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https://www.vtupulse.com/cbcs-ece-notes/17ec834-machine-learning-ml-vtu-notes/ | 1,702,239,402,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00652.warc.gz | 1,129,060,738 | 12,600 | # 17EC834 Machine learning – ML VTU Notes
[wptelegram-join-channel]
### 17EC834 Machine learning – ML VTU CBCS Notes
Here you can download the VTU CBCS 2017 Scheme notes, and Study materials of Machine learning – ML VTU of the Electronics and Communication Engineering department.
University Name: Visvesvaraya Technological University (VTU), Belagavi
Branch Name: Electronics and Communication Engineering – ECE
Semester: 8 (4th Year)
Subject Code and Subject Name: 17EC834 Machine learning – ML
Scheme of Examination: 2017 Scheme
## Machine Learning Video Tutorials
Important Concepts discussed:
Following are the contents of module 1 – Introduction to Machine Learning and Concept Learning
Introduction to Machine Learning. “Learning problems and Designing a Learning system. Different Perspectives and Machine Learning issues.”
Introduction to Concept Learning and Concept learning. “Concept learning as a search of a hypothesis. Find-S and Candidate Elimination algorithm. Version space, Inductive Bias of Find-S, and Candidate Elimination algorithm.”
Following are the contents of module 2 – Decision Tree Learning
Introduction to Decision Tree Learning Algorithm. “Decision tree representation and appropriate problems for
decision tree learning. The Decision Tree Learning Hypothesis space search, Inductive bias, and Issues in decision tree learning algorithm.”
Following are the contents of module 3 – Artificial Neural Networks
Introduction to Artificial Neural Networks. “Artificial Neural Network representation, appropriate problems Artificial Neural Network, Perceptrons, a sigmoid function, Back-propagation algorithm, and its derivation.”
Following are the contents of module 4 – Bayesian Learning
“Introduction to Bayesian Learning. Bayes theorem and its concept learning, Minimum Description Length principle. Introduction to Naive Bayes classifier and numerical example, Bayesian belief networks, and EM, K-means algorithm.”
Following are the contents of module 5 – Evaluating Hypothesis, Instance-Based, and Reinforcement Learning
Introduction to Evaluating Hypothesis. “Basics of the sampling theorem, General approach for deriving confidence intervals, calculating the difference in the error of two hypotheses, paired t-Tests, Comparing two learning algorithms.”
Click the below link to download the 2017 Scheme VTU CBCS Notes of 17EC834 Machine learning – ML
M1, M2, M3, M4, and M5 Another Seet M2, M3, M4, and M5 | 516 | 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} | 2.5625 | 3 | CC-MAIN-2023-50 | latest | en | 0.811705 |
https://www.aqua-calc.com/one-to-all/volume/preset/cubic-yard/16 | 1,642,415,916,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00352.warc.gz | 670,225,372 | 7,525 | # Convert cubic yards [yd³] to other units of volume
## yards³ [yd³] volume conversions
16 yd³ = 0.01 acre-foot yd³ to ac-ft 16 yd³ = 17 989 526.08 capsules 0 yd³ to cap 0 16 yd³ = 13 592 086.37 capsules 00 yd³ to cap 00 16 yd³ = 8 929 107.82 capsules 000 yd³ to cap 000 16 yd³ = 12 232 877.73 capsules 00E yd³ to cap 00E 16 yd³ = 15 683 176.58 capsules 0E yd³ to cap 0E 16 yd³ = 25 485 161.92 capsules 1 yd³ to cap 1 16 yd³ = 679 604.32 capsules 10 yd³ to cap 10 16 yd³ = 1 223 287.77 capsules 11 yd³ to cap 11 16 yd³ = 2 446 575.55 capsules 12 yd³ to cap 12 16 yd³ = 1 631 050.37 capsules 12el yd³ to cap 12el 16 yd³ = 3 822 774.29 capsules 13 yd³ to cap 13 16 yd³ = 33 980 215.84 capsules 2 yd³ to cap 2 16 yd³ = 45 306 954.56 capsules 3 yd³ to cap 3 16 yd³ = 61 164 388.64 capsules 4 yd³ to cap 4 16 yd³ = 94 099 059.52 capsules 5 yd³ to cap 5 16 yd³ = 509 703.24 capsules 7 yd³ to cap 7 16 yd³ = 436 888.49 capsules Su7 yd³ to cap Su7 16 yd³ = 1.22 × 10+31 cubic angstroms yd³ to ų 16 yd³ = 3.65 × 10-33 cubic astronomical unit yd³ to au³ 16 yd³ = 12 232 877.73 cubic centimeters yd³ to cm³ 16 yd³ = 0.002 cubic chain yd³ to ch³ 16 yd³ = 12 232.88 cubic decimeters yd³ to dm³ 16 yd³ = 0.01 cubic dekameter yd³ to dam³ 16 yd³ = 2 cubic fathoms yd³ to ftm³ 16 yd³ = 432 cubic feet yd³ to ft³ 16 yd³ = 1.5 × 10-6 cubic furlong yd³ to fur³ 16 yd³ = 1.22 × 10-5 cubic hectometer yd³ to hm³ 16 yd³ = 746 496 cubic inches yd³ to in³ 16 yd³ = 1.22 × 10-8 cubic kilometer yd³ to km³ 16 yd³ = 1.44 × 10-47 cubic light year yd³ to ly³ 16 yd³ = 12.23 cubic meters yd³ to m³ 16 yd³ = 7.46 × 10+23 cubic microinches yd³ to µin³ 16 yd³ = 1.22 × 10+19 cubic micrometers yd³ to µm³ 16 yd³ = 1.22 × 10+19 cubic microns yd³ to µ³ 16 yd³ = 7.46 × 10+14 cubic mils yd³ to mil³ 16 yd³ = 2.93 × 10-9 cubic mile yd³ to mi³ 16 yd³ = 12 232 877 727.74 cubic millimeters yd³ to mm³ 16 yd³ = 1.22 × 10+28 cubic nanometers yd³ to nm³ 16 yd³ = 1.93 × 10-9 cubic nautical mile yd³ to nmi³ 16 yd³ = 4.16 × 10-49 cubic parsec yd³ to pc³ 16 yd³ = 1.22 × 10+37 cubic picometers yd³ to pm³ 16 yd³ = 7.46 × 10+14 cubic thous yd³ to thou³ 16 yd³ = 122 328.78 deciliters yd³ to dl 16 yd³ = 1 223.29 dekaliters yd³ to dal 16 yd³ = 244 657 555.2 drops yd³ to gt 16 yd³ = 1.22 × 10-5 gigaliter yd³ to Gl 16 yd³ = 43 053.69 Imperial cups yd³ to imperial c 16 yd³ = 430 536.93 Imperial fluid ounces yd³ to imperial fl.oz 16 yd³ = 2 690.86 Imperial gallons yd³ to UK gal 16 yd³ = 21 526.85 Imperial pints yd³ to imperial pt 16 yd³ = 10 763.42 Imperial quarts yd³ to UK qt 16 yd³ = 12.23 kiloliters yd³ to kl 16 yd³ = 12 232.88 liters yd³ to l 16 yd³ = 0.01 megaliter yd³ to Ml 16 yd³ = 48 931.51 metric cups yd³ to metric c 16 yd³ = 1 223 287.78 metric dessertspoons yd³ to metric dstspn 16 yd³ = 815 525.18 metric tablespoons yd³ to metric tbsp 16 yd³ = 2 446 575.55 metric teaspoons yd³ to metric tsp 16 yd³ = 12 232 877.73 milliliters yd³ to ml 16 yd³ = 76.94 oil barrels yd³ to bbl 16 yd³ = 1.22 × 10-8 teraliter yd³ to Tl 16 yd³ = 51 705.35 US cups yd³ to US c 16 yd³ = 1 654 571.22 US dessertspoons yd³ to US dstspn 16 yd³ = 413 642.81 US fluid ounces yd³ to fl.oz 16 yd³ = 3 231.58 US gallons yd³ to US gal 16 yd³ = 25 852.68 US pints yd³ to pt 16 yd³ = 12 926.34 US quarts yd³ to US qt 16 yd³ = 827 285.61 US tablespoons yd³ to US tbsp 16 yd³ = 2 481 856.83 US teaspoons yd³ to US tsp
#### Foods, Nutrients and Calories
GOURMET SEAFOOD SALAD, UPC: 018894329978 weigh(s) 224 grams per metric cup or 7.5 ounces per US cup, and contain(s) 236 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
23525 foods that contain Magnesium, Mg. List of these foods starting with the highest contents of Magnesium, Mg and the lowest contents of Magnesium, Mg, and Recommended Dietary Allowances (RDAs) for Magnesium
#### Gravels, Substances and Oils
Substrate, Mix weighs 1 153 kg/m³ (71.97944 lb/ft³) with specific gravity of 1.153 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
OrganoClay MCM-830P weighs 705 kg/m³ (44.01171 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-32, liquid (R32) with temperature in the range of -40°C (-40°F) to 65.56°C (150.008°F)
#### Weights and Measurements
A yoctoohm is a SI-multiple (see prefix yocto) of the electric resistance unit ohm and equal to equal to 1.0 × 10-24 ohm
The length measurement was introduced to measure distance between any two objects.
dwt/metric c to kg/US qt conversion table, dwt/metric c to kg/US qt unit converter or convert between all units of density measurement.
#### Calculators
Ellipse calculator, equations, area, vertices and circumference | 2,009 | 4,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.383422 |
https://oeis.org/A335645 | 1,643,427,333,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00218.warc.gz | 477,745,418 | 4,179 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A335645 Smallest palindrome with exactly n distinct prime factors. 2
1, 2, 6, 66, 858, 6006, 222222, 20522502, 244868442, 6172882716, 231645546132, 49795711759794, 2415957997595142, 495677121121776594, 22181673755737618122 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS max{A002110(n), A076886(n), A239696(n)} <= a(n) <= A046399(n). No more terms with less than 17 digits. Next term: 10^16 <= a(13) <= 495677121121776594. LINKS EXAMPLE a(3)=66 because 66 is the smallest palindromic number with 3 distinct prime factors, 2*3*11. PROG (Python) from sympy import factorint def A335645(n): d = 1 while True: half = (d+1)//2 for left in range(10**(half-1), 10**half): strleft = str(left) if d%2 == 0: m = int(strleft + strleft[::-1]) else: m = int(strleft + (strleft[:-1])[::-1]) if len(factorint(m)) == n: return m d += 1 print([A335645(n) for n in range(8)]) # Michael S. Branicky, Oct 02 2020 CROSSREFS Subsequence of A002113. Cf. A002110, A046399, A076886, A239696. Sequence in context: A262047 A280393 A239696 * A082619 A046399 A082617 Adjacent sequences: A335642 A335643 A335644 * A335646 A335647 A335648 KEYWORD nonn,base,hard,more AUTHOR Michael S. Branicky, Oct 02 2020 EXTENSIONS a(13) from Michael S. Branicky and David A. Corneth, Oct 03 2020 a(14) from David A. Corneth, Oct 03 2020 STATUS approved
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Last modified January 28 22:30 EST 2022. Contains 350669 sequences. (Running on oeis4.) | 627 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-05 | latest | en | 0.535631 |
https://www.convertunits.com/from/megameter/to/point | 1,643,410,268,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00704.warc.gz | 768,848,349 | 17,324 | ## ››Convert megametre to point [Adobe]
megameter point
Did you mean to convert megameter to point [Adobe] point [Britain, US] point [Didot] point [TeX]
How many megameter in 1 point? The answer is 3.5277777777778E-10.
