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https://www.slideserve.com/rmisty/structure-from-motion-powerpoint-ppt-presentation | 1,623,755,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00377.warc.gz | 907,040,085 | 22,255 | Structure from motion
# Structure from motion
## Structure from motion
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##### Presentation Transcript
1. Structure from motion
2. Multiple-view geometry questions • Scene geometry (structure): Given 2D point matches in two or more images, where are the corresponding points in 3D? • Correspondence (stereo matching): Given a point in just one image, how does it constrain the position of the corresponding point in another image? • Camera geometry (motion): Given a set of corresponding points in two or more images, what are the camera matrices for these views?
3. Xj x1j x3j x2j P1 P3 P2 Structure from motion • Given: m images of n fixed 3D points • xij = Pi Xj , i = 1, … , m, j = 1, … , n • Problem: estimate m projection matrices Piand n 3D points Xjfrom the mn correspondences xij
4. Structure from motion ambiguity • If we scale the entire scene by some factor k and, at the same time, scale the camera matrices by the factor of 1/k, the projections of the scene points in the image remain exactly the same: • It is impossible to recover the absolute scale of the scene!
5. Structure from motion ambiguity • If we scale the entire scene by some factor k and, at the same time, scale the camera matrices by the factor of 1/k, the projections of the scene points in the image remain exactly the same • More generally: if we transform the scene using a transformation Q and apply the inverse transformation to the camera matrices, then the images do not change
6. Reconstruction ambiguity: Similarity
7. Reconstruction ambiguity: Projective
8. Projective ambiguity
9. From projective to affine
10. From affine to similarity
11. Hierarchy of 3D transformations • With no constraints on the camera calibration matrix or on the scene, we get a projective reconstruction • Need additional information to upgrade the reconstruction to affine, similarity, or Euclidean Projective 15dof Preserves intersection and tangency Preserves parallellism, volume ratios Affine 12dof Similarity 7dof Preserves angles, ratios of length Euclidean 6dof Preserves angles, lengths
12. Structure from motion • Let’s start with affine cameras (the math is easier) center atinfinity
13. Recall: Orthographic Projection • Special case of perspective projection • Distance from center of projection to image plane is infinite • Projection matrix: Image World Slide by Steve Seitz
14. Affine cameras Orthographic Projection Parallel Projection
15. x a2 X a1 Affine cameras • A general affine camera combines the effects of an affine transformation of the 3D space, orthographic projection, and an affine transformation of the image: • Affine projection is a linear mapping + translation in inhomogeneous coordinates Projection ofworld origin
16. Affine structure from motion • Given: m images of n fixed 3D points: • xij = Ai Xj + bi , i = 1,… , m, j = 1, … , n • Problem: use the mn correspondences xij to estimate m projection matrices Ai and translation vectors bi, and n points Xj • The reconstruction is defined up to an arbitrary affine transformation Q (12 degrees of freedom): • We have 2mn knowns and 8m + 3n unknowns (minus 12 dof for affine ambiguity) • Thus, we must have 2mn >= 8m + 3n – 12 • For two views, we need four point correspondences
17. Affine structure from motion • Centering: subtract the centroid of the image points • For simplicity, assume that the origin of the world coordinate system is at the centroid of the 3D points • After centering, each normalized point xij is related to the 3D point Xi by
18. Affine structure from motion • Let’s create a 2m×n data (measurement) matrix: cameras(2m) points (n) C. Tomasi and T. Kanade. Shape and motion from image streams under orthography: A factorization method.IJCV, 9(2):137-154, November 1992.
19. Affine structure from motion • Let’s create a 2m×n data (measurement) matrix: points (3× n) cameras(2m ×3) The measurement matrix D = MS must have rank 3! C. Tomasi and T. Kanade. Shape and motion from image streams under orthography: A factorization method.IJCV, 9(2):137-154, November 1992.
20. Factorizing the measurement matrix Source: M. Hebert
21. Factorizing the measurement matrix • Singular value decomposition of D: Source: M. Hebert
22. Factorizing the measurement matrix • Singular value decomposition of D: Source: M. Hebert
23. Factorizing the measurement matrix • Obtaining a factorization from SVD: Source: M. Hebert
24. Factorizing the measurement matrix • Obtaining a factorization from SVD: This decomposition minimizes|D-MS|2 Source: M. Hebert
25. Affine ambiguity • The decomposition is not unique. We get the same D by using any 3×3 matrix C and applying the transformations M → MC, S →C-1S • That is because we have only an affine transformation and we have not enforced any Euclidean constraints (like forcing the image axis to be perpendicular, for example) Source: M. Hebert
26. Eliminating the affine ambiguity • Orthographic: image axes are perpendicular and scale is 1 • This translates into 3m equations in L = CCT: • Ai L AiT = Id, i = 1, …, m • Solve for L • Recover C from L by Cholesky decomposition: L = CCT • Update M and S: M = MC, S = C-1S a1 · a2 = 0 x |a1|2 = |a2|2= 1 a2 X a1 Source: M. Hebert
27. Algorithm summary • Given: m images and n features xij • For each image i, center the feature coordinates • Construct a 2m ×n measurement matrix D: • Column j contains the projection of point j in all views • Row i contains one coordinate of the projections of all the n points in image i • Factorize D: • Compute SVD: D = U W VT • Create U3 by taking the first 3 columns of U • Create V3 by taking the first 3 columns of V • Create W3 by taking the upper left 3 × 3 block ofW • Create the motion and shape matrices: • M = U3W3½ and S = W3½V3T(or M = U3 and S = W3V3T) • Eliminate affine ambiguity Source: M. Hebert
28. Reconstruction results C. Tomasi and T. Kanade. Shape and motion from image streams under orthography: A factorization method.IJCV, 9(2):137-154, November 1992.
29. Dealing with missing data • So far, we have assumed that all points are visible in all views • In reality, the measurement matrix typically looks something like this: cameras points
30. Dealing with missing data • Possible solution: decompose matrix into dense sub-blocks, factorize each sub-block, and fuse the results • Finding dense maximal sub-blocks of the matrix is NP-complete (equivalent to finding maximal cliques in a graph) • Incremental bilinear refinement • Perform factorization on a dense sub-block (2) Solve for a new 3D point visible by at least two known cameras (linear least squares) (3) Solve for a new camera that sees at least three known 3D points (linear least squares) F. Rothganger, S. Lazebnik, C. Schmid, and J. Ponce. Segmenting, Modeling, and Matching Video Clips Containing Multiple Moving Objects. PAMI 2007.
31. Xj x1j x3j x2j P1 P3 P2 Projective structure from motion • Given: m images of n fixed 3D points • zijxij = Pi Xj , i = 1,… , m, j = 1, … , n • Problem: estimate m projection matrices Pi and n 3D points Xj from the mn correspondences xij
32. Projective structure from motion • Given: m images of n fixed 3D points • zijxij = Pi Xj , i = 1,… , m, j = 1, … , n • Problem: estimate m projection matrices Pi and n 3D points Xj from the mn correspondences xij • With no calibration info, cameras and points can only be recovered up to a 4x4 projective transformation Q: • X → QX, P → PQ-1 • We can solve for structure and motion when • 2mn >= 11m +3n – 15 • For two cameras, at least 7 points are needed
33. Projective SFM: Two-camera case • Compute fundamental matrixF between the two views • First camera matrix: [I|0]Q-1 • Second camera matrix: [A|b]Q-1 • Let • Then b: epipole (FTb = 0), A = –[b×]F F&P sec. 13.3.1
34. Projective factorization • If we knew the depths z, we could factorize D to estimate M and S • If we knew M and S, we could solve for z • Solution: iterative approach (alternate between above two steps) points (4× n) cameras(3m ×4) D = MS has rank 4
35. Sequential structure from motion • Initialize motion from two images using fundamental matrix • Initialize structure • For each additional view: • Determine projection matrix of new camera using all the known 3D points that are visible in its image – calibration points cameras
36. Sequential structure from motion • Initialize motion from two images using fundamental matrix • Initialize structure • For each additional view: • Determine projection matrix of new camera using all the known 3D points that are visible in its image – calibration • Refine and extend structure: compute new 3D points, re-optimize existing points that are also seen by this camera – triangulation points cameras
37. Sequential structure from motion • Initialize motion from two images using fundamental matrix • Initialize structure • For each additional view: • Determine projection matrix of new camera using all the known 3D points that are visible in its image – calibration • Refine and extend structure: compute new 3D points, re-optimize existing points that are also seen by this camera – triangulation • Refine structure and motion: bundle adjustment points cameras
38. Bundle adjustment • Non-linear method for refining structure and motion • Minimizing reprojection error Xj P1Xj x3j x1j P3Xj P2Xj x2j P1 P3 P2
39. Self-calibration • Self-calibration (auto-calibration) is the process of determining intrinsic camera parameters directly from uncalibrated images • For example, when the images are acquired by a single moving camera, we can use the constraint that the intrinsic parameter matrix remains fixed for all the images • Compute initial projective reconstruction and find 3D projective transformation matrix Q such that all camera matrices are in the form Pi = K [Ri | ti] • Can use constraints on the form of the calibration matrix: zero skew
40. Summary: Structure from motion • Ambiguity • Affine structure from motion: factorization • Dealing with missing data • Projective structure from motion: two views • Projective structure from motion: iterative factorization • Bundle adjustment • Self-calibration
41. Summary: 3D geometric vision • Single-view geometry • The pinhole camera model • Variation: orthographic projection • The perspective projection matrix • Intrinsic parameters • Extrinsic parameters • Calibration • Multiple-view geometry • Triangulation • The epipolar constraint • Essential matrix and fundamental matrix • Stereo • Binocular, multi-view • Structure from motion • Reconstruction ambiguity • Affine SFM • Projective SFM | 2,653 | 10,700 | {"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-2021-25 | latest | en | 0.78068 |
https://au.mathworks.com/matlabcentral/cody/problems/594-look-and-say-sequence/solutions/1466815 | 1,606,680,677,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141202590.44/warc/CC-MAIN-20201129184455-20201129214455-00527.warc.gz | 205,907,010 | 16,964 | Cody
# Problem 594. "Look and say" sequence
Solution 1466815
Submitted on 20 Mar 2018 by Naimul Hassan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
assert(isequal(look_and_say([1]),[1 1]))
NEXT = 1 1
2 Pass
assert(isequal(look_and_say([1 1 1 1 1]),[5 1]))
NEXT = 5 1
3 Pass
assert(isequal(look_and_say([1 3 3 1 5 2 2]),[1 1 2 3 1 1 1 5 2 2]))
NEXT = 1 1 2 3 1 1 1 5 2 2
4 Pass
assert(isequal(look_and_say([8 6 7 5 3 0 9]),[1 8 1 6 1 7 1 5 1 3 1 0 1 9]))
NEXT = 1 8 1 6 1 7 1 5 1 3 1 0 1 9
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https://palmettoinsider.com/isobars-are-lines-on-a-weather-map-representing-what.html | 1,620,436,157,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00613.warc.gz | 461,183,289 | 8,607 | # Isobars Are Lines On A Weather Map Representing What?
What data do you no longer in finding on a weather map? solution choices . cold entrance. heat entrance. prime or low force. seasons. Tags: Question 8 . SURVEY . 30 seconds . Q. What are lines on a weather map connecting points of equal pressure? answer possible choices Which map absolute best represents a low force system in the NORTHERN Hemisphere? answer alternativesMany nationwide research institutes supply a vast variety numerically modelled weather forecast data. Weather services like WeatherOnline or the Met Office are the usage of the knowledge for plotting weather maps, such because the mean-sea-level stress chart, which may be identified simply as floor chart.The main characteristic of a surface map are the isobars, lines which join points of equivalent mean-sea-level-pressure.Isobars are lines on a weather map joining together puts of equal atmospheric power. On the map the isobar marked 1004 represents a space of high power, whilst the isobar marked 976...Isobars are lines of equal atmospheric power drawn on a meteorological map. Each line passes through a pressure of a given price, provided positive regulations are adopted.Isobar Weather Maps Explained Isobars are lines linking points on the earth's floor (or at fastened heights above the earth's surface) where the air drive remains constant. Isobars are like contour lines on an ordnance survey map that mark points on the earth's floor that are at the similar peak above a datum point (typically sea degree).
## Weather Facts: Isobars on surface maps | weatheronline.co.uk
Isobars and Isotherms Isobars and isotherms are lines on weather maps which represent patterns of force and temperature, respectively. They show how temperature and pressure are changing over space and so help describe the large-scale weather patterns throughout a area within the map. Why do I care?weather maps, and have focused on drawing isotherms (lines of equivalent temperature) and especially importantly isobars (lines of equal drive). It is important to remember that surface weather maps (as opposed to upper atmospheric weather maps) plot isobars at four mb durations. Be sure you apply this convention in drawing your weather maps on yourIsobars are lines on a weather map representing what? lines of equivalent air drive. What direction is a south east wind coming from? Where is the Coriolis power greatest. top latittude. Closely spaced isobars indicate. top pressure gradient. What would the drive tendency be if stormy weather is approaching. Low. A land breeze generallyisobar definition: 1. a line drawn on a weather map joining all the puts that have the same air pressure 2. a lineā¦. Learn extra.
### Isobars and fronts - Weather systems - GCSE Geography
An isobar is a line connecting puts of equal _____. drive. rain and top winds. When a tropical cyclone passes an area, on the other hand, there are in most cases several days of clear, dry weather. Why would possibly this occur? The cyclone must have an outflow aloft, which feeds descending air surrounding the storm. Isobars are lines on a weather mapIsobars lines of continuous pressureA line drawn on a weather map connecting points of equal drive is named an "isobar". Isobars are generated from imply sea-level drive stories and are given in millibars. The diagram under depicts a pair of pattern isobars.The term "isobar" originates from the Greek, isos (equal) and baros (weight). The lines are drawn using data from mean sea-level power studies. Because many of the weather stations are now not located at sea point, however at a certain elevation, the power measured at each location has to be transformed into sea level power before the isobars are drawn.Using a black coloured pencil, frivolously draw lines connecting identical values of sea-level pressure. Remember, those lines, referred to as isobars, don't cross every other. Isobars are normally drawn for every 4 millibars, the usage of a thousand millibars as the place to begin.To learn air drive on a floor analysis weather map, check for isobars (iso = equivalent, bar = drive) - plain, curved lines that point out spaces of equivalent air power. Isobars play a main position in figuring out the rate and path of wind.
### Isobars: lines of constant pressure
Isobars lines of continuous power
A line drawn on a weather map connecting points of equivalent force is called an "isobar". Isobars are generated from mean sea-level power studies and are given in millibars.
The diagram under depicts a pair of pattern isobars. At each and every level along the highest isobar, the force is 996 mb while at every level along the ground isobar, the pressure is 1000 mb. Points above the a thousand mb isobar have a lower force and issues beneath that isobar have a upper drive.
Any point mendacity in between those two isobars will have to have a force somewhere between 996 mb and a thousand mb. Point A, for instance, has a force of 998 mb and is due to this fact located between the 996 mb isobar and the one thousand mb isobar.
Sea-level pressure studies are to be had every hour, which means that isobar maps are likewise available each hour. The solid blue contours (within the map beneath) represent isobars and the numbers along decided on contours point out the force price of that specific isobar.
Such maps are helpful for locating areas of low and high force, which correspond to the positions of surface cyclones and anticyclones. A map of isobars may be useful for finding robust force gradients, which are identifiable by way of a tight packing of the isobars. Stronger winds are associated with more potent drive gradients.
variation with height pressure surfaces | 1,239 | 5,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.889518 |
http://www.numbersaplenty.com/40401131131 | 1,601,186,147,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400265461.58/warc/CC-MAIN-20200927054550-20200927084550-00134.warc.gz | 208,957,236 | 3,557 | Search a number
40401131131 = 371091922463
BaseRepresentation
bin100101101000000110…
…000101011001111011
310212021121212002222001
4211220012011121323
51130220142144011
630320544241431
72630114643214
oct455006053173
9125247762861
1040401131131
1116152426879
1279b6310277
133a6b1cb96a
141d5398840b
1510b6d207c1
hex96818567b
40401131131 has 4 divisors (see below), whose sum is σ = 41493053632. Its totient is φ = 39309208632.
The previous prime is 40401131123. The next prime is 40401131167. The reversal of 40401131131 is 13113110404.
Adding to 40401131131 its reverse (13113110404), we get a palindrome (53514241535).
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 40401131131 - 23 = 40401131123 is a prime.
It is a super-2 number, since 2×404011311312 (a number of 22 digits) contains 22 as substring.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 40401131099 and 40401131108.
It is not an unprimeable number, because it can be changed into a prime (40401131101) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 545961195 + ... + 545961268.
It is an arithmetic number, because the mean of its divisors is an integer number (10373263408).
Almost surely, 240401131131 is an apocalyptic number.
40401131131 is a deficient number, since it is larger than the sum of its proper divisors (1091922501).
40401131131 is a wasteful number, since it uses less digits than its factorization.
40401131131 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1091922500.
The product of its (nonzero) digits is 144, while the sum is 19.
The spelling of 40401131131 in words is "forty billion, four hundred one million, one hundred thirty-one thousand, one hundred thirty-one".
Divisors: 1 37 1091922463 40401131131 | 624 | 2,061 | {"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} | 3.21875 | 3 | CC-MAIN-2020-40 | latest | en | 0.848565 |
https://rdrr.io/cran/VGAM/man/inv.paralogistic.html | 1,493,571,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125719.13/warc/CC-MAIN-20170423031205-00213-ip-10-145-167-34.ec2.internal.warc.gz | 831,379,478 | 33,442 | # inv.paralogistic: Inverse Paralogistic Distribution Family Function
### Description
Maximum likelihood estimation of the 2-parameter inverse paralogistic distribution.
### Usage
```1 2 3 4``` ```inv.paralogistic(lscale = "loge", lshape1.a = "loge", iscale = NULL, ishape1.a = NULL, imethod = 1, lss = TRUE, gscale = exp(-5:5), gshape1.a = seq(0.75, 4, by = 0.25), probs.y = c(0.25, 0.5, 0.75), zero = "shape") ```
### Arguments
`lss` See `CommonVGAMffArguments` for important information. `lshape1.a, lscale` Parameter link functions applied to the (positive) parameters `a` and `scale`. See `Links` for more choices. `iscale, ishape1.a, imethod, zero` See `CommonVGAMffArguments` for information. For `imethod = 2` a good initial value for `ishape1.a` is needed to obtain a good estimate for the other parameter. `gscale, gshape1.a` See `CommonVGAMffArguments` for information. `probs.y` See `CommonVGAMffArguments` for information.
### Details
The 2-parameter inverse paralogistic distribution is the 4-parameter generalized beta II distribution with shape parameter q=1 and a=p. It is the 3-parameter Dagum distribution with a=p. More details can be found in Kleiber and Kotz (2003).
The inverse paralogistic distribution has density
f(y) = a^2 y^(a^2-1) / [b^(a^2) (1 + (y/b)^a)^(a+1)]
for a > 0, b > 0, y >= 0. Here, b is the scale parameter `scale`, and a is the shape parameter. The mean is
E(Y) = b gamma(a + 1/a) gamma(1 - 1/a) / gamma(a)
provided a > 1; these are returned as the fitted values. This family function handles multiple responses.
### Value
An object of class `"vglmff"` (see `vglmff-class`). The object is used by modelling functions such as `vglm`, and `vgam`.
### Note
See the notes in `genbetaII`.
T. W. Yee
### References
Kleiber, C. and Kotz, S. (2003) Statistical Size Distributions in Economics and Actuarial Sciences, Hoboken, NJ, USA: Wiley-Interscience.
`Inv.paralogistic`, `genbetaII`, `betaII`, `dagum`, `sinmad`, `fisk`, `inv.lomax`, `lomax`, `paralogistic`, `simulate.vlm`.
### Examples
```1 2 3 4 5 6 7``` ```idata <- data.frame(y = rinv.paralogistic(n = 3000, exp(1), scale = exp(2))) fit <- vglm(y ~ 1, inv.paralogistic(lss = FALSE), data = idata, trace = TRUE) fit <- vglm(y ~ 1, inv.paralogistic(imethod = 2, ishape1.a = 4), data = idata, trace = TRUE, crit = "coef") coef(fit, matrix = TRUE) Coef(fit) summary(fit) ```
Search within the VGAM package
Search all R packages, documentation and source code
Questions? Problems? Suggestions? or email at ian@mutexlabs.com.
Please suggest features or report bugs with the GitHub issue tracker.
All documentation is copyright its authors; we didn't write any of that. | 819 | 2,686 | {"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-2017-17 | longest | en | 0.602439 |
http://www.padipa.org.uk/767715-physics-help-power-problem.php | 1,563,507,070,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00021.warc.gz | 247,144,184 | 7,800 | 5748 Shares
# Physics Help: Power Problem?
Topic: Physics Help: Power Problem?
July 19, 2019 / By Rose
Question: I solved part A but cant solve part B. Thanks for help A skier of mass 70.8 kg is pulled up a slope by a motor-driven cable. (a) How much work is required for him to be pulled a distance of 59.2 m up a 30.4° slope (assumed frictionless) at a constant speed of 1.94 m/s? This I found to be 20.785 kJ b) A motor of what power is required to perform this task Answer in hp
## Best Answers: Physics Help: Power Problem?
Muriel | 6 days ago
If it pulls him 59.2 meters at 1.94 meters per second, it takes 59.2 / 1.94 = 30.5 seconds 20.8 kjoules in 30.5 seconds is 0.681 kwatts 0r 681 watts. 1 hp = 746 watts, to this is 681/746 = 0.91 HP .
👍 120 | 👎 6
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Work: force exerted over a distance: F x d Power: how much work per second since work is an energy we can say W = -U (where U is potential energy) U = mgh, at the top of the hill U = (56kg)(-9.8m/s2)(30.5m) = -16738.4 Joules so W = 16738.4 Joules now since you know how long it takes to get up to the top (10s) divide work by time to get power P = W/t = 16738.4 Joules/10 seconds = 1673.84 Watts or approximately 1.7 kiloWatts
Lolicia
Power is work divided by time. Your time is ( 59.2 m ) / ( 1.94 m/s ) = 30.5 seconds, so the power is ( 20.785 kJ ) / ( 30.5 sec ) = 681 Watts. Now convert that to horsepower: 1 hp = 746 Watts.
👍 40 | 👎 -1
Karyn
Power = Work/Time You can find time from part (a) => Distance = speed*time => time = distance/speed = 59.2/1.94 = 30.52 s => Power = Work/time = (20.785*1000)/30.52.......(I multiplied work by 1000 to convert from kJ to J) => 681.089 W => now 1W = 0.001341 hp => Power = 681.089*0.001341 = 0.913335 hp
👍 35 | 👎 -8
Haydee
Power is simply the total energy used divised by the time. T = time, X = distance moved, V=Velocity, E=total Energy P = Power = E.V / X
👍 30 | 👎 -15
Originally Answered: PHYSICS: How large a collecting area of solar cells would be necessary to produce a 1-gigawatt power plant?
The energy received by 1 m² = 1400 watt Energy required to be produced = 1 giga-watt = 1 x 10^12 The area reqd. = 1x10^12 / 1400 = 714285714 m² .................. ....................... ...........= 714 .28 Km² ................ ....................... ...............===========
If you have your own answer to the question Physics Help: Power Problem?, then you can write your own version, using the form below for an extended answer. | 822 | 2,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-30 | latest | en | 0.912925 |
http://www.conversion-website.com/volume/cubic-centimeter-to-deciliter.html | 1,511,205,248,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806124.52/warc/CC-MAIN-20171120183849-20171120203849-00081.warc.gz | 373,502,745 | 4,380 | # Cubic centimeters to deciliters (cm³ to dL)
## Convert cubic centimeters to deciliters
Cubic centimeters to deciliters conversion calculator shown above calculates how many deciliters are in 'X' cubic centimeters (where 'X' is the number of cubic centimeters to convert to deciliters). In order to convert a value from cubic centimeters to deciliters (from cm³ to dL) type the number of cm³ to be converted to dL and then click on the 'convert' button.
## Cubic centimeters to deciliters conversion factor
1 cubic centimeter is equal to 0.01 deciliters
## Cubic centimeters to deciliters conversion formula
Volume(dL) = Volume (cm³) × 0.01
Example: Assume there are 589 cubic centimeters. Shown below are the steps to express them in deciliters.
Volume(dL) = 589 ( cm³ ) × 0.01 ( dL / cm³ )
Volume(dL) = 5.89 dL or
589 cm³ = 5.89 dL
589 cubic centimeters equals 5.89 deciliters
## Cubic centimeters to deciliters conversion table
cubic centimeters (cm³)deciliters (dL)
70.07
90.09
110.11
130.13
150.15
170.17
190.19
210.21
230.23
250.25
270.27
290.29
310.31
330.33
350.35
370.37
390.39
410.41
430.43
450.45
cubic centimeters (cm³)deciliters (dL)
1501.5
2002
2502.5
3003
3503.5
4004
4504.5
5005
5505.5
6006
6506.5
7007
7507.5
8008
8508.5
9009
9509.5
100010
105010.5
110011
Versions of the cubic centimeters to deciliters conversion table. To create a cubic centimeters to deciliters conversion table for different values, click on the "Create a customized volume conversion table" button.
## Related volume conversions
Back to cubic centimeters to deciliters conversion
TableFormulaFactorConverterTop | 521 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-47 | longest | en | 0.465219 |
http://www.flowerbrackets.com/adding-two-integers-without-using-arithmetic-operator/ | 1,539,661,297,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509996.54/warc/CC-MAIN-20181016030831-20181016052331-00271.warc.gz | 469,410,123 | 9,796 | # Adding Two Integers Without Using Arithmetic Operator
Hey guys!!! Welcome to flower brackets blog. In this post we are going to learn adding two integers without using arithmetic operator.
Since we are not using arithmetic operators in the below example, addTwoNumbers function is used.
The following program demonstrates the example on how to add using bitwise operators,
#### Example: adding two integers without using arithmetic operator
```import java.util.Scanner;
{
static int addTwoNumbers(int a, int b)
{
int carryNum;
while(b ! = 0)
{
carryNum = a & b;
a = a ^ b;
b = carryNum << 1;
}
return a;
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int a = sc.nextInt(); // Scans the next token of the input as an int.
int b = sc.nextInt(); // Scans the next token of the input as an int.
System.out.println("The output is: " + addTwoNumbers(a, b));
sc.close();
}
}
```
Output:
#### Using recursion
Here let’s see how to add two integers without using arithmetic operators in java using recursion method,
```import java.util.Scanner;
{
static int addTwoNumbers(int a, int b)
{
if (b == 0)
return a;
int total = a ^ b;
int carryNum = (a & b) << 1;
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
int b = sc.nextInt();
System.out.println("The output is: " + addTwoNumbers(a,b));
sc.close();
}
}
```
Output:
The output is: 50
##### conclusion
That’s it. So this is all about how to add two integers without using arithmetic operators in java.
I hope you have understood the concept of add two numbers without using + operator.
Do subscribe to my blog flower brackets if you haven’t already. | 408 | 1,706 | {"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-2018-43 | latest | en | 0.579865 |
https://wiki.ogre3d.org/tiki-print.php?q=%2FNearest+point+on+a+line&page=Nearest+point+on+a+line | 1,680,380,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00434.warc.gz | 690,030,088 | 18,731 | Keep in mind that pt1 and pt2 define the line. So the point returned may not be between pt1 and pt2.
```Ogre::Vector3 nearestPoint(Ogre::Vector3 pt1, Ogre::Vector3 pt2, Ogre::Vector3 testPoint)
{
// Find the point on a line defined by pt1 and pt2 that
// is nearest to a given point tp
// tp
// /|
// A / |
// / |
// / |
// pt1---o--------------pt2
// B' B
// Get the vectors between the points
Ogre::Vector3 A = testPoint - pt1;
Ogre::Vector3 B = pt2 - pt1;
// Find the cos of the angle between the vectors
float cosTheta = A.dotProduct(B) / (A.length() * B.length());
// Use that to calculate the length of B'
float BPrimeLength = A.length() * cosTheta;
// Find the ratio of the length of B' and B
float scale = BPrimeLength / B.length();
// Scale B by that ratio
B *= scale;
// Translate p1 by B, this puts it at o
Ogre::Vector3 C = pt1 + B;
return C;
}```
A smaller and faster version:
```inline Ogre::Vector3 nearestPoint(const Ogre::Vector3 &pt1, const Ogre::Vector3 &pt2, const Ogre::Vector3 &testPoint)
{
const Ogre::Vector3 A = testPoint - pt1;
const Ogre::Vector3 u = (pt2-pt1).normalisedCopy();
return pt1 + (A.dotProduct(u)) * u;
};```
If you wish to find the closest point that exists between the 2 points in the line
```//Give the spot on the line, thats closest to testPoint
inline Ogre::Vector3 nearestPointInBetween(const Ogre::Vector3 &pt1, const Ogre::Vector3 &pt2, const Ogre::Vector3 &testPoint)
{
//probably not most efficient(?)
const Ogre::Vector3 result = nearestPoint(pt1,pt2,testPoint);
const Ogre::Real lineLength = pt1.squaredDistance(pt2);
const Ogre::Real p1R = pt1.squaredDistance(result);
const Ogre::Real p2R = pt2.squaredDistance(result);
//R R p1 R p2 R R
if ( p1R > lineLength )
{
if ( p2R > p1R )
return pt1; //pt 1 is closer to result
else
return pt2;
}
else if ( p2R > lineLength )
return pt1;
return result;
};```
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https://physics.stackexchange.com/questions/627519/when-and-why-can-the-spin-connection-term-of-the-dirac-operator-be-omitted | 1,653,136,453,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00082.warc.gz | 496,406,956 | 66,915 | # When and why can the spin connection term of the Dirac Operator be omitted?
The Dirac Operator $$D$$ is defined by $$$$\tag{1} D=i\gamma^a\nabla_a=i\gamma^a\nabla_{e_a}=i\underbrace{\gamma^a{e_a}^\mu}_{=\gamma^\mu}\nabla_{\partial_\mu}=i\gamma^\mu\nabla_\mu=i\gamma^\mu(\partial_\mu+\omega_\mu+A_\mu)$$$$ and $$$$\omega_a=-\frac{1}{4}\omega_{abc}\gamma^{bc}\psi={e_a}^\mu\omega_\mu.$$$$ However, in his derivation of the Atiyah Singer Index theorem$$^1$$, Fujikawa (chapter $$5.5$$ of Path Integrals and Quantum Anomalies) assumes $$$$D=i\gamma^\mu\nabla_\mu=i\gamma^\mu(\partial_\mu+A_\mu).$$$$ One might think that Fujikawa's $$D$$ is simply another operator, but Nakahara - who derives the same equation in section $$13.2.1$$ (Fujikawa's method) - says that the spin connection "plays no role" under certain assumptions:
We compactify the space in such a way that the geometry (the spin connection) plays no role. For example, this can be achieved by compactifying $$\mathbf{R}^4$$ to $$S^4=\mathbf{R}^4\cup\{\infty\}$$, for which the Dirac genus $$\hat{A}(TM)$$ is trivial.
I know that $$S^n\cong\mathbf{R}^n\cup\{\infty\}$$, but I don't see why this implies that the spin connection "plays no role".
$$^1$$ By the "Atiyah Singer Index theorem" I mean the equation $$$$\mathrm{ind}\,D_+=-\frac{1}{8\pi^2}\int_M\mathrm{tr}(F_{ij}F_{kl})\epsilon^{ijkl}\cdot\omega,$$$$ where $$M$$ is a $$4$$-dimensional Riemannian manifold with euclidean signature and $$\omega$$ is the volume form. (I am aware of the fact that $$D\colon\Gamma(M,S\otimes E)\to\Gamma(M,S\otimes E)$$ is a Fredholm operator and that $$(1)$$ is only valid after the choice of a local gauge - here we assume that $$E$$ is the the associated vector bundle induced by the adjoint representation.)
If you compacify to a torus then, in Cartesian coordinates, the spin connection vanishes and so is irrelevent. If you compactify to a sphere, as Fujikawa suggests, it is far less obvious that $$\hat A(TM)$$ is not needed. That it plays no role is because $$\hat A(TM)$$ is a genus and so cobordism invariant. This means that the curvature contribution to the index is zero when $$M$$ is a boundary: $$M=\partial N$$, and the spin connection can be extended through $$N$$. This is true in the case of a sphere. I do not know a simple proof of the cobordism property though.
• To prove the ASIT, it doesn't suffice to prove $\mathrm{ind}\,D_+=-\frac{1}{8\pi^2}\int_M\mathrm{tr}(F_{ij}F_{kl})\epsilon^{ijkl}\cdot\omega$ like I did, but one needs to show that RHS is equal to the topological index and that's where the A hat genus joins the party. Considering the operator $i\gamma^\mu(\partial_\mu+A_\mu)$ is useful to prove $\mathrm{ind}\,D_+=-\frac{1}{8\pi^2}\int_M\mathrm{tr}(F_{ij}F_{kl})\epsilon^{ijkl}\cdot\omega$, but $i\gamma^\mu(\partial_\mu+A_\mu)$ is not the Dirac operator. (A confirmation would be appreciated.) Apr 7, 2021 at 19:10
• Yes. There are curvature terms coming from $\hat A$ in in the expression for the index general spin manifolds. Even when $\hat A$ makes not contribution you will need the spin connection to makes ense of the Dirac operator. Apr 7, 2021 at 19:29 | 966 | 3,158 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | longest | en | 0.850997 |
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Topic: Matheology § 222 Back to the roots
Replies: 606 Last Post: Apr 18, 2013 2:04 AM
Messages: [ Previous | Next ]
Virgil Posts: 8,833 Registered: 1/6/11
Re: WMytheology � 222 Rejects Induction
Posted: Feb 18, 2013 4:21 PM
In article
WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 18 Feb., 14:11, William Hughes <wpihug...@gmail.com> wrote:
> > On Feb 18, 1:38 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 18 Feb., 11:53, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > Is there a potentially infinite sequence,
> > > > x, such that the nth FIS of x consists of
> > > > n 1's
> >
> > > Yes, of course
> >
> > Let y be a potentially infinite process
>
> There are no processes with respect to numbers and lists. They are
> existing or are not existing.
Then ther is no such thing a potentially infinite, as that would require
existence of a process that does not end.
> Potential infinity with respect to natural numbers means: You can
> consider every natural number you like. There is no upper threshold.
> So name any set of natural numbers - except using naive and
> couterfactual "all" of some kind like "all prime numbers" or all "even
> numbers".
So for any set of naturals you can name, there is a process for finding
a superset of that set, even in Wolkenmuekenheim.
If one cannot ever have something true for ALL natural numbers, how can
one ever use inductive proofs?
> Now realize what potential infinity means: There are no processes in
> above list.
It also means that induction can never conclude that anything is true
for all natural numbers.
In Wolkenmuekenheim one cannot say
"For all n, if n is a natural then n+1 is a natural."
because on cannot ever say "For all n"
> We simply write some lines and stop at some point.
Thus WM can never prove that anything by induction.
--
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# Hash Tables
SDP-4
## 2. Dictionary
Dictionary:
Dynamic-set data structure for storing items indexed using keys.
Supports operations Insert, Search, and Delete.
Applications:
Symbol table of a compiler.
Memory-management tables in operating systems.
Large-scale distributed systems.
Hash Tables:
Effective way of implementing dictionaries.
Generalization of ordinary arrays.
Applicable when we can afford to allocate an array with one
position for every possible key.
Element whose key is k is obtained by indexing into the kth position
of the array.
i.e. when the universe of keys U is small.
Dictionary operations can be implemented to take O(1) time.
Details in Sec. 11.1.
## 4. Hash Tables
Notation:
When U is very large,
U – Universe of all possible keys.
K – Set of keys actually stored in the dictionary.
|K| = n.
Arrays are not practical.
|K| << |U|.
Use a table of size proportional to |K| – The hash tables.
However, we lose the direct-addressing ability.
Define functions that map keys to slots of the hash table.
## 5. Hashing
Hash function h: Mapping from U to the slots of a hash table
T[0..m–1].
h : U {0,1,…, m–1}
With arrays, key k maps to slot A[k].
With hash tables, key k maps or “hashes” to slot T[h[k]].
h[k] is the hash value of key k.
## 6. Hashing
0
U
(universe of keys)
h(k1)
h(k4)
k1
K
(actual k2
keys)
k4
collision
k5
h(k2)=h(k5)
k3
h(k3)
m–1
## 7. Issues with Hashing
Multiple keys can hash to the same slot – collisions are
possible.
Design hash functions such that collisions are minimized.
But avoiding collisions is impossible.
Design collision-resolution techniques.
Search will cost Ө(n) time in the worst case.
However, all operations can be made to have an expected
complexity of Ө(1).
## 8. Methods of Resolution
Chaining:
Store all elements that hash to the same slot in a
the hash table slot.
All elements stored in hash table itself.
When collisions occur, use a systematic
(consistent) procedure to store elements in free
slots of the table.
0
k1
k4
k5
k2
k7
k3
k8
m–1
k6
## 9. Collision Resolution by Chaining
0
U
(universe of keys)
X
h(k1)=h(k4)
k1
k4
K
(actual
keys)
k2
k6
k5
X
h(k2)=h(k5)=h(k6)
k7
k8
k3
X
h(k3)=h(k7)
h(k8)
m–1
## 10. Collision Resolution by Chaining
0
U
(universe of keys)
k1
k4
k5
k2
k7
k3
k1
k4
K
(actual
keys)
k2
k6
k5
k7
k8
k3
k8
m–1
k6
## 11. Hashing with Chaining
Dictionary Operations:
Chained-Hash-Insert (T, x)
Chained-Hash-Delete (T, x)
Insert x at the head of list T[h(key[x])].
Worst-case complexity – O(1).
Delete x from the list T[h(key[x])].
Worst-case complexity – proportional to length of list with
Chained-Hash-Search (T, k)
Search an element with key k in list T[h(k)].
Worst-case complexity – proportional to length of list.
## 12. Analysis on Chained-Hash-Search
Load factor =n/m = average keys per slot.
m – number of slots.
n – number of elements stored in the hash table.
Worst-case complexity: (n) + time to compute h(k).
Average depends on how h distributes keys among m slots.
Assume
Simple uniform hashing.
Any key is equally likely to hash into any of the m slots, independent of
where any other key hashes to.
O(1) time to compute h(k).
Time to search for an element with key k is (|T[h(k)]|).
Expected length of a linked list = load factor = = n/m.
## 13. Expected Cost of an Unsuccessful Search
Theorem:
An unsuccessful search takes expected time Θ(1+α).
Proof:
Any key not already in the table is equally likely to hash to
any of the m slots.
To search unsuccessfully for any key k, need to search to
the end of the list T[h(k)], whose expected length is α.
Adding the time to compute the hash function, the total
time required is Θ(1+α).
## 14. Expected Cost of a Successful Search
Theorem:
A successful search takes expected time Θ(1+α).
Proof:
The probability that a list is searched is proportional to the number of
elements it contains.
Assume that the element being searched for is equally likely to be any of
the n elements in the table.
The number of elements examined during a successful search for an
element x is 1 more than the number of elements that appear before x in
x’s list.
These are the elements inserted after x was inserted.
Goal:
Find the average, over the n elements x in the table, of how many elements were
inserted into x’s list after x was inserted.
## 15. Expected Cost of a Successful Search
Theorem:
A successful search takes expected time Θ(1+α).
Proof (contd):
Let xi be the ith element inserted into the table, and let ki = key[xi].
Define indicator random variables Xij = I{h(ki) = h(kj)}, for all i, j.
Simple uniform hashing Pr{h(ki) = h(kj)} = 1/m
E[Xij] = 1/m.
Expected number of elements examined in a successful search is:
n
1 n
E 1 X ij
n i 1 j i 1
No. of elements inserted after xi into the same slot as xi.
## 16. Proof – Contd.
1 n
E 1
n i 1
X
ij
j i 1
1 n
1
n i 1
n
E
[
X
]
ij
j i 1
1 n
1
n i 1
1
j i 1m
n
(linearity of expectation)
n
1 n
1
(n i )
nm i 1
n
1 n
1
n i
nm i 1
i 1
1 2 n(n 1)
n
nm
2
n 1
1
2m
1
1
2
2n
Expected total time for a successful search
= Time to compute hash function + Time
to search
= O(2+ /2 – /2n) = O(1+ ).
## 17. Expected Cost – Interpretation
If n = O(m), then =n/m = O(m)/m = O(1).
Searching takes constant time on average.
Insertion is O(1) in the worst case.
Deletion takes O(1) worst-case time when lists are
Hence, all dictionary operations take O(1) time on
average with hash tables with chaining.
## 18. Good Hash Functions
Satisfy the assumption of simple uniform hashing.
Not possible to satisfy the assumption in practice.
Often use heuristics, based on the domain of the keys, to
create a hash function that performs well.
Regularity in key distribution should not affect uniformity.
Hash value should be independent of any patterns that might
exist in the data.
E.g. Each key is drawn independently from U according to a
probability distribution P:
k:h(k) = j P(k) = 1/m
for j = 0, 1, … , m–1.
An example is the division method.
## 19. Keys as Natural Numbers
Hash functions assume that the keys are natural numbers.
When they are not, have to interpret them as natural
numbers.
Example: Interpret a character string as an integer expressed
in some radix notation. Suppose the string is CLRS:
ASCII values: C=67, L=76, R=82, S=83.
There are 128 basic ASCII values.
So, CLRS = 67·1283+76 ·1282+ 82·1281+ 83·1280
= 141,764,947.
## 20. Division Method
Map a key k into one of the m slots by taking the
remainder of k divided by m. That is,
h(k) = k mod m
Example: m = 31 and k = 78 h(k) = 16.
Advantage: Fast, since requires just one division
operation.
Disadvantage: Have to avoid certain values of m.
Don’t pick certain values, such as m=2p
Or hash won’t depend on all bits of k.
Good choice for m:
Primes, not too close to power of 2 (or 10) are good.
## 21. Multiplication Method
If 0 < A < 1, h(k) = m (kA mod 1) = m (kA – kA )
where kA mod 1 means the fractional part of kA, i.e., kA –
kA .
Disadvantage: Slower than the division method.
Advantage:Value of m is not critical.
Typically chosen as a power of 2, i.e., m = 2p, which makes
implementation easy.
Example: m = 1000, k = 123, A 0.6180339887…
h(k) = 1000(123 · 0.6180339887 mod 1)
= 1000 · 0.018169... = 18.
## 22. Multiplication Mthd. – Implementation
Choose m = 2p, for some integer p.
Let the word size of the machine be w bits.
Assume that k fits into a single word. (k takes w bits.)
Let 0 < s < 2w. (s takes w bits.)
Restrict A to be of the form s/2w.
Let k s = r1 ·2w+ r0 .
r1 holds the integer part of kA ( kA ) and r0 holds the fractional part
of kA (kA mod 1 = kA – kA ).
We don’t care about the integer part of kA.
So, just use r0, and forget about r1.
## 23. Multiplication Mthd – Implementation
w bits
k
binary point
r1
s = A·2w
·
r0
extract p bits
h(k)
We want m (kA mod 1) . We could get that by shifting r0 to the
left by p = lg m bits and then taking the p bits that were shifted to
the left of the binary point.
But, we don’t need to shift. Just take the p most significant bits of
r0.
## 24. How to choose A?
Another example: On board.
How to choose A?
The multiplication method works with any legal value of A.
But it works better with some values than with others, depending
on the keys being hashed.
Knuth suggests using A ( 5 – 1)/2. | 2,517 | 8,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.822709 |
https://www.jove.com/de/science-education/15542/equation-of-motion-rotation-about-a-fixed-axis | 1,721,243,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00016.warc.gz | 756,080,228 | 26,397 | Trial ends in
# 15.11: Equation of Motion: Rotation About a Fixed Axis
TABLE OF
CONTENTS
### 15.11: Equation of Motion: Rotation About a Fixed Axis
Consider a flywheel, having an uneven mass distribution, rotating steadily around a fixed axis. As this rotation occurs, the center of mass of the flywheel traces a circular path. Understanding the acceleration of this center of mass requires observing both its tangential and normal components.
The tangential component is dependent on the direction of the angular acceleration of the flywheel. The tangential component of the acceleration propels the flywheel along its path. On the other hand, the normal component is always directed along the radius towards point O. Point O lies on the axis of rotation along which the flywheel spins.
A crucial aspect of this scenario is the moment applied to the flywheel's center of mass. This is calculated by multiplying the moment of inertia of the center of mass by its angular acceleration. The equation for this moment can be articulated in terms of the moment about point O, which effectively eliminates any unknown forces acting on the body. Interestingly, the moment resulting from the normal component of acceleration is not taken into account in these calculations. The reason for this exclusion is that the normal component of the acceleration passes through point O and is parallel to the radial vector, resulting in no moment.
To further refine this understanding, one could employ the parallel axis theorem. This allows the moment equation to be expressed in terms of the moment of inertia about point O, providing a more detailed view of the flywheel's motion.
#### Tags
Keywords: Rotation Fixed Axis Flywheel Mass Distribution Center Of Mass Acceleration Tangential Component Normal Component Angular Acceleration Moment Of Inertia Moment About Point O Parallel Axis Theorem
X | 364 | 1,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.893688 |
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Flashcards - Rotational Motion
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Inertia
Inertia is the resistance of any physical object to any change in its state of motion; this includes changes to its speed, direction or state of rest
Got it
angle
In planar geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle
Got it
torque
Torque, moment, or moment of force is the tendency of a force to rotate an object about an axis, fulcrum, or pivot
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or choose a specific lesson: See all lessons in this chapter
7 cards in set
Front
Back
torque
Torque, moment, or moment of force is the tendency of a force to rotate an object about an axis, fulcrum, or pivot
angle
In planar geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle
Inertia
Inertia is the resistance of any physical object to any change in its state of motion; this includes changes to its speed, direction or state of rest
Power
Power is an American crime drama television series airing on Starz
Rotational motion
Rotation around a fixed axis is a special case of rotational motion
Kinetic energy of rotation
Rotational energy or angular kinetic energy is kinetic energy due to the rotation of an object and is part of its total kinetic energy
The radian is the standard unit of angular measure, used in many areas of mathematics
To unlock this flashcard set you must be a Study.com Member. | 426 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-18 | latest | en | 0.921119 |
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I had written a code in SCN Discussion started by Kiran Premlal to calculate all possible permutations.
The code in this document has certain improvements like:
1. Choose permutation or combination using radio-button
2. Separation of data from logic
3. Minimum and maximum set size can be specified
## Why was data separated from logic?
Let us look at some example sets.
SET 1 SET 2 SET 3 54 apple 1 23 orange 2 101 banana 3
As we can see, sets can be of integers, strings, mixed data types and data can be unsorted.
However, for calculation of output, all we need is a set of indexes.
For index combination {2,3}, subset of SET 1 would be {23, 101}, whereas subset of SET 2 would be {orange, banana}
The code first calculates a list of index combinations, which can then be used to build data sets.
## Example Explanation
We look at SET 2 and see how output is being calculated.
INDEX DATA 1 apple 2 orange 3 banana
### Permutation of all pairs
We have 2 slots to fill, and first element can reserve a slot for starters.
Data sets are {apple, orange} and {apple, banana}.
Index set equivalents would be {1, 2} and {1, 3}.
Program would calculate index sets as:
1 2 1 3 2 1 2 3 3 1 3 2
When these indexes are replaced by their data equivalents, result would be:
apple orange apple banana orange apple orange banana banana apple banana orange
### Combinations of all singles and pairs
Keeping above SET 2 as input data, we see how output is calculated.
For calculating singles and pairs, we have minimum set size = 1 and maximum set size = 2.
We first find all sets of size 1, and of size 2.
Output in index as well data form would be:
1 apple 2 orange 3 banana 1 2 apple orange 1 3 apple banana 2 3 orange banana
For same input data, 6 permutation of pairs was calculated, whereas only 3 combination of pairs.
## Screenshots from program
The dataset in program is build using characters starting from ‘B’.
For dataset of size 3, program would build internal table of characters B, C and D.
INDEX DATA 1 B 2 C 3 D
We specify in program selection the size of dataset, max/min size of set, and option to find permutation or combination.
Program would store resulting sets in internal table GT_SETS, whose record is capable of storing table of integers.
Although this program prints the resulting sets, minor changes can be done to return the results to another routine.
## Code Snippet
1. TYPES: tty TYPE TABLE OF i.
2. * store superset
3. DATA gt_superset_index TYPE TABLE OF i.
4. DATA gt_superset_data TYPE TABLE OF c.
5. * temporarily store set
6. DATA gt_set TYPE TABLE OF i.
7. * gt_sets to store list of sets
8. DATA gt_sets LIKE TABLE OF gt_set.
9. DATA gv_setsize TYPE i.
10. PARAMETERS: p_total TYPE i DEFAULT 5, “total number of elements in superset
13. p_max TYPE i DEFAULT 3, “maximum size of set
14. p_min TYPE i DEFAULT 1. “minimum size of set
15. CHECK p_total GT 0 AND
16. p_max GT 0 AND
17. p_min GT 0 AND
18. p_total GE p_max AND
19. p_max GE p_min.
20. * populate superset index and data
21. DO p_total TIMES.
22. APPEND syindex TO gt_superset_index.
23. APPEND syabcde+syindex(1) TO gt_superset_data.
24. ENDDO.
25. * populate gt_sets table with various combinations of indexes
26. gv_setsize = p_min.
27. WHILE gv_setsize LE p_max.
28. PERFORM build_sets USING gt_superset_index
29. gv_setsize.
30. gv_setsize = gv_setsize + 1.
31. ENDWHILE.
32. * display sets by using data from gt_superset_data and index from gt_sets
33. PERFORM display.
34. *&———————————————————————*
35. *& Form BUILD_SETS
36. *&———————————————————————*
37. FORM build_sets USING pt TYPE tty
38. p_setsize TYPE i.
39. DATA lv TYPE i.
40. DATA lv_lines TYPE i.
41. DATA lt TYPE tty.
42. DATA lv_tabix TYPE i.
43. LOOP AT pt INTO lv.
44. lv_tabix = sytabix.
45. * PUSH element to stack and pass remaining to recursive call
46. APPEND lv TO gt_set.
47. * save set if set size is achieved
48. DESCRIBE TABLE gt_set LINES lv_lines.
49. IF lv_lines EQ p_setsize.
50. APPEND gt_set TO gt_sets.
51. DELETE gt_set INDEX lv_lines.
52. CONTINUE.
53. ENDIF.
54. * recursive call for subset
55. lt = pt.
56. CASE ‘X’.
57. WHEN p_perm.
58. “permutation – calculate for all elements excluding current
59. DELETE lt INDEX lv_tabix.
60. WHEN p_comb.
61. “combination – calculate for all elements following current
62. DELETE lt TO lv_tabix.
63. ENDCASE.
64. IF lt IS NOT INITIAL.
65. PERFORM build_sets USING lt
66. p_setsize.
67. ENDIF.
68. * POP the element from stack once all combinations are tried
69. DESCRIBE TABLE gt_set LINES lv_lines.
70. DELETE gt_set INDEX lv_lines.
71. ENDLOOP.
72. ENDFORM. ” BUILD_SETS
73. *&———————————————————————*
74. *& Form DISPLAY
75. *&———————————————————————*
76. FORM display .
77. FIELD-SYMBOLS: <fs_set> TYPE ANY TABLE,
78. <fs_index> TYPE i,
79. <fs_data> TYPE c.
80. LOOP AT gt_sets ASSIGNING <fs_set>.
81. NEW-LINE.
82. LOOP AT <fs_set> ASSIGNING <fs_index>.
83. READ TABLE gt_superset_data INDEX <fs_index> ASSIGNING <fs_data>.
84. IF sysubrc EQ 0.
85. WRITE <fs_data>.
86. ENDIF.
87. ENDLOOP.
88. ENDLOOP.
89. ENDFORM. ” DISPLAY | 1,504 | 5,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-50 | latest | en | 0.887344 |
https://site.chimpvine.com/article/what-is-divisor/ | 1,725,799,501,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00178.warc.gz | 515,025,671 | 40,278 | # Understanding the Role of Divisor in Division
## Introduction
### Divisors
In the realm of mathematics, the concept of division plays a crucial role in breaking down numbers into equal parts. At the heart of division lies the divisor, an essential component that determines the number of equal parts a given quantity is divided into. Let’s explore the significance of the divisor and its role in the division process.
## Analogy of Definition
### What are Divisors?
The divisor is a number by which another number is divided. In a division equation, the divisor is the quantity that specifies the number of equal parts the dividend is divided into. It is an integral part of the division process and is essential for determining the quotient and remainder. Understanding divisors is quite straightforward. Suppose we want to divide 42 by 6. This division is represented as 8 ÷ 4 = 2. Here, 8 is the dividend, 4 is the divisor, and 2 is the quotient.
### Parts of Division
In any division operation, we encounter four key components:
Dividend: This is the number or quantity that is being divided.
Divisor: The divisor is the number by which the dividend is divided.
Quotient: It is the result of the division operation, representing the number obtained by dividing the dividend by the divisor.
Remainder: This is the number that remains after the division process, which may occur when the division is not exact.
## Method
### Understanding Divisor
Let’s assume that we have 10 balls in total, and we need to divide them into groups of 5. In this case, 5 is the divisor because we need to divide the whole of 10 into parts of 5.
### Finding the Divisor
The divisor can be found using the divisor formula, which is expressed as:
This formula helps in calculating the divisor when the dividend and quotient are known. Additionally, the divisor can be determined by identifying the factors of the dividend and understanding their relationship with the quotient.
## Examples
Consider the division problem: 24 ÷ 6 = 4
In this example, 24 is the dividend, 6 is the divisor, 4 is the quotient, and 0 is the remainder. The divisor, 6, determines the number of equal parts into which 24 is divided, resulting in 4 as the quotient.
Consider the division problem: 15 ÷ 3 = 5
In this example, 15 is the dividend, 3 is the divisor, 5 is the quotient, and 0 is the remainder. The divisor, 3, determines the number of equal parts into which 15 is divided, resulting in 5 as the quotient.
Consider the division problem:36 ÷ 9 = 4
In this example, 36 is the dividend, 9 is the divisor, 4 is the quotient, and 0 is the remainder. The divisor, 9, determines the number of equal parts into which 36 is divided, resulting in 4 as the quotient.
Consider the division problem: 48 ÷ 12 = 4
In this example, 48 is the dividend, 12 is the divisor, 4 is the quotient, and 0 is the remainder. The divisor, 12, determines the number of equal parts into which 48 is divided, resulting in 4 as the quotient.
Consider the division problem: 27 ÷ 9 = 3
In this example, 27 is the dividend, 9 is the divisor, 3 is the quotient, and 0 is the remainder. The divisor, 9, determines the number of equal parts into which 27 is divided, resulting in 3 as the quotient.
## Tips and Tricks
1. Divisor in Division
Tip: The divisor is the number by which the dividend is divided to obtain the quotient and remainder in a division operation.
2. Divisor Formula
Tip: Use the formula Divisor = Dividend ÷ Quotient to calculate the divisor when the dividend and quotient are known.
3. Properties of Divisor
Tip: The divisor has the property of determining the number of equal parts the dividend is divided into, influencing the resulting quotient and remainder.
4. Divisor vs. Dividend
Tip: Distinguish between the divisor and dividend by understanding their roles in the division process. The divisor is the number being divided, while the dividend is the number by which the division is performed.
5. Parts of Division
Tip: Familiarize yourself with the three main parts of a division operation: dividend, divisor, and quotient, each playing a distinct role in the division process.
## Real life application
Story: “The Divisor Chronicles”
The concept of the divisor found its way into the real world, impacting various scenarios and everyday situations.
Scenario 1: Sharing Treats
In a classroom, a teacher distributed 24 candies among 6 students. The divisor, 6, determined the number of candies each student received, resulting in an equal distribution of treats.
