cost-effective aesthetic appeal and sustainable sources.[9] Tropical hardwoods are found within the equatorial belt, including Africa, Asia, and South America. Hardwoods flaunt a higher density, around 65lb/cu ft as a result of slower growing rates and is more stable when drying.[9] As a result of its high density, hardwoods are typically heavier than softwoods but can also be more brittle.[9] While there are an abundant number of hardwood species, only 200 are common enough and pliable enough to be used for woodworking.[11] Hardwoods have a wide variety of properties, making it easy to find a hardwood to suit nearly any purpose, but they are especially suitable for outdoor use due to their strength and resilience to rot and decay.[9] The coloring of hardwoods ranges from light to very dark, making it especially versatile for aesthetic purposes. However, because hardwoods are more closely grained, they are typically harder to work than softwoods. They are also harder to acquire in the United States and, as a result, are more expensive.[9]
From the source tutorial, you can get illustrates to the instruction about the plan. Everything is fairly described as diagrams, images, the list of supplies and tools need etc. The process to this plan is very easy to understand and follow for if you are having some basic woodworking knowledge. Make sure to collect all the supplies you need before you start with the project. You may even ask any question directly in the comment section of the tutorial post and also comment the images of your final product if you have completed it. Either way, I hope that you will manage to build this one nicely.
Finding affordable lumber has always been a mainstay for woodworkers, and when you tie our dwindling natural resources into the conversation the time is right to look at milling your own lumber. This seven-part weekly video series takes you through how to find lumber, how to operate a sawmill, details on types of sawing methods, stickering and drying and ultimately advice on using a mill as part of a business. Learn what you need to know to understand Milling Your Own Lumber.
```A small oak table is a very useful wooden item for every household. You can yourself make a nice, strong and beautiful oak table suitable for any purpose. See the image below. As you can see, it is a small, yet good enough table to be used as a coffee table, lamp stand, breakfast table, etc. You can also find many other design variants on the internet. Choose the one you want for yourself and start making it now.
```
We have a small dining room area in our farmhouse that is separate from the living room and kitchen. The area is much smaller in space than our last house. I was little confused that our typical rectangular farmhouse table was not going to cut it. So, I walked in I came to know that we needed to build a round dining table. So, I searched for a plan design idea and build a very own round farmhouse dining table. I was an amazing DIY plan, I just love it!
Hardwoods, botanically known as angiosperms, are deciduous and shed their leaves annually with temperature changes.[8] Softwoods come from trees botanically known as gymnosperms, which are coniferous, cone-bearing, and stay green year round.[8] Although a general pattern, softwoods are not necessarily always “softer” than hardwoods, and vice versa.[9] | 5,541 | 25,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-13 | latest | en | 0.954557 |
https://ebrary.net/135732/engineering/ultimate_compressive_resistance_static_load_tests | 1,642,895,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00111.warc.gz | 287,050,733 | 8,726 | # Ultimate compressive resistance from static load tests
If static load tests are performed on several piles with similar characteristics, the characteristic compressive resistance may be computed as:
where £, and £2 are correlation factors applied to the mean and the lowest of the measured resistances, Rc;m, respectively. The objective of these correlation factors is to introduce the variability of the ground conditions into the analysis. Their recommended values are presented in Table 7.11.
# Ultimate compressive resistance from ground test results
Two approaches may be adopted: the “model pile” and the “alternative” or “ground model” approach. In the “model pile” approach, a pile compressive resistance is calculated for each
Table 7.11 Correlation factors £ to derive characteristic values from static pile load tests (EN 1997-1:2004).
/ 2 3 4 >5 1.40 1.3 1.2 1.1 1.00 1.40 1.2 1.05 1 1.00
n = number of tests
test profile, using a semi-empirical method based on in situ or laboratory test results. After calculating one resistance, Rc.cah per profile, the characteristic value is computed as:
where £s and are correlation factors applied to the mean and the lowest of the calculated resistances, Rc.cai, respectively, and depend on the number of test profiles, n, and are applied, respectively:
• to the mean values:
• to the lowest values:
The values of the proposed correlation factors are given in Table 7.12.
In the “alternative” or “ground model” procedure, the characteristic values of the base and shaft resistances are computed, based on ground parameters:
Note that, for calculation of the characteristic values of the unit base and shaft resistances, no partial factor is applied to the ground parameters (except for Design Approach 3) and no correlation factors £ are applied to the resistances. In this “ground model” alternative procedure, a model factor, yRd, greater than 1, defined in the National Annex, is used to
Table 7.12 Correlation factors £ to derive characteristic values from ground test results (EN 1997-1:2004).
/ 2 3 4 5 7 10 1.40 1.35 1.33 1.31 1.29 1.27 1.25 1.40 1.27 1.23 1.2 1.15 1.12 1.08
n = number of test profiles increase the partial factors for base and shaft resistances. In this case, Equation 7.100 can be rewritten as:
# Ultimate compressive resistance from dynamic impact tests
When dynamic impact tests are used to determine pile resistance, the characteristic value is given by:
The correlation factors £s and £6 are listed in Table 7.13.
Table 7.13 Correlation factors g to derive characteristic values from dynamic impact tests (EN 1997-1:2004).
>2 >5 >10 >15 >20 1.60 1.50 1.45 1.42 1.40 1.50 1.35 1.30 1.25 1.25
• * n = number of tested piles
• 1 The ^-values in the table are valid for dynamic impact tests.
b The ^-values may be multiplied by a model factor of 0.8S, when using dynamic impact tests with signal matching.
c The ^-values should be multiplied by a model factor of 1.10, when using a pile driving formula with measurements of the quasi-elastic pile head displacement during the impact.
d The ^-values should be multiplied by a model factor of 1.20, when using a pile driving formula without measurements of the quasi-elastic pile head displacement during the impact.
' If different piles exist in the foundation, groups of similar piles should be considered separately when selecting the number n of test piles.
## Load and Resistance-Factored Design (LRFD)
The Load and Resistance-Factored Design has been presented in Chapter 3. The verification of safety is performed under the following conditions:
where Pu is the factored normal load, computed as the sum of factored overall load effects, for a given load combination, Ф is the resistance factor, and R„ is the nominal normal load capacity. | 933 | 3,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-05 | latest | en | 0.873437 |
https://stats.stackexchange.com/questions/tagged/inference?tab=newest&page=4 | 1,660,816,757,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00167.warc.gz | 498,968,356 | 76,078 | # Questions tagged [inference]
Drawing conclusions about population parameters from sample data. See https://en.wikipedia.org/wiki/Inference and https://en.wikipedia.org/wiki/Statistical_inference
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### Motivating use of Bayesian splines in excess mortality estimation
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### Is there a standard measure of the sufficiency of a statistic?
Given a parametrical model $f_\theta$ and a random sample $X = (X_1, \cdots, X_n)$ from this model, a statistic $T(X)$ is sufficient if the distribution of $X$ given $T(X)$ doesn't depend on $\theta$. ...
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### Using Pairwise Differences Between Two Conditions as Data (Bayesian)?
Suppose I have measurements for the expression-level of a "gene" from two groups of arbitrary (possibly different) sizes. Maybe one group is a control and the other treated. $x$ = <4.5, 5....
1 vote
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### Why is "proving" alternative hypothesis true, harder than "proving" Null hypothesis false
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### Infer (supposed) Poissonian probability from data
Suppose to count the drops of rain in a square meter in 15 seconds, producing 16 observations: 40, 20, 24, 15, 23, 12, 39, 26, 29, 33, 16, 36, 17, 32, 40, 15. What is the probability of counting 28 ...
• 153
1 vote | 2,399 | 10,322 | {"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} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.899934 |
https://edurev.in/course/quiz/attempt/-1_Test-Viscous-Flow-of-Incompressible-Fluids--Flow-L/15579054-8f0e-4e95-aaf5-b7fbea3e2713 | 1,632,616,844,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00573.warc.gz | 286,194,022 | 47,963 | Courses
# Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2
## 20 Questions MCQ Test Engineering Mechanics | Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2
Description
This mock test of Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2 for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 20 Multiple Choice Questions for Civil Engineering (CE) Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2 quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2 exercise for a better result in the exam. You can find other Test: Viscous Flow of Incompressible Fluids, Flow Losses Level - 2 extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.
QUESTION: 1
### A piping system consists of three pipes arranged in series; the lengths of the pipes are 1200 m, 750 m and 600 m and diameters 750 mm, 600 mm and 450 mm respectively.Transform the system to an equivalent 450 mm diameter pipe
Solution: l1 = 1200 m, l2 = 750 m, l3 = 600 m
d1 = 750 mm d2 = 600 mm d3 = 450 mm
de = 450 mm
For pipes in series
le = 871 m
QUESTION: 2
### A fluid jet is discharging from a 100 mm nozzle and the vena contracta formed has a diameter of 90 mm. If the coefficient of velocity is 0.95, then the coefficient of discharge for the nozzle is
Solution:
CV = Vcontracta actual / Vcontracta Theory = 0.95
Q actual = Vcontracta actual × Acontracta
= VC,Theory × CV × AC / Anozzle × Anozzle
Qactual = VC,Theory × Anozzle × CV × AC / Anozzle
Now CV × AC / AN = Cd
where Cd = Coefficient of discharge
So Cd = 0.95 × 902 / 1002
Cd = 0.7695
QUESTION: 3
### Oil at 20°C (ρ = 888 kg⁄m3 and μ = 0.800 kg ⁄ ms) is flowing steadily through a 5-cm-diameter 40-m-long pipe as shown in the below figure. The pressure at the pipe inlet and outlet are measured to be 745 and 97 kPa, respectively. Determine the flow rate of oil through the pipe (in m3/s) assuming the pipe is inclined 15° upward.
Solution: In this question the total pressure drop will be the addition of static pressure drop and the piezometric pressure drop.
QUESTION: 4
Consider the diagram below.
ρ = 1000 kg/m3 = constant
If pressure at 1 = 150 kPa and pressure at 2 = 100 kPa, the loss coefficient = 0.5 and discharge = 0.01 m3 /s. Calculate the major head loss, correct upto 2 decimal places (in m)
Solution:
QUESTION: 5
Where does the point of maximum instability exist for laminar flow through a pipe of radius R?
Solution: The velocity gradient for laminar flow through a pipeline is
Where the radial distance r from the pipe axis is r = R − y
The dimensionless parameter x is
Hence the point of maximum instability exists at a distance two-third of pipe radius from the wall.
QUESTION: 6
The value of kinetic energy correction factor for laminar flow through a circular pipe is approximately equal to
Solution: Kinetic energy correction factor,
⇒ α = 2
QUESTION: 7
Water is discharged from a tank maintained at a constant head of 5 m above the exit of a straight pipe 100 m long and 15 cm in diameter. If the friction coefficient for the pipe is 0.01, the rate of flow will be nearly
Solution:
Applying Bernoulli’s equation at 1 and 2
Note: f = friction coefficient (CF) not Darcy friction factor.
QUESTION: 8
An oil of viscosity 8 poise flows between two parallel fixed plates, which are kept at a distance of 30 mm apart. If the drop of pressure for a length of 1 m is 0.3 × 104 N⁄m2 and width of the plates is 500 mm, the rate of oil flow between the plates will be
Solution: μ = 0.8; h = 30 mm; L = 1 m Δp = −0.3 × 104 N⁄m2 w = 500 mm
Vavg = 0.281 m/s
Q = h × w × Vavg
= 0.03 × 0.5 × 0.281
= 0.00421m
= 4.21 × 10−3m
QUESTION: 9
With reference to section ① and ② consider the following statements, neglecting all losses.
Solution: HGL joins the piezometric heads at all points in flow while TEL joins the total energy head at all points in flow.
∴ TEL = HGL + V2 / 2g at a particular point
!t section ① due to sudden contraction, velocity head increases but the total energy remains constant, (as all loses are neglected) therefore HGL will drop but TEL will remain constant. !t section ② external energy is supplied to the fluid with the help of a pump, due to which both HGL and TEL will rise.
Note: Since the diameter of the pipe is the same on either side of the pump, velocity head remains constant on either side of the pump.
QUESTION: 10
Considering all the major and minor losses in the pipe network, the total head loss is (Assume Q and Darcy friction factor, f is a constant) __________ m.
1: Length = 200 cm, diameter = 10 cm
2: Length = 2.5 m, diameter = 20 cm
3: Length = 1 m, diameter = 15 cm Loss coefficient = 0.5
4: Mean length = 500 cm, diameter = 15 cm Loss coefficient = 0.33
Q = 0.1 m3 /s, f = 0.05. Assume flow is laminar everywhere.
Solution: Total head loss = Σ Hmajor + Σ Hminor
QUESTION: 11
Two reservoirs connected by two pipelines in parallel of the same diameter D and length l. It is proposed to replace the two pipelines by a single pipeline of the same length l, without affecting the total discharge and loss of head due to friction. The diameter of the equivalent pipe De in terms of the diameter of the existing pipeline, De / D is
Solution: Given that hf is same for both cases QTotal = same for both cases Length of pipe is same for both cases For parallel pipes we know hf1 = hf2 So,
QUESTION: 12
Consider fully developed laminar flow in a circular pipe of a fixed length
1. The friction factor is inversely proportional to Reynolds number
2. The pressure drop in the pipe is proportional to the average velocity of the flow in the pipe
3. The friction factor is higher for a rough pipe as compared to a smooth pipe
4. The pressure drop in the pipe is proportional to the square of average of flow in the pipe
What of the above statements are correct?