We assume you are converting between megametre and point [Adobe].
You can view more details on each measurement unit:
megameter or point
The SI base unit for length is the metre.
1 metre is equal to 1.0E-6 megameter, or 2834.6456692913 point.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between megameters and points.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of megameter to point
1 megameter to point = 2834645669.2913 point
2 megameter to point = 5669291338.5827 point
3 megameter to point = 8503937007.874 point
4 megameter to point = 11338582677.165 point
5 megameter to point = 14173228346.457 point
6 megameter to point = 17007874015.748 point
7 megameter to point = 19842519685.039 point
8 megameter to point = 22677165354.331 point
9 megameter to point = 25511811023.622 point
10 megameter to point = 28346456692.913 point
## ››Want other units?
You can do the reverse unit conversion from point to megameter, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Megameter
A megametre (American spelling: megameter, symbol: Mm) is a unit of length equal to 10^6 metres (from the Greek words megas = big and metro = count/measure). Its customary equivalent is 621.37 miles.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 572 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.745893 |
https://www.airmilescalculator.com/distance/lbs-to-vbv/ | 1,675,532,758,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00828.warc.gz | 634,441,351 | 9,750 | # How far is Vanua Balavu from Labasa?
The distance between Labasa (Labasa Airport) and Vanua Balavu (Vanuabalavu Airport) is 124 miles / 200 kilometers / 108 nautical miles. The estimated flight time is 44 minutes.
124
Miles
200
Kilometers
108
Nautical miles
## Distance from Labasa to Vanua Balavu
There are several ways to calculate the distance from Labasa to Vanua Balavu. Here are two standard methods:
Vincenty's formula (applied above)
• 124.405 miles
• 200.210 kilometers
• 108.105 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 124.382 miles
• 200.173 kilometers
• 108.085 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 Labasa to Vanua Balavu?
The estimated flight time from Labasa Airport to Vanuabalavu Airport is 44 minutes.
## What is the time difference between Labasa and Vanua Balavu?
There is no time difference between Labasa and Vanua Balavu.
## Flight carbon footprint between Labasa Airport (LBS) and Vanuabalavu Airport (VBV)
On average, flying from Labasa to Vanua Balavu generates about 43 kg of CO2 per passenger, and 43 kilograms equals 95 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path from Labasa to Vanua Balavu
Shortest flight path between Labasa Airport (LBS) and Vanuabalavu Airport (VBV).
## Airport information
Origin Labasa Airport
City: Labasa
Country: Fiji
IATA Code: LBS
ICAO Code: NFNL
Coordinates: 16°28′0″S, 179°20′23″E
Destination Vanuabalavu Airport
City: Vanua Balavu
Country: Fiji
IATA Code: VBV
ICAO Code: NFVB
Coordinates: 17°16′8″S, 178°58′33″W | 518 | 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} | 2.703125 | 3 | CC-MAIN-2023-06 | latest | en | 0.84097 |
http://stackoverflow.com/questions/10633245/cannot-traverse-a-tree-nodes-are-re-traversed-a-huge-amount-of-times-but-still | 1,427,850,488,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131302428.75/warc/CC-MAIN-20150323172142-00274-ip-10-168-14-71.ec2.internal.warc.gz | 247,674,357 | 15,363 | # Cannot traverse a tree- nodes are re-traversed a huge amount of times but still acyclical
I've decided to admit defeat. Here is my code. It should build a tree of the execution of the SHA-2 algorithm for a single chunk 512bit input without preprocessing. However, I turned it down from 64 iterations to 4 iterations because traversing the tree never finished for 64 iterations. Even at 4 iterations, it traverses 10,000 nodes- even when only 1000 have been allocated, including a bunch of constants which are leaves and not even counted in the traversal count. In addition, the assertions guarantee acyclical, and if it was cyclical, it would simply never return, not take forever and then return.
What on earth have I done with this code to cause it to take forever to traverse for such a tiny acyclical tree?
-
Well, that is to be expected. The tree may be stored in a very compact manner, but it has lots and lots of repeated portions. It is massively big.
As an example, consider this sequence:
``````auto s0 = la.rotate(2) ^ la.rotate(13) ^ la.rotate(22);
``````
Now the `s0` tree has three of the original `la` expressions below it.
``````auto t2 = s0 + maj;
``````
... and that tree goes in `t2`...
``````la = t1 + t2;
``````
... which ends up in `la` again at the end of the loop. So, `la` has now (at least) three references to the `la` tree of the previous iteration. The next iteration will again triplicate this. And over and over. From this we can derive a lower bound of 3^64 instances of the original `la` in the final tree (even though only one will exist in storage). 3^64 is 3.43368382 × 10^30, i.e., about three gazillions.
The generation of the tree takes the easy route and just reuses the subtrees. The traversal function on the other hand, will end up traversing the same subtree many, many times.
- | 469 | 1,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-14 | latest | en | 0.940938 |
https://www.coursehero.com/file/5802402/Final-Fall-1999-2000-Abu-Khuzam-Lyzzaik/ | 1,521,783,308,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648178.42/warc/CC-MAIN-20180323044127-20180323064127-00043.warc.gz | 768,687,135 | 29,001 | {[ promptMessage ]}
Bookmark it
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Final_Fall-1999-2000_Abu-Khuzam_Lyzzaik
# Final_Fall-1999-2000_Abu-Khuzam_Lyzzaik - January 27 2000...
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Unformatted text preview: January 27 , 2000 FINAL EXAMINATION VERSION II MATHEMATICS 201 FALL 1999‘ 2000 Time : 2 Hours Circle out your instructor’s name and your section number Prof. H.Abu-Khuzam ............................................. Section: 6 and 12 Prof. A. Lyzzaik ................................................... Section: 5 and 7 NOTE: Make sure that you have 6 pages and twenty questions Circle the correct answer in each of the followin roblems 5 oints for each correct answer, and 0 f0!“ no 01' more than one answer! I 1 no 2" . _ 3 H . 1. The interval of convergence of the followmg series 2 —-—92—) IS nil H 5 7 5 7 a A —<~<—— B —S.S-— C ls SS ( ) 2 x 2 ( ) 2 x 2 ( ) x (D) 1< x < 5 (E) none of the above 2. The surface £- y_‘ = 5 is a 36 6 (A) elliptic paraboloid (B) elliptic cone (C) circular paraboloid (D) hyperbolic paraboloid (E) none of the above 3. The sum of the series 2nd I 2)”1 is "=1 (A) 6 (B) .32 (C) 4 (D) 25/4 (E) noncofthe above 4. Which one of the following series converges: (A) ELM - JR] (B) i W” (C) in ”=1 ":1 (3”)! ”=1 (D) icosmr (E) ii n=l rr=l n S. An estimate of the magnitude of the error obtained by taking the first four terms of co . 1 H the series 2 (—1)"+1 (0 0 ) 3 "=1 ['1 to appmximate its value is (A) 1.56 x 10‘12 (B) s x 10‘” (C) s x 10'” (D) 1.56 x 10'”J (E) none of the above . _ 7r ,1 6. The sequence whose n-th term 18 an = (1 ~ ——)2 n (A) Converges to e"r (B) Diverges (C) Converges to 62” (D) Converges to e‘?”r (E) none of the above 7. The series GOSH” ”:1 n Inn (A) is not alternating (B) diverges (C) converges conditionally (D) converges absolutely (E) none of the above 8. The spherical coordinate equation for the sphere x2 + y2 + (z — 4) 3 = 16 is (A)p=4cosB (B) p=4cos¢ ,(C)p=8cos<|) (D) p = 8 cos 8 (E) none of the above l 9. The value of the double integral Evaluate j Jay: dydx is 0.1:]? (A) l - lie (B) l + l/e (C) l (D) e (E) none ofthe above 10. If f(x,y) = ln xy +ln yz +1n x2, then the derivative of f at P(1,l,l) in the direction where f increases most rapidly is (A) J3 (B) 2% (C) 0 (D) 3J3— (E) none of the above — 4 11. Using the Maclaurin series for , the Maclaurin series of ——:£3—7 is 1+ x (1 + x )' (A) 25HXSH U3) :‘(--1)"5rtt5"_1 (C) 2(— 1) "+1 Sims"—1 “:1 ”:1 ”:1 (D) Z — SHXS’H rt=l (E) none of the above 4-4] A )" fbr(x,_v)¢(0,0) Then 12. Consider the function f(x,y) = x6 + y" k for(x, y) = (0,0) (A) f is discontinuous at (0,0) [or all values of k. (B) f is continuous at (0,0) provided that k:() (C) f is continuous at (0,0), provided that k=l (D) f is continuous at (0,0) , provided that kz—l. (E) f is continuous at (0,0) for all values of k. 13. The function f(x, y) =—.):3 — y3 —5xy “1. has (A) saddle point at (0,0) and local maximum at (—§,— §) . (B) saddle point at (0,0) and local minimum at (—§,-— %) . (C) saddle point at (—%,— g) and local minimum at (0,0). (D) saddle point at (—§,—%) and local maximum at (0,0). . 5 5 (E) local maxrmum at (—§,—- :) . J 14. The function f (x, y) =3xy — 6.x—3y+7 defined on the triangular plate R with vertices (0,0), (3,0), and (0,5) has 9 . . (A) an absolute maximum g and absolute mmlmum e8. . 9 . . (B) an absolute maxrmum g and absolute minimum —1 l. (C) an absolute maximum 7 and absolute minimum 1. (D) an absolute maximum 8 and absolute minimum —l l. (E) none of the above. 15. The area of the region that is inside the cardioid r = 4 + 4 cos 0 and outside the circle r = 6 is (A) 18J§ +2n (B) 18J§ — 411 (C) 51: (D) 5 45 +10 (E) none of the above 16. Using triple integration in cylindrical coordinates, the volume of the solid bounded above by the hemisphere z = 1125 --x2 - y2 , below by the xy—plane, and laterally by the cylinder x2 + yz :9 is (A) 401t/3 (B) 351113 (C) 1221:13 (D) 4611/3 (E) none of the above ...