Scenario 2: Budget Allocation
A group of friends decided to divide their expenses equally for a trip. The total cost of the trip, the dividend, was divided by the number of friends, the divisor, to determine the individual share of each friend.
Scenario 3: Time Management
A project manager allocated tasks to team members based on the divisor of the total workload. By dividing the tasks equally, the team members were able to manage their time effectively and complete the project on schedule.
## FAQ's
The divisor plays a crucial role in division by determining the number of equal parts the dividend is divided into, leading to the calculation of the quotient and remainder.
The divisor has the property of influencing the resulting quotient and remainder in a division operation. It determines the number of equal parts the dividend is divided into.
The divisor formula, Divisor = Dividend ÷ Quotient, provides a method to calculate the divisor when the dividend and quotient are known, aiding in the division process.
The main parts of a division operation are the dividend, divisor, and quotient, each playing a distinct role in the division process.
Yes, negative numbers can be divisors just like positive numbers. However, when dealing with negative divisors, it’s essential to consider the sign conventions and rules for operations involving negative numbers.
Like? Share it with your friends | 1,309 | 5,966 | {"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} | 5 | 5 | CC-MAIN-2024-38 | latest | en | 0.930514 |
http://mathhelpforum.com/calculus/133628-differential-arc-trajectory.html | 1,527,453,655,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00087.warc.gz | 184,226,698 | 9,540 | ## Differential arc of a trajectory
The trajectory of a particle is given by $\displaystyle \vec r(t)$ and its velocity by $\displaystyle \vec v(t)$.
Calculate $\displaystyle d\vec s = dx \hat i + dy \hat j + dz \hat k$ and $\displaystyle ds^2$ in cylindrical and spherical coordinates.
So far I have converted $\displaystyle \hat i$, $\displaystyle \hat j$ and $\displaystyle \hat k$ into a combination of unit vectors in cylindrical coordinates. I also done the opposite.
I just don't know how to convert dx into a differential depending on $\displaystyle r$, $\displaystyle \varphi$ and z. I realize it will only depend on $\displaystyle r$ and $\displaystyle \varphi$.
I know that $\displaystyle x=r\cos \varphi$. I'm not really sure how to get dx. I don't think I can call "dx" a total derivative. | 206 | 803 | {"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} | 3.21875 | 3 | CC-MAIN-2018-22 | latest | en | 0.830411 |
https://work-at-home-on-your-computer.com/qa/quick-answer-what-is-the-missing-factor-of-36.html | 1,603,622,727,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888931.67/warc/CC-MAIN-20201025100059-20201025130059-00140.warc.gz | 582,095,736 | 8,041 | # Quick Answer: What Is The Missing Factor Of 36?
## What are the missing factors of 36 and 42?
The gcf of 36 and 42 can be obtained like this: The factors of 36 are 36, 18, 12, 9, 6, 4, 3, 2, 1.
The factors of 42 are 42, 21, 14, 7, 6, 3, 2, 1.
The common factors of 36 and 42 are 6, 3, 2, 1, intersecting the two sets above..
## What is the HCF of 36 48?
12Greatest common factor (GCF) of 36 and 48 is 12. We will now calculate the prime factors of 36 and 48, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 36 and 48.
## What is the missing factor of 30?
30 = 1 x 30, 2 x 15, 3 x 10, or 5 x 6. Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Prime factorization: 30 = 2 x 3 x 5.
## What is the GCF of 36 63?
9To sum up, the gcf of 36 and 63 is 9. In common notation: gcf (36,63) = 9.
## What is a common factor for 36?
1, 2, 3, 4, 6, 9, 12, 18, and 36. 1, 2, 3, 6, 9, 18, 27, and 54. Although the numbers in bold are all common factors of both 36 and 54, 18 is the greatest common factor.
## What are 5 factors of 36?
36 is a composite number, and it is 6 squared. 36 = 1 x 36, 2 x 18, 3 x 12, 4 x 9, or 6 x 6. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. Prime factorization: 36 = 2 x 2 x 3 x 3, which can also be written 36 = 2² x 3².
## Is 36 a perfect number?
-perfect numbers are 36, 1800, 2700, 17424, … (OEIS A054980).
## What is the HCF of 36 and 42?
We found the factors and prime factorization of 36 and 42. The biggest common factor number is the GCF number. So the greatest common factor 36 and 42 is 6.
## What is the GCF of 60 and 24?
Greatest Common Factor We found the factors and prime factorization of 24 and 60. The biggest common factor number is the GCF number. So the greatest common factor 24 and 60 is 12.
## What is the GCF of 24 and 36?
The common factors of 24 and 36 are 1, 2, 3, 4, 6 and 12. The greatest common factor of 24 and 36 is 12. Example 5: Find the GCF of 56 and 63.
## How do you find a factor?
“Factors” are the numbers you multiply to get another number. For instance, factors of 15 are 3 and 5, because 3×5 = 15. Some numbers have more than one factorization (more than one way of being factored). For instance, 12 can be factored as 1×12, 2×6, or 3×4.
## How many factors does 36 square have?
Factors of 36 are 1,2,3,4,6,9,12,18 & 36.
## What is a missing factor?
A factor is a number multiplied by another number to make a product. A product is the answer we get when we multiply two or more factors. … Using a Calculator to divide: Product / Factor = Missing Factor.
## How many ways can you make 36?
9 different waysso, a+c=2 ; b+d=2; so, 3c1*3c1 = 9. So. 9 different ways we can write 36 as a product of two natural numbers.
## What 4 numbers add up to 36?
1 + 35 = 36, 2 + 34 = 36, 3 + 33 = 36, 4 _ 32 = 36, 5 + 31 = 36, 6 + 30 = 36, 7 + 29 = 36, 8 + 28 = 36, 9 + 27 = 36, 10 + 26 = 36, 11 + 25 = 36, 12 + 24 = 36, 13 + 23 = 36, 14 + 22 = 36, 15 + 21 = 36, 16 + 20 = 36, 17 + 19 = 36, 18 + 18 = 36.
## What is the HCF of 24 36 and 60?
4H.C.F of 24, 36 and 60 is 4.
## What is the HCF of 24 36 and 48?
find gcf (24, 36 and 48) = ? step 3The common prime factor or the product of all common prime factors between all integers is the highest common factor of this group of integers (24, 36 & 48). | 1,263 | 3,351 | {"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-2020-45 | latest | en | 0.919505 |
https://cg-tower.com/what-equals-3-4-teaspoon/ | 1,638,241,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00490.warc.gz | 236,930,694 | 6,382 | What Is 3/4 that a Teaspoon? when measuring materials in the kitchen, ¾ the a teaspoon is equal to ¼ the a tablespoon or around 4 milliliters. This is measure up by using a ½ teaspoon and also a ¼ teaspoon. Many sets of measure spoons do not incorporate a ¾ teaspoon option, so combining smaller choices is normally easiest.
You are watching: What equals 3/4 teaspoon
Click come see full answer. Also, how can I get fifty percent of 3/4 teaspoon?
One half of 3/4 teaspoon is same to 3/8 or 0.375 teaspoons. One technique of finding the price in the fraction type of 3/8 is to main point the fraction 1/2 by 3/4.
Additionally, what is 3/4 that a teaspoon in ML? measure up of drugs 1/4 teaspoon 1.25 ml
3/4 teaspoon 3.75 ml
1 teaspoon 5 ml
1-1/2 teaspoon 7.5 ml
1 tablespoon 15 ml
Besides, what 3/4 tablespoon doubled?
Scale, fifty percent and double Quantity quantities in a cooking recipes (Chart) original Recipe Measure fifty percent Scaled Measure twin Scaled measure up
1/3 cup 2 tbsp. + 2 tsp. 2/3 cup
1/2 cup (4 fl. Oz.) 1/4 cup 1 cup
2/3 cup 1/3 cup 1 1/3 cups
3/4 cup 3 tbsp. 1 1/2 cups
How can I measure 3/4 cup?
The various other simple way to measure 3/4 cup is as follows : fill a cup through the point you desire to measure. Pour or take it out fifty percent of that into another cup(this is 1/2 cup,). Now from either of the cups take it out half of the thing you space measuring(it is 1/4 cup) .
Yeny CondalProfessional
## How deserve to I measure up 3/4 teaspoon?
What Is 3/4 of a Teaspoon? when measuring products in the kitchen, ¾ that a teaspoon is equal to ¼ of a tablespoon or about 4 milliliters. This is measured by making use of a ½ teaspoon and a ¼ teaspoon. Many sets the measuring spoons perform not incorporate a ¾ teaspoon option, so combine smaller options is typically easiest.
Lelah HomedesProfessional
## What is fifty percent of 2/3 cup of flour?
reducing the size of Recipes
1/3 cup 2 tablespoons + 2 teaspoons
1/2 cup 1/4 cup
2/3 cup 1/3 cup
3/4 cup 6 tablespoons
Azzeddine HoyleProfessional
## What is half of 3/4 cup of sugar?
Measure half that 3/4 cup sugar by utilizing a tablespoon. The variety of tablespoons the adds approximately 3/4 cup is 12, so division 12 in half and include 6 tablespoons that sugar to your recipe because that half the 3/4 cup.
Charifa SiscarExplainer
## What is fifty percent of 3/4 in fraction?
Half of 3/4: may be composed as 3/4/2 or as 3/4*1/2 (which is additionally 3/8). Conceptually, what is half that 3 parts out the 4 total parts. It does do sense, visually, the the visualized size of half of 3/4 would certainly be 3/8.
Magencia VicariaExplainer
## How have the right to I measure up 2/3 cup?
Use a 1/3 of a cup and also fill it double if girlfriend don"t own or can"t uncover your 2/3 measure up cup. You can additionally use 10 tablespoons to add 2 teaspoons in a pinch together a conversion for 2/3 that a cup.
Hendrik AlierExplainer
## What is .375 together a fraction?
Fractions to decimals to Inches come MM conversion Chart
fractions Decimal millimeter
11/32 .3437 8.731
23/64 .3594 9.128
3/8 .375 9.525
25/64 .3906 9.921
Massinissa CoronhoPundit
## What is fifty percent of 2/3 in fraction form?
Since 4/12 = 1/3, we see that half of 2/3 is 1/3. One more option is to main point 2/3 by 1/2: 2/3 x 1/2 = 2/6 = 1/3.
Onelia MambrillaPundit
## How many ounces is 2/3 cup that water?
5.37
Sheyla BotelhoPundit
## How much is 1/4 tespoon doubled?
Doubling Ingredients
A B
ingredient: 2 pinches doubled: 4 pinches
ingredient: 2/5 cup doubled: 4/5 cup
ingredient: 1/3 cup doubled: 2/3 cup
ingredient: 1/4 teaspoon doubled: 1/2 teaspoon
Brett GrothePundit
## How countless 2/3 cup does it require to make 1 cup?
One means would be to have actually two cups one the is 2/3 and one that is exactly 1 cup. Fill the 2/3 cup up v what ever before you want to measure.. Favor water and put the in the empty cup. Fill the 2/3 cup again and also fill the partially filled cup through water again.
Leyda FanjulPundit
## How plenty of 3rds room in a cup?
Explanation: 13 rd that a cup means there space three 13 the a cup, every cup.
Talwinder GaruzTeacher
## What is half of 3rd cup?
When dividing by a fraction, invert that portion (the denominator) and also multiply. As soon as multiplying 2 fractions, main point the 2 numerators. Climate multiply the two denominators. So, half of a third cup = 1/6 (one sixth).
Germaine RamiloSupporter
## How numerous milligrams room in a tespoon of liquid?
Teaspoon: that is a unit of measure up of volume the a medication or dosage i m sorry is equal to 5 milliliters. The unit is abbreviated together tsp. Transform Milligrams (mg) to Teaspoons (tsp): 1 mg is approximately equal to 0.0002 tsps.
Georgeanna ArochaSupporter
## How numerous TSP is 7.5 mL?
7.5ml equals to 1.52 teaspoons or there space 1.52 tsp in 7.5 milliliters.
Syuleyman BauerrichterSupporter
## Is 0.5 mL the same as 5 ml?
0.5ml is no the same as 5ml. 5ml is 10 times as much as 0.5ml.
Piotr ArangoitiBeginner
## What is 10 mL identical to in teaspoons?
10 ml to tsp. How plenty of Teaspoons is 10ml? - 10 ml is equal come 2.03 teaspoons.
## How do you measure up 10 mL the medicine?
by Drugs.com
10mL amounts to two teaspoons (2tsp). A tablespoon is 3 times bigger 보다 a teaspoon and also three teaspoons same one tablespoon (1Tbsp or 1Tb). One tablespoon also equals 15mL.
See more: It Was Nice To Meet You In Japanese ? How To Say Nice To Meet You In Japanese
Huifang LikhodeiBeginner
## How lot is one mL in a dropper?
A conventional dropper produces 20 drops per milliliter (20 drops = 1ML = 7 MG) yet dropper sizes deserve to differ. You can measure the number of drops in a milliliter making use of your dropper and find charts to readjust the variety of drops/ML if your dropper is different. | 1,754 | 5,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-49 | longest | en | 0.888266 |
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# Equal volumes of 1 M HCl and 1 M ${{H}_{2}}S{{O}_{4}}$ are neutralized by 1M NaOH solution and x and y kJ equivalent heat are liberated respectively. Which of the following relations is correct.A. x = 2yB. x = 3yC. x = 4yD. x = 0.5y
Last updated date: 12th Sep 2024
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Hint: When acid and base react they form a salt and release some amount of energy also. These types of reactions are called exothermic reactions and neutralization reactions due to the formation of a neutral compound (salt). The amount of heat liberated in a reaction depends on the number of protons released during the reaction.
Complete step by step solution:
In the question it is given that 1 M HCL and 1 M ${{H}_{2}}S{{O}_{4}}$reacts with 1 M NaOH solution and liberates x and y kJ equivalents of heat.
HCl dissociates and gives the following products.
$HCl\to {{H}^{+}}+C{{l}^{-}},\text{ }\Delta H=x$
${{H}_{2}}S{{O}_{4}}$ dissociates and gives the following products.
${{H}_{2}}S{{O}_{4}}\to 2{{H}^{+}}+SO_{4}^{2-},\text{ }\Delta H=y$
From the above equations we can say that HCl gives one mole of ${{H}^{+}}$and ${{H}_{2}}S{{O}_{4}}$gives two moles of${{H}^{+}}$, means ${{H}_{2}}S{{O}_{4}}$releases double amount of heat when compared to HCl in neutralization reaction with 1 M NaOH solution.
Therefore, $x=\dfrac{y}{2}\text{or }y=2x\text{ or }x=0.5y$.
So, the correct option is D.
Note: Hydrochloric acid liberates less amount of heat when compared to sulphuric acid during neutralisation reaction because sulphuric acid contains two protons and hydrochloric acid contains only one proton. Coming to sulphuric acid and nitric acid, nitric acid releases more heat compared to sulphuric acid, because nitric acid contains three protons and sulphuric acid contains two protons. | 572 | 1,890 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.871568 |
https://www.onemathematicalcat.org/algebra_book/online_problems/fac_tri_one_cneg.htm | 1,642,449,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00120.warc.gz | 949,864,740 | 10,904 | Factoring Trinomials (coefficient of x^2 term is 1, constant term is negative)
FACTORING TRINOMIALS OF THE FORM $\,x^2 + bx + c\,$, WHERE $\,c\lt 0$
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Here, you will practice factoring trinomials of the form $\,x^2 + bx + c\,$,
where $\,b\,$ and $\,c\,$ are integers, and $\,c\lt 0\,$.
That is, the constant term is negative.
Recall that the integers are: $\,\ldots,-3,-2,-1,0,1,2,3,\ldots$
As discussed in Basic Concepts Involved in Factoring Trinomials,
you must first find two numbers that add to $\,b\,$ and that multiply to $\,c\,$, since then: $$\,\cssId{s11}{x^2 + bx + c} \ \ \cssId{s12}{=\ \ x^2 + (\overset{=\ b}{\overbrace{f+g}})x + \overset{=\ c}{\overbrace{\ fg\ }}} \ \ \cssId{s13}{=\ \ (x + f)(x + g)}$$
Since $\,c\,$ is negative in this exercise, one number will be positive, and the other will be negative.
(How can two numbers multiply to give a negative result? One must be positive, and the other negative.)
That is, the numbers will have different signs.
For example, to mentally add $\,(-5) + 3\,$, in your head you would compute $\,5 - 3\,$,
Think of it this way: Start at zero on a number line. Walk $\,5\,$ units to the left, and $\,3\,$ units to the right. You end up at $\,-2\,$.
You walked farther to the left than you did to the right, so your final answer is negative.
The sign of $\,b\,$ (the coefficient of the $\,x\,$ term) determines which number will be positive, and which will be negative:
If $\,b\gt 0\,$, then the bigger number (the one farthest from zero) will be positive.
If $\,b\lt 0\,$, then the bigger number (the one farthest from zero) will be negative.
In other words, the biggest number takes the sign (plus or minus) of $\,b\,$.
These results are summarized below:
FACTORING TRINOMIALS OF THE FORM $\,x^2 + bx + c\,$, $\,c\lt 0$
• Check that the coefficient of the square term is $\,1\,$.
• Check that the constant term ($\,c\,$) is negative.
• It's easier to do mental computations involving only positive numbers.
So, you will initially ignore all minus signs and just work with the numbers $\,|b|\,$ and $\,|c|\,$.
• Find two numbers whose DIFFERENCE is $\,|b|\,$ and whose PRODUCT is $\,|c|\,$.
That is, find two numbers that subtract to give $\,|b|\,$ and that multiply to give $\,|c|\,$.
• Now (and only now), you'll use the actual plus-or-minus sign of $\,b\,$.
If $\,b\gt 0\,$, then the bigger of your two numbers is positive; the other is negative.
If $\,b\lt 0\,$, then the bigger of your two numbers is negative; the other is positive.
That is, the biggest number takes the sign (plus or minus) of $\,b\,$.
• Use these two numbers to factor the trinomial, as illustrated in the examples below.
EXAMPLES:
Question: Factor: $x^2 + 5x - 6$
Solution: Thought process:
Is the coefficient of the $\,x^2\,$ term equal to $\,1\,$? Check!
Is the constant term negative? Check!
Find two numbers whose difference is $\,5\,$ and whose product is $\,6\,$.
That is, find two numbers that subtract to give $\,5\,$ and that multiply to give $\,6\,$.
The numbers $\,1\,$ and $\,6\,$ work, since $\,6 - 1 = 5\,$ and $\,6\cdot 1 = 6\,$.
Since the coefficient of $\,x\,$ is positive, the bigger number ($\,6\,$) will be positive, and the other will be negative.
The desired numbers are therefore $\,6\,$ and $\,-1\,$.
Then, $$\,\cssId{s64}{x^2 + 5x - 6} \ \ \cssId{s65}{=\ \ x^2 + (\overset{=\ 5}{\overbrace{6+(-1)}})x + \overset{=\ -6}{\overbrace{\ 6\cdot (-1)\ }}} \ \ \cssId{s66}{=\ \ (x + 6)(x - 1)}$$ Check: $\cssId{s68}{(x+6)(x-1)} \cssId{s69}{= x^2 - x + 6x - 6} \cssId{s70}{= x^2 + 5x - 6}$
Question: Factor: $x^2 - 5x - 6$
Solution: Thought process:
Is the coefficient of the $\,x^2\,$ term equal to $\,1\,$? Check!
Is the constant term negative? Check!
Find two numbers whose difference is $\,5\,$ and whose product is $\,6\,$.
That is, find two numbers that subtract to give $\,5\,$ and that multiply to give $\,6\,$.
The numbers $\,1\,$ and $\,6\,$ work, since $\,6 - 1 = 5\,$ and $\,6\cdot 1 = 6\,$.
Since the coefficient of $\,x\,$ is negative, the bigger number ($\,6\,$) will be negative, and the other will be positive.
The desired numbers are therefore $\,-6\,$ and $\,1\,$.
Then, $$\,\cssId{s85}{x^2 - 5x - 6} \ \ \cssId{s86}{=\ \ x^2 + (\overset{=\ -5}{\overbrace{(-6)+1}})x + \overset{=\ -6}{\overbrace{\ (-6)\cdot 1\ }}} \ \ \cssId{s87}{=\ \ (x - 6)(x + 1)}$$ Check: $\cssId{s89}{(x-6)(x+1)} \cssId{s90}{= x^2 + x - 6x - 6} \cssId{s91}{= x^2 - 5x - 6}$
Question: Factor: $x^2 + x - 1$
Solution:
There are no integers whose difference and product are both $\,1\,$.
Thus, $\,x^2 + x - 1\,$ is not factorable over the integers.
Master the ideas from this section
When you're done practicing, move on to:
Factoring Trinomials, All Mixed Up
CONCEPT QUESTIONS EXERCISE: | 1,625 | 4,855 | {"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} | 4.90625 | 5 | CC-MAIN-2022-05 | longest | en | 0.807616 |
http://math.stackexchange.com/questions/72729/proving-sets-are-measurable/72768 | 1,469,516,093,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824756.90/warc/CC-MAIN-20160723071024-00148-ip-10-185-27-174.ec2.internal.warc.gz | 156,447,143 | 17,694 | # Proving sets are measurable
The problem statement, all variables and given/known data
The question is from Stein and Shakarchi, Real Analysis 2, Chapter 1, Problem 5:
Suppose $E$ is measurable with $m(E) < \infty$, and $E=E_1\cup E_2$, $E_1\cap E_2=\emptyset$.
Prove:
a) If $m(E) = m^{*}(E_1) + m^{*}(E_2)$, then $E_1$ and $E_2$ are measurable.
b) In particular, if $E \subset Q$, where $Q$ is a finite cube, then $E$ is measurable if and only if $m(Q) = m^{*}(E) + m^{*}(Q − E)$.
The definition of a 'measurable set' given in the book is that for any $\epsilon > 0$ there exists an open set $O$ with $E \subset O$ and $m^{*}(O − E) \leq \epsilon$, so I'm looking for a set of implications that lead me back to this definition.
all i could prove is that if $E$ measurable from my definition up, iff $m(A) = m( A \cap E) + m(A \cap E^{c})$
Thanks in advance for any help you can give me - it's very much appreciated.
-
We define the inner measure $m_*$ of a set $X$ as $$m_*(X)=\sup_{F\in\mathcal{C}}\ m(F),$$ where $\mathcal{C}$ is the family of closed subsets of $X$.
Then you can prove the following lemmas:
Lemma 1 For all $E$:
$i)$ $m_{\star}(E)\leq m^{\star}(E)$
$ii)$ If $E$ is measurable then $m_*(E)=m^*(E)$. If $m_*(E)=m^*(E)\lt \infty$ then $E$ is measurable.
Lemma 2 If $E$ is measurable and $A$ is any subset of $E$, then $$m(E)=m_*(A)+m^*(E\setminus A).$$
Now, note that if $E_1\cap E_2=\emptyset$ and $E=E_1\cup E_2$ then \begin{align*} E\setminus E_2&= (E_1\cup E_2)\setminus E_2\\ &= E_1\setminus E_2\\ &= E_1\setminus (E_1\cap E_2)\\ &= E_1. \end{align*}
Also note that is enough to show that $E_1$ is measurable. Since $E_2\subseteq E$, $m^*(E_2)\lt \infty$. By your hypothesis and the lemma 2 you have $$m(E)=m^*(E_1)+m^*(E_2)$$ and $$m(E)=m_*(E_1)+m^*(E_2).$$
I think you can conclude the proof from this point.
-
Thanks!! got it :) – alice Oct 15 '11 at 15:32 | 712 | 1,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-30 | latest | en | 0.818232 |
https://www.advancedfootballanalytics.com/2007/01/luck.html | 1,726,124,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00144.warc.gz | 580,553,371 | 18,133 | Which teams were luckiest in 2006?
When we are engaged in any endeavor, we know we shouldn't think about how much luck has to play in the outcome. It is usually counterproductive to dwell on luck because that tends to reduce our efforts towards our goal. Think about it--Why should I try so hard if it comes down to luck anyway?
In the NFL, if teams thought that way they'd probably be dead meat. But luck is a factor in all sports. Think about a very simple example game. Assume both PIT and CLE each get 12 1st downs in a game against each other. PIT's 1st downs come as 6 separate bunches of 2 consecutive 1st downs followed by a punt. CLE's 1st downs come as 2 bunches of 6 1st downs resulting in 2 TDs. CLE's remaining drives are all 3-and-outs followed by a solid punt. Each team performed equally well, but the random "bunching" of successful events gave CLE a 14-0 shutout.
The linear efficiency model I've began using early in the season has an r-squared value of almost 0.75. That means that 75% of the variance in the outcome (season wins) can be explained by the model's variables. But by including additional variables such as penalty yards or special teams the r-squared only marginally increases and they are largely insignificant. Those factors are fairly random and chaotic anyway, which is one way to define luck. So we could conjecture that a measurable part of a team's win-loss record, but something less than 20% is due to luck.
So how can we determine how lucky a team is? By using the model and estimating the number of wins a team "should" have based on its stats, the number of "expected wins" is calculated. The difference between a team's actual wins and its expected wins reveals how lucky a team is.
Seattle appears to have been the luckiest team this year, winning about 4 more games than we would expect given their stats. They squeaked out 1 game over .500 in a very weak NFC-West to make the playoffs. If they were in the AFC they probably would not have made the post season at all.
Minnesota appears to be the unluckiest. This is probably due to their league-leading run defense, but they didn't have the wins to show for it.
Notice Pittsburgh's expected win number--10.24. They played well enough this year to win 10 games but, according to most analysts "collapsed" and didn't make the playoffs a year after winning the Super Bowl. But their regular season record last year was 11-5. One could make the case that the Steelers played only slightly worse than they did last year, but just got unlucky.
Jim Mora, Jr., former coach of the Falcons, or other fired coaches might have used this to save his job. "Mr. Blank, look, we actually played well enough to win 9 or 10 games and would have made the playoffs!" But somehow I don't think that would fly, no matter how sound the math.
2 Responses to “Which teams were luckiest in 2006?”
1. Anonymous says:
"Notice Pittsburgh's expected win number--10.24. They played well enough this year to win 10 games but, according to most analysts "collapsed" and didn't make the playoffs a year after winning the Super Bowl. But their regular season record last year was 10-6."
I think their regular season record last year was 11-5.
2. Brian Burke says:
That's true. I'll make the correction in the post. | 767 | 3,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.983631 |
http://www.speedsolving.com/wiki/index.php/LBL | 1,411,047,690,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657127503.54/warc/CC-MAIN-20140914011207-00168-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 819,392,699 | 7,049 | Layer by layer
(Redirected from LBL)
LBL (3x3x3) Information Proposer(s): none Proposed: 1980s Alt Names: none Variants: CFOP, 8355 No. Steps: 5+ No. Algs: 6 to 16 (depending on LL) Avg Moves: unknown Purpose(s):
Layer-By-Layer, or normally only LBL is a group of methods that solves the cube in layers. In the basic, beginner LBL method, the solver finishes the layers one at a time: the first layer edges, then corners, then the second layer edges, and finally the last layer. This is the most common method for new cubers to discover on their own. In more advanced LBL methods, you solve layers more efficiently or solve two layers at once. For example, in the Fridrich method, one solves the first two layers simultaneously by forming a cross of the first layer edges, and then filling in four pairs of a corner and an edge into the so-called slots.
3x3x3 LBL Method
Also known as LBL. Solves the cube in layers: First the bottom layer, then the middle layer, and finally the top layer or LL. As a slight advancement to basic Layer By Layer, the Keyhole method can be used to enable solving of the middle layer in fewer moves.
2x2x2 LBL Method
LBL (2x2x2) Information Proposer(s): none Proposed: 1980s Alt Names: none Variants: CLL No. Steps: 3 No. Algs: 2 to 9 (depending on LL) Avg Moves: unknown Purpose(s):
Done as if solving the 3x3 corners using the Layer-By-Layer method. You can use shorter last layer algs because there are no edges that need to be kept in position on a 2x2. You can use any OLL of your choice as long as the corners are in the correct position. There are two options for the last layer in a beginner 2x2 solve:
• Orientation followed by Permutation
• Permutation followed by Orientation | 446 | 1,727 | {"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} | 3.21875 | 3 | CC-MAIN-2014-41 | latest | en | 0.898654 |
https://www.physicsforums.com/threads/neophyte-with-basic-questions.550676/ | 1,653,576,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00294.warc.gz | 1,068,917,774 | 23,643 | # Neophyte with basic questions
By way of introduction, I am a database engineer looking at returning to graduate school to focus on the overlaps in information theory, quantum theory and relativity. So I'm working my way through undergraduate texts on the foundations of said theories, attempting to resurrect long dormant math skills, etc.
Needless to say, I have many questions, and as I'm not at University, few resources for getting answers.
Many of my questions are at this point conceptual not mathematical, so I apologize for my lack of rigor. But this is important to me as I need to build appropriate visualizations before the math makes sense to me.
So two very basic questions:
Does it make sense to talk about the "shape" of the universe at the Plank scale? I mean this in the same sense that at the macro scale the universe is thought to be spherical, flat or hyperbolic.
I understand that the position of an electron in its orbital cloud is essentially probabilistic and you can't talk about where it is at one moment and then predict where it will be in the next moment. It's next position could be anywhere within the cloud based on the probability function. So my question: Does it make sense to describe the motion of the electron as discontinuous? E.G. in classical mechanic, you can't get from point A to point C without passing through point B. But in a probabilistic model, there is a non-zero chance of "appearing" at point C.
Bob
ZapperZ
Staff Emeritus
By way of introduction, I am a database engineer looking at returning to graduate school to focus on the overlaps in information theory, quantum theory and relativity. So I'm working my way through undergraduate texts on the foundations of said theories, attempting to resurrect long dormant math skills, etc.
Needless to say, I have many questions, and as I'm not at University, few resources for getting answers.
Many of my questions are at this point conceptual not mathematical, so I apologize for my lack of rigor. But this is important to me as I need to build appropriate visualizations before the math makes sense to me.
So two very basic questions:
Does it make sense to talk about the "shape" of the universe at the Plank scale? I mean this in the same sense that at the macro scale the universe is thought to be spherical, flat or hyperbolic.
What exactly is meant by the "shape of the universe" at that scale? In fact, what does it mean at ANY scale? The "shape" of something means that there's a boundary signifying when the object has an abrupt end. How do you ask that at a microscopic scale that is inside the object itself? Does this makes any sense to you?
I understand that the position of an electron in its orbital cloud is essentially probabilistic and you can't talk about where it is at one moment and then predict where it will be in the next moment. It's next position could be anywhere within the cloud based on the probability function. So my question: Does it make sense to describe the motion of the electron as discontinuous? E.G. in classical mechanic, you can't get from point A to point C without passing through point B. But in a probabilistic model, there is a non-zero chance of "appearing" at point C.
Bob
You have to be very careful here. Once you have made a measurement, the description of the system is now different than what it was before. The act of measurement has put the system into a particular state, and the original description is no longer valid.
The "motion" of the electron isn't continuous, simply because we cannot track such a movement. If we simply go by the wavefunction, then we can already see that it isn't discontinuous based on such a description.
While it is tempting to do quantum mechanics before understanding the mathematics, relying purely on physical intuition isn't something I highly recommend. What grounds QM to what we already know IS the mathematics. The physical description of it is discontinuous (since you are interested in such a concept) from our classical physical intuition. So to start with that ahead of the mathematics will often result in a misunderstanding of what QM is.
Zz.
What exactly is meant by the "shape of the universe" at that scale? In fact, what does it mean at ANY scale? The "shape" of something means that there's a boundary signifying when the object has an abrupt end. How do you ask that at a microscopic scale that is inside the object itself? Does this makes any sense to you?
Perhaps "shape" is the wrong word. Maybe the correct word is geometry? Parallel lines behave differently depending on the geometry of the space in which they reside.
"Does such a geometry exist at the Planck scale?" might be the better question.
You have to be very careful here. Once you have made a measurement, the description of the system is now different than what it was before. The act of measurement has put the system into a particular state, and the original description is no longer valid.
Having made a measurement, can we say with any certainty what the next measurement will be? This "feels" a little like a Markov Process - the measurement "erases" any previous description (a Markov Process requires no knowledge of previous state) and the next state is wholly independent of that description. But the next state is still probabilistic.
The "motion" of the electron isn't continuous, simply because we cannot track such a movement. If we simply go by the wavefunction, then we can already see that it isn't discontinuous based on such a description.
(hopeless naivete ahead) If our ACTUAL measurements are essentially discrete then how can we verify that ANY description of continuous motion is valid? The wave function is a model that suggests continuity but that continuity cannot be verified with actual measurements?
While it is tempting to do quantum mechanics before understanding the mathematics, relying purely on physical intuition isn't something I highly recommend. What grounds QM to what we already know IS the mathematics. The physical description of it is discontinuous (since you are interested in such a concept) from our classical physical intuition. So to start with that ahead of the mathematics will often result in a misunderstanding of what QM is.
Zz.
I am in no way avoiding the mathematics. I'm neck deep in it. I just work on multiple tracks and one of those is not so much developing an intuitive model so much as correcting the intuitive (or abandoning) model as needed.
ZapperZ
Staff Emeritus
Perhaps "shape" is the wrong word. Maybe the correct word is geometry? Parallel lines behave differently depending on the geometry of the space in which they reside.
"Does such a geometry exist at the Planck scale?" might be the better question.
No one knows.
Having made a measurement, can we say with any certainty what the next measurement will be? This "feels" a little like a Markov Process - the measurement "erases" any previous description (a Markov Process requires no knowledge of previous state) and the next state is wholly independent of that description. But the next state is still probabilistic.
That's the whole point of "wavefunction collapse"! You have put the system in a particular state. If I measure the spin of an electron, and finds that it is in a particular direction, and I continue to isolate it from everything else, what do you think is the outcome of my next, identical measurement of it?
[qutoe](hopeless naivete ahead) If our ACTUAL measurements are essentially discrete then how can we verify that ANY description of continuous motion is valid? The wave function is a model that suggests continuity but that continuity cannot be verified with actual measurements?[/quote]
But just because your measurement is discrete also doesn't make the system discrete! So the same argument can be applied to you.
I have a system that is described by, say, a plane wave. Can you tell me how, from that description, that I have a discrete system for, say, it's position? Again, unless you can argue this from the point of view of the mathematics, you really have nothing to stand on if you simply want to argue this based on "intuition".
I am in no way avoiding the mathematics. I'm neck deep in it. I just work on multiple tracks and one of those is not so much developing an intuitive model so much as correcting the intuitive (or abandoning) model as needed.
Read above. You simply cannot put the cart before the horse. Rather than looking at the actual object itself, you are already making guesses on what it is based on simply looking at the shadow of the object.
Zz.
No one knows.
Seems like an obvious question. Is the geometry at Planck scale irrelevant or just not yet determined?
(from your previous response - emphasis mine)
If we simply go by the wavefunction, then we can already see that it isn't discontinuous based on such a description.
But just because your measurement is discrete also doesn't make the system discrete! So the same argument can be applied to you.
If measurement is discrete how does one verify actual continuity? Seems that even if the wave function describes continuous motion, there is no form of measurement that allows us to verify that.
Ken G
Gold Member
Seems like an obvious question. Is the geometry at Planck scale irrelevant or just not yet determined?
Not yet determined, in the worst way. The best theory we have now (GR) isn't just moot on the question, it gets what must be the wrong answer to the question-- it's not consistent with quantum mechanics. This question awaits unification of QM and GR.
If measurement is discrete how does one verify actual continuity? Seems that even if the wave function describes continuous motion, there is no form of measurement that allows us to verify that.
Yes, I would say an important lesson of quantum mechanics is that we are not allowed to "mentally fill in the gaps" between our measurements, like we thought we could do in classical mechanics (notice what happens, for example, in the double slit experiment if you try to do that). Instead, quantum mechanics introduces a fundamental role for indeterminacy-- instead of what is determined and what is not determined being an issue of little importance, as it was in classical mechanics, in quantum mechanics it is a very important part of the complete picture. One might even say that quantum mechanics involves the union of what is determined from what is indeterminate, and both occupy essential places in the patchwork. Continuous motion was always a kind of fantasy of classical mechanics that we got away with but could never really verify. In quantum mechanics, it is part of the laws that we cannot verify it.
ZapperZ
Staff Emeritus
Seems like an obvious question. Is the geometry at Planck scale irrelevant or just not yet determined?
Why is this an "obvious" question? Obvious to whom? And what's with this Planck scale stuff when you're not even anywhere close to understanding QM, much less, QFT?
If measurement is discrete how does one verify actual continuity? Seems that even if the wave function describes continuous motion, there is no form of measurement that allows us to verify that.
Think of classical physics. Applying what you just said, even classical physics, using your logic, is "discrete". Is this something you are arguing for?
Zz.
Why is this an "obvious" question? Obvious to whom? And what's with this Planck scale stuff when you're not even anywhere close to understanding QM, much less, QFT?
Well it was obvious to me. And apparently not irrelevant based on Ken G's response.
Think of classical physics. Applying what you just said, even classical physics, using your logic, is "discrete". Is this something you are arguing for?
Measurement is discrete? No? At some point any form of ACTUAL measurement has some unit value that beyond which measurement has no meaning.
I'm not arguing for anything. I'm just asking questions. If measurement is the limiting factor in the validation of ANY physical theory, then anything that cannot be measured cannot be verified. Thus continuity cannot be verified. This does not argue that the universe is continuous or discontinuous, only that we cannot determine it to be one way or the other.
Does it make sense to talk about the "shape" of the universe at the Plank scale? I mean this in the same sense that at the macro scale the universe is thought to be spherical, flat or hyperbolic.
Hi Bob, I think this is a very fair question. I've also heard of things like this from TV documentaries on string theory. I think those who have already commented are much more qualified to talk about this since it is something I know very little about, but I think I can offer a simple suggestion for a physical picture that explains what other people have already said, which is that nobody knows.
If you think of usual X-ray crystallography: it is possible to study structure of a crystalline material because the wavelength is comparable to the lattice spacing. Now if instead of X-rays you shine physical light, you clearly see nothing of the internal structure, because the wavelength is much larger than the distance between atoms (the internal structure is 'smeared out'). If you care about long-wavelength behaviour, for example the long-range wavefunction of an electron, the electron appears to move in free space, and the only effect of the crystal is to give the electron a different mass (in condensed matter we call this the effective mass approximation). This is valid when you are dealing with distance scales much larger than the distance between atoms.
Now it turns out that you can do calculations in this approximation by writing the Hamiltonian for a particle that moves in vacuum with some effective mass. Instead of looking at the crystal, all information about the electron's interaction with the atomic cores and other electrons is buried in the new "mass" of the electron (the renormalized mass).
For a single electron, this picture is good, but as soon as you go to field theory, you find that as soon as you calculate anything, you get an infinite answer, because you must integrate over all momenta. Usually your integral diverges as momentum becomes infinite. There is a clear physical interpretation: when momentum becomes large, this corresponds to a distance scale (via the de Broglie relation) which is smaller than the spacing between atoms. So you have to introduce a cutoff (the ultraviolet cutoff), p_max at some "high" value of momentum to prevent your answer from becoming infinite. The interpretation is that in the theory you do not probe at distances less than h-bar/p_max.
Now for a crystal this is very sensible. However it turns out that when you do field theory in a vacuum, you also get similar infinities. "Common sense" would say that you can go to arbitrarily small distance scales and the vacuum should not have any structure there, there should be no problem when momentum becomes very large, so people do not understand the origin of these infinities. So people introduce a bit of mathematical machinery to get around these infinities. You say that there is a cutoff momentum beyond which you do not integrate, and then you always get a finite answer. However if you decide to interpret this physically, you can say that the momentum corresponds to a length scale which is the scale on which the gravitational force dominates, and possibly something disruptive is happening there, and the momenta over which you do integrate do not account for gravity at all.
However it turns out that when you do field theory in a vacuum, you also get similar infinities. "Common sense" would say that you can go to arbitrarily small distance scales and the vacuum should not have any structure there, there should be no problem when momentum becomes very large, so people do not understand the origin of these infinities. So people introduce a bit of mathematical machinery to get around these infinities.
Your entire response was helpful, but this part jumped out at me. The limits to what we can probe and the mathematical machinery to deal with the infinities is very interesting.
FWIW, my motivation behind these questions lies in information systems and how they map to physics. Our computer representations are fully discrete, our measurements have "granularity" that is easily modeled as discrete and even our mathematics are discrete strings of symbols. My interest in Planck scale is that it is essentially a lower limit to what can be validly symbolically represented in something like a Turing Machine or a Deterministic Finite Automata.
ZapperZ
Staff Emeritus
Why is this an "obvious" question? Obvious to whom? And what's with this Planck scale stuff when you're not even anywhere close to understanding QM, much less, QFT?
Well it was obvious to me. And apparently not irrelevant based on Ken G's response.
Measurement is discrete? No? At some point any form of ACTUAL measurement has some unit value that beyond which measurement has no meaning.
No, your measurement, ANY measurement, is "discrete". Show me what you think is a continuous measurement, and I'll show you a discrete one.
I'm not arguing for anything. I'm just asking questions. If measurement is the limiting factor in the validation of ANY physical theory, then anything that cannot be measured cannot be verified. Thus continuity cannot be verified. This does not argue that the universe is continuous or discontinuous, only that we cannot determine it to be one way or the other.
But unless you argue that nature puts something very strange in between our measurements, there is a physical description of a continuous evolution from one to the other.
For example, in a plane wave state, the particle has a continuous probability to be in all those places. Now, while measurements will reveal a particle being at x1, x2, x3, etc... unless you are arguing that there's something strange going on in between x2 and x2, then from the description that we have (the wavefunction), one can say that the position probability is continuous between x1 and x2, and so on, even when our measurements produces values at discrete locations.
Again, this is also what is done in classical systems. It is just that it is so closed together, it APPEARS as if it is a continuous set. It isn't!
And no, we DO know when there's a discontinuity. The existence of phase transition is a clear example. So it is not as if we don't know if there is a breakdown in the continuous evolution of a parameter. We do, and know if one occurs!
Again, show me an example of what you think is continuous. What you claim is not supported by the physics. That is the main problem that I have with what you are claiming.
Zz.
Again, show me an example of what you think is continuous. What you claim is not supported by the physics. That is the main problem that I have with what you are claiming.
Zz.
Maybe I've given the wrong impression. I'm not trying to claim that the universe is or isn't continuous.
I'm concerned with computer representations of physics. Continuity CANNOT be represented in a computer. And it seems to go deeper - that continuity is an abstraction that cannot be verified by ANY measurement. So while continuous functions are extraordinarily powerful, both descriptively and predictively, they cannot be linked to ACTUAL observation.
Ken G
Gold Member
I actually think you are making a very valid point, which is that in classical physics, we had discrete measurements but imagined that they were just tracing out something fundamentally continuous. This never created a problem in classical physics (though encounters subtleties in nonlinear dynamics), but it does in quantum physics-- what is continuous in quantum mechanics is the wave function, but the wave function implies certain inherent indeterminacies. So in a sense, it is what is indeterminate that is continuous, but what is determinate cannot be demonstrated as being continuous. The closer you look for continuity, the more you mess up the result! Verifying continuity is a bit like opening a book to verify that it was closed.
So we can address this in two ways-- rationalistically, we say that if the wave function is evolving continuously, then so is "nature", and it is only our ability to verify this fact that is the problem. But empirically, we say that what we can verify is all we can claim to be true, so if we cannot verify continuity, then we cannot claim it either. That is more along the lines of the way you are thinking, and I think it's a fine angle of approach. We really don't know at this stage which angle will bear the most fruit down the road (such as in investigations of the vacuum like tommyli was insightfully describing), so exploring every avenue is the rational course.
Indeed, it sounds like you are interested in how we manipulate finite amounts of (therefore discrete) information, so even if nature is "really" continuous, it is never going to matter to us that it is-- what will matter to us will always be the finite information we actually have access to. So it makes sense to model information as discrete, so that we are talking about the same kinds of information as what we actually get access to. Underlying that information might be a continuous theory, just as it was in classical mechanics, but that was never what we were actually dealing with-- which has important implications even in classical physics, in the context of chaotic systems.
I think what ZapperZ is saying is that quantum mechanics by itself really doesn't have answers to your questions, it is a mathematical structure that is fully self-contained and answers what it answers (basically, statistical predictions), but leaves open what it leaves open (what vacuum is, whether finite information is an inherent aspect of how nature works or if it is just a limitation we face in exploring nature, etc.). Since you are talking about going beyond quantum mechanics, it might be a good idea to master the mathematics of quantum mechanics first, while perhaps keeping your eye out for potential insights into the questions that really interest you. I think developing a working philosophy, while you learn the math, is a good idea-- it can really help channel your learning into fruitful directions, and increase your pleasure in a process that can otherwise be a bit dry if not daunting.
Last edited:
I actually think you are making a very valid point, which is that in classical physics, we had discrete measurements but imagined that they were just tracing out something fundamentally continuous. This never created a problem in classical physics (though encounters subtleties in nonlinear dynamics), but it does in quantum physics-- what is continuous in quantum mechanics is the wave function, but the wave function implies certain inherent indeterminacies. So in a sense, it is what is indeterminate that is continuous, but what is determinate cannot be demonstrated as being continuous. The closer you look for continuity, the more you mess up the result! Verifying continuity is a bit like opening a book to verify that it was closed.
So we can address this in two ways-- rationalistically, we say that if the wave function is evolving continuously, then so is "nature", and it is only our ability to verify this fact that is the problem. But empirically, we say that what we can verify is all we can claim to be true, so if we cannot verify continuity, then we cannot claim it either. That is more along the lines of the way you are thinking, and I think it's a fine angle of approach. We really don't know at this stage which angle will bear the most fruit down the road (such as in investigations of the vacuum like tommyli was insightfully describing), so exploring every avenue is the rational course.
Indeed, it sounds like you are interested in how we manipulate finite amounts of (therefore discrete) information, so even if nature is "really" continuous, it is never going to matter to us that it is-- what will matter to us will always be the finite information we actually have access to. So it makes sense to model information as discrete, so that we are talking about the same kinds of information as what we actually get access to. Underlying that information might be a continuous theory, just as it was in classical mechanics, but that was never what we were actually dealing with-- which has important implications even in classical physics, in the context of chaotic systems.
I think what ZapperZ is saying is that quantum mechanics by itself really doesn't have answers to your questions, it is a mathematical structure that is fully self-contained and answers what it answers (basically, statistical predictions), but leaves open what it leaves open (what vacuum is, whether finite information is an inherent aspect of how nature works or if it is just a limitation we face in exploring nature, etc.). Since you are talking about going beyond quantum mechanics, it might be a good idea to master the mathematics of quantum mechanics first, while perhaps keeping your eye out for potential insights into the questions that really interest you. I think developing a working philosophy, while you learn the math, is a good idea-- it can really help channel your learning into fruitful directions, and increase your pleasure in a process that can otherwise be a bit dry if not daunting.
You captured the gist of my thinking here. I bolded the parts that were particularly salient.
As for mastering the math of quantum mechanics, I fully understand this as a necessity, else everything is so much speculative blather. I just learn on parallel tracks. The math has to grow along side the conceptualization. The math tells me which parts of the conceptualizations need adjusting or must be abandoned, but I need both to progress.
Ken G
Gold Member
Yes, this is why many of us chose to be physicists, rather than mathematicians-- that "parallel track" that is so important to many physicists. | 5,320 | 26,116 | {"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 | 0.966801 |
https://www.ecologycenter.us/rain-forests-2/info-yol.html | 1,590,784,890,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406365.40/warc/CC-MAIN-20200529183529-20200529213529-00108.warc.gz | 710,863,669 | 6,914 | # Info
Step 1 Determine the x-axis and y-axis variables. Time varies independently of distance and is plotted on the x-axis. Distance is dependent on time and is plotted on the y-axis.
Step 2 Determine the scale of each axis.The x-axis data ranges from 0 to 5.The y-axis data ranges from 0 to 40.
Step 3 Using graph paper, draw and label the axes. Include units in the labels.
Step 4 Draw a point at the intersection of the time value on the x-axis and corresponding distance value on the y-axis.Connect the points and label the graph with a title, as shown in Figure 20.
Distance v. Time
Figure 20 This line graph shows the relationship between distance and time during a bicycle ride.
Figure 20 This line graph shows the relationship between distance and time during a bicycle ride.
Practice Problem A puppy's shoulder height is measured during the first year of her life.The following measurements were collected: (3 mo, 52 cm), (6 mo,72 cm), (9 mo,83 cm),(12 mo,86 cm).Graph this data.
Find a Slope The slope of a straight line is the ratio of the vertical change, rise, to the horizontal change, run.
g _ verticalchange (rise) _ change in y p horizontal change (run) change in x
Example Find the slope of the graph in Figure 20.
Step 1 You know that the slope is the change in y divided by the change in x.
Slope _ changein y change in x
Step 2 Determine the data points you will be using. For a straight line, choose the two sets of points that are the farthest apart. (40-0) km
Slope _
Step 3 Find the change in y and x. 40km
Slope _ "5iT
Step 4 Divide the change in y by the change in x. 8 km
Slope _
The slope of the graph is 8 km/h.
Bar Graph To compare data that does not change continuously you might choose a bar graph. A bar graph uses bars to show the relationships between variables. The x-axis variable is divided into parts. The parts can be numbers such as years, or a category such as a type of animal. The y-axis is a number and increases continuously along the axis.
Example A recycling center collects 4.0 kg of aluminum on Monday, 1.0 kg on Wednesday,and 2.0 kg on Friday.Create a bar graph of this data.
Step 1 Select the x-axis and y-axis variables.The measured numbers (the masses of aluminum) should be placed on the y-axis.The variable divided into parts (collection days) is placed on the x-axis.
Step 2 Create a graph grid like you would for a line graph. Include labels and units.
Step 3 For each measured number,draw a vertical bar above the x-axis value up to the y-axis value. For the first data point, draw a vertical bar above Monday up to 4.0 kg.
Aluminum Collected During Week
Monday Wednesday Friday Day of collection
Practice Problem Draw a bar graph of the gases in air: 78% nitrogen, 21% oxygen, 1% other gases.
Circle Graph To display data as parts of a whole, you might use a circle graph. A circle graph is a circle divided into sections that represent the relative size of each piece of data. The entire circle represents 100%, half represents 50%, and so on.
Example Air is made up of 78% nitrogen, 21% oxygen, and 1% other gases. Display the composition of air in a circle graph.
Step 1
Multiply each percent by 360° and divide by 100 to find the angle of each section in the circle.
Step 2 Use a compass to draw a circle and to mark the center of the circle. Draw a straight line from the center to the edge of the circle.
### Step 3
Use a protractor and the angles you calculated to divide the circle into parts. Place the center of the protractor over the center of the circle and line the base of the protractor over the straight line.
Other
Nitrogen 78%
Practice Problem Draw a circle graph to represent the amount of aluminum collected during the week shown in the bar graph to the left. | 898 | 3,780 | {"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-2020-24 | latest | en | 0.914141 |
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A059276 Row sums of triangle in A059274. 1
%I
%S 1,3,14,79,482,3088,20502,139882,975084,6915354,49742974,362054512,
%T 2661657346,19735338206,147418892116,1108339796742,8380510594346,
%U 63689530713888,486218971691294,3727035449588930,28674306373680492
%N Row sums of triangle in A059274.
%H G. C. Greubel, <a href="/A059276/b059276.txt">Table of n, a(n) for n = 0..250</a>
%t T[i_, j_]:= T[i, j] = Module[{r, s, t1}, If[i == 0 && j == 0, Return[1]]; If[j == 0, Return[1]]; t1 = T[i, j - 1]; For[r = 0, r <= i - j, r++,
%t For[s = 0, s <= j, s++, If[r + s != i, t1 = t1 + T[r + s, s]]]];
%t Return[t1]]; Table[Sum[T[n, k], {k, 0, n}], {n, 0, 10}] (* _G. C. Greubel_, Jan 04 2017 *)
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Jan 24 2001
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Last modified June 13 09:54 EDT 2021. Contains 344981 sequences. (Running on oeis4.) | 477 | 1,257 | {"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-2021-25 | latest | en | 0.558756 |
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How to calculate total sum of these table Rate Topic Display Mode Topic Options
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Posted Friday, July 11, 2014 12:13 AM
SSC-Enthusiastic Group: General Forum Members Last Login: Wednesday, July 27, 2016 4:25 AM Points: 186, Visits: 473
Hi Friends i ve the table like create table accutn_det( fs_locn char(50), fs_accno varchar(100),fs_cost_center varchar(100), fs_tran_type char(50) fs_amount numeric(50), fs_trans_date datetime,)insert into accutn_det values('CHN','E-Sw-2100','205produ','Cr','5000','2014-05-01')values('CHN','E-Sw-2100','205produ','Dr','15000','2014-05-06')values('HYD','E-Sw-2100','206produ','Dr','8000','2014-05-03')values('BANG','E-Sw-2100','208produ','Dr','25000','2014-05-01')values('BANG','E-Sw-2100','208produ','Cr','5000','2014-05-06')like all loctaion details stored from all months in these table here Dr=debit,Cr=Credit Formula= 'Dr-Cr' to find the salary wavges of amountso i made the query to find the amount for mayselect fs_locn, fs_accno, amount=sum(case when fs_accno like 'E%' and fs_tran_type='Dr' then fs_amount when fs_accno like 'E%' and fs_tran_type='Cr' then fs_amount * -1 end )from accutn_det where fs_trans_date between '01-may-2014' and '31-may-2014'groupby fs_locn,fs_accnonow i need the sum values of all costcenter for the particular accounthow to do that?
Post #1591438
Posted Friday, July 11, 2014 12:36 AM
SSCoach Group: General Forum Members Last Login: Thursday, September 22, 2016 2:39 AM Points: 15,371, Visits: 13,116
Add the cost center in the GROUP BY? How to post forum questions.Need an answer? No, you need a question.What’s the deal with Excel & SSIS?Member of LinkedIn. My blog at SQLKover. MCSA SQL Server 2012 - MCSE Business Intelligence
Post #1591447
Posted Friday, July 11, 2014 12:45 AM
Ten Centuries Group: General Forum Members Last Login: Tuesday, August 23, 2016 12:03 AM Points: 1,134, Visits: 1,399
Add Cost center (fs_cost_center) in Group by clause. Thanks
Post #1591459
Posted Friday, July 11, 2014 3:07 AM
SSC-Enthusiastic Group: General Forum Members Last Login: Wednesday, July 27, 2016 4:25 AM Points: 186, Visits: 473
Hi Friends i need out put likefor exmaple :account no 205prod 206prod totalE-SW-100 10000 3000 13000how make code?