Solution: 1 is true, 2 is true, 3 is false as friction factor only depends on Reynolds number of the pipe and has nothing to do with roughness of pipe. 4 is false as pressure drop is directly proportional U and U × !rea = Q. So, if area is constant then U is proportional to Q and hence pressure drop is also proportional to Q.
*Answer can only contain numeric values
QUESTION: 13
The space between two plates (20cm*20cm*1cm), 1 cm apart, is filled with a liquid of viscosity 1 Poise. The upper plate is dragged to the right with a force of 5N keeping the lower plate stationary.
What will be the velocity in m/s of flow at a point 0.5 cm below the lower surface of the upper plate if linear velocity profile is assumed for the flow?
Solution:
where Fν = viscous force, A = area, du ⁄ dx = velocity gradient, μ = coefficient of viscosity. If linear velocity profile is assumed, du⁄dx = U/x, where U = velocity of the upper plate and x = distance between the two plates. Now, the viscous force Fv = -F= -5N. Substituting all the values in the equation, U becomes 12.5 m/s.
QUESTION: 14
Consider the case of minor head loss due to sudden contraction
(here C is the point of vena contracta)
Which of the following is correct?
Solution: The minor head loss due to sudden contraction occurs downstream of C because of the presence of adverse pressure gradient (dp /dx > 0) which causes flow separation.
QUESTION: 15
A 0.20 m diameter pipe 20 km long transports oil at a flow rate of 0.01 m3⁄s. Calculate power required to maintain flow if dynamic viscosity and density of oil is 0.08 Pas and 900 kg⁄m3 respectively.
Solution:
QUESTION: 16
Consider flow of oil and water through a channel; the boundary conditions at the interface are
Solution: Since the interface will have a defined velocity (Vi). The velocity of a particle just below the interface (in water) has to be the same as the velocity of the particle just above the interface (in oil), otherwise the interface will deform due to uneven velocity. Assuming the flow is fully developed. The shear stress (τ) just above and below the interface must be the same, or there will be a net force on the interface which will accelerate the interface. Consider an element at the interface.
For uniform velocity (τoil = τwater ) at interface Since τ is equal at interface in both the liquids, shear stress is continuous.
Since the interface has a defined velocity, velocity will be continuous.
QUESTION: 17
Consider the image below.
Codes: (i) (ii) (iii) (iv)
Solution: The general velocity profile for Couette flow is
then V = 0 (Parabolic)
V < 0="" gives="" (iii)="" and="" v="" /> 0 gives (iv)
QUESTION: 18
Assertion (A): The power transmitted through a pipe is maximum when the loss of head due to friction is equal to one-third of total head at the inlet.
Reason (R): Velocity is maximum when the friction loss is one-third of the total head at the inlet.
Solution: Maximum velocity can only be obtained when head loss is zero and the flow is ideal (no head loss). In viscous flow through pipes, head loss = initial head/3 is a condition for maximum power and not for maximum velocity. Assertion is true, Reason is false.
QUESTION: 19
Which one of the following is incorrect for the developing region of the pipe flow?
Solution: Assuming constant discharge and incompressible flow
Clearly, the velocity profile is more parabolic downstream. The velocity near the centre is increasing, so the fluid has convective acceleration.
By continuity equation Q entry = Q developing region = Q developed region So ! × U = A × Uavg,developing = A × Uavg Since A is constant, as diameter of pipe is constant.
U∞ = Uavg,developing = Uavg = Umax / 2
So the average velocity stays constant, and velocity at the centre line is the maximum velocity (Umax ) which is always greater than free stream velocity (U).
QUESTION: 20
For viscous, incompressible flow between two flat plates as shown, the velocity at y = h / 2 is
Solution: For Couette flow, the general velocity profile is given as | 2,646 | 10,322 | {"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} | 3.53125 | 4 | CC-MAIN-2021-39 | latest | en | 0.796846 |
https://www.physicsforums.com/threads/friction-at-an-angle-verification.232415/ | 1,695,752,013,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510219.5/warc/CC-MAIN-20230926175325-20230926205325-00736.warc.gz | 1,022,788,955 | 14,226 | # Friction at an angle verification
• warmfire540
#### warmfire540
Hey, i just want to make sure I'm doing this right:
At the end of a factory production line, boxes start from rest and slide down a 30o ramp 5.4 m long. If the slide is to take no more than 3.3 s, what is the maximum coefficient of friction between the box and the slide?
I first found out the acceleration:
5.4=0.5a(3.3)^2
a=.99m/s^2
so i then did
mgsinTHETA-umgcosTHETA=ma (mass cancels out)
4.9-8.5u=0.99
-8.5u=-3.91
u=.46...the coefficent of friction, it seems to work..
however i still need help on my post "Tension and centripital force revisited"
THANKS for alll ya'lls help
Hey, i just want to make sure I'm doing this right:
At the end of a factory production line, boxes start from rest and slide down a 30o ramp 5.4 m long. If the slide is to take no more than 3.3 s, what is the maximum coefficient of friction between the box and the slide?
I first found out the acceleration:
5.4=0.5a(3.3)^2
a=.99m/s^2
so i then did
mgsinTHETA-umgcosTHETA=ma (mass cancels out)
4.9-8.5u=0.99
-8.5u=-3.91
u=.46...the coefficent of friction, it seems to work..
however i still need help on my post "Tension and centripital force revisited"
THANKS for alll ya'lls help
I didn't check the numbers, but your method seems correct. | 420 | 1,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.895881 |
https://testroom.com/tests/255935/fraction-word-problems-addition-and-subtraction | 1,547,910,437,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583668324.55/warc/CC-MAIN-20190119135934-20190119161934-00149.warc.gz | 657,693,941 | 6,867 | ##### Notes
Supports Common Core Mathematics Standard CCSS 4.NF.B.3d
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## Fraction Word Problems - Addition and Subtraction
1.
In a pond, 1/8 of the animals are amphibians, 1/8 are mammals, 2/8 are fish, and 3/8 are insects. The remaining animals are reptiles. Write a subtraction equation that shows the fraction of animals in the pond that are reptiles.
2.
Josie and Xavier are lab partners in science class. They work together to write their lab report by each writing the parts of the report shown below.
Introduction - Josie
Materials - Xavier
Procedure - Xavier
Data - Josie
Conclusions - Xavier
Write an addition fraction equation the shows how Josie and Xavier shared writing the report.
3.
Write a word problem for the fraction model equation shown below.
$+$$=$
4.
Write a word problem for the fraction model equation shown below.
$-$$-$$=$
You need to be a HelpTeaching.com member to access free printables. | 280 | 1,216 | {"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} | 2.875 | 3 | CC-MAIN-2019-04 | latest | en | 0.834556 |
https://www.physicsforums.com/threads/thickness-of-lenses.780646/ | 1,532,306,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594675.66/warc/CC-MAIN-20180722233159-20180723013159-00625.warc.gz | 956,988,924 | 15,472 | # Homework Help: Thickness of lenses
1. Nov 7, 2014
### toothpaste666
1. The problem statement, all variables and given/known data
A highly reflective mirror can be made for a particular wavelength at normal incidence by using two thin layers of transparent materials of indices of refraction n1 and n2 (1 < n1<n2 ) on the surface of the glass ( n>n2 ).(Figure 1)
A)What should be the minimum thicknesses d1 in the figure in terms of the incident wavelength λ, to maximize reflection?Express your answer in terms of the variables n1, n2, and λ.
B)
What should be the minimum thickness d2 in the figure in terms of the incident wavelength λ, to maximize reflection?
Express your answer in terms of the variables n1, n2, and λ.
2. Relevant equations
2t = (lamda n)
(lamda n) = lamda/n
3. The attempt at a solution
they want us to maximize reflection so that means they want constructive interference so we use
$2d_1 = (lambda_n) = lamda/n_1$
$d_1= lambda / 2n_1$
i put this as my answer for part a but it was wrong. I am guessing I need to include $n_2$
in my equation but i am not quite sure how to relate them. can someone point me in the right direction?
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Views:
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2. Nov 8, 2014
### toothpaste666
I think the main reason i am confused is that i dont see how the index of refraction n2 would affect the thickness d1. it seems that n1 and n2 would affect d2 but only n1 would affect d1
3. Nov 8, 2014
### toothpaste666
oh wait i accidently entered it into mastering physics as n1(lambda/2) -__- i turned out to be right about that one
4. Nov 8, 2014
### toothpaste666
so for the part B)
the wavelength of the light entering d2 is lambda/n1
$2d_2 = \frac{\frac{lambda}{n_1}}{n_2}$
$d_2 = \frac{lambda}{2n_1n_2}$
is this correct?
5. Nov 8, 2014
### toothpaste666
or would it undergo a 180 degree phase shift and be
d2 = lambda/4n1n2 | 543 | 1,872 | {"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} | 3.71875 | 4 | CC-MAIN-2018-30 | latest | en | 0.905693 |
https://www.experts-exchange.com/questions/21096400/Numerical-value-from-pointArray-i.html | 1,488,174,815,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172447.23/warc/CC-MAIN-20170219104612-00635-ip-10-171-10-108.ec2.internal.warc.gz | 819,952,959 | 26,045 | Solved
# Numerical value from pointArray[ i ]
Posted on 2004-08-17
152 Views
I need the numerical values of the x-coordinates of the points on the screen where I have clicked with the mouse. Briefly I do the following (say, clicking twice):
// add mouse position to vertex list
points->Add( __box( Point( e->X, e->Y ) ) );
// if I clicked 2 times
if ( points->Count == 2 ) {
// create array of points
Point pointArray[] = new Point[ points->Count ];
// add each point to the array
for( int i = 0; i < points->Count; i++ )
pointArray[ i ] = *(
dynamic_cast< Point* >( points->Item[ i ] ) );
}
As a result, I see in the messagebox MessageBox::Show( pointArray[ 0 ].ToString(), pointArray[ 1 ].ToString() ); the following:
{X=124,Y=231} and {X=67,Y=96}
How do I extract the numerical values of the x-coordinates from these data (from the array pointArray[ i ])?
0
Question by:judico
• 2
Author Comment
ID: 11822217
I got that:
X1 = pointArray[ 0 ].X;
X2 = pointArray[ 1 ].X;
but how does one first plot
void OnPaint( PaintEventArgs *paintEvent )
{
<some code>
}
and then, after clicking twice and assigning non-zero values of X1 and X2, plots again
graphicsObject->DrawEllipse( pen, X1, X2, 7, 7 ); ?
0
Author Comment
ID: 11822789
I got that too. After values X1 and X2 are made non-zero the graphics should be repainted through Refresh();
0
LVL 5
Accepted Solution
Netminder earned 0 total points
ID: 11859253
User resolved; closed, 50 points refunded.
Netminder
0
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https://learn.careers360.com/school/question-please-solve-r-d-sharma-class-12-chapter-5-determinants-exercise-5-point-4-question-3-maths-textbook-solution/?question_number=3.0 | 1,679,993,867,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00366.warc.gz | 412,944,958 | 115,531 | # Get Answers to all your Questions
### Answers (1)
Answer: $\mathrm{x}=7 \text { and } \mathrm{y}=-3$
Hint: Use Cramer’s rule to solve a system of two equations in two variables.
Given: \begin{aligned} &2 x-y=17 \\ &3 x+5 y=6 \end{aligned}
Solution:
First D: determinant of the coefficient matrix
$\mathrm{D}=\left|\begin{array}{cc} 2 & -1 \\ 3 & 5 \end{array}\right| \quad \because\left|\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}\right|=\left(a_{1} b_{2}-a_{2} b_{1}\right)$
\begin{aligned} &=(2)(5)-(3)(-1) \\ &=10+3 \\ &=13 \end{aligned}
Now, $D\neq 0$. If we are solving for x, the x column is replaced with constant column i.e.
$\mathrm{D}_{1}=\left|\begin{array}{cc} 17 & -1 \\ 6 & 5 \end{array}\right|$
\begin{aligned} &=(2)(6)-(17)(3) \\ &=12-51 \\ &=-39 \\ &\text { Now, } x=\frac{D_{1}}{D}=\frac{91}{13}=7 \end{aligned}
$\mathrm{y}=\frac{D_{2}}{D}=\frac{-39}{13}=-3$
Hence, x = 7 and y=-3
Concept: Cramer’s rule for system of two equations.
Note: Cramer’s rule will give us unique solution to a system of equations, if it exists. However, if the system has no solution or an infinitive number of solutions that is determinant is zero.
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• Faculty Support | 450 | 1,302 | {"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-14 | longest | en | 0.687655 |
https://www.reference.com/geography/can-calculate-gas-cost-between-cities-mileage-8a5a943d2a6b76d1 | 1,480,832,672,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541214.23/warc/CC-MAIN-20161202170901-00301-ip-10-31-129-80.ec2.internal.warc.gz | 1,026,773,817 | 21,536 | Q:
Can you calculate the gas cost between cities if you know the mileage?