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Topic: Help Needed in math
Replies: 25 Last Post: May 2, 2014 6:01 PM
Messages: [ Previous | Next ]
William Elliot Posts: 2,637 Registered: 1/8/12
Re: Help Needed in math
Posted: Apr 30, 2014 11:59 PM
> > > > > > Find the largest values of the integer a for which ax^2-5x-3 is
> > > > > > negative for all values of x.
> > > > > set a = (5x + 3) / x^2 and find the minima of this function. This
> > > > > will, with your constraints, give the answer: a = -3
> > > > 5/x + 3/x^2; -5/x^2 - 6/x^3 = 0; -5x + -6 = 0; x = -6/5
> > > > a = (-6 + 3)/(36/25) = -25/12
> > > > Why does the minimum work?
> a < f(x) = (5x + 3) / x^2, x <> 0
What's f?
> If a < f(x) for all x, then a has to be less than the global minimum of f(x).
> The global minimum of f(x) = -25/12. The largest integer less than this is -3.
Date Subject Author
4/28/14 punisher
4/28/14 quasi
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/28/14 Peter Percival
4/29/14 William Elliot
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
5/1/14 Karl-Olav Nyberg
5/1/14 William Elliot
5/1/14 quasi
5/2/14 William Elliot
5/2/14 quasi
5/2/14 William Elliot
5/2/14 snmpprotocol@gmail.com
5/2/14 quasi
5/2/14 Karl-Olav Nyberg
5/2/14 quasi | 604 | 1,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-09 | longest | en | 0.708793 |
https://www.physicsforums.com/threads/how-do-you-intergrate-lnx.303696/ | 1,513,265,431,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544677.45/warc/CC-MAIN-20171214144324-20171214164324-00699.warc.gz | 793,756,618 | 13,835 | # How do you intergrate lnx?
1. Mar 30, 2009
### matt_crouch
how do you intergrate lnx?
2. Mar 30, 2009
### Count Iblis
Re: intergratation
There are two easy ways:
1) Partial integration
2) Differentiation w.r.t. a parameter.
Method 1):
Leibnitz's rule can be written as:
d(fg) = f dg + g df
We can rewrite this as:
f dg = d(fg) - g df (1)
So, if we want to integrate ln(x), we can write according to Eq. (1):
ln(x) dx = d[x ln(x)] - x d[ln(x)] = d[xln(x)] - dx
So, the integral of ln(x) dx is the integral of d[xln(x)] minus the integral of dx, which is x ln(x) - x plus an arbitrary constant.
Method 2):
Consider integrating the function x^p. Then differentiate both sides w.r.t. the parameter p. Then set p equal to zero. Try it! | 242 | 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} | 3.9375 | 4 | CC-MAIN-2017-51 | longest | en | 0.682248 |
http://mathhelpforum.com/statistics/89027-probability-problem.html | 1,481,043,799,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541910.39/warc/CC-MAIN-20161202170901-00226-ip-10-31-129-80.ec2.internal.warc.gz | 174,304,475 | 9,979 | # Thread: probability problem
1. ## probability problem
Hi I was wondering if anyone could help me with this problem.
Only about 14% of senior citizens get the flu each year. However, about 24% of people under 65 years old get the flu each year. The general population consists of 12.5% senior citizens.
a) What is the probability that a person selected at random from the general population is a senior citizen who will get the flu this year?
b) What is the probability that a person selected at random from the general population is a person under age 65 who will get the flu this year?
Any help would be greatly appreciated.
2. Originally Posted by Klynn
Hi I was wondering if anyone could help me with this problem.
Only about 14% of senior citizens get the flu each year. However, about 24% of people under 65 years old get the flu each year. The general population consists of 12.5% senior citizens.
a) What is the probability that a person selected at random from the general population is a senior citizen who will get the flu this year?
b) What is the probability that a person selected at random from the general population is a person under age 65 who will get the flu this year?
Any help would be greatly appreciated.
a) $(\frac{12.5}{100})(\frac{14}{100})$
b) $(\frac{88.5}{100})(\frac{24}{100})$
3. Originally Posted by noob mathematician
a) $(\frac{12.5}{100})(\frac{14}{100})$
b) $(\frac{88.5}{100})(\frac{24}{100})$
Thanks so much! | 366 | 1,462 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2016-50 | longest | en | 0.95495 |
http://www.slideserve.com/naeva/sequential-sums-of-squares-1326315 | 1,505,952,107,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687582.7/warc/CC-MAIN-20170920232245-20170921012245-00019.warc.gz | 561,821,271 | 16,290 | 1 / 30
# Sequential sums of squares - PowerPoint PPT Presentation
Sequential sums of squares. … or … extra sums of squares. Sequential sums of squares: what are they?. The reduction in the error sum of squares when one or more predictor variables are added to the regression model.
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### Sequential sums of squares
… or … extra sums of squares
Sequential sums of squares: what are they?
• The reduction in the error sum of squares when one or more predictor variables are added to the regression model.
• Or, the increase in the regression sum of squares when one or more predictor variables are added to the regression model.
• They can be used to test whether one slope parameter is 0.
• They can be used to test whether a subset (more than two, but less than all) of the slope parameters are 0.
• Sample of n = 38 college students
• Response (Y): intelligence based on the PIQ (performance) scores from the (revised) Wechsler Adult Intelligence Scale.
• Predictor (X1): Brain size based on MRI scans (given as count/10,000)
• Predictor (X2): Height in inches
• Predictor (X3): Weight in pounds
The regression equation is PIQ = 4.7 + 1.18 MRI
Predictor Coef SE Coef T P
Constant 4.65 43.71 0.11 0.916
MRI 1.1766 0.4806 2.45 0.019
Analysis of Variance
Source DF SS MS F P
Regression 1 2697.1 2697.1 5.99 0.019
Error 36 16197.5 449.9
Total 37 18894.6
The regression equation is
PIQ = 111 + 2.06 MRI - 2.73 Height
Predictor Coef SE Coef T P
Constant 111.28 55.87 1.99 0.054
MRI 2.0606 0.5466 3.77 0.001
Height -2.7299 0.9932 -2.75 0.009
Analysis of Variance
Source DF SS MS F P
Regression 2 5572.7 2786.4 7.32 0.002
Residual 35 13321.8 380.6
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Height 1 2875.6
The regression equation is
PIQ = 111 + 2.06 MRI - 2.73 Height + 0.001 Weight
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
MRI 2.0604 0.5634 3.66 0.001
Height -2.732 1.229 -2.22 0.033
Weight 0.0006 0.1971 0.00 0.998
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Height 1 2875.6
Weight 1 0.0
Sequential sums of squares: definition using SSE notation
• SSR(X2|X1) = SSE(X1) - SSE(X1,X2)
• In general, you subtract the error sum of squares due to all of the predictors both left and right of the bar from the error sum of squares due to the predictor to the right of the bar.
• SSR(X2,X3|X1) = SSE(X1) - SSE(X1,X2,X3)
Sequential sums of squares: definition using SSR notation
• SSR(X2|X1) = SSR(X1,X2) – SSR(X1)
• In general, you subtract the regression sum of squares due to the predictor to the right of the bar from the regression sum of squares due to all of the predictors both left and right of the bar.
• SSR(X2,X3|X1) = SSR(X1,X2,X3)-SSR(X1)
In multiple regression, there is more than one way to decompose the regression sum of squares. For example:
The regression equation is
PIQ = 111 + 2.06 MRI - 2.73 Height
Predictor Coef SE Coef T P
Constant 111.28 55.87 1.99 0.054
MRI 2.0606 0.5466 3.77 0.001
Height -2.7299 0.9932 -2.75 0.009
Analysis of Variance
Source DF SS MS F P
Regression 2 5572.7 2786.4 7.32 0.002
Residual 35 13321.8 380.6
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Height 1 2875.6
The regression equation is
PIQ = 111 - 2.73 Height + 2.06 MRI
Predictor Coef SE Coef T P
Constant 111.28 55.87 1.99 0.054
Height -2.7299 0.9932 -2.75 0.009
MRI 2.0606 0.5466 3.77 0.00
Analysis of Variance
Source DF SS MS F P
Regression 2 5572.7 2786.4 7.32 0.002
Error 35 13321.8 380.6
Total 37 18894.6
Source DF Seq SS
Height 1 164.0
MRI 1 5408.8
A sequential sum of squares involving one extra predictor variable has one degree of freedom associated with it:
A sequential sum of squares involving two extra predictor variables has two degrees of freedom associated with it:
Degrees of freedom and regression mean squares
• The SSR is automatically decomposed into one-degree-of-freedom sequential sums of squares, in the order in which the predictor variables are entered into the model.
• To get sequential sum of squares involving two or more predictor variables, sum the appropriate one-degree-of-freedom sequential sums of squares.
The regression equation is
PIQ = 111 + 2.06 MRI - 2.73 Height + 0.001 Weight
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
MRI 2.0604 0.5634 3.66 0.001
Height -2.732 1.229 -2.22 0.033
Weight 0.0006 0.1971 0.00 0.998
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Height 1 2875.6
Weight 1 0.0
The regression equation is
PIQ = 111 - 2.73 Height + 0.001 Weight + 2.06 MRI
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
Height -2.732 1.229 -2.22 0.033
Weight 0.0006 0.1971 0.00 0.998
MRI 2.0604 0.5634 3.66 0.001
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
Height 1 164.0
Weight 1 169.5
MRI 1 5239.2
Testing one slope β1= βMRI is 0
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
Height -2.732 1.229 -2.22 0.033
Weight 0.0006 0.1971 0.00 0.998
MRI 2.0604 0.5634 3.66 0.001
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
Height 1 164.0
Weight 1 169.5
MRI 1 5239.2
Testing one slope β2= βHT is 0
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
MRI 2.0604 0.5634 3.66 0.001
Weight 0.0006 0.1971 0.00 0.998
Height -2.732 1.229 -2.22 0.033
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Weight 1 940.9
Height 1 1934.7
Testing one slope β3= βWT is 0
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
MRI 2.0604 0.5634 3.66 0.001
Height -2.732 1.229 -2.22 0.033
Weight 0.0006 0.1971 0.00 0.998
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Height 1 2875.6
Weight 1 0.0
Reduced model:
Testing one slope βk is 0: why it works?
becomes:
Testing one slope βk is 0: why it works? (cont’d)
Reduced model:
Testing whether β2 = β3 = 0
becomes:
Testing whether β2 = β3 = 0 (cont’d)
The regression equation is
PIQ = 111 + 2.06 MRI - 2.73 Height + 0.001 Weight
Predictor Coef SE Coef T P
Constant 111.35 62.97 1.77 0.086
MRI 2.0604 0.5634 3.66 0.001
Height -2.732 1.229 -2.22 0.033
Weight 0.0006 0.1971 0.00 0.998
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
MRI 1 2697.1
Height 1 2875.6
Weight 1 0.0
Cumulative Distribution Function
F distribution with 2 DF in numerator and 34 DF in denominator
x P( X <= x )
3.6700 0.9640
• Select Calc >> Probability Distributions >> F…
• Select Cumulative Probability. Use default noncentrality parameter of 0.