Post #1591493
Posted Friday, July 11, 2014 3:17 AM
SSCoach Group: General Forum Members Last Login: Thursday, September 22, 2016 2:39 AM Points: 15,371, Visits: 13,116
raghuldrag (7/11/2014)Hi Friends i need out put likeaccount no 205prod 206prod totalE-SW-100 10000 3000 13000how make code?How is the 3000 calculated? How to post forum questions.Need an answer? No, you need a question.What’s the deal with Excel & SSIS?Member of LinkedIn. My blog at SQLKover. MCSA SQL Server 2012 - MCSE Business Intelligence
Post #1591500
Posted Friday, July 11, 2014 3:35 AM
SSC-Enthusiastic Group: General Forum Members Last Login: Wednesday, July 27, 2016 4:25 AM Points: 186, Visits: 473
Hi friends,i need the output like Accno 205produ 206produ 208produ totalE-SW-2100 15000 8000 20000 42000after find the sum of account how to calculate the total value of each costcenter for that account
Post #1591503
Posted Friday, July 11, 2014 3:55 AM
SSCoach Group: General Forum Members Last Login: Thursday, September 22, 2016 2:39 AM Points: 15,371, Visits: 13,116
Is it always 205, 206 and 208, or are there more columns possible? How to post forum questions.Need an answer? No, you need a question.What’s the deal with Excel & SSIS?Member of LinkedIn. My blog at SQLKover. MCSA SQL Server 2012 - MCSE Business Intelligence
Post #1591531
Posted Friday, July 11, 2014 4:22 AM
SSC-Enthusiastic Group: General Forum Members Last Login: Wednesday, July 27, 2016 4:25 AM Points: 186, Visits: 473
Hi Friends, The costcneter account wont be changed always same account and unique
Post #1591537
Posted Friday, July 11, 2014 4:52 AM
SSC-Enthusiastic Group: General Forum Members Last Login: Wednesday, July 27, 2016 4:25 AM Points: 186, Visits: 473
Hi Friends , i just added my code with rollupselect fs_locn, fs_accno,fs_cost_center,amount=sum(case when fs_accno like 'E%' and fs_tran_type='Dr' then fs_amountwhen fs_accno like 'E%' and fs_tran_type='Cr' then fs_amount * -1end)from accutn_det where fs_trans_date between '01-may-2014' and '31-may-2014'groupby fs_locn,fs_accno,fs_cost_center with rollupits giving the sum total value with NULLhow to avoid "Null" replace on TOTAL In That???
Post #1591551
Posted Friday, July 11, 2014 5:04 AM
Ten Centuries Group: General Forum Members Last Login: Tuesday, August 23, 2016 12:03 AM Points: 1,134, Visits: 1,399
Give it a try:`; WITH AccountCTE AS (SELECT fs_accno, fs_cost_center, fs_amount from accutn_det)SELECT fs_accno,[205produ],[206produ],[208produ], COALESCE([205produ], 0)+ COALESCE([206produ], 0)+ COALESCE([208produ],0) AS Total FROM AccountCTEPIVOT (SUM(fs_amount) FOR fs_cost_center IN ([205produ],[206produ],[208produ])) p` Thanks
Post #1591558
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Kinetic Friction Formula
Kinetic friction is a force that acts between moving surfaces. An object that is being moved over a surface will experience a force in the opposite direction as its movement. The magnitude of the force depends on the coefficient of kinetic friction between the two kinds of material. Every combination is different. The coefficient of kinetic friction is assigned the Greek letter "mu" (μ), with a subscript "k". The force of kinetic friction is μk times the normal force on an object, and is expressed in units of Newtons (N).
force of kinetic friction = (coefficient of kinetic friction)(normal force)
Fk = μk η
Fk = force of kinetic friction
μk = coefficient of kinetic friction
η = normal force (Greek letter "eta")
Kinetic Friction Formula Questions:
1) A worker in a stock room pushes a large cardboard box across the floor. The coefficient of kinetic friction is μk = 0.520. The box has a mass of 75.0 kg, and the worker is exerting a force of 400.0 Nforward. What is the magnitude of the force of friction, and what is the net force moving the box?
Answer: On a flat surface, the normal force on an object is η = mg. Using this, the formula can be used to find the force of friction:
Fk = μk η
Fk = μk mg
Fk = (0.520)(75.0 kg)(9.80 m/s2)
Fk = 382.2 kg∙m/s2
Fk = 382.2 N
The force of kinetic friction acting in the opposite direction as the motion of the box is 382.2 N. The net force acting on the box is the sum of forces. The two forces to consider are the force of kinetic friction acting in the opposite direction as the box's motion, and the force exerted by the worker, which is 400 N forward. If we define "forward" as the positive direction, the net force is:
Fnet=Fworker-Fk
Fnet=400.0 N-382.2N
Fnet = 17.8 N
The net force acting on the box is 17.8 N forward.
2) A woman is skiing straight down a snow-covered hill. The coefficient of kinetic friction between the skis and the snow is μk = 0.0800. The hill is at a 60.0° angle from the horizontal. The skier's mass is 55.00 kg. What is the magnitude of the force of kinetic friction, and what is the net force along the skier's direction of motion?
Answer: On a flat surface, the normal force on an object is η = mg. On a surface that is at an angle relative to the horizontal axis, The total force due to gravity, F = mg, must be broken in to components. The normal force is the component that is perpendicular to the angled surface, and the remaining force is parallel to the angled surface. The normal force is η = mg cosθ, and the remaining force component is F = mg sinθ. Using this, the formula can be used to find the magnitude of the force of kinetic friction:
Fk = μk η
Fk = μk mg cosθ
Fk = (0.0800)(55.00 kg)(9.80 m/s2)cos(60°)
Fk = 21.6 kg∙m/s2
Fk = 21.6 N
The force of kinetic friction opposes the motion of the skier. The force that moves the skier down the hill is the remaining component of the force of gravity, Fθ = mg sinθ. This force is:
F = mg sinθ
F =(55.0 kg)(9.80 m/s2)sin(60°)
F = 466.8kg∙m/s2
F = 466.8N
The net force acting on the skier is the sum of forces. The two forces to consider are the force directed down the hill and the force of kinetic friction directed up the hill. If we define the positive direction as down the hill, which is the skier's direction of motion, the net force is:
Fnet = F-Fk
Fnet = 466.8 N-21.6 N
Fnet = 445.2 N
The net force acting on the skier in her direction of motion is on the box is 445.2 N.
Related Links: Potential & Kinetic Energy Quiz Chemical Kinetics and Rates Quiz Kinetic Energy Examples Kinetic Energy Formula Inertia Examples Rolling Friction Examples Static Friction Formula Stopping Distance Formula | 991 | 3,730 | {"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} | 4.34375 | 4 | CC-MAIN-2020-05 | latest | en | 0.926939 |
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[SOLVED] Quaternion - Vector multiplication? (Read 4497 times) 0 Members and 1 Guest are viewing this topic.
theagentd
« JGO Bitwise Duke »
Medals: 437
Projects: 2
Exp: 8 years
« Posted 2012-01-14 19:40:37 »
I'm still trying to get skeleton animation working, but first I need to be able to load a model.
At first I was like "WHAT THE HELL IS THIS?!". Then I was like "Oh, this actually kind of makes sense!". Then I was like "WHAT THE HECK IS A QUATERNION?!". Then I was like "Haha! There's a Quaternion class in LWJGL!".
Then my example code was like
1 `vec3 rotPos = joint.m_Orient * weight.m_Pos; //m_Orient is a quaternion`
and I was like "NO WAI YOU DID NOT JUST DO THAT" and it was like "Oh, yes, I did." and I was like "But there is no quaternion-vector multiplication function in the LWJGL classes!" and it was like "Well, screw you." Then Google was like "Did ya mean quaternion-quaternion multiplication X 1 000 000?" and I was like "NO." and then, well, here I am.
So, how do I multiply a quaternion and a vector to get a new magical vector?
Myomyomyo.
Roquen
« Reply #1 - Posted 2012-01-14 19:55:08 »
I assume that your question is "how do I rotate a vector by a quaternion that represents a rotation?" Because there's more that one meaning of multiple a vector by a quaternion. If so, the the answer depends on how many vectors you want to rotate since there are multiple formulations.
sproingie
JGO Kernel
Medals: 202
« Reply #2 - Posted 2012-01-14 22:33:46 »
LWJGL's Quaternion class is ... well it just isn't very good. I'm not sure it even has a method to transform a Vector.
Java3D has a much more fully-featured quaternion class. You can find it here though you might want to find an alternative implementation with source both for licensing and documentation reasons (that one doesn't even have a javadoc artifact)
theagentd
« JGO Bitwise Duke »
Medals: 437
Projects: 2
Exp: 8 years
« Reply #3 - Posted 2012-01-14 22:35:24 »
I assume that your question is "how do I rotate a vector by a quaternion that represents a rotation?" Because there's more that one meaning of multiple a vector by a quaternion. If so, the the answer depends on how many vectors you want to rotate since there are multiple formulations.
You're exactly right. I want to rotate my position vector by the orientation quaternion. What do you mean by multiple formulations?
Also, this seems to be the relevant code in the C-something file:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 ` inline detail::tvec3 operator* ( detail::tquat const & q, detail::tvec3 const & v ) { typename detail::tquat::value_type Two(2); detail::tvec3 uv, uuv; detail::tvec3 QuatVector(q.x, q.y, q.z); uv = glm::cross(QuatVector, v); uuv = glm::cross(QuatVector, uv); uv *= (Two * q.w); uuv *= Two; return v + uv + uuv; }`
I have no idea what this means though...
LWJGL's Quaternion class is ... well it just isn't very good. I'm not sure it even has a method to transform a Vector.
Java3D has a much more fully-featured quaternion class. You can find it here though you might want to find an alternative implementation with source both for licensing and documentation reasons (that one doesn't even have a javadoc artifact)
I looked at the javadoc for that library, but I couldn't find a function for that either...
Myomyomyo.
Roquen
« Reply #4 - Posted 2012-01-15 09:33:08 »
I'm not at home, so here's a quick answer off the top of my head. (In other words, it probably doesn't work so verify it)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 `public final Vect3f rot(Vect3f v){ float k0 = w*w - 0.5f; float k1; float rx,ry,rz; // k1 = Q.V k1 = v.x*x; k1 += v.y*y; k1 += v.z*z; // (qq-1/2)V+(Q.V)Q rx = v.x*k0 + x*k1; ry = v.y*k0 + y*k1; rz = v.z*k0 + z*k1; // (Q.V)Q+(qq-1/2)V+q(QxV) rx += w*(y*v.z-z*v.y); ry += w*(z*v.x-x*v.z); rz += w*(x*v.y-y*v.x); // 2((Q.V)Q+(qq-1/2)V+q(QxV)) rx += rx; ry += ry; rz += rz; return new Vect3f(rx,ry,rz);}`
Quote
I want to rotate my position vector by the orientation quaternion. What do you mean by multiple formulations?
There are three main ways to have a quaternion represent a rotation. Each of these can be "reversed" yielding six formulations with different properties (they are all related). Given a specific mathematical formulation, each of may be used directly as if one was performing the operation in quaternions or it can be converted to some other mathematical type (almost always a matrix). The above (assuming I didn't screw up) is a direct formulation where the quaternion is assume to be of unit length (otherwise it will yield a scaled rotation, where the scaling factor is the magnitude of the quaternion squared). This question and your previous are really the same "in wolves clothing". If you were to pug the vectors (1,0,0), (0,1,0), & (0,0,1) into the above it will yield three vectors (really every where we're saying vectors should be bivectors, but that's a different story). Shove these three vectors into either a row or column of a matrix and you have the (misnamed) quaternion to matrix conversion. Make it a 4x4 matrix and you can shove the translation vector in the appropriate spot and bingo...you're done.
Quote
LWJGL's Quaternion class is ... well it just isn't very good.
It doesn't really do anything and was obviously thrown together by someone with no understanding of quaternions (no offense intended if said person is reading this...you're far from alone.)
theagentd
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« Reply #5 - Posted 2012-01-15 11:50:34 »
That was exactly what I needed! Thanks a lot! I also look forward to actually understanding that code when I get to college... >_>
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« Reply #6 - Posted 2012-01-15 18:20:24 »
Roquen
« Reply #7 - Posted 2012-01-16 05:18:22 »
theagentd
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« Reply #8 - Posted 2012-01-16 07:22:23 »
Those tmp variables looked really bad, but were they actually used somewhere in the code?
Myomyomyo.
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« Reply #9 - Posted 2012-01-16 07:59:46 »
I only use quaternion to interpolate rotation. When I have I have compute all rotations, I transform quaternions to matrices. Then I only use matrix. I have never use quaternion with vector directly.
Roquen
« Reply #10 - Posted 2012-01-16 08:51:09 »
The sweet point between direct computation and conversion to a matrix is hard to qualify. For scalar computation converting to matrices generally be a win past some very small number of conversions. For SIMD it really depends on the low level ops exposed.
@theagentd: I just skimmed...didn't look that closely.
Roquen
« Reply #11 - Posted 2012-01-16 14:50:25 »
Quote
Quote
LWJGL's Quaternion class is ... well it just isn't very good.
It doesn't really do anything and was obviously thrown together by someone with no understanding of quaternions (no offense intended if said person is reading this...you're far from alone.)
Opps..I was actually thinking of some other library...I guess I should check my facts before opening my mouth.
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## October 31, 2011
Study for properties quiz on Tuesday. Study for nine weeks test to be given on Friday. Study entire 1st nine weeks notebook. All of these notes will appear on the nine weeks test. Also, study the blue review booklet.
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## October 28, 2011
All classes: Study your interactive notebooks and your blue review booklet. The nine weeks test is next Friday.
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## October 27, 2011
ALL BLOCKS: Complete Ch. 6 Pretest Worksheet that was begun in class today. ANNOUNCEMENTS: All students will need a new Math notebook for the second nine weeks. Please bring them in on Wednesday, November 9th. Tomorrow (Friday) is Hat Day. … Continue reading
Posted in Uncategorized | Comments Off on October 27, 2011
## October 26, 2011
ALL BLOCKS: Red Book p. 608 Lesson 7-3 #1,2,3,4,5 and Lesson 7-5 #1,2,3,4,5 Quizzes on Thursday and Friday on Fractions and word problems
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## October 25,2011
All Blocks are to complete the 7 problems copied from the board into their interactive notebooks. The problems are on adding and subtracting mixed numbers.
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## October 24, 2011
All students received a copy of our red math book today to keep at home. From now on the homework will say if it is in the red book or the green book. This book is to be returned in June … Continue reading
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## October 21, 2011
Blocks 1 and 3: Green Math Book: p. 111 # 11, 13, 14 and p. 115 # 3, 9 Block 2: Green Math Book: p. 12 # 23, 28, 30 and p. 15 # 12a and 12 b Nine weeks … Continue reading
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## October 20, 2011
Blocks 1 : Using the green math book, complete p. 103 # 44, 45A, 45B, 50 and p. 108 #26. Block 2: Study all notes on fractions. Block 3: Using the green math book, complete p. 103 #44, 45A, 45B,p. … Continue reading
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## October 19, 2011
Blocks 1 and 3: Green math book: p. 95 #6, p. 96 # 22, 23, 24 and p. 101 #4 Block 2: Complete worksheet on modeling division of fractions.
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## October 17, 2011
Block 1 and 3 complete the front side of the modeling fraction multiplication problems worksheet that we marked as homework in class today. DO NOT do the other side. Block 2 complete INp. 34 Reviewing decimals.
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4. The sum to infinity of a geometric series if four times its first term. Find the common ratio.
3. How many terms of the arithmetic series 15+13+11... are required to make a total of -36?
3) a = 15 d = -2
Sn = n/2 (a + (n - 1) d)
-36 = n/2 (15 + (n - 1) -2)
-72 = n (15 - 2n + 2)
2n^2 - 17n + 72 = 0
(2n - 9)(n - 8) = 0
n = 8 or n = 9/2
FUCK YOU! GET PUMPED!
Quote by Craigo
Alright, these are messing me up, so...
4. The sum to infinity of a geometric series if four times its first term. Find the common ratio.
3. How many terms of the arithmetic series 15+13+11... are required to make a total of -36?
4.
S-inf = a/1-r
a/1-r=4a
a = 4a-4ar
1=4-4r
r=3
Edit:
Damn;
4r=3
r=3/4
Last edited by bequickorbedead at May 30, 2008,
4.
S = u/(1-r)
in this case becomes
4u = u/(1-r)
4u(1-r) = u
=> 1-r = 1/4
r = 3/4
4.
S-inf = a/1-r
a/1-r=4a
a = 4a-4ar
1=4-4r
r=3
Dude, condition for Sum to infinity to hold is that r < 1.
EDIT again, sorry, seems you realised yourself (-:
Last edited by Mad_BOB at May 30, 2008,
haha brain hurts from reading the questions
i took the exam 2 weeks ago pheeew
EDIT: core 2?
Quote by SuperBlob
3) a = 15 d = -2
Sn = n/2 (a + (n - 1) d)
-36 = n/2 (15 + (n - 1) -2)
-72 = n (15 - 2n + 2)
2n^2 - 17n + 72 = 0
(2n - 9)(n - 8) = 0
n = 8 or n = 9/2
i would like to correct this one
3) a = 15 d = -2
Sn = n/2 (2a + (n - 1) d)
-36 = n/2 (30 + (n - 1) -2)
-72 = n (30 - 2n + 2)
-72 = 32n -2n^2
2n^2 - 32n - 72 = 0
(2n + 4)(n - 18) = 0
n = 18 or (n = -2 which it cannot be)
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Ah crap.
I knew it was off with it giving 2 answers
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# Certain oil companies have been called poor corporate citizens because
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Certain oil companies have been called poor corporate citizens because [#permalink]
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27 Jul 2009, 22:39
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Certain oil companies have been called poor corporate citizens because they have opposed government action to limit global warming by undermining scientific research that characterizes the issue as severe. However, these same oil companies have also invested millions of dollars in scientific research to address the long term effects of climate change.
Which of the following best explains the apparent discrepancy in the situation described above?
A.) The oil companies only recently began investing in scientific research to address climate change issues.
B.) The research dollars invested by the oil companies are specifically earmarked for developing practical technologies that might be used to combat global warming.
C.) The government action opposed by the oil companies would negatively impact their profits.
D.) The scientific research that characterizes global warming as a severe problem has not been definitively proven.
E.) The oil companies don't believe that any scientific research related to climate change will ultimately serve their interests.
I can't for the life of me figure out how the OA makes any sense - just want to see if anyone else will get it.
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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27 Jul 2009, 23:30
IMO D
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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29 Jul 2009, 07:15
A,C and E>>i think are irrelevant.
was stuck b/w B and D....
i go for B...OA plzzz..
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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29 Jul 2009, 07:24
understudy wrote:
Certain oil companies have been called poor corporate citizens because they have opposed government action to limit global warming by undermining scientific research that characterizes the issue as severe. However, these same oil companies have also invested millions of dollars in scientific research to address the long term effects of climate change.
Which of the following best explains the apparent discrepancy in the situation described above?
A.) The oil companies only recently began investing in scientific research to address climate change issues.
B.) The research dollars invested by the oil companies are specifically earmarked for developing practical technologies that might be used to combat global warming.
C.) The government action opposed by the oil companies would negatively impact their profits.
D.) The scientific research that characterizes global warming as a severe problem has not been definitively proven.
E.) The oil companies don't believe that any scientific research related to climate change will ultimately serve their interests.
I can't for the life of me figure out how the OA makes any sense - just want to see if anyone else will get it.
A) MAYBE. The companies might have started investing due to the criticism.
B) MAYBE. In a way if they are doing "just enough" to not make the problem as severe so they can show people that due to their "advancement" the problem is not that severe
C) irrelevant. does not explain the discrepancy
D) prob not. I dont think they are undermining the research of others if they are doing similar research?
E) no. what's the point of them doing research if they dont believe it?
I think B. If the oil companies are "countering" claims made by researches by researching solutions then the overall problem will be less severe. In a way it is like putting simple patches on a sinking boat - it will make it sink much slower and whenever there is another hole it will be patched up but eventually the patches will not hold for long.
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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29 Jul 2009, 07:27
While D might be tempting, it will not help answer both the contradictory premises. 1) Undermining Government by saying that the global warming is severe 2) spending millions of dollars in research in long term effect of climate change
B will answer both of these discripiencies, since companies still do not change their position from undermining govt, at the same time invest money on further research.
The research dollars invested by the oil companies are specifically earmarked for developing practical technologies that might be used to combat global warming.
D) The scientific research that characterizes global warming as a severe problem has not been definitively proven. - this will give us the reason for the "undermining of govt" but doesnt give a reason for companies to invest.
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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31 Jul 2009, 09:14
B
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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31 Jul 2009, 10:12
OA is B. Good job guys.
Still seems that B simply restates what is already said in the paragraph.
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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02 Aug 2009, 10:47
1
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my choice too was B
To begin its easy to boil down to two options B & E. in case of E its says that companies believe that NO scientific research wold be beneficial. this is too extreme poison which does not correspond with argument that they invest in research for long term effects . . else why would they invest. so this option does not work.
now we are left with B and that's the answer. (though justifying B is a bit difficult, process of elimination worked for me)
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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03 Aug 2009, 10:39
[EDIT - REMOVED BY THE USER]
Last edited by nplaneta on 17 Jun 2012, 19:12, edited 1 time in total.
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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03 Aug 2009, 12:14
+1 for B
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Certain oil companies have been called poor corporate [#permalink]
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10 Oct 2009, 21:02
Certain oil companies have been called poor corporate citizens because they have opposed government action to limit global warming by undermining scientific research that characterizes the issue as severe. However, these same oil companies have also invested millions of dollars in scientific research to address the long term effects of climate change.
Which of the following best explains the apparent discrepancy in the situation described above?
A. The oil companies only recently began investing in scientific research to address climate change issues.
B. The research dollars invested by the oil companies are specifically earmarked for developing practical technologies that might be used to combat global warming.
C. The government action opposed by the oil companies would negatively impact their profits.
D. The scientific research that characterizes global warming as a severe problem has not been definitively proven.
E. The oil companies don't believe that any scientific research related to climate change will ultimately serve their interests.
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10 Oct 2009, 21:29
IMO D.
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10 Oct 2009, 21:32
Only D looks reasonable.
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11 Oct 2009, 10:23
IMO also D..OA pls..
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11 Oct 2009, 14:43
I went with D as well - this one is a toughie
OA is B
Explained by -
This choice provides a motive for companies to invest in global warming research. If the research is specifically geared at developing technologies used to combat global warming, this research represents a long-term strategy for finding new ways for the companies to profit. This is in contrast with the research that the oil companies oppose - research focused on the severity and immediacy of the issue, a focus which might lead to government action that could have a negative impact on the profitability of the oil companies in the short term.
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Re: Certain oil companies have been called poor corporate citizens because [#permalink]
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02 Nov 2010, 05:24
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Wats the OA?
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Certain oil companies have been called poor corporate [#permalink]
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08 Oct 2012, 07:47
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Certain oil companies have been called poor corporate citizens because they have opposed government action to
limit global warming by undermining scientific research that characterizes the issue as severe. However, these
same oil companies have also invested millions of dollars in scientific research to address the long term effects of
climate change. Which of the following best explains the apparent discrepancy in the situation described above?
A The oil companies only recently began investing in scientific research to address climate change issues.
B The research dollars invested by the oil companies are specifically earmarked for developing practical technologies
that might be used to combat global warming.
C The government action opposed by the oil companies would negatively impact their profits.
D
The scientific research that characterizes global warming as a severe problem has not been definitively proven.
E The oil companies don’t believe that any scientific research related to climate change will ultimately serve their
interests.
[Reveal] Spoiler:
B
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Re: Certain oil companies have been called poor corporate... [#permalink]
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08 Oct 2012, 09:26
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Premise 1- Oil Companies - Poor Corporate Citizen because they oppose Govt Action on Global warming which is based on scientific studies that term Global Warming as severe.
Premise 2 - Same companies spend dollars in scientific study to study effects of climatic change.
To solve this discrepancy that Oil Companies are on one hand opposing govt action (to oppose global warming, based on scientific studies) and on the other hand same companies are spending dollars to address Climatic change.
We need some info which can justify this disconnect between money spent by oil companies and where it is spent.
(A) Time duration not sufficient.
(B) Very finely proves that if the Oil companies are putting dollars to find out effects of climatic change then they very well believe that global warming can bring climatic change, thus they cannot refute Govt's action to oppose global warming by undermining the scientific study
(C) Profits. Not related here.
(D) Incorrect because if that scientific study is not proven yet then there is no discrepancy.
(E) What oil companies believe or not does not help solve the discrepancy.
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Re: Certain oil companies have been called poor corporate... [#permalink]
### Show Tags
09 Oct 2012, 00:44
Certain oil companies have been called poor corporate citizens because they have opposed government action to
limit global warming by undermining scientific research that characterizes the issue as severe. However, these
same oil companies have also invested millions of dollars in scientific research to address the long term effects of
climate change
. Which of the following best explains the apparent discrepancy in the situation described above?
• The oil companies only recently began investing in scientific research to address climate change issues.
• The research dollars invested by the oil companies are specifically earmarked for developing practical technologies
that might be used to combat global warming.
• The government action opposed by the oil companies would negatively impact their profits.
• The scientific research that characterizes global warming as a severe problem has not been definitively proven.
• The oil companies don’t believe that any scientific research related to climate change will ultimately serve their
interests.
[Reveal] Spoiler:
B
aditi1903 - they key here is the bold part. So option B right.
C oppose G's scientific research because they are not finding sols to the problems but seeing how big the problem is, which is already sever and the results would affect the company further. But the company is finding sol to the problem using their own research that will resolve the problem.
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Re: Certain oil companies have been called poor corporate... [#permalink]
### Show Tags
09 Oct 2012, 09:12
Certain oil companies have been called poor corporate citizens because they have opposed government action to
limit global warming by undermining scientific research that characterizes the issue as severe. However, these
same oil companies have also invested millions of dollars in scientific research to address the long term effects of
climate change. Which of the following best explains the apparent discrepancy in the situation described above?
• The oil companies only recently began investing in scientific research to address climate change issues.
• The research dollars invested by the oil companies are specifically earmarked for developing practical technologies
that might be used to combat global warming.
• The government action opposed by the oil companies would negatively impact their profits.
• The scientific research that characterizes global warming as a severe problem has not been definitively proven.
• The oil companies don’t believe that any scientific research related to climate change will ultimately serve their
interests.
[Reveal] Spoiler:
B
I had to chose b/w B and D, and I think D is better, any thoughts, anyone?
How do I eliminate D?
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Re: Certain oil companies have been called poor corporate... [#permalink] 09 Oct 2012, 09:12
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Display posts from previous: Sort by | 4,417 | 18,430 | {"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-2017-43 | latest | en | 0.947136 |
http://thetopsites.net/article/58118579.shtml | 1,600,424,881,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400187390.18/warc/CC-MAIN-20200918092913-20200918122913-00204.warc.gz | 166,593,224 | 5,329 | Find array index inside another array
matlab find position of value in array
matlab find array in array
matlab ismember
matlab find first instance of value in array
matlab intersect
array of objects javascript
javascript array
array in array javascript
Lets say i have two numpy arrays or lists example
```a = ['P', 'P' ,'Q', 'Q','P' ,'P','Q', 'Q']
b = ['Q', 'Q','P' ,'P']
```
and the result that i expect would be that it starts at
```start = 2
end = 6
```
Example below without converting to string
```a = ['P', 'P' ,'Q', 'Q','P' ,'P','Q', 'Q']
b = ['Q', 'Q','P' ,'P']
for i in range(len(a)):
if a[i:i+len(b)] == b:
print i,i+len(b)
break
```
Finding the indices of the elements of one array in another , Given two vectors A and B, find the index, idx into A of the element of B so that. A(idx)=B. Now I know there must be many ways it can be done, but is there a FindIndex<T> (T [], Int32, Predicate<T>) Searches for an element that matches the conditions defined by the specified predicate, and returns the zero-based index of the first occurrence within the range of elements in the Array that extends from the specified index to the last element.
It is easiest, if you convert the lists to a string:
```a1 = ''.join(a)
b1 = ''.join(b)
```
then you can use the `str.find()` function that returns the lowest index of the substring if it is found in given string. If its is not found then it returns -1.
Find indices of one array in another array., Find indices of one array in another array.. Learn more about find, index, array. How to search for an array inside another array?. Learn more about matlab, arrays, find, indexing, matrix manipulation MATLAB
thanks for you help, based on commentary discussion i was able to find this
```a = ['P', 'P' ,'Q', 'Q','P' ,'P','Q', 'Q']
b = ['Q', 'Q','P' ,'P']
start = '|'.join(a).split('|'.join(b))[0].count('|')
end = start + len(b)
if start == len(a)-1:
print('the subset wasnt found')
else:
print('the subset was found, start: ', start, 'end: ',end)
```
Data Structures: Objects and Arrays :: Eloquent JavaScript, The notation for getting at the elements inside an array also uses square brackets. expression, with another expression inside of them, will look up the element To find out what properties an object has, you can use the Object.keys function Binary search: Binary search can also be used to find the index of the array element in an array. But the binary search can only be used if the array is sorted. Java provides us with an inbuilt function which can be found in the Arrays library of Java which will rreturn the index if the element is present, else it returns -1.
5. Working with Arrays and Loops, Arrays in JavaScript are zero-based, which means the first element index is zero, and you might want to traverse an array until you find either a specific element, are managed by creating a new array as an element within an existing array. Thanks for the three solutions. Here is a function to test all three. "method_1" wins as the most elegant and fastest, but the other two taught other useful ways of looking at the problem.
jQuery.inArray(), Description: Search for a specified value within an array and return its index (or -1 if not indexOf() method in that it returns -1 when it doesn't find a match. If you need the index of the found element in the array, use findIndex (). If you need to find the index of a value, use Array.prototype.indexOf (). (It’s similar to findIndex (), but checks each element for equality with the value instead of using a testing function.)
jQuery.map(), The \$.map() method applies a function to each item in an array or object and arguments: The element's value and its index or key within the array or object. Let's say I have 2 arrays of double, call then A and B. If both have unique entries and I want to find the position of each element of A in array B I can do: | 962 | 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} | 3.140625 | 3 | CC-MAIN-2020-40 | latest | en | 0.761624 |
https://www.futurelearn.com/courses/data-science/0/steps/25963 | 1,603,125,998,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107863364.0/warc/CC-MAIN-20201019145901-20201019175901-00343.warc.gz | 719,645,994 | 25,269 | ## Want to keep learning?
This content is taken from the Purdue University & The Center for Science of Information's online course, Introduction to R for Data Science. Join the course to learn more.
2.12
## Purdue University
Skip to 0 minutes and 8 seconds I wanna extend the example that I just did, where we were looking at the expected arrival delay on each day. And restricting the flights that were arriving to Indianapolis, okay? We can extend that example, that the destination airport is IND, and that the to now specify is ORD, which is O’Hare. There’s a lot of flights like that, okay? So, if I could look at the head of this vector, I just see that this vector contains trues and falses. Similarly, I could go look at another vector that tells me if the origin is O’hare, another vector of trues and falses. One thing I can do is I can go put those together with an ampersand.
Skip to 1 minute and 1 second Let’s go get rid of all our parentheses. You can keep the parenthesie in there if you want to. But this is a vector that identifies the flights, which have IND as the Dest and have ORD as the Origin. It is a vector of 7 million TRUE’s and FALSE’s. We can go take a look at the head of that vector. Again, trues and falses. You can take a look at the length of that vector, 7 million. And if you like, so that you don’t have to put all of that long string into your query above, you can even go save this. Let’s say we call ORD to IND. You can call it anything you want.
Skip to 1 minute and 45 seconds So now I’m gonna go take the code we had before. Say we find the expected arrival delay for flights from ORD to IND, categorized according to the day. So I’m no longer just gonna look for flights that have Indy as the destination. Gonna put in there whether the flight was from O’Hare to Indy or not. And if the flight was from O’Hare to Indy, we count it. And if it wasn’t, we don’t count it. There you go. So for instance, on December 23rd, there must have been some long kind of delays there. Because you had a three hour delay, and on February 12th, 142 minutes delay on average, all right?
Skip to 2 minutes and 29 seconds Were there’s some days in which the expected arrival time was negative, in which you even expected to arrive early? There sure were. Can insert [LAUGH] lots and lots and lots and questions with these t apply functions. It’s good to be practicing. I hope you’re practicing as we’re going through some of these examples. And just trying some things and sort of doing some kind of checks to make sure that you’re getting reasonable results, right? And I’m not surprised at all that the longest expected arrival delay was right in the dead of the Winter. And there may have been very bad weather that day.
Skip to 3 minutes and 3 seconds That’s not entirely surprising, that that’s gonna happen at the end of the year when there could be snow or there could be bad weather. | 689 | 2,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-45 | latest | en | 0.960921 |
http://lbartman.com/worksheet/math-worksheets-for-fourth-grade.php | 1,566,367,036,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315809.69/warc/CC-MAIN-20190821043107-20190821065107-00100.warc.gz | 116,816,136 | 12,826 | ## lbartman.com - the pro math teacher
• Subtraction
• Multiplication
• Division
• Decimal
• Time
• Line Number
• Fractions
• Math Word Problem
• Kindergarten
• a + b + c
a - b - c
a x b x c
a : b : c
# Math Worksheets For Fourth Grade
Public on 08 Oct, 2016 by Cyun Lee
### mountain math worksheet fourth grade educational math activities
Name : __________________
Seat Num. : __________________
Date : __________________
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free printable math worksheets for 3rd grade word problems | 1,776 | 4,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-35 | longest | en | 0.17023 |
https://unseenpassage.com/class-12-physics-sample-paper-term-2-set-b/ | 1,713,000,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00039.warc.gz | 553,120,123 | 26,602 | # Unseen Passage
For Class 4 to Class 12
# Class 12 Physics Sample Paper Term 1 Set B
Please see below Class 12 Physics Sample Paper Term 1 Set B with solutions. We have provided Class 12 Physics Sample Papers with solutions designed by Physics teachers for Class 12 based on the latest examination pattern issued by CBSE. We have provided the following sample paper for Term 1 Class 12 Physics with answers. You will be able to understand the type of questions which can come in the upcoming exams.
## CBSE Sample Paper for Class 12 Physics Term 1 Set B
Section – A
This section consists of 25 multiple choice questions with overall choice to attempt any 20 questions. In case more than desirable number of questions are attempted, ONLY first 20 will be considered for evaluation.
Q. 1. Suppose the charge of a proton and an electron differ slightly. One of them is – e, the other is (e+e).
If the net f electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of : [Given mass of hydrogen mH = 1.67 × 10-27 kg]
(A) 10–23 C
(B) 10–37 C
(C) 10–47 C
(D) 10–20 C
B
Q.2. The electrostatic potential on the surface of a charged conducting sphere is 100 V.
Two statements are made in this regard :
S1 : At any point inside the sphere, electric intensity is zero.
S2 : At any point inside the sphere, the electrostatic potential is 100 V.
Which of the following is a correct statement?
(A) S1 is true, but S2 is false.
(B) Both S1 and S2 are false.
(C) S1 is true, S2 is also true and S1 is the cause of S2.
(D) S1 is true, S2 is also true but the statements are independent.
C
Q. 3. Equipotential at a great distance from a collection of charges whose total sum is not zero are approximately
(A) spheres.
(B) planes.
(C) paraboloids.
(D) ellipsoids.
A
Q. 4. In a region the potential is represented by V(x, y, z) = 6x – 8xy – 8y + 6yz, where V is in volts and x, y, z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1, 1, 1) is :
(A) 6√5 N
(B) 30 N
(C) 24 N
(D) 4√35
D
Q. 5. Which of the following characteristics of electrons determines the current in a conductor?
(A) Drift velocity alone
(B) Thermal velocity alone
(C) Both drift velocity and thermal velocity
(D) Neither drift nor thermal velocity.
A
Q. 6. A metal rod of length 10 cm and a rectangular cross-section of 1 cm × 1/2cm is connected to a battery across opposite faces. The resistance will be
(A) maximum when the battery is connected across 1 cm × 1/2 cm faces.
(B) maximum when the battery is connected across 10 cm × 1 cm faces.
(C) maximum when the battery is connected across 10 cm × 1/2 cm faces.
(D) same irrespective of the three faces.
A
Q. 7. Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron) :
C
Q. 8. A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part AKB of the ring is E. The electric field at the centre due to the charge on the part ACDB of the ring is :
(A) 3 E along OK
(B) 3 E along KO
(C) E along OK
(D) E along KO
C
Q. 9. Which of the following I-V characteristic represent the characteristic of a Ohmic conductor?
D
Q. 10. What is the potential difference between points A and B in the following circuit?
(A) 10 V
(B) 5 V
(C) 2.5 V
(D) 20 V
B
Q. 11. The acceleration of an electron in an electric field of magnitude 50 V/cm, if e/m value of the electron is 1.76 × 1011 C/kg, is :
(A) 8.8×1014 m/sec2
(B) 6.2×1013 m/sec2
(C) 5.4×1012 m/sec2
(D) Zero
A
Q. 12. A toroid of n turns, mean radius R and cross-sectional radius a carries current I. It is placed on a horizontal table taken as x-y plane. Its magnetic moment m
(A) is non-zero and points in the z-direction by symmetry.
(B) points along the axis of the toroid (m = mφ).
(C) is zero, otherwise there would be a field falling as 1/r3 at large distances outside the toroid.
C
Q. 13. A square of side L meters lies in the x-y plane in a region where the magnetic field is given by B = Bo (2ˆi + 3ˆj + 4kˆ) Tesla, where B0 is constant. The magnitude of flux passing through the square is
(A) 2BoL2Wb
(B) 3BoL2Wb
(C) 4BoL2Wb
(D) √229B0L2Wb
C
Q. 14. A loop, made of straight edges has six corners at A(0, 0, 0), B(L, 0, 0) C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0, 0, L). A magnetic field B=B0 (iˆ + kˆ) Tesla is present in the region. The flux passing through the loop ABCDEFA (in that order) is
(A) BoL2 Wb.
(B) 2BoL2 Wb.
(C) √2BoL2 Wb .
(D) 4BoL2Wb.
B
Q. 15. A capacitor of 20 μF is charged to 500 volts and connected in parallel with another capacitor of 10 μF and charged to 200 volts. The common potential is :
(A) 200 volts
(B) 300 volts
(C) 400 volts
(D) 500 volts
C
Q. 16. The capacity of a parallel plate condenser is 5 μF. When a glass plate is placed between the plates of the conductor, its potential becomes 1/8th of the original value. The value of dielectric constant will be :
(A) 1.6
(B) 5
(C) 8
(D) 40
C
Q. 17. A condenser of capacity 50 μF is charged to 10 volts. Its energy is equal to :
(A) 2.5×10−3 joule
(B) 2.5×10−4 joule
(C) 5×10−2 joule
(D) 1.2×10−8 joule
A
Q. 18. An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg and a reactance XL. For maximum power to be delivered from the generator to the load, the value of XL is equal to
(A) zero
(B) Xg
(C) −Xg
(D) Rg
C
Q. 19. When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. this means
(A) input voltage cannot be AC voltage, but a DC voltage.
(B) maximum input voltage is 220 V.
(C) The meter reads not v but (v2) and is calibrated to read √(v2 ).
(D) The pointer of the meter is stuck by some mechanical defect.
C
Q. 20. In a charged capacitor, the energy resides :
(A) the positive charges.
(B) both the positive and negative charges
(C) the field between the plates.
(D) around the edge of the capacitor plates.
C
Q. 21. Electric field inside a hollow conducting sphere is
(A) zero
(B) infinite
(C) depends on the magnitude of charge
(D) depends on the radius of the sphere
A
Q. 22. Fuse wire has high resistance and low melting point.
(A) should have high resistance
(B) should have low melting point
(C) should have low resistance
(D) both (A) and (B)
D
Q. 23. The resistance of superconductor is
(A) too large
(B) too small
(C) Negative
(D) zero
D
Q. 24. An electron and a proton moving with same velocity enters a magnetic field.
(A) Proton will experience more force
(B) Electron will experience more force
(C) Both will experience same force
(D) Force is mass dependent
C
Q. 25. Faraday’s laws of electromagnetic induction are consequences of law of conservation of
(A) Energy
(B) Momentum
(C) Mass
(D) None of the above
A
Section – B
This section consists of 24 multiple choice questions with overall choice to attempt any 20 questions. In case more than desirable number of questions are attempted, ONLY first 20 will be considered for evaluation.
Q. 26. The average value of current over a complete cycle is
(A) 0
(B) I0
(C) I0/2
(D) I0/√2
A
Q. 27. Which one of the following has dimension of time?
(A) LR
(B) LC
(C) 1/RC
(D) RC
D
Q. 28. In a LCR circuit energy is dissipated by
(A) L
(B) C
(C) R
(D) All of the above
C
Q. 29. The ratio of XC and XL in a A.C. circuit is
(A) 1
(B) 1/ω2LC
(C) ω2L
(D) None of the above
B
Q. 30. In series resonant circuit, the total reactance is
(A) 0
(B) ∞
(C) Maximum
(D) Minimum
A
Q. 31. A metal plate can be heated by
(A) placing in a time varying magnetic field.
(B) passing either a direct or alternating current.
(C) placing in a space varying magnetic field, but does not vary with time.
(D) both (A) and (B) are correct.
D
Q. 32. Which of the following is not an application of eddy current?
(A) Electric power meters
(B) Induction furnace
(C) Magnetic brakes in trains
(D) LED lights
D
Q. 33. 1 Henry is equal to
(A) Weber / Volt
(B) Weber / ampere
(C) Weber Ampere
(D) Weber Volt
B
Q. 34. 1 Farad is equivalent to
(A) 1 C/J
(B) 1 J/C
(C) 1 C2/J
(D) 1 C2/J2
C
Q. 35. The role of inductance is equivalent to
(A) Force
(B) Inertia
(C) Moment
(D) Energy
B
Q. 36. Which of the following represents Ampere’s Circuital law?
B
Q. 37. The nature of parallel and anti-parallel current carrying wires are
(A) parallel current carrying wires repel and anti-parallel current carrying wires attract.
(B) parallel current carrying wires attract and anti-parallel currents carrying wires repel.
(C) both attract.
(D) both repel.
B
Q. 38. The defection of pointer of a moving coil galvanometer is given by
B
Q. 39. A moving coil galvanometer can be converted into an ammeter by
(A) putting a shunt in series.
(B) putting a shunt in parallel.
(C) putting a high value resistance in series.
(D) Putting a high value resistance in parallel.
B
Q. 40. The conversion of a moving coil galvanometer into a voltmeter is done by
(A) putting a shunt in series.
(B) putting a shunt in parallel.
(C) putting a high value resistance in series.
(D) Putting a high value resistance in parallel.
C
Q. 41. Two thin long parallel wires carrying current I ampere is separated by a distance r. The magnitude of the force per unit length applied by one wire on the other is
(A) μ0I2 / 2πr
(B) μ0I2 / r2
(C) μ0I2 / 4πr
(D) μ0I2 / 2πr2
A
Q. 42. The force on a current carrying conductor in a magnetic field is maximum, when angle between the length of the conductor and the magnetic field is:
(A) π/4
(B) π/2
(C) π
(D) 0
B
Q. 43. You are given three voltmeters A, B and C having internal resistance 50 Ω, 50 kΩ and 1 MΩ. Which one will you select for better accuracy?
(A) A
(B) B
(C) C
(D) Any one
C
Q. 44. A current carrying coil is placed in a uniform magnetic field. If θ is the angle between the axis of the coil and the direction of magnetic field, then the torque is directly proportional to:
(A) sin θ
(B) cos θ
(C) cosec θ
(D) sec θ
A
Given below are two statements labelled as Assertion (A) and Reason (R)
Directions: In the following questions, a statement of Assertion (A) and is followed by a statement of
Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason(R) is the correct explanation of (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
A
Q. 45. Assertion (A) : In a non-uniform electric field, a dipole will have translatory as well as rotational motion.
Reason (R) : In a non-uniform electric field, a dipole experiences a force as well as torque.
A
Q.46. Assertion (A) : Electric field is always normal to equipotential surfaces and along the direction of decreasing order of potential.
Reason (R) : Negative gradient of electric potential is electric field.
A
Q. 47. Assertion (A) : Electric appliance with metal body has three electrical connections. But an electrical bulb has two electrical connection.
Reason (R) : Three pin connection reduces chances of electrical shocks.
A
Q. 48. Assertion (A) : The magnetic field at the ends of a very long current carrying solenoid is half of that at the centre.
Reason (R) : Magnetic field within a sufficiently long solenoid is uniform.
B
Q.49. Assertion (A) : The magnetic field configuration with 3 poles is not possible.
Reason (R) : No torque acts on a bar magnet itself due to its own field.
B
Section – C
This section consists of 6 multiple choice questions with an overall choice to attempt any 5. In case more than desirable number of questions are attempted, ONLY first 5 will be considered for evaluation.
Q. 50. The angle of dip where the horizontal component of earth’s magnetic field is equal to the vertical component is
(A) 30°
(B) 60°
(C) 45°
(D) 90°
C
Q. 51. The magnetic lines of force inside a bar magnet is
(A) South to North pole
(B) North to South pole
(C) There is no lines of force inside a bar magnet
(D) Direction of lines of force depends on the pole strength
A
CASE STUDY
Read the following text and answer the following questions on the basis of the same:
Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor μ. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 × 10–4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable. With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5–1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,…
Q. 52. Magnetic effects:
(A) are equal to electric effects.
(B) are greater than electric effects.
(C) are smaller than electric effects.
(D) cannot be compared with electric effects.
C
Q. 53. The force 10–3 N,is equivalent to:
(A) 100 mg
(B) 100 g
(C) 10 g
(D) 10 mg
A
Q. 54. Why the spring shrinks in Roget’s spiral ?
(A) The spring functions as a solenoid
(B) Due to force acting between two current carrying wires
(C) Due to magnetic effect of current
(D) Since the spring is soft
B
Q. 55. What are the main 3 components in a Roget’s spiral?
(A) Mercury, AC voltage source
(B) Mercury, DC voltage source
(C) Mercury, DC voltage source, key
(D) Mercury, AC voltage source, key | 4,251 | 14,956 | {"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} | 3.21875 | 3 | CC-MAIN-2024-18 | longest | en | 0.917955 |
https://www.reddit.com/r/explainlikeimfive/comments/qadv7/eli5_evidence_for_other_dimensions_beyond_3/ | 1,508,193,091,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820466.2/warc/CC-MAIN-20171016214209-20171016234209-00017.warc.gz | 1,247,699,083 | 21,936 | This is an archived post. You won't be able to vote or comment.
[–] 2 points3 points * (2 children)
From a mathematical standpoint, you don't need evidence for other dimensions. You can just have them for free. If you're familiar with a 2d coordinate system, where a point is described by a list of 2 numbers, say (2,3), then you can understand this idea fully. On a 2d coordinate system, for instance, we can define a circle with its center at the origin and radius 2 like this: "the set of all points whose distance from (0,0) is exactly 2". This describes the circle perfectly. Moving up to three dimensions is no problem—you just add another number, so that points look like (x,y,z). Then instead of a circle, we can define a sphere, which is kind of a circle generalized to three dimensions. A sphere with its center at (0,0,0) and radius 2 would be "the set of points with distance exactly 2 from (0,0,0)".
Now, there's nothing stopping you from moving from 3 dimensions to 4 dimensions by these definitions. As long as you don't let the fact that this is impossible to imagine visually bother you, there's nothing conceptually stopping you from saying that a point in 4d space looks like (x,y,z,w) and a hypersphere with radius 2 and center (0,0,0,0) is the set of all points exactly 2 away from (0,0,0,0).
In a lot of applications you might be dealing with 4 or 5 or whatever dimensional space—by which I mean lists of 4 or 5 or whatever numbers—and solving problems with these lists will have geometric analogies. Solving the system of equations:
``````5x + 3y + 2z + 5w = 5
3x + 6y + 4z + 2w = 3
``````
is just like finding the set of points in 4 dimensional space where two three dimensional pieces of four dimensional space intersect with each other. I know you can't picture that, (neither can I), but conceptually it's not different from the fact that
``````2x + y = 3
5x - 2 = 5
``````
Is finding the set of points where two one dimensional pieces of two dimensional space (lines) intersect.
[–][S] 0 points1 point (1 child)
i have heard of people talk about something like up to 11 dimensions. It seems like , by your explanation, there can be infinite dimensions...?
[–] 0 points1 point (0 children)
Yep. Lots of important problems are in infinite-dimensional space.
[–] 15 points16 points * (2 children)
At the simplest level, a dimension is a number you need to describe something fully. For instance, in our 3d space, you need 3 numbers to describe where something is...up/down, left/right, forwards/backwards (or any one of an infinity of equivalent systems).
But actually, it's immediately clear that 3 numbers isn't enough to describe something. You need another one to describe when something is. Therefore, in this sense, time is a dimension. General relativity, our excellent theory of how space and time works, describes time in exactly this way - it gets more or less the same treatment as a spatial dimension, though it's mathematically a slightly different type of thing so it ends up not behaving quite the same...such as only going one way for us.
Now, you're probably actually thinking about extra dimensions in the popular science way, where people throw terms like '11 dimensions' around like confetti. These dimensions are theoretical, and arise from making a set of assumptions and seeing that these assumptions necessitate those dimensions. The central idea to this area of science (and all other areas, though the other ones often have existing practical successes) is trying to make a unique prediction from these assumptions. If that prediction could be verified, the assumptions would have strong evidence for being correct, and so would the dimensions. We do not currently have good scientific evidence for these theories, and the existence of more than 4 dimensions is not well accepted scientific theory at the moment.
To elaborate a little further, the extra dimensions discussed here aren't quite like the dimensions you think of as up, down, left and right. A good (and common) example is of an ant on a telephone wire. From some way away, this system seems 'one dimensional' to you because you can describe where the ant is with a single number - how far along the wire it is. However if you look closely, as the ant does, it's actually a two dimensional system; the ant has both a distance along the wire and a distance around the wire. This 'around' is still a dimension because it's a number that you need to describe where the ant is, but you couldn't see that dimension from far away, because it's so small as to seem irrelevant to you. The idea in string theory (or whatever) is that there may be analogous cases on quantum scales which may govern how particles behave without being obvious to us.
[–] 6 points7 points (1 child)
You must watch this. Watch the entire thing. It is crystal clear.
[–] 1 point2 points (0 children)
Here is a pretty ELI5 explanation, however I urge you not to accept this as fact.
EDIT: I just realized this isn't really evidence for it. Technically there is no evidence for it, but more dimensions would be necessary for things like string theory to be true, and it would explain why some matter seems to disappear from existence during particle collision experiments.
[–] -2 points-1 points (5 children)
Because it makes the math work.
[–][S] 0 points1 point (4 children)
um, what?
[–] 4 points5 points (3 children)
Well, for one, gravity seems to be a lot weaker than it should. For example, you can pull a paperclip up with a small magnet, even when the opposite force is the pull of the entire Earth. Why is it so much weaker? Some people think it is because the gravity is leaking into other the dimensions. When we try to run the equations with 10 or 11 dimensions instead of 3 or 4, the math works out much better.
But, a good theory can also make predictions based on the math, and we're currently having a bit of trouble with that part; the predictions that current multi-dimensional theories are making require a lot of energy to see properly, so that makes them hard to test.
[–][S] 0 points1 point (2 children)
So are you saying there is not any good evidence of these dimensions?
[–] 5 points6 points (0 children)
Besides that it makes the math work? Not yet.
[–] 0 points1 point (0 children)
Good physical evidence, no. Mathematical and logical evidence, yes.
[–] -3 points-2 points (0 children)
Aliens | 1,530 | 6,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-43 | latest | en | 0.940651 |
https://www.office-forums.com/threads/tricky-one.2338706/ | 1,611,328,499,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00513.warc.gz | 905,812,954 | 12,216 | # Tricky one
L
#### LondonJames
Hi all,
I have the following example set of data
n|A|B
1|1|10
2|2|55
3|2|34
4|1|12
5|3|45
6|2|344
7|3|32
8|3|56
9|2|23
Each row needs to have a sum formula in row C, which needs to sum th
"last n rows" (say n is 3 for this example) where the value in A is NO
3
So:
In row 3, it would sum and return rows 1 2 3.
In row 4, it would sum and return rows 2 3 4
In row 6, it would sum and return rows 6 4 3 (skipping row 5)
In row 9, it would sum and return rows 9 6 5 (skipping rows 8 and 7)
I cant seem to work this out, even using fancy sumproduct/matc
combinations. Don't want to use VBA
Many thanks!
Jame
P
#### plinius
LondonJames said:
Hi all,
I have the following example set of data
n|A|B
1|1|10
2|2|55
3|2|34
4|1|12
5|3|45
6|2|344
7|3|32
8|3|56
9|2|23
Each row needs to have a sum formula in row C, which needs to sum the
"last n rows" (say n is 3 for this example) where the value in A is NOT
3
So:
In row 3, it would sum and return rows 1 2 3.
In row 4, it would sum and return rows 2 3 4
In row 6, it would sum and return rows 6 4 3 (skipping row 5)
In row 9, it would sum and return rows 9 6 5 (skipping rows 8 and 7)
I cant seem to work this out, even using fancy sumproduct/match
combinations. Don't want to use VBA
Many thanks!
James
Put in cell I1 the value escluded (3)
In C4 this matricial formula:
=SUM((OFFSET(A4,,,-MATCH(\$I\$1,COUNTIF(OFFSET(A4,,,-ROW(INDIRECT("\$A\$1:A"&ROW())),),"<>"&\$I\$1)),)<>\$I\$1)*OFFSET(A4,,1,-MATCH(\$I\$1,COUNTIF(OFFSET(A4,,,-ROW(INDIRECT("\$A\$1:A"&ROW())),),"<>"&\$I\$1)),))
and fill down.
Bye,
E. | 622 | 1,599 | {"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-2021-04 | latest | en | 0.803544 |
https://bestbinaryyptv.web.app/peightell66424ky/calculate-annualized-rate-of-return-in-excel-boci.html | 1,627,864,260,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00158.warc.gz | 140,619,180 | 7,056 | ## Calculate annualized rate of return in excel
An annualized rate of return is, essentially, the average return an investor receives over a given period, scaled down to a period of one year. It is not a simple
The compound annual growth rate (CAGR) shows the rate of return of an is an overview of how to calculate it both by hand and by using Microsoft Excel. The calculation of your annualized portfolio return answers one question: what is the compound rate of return earned on the portfolio for the period of investment? While the various Know the Excel formulas for these calculations. The formula Excel 2013 provides the DATE and XIRR functions, which derive the final annual interest rate from a series of payments on known dates. 1. Enter your list of An annualized rate of return is, essentially, the average return an investor receives over a given period, scaled down to a period of one year. It is not a simple Average all annual growth rate with entering Kutools for Excel includes more than 300 buy a share at when I have a total expected return. Excel's Internal Rate of Return (IRR) function is an annual growth rate formula for investments that pay out at regular intervals. It takes a list of dates and
## 1 Apr 2011 Note: I assumed your interest rate is annual and divided it by 12. How to use excel to calculate real rate of return factor for 5 & 10 years?
Use this calculator to determine the annual return of a known initial amount, a stream of deposits, plus a known final future value. Excel calculates the average annual rate of return as 9.52%. Remember that when you enter formulas in Excel, you double-click on the cell and put it in formula mode by pressing the equals key (=). When Excel is in formula mode, type in the formula. Related Articles Step 1. Open Excel by double-clicking its icon. Step 2. Put all of your cash flows in column A. Step 3. Enter the date of each cash flow in the same row as the cash flow in column B using Excel's Step 4. Below your two tables of cash flows and dates, type "=XIRR (" without the In the case of investment #2, with an investment of \$1,000 in 2013, the yield will bring an annual return of 80%. If no parameters are entered, Excel starts testing IRR values differently for the entered series of cash flows and stops as soon as a rate is selected that brings the NPV to zero. Annualized Rate of Return = 25% So, Annualize Rate of return on shares is 25%. Now, let us calculate the rate of return on shares. Rate of Return = (Current Value – Original Value) * 100 / Original Value Annualized Rate of Return = [(\$990 + \$600) / \$990 ] 1 / 10 – 1; Annualized Rate of Return = 4.85%; Therefore, the investor earned annualized rate of return of 4.85% from the bond investment over the 10-year holding period. Explanation. The formula for the Annualized Rate of Return can be calculated by using the following steps:
### I am attaching one of my excel files for reference. Thanks in advance. Kumar.
The way to set this up in Excel is to have all the data in one table, then break out the calculations line by line. For example, let's derive the compound annual growth rate of a company's sales over 10 years: The CAGR of sales for the decade is 5.43%. You can do as follows: 1 . Besides the original table, enter the below formula into the blank Cell C3 and, 2 . Select the Range D4:D12, click the Percent Style button on the Home tab, 3 . Average all annual growth rate with entering below formula into Cell F4, and press the Enter key. The internal rate of return (IRR) is a core component of capital budgeting and corporate finance. Businesses use it to determine which discount rate makes the present value of future after-tax One advantage of using IRR, which is expressed as a percentage, is that it normalizes returns: everyone understands what a 25% rate means, compared to a hypothetical dollar equivalent (the way the NPV is expressed). Unfortunately, there are also several critical disadvantages with using the IRR to value projects. Microsoft Excel: 3 ways to calculate internal rate of return in Excel 1. Negative and positive cash flow values required. 2. Monthly versus annual yields. When calculating the IRR or MIRR of monthly cash flows, 3. Guess. The IRR and XIRR functions allow you to enter a guess as the beginning The real rate of return is the actual annual rate of return after taking into consideration the factors that affect the rate like inflation and this formula is calculated by one plus nominal rate divided by one plus inflation rate minus one and inflation rate can be taken from consumer price index or GDP deflator. I am looking for a formula that will calculate annualized return. Specifically, I have the annual returns for the S&P 500 from 1976 through 2005. I would like to be able to calculate the annualized (compounded or CAGR) rate of return over various time frames. There is no formula listed under tools for Annualized return or CAGR in excel.