A:
Gas cost can be calculated using the miles per gallon and fuel cost at stations along the road, should a refuel be necessary in the middle of the trip, using the mileage as a basis. Websites like FuelEconomy.gov help with calculating this cost without any lengthy research.
Keep Learning
Credit: YinYang Vetta Getty Images
Mileage is simply the distance between two points as measured in miles. If a vehicle's fuel efficiency is calculated alongside the mileage between two cities, it is possible to know when refuels will be necessary and how much it will cost to refuel at a given location. Each trip has a high likelihood of having a different cost than the last, due to the range of variables that go into fuel costs.
It is always necessary to use the miles per gallon of a given vehicle to get an accurate estimation of fuel costs, as each vehicle can vary in fuel cost due to age, model and parts that vary from the factory model. Once this is accounted for, a state's average fuel prices must be added into the equation, if a vehicle needs to stop at a station during a long trip. These prices vary from state to state and city to city, and can even be different in stations a couple of minutes from each other. Due to this variance, only a rough estimate can be made of overall fuel costs.
Sources:
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PEOPLE SEARCH FOR | 462 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-50 | longest | en | 0.951649 |
https://simplewebtool.com/converters/area/squarerodstosquareparsecs/squarerodstosquareparsecs.html | 1,720,986,043,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00460.warc.gz | 495,547,045 | 12,527 | Square Rods to Square Parsecs conversion
Square Rods to Square Parsecs conversion calculator and how to convert.
How to convert Square Rods to Square Parsecs: 1 square rod is approximately equal to 2.66e-32 square parsecs. 1 square rod ≈ (2.66 * 10-32) ≈ 2.66e-32 square parsecs The areaQ(sp) in square parsecs is approximately equal to the area Q(sr) in square rods multiplied by 2.66e-32.
Formula: Q(sp) ≈ Q(sr) * 2.66e-32
Square Rods to Square Parsecs conversion table
Square Rods Square Parsecs
1 square rod 2.66e-32 square parsecs
2 square rods 5.31e-32 square parsecs
3 square rods 7.97e-32 square parsecs
4 square rods 1.06e-31 square parsecs
5 square rods 1.33e-31 square parsecs
6 square rods 1.59e-31 square parsecs
7 square rods 1.86e-31 square parsecs
8 square rods 2.13e-31 square parsecs
9 square rods 2.39e-31 square parsecs
10 square rods 2.66e-31 square parsecs | 285 | 885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.700697 |
https://www.airmilescalculator.com/distance/cok-to-pgh/ | 1,619,134,291,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00199.warc.gz | 708,764,932 | 72,333 | # Distance between Cochin (COK) and Pantnagar (PGH)
Flight distance from Cochin to Pantnagar (Cochin International Airport – Pantnagar Airport) is 1314 miles / 2115 kilometers / 1142 nautical miles. Estimated flight time is 2 hours 59 minutes.
Driving distance from Cochin (COK) to Pantnagar (PGH) is 1617 miles / 2603 kilometers and travel time by car is about 30 hours 49 minutes.
## Map of flight path and driving directions from Cochin to Pantnagar.
Shortest flight path between Cochin International Airport (COK) and Pantnagar Airport (PGH).
## How far is Pantnagar from Cochin?
There are several ways to calculate distances between Cochin and Pantnagar. Here are two common methods:
Vincenty's formula (applied above)
• 1314.010 miles
• 2114.694 kilometers
• 1141.843 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1319.624 miles
• 2123.730 kilometers
• 1146.722 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Cochin International Airport
City: Cochin
Country: India
IATA Code: COK
ICAO Code: VOCI
Coordinates: 10°9′7″N, 76°24′6″E
B Pantnagar Airport
City: Pantnagar
Country: India
IATA Code: PGH
ICAO Code: VIPT
Coordinates: 29°2′0″N, 79°28′25″E
## Time difference and current local times
There is no time difference between Cochin and Pantnagar.
IST
IST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 168 kg (370 pounds).
## Frequent Flyer Miles Calculator
Cochin (COK) → Pantnagar (PGH).
Distance:
1314
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1314
Round trip? | 502 | 1,858 | {"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-2021-17 | latest | en | 0.813304 |
https://discussions.unity.com/t/leantween-reverse-tween/232850 | 1,716,969,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00634.warc.gz | 164,658,037 | 5,533 | # LeanTween reverse tween
Is it possible to reverse a tween? I am working on a wall running script and I can not make the camera go back smoothly it just snaps back.
Here is the code.
`````` if (isWallRunning)
{
if (leftRight.Equals("left") || leftRight.Equals("right"))
{
return;
}
if (leftHit.collider)
{
float direction = -1f;
LTDescr rotateOut = LeanTween.rotateLocal(playerCam.gameObject, new Vector3(0, 0, 12.5f * direction), timeToSnap);
leftRight = "left";
_rb.useGravity = false;
}
if (rightHit.collider)
{
float direction = 1f;
LTDescr rotateOut = LeanTween.rotateLocal(playerCam.gameObject, new Vector3(0, 0, 12.5f * direction), timeToSnap);
leftRight = "right";
_rb.useGravity = false;
}
}
else
{
if (leftRight.Equals("none"))
{
return;
}
if (leftRight.Equals("left"))
{
float direction = 1f;
LTDescr rotateOut = LeanTween.rotateLocal(playerCam.gameObject, new Vector3(0, 0, 12.5f * direction), timeToSnap);
//LeanTween.cancel(playerCam.gameObject);
_rb.useGravity = true;
Debug.Log("test left");
}
if (leftRight.Equals("right"))
{
float direction = -1f;
LTDescr rotateOut = LeanTween.rotateLocal(playerCam.gameObject, new Vector3(0, 0, 12.5f * direction), -timeToSnap);
//LeanTween.cancel(playerCam.gameObject);
_rb.useGravity = true;
Debug.Log("test right");
}
leftRight = "none";
}
``````
Can someone help?
Tweening forward and backward is possible with Ext.LeanTween.Framework.LTDescr.setLoopPingPong with argument 1.
Like:
LeanTween.value(target)
.setLoopPingPong(1); | 415 | 1,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.539079 |
https://www.cfd-online.com/Forums/openfoam/78991-why-icofoam-solver-results-not-true-cavity-10000-re-5000-a-print.html | 1,498,351,388,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320368.57/warc/CC-MAIN-20170624235551-20170625015551-00113.warc.gz | 843,538,486 | 3,571 | CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
- OpenFOAM (https://www.cfd-online.com/Forums/openfoam/)
- - Why icoFoam solver results are not true for cavity (10000>Re>5000) ? (https://www.cfd-online.com/Forums/openfoam/78991-why-icofoam-solver-results-not-true-cavity-10000-re-5000-a.html)
maysmech August 9, 2010 11:57
Why icoFoam solver results are not true for cavity (10000>Re>5000) ?
Dear Foamers,
i used icoFoam solver for cavity. when the reynolds number is below 5000 the results are good. but for reynolds 5000 and 10000 results are different from Ghia benchmark. i promoted orders of discretisations in fvSchemes but results were same.
it is wondering because for reynolds below Re=10000 the cavity case is laminar and icoFoam should leads to true results.
:confused:WHAT IS THE PROBLEM?
stevenvanharen August 10, 2010 10:44
Hi maysam,
I am assuming you are referring to the lid-driven cavity flow.
i think both Reynolds number will not result in steady laminar flow. The OpenFoam user guide switches to pisoFoam for Re = 10^4. Botella and Peyret found unsteady behaviour at least for Re=9000.
What is your source for assuming laminar flow for Re < 10^4?
maysmech August 10, 2010 15:25
???
I examined cavity lid driven with pisoFoam RAS but results are same as icoFoam. for Re= 10000 and 5000 results are far from Ghia benchmark:confused:
stevenvanharen August 11, 2010 04:48
Ok, I will look at the Ghia paper this afternoon.
In the mean time, in what way do your results differ from the Ghia paper? In what way are velocity plots and pressure plots different?
maysmech August 11, 2010 05:02
Results for Re=10000.
velocity diagram over horizontal and vertical centreline.
U=1, D=1, nu=10^-4, time 100 sec
stevenvanharen August 11, 2010 08:57
Ok, I am not a 100 percent sure about what it is but I have some thoughts:
- Ghai et al use upwinding, are you using upwinding too in icoFoam? Could it be that your numerical diffusion is less strong than theirs? The fact that your velocity profile is less smooth suggests this.
- the fact that you use upwinding could explain the laminar solution for Re=10000.
- numerical diffusion is less dominant at low Reynolds numbers, therefore your error increases for large Reynolds number
- I suggest you use the paper by Botella & Peyret (1998), their method is much more accurate (spectral) and they also compare to Ghai et. al., unfortunately only up to Re = 1000.
Hope this helps!
maysmech August 12, 2010 04:10
Quote:
Originally Posted by stevenvanharen (Post 271145) Ok, I am not a 100 percent sure about what it is but I have some thoughts: - Ghai et al use upwinding, are you using upwinding too in icoFoam? Could it be that your numerical diffusion is less strong than theirs? The fact that your velocity profile is less smooth suggests this. - the fact that you use upwinding could explain the laminar solution for Re=10000. - numerical diffusion is less dominant at low Reynolds numbers, therefore your error increases for large Reynolds number - I suggest you use the paper by Botella & Peyret (1998), their method is much more accurate (spectral) and they also compare to Ghai et. al., unfortunately only up to Re = 1000. Hope this helps!
Dear Steven,
Thanks for your suggestions.
i examined icoFoam for Re=100, 400 and 1000 and there was no difference between them and Ghia et al.
but there was problem in 5000 and 10000 and it is solved by considering two things::)
- I should use higher order discretisation method i used vanleer.
- I shouldn't use initial value u=0 for all domain. i used steady state velocity and pressure of Re=1000 for Re= 5000 and also used steady satate values of Re=5000 for 0 folder of 10000. i think it is an important work for high reynolds problems.:cool:
Best wishes,
Maysam
stevenvanharen August 12, 2010 16:12
ok, first of all congratulations on solving the problem!
But your solution should be independent of the initial velocity field!
In the previous simulation which yielded the erroneous results how did you stop the simulation? A simulation with a higher Reynolds number should take longer to converge to a steady state solution (in physical time).
maysmech August 12, 2010 16:42
for Re=1000 after 20 sec the problem became steady and i let 50 sec for 5000 and 100sec for 10000. i thought the time was sufficient to become steady because i didn't see any changes in several last time steps by eye.
maybe it was not enough i will try it with more time because a sudden shock in high reynolds leads to late steady and putting steady state values of lower Re to 0 values of higher Re makes less run time.
All times are GMT -4. The time now is 20:43. | 1,253 | 4,712 | {"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-2017-26 | latest | en | 0.876472 |
https://plainmath.org/quantum-mechanics/102678-the-position-of-an-object-movi | 1,709,174,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00629.warc.gz | 454,201,519 | 21,183 | Jamari Bowman
2023-02-21
The position of an object moving along a line is given by . What is the speed of the object at ?
Bobby Espinoza
Solution
The speed of an object is the magnitude of the object's velocity , which is the derivative of displacement (magnitude is position).
To find the speed of the object at time t, we can begin by taking the derivation of the provided equation for position:
$p\left(t\right)=2{t}^{3}-2{t}^{2}+1$
$⇒p\prime \left(t\right)=v\left(t\right)=6{t}^{2}-4t$
At t=3, we have:
$v\left(t\right)=6{\left(3\right)}^{2}-4\left(3\right)$
$=54-12$
$=42$
At t=3, the object has a speed of 42 units.
Do you have a similar question? | 212 | 658 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.839212 |
https://salttheoats.wordpress.com/2011/10/06/modeling-unit-6/ | 1,529,381,924,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861899.65/warc/CC-MAIN-20180619041206-20180619061206-00291.warc.gz | 703,873,415 | 18,307 | # Modeling Unit 6
Unit VI: 2-D Particle Model
Our introductory/paradigm demo was Jon and Chris tossing a ball back and forth. Jon then asked questions like: “Once it leaves my hand, where will it go?””Does the ball have a choice as to where it goes after it leaves my hand?”
One thing that come to mind during this demo was the following video from Veritasium.com:
What do you notice?
What can you measure?
{at this point, Jon showed us the equipment that we would be using}
{Jon build hold that converts dynamic cart w/ spring into ball launcher}
Here’s a rough sketch:
Where the blue shape is the dynamics cart with the spring plunger extended, the silver circle is the ball to be launched, and the brown shape is the holder Jon built out of wood. He also cut/routed a groove for the ball to roll in on the top of the “shelf.”
After being shown equipment What can you manipulate?