• Type in numerator DF and denominator DF.
• Select Input constant. Type in F-statistic. Answer appears in session window.
• P-value is 1 minus the number that appears.
Test whether β1 = β3 = 0
Analysis of Variance
Source DF SS MS F P
Regression 3 5572.7 1857.6 4.74 0.007
Error 34 13321.8 391.8
Total 37 18894.6
Source DF Seq SS
Height 1 164.0
Weight 1 169.5
MRI 1 5239.2 | 3,009 | 8,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-39 | latest | en | 0.840117 |
http://www.jiskha.com/display.cgi?id=1316802905 | 1,495,771,929,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608633.44/warc/CC-MAIN-20170526032426-20170526052426-00067.warc.gz | 673,348,679 | 4,281 | # binomial probabbility
posted by on .
Assume that a procedure yields a binomial distribution with a trial repeated n times. use the binomial probability formula to find the probability of x success given the probability p of success on a single trial.
1. by formula- n=9, x=2, p=0.35
2. by PDF- n=15, x=13, p=1/3
• binomial probabbility - ,
C(9,2) = 9!/[2!(7!)]
= 9*8/2 = 9*4 = 36
so 36(.35)^2(1-.35)^7
= 36 * .1225 * .049
= .216
I do not know why by PDF means (By probability distribution function?)
• binomial probabbility - ,
yes thank you I believe that's what it stands for...
• binomial probabbility - ,
Well, I suppose I could look it up in a table of the distribution for n = 15, x =13, p =.3333... by symmetry
it is the same as for n = 15, x = 2, p = .33333....
but I do not have a good table
I have only up to n = 10
so have o do it the same old way
C(15,13) = 105
so
105 * (1/3)^13 * (2/3)^2
= 2.927*10^-5
c(15,13)
• binomial probabbility - ,
Now maybe they mean using the normal function which is the limit of binomial for large n
u = n p = 15*1/3 = 5
s^2 = 5(2/3) = 10/3 = 3.333...
s = 1.83
z = (x-u)/s = (13-5)/1.83 = 4.38
f(z) = (1/sqrt(2pi) e^-(4.38^2/2)
= 2.72 * 10^-5
gee, not too far off
• binomial probabbility - ,
For a binomial distribution with parameters
n = 5 , p = 0.3 . Find the probabilities of
getting :
(i) Atleast 3 successes.
(ii) Atmost 3 successes.
n = 5 , p = 0.3 | 527 | 1,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-22 | latest | en | 0.838885 |
https://lexique.netmath.ca/en/system-of-equations/ | 1,680,390,549,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00485.warc.gz | 402,879,106 | 19,711 | # System of Equations
## System of Equations
A finite set of equations that can be verified simultaneously.
A system of equations in two variables can be represented in a Cartesian plane using the graph of each equation. The points of intersection of these graphs are the solutions to the system.
### Example
Consider the following system of equations:
$$\left\{\begin{matrix} y=2x\\ y=4x^2-1 \end{matrix}\right.$$
The solutions to this system are the ordered pairs of coordinates ($$\frac{(1+\sqrt{5})}{4}$$, $$\frac{(1+\sqrt{5})}{2}$$)
and ($$\frac{(1-\sqrt{5})}{4}$$, $$\frac{(1-\sqrt{5})}{2}$$). | 171 | 606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-14 | longest | en | 0.799193 |
https://talesofpiglingbland.com/boil/what-temperature-is-medium-high-on-a-charcoal-grill.html | 1,656,520,462,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103640328.37/warc/CC-MAIN-20220629150145-20220629180145-00437.warc.gz | 584,032,176 | 19,174 | # What temperature is medium high on a charcoal grill?
Contents
## What is medium high on charcoal grill?
Telling the Temperature
2 to 3 seconds: the fire is medium-high heat, about 375°F, and coals have a faint coat of ash. 3 to 4 seconds: the fire is medium heat, about 350°F, and coals have a light coat of ash. 4 to 5 seconds: the fire is medium-low heat, about 325°F, and coals have a moderate coat of ash.
## What is medium heat on charcoal?
The number of seconds you can keep your hand there gives you an indication of how hot the coals are at the grate. High heat450°F to 550°F2 to 4 seconds. Medium heat350°F to 450°F5 to 6 seconds. Low heat250°F to 350°F8 to 10 seconds.
## What temperature should your charcoal grill be at?
When it comes to smoked meats, 225 is the magic number—well, 225°F, that is. That’s the sweet spot of your grill, the temperature that, with a little finessing and some patience, yields tender, juicy, pulls-apart-without-even-trying barbecued meat.
## How do you keep a charcoal grill at 250 degrees?
Direct Cooking:
1. Light enough coals to bring the grill up to the desired temperature. Usually around five lit coals will get you up to 225-250°F.
2. Spread unlit coals evenly on one side of the coal grate.
3. Place the lit coals evenly amongst the unlit coals.
4. Place your food above the coals and put on the lid.
IT\'S FUNNING: Can you cook in a glass pan on the stove?
## How many briquettes is 350 degrees?
So, in the case of an 8” Dutch oven, to get a temperature of 350° you need a total of 16 briquettes. Below the 16 you will notice the numbers 11/5. 11 is the number of briquettes for the top of the oven. 5 is the number of briquettes to go under the oven.
## What temperature is a medium grill Celsius?
Oven Temperature Conversions
Cooking Instructions Fahrenheit Celsius
Moderately Hot 375 190
Moderate/Medium 350 175
Warm 325 165
Slow/Low 300 150
## What temp should I cook steak on grill?
The best temperature for steaks is 450°F to 500°F. 4. Put your steaks on the grill, close the lid, and set your timer for 2 to 3 minutes, depending on the thickness of your steak.
## Is 350 degrees medium heat?
In this way, is 350 degrees medium heat? I’ve discovered that, with an IR thermometer to see pan bottom temperature after 5 minutes of heat (the weather is no more climbing ), high contrasts to 375 degrees F, moderate-high to 330 degrees, moderate to 300 levels, and reduced to approximately 275 degrees.
## How high should grill be above charcoal?
REASON: Manufacturers design their grills for 5″ between the top of the grill and the top of the layer of charcoal briquettes, so that the temperature will be about 370°. A temperature of 370° is ideal for cooking hamburgers all the way through.
## What temperature does coal burn at?
The burning of coal can produce combustion gases as hot as 2,500 °C (4,500 °F), but the lack of materials that can withstand such heat forces even modern power plants to limit steam temperatures to about 540 °C (1,000 °F)—even though the thermal efficiency of a power plant increases with increasing operating fluid ( …
IT\'S FUNNING: How do you clean Weber porcelain enameled cast iron cooking grates?
## Does more charcoal mean more heat?
When grilling on high heat, create a two-fire zone: Stack more coals on one side of the grill for higher-temperature cooking, and the other side of the grill should have less charcoal for lower-temperature cooking.
## What is the vent for on a charcoal grill?
Those are the vents! Both control the flow of air inside of the grill, which changes the heat level and direction. … Closed vents mean less oxygen, which in turn means less heat and slower-burning charcoal.
## How do I know if my grill is hot enough?
If you need to pull your hand away after 2 to 4 seconds, the heat is high. If you need to pull your hand away after 5 to 7 seconds, the heat is medium. If you need to pull your hand away after 8 to 10 seconds, the heat is low. | 1,003 | 3,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.868367 |
http://docplayer.net/885293-Chapter-15-current-liabilities-management.html | 1,521,783,273,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648178.42/warc/CC-MAIN-20180323044127-20180323064127-00565.warc.gz | 71,981,754 | 25,059 | # Chapter 15 Current Liabilities Management
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1 Chapter 15 Current Liabilities Management Solutions to Problems P15-1. LG 1: Payment Dates December 25 (b) December 30 (c) January 9 (d) January 30 P15-2. LG 1: Cost of Giving Up Cash Discount ( ) (365 20) = 37.24% (b) ( ) (365 20) = 18.43% (c) ( ) (365 35) = 21.28% (d) ( ) (365 35) = 32.25% (e) ( ) (365 50) = 7.37% (f) ( ) (365 20) = 56.44% (g) ( ) ( ) = 8.95% P15-3. LG 1: Credit Terms 1/15 net 45 date of invoice 2/10 net 30 EOM 2/7 net 28 date of invoice 1/10 net 60 EOM (b) 45 days 50 days 28 days 80 days
2 (c) CD % CD N 1% % 1% 30 Cost of giving up cash discount = = = 12.29% 2% % 2% 20 Cost of giving up cash discount = = = 37.23% 2% % 2% 21 Cost of giving up cash discount = = = 36.46% 1% % 1% 50 Cost of giving up cash discount = = = 14.89% (d) In all four cases the firm would be better off to borrow the funds and take the discount. The annual cost of not taking the discount is greater than the firm s 8% cost of capital. P15-4. LG 1: Cash Discount versus Loan Cost of giving up cash discount = ( ) (365 35) = 32.25% Since the cost of giving up the discount is higher than the cost of borrowing for a short-term loan, Erica is correct; her boss is incorrect. P15-5. LG 1, 2: Cash Discount Decisions (b) Supplier Cost of Forgoing Discount Decision J ( ) (365 20) = 18.43% Borrow K ( ) (365 60) = 12.42% Give up L ( ) (365 40) = 9.22% Give up M ( ) (365 45) = 25.09% Borrow Prairie would have lower financing costs by giving up Ks and Ls discount since the cost of forgoing the discount is lower than the 16% cost of borrowing. (c) Cost of giving up discount from Supplier M = ( ) (365 75) = 15.05% In this case the firm should give up the discount and pay at the end of the extended period. P15-6. LG 2: Changing Payment Cycle Annual Savings = (\$10,000,000) (0.13) = \$1,300,000
3 P15-7. LG 2: Spontaneous Sources of Funds, Accruals Annual savings = \$750, = \$82,500 P15-8. LG 3: Cost of Bank Loan Interest = (\$10, ) (90 365) = \$ \$375 (b) Effective 90 day rate = = 3.75% \$10,000 (c) Effective annual rate = ( ) 4 1 = 15.87% P15-9. LG 3: Effective Annual Rate of Interest \$10, Effective interest = = 14.29% [\$10,000 ( )] P LG 3: Compensating Balances and Effective Annual Rates Compensating balance requirement = \$800,000 borrowed 15% = \$120,000 Amount of loan available for use = \$800,000 \$120,000 = \$680,000 Interest paid = \$800, % = \$88,000 \$88, 000 Effective interest rate = = 12.94% \$680, 000 (b) Additional balances required = \$120,000 \$70,000 = \$50,000 \$88, 000 Effective interest rate = = 11.73% \$800, 000 \$50, 000 (c) Effective interest rate = 11% (None of the \$800,000 borrowed is required to satisfy the compensating balance requirement.) (d) The lowest effective interest rate occurs in situation (c), when Lincoln has \$150,000 on deposit. In situations and (b), the need to use a portion of the loan proceeds for compensating balances raises the borrowing cost.