### Annualized Return Formula APY = ((principal + gain) / principal) ^ (365/days) - 1 So, for example, suppose our initial investment (ie. principal) is \$10,000, and after 2.5 years we are sitting on \$14,000. What is our annual return?
In this section we will see how to calculate the rate of return on a bond so it measures the expected compound average annual rate of return if the bond is You want to calculate your portfolio's annual rate of return to compare it to a benchmark index's return or to a friend's Excel or an online calculator. If it does not Learn how to to calculate the Compound Annual Growth Rate (CAGR) in Excel the return would have been assuming a constant growth rate over the period.
## To calculate rates of return for any given period of time or to determine com- pound annual returns, follow the instructions in this Fact Sheet. Period Returns.
The way to set this up in Excel is to have all the data in one table, then break out the calculations line by line. For example, let's derive the compound annual growth rate of a company's sales over 10 years: The CAGR of sales for the decade is 5.43%. You can do as follows: 1 . Besides the original table, enter the below formula into the blank Cell C3 and, 2 . Select the Range D4:D12, click the Percent Style button on the Home tab, 3 . Average all annual growth rate with entering below formula into Cell F4, and press the Enter key. The internal rate of return (IRR) is a core component of capital budgeting and corporate finance. Businesses use it to determine which discount rate makes the present value of future after-tax One advantage of using IRR, which is expressed as a percentage, is that it normalizes returns: everyone understands what a 25% rate means, compared to a hypothetical dollar equivalent (the way the NPV is expressed). Unfortunately, there are also several critical disadvantages with using the IRR to value projects. Microsoft Excel: 3 ways to calculate internal rate of return in Excel 1. Negative and positive cash flow values required. 2. Monthly versus annual yields. When calculating the IRR or MIRR of monthly cash flows, 3. Guess. The IRR and XIRR functions allow you to enter a guess as the beginning The real rate of return is the actual annual rate of return after taking into consideration the factors that affect the rate like inflation and this formula is calculated by one plus nominal rate divided by one plus inflation rate minus one and inflation rate can be taken from consumer price index or GDP deflator. I am looking for a formula that will calculate annualized return. Specifically, I have the annual returns for the S&P 500 from 1976 through 2005. I would like to be able to calculate the annualized (compounded or CAGR) rate of return over various time frames. There is no formula listed under tools for Annualized return or CAGR in excel.
Multiple the result by 100 to calculate the annualized return expressed as a percentage. Completing the example, multiply 0.0619 by 100 to get 6.19 percent. The calculation of the effective rate on the loan in Excel «Nominal rate» - is the annual rate of interest on the credit, which is designated in the monthly rate, we need use the IRR function (return to the internal rate of return for cash flow):. 6 Jun 2019 When it comes to compounding annual growth rates, there's more than meets the eye. Discover how to calculate CAGR while avoiding common pitfalls! Although average annual return is a common measure for mutual funds, Calculating Internal Rate of Return Using Excel or a Financial Calculator. To calculate rates of return for any given period of time or to determine com- pound annual returns, follow the instructions in this Fact Sheet. Period Returns. The Sharpe ratio is calculated by subtracting the risk-free rate - such as that of the 10-year U.S. Treasury bond - from the rate of return for a portfolio and dividing the geometric mean, we multiply by 365 to get the annualized portfolio return. | 1,980 | 8,964 | {"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-2021-31 | latest | en | 0.92477 |
https://fr.slideserve.com/cyma/robotic-rational-reasoning-lecture-2-powerpoint-ppt-presentation | 1,627,542,025,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00063.warc.gz | 265,440,347 | 26,368 | Robotic Rational Reasoning! Lecture 2
# Robotic Rational Reasoning! Lecture 2
## Robotic Rational Reasoning! Lecture 2
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##### Presentation Transcript
1. Robotic Rational Reasoning! Lecture 2 P.H.S. Torr
2. Summary of last lecture • We talked about the problem of induction and how it might relate to AI. • We saw probability might be the basis of reasoning about the world. • We saw there were problems with the attempts to make objective interpretations of probability. • Venn • Von Mises (frequentists) • Popper (propensity)
3. This lecture • We will develop a system of probability that can be used within AI. • We shall show how this can be used to reason about events and decide actions. • How to represent our belief in propositions by numbers • How to make action on the basis of those beliefs • Once we have done this we want to show how to act rationally, so as to optimize utility.
4. Belief • Why is belief important? • Consider again the example of the coin hidden in one of my hands. • Your belief in which hand it is in, is different from mine. • Can we make this more formal so that we can use belief in practise?
5. We talked about ‘stronger’ and ‘weaker’ belief; and being ‘more or less confident’. Can we make sense of this talk? That is, can we quantify partial belief, and give a numerical basis to these comparative judgements? Assign numbers to measure the strength of a belief; Prove that those numbers must be, formally, probabilities; and Provide a rule for updating beliefs given new evidence. These three achievements constitute Bayesian epistemology. Quantifying Belief
6. Task 1: Measuring Belief • We turn first to the task of assigning numbers to measure belief. • Is it possible to measure belief in some rational way?
7. Theories of Belief • Logical • Keynes, Carnap, Cox, Jaynes • Preference over actions • Ramsey, Savage • Betting • De Finetti
8. Logical theory of belief • Here logic is extended from binary valued to include uncertainty over propositions. • This theory was first put forward by economist Keynes and then given rigour by R.T. Cox. • It is convincing but difficult maths.
9. Subjective theory of belief based on Actions • By Ramsey and Savage. • From a person expressing preferences between actions one can derive the whole of probability theory, a remarkable feat. • Again highly mathematical.
10. Gambles and Degrees of Belief • Probability has a long association with games of chance. • One way that we have talked about probabilities was in terms of rolls of die and tosses of coins. • A better way to represent degrees of belief will be in terms of gambles. • A gamble is simply a choice between two possibilities or options. • The idea is to: • Assign cash payoffs to options, and then • Assign numbers to which of options we think is more likely.
11. Example of a Gamble • Setup: • First we need to list our options: • B = The US will bomb North Korea by June 30, 2007. • ~B = The US will not bomb North Korea by June 30, 2007. • Second we need to think of some sort of payoff for when one of these options obtains; the payoffs will be understood in terms of gambles, that is, taking one option as more likely than the other.(Here think of a reward for yourself, say getting \$10) • Gamble 1: Win \$10, if B occurs, otherwise nothing • Gamble 2: Win \$10, if ~B occurs, otherwise nothing
12. There are three possibilities here • Indifference — You might treat the two options as equally likely, thus you have no reason to accept Gamble 1 over Gamble 2. • You want to take Gamble 1— If you choose Gamble 1, then you feel that B is more likely to occur than ~B. • You want to take Gamble 2— If you choose Gamble 2, then you feel that ~B is more likely to occur than B.
13. Betting Rates and Payoff Matrixes: • The examples thus far did not involve taking a risk—in that, you lose nothing if the gamble you chose above did not pay out. • Another way of assigning numbers to confidence levels is to make the decision a risky one, that is, one where the participants stand to lose something by choosing incorrectly.
14. Betting Rates and Payoff Matrixes • Notation: • Suppose two people (J and K) make a bet that some event A will turn out. • J bets amount \$10on A, and • K bets amount \$10 against A. (Betting against A is the same as betting on ¬A.) • What is the total amount of money at stake between J and K? • \$20. • The total amount is the sum of the individual bets (i.e., \$(10 + 10) ). • The total amount here is called the stake.
15. Betting Rates and Payoff Matrixes • In some sense the previous example is fair if the probability is 0.5 • Consider, what would you pay to enter a lottery in which the stake is 10,000 and the chance of winning is 0.1?
16. Betting Rate • What we want to know is what a person’s betting rate is. This will be used as a proxy for their personal probability. • The probability that you believe in some hypothesis will be equivalent to your betting on the hypothesis being true. • The betting rate is calculated by taking your individual bet and dividing it by the stake: • Your Betting Rate= Your Bet/Stake • NOTE: We can also determine what a person’s bet is if we know the stake and their betting rate, by the following equation: • Bet = (Personal Betting Rate) * (Stake) • All that we have done here is solved the previous equation for the Bet, which involves multiplying both sides by the Stake.
17. Bets • A bet involves two bettors X & Y • X bets £x on H, • Y bets £y on ¬H (i.e. against H). • The winner takes £(x+y) • Thus the odds against H are y:x • The betting rate of X on H is b(X) = x/(x+y)
18. Definition: Fair Bet • A bet on H is fair for X iff X is indifferent between the two options of • Betting on H with wager £x; or • Betting against H with wager £y. • A fair bet is one where X ’s betting rates £x/(x+y) and £y/(x+y) are such that X does not prefer to bet either on H or on ¬H at those rates (or, equivalently, X wouldn’t prefer either side at those odds). • A fair bet is not one we would take: in betting we usually wish to win, and hence to bet at unfair rates—unfair in our favour!
19. Betting Rates and degrees of belief. • If agent X has betting rate b(H) in what X regards as a fair bet on H, then X ’s degree of belief in H is equal to b(H). • As we define them, X need not actually bet to have a belief: X must simply have preferences which would make him indifferent between the two betting options, whether or not X ever acts on those preferences (or is ever faced by just those options). • We have completed task one: devised a method for assigning numerical values to propositions.
20. Pay off Matrix • Assume X bets on A, where the Stake = £S and X’s betting rate = p. • Then X’s bet is £pS, which is what he stands to lose, he stands to gain • £S(1-p), this can be shown in a pay off matrix: • If you bet for A, then: • (i) If A occurs, you get S(1 - p), or • (ii)If ~A occurs, you get –pS. • If you bet against A, then: • (i) If A occurs, you get - S(1 - p); or • (ii) If ~A occurs, you get pS.
21. Subjective Belief • So now we have assigned numbers representing degrees of belief to propositions. • Note these are so far entirely personal (subjective). • Are there any rules governing these beliefs? • Next we shall show that, remarkably, they obey the rules of probability.
22. Set of Propositions • First we consider (rather informally) the set of propositions we are dealing with. • These can be any set of propositions, P, providing they satisfy a Boolean algebra: • H in P; implies not H in P • H1 in P and H2 in P; implies (H1 V H2) in P
23. Recall standard rules of probability
24. Probability Function over propositions
25. Ramsey De Finetti Theorem • We are now in a position to embark on our second task: proving that rational degrees of belief obey the rules of the probability calculus. • That is, we must prove • Theorem (Ramsey De Finetti) If X ’s degrees of belief are rational, then X ’s degrees of belief function defined by fair betting rates is (formally) a probability function. • We begin by settling what it means for degrees of belief to be rational.
26. Rationality • Defining ‘rationality’ in general is a huge task. It is extremely difficult, if not impossible, to give an uncontroversial account; it would be unfortunate if we had to give such an account before we could prove our theorem. • Thankfully, we don’t have to. We do accept the following: • If we can show that nonprobabilistic belief would lead to sure betting losses, we can infer from this Commonsense Principle that such belief are not rational.
27. Sure-Loss Contracts: • The idea is that when your betting rates are inconsistent, we could think of a crafty person being able to develop a betting contract, by taking your betting rates, where you lose no matter what happens. When you can only be put in a situation where you are sure to lose if you betting rates are inconsistent, and this (as we will see) only happens when your betting rates violate the rules of probability.
28. Sure loss contract or Dutch Book • BETTING CONTRACT DEFINED: A betting contract is a contract between two people to settle a bet at some agreed upon set of betting rates. • BOOKMAKER DEFINED: A bookmaker is someone who makes betting contracts. He pays you if you win, and collects from you if you lose. • SURE-LOSS CONTRACT DEFINED: A sure-loss contract is a betting contract in which you will lose no matter what outcome occurs.
29. Example of Sure Loss Contract: • Recall from above: • B = The US will bomb North Korea by June 30, 2005. • Your betting rate on B is 5/8. • ~B = The US will not bomb North Korea by June 30, 2005. • Your betting rate on ~B is 3/4 • The bookmaker chooses which way to run the bet, given the betting rates that you have already selected. The bookmaker is a crafty fellow so he will take advantage of both your odds, so you will place independent bets for B and ~B.
30. The Bets • Bets on B (at a rate of 5/8): • You bet \$5 and the bookmaker bets \$3. • Bets on ~B (at a rate of 3/4): • You bet \$6 and the bookmaker bets \$2.
31. Result • What happens if US bomb Korea (i.e., B occurs)? • What do you win? • You would win \$3. This is what the bookmaker put up against your bet on B. • What do you lose? • You would lose \$6. This is what you bet on ~B. • What is the net payoff if B occurs? • YOU WOULD SUFFER A NET LOSS OF \$3. • What happens if US don’t bomb Korea (i.e., ~B occurs)? • What do you win? • You would win \$2. This is what the bookmaker put up against your bet on ~B.. • What do you lose? • You would lose \$5. This is what you be on B. • What is the net payoff if ~B occurs? • YOU AGAIN WOULD SUFFER A NET LOSS OF \$3.
32. Pay Off Matrix The basic idea is that, no matter how things turn out, you lose \$3. This is what a sure-loss contract looks like, since these were your own betting rates you cannot cry foul that you were mislead by the bookmaker. The notion of a sure-loss contract gives us a way to define coherence among betting rates.
33. The Argument for Coherence from the Basic Rules of Probability • Outline: • What we are aiming for is a notion of a set of betting rates being coherent if and only if the set of betting rates satisfies the basic rules of probability. • What we will do is show that when we develop betting rates that violate the rules of probability the result will be that a bookmaker can construct a sure-loss contract from our betting rates. • First • First we show all beliefs must lie within 0 and 1.
34. Normal Beliefs • If b(H) is between 0 and 1 it is said to be normal. • We can show that if we do not have belief that is normal then we are vulnerable to a sure loss contract. • It is sufficient to show this for b(H) > 1 as • b(H) < 0 implies b(¬H) > 1.
35. Normal Beliefs • If b(H) > 1 then our betting rate is expressible as x/(x+y) where x > (x+y) • hence y is negatitve. • Hence we are indifferent between • Betting x on H and either losing x is H is false; or winning (x+y) if H is true; but (x+y) < x so we lose either way: a sure loss • Betting y on ¬H and either losing y (a win); or winning x+y; either way a sure win. • Being indifferent between a sure loss and a sure win is irrational and the book maker can always make us lose.
36. Argument for Rule of Addition: • When A and B are mutually exclusive events, • then Pr(A v B) = Pr(A) + Pr(B) • If our betting rates are to meet this, then our betting for (AvB) should equal the sum of our betting rate for (A) and our betting rate for (B). • Thus, if A and B are mutually exclusive, then the Addition Rule requires: • Betting rate on (A v B) = betting rate on A + betting rate on B
37. Strategy • What we will do is see if a sure-loss contract can arise when we violate this rule. • Consider three betting rates: • Betting rate on A = p • Betting rate on B = q • Betting rate on A v B = r • Assume that the restricted rule of addition is violated here. • Thus: r < p + q. That is, assume that we assign a betting rate to r that is less than the sum of the betting rates for p and q. Thus, the sum of p and q will always be greater than r.
38. Layout of The Bets: • Suppose a bookmaker asks you to make bets on your rates simultaneously, where the stake for each bet is \$1. • We bet \$1 dollar so we can use the betting rates as proxies for the bets themselves. • Bet 1: • Here we bet p on A. • What do we win if A occurs? • We win the remainder of the stake, which is (1 – p). • What do we lose if ~A occurs? • We lose our original bet of p. • Bet 2: • Here we bet q on B. • What do we win if B occurs? • Here we win the remainder of the stake, which is (1 – q). • What do we lose if ~B occurs? • Here we lose our original bet q.
39. Layout of The Bets: • Bet 3: • Here we bet (1 – r) against (A v B). • Since this is against either A or B, we will only win if neither A nor B occurs. • What do we win if neither A or B occurs? • Here we would win r. • What do we lose if either A or B occurs? • Here we would lose our original bet of (1 – r).
40. Setting up the Payoff Table: • Recall that we are assuming that r is less that the sum of p and q, that is, r < p + q. • If that is so, then r – (p + q) is always negative. That is, regardless of what p and q sum to, r must be less than that. Thus, we would lose given these better rates no matter which scenario took place. • Payoff (iii) is the easiest to see: Neither A or B are true. • Here you win r, less p and q. • Therefore, the net payoff is r – p – q. • But since r is less than p and q, this means that regardless of what r is, you lose [r - (p + q)].
41. Simple Numerical example • Suppose I have b(A) = 0.2, b(B) = 0.3, b(A v B) = 0.4, hence b¬(A v B) = 0.6. • The book keeper offers a £1 stake on each proposition ¬A and ¬B and (A v B) • If A & ¬B my gain is -0.1 • +0.8 - 0.3 - 0.6 • If B & ¬A my gain is -0.1 • -0.2 + 0.7 -0.6 • If neither A nor B my gain is -0.1 • -0.2 – 0.3 + 0.4
42. Simple Numerical example • Suppose I have b(A) = 0.2, b(B) = 0.3, b(A v B) = 0.6 hence b¬(A v B) = 0.4. • The book keeper offers £1 with me taking the bets ¬A and ¬B and (A v B) • If A my gain is -0.1 • -0.2 + 0.7 - 0.6 • If B my gain is -0.1 • +0.8 - 0.3 – 0.6 • If neither A nor B my gain is -0.1 • -0.2 – 0.3 + 0.4
43. Belief = Betting Rates • Thus, we have a sure-loss contract, and this means that by violating the restricted rule of addition we end up with inconsistent betting rates. • If our betting rates here had followed the restricted rule of addition, that is, if r = p + q, then there would have been no way to create a sure loss contract involving A, B, and A v B. • LARGER POINT: It is a necessary and sufficient condition that a set of betting rates, including conditional betting rates, should be coherent is that they should satisfy the basic rules of probability.
44. Conditional Probability • We have left out conditional probability but this can be easily demonstrated using similar arguments.
45. So now we have completed our first two tasks: Assign numbers to measure the strength of a belief; Prove that those numbers must be, formally, probabilities; and Next we need to Provide a rule for updating beliefs given new evidence. Quantifying Belief
46. Conditional Probability • We shall next do a little revision on conditional probability as it is necessary to know about this to define Bayes’ rule.
47. Conditional Probability • Many times, however, we take the probability of one event to be dependent on another event occurring (or proposition being true). • Example of Conditional Probability Statement: • P2: Given that she starts the chemotherapy immediately, Sarah has a 70% of stopping the spread of the cancer. • Conditional probabilities are complex statements—they are composed of two constituent statements. • In each case here, there are two sentences (or events) that are taken to be related to one another: • The probability of one of the statements is dependent on the other statement being true.
48. Conditional Probability • NOTATION FOR CONDITIONAL PROBABILITY: Pr(•/•); • e.g., Pr(A/B) • We can read conditional probabilities in the following ways: • The probability of A given B. • The probability of A on B. • The probability of A on the condition that B. • Definition of Conditional Probability
49. Conditional Probability • Definition of Conditional Probability • Read: • When the probability of B is greater than 0, then the probability of A given B is equal to the probability of A and B jointly occurring, divided by the probability of B occurring without A. • Why must B be greater than 0? • Because we cannot divide by 0.
50. So now we have completed our first two tasks: Assign numbers to measure the strength of a belief; Prove that those numbers must be, formally, probabilities; and Next we need to Provide a rule for updating beliefs given new evidence. Quantifying Belief | 4,603 | 18,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-31 | latest | en | 0.879146 |
http://mathtutorkk1969.blogspot.com/2014/05/trigonometry-measurement-of-triangles.html | 1,532,026,182,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591216.51/warc/CC-MAIN-20180719183926-20180719203926-00163.warc.gz | 238,255,378 | 13,656 | ## Friday, May 2, 2014
Trigonometry, measurement of triangles, is used in various occupations including engineering, architecture and navigation. We start with the study of trigonometry by examining six functions. These trigonometric functions are determined by inputs that are he measures of acute angles of a right triangle. The outputs are the ratios of the length of the sides of right triangles.
If you draw a right triangle, that is a triangle with one right (90 degree) angle and two acute angles, note a base angle as Ө. We note the sides of the triangle based off this angle and the right angle. The side opposite the right angle, which is the longest side of a right triangle is the hypotenuse. The other sides are opposite of Ө and adjacent to Ө. It's important to know the distinction between the sides because the six trigonometric functions are based on the ratio between these sides.
The trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. If you have a scientific calculator, you may have seen buttons with sin, cos and tan on them. Those are the trigonometric functions sine, cosine and tangent. Sine is the length of the side opposite Ө divided by the length of the hypotenuse. Cosine is the length of the side adjacent to Ө divided by the length of the hypotenuse. Tangent is the length of the side opposite of Ө divided by the length of the side adjacent to Ө. To figure out cotangent, secant and cosecant, they are simply the reciprocals of tangent, cosine and sine, respectively.
An important thing to note is that the value of the trigonometric functions depend only of the size of the angle Ө and not on the length of the sides. For example, a triangle with opposite side 2 and adjacent side 1 has the same trigonometric functions as a triangle with opposite side 30 and adjacent side 15, since the ratio between the two is 2:1 in both cases.
Now that we know the basic trigonometric functions, let's evaluate them in an example. Suppose we have a right triangle with the length of the opposite side of angle Ө to be 4 and the length of the adjacent side to angle Ө to be 3. We can find the values of the six trigonometric functions once we know the length of the hypotenuse. We can get that by using the Pythagorean Theorem, which is "a squared" plus "b squared" equals "c squared", if a, b and c are the sides of the triangle, c being the hypotenuse. Using the equation, we get c = 5. Therefore sine is opposite divided by hypotenuse, which is 4/5. Cosine is adjacent divided by hypotenuse, which is 3/5 and tangent is opposite over adjacent, which is 4/3. Cotangent is the reciprocal of tangent, so cotangent is 3/4. Secant is reciprocal of cosine, so secant is 5/3. Cosecant is reciprocal of sine, so cosecant is 5/4.
We can also find the angle measures of a right triangle using trigonometric functions. If you've ever noticed sin-1 , cos-1 , and tan-1 on your calculator those are the inverse sine, inverse cosine and inverse tangent buttons and are used to find the angle given a the trigonometric ratio. For example, in our above problem we know sine = 4/5, cosine = 3/5 and tangent = 4/3. Knowing these ratios we can get the value of Ө and therefore the other angles of the triangle. Take inverse sine of 4/5 and we get 53.1 degrees, which will be the same as taking inverse cosine of 3/5 and inverse tangent of 4/3. All will give the value of Ө at 53.1 degrees. Since we have a right triangle, we know one angle is 90 degrees and since the angles of a triangle add to 180 degrees, the third angle is 36.9 degrees.
These are just some of the basics with right triangle trigonometry. There is much more to learn, but understanding these basics are key to progressing to more difficult topics and applications of trigonometry. | 922 | 3,795 | {"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} | 4.8125 | 5 | CC-MAIN-2018-30 | latest | en | 0.934667 |
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# What is 1 over 2 minus 3 over 4?
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3 and 2/5 minus 1 and 1/3 = 2 and 1/15
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1 over 2 | 336 | 755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.85256 |
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Profit and Loss Formulaes
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A. Rs.10000 B. Rs.10074.25 C. Rs.10100 D. Rs.12000
103. If a person sells a ‘sari’ for Rs.5200, making a profit of 30%, then the cost price of the sari is
A. Rs.4420 B. Rs.4000 C. Rs.3900 D. Rs.3800
104. If the selling price of 40 articles is equals to the cost price of 50 articles, then the loss or gain per cent is
A. 25% loss B. 20%loss C. 25% gain D. 20% gain
105. Oranges are bought at the rate of 10 for Rs.25 and sold at the rate of 9 for Rs.25. The profit is
A. 9 1/11% B. 10% C. 11 1/9% D. 12 ½%
106. Nikhil bought a painting at a certain price and sold it at Rs.462000 which was 10% more than the original price. What was the price at which Nikhil bought the painting?
A. Rs.420000 B. Rs.440000 C. Rs.402000 D. Cannot be determined
107. The labeled price of a product is Rs.620. If it is sold at 15% discount and the dealers earn a 20% profit, what is the cost price?
A. Rs.439.16 B. Rs.423.16 C. Rs.412.16 D. Rs.436.11
108. If the cost of 12 pencils is equal to selling price of 10 pencils, the profit per cent in the transaction is
A. 16 2/3% B. 18% C. 20% D. 25%
109. When a discount of 12% on the marked price of an article is allowed, the article is sold for Rs.264. The marked price of the article is
A. Rs.300 B. Rs.276 C. Rs.312 D. Rs.325
110. Successive discounts of 20% and 15% are equivalent to a single discount of
A. 35% B. 32% C. 17.5% D. 17%
Related Topics
Who are all can get the advantages from this Profit and Loss Question and Answers section?
Those are all planning for any competitive examinations can use this segment to enhance their abilities.
• L.I.C/ G. I.C Competitive Exams
• Career Aptitude Test (IT Companies) and etc.
• Defence Competitive Exams
• Common Aptitude Test (CAT)
• UPSC Competitive Exams
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• University Grants Commission (UGC) | 715 | 2,263 | {"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} | 4.09375 | 4 | CC-MAIN-2017-34 | latest | en | 0.926577 |
https://www.iannix.org/fr/help/topic/smoothing-a-curve/?bbp_reply_to=2499&_wpnonce=1772819724 | 1,638,969,152,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363510.40/warc/CC-MAIN-20211208114112-20211208144112-00502.warc.gz | 839,279,172 | 8,418 | ## Smoothing a curve
Accueil Forums Wanted! Smoothing a curve
• Ce sujet contient 4 réponses, 3 participants et a été mis à jour pour la dernière fois par Anonyme, le il y a 10 années et 2 mois.
5 sujets de 1 à 5 (sur un total de 5)
• Auteur
Messages
• #2169 Répondre
Anonyme
Inactif
IanniX can manage curves composed of line segments. But we can also manage bezier curves.
When you draw curve with the freehand tool, we sample the mouse and construct a curve with line segments. But we’re sure there is a mathematical tip to make a smooth curve (thanks to bezier curves).
Input : a set of 2D-points
Output : a set of 2D-points with cubic-bezier control (so one 2D-point for point position and two 2D-points for bezier control).
Does somebody have an idea?
#2496 Répondre
Anonyme
Inactif
I am not a math-guy, but I sure enjoyed viewing the animations on this page: Bézier curves. The External Links section seems to point to possible sources of algorithms, some of which might be in code-form. I don’t know if code is what you’re asking for, Guillaume, but that jsDraw2d library might be useful for design ideas, even though it’s not C/C++.
I suppose a C/C++ library that wraps OpenGL would be what you’re looking for, optimally. Given that drawing smooth curves is such a core problem, there must be a GPL’ed library out there already.
Cheers,
–Bob
#2497 Répondre
Anonyme
Inactif
Do you still need a way to smoothly connect Bezier curves?
Here is an article on how to do this, along with open source code implementing it.
http://www.codeproject.com/KB/graphics/ClosedBezierSpline.aspx
I hope this helps.
#2498 Répondre
Anonyme
Inactif
Hi!
Yeah I still need these kind of stuff. The link of AvantGuy was cool, but the translation from JS to C++ was not easy (and need timmmmmme :)).
Thanks a lot!
#2499 Répondre
Anonyme
Inactif
No problem. I found that article while researching for my own path-following music project.
I just remembered that the above article is about creating closed curves. You probably also want to be able to make open ended smooth Bezier curves. Here is the companion article that covers open curves:
http://www.codeproject.com/KB/graphics/BezierSpline.aspx
5 sujets de 1 à 5 (sur un total de 5)
Répondre à : Répondre #2499 dans Smoothing a curve
Vos informations : | 612 | 2,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.782513 |
gbraddock.blogspot.com | 1,531,952,072,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00457.warc.gz | 151,260,805 | 20,040 | Sunday, March 1, 2009
Websites I found interesting (weekly)
• Math is Fun starts with the concept of squares. "To square a number, just multiply it by itself." Then introduces square roots, perfect squares, and the radical sign. " This is the special symbol that means ‘square root', it is sort of like a tick, and actually started hundreds of years ago as a dot with a flick upwards. It is called the radical, and always makes math look important!" The lesson includes lots of examples and illustrations, and concludes by explaining the guess, divide and average method of solving square roots without a calculator.
• 3 stars
"Finding the square root of a number is the inverse operation of squaring that number. Remember, the square of a number is that number times itself." In four pages (called steps one through four) Math.com introduces perfect squares and how to use estimation to calculate the square root for numbers that are not perfect squares. The last step is an interactive quiz which worked for me in Internet Explorer, but not Firefox or Safari.
• Eureka! We've found the mother lode of printable square root problem sheets with this worksheet generator from Homeschool Math. When creating your worksheet and answer key, make choices such as only perfect squares, the number of decimals to round the answers to, the range of radicands, and the number of problems on the page. When you've found the combination you like, you can simply reload your worksheet to generate another one with the same specs. To make it even easier, six sample worksheets are provided.
• 4 stars
Dr. Math explains two methods fo finding a square root by hand. The first includes three steps: guess, divide and average. Of course depending on your guess in step one, steps two and three might need repeating. Dr. Math demonstrates by showings all the steps involved in finding the square root of twelve assuming your first guess was two. The second method uses the Binomial Theorem and infinite series, and is probably not for most middle school students.
• Aplusmath.com: Square Root Flashcards
5 stars
Because math practice is so vital to learning, this online flashcard game is a sure hit. Each quiz offers ten perfect square roots to solve, randomly choosing radicands (the number under the square root symbol) from 1 to 144. Because the problems are randomized, you can play this game as many times as you like. For more math games, look in both the Flashcards and Games sections.
• Protected Glogster environment for schools
• Free site
• Feb. 23
- Light and the Reflectance Spectrometer - The Reflectance Spectrometer is a hand-held device that measures the light reflected off a surface. It’s easy to use, and students can use data gleaned from it to construct specific line graphs of leaves, colored paper, grass, or the like. It’s a “remote-sensing” light meter....student friendly! Lausanne - Dr. John Frassinelli (jfrassinelli@lausanneschool.com )
Feb. 24
- Google Earth Goodies - Captivate your students' attention while teaching content with Google Earth. Standard program features and creating custom files will be discussed. Gain ideas for lessons with this powerful software for multiple subject areas K-12. St. George’s Memphis - Gail Braddock (lgbraddock@gmail.com)
Feb. 25
- Flip Videos - Smaller than most cell phones, Flip Video cameras fit into a pocket and can capture student stories wherever they happen - in the classroom, on field trips, even on the playground. These modestly priced, yet powerful camcorders allow you to simply plug into any computer for immediate video editing and sharing. MUS - Terry Balton (terry.balton@musowls.org )
Feb. 26
- Show and Tell: The Digital Way - Bring Your Photos to Life with PhotoStory 3 - Photo Story 3 is a program you can use to create media presentations from your digital photos. With a single clic, you can tough-up, crop, or rotate pictures. Add special effects, music, and your own voice narration to your photo stories. Small file sizes make it easy to sent your photo stories in an email. All this and it is FREE! Hutchison - Whitney Miller (wmiller@hutchisonschool.org)
March 3
- Doc It! Google Docs Can Simplify Your Teaching Life - This session will demonstrate how to create a Google account and the process of creating and using Google Docs. We will also share ideas of how to use this to get organized and save time. A fourth grade teacher will demonstrate how she uses Google Docs with her fourth grade team for planning purposes. Jennifer Stover (jstover@hutchisonschoo
• Posted: 19 Feb 2009 04:01 PM PST
Finding kid safe videos on YouTube can be a time consuming process. Kideo Player makes it easier to locate videos appropriate for students in pre-K through second grade. Kideo Player plays a continuous stream of videos for pre-K through second grade students. When a video you like is playing, click on the the YouTube icon and you will be taken to the original YouTube source where you can find out more about each video and grab the embed code. Clicking your keyboard's space bar allows you to skip ahead in the video clips.
Applications for Education
As mentioned above, Kideo Player is a good resource for pre-K through second grade videos. Some of the videos have educational value like the "ABC's" video while other videos are more of a children's entertai
• The Forward Thinking Museum is a virtual art museum containing more than 100 exhibitions and videos. Visitors can view image exhibitions or watch video exhibitions. The content ranges from subjects such as wildlife to architecture to people at work. The Forward Thinking Museum could keep you occupied for hours as you explore all of the exhibits. Visitors navigate through the Forward Thinking Museum by using their mouse or keyboard arrow keys.
Applications for Education
The Forward Thinking Museum could be a great resource for art teachers and art students. The museum could be used as an introduction to photography and videography styles. The Forward Thinking Museum could also be used as the centerpiece of a pairing and sharing activity.
• This site was created by a voter registration organization who
wanted to keep the young people they registered involved and engaged.
To do this, they provided them with free online tools and raw
materials through “America Now” and “America Then” playlists. Remix
America encourages students to draw parallels between the present and
the past. They hope that viewing seminal speeches and events from
American History will inspire young people to express themselves and
take action on the issues that matter to them.
Teachers around America have stumbled upon the software and
incorporated into their classroom. One teacher asked her students to
take a quote from Shakespeare’s Julius Caesar and apply it to the 2008
election. Another asked her students to create PSAs on the issues that
matter most to them – censorship, war, civil rights. You can browse
through “Favorite Remixes” section to see some of these great remixes!
• NASA has done something similar to engage students in science,
technology, engineering and mathematics. The NASA's Do-It-Yourself
Podcast activity provides students with audio clips, video, and photos
related to space. Students can use the NASA materials produce their
own audio or video productions.
• PrimaryAccess is a web-based tool that offers teachers and
to construct movies using tools provided by the web site.
Althought many of the primary source materials are photograph and
still images, the tools provided on the website allows students to add
motions to create a movie effect. I fist learned about Primary Access
while listening to Glen Bull's presentation during the 2008 K-12
online conference.
• rg
This project is slightly different in that it not only provides
the raw materials for students to produce a video, but also complete
an advocacy event. The project requires schools to register and the
topic is more focused. According to the project web site “Each year,
Take 2 shoots 2-3 months of high definition footage in a different
conflict region and creates extensive supporting and background
documentation then licenses the package free of charge to qualified
educational institutions. Participating schools will complete one
small task to help grow Take 2’s infrastructure and undertake at least
one advocacy event upon completion of their projects
• This website is not yet populated with lots of materials, but
has promise in offering students free, educational, copyright-friendly
media resources. According to the project website “Students and
teachers around the world can access pre-made collections, or "kits,"
of various digital assets - still images, background music,
narratives, video and text. Each kit is built around a common theme,
or curricular topic. For students, this becomes the construction paper
of the 21st century --allowing them to create reports and projects
filled with rich, immersive media for communicating their vision of
whatever subjects they chose. AS they master the technology, they will
progress from building projects with supplied materials to projects
where they find or create their own resources -- a strategy that
results in truly authentic assessment as measured by the projects
produced."
Posted from Diigo. The rest of my favorite links are here. | 1,939 | 9,370 | {"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-2018-30 | latest | en | 0.935372 |
https://codegolf.stackexchange.com/questions/201789/convert-planck-unit-to-si-metric/201807 | 1,620,294,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00151.warc.gz | 212,656,377 | 56,729 | # Objective
Given a dimension of an SI unit, convert the Lorentz-Heaviside version of a Planck unit $$\1\$$ into SI metric.
# What is a Planck unit?
Planck units are a set of units of measurement. It defines five fundamental constants of the universe as dimensionless $$\1\$$.
# What is a dimension?
There are five types of fundamental dimension: L, M, T, Q, and Θ (U+0398; Greek Capital Letter Theta).
L stands for length and corresponds to SI unit m (meter).
M stands for mass and corresponds to SI unit kg (kilogram).
T stands for time and corresponds to SI unit s (second).
Q stands for electric charge and corresponds to SI unit C (Coulomb).
Θ stands for temperature and corresponds to SI unit K (Kelvin).
A dimension is a multiplicative combination of these. For example, SI unit V (volt) is same as kg·m²·C/s⁴ and thus corresponds to dimension L²MQ/T⁴.
# Planck to SI
$$\1\$$ as Planck unit can be converted to SI metric as follows:
$$1 = 5.72938×10^{−35} \space [\text{m}] = 6.13971×10^{−9} \space [\text{kg}] = 1.91112×10^{−43} \space [\text{s}] = 5.29082×10^{−19} \space [\text{C}] = 3.99674×10^{31} \space [\text{K}]$$
# Input and Output
A dimension is given as the input. Its type and format doesn't matter. In particular, it can be an size-5 array of signed integers, each integer representing the exponent of a fundamental dimension.
The Planck unit $$\1\$$ is to be converted to the SI unit that corresponds to the inputted dimension, and then outputted. The output type and format doesn't matter.
# Examples
Let's say the input format is a tuple of five integers, representing L, M, T, Q, and Θ, respectively.
For example, If the input is $$\(2,1,-1,-2,0)\$$, it corresponds to SI unit Ohm, and thus:
$$1 = \frac{(5.72938×10^{−35})^2 × (6.13971×10^{−9})}{(1.91112×10^{−43})×(5.29082×10^{−19})^2} \space [\text{Ω}]$$
So the output is approximately $$\376.730\$$.
For another example, if the input is $$\(-2,-1,3,0,1)\$$, it corresponds to SI unit K/W, and thus:
$$1 = \frac{(1.91112×10^{−43})^3 × (3.99674×10^{31})}{(5.72938×10^{−35})^2 × (6.13971×10^{−9})} \space [\text{K/W}]$$
So the output is approximately $$\1.38424×10^{−20}\$$.
Note that, if the input is $$\(0,0,0,0,0)\$$, the output must be $$\1\$$.
• How much precision do we need in the constants? I assume that if we use them as is, it's five digits as given. But, if our solution computes them via their logs or inverses, how accurate do we need to get? – xnor Mar 28 '20 at 8:28
• @xnor At least 5 digits, as you say. (This is why I hate floating-point numbers.) – Dannyu NDos Mar 28 '20 at 8:36
• Yeah, floats are that worst. Do you mean give digits in the numbers we use, or five digits in the result in produces? for example, if my answer computes $5.72938×10^{−35}$ as 10**-34.241892, where can I cut off the 34.241892? If I cut it off to five decimal places as 34.24189, then the result of 5.72941129637e-35 isn't quite accurate to five places. Sorry to be so finicky here, but this does seem to matter for answers I'm trying. – xnor Mar 28 '20 at 9:58
• @xnor Rounding to six significant digits is acceptable. – Dannyu NDos Mar 28 '20 at 20:02
• Regarding (0,0,0,0,0) producing 1: Literally "1", or a floating point representation of "1", e.g., "1.00000"? – Eric Towers Mar 30 '20 at 21:55
# 05AB1E, 31 29 bytes
•¿NkˆSSÀ¯j5ÄΣDt₁ñµ•8ô6°/₃-*O°
Try it online!
-2 thanks to @Kevin Cruijssen
inspired by @xnor.
explanation:
•¿NkˆSSÀ¯j5ÄΣDt₁ñµ•8ô6°/₃- compressed list [-34.241893, -8.211853000000005, -42.718713, -18.276477999999997, -82.33983, -94.999995]
* multiply by the implicit input (e.g. [-68.48378600000001, -8.211852999999998, 42.718713, 36.552955999999995, 0.0])
O sum (e.g. 2.576029999999996)
° raise 10 to the power of that (e.g. 376.729821659)
# Old Solution, 3635 34 bytes
"7Q/Gy"Ç₃-*O°•7“µ$₁$´*h$Ć²E•6ôImP* -1 thanks to @Expired Data Try it online! explanation: "7Q/Gy"Ç₃- compressed list [-40,-14,-48,-24,26] * multiply (vectorize) with the implicit input - [-80, -14, 48, 48, 0] O sum - 2 ° 10^ - 100 •7“µ$₁$´*h$Ć²E• compressed number 572938613971191112529082399674
6ô split into groups of length 6 - [572938,613971,191112,529082,399674]
I push input to the stack, to get [2, 1, -1, -2, 0] [572938,613971,191112,529082,399674] 100
m power (vectorize) - [328257951844, 613971, 5.232533802168362e-6, 3.5723502030270884e-12, 1]
P product - 3.7672911312918473
* multiply - 376.72911312918473, and then implicitly output
• •;sê •2ô58- can be "7Q/Gy"Ç₃- for -1 – Expired Data Mar 28 '20 at 16:05
• thanks! how did you find it? I also need to update to TIO link – Command Master Mar 28 '20 at 16:44
• Well 95 is a 1 byte built in as opposed to 58 which is two and integer lists can be represented as strings and converted from ASCII if they're all in the range you need – Expired Data Mar 28 '20 at 19:24
• You can remove the s before the * in your 31-byte version. – Kevin Cruijssen Mar 30 '20 at 8:16
• 29 bytes by also using ₃- instead of 53- with a different compressed integer. – Kevin Cruijssen Mar 30 '20 at 8:28
# Python 3, 118 $$\\cdots\$$ 85 80 bytes
Saved a whopping 29 bytes thanks to Surculose Sputum!!!
Saved 4 bytes thanks to Arnauld!!!
Saved 5 bytes thanks to xnor!!!
lambda a,b,c,d,e:399674e26**e/174539e29**a/162874e3**b/523253e37**c/189007e13**d
Try it online!
• 116 bytes haha – RGS Mar 28 '20 at 2:47
• 89 bytes by doing it the vanilla way :) – Surculose Sputum Mar 28 '20 at 4:52
• Not as short as Surculose Sputum's direct solution, but another idea is to convert the problem to addition by working in log-space: Try it online! I don't know though how many digits of precision are needed in the list since the challenge doesn't say. Along similar lines, we could probably save bytes by changing some multiplications to divisions in the direct solution. – xnor Mar 28 '20 at 8:24
• @xnor Wow! You've totally blown my mind. Have to have way more coffee and see if I can figure out what's going there. :D – Noodle9 Mar 28 '20 at 11:02
• @SurculoseSputum The beauty (and oft shortness) of simplicity. Sweet - thanks! :-) – Noodle9 Mar 28 '20 at 11:03
## Frink, 76 bytes
f[a,b,z,d,e]:=l_P^a m_P^b(4π)^((a-b+z-e)/2)t_P^z(ℏ c epsilon0)^(d/2)T_P^e
Example usage:
f[2,1,-1,-2,0]
376.73031366686987155 m^2 s^-3 kg A^-2 (electric_resistance)
f[-2,-1,3,0,1]
1.3842382138391631282e-20 m^-2 s^3 kg^-1 K (thermal_resistance)
(As a bonus, it outputs a description of the unit of the result!)
Frink has the constants plancklength, planckmass, plancktime, and plancktemperature built in, with abbreviations l_P, m_P, t_P, and T_P respectively. Unfortunately, these are the Gaussian versions of the units, and the challenge specifies that we must use the Lorentz-Heaviside versions. This is a factor of $$\\sqrt{4\pi}\$$ conversion for each unit, which is consolidated into the (4π)^((a-b+z-e)/2) expression in the solution. (This expression is placed in the middle of the code to save a byte by letting us remove the space between m_P^b and t_P^z.)
Incidentally, is exactly as long as 4pi in UTF-8, so it doesn't actually save any bytes -- it just looks fancier. :P
Finally, Frink inexplicably lacks the built-in unit planckcharge, so we calculate the Lorentz-Heaviside version directly ourselves as (ℏ c epsilon0)^(1/2) (this time, actually saves a byte over hbar!).
# Fortran (2008), 133 bytes
At least for this challenge there should be a Fortran solution.
Takes input as integer array.
real(real64)function h(I)
use iso_fortran_env
integer I(5)
h=product([572938d-40,613971d-14,191112d-48,529082d-24,399674d26]**I)
end
A slightly longer and less accurate function uses the logarithm (stolen from @xnor):
real(real64)function f(I)
use iso_fortran_env
integer I(5)
f=exp(sum([-7884487d-5,-1890850d-5,-9836347d-5,-42083140d-5,7276562d-5]*I))
end
• Would be awesome if you could solve this one with Fortran! :D – Noodle9 Mar 28 '20 at 11:11
• Thank you for the tip – mcocdawc Mar 28 '20 at 11:53
• You can save a few bytes by getting rid of the decimal points (e.g. 5.72938d-35 ~> 572938d-40). – Arnauld Mar 28 '20 at 15:41
• Thank you very much – mcocdawc Mar 28 '20 at 16:08
# Jelly, 35 34 bytes
“~VHð¡ȷ‘I⁵*ד®ƬØʋ¥4ẋİ8nCaḌ’bȷ6¤*⁸P
A monadic Link accepting a list of the five exponents, [L, M, T, Q, Θ], which yields a floating-point number.
Try it online!
### How?
“~VHð¡ȷ‘I⁵*ד®ƬØʋ¥4ẋİ8nCaḌ’bȷ6¤*⁸P - Link: list of numbers [L, M, T, Q, Θ]
“~VHð¡ȷ‘ - list of code-page indices [126, 86, 72, 24, 0, 26]
I - differences [-40,-14,-48,-24,26]
⁵ - ten 10
* - exponentiate [1e-40, 1e-14, 1e-48, 1e-24, 1e26]
“®ƬØʋ¥4ẋİ8nCaḌ’ - base 250 number 572938613971191112529082399674
ȷ6 - 10^6 1000000
b - convert from base [572938,613971,191112,529082,399674]
× - multiply (vectorises) [5.72938e-35,6.13971e-09,1.91112e-43,5.29082e-19,3.99674e31]
⁸ - chain's left argument [L, M, T, Q, Θ]
* - exponentiate [5.72938e-35^L,6.13971e-09^M,1.91112e-43^T,5.29082e-19^Q,3.99674e31^Θ]
P - product 5.72938e-35^L×6.13971e-09^M×1.91112e-43^T×5.29082e-19^Q×3.99674e31^Θ
• 34 bytes by using STDIN Ɠ instead of first argument , enabling you to use for -100 instead of the _48. – Kevin Cruijssen Mar 30 '20 at 19:54
• @KevinCruijssen Nice & thanks; I'm going to do 34 a different way though... – Jonathan Allan Mar 30 '20 at 22:02
• Ah, using differences. Yeah, that's also a way. :) – Kevin Cruijssen Mar 31 '20 at 6:28
# Perl 5-pa, 84 80 bytes
@Arnauld and I both realized that you can save bytes by removing the decimal points.
$\=1;map$\*=$_**$F[$i++],572938E-40,6.13971E-9,191112E-48,529082E-24,399674E26}{ Try it online! • I was actually just working on that when you posted. – Xcali Mar 28 '20 at 15:41 # dc, 85817875 70 bytes 16iFFk1F875AI15^*?^35C8BI18^*?^/9B5431E?^/3C1109I1E^*?^/1A3ADDIA^*?^/p Try it online! Or try a test suite that shows the sample cases listed in the challenge. (This is a bash script that runs the dc program on each test data set in turn.) Saved 4 bytes by removing the decimal points, thanks to @Arnauld. Had to add 1 byte also though, because I had inadvertently omitted a digit in one of the constants. Saved 3 more bytes by changing constants to hexadecimal. Thanks to @Noodle9 for pointing out that using division instead of multiplication can shorten this. That, plus careful attention to how many significant digits were needed throughout, yielded a savings of 5 additional bytes. Input on stdin, one number per line, in base 16, in the following order: Θ L M T Q This is the OP's order except the last one there is moved to the first position instead. (The challenge says, "A dimension is given as the input. Its type and format doesn't matter." Also, hexadecimal input won't make a difference in practice given the problem domain, because we're not going to have a dimensional exponent outside the range -9 to 9. And this is allowed anyway by "Yes, I/O in unary, binary, octal, decimal or hexadecimal should be acceptable by default", which got the required support of at least 5 net votes, and at least twice as many upvotes as downvotes -- in this case, 20 upvotes, 8 downvotes, and 12 net votes.) Output on stdout, in base 10 as usual, with lots of decimal places :) . If you try this out, note that when you enter negative numbers in dc, they're written with an initial _ character instead of -, since - is reserved for the binary subtraction operator. And don't forget to put Θ first, and to enter them all in hexadecimal! (Output is still in decimal.) # JavaScript (ES7), 81 76 bytes Saved 5 bytes thanks to @Noodle9 Seems like the direct formula is the shortest way... ¯\_(ツ)_/¯ Takes input as 5 distinct arguments. (a,b,c,d,e)=>399674e26**e/174539e29**a/162874e3**b/523253e37**c/189007e13**d Try it online! # JavaScript (ES7), 86 bytes Using reduce() adds 5 10 bytes. Takes input as an array. a=>a.reduce((p,c,i)=>p*([572938e8,613971e34,191112,529082e24,399674e74][i]/1e48)**c,1) Try it online! • – Noodle9 Mar 29 '20 at 0:13 # Mathematica, 260 bytes Using a different method than the OP's. In particular, Table 2 at https://en.wikipedia.org/wiki/Planck_units#Definition Many opportunities for cleanup (e.g., shrink name to c, saving 6 bytes), but seems pointless. Newlines and indenting are inessential and are only present for presentation. Two spaces are required, at the end of the BoltzmannConstant line and at the end of the ElectricConstant line. convert=Function[{L,M,T,Q,t}, N[UnitConvert[ Quantity["BoltzmannConstant"]^-t Sqrt[ (4Pi)^(L-M+T-t) Quantity["SpeedOfLight"]^(-3L+M-5T+Q+5t) Quantity["ElectricConstant"]^Q Quantity["GravitationalConstant"]^(L-M+T-t) Quantity["ReducedPlanckConstant"]^(L+M+T+Q+t) ] ],6] ] Checking: convert[2, 1, -1, -2, 0] (* 376.730 kg m^2/(s^3A^2) *) (* FullForm: Quantity[376.730, ("Kilograms" ("Meters")^2)/(("Amperes")^2 ("Seconds")^3)] *) convert[-2, -1, 3, 0, 1] (* 1.3842\[Times]10^-20 s^3K/(kg m^2) *) (* FullForm: Quantity[1.3842\[Times]10^-20, ("Kelvins"*"Seconds"^3)/("Kilograms"*"Meters"^2)] *) convert[0, 0, 0, 0, 0] (* 1.00000 *) Notice that this code uses a curated source of consensus scientific values of the constants, and the output will change as those values change. Of relevance to the Challenge (and exemplified in the above samples), the gravitational constant is only 6.674 30(15) *10^(-11) m^3 /(kg s^2), so is only known to four significant digits. This impacts the precision of length, mass, and time (except in combinations where this term cancels) in the Challenge, although the Challenge does not acknowledge this limitation of precision in those constants. # PowerShell, 124 101 bytes -23 bytes thanks to mazzy$t=1
$args|%{$t/=[math]::pow((174539e29,162874e3,523253e37,189007e13, 2502039161917e-44)[$i++],$_)}
\$t
• – mazzy Apr 2 '20 at 3:55
• It looks nice. I don't see any good reason to use 1/const and const in the same expression at the same time, unless it shortens the expression. – mazzy Apr 2 '20 at 10:30
# Java 8, 118 bytes
Math M=null;(a,b,c,d,e)->M.pow(399674e26,e)/M.pow(174539e29,a)/M.pow(162874e3,b)/M.pow(523253e37,c)/M.pow(189007e13,d)
Port of @Noodle9's Python answer, so make sure to upvote him!!
Try it online.
# TI-83 Basic, 62 tokens
Using list operations:
prod({5.72938ᴇ−35,6.13971ᴇ−9,1.91112ᴇ−43,5.29082ᴇ−19,3.99674ᴇ31}^Ans
where
prod( Multiply all list elements together (1 token)
{ ... } Constants in list (59 tokens)
^ Power using 2 lists as arguments (1 token)
Ans Input list stored in answer variable (1 token)
Example:
| {2,1,-1,-2,0} | Give input
| {2 1 -1 -2 0} |
| prgmPLANCK | Run program
| 376.7291131 | Result
Second example unfortunately returns 0, because of insufficient precision on calculator. | 5,025 | 14,949 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-21 | latest | en | 0.718613 |
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# Risk management using risk+ (v5)
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Employing Monte Carlo Simulation for the assessment of schedule and cost risk in the Integrated Master Schedule (IMS).
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### Risk management using risk+ (v5)
1. 1. 1 INCREASING THE PROBABILITY OF PROGRAM SUCCESS USING RISK+Glen B. Alleman A workshop on the principles and practices of Risk+ andNiwot Ridge LLC increasing the Probability of Program Success
2. 2. A Warning We’re going to cover a lot of material in 3 hours2
3. 3. Risk involves uncertainty. Uncertainty involves probability.3
4. 4. 4 Douglas Adams, Hitchhikers Guide to the Galaxy
5. 5. MOTIVATION? Your motivation? Your motivation is your pay packet on Friday. Now get on with it. – Noel Coward, English actor, dramatist, & songwriter (1899 – 1973)5
6. 6. We have to know the underlying statistical behavior of the processes driving the project This means cost, schedule, and technical performance measures with probabilistic models We need to know how these three statistical drivers are coupled What drives what? What are the multipliers between each random6 variable?
8. 8. Remember High School Statistics8
9. 9. The IMS is a collection of probabilistic processes all coupled together9
10. 10. What does this really mean?10 In building a risk tolerant IMS, we’re interested in the probability of a successful outcome… “What is the probability of making a desired completion date?” But the underlying statistics of the tasks influence this probability The statistics of the tasks, their arrangement in a network of tasks and correlation define how this probability based estimated developed.
11. 11. There are real problems with those pesky Unknowns that get in the way of progress Imprint of a bird on our west facing family room second story window on a bright afternoon11 The Bird survived
12. 12. The “units of measure” of Risk12 These classifications can be used to avoid asking the “3 point” question for each task. Anchoring and Adjustment† of all estimating processes produces a bias. Knowing this is necessary for credible estimates. Classification Uncertainty Overrun 1 Routine, been done before Low 0% to 2% 2 Routine, but possible difficulties Medium to Low 2% to 5% 3 Development, with little technical difficulty Medium 5% to 10% 4 Development, but some technical difficulty Medium High 10% to 15% 5 Significant effort, technical challenge High 15% to 25% 6 No experience in this area Very High 25% to 50% † Tversky and Khanemann Anchoring and Adjustment
13. 13. We’re looking for knowledge of what is going to happen in he future, with a known level of confidence13 Harvard main library
14. 14. 14
15. 15. 15 What is Monte Carlo Simulation? With some principals behind us, let’s see how to use Risk+ to address the problem of forecasting the future of schedule and cost performance.