{If you don’t have time to build this and have students tape it, use videos in loggerpro}
Open logger pro
Click Insert – Movie
Click “expand menu” in bottom right corner
Click scale icon (looks like a ruler)
Make sure you have scale (meter stick) in the movie
Click and trace standard length in screen & define length
Click on track (find name) button and click on specific point on object
Continue clicking on the same spot of the object (vernier advances to next frame)
Jon and Chris then tried to show us the classic Monkey-Blow gun demo using the Pasco equipment:
Since this is quite expensive, Jon explained how he made a “homemade version” of this:
Materials:
Electric conduit (1/2 inch? Metal)
Nail with cone of paper hot glued in
Electromagnet
Wire
12 V power source (3 or 6 V should also work)
Target – Balloon with brass mass inside, washer stretching the opening
Stuffed animal with metal screw in its head
He attaches the Electromagnet to the ceiling in the back of his room and runs the ingoing and outgoing wires above is ceiling (drop-down I’m guessing) to the front of his room. He uses the conduit as the blow gun and makes darts by gluing cones of paper to the head of the nail. Have the two wires run up the side of the conduit and each extent the bare wires beyond the opening of the conduit. Bend the wires so they touch in the middle of the opening. As the dart shoots out, it will separate the wires, breaking the connection. Here’s his sketch:
(click to embiggen)
I missed this day of the workshop, however, a few of my cohort were gracious enough to take notes. I’m doing my best to take what they gave me. Any help to clarify things would be greatly appreciated.
The day began with everyone working on Unit VI worksheet. Everyone worked individually, and then the groups met to create whiteboards.
– Useful to separate horizontal and vertical givens in table:
-Good to explicitly show + state that t is the same for horizontal and vertical motion
-Good to keep algebra in variable until the last step – then plug in number
#4 Would be interesting in adding a horizontal & vertical motion map for car and ball
-stress constant velocity in horizontal direction
– ESL students have difficulty with “how long” thinking it means distance
LoggerPro basket ball shot analysis follow up
– After students have generated data, insert 3 graphs + auto arrange
• x vs t
• y vs t
• $v_x$ vs t
• $v_y$ vs t
-Highlight first 1.5 second to analyze
• compare slope of x vs t and average value of $v_x$ from $v_x$ vs t graph
• lead students to see that $v_x$ is constant but $v_y$ is changing: slope is 9.8 $m/s^2$
• If you want, have students insert a quadratic fit onto y vs t graph and lead them to find what the meaning of the constants are in the regressed equation.
Next on the agenda was to split up an article to have summarized on whiteboards by the groups.
After lunch, Jon and Chris asked for feedback for Unit VI
What worked:
Video analysis lab
Plan for Dart Gun for Classic Monkey Problem
Worksheet #3
Wells Reading
Hammer article about Lisa & Ellen
Group Work
Adaptability of labs to every level of student (*response to comment on what didn’t work)
What didn’t work:
Transition from 1D to 2D – we would like to see the process
Time constraints
Simplicity of labs
Advertisements | 992 | 4,216 | {"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": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-26 | latest | en | 0.921236 |
https://www.learn-with-math-games.com/fraction-math-games.html | 1,685,707,977,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00187.warc.gz | 923,389,207 | 20,326 | # Fraction Math Games
Fraction math games make learning fractions fun and exciting for kids and also boost confidence. What if you could get kids captivated by math so much they don't even realize they are learning? Wouldn't that be great?
Four-Sum Fractions
Skill: Adding and simplifying fractions with common denominators
Number of players: 2-4
Object of the game: Be the first player to score four in a row.
Supplies:
• A game board for each player
• List of possible sums
• Markers to cover the game board spaces
• Paper and pencil for each player
Preparation:
To Play:
• Each player writes any of the possible sums in the spaces on his blank game card. He may use all of the sums just once, or use some more than once and not use others at all. Each space of the game card should be filled. Players should know that many of the sums are used for more than one fraction combination. For example, 1/2 is the answer for 1/8 + 3/8 and 1/4 + 1/4.
• The first player to complete step one starts the game.
• The first player takes a fraction combination card from the top of the pile. He calls out the combination. Each player writes this on his paper, adds the fractions, and writes the answer in lowest terms. Any player who has this sum on his game card covers the space with a marker.Note: Players may only cover one space per turn. For example, suppose Jim has written the sum of 3/4 on his game board in two places. He takes the card that says 7/8 + 7/8. He adds and finds the answer is 3/4. He may cover only one of the spaces on the game board with that answer.
• The next player takes a fraction card, reads it to the players as in Step 3.
• Play continues until one player has covered 4 spaces in a row, vertically, horizontally or diagonally. He calls out, “Four-Sum,” and play stops.
• The other players check his answers with the used fraction cards to be sure he marked the correct spaces. If the answers are correct, that player wins the round. If 1 or more of his answers were incorrect, he must clear his board of all markers, and the round continues until a player wins.
• Variations:
1. Use more fraction cards and 5 x 5 game boards for a longer game.
2. Use harder fraction cards (with unlike denominators, three addends, etc.) for older players.
Have Fun!!!
© 2009 www.learn-with-math-games.com All rights reserved.Now, if you had tons of different fun fraction math games, you might be well on your way to turning them into math maniacs. I'm sure you might not mind so much if you see better grades and their confidence in math skyrocket.
Fraction math games should really be a part of a child's overall math learning experience. More and more teachers and parents are realizing the value of making math fun for kids. If you can capture their attention, that's most of the battle right there!
Go to main Fraction Games page
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http://www.utdallas.edu/~cantrell/ee4301/MT2_topics_13.html | 1,369,134,765,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699899882/warc/CC-MAIN-20130516102459-00004-ip-10-60-113-184.ec2.internal.warc.gz | 782,275,951 | 1,791 | ## Topics for Midterm Exam 2
EMAG CONCEPTS
Ampere's circuital law
differential form
integral form
Examples:
H inside a solenoid
H in a coaxial cable
Current densities
surface
volume
Displacement current
examples: capacitor
use integral form to get EMF
differential form
Maxwell's equations
differential and integral form
Test a proposed field for consistency with Maxwell's equations
Recover the missing field given either E or H
Find the surface charge density given E in an infinite parallel-plate
waveguide
Find the surface current density given H in an infinite parallel-plate
waveguide
Material properties
\epsilon and \mu
Give examples of materials that are:
good dielectrics
good conductors
semiconductors
Use appropriate boundary conditions on fields at interfaces
Wave equation in 1 dimension
General solution
Application to plane waves
Test whether a function satisfies the 1-d wave equation
Physical interpretation
Phase velocity
Wave equation in 1 dimension
Single-frequency solutions
Temporal frequency
Spatial frequency
Boundary conditions on B_n
Boundary conditions on H_t
Poynting's theorem
Calculate magnitude and direction of power flux given E and H
Wave equation in 1 dimension
Single-frequency solutions
Temporal frequency
Spatial frequency
Wave equation in 3 dimensions
Plane waves in conducting media
Skin depth
Surface resistance
Polarization of electromagnetic waves
Plane and circular polarization
Wire-grid polarizer
Dichroic materials
Malus' law
Parallel-plate waveguide
Application of boundary conditions to determine the directions of
electric and magnetic fields
Parallel-plate waveguide as a polarization-sensitive
device
VECTOR CALCULUS
curl of a vector field
Cartesian
cylindrical
Stokes' theorem
All vector calculus topics from MT 1 | 390 | 1,787 | {"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} | 2.8125 | 3 | CC-MAIN-2013-20 | latest | en | 0.785793 |
https://issuu.com/cooladityabhardwaj/docs/44.x-rays | 1,508,491,005,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823997.21/warc/CC-MAIN-20171020082720-20171020102720-00770.warc.gz | 748,138,242 | 20,320 | X - RAYS CHAPTER 44 1.
= 0.1 nm a) Energy =
hc 1242 ev.nm 0.1 nm
= 12420 ev = 12.42 Kev = 12.4 kev. b) Frequency =
C 3 108 3 108 3 1018 Hz 10 0.1 10 9 10
c) Momentum = E/C = 2.
12.4 103 1.6 10 19 3
3 108
= 6.613 10
–24
kg-m/s = 6.62 10
–24
kg-m/s.
Distance = 3 km = 3 10 m 8 C = 3 10 m/s
Dist 3 103 10 5 sec. Speed 3 108 –8 10 10 sec = 10 s in both case. V = 30 KV hc hc 1242 ev nm –4 = = 414 10 nm = 41.4 Pm. 3 E eV e 30 10 –10 –34 = 0.10 nm = 10 m ; h = 6.63 10 J-s 8 –19 C = 3 10 m/s ; e = 1.6 10 C hc hc min = or V= eV e t=
3.
4.
=
6.63 1034 3 108 1.6 10 19 10 10
3
= 12.43 10 V = 12.4 KV.
hc 6.63 10 34 3 108 –18 –15 –15 = 19.89 10 = 1.989 10 = 2 10 J. 10 10 hc 1242 3 = 80 pm, E = = 15.525 10 eV = 15.5 KeV 80 10 3 hc We know = V hc Now = 1.01V 1.01 0.01 – = . 1.01 0.01 1 = 0.9900 = 1%. % change of wave length = 100 1.01 1.01 –3 d = 1.5 m, = 30 pm = 30 10 nm hc 1242 3 = 41.4 10 eV E= 30 10 3 Max. Energy =
5. 6.
7.
V 41.4 103 3 = 27.6 10 V/m = 27.6 KV/m. d 1.5 Given = – 26 pm, V = 1.5 V hc hc , = Now, = ev ev or V = V –12 V = ( – 26 10 ) 1.5 V Electric field =
8.
44.1
X-Rays = 1.5 – 1.5 26 10 =
–12
39 10 12 –12 = 78 10 m 0.5
hc 6.63 3 10 34 108 5 3 = 0.15937 10 = 15.93 10 V = 15.93 KV. 19 12 e 1.6 10 78 10 3 9. V = 32 KV = 32 10 V When accelerated through 32 KV 3 E = 32 10 eV hc 1242 –3 = = 38.8 10 nm = 38.8 pm. E 32 103 hc 18 10. = ; V = 40 kV, f = 9.7 10 Hz eV eV h h i h or, ; or, ; or h V s f c eV f eV V=
eV 40 103 –15 Vs = 4.12 10 eV-s. f 9.7 1018 3 11. V = 40 KV = 40 10 V 3 Energy = 40 10 eV 70 3 Energy utilized = 40 103 = 28 10 eV 100 hc 1242 ev nm –3 = 44.35 10 nm = 44.35 pm. 3 E 28 10 ev =
For other wavelengths, E = 70% (left over energy) =
70 2 (40 28)103 = 84 10 . 100
hc 1242 –3 = 147.86 10 nm = 147.86 pm = 148 pm. 3 E 8.4 10 For third wavelength, 70 3 2 2 = (12 – 8.4) 10 = 7 3.6 10 = 25.2 10 E= 100 hc 1242 –2 = = 49.2857 10 nm = 493 pm. E 25.2 102 1242 –12 12. K = 21.3 10 pm, Now, EK – EL = = 58.309 kev 21.3 10 3 EK = 58.309 + 11.3 = 69.609 kev EL = 11.3 kev, Now, Ve = 69.609 KeV, or V = 69.609 KV. 13. = 0.36 nm 1242 = 3450 eV (EM – EK) E= 0.36 Energy needed to ionize an organ atom = 16 eV Energy needed to knock out an electron from K-shell = (3450 + 16) eV = 3466 eV = 3.466 KeV. 14. 1 = 887 pm =
C 3 108 7 16 8 = = 3.382 10 = 33.82 10 = 5.815 10 887 10 12 2 = 146 pm v=
v=
3 108 146 10 12
= 0.02054 10
20
= 2.054 10
18
9
= 1.4331 10 .
44.2
X-Rays
v a(z b)
We know,
5.815 108 a(13 b) 1.4331 109 a(30 b)
13 b 5.815 10 1 = 0.4057. 30 b 1.4331 30 0.4057 – 0.4057 b = 13 – b 12.171 – 0.4.57 b + b = 13 0.829 b= = 1.39491 0.5943
5.815 108 7 0.51323 108 = 5 10 . 11.33 For ‘Fe’, a=
7
7
7
v = 5 10 (26 – 1.39) = 5 24.61 10 = 123.05 10 14 c/ = 15141.3 10
3 108
–6 –12 = 0.000198 10 m = 198 10 = 198 pm. 15141.3 1014 15. E = 3.69 kev = 3690 eV hc 1242 = = 0.33658 nm E 3690
==
c / a(z – b);
a = 5 10
7
Hz , b = 1.37 (from previous problem)
3 108
5 107 (Z 1.37) 8.82 1017 5 107 (Z 1.37) 0.34 10 9 8 7 9.39 10 = 5 10 (Z – 1.37) 93.9 / 5 = Z – 1.37 Z = 20.15 = 20 The element is calcium. 16. KB radiation is when the e jumps from n = 3 to n = 1 (here n is principal quantum no) 1 2 1 E = h = Rhc (z – h) 2 2 3 2
v
9RC (z h) 8
v z Second method : We can directly get value of v by ` hv = Energy Energy(in kev) v= h This we have to find out 17. b = 1 For a (57)
10 20 30 40
v and draw the same graph as above.