4 P LG 3: Compensating Balance vs. Discount Loan \$150, \$13, 500 State Bank interest = = = 10.0% \$150,000 (\$150, ) \$135, 000 This calculation assumes that Weathers does not maintain any normal account balances at State Bank. \$150, \$13, 500 Frost Finance interest = = = 9.89% \$150,000 (\$150, ) \$136, 500 (b) If Weathers became a regular customer of State Bank and kept its normal deposits at the bank, then the additional deposit required for the compensating balance would be reduced and the cost would be lowered. P LG 3: Integrative Comparison of Loan Terms Challenge ( ) 0.80 = % (b) Effective annual interest rate = [\$2,000,000 ( ) + (0.005 \$2,000,000)] = % (\$2,000, ) (c) The revolving credit account seems better, since the cost of the two arrangements is the same; with a revolving loan arrangement, the loan is committed. P LG 4: Cost of Commercial Paper \$1,000,000 \$978,000 Effective 90-day rate = = 2.25% \$978,000 Effective annual rate = ( ) 365/90 1 = 9.44% [\$1,000,000 \$978,000 + \$9,612] (b) Effective 90-day rate = = 3.26% (\$978,000 - \$9,612) Effective annual rate = ( ) 365/90 1 = 13.89% P LG 5: Accounts Receivable as Collateral Acceptable Accounts Receivable Customer Amount D \$8,000 E 50,000 F 12,000 H 46,000 J 22,000 K 62,000 Total Collateral \$200,000
5 (b) Adjustments: 5% returns/allowances, 80% advance percentage. Level of available funds = [\$200,000 (1 0.05)] 0.80 = \$152,000 P LG 5: Accounts Receivable as Collateral Customer Amount A \$20,000 E 2,000 F 12,000 G 27,000 H 19,000 Total Collateral \$80,000 (b) \$80,000 (1 0.1) = \$72,000 (c) \$72,000 (0.75) = \$54,000 P LG 3, 5: Accounts Receivable as Collateral, Cost of Borrowing Challenge [\$134,000 (\$134, )] 0.85 = \$102,510 (b) (\$100, ) + (\$100, ) = \$2,000 + \$11,500 = \$13,500 \$13,500 Interest cost = = 13.5% for 12 months \$100, (\$100, ) + \$100, = \$2,000 + \$5,750 = \$7,750 \$7,750 Interest cost = = 7.75% for 6 months \$100,000 Effective annual rate = ( ) 2 1 = 16.1% (\$100, ) + \$100,000 4 = \$2,000 + \$2,875 = \$4,875 \$4,875 Interest cost = = 4.88% for 3 months \$100,000 Effective annual rate = ( ) 4 1 = 21.0%
6 P LG 5: Factoring Holder Company Factored Accounts May 30 Status on Accounts Amount Date Due May 30 Amount Remitted Date of Remittance A \$200,000 5/30 C 5/15 \$196,000 5/15 B 90,000 5/30 U 88,200 5/30 C 110,000 5/30 U 107,800 5/30 D 85,000 6/15 C 5/30 83,300 5/30 E 120,000 5/30 C 5/27 117,600 5/27 F 180,000 6/15 C 5/30 176,400 5/30 G 90,000 5/15 U 88,200 5/15 H 30,000 6/30 C 5/30 29,400 5/30 The factor purchases all acceptable accounts receivable on a nonrecourse basis, so remittance is made on uncollected as well as collected accounts. P LG 1, 6: Inventory Financing Challenge City-Wide Bank: [\$75,000 ( )] +( \$100,000) = \$1,000 Sun State Bank: \$100,000 ( ) = \$1,083 Citizens Bank and Trust: [\$60,000 ( )] + (0.005 \$60,000) = \$1,050 (b) City-Wide Bank is the best alternative, since it has the lowest cost. (c) Cost of giving up cash discount = ( ) (365 20) = 37.24% The effective cost of taking a loan = (\$1,000 \$75,000) 12 = 16.00% Since the cost of giving up the discount (37.24%) is higher than borrowing at Citywide Bank (16%), the firm should borrow to take the discount. P Ethics Problem Management should point out that what it is doing shows integrity, as it is honest, just and fair. The ethics reasoning portrayed in the ethics focus box could be used.
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### Cash Flow. Summary. Cash Flow. Louise Söderberg, 2010-10-15
Cash Flow Louise Söderberg, 2010-10-15 Summary The statement of cash flow reports the cash generated and used during the time interval specified in its headings. A cash flow analysis is a method of checking | 7,956 | 31,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.87928 |
https://studyadda.com/question-bank/mock-test-mathematical-reasoning_q16/5489/427617 | 1,716,424,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00374.warc.gz | 478,852,418 | 21,293 | • # question_answer If p is false and q is true, then A) $p\wedge q$ is true B) $p\,\vee \tilde{\ }q$is true C) $q\wedge q$is true D) $p\Rightarrow q$is true
Solution :
[d] When p is false and q is true, then $p\wedge q$ is false$p\vee \sim q$ is false, ($\therefore$ both p and $\sim q$ are false) and $q\Rightarrow p$is also false, Only$p\Rightarrow q$ is true.
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You will be redirected in 3 sec | 150 | 458 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.613137 |
https://math.stackexchange.com/questions/1088796/proof-that-difference-of-compact-and-closed-sets-is-also-closed | 1,566,739,992,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330233.1/warc/CC-MAIN-20190825130849-20190825152849-00490.warc.gz | 542,630,428 | 27,472 | # Proof that difference of compact and closed sets is also closed [duplicate]
I am trying to prove that for any $A$ compact, $B$ closed sets $\Rightarrow A-B = \{a-b | a\in A, b\in B\}$ is also closed, where A and B are subsets of a topological vector space $X$.
## marked as duplicate by saz, user147263, Mark Bennet, 6005, André NicolasJan 2 '15 at 19:53
This result holds for any Hausdorff topological group $G$. Indeed, let $c\in G\setminus AB^{-1}$ be an arbitrary point. It follows that the sets $cB$ and $A$ are disjoint. For every point $x\in A$ there exists a neighborhood $V(x)$ of the unit such that $V(x)x\cap cB=\varnothing$. Let $U(x)$ be a neighborhood of the unit such that $U(x)^2\subset V(x)$. There exists a finite family $\mathcal X$ such that $A\subset\bigcup\{U(x)x:x\in\mathcal X\}$. Put $U=\bigcap\{U(x):x\in\mathcal X\}$. Then $$UA\cap cB\subset U\bigcup\{U(x)x:x\in\mathcal X\}\cap cB \subset \bigcup\{U(x)^2x:x\in\mathcal X\}\cap cB=\varnothing.$$ Then $U^{-1}c$ is a neighborhood of the point $c$, which does not intersect with the set $AB^{-1}$ | 359 | 1,076 | {"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} | 3.109375 | 3 | CC-MAIN-2019-35 | latest | en | 0.835207 |
https://stats.stackexchange.com/questions/439451/one-sample-t-test-or-one-sample-z-test/439563#439563 | 1,642,631,296,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301592.29/warc/CC-MAIN-20220119215632-20220120005632-00040.warc.gz | 599,011,610 | 36,203 | # One sample t test or one sample z-test
I have a question as to whether the one-sample t test or the one-sample z test is appropriate in a sample of n=17 BUT with a known population variance.
I would usually use a one-sample t-test for such small samples. However, since I have a reference variance from another independent sample I was wondering if maybe the one-sample z-test would be appropriate.
I have conducted both and have obtained some significant effects. I thought that the tests would be underpowered given the small sample size. How should I interpret the results now? Are the results still taken to be seriously?
Thank you everyone!
• It sounds like you're using the variance from another sample when calculating your test statistic. That's not "a known population variance" it's a sample variance (albeit from a different sample). Can you clarify your question to be consistent with whatever the actual situation is? (to either be talking about a known population variance or about a sample variance from another sample) Dec 5 '19 at 12:56
• Our statistics professor has always been telling us that if we have a variance from another independent sample we can treat it as a known "population" variance. That's why I was talking about the population variance. Dec 5 '19 at 17:03
• Do you get different results with t-test and z-test? 17 is not a such a small sample. Dec 5 '19 at 18:07
• @Milkovena Interesting. Outside of the circumstances where it would be reasonable to treat an insample standard deviation as the population s.d. (i.e. very large sample size), your professor is wrong. I'll come back and write up an answer explaining how to correctly use a sample variance from a different sample (given suitable assumptions) Dec 6 '19 at 1:48
Imagine we have the following situation:
A previous sample of size $$m$$, from which we have an estimate $$s^2_m$$ of its population variance $$\sigma^2$$. (Here the $$m$$ subscript is simply denoting the size of the first sample, rather than the denominator in the variance calculation. It's just there to remind us that it's not calculated on the current sample.)
A current sample of size $$n$$ from which we have an estimate $$\bar{y}$$ of its population mean $$\mu$$. We assume here that the variance of the population from which the current sample was drawn is also $$\sigma^2$$.
We assume both populations are normally distributed. We assume independence within and between groups.
For a one sample test of $$H_0$$: $$\mu = \mu_0$$ (against one or two sided alternatives in the usual fashion), we can form the statistic $$t=\frac{\bar{y}-\mu_0}{s_m/\sqrt{n}}$$. This will have a t-distribution with $$m-1$$ d.f. (NB $$n$$ doesn't come into the df at all)
For a two sample test of equality of the population means of the two samples described above, we can form a test statistic whose numerator is the difference in their sample means and whose denominator is $$s_m \sqrt{\frac{1}{m}+\frac{1}{n}}$$, and again the d.f. is $$m-1$$.