16. 16. A Quick Look At Monte Carlo16 George Louis Leclerc, Comte de Buffon, asked what was the probability that the needle would fall across one of the lines, marked here in green. That outcome will occur only if A l sin
17. 17. 17 Los Alamos Science, Special Issue 1987
18. 18. Monte Carlo Simulation18 Monte Carlo Simulation is named after the city, in Monaco, of casinos on the French Rivera. Monte Carlo … Examines all paths not just the critical path. Provides an accurate (true) estimate of completion: Overall duration distribution Confidence interval (accuracy range) Sensitivity analysis of interacting tasks Varied activity distribution types – not restricted to a single distribution Schedule logic can include branching – both probabilistic and conditional When resource loaded schedules are used – provides integrated cost and schedule probabilistic model.
19. 19. 19
20. 20. 20 What Are We Really After? We need to answer the question … What is the confidence we will complete “on or before” at date and “at or below” at cost? This is the question that should be asked and answered on a periodic basis. We need to have Schedule and Cost margin to protect the deliverables and our Budget At Completion.
21. 21. Here is some advice on how to depict this margin and where to place this margin. No matter how we show manage these two elements in the IMS, if we don’t have margin we are late and over budget before we start. http://www.ndia.org/Divisions/Divisions/Procurement/Documents/PMSCommittee/CommitteeDocuments/21 WhitePapers/NDIAScheduleMarginWhitePaperFinal-2010(2).pdf
22. 22. Confidence levels for margin change as the program proceeds22 As the program proceeds we want to have Increased accuracy Reduced schedule risk Increasing visual confirmation Current Estimate Confidence that success can be reached
23. 23. Our REAL goal here is to Manage Margin using probabilistic models23 Programmatic Margin is Margin that is not used in the added between Development, IMS for risk mitigation will be Production and Integration & moved to the next sequence Test phases of risk alternatives Risk Margin is added to the This enables us to buy back IMS where risk alternatives schedule margin for activities are identified further downstream This enables us to control the Downstream ripple effect of schedule shifts Plan B Duration of Plan B < Plan A + Margin Activities shifted to left 2 days on Margin activities Plan B 3 Days Margin Used Plan A 5 Days Margin First Identified Risk Alternative in IMS Plan A 5 Days Margin Second Identified Risk 2 days will be added to this margin task Alternative in IMS to bring schedule back on track
24. 24. Sensitivity Analysis24 The schedule sensitivity of a task measures the closeness with which change in the task duration matches change in the project duration over the simulation. This closeness is the correlation between changes in individual activities and their impacts on other activities. A task with high schedule sensitivity is more likely to be a major driver of the project duration than a lower ranked task. : Models of the Schedule
25. 25. Task Criticality Analysis25 A measure of the frequency that an activity in the project schedule is critical (Total Float = 0) in a simulation If a task is critical in 500 of the 1,000 iterations of the simulation, it has a Criticality Index of 0.5 The higher the criticality index, the more certain it is that the task will always be critical in the project : Models of the Schedule
26. 26. Cruciality shows each task’s tolerance to risk26 Cruciality = Schedule Sensitivity x Criticality Schedule Sensitivity can be statistically misleading: A task with high sensitivity may not be on or near the critical path. Thus a reduction in that task’s duration may have little effect on the project duration. Cruciality sharpens the analytical focus: It highlights critical or near–critical activities with high. Schedule Sensitivity These tasks are most likely to drive project duration. : Models of the Schedule
27. 27. Guiding the Risk Factor Process means weighting each level of risk27 For tasks marked “Low” a reasonable approach is to score the maximum Min Most Max 10% greater than the minimum. Likely The “Most Likely” is then scored as a geometric progression for the Low 1.0 1.04 1.10 remaining categories with a common Low+ 1.0 1.06 1.15 ratio of 1.5 Moderate 1.0 1.09 1.24 Tasks marked “Very High” are bound Moderate+ 1.0 1.14 1.36 at 200% of minimum. High 1.0 1.20 1.55 No viable project manager would like a task grow to three times the planned High+ 1.0 1.30 1.85 duration without intervention Very High 1.0 1.46 2.30 The geometric progress is somewhat Very High+ 1.0 1.68 3.00 arbitrary but it should be used instead of a linear progression : Examples of Monte Carlo
28. 28. Progressive Risk Factors28 A geometric progression (1.534) of risk can be used. The phrases associated with increasing risk have been shown at the Naval Research Laboratory to correlate with an engineers “sense” of increasing risk. : Examples of Monte Carlo
29. 29. Risk Factor Attributes29 The “narrative” for each risk factor needs to be developed. Each description is dependent on… Discipline Program stage Complexity Historical data Current “risk state” of the program This is currently missing from our efforts to quantify schedule and cost risk. : Examples of Monte Carlo
30. 30. Accuracy30 Given a specified final cost or project duration, what is the probability of achieving this cost or duration? Frequentist approach Over many different projects, four out of five will cost less or be completed in less time than the specified cost or duration. Bayesian approach We would be willing to bet at 4 to 1 odds that the project will be under the 80% point in cost or duration. Accuracy is needed to plan reserves. Accuracy is needed when comparing competing proposals. : What is the Purpose of Project Risk Analysis?
31. 31. Structured Thinking31 All estimates will be in error to some degree of variance. Trying to quantify these errors will result in bounds too wide to be useful for decision making. Risk analysis should be used to Think about different aspects of the project Try to put numbers against probabilities and impacts Discuss with colleagues the different ideas and perceptions Thinking things through carefully results in Which elements of the programmatic and technical risk are represented in the IMS. The process becomes more valuable than the numbers. : What is the Purpose of Project Risk Analysis?
32. 32. To properly use Schedule Margin†32 Work must be represented in single units – either task or work packages. The overall schedule margin must be related to the variation of individual units of work. The importance of the units of work must be shared among all participants (ordinal ranking of work and its risk). The schedule must be reasonable in some units of measure shared by all the participants. † “Protecting Earned Value Schedules with Schedule Margin,” Newbold, Budd, and Budd, http://www.prochain.com/pm/articles/ProtectingEVSchedules.pdf
33. 33. Let’s Apply a Monte Carlo Simulation Tool The Monte Carlo trolley, or FERMIAC, was invented by Enrico Fermi and constructed by Percy King. The drums on the trolley were set according to the material being traversed and a random choice between fast and slow neutrons. Another random digit was used to determine the direction of motion, and a third was selected to give the distance to the next collision. The trolley was then operated by moving it across a two dimensional scale drawing of the nuclear device or reactor assembly being studied. The trolley drew a path as it rolled, stopping for changes in drum settings whenever a material boundary was crossed. This infant computer was used for about two years to determine, among other things, the change in neutron population with33 time in numerous types of nuclear systems.
34. 34. 34
35. 35. 35 A Small Diversion Most Likely Isn’t Likely to be the Most Likely When we say “most likely” what do we think this actually means? If you pick the wrong meaning, your Monte Carlo model will be seriously flawed.
36. 36. The problem with “Most Likelies”36 For each activity the “best” estimate is … The “most likely” duration – the mode of the distribution of durations? (Mode is the number that appears most often) It’s 50th percentile duration – the median of the distribution? (Median is the number in the middle of all the numbers) It’s expected duration – the mean of the distribution? (Mean is the average of all the numbers) These definitions lead to values that are almost always different from each other. Rolling up the “best” estimate of completion is almost never one of these.
37. 37. Durations are Probability Estimates not Single Point Values37 We know this because… “Best” estimate is not the only possible estimate, so other estimates must be considered “worse.” Common use of the phrase “most likely duration” assumes that other possible durations are “less likely.” “Mean,” “median,” and “mode” are statistical terms characteristic of probability distributions. This implies activity distributions have probability distributions They are random variables drawn from the probability distribution function (pdf). “Actual” project duration is an uncertain quality that can be modeled as a sum of random variables The pdf may be known or unknown.
38. 38. 3 Task Most Likely ≠ Project Most Likely38 PERT assumes probability distribution of the project times is the same as the tasks on the critical path. Because other paths can become critical paths, PERT consistently underestimates the project completion time. 1+1=3 : Managing Uncertainty in the IMS
39. 39. Probability Distribution Function is the Lifeblood of good planning39 Probability of occurrence as a function of the number of samples. “The number of times a task duration appears in a Monte Carlo simulation.” : Managing Uncertainty in the IMS
40. 40. Remember the quote about statistics40 Lies, Damn Lies, and Statistics – Benjamin Disraeli But we know better, we know that any estimate without a variance is not trustworthy. We know that the variances have to be calibrated from past performance to be credible
41. 41. 41
42. 42. 42 A “Real World” Schedule Analysis One should expect that the expected can be prevented, but the unexpected should have been expected. — Augustine Law XLVThis is a must own book for everyone in our business. It definesfundamental Laws of program and business management, whichare many times ignored – like the one above
43. 43. Our Starting Point43 Risk+ Installed Let’s define the needed fields These are used by Risk+ to hold information and run the application. If there are conflicts, you can make changes in Risk+ to work around your fields.
44. 44. A Simple IMS44 By simple it means serial cascaded work efforts.
45. 45. Initial Field Usage45 Minimum Remaining Duration The duration that is least you’d expect this task to complete in Most Likely Remaining Duration The ML (Mode) of the duration Maximum Remaining Duration The duration that is the most you’d expect this task to complete in Task Reporting ID The tasks we want to watch
46. 46. Define a View and Table for Risk+46 Start with the Gantt View and Entry Table Set up both to match the Risk+ field usage Use the default if there are no field conflicts
47. 47. Fields used for Risk+ example47
48. 48. Let’s actually “doing something”48 Initialize the Most Likely. This sets the Most Likely duration to the same value that is in the “Duration” field of your IMS. The “planned duration” now becomes the ML duration. If this “planned duration” is bogus then your model will be as well. Choose wisely.
49. 49. Now the ML = DURATION step49 All the DURATION values have been moved to the ML field. But remember our discussion of the ML’s Choose them carefully The next we’ll set the upper and lower limits of that ML value Using risk factors. OK, 3 point estimates if you have to.
50. 50. Let’s do this the simple way50 Let’s pick MEDIUM confidence. MEDIUM means –25% +25% And a NORMAL (Gaussian) curve
51. 51. Let’s have Risk+ do something for us51 Enter a “1” in the RPT field (Number 1) This marks that ROW in the schedule as a work activity we want to see the Monte Carlo output for
52. 52. Now we’re ready to run52 The RISK ANALYSIS command starts the process going. Let’s make 200 iteration and look at the DURARTION ANALYSIS for the activities we are watching.
53. 53. This is nice but what actually is Risk + doing?53 Risk+ is picking a random number from under the normal distribution within the range of the Least remaining and most remaining This is not some ordinary random number it is chosen through an algorithm called the Latin Hypercube - more on that later. Risk+ then plugs that number into the “real” DURATION field and does that for all the DURATIONS in the schedule Then the F9 key is pressed and the date is recorded for the finish of UID 41. This is done 200 times and a histogram of all the dates that appeared for those 200 time is recorded.
54. 54. And We Get54 Date: 11/29/2011 4:32:17 PM Completion Std Deviation: 2.06 days Samples: 500 95% Confidence Interval: 0.18 days Unique ID: 19 Each bar represents 1 day Name: End Work Package 3 0.20 1.0 Completion Probability Table 0.18 0.9 Cumulative Probability Prob Date Prob Date 0.16 0.8 0.05 Wed 3/7/12 0.55 Tue 3/13/12 0.14 0.7 0.10 Thu 3/8/12 0.60 Tue 3/13/12 Frequency 0.12 0.6 0.15 Thu 3/8/12 0.65 Tue 3/13/12 0.10 0.5 0.20 Fri 3/9/12 0.70 Wed 3/14/12 0.08 0.4 0.25 Fri 3/9/12 0.75 Wed 3/14/12 0.06 0.3 0.30 Fri 3/9/12 0.80 Wed 3/14/12 0.35 Mon 3/12/12 0.85 Thu 3/15/12 0.04 0.2 0.40 Mon 3/12/12 0.90 Thu 3/15/12 0.02 0.1 0.45 Mon 3/12/12 0.95 Fri 3/16/12 Fri 3/2/12 Mon 3/12/12 Tue 3/20/12 0.50 Mon 3/12/12 1.00 Tue 3/20/12 Completion Date
55. 55. Learning to Speak in Risk+55 Risk +shows use the probability of finish “on or before” a date It does NOT show the probability of success. But even the “on or before” term is loaded with special meaning. It means for the 500 iterations of Risk+ using the upper and lower bounds of the duration, drawn from the probability density function (pdf) with the Normal (Gaussian) shape, 60% of the finish dates were recorded to be on or before 3/12/12.
56. 56. Medium confidence for a large project56 Date: 11/30/2011 6:05:35 PM Completion Std Deviation: 4.49 days Samples: 200 95% Confidence Interval: 0.62 days Unique ID: 17 Each bar represents 2 days Name: (SA) Systems Requirements Completed 0.22 1.0 Completion Probability Table 0.20 0.9 Cumulative Probability Prob Date Prob Date 0.17 0.8 0.05 Fri 5/4/12 0.55 Thu 5/17/12 0.7 0.10 Wed 5/9/12 0.60 Thu 5/17/12 Frequency 0.15 0.6 0.15 Thu 5/10/12 0.65 Fri 5/18/12 0.13 0.5 0.20 Fri 5/11/12 0.70 Mon 5/21/12 0.10 0.25 Mon 5/14/12 0.75 Mon 5/21/12 0.4 0.08 0.3 0.30 Mon 5/14/12 0.80 Tue 5/22/12 0.05 0.35 Tue 5/15/12 0.85 Wed 5/23/12 0.2 0.40 Tue 5/15/12 0.90 Thu 5/24/12 0.03 0.1 0.45 Wed 5/16/12 0.95 Mon 5/28/12 Wed 5/2/12 Wed 5/16/12 Mon 6/4/12 0.50 Wed 5/16/12 1.00 Mon 6/4/12 Completion Date
57. 57. Low confidence for a large project57 Date: 11/30/2011 10:30:05 PM Completion Std Deviation: 9.14 days Samples: 200 95% Confidence Interval: 1.26 days Unique ID: 17 Each bar represents 3 days Name: (SA) Systems Requirements Completed 0.16 1.0 Completion Probability Table 0.9 0.14 Cumulative Probability Prob Date Prob Date 0.8 0.12 0.05 Thu 5/3/12 0.55 Fri 5/25/12 0.7 0.10 Tue 5/8/12 0.60 Mon 5/28/12 Frequency 0.10 0.6 0.15 Wed 5/9/12 0.65 Wed 5/30/12 0.08 0.5 0.20 Mon 5/14/12 0.70 Wed 5/30/12 0.4 0.25 Tue 5/15/12 0.75 Fri 6/1/12 0.06 0.3 0.30 Thu 5/17/12 0.80 Mon 6/4/12 0.04 0.35 Fri 5/18/12 0.85 Wed 6/6/12 0.2 0.02 0.40 Mon 5/21/12 0.90 Fri 6/8/12 0.1 0.45 Wed 5/23/12 0.95 Thu 6/14/12 Tue 4/24/12 Thu 5/24/12 Wed 6/27/12 0.50 Wed 5/23/12 1.00 Wed 6/27/12 Completion Date
58. 58. Let’s run some simulations58
59. 59. 59
60. 60. 60 Basic Principles of Probabilistic Cost Now that the schedule can be produced using probabilistic methods, it’s time to talk about the cost. Cost does not have a linear relationship with schedule unfortunately. : Basic Principles of Probabilistic Cost
61. 61. Basic Principles with Probabilistic Cost Estimating are coupled with scheduling61 Cost estimates usually involve many CERs Each of these CERs has uncertainty (standard error) CER input variables have uncertainty (technical uncertainty) Must combine CER uncertainty with technical uncertainty for many CERs in an estimate Usually cannot be done arithmetically; must use simulation to roll up costs derived from Monte Carlo samples Add and multiply probability distributions rather than numbers Statistically combining many uncertain, or randomly varying, numbers Monte Carlo simulation Take random sample from each CER and input parameter, add and multiply as necessary, then record total system cost as a single sample Repeat the procedure thousands of times to develop a frequency histogram of the total system cost samples This becomes the probability distribution of total system cost : Basic Principles of Probabilistic Cost
62. 62. The Cost Probability Distributions as a function of the weighted cost drivers62 Combined Cost Modeling and Technical Uncertainty Cost = a + bXc Cost Modeling Uncertainty Cost Estimate Historical data point \$ Cost estimating relationship Technical Uncertainty Standard percent error bounds Cost Driver (Weight) : Basic Principles of Probabilistic Cost
63. 63. The Risk Adjusted Cost Estimate Connected To The IMS63 In the risk–adjusted cost estimate, we now combine discrete risk events and the uncertainty of the input distributions with the uncertainty of the CERs Since the input distributions tend to be right–skewed, the expected cost tends to be larger than the baseline estimate In addition, the risk–adjusted cost distribution tends to be wider than the baseline estimate The difference between the expected cost of the risk– adjusted estimate and the expected cost of the baseline estimate is, by definition, the amount of RISK dollars included in the risk–adjusted estimate : Basic Principles of Probabilistic Cost
64. 64. Baseline versus Risk Adjusted Cost Estimates Usually Show a Cost Increase64 Baseline vs. Risk-Adjusted Estimates Baseline: Mean = \$102.6M Std Dev = \$29.8M Risk–Adjusted: Mean = \$122.6MLikelihood Std Dev = \$42.8M 0 50 100 150 200 250 300 350 FY\$M : Basic Principles of Probabilistic Cost
65. 65. The S–Curve for Cost Modeling65 Cumulative Distribution Function 100% 90% 80% 80th percentileCumulative Probability 70% 50th percentile \$153.5M 60% \$114.7M 50% Baseline Estimate Mean \$102.6M Risk–adjusted 40% Estimate Mean \$122.6M 30% 20% 10% 0% \$60 \$80 \$100 \$120 \$140 \$160 \$180 \$200 FY00\$M : Basic Principles of Probabilistic Cost
66. 66. The Real Question Always Returns to… “But How Much Does It Cost? Really?”66 This is impossible to answer precisely Decision–makers and cost analysts should always think of a cost estimate as a probability distribution, NOT as a deterministic number The best we can provide is the probability distribution – If we think we can be any more precise, we’re fooling ourselves It is up to the decision–maker to decide where he/she wants to set the budget The probability distribution provides a quantitative basis for making this determination Low budget = high probability of overrun High budget = low probability of overrun : Basic Principles of Probabilistic Cost
67. 67. 67
68. 68. 68 Some More Parts to using Risk+ Just having the pictures is necessary, but knowing what they mean is required. Making changes to the IMS to increase the Probability of Program Success is the primary outcome from Monte Carlo Simulation.
69. 69. Without Integrating \$, Time, and TPM you’re driving in the rearview mirror69 Technical Performance (TPM)
70. 70. Risk Management Demands a Well Defined Process
71. 71. Statistics of a Triangle Distribution71 50% of all possible values are under this area of the curve. This is the definition of the median Minimum Maximum 1000 hrs 6830 hrs Mode = 2000 hrs Mean = 3879 hrs Median = 3415 hrs Basic Statistics
72. 72. TPM Trends & Responses directly impact risk and credibility of the IMS72 Design ModelROM in Proposal Detailed Design Model Bench Scale Model Measurement Technical Performance Measure 28kg Prototype Measurement Vehicle Weight Flight 1st Article 26kg 25kg 23kg CA SFR SRR PDR CDR TRR Dr. Falk Chart – modified
73. 73. Not A Mitigation Plan Mitigation is too late, the risk has turned into an issue. The money has been spent, and the time has passed.73
74. 74. Ordinal versus Cardinal74 Ordinal Cardinal A variable is ordinally measurable A variable is cardinally measurable if ranking is possible for values of if a given interval between the variable. For example, a gold measures has a consistent meaning, medal reflects superior performance i.e., if the measure corresponds to to a silver or bronze medal in the points along a straight line. For Olympics, or you may prefer French example, height, output, and income toast to waffles, and waffles to oat are cardinally measurable. bran muffins. All variables that are cardinally measurable are also ordinally measurable, although the reverse may not be true.
75. 75. Correcting Ordinal Risk Scales75 Classify and calibrate risk ranking in units meaningful to the decision makers Risk rank 1, 2, 3, 4, is NOT sufficient The Risk Rank must have a measurable value connected to the actual behavior of the system being assessed Calibration coefficients between ordinal probability and consequences should also be used. Ordinal analysis assumes ordering of the risks. Cardinal analysis provides objective measures of probability and consequential impact.
76. 76. Level Likelihood ValueNever multiply Likelihood E E Near Certainty E ≥ 90%by outcome. They are not“numbers,” they a D Highly Likely 74% ≤ D ≤ 90% Dprobability distributions. C Likely 40% ≤ C ≤ 60% COnly convolution is B Low Likelihood 20% ≤ B ≤ 40% Bpossible A Not Likely A ≤ 20% AThese are Cardinal measures of probability of occurrence and A B C D Econsequential impactLevel Technical Performance Schedule Cost Minimal or no consequence to A Minimal or no impact Minimal or no impact technical performance. Budget increase or unit Minor reduction in technical B Able to meet key dates production cost increases. performance or supportability. < (1% of Budget) Moderate reduction in Minor schedule slip. Budget increase or unit technical performance or Able to meet key C production cost increase supportability with limited milestones with no < (5% of Budget) impact on program objectives. schedule float. Significant degradation in Budget increase or unit technical performance or Program critical path D production cost increase major shortfall in affected < (10% of Budget) supportability. Cannot meet key Exceeds budget increase or Severe degradation in technical E program milestones. unit production cost performance. 76 Slip > X months threshold
77. 77. Example of Ordinal Probability Complexity Scale†77 Definition of the Ordinal Scale Ranking Scale Level Greater than 20% of the interface design has been E altered because of modifications to the ICD’s. Greater than 15% but less than 20% of the interface design has been altered because of modifications of the D ICD’s. Greater than 10% but less than 15% of the interface design has been altered because of modifications of the C ICD’s. At least 5% but less than 10% of the interface design has B been altered because of modifications of the ICD’s. At least 5% of the interface design has been altered A because of modifications of the ICD’s. † Effective Risk Management: Some Keys to Success, Edmund Conrow, AIAA Press, 2003
78. 78. A “real” risk Ordinal Ranking Table78Risk Percent Variance Interpretation of Risk RankingRank Normal business, technical & manufacturing A – 5% ≤ A ≤ 10% processes are applied Normal business & technical processes are B – 5% ≤ B≤ 15% applied; new or innovative manufacturing processes Flight software development & certification C – 5% ≤ C ≤ 35% processes Build & qualification of flight components, D – 10% ≤ D ≤ 25% subsystems & systems E – 10% ≤ E ≤ 35% Flight software qualification F – 5% ≤ F ≤ 175% ISS thermal vacuum acceptance testing
79. 79. Project Train Wrecks Occur When There is… Inattention to budgetary responsibilities Work authorizations that are not always followed Issues with Budget and data reconciliation Lack of an integrated management system Baseline fluctuations and frequent replanning Untimely and unrealistic Latest Revised Current period and retroactive Estimates (LRE) changes Progress not monitored in a regular and Improper use of management consistent manner reserve Lack of vertical and horizontal traceability EV techniques that do not cost and schedule data for corrective action reflect actual performance Lack of internal surveillance and controls Lack of predictive variance Managerial actions not demonstrated using analysis Earned Value 79
80. 80. Our Final Check List80 Set up the Risk+ fields, flags, views, and tables for the program standard IMS. Build an IMS that passes the DCMA 14 Point Assessment with all GREEN. Build the Ordinal Risk Ranking table for the various risk categories on the program. Assign risk ranking to each activities in the IMS, with the variances defined in the Ordinal Table. Run Risk+ to see the confidence in the deliverables. Develop the needed schedule margin to protect the delivery to at least the 80% confidence level.
81. 81. Advice from the school of hard knocks81 Put margin in front of critical deliverables. Build a margin burn down chart and allocate schedule margin just like you do MR for the PMB. This real world advice is counter to the current DCMA guidance.
82. 82. 82
83. 83. 83 Putting This New Knowledge To Work
84. 84. Managing margin is what Risk+ is all about84 CP Total Float Float Erosion: Critical Path Time Usage Acceptable Rate of Float Erosion Linear (CP Total Float ) 100 80 Time Now October 31, 2005 Critical Path - Time Reserve 60 40 Spacecraft Contract Delivery December 10, 2007 20 0 -20 -40
85. 85. How much margin do we need?85 The Missing Link: Schedule Margin Management, Rick Price, PS–10, PMI–CPM EVM World 2008
86. 86. Deterministic versus Probabilistic86 Baseline Plan Sep 2011 Oct 2011 Current Plan with risks is the Ready deterministic schedule Nov 2001 Early Plan Margin Dec 2011 Launch 20% Risk Period Margin Jan 2012 Mean Current Plan with risks is the Feb 2012 stochastic schedule The probability distribution can 80% Mar 2012 vary as a Missed function of time Launch ATLO Period CDR PDR FRR SRR Apr 2012
87. 87. 87
88. 88. 88 References
89. 89. References89 “Protecting Earned Value with Schedule Margin,” http://www.prochain.com/pm/articles/ProtectingEVSchedules.pdf Depicting Schedule Margin in the Integrated Master Schedule, http://www.ndia.org/Divisions/Divisions/Procurement/Documents/PMSCommittee/C ommitteeDocuments/WhitePapers/NDIAScheduleMarginWhitePaperFinal- 2010(2).pdf Effective Risk Management: Some Keys to Success, Second Edition, Edmund Conrow, AIAA Press. How to Lie with Statistics, Darrell Huff, Norton, 1954 (Available in paper back at any good book store) DID DI–MGMT–81650 “A management method for accommodating schedule contingencies. It is a designated buffer and shall be identified separately and considered part of the baseline.
90. 90. References90 Interfacing Risk and Earned Value Management, Association for Project Management, 150 West Wycombe Road, High Wycombe, Buckinghamshire, HP12 3AE, United Kingdom. Practice Standard for Earned Value Management, Second Edition, Project Management Institute, 2011. Effective Opportunity Management for Projects, David Hillson, Taylor and Francis, 2004. Measuring Time: Improving Project Performance Using Earned Value, Mario Vanhoucke, Springer, 2009. Performance Based Earned Value, Paul Solomon and Ralph Young, Wiley, 2007. Effective Risk Management: Some Keys to Success, Edmund Conrow, AIAA Press, 2003.
91. 91. Niwot Ridge LLC (: 303.241.9633 4347 Pebble Beach Drive -: glen.alleman@niwotridge.com Niwot, Colorado 8050391 | 8,229 | 32,150 | {"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-2019-43 | latest | en | 0.7562 |
https://www.dataunitconverter.com/megabyte-to-zettabyte | 1,708,911,009,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00302.warc.gz | 739,389,111 | 16,909 | # MB to ZB → CONVERT Megabytes to Zettabytes
expand_more
info 1 MB is equal to 0.000000000000001 ZB
Input Megabyte (MB) - and press Enter.
## Megabyte (MB) Versus Zettabyte (ZB) - Comparison
Megabytes and Zettabytes are units of digital information used to measure storage capacity and data transfer rate.
Both Megabytes and Zettabytes are the "decimal" units. One Megabyte is equal to 1000^2 bytes. One Zettabyte is equal to 1000^7 bytes. There are 1,000,000,000,000,000 Megabyte in one Zettabyte. Find more details on below table.
Unit Name Megabyte Zettabyte
Unit Symbol MB ZB
Standard decimal decimal
Defined Value 10^6 or 1000^2 Bytes 10^21 or 1000^7 Bytes
Value in Bits 8,000,000 8,000,000,000,000,000,000,000
Value in Bytes 1,000,000 1,000,000,000,000,000,000,000
## Megabyte (MB) to Zettabyte (ZB) Conversion - Formula & Steps
The MB to ZB Calculator Tool provides a convenient solution for effortlessly converting data units from Megabyte (MB) to Zettabyte (ZB). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Megabyte) and target (Zettabyte) data units.
Source Data Unit Target Data Unit
Equal to 1000^2 bytes
(Decimal Unit)
Equal to 1000^7 bytes
(Decimal Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Megabyte to Zettabyte in a simplified manner.
÷ 8
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
x 8
x 1000
x 1000
x 1000
x 1000
x 1000
x 1000
x 1000
x 1000
Based on the provided diagram and steps outlined earlier, the formula for converting the Megabyte (MB) to Zettabyte (ZB) can be expressed as follows:
diamond CONVERSION FORMULA ZB = MB ÷ 10005
Now, let's apply the aforementioned formula and explore the manual conversion process from Megabyte (MB) to Zettabyte (ZB). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Zettabytes = Megabytes ÷ 10005
STEP 1
Zettabytes = Megabytes ÷ (1000x1000x1000x1000x1000)
STEP 2
Zettabytes = Megabytes ÷ 1000000000000000
STEP 3
Zettabytes = Megabytes x (1 ÷ 1000000000000000)
STEP 4
Zettabytes = Megabytes x 0.000000000000001
Example : By applying the previously mentioned formula and steps, the conversion from 1 Megabyte (MB) to Zettabyte (ZB) can be processed as outlined below.
1. = 1 ÷ 10005
2. = 1 ÷ (1000x1000x1000x1000x1000)
3. = 1 ÷ 1000000000000000
4. = 1 x (1 ÷ 1000000000000000)
5. = 1 x 0.000000000000001
6. = 0.000000000000001
7. i.e. 1 MB is equal to 0.000000000000001 ZB.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Megabytes to Zettabytes using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Megabyte ?
A Megabyte (MB) is a decimal unit of digital information that is equal to 1,000,000 bytes (or 8,000,000 bits) and commonly used to express the size of a file or the amount of memory used by a program. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of mebibyte (MiB) is used instead.
arrow_downward
#### What is Zettabyte ?
A Zettabyte (ZB) is a decimal unit of measurement for digital information storage. It is equal to 1,000,000,000,000,000,000,000 (one sextillion) bytes. It is commonly used to measure the storage capacity of large data centers, computer hard drives, flash drives, and other digital storage devices.
## Excel Formula to convert from Megabyte (MB) to Zettabyte (ZB)
Apply the formula as shown below to convert from 1 Megabyte (MB) to Zettabyte (ZB).
A B C
1 Megabyte (MB) Zettabyte (ZB)
2 1 =A2 * 0.000000000000001
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Megabyte (MB) to Zettabyte (ZB) Conversion
You can use below code to convert any value in Megabyte (MB) to Megabyte (MB) in Python.
megabytes = int(input("Enter Megabytes: "))
zettabytes = megabytes / (1000*1000*1000*1000*1000)
print("{} Megabytes = {} Zettabytes".format(megabytes,zettabytes))
The first line of code will prompt the user to enter the Megabyte (MB) as an input. The value of Zettabyte (ZB) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Zettabytes(ZB) are there in a Megabyte(MB)?expand_more
There are 0.000000000000001 Zettabytes in a Megabyte.
#### What is the formula to convert Megabyte(MB) to Zettabyte(ZB)?expand_more
Use the formula ZB = MB / 10005 to convert Megabyte to Zettabyte.
#### How many Megabytes(MB) are there in a Zettabyte(ZB)?expand_more
There are 1000000000000000 Megabytes in a Zettabyte.
#### What is the formula to convert Zettabyte(ZB) to Megabyte(MB)?expand_more
Use the formula MB = ZB x 10005 to convert Zettabyte to Megabyte.
#### Which is bigger, Zettabyte(ZB) or Megabyte(MB)?expand_more
Zettabyte is bigger than Megabyte. One Zettabyte contains 1000000000000000 Megabytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,546 | 5,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-10 | latest | en | 0.710104 |
https://www.physicsforums.com/threads/charging-a-capacitor.813032/ | 1,519,560,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00054.warc.gz | 913,381,074 | 15,840 | # Charging a capacitor
1. May 9, 2015
### henry3369
I need some clarification on batteries and capacitors.
Before a battery is connected to a capacitor, the capacitor is uncharged. When a battery is connected, an electron from one conductor moves toward the positive terminal of the battery, leaving this conductor positively charged. What causes the other conductor to become negatively charged? At this point, the positive terminal loses charge because it has an additional electron. The other conductor is still neutral at this point and something has to cause an electron from the negative terminal to move to the neutral conductor.
2. May 9, 2015
### Simon Bridge
The battery moves electrons from one side of the capacitor to the other...
... electrons get added to it.
... of the battery? The + terminal loses electrons, they get transferred to the - terminal.
The neg terminal is negatively charged, so it repels electrons.
Like charges on a conductor - which is the neg terminal, the wire, and the capacitor plate, all together - try to get as far away from each other as possible.
3. May 9, 2015
### henry3369
Doesn't something prevent the electrons from moving from the positive terminal to the negative terminal? If not, how would the two terminals have opposite charges to begin with?
4. May 9, 2015
### Simon Bridge
The battery uses a chemical energy store to move charges from the positive side to the negative side.
The exact mechanism varies from battery to battery
Think of the electrons being like water and the battery is a pump moving water from a low reservoir to a high one. What prevents the water from flowing back down?
5. May 9, 2015
### OmegaKV
A battery is not a gun that can just shoot electrons. It is important that you understand that for every single electron that comes out of the negative terminal of the battery, an electron must go into the positive terminal at the same time (at least for an ideal battery). If you were to look at a sketch the electric field lines for this circuit, you would see why. The positive end of the battery pulls electrons just as much as the negative end pushes electrons. Remember, there are always electrons at all parts of the circuit, since everything is made of atoms which contain both protons and electrons.
When a battery is connected to a cap, electrons are added to one plate, making that plate negative, and removed from the other plate, making that plate positive. These happen concurrently.
6. May 10, 2015
### davenn
yes, and to follow on from that, to the OP, the voltage source doesnt charge a capacitor, it energises a capacitor
The net charge in a capacitor is always zero
Dave | 575 | 2,685 | {"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-2018-09 | longest | en | 0.944745 |
https://wikivisually.com/wiki/Reflection_coefficient | 1,566,302,074,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315329.55/warc/CC-MAIN-20190820113425-20190820135425-00001.warc.gz | 688,723,201 | 40,696 | # Reflection coefficient
In physics and electrical engineering the reflection coefficient is a parameter that describes how much of an electromagnetic wave is reflected by an impedance discontinuity in the transmission medium. It is equal to the ratio of the amplitude of the reflected wave to the incident wave, with each expressed as phasors. For example, it is used in optics to calculate the amount of light that is reflected from a surface with a different index of refraction, such as a glass surface, or in an electrical transmission line to calculate how much of the electromagnetic wave is reflected by an impedance; the reflection coefficient is closely related to the transmission coefficient. The reflectance of a system is also sometimes called a "reflection coefficient".
A wave experiences partial transmittance and partial reflectance when the medium through which it travels suddenly changes. The reflection coefficient determines the ratio of the reflected wave amplitude to the incident wave amplitude.
Different specialties have different applications for the term.
## Telecommunications
In telecommunications, the reflection coefficient is the ratio of the complex amplitude of the reflected wave to that of the incident wave. In particular, at a discontinuity in a transmission line, it is the complex ratio of the electric field strength of the reflected wave (${\displaystyle E^{-}}$) to that of the incident wave (${\displaystyle E^{+}}$). This is typically represented with a ${\displaystyle \Gamma }$ (capital gamma) and can be written as:
${\displaystyle \Gamma ={\frac {E^{-}}{E^{+}}}}$
The reflection coefficient may also be established using other field or circuit quantities.
The reflection coefficient of a load is determined by its impedance ${\displaystyle Z_{L}\,}$(load impedance) and the impedance toward the source ${\displaystyle Z_{S}\,}$(source impedance).
Simple circuit configuration showing measurement location of reflection coefficient.
${\displaystyle \Gamma ={Z_{L}-Z_{S} \over Z_{L}+Z_{S}}}$
Notice that a negative reflection coefficient means that the reflected wave receives a 180°, or ${\displaystyle \pi }$, phase shift.
The magnitude (designated by vertical bars) of the reflection coefficient can be calculated from the standing wave ratio, ${\displaystyle SWR}$:
${\displaystyle |\Gamma |={SWR-1 \over SWR+1}}$
The reflection coefficient is displayed graphically using a Smith chart.
## Seismology
Reflection coefficient is used in feeder testing for reliability of medium.
## Optics and microwaves
In optics and electromagnetics in general, "reflection coefficient" can refer to either the amplitude reflection coefficient described here, or the reflectance, depending on context. Typically, the reflectance is represented by a capital R, while the amplitude reflection coefficient is represented by a lower-case r.These related concepts are covered by Fresnel equations in classical optics.
## Acoustics
Acousticians use reflection coefficients to understand the effect of different materials on their acoustic environments. | 619 | 3,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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-2019-35 | longest | en | 0.881716 |
https://www.fundsindia.com/blog/mf-research/demystifying-home-loan-emis-what-they-dont-tell-you/27813 | 1,721,392,651,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00557.warc.gz | 654,117,333 | 20,469 | # Demystifying Home Loan EMIs: What They Don’t Tell You
Do you have a home loan outstanding or are you planning to take a home loan?
If ‘Yes’, do you really understand the nuances of how a home loan EMI works?
Let’s find out.
Try and answer these 3 questions.
Assume you have taken a home loan of Rs 50 lakhs at 8.50% interest for a tenure of 20 years with an EMI of Rs 43,391.
Question 1: In the first 5 years you would have paid a total EMI of ~Rs 26 lakhs (read as more than 50% of your original loan amount). How much of your principal loan amount have you repaid?
Option A – 30% to 40%
Option B – 20% to 30%
Option C – less than 15%
Question 2: For the 20 year home loan, how long does it take to repay 50% of the loan amount (principal)?
Option A – 10 years
Option B – 12 years
Option C – 14 years
Question 3: For the loan of Rs 50 lakhs, what is the total EMI amount that you pay over 20 years?
Option A – Rs 70 lakhs to Rs 80 lakhs
Option B – Rs 80 lakhs to Rs 90 lakhs
Option C – more than Rs 1 cr
Now let’s check if you got them right!
Question 1: In the first 5 years you would have paid a total EMI of ~Rs 26 lakhs (read as more than 50% of your original loan amount). How much of your principal loan amount have you repaid?
Correct AnswerOption C (less than 15%) – its actually 12%!!
Question 2: For the 20 year home loan, how long does it take to repay 50% of the loan amount (principal)?
Correct AnswerOption C (14 to 15 years)
Question 3: For Rs 50 lakhs home loan, what is the total EMI amount that you pay over 20 years?
Correct AnswerOption C (more than 1 cr) – its 1.04 crs
Surprised!
Here is the evidence – the Detailed Home Loan EMI table which shows the 20 year journey
*You can refer to the annexure section of the blog to understand the various columns
### How does a home loan really work: 3 Surprising Insights!
#### INSIGHT 1: During the initial years, most of your EMI goes only for interest payments!
Sample this.
For a Rs 50 lakhs home loan for 20 years at 8.5% interest rate…
• In the first year, out of Rs 5.20 lakhs that you paid as EMI, Rs 4.2 lakhs goes only towards Interest – a massive 81% of your yearly EMI!.
• 5 years later, the total cumulative EMI amount is Rs 26 lakh, out of which Rs 20 lakhs (77% of cumulative EMI) were only interest payments!
Why does this happen?
As seen from the chart below, a large percentage of your EMI in the initial years goes only towards Interest.
Does this hold true for different loan rates?
Yes it does. Historically, in India interest rates have been around 7% to 9%.
Assuming 7-9% home loan rates, around 70%-80% of the EMI that you pay in the first 5 years goes only towards Interest!
#### INSIGHT 2: While you pay off almost HALF of the loan amount as EMIs in the first five years, only 10-15% of the loan is paid off!
During the initial years of the loan tenure the contribution of EMI towards the Principal is low which means the loan amount (principal) repaid is also low.
In the chart below, for the same example of a Rs 50 lakhs home loan for 20 years at 8.5% interest rate, you can see how much of the original loan gets repaid cumulatively after every year.
Here comes the shocker…
In the first 5 years the principal repaid is only 12% despite paying off 50% of the home loan as EMIs!
Let’s check if this holds true for different interest rates (7% to 9%).
As seen above, this holds true across different home loan rates between 7%-9%.
Only 10-15% of the loan gets paid off the first 5 years despite paying off almost half the loan amount (45%-55%) as EMIs.
#### INSIGHT 3: You almost end up paying TWICE the original loan amount as EMIs for a 20-year home loan
While we regularly track the EMIs, Interest and Principal, what we usually overlook is the total amount that we have to pay for the home loan over the entire tenure.
For a Rs 50 lakhs home loan at 8.5% interest, you end up paying Rs 1.04 cr over 20 years – this is almost 2 times the loan amount!
Interest is more than the loan amount i.e. Rs 54 lakhs!
In the table below you can see that even at different home loan rates (7-9%), you still end up paying almost 2 times the original loan amount.
Understanding all the above 3 nuances of how a home loan really works, is important to ensure that you don’t get frustrated in the initial years.
### 3 Ideas to manage your home loan better
#### IDEA 1: Use 5 year cumulative blocks to understand how your home loan EMI is split across Interest and Principal
Assume you have a loan tenure of 20 years. To make it simpler, divide this into 5 year blocks (4 in this case) and summarize the cumulative totals.
This makes it easier and simple to understand the proportion of EMI that goes towards Principal vs Interest.
To calculate this, you can use the home loan EMI calculator here
#### IDEA 2: Try to prepay in early years and increase your EMI every year in line with your salary increase
Since in the early years of loan tenure the majority of EMI goes towards interest, it is better to prepay some of your home loan in the initial years of the loan tenure which will help reduce the total amount paid (over the tenure for the loan) and shorten the loan tenure. Home loan prepayments simply mean you pay a certain portion of your loan amount earlier than the planned repayment period.
This can be done in two ways
2. Prepay whenever you receive any lumpsum amount or bonus
How much of a difference does it really make?
• If you prepay 1 extra EMI every year, then your total EMI payments (over the loan tenure) reduce by almost 20% of the original loan amount.
• If you prepay 1 extra EMI and also increase this by 5% every year, then your total EMI payments reduce by almost 25% of the original loan amount.
• This gets even better if you are able to prepay more/increase the EMI.
In the table below we have compared the Rs 50 lakhs home loan assuming no prepayment, with prepayment and with yearly increase in prepayment.
#### IDEA 3: If home loan rates go up, don’t forget to increase EMI or Prepay to keep tenure constant
While taking a home loan we usually keep the prevailing home loan rate in mind and do not plan for situations like an increase in home loan rates. When interest rates go up, while your EMI remains the same, the banks increase the tenure of your loan.
So, whenever your home loan rates increase, don’t forget to increase your EMI or prepay – to keep your loan tenure the same.
#### Summing it up
1. Understand these 3 nuances of a home loan EMI
• During the initial years, most of your EMI goes only for interest payments
• While you pay off almost HALF of the loan amount as EMIs in the first five years, only 10-15% of the loan is paid off
• You almost end up paying TWICE the original loan amount as EMIs for a 20-year home loan
1. Use these 3 ideas to manage your home loan better
• Use 5 year cumulative blocks to simplify and understand how your home loan EMI is split across Interest and Principal
• Try to prepay in early years and increase your EMI every year in line with your salary increase
• If home loan rates go up, don’t forget to increase EMI or Prepay to keep tenure constant
Annexure:
Home Loan EMI table Glossary:
Opening Balance = loan outstanding at the start of the year
EMI paid yearly = yearly EMIs paid (monthly EMI * 12)
Interest paid yearly = from the yearly EMI, the amount that goes towards interest
Principal paid yearly = from the yearly EMI, the amount that goes towards principal
% of interest and % of principal = the proportion of EMI that goes towards interest and principal
Closing Balance = loan outstanding at the end of the year
EMI paid cumulative = total EMI paid till date
Interest paid yearly = total interest paid till date
Principal paid yearly = total principal paid till date
% of Principal paid = total principal paid till date as a proportion of loan outstanding
Cumulative EMI paid as % of total loan = total amount paid till date as a proportion of loan outstanding
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This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2,021 | 8,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.95358 |
https://www.physicsforums.com/threads/two-simple-magnetism-questions.703239/ | 1,526,920,501,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00210.warc.gz | 800,346,590 | 17,003 | # Homework Help: Two Simple Magnetism Questions
1. Jul 28, 2013
### Sunturday
1. The problem statement, all variables and given/known data
2. Relevant equations
F=ILB
3. The attempt at a solution
F1 - F2 = FT = 5*10^-5
Second one no clue.
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2. Jul 28, 2013
### Sunturday
And why is the current in this one clockwise as viewed from the top?
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3. Jul 28, 2013
### siddharth23
I think the answer to qn 4 is the current in the loop will flow in such a direction that the magnetic field produced by it will oppose the magnetic field of the magnet (law of conservation of energy). Hence, clockwise.
4. Jul 28, 2013
### siddharth23
For the first one, the current will flow in such a direction that the magnetic field produced by that current will oppose the motion of the loop to the right by applying a force to the left
F=ILB
5. Jul 28, 2013
### Sunturday
But inside the loop the magnetic field would be the other way.
6. Jul 28, 2013
### siddharth23
Yup. Basically it tries to stay in it's original state of rest and oppose change. | 388 | 1,277 | {"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} | 3.34375 | 3 | CC-MAIN-2018-22 | latest | en | 0.88583 |
http://betterlesson.com/lesson/reflection/5126/student-responses | 1,487,822,255,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171070.80/warc/CC-MAIN-20170219104611-00449-ip-10-171-10-108.ec2.internal.warc.gz | 28,764,935 | 20,665 | Reflection: Checks for Understanding Plus Ten Minus Ten - Section 4: Closing
I received a wide range of responses when I asked students what happens when they add 10 to a number. Several students said something like, "when I add 10 I just change the tens place." I also had several who said they just go up one on the number grid. Two children just wrote a number sentence like 38 + 10 = 48. Others mentioned counting up 10.
When I look at the different responses, I see that students are not all in the same place with their understanding, but each one has some sort of strategy for approaching the adding of 10 to a number. Some are still relying on counting while others are moving to a more mental math response to adding 10 to a number. There are still children who are counting up who need continued support with place value understanding.
Student Responses
Checks for Understanding: Student Responses
Plus Ten Minus Ten
Unit 2: Adding and Subtracting the Basics
Lesson 11 of 18
Big Idea: To help students develop fluency skills with adding and subtracting numbers within 100. A variety of strategies need to be presented to encourage students to think mathematically.
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Environment: Rural | 410 | 1,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-09 | latest | en | 0.942026 |
https://www.iieta.org/journals/mmep/paper/10.18280/mmep.070409 | 1,722,815,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00530.warc.gz | 642,197,750 | 24,430 | # Numerical Study of Fractional Mathieu Differential Equation Using Radial Basis Functions
Numerical Study of Fractional Mathieu Differential Equation Using Radial Basis Functions
Hojjat Ghorbani Yaghoub MahmoudiFarhad Dastmalchi Saei
Department of Mathematics, Tabriz Branch, Islamic Azad University, Tabriz, Iran
Corresponding Author Email:
mahmoudi@iaut.ac.ir
Page:
568-576
|
DOI:
https://doi.org/10.18280/mmep.070409
2 September 2020
|
Revised:
22 November 2020
|
Accepted:
3 December 2020
|
Available online:
18 December 2020
| Citation
OPEN ACCESS
Abstract:
In this paper, we introduce a method based on Radial Basis Functions (RBFs) for the numerical approximation of Mathieu differential equation with two fractional derivatives in the Caputo sense. For this, we suggest a numerical integration method for approximating the improper integrals with a singularity point at the right end of the integration domain, which appear in the fractional computations. We study numerically the affects of characteristic parameters and damping factor on the behavior of solution for fractional Mathieu differential equation. Some examples are presented to illustrate applicability and accuracy of the proposed method. The fractional derivatives order and the parameters of the Mathieu equation are changed to study the convergency of the numerical solutions.
Keywords:
radial basis functions, fractional Caputo derivative, Mathieu differential equation
1. Introduction
Differential equations explain mathematically most of physical phenomena such as electromagnetic, elastic wave equations, frequency modulation, parametric oscillators, mirror trap for neutral particles, motion of a quantum particles in a periodic potential [1], and a lot more. Mathieu differential equation
${{D}^{2}}y(t)+[a-2qcos(2t)]y(t)=0, y(0)=\underline{y}, y'(0)=\overline{y}$ (1)
is a famous differential equation which appears to describe such phenomena [2]. In Eq. (1) $a$ is the characteristic number and q is the characteristic parameter of the equation. The French mathematician Mile Lonard Mathieu introduced this equation for the first time and lately this equation was named in his honor the Mathieu equation [3]. The equation was formulated to describe the vibration modes of an elliptical membrane [2], but it has been applied to quadrupole ion traps theory in chemistry [4, 5], ultra cold atoms [6], models of quantum rotor [7], inverted pendulum, vibrations in an elliptic drum, stability of a floating body and Paul trap for charged particles [8, 9].
A simple harmonic oscillator obtains in Eq. (1) when we put q=0, It is well known that this oscillator performs free vibrations around the stable equilibrium position y=0. For $q\ne 0$ the Mathieu Eq. (1) may have a stable solution (when the motion is bounded) or unstable solution (when the motion is unbounded). The occurrence of one of these two outcomes depends on the combination of the parameters q and $a$. When presented graphically, this gives the so-called stability chart with regions of stability and regions of instability separated by the so-called transition curves, enabling one to clearly determine the resulting behavior and the stability property mentioned. This was extensively studied by Rand et al. [10] for classical damped Mathieu equation and fractional Mathieu equation with one fractional derivative.
The asymptotic solutions and transition curves for the generalized form of the non-homogeneous Mathieu differential equation are investigated in the paper [11]. An efficient numerical scheme is proposed for obtaining the stability charts for Mathieu equation by authors [12]. The existence of periodic and quasiperiodic solutions for generalized Van der Pol-Mathieu differential equation is proved using the averaging method in the paper [13]. Kovacic et al. provided a systematic overview of the methods to determine the corresponding stability chart of the classical Mathieu’s equation [14].
Mathieu equation is especially important to the theory of Josephson junctions, where it is equivalent to Schrödinger’s equation. Recently Wilkinson et al. [15] collected various approximations which appear throughout the physics and mathematics literature and examined their accuracy and regimes of applicability.
The fractional calculus is very popular and applicable tool to describe different phenomena these days. The memory and hereditary properties of various materials and processes were considered by the models based on fractional derivatives, where as in integer order models, such types of aspects are not considered. These leads fractional calculus to become an increasingly important topic in the literature of applied mathematics, engineering and other fields. For detailed definitions and theorems on fractional calculus and applications, the readers are referred to the Ref. [16].
Adding the term $^{C}{{D}^{\alpha }}y(t)$ in (1), which denotes the Caputo fractional derivative of function y of order $\alpha$, we obtain the fractional Mathieu differential equation as follows [10]:
${{D}^{2}}y(t)+\kappa {{ }^{C}}{{D}^{\alpha }}y(t)+[a-2qcos(2t)]y(t)=0, 0<\alpha \le 1,$ (2)
where, $\kappa$ is a constant. The general fractional form of Mathieu Eq. (2) is as follows:
$^{C}{{D}^{\beta }}y(t)+{{\kappa }^{C}}{{D}^{\alpha }}y(t)+[a-2qcos(2t)]y(t)=f(t),$ (3)
where,$0<\alpha \le 1$ and $1<\beta \le 2$, are fractional derivatives. Setting $\alpha =1$ and $\beta =2$ in (3), leads to the familiar damped Mathieu equation.
Fractional Mathieu equation is solved numerically by many scientists. For example, authors [10] have used the harmonic balance method. The Adomian decomposition and the series method were used in the paper [17] for fractional Mathieu equation with damped term. Najafi et al. have used the generalized differential transform method [18]. The dynamics of Mathieu equation with two kinds of van der Pol (VDP) fractional-order terms is investigated by Wen et al. [19]. Recently the Block-Pulse wavelets approximation method [20] was used by Pirmohabbati et al. for numerical solution of fractional Mathieu equation.
The Radial Basis Functions (RBF) method was introduced by Hardy [21] in 1971. At first, it was popular in multivariate interpolation [22]. In 1990, Kansa introduced a new method to use RBFs for solving parabolic, hyperbolic, and elliptic partial differential equations [23]. After that, radial basis functions have been widely applied in different fields of computational science.
In this paper, we use Radial Basis Functions (RBFs) to solve numerically the general fractional Mathieu differential Eq. (3). The results of this paper are compatible with those in the papers [10, 20] and other literatures. The rest of this paper is organized as follows. In Section 2, we review some basic definitions of fractional calculus and Radial Basis Functions. We also introduce a composite Simpson's numerical scheme for the improper integrals with a singularity at the right endpoint of the integration domain. In Section 3, we present the new numerical algorithm for solving fractional Mathieu equation. In Section 4, some examples are presented to illustrate the numerical method. Finally, we give conclusions in Section 5.
2. Preliminaries
2.1 Fractional calculus
In this part a brief description of Riemann-Liouville and Caputo fractional integrals and derivatives are presented.
Definition 2.1: 1 The Riemann-Liouville fractional integration of order $\alpha (\alpha \in {{R}^{+}})$ is expressed by [16],
$\left\{ \begin{array}{*{35}{l}} _{0}I_{x}^{\alpha }f(x)=\frac{1}{\Gamma (\alpha )}\int_{0}^{x}{}{{(x-\tau )}^{\alpha -1}}f(\tau )d\tau ,\quad \quad & \alpha >0, \\ {} & {} \\ _{0}I_{x}^{0}f(x)=f(x), & \alpha =0, \\\end{array} \right.$ (4)
where, $\Gamma (\alpha )=\int_{0}^{\infty }{}{{x}^{\alpha -1}}{{\text{e}}^{-x}}\ dx$ is the well-known gamma function. This to end for simplicity we omit 0 and x in the fractional integration notation.
The operator ${{I}^{\alpha }}$ satisfies the following relations,
${{I}^{\alpha }}{{I}^{\gamma }}f(x)={{I}^{\alpha +\gamma }}f(x),$
${{I}^{\alpha }}{{I}^{\gamma }}f(x)={{I}^{\gamma }}{{I}^{\alpha }}f(x),$
$_{0}I_{x}^{\alpha }{{x}^{b}}=\frac{\Gamma (b+1)}{\Gamma (b+1+\alpha )}{{x}^{b+\alpha }}.$
Definition 2.2: 2 The Riemann-Liouville fractional derivative of order $\alpha$ of f(x) is defined as [16],
$_{0}^{RL}D_{x}^{\alpha }f(x)=\frac{1}{\Gamma (n-\alpha )}\frac{{{d}^{n}}}{d{{x}^{n}}}\int_{0}^{x}{}\frac{f(\tau )}{{{(x-\tau )}^{\alpha -n+1}}}d\tau ,$ (5)
where, $n-1<\alpha \le n$, $n\in N$.
Definition 2.3: 3 The Caputo fractional derivative so that $n-1<\alpha <n$ are expressed as [16],
$_{0}^{C}D_{x}^{\alpha }f(x)=\frac{1}{\Gamma (n-\alpha )}\int_{0}^{x}{}\frac{{{f}^{(n)}}(\tau )}{{{(x-\tau )}^{\alpha -n+1}}}d\tau .$ (6)
The main properties of the operator $_{0}^{C}D_{x}^{\alpha }$ are [16],
$_{0}^{C}D{{_{x}^{\alpha }}_{0}}I_{x}^{\alpha }f(x)=f(x),$ (7)
${ }_{0} I_{x}^{\alpha}{ }_{0}^{C} D_{x}^{\alpha} f(x)=f(x)-\sum_{j=0}^{n-1} f^{(j)}\left(0^{+}\right) \frac{x^{j}}{j !}$, (8)
$_{0}^{C}D_{x}^{\alpha }{{x}^{j}}=\left\{ \begin{array}{*{35}{l}} 0 & j\in {{N}_{0}}\ and\ j<n, \\ {} & {} \\ \frac{\Gamma (j+1)}{\Gamma (j+1-\alpha )}{{x}^{j-\alpha }} & j\notin {{N}_{0}}\ or\ j\ge n, \\\end{array} \right.$ (9)
where, ${{N}_{0}}=\{0,1,2,...\}$. The Caputo fractional derivative (6) is determined by Riemann-Liouville fractional derivative (5) as follows [16],
$_{0}^{C}D_{x}^{\alpha }f(x)=_{0}^{RL}D_{x}^{\alpha }f(x)-\sum\limits_{i=0}^{n-1}{}\frac{{{f}^{(i)}}(0)}{\Gamma (i+1-\alpha )}{{x}^{i-\alpha }},$ (10)
Accordingly, if ${{f}^{(i)}}(0)=0,\ i=0,1,...,n-1,$ then $_{0}^{C}D_{x}^{\alpha }f(x)$ and $_{0}^{RL}D_{x}^{\alpha }f(x)$ are equivalent.