v = a (Z – b) v = a (57 – 1) = a 56 For Cu(29)
v
…(1)
1.88 1078 = a(29 –1) = 28 a …(2) dividing (1) and (2)
44.3
50 60
Z
X-Rays
v a 56 = 2. 1.88 1018 a 28 18 2 18 8 v = 1.88 10 (2) = 4 1.88 10 = 7.52 10 Hz. 18. K = EK – EL ,,,(1) K = 0.71 A° K = EK – EM ,,,(2) K = 0.63 A° L = E L – EM ,,,(3) Subtracting (2) from (1) K – K = EM – EL = –L or, L = K – K =
3 108 10
L K
K
E3
E2 L
3 108
0.63 10 0.71 10 10 18 18 = 4.761 10 – 4.225 10 = 0.536 10 Hz. 18
Again =
3 108
–10
= 5.6 10 = 5.6 A°. 0.536 1018 1242 3 19. E1 = = 58.309 10 ev 3 21.3 10 1242 3 = 8.8085 10 ev E2 = 3 141 10 3 E3 = E1 + E2 (58.309 + 8.809) ev = 67.118 10 ev hc 1242 –3 = = 18.5 10 nm = 18.5 pm. E3 67.118 103
L E1 K
K
20. EK = 25.31 KeV, EL = 3.56 KeV, EM = 0.530 KeV K = EK – KL = hv E EL 25.31 3.56 15 v= K 103 = 5.25 10 Hz h 4.14 10 15 K = EK – KM = hv E EM 25.31 0.53 18 v= K 103 = 5.985 10 Hz. h 4.14 10 15 21. Let for, k series emission the potential required = v Energy of electrons = ev This amount of energy ev = energy of L shell The maximum potential difference that can be applied without emitting any electron is 11.3 ev. 22. V = 40 KV, i = 10 mA 1% of TKE (Total Kinetic Energy) = X ray i = ne
or n =
102
17
= 0.625 10
no.of electrons. 1.6 10 –19 3 –15 KE of one electron = eV = 1.6 10 40 10 = 6.4 10 J 17 –15 2 TKE = 0.625 6.4 10 10 = 4 10 J. 2 a) Power emitted in X-ray = 4 10 (–1/100) = 4w b) Heat produced in target per second = 400 – 4 = 396 J. 23. Heat produced/sec = 200 w neV = 200 (ne/t)V = 200 t i = 200 /V = 10 mA. 14 2 24. Given : v = (25 10 Hz)(Z – 1) 14 2 Or C/ = 25 10 (Z – 1) 19
3 108
(Z 1)2 78.9 10 12 25 1014 2 6 or, (Z – 1) = 0.001520 10 = 1520 Z – 1 = 38.98 or Z = 39.98 = 40. It is (Zr) a)
44.4
M
K
X-Rays 8
3 10 (Z 1)2 146 10 12 25 1014 2 6 or, (Z – 1) = 0.0008219 10 Z – 1 = 28.669 or Z = 29.669 = 30. It is (Zn). 3 108
c)
12
14
Intensity
b)
(Z 1)2
158 10 25 10 2 6 or, (Z – 1) = 0.0007594 10 Z – 1 = 27.5589 or Z = 28.5589 = 29. It is (Cu).
78.9 146 158 198 Wavelength (in pm)
3 108
(Z 1)2 198 10 12 25 1014 2 6 or, (Z – 1) = 0.000606 10 Z – 1 = 24.6182 or Z = 25.6182 = 26. It is (Fe). 25. Here energy of photon = E 3 E = 6.4 KeV = 6.4 10 ev d)
Momentum of Photon = E/C =
6.4 103
= 3.41 10
–24
m/sec. 3 10 According to collision theory of momentum of photon = momentum of atom –24 Momentum of Atom = P = 3.41 10 m/sec 2 Recoil K.E. of atom = P / 2m
(3.41 10 24 )2 eV (2)(9.3 1026 1.6 10 19 )
8
3.9 eV [1 Joule = 1.6 10–19 ev]
26. V0 Stopping Potential, Wavelength, eV0 = hv – hv0 eV0 = hc/ V0 = hc/e V Potential difference across X-ray tube, Cut of wavelength = hc / eV or V = hc / e Slopes are same i.e. V0 = V
hc 6.63 10 34 3 108 –6 = 1.242 10 Vm e 1.6 10 19 –12 27. = 10 pm = 100 10 m –2 D = 40 cm = 40 10 m –3 = 0.1 mm = 0.1 10 m d=
V
V0
1/
D d
D 100 10 12 40 10 2 –7 = 4 10 m. 3 10 0.1
44.5
1/
44.X-RAYS
hc 1242 hc 1242 Now, = hc ev , = hc ev or V = V V = ( – 26 10 6.63 10 3 10 1.6 10 10 30 10 = 41.4 10 3 eV 34 8 34... | 4,620 | 7,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-43 | latest | en | 0.357384 |
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# A store purchased 20 coats that each cost an equal amount
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A store purchased 20 coats that each cost an equal amount and then sold each of the 20 coats at an equal price. What was the store's gross profit on the 20 coats?
1) If the selling price per coat had been twice as much, the store's gross profit on the 20 coats would have benn $2400 2) If the selling price per coat had been$2 more, the store's gross profit on the 20 coats would have been \$440
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19 Dec 2005, 11:52
B
i gives you 20(SP) + Profit = 2400 - 2 variables - insuff
ii gives you SP-CP = 400 - sufficicient to find 20(SP- CP)
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26 Dec 2005, 17:28
huh?
what am i missing?
gross profit (GP) = revenue (SP) - cost of merchandise (CP)
S1 gives:
GP =2400 and SP = 2xCP
so GP = 2xCP - CP = 2400 = CP
I must be brain dead tonight!
BTW I started looking at this site a few weeks ago -- It is very cool and very helpful.
Shoe
26 Dec 2005, 17:28
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bretting (nmb2278) – HW09 – meth – (54160) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points From the contour map of f shown below decide whether f x , f y are positive, negative, or zero at P . 0 0 -2 -2 -4 -4 -6 -6 P x y 1. f x > 0 , f y = 0 2. f x < 0 , f y = 0 3. f x < 0 , f y > 0 4. f x < 0 , f y < 0 correct 5. f x > 0 , f y > 0 6. f x > 0 , f y < 0 002 10.0points Determine whether the partial derivatives f x , f y of f are positive, negative or zero at the point P on the graph of f shown in P x z y 1. f x < 0 , f y < 0 2. f x = 0 , f y < 0 3. f x > 0 , f y = 0 4. f x < 0 , f y > 0 correct 5. f x = 0 , f y = 0 6. f x = 0 , f y > 0 7. f x > 0 , f y > 0 8. f x < 0 , f y = 0
bretting (nmb2278) – HW09 – meth – (54160) 2 and so the sign of f y indicates whether f is increasing or decreasing in the y -direction, or whether the tangent line in that direction at P is horizontal. From the graph it thus follows that at P f x < 0 , f y > 0 . keywords: surface, partial derivative, first or- der partial derivative, graphical interpreta- tion 003 10.0points Determine f x when f ( x, y ) = 2 x - y 2 x + y . 1. f x = - 4 x (2 x + y ) 2 2. f x = - 5 x (2 x + y ) 2 3. f x = 5 y (2 x + y ) 2 4. f x = 3 x (2 x + y ) 2 5. f x = 4 y (2 x + y ) 2 correct 6. f x = - 3 y (2 x + y ) 2 Explanation: From the Quotient Rule we see that f x = 2(2 x + y ) - 2(2 x - y ) (2 x + y ) 2 . Consequently, f x = 4 y (2 x + y ) 2 . 004 10.0points Determine f x when f ( x, y ) = cos(2 y - x ) - x sin(2 y - x ) . 1. f x = - 2 sin(2 y - x ) - x cos(2 y - x ) 2. f x = - x cos(2 y - x ) 3. f x = - cos(2 y - x ) - x sin(2 y - x ) 4. f x = x cos(2 y - x ) - sin(2 y - x ) 5. f x = - x sin(2 y - x ) 6. f x = x cos(2 y - x ) correct 7. f x = 2 sin(2 y - x ) - x cos(2 y - x ) 8. f x = x sin(2 y - x ) Explanation: From the Product Rule we see that f x = sin(2 y - x ) - sin(2 y - x )+ x cos(2 y - x ) . Consequently, f x = x cos(2 y - x ) . 005 10.0points Determine f x when f ( x, y ) = (1 - 3 xy ) e - xy . 1. f x = y (3 xy - 4) e - xy correct 2. f x = x (2 - xy ) e - xy 3. f x = x (4 - xy ) e - xy 4. f x = x ( xy - 4) e - xy 5. f x = x (2 - 3 xy ) e - xy 6. f x = y (2 + 3 xy ) e - xy
bretting (nmb2278) – HW09 – meth – (54160) 3 | 1,069 | 2,513 | {"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-2022-05 | latest | en | 0.515898 |
https://de.zxc.wiki/wiki/Cauchyscher_Integralsatz | 1,718,513,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861643.92/warc/CC-MAIN-20240616043719-20240616073719-00293.warc.gz | 171,707,174 | 20,962 | # Cauchy's integral theorem
The Cauchy's integral theorem (after Augustin Louis Cauchy ) is one of the key sets of function theory . It deals with curve integrals for holomorphic (complex-differentiable on an open set) functions. In essence, it says that two paths connecting the same points have the same path integral if the function is holomorphic everywhere between the two paths. The theorem derives its meaning from the fact that it is used to prove Cauchy's integral formula and the residual theorem .
The first formulation of the theorem dates from 1814 , when Cauchy proved it for rectangular areas. He generalized this over the next few years, although he took the Jordanian curve set for granted. Thanks to Goursat's lemma, modern proofs get by without this profound statement from the topology .
## The sentence
The integral theorem was formulated in numerous versions.
### Cauchy's integral theorem for elementary domains
Let be an elementary domain, i.e. a domain in which every holomorphic function has an antiderivative . Star regions are, for example, elementary regions . Cauchy's integral theorem now says that ${\ displaystyle D \ subseteq \ mathbb {C}}$ ${\ displaystyle f \ colon D \ to \ mathbb {C}}$
${\ displaystyle \ oint _ {\ gamma} f (z) \, dz = 0}$
for each closed curve (where and ). For the integral sign with circle see notation for curve integrals of closed curves . ${\ displaystyle \ gamma \ colon [a, b] \ to D}$${\ displaystyle a, b \ in \ mathbb {R}}$${\ displaystyle a
If there is no elementary area, the statement is false. For example, the field is holomorphic, but it does not vanish over every closed curve. For example ${\ displaystyle D}$${\ displaystyle f \ colon z \ mapsto {\ tfrac {1} {z}}}$${\ displaystyle \ mathbb {C} \ setminus \ {0 \}}$${\ displaystyle \ textstyle \ oint _ {\ gamma} f (z) \, dz}$
${\ displaystyle \ oint _ {\ partial U_ {r} (0)} {\ frac {1} {z}} \ mathrm {d} z = 2 \ pi \ mathrm {i} \ neq 0}$
for the simply traversed edge curve of a circular disk with a positive radius . ${\ displaystyle 0}$${\ displaystyle r}$
### Cauchy's integral theorem (homotopy version)
Is open and two homotopic curves are in , then is ${\ displaystyle D \ subseteq \ mathbb {C}}$${\ displaystyle \ alpha, \ beta \ colon [0,1] \ to D}$${\ displaystyle D}$
${\ displaystyle \ int \ limits _ {\ alpha} f (z) \, dz = \ int \ limits _ {\ beta} f (z) \, dz}$
for every holomorphic function . ${\ displaystyle f \ colon D \ to \ mathbb {C}}$
Is a simply connected region, then the integral vanishes after homotopies version for each closed curve d. H. is an elementary area . ${\ displaystyle D}$${\ displaystyle D}$
If you look back at the example above, you notice that is not simply connected. ${\ displaystyle \ mathbb {C} \ setminus \ {0 \}}$
### Cauchy's integral theorem (homology version)
If an area and a cycle is in , then it disappears ${\ displaystyle D \ subseteq \ mathbb {C}}$${\ displaystyle \ Gamma}$ ${\ displaystyle D}$
${\ displaystyle \ int \ limits _ {\ Gamma} f (z) \, dz}$
for every holomorphic function if and only if is zero homolog in . ${\ displaystyle f \ colon D \ to \ mathbb {C}}$${\ displaystyle \ Gamma}$ ${\ displaystyle D}$
## Isolated singularities
### Number of turns of the integration path
Let it be an area, an interior point, and holomorphic. Let be a dotted neighborhood that is holomorphic. Furthermore, let it be a completely closed curve that revolves exactly once with a positive orientation, i.e. H. for the circulation rate applies (in particular is not on ). With the integral theorem, we now have ${\ displaystyle D \ subseteq \ mathbb {C}}$${\ displaystyle a \ in D}$${\ displaystyle f \ colon D \ setminus \ {a \} \ to \ mathbb {C}}$${\ displaystyle U: = U_ {r} (a) \ setminus \ {a \} \ subset D}$${\ displaystyle f}$${\ displaystyle \ gamma}$${\ displaystyle D}$${\ displaystyle a}$${\ displaystyle \ operatorname {ind} _ {\ gamma} (a) = 1}$${\ displaystyle a}$${\ displaystyle \ gamma}$
${\ displaystyle \ oint _ {\ gamma} f (z) \, dz = \ oint _ {\ partial U} f (z) \, dz.}$
By generalizing to any number of circulation , one obtains ${\ displaystyle \ gamma}$
${\ displaystyle \ oint _ {\ gamma} f (z) \, dz = \ operatorname {ind} _ {\ gamma} (a) \ oint _ {\ partial U} f (z) \, dz.}$
With the help of the definition of the residual it even results
${\ displaystyle {\ frac {1} {2 \ pi \ mathrm {i}}} \ oint _ {\ gamma} f (z) \, dz = \ operatorname {ind} _ {\ gamma} (a) \ operatorname {Res} _ {a} f (z).}$
The residual theorem is a generalization of this approach to several isolated singularities and to cycles.