This result is relatively straightforward mathematically, but we can check it via simulation. Here's an example done in R. Assume $$m=10$$ and $$n=5$$ (I choose small samples so that we can easily distinguish the distributions). Let's demonstrate that the null distribution of the statistic for the first case above is consistent with the claim that the distribution is $$t_9$$ (i.e. $$t$$ with $$m-1$$ df). We simulate the t-test statistic and compare the distribution of under the two claims (your professor's and mine). However, it can be hard to visually distinguish a t and a normal distribution. Consequently, I will convert the distribution of the t-statistic into the corresponding p-values under the assumption that the statistic has a normal distribution and the assumption it has a $$t$$ distribution with 9 df. If one of us is correct, the p-values from our claimed distribution should look uniformly distributed between 0 and 1. I will also look at the achieved type I error rate (i.e. actual significance level) for a 5% two tailed test under both approaches.
Here's the R code to do that for the one-sample test:
nsim = 100000
sigma = 1 # without loss of generality but feel free to change things
m = 10
n = 5
mux = 20 # ditto
muy = 10 # ditto, demonstrating these don't have to be the same
tsim = replicate(nsim,{
x = rnorm(m,mux,sigma)
sm = sd(x)
y = rnorm(n,muy,sigma)
ybar = mean(y)
(ybar-muy)/(sm/sqrt(n))
})
opar=par()
par(mfrow=c(1,2))
hist(pt(tsim,9),n=100) # transforming test stat. to p-values if it's t9
hist(pnorm(tsim),n=100) # ditto, if it's normal
par(opar)
Which gives:
As you see, my p-values (on the left) are plausibly close to a uniform while your professor's p-values (on the right) are definitely not. This doesn't prove it's t9, but it does demonstrate that it's not normal, and that it can't be very far from a t9. My claim remains plausible.
Let's seem how much this would affect the significance level of a two-tailed t-test that we wish to be at the 5% level:
mean(pt(tsim,9)<.025|pt(tsim,9)>.9755) # my claim
[1] 0.05011 # about 5%; within sampling variation
[1] 0.08228 # 8.2% - this is quite far out
So you should do the t-test but use as the df that for the sample standard deviation you used in the denominator (i.e. one less than the sample size in the original sample your variance estimate came from).
We could do a simulation for the two sample case in similar fashion; again it's $$m-1$$ df. You'd need to make the population means the same, calculate both sample means and then set the statistic to
(xbar-ybar)/(sm*sqrt(1/m+1/n)).
So the problem with doing a z-test instead of a t-test in this situation is that your significance level is inflated (just as it is if you did a z-test instead of a t-test with the s.d. calculated from the current sample). This lifts both power and the type I error rate (potentially a lot, as we saw in the simulation).
Naturally, if $$m$$ is quite large, you can get a reasonable approximation from treating the t-statistic as normal, but there's no reason to avoid doing the t-test if the assumptions are reasonable (why approximate what is easily done exactly?)
• Thank you so much! :) Dec 16 '19 at 5:52
• What to do when someone answers your question -- in particular note that if you follow the other suggestions, there's no need to say thank you Dec 16 '19 at 6:19 | 1,578 | 6,387 | {"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": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-05 | latest | en | 0.960443 |
https://bedtimemath.org/users/172339702987f046bf/ | 1,726,058,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00525.warc.gz | 113,268,243 | 17,702 | # rose woods’s Dinner in Space
You’re super excited for dinner tonight: rootbeer is on the menu! But oh no: rootbeer comes topped with fish. You don’t want to hurt the chef’s feelings, so you sneak every bit of fish off the plate and launch it into the sky on a light pink hot air balloon. But fish doesn’t fly far before a hungry seagull gobbles it up. Problem solved, dinner served!
Wee Ones:
Can you find 3 light pink objects near you? Which item is largest?
Little Kids:
If you’re at a restaurant sharing 10 plates of fish with a seagull, how many different ways could you and the seagull sort those plates into 2 piles?
Bonus:
If there are twice as many fish-topped cupcakes as rootbeer-topped ones, and there are 7 rootbeer-topped ones, how many fish-topped cupcakes are there?
Big Kids:
You’re opening your own restaurant and order a big box of 72 onions! But 1/2 of the onions are light pink – that can’t be good. How many onions do you need to toss in the compost?
Bonus:
Oh no! The onion people also keep delivering 4 1/2 pound boxes of fish every Wednesday this May. How many pounds does that add up to for the month?
Wee Ones:
Different for everyone!
Little Kids:
There are 5 ways that you and the birds could split 10 plates into 2 piles: 1+9, 2+8, 3+7, 4+6, 5+5.
Bonus:
14 fish-topped cupcakes.
Big Kids:
36 onions.
Bonus:
22 1/2 pounds, because there are 5 Wednesdays in May this year. | 372 | 1,421 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.917496 |
https://onlinecourses.nptel.ac.in/noc22_ma18/preview | 1,721,898,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00185.warc.gz | 362,825,970 | 14,457 | # Numerical Linear Algebra
By Prof. P. N. Agarwal, Prof. D. N. Pandey | IIT Roorkee
Learners enrolled: 823
This course is a basic course offered to UG/PG students of Engineering/Science background. It contains basics of matrix algebra, computer arithmetic, conditioning and condition number, stability of numerical algorithms, vector and matrix norms, convergent matrices, stability of non-linear systems, sensitivity analysis, singular value decomposition (SVD), algebraic and geometric properties of SVD, least square solutions, Householder matrices and applications, QR method, Power method and applications, Jacobi method for finding the eigenvalues of a given matrix. This course has tremendous applications in diverse fields of Engineering and Sciences such as control theory, image processing, numerical analysis and dynamical systems etc.
INTENDED AUDIENCE : UG and PG students of technical institutions/ universities/colleges
PRE-REQUISITES : None
INDUSTRY SUPPORT : None
Summary
Course Status : Completed Course Type : Elective Duration : 12 weeks Category : Mathematics Credit Points : 3 Level : Undergraduate/Postgraduate Start Date : 24 Jan 2022 End Date : 15 Apr 2022 Enrollment Ends : 07 Feb 2022 Exam Date : 24 Apr 2022 IST
Note: This exam date is subjected to change based on seat availability. You can check final exam date on your hall ticket.
### Course layout
Week 1: Matrix operations and type of matrices, Determinant of a Matrix, Rank of a matrix, Vector Space-I, Vector Space-II
Week 2: Linear dependence and independence, Bases and Dimensions – I, Bases and Dimension - II, Linear Transformation - I,
Linear Transformation - II
Week 3: Orthogonal subspaces, Row space, column space and null Space, Eigenvalues and Eigenvectors-I, Eigenvalues and
Eigenvectors-II, Diagonalizable Matrices
Week 4: Orthogonal Sets, Gram Schmidt orthogonalization and orthonormal bases, Introduction to Matlab, Sign integer representation
Computer representation of numbers
Week 5: Floating point representation, Round-off error, Error propagation in computer arithmetic, Addition and multiplication of floating
point numbers, Conditioning and condition numbers-I
Week 6: Conditioning and condition numbers-II, Stability of numerical algorithms-I, Stability of numerical algorithms-II, Vector
norms - I, Vector norms - II
Week 7: Matrix Norms - I, Matrix Norms-II, Convergent Matrices - I, Convergent Matrices - II, Stability of non-linear system
Week 8: Condition number of a matrix: Elementary properties, Sensitivity analysis-I, Sensitivity analysis-II, Residual theorem, Nearness to singularity
Week 9: Estimation of the condition number, Singular value decomposition of a matrix – I, Singular value decomposition of a
matrix - II, Orthogonal Projections, Algebraic and geometric properties of matrices using SVD
Week 10: SVD and their applications, Perturbation theorem for singular values, Outer product expansion of a matrix, Least
square solutions-I, Least square solutions-II
Week 11: Psudeo - inverse and least square solution, Householder matrices and their applications, Householder QR factorization –I,
Householder QR factorization –II, Basic theorems on eigenvalues and QR method
Week 12: Power method, Rate of convergence of Power method, Applications of Power method with shift, Jacobi method-I, Jacobi method-II
### Books and references
1. V. Sundarapandian, Numerical Linear Algebra, PHI, 2008.
2. Biswa Nath Dutta, Numerical Linear Algebra and Applications, SIAM, 2010.
3. Roger A. Horn and Charles R. Johnson, Matrix Analysis, Cambridge University Press, 1994.
4. William Ford, Numerical Linear Algebra with Applications, Academic Press, 2014..
### Prof. P. N. Agarwal
IIT Roorkee
Dr. P. N. Agarwal is a Professor in the Department of Mathematics, IIT Roorkee. His area of research includes approximation Theory and Complex Analysis. He delivered 13 video lectures on Engineering Mathematics in NPTEL Phase I and recently completed Pedagogy project on Engineering Mathematics jointly with Dr. Uaday Singh in the same Department. Further he has completed online certification course “Mathematical methods and its applications” jointly with Dr. S.K. Gupta of the same department. He taught the course on “Integral equations and calculus of variations” several times to MSc (Industrial Mathematics and Informatics) students. He has supervised nine Ph.D. theses and has published more than 187 research papers in reputed international journals of the world. Currently, he is supervising eight research students.
### Prof. D. N. Pandey
Dr. D. N. Pandey is an Associate Professor in the Department of Mathematics, IIT Roorkee. Before joining IIT Roorkee, he worked as a faculty member in BITS-Pilani Goa campus and LNMIIT Jaipur. His area of expertise includes semigroup theory and functional differential equations of fractional and integral orders. He has already prepared e-notes for the course titled “Ordinary Differential Equations and Special Functions” under e-Pathshala funded by UGC. Also, he has published a book titled “Nonlocal Functional Evolution Equations: Integral and fractional orders, LAP LAMBERT Academic Publishing AG Germany”. He has delivered several invited talks at reputed institutions in India and abroad. He has guided three PhD theses and has published more than 60 papers in various international journals of repute. Currently, he is supervising five research students.
### Course certificate
The course is free to enroll and learn from. But if you want a certificate, you have to register and write the proctored exam conducted by us in person at any of the designated exam centres.
The exam is optional for a fee of Rs 1000/- (Rupees one thousand only).
Date and Time of Exams: 24 April 2022 Morning session 9am to 12 noon; Afternoon Session 2pm to 5pm.
Registration url: Announcements will be made when the registration form is open for registrations.
The online registration form has to be filled and the certification exam fee needs to be paid. More details will be made available when the exam registration form is published. If there are any changes, it will be mentioned then.
Please check the form for more details on the cities where the exams will be held, the conditions you agree to when you fill the form etc.
CRITERIA TO GET A CERTIFICATE
Average assignment score = 25% of average of best 8 assignments out of the total 12 assignments given in the course.