2.2 Review of interpolation by RBFs
In this subsection, we briefly introduce the Radial Basis Functions. We state the relevant basic definitions and refer to the literatures (for example [24, 25]) for more details.
A function $K:{{R}^{d}}\times {{R}^{d}}\to R$ is symmetric, if$K(x,y)=K(y,x)$ holds for all $x, y\in {{R}^{d}}$. For scattered nodes ${{x}_{1}},\ldots ,{{x}_{n}}\in {{R}^{d}}$, the translates ${{K}_{j}}(x)=K({{x}_{j}},x)$ are called the trial functions.
$K(x, y)=\phi(r), r=\|x-y\|, x, y \in R^{d}$, (11)
where, $\phi$ :[0,\infty )\to R,$is a scalar function. The function$\varphi $is called a radial basis function. One can scale the kernels on Rd by a positive factor c which is called shape parameter. The accuracy of numerical solution and the condition number of collocation matrix are affected by the shape parameter c. There are some numerical methods for finding an appropriate parameter c to control the accuracy of the solution as well as conditioning of the collocation matrix. By importing the shape parameter c, the new scaled kernel is given by:${{K}_{c}}(x,y)=K(\frac{x}{c},\frac{y}{c}), \forall x,y\in {{R}^{d}}.$(12) The scaled radial kernels on Rd is defined as${{K}_{c}}(x,y)=\varphi (\frac{r}{c}), r=\left\| x-y \right\|, x,y\in {{R}^{d}}.$(13) Table 1 represents the most commonly used global RBFs$\varphi (r)$, where$n, \beta $are RBF parameters and c is shape parameter. Table 1. Global RBFs Name$\varphi (r)$Condition Gaussian(GS)${{e}^{-{{(cr)}^{2}}}}$Multi quadric(MQ)${{(1+{{(cr)}^{2}})}^{\beta /2}}\beta \ne 0, \beta \notin 2N$Inverse quadric(IQ)${{(1+{{(cr)}^{2}})}^{-1}}$Powers${{r}^{\beta }}0\le \beta \notin 2N$Thin-plate splines${{r}^{2n}}ln(r)n\in N$Yoon [24] showed that applying RBFs in Sobolov space has exponential convergence. Fornberg et al. [25] proved the spectral convergence of the method in the limit of flat RBFs. Madych has proved in the paper [26] that under certain conditions the interpolation error is$\varepsilon =O({{\lambda }^{c/h}})$where h is the mesh size, and$0<\lambda <1$is a constant. He presented that, there are two ways to improve the approximation, by reducing the size of h and by increasing the size of c. It means that if$c\to \infty $then$\varepsilon \to 0$. While reducing h leads to the heavy computations, increasing c is without the extra computational cost, but it was proven in the paper [26] that as the error becomes smaller, the matrix becomes more ill-conditioned; hence the solution will break down as$c$becomes too large. So, if the ill-conditioned system could be solved, h could be increased to obtain the best approximation. Selecting the appropriate c affects the accuracy and convergence properties of RBFs. Optimal choice of c is still an open problem and lots of papers have been published in this area. We refer the readers to the papers [26-30] for more details. In this study, we select$c$through our numerical observation such that the stability of the proposed method is achieved. 2.3 Composite rule for singular integrals In this subsection a composite Simpson's numerical scheme is provided for integrals with a singularity at the right endpoint of the integration domain [31]. The improper integral with a singularity at the right endpoint,$\int_{a}^{b}{}\frac{dt}{{{(b-t)}^{q}}},$(14) converges if and only if$0<q<1$, and in this case, one defines,$\int_{a}^{b}{}\frac{dt}{{{(b-t)}^{q}}}=\frac{{{(b-a)}^{1-q}}}{1-q}.$(15) Let g is continuous on$[a,b]$, then the improper integral$\int_{a}^{b}{}\frac{g(t)}{{{(b-t)}^{q}}}dt,$(16) exists, where$0<q<1$. This integral can be approximated with Composite Simpson’s rule. For this we assume that$g\in {{C}^{5}}[a,b]$. In this case, the fourth order Taylor polynomial,${{P}_{4}}(t)$, for g about b is constructed as:${{P}_{4}}(t)=g(b)-{g}'(b)(b-t)+\frac{{{g}'}'(b)}{2!}{{(b-t)}^{2}}-\frac{{{g}^{(3)}}(b)}{3!}{{(b-t)}^{3}}+\frac{{{g}^{(4)}}(b)}{4!}{{(b-t)}^{4}}.$(17) Then we can write,$\int_{a}^{b}{}\frac{g(t)}{{{(b-t)}^{q}}}dt=\int_{a}^{b}{}\frac{g(t)-{{P}_{4}}(t)}{{{(b-t)}^{q}}}dt+\int_{a}^{b}{}\frac{{{P}_{4}}(t)}{{{(b-t)}^{q}}}dt.$(18) P4(x) is a polynomial, then the first integral in (18) can be exactly determined by:$\begin{array}{*{35}{l}} \int_{a}^{b}{}\frac{{{P}_{4}}(t)}{{{(b-t)}^{q}}}dt & =\sum\limits_{k=0}^{4}{}{{(-1)}^{k}}\int_{a}^{b}{}\frac{{{g}^{(k)}}(b)}{k!}{{(b-t)}^{k-q}}dt \\ {} & =\sum\limits_{k=0}^{4}{}{{(-1)}^{k}}\frac{{{g}^{(k)}}(b)}{k!(k+1-q)}{{(b-a)}^{k+1-q}}. \\\end{array}$(19) When the Taylor polynomial P4(t) is too close to g(x) throughout the interval$[a,b]$, (19) is the dominant part of the approximation. We must add to this value the following approximation,$\int_{a}^{b}{}\frac{g(t)-{{P}_{4}}(t)}{{{(b-t)}^{q}}}dt.$To this, we first define the function G(t) as follows$G(t)=\left\{ \begin{matrix} \frac{g(t)-{{P}_{4}}(t)}{{{(b-t)}^{q}}} & a\le t<b \\ 0 & t=b \\\end{matrix} \right.$G(t) is a continuous function on$[a,b]$. Because,$0<q<1$and$P_{4}^{(k)}(b)={{g}^{(k)}}(b)$for$k=0,1,2,3,4$, so$G\in {{C}^{4}}[a,b]$. Therefore, we can use the Composite Simpson’s rule to approximate the integral of G on$[a,b]$. Adding this to the value in Eq. (19) gives an approximation to the improper integral (18), with the accuracy of the Composite Simpson’s rule approximation (which is of order$O({{h}^{4}})$for any given step length (h). The function$SINGSIMP(g,a,b,q,\varepsilon )$approximates (16) with the accuracy of$\varepsilon $(see Appendix 1). 3. Method of Solution To solve the general Mathieu Eq. (3) with the initial conditions$y(0)=\underline{y}, {y}'(0)=\overline{y}$on the interval$[0,T]$, we set:$y(t)=\sum\limits_{j=1}^{N}{}{{y}_{j}} {{\varphi }_{j}}(t)={{\Phi }^{T}}(t)Y,$(20) where,$\begin{align} & \Phi (t)={{\left[ {{\varphi }_{1}}(t),{{\varphi }_{2}}(t),\ldots ,{{\varphi }_{N}}(t) \right]}^{T}}, \\ & Y=[{{y}_{1}},{{y}_{2}},\ldots ,{{y}_{N}}{{]}^{T}}, \\ \end{align}$(21) are RBF basis vector and coefficients vector respectively and${{\varphi }_{j}}(t)=\varphi (|t-{{t}_{j}}|), j=1,2,\ldots ,N$are RBF basis functions and${{t}_{1}},{{t}_{2}},\ldots ,{{t}_{N}}$are arbitrary (not necessary equally spaced) points of$[0,T]$. Differentiating (20) of order$\alpha $and β in the fractional Caputo scene, we get$_{0}^{C}D_{t}^{\alpha }y(t)=\sum\limits_{j=1}^{N}{}{{y}_{j}} _{0}^{C}D_{t}^{\alpha }{{\varphi }_{j}}(t)=\Phi _{\alpha }^{T}(t)Y,$(22)$_{0}^{C}D_{t}^{\beta }y(t)=\sum\limits_{j=1}^{N}{}{{y}_{j}} _{0}^{C}D_{t}^{\beta }{{\varphi }_{j}}(t)=\Phi _{\beta }^{T}(t)Y,$(23) where,${{\Phi }_{\alpha }}(t)={{\left[ _{0}^{C}D_{t}^{\alpha }{{\varphi }_{1}}(t),_{0}^{C}D_{t}^{\alpha }{{\varphi }_{2}}(t),\ldots ,_{0}^{C}D_{t}^{\alpha }{{\varphi }_{N}}(t) \right]}^{T}},$and${{\Phi }_{\beta }}(t)={{\left[ _{0}^{C}D_{t}^{\beta }{{\varphi }_{1}}(t),_{0}^{C}D_{t}^{\beta }{{\varphi }_{2}}(t),\ldots ,_{0}^{C}D_{t}^{\beta }{{\varphi }_{N}}(t) \right]}^{T}}.$Substituting (20), (22) and (23) in (3) we get:$\left( \Phi _{\beta }^{T}(t)+\kappa \ \Phi _{\alpha }^{T}(t)+(a-2qcos2t)\,{{\Phi }^{T}}(t) \right)\,Y=f(t).$(24) Now we choose$N-2$distinct collocation points${{x}_{1}},{{x}_{2}},\ldots ,{{x}_{N-2}}\in [0,T]$and collocate (24) as follows:$\left( \Phi _{\beta }^{T}({{x}_{i}})+\kappa \,\Phi _{\alpha }^{T}({{x}_{i}})+(a-2qcos2{{x}_{i}}){{\Phi }^{T}}({{x}_{i}}) \right)Y=f({{x}_{i}}), i=1,2,\ldots ,N-2.$(25) Finally, we summarize (25) as the following matrix form:$\left( A+\kappa \,B+C \right)\,Y=F,$(26) where,$\begin{matrix} {{A}_{ij}}=_{0}^{C}D_{t}^{\beta }{{\varphi }_{j}}(t{{)|}_{t={{x}_{i}}}}, \\ {{B}_{ij}}=_{0}^{C}D_{t}^{\alpha }{{\varphi }_{j}}(t{{)|}_{t={{x}_{i}}}}, \\ {{C}_{ij}}=(a-2qcos2{{x}_{i}}){{\varphi }_{j}}({{x}_{i}}), \\ {{F}_{i}}=f({{x}_{i}}), \\\end{matrix}$(27) for$i=1,2,\ldots ,N-2$,$j=1,2,\ldots ,N$. (26) is a system of$(N-2)\times N$linear equations. Together with two initial conditions,$y(0)=\underline{y}$(28)${y}'(0)=\overline{y}$(29) we get a system of N linear equations of N unknown coefficients${{y}_{1}},{{y}_{2}},\ldots ,{{y}_{N}}$. In (29)${{\Phi }_{1}}(t)$denotes the first derivative of vector function$\Phi (t).$Now we write the final linear equation as follows:$M\,Y=Z,$(30) where,$\begin{align} & {{M}_{ij}}={{A}_{ij}}+\kappa \,{{B}_{ij}}+{{C}_{ij}},\quad i=1,2,...,N-2\text{ , } \\ & {{M}_{N-1\,j}}={{\varphi }_{j}}(0),\quad \quad \quad \ j=1,2,...N,\quad \\ & {{M}_{Nj}}=\varphi {{'}_{j}}(0), \\ \end{align}$(31) and$\begin{align} & {{Z}_{i}}={{F}_{i}},\quad \quad i=1,2,...,N-2\text{ , } \\ & {{Z}_{N-1}}=\underline{y},\quad \\ & {{Z}_{N}}=\overline{y}. \\ \end{align}$(32) Here we must emphasize that the matrix components${{A}_{ij}}$and${{B}_{ij}}$are as follows$\begin{align} & {{A}_{ij}}=_{0}^{C}D_{t}^{\beta }{{\varphi }_{j}}(t{{)|}_{t={{x}_{i}}}} \\ & \quad \ =\frac{1}{\Gamma (2-\beta )}\int_{0}^{{{x}_{i}}}{}{{({{x}_{i}}-\tau )}^{1-\beta }}{{{{{\varphi }'}'}}_{j}}(\tau )d\tau . \\ \end{align}$(33)$\begin{align} & {{B}_{ij}}=_{0}^{C}D_{t}^{\alpha }{{\varphi }_{j}}(t{{)|}_{t={{x}_{i}}}} \\ & \quad \,\,\,=\frac{1}{\Gamma (1-\alpha )}\int_{0}^{{{x}_{i}}}{}{{({{x}_{i}}-\tau )}^{-\alpha }}{{{{\varphi }'}}_{j}}(\tau )d\tau . \\ \end{align}$(34) Since$0<\alpha <1$and$1<\beta <2$then both improper integrals (33) and (34) are convergent and can be approximated by composite Simpson's method as mentioned in section 2. The algorithm of this method is given as follows: Algorithm 1. 1. choose$\ \phi (t),\ T,N,\varepsilon $2. set$h=T/(N-1)$3. set${{t}_{j}}=(j-1)h,\ \ j=1,2,...,N$4. set${{\phi }_{j}}(t)=\phi (|t-{{t}_{j}}|),\ j=1,2,...,N$5. set${{x}_{k}}=(k-1/2)h,\ k=1,2,...,N-1$6. for$i=1:N-2$for$j=1:NA_{ij}^{{}}=SINGSIMP(\phi '{{'}_{j}}(t),0,{{x}_{i}},\beta -1,\varepsilon )/\Gamma (2-\beta )B_{ij}^{{}}=SINGSIMP(\phi {{'}_{j}}(t),0,{{x}_{i}},\alpha ,\varepsilon )/\Gamma (1-\alpha ){{C}_{ij}}=(a-2qcos2{{x}_{i}}){{\varphi }_{j}}({{x}_{i}}){{M}_{ij}}={{A}_{ij}}+\kappa \,{{B}_{ij}}+{{C}_{ij}}$end${{Z}_{i}}=f({{x}_{i}})$end 7. for$j=1:N{{M}_{N-1\,j}}={{\varphi }_{j}}(0){{M}_{Nj}}=\varphi {{'}_{j}}(0)$end 8.${{Z}_{N-1}}=\underline{y}\quad $9. solve MY=Z 10. set$y(t)=\sum\limits_{j=1}^{N}{}{{y}_{j}} {{\varphi }_{j}}(t)$4. Illustrations In this section, in order to test the numerical method, we present some illustrative examples. We use different values of$\alpha , \beta , \kappa , q, a$and f and plot the achieved numerical solutions and compare them with each other and the results obtained by Pirmohabbati et al. [20]. When$0<\alpha <1$and$1<\beta <2$, the Mathieu equation is of fractional order. In this case there is no known analytic solution for Mathieu equation so the results are compared with the case that the orders are integer. By numerical experiments we choose the shape parameter c=h, where c is RBF parameter and hare the mesh size. The computations are performed by Maple 16 with 40 digits of computation. The Condition number of coefficient matrix are often in the domain${{[10}^{10}}{{,10}^{20}}]$. We study (3) in three different cases. Case A: In (3), we set$\kappa =1$,$a=1$and q=0, with initial values$y(0)=1$and${y}'(0)=-1$and$f(t)={{e}^{-t}}$where the exact solution is$y(t)={{e}^{-t}}$. Table 2 presents the numerical results for different RBFs with$T=1,\ \beta =2$,$\alpha =1$and$c=h$. As we see in Table 2, the error norms highly affected by increasing N. In Table 3 we fixed h=0.1 and changed c=kh. The results for different RBFs show that L2 norm decreases as c increases. The numerical solution with$\beta =2$,$\alpha =1$and the exact solution$y(t)={{e}^{-t}}$are plotted in Figure 1. The graph of absolute error with$\beta =2$,$\alpha =1$is represented in Figure 2. Table 2. The${{L}_{\infty }}$and L2 norm for different values of N RBFs N${{L}_{\infty }}$L2 GS 5 1.2367e-2 2.6761e-3 10 3.1366e-8 3.1028e-9 20 1.7318e-14 2.4091e-16 40 5.4396e-21 2.7160e-23 MQ 5 4.7986e-3 1.0433e-3 10 8.0846e-9 7.9907e-10 20 1.1441e-20 4.7918e-22 40 4.7918e-25 4.3043e-27 IQ 5 1.0182e-2 2.1988e-3 10 1.5795e-8 1.5603e-9 20 1.6640e-21 7.2423e-23 40 4.2156e-26 2.2128e-28 Table 3. The affects of c=kh on L2for h=0.1 $k$GS MQ IQ 8 2.8127e-6 9.1934e-5 8.5498e-4 4 2.1802e-7 1.1431e-6 6.1748e-6 2 2.1937e-8 1.8897e-8 5.9586e-8 1 3.1028e-9 7.9907e-10 1.5603e-9 1/2 6.5205e-10 1.1637e-10 1.5880e-10 1/4 2.0685e-10 4.9421e-11 5.5103e-11 1/8 9.5615e-11 3.7367e-11 3.8537e-11 1/16 5.9532e-11 3.4397e-11 3.4898e-11. ## 1.png Figure 1. The graph of exact and approximate solution ($\beta =2, \alpha =1$) with$N=20$for Case A ## 2.png Figure 2. The graph of absolute error ($\beta =2, \alpha =1$) with$N=20$for Case A Now we keep$\beta =2$fixed and change$\alpha =1$to$\alpha =0.75$and$\alpha =0.5$. Then we keep$\alpha =1$and change$\beta =2$to$\beta =1.9$and$\beta =1.75$respectively, which the results are plotted in Figures 3 and 4. The results show that the numerical solution converges in all the cases when$\beta $and$\alpha $are integer or rational numbers. To show the convergency of the approximate solution we plotted the exact and numerical solution with$\beta =2$and$\alpha =1$on the interval [0,50] in Figure 5. ## 3.png Figure 3. The graph of approximate solution for$\beta =2$and$\alpha =1, 0.75, 0.5$with$N=20$for Case A. ## 4.png Figure 4. The graph of approximate solution for$\beta =2, 1.9, 1.75$and$\alpha =1$with$N=20$for Case A. ## 5.png Figure 5. The graph of exact solution$y(t)={{e}^{-t}}$and approximate solution with$N=200$for Case A. Case B: In (3), suppose$f(t)=0$,$\kappa =0$,$a=1$and$q=0.1$, with initial values$y(0)=0$and${y}'(0)=0.5$. The numerical solution with$\alpha =1$,$\beta =2$is plotted as follows as Figure 6. In fact, in this case we solve the Mathieu equation without damping term ($\kappa =0$). It is clear from Figure 6 that the solution diverges in this case. Our numerical solution is compatible with the results of the paper [20]. ## 6.png Figure 6. The graph of approximate solution for$\beta =2$,$\alpha =1$and$N=200$for Case B. ## 7.png Figure 7. The graph of approximate solution for$\beta =2, 1.97, 1.95$and$\alpha =1$for Case B. ## 8.png Figure 8. The graph of approximate solution for$\beta =1.9, 1.8, 1.7$and$\alpha =1$for Case B. For the next attempt we keep$\alpha =1$and the other coefficients fixed and then change$\beta $from 2 to 1.97 and 1.95 (see Figure 7). The numerical results show that the solution is divergent. It is obvious from Figure 7 that when the derivatives occur with fractional order, the solution diverges faster and the divergency rate decreases as$\beta $decreases. Now we continue with$\alpha =1$and other coefficients fixed and change$\beta $from 2 to 1.9, 1.8 and 1.7. The results which were plotted in Figure 8 show that for these values of$\beta $the behavior of numerical solution changes to convergency. Then we conclude that in the absence of damping term the solution of the Mathieu equations diverges for large values of$\beta $($\beta $near to 2) and converges for smaller values of$\beta $. Case C: In (3), suppose$f(t)=0$,$\kappa =0.15$$a=1 and q=0.1, with initial values y(0)=0 and {y}'(0)=0.5. The numerical solution with \alpha =1, \beta =2 is plotted in Figure 9. The numerical results of our method, which are compatible with those on the paper [20], show the convergency of the solution in this case. ## 9.png Figure 9. The graph of approximate solution (\beta =2,\ \alpha =1) for Case C. ## 10.png Figure 10. The graph of approximate solution for \beta =2 and \alpha =0.9, 0.8 for Case C. ## 11.png Figure 11. The graph of approximate solution for \beta =2, 1.97, 1.95 and \alpha =1 for Case C. ## 12.png Figure 12. The graph of approximate solution with fractional order of derivatives for Case C. We keep \beta =2 and fix the other coefficients and change \alpha from 1 to 0.9and 0.8. The results plotted in Figure 10 show that the solution converges faster than the case that \alpha =1. If we keep \alpha =1 fixed and change \beta from 2 to 1.97 and then 1.95, the plots of results in Figure 11 show the convergency of the results which decays with decreasing of \alpha . For the last attempt we change \beta =2 and \alpha =1 to \beta =1.95 and \alpha =0.8 first and then to \beta =1.9 and \alpha =0.7 which the graph is plotted in Figure 12. 5. Conclusion In this work, the fractional Mathieu differential Eq. (3), including fractional order damping factor, was solved by using radial basis functions. Three classes of radial basis functions, the Gaussian (GS), multi quadric (MQ) and inverse quadric (IQ), have been used. Numerical results in all three classes converge to the solution of the equation. But the MQ and IQ classes are a little more accurate than GS. As N the number of radial basis functions increases, the approximation error decreases rapidly in all three cases, but the condition number of the required linear system increases rapidly. This requires that double accuracy of computation to be used. By changing the characteristic parameters of the equation, the resulting numerical solutions in both stability and instability states converge to the exact solution of the Mathieu equation, which are compatible with the analytical and numerical results of other studies. Appendix The function SINGSIMP(g, a, b, q, ε) approximates the singular integral, I=\int_{a}^{b}{}\frac{g(t)}{{{(b-t)}^{q}}}dt, with absolute error less than or equal to \varepsilon , where 0<q<1 and g is a smooth function. function\quad SINGSIMP(g,a,b,q,\varepsilon ) {{P}_{4}}(t)=\sum\limits_{k=0}^{4}{\frac{{{g}^{(k)}}(b)}{k!}{{(b-t)}^{k}}}, {{I}_{1}}=\sum\limits_{k=0}^{4}{}{{(-1)}^{k}}\frac{{{g}^{(k)}}(b)}{k!(k+1-q)}{{(b-a)}^{k+1-q}} G(t)=\left\{ \begin{matrix}\frac{g(t)-{{P}_{4}}(t)}{{{(b-t)}^{q}}} & a\le t<b \\ 0 & t=b \\\end{matrix} \right. {{M}_{4}}=\underset{a\le t\le b}{\mathop{\max }}\,\left| {{G}^{(4)}}(t) \right|, {{h}_{1}}={{\left( \frac{180\varepsilon }{(b-a){{M}_{4}}} \right)}^{1/4}}, N=2\left[ \frac{b-a}{2{{h}_{1}}} \right]+2, {{h}_{2}}:=\frac{b-a}{N}, {{x}_{k}}=a+k{{h}_{2}},\quad \quad k=0,1,2,...,N$${{I}_{2}}=\frac{{{h}_{2}}}{3}\left( G({{x}_{0}})+4\sum\limits_{k=0}^{N/2-1}{G({{x}_{2k+1}})}+2\sum\limits_{k=1}^{N/2-2}{G({{x}_{2k}})}+G({{x}_{N}}) \right),return\quad I={{I}_{1}}+{{I}_{2}},end.\$
References
[1] Slater, J. (1952). A soluble problem in energy bands. Phys. Rev., 87(5): 807.
[2] Mathieu, E. (1868). Memoire sur le movement vibratoire dune membrance de forme elliptique. Journal de Mathematiques Pures Appliqués, 13: 137-203.
[3] Duhem, P. (1892). Emile Mathieu, his life and works. Bull. Amer. Math., Soc., 1(7): 156-168. https://projecteuclid.org/euclid.bams/1183407338
[4] March, R. (1997). An introduction to quadrupole ion trap mass spectrometry. Journal of Mass Spectrometry, 1(7): 351-369. arXiv:1211.0050.
[5] Baranov, V. (2003). Analytical approach for description of ion motion in quadrupole mass spectrometer. Journal of the American Society for Mass Spectrometry, 14(8): 818-824. https://doi.org/10.1016/S1044-0305(03)00325-8
[6] Rey, A., Pupillo, G., Clark, C., Williams, C. (2005). Ultracold atoms confined in an optical lattice plus parabolic potential: A closed form approach. Phys. Rev., 72(3): 033616. https://doi.org/10.1103/PhysRevA.72.033616
[7] Ayub, M., Naseer, K., Ali, M., Saif, F. (2009). Atom optics quantum pendulum. J. Russ. Laser Res., 30(3): 205-223. https://doi.org/10.1007/s10946-009-9078-x
[8] Nwamba, J. (2013). Delayed Mathieu equation with fractional order damping an approximate analytical solution. International Journal of Mechanics and Applications, 3(4): 70-75. https://doi.org/10.5923/j.mechanics.20130304.02
[9] Stoker, J. (1950). Nonlinear Vibrations in Mechanical and Electrical Systems. Interscience Publishers, New York.
[10] Rand, R., Sah, S., Suchorsky, M. (2010). Fractional mathieu equation. Communications in Nonlinear Science and Numerical Simulation, 15(11): 3254-3262. https://doi.org/10.1016/j.cnsns.2009.12.009
[11] Younesian, D., Esmailzadeh, E., Sedaghati, R. (2007). Asymptotic solutions and stability analysis for generalized non-homogeneous Mathieu equation. Communications in Nonlinear Science and Numerical Simulation, 12(1): 58-71. https://doi.org/10.1016/j.cnsns.2006.01.005
[12] Bobryk, R.V., Chrzeszczyk, A. (2009). Stability regions for Mathieu equation with imperfect periodicity. Physics Letters A, 373(39): 3532-3535. https://doi.org/10.1016/j.physleta.2009.07.069
[13] Kalas, J., Kaderabek, K. (2014). Periodic solutions of a generalized Van der Pol-Mathieu differential equation. Applied Mathematics and Computation, 234: 192-202. https://doi.org/10.1016/j.amc.2014.01.161
[14] Kovacic, I., Rand, R., Sah, S.M. (2018). Mathieu’s equation and its generalizations: Overview of stability charts and their features. Applied Mechanics Reviews, 70(2): 020802. https://doi.org/10.1115/1.4039144
[15] Wilkinson, S.A., Vogt, N., Golubev, D.S., Cole, J.H. (2018). Approximate solutions to Mathieu’s equation. Physica E: Low-dimensional Systems and Nanostructures, 100: 24-30. https://doi.org/10.1016/j.physe.2018.02.019
[16] Podlubny, I. (1990). Fractional Differential Equations. San Diego, Academic Press.
[17] Abdelhalim, E., Elsayed, D.M., Aljoufi, M.D. (2012). Fractional calculus model for damped Mathieu equation: Approximate analytical solution. Applied Mathematical Science, 6(82): 4075-4080.
[18] Najafi, H., Mirshafaei, S., Toroqi, E. (2012). An approximate solution of the Mathieu fractional equation by using the generalized differential transform method (GDTM). Applications and Applied Mathematics, 7(1): 347-384.
[19] Wen, S., Shen, Y., Li, X., Yang, S. (2016). Dynamical analysis of Mathieu equation with two kinds of van der Pol fractional-order terms. International Journal of Non-Linear Mechanics, 84: 130-138. https://doi.org/10.1016/j.ijnonlinmec.2016.05.001
[20] Pirmohabbati, P., Sheikhani, A.R., Najafi, H.S., Ziabari, A.A. (2019). Numerical solution of fractional mathieu equations by using block-pulse wavelets. Journal of Ocean Engineering and Science, 4(4): 299-307. https://doi.org/10.1016/j.joes.2019.05.005
[21] Hardy, R. (1971). Multiquadratic equation of topology and other irregular surface. Journal of Geophysical Research, 76(8): 1905-1915. https://doi.org/10.1029/JB076i008p01905
[22] Buhmann, M. (1990). Multivariate interpolation in odd-dimensional Euclidean spaces using multiquadrics. Constructive Approximation, 6(1): 21-34. https://doi.org/10.1007/BF01891407
[23] Kansa, E. (1990). Multiquadricsa scattered data approximation scheme with applications to computational fluid-dynamics-II solutions to parabolic, hyperbolic and elliptic partial differential equations. Computers & Mathematics with Applications, 19(8-9): 147-161. https://doi.org/10.1016/0898-1221(90)90271-K
[24] Yoon, J. (1999). Spectral approximation orders of radial basis function interpolation on the Sobolov space. SIAM J. Math. Anal., 33(4): 946-958. https://doi.org/10.1137/S0036141000373811
[25] Fornberg, B., Wright, G., Larsson, E. (2004). Some observations regarding interpolants in the limit of flat radial basis functions. Comput. Math. Appl., 47(1): 37-55. https://doi.org/10.1016/S0898-1221(04)90004-1
[26] Madych, W. (1992). Miscellaneous error bounds for multiquadric and related interpolators. Computers & Mathematics with Applications, 24(12): 121-138. https://doi.org/10.1016/0898-1221(92)90175-H
[27] Haq, S., Hussain, M. (2018). Selection of shape parameter in radial basis functions for solution of time-fractional Black-Sholes models. Appl. Math. Comput., 335: 248-263. https://doi.org/10.1016/j.amc.2018.04.045
[28] Mongillo, M. (2011). Choosing basis functions and shape parameters for radial basis function methods. SIAM Undergrad. Res. Online, 190-209.
[29] Fallah, A., Jabbari, E., Babaee, R. (2019). Development of the Kansa method for solving seepage problems using a new algorithm for the shape parameter optimization. Comput. Math. Appl., 77(3): 815-829. https://doi.org/10.1016/j.camwa.2018.10.021
[30] Golbabai, A., Mohebianfar, E., Rabiei, H. (2015). On the new variable shape parameter strategies for radial basis functions. Computational and Applied Mathematics, 34(2): 691-704. https://doi.org/10.1007/s40314-014-0132-0
[31] Burden, R.L., Faires, J.D. (2010). Numerical Analysis. Richard Stratton Ninth Edition. | 11,397 | 34,012 | {"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": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.825943 |
https://gateoverflow.in/312339/made-easy-test-series-algorithm-time-complexity | 1,568,588,032,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00113.warc.gz | 499,965,243 | 19,061 | +1 vote
111 views
Consider a procedure $find()$ which take array of $n$ integers as input, and produce pair of element of array whose difference is not greater than the difference of any other pair of element of that array. Which of the following represent worst case time complexity of $find()$ procedure??
$A)O(n)$ $B)O(nlogn)$ $C)O(n^{2})$ $D)O(n^{2}log n)$
Here we need to sort first and then need to compare adjacent element
right??
Then what will be complexity??
| 111 views
0
yes...Time Complexity will be $O(nlgn + n) =O(nlgn)$
0
@ankitgupta.1729
which algo will be good here??Merge Sort or quick sort??
Can u tell me the algo, from where r u getting T.C. nlogn?? and not $n^{2}$ or $n^{2}log n??$
It is asking for worst case
right??
+2
mam, either apply merge sort or heap sort.. both will give $O(nlgn)$ running time in worst case whereas worst case running time of quick sort is $O(n^2).$
Here, We have to find a pair of elements whose absolute difference is minimum.
Algo :-
1) First sort the elements in O(nlgn) time.
2) scan the whole array and check the absolute difference between each pair of elements. So, comparison should be like this A[0] with A[1] then A[1] with A[2] and so on. We are doing like this because if elements are sorted then only adjacent will give least difference if we scan array from left to right and elements are sorted in non-decreasing order.
This whole process of scanning takes O(n) time to find the least difference pair of elements.
So, Total running time will be O(nlgn +n) =(nlgn)
0
@ankitgupta.1729
but why u have taken minimum among all worst case??
It is not told
rt??
0
After sorting program will be like this
max=0;
for(i=0;i<n;i++)
{
temp=a[i+1]-a[i];
if(temp>max){
max=temp;
}
}
right??
0
0
+1
but why u have taken minimum among all worst case??
I have not taken minimum..Answer can be $O(n^2)$.. I have given one algo which works better than $O(n^2)$..There might be some better algo than $O(nlgn)$ worst case running time also.
After sorting program will be like this
here we are finding min. difference.. first store the difference of first 2 elements in a variable then compare this value by the difference of further consecutive elements of this array. If you get less than that at any time then update the value of that variable..
+1
I have written the program
first store the difference of first 2 elements in a variable then compare this value by the difference of further consecutive elements of this array
I have taken it as max.
But question was O(nlogn)*O(n)
or
O(nlogn)+O(n)??
But here first we sort , then find max. difference , i.e. O(nlogn)+O(n)
right??
0
yes right.. it should be O(nlogn) +O(n) not O(nlogn)*O(n)
1
2 | 760 | 2,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-39 | longest | en | 0.822548 |
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# Right answer, incorrect solution
I'm curious to hear from the folks at Brilliant.org what fraction of submitted solutions are seriously flawed or incorrect. This past week was the first I was invited to submit solutions, and it forced me to notice when I took steps I couldn't fully justify. It would be interesting to see what the general gap is between a solver's intuition and his/her rigor, whether this changes with level, and whether it is different for math vs. physics.
Note by Noah Segal
4 years, 11 months ago
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It heavily depends on the question. For the straight-forward questions, the percentage of correct solution tends to be extremely high, especially since we only solicit solutions from correct numerical answers, and students only submit solutions when they are certain of what they are writing.
However, there are questions in which the 'obvious' step is false, and students apply a simplistic logic. 2 questions come to mind from the Jan 14 set. In Number of Divisors, every solution made the fatal error of assuming that the smallest value of $$N$$ must result from the smallest value of $$A$$, which must result from the smallest value of $$B$$, so on and so forth. However, while this logic is flawed, few students were able to spot the error in their thinking.
The other example is Product of Values of Roots, where students forget that in taking square roots they need to be clear about the positive and negative possibilities, which could result in wrong solutions as in the Proof that -1 = 1. While lots of students could see the fallacy in the blog proof and quickly criticize it, they fail to see the error of their own thinking just because they think they had the solution.
Staff - 4 years, 11 months ago | 681 | 2,619 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-05 | longest | en | 0.903295 |
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# Double Integral and Change of Order of Integration
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I) Evaluate the integral....
ii) Change the order of integration and verify the answer is the same by evaluating the resulting integral.
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1. The integral is
...
#### Solution Summary
A double integral is evaluated and changing the of order of integration is investigated. The solution is detailed and well presented.
\$2.49 | 172 | 850 | {"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-2022-21 | longest | en | 0.879672 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=13&t=16443&p=42798 | 1,585,723,054,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505366.8/warc/CC-MAIN-20200401034127-20200401064127-00374.warc.gz | 560,300,150 | 11,311 | ## Percent Yield
Katie McCombs 3G
Posts: 19
Joined: Wed Sep 21, 2016 2:57 pm
### Percent Yield
Can someone remind me of the formula for percent yield in the event of a limiting reactant?
nikita bhat 2D
Posts: 29
Joined: Wed Sep 21, 2016 2:59 pm
### Re: Percent Yield
When solving for a limiting reactant problem that asks for the percent yield of the product these are the steps to follow:
1. convert the moles/grams of the reactants to the moles/ grams of the product that you have to find the percent yield of. Depending on whether the product amount given is in moles or grams is what you convert to. Compare the the results of the product of the reactants.
2. The reactant with the smallest product is the limiting reactant.
3. percent yield = ( actual yield / theoretical yield ) x 100
Alyssa_Hsu_2K
Posts: 55
Joined: Wed Sep 21, 2016 3:00 pm
### Re: Percent Yield
Percent yield is actual yield/theoretical yield.
Actual yield is determined experimentally and will be given to you in the question.
You can calculate theoretical yield by first determining the limiting reactant, then use its molar ratio to calculate the amount in grams for a final product.
Then, you divide the actual yield(which again, is given to you) by the theoretical yield.
Jessica_Nakahira_1G
Posts: 16
Joined: Wed Sep 21, 2016 2:57 pm
### Re: Percent Yield
Also when providing your answer there are no distinct/measurable units that must be accounted for. Just include the percent sign. | 380 | 1,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-16 | latest | en | 0.882312 |
https://alumniagri.in/task/name-the-different-kind-of-resistance-to-vehicle-motion-3736069 | 1,695,920,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510427.16/warc/CC-MAIN-20230928162907-20230928192907-00864.warc.gz | 100,064,609 | 5,695 | asked by nazirhaq6856, 10 days ago
# Name the different kind of resistance to vehicle motion
1
Resistance to motion
This is the resistance a vehicle faces while attempting to move from a stall condition or while accelerating. This resistance must be overcome by the powerplant of the engine in order to sustain motion. When the power produced is smaller than the resistance to motion, the vehicle will gradually slow down. We must have experienced the slowing down of bicycles if we stop pedaling. The bicycle also slows down if we go uphill or if wind blows from front. A poorly inflated tire also causes the vehicle to groan more and slow down. These are the resistances that force the vehicle to slow down under their effect.
Broadly the resistances can be categorized into the following categories:
Aerodynamic drag
Rolling resistance
Inertia
All the above produce a restraining force working against the tractive force. The tractive force must be greater than or equal to the resistive forces in order to maintain a sustainable motion. We can balance them as
F = F req = FA + FG + FR + FI
0
## Vehicle Motion
Explanation:
1. When a vehicle is moving, at that point .There are three protections following up on a vehicle moving.
2. A vehicle proceeding onward plane street encounters just air opposition and moving obstruction though the one proceeding onward a slant or inclination encounters all the three protections.
3. This is the opposition a vehicle faces while endeavoring to move from a slow down condition or while quickening.
4. These are the protections that power the vehicle to back off under their impact. Extensively the protections can be ordered into the accompanying classes Aerodynamic drag.
5. The powers following up on the toy vehicle as it moves down are gravity, grinding and the ordinary power.
6. The segment of gravity along the course of the slant is more prominent than any frictional powers and gives a speeding up down the slope.
7. Four Fundamental Forces of Nature are Gravitational power, Weak Nuclear power, Electromagnetic power and Strong Nuclear power.
8. The powerless and solid powers are compelling just over a short range and overwhelm just at the degree of subatomic particles.
9. Gravity and Electromagnetic power have boundless range.
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History, 1 month ago | 528 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-40 | latest | en | 0.928265 |
https://thesuntrapp.com/blog/how-many-cups-are-in-a-gallon-of-milk/ | 1,696,318,288,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00363.warc.gz | 597,056,344 | 55,237 | # How many cups are in a gallon of milk? | easy measuring tips
If you’ve ever gone to the store and been taken aback by how expensive milk can be, you may have found yourself wondering just how much a gallon of milk really contains. It’s such an essential part of our diets and so many recipes call for it, yet knowing exactly what measurements make up a gallon can feel like quite the mystery! In this blog post, we’ll take a look at the answer to the age-old questions: How many cups are in a gallon of milk? We’ll discuss some useful conversions from gallons to cups, along with providing tips on ways to save money when buying milk. So let’s dive right into it.
## What is a gallon?
If you’ve ever found yourself stumped by a recipe that calls for gallons, fear not– each gallon equals 8 cups! In other words, 1 cup is equivalent to one-eighth of a gallon. For larger servings like soup or cake batter? Don’t worry about doing the math – just double it and remember 4 gallons makes 32 cups while 8 buckets of liquid come out to 64 cuppas deliciousness.
## What is cup?
A cup is a unit of measurement used to measure both liquid and dry ingredients. It can be abbreviated as c, c., or C (capitalized). A cup is equal to 8 fluid ounces (fl oz), 1/16 of a gallon, or about 237 milliliters.
## Difference between cups and gallons
Let’s look a bit more closely at the differences between cups and gallons. A gallon is used to measure volume whereas a cup measures both weight and volume. This means that the same quantity in one may not be the same as in another. For example, 8 ounces of water would be 1 cup if you were measuring by volume but could weigh differently if you were measuring by weight.
## Benefits of knowing how many cups are in a gallon of milk
Knowing how many cups are in a gallon of milk can be incredibly useful when you’re shopping for the right amount to buy. It’s also helpful when it comes to measuring out portions, as knowing exactly how much one cup is can help ensure that you don’t get too much or too little. Not only that, but this knowledge can help you save money. Milk is expensive, so being able to accurately measure out just the right amount can help you avoid overspending on more than you need.
## How many cups are in a gallon of milk?
So, How many cups are in a gallon of milk? The short answer is that 8 cups make up 1 gallon of milk. This means that if you need to buy a gallon of milk for a recipe or if you’re just stocking up on the essentials, all you have to do is measure out eight cups and you’ve got yourself one gallon. It’s that simple.
## What to use to convert cups in a gallon of milk?
If you’re looking for an easier way to convert cups into gallons, then there are plenty of online conversion tools available. These sites can be used to quickly and accurately calculate the number of cups that equal a gallon, as well as other measurements such as teaspoons, tablespoons, quarts, and liters. It’s also helpful to have a kitchen scale on hand to measure out the right amount of liquid as some recipes might call for a specific weight. With the right tools and knowledge, you’ll have no trouble converting cups into gallons.
## How to convert cups in a gallon of milk?
The easiest way to convert cups into gallons (or vice versa) is using a simple online calculator. Enter in the amount of milk you need in either cups or gallons and the calculator will give you a precise measurement of how much liquid there is. It’s important to note, however, that these measurements are approximate since different types of liquid can weigh different amounts. For example, one cup of oil may be slightly heavier than one cup of water.
## Cups in a gallon of milk conversion table
To help you out, here’s a handy cups in a gallon of milk conversion table. This should make figuring out how many cups are in a gallon of milk much easier.
| Cups | Gallons |
|—|—|
| 8 | 1 |
| 16 | 2 |
| 24 | 3 |
| 32 | 4 |
| 40 | 5 |
| 48 | 6 |
| 56 | 7 |
| 64 | 8 |
By using this table, you can quickly and easily determine how many cups are in a gallon of milk. All you have to do is find the corresponding number of cups and you’ll have your answer.
## Notes when converting cups in a gallon of milk
It’s important to note when converting cups into gallons that the measurements are approximate. Different types of liquids may weigh differently, so it’s best to use a kitchen scale or online conversion calculator for precise measurements. Additionally, if you’re looking to save money on milk, try buying in bulk and portioning out the liquid with measuring cups.
## Cups in a gallon of milk conversion tips
When converting cups into gallons, it’s important to use the right tools. A kitchen scale is ideal for measuring out precise amounts of liquid, as well as a calculator or conversion website. Additionally, when buying milk in bulk, try portioning out the liquid with measuring cups and storing it in airtight containers so that you don’t have to worry about wasting any of it. By taking these extra steps, you can save money and make sure that you always have the right amount of milk for any recipe or cup of tea.
## How to apply the cups in a gallon of milk transformation to your life?
The knowledge of how many cups are in a gallon of milk can be applied to your life in a variety of ways. Whether you’re shopping for groceries or just measuring out the right amount for your recipe, this information can come in handy. Additionally, it can help you save money when buying milk as portioning out liquid with measuring cups and storing it in airtight containers means you don’t have to worry about wasting any of it. Finally, being able to accurately measure out your ingredients can help make sure that your recipes turn out just the way you want them.
## Conclusion: how many cups are in a gallon of milk
To wrap it up, 8 cups are equal to 1 gallon of milk. This means that when shopping for groceries, all you have to do is measure out eight cups and you’ve got yourself one gallon. Additionally, there are a variety of online conversion tools available that can help you quickly and accurately calculate how many cups make up a gallon of milk. Finally, using a kitchen scale and measuring cups can help you accurately portion out your ingredients so that recipes turn out just the way you want them. Now that you know how many cups are in a gallon of milk, you have the confidence to shop for groceries or measure out recipes with ease.
Related: how many gallons is 12 cups
## FAQs: cup in a gallon of milk
### Are there 2 cups in a gallon of milk?
Did you know that one cup is equivalent to 8 fluid ounces, or 16 tablespoons? And in the United States, a gallon can be broken down into 128 fluid ounces, or 16 cups.
### Are there 5 cups in a gallon of milk?
Contrary to popular belief, there are not 5 cups in a gallon of milk. In fact, there are 8 cups in a single gallon of milk. This means that 5 cups would actually only amount to 2/3 of a gallon of milk.
### Is 10 cups equal to a gallon of milk?
Did you know that 1 gallon is equivalent to 16 cups? And if you double that to 2 gallons, you’ll have 32 cups. But wait, it gets even better – with 3 gallons, you’ll have a whopping 48 cups! And if you’re really looking to stock up, 4 gallons will give you a staggering 64 cups. Now that’s a lot of cups.
### Are there more than 20 cups in a gallon of milk?
Discover the milk measurement breakdown: 1 gallon = 16 cups. Digging deeper, each cup holds 8 ounces, while a gallon boasts a whopping 128 ounces. Unveil the precise measures of milk with ease.
### Does 16 cups in a gallon of milk?
Did you know that a gallon of milk is equal to 16 cups? To put it simply, there are 8 ounces in one cup and 128 ounces in a gallon. When you do the math, 8 multiplied by 16 equals 128, which is the total number of ounces in a gallon. With each cup holding 8 ounces, it’s no surprise that there are 16 cups in a gallon.
### How many cups are in a gallon of milk in uk?
A gallon is the same measurement in both the UK and the US. As a result, one gallon of milk will also be equal to 16 cups in both countries. The only difference is that while Americans measure their milk using customary units (ounces, quarts, pints), those in the UK use metric measurements (liters, milliliters). But no matter where you are, 8 ounces is equal to 1 cup and 128 ounces is equal to 1 gallon. So no matter whether you’re a US citizen or a UK resident, 16 cups will always make up one gallon of milk.
### How many cups are in a gallon of milk in us?
In the United States, a gallon of milk is equal to 16 cups. This is because 1 cup holds 8 ounces and 1 gallon contains 128 ounces. When you do the math, 8 multiplied by 16 equals 128, which is the total number of ounces in a gallon. With each cup holding 8 ounces, it’s no surprise that there are 16 cups in a gallon. So the next time you’re measuring out milk for a recipe, you know that 16 cups equals one gallon of milk.
### Is it difficult to convert cups in a gallon of milk?
No, it’s not difficult to convert cups into gallons. All you have to do is find the corresponding number of cups and you’ll have your answer. It’s important to note when converting from cups into gallons that measurements are approximate, so it’s best to use a kitchen scale or online conversion calculator for precise measurements.
### Is 1/4 cup equal to 1 gallon of milk?
Quickly calculate the quantity of milk you need with this simple conversion: in the US, one gallon is equivalent to 16 cups. All you have to do is divide the total fluid ounces (fl oz) by 8.
### 1 cup of 500ml equals how many gallons of milk?
One cup of 500ml equals 0.56 gallons of milk, as 1 gallon is equivalent to 128 ounces and 500ml is equal to 17 ounces. To calculate this quickly, divide the total fluid ounces (fl oz) by 8. In this case, you would divide 17 by 8 which gives you 2.125 cups in a gallon or 0.56 gallons in one cup. So if you need 500ml of milk, that would equal 0.56 gallons or 2.125 cups. This information can help you accurately measure out the right amount for a recipe or shopping trip and save you money in the process.
### How many 12oz cups in a gallon of milk?
A gallon of milk contains 16 cups that are 8 ounces each, or 32 cups that are 4 ounces each. Therefore, a gallon of milk will contain 16 12-ounce cups of liquid.
### Is a gallon of milk the same as a cup?
Is a gallon of milk equivalent to a cup? Not quite. A gallon of milk contains 128 ounces of liquid, whereas a cup holds only 8 ounces. So, in other words, one gallon comprises a total of 16 cups.
### Is a 250ml cups equal a gallon of milk?
No, a 250ml cup is not the same as a gallon of milk. In fact, one gallon of milk is equivalent to 3785ml, which is approximately 15 times more than 250ml. | 2,549 | 10,883 | {"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} | 3.953125 | 4 | CC-MAIN-2023-40 | latest | en | 0.960854 |
http://www.wikihow.com/Put-a-Rubik's-Cube-Back-Together | 1,462,478,671,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860127983.53/warc/CC-MAIN-20160428161527-00110-ip-10-239-7-51.ec2.internal.warc.gz | 903,764,367 | 39,889 | Edit Article
# How to Put a Rubik's Cube Back Together
Community Q&A
This is a tutorial on how to re-assemble a Rubik's cube.
## Steps
1. 1
If not already, disassemble the entire cube but do not remove the center pieces, as they are a support to the entire cube..
2. 2
Separate the edge pieces from the corner pieces. Corner pieces Have three coloured sides, edge pieces have two.
3. 3
Choose a colour and then insert all the edge pieces of that colour into their correct position. Do this by simply sliding the edges between the 2 centers of matching colours.
4. 4
Place the corners of your chosen colour into their correct positions, this should just fall into place.
5. 5
You should now have placed one layer of the Rubik's Cube into it's correct position. Now we start on the next layer. There are only 4 edges to place in this step; these are the edges that go directly above the corners you placed in the previous step. They should slot in like the edges in step 3.
6. 6
Now we need to do the top layer. When inserting an edge in the last layer, twist the top of the cube 45 degrees to make it easier.
7. 7
Place the pieces in this order. Edge, the two adjacent corners, the 2 adjacent edges, the last two corners, the last edge.
8. 8
Finished.
## Tips
• Don't worry if you need to be quite forceful with the last layer edges. The best way to do them is to hook the piece under the top center and push into the cube.
• Don't use this as a way of solving the cube. Learn it properly.
## Article Info
Categories: Stub | Rubik's Cube
In other languages:
Español: reensamblar un cubo de Rubik, Italiano: Riassemblare un Cubo di Rubik, Português: Remontar um Cubo Mágico, Русский: снова собрать Кубик Рубика
Thanks to all authors for creating a page that has been read 59,762 times. | 469 | 1,795 | {"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-2016-18 | longest | en | 0.839887 |
https://www.univerkov.com/based-on-the-ecological-pyramid-rule-determine-how-much-plankton/ | 1,716,903,674,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00628.warc.gz | 890,619,286 | 6,287 | # Based on the ecological pyramid rule, determine how much plankton
Based on the ecological pyramid rule, determine how much plankton is needed for one dolphin weighing 400 kg to grow and exist in the Black Sea.
To solve this problem, it is necessary to create a food chain, on the basis of which a solution can be proposed. In the case of drawing up different food chains, the solution will be different.
For example: plankton – herbivorous fish – dolphin.
According to the rule of the ecological pyramid, only 10% of the matter and energy from the previous one goes to each next level (trophic).
Then for the existence of a dolphin weighing 400 kg, you need 400 * 10 = 4000 kg of herbivorous fish, and for their existence 4000 * 10 = 40,000 kg of plankton.
Once again, I draw your attention to the fact that this food chain will require 40,000 kg of plankton.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 260 | 1,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.919074 |
http://en.allexperts.com/q/Geometry-2060/2012/11/math-help-1.htm | 1,487,997,933,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171664.76/warc/CC-MAIN-20170219104611-00420-ip-10-171-10-108.ec2.internal.warc.gz | 83,563,976 | 6,821 | You are here:
# Geometry/Math help
Question
math
"I am having some trouble proving a Triangle is Isosceles. There is a picture of a triangle, with points A-D. The top point is A, bottom Left is B, bottom right is C. A line is in between (perpendicular) A and D. Point D is the bottom middle of the triangle. I need to have a statement and reason.
Attached is the pdf.
My question is that I need another statement/reason for angles. What am I missing???
Thank You!!!"
I could work with your method, but this one is obvious to me.
Since the angles at B and C are equal and angles BDA and CDA are right triangles,
angle BAD and angle DAC are also both equal since the sum of angles in a triangle
is always 180°.
Given the angles are always equal and both share the side DA,
this says that the other sides are equal.
That is, the line BC is cut in half and the line BA has the same length at line CA.
From here, it can be shown that thee two triamgles are relected on each other.
Once this has been done, anything else can be found.
Is this what the problem is suppose to prove?
Geometry
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#### Scott A Wilson
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https://askanydifference.com/sales-tax-calculator/ | 1,726,039,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00882.warc.gz | 99,512,453 | 55,295 | # Sales Tax Calculator
Sharing is caring!
Instructions:
• Enter the names and quantities of items you want to calculate tax for.
• Click "Add Item" to add each item and quantity.
• Enter the tax rate in percentage (%).
• Click "Calculate Tax" to calculate the tax amount for the entered items.
• Click "Clear Results" to reset the items and tax calculation.
• Click "Copy Results" to copy the tax calculation results to the clipboard.
Calculation History
## Introduction to Sales Tax Calculator
### Understanding Sales Tax
Sales tax is a consumption tax imposed by the government on the sale of goods and services. The tax is a percentage of the price of the goods or services sold. As sales tax rates vary by country, state, or even city, the process of calculating the total cost including tax can become complicated, especially for businesses that operate in multiple jurisdictions.
### The Concept of a Sales Tax Calculator
A Sales Tax Calculator is a digital tool designed to automate the calculation of sales tax. Users input the net amount (price before tax) and the sales tax rate; the calculator then computes the total cost (net amount plus sales tax) and, often, the amount of tax itself. This tool is invaluable for both consumers and business owners, simplifying budgeting, accounting, and financial planning processes.
Also Read: Alpha vs Omega: Difference and Comparison
## Formulae for Sales Tax Calculation
### Basic Sales Tax Calculation
The most fundamental formula used in a Sales Tax Calculator is for determining the total cost of a purchase. The formula is:
`Total Cost = Net Amount + (Net Amount * Sales Tax Rate)`
### Reverse Sales Tax Calculation
Sometimes, users may need to determine the net amount and sales tax amount from the total cost. This is known as reverse sales tax calculation, and the formulae are:
`Net Amount = Total Cost / (1 + Sales Tax Rate) Sales Tax Amount = Total Cost - Net Amount`
### Compound Sales Tax Calculation
In some jurisdictions, sales tax is applied on top of another tax (e.g., municipal tax on top of state tax). The compound tax can be calculated as follows:
`Total Tax Rate = 1 + (First Tax Rate + Second Tax Rate + (First Tax Rate * Second Tax Rate)) Total Cost = Net Amount * Total Tax Rate`
## Benefits of Sales Tax Calculator
### Accuracy
Manual calculations, especially when dealing with percentages and large numbers, are prone to errors. A sales tax calculator minimizes these errors, ensuring that the calculations are precise.
### Time Efficiency
Automating the calculation process saves time, particularly for businesses that process numerous transactions daily.
### Financial Planning
Understanding the exact cost including tax helps individuals and businesses in budgeting and financial planning.
### Compliance
For businesses, accurate calculation of sales tax is crucial for tax compliance. Underestimating sales tax can lead to penalties, while overestimating can affect profitability.
## Facts About Sales Tax
### Variability
Sales tax rates can vary significantly, not just between countries but within regions of the same country. For instance, in the United States, some states have no sales tax, while others have rates higher than 10%.
Also Read: P vs P Hat: Difference and Comparison
### History
Sales tax has a long history, with evidence of its use in ancient civilizations. It was a primary form of revenue for states and was used to fund wars, build infrastructure, and cover administrative costs.