### example
The integral is below with determined. As integration path Choose a circle with radius to , so ${\ displaystyle \ oint _ {\ partial U (a)} {\ frac {1} {(za) ^ {n}}} \ mathrm {d} z}$${\ displaystyle n \ in \ mathbb {Z}}$${\ displaystyle \ partial U (a) = \ partial U_ {r} (a)}$${\ displaystyle r}$${\ displaystyle a}$
${\ displaystyle z = \ gamma (t) = a + re ^ {2 \ pi \ mathrm {i} t} \ quad \ Rightarrow \ quad \ mathrm {d} z = {\ frac {\ partial \ gamma} {\ partial t}} \ mathrm {d} t = 2 \ pi ire ^ {2 \ pi \ mathrm {i} t} \ mathrm {d} t}$
When used, results in:
{\ displaystyle {\ begin {aligned} \ oint _ {\ partial U_ {r} (a)} {\ frac {1} {(za) ^ {n}}} \ mathrm {d} z & = \ int _ { 0} ^ {1} {\ frac {2 \ pi \ mathrm {i} re ^ {2 \ pi \ mathrm {i} t}} {r ^ {n} e ^ {2 \ pi n \ mathrm {i} t}}} \ mathrm {d} t = 2 \ pi \ mathrm {i} r ^ {1-n} \ int _ {0} ^ {1} e ^ {2 \ pi \ mathrm {i} t (1 -n)} \ mathrm {d} t = {\ begin {cases} 2 \ pi \ mathrm {i} [t] _ {0} ^ {1} & {\ mbox {for}} \ n = 1 \\ {\ frac {r ^ {1-n}} {1-n}} [e ^ {2 \ pi \ mathrm {i} t (1-n)}] _ {0} ^ {1} & {\ mbox {for}} \ n \ neq 1 \ end {cases}} \\ & = {\ begin {cases} 2 \ pi \ mathrm {i} & {\ mbox {for}} \ n = 1 \\ 0 & {\ mbox {for}} \ n \ neq 1 \ end {cases}} = 2 \ pi \ mathrm {i} \ delta _ {n, 1} \ end {aligned}}}
Since every function that is holomorphic on a circular ring around can be expanded into a Laurent series , the integration results around : ${\ displaystyle f (z)}$${\ displaystyle a}$ ${\ displaystyle f (z) = \ sum _ {n = - \ infty} ^ {\ infty} c_ {n} (za) ^ {n}}$${\ displaystyle a}$
${\ displaystyle \ oint _ {\ partial U (a)} f (z) \ mathrm {d} z = \ oint _ {\ partial U (a)} \ sum _ {n = - \ infty} ^ {\ infty } c_ {n} (za) ^ {n} \ mathrm {d} z = \ sum _ {n = - \ infty} ^ {\ infty} c_ {n} \ oint _ {\ partial U (a)} ( za) ^ {n} \ mathrm {d} z}$
The above result can now be used: ${\ displaystyle \ oint _ {\ partial U_ {r} (a)} (za) ^ {n} \ mathrm {d} z = 2 \ pi \ mathrm {i} \ delta _ {n, -1}}$
${\ displaystyle \ oint _ {\ partial U (a)} f (z) \ mathrm {d} z = \ sum _ {n = - \ infty} ^ {\ infty} c_ {n} 2 \ pi \ mathrm { i} \ delta _ {n, -1} = 2 \ pi \ mathrm {i} \, c _ {- 1} = 2 \ pi \ mathrm {i} \, {\ text {Res}} _ {a} ( f)}$,
where the expansion coefficient was called the residual . ${\ displaystyle c _ {- 1}}$
## Derivation
The following derivation, which presupposes the continuous complex differentiability, leads the complex integral back to real two-dimensional integrals.
Be with and with . Then applies to the integral along the curve in the complex plane, or to the equivalent line integral along the curve ${\ displaystyle z = x + iy \ in \ mathbb {C}}$${\ displaystyle x, y \ in \ mathbb {R}}$${\ displaystyle f (z) = f (x, y) = u (x, y) + iv (x, y) \ in \ mathbb {C}}$${\ displaystyle u, v \ in \ mathbb {R}}$${\ displaystyle \ gamma (z)}$
${\ displaystyle C (x, y) = {\ begin {pmatrix} \ Re (\ gamma (z)) \\\ Im (\ gamma (z)) \ end {pmatrix}} = {\ begin {pmatrix} \ Re (\ gamma (x, y)) \\\ Im (\ gamma (x, y)) \ end {pmatrix}}}$
in the real plane ${\ displaystyle \ mathbb {R} ^ {2}}$
{\ displaystyle {\ begin {aligned} {\ underset {\ gamma \ subset \ mathbb {C}} {\ int}} f (z) \, dz & = {\ underset {C \ subset \ mathbb {R} ^ { 2}} {\ int}} f (x, y) \, (dx + idy) = {\ underset {C \ subset \ mathbb {R} ^ {2}} {\ int}} {\ begin {pmatrix} f (x, y) \\ if (x, y) \ end {pmatrix}} \ cdot {\ begin {pmatrix} dx \\ dy \ end {pmatrix}} \\ & = {\ underset {C \ subset \ mathbb {R} ^ {2}} {\ int}} {\ begin {pmatrix} u (x, y) \\ - v (x, y) \ end {pmatrix}} \ cdot {\ begin {pmatrix} dx \\ dy \ end {pmatrix}} + i {\ underset {C \ subset \ mathbb {R} ^ {2}} {\ int}} {\ begin {pmatrix} v (x, y) \\ u (x , y) \ end {pmatrix}} \ cdot {\ begin {pmatrix} dx \\ dy \ end {pmatrix}} \ end {aligned}}}
The complex curve integral was thus expressed by two real curve integrals.
For a closed curve that borders a simply connected area S, Gauss's theorem (here the continuity of the partial derivatives is used) can be applied ${\ displaystyle C = \ partial S}$
{\ displaystyle {\ begin {aligned} {\ underset {\ gamma \ subset \ mathbb {C}} {\ oint}} f (z) \, dz & = {\ underset {S \ subset \ mathbb {R} ^ { 2}} {\ int}} {\ begin {pmatrix} \ partial _ {x} \\\ partial _ {y} \ end {pmatrix}} \ cdot {\ begin {pmatrix} u \\ - v \ end { pmatrix}} dxdy + i {\ underset {S \ subset \ mathbb {R} ^ {2}} {\ int}} {\ begin {pmatrix} \ partial _ {x} \\\ partial _ {y} \ end {pmatrix}} \ cdot {\ begin {pmatrix} v \\ u \ end {pmatrix}} dxdy \\ & = {\ underset {S \ subset \ mathbb {R} ^ {2}} {\ int}} \ left \ {\ partial _ {x} u- \ partial _ {y} v \ right \} dxdy + i {\ underset {S \ subset \ mathbb {R} ^ {2}} {\ int}} \ left \ {\ partial _ {x} v + \ partial _ {y} u \ right \} dxdy \ end {aligned}}}
or alternatively the Stokes theorem
{\ displaystyle {\ begin {aligned} {\ underset {\ gamma \ subset \ mathbb {C}} {\ oint}} f (z) \, dz & = {\ underset {S \ subset \ mathbb {R} ^ { 2}} {\ int}} \ left [{\ begin {pmatrix} \ partial _ {x} \\\ partial _ {y} \\ 0 \ end {pmatrix}} \ times {\ begin {pmatrix} u \ \ -v \\ 0 \ end {pmatrix}} \ right] _ {3} dxdy + i {\ underset {S \ subset \ mathbb {R} ^ {2}} {\ int}} \ left [{\ begin {pmatrix} \ partial _ {x} \\\ partial _ {y} \\ 0 \ end {pmatrix}} \ times {\ begin {pmatrix} v \\ u \\ 0 \ end {pmatrix}} \ right] _ {3} dxdy \\ & = {\ underset {S \ subset \ mathbb {R} ^ {2}} {\ int}} \ left \ {- \ partial _ {x} v- \ partial _ {y} u \ right \} dxdy + i {\ underset {S \ subset \ mathbb {R} ^ {2}} {\ int}} \ left \ {\ partial _ {x} u- \ partial _ {y} v \ right \} dxdy \ end {aligned}}}
If the function in S is complex differentiable , the Cauchy-Riemann differential equations must there${\ displaystyle f (z)}$
${\ displaystyle \ partial _ {x} u = \ partial _ {y} v}$ and ${\ displaystyle \ partial _ {x} v = - \ partial _ {y} u}$
hold, so that the integrands above (regardless of whether in the Gauss or Stokes version) vanish:
${\ displaystyle {\ underset {\ gamma} {\ oint}} f (z) \, dz = 0}$
Thus, Cauchy's integral theorem for holomorphic functions on simply connected domains is proven.
### Cauchy's integral theorem with Wirtinger calculus and Stokes' theorem
The Cauchy integral theorem is obtained as slight consequence of the set of Stokes when the Wirtinger derivatives can bring to bear. To prove the integral theorem, the calculation of the curve integral is understood as the integration of the complex-valued differential form
${\ displaystyle \ omega = f (z) dz}$
via the closed curve that runs around the simply contiguous and bordered area . ${\ displaystyle C}$${\ displaystyle C = \ partial S}$ ${\ displaystyle S}$
The Wirtinger calculus says that the differential is the representation ${\ displaystyle df}$
${\ displaystyle df = {\ frac {\ partial f} {\ partial z}} dz + {\ frac {\ partial f} {\ partial {\ bar {z}}}} {d {\ bar {z}}} }$
has what immediately
${\ displaystyle d {\ omega} = df \ wedge dz = {{\ frac {\ partial f} {\ partial z}} dz} \ wedge dz + {{\ frac {\ partial f} {\ partial {\ bar { z}}}} {d {\ bar {z}}}} \ wedge dz}$
follows.
Well first is fundamental
${\ displaystyle dz \ wedge dz = 0}$
Furthermore, the assumed means Holomorphiebedingung for after Wirtinger derivatives nothing more than ${\ displaystyle f}$
${\ displaystyle {\ frac {\ partial f} {\ partial {\ bar {z}}}} = 0}$ ,
what immediately
${\ displaystyle {{\ frac {\ partial f} {\ partial {\ bar {z}}}} {d {\ bar {z}}}} \ wedge dz = 0}$
entails.
So the overall result is:
${\ displaystyle d {\ omega} = 0}$
and finally by means of Stokes' theorem :
${\ displaystyle \ int _ {C} f (z) dz = \ int _ {\ partial S} \ omega = \ int _ {S} \ mathrm {d} \ omega = \ int _ {S} \ mathrm {0 } = 0}$
#### annotation
With the help of Goursat's integral lemma , it can be shown that complex differentiability alone - i.e. without the additional assumption of continuity of the derivatives! - Cauchy's integral theorem and then also the existence of all higher derivatives result. This approach to Cauchy's integral theorem bypasses Stokes' theorem and is preferable from a didactic point of view.
## literature
• Kurt Endl, Wolfgang Luh : Analysis. Volume 3: Function Theory, Differential Equations. 6th revised edition. Aula-Verlag, Wiesbaden 1987, ISBN 3-89104-456-9 , p. 143, sentence 4.7.3
• Wolfgang Fischer, Ingo Lieb : Function theory. 7th improved edition. Vieweg, Braunschweig a. a. 1994, ISBN 3-528-67247-1 , p. 57, chapter 3, sentence 1.4 ( Vieweg study. Advanced course in mathematics 47).
• Günter Bärwolff : Higher Mathematics for Natural Scientists and Engineers. 2nd edition, 1st corrected reprint. Spectrum Academic Publishing House, Munich a. a. 2009, ISBN 978-3-8274-1688-9 .
• Klaus Jänich : Introduction to Function Theory . 2nd Edition. Springer-Verlag, Berlin (inter alia) 1980, ISBN 3-540-10032-6 .
## Individual evidence
1. ^ Klaus Jänich : Introduction to Function Theory . 2nd Edition. Springer-Verlag, Berlin (inter alia) 1980, ISBN 3-540-10032-6 , pp. 19-20 .
2. ^ Klaus Jänich : Introduction to Function Theory . 2nd Edition. Springer-Verlag, Berlin (inter alia) 1980, ISBN 3-540-10032-6 , pp. 15, 20 .