Exam score = 75% of the proctored certification exam score out of 100
Final score = Average assignment score + Exam score
YOU WILL BE ELIGIBLE FOR A CERTIFICATE ONLY IF AVERAGE ASSIGNMENT SCORE >=10/25 AND EXAM SCORE >= 30/75. If one of the 2 criteria is not met, you will not get the certificate even if the Final score >= 40/100.
Certificate will have your name, photograph and the score in the final exam with the breakup.It will have the logos of NPTEL and IIT Roorkee.It will be e-verifiable at nptel.ac.in/noc.
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- NPTEL team | 1,606 | 7,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.752201 |
https://metanumbers.com/65029 | 1,638,056,739,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358323.91/warc/CC-MAIN-20211127223710-20211128013710-00369.warc.gz | 461,216,979 | 7,287 | # 65029 (number)
65,029 (sixty-five thousand twenty-nine) is an odd five-digits prime number following 65028 and preceding 65030. In scientific notation, it is written as 6.5029 × 104. The sum of its digits is 22. It has a total of 1 prime factor and 2 positive divisors. There are 65,028 positive integers (up to 65029) that are relatively prime to 65029.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 65 thousand 29 sixty-five thousand twenty-nine
## Notation
Scientific notation 6.5029 × 104 65.029 × 103
## Prime Factorization of 65029
Prime Factorization 65029
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 65029 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 11.0826 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 65,029 is 65029. Since it has a total of 1 prime factor, 65,029 is a prime number.
## Divisors of 65029
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 65030 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 32515 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 255.008 Returns the nth root of the product of n divisors H(n) 1.99997 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 65,029 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 65,029) is 65,030, the average is 32,515.
## Other Arithmetic Functions (n = 65029)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 65028 Total number of positive integers not greater than n that are coprime to n λ(n) 65028 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6497 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 65,028 positive integers (less than 65,029) that are coprime with 65,029. And there are approximately 6,497 prime numbers less than or equal to 65,029.
## Divisibility of 65029
m n mod m 2 3 4 5 6 7 8 9 1 1 1 4 1 6 5 4
65,029 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (65029)
Base System Value
2 Binary 1111111000000101
3 Ternary 10022012111
4 Quaternary 33320011
5 Quinary 4040104
6 Senary 1221021
8 Octal 177005
10 Decimal 65029
12 Duodecimal 31771
20 Vigesimal 82b9
36 Base36 1e6d
## Basic calculations (n = 65029)
### Multiplication
n×y
n×2 130058 195087 260116 325145
### Division
n÷y
n÷2 32514.5 21676.3 16257.2 13005.8
### Exponentiation
ny
n2 4228770841 274992739019389 17882502825691847281 1162881276251915136836149
### Nth Root
y√n
2√n 255.008 40.2132 15.969 9.17532
## 65029 as geometric shapes
### Circle
Diameter 130058 408589 1.32851e+10
### Sphere
Volume 1.15189e+15 5.31403e+10 408589
### Square
Length = n
Perimeter 260116 4.22877e+09 91964.9
### Cube
Length = n
Surface area 2.53726e+10 2.74993e+14 112634
### Equilateral Triangle
Length = n
Perimeter 195087 1.83111e+09 56316.8
### Triangular Pyramid
Length = n
Surface area 7.32445e+09 3.24082e+13 53096
## Cryptographic Hash Functions
md5 ab07090dd535ddbe692e95ec26087d4f 9ee122f6a637fb6482839c1a10930e386def4e3c 0ced29bb96f2baa9734c9b55a037ecef9f68dde95e52d69999aa0f97c805f709 f19fed992c5962fe05ee03c2ac8d7afed62f34085f92d7df40a52b84f776dab230d6b32eaae58062d3800dc3bf1ebfddfec06160d4eedb0e970985a2fe5f4127 db2d7c9b169c8fbaac26d4a3425d4dd92cb19a06 | 1,426 | 4,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-49 | latest | en | 0.813968 |
http://www.gabormelli.com/RKB/N0_Natural_Number_Sequence | 1,726,537,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00712.warc.gz | 44,616,878 | 22,949 | # The N0 Natural Number Sequence
(Redirected from N0 Natural Number Sequence)
The N0 Natural Number Sequence is an Infinite Number Sequence defined in terms the Successor Function, Peano's Axioms, and the starting element of Zero.
## References
### 2009
• (Wikipedia, 2009) ⇒ http://en.wikipedia.org/wiki/Natural_number
• In mathematics, a natural number means either an element of the set {1, 2, 3, ...} (the positive integers) or an element of the set {0, 1, 2, 3, ...} (the non-negative integers).
• Natural numbers have two main purposes: counting ("there are 3 apples on the table") and ordering ("this is the 3rd largest city in the country").
• Properties of the natural numbers related to divisibility, such as the distribution of prime numbers, are studied in number theory. Problems concerning counting, such as Ramsey theory, are studied in combinatorics.
• Mathematicians use N or \mathbb{N} (an N in blackboard bold, displayed as ℕ in Unicode) to refer to the set of all natural numbers. This set is countably infinite: it is infinite but countable by definition. This is also expressed by saying that the cardinal number of the set is aleph-null (\aleph_0).
• To be unambiguous about whether zero is included or not, sometimes an index "0" is added in the former case, and a superscript "*" or subscript "1" is added in the latter case:
• \mathbb{N}_0 = \{ 0, 1, 2, \ldots \}; \quad \mathbb{N}^* = \mathbb{N}_1 = \{ 1, 2, \ldots \}.
• WordNet http://wordnet.princeton.edu/perl/webwn?s=natural%20number
• the number 1 and any other number obtained by adding 1 to it repeatedly
• (Wikipedia, 2009) ⇒ http://en.wikipedia.org/wiki/Natural_number#Peano_axioms
• The Peano axioms give a formal theory of the natural numbers starting with 0. The axioms are:
• There is a natural number 0.
• Every natural number a has a natural number successor, denoted by S(a). Intuitively, S(a) is a+1.
• There is no natural number whose successor is 0.
• Distinct natural numbers have distinct successors: if a ≠ b, then S(a) ≠ S(b).
• If a property is possessed by 0 and also by the successor of every natural number which possesses it, then it is possessed by all natural numbers. (This postulate ensures that the proof technique of mathematical induction is valid.)
• It should be noted that the "0" in the above definition need not correspond to what we normally consider to be the number zero. “0" simply means some object that when combined with an appropriate successor function, satisfies the Peano axioms. | 643 | 2,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-38 | latest | en | 0.879404 |
https://www.chegg.com/homework-help/elementary-algebra-early-graphing-for-college-students-3rd-edition-chapter-10-solutions-9780136134169 | 1,547,699,341,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658702.9/warc/CC-MAIN-20190117041621-20190117063621-00372.warc.gz | 742,550,789 | 23,031 | # Elementary Algebra Early Graphing for College Students (3rd Edition) View more editions Solutions for Chapter 10
• 7557 step-by-step solutions
• Solved by professors & experts
• iOS, Android, & web
Chapter: Problem:
Sample Solution
Chapter: Problem:
• Step 1 of 2
We wish to evaluate the given expression whenis equal toandis equal to
We will substituteand in the expression and simplify it.
Given expression:
Substituteand:
And:
Multiply:
Multiply:
Combine like terms:
• Step 2 of 2
Hence the value of expression is
Corresponding Textbook
Elementary Algebra Early Graphing for College Students | 3rd Edition
9780136134169ISBN-13: 0136134165ISBN: Authors:
Alternate ISBN: 9780131580671, 9780131580718, 9780131580732, 9780132348867, 9780135131954, 9780321747198 | 211 | 775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-04 | latest | en | 0.67316 |
http://math.stackexchange.com/questions/658224/the-kernel-of-homomorphism-of-a-local-ring-into-a-field-is-its-maximal-ideal | 1,469,315,950,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823802.12/warc/CC-MAIN-20160723071023-00189-ip-10-185-27-174.ec2.internal.warc.gz | 161,365,293 | 18,545 | # The kernel of homomorphism of a local ring into a field is its maximal ideal?
I have a question about the proof of Theorem 3.2. of Algebra by Serge Lang.
In the theorem $A$ is a subring of a field $K$ and $\phi:A \rightarrow L$ is a homomorphism of $A$ into an algebraically closed field $L$.
In the beginning of the proof
We may first extend $\phi$ to a homomorphism of the local ring $A_\mathfrak{p}$, where $\mathfrak{p}$ is the kernel of $\phi$. Thus without loss of generality, we may assum that $A$ is a local ring with maximal ideal $\mathfrak{m}$.
Then, later
Since $\phi$ and the canonical map $A \rightarrow A/\mathfrak{m}$ have the same kernel,
I understand that $\mathfrak{m}$ is the kernel of the extension of $\phi$ to $A_\mathfrak{p}$. But in the proof, $\mathfrak{m}$ seems to be assumed to be the kernel of $\phi$ even if $A$ itself is a local ring.
In general, is the kernel of homomorphism of a local ring to a field its maximal ideal? If so, why is that ?
-
Is $\phi$ surjective? Otherwise pick your favorite discrete valuation ring and look at the inclusion mapping into its field of fractions... – JSchlather Jan 31 '14 at 8:05
Not generally true. You have to assume (wlog) that $\mathfrak m$ is the kernel. – Karl Kronenfeld Jan 31 '14 at 8:08
Bah. ${}{}{}{}$ – anon Jan 31 '14 at 8:16
No, in general it is not true that a homomorphism from a local ring into a field has the maximal ideal as kernel. Just consider the inclusion ${\mathbb Z}_{(p)} \to {\mathbb Q}$ for some prime number $p$.
What is happening here is the following. Suppose you start out with a ring $A$ that is already local with maximal ideal ${\frak m}$ and a homomorphism $\phi \colon A \to L$. The kernel ${\frak p}$ of $\phi$ doesn't have to be ${\frak m}$, but you can still localize at ${\frak p}$, obtaining an other local ring $A_{\frak p}$.
So the claim by Lang that "$\phi$ and the canonical map $A \to A/{\frak m}$ have the same kernel" is true, but that's because earlier he localized exactly at the kernel. (It might have been clearer if Lang had written "we may assume that $A$ is a local ring with maximal ideal ${\frak p}$" (or "with maximal ideal ${\frak m} = {\frak p}$")).
-
I believe, Lang meant that one can consider only the case of a local ring and the kernel being $\mathfrak{m}$ (i.e. $A:=A_{\mathfrak{p}}$). For the kernel: let $R$ be a ring of germs of infintely-differentiable functions, and take a homomorphism of $R$ to formal power series. That is embeddeble into the field of Laurent series, but the kernel is obviously not maximal ideal.
-
We can only infer $\mathfrak m=\ker\varphi$ when $\mathfrak m$ is the only prime ideal of $A$. In this case $A/\ker\varphi$ is a subring of $L$, which is necessarily an integral domain.