In the modern era, sales tax has adapted to include digital goods and services, with many jurisdictions implementing specific digital sales taxes to keep up with the evolving market.
## Conclusion
A Sales Tax Calculator is more than just a simple tool; it is a necessary asset for individuals and businesses alike. Its ability to provide quick, accurate calculations not only saves time but also aids in financial planning and compliance. As the economic landscape becomes more complex and interconnected, the importance of accurate tax calculation tools like the Sales Tax Calculator becomes ever more apparent. Utilizing such tools effectively can lead to significant benefits, ensuring both compliance and financial efficiency.
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#### By Emma Smith
Emma Smith holds an MA degree in English from Irvine Valley College. She has been a Journalist since 2002, writing articles on the English language, Sports, and Law. Read more about me on her bio page. | 950 | 4,798 | {"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} | 3.15625 | 3 | CC-MAIN-2024-38 | latest | en | 0.916429 |
https://newsletters.naavi.com/i/xNvogdJ/partners-in-learning/page/3 | 1,643,460,887,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00680.warc.gz | 463,219,866 | 13,744 | # Numeracy
WHAT DO WE LEARN IN MATHEMATICS?
Year Three
Dear Parents and Carers,
Over the coming weeks I will continue to outline what the teaching and learning Mathematics curriculum encompasses for each year level. In this article I will focus on Year Three. In the classroom children may be working at level, below level or above level. Teachers differentiate the teaching and tasks to cater for the learning each child needs to access.
In level 3, students increasingly use mathematical terms and symbols to describe computations. There are three major areas Number and Algebra, Measurement and Geometry, Statistics and Probability.
Number and Algebra
Number and place value
• Investigate the conditions required for a number to be odd or even and identify odd and even numbers
• Recognise, model, represent and order numbers to at least 10 000
• Apply place value to partition, rearrange and regroup numbers to at least 10 000 to assist calculations and solve problems
• Recognise and explain the connection between addition and subtraction
• Recall addition facts for single-digit numbers and related subtraction facts to develop increasingly efficient mental strategies for computation
• Recall multiplication facts of two, three, five and ten and related division facts
• Represent and solve problems involving multiplication using efficient mental and written strategies and appropriate digital technologies.
Money and financial mathematics
• Represent money values in multiple ways and count the change required for simple
Fractions and Decimals
• Model and represent unit fractions including 1/2, 1/4, 1/3, 1/5 and their multiples to a complete whole.
Patterns and algebra
• Describe, continue, and create number patterns resulting from performing addition or subtraction
• Use a function machine and the inverse machine as a model to apply mathematical rules to numbers or shapes.
Measurement and Geometry
Using units of measurement
• Measure, order and compare objects using familiar metric units of length, area, mass and capacity
• Tell time to the minute and investigate the relationship between units of time.
Shape
• Make models of three-dimensional objects and describe key features.
Location and transformation
• Create and interpret simple grid maps to show position and pathways
• Identify symmetry in the environment
• Identify and describe slides and turns found in the natural and built environment.
Geometric reasoning
• Identify angles as measures of turn and compare angle sizes in everyday situations
Statistics and Probability
Chance
• Conduct chance experiments, identify and describe possible outcomes and recognise variation in results.
Data representation and interpretation
• Identify questions or issues for categorical variables. Identify data sources and plan methods of data collection and recording
• Collect data, organise into categories and create displays using lists, tables, picture graphs and simple column graphs, with and without the use of digital technologies
• Interpret and compare data displays.
Kind regards,
Jennifer O’Connor
Assistant Principal | 579 | 3,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-05 | latest | en | 0.869002 |
http://openstudy.com/updates/4dc6aa68a5918b0b9ea58420 | 1,519,134,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812959.48/warc/CC-MAIN-20180220125836-20180220145836-00762.warc.gz | 279,239,184 | 8,396 | • anonymous
Can someone set these word problems up so that I can solve for the answer. 1. One canned juiced drink is 20% orange juice; another is 10% orange juice. How much many liters of each should be mixed together in order to get 10L that is 18% orange juice? How many liters of the 20% orange juice should be in the mixture? How many liters of the 10% orange juice should be in the mixture?
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 343 | 1,325 | {"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-2018-09 | latest | en | 0.531016 |
https://marknelson.us/posts/1996/04/01/a-floating-point-problem | 1,713,685,021,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00047.warc.gz | 351,976,589 | 5,100 | Probably the scariest thing I have to do at work each day is to walk past Marc Leger’s office. Since Marc’s office is on the way to the bathroom, I have a tough time avoiding it. So several times a day, I have to face the possibility that Marc will accost me at a time when I am really in a hurry to be elsewhere.
Today was not going to be one of those lucky days when I got off the hook. I slipped down the hallway as quietly as I could, but I wasn’t quiet enough. Just as I reached Marc’s office, I heard the familiar demand. “Mark! Mark! Come here, I need help!”
#### Into the breach
With a sigh of sympathy for my bladder, I pulled up a chair behind Marc’s desk. He was running Turbo Debugger, and had a program similar to that shown in fp1.cpp.
Marc was working on a customer problem. A Greenleaf Database Library user was attempting to insert floating point data into a database, and our library was returning an error code indicating a loss of precision. Marc was in his usual state of consternation for two reasons. First, he hadn’t seen this problem occur before. Second, he couldn’t reproduce the problem when using doubles, just floats.
#### The smoking gun
The offending code is fairly simple. It converts a C++ double to ASCII representation for storage in a database. After converting and storing the result, the code wants to see if any loss of precision was caused during the conversion process. For example, if I’m using 2 digits after the decimal point, I should get an error if I try to store 3.141. The trailing 1 would be lopped off by the binary to ASCII conversion done by sprintf(). Marc’s test program from Listing 1 produced the following output:
`````` add_double( 1.12 ) returns success
add_double( 1.125 ) returns failure
add_double( d ) returns success
add_double( f ) returns failure
``````
As expected, when using two digits of precision, 1.12 is added to the database without error. 1.125 fails, because it will be stored in ASCII as 1.12. Using a double that has been assigned the value of 1.12 works properly. But using a float with a value of 1.12 fails!
#### Seeing double
The thing that was really bothering Marc was the display on his debugger. He was at the end of the add_double() routine, and had all three local variables in his watch window. They all looked perfect. The two floating point numbers had values of 1.12, and the character buffer contained a “1.12”. Even so, the comparison “test == d” had just failed, causing the routine to return an error. Was the compiler somehow at fault? Marc was convinced it was a conspiracy on Borland’s part to make his life hell.
Mark’s problem was fairly simple, once you looked at it from the right perspective. Numbers such as 1.12 can’t be represented exactly in a float, or even in a double. Thus, there are rounding and truncation errors when the conversion takes place. The specific problem in fp1.cpp is the loss of precision found when placing 1.12 in a float. Assigning that value to a double later on doesn’t fix the problem, it only perpetuates it.
In the add_double() routine, the actual value passed using a float is just a little bit off from the desired value of 1.12. But printing the value out to a buffer using just two digits of precision manages to throw out the error. Thus, when atof() converts the buffer back to a double, the error is gone. This means that the input value and the test value are going to differ by a small version.
fp2.cpp is a program that demonstrates how this works. d1 is a double that has been initialized with the value of 1.12. d2 has the same value after being initialized using a float. The output of the program is shown below:
``````f = 1.12 1.120000005 dump: 29 5c 8f 3f
d1 = 1.12 1.12 dump: ec 51 b8 1e 85 eb f1 3f
d2 = 1.12 1.120000005 dump: 00 00 00 20 85 eb f1 3f
``````
As you can see, looking at all three values with just 3 digits of precision makes them look identical. However, when you use a greater precision, the representation error of a float rears its ugly head. The binary dump of the second double shows exactly where the problem lies. The lower half of the double contains all zeros, a result of having the smaller float value copied into it. Clearly, when comparing (test == d ) in fp1.cpp, the compiler will notice the difference.
#### A Happy Camper
Marc wasn’t particularly happy with my explanation. First, it didn’t offer an immediate solution to his problem. And second, he resented my glib advice: “Never use floating point numbers if you expect your program to work.”
The best answer for our customer would have been to keep all floating point data in doubles at all times. Unfortunately, this solution was impractical. Instead, we created a short routine for her that simply rounded off doubles to 2 significant digits, throwing out anything beyond that. Life would have been much easier if we could have convinced her to simply ignore the error return!
My advice to you is the same: avoid floating point anytime it isn’t really necessary. If you do find you need to use it, try to avoid mixing doubles and floats. Stick to the preferred C++ floating format of double for all your data and calculations, and maybe you’ll avoid those moments of hysteria that Marc has to constantly deal with. | 1,246 | 5,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.940289 |
http://www.algebra.com/algebra/homework/equations/Equations.faq.question.251244.html | 1,369,376,251,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704234586/warc/CC-MAIN-20130516113714-00028-ip-10-60-113-184.ec2.internal.warc.gz | 311,571,937 | 4,509 | # SOLUTION: What is the answer to this? y=(1.023*-(a))-(1.023*b)
Algebra -> Algebra -> Equations -> SOLUTION: What is the answer to this? y=(1.023*-(a))-(1.023*b) Log On
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Algebra: Equations Solvers Lessons Answers archive Quiz In Depth
Question 251244: What is the answer to this? y=(1.023*-(a))-(1.023*b)Answer by stanbon(57361) (Show Source): You can put this solution on YOUR website!What is the answer to this? y=(1.023*-(a))-(1.023*b) ------------- It cannot be solved. There are too many variables. Cheers, Stan H. | 206 | 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} | 2.96875 | 3 | CC-MAIN-2013-20 | latest | en | 0.843735 |
https://onews.info/how-long-is-a-day-on-uranus-compared-to-earth/ | 1,713,962,212,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00433.warc.gz | 386,483,205 | 15,294 | # How long is a day on Uranus compared to Earth?
## How long is a day on Uranus compared to Earth?
about 17 hours One day on Uranus takes about 17 hours (the time it takes for Uranus to rotate or spin once). And Uranus makes a complete orbit around the Sun (a year in Uranian time) in about 84 Earth years (30,687 Earth days).
## How long is one hour on Uranus?
Uranus spins on its axis about every 17 hours 14 minutes. So its day-night cycle lasts that long. So for much of the planet's spring and fall a large percentage of the planet has day and night about every 17 hours.
## How long is one night on Uranus?
Uranus spins on its axis about every 17 hours, 14 minutes. So its day-night cycle lasts that long. So, for much of the planet's spring and fall, a large percentage of the planet has day and night about every 17 hours.
## How long is one day on each planet?
The Earth is the only planet with an approximately 24-hour day
Planet Length of Day
Venus 243 Earth days
Earth 23 hours, 56 minutes
Mars 24 hours, 37 minutes
Jupiter 9 hours, 55 minutes
•Jul 18, 2019
## Why is a day on Uranus 17 hours?
A planet's day is the time it takes the planet to rotate or spin once on its axis. Uranus rotates faster than Earth so a day on Uranus is shorter than a day on Earth. A day on Uranus is 17.24 Earth hours while a day on Earth is 23.934 hours.
## What would a day on Uranus look like?
What this means is that a day on Uranus is the same as an entire season on Uranus. Even though the planet is rotating on its axis, the Sun will just spiral around in the sky until the planet has gone far enough around the Sun for it to be obscured.
## Which planet has 48 hours a day?
This is also referred to as its rotational period. So, Venus has the longest day of any planet in our solar system.
## What planet has the longest day?
Venus It was already known that Venus has the longest day – the time the planet takes for a single rotation on its axis – of any planet in our solar system, though there were discrepancies among previous estimates.
## Do u age slower in space?
Scientists have recently observed for the first time that, on an epigenetic level, astronauts age more slowly during long-term simulated space travel than they would have if their feet had been planted on Planet Earth.
## What planet has a 30 hour day?
This is also referred to as its rotational period. So, Venus has the longest day of any planet in our solar system. It completes one rotation every 243 Earth days. Its day lasts longer than its orbit.
## Does it rain on Uranus?
Two ice giants in our solar system experience diamond rain. Nope, we are not making this up. On Neptune and Uranus, it rains diamonds! While this is a striking reminder of how ordinary diamonds are in the greater scheme of this universe, it also sheds light on how elusive most terrains remain for humanity.
## How is 1 hour in space 7 years on Earth?
The first planet they land on is close to a supermassive black hole, dubbed Gargantuan, whose gravitational pull causes massive waves on the planet that toss their spacecraft about. Its proximity to the black hole also causes an extreme time dilation, where one hour on the distant planet equals 7 years on Earth.
## Which planet is the longest day?
Venus It was already known that Venus has the longest day – the time the planet takes for a single rotation on its axis – of any planet in our solar system, though there were discrepancies among previous estimates.
## What planet takes 7 years to get to?
Saturn FAQ – Spacecraft
Spacecraft Target Time
Messenger Mercury 6.5 years
Cassini Saturn 7 years
Voyager 1 & 2 Jupiter; Saturn; Uranus; Neptune 13,23 months; 3,4 years; 8.5 years; 12 years
New Horizons Pluto 9.5 years
## What planet has shortest day?
Jupiter Jupiter is the fastest spinning planet in our Solar System, rotating on average once in just under 10 hours. That is very fast, especially considering how large Jupiter is. This means that Jupiter has the shortest day of all the planets in the Solar System.
## Is an hour in space 7 years on Earth?
The first planet they land on is close to a supermassive black hole, dubbed Gargantuan, whose gravitational pull causes massive waves on the planet that toss their spacecraft about. Its proximity to the black hole also causes an extreme time dilation, where one hour on the distant planet equals 7 years on Earth.
## Do astronauts get paid for life?
So, while they may not be paid for life, astronauts do receive many benefits that help offset the risks of their profession. However, astronauts may enjoy many potential financial rewards once they retire.
## How is 1 hour in Space 7 years on Earth?
The first planet they land on is close to a supermassive black hole, dubbed Gargantuan, whose gravitational pull causes massive waves on the planet that toss their spacecraft about. Its proximity to the black hole also causes an extreme time dilation, where one hour on the distant planet equals 7 years on Earth.
## What planet is made of diamonds?
In 2012, Yale University scientists published a study announcing the identification of a planet rich in diamonds. Called 55 Cancri e, the planet is "possibly covered in diamond, rather than water and granite," scientists explained at the time. The exoplanet is twice the size of Earth but has eight times its mass.
## Where does it rain diamonds?
Deep within Neptune and Uranus, it rains diamonds—or so astronomers and physicists have suspected for nearly 40 years. The outer planets of our Solar System are hard to study, however. Only a single space mission, Voyager 2, has flown by to reveal some of their secrets, so diamond rain has remained only a hypothesis.
## Do you age slower in space?
Scientists have recently observed for the first time that, on an epigenetic level, astronauts age more slowly during long-term simulated space travel than they would have if their feet had been planted on Planet Earth.
## Do you age in space?
Flying through outer space has dramatic effects on the body, and people in space experience aging at a faster rate than people on Earth.
## Which planet has shortest day?
Jupiter is the fastest spinning planet in our Solar System, rotating on average once in just under 10 hours. That is very fast, especially considering how large Jupiter is. This means that Jupiter has the shortest day of all the planets in the Solar System.
## What planet is the hottest?
Venus It has a strong greenhouse effect, similar to the one we experience on Earth. Because of this, Venus is the hottest planet in the solar system. The surface of Venus is approximately 465°C!
## How old would I be on Mars if I was 12?
A year on Mars is longer than a year on Earth—almost twice as long at 687 days. This is roughly 1.88 times the length of a year on Earth, so to calculate your age on Mars we simply have to divide your Earth age by 1.88.
## Which planet has a longest day?
Venus It was already known that Venus has the longest day – the time the planet takes for a single rotation on its axis – of any planet in our solar system, though there were discrepancies among previous estimates.
## Do you age less in space?
Training on the simulated martian terrain of Mars-500. Scientists have recently observed for the first time that, on an epigenetic level, astronauts age more slowly during long-term simulated space travel than they would have if their feet had been planted on Planet Earth.
## Do you stop aging in space?
In space, people usually experience environmental stressors like microgravity, cosmic radiation, and social isolation, which can all impact aging. Studies on long-term space travel often measure aging biomarkers such as telomere length and heartbeat rates, not epigenetic aging.
## What happens if an astronaut gets pregnant in space?
For one, without the stress of Earth's gravity, her bones lose density. Studies show that astronauts, for instance, lose 1% to 2% of their bone density for every month spent in space, and that would be especially concerning for giving birth because the pelvis could fracture in the process.
## Can it rain glass?
The planet, known as HD 189733b, is a gas giant with a daytime temperature of 2,000 degrees Fahrenheit where it possibly rains liquid glass sideways amid 4,500 mph winds, NASA says. | 1,865 | 8,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.950089 |
https://www.airmilescalculator.com/distance/mub-to-tld/ | 1,653,331,815,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00007.warc.gz | 698,534,014 | 27,520 | # Distance between Maun (MUB) and Tuli Lodge (TLD)
Flight distance from Maun to Tuli Lodge (Maun Airport – Tuli Lodge Airport) is 398 miles / 641 kilometers / 346 nautical miles. Estimated flight time is 1 hour 15 minutes.
Driving distance from Maun (MUB) to Tuli Lodge (TLD) is 521 miles / 838 kilometers and travel time by car is about 12 hours 1 minutes.
398
Miles
641
Kilometers
346
Nautical miles
1 h 15 min
## How far is Tuli Lodge from Maun?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 398.088 miles
• 640.660 kilometers
• 345.929 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
• 397.814 miles
• 640.219 kilometers
• 345.690 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).
## How long does it take to fly from Maun to Tuli Lodge?
Estimated flight time from Maun Airport to Tuli Lodge Airport is 1 hour 15 minutes.
## What is the time difference between Maun and Tuli Lodge?
There is no time difference between Maun and Tuli Lodge.
## Flight carbon footprint between Maun Airport (MUB) and Tuli Lodge Airport (TLD)
On average flying from Maun to Tuli Lodge generates about 84 kg of CO2 per passenger, 84 kilograms is equal to 184 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Maun to Tuli Lodge
Shortest flight path between Maun Airport (MUB) and Tuli Lodge Airport (TLD).
## Airport information
Origin Maun Airport
City: Maun
Country: Botswana
IATA Code: MUB
ICAO Code: FBMN
Coordinates: 19°58′21″S, 23°25′51″E
Destination Tuli Lodge Airport
City: Tuli Lodge
Country: Botswana
IATA Code: TLD
ICAO Code: FBTL
Coordinates: 22°11′21″S, 29°7′36″E | 546 | 2,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.856239 |
https://mathhelpboards.com/threads/2-2-18-initial-value-differentual-eq-with-complete-the-square.25422/ | 1,620,684,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989749.3/warc/CC-MAIN-20210510204511-20210510234511-00498.warc.gz | 382,013,439 | 15,176 | # [SOLVED]2..2.18 initial value differentual eq with complete the square??
#### karush
##### Well-known member
$\quad\displaystyle y^{\prime}= \frac{e^{-x}-e^x}{3+4y}, \quad y(0)=1$
rewrite
$\frac{dy}{dx}=\frac{e^{-x}-e^x}{3+4y}$
separate
$3+4y \, dy = e^{-x}-e^x \, dx$
integrate
$2y^2+3y=-e^{-x}-e^x+c$
well if so far ok presume complete the square ???
$(a)\quad y=-\frac{3}{4}+\frac{1}{4} +\sqrt{65-8e^x-8e^{-x}}$\\
$(c)\quad|x|<2.0794\textit{ approximately}$
#### skeeter
##### Well-known member
MHB Math Helper
$\quad\displaystyle y^{\prime}= \frac{e^{-x}-e^x}{3+4y}, \quad y(0)=1$
rewrite
$\frac{dy}{dx}=\frac{e^{-x}-e^x}{3+4y}$
separate
$3+4y \, dy = e^{-x}-e^x \, dx$
integrate
$2y^2+3y=-e^{-x}-e^x+c$
well if so far ok presume complete the square ???
$(a)\quad y=-\frac{3}{4}+\frac{1}{4} +\sqrt{65-8e^x-8e^{-x}}$\\
$(c)\quad|x|<2.0794\textit{ approximately}$
$y(0) = 1 \implies 5 = -2+c \implies c = 7$
$y^2 + \dfrac{3}{2}y = \dfrac{-e^{-x} -e^x + 7}{2}$
$y^2 + \dfrac{3}{2}y + \dfrac{9}{16} = \dfrac{-e^{-x} -e^x + 7}{2} + \dfrac{9}{16}$
$\left(y+\dfrac{3}{4}\right)^2 = \dfrac{-8e^{-x}-8e^x+65}{16}$
$y+\dfrac{3}{4} = \dfrac{\sqrt{-8e^{-x}-8e^x+65}}{4}$
check the book "answer" ...
$y = \dfrac{-3 + \sqrt{-8e^{-x}-8e^x+65}}{4}$
#### karush
##### Well-known member
thanks for all the steps
i get lost on the initial value thing
I dont see where the 5 comes from?
Last edited:
#### skeeter
##### Well-known member
MHB Math Helper
thanks for all the steps
i get lost on the initial value thing
I dont see where the 5 comes from?
$2y^2+3y=-e^{-x}-e^x+C$
note $y(0)=1 \implies y = 1 \, \text{when} \, x=0$
$2(1)^2 + 3(1) = -e^{-0}-e^0+C$
... see it now? | 751 | 1,685 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-21 | latest | en | 0.453305 |
https://news.ycombinator.com/item?id=6685951 | 1,550,326,822,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480472.38/warc/CC-MAIN-20190216125709-20190216151709-00432.warc.gz | 642,370,116 | 3,139 | There's a bit of a mischaracterization of the way 2 dimensional Fourier transforms operate on images in this post. The 2D DFT (or DCT rather) doesn't deal with anything as complex as tracing shapes on a 2D plane like seen in the video. What it does is treat each 1 dimensional line of the image as a signal/waveform, with the pixel intensity as amplitude. Fourier-family transforms are separable, so the 2D (and ND) case is equivalent to the transform of each line followed by a transform along the resulting columns. So the actual representation that arises from this looks like so, with each image representing the corresponding cosine component: http://0x09.net/img/dct32.png (here grey is 0)Visually most of the sinusoidal components here are zero or nearly so. However if we scale them logarithmically, we'll see that it's actually not so: http://0x09.net/img/dct32log.pngWhat transform coders like JPEG do is reduce the precision of these components, causing many of them to become zero. Which is good for the entropy coder, and mostly imperceptible to us. Of course JPEG operates on 8x8 blocks only * rather than a whole image like here.It's hard to imagine this as an image, so here's a progressive sum starting from the second term, which essentially demonstrates an inverse DCT: http://0x09.net/img/idct32.pngmind that 0 is adjusted to grey in this rendering, and the brightness of the result is not an artifact of the transform.It's easier to understand what goes on with these transforms if you can visualize things in terms of the basis functions. Which in the case of a 32x32 image like above would be http://0x09.net/img/basis.png (warning: eye strain).All the examples above pertain to the DCT, partly because of JPEG and partly so I could avoid getting phase involved, but the principles apply equally to the other transforms in the family.* although recent versions of libjpeg can use other sizes
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http://www.ebdigest.org/search/shsat_practice_test_admission | 1,550,464,691,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484648.28/warc/CC-MAIN-20190218033722-20190218055722-00187.warc.gz | 330,199,765 | 13,389 | #### Search result: shsat practice test admission
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More | 3,121 | 10,827 | {"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} | 3.15625 | 3 | CC-MAIN-2019-09 | latest | en | 0.677091 |
https://answerprime.com/how-does-log55-4-2/ | 1,713,182,917,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00519.warc.gz | 83,535,353 | 40,648 | # How does Log(55) = 4?
Its on a math clock in the position of the 4pm. I haven’t been able to see how log(55) = 4. When I type it in the calculator I get around 1.7…
Log(55)=log(5*11)=log5+log11=1.74.
Log 55
In Calculus,Log=ln,ln(55)~4.00733
As u said, log(55)=1.7, from a calculator,
bt ln(55)=4, check it if u want in a calculator…
log is of base 10 whereas ln is of base ‘e’….
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Department of Physics Lior Burko University of Alabama in Huntsville Fall semester, 2010 PH 251: Special Relativity Homework Assignment No. 1 Due date: 9/3/2010 All coded problems are from the textbook: 1. Take the solution for the wave equation in the primed reference frame to be f ʹ x , ʹ t ( ) = sin ʹ x c ʹ t ( ) . a) Show that under Galilean transformations x ʹ x x , t ( ) = x ut t ʹ t x , t ( ) = t this solution is transformed to
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Unformatted text preview: f x , t ( ) = sin x − c + u ( ) t [ ] . b) Show that this solution does not satisfy the wave equation in its form in the primed frame, i.e., ˙ ˙ f − c 2 ʹ ʹ f ≠ . c) Show that this solution does satisfy the transformed wave equation ˙ ˙ f − c 2 − u 2 ( ) ʹ ʹ f + 2 u ˙ ʹ f = . R1B.2 R1B.3 R1S.2 R1S.5 R1S.9 R1R.2 R1A.1...
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## This note was uploaded on 07/19/2011 for the course PH 251 taught by Professor Staff during the Fall '10 term at University of Alabama - Huntsville.
Ask a homework question - tutors are online | 397 | 1,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-17 | longest | en | 0.829209 |
https://www.happyzebra.com/timezones-worldclock/difference-between-Suva-and-Santo%20Domingo.php | 1,558,861,661,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259015.92/warc/CC-MAIN-20190526085156-20190526111156-00431.warc.gz | 782,259,518 | 5,179 | # Suva to Santo Domingo time difference
## Suva time vs. Santo Domingo time:
SuvatoSanto Domingo
12:00 am=8:00 am previous day
12:30 am=8:30 am previous day
1:00 am=9:00 am previous day
1:30 am=9:30 am previous day
2:00 am=10:00 am previous day
2:30 am=10:30 am previous day
3:00 am=11:00 am previous day
3:30 am=11:30 am previous day
4:00 am=12:00 pm previous day
4:30 am=12:30 pm previous day
5:00 am=1:00 pm previous day
5:30 am=1:30 pm previous day
6:00 am=2:00 pm previous day
6:30 am=2:30 pm previous day
7:00 am=3:00 pm previous day
7:30 am=3:30 pm previous day
8:00 am=4:00 pm previous day
8:30 am=4:30 pm previous day
9:00 am=5:00 pm previous day
9:30 am=5:30 pm previous day
10:00 am=6:00 pm previous day
10:30 am=6:30 pm previous day
11:00 am=7:00 pm previous day
11:30 am=7:30 pm previous day
12:00 pm=8:00 pm previous day
12:30 pm=8:30 pm previous day
1:00 pm=9:00 pm previous day
1:30 pm=9:30 pm previous day
2:00 pm=10:00 pm previous day
2:30 pm=10:30 pm previous day
3:00 pm=11:00 pm previous day
3:30 pm=11:30 pm previous day
4:00 pm=12:00 am
4:30 pm=12:30 am
5:00 pm=1:00 am
5:30 pm=1:30 am
6:00 pm=2:00 am
6:30 pm=2:30 am
7:00 pm=3:00 am
7:30 pm=3:30 am
8:00 pm=4:00 am
8:30 pm=4:30 am
9:00 pm=5:00 am
9:30 pm=5:30 am
10:00 pm=6:00 am
10:30 pm=6:30 am
11:00 pm=7:00 am
11:30 pm=7:30 am
Santo Domingo ↔ Suva time
Convert time to other cities:
Suva Santo Domingo
## Time difference between Suva and Santo Domingo, Dominican Republic is -16:0 hours
Santo Domingo doesn't observe daylight saving time but Suva does. DST in Suva will start on 03 November 2019 and will end on 19 January 2020. Once DST starts in Suva the time difference between Suva and Santo Domingo will be -17:0 hours.
Suva, Fiji:
Time zone: FJT - Fiji Time (UTC/GMT +12)
DST will start on 03 November 2019
DST will end on 19 January 2020
Time zone difference / Flight distance
Santo Domingo, Dominican Republic:
Time zone: AST - Atlantic Standard Time (UTC/GMT -4)
Time zone difference / Flight distance
FJT to AST
Suva time to AST
FJT to Santo Domingo time | 776 | 2,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-22 | latest | en | 0.61168 |
https://exoplanetmusings.wordpress.com/category/orbital-mechanics/ | 1,550,288,600,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479838.37/warc/CC-MAIN-20190216024809-20190216050809-00240.warc.gz | 559,073,337 | 27,234 | ## Tidal Theory
Tidal Wave (Source)
We’re used to thinking of various astronomical bodies having gravity and gravitationally interacting with each other, and it’s certainly true that they do, and that these interactions can be exploited to detect extrasolar planets through their pull on their parent stars (either by astrometry or Doppler spectroscopy). So far in our study of extrasolar planetary systems, we have treated astronomical bodies as point masses — that is, the entirety of their gravity is assumed to come from a single point, and that the entirety of their matter exists at that point. But this is, of course, an oversimplification. The consideration of astronomical bodies as extended objects introduces a range of new effects that are collectively called “tides.”
We may be familiar with the concept of the tides in the ocean, where the gravitational pull of the Moon on the oceans of Earth causes them to be raised twice a day, and so this is where we will start our consideration of tides. The gravity of Earth’s moon pulls on Earth, causing it to stretch into a more elliptical shape. A diagram illustrating this is shown below.
We see then that the rise over the side of Earth opposite the Moon is caused by the distortion of the shape of Earth. It’s important to understand that this is a very general phenomenon: It is not exclusive to the Earth-moon system, but any two-body system can be represened this way. Furthermore, while it is not shown in the above image for simplicity, the primary body (Earth in the case of the Earth-Moon system, the star in the case of a hot Jupiter system) will also raise a tide on the secondary mass.
The tidal deformation of the primary body is not an instantaneous process. There is inertia to overcome and so if the primary mass is rotating, the tidal bulge will be dragged away from the line connecting the two bodies. A diagram of this is shown below.
In the above image, we assume a prograde orbit for the satellite, and $r$ is a line joining the centre of the two bodies. Note that the rotation of the primary forces its long axis to lead ahead of the orbiting body. For a circular orbit, the long axis of the planet will lead ahead of $r$ by some constant angle, even while the planet rotates. This is why there are two lunar tides each day. Aside from causing boats to rise and descend in the water, relative to fixed points on land, these tidal effects have some important implications.
Firstly, it affects the orbits of the two bodies. In the case of the Earth-Moon system, the oceanic tidal bulge facing the Moon has a stronger gravitational attraction on the Moon than the tidal buge on the opposite hemisphere on Earth, simply because it is closer (and gravitational attraction scales inversely with $r^2$). Since the nearer tidal bulge is pushed ahead of the $r$ line, the net gravitational attraction of the Moon toward Earth is also displaced from $r$. Since the tidal bulge is pushed ahead of the Moon in its orbit, the gravitational attraction of the Moon toward Earth has a small component in the direction of its orbital motion. This causes the Moon to accelerate (it’s orbital velocity increases), which then causes its orbital altitude (the semi-major axis) to increase. Any readers casually aware of orbital mechanics is familiar with this effect. If you’re unfamiliar with orbital mechanics, a full description of it is beyond the scope of this writing, but I would encourage you to try Kerbal Space Program.
Conversely, if the tidal bulge lagged behind the $r$ line, the orbit of the Moon would decay. This can be done either by having the Moon in a retrograde orbit as is the case for Neptune’s moon Triton, or an orbit that is closer to the planet than the geosynchronous orbit radius, as is the case for Mars’ moon Phobos. Triton and Phobos are thus doomed.
Let’s define two factors (given by Dobbs-Dixon et al. 2004, Ferraz-Mello, et al. 2008), $\hat{s}$ and $\hat{p}$, as the strength of the stellar and planetary tides, respectively,
$\displaystyle \hat{s}\equiv\frac{9}{4}\frac{k_0}{Q_0}\frac{m_1}{m_0}R_0^5,\qquad\hat{p}\equiv\frac{9}{2}\frac{k_1}{Q_1}\frac{m_0}{m_1}R_1^5$
Where $k_0$ and $k_1$ is the Love number of the primary and secondary, respectively, $Q_0$ and $Q_1$ is the tidal dissipation factor of the primary and secondary (a quantity that encapsulates how a body responds to tidal deformation, but it is very hard to measure, and usually poorly known), respectively, $m_0$ and $m_1$ as the masses of the two bodies, and $R$ as their radii. The Love number quantifies the mass concentration of the body and will not be expanded on significantly here. The separation between the two bodies will change with a rate of
$\displaystyle \dot{a}=-\frac{4}{3}na^{-4}\hat{s}\left[(1+23e^2)+7e^2(\hat{p}/2\hat{s})\right]$
Where $a$ is the semi-major axis of the orbit, $e$ is the eccentricity of the orbit, and $n$ is the
Secondly, the tidal deformation of a body will be resisted by inertia, causing its rotation to slow (and its synchronous orbit radius to expand outward). This will continue until either the body’s spin period mathes its orbital period (as is the case for the Moon), or is some integer ratio of the orbital period other than 1:1 (as is the case for Mercury), as is often the case for bodies in eccentric orbits.
For fluid bodies (such as gas giant planets or stars) in circular orbits, the end result of tidal spindown will result in a perfectly synchronous rotation, via a process called tidal synchronisation. In eccentric orbits, the planet moves around the star at varying speeds, making a synchronous rotation impossible. For such planets, the end result of tidal spindown is “pseudosynchronisation,” with a planet in a pseudosynchronous rotation such that the rotation rate of the planet matches the angular velocity around the star at periapsis (where the planet moves fastest), but with the rotation of the planet being faster than the angular velocity elsewhere. In rigid bodies (such as the terrestrial planets), tidal synchronisation is achieved in the form of a spin-orbit resonance, such that the long axis of the body is aligned with $r$ at each periapsis. It doesn’t need to be the same hemisphere of the body, as we see in the case of Mercury, which presents alternating hemispheres toward the sun at each periapsis.
Thirdly, tidal deformation of a body in a non-synchronous, eccentric orbit will cause the orbital eccentricity to tend to zero. This process, called “tidal circularisation,” is responsible for the circular orbits we see for the hot Jupiter population. As a body swings around periapsis, the tidal deformation intensifies and inertia resists this motion. In addition to causing heating to the object, it also saps some of its orbital energy, causing it to slow down. As this happens at periapsis more strongly than anywhere else, the net result is that the eccentricity of the orbit tends to zero.
The rate of the change in the eccentricity is given by
$\displaystyle \dot{e}=-\frac{2}{3}nea^{-5}\hat{s}\left[9+7(\hat{p}/2\hat{s})\right]$
Tidal deformations are a significant source of heating of a form that is collectively called “tidal heating.” An extreme example of it is seen at Io, where Jupiter’s other large moons perturb Io, pumping up its eccentricity, which is subsequently damped by tidal interaction with Jupiter, causing heating, and some rather spectacular volcanism.
Understanding the influences of tides on planetary systems is a vital component of understanding their evolution, formation and habitability. Tidal interactions between moons and planets can completely melt those moons, and tidal interaction between planets can tidally heat those planets and potentially completely melt them, too. Tides can also dominate the final rotational behaviour of a planet, with effects for its climate that are still being examined.
## RV Fits: The Newtonian Solution
Isaac Newton
In the last post we looked at modelling the radial velocity of a star as the reflex motion due to an orbiting planet’s gravitational influence. This is the most prominent way used to model extrasolar planets to fit RV data. The Keplerian solution offers a lot in its simplicity, but it has a major shortcoming in that it is not particularly dynamic, and really only approximates any system with more than two bodies.
A planetary system may be thought of as an n-body problem, where you have n gravitating objects all influencing each other. This is different from the Keplerian models that take into account only the star-planet gravitational interaction. This will be a rather rudimentary look at simulating a gravitational n-body model.
First let’s look at vectors. In this case, we’ll consider a velocity vector $\vec{v}$ defined simply with an x and a y component.
$\displaystyle \vec{v} = [v_x, v_y]$
This simply allows us to get some sense of where our velocity is toward in a two-dimensional cartesian plane. We’re used to saying that an object has a velocity of, for example, 5 m s-1, but this alone tells you little useful in the context of modelling a system – 5 m s-1 in what direction? By defining a vector with an x and y component, we can specify the direction that the velocity is in. For example, $\vec{v} = [1, 0]$, then the velocity of the object is to the right, whereas $\vec{v} = [0, -1]$ implies moving down. If $\vec{v} = [1, 1]$, then the object is moving both to the right and up, with the actual total velocity being determined by Pythagorean Theorem.
$\displaystyle v = \sqrt{v_x^2 + v_y^2}$
This is called the magnitude of the velocity vector.
Now each particle will have its own gravitational field. To determine the attraction to a particle from a given coordinate, we can use Newton’s Law of Universal Gravitation.
$\displaystyle \vec{F_g} = -G\frac{m}{r^2}$
Where m is the mass of the particle we’re considering, let’s call it $p_0$, and r is the distance between it and the other particle, let’s call it $p_1$. Since we’re considering in this context only the gravitational attraction of $p_1$ on $p_0$ (not both particles on each other), we need not take into account the mass of both particles at this time.
Once you have determined the gravitational attraction of $p_0$ toward $p_1$, you will need to determine which direction the gravitational attraction is toward. The angle between two points is simply given by
$\displaystyle \theta = \tan^{-1}{\frac{y_0-y_1}{x_0-x_1}}$
The x and y components of the gravitational attraction may be determined with $F_g \cos{\theta}$ and $F_g \sin{\theta}$, respectively. These x and y components may be added to the x and y components of the velocity vector to determine the new $\vec{v}$.
This will have to be done for every particle. You will have to calculate the gravitational influence of all particles on each other particle. For i particles, the new velocity of any given particle j, $\vec{v}'_j$, is given by
$\displaystyle \vec{v}_j' = \vec{v}_j - \sum_{i\neq j}^i G\frac{m_i}{r^2_{ij}}$
Where r is the separaton between the two particles. For most chosen values for $\vec{v}$, the system will have a systemic velocity (corresponding to the γ term in the Keplerian radial velocity equation). If you want to determine $\vec{\gamma}$, you can simply sum up all of the initial velocity vectors and divide it by the total system mass M.
$\displaystyle \vec{\gamma} = M^{-1} \sum_i m_i\vec{v}_i$
N-body simulations are calculation-intensive and are typically run on computers, where gravitational attractions and their modifications of velocity vectors may be calculated over and over for each new position. Particles are typically assigned a mass, a starting coordinate and a starting velocity vector, and a computer will calculate new positions as dictated by Newtonian physics for as long as the user wants.
Let’s look at a specific example. For simplicity, I set the gravitational constant to unity and use no units. I simulate three particles, a star and two planets, with positions of (0,0), (-30, 0), (60, 0), starting velocity vectors of (0.1, 0.1), (0, 0.5), (0, -0.22), and masses of 10, 0.1 and 0.1, respectively. Letting the simulation run for 3,000 time steps and plotting the y-axis component of the star’s velocity vector provides us with the stellar radial velocity.
Simulated RV data for chaotic system
What is going on? The RV behaviour is decidedly non-Keplerian – there are no Keplerian models that would be able to fit this data. If we look at a diagram of the system, it becomes clear what is happening here.
Planet Scattering
Our radial velocity plot is showing us a nightmare in planetary system architectures: planet-planet scattering. The outer planet (whose positions are shown in green) is clearly interacting very strongly with the inner planet (whose positions are marked in red).
Not all dynamically interacting multi-planet systems are so chaotic. Systems often have planets in close but stable orbits where they gravitationally tug on each other, causing the stellar radial velocity profile to deviate from a simple Keplerian model.
While Keplerian models are not capable of fitting to planetary systems with interacting planets, this Newtonian fit can. It is dynamical in that it simulates the influences of the planets on each other, however it can be particularly computer-intensive. Newtonian fits are often run of systems where Keplerian fits suffice simply to probe the dynamical behaviour of a system, such as its long-term stability or possible resonances between the orbits. Knowing whether to use Keplerian or Newtonian models for planets will depend on the planetary system architecture, but it is clear that Newtonian models are powerful tools for predicting the behaviour of planets in multi-planet systems.
## RV Fits: The Keplerian Solution
Johannes Kepler
We looked at deriving the radial velocity equation a little here but for the sake of this discussion, we will recover some of the material and expand on it. Furthermore, while modelling the radial velocity curve was described here, it was done so from the perspective of statistics and data, as opposed to actually modelling the stellar reflex velocity from of the gavitational influence of a planet. In hind sight, maybe it is this article which should have been titled “making waves.” I will seek to clarify the physical modelling of the radial velocity curve itself, and how the orbital orientation of the planet plays into how the Keplerian radial velocity curve is shaped.
The radial velocity equation is heavily reliant on Kepler’s Laws of Planetary Motion which govern the shape and periods of the orbits of planets and stars. As such, any fits to radial velocity data using the radial velocity equation is called a Keplerian fit. A Keplerian fit is therefore a fit to RV data that is effectively a model of the motion of the star as expected from Kepler’s Laws.
If you recall, the radial velocity equation is given by
$\displaystyle V_r(t) = K[\cos(\epsilon+\omega)+e\cos(\omega)]+\gamma+\dot{\gamma}+\ddot{\gamma}$
Where ω is the longitude of perihelion of the orbit, e is the eccentricity of the orbit, $\gamma$ is a term used to represent the radial velocity offset of the system (the radial velocity of the system arising simply from its motion through space), $\dot{\gamma}$ is any linear radial velocity trend in the data, measured in m s-1, and $\ddot{\gamma}$ is a term for a quadratic trend in the RV data, where the RV data may be curved but not in such a way that it permits the period to be readily known. This is typically observed when the aforementioned trend has been observed long enough that it begins to show some curvature in the RV data. If there is no need for any of the $\gamma$ terms, they may be set to zero. Lastly, K is the amplitude of the RV curve and directly corresponds to the mass of the planet through the semi-major axis of the star around the system barycentre, as well as being dependent on the unknown inclination of the system and the eccentricity of the orbit.
$\displaystyle K=\frac{2\pi a_*\sin{i}}{P\sqrt{1-e^2}}$
The nature of $\epsilon$ was not expanded on in the previous aforementioned page. It represents the “eccentric anomaly” and can be understood to be representative of where the planet is in its orbit at a given time. For circular orbits such that e = 0, the eccentric anomaly is simply equal to the mean anomaly, and is understood to be 0 at the start and 2π = 360o after a full orbit. For non-circular orbits such that e > 0, then it must be calculated.
Eccentric Anomaly (source)
The relationship between the mean anomaly and the eccentric anomaly is given by
$\displaystyle M = \epsilon - e \sin{\epsilon}$
It turns out that it’s impossible to solve for $\epsilon$ mathematically (in the same sense that it is impossible to solve for π) so a recursive algorithm will be needed to approximate it. This algorithm is called Newton’s Method and is given as
$\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
An initial guess, x0 is plugged into the algorithm and is used to determine x1, which is plugged back into the algorithm, ad infinitum. This algorithm will converge on the actual value for x regardless of the initial guess, with it returning the exact value at x, though of course after a handfull of iterations, a sufficiently close guess is arrived at to avoid the need for excessively redundant calculations. The initial guess is plugged into the original equation being solved for, divided by the initial guess plugged into the derivative of the original equation. As you might imagine from the orbit diagram above, the mean anomaly itself is a fairly decent guess at the the eccentric anomaly, so one is safe using this for x0. In the case of the expression for the eccentric anomaly, we first set the expression equal to zero: $0 = \epsilon-e\sin{\epsilon}-M$ and then with Newton’s Method
$\displaystyle \epsilon_{n+1} = \epsilon_n - \frac{\epsilon -e\sin{\epsilon}-M}{1-e\cos{\epsilon_n}}$
This may be done until $\epsilon$ is determined to the desired precision.
Graphing the RV equation is rather straightforwad with some computer scripting, though it can be somewhat challenging with a graphing calculating given the difficulty in determining $\epsilon$ to sufficient precision throughout the orbit (it turns out that astronomy uses both math and computers!). Let us examine how the shape of the RV curve varies with the orbital parameters (for a description of the various orbital parameters, see here). Obviously changing the period widens the the wavelength of the curve, and obviously varying K affects the amplitude. Changing the eccentricity will change the shape of the orbit significantly and will therefore affect the y-axis symmetry of the RV curve.
Manifestation of e on the RV curve
The “jagged” appearance of RV curves arising from eccentric planets is distinctive and difficult to replicate through other phenomena such as pulsations. As such, the RV signals from eccentric planets is less ambiguously of planetary origin than for planets in circular orbits.
Changes in the mean anomaly of the orbit simply effect a phase shift and move the RV curve of the planet along the x-axis (time axis). Changing the longitude of perihelion, ω, rotates the planetary orbit about its plane in space and therefore will cause what appears to be a phase offset to the orbit for a circular orbit, but will cause noticeable changes to the RV curve shape for eccentric planets. Changing ω for eccentric orbits causes a loss of x-axis symmetry.
Manifestation of ω on the RV curve
As before, the shape of the RV curve here is hard to replicate through processes intrinsic to the star. This “sawtooth” shape of the RV curve is much more reason to believe the RV curve is due to an orbiting companion rather than other issues such as starspots or pulsations.
Now multiple planets can be modeled by simply adding multiple RV curves over each other. For a system of n planets, the RV curve is given by
$\displaystyle V_r(t) = \gamma + \dot{\gamma} + \ddot{\gamma} + \sum_{i=1}^n K_i[\cos(\epsilon_i + \omega_i)+e_i\cos(\omega_i)]$
Keplerian modelling of RV data is simple, fast, and computationally easy. However it leaves out very important aspects which we will look at later, such as planet-planet interactions. That being said, Keplerian fitting is still the most widely used method to deriving extrasolar planet properties from RV data.
## The Path of the Wanderers
The Heliocentric Model of the Solar System
While the ancients did not understand the motions of the “wandering stars,” Newton and Kepler made great strides toward understanding them.
Assuming the simple case of a single body, A, orbiting another, B, the two bodies will orbit around a common barycentre that represents the centre of mass of the system. Thus while it is acceptable to say that one body orbits another, it is more physically accurate to say that both bodies are in a mutual orbit about each other.
Assume the existence of two objects, A and B, with masses MA and MB, respectively. Let aA be the distance from A to the barycentre of a system, and let aB be the distance from B to the barycentre. The ratio of aA / aB is determined purely by inverse ratio of the two masses. For example if MA = MB, then aA = aB, and if MA = 3MB, then aA = (1/3)MB. Physically, this is because A is more massive than B, and so it has more gravity and is thus B needs to be farther from the centre of mass to balance the system.
In the latter case, the gravitational acceleration of A on B is also greater, causing B to have a higher velocity than A. Thus despite B having a wide orbit, it will complete the orbit at the same time as A, and thus, the orbital period, P, of the two will be equal. The fact that is true for all values of is important for constraining the orbital period of one object if the other is not visible for whatever reason. The orbital period of both bodies may be defined as
$\displaystyle P = 2\pi \sqrt{a^3 / \mu}$
Where a is the semi-major axis, roughly the average distance between A and B (and is defined as half the length of the orbits longest axis, the “major axis”), and μ is defined as μ = GM, the product of the gravitational constant, G, multiplied by the total masses of the two bodies, M. It is necessary to consider the sum of the masses because the gravitational pull of both objects on each other will contribute to the orbital period. In cases where MA > MB, the contribution of B may be considered negligible and the orbital period determined more or less accurately without taking it into account (for example, calculating the orbital period of Earth about the sun may be done reasonably well without taking Earth’s gravitational pull on the sun into account).
Not all orbits are perfect circles. Some may be eccentric. Mathematically, an orbit should never be thought of as a circle, but rather as an ellipse with the barycentre at one focus. However an orbit with an eccentricity of e = 0 is physically indistinguishable from circular.
For an eccentric orbit (any orbit for which e > 0), there exists a point in the orbit where the bodies are closest to each other, this is referred to as the periapsis, ap. The point in the orbit where the two bodies are most separated is the apoapsis aa, though these terms vary depending on which body is orbited. Aphelion and perihelion are used for bodies orbiting the sun, and apastron and periastron are used for bodies orbiting other stars.
If e = 0, the orbit may be considered circular. If 0 < e < 1, then the orbit is eccentric. If e = 1, then the orbit is parabolic, and not bound to the orbited body. If e > 1, then the orbit is hyperbolic and is not bound to the orbited body. In the case of e ≥ 1, there exists a periapsis, but no apoapsis.
Regardless of the eccentricity of an orbit, the orbital period will remain constant for a constant value of a. However the orbital velocity will change in such a way so that a line drawn from the planet to the star will cover a constant amount of area per unit time. As such, a planet will orbit faster at periastron, and slower at apastron.
Conveniently, the eccentricity of a planet around the barycenter will be equal to the eccentricity of the star’s orbit around the barycentre. Note that if a body in a circular orbit accelerates (be it by its own power like a spacecraft, or by a gravitational pull from another body), its orbit will become more eccentric.
Inclination for the planets in our solar system is defined as the angle between the plane of the planet’s orbit and the plane of Earth’s orbit, which defines the “ecliptic.” However this is invalid for other planetary systems for which Earth is not presently orbiting to provide such a reference. As such, we use a different reference plane upon which the sky is. We view the sky as if in the centre of an all-encompassing sphere. When looking at a planetary system from the viewpoint of Earth, we may consider the background sky to be a perpendicular plane, and determine the inclination of an orbit as the angle between that body’s orbital plane and the plane of the sky.
Inclination of a planetary system
Thus an edge-on orbit around some distant star would be perpendicular to the plane of the sky, and would be considered to have an inclination of 90°, and face-on orbits are parallel to the plane of the sky and have an inclination of 0° or 180°. As with the orbital period and eccentricity, both the planet and the star will have an equal inclination (and indeed their orbital planes are identical).
One may rotate an orbit about the orbited body within the orbit plane, such that the rotation axis is perpendicular to the orbit plane. This is called the longitude of perihelion, ω, because it rotates the perihelion longitudinally around the orbit, and is measured in degrees. For an orbit with e = 0, this does not make a noticeable change in the orbit, and in such cases, it is customary to set ω = 0°. The reference angle for ω is the line between the observer and the star. There is a 180° difference between ωplanet and ωstar.
The longitude of the ascending node, Ω, is the angle around the reference plane where the orbital plane of the planet in an inclined orbit crosses the reference plane. For the specific case of the ascending node, the planet rises above the reference plane (more northward). For the specific case of when the planet dips below the reference plane, this is called the longitude of the descending node. From visual observation alone, it is impossible to discern the ascending node from the descending node for a distant planetary system. The value for Ω is equal for both the planet and the star. For non-inclined orbits we let Ω = 0°.
The mean anomaly of an orbit, M, is the position of the planet around the orbit, in degrees, and is unrelated to the shape of the orbit. It is typically given with an ephemeris, or a date (or more scientifically, ‘epoch’) for which the value of M is true. There is a 180° offset between the values of M for the star and the planet for the same epoch.
The elements of an orbit | 6,166 | 27,208 | {"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": 59, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.92974 |
http://uk.mathworks.com/help/matlab/ref/rcond.html?nocookie=true | 1,430,785,307,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430455162832.10/warc/CC-MAIN-20150501043922-00013-ip-10-235-10-82.ec2.internal.warc.gz | 197,583,397 | 10,028 | rcond
Reciprocal condition number
Description
example
````C = rcond(A)` returns an estimate for the reciprocal condition of `A` in 1-norm. If `A` is well conditioned, `rcond(A)` is near 1.0. If `A` is badly conditioned, `rcond(A)` is near 0. ```
Examples
collapse all
Examine the sensitivity of a badly conditioned matrix.
A notable matrix that is symmetric and positive definite, but badly conditioned, is the Hilbert matrix. The elements of the Hilbert matrix are H(i,j) = 1/(i + j -1).
Create a 10-by-10 Hilbert matrix.
```A = hilb(10); ```
Find the reciprocal condition number of the matrix.
```C = rcond(A) ```
```C = 2.8286e-14```
The reciprocal condition number is small, so `A` is badly conditioned.
The condition of `A` has an effect on the solutions of similar linear systems of equations. To see this, compare the solution of Ax = b to that of the perturbed system, Ax = b + 0.01.
Create a column vector of ones and solve Ax = b.
```b = ones(10,1); x = A\b;```
Now change b by `0.01` and solve the perturbed system.
```b1 = b + 0.01; x1 = A\b1;```
Compare the solutions, `x` and `x1`.
```norm(x-x1) ```
```ans = 1.1250e+05```
Since `A` is badly conditioned, a small change in `b` produces a very large change (on the order of 1e5) in the solution to x = A\b. The system is sensitive to perturbations.
Find Condition of Identity Matrix
Examine why the reciprocal condition number is a more accurate measure of singularity than the determinant.
Create a 5-by-5 multiple of the identity matrix.
`A = eye(5)*0.01;`
This matrix is full rank and has five equal singular values, which you can confirm by calculating `svd(A)`.
Calculate the determinant of `A`.
`det(A)`
```ans = 1.0000e-10```
Although the determinant of the matrix is close to zero, `A` is actually very well conditioned and not close to being singular.
Calculate the reciprocal condition number of `A`.
`rcond(A)`
```ans = 1 ```
The matrix has a reciprocal condition number of `1` and is, therefore, very well conditioned. Use `rcond(A)` or `cond(A)` rather than `det(A)` to confirm singularity of a matrix.
Input Arguments
collapse all
`A` — Input matrixsquare numeric matrix
Input matrix, specified as a square numeric matrix.
Data Types: `single` | `double`
Output Arguments
collapse all
`C` — Reciprocal condition numberscalar
Reciprocal condition number, returned as a scalar. The data type of `C` is the same as `A`.
The reciprocal condition number is a scale-invariant measure of how close a given matrix is to the set of singular matrices.
• If `C` is near 0, the matrix is nearly singular and badly conditioned.
• If `C` is near 1.0, the matrix is well conditioned.
collapse all
Tips
• `rcond` is a more efficient but less reliable method of estimating the condition of a matrix compared to the condition number, `cond`. | 756 | 2,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2015-18 | latest | en | 0.895614 |
http://www.chegg.com/homework-help/questions-and-answers/spheres-mass-m-slide-freely-africtionless-horizontal-surface-sphere-moving-speed-48-m-s-st-q56444 | 1,387,590,383,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345774525/warc/CC-MAIN-20131218054934-00053-ip-10-33-133-15.ec2.internal.warc.gz | 334,407,352 | 7,864 | 0 pts pending
Two spheres, each of mass m, can slide freely on africtionless, horizontal surface. Sphere A is moving at a speed= 4.8 m/s when it strikes sphere B which is atrest and the impact causes sphere B to break into two pieces, eachof mass m/2. Knowing that 0.7 seconds after the collision one piecereaches point C and 0.9 seconds after the collision the other piecereaches point D, determine (a) the velocity of sphere A after thecollision, (b) the angle θ and the speeds of the two piecesafter the collision. | 131 | 517 | {"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-2013-48 | latest | en | 0.865621 |
https://www.powershow.com/view4/508f13-M2Y4Y/The_Study_of_Modulation_Schemes_powerpoint_ppt_presentation | 1,540,303,731,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516194.98/warc/CC-MAIN-20181023132213-20181023153713-00354.warc.gz | 1,011,092,637 | 19,454 | # The Study of Modulation Schemes ?????????? - PowerPoint PPT Presentation
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Title: The Study of Modulation Schemes ??????????
1
The Study of Modulation Schemes??????????