3. ^ Klaus Jänich : Introduction to Function Theory . 2nd Edition. Springer-Verlag, Berlin (inter alia) 1980, ISBN 3-540-10032-6 , pp. 16, 20 . | 4,425 | 14,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 86, "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} | 3.9375 | 4 | CC-MAIN-2024-26 | latest | en | 0.85996 |
https://www.slideserve.com/moeshe/csce-3110-data-structures-algorithm-analysis | 1,513,585,080,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948609934.85/warc/CC-MAIN-20171218063927-20171218085927-00356.warc.gz | 812,111,715 | 15,720 | 1 / 28
# CSCE 3110 Data Structures & Algorithm Analysis - PowerPoint PPT Presentation
CSCE 3110 Data Structures & Algorithm Analysis. Rada Mihalcea http://www.cs.unt.edu/~rada/CSCE3110 Binary Search Trees Reading: Chap. 4 (4.3) Weiss. A Taxonomy of Trees. General Trees – any number of children / node Binary Trees – max 2 children / node Heaps – parent < (>) children
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### CSCE 3110Data Structures & Algorithm Analysis
Binary Search Trees
• General Trees – any number of children / node
• Binary Trees – max 2 children / node
• Heaps – parent < (>) children
• Binary Search Trees
• Binary search tree
• Every element has a unique key.
• The keys in a nonempty left subtree (right subtree) are smaller (larger) than the key in the root of subtree.
• The left and right subtrees are also binary search trees.
Binary Search Trees (BST) are a type of Binary Trees with a special organization of data.
This data organization leads to O(log n) complexity for searches, insertions and deletions in certain types of the BST (balanced trees).
O(h) in general
Binary Search Trees
Binary Search Algorithm special organization of data.
Binary Search algorithm of an array of sorted items reduces the search space by one half after each comparison
34
41
56
63
72
89
95
0 1 2 3 4 5 6
72
89
95
34
41
56
4 5 6
0 1 2
72
95
34
56
4 6
0 2
Organization Rule for BST special organization of data.
• the values in all nodes in the left subtree of a node are less than
• the node value
• the values in all nodes in the right subtree of a node are greater
• than the node values
63
89
41
72
95
34
56
Binary Tree special organization of data.
typedef struct tnode *ptnode;typedef struct node { short int key; ptnode right, left;
} ;
BST Operations: Search special organization of data.
• Searching in the BST
• method search(key)
• implements the binary search based on comparison of the items
• in the tree
• the items in the BST must be comparable (e.g integers, string, etc.)
• The search starts at the root. It probes down, comparing the
• values in each node with the target, till it finds the first item equal
• to the target. Returns this item or null if there is none.
Search in BST - Pseudocode special organization of data.
if the tree is empty
return NULL
else if the item in the node equals the target
return the node value
else if the item in the node is greater than the target
return the result of searching the left subtree
else if the item in the node is smaller than the target
return the result of searching the right subtree
Search in a BST: C code special organization of data.
Ptnode search(ptnode root, int key)
{
/* return a pointer to the node that
contains key. If there is no such node, return NULL */
if (!root) return NULL;
if (key == root->key) return root;
if (key < root->key)
return search(root->left,key);
return search(root->right,key);
}
BST Operations: Insertion special organization of data.
• method insert(key)
• places a new item near the frontier of the BST while retaining its organization of data:
• starting at the root it probes down the tree till it finds a node whose left or right pointer is empty and is a logical place for the new value
• uses a binary search to locate the insertion point
• is based on comparisons of the new item and values of nodes in the BST
• Elements in nodes must be comparable!
Case 1: special organization of data. The Tree is Empty
• Set the root to a new node containing the item
Case 2: The Tree is Not Empty
• Call a recursive helper method to insert the item
10 > 7
10 > 9
10
7
5
9
4
6
8
10
Insertion in BST - Pseudocode special organization of data.
if tree is empty
create a root node with the new key
else
compare key with the top node
ifkey = node key
replace the node with the new value
else if key > node key
compare key with the right subtree:
if subtree is empty create a leaf node
else key < node key
compare key with the left subtree:
if the subtree is empty create a leaf node
else add key to the left subtree
Insertion into a BST: C code special organization of data.
void insert (ptnode *node, int key)
{
ptnode ptr,
temp = search(*node, key);
if (temp || !(*node)) {
ptr = (ptnode) malloc(sizeof(tnode));
if (IS_FULL(ptr)) {
fprintf(stderr, “The memory is full\n”);
exit(1);
}
ptr->key = key;
ptr->left = ptr->right = NULL;
if (*node)
if (key<temp->key) temp->left=ptr;
else temp->right = ptr;
else *node = ptr;
}
}
BST Shapes special organization of data.
• The order of supplying the data determines where it is placed in the BST , which determines the shape of the BST
• Create BSTs from the same set of data presented each time in a different order:
• a) 17 4 14 19 15 7 9 3 16 10
• b) 9 10 17 4 3 7 14 16 15 19
• c) 19 17 16 15 14 10 9 7 4 3 can you guess this shape?
BST Operations: Removal special organization of data.
• removes a specified item from the BST and adjusts the tree
• uses a binary search to locate the target item:
• starting at the root it probes down the tree till it finds the target or reaches a leaf node (target not in the tree)
• removal of a node must not leave a ‘gap’ in the tree,
Removal in BST - Pseudocode special organization of data.
method remove (key)
I if the tree is empty return false
II Attempt to locate the node containing the target using the binary search algorithm
else the target is found, so remove its node:
Case 1:if the node has 2 empty subtrees
replace the link in the parent with null
Case 2: if the node has a left and a right subtree
- replace the node's value with the max value in the
left subtree
- delete the max node in the left subtree
Removal in BST - Pseudocode special organization of data.
Case 3: if the node has no left child
- link the parent of the node
- to the right (non-empty) subtree
Case 4:if the node has no right child
- link the parent of the target
- to the left (non-empty) subtree
Removal in BST: Example special organization of data.
Case 1: removing a node with 2 EMPTY SUBTREES
parent
7
cursor
5
9
4
6
8
10
7
Removing 4
replace the link in the parent with null
5
9
6
8
10
Removal in BST: Example special organization of data.
Case 2: removing a node with 2 SUBTREES
- replace the node's value with the max value in the left subtree
- delete the max node in the left subtree
What other element
can be used as
replacement?
Removing 7
cursor
cursor
7
6
5
9
5
9
4
6
8
10
4
8
10
Removal in BST: Example special organization of data.
Case3: removing a node with 1 EMPTY SUBTREE
the node has no left child:
link the parent of the node to the right (non-empty) subtree
parent
parent
7
7
cursor
5
9
5
9
cursor
6
8
10
6
8
10
Removal in BST: Example special organization of data.
Case4: removing a node with 1 EMPTY SUBTREE
the node has no right child:
link the parent of the node to the left (non-empty) subtree
Removing 5
parent
parent
cursor
7
7
cursor
5
9
5
9
4
8
10
8
10
4
Analysis of BST Operations special organization of data.
• The complexity of operationsget, insert and remove in BST is O(h) , where h is the height.
• O(log n) when the tree is balanced. The updating operations cause the tree to become unbalanced.
• The tree can degenerate to a linear shape and the operations will become O (n)
Best Case special organization of data.
BST tree = new BST();
tree.insert ("E");
tree.insert ("C");
tree.insert ("D");
tree.insert ("A");
tree.insert ("H");
tree.insert ("F");
tree.insert ("K");
K
H
F
E
D
C
A
Output:
Worst Case special organization of data.
BST tree = new BST();
for (int i = 1; i <= 8; i++)
tree.insert (i);
>>>> Items in worst order:
8
7
6
5
4
3
2
1
Output:
Random Case special organization of data.
tree = new BST ();
for (int i = 1; i <= 8; i++)
tree.insert(random());
Output:
>>>> Items in random order:
X
U
P
O
H
F
B
Applications for BST special organization of data.
• Sorting with binary search trees
• Input: unsorted array
• Output: sorted array
• Algorithm ?
• Running time ?
Better Search Trees special organization of data.
Prevent the degeneration of the BST :
• A BST can be set up to maintain balance during updating operations (insertions and removals)
• Types of ST which maintain the optimal performance:
• splay trees
• AVL trees
• 2-4 Trees
• Red-Black trees
• B-trees | 2,396 | 9,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-51 | latest | en | 0.78917 |
https://pynbody.github.io/pynbody/tutorials/profile.html | 1,603,671,829,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00391.warc.gz | 486,887,608 | 7,513 | # Profiles in Pynbody¶
Making profiles of all kinds of quantities is easy – here’s a simple example that shows how to plot a density profile:
import pynbody
import matplotlib.pylab as plt
# load the snapshot and set to physical units
h = s.halos()
# center on the largest halo and align the disk
pynbody.analysis.angmom.faceon(h[1])
# convert all units to something reasonable (kpc, Msol etc)
s.physical_units()
# create a profile object for the stars (by default this is a 2D profile)
p = pynbody.analysis.profile.Profile(h[1].s, vmin =.01, max=50)
# make the figure and sub plots
f, axs = plt.subplots(1,2,figsize=(14,6))
# make the plot
axs[0].plot(p['rbins'],p['density'], 'k')
axs[0].semilogy()
axs[0].set_xlabel('R [kpc]')
axs[0].set_ylabel(r'$\Sigma_{\star}$ [M$_{\odot}$ kpc$^{-2}$]')
# make a 3D density plot of the dark matter (note ndim=3 in the constructor below)
p = pynbody.analysis.profile.Profile(h[1].d,min=.01,max=50,ndim=3)
axs[1].plot(p['rbins'],p['density'], 'k')
axs[1].semilogy()
axs[1].set_xlabel('R [kpc]')
axs[1].set_ylabel(r'$\rho_{DM}$ [M$_{\odot}$ kpc$^{-3}$]')
Below is a more extended description of the profile module.
## The Profile Class¶
The Profile class is meant to be a general-purpose class to satisfy (almost) all simulation profiling needs. Profiles are the most elementary ways to begin analyzing a simulation, so the Profile class is designed to be an extension of the syntax implemented in the SimSnap class and its derivatives.
Creating a Profile instance simply defines the bins etc. Importantly, it also stores lists of particle indices corresponding to each bin, so you can easily identify where the particles belong.
In [1]: import pynbody; from pynbody.analysis import profile; import matplotlib.pylab as plt
In [2]: s = pynbody.load('testdata/g15784.lr.01024'); s.physical_units()
In [3]: h = s.halos()
In [4]: pynbody.analysis.angmom.faceon(h[1])
Out[4]: <pynbody.transformation.GenericRotation at 0x7ff6a95f1040>
We’re centered on the main halo (as in the cookbook example above) and we make the Profile instance:
In [5]: p = profile.Profile(h[1].s, rmin='.01 kpc', rmax='50 kpc')
With the default parameters, the profile is made in the xy-plane. To make a spherically-symmetric 3D profile, specify ndim=3 when creating the profile.
In [6]: pdm_sph = profile.Profile(s.d, rmin = '.01 kpc', rmax = '250 kpc')
Even though we use s.d here (i.e. the full snapshot, not just halo 1), the whole snapshot is still centered on halo 1.
Note
You can pass unit strings to rmin and rmax and the conversion will be done automatically into whatever the current units of the snapshot are so you don’t have to explicitly do any unit conversions.
## Automatically-generated profiles¶
Many profiling functions are already implemented – see the Profile documentation for a full list with brief descriptions. You can also check the available profiles in your session using derivable_keys() just like you would for a SimSnap:
In [7]: p.derivable_keys()
Out[7]:
['weight_fn',
'mass',
'density',
'fourier',
'pattern_frequency',
'mass_enc',
'density_enc',
'dyntime',
'g_spherical',
'rotation_curve_spherical',
'j_circ',
'v_circ',
'E_circ',
'pot',
'omega',
'kappa',
'beta',
'magnitudes',
'sb',
'Q',
'X',
'jtot',
'j_theta',
'j_phi']
Additionally, any existing array can be ‘profiled’. For example, if the metallicity [Fe/H] is a derived field stored under ‘feh’, then plotting a metallicity profile is as simple as:
In [8]: plt.plot(p['rbins'].in_units('kpc'),p['feh'],'k')
Out[8]: [<matplotlib.lines.Line2D at 0x7ff668e3f160>]
In [9]: plt.xlabel('$R$ [kpc]'); plt.ylabel('[Fe/H]')
Out[9]: Text(0, 0.5, '[Fe/H]')
If the array doesn’t exist but is deriveable (check with s.derivable_keys()), it is automatically calculated.
In addition, you can define your own profiling functions in your code by using the Profile.profile_property decorator:
@profile.Profile.profile_property
def random(self):
import numpy as np
return np.random.rand(self.nbins)
Now this will be automatically derivable for any newly-created profile as 'random'.