2
Elliptical Modulated Signal
Current Schemes
Sine Wave Signal
Circle
Proposed Schemes
Elliptical Modulated Signal
Ellipse
3
Flexibility of Elliptical Modulation
• AM, FM, PM
• Transformation of Carrier Signal
• in Amplitude, Phase and Frequency
Proposed Modulation Schemes
Amplitude, Phase , Frequency
and Eccentricity,
Inclination Angle, Rotation Frequency, Rotation
Direction
4
Concept Theory
P
R 2
a
R 1
r
?
o
a
F
-F
-b
Semi-major axis (a)
? Revolution Angle a Offset Inclination
Angle ?r the revolution frequency of the
carrier eccentricity
5
Concept Theory
P
R 2
a
R 1
r
?
o
a
F
-F
-b
Semi-major axis (a)
? Revolution Angle a Offset Inclination
Angle ?r is the revolution frequency of the
carrier ?i is the rotation frequency of the
carrier
6
Elliptical Modulation Schemes
Basic Schemes
• Eccentricity Shift Keying (ESK)
• Inclination Angle Shift Keying (IASK)
• Rotation Frequency Shift Keying (RFSK)
7
Inclination Angle Shift Keying
Binary 0
Binary 1
8
Inclination Angle Shift Keying Signal (
,ec 0.3, 0.6, 0.9)
9
Rotation Frequency Shift Keying (RFSK)
where wi is modulation variable
Binary 0
Binary 1
10
4-ary EPSK(Elliptical Phase Shift Keying)
Offset inclination angle phase
where ?i and ?j can have N and K discrete
values respectively. M NK
4-ary Elliptical Phase Shift Keying (N2K2,
(?i ,?j) (p/4, p/4) (3p/4, 3p/4)
(p/4, 5p/4) (3p/4, 7p/4)
where integer i 0,1,2,3
11
Correlation Coefficient
where integer i 0,1,2,3, and corresponding
signals are defined as S1, S2, S3, S4.
CS1S2_SIN CS2S3_COS CS3S4_SIN CS4S1_COS
- CS1S2_COS CS2S3_SIN CS3S4_COS CS4S1_SIN
ec 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
? 0.005 0.020 0.047 0.087 0.144 0.222 0.334 0.500 0.786
12
Mathematical Analysis of BER
ec 0.9 0.8 0.7 0.6 0.5 0.4
a 1.51A 1.29A 1.18A 1.11A 1.07A 1.04A
BER comparison
13
Modulation and Demodulation
Carrier recovery circuit
LPF
block
Advantage can be strenthened by increasing
eccentricity
14
Simulation Results on BER Performance
Under AWGN
• 4-EPSK outperforms QPSK
• error performance is directly proportional to the
value of eccentricity
• Both results are consistent with conclusions
achieved from mathematical
• analysis and demodulation illustration.
15
Proposal of 8-ary EPSK
• 8-ary Elliptical Phase Shift Keying ( N2K4)
where i 0,1,2,3 j0,1
16
Mathematical Analysis of BER
Euclidean distance comparison
BER comparison
ec 0.9 0.8 0.7 0.6 0.5 0.4
a 1.51A 1.29A 1.18A 1.11A 1.07A 1.04A
rmin 0.47A 0.55A 0.59A 0.63A 0.65A 0.67A
17
18
Effect of Eccentricity to Performance of 8-EPSK
Under AWGN
19
Performance Comparison between 8PSK 8-EPSK at
eccentricity of larger than 0.5
20
New Concept of Waveform
inclination angle
frequency
eccentricity | 1,199 | 3,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-43 | latest | en | 0.68555 |
https://discourse.slicer.org/t/radius-extraction-from-centerline/34066 | 1,709,139,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00434.warc.gz | 217,050,896 | 7,358 | Hi to everyone. I’m working with centerlines, which I made with a external python code. From this code I return the centerlines as Markups nodes, and now I would like to extract radius info. For this purpose I use the Cross Section Analysis. This module works well more of the times but fails in bifurcations, where minimum sphere and cross section methods can’t find the proper area…
I noticed that centerlines made with Extract Centerlines module can give radius data when you use the centerline model (see code). I know that Extract Centerlines workflow is based on Voronoi Diagrams, but the radius extraction is better than Cross Section Analysis?
``````>>> c = getNode('Centerline model')
>>> points = slicer.util.arrayFromModelPoints(c)
``````
In that case, could I use my centerlines with the same code?
If there is no solution for this, you have any idea to perform radius extraction in bifurcations ?
Thanks a lot!
Can you draw by hand or with a mouse a bifurcation to illustrate what you mean by ‘proper area’, from which point or circumference to which point or circumference you idealize measurements in a bifurcation?
The minimum sphere radius is more accurate in a bifurcation than the circular-equivalent diameter since the latter is derived from a surface area. This area will include all branches in a bifurcation. You could cut the segment so as to isolate single tubes of interest and regenerate everything.
Hi Chir.set, thanks for your help. The goals is :
Where clearly both structures are well defined.
At this moment, I’m using the CSA process and then correct the radius for nodes in bifurcation zones using MIS by myself.¡, but this work well. MIS strategy used is:
Create a sphere with radius closer to zero and start increasing the radius until the sphere contains at least 30-40 edges of the segmentation…
Did you see any inconvenience on this procedure?
Is there any other way to attack this problem?
I can’t really have an authoritative opinion on the matter.
According to the above image: you might process your bifurcated centerline model with the ‘Centerline disassembly’ module in SlicerVMTK. You can thus get a complete centerline model from the first endpoint to each other endpoint, when selecting the ‘Centerlines’ component in the UI.
Each new centerline model will have a ‘Radius array’, which is the MIS radius. This may help you generate the tubes you are pointing to, probably using vtkTubeFilter.
Okay, but MIS would fail when the segment surface is non-smooth or has any stenosis… | 546 | 2,542 | {"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-2024-10 | latest | en | 0.912631 |
https://www.unitconverters.net/power/erg-second-to-foot-pound-force-hour.htm | 1,718,632,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00057.warc.gz | 911,443,156 | 3,615 | Home / Power Conversion / Convert Erg/second to Foot Pound-force/hour
# Convert Erg/second to Foot Pound-force/hour
Please provide values below to convert erg/second [erg/s] to foot pound-force/hour, or vice versa.
From: erg/second To: foot pound-force/hour
### Erg/second to Foot Pound-force/hour Conversion Table
Erg/second [erg/s]Foot Pound-force/hour
0.01 erg/s2.655223737402E-6 foot pound-force/hour
0.1 erg/s2.65522E-5 foot pound-force/hour
1 erg/s0.0002655224 foot pound-force/hour
2 erg/s0.0005310447 foot pound-force/hour
3 erg/s0.0007965671 foot pound-force/hour
5 erg/s0.0013276119 foot pound-force/hour
10 erg/s0.0026552237 foot pound-force/hour
20 erg/s0.0053104475 foot pound-force/hour
50 erg/s0.0132761187 foot pound-force/hour
100 erg/s0.0265522374 foot pound-force/hour
1000 erg/s0.2655223737 foot pound-force/hour
### How to Convert Erg/second to Foot Pound-force/hour
1 erg/s = 0.0002655224 foot pound-force/hour
1 foot pound-force/hour = 3766.1609675818 erg/s
Example: convert 15 erg/s to foot pound-force/hour:
15 erg/s = 15 × 0.0002655224 foot pound-force/hour = 0.0039828356 foot pound-force/hour | 362 | 1,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.552759 |
https://py.checkio.org/forum/post/13144/what-is-wrong/ | 1,713,240,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00205.warc.gz | 442,161,827 | 6,805 | • # what is wrong?
## Question related to mission The Stones
``` def stones(pile,moves):
results=[False]+[True]*pile
for count in range(1,pile+1):
possible=[x for x in moves if count-x>=0]
results[count]= not(all(results[count-i] for i in possible))
if results[pile]:
return 2
else:
return 1
```
why does it not work? | 94 | 323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.696287 |
https://rdrr.io/rforge/arf3DS4/src/R/displayFunctions.R | 1,576,142,936,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00010.warc.gz | 509,957,457 | 16,247 | # R/displayFunctions.R In arf3DS4: Activated Region Fitting, fMRI data analysis (3D).
```#############################################
# arf3DS4 S4 DISPLAY FUNCTIONS #
# Wouter D. Weeda #
# University of Amsterdam #
#############################################
#[CONTAINS]
#makeColors
#sliceColor
#makeDiscreteImage
#newprogressElement
#writeProgress
makeDiscreteImage <-
function(datavec,zerotol=1e-03)
#make a discritezed image of a datavector (divide into steps relative to zero-point)
{
#define maxsteps
maxsteps = 64
datavec[abs(datavec)<zerotol]=0
wzero = datavec[-which(datavec==0)]
if(length(which(wzero[wzero>0]>0))) pos_small = min(wzero[wzero>0]) else pos_small = 0
if(length(which(wzero[wzero<0]>0))) neg_small = min(abs(wzero[wzero<0])) else neg_small = 0
max_dat = max(datavec)
min_dat = min(datavec)
total = abs(min_dat)+abs(max_dat)
possteps = round(maxsteps*(abs(max_dat)/total))
negsteps = round(maxsteps*(abs(min_dat)/total))
pos_data = datavec[datavec>0]
neg_data = datavec[datavec<0]
if(max_dat>0 & min_dat<0) {
pq = quantile(pos_data,probs=seq(0,1,1/possteps)[-1])
nq = quantile(neg_data,probs=seq(1,0,-1/negsteps)[-1])
}
if(max_dat>0 & min_dat>=0) {
possteps = maxsteps
pq = quantile(pos_data,probs=seq(0,1,1/possteps)[-1])
nq = numeric(0)
}
if(max_dat<=0 & min_dat<0) {
negsteps = maxsteps
nq = quantile(neg_data,probs=seq(1,0,-1/maxsteps)[-1])
pq = numeric(0)
}
if(max_dat==0 & min_dat==0) {
nq = numeric(0)
pq = numeric(0)
}
newdata=rep(NA,length(datavec))
if(length(pq)>0) newdata[datavec>0 & datavec<pq[1]]=1
if(length(nq)>0) newdata[datavec<0 & datavec>nq[1]]=-1
if(length(pq)>0) for(i in 1:possteps) newdata[datavec>=pq[i]]=i+1
if(length(nq)>0) for(i in 1:negsteps) newdata[datavec<=nq[i]]=-i-1
newdata[datavec==0]=0
return(list(newdata=newdata,minmax=c(min_dat,max_dat),small=c(pos_small,neg_small)))
}
makeColors <-
function(datavec,gray=FALSE)
## make colors for overlay images, input is a discretized image
{
datasort = sort(unique(datavec))
neg_dat = datasort[datasort<0]
pos_dat = datasort[datasort>0]
if(gray) {
pos_col = gray(seq(0,1,1/length(pos_dat))[-1])
neg_col = gray(seq(1,0,-1/length(neg_dat))[-length(seq(1,0,-1/length(neg_dat)))])
zero_col = gray(0)
} else {
pos_col <- rgb(1,seq(0,1,1/length(pos_dat))[-1],0)
neg_col <- rgb(seq(.5,0,-.5/length(neg_dat))[-1],seq(.5,0,-.5/length(neg_dat))[-1],1)
zero_col <- rgb(0,0,0)
}
colvec <-c(neg_col,zero_col,pos_col)
neg = matrix(NA,2,length(neg_col))
pos = matrix(NA,2,length(pos_col))
neg[1,]=neg_dat
neg[2,]=neg_col
pos[1,]=pos_dat
pos[2,]=pos_col
return(list(pos=pos,neg=neg,colvec=colvec,data=c(neg_dat,0,pos_dat)))
}
sliceColor <-
function(slicedata,colors)
## calculate the colorvector for the discretized slice based on an makeColor object.
{
slice_max = max(slicedata)
slice_min = min(slicedata)
mp = which(as.numeric(colors\$pos[1,])<slice_max)
mn = which(as.numeric(colors\$neg[1,])<=slice_min)
colvec_pos = colors\$pos[2,mp]
colvec_neg = colors\$neg[2,-mn]
colvec=c(colvec_neg,rgb(0,0,0),colvec_pos)
return(colvec)
}
newProgressElement <-
function(arfmodel,options,lower,upper)
#make a new Progress Window, return an object of class progress (S3)
{
if(.options.output.mode(options)=='none') {
disabled = T
} else {
disabled = F
##library(tcltk)
}
if(!disabled) {
tt <- tktoplevel()
mt = .model.modeltype(arfmodel)
nr = .model.regions(arfmodel)
tktitle(tt) <- paste('ARF Progress [ ',mt,' @ ',nr,' ]',sep='')
scr <- tkscrollbar(tt, repeatinterval=5,command=function(...)tkyview(txt,...))
txt <- tktext(tt,bg="white",font="courier",yscrollcommand=function(...)tkset(scr,...),height=50,width=45)
tkgrid(txt,scr)
tkgrid.configure(scr,sticky="ns")
tkinsert(txt,"end",paste('ARF [',.model.name(arfmodel),'] ',as.character(Sys.time()),sep=''))
tkconfigure(txt, state="disabled")
tkfocus(txt)
}
#make progress object (S3)
if(!disabled) {
progress = list(disabled=disabled,tt=tt,txt=txt,lower=lower,upper=upper,iterlim=.options.min.iterlim(options),perslim=.options.min.boundlim(options))
} else {
progress = list(disabled=disabled,tt=NULL,txt=NULL,lower=lower,upper=upper,iterlim=.options.min.iterlim(options),perslim=.options.min.boundlim(options))
}
attr(progress,'class') <- 'progress'
#assign global counters
assign('.oldobj',0,envir=.arfInternal)
assign('.objit',1,envir=.arfInternal)
assign('.bounded',rep(0,.model.regions(arfmodel)),envir=.arfInternal)
return(progress)
}
writeProgress <-
#write down the progress of the iterations
{
txt = progress\$txt
tkconfigure(txt, state="normal")
tkdelete(txt,"1.0","end")
tkinsert(txt,"end",paste(as.character(Sys.time()),'\n',sep=''))
tkinsert(txt,"end",paste("\n"))
tkinsert(txt,"end",sprintf("Iteration obj. : %10.0f\n",objit))
tkinsert(txt,"end",sprintf("Iteration limit : %10.0f\n",progress\$iterlim))
tkinsert(txt,"end",sprintf("Boundary limit : %10.0f\n",progress\$perslim))
tkinsert(txt,"end",sprintf("Objective value : %10.0f\n",round(ssqdat)))
tkinsert(txt,"end",paste("\n"))
tkinsert(txt,"end",paste("Region Information\n"))
tkinsert(txt,"end",paste("Bounded regions (",paste(which(bounded>0),collapse=','),")\n",sep=''))
tkinsert(txt,"end",paste("\n"))
estvec = matrix(theta,10)
svec = sprintf(' [%3.0f] (%5.2f %5.2f %5.2f) |%8.0f|',1,estvec[7,1],estvec[8,1],estvec[9,1],sqrt(sum(gradmat[,1]^2)))
if(bounded[1]>0) svec=paste(svec,'*',sep='')
tkinsert(txt,"end",paste(svec,"\n"))
svec = sprintf(' [%3.0f] (%5.2f %5.2f %5.2f) |%8.0f|',i,estvec[7,i],estvec[8,i],estvec[9,i],sqrt(sum(gradmat[,i]^2)))
if(bounded[i]>0) svec=paste(svec,'*',sep='')
tkinsert(txt,"end",paste(svec,"\n"))
}
}
tkconfigure(txt, state="disabled")
tkfocus(txt)
}
```
## Try the arf3DS4 package in your browser
Any scripts or data that you put into this service are public.
arf3DS4 documentation built on May 2, 2019, 5:16 p.m. | 1,938 | 5,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-51 | latest | en | 0.157979 |
https://www.skillsworkshop.org/functional-maths-e3.23 | 1,723,424,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641023489.70/warc/CC-MAIN-20240811235403-20240812025403-00226.warc.gz | 768,758,729 | 8,604 | # FM E3.23 Organise & represent information in appropriate ways inc. tables, diagrams, simple line graphs & bar charts
3.23 Organise and represent information in appropriate ways including tables, diagrams, simple line graphs and bar charts
Subject content [reformed] functional skills: Mathematics (DfE, Feb 2018):
https://www.gov.uk/government/publications/functional-skills-subject-content-mathematics
## Earn and Stash the Cash - Australian version
An Australian version (i.e. uses A\$, Australian vocabulary and minimum wage details, etc.) of the popular Stash the Cash written by Andy Morrell in 2010. In addition to currency changes this version also includes an extra version of the calculations table and other changes. For example, the table on page 3 has extra columns to make the mathematical operations explicit and to allow students to practise writing mathematical “sentences”.
Level
Entry Level 3
Level 1
Level 2
Maths
FM E3.8 Read, write & use decimals up to two decimal places
FM L1.2 Recognise and use positive and negative numbers
FM L1.11 Add, subtract, multiply & divide decimals up to 2 decimal places
FM L2.1 Read, write, order & compare positive & negative numbers of any size
FM E3.10 Calculate with money using decimal notation & express money correctly in writing in pounds and pence
FM E3.21 Extract information from lists, tables, diagrams, charts; create frequency tables
FM E3.23 Organise & represent information in appropriate ways inc. tables, diagrams, simple line graphs & bar charts
FM L1.29 Find the mean & range of a set of quantities
FM L2.23 Calculate the median & mode of a set of quantities
Context
Independent living
## Valentine Meal Deal - E3, L1 & L2 Functional Maths
A follow-on Valentine's Day resource from the E1-E3 worksheet of the same name. It continues the general problem solving of finding suitable meal deals for various couples and calculating how much they save by purchasing a Valentine's meal deal. For L2 learners it also delves into compound measures with challenging multiple questions on on unit prices (e.g. £/kg, £/litre).
Level
Entry Level 3
Level 1
Level 2
Maths
FM Straightforward problem(s) with more than 1 step
FM Complex multi-step problem(s)
FM E3.10 Calculate with money using decimal notation & express money correctly in writing in pounds and pence
FM E3.21 Extract information from lists, tables, diagrams, charts; create frequency tables
FM E3.23 Organise & represent information in appropriate ways inc. tables, diagrams, simple line graphs & bar charts
FM L1.20 Convert between units of length, weight, capacity, money and time, in the same system
FM L2.15 Calculate using compound measures inc. speed, density & rates of pay
Context
Retail Hospitality Customer service
Catering Food Nutrition
## The 100 Metres Dash probability investigation
A practical investigation that can be enjoyed in mixed level classes. There's something for all learners from Entry Level 1 to Level 2.
Editor's note. One of my favourite resources that somehow got lost when the site was redeveloped in 2011. Now fully updated to reflect the new 2018 Functional Skills content that come into effect September 2019. Pages 7-8 show detailed FS mapping to content descriptors and problem solving skills.
Level
Entry Level 1
Entry Level 2
Entry Level 3
Level 1
Level 2
Maths
FM E1.3 Add numbers which total up to 20, & subtract numbers from numbers up to 20
FM E1.13 Read & draw simple charts & diagrams inc. a tally chart, block diagram/graph
FM E2.25 Take information from one format and represent the information in another format including use of bar charts
FM E3.23 Organise & represent information in appropriate ways inc. tables, diagrams, simple line graphs & bar charts
FM L1.8 Read, write, order & compare common fractions & mixed numbers
FM L2.27 Express probabilities as fractions, decimals & percentages
Context
Sport and fitness | 911 | 3,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.794044 |
https://www.physicsforums.com/threads/electron-scattering-nuclear-radius-to-nucleon-number-relationship.710556/ | 1,695,530,016,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00716.warc.gz | 1,031,449,836 | 16,752 | # Electron scattering: nuclear radius to nucleon number relationship
• AlexCdeP
#### AlexCdeP
Hi! I am extremely confused on what seems to be quite a simple question. The question contains a graph of root mean square radius <r$^{2}$> plotted against A$^{1/3}$ where A is the nucleon number. In the lecture notes he specifies that <r$^{2}$> is not the same as R but does not really say specifically what the difference is. It is given that the gradient of the line in this graph is 0.96. I am meant to find the value of R$_{0}$ in the relationship R=R$_{0}$A$^{1/3}$ but I'm not certain how to find it because I don't know what the relationship between the rms r and R is.
Any help would be amazing.
I have worked through trying to use the charge radius but without success, he never mentioned the charge radius in the lecture so I'm not sure what to do. http://en.wikipedia.org/wiki/Charge_radius
#### Attachments
• PC3232_assignment_2_1314s1.pdf
113.7 KB · Views: 653
If you have a uniform sphere of radius R, what is the (squared) rms radius <r2> of its volume? Hint: You'll need an integral (or a formula).
1 person
Thanks so much I think I'm on the right track now. So for a uniform sphere you use some spherical coordinate integrals to find that the rms radius is related to the radius by $\frac{3}{5}$R$^{2}$. Is this the correct formula to get?
Also am I right in saying that I am effectively finding the expectation value of r$^{2}$?
Finally got the answer! Thanks again man couldn't have done it without you!
Thanks so much I think I'm on the right track now. So for a uniform sphere you use some spherical coordinate integrals to find that the rms radius is related to the radius by $\frac{3}{5}$R$^{2}$. Is this the correct formula to get?
Also am I right in saying that I am effectively finding the expectation value of r$^{2}$?
Yup. | 473 | 1,857 | {"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} | 3.28125 | 3 | CC-MAIN-2023-40 | latest | en | 0.935864 |
https://ask.sagemath.org/answers/26093/revisions/ | 1,695,395,820,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00082.warc.gz | 136,373,820 | 6,432 | # Revision history [back]
A small edit to tmonteil's first anser does the trick. Need to define reverse multiplication and division (http://www.rafekettler.com/magicmethods.pdf):
class MYExp(sage.symbolic.expression.Expression):
def __sub__(self,other):
if self.convert() == other.convert():
return self
else:
raise TypeError('Subtraction of non-matching units')
if self.convert() == other.convert():
return self
else:
raise TypeError('Sum of non-matching units')
def __mul__(self,other):
return MYExp(SR, super(MYExp, self).__mul__(other))
def __rmul__(self,other):
return MYExp(SR, super(MYExp, self).__mul__(other))
def __div__(self,other):
return MYExp(SR, super(MYExp, self).__div__(other))
def __rdiv__(self,other):
return MYExp(SR, super(MYExp, self).__div__(other))
def __pow__(self,other):
return MYExp(SR, super(MYExp, self).__pow__(other))
U = lambda x: MYExp(SR,x)
for s in units.length.trait_names():
globals()[s] = U(getattr(units.length,s))
Now we can write:
meter - meter
meter
1*meter - 1*meter
meter
meter^2 - meter^2
meter^2
A small edit to tmonteil's first anser does the trick. Need to define reverse multiplication and division (http://www.rafekettler.com/magicmethods.pdf):
class MYExp(sage.symbolic.expression.Expression):
def __sub__(self,other):
if self.convert() == other.convert():
return self
else:
raise TypeError('Subtraction of non-matching units')
if self.convert() == other.convert():
return self
else:
raise TypeError('Sum of non-matching units')
def __mul__(self,other):
return MYExp(SR, super(MYExp, self).__mul__(other))
def __rmul__(self,other):
return MYExp(SR, super(MYExp, self).__mul__(other))
def __div__(self,other):
return MYExp(SR, super(MYExp, self).__div__(other))
def __rdiv__(self,other):
return MYExp(SR, super(MYExp, self).__div__(other))
def __pow__(self,other):
return MYExp(SR, super(MYExp, self).__pow__(other))
U = lambda x: MYExp(SR,x)
for s in units.length.trait_names():
globals()[s] = U(getattr(units.length,s))
Now we can write:
meter - meter
meter
1*meter - 1*meter
meter
meter^2 - meter^2
meter^2
However, note that external functions can still result in problems, e.g.
sqrt(meter) - sqrt(meter)
0
sin(meter) - sin(meter)
0
A small edit to tmonteil's first anser does answer allows handling multiplications and divisions in the trick. above framework. Need to define reverse multiplication and division (http://www.rafekettler.com/magicmethods.pdf):
class MYExp(sage.symbolic.expression.Expression):
def __sub__(self,other):
if self.convert() == other.convert():
return self
else:
raise TypeError('Subtraction of non-matching units')
if self.convert() == other.convert():
return self
else:
raise TypeError('Sum of non-matching units')
def __mul__(self,other):
return MYExp(SR, super(MYExp, self).__mul__(other))
def __rmul__(self,other):
return MYExp(SR, super(MYExp, self).__mul__(other))
def __div__(self,other):
return MYExp(SR, super(MYExp, self).__div__(other))
def __rdiv__(self,other):
return MYExp(SR, super(MYExp, self).__div__(other))
def __pow__(self,other):
return MYExp(SR, super(MYExp, self).__pow__(other))
U = lambda x: MYExp(SR,x)
for s in units.length.trait_names():
globals()[s] = U(getattr(units.length,s))
Now we can write:
meter - meter
meter
1*meter - 1*meter
meter
meter^2 - meter^2
meter^2
However, note that external functions can still result in problems, e.g.
sqrt(meter) - sqrt(meter)
0
sin(meter) - sin(meter)
0
Actually, I found a more practical way to automatically check for consistent units. Provided a dictionary containing all variables of eq as keys and each variable multiplied by its units as entries:
def units_check(eq, i0 = 1):
'''
Checks whether all arguments are keys in udict and returns simplified units
If units cancel out on one side while they should not, set i0=2 or something other than 1.
Example:
sage: var('a b c d')
sage: udict = {a: a*units.length.meter^3, b: b*units.length.meter, c: c*units.length.meter, d: d*units.length.meter}
sage: eq = a == b*c*d
sage: units_check(eq)
meter^3 == meter^3
'''
valdict = {}
i = i0
for arg in eq.arguments():
valdict[arg] = i
i = i + 1
udict[arg]
eq_val = eq.subs(valdict)
return (eq.subs(udict).subs(valdict))/eq_val | 1,199 | 4,223 | {"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-2023-40 | latest | en | 0.502198 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=11650&CurriculumID=49&Method=Worksheet&NQ=10&Num=6.7&Type=D | 1,566,640,079,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00351.warc.gz | 114,662,336 | 3,457 | Name: ___________________Date:___________________
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### High School Mathematics - 26.7 Inverse Matrix
Method:- 1. Check whether the given matrix 'A' is a singular or non-singular matrix. That is |A| ¹ 0. 2. If A is singular, A-1 does not exit. If A is non-singular proceed to find the inverse of the gives matrix. 3. Interchanging the elements of the principal diagonal of A. 4. Change the sign of the other two elements in A. 5. Multiply the resultant matrix with the scalar 1/(ad-bc). Directions: Find the inverse of the given matrix. Also write at least ten examples of your own.
Name: ___________________Date:___________________
### High School Mathematics - 26.7 Inverse Matrix
Q 1: dabc Q 2: dcab Q 3: abcd Q 4: cadb Q 5: dcba Q 6: dcba Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only! | 255 | 968 | {"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-2019-35 | latest | en | 0.780189 |
http://clay6.com/qa/28301/a-ray-of-light-passes-through-four-transparent-media-with-refractive-indice | 1,527,342,866,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00152.warc.gz | 66,663,833 | 26,396 | # A ray of light passes through four transparent media with refractive indices $\mu_1,\mu_2,\mu_3$ of all media are parallel, If the emergent ray CD is parallel ti incident ray AB. We must have
$(a)\;\mu_1 = \mu_2 \\ (b)\;\mu_2=\mu_3 \\ (c)\;\mu_3=\mu_4 \\ (d)\;\mu_4 =\mu_1$
According to Snell's law,
$\mu \sin \theta= constant$
Since the emergent ray is parallel to the incident ,the emergent angle = incident angle
Which gives, $\mu_1=\mu_4$
Hence d is the correct answer.
edited Jul 23, 2014 | 164 | 498 | {"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} | 3.390625 | 3 | CC-MAIN-2018-22 | latest | en | 0.759739 |
https://socratic.org/questions/how-do-you-solve-the-exponential-equation-9-2x-1-3-5x-1 | 1,685,328,284,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00033.warc.gz | 584,384,176 | 5,952 | How do you solve the exponential equation 9^(2x+1)=3^(5x-1)?
Mar 5, 2017
$x = 3$
Explanation:
Express the left side with a base of 3
$\Rightarrow {\left({3}^{2}\right)}^{2 x + 1} = {3}^{5 x - 1}$
$\Rightarrow {3}^{4 x + 2} = {3}^{5 x - 1}$
Since the bases are equal, we can equate the exponents.
$\Rightarrow 5 x - 1 = 4 x + 2$
$\Rightarrow x = 2 + 1 = 3$ | 147 | 364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.3125 | 4 | CC-MAIN-2023-23 | latest | en | 0.62194 |
http://www.uff.br/trianglecenters/X3070.html | 1,508,370,081,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823168.74/warc/CC-MAIN-20171018233539-20171019013539-00030.warc.gz | 588,196,813 | 2,304 | ## X(3070) (INTERSECTION OF X(4)X(6) AND X(5)X(372))
Information from Kimberling's Encyclopedia of Triangle Centers
Trilinears sin A + 2 cos B cos C : sin B + 2 cos C cos A : sin C + 2 cos A cos B
Barycentrics (sin A)(sin A + 2 cos B cos C) : (sin B)(sin B + 2 cos C cos A) : (sin C)(sin C + 2 cos A cos B)
The points X(3070) and X(3071) are on the Evans conic.
X(3070) lies on these lines:
2,490 3,485 4,6 5,372 20,1151 30,371 487,1991 642,2482 1124,1479 1335,1478 1837,2362
X(3070) = complement of X(490)
X(3070) = crosspoint of X(4) and X(485)
X(3070) = crosssum of X(3) and X(371)
X(3070) = {X(5),X(372)}-harmonic conjugate of X(615)
This is a joint work of
Humberto José Bortolossi, Lis Ingrid Roque Lopes Custódio and Suely Machado Meireles Dias. | 326 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | latest | en | 0.356052 |
https://www.vedantu.com/iit-jee/keplers-laws-of-planetary-motion | 1,679,548,256,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00648.warc.gz | 1,156,985,742 | 34,231 | Courses
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# Kepler’s Laws of Planetary Motion
## Kepler's Laws of Motion
Last updated date: 16th Mar 2023
Total views: 273.9k
Views today: 6.52k
The Kepler’s Law of planetary motion typically revolves around the periods, orbits, and areas with regard to planet and sun. The principle thus states that the planets revolve around the sun in elliptical orbits with the sun at one focus in a sun-centred solar system.
Owing to this evaluation of the motions of the planets, Kepler introduced a series of laws, now popularly known as Kepler’s three laws. The three laws by Kepler described the behaviour of planets depending on their paths through space. The first two Kepler's laws of motion were published in 1609 in The New Astronomy. Their discovery turned out to be an earnest approach to the development of modern science.
### Kepler's First Law
Referred to as the law of ellipses, this law describes that planets are orbiting around the sun in a path explained as an ellipse. This Kepler's first law is also popularly known as the law of orbits. The law suggests that the orbit of any planet is an ellipse along with the Sun with Sun at one of the two focal points of an ellipse. We are already acquainted with the fact that planets move around the Sun in a circular orbit. However, as per Kepler, it is beyond doubt that planets revolve around the Sun, but not in a circular orbit. However, it revolves around an ellipse. In an ellipse, we have two foci. Sun is situated at one of the focus points of the ellipse.
### Kepler's Second Law
Known as the law of equal areas- this law defines the speed at which any given planet will revolve while orbiting the sun. The speed at which any planet revolves through space changes constantly. A planet revolves the fastest when it is closest to the sun and slowest when it is farthest from the sun. Still, if an imaginary line is drawn from the centre of the planet to the centre of the sun that line would blow off the similar area in equal periods. For example, if an imaginary line is drawn from the earth to the sun, then the area swept aside by the line every 31-day month will be the same. This is depicted in the diagram below.
### Diagrammatic Representation Of Kepler's Second Law
As can be seen from the diagram below, the areas developed when the earth is closest to the sun can be estimated as a broad but short triangle. On the other hand, the areas developed when the earth is farthest from the sun can be estimated as a slender but long triangle. These areas are of similar size. Seeing that the base of these triangles are shortest when the earth is farthest from the sun. It is observed that the earth would have to be in motion more slowly for the sake of this imaginary area to be the same size as when the earth is closest to the sun.
Diagrammatic Representation Of Kepler's Second Law
### Kepler’s Third Law
Referred to as the law of harmonies, Kepler’s third law sets a comparison for the orbital period and radius of orbit of a planet against other planets. Distinct from Kepler's 1st and 2nd laws that explain the motion attribute of a single planet, the 3rd law compares the motion characteristics of different planets. The comparison established is that the ratio of the squares of the periods to the cubes of their average distances from the sun is similar for every one of the planets.
### Solved Examples for Kepler’s Laws of Motion
Problem 1
a. Which scientist is attributed to the gathering of the data relevant to support the planet's elliptical motion?
b. Which scientist is credited with the correct description of the data?
c. Which scientist is acknowledged with the long and complex task of evaluating the data?
1. Tycho Brahe collected the data.
2. Johannes Isaac Newton appropriately described the data.
3. Kepler evaluated the data.
Problem 2
The average orbital distance of Mars is 1.52 times to that of the average orbital distance of the Earth. Having known that the Earth orbits around the sun in about 365 days, use Kepler’s third law (law of harmonies) to anticipate the time that Mars would require to orbit the sun.
Given that, the ratio of Rmars / Rearth is 1.52
= Tmars will be 684 days
Problem 3
Employ the graphing competencies of your TI calculator to trace out T2 vs. R3 (T2 must be plotted about the vertical axis) and to identify the equation of the line. Express the equation in slope-intercept form below.
T2 = [3.03 ×10-16] × R3 - 4.62 × 10+9
Given the unpredictability in the y-intercept value, it can be estimated as 0.
Hence, T2 = (3.03 × 10-16) × R3
## FAQs on Kepler’s Laws of Planetary Motion
1. What are Kepler’s Three Laws?
Ans. Kepler’s 3 laws of planetary motion can be explained as follows:
1. The Law of Ellipses: By this postulate, the path of the planets along the sun is elliptical, with the nucleus (centre) of the sun being situated at one focus.
2. The Law of Harmonies: The proportion of the squares of the periods of any two planets is equivalent to the proportion of the cubes of their mean distances from the sun.
3. The Law of Equal Areas: An imaginary segment drawn from the nucleus of the sun to the nucleus of the planet will eradicate equal areas at equal intervals of time.
2. How to manually draw an Ellipse?
Ans. You can easily construct an ellipse using a pencil, string, sheet of paper, piece of cardboard and two tacks and wrap the string tightly around the tacks. Tack the paper sheet to the cardboard with two tacks. Then tie the string into a loop and around the two tacks. Take your pencil and tug the string until the two tacks and pencil form a triangle. Then start to draw a path with the pencil. The resulting shape is an ellipse.
An ellipse is a unique curve in which the sum of the total distances from each point on the curve to two other points is constant. For proper understanding, refer to the illustrative image below. | 1,338 | 5,968 | {"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} | 3.078125 | 3 | CC-MAIN-2023-14 | latest | en | 0.952984 |
https://www.wishfin.com/business-loan/saraswat-bank-business-loan-1955/ | 1,726,650,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00134.warc.gz | 1,005,833,539 | 16,014 | Last Updated : May 28, 2021, 9:12 p.m.
In search of a bank that funds your business as per your requirement? Saraswat bank is the right choice, as the bank proffers several loan schemes targeting different requirements of the diverse categories. The bank provides loan for the maximum tenure of 7 years. The loan can be used to manage various business needs such as buying property, working capital and purchase of assets. No processing fee is charged for the credit limit up to Rs. 5,00,000. However, for the loan amount exceeding Rs. 5,00,000 minimal processing fee is charged.
Particulars Details
Interest Rate 14.25% p.a. & above
Loan Amount Up to Rs. 1 Crore
Tenure 7 years
Processing Fee
No processing fee up to loan amount of Rs. 5,00,000
How To Apply
A business loan can be availed from Saraswat Bank by applying online. The customer can even locate a branch office in nearby vicinity to apply for the business loan.
Saraswat Bank Business Loan EMI Calculator
The Business Loan EMI Calculator is a calculator designed to make the calculation of EMI quick and simple. It computes the real results in the shortest time. The calculator helps the user to know his/ her EMI value and plan for the future payments accordingly. The user can also learn about the amount of interest he/ she has to pay on the loan amount being taken. The calculator works on a simple principle. The user only needs to enter the loan amount, the rate of interest and the tenure of the loan.
For Example – If a borrower avails a loan of Rs. 50,00,000 for the period of 7 years at the prevailing rate of interest which is 14.25% p.a then the EMI will be
Table Showing EMI, Total Interest Amount, Total Repayment Amount
Loan Amount (in Rs.) Interest Rate (p.a) Tenure (in years) EMI (in Rs.) Total Interest Amount (in Rs.) Total Repayment Amount (Principal + Interest) (in Rs.)
50,00,000 14.25% 1 4,49,524 3,94,287 53,94,287
50,00,000 14.25% 2 2,40,655 7,75,729 57,75,729
50,00,000 14.25% 3 1,71,496 11,73,852 61,73,852
50,00,000 14.25% 4 1,37,260 15,88,492 65,88,492
50,00,000 14.25% 5 1,16,990 20,19,419 70,19,419
50,00,000 14.25% 6 1,03,699 24,66,344 74,66,344
50,00,000 14.25% 7 94,392 29,28,923 79,28,923
Amortization Table
Year Principal (in Rs.) Interest (in Rs.) Balance Amount (in Rs.)
1 448,765 683,939 4,551,235
2 517,060 615,644 4,034,175
3 595,749 536,955 3,438,426
4 686,416 446,288 2,752,010
5 790,875 341,829 1,961,135
6 911,237 221,467 1,049,898
7 1,051,006 82,790 0
Features And Benefits
• Loan available to meet various business requirements
• Higher credit limit
• Quick loan approval
• Simplified documentation
Saraswat Bank Loan Schemes :
The bank provides business loan to several beneficiaries under various schemes that are as follows :
1. Udyogini (Business Loan for Women Entrepreneurs)
This scheme is introduced to benefit the women entrepreneurs who want to expand their business. The bank offers a loan up to Rs. 50,00,000 for the maximum period of 7 years. The loan is available against the security and two guarantors are required. The processing fee is 0.1% of the sanctioned loan amount.
2. Micro Finance -Self Help Groups Business Loan
A self- help group of 10 to 20 members can avail a loan of up to 4 times the amount available in the savings account or Rs. 2,00,000. The loan is available at a rate of interest 1% lower than the PLR. The loan is sanctioned for the maximum tenor of 60 months. The guarantee of all group members and the third party guarantor is required as security. The bank provides loan to self- help groups involved in following activities –
a) selling fruits, milk, vegetables, fish
b) making of brooms, baskets, and other bamboo products
c) Other professional activities such as hairdressing, plumbing, electrician etc.
The bank provides the credit facilities to retail traders in the form of working capital and property loan. The loan is available for the amount ranging between Rs. 2,00,000 to Rs. 50,00,000. The repayment has to be made in 60 installments. The rate of interest is linked to PLR which is 14.5% p.a. at present. The benefit of this scheme can be availed by the following:
Eligibility Criteria
-Salaried employees with a minimum net salary of Rs. 10,000 per month
-Professionals, self - employed and others who are income tax assessee having a net annual taxable income of Rs. 1,50,000/- for at least 3 years continuously
-Firms / companies having a net annual taxable income of Rs. 1,50,000 per annum and in operation for last 3 years making cash profit for last 3 years
4. Small and Medium Enterprises Business Loan
Saraswat Bank provides a business loan to the SME units for the expansion and growth. The bank aids the units by providing funds to meet the working capital requirements, term loans, and property loans. The loan is available for the amount of & above Rs. 25,00,000. No processing fee is charged on the loan of up to Rs. 5,00,000 .
Saraswat Bank Business Loan Eligibility Criteria :
The loan is available for the candidates running business under following constitutions :
• Proprietary
• Partnership Firm
• Private Limited Company
• Limited Company
• Co-operative Society
Saraswat Bank Business Loan Application form
The application form is to be mandatorily submitted in order to avail the business loan from the bank. The application form comprises of the following sections:
• Name of the applicant and PAN details
• Date of establishment
• Details of promoters, proprietor, partners or directors
• Loan details
• Existing account and credit facility details
• Reason for taking loan
Documents Required
Documents Required
1. Proof of identity - Voter’s ID Card/ Passport/ Driving License/PAN Card/ Aadhar Card/ signature identification from present bankers of proprietor, partner or Director (if a company)
2. Proof of residence - Recent telephone / electricity bill, property tax receipt / passport / Aadhar Card/ voter’s ID card of proprietor, partner or director ( if a company)
4. Proof of SC/ ST/ Minority section, if applicable
5. Certificates of incorporation/registration etc
6. Memorandum and Articles of association of the company/Registered partnership / Trust Deed etc
7. Assets and liabilities statement of promoters and guarantors along with latest income tax returns should be certified by C.A.
8. Personal IT returns of Promoters/Partners/Directors etc & Guarantors (other than these)
9. Rent Agreement/s (if business premises are rented)
11. Last three year's balance sheet and Profit & Loss (with schedules) of the unit (For all applications for Rs.2 lakhs and above) However, for fund based limits applied below Rs. 50 lakhs if audited Accounts are not available, then Unaudited balance sheets and Profit & Loss (with schedules) are acceptable. For cases of Rs.50 lacs and above, the audited balance sheets and Profit & Loss (with schedules) are mandatory
12. Income Tax / Sales Tax returns of the applicant
13. Projected balance sheets for the next two years in case of working capital limits and for the period of the loan in case of term loan
14. In case of takeover of advances, sanction letters of all facilities being availed from existing bankers/ Financial Institution along with detailed terms and conditions
15. Last three balance sheets of the Associate/ Group Companies, if any (Applicable For Cases With Exposure Above 50 Lakhs)
16. Photocopies of lease deeds/ title deeds and valuation reports, if available, of all the properties being offered as primary and collateral securities
17. Bank statement for last 6 months if banking with other Bank
18. Statement of Loans in individual names of Promoters/ Partners / Directors, etc.
19. Statement of Current/ Savings Account of Promoters/ Partners / Directors etc for last 6 months
Saraswat Bank Business Loan Interest Rate
The rate of interest charged on the business loans is linked to the PLR (Prime Lending Rate) . It is the rate at which the company charges interest to it’s most efficient and creditworthy customer. At present, the PLR is 14.5% p.a.
Saraswat Bank Customer Care Number
Want to know more about the product or have any query regarding the product call at the customer care number. The toll-free number is active from Monday to Saturday. Place a call on Toll- Free No.: 1800229999 between 8:00 AM – 8:00 PM and get your queries resolved. | 2,105 | 8,355 | {"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} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.931796 |
https://strangeherring.com/what-is-a-pivot-chord-music-theory/ | 1,708,728,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00653.warc.gz | 562,483,329 | 11,078 | If you are an aspiring musician or a music enthusiast, you might have come across the term pivot chord in music theory. A pivot chord is a chord that is shared between two different keys and acts as a connection point between them. In this article, we will explore the concept of pivot chords in music theory in detail.
## What is a Pivot Chord?
A pivot chord is a chord that appears in both the old key and the new key. It acts as a harmonic bridge between the two keys and helps to smooth the transition from one key to another.
### How does it work?
The simplest way to explain how pivot chords work is through an example. Let’s take a look at the following chord progression:
C – Am – F – G
In this progression, we start with the key of C major and end with G major. However, we can also view this progression as starting in A minor and ending in F major. The pivot chord in this case is F, which appears both as IV (fourth) chord in C major and as VI (sixth) chord in A minor.
• C major: C – Am – F – G
• A minor: F – G – C – F
• F major: IV – V – I
• G major: V – vi – IV
As you can see, by using the same pivot chord (F), we are able to move seamlessly from one key to another.
### Types of Pivot Chords
There are two types of pivot chords: diatonic and chromatic.
Diatonic Pivot Chords: Diatonic pivot chords are chords that belong to both the old and new keys. For example, in the key of C major, the diatonic pivot chords for the key of G major would be C, G, Am, and Em.
Chromatic Pivot Chords: Chromatic pivot chords, on the other hand, are chords that do not belong to both keys but are altered so that they can fit into both keys. For example, in the key of C major moving to D major, we could use a Bb chord as a chromatic pivot chord.
### Why use Pivot Chords?
The use of pivot chords is a common technique in music composition. It helps to create smooth transitions between different sections of a song or between different keys. It also adds variety and interest to a song by allowing for key changes without abrupt modulation.
## Conclusion
In conclusion, pivot chords are an essential tool for any composer or songwriter. They provide a harmonically smooth transition between two different keys and add interest and variety to a song. By understanding how pivot chords work and how to use them effectively, you can take your music composition skills to the next level. | 555 | 2,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-10 | latest | en | 0.959382 |
https://fdocuments.us/document/cosmic-brownies.html | 1,631,951,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00618.warc.gz | 299,137,446 | 15,895 | of 27
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Cosmic Brownies. Attack of the Demon Sun. Not To Scale. 1.5*10 8 km. 0.56m. 2.9m. 2.9m. Φ = π -2 ϴ. ϴ = tan -1 (2.9/.56) = 1.38 rad. Φ = 0.38 rad. Φ. 2.9m. ϴ ϴ. 0.56 m. ϴ = 1.38 rad. Φ = 0.38 rad. Night Fall @ 18:30. Day Break @ 5:30. 0 .38*780 π. 11:13 – 12:47. - PowerPoint PPT Presentation
### Transcript of Cosmic Brownies
Cosmic Brownies
Cosmic BrowniesAttack of the Demon SunNot To Scale1.5*108 km2.9m2.9m0.56m2.9m0.56m = -2 = tan-1(2.9/.56) = 1.38 rad = 0.38 rad = 1.38 rad = 0.38 radDay Break @ 5:30Night Fall @ 18:30780 min in the dayHorizon1.38*7801.38*7800.38*780343 min343 min94 min11:13 12:475:30 11:1312:47 18:30So during the time of 11:13 and 12:47 We should be getting the most hits
Computer
Top view
Overall View #1
Overall view #2
Bottom Bottom veiw
TOP- Bottom view
TOP VIEW
Data
The time we detect the maximum number of muons should be at precisely the peak of the solar day. However is seems to more often be around 1pm.
The amount of muons hitting our detector should increase and decrease in a 24 hour cycle. This does not mean however that at noon there will be a peak in the number of muons produced. Assuming there is a consistent stream of cosmic rays coming in from the sun, then at noon it should peak, although the particles creating the muons in the atmosphere will be much older than the light we see because the particles travel slower. There are other factors in the amount of muons, for example as the distance between the earth and sun increases or decreases.list of causes of changes in # of cosmic rayssolar day (roughly 24 hour cycle what we were looking at)11 years Schwabe cycle(increase/decrease in sunspots)22 years Hale cycle(suns poles full revolution of suns poles) | 589 | 1,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-39 | latest | en | 0.837543 |
https://www.sscwill.in/2021/09/exercise-13-2-class-10-hindi-pdf.html | 1,686,030,611,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00012.warc.gz | 1,091,291,191 | 30,081 | # Exercise 13.2 Class 10 NCERT Solutions in Hindi
Get Free NCERT Solutions for Class 10 Maths Chapter 13 Ex 13.2 PDF. Surface Areas and Volumes Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 13.2 Class 10 Maths NCERT Solutions were prepared by Experienced sscwill.in Teachers. Detailed answers of all the questions in Chapter 13 Maths Class 10 Surface Areas and Volumes Exercise 13.2 provided in NCERT TextBook.
Topics and Sub Topics in Class 10 Maths Chapter 13 Surface Areas and Volumes:
Section Name Topic Name 13 Surface Areas And Volumes 13.1 Introduction 13.2 Surface Area Of A Combination Of Solids 13.3 Volume Of A Combination Of Solids 13.4 Conversion Of Solid From One Shape To Another 13.5 Frustum Of A Cone
You can also download the free PDF of Ex 13.2 Class 10 Surface Areas and Volumes NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.
Exercise 13.1 Class 10 Exercise 13.2 Class 10 Exercise 13.3 Class 10 Exercise 13.4 Class 10 Exercise 13.5 Class 10
Board CBSE Textbook NCERT Class Class 10th Subject Maths Chapter Chapter 13 Chapter Name Surface Areas and Volumes Exercise Exercise 13.2 Number of Questions Solved 5
## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 in Hindi
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2
### Exercise 13.2 Class 10 in Hindi Download PDF
Exercise 13.1
Exercise 13.3
Exercise 13.4
Exercise 13.5
We hope the NCERT Solutions in hindi for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, help you. If you have any query regarding NCERT Solutions in hindi for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 in hindi, drop a comment below and we will get back to you at the earliest.
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Tagged With: Maths NCERT Solutions For Class 10 in hindi, Maths Solutions For Class 10 NCERT, NCERT Maths Solutions in hindi For Class 10, NCERT Solutions for Class 10 Maths Chapter 13 in hindi, NCERT Solutions For Class 10 Maths Pdf in hindi, NCERT Solutions in hindi For Maths Class 10, NCERT Solutions Of Maths in hindi For Class 10, Solutions in hindi For NCERT Maths Class 10 in hindi | 619 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.798006 |
https://www.mcqslearn.com/business-statistics/statistical-techniques.php | 1,561,551,797,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00357.warc.gz | 821,441,742 | 7,116 | # Statistical Techniques MCQs Questions and Answers PDF Book Download
Statistical techniques multiple choice questions (MCQs), statistical techniques quiz answers to learn business analytics online courses. Introduction to statistics MCQs, statistical techniques quiz questions and answers for online bachelor degree. Learn sources of data, data measurement in statistics, statistical analysis methods, introduction to statistics, statistical techniques test prep for business analyst certifications.
Learn introduction to statistics test MCQs: analysis of labor turnover rates, performance appraisal, training programs and planning of incentives are examples of role of, with choices statistics in personnel management, statistics in finance, statistics in marketing, and statistics in production for online bachelor degree. Practice merit scholarships assessment test, online learning statistical techniques quiz questions for competitive assessment in business majors for business administration certifications.
## MCQ on Statistical TechniquesQuiz Book Download
MCQ: Analysis of labor turnover rates, performance appraisal, training programs and planning of incentives are examples of role of
1. statistics in personnel management
2. statistics in finance
3. statistics in marketing
4. statistics in production
A
MCQ: Branch of statistics which considers ratio scale and interval scale is considered as
1. parametric statistics
2. non-parametric statistics
3. distribution statistics
4. sampling statistics
A
MCQ: Process of converting inputs into outputs in presence of repeatedly same conditions is classified as
1. sampler
2. parameters
3. process
4. mixer
C
MCQ: Branch of statistics in which data is collected according to ordinal scale or nominal scale is classified as
1. distribution statistics
2. sampling statistics
3. parametric statistics
4. non-parametric statistics
D
MCQ: Time frame to complete a transaction in bank is classified as
1. parameters
2. process
3. mixer
4. sampler
B | 377 | 2,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-26 | latest | en | 0.870824 |
https://studylib.net/doc/5894727/physics-6b-practice-questions | 1,560,813,749,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998581.65/warc/CC-MAIN-20190617223249-20190618005249-00442.warc.gz | 591,878,393 | 42,778 | # Physics 6B Practice Questions
```Physics 6B Practice Midterm
A block of plastic with a density of 920 kg/m3 floats at the interface between oil of density 850
kg/m3 and water of density 1000 kg/m3, as shown. Calculate the percentage of the plastic which is
submerged in the water.
1.
a)
b)
c)
d)
23%
47%
72%
85%
oil
interface
plastic
water
A U-shaped tube contains 10 cm of alcohol (ρa=800 kg/m3), 12 cm of oil (ρo=850 kg/m3), and
water (ρw=1000 kg/m3) as shown. Find the distance h between the surface of the alcohol and the
surface of the water.
2.
a)
b)
c)
d)
2.0 cm
3.8 cm
8.5 cm
12.0 cm
10 cm
12 cm
alcohol
h=?
oil
water
3.
An incompressible fluid is flowing in a horizontal pipe. The radius of the pipe gradually decreases
from 10 cm to 5 cm. The velocity of the fluid in the narrow portion will be:
a)
b)
c)
d)
twice as large as in the wide section
half as large as in the wide section
the same as in the wide section
four times as large as in the wide section
4.
A 750g mass attached to a spring of spring constant k=124 N/m is at rest on a horizontal
frictionless surface. It is struck by a hammer which gives it a speed of 2.76 m/s directly toward
the spring. It does not move any significant distance while being struck. Which of these
statements is true?
a)
b)
c)
It will oscillate with a frequency of 12.86 Hz and an amplitude of 0.215 m.
It will oscillate with a period of 0.489 seconds and an amplitude of 0.304 m.
It will move a maximum distance of 21.5 cm from its equilibrium position, and take 0.244 seconds
It will move a total distance of 86 cm before it first returns to the equilibrium position, and this
will take 0.489 seconds.
d)
5.
a)
b)
c)
d)
What happens to a simple pendulum’s frequency if both the mass and length are increased?
The frequency increases.
The frequency decreases.
The frequency does not change.
The frequency may increase or decrease; it depends on the length to mass ratio.
6.
After plunging from a bridge, a 60kg bungee-jumper oscillates up and down, completing one cycle
every 2 seconds. If the bungee cord is assumed to be massless and has an unstretched length of 10m, find
the spring constant of the bungee cord.
a)
865 N/m
b)
243 N/m
c)
592 N/m
d)
1207 N/m
7.
A 50kg block is placed at the top of a cylindrical steel pole with a radius of 0.1m and a height of
1m. Determine the change in the length of the pole. The elastic (Young’s) modulus for steel is 200x109
N/m2.
a)
78 nm
b)
23 nm
c)
156 nm
d)
47 nm
8.
a)
b)
c)
d)
9.
a)
b)
c)
d)
10.
a)
b)
c)
d)
11.
a)
b)
c)
d)
You are in the front row of the Boombox Orchestra concert, standing 1m away from the speakers.
The sound intensity level is an earsplitting 120 db, so you decide to move away to a quieter
position. How far away from the speaker do you need to be so that the level is only 80 db?
10m
40m
100m
200m
A loudspeaker playing a constant frequency tone is dropped off a cliff. As it accelerates
downward, a person standing at the bottom of the cliff will hear a sound of:
increasing frequency and decreasing amplitude
constant frequency and increasing amplitude
increasing frequency and increasing amplitude
decreasing frequency and constant amplitude
A “boat” consists of a massless hollow cube of side length 50cm, floating in a freshwater lake.
When a person steps onto the “boat” it sinks down 30cm. Find the weight of the person.
500 N
750 N
1000 N
1500 N
Two strings on the same guitar (same length) are tuned so that string B is one octave higher
frequency than string A. Given that string A has 4 times the mass of string B, what is the ratio of
the tensions in the strings? (Hint: One octave higher frequency means the frequency is twice as
high).
String A has 4 times the tension of String B
String A has 2 times the tension of String B
The tensions in the strings are equal
String A has half the tension of String B
12.
One of the harmonics on a string that is 1.30 m long has a frequency of 15.60 Hz. The next higher
harmonic has frequency 23.40 Hz. Find the fundamental frequency and the speed of the waves on
the string.
a) 7.80 Hz, 20.3 m/s
b) 15.6 Hz, 24.3 m/s
c) 23.4 Hz, 9.8 m/s
d) 39.0 Hz, 20.3 m/s
13.
Two protons are placed as follows: Proton A is placed on the X-axis, +5 cm from the origin.
Proton B is placed on the Y-axis, +10 cm from the origin. Find the electric field at the origin
(magnitude and direction). The charge on a proton is 1.6 x 10 -19C.
7.2 x 10-7 N/C, 14° below negative x-axis
5.9 x 10-7 N/C, 14° above positive x-axis
7.2 x 10-7 N/C, 76° above positive x-axis
5.9 x 10-7 N/C, 14° below negative x-axis
a)
b)
c)
d)
Answers: 1)b 2)b 3)d 4)c 5)b 6)c 7)a 8)c 9)c 10)b 11)c 12)a 13)d
```
Nuclear physics
42 Cards
Mass
15 Cards | 1,404 | 4,684 | {"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} | 4.03125 | 4 | CC-MAIN-2019-26 | latest | en | 0.909384 |
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