## Calculating Derivatives and Dispersions¶
You can calculate derivatives of profiles automatically. For instance, you might be interested in d phi / dr if you’re looking at a disk. This is as easy as attaching a d_ to the profile name. For example:
In [10]: p_all = profile.Profile(s, rmin='.01 kpc', rmax='250 kpc')
In [11]: p_all['pot'][0:10] # returns the potential profile
Out[11]:
SimArray([-1883725.20605628, -1774955.4230129 , -1722218.33725498,
-1690633.68613091, -1668854.69876004, -1652878.17689281,
-1640433.23593114, -1630243.68972168, -1621605.13215674,
-1614220.30221359], 'km**2 s**-2')
In [12]: p_all['d_pot'][0:10] # returns d phi / dr from p["phi"]
Out[12]:
SimArray([43509.6536035 , 32302.66586689, 16865.02197728, 10673.15462517,
7551.40390378, 5684.51994658, 4527.07851737, 3765.77138574,
3204.80569384, 2767.54060378], 'km**2 kpc**-1 s**-2')
Similarly straightforward is the calculation of dispersions and root-mean-square values. You simply need to attach a _disp or _rms as a suffix to the profile name. To get the stellar velocity dispersion:
In [13]: plt.clf(); plt.plot(p['rbins'].in_units('kpc'),p['vr_disp'].in_units('km s^-1'),'k')
Out[13]: [<matplotlib.lines.Line2D at 0x7ff6a9f373d0>]
In [14]: plt.xlabel('$R$ [kpc]'); plt.ylabel('$\sigma_{r}$')
Out[14]: Text(0, 0.5, '$\\sigma_{r}$')
In addition to doing this by hand, you can make a QuantileProfile that can return any desired quantile range. By default, this is the mean +/- 1-sigma:
In [15]: p_quant = profile.QuantileProfile(h[1].s, rmin = '0.1 kpc', rmax = '50 kpc')
In [16]: plt.clf(); plt.plot(p_quant['rbins'], p_quant['feh'][:,1], 'k')
Out[16]: [<matplotlib.lines.Line2D at 0x7ff6a9f88dc0>]
In [17]: plt.fill_between(p_quant['rbins'], p_quant['feh'][:,0], p_quant['feh'][:,2], color = 'Grey', alpha=0.5)
Out[17]: <matplotlib.collections.PolyCollection at 0x7ff6695a4340>
In [18]: plt.xlabel('$R$ [kpc]'); plt.ylabel('[Fe/H]')
Out[18]: Text(0, 0.5, '[Fe/H]')
## Making a profile using a different quantity¶
Radial profiles are nice, but sometimes we want a “profile” using a different quantity. We might want to know, for example, how the mean metallicity varies as a function of age in the stars. Profile calls the function _calculate_x() by default and this simply returns the 3D or xy-plane radial distance, depending on the value of ndim. We can specify a different function using the calc_x keyword. Often these are simple so a lambda function can be used (e.g. if we just want to return an array) or can also be more complicated functions. For example, to make the profile of stars in halo 1 according to their age:
In [19]: s.s['age'].convert_units('Gyr')
In [20]: p_age = profile.Profile(h[1].s, calc_x = lambda x: x.s['age'], rmax = '10 Gyr')
In [21]: plt.clf(); plt.plot(p_age['rbins'], p_age['feh'], 'k', label = 'mean [Fe/H]')
Out[21]: [<matplotlib.lines.Line2D at 0x7ff6695ca8e0>]
In [22]: plt.plot(p_age['rbins'], p_age['feh_disp'], 'k--', label = 'dispersion')
Out[22]: [<matplotlib.lines.Line2D at 0x7ff6a9f40220>]
In [23]: plt.xlabel('Age [Gyr]'); plt.ylabel('[Fe/H]')
Out[23]: Text(0, 0.5, '[Fe/H]')
In [24]: plt.legend()
Out[24]: <matplotlib.legend.Legend at 0x7ff6695cac10>
## Vertical Profiles and Inclined Profiles¶
For analyzing disk structure, it is frequently useful to have a profile in the z-direction. This is done with the VerticalProfile which behaves in the same way as the Profile. Unlike in the basic class, you must specify the radial range and maximum z to be used:
In [25]: p_vert = profile.VerticalProfile(h[1].s, '3 kpc', '5 kpc', '5 kpc')
In [26]: plt.clf(); plt.plot(p_vert['rbins'].in_units('pc'), p_vert['density'].in_units('Msol pc^-3'),'k')
Out[26]: [<matplotlib.lines.Line2D at 0x7ff6695edd00>]
In [27]: plt.xlabel('$z$ [pc]'); plt.ylabel(r'$\rho_{\star}$ [M$_{\odot}$ pc$^{-3}$]')
Out[27]: Text(0, 0.5, '$\\rho_{\\star}$ [M$_{\\odot}$ pc$^{-3}$]')
Similarly, one can make inclined profiles using the VerticalProfile, but the snapshot needs to be rotated first:
In [28]: s.rotate_x(60) # rotate the snapshot by 60-degrees
Out[28]: <pynbody.transformation.GenericRotation at 0x7ff699623550>
In [29]: p_inc = profile.InclinedProfile(h[1].s, 60, rmin = '0.1 kpc', rmax = '50 kpc') | 2,532 | 8,341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.653882 |
http://guides.wkbw.com/Approach_to_Solving_Math_Problems_Utica_NY-r992520-Utica_NY.html | 1,519,023,239,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812405.3/warc/CC-MAIN-20180219052241-20180219072241-00355.warc.gz | 142,038,482 | 6,823 | # Approach to Solving Math Problems Utica NY
When learning math, getting stumped without knowing what next is very common. However, before you let go of that pen and paper and scream I hate math to the top of your lungs, take time to read this article and get more information about learning math in Utica.
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When learning math, getting stumped without knowing what next is very common. However, before you let go of that pen and paper and scream I hate math to the top of your lungs, take time to read this article.
If you have read other articles or books on how to solve math problems, this article will not really provide anything new. However, this provides you with a simple step by step approach to help you through each math problem you encounter.
How do you go about solving a difficult math problem?
Before discussing the steps, you must realize how keeping an open mind about math will improve your performance greatly. You may dislike math, but reminding yourself over and over will make you worse off. So, try assuming even just for a few minutes, that math is manageable. As it actually is.
So, here it goes.
In every math problem, you have to ask yourself 4 very important questions, which are:
1. What information am I given?In this part, you can do any one or more of the following:
• List down the information and facts given in the word problem, along with units and other details
• Point out key words and underline or circle them. There are hundreds of key words possible such as area, speed, factor, per, for every, among many others
• Watch out for units that are mixed. Units are important information, so never neglect them.
• Express the information in the form of math symbols, for example speed is distance over time.
2. What information is being asked?On this step, make sure you do the following:
• Know what the problem expects you to find. Check if it has to be expressed in certain units.
• Express your goal in terms of math symbols, if possible. If you are being asked for the area of a circle, express it as A = pi*r*r
3. What can I do with the given information?
• Play around with the facts you are given. Check what information you can get with the given and see whether it brings you closer to the information you have to solve for.
• Don t hesitate to manipulate the information one step at a time. Simplify the problem if it is relatively complicated.
• Make a picture or diagram, or even a table or chart. if the situation calls for it.
• Look for certain distinct pattern if applicable.
• If the problem seems difficult to understand, find a way to restate the problem which might be easier to grasp.
• Keep manipulating until you arrive at an answer.
4. Does my answer make sense?
• Once you have successfully found a way to solve the problem, check if it makes sense.
• Read the problem again and understand it once more.
• Check if you have the right units, as asked.
• Think of a way to confirm if your answer is right, if there is at all.
Solving math problems should not be tough if you keep your mind and heart open. Start with these 4 questions and see how you ll fare.
John has a site called http://Mathtrench.com , that offers thousands of solved math problems
Provided by ZingArticles.com
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- LSAT Prep Courses Utica NY | 1,096 | 4,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-09 | latest | en | 0.897443 |
https://goobi.tib.eu/viewer/!fulltext/856671290/242/ | 1,718,421,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00405.warc.gz | 256,050,308 | 34,177 | # Full text: Proceedings International Workshop on Mobile Mapping Technology
```A METHOD OF ROAD REPRESENTATION IN 3D MAPPING TECHNOLOGY
Tsukasa Hosomura
Kanazawa Institute of Technology
7-1 Ohgigaoka, Nonoichi-machi, Ishikawa-gun, Ishikawa 921-8501
Japan
E-mail: hosomura@neptune.kanazawa-it.ac.jp
KEY WORDS : Map, 3D-CG, Road
ABSTRACT
We are trying to make the map that everyone can understand easy, and we use approximate shape
of objects. There appeared some problems in previous study. One of the problems is a representation
method of road on map. We suppose the road is an important object to represent the map. But it is
difficult to represent all these objects correctly by using 3DCG. We describe the efficient
representation method of real road for 3D model. We used the data of road as raster for making the
map. It took a lot of time to do the work. So, we managed a road as a vector data in this study. This
realized the reduction of data and made the work efficient.
1. INTRODUCTION
Recently, various maps are made by digital
techniques. Digital map is generally used in
some systems such as car navigation system, as
multimedia spreads out many fields. So, we are
studying about the digital mapping that is
useful and can represent real view to the next
future. Trough the study of last year, we
recognized that we could make the map
beautiful and easy to understand for urban area
by using a representation method of 3DCG. But
some problems are coming up.
1.1 Requirement and Counter Plan
One of the proposals is how to represent the
road. In making a road by 3DCG, we take at
least four points of coordinates in previous
study. As the advantage of this method,
representing the outline of road correctly is
raised. But in case of representing the elements
of the road as centerline, sidewalk and so on,
this method is not efficient because it requires a
lot of coordinates data. We try to represent
centerline and sidewalk by managing the road
as a vector data to improve this. And we
consider the road through the mountain for
adding the elevation value to coordinates
consists of road from numerical map of
elevation published by
Geographical Survey Institute.
Representation of mountain, river and other
object will be needed when the scale of map
expands and we approach the real
representation. This time, we try to represent
only mountain from numerical map of elevation.
Scale of map for road is about 1.5km east to
west, 3km south to north area around the
```
## Note to user
Dear user,
In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser.
Please use one of the following browsers to display this page correctly.
Thank you. | 644 | 2,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-26 | latest | en | 0.93878 |
https://www.sourcecodeexamples.net/2024/01/dutch-national-flag-problem-cpp-solution.html | 1,726,851,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00110.warc.gz | 905,342,634 | 28,444 | # 1. Introduction
The Dutch National Flag Problem is a classic algorithm problem that involves sorting an array containing only three distinct elements. The challenge is to accomplish this in linear time and using constant space, which tests a programmer's ability to implement efficient sorting algorithms with space constraints.
## Problem
Given an array containing only elements 0, 1, and 2, the task is to sort the array in ascending order. The solution should be implemented in-place, i.e., without using extra space, and it should have linear time complexity.
# 2. Solution Steps
1. Initialize three pointers: low, mid, and high. low and mid are set to the beginning of the array, and high is set to the end.
2. Traverse the array using the mid pointer.
3. If mid points to 0, swap it with the element at low and increment both low and mid.
4. If mid points to 1, just increment mid.
5. If mid points to 2, swap it with the element at high and decrement high.
6. Continue this process until mid crosses high.
# 3. Code Program
``````#include <iostream>
#include <vector>
using namespace std;
// Utility function to swap two elements in the array
void swap(int &a, int &b) {
int temp = a;
a = b;
b = temp;
}
// Function to sort the array containing 0s, 1s, and 2s
void sortArray(vector<int>& nums) {
int low = 0, mid = 0, high = nums.size() - 1;
while (mid <= high) {
switch(nums[mid]) {
case 0:
swap(nums[low++], nums[mid++]);
break;
case 1:
mid++;
break;
case 2:
swap(nums[mid], nums[high--]);
break;
}
}
}
int main() {
vector<int> nums = {0, 1, 2, 2, 1, 0, 0, 2, 0, 1, 1, 0};
sortArray(nums);
// Print the sorted array
for (int num : nums) {
cout << num << " ";
}
cout << endl;
return 0;
}
``````
### Output:
0 0 0 0 0 1 1 1 1 2 2 2
### Explanation:
The program implements the Dutch National Flag algorithm to sort the array. It uses three pointers: low for 0s, mid for 1s, and high for 2s.
By swapping elements based on the value at the mid pointer and adjusting the pointers accordingly, the array is sorted in-place.
This method ensures linear time complexity and constant space usage, making it an efficient solution for sorting arrays with a limited set of distinct elements. | 590 | 2,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.754638 |
https://learn.saylor.org/mod/book/tool/print/index.php?id=27252&chapterid=3262 | 1,675,468,262,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00128.warc.gz | 378,664,694 | 9,938 | # Try It Now
## Exercises
1. List all partitions of the set A = {a, b, c}.
2. A student, on an exam paper, defined the term partition the following way: “Let A be a set. A partition of A is any set of nonempty subsets A1, A2, A3, . . . of A such that each element of A is in one of the subsets.” Is this definition correct? Why?
3. Show that {{2n| n ∈ ℤ},{2n+ 1 | n ∈ ℤ}} is a partition of ℤ. Describe this partition using only words.
4. A survey of 90 people, 47 of them played tennis and 42 of them swam. If 17 of them participated in both activities, how many of them participated in neither?
5. Regarding the Theorem 2.3.9,
1. (a) Use the two set inclusion-exclusion law to derive the three set inclusion- exclusion law. Note: A knowledge of basic set laws is needed for this exercise.
2. (b) State and derive the inclusion-exclusion law for four sets.
6. The definition of ℚ = {a/b| a, b ∈ ℤ, b ≠ 0} given in Chapter 1 is awkward. If we use the definition to list elements in ℚ, we will have duplications such as $\frac{1}{2}$$\frac{-2}{-4}$, and $\frac{300}{600}$. Try to write a more precise definition of the rational numbers so that there is no duplication of elements. | 346 | 1,185 | {"found_math": true, "script_math_tex": 3, "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} | 3.6875 | 4 | CC-MAIN-2023-06 | latest | en | 0.886203 |
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