However, $A$ is assumed to be a subring of a field! So, $(0)\subset A$ is prime, and we can always consider the case of $\varphi$ being the canonical embedding of $A$ in the algebraic closure of its field of fractions. With this choice of $\varphi$ if $A$ is not a field, then $\ker\varphi$ is not the maximal ideal $\mathfrak m$.
- | 895 | 3,102 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-30 | latest | en | 0.905802 |
http://establish-fp7.eu/field-collection/field-item/189.html | 1,708,804,777,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00423.warc.gz | 12,827,838 | 4,677 | Learning Aims:
Using their knowledge of physics, the pupils should be able to explain how/why some disability aids function and which movements are made easier or possible.
Subjects:
Biology, physics and technology
Materials:
Everyday aids for disabled people. Products of various types to assist in everyday tasks such as eating and drinking, getting dressed or opening packages; these can be bought in a shop or on the internet. Make available too some different packages that can be difficult to open, for example plastic bottles, glass jars or tins of food and some paper cartons. It is important to use real packages so that the pupils can test them and understand for themselves the functions and describe the physics behind the activity with their own words.
Suggestions for use:
Divide the class in groups of 3-4 pupils. The task is in three parts.
• The first part is about the disability aids
• The second part is about opening a common package
• The third part is about describing the underlying physical principles
Distribute to each group some aids for people with decreased motor skills. The pictures show examples of some aids that can be used:
Ask the pupils to explain what they are and what they are to be used for. What functions do they have? Explain how movement occurs!
After this, each group is given a common package to open. The task is to discover how a person with reduced hand strength can open the package with the help of one or other of the objects. Examples of the packages that can be used are:
To conclude the Activity, the pupils are asked to explain and demonstrate the principles for increased hand strength. Theoretically, most cases concern turning around a fixed point. What is to be noticed is that the force used does not work directly at the point at which the result is desired but at some distance from it. The result is dependent upon the degree of force and the distance between the force and the point where the result takes place. This distance is called torque or leverage and is measured at right angles to the force. Note that torque also has a direction – one can turn clockwise or anti-clockwise. This is, therefore, the same principle one uses when a pole or bar is used to lift a stone from a hollow in the ground. The pole functions as a lever; see picture.
Note that the lever always has a fixed point around which the turn occurs.
Using a lever is a discovery made in ancient times and is counted – together with the sloping plane, the wedge, the screw and the wheel – amongst the most simple machines. It can be interesting to know that scientists and philosophers in ancient times speculated a great deal about the properties of these simple machines and their use in increasing force.
In order to demonstrate the idea of a lever, you can do the following: Open a door only about 10cm wide and try pressing against the door close to the hinge. Note the amount of force required to open the door further. Then try pressing against the edge of the door and compare how much force is required in these two situations. When is least force required? The correct answer is, that least force is required when one presses against the edge of the door. The lever in this case is the complete breadth of the door. | 658 | 3,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.958158 |
https://fr.mathworks.com/matlabcentral/answers/493331-how-do-i-determine-which-row-my-value-change-significantly | 1,716,133,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00563.warc.gz | 232,485,704 | 26,657 | # How do I determine which row my value change significantly
2 vues (au cours des 30 derniers jours)
new2matlab le 26 Nov 2019
Commenté : new2matlab le 26 Nov 2019
I'm looking to automate the process of finding where my column data changes from 375 (about the first 7 rows right now), as it starts decreasing constantly after that. Depending on the data loaded in, the change from a constant 375 can start anywhere from row 3 to row 8, so I'd like to detect where it begins. Thanks!
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### Réponse acceptée
Turlough Hughes le 26 Nov 2019
Modifié(e) : Turlough Hughes le 26 Nov 2019
You could find where data is not equal to 375 and then take the first index where this condition is true:
rowstart=min(find(data~=375))
Then your new data would be:
data2=data(rowstart:end);
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new2matlab le 26 Nov 2019
This is exactly what I needed. Thank you!
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### Plus de réponses (1)
TADA le 26 Nov 2019
Im assuming its roughly 375 and not exactly that value, otherwise go with Turlough Hughes' solution which is simpler.
The easiest method is to compare the values with tolerance to changes. You can define a threshold value that you would concider as significant reduction
% for the sake of this example I'm using 5% of the initial value (should be ~5% of 375)
threshold = 0.05*data(1);
sigDecMask = data < (data(1) - threshold);
startDecIdx = find(sigDecMask, 1, 'first');
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Translated by | 510 | 1,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 3 | CC-MAIN-2024-22 | latest | en | 0.585314 |
https://essayfount.com/geometry-sketch/ | 1,670,516,682,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00415.warc.gz | 282,078,269 | 15,449 | # Geometry Sketch
Table of Contents
## Unit 6 Assignment
##### Assignment Instructions
Measure and draw a two-dimensional sketch of your classroom or a room in your house, listing the exact dimensions in feet or meters. Then, draw a miniature version of the same room using the scale factor of 1/50. Explain why the two pictures are “similar” using the terms: angles, lengths, and proportions.
Note
• The first drawing is a rough sketch: For this drawing, put the actual dimensions of your classroom or room in your house (i.e. depending on the one you are drawing).
• The second drawing needs to be drawn using a scale factor of 1/50: For this drawing, put the dimensions using the scale factor of 1/50 of the actual dimensions of your classroom or room in your house.
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"This company is the best there is. They saved me so many times, I cannot even keep count. Now I recommend it to all my friends, and none of them have complained about it. The writers here are excellent." | 325 | 1,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-49 | longest | en | 0.928186 |
https://metric-calculator.com/convert-grain-to-tonne.htm | 1,701,591,221,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00032.warc.gz | 463,238,361 | 7,010 | # Grains to Tonnes Converter
Select conversion type:
Rounding options:
Convert Tonnes to Grains (tonne to gr) ▶
## Conversion Table
grains to tonnes gr tonne 10000000 gr 0.648 tonne 20000000 gr 1.296 tonne 30000000 gr 1.944 tonne 40000000 gr 2.592 tonne 50000000 gr 3.2399 tonne 60000000 gr 3.8879 tonne 70000000 gr 4.5359 tonne 80000000 gr 5.1839 tonne 90000000 gr 5.8319 tonne 100000000 gr 6.4799 tonne 110000000 gr 7.1279 tonne 120000000 gr 7.7759 tonne 130000000 gr 8.4239 tonne 140000000 gr 9.0718 tonne 150000000 gr 9.7198 tonne 160000000 gr 10.3678 tonne 170000000 gr 11.0158 tonne 180000000 gr 11.6638 tonne 190000000 gr 12.3118 tonne 200000000 gr 12.9598 tonne
## How to convert
1 grain (gr) = 6.47989E-08 tonne (tonne). Grain (gr) is a unit of Weight used in Standard system. Tonne (tonne) is a unit of Weight used in Metric system.
## Grains: A Unit of Weight
Grains are a unit of weight that are used for measuring small masses, such as bullets, arrows, and seeds. Grains are derived from the English word grain, which was the name of a seed or a cereal plant. The symbol for grain is gr.
## Definition of the Grain
The grain is defined as one seven-thousandth of an avoirdupois pound, which is the common unit of weight in the US customary system and the British imperial system. The grain is equal to the troy grain, which is the smallest unit of the troy weight system, which is used for measuring precious metals. The grain is also equal to the apothecaries’ grain, which is the smallest unit of the apothecaries’ weight system, which is used for measuring medicines.
The grain is equal to about 0.0023 avoirdupois ounces or 0.0648 grams. The grain is also equal to about 0.00014 troy ounces or 0.0417 pennyweights.
## How to Convert Grains
Grains can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert grains to other units of weight in the US customary system and the SI system:
• To convert grains to avoirdupois ounces, divide by 437.5. For example, 10 gr = 10 / 437.5 = 0.0229 oz.
• To convert grains to avoirdupois pounds, divide by 7000. For example, 5 gr = 5 / 7000 = 0.0007 lb.
• To convert grains to tons (short), divide by 14000000. For example, 20 gr = 20 / 14000000 = 0.0000014 ton.
• To convert grains to kilograms, divide by 15432.3584. For example, 15 gr = 15 / 15432.3584 = 0.000972 kg.
• To convert grains to grams, divide by 15.4323584. For example, 25 gr = 25 / 15.4323584 = 1.6218 g.
• To convert grains to milligrams, multiply by 64.79891. For example, 30 gr = 30 x 64.79891 = 1943.9673 mg.
## Where Grains are Used
Grains are used in different countries and regions for different applications and purposes. Here are some examples of where grains are used:
• In most countries that use the SI system, grains are not used for measuring weight, but only for measuring small masses, such as bullets, arrows, and seeds.
• In the United States, grains are sometimes used for measuring small masses, such as bullets, arrows, and seeds.
• In Canada, Australia and New Zealand, grains are sometimes used for measuring small masses, such as bullets, arrows, and seeds.
• In the United Kingdom, grains are sometimes used for measuring small masses, such as bullets, arrows, and seeds.
• In India, grains are sometimes used for measuring small masses, such as bullets, arrows, and seeds.
## History of Grains
Grains have a long history that dates back to ancient times. Here are some highlights of the history of grains:
• The grain was originally based on the weight of a single seed of wheat or barley, which was used as a unit of mass in ancient civilizations such as Egypt, Mesopotamia and Greece.
• The grain was part of different systems of measurement, such as the Tower system, the Troy system, the Avoirdupois system and the Apothecaries’ system. It varied from about 40 milligrams to about 80 milligrams depending on the region and the time period.
• The grain was standardized by royal statutes and international agreements in different periods of history. For example, in 1527 an act of Henry VIII fixed the grain at exactly one seven-thousandth of an avoirdupois pound; in 1828 an act of Congress adopted the troy ounce as the official unit of weight for coinage in the United States; in 1959 an international agreement defined the international avoirdupois ounce as exactly 28.349523125 grams.
## Example Conversions of Grains to Other Units
Here are some examples of conversions of grains to other units of weight:
• 1 gr = 0.0023 oz
• 1 gr = 0.0001 lb
• 1 gr = 0.00000007 ton
• 1 gr = 0.000065 kg
• 1 gr = 0.0648 g
• 1 gr = 64.79891 mg
• 1 gr = 0.0042 oz t
• 1 gr = 0.0417 dwt
## Tonnes: A Unit of Weight
Tonnes are a unit of weight that are used for measuring large masses, such as vehicles, ships, and buildings. Tonnes are also known as metric tons or megagrams. The symbol for tonne is t.
## Definition of the Tonne
The tonne is defined as 1000 kilograms, which is the base unit of mass in the International System of Units (SI). The tonne is also equal to one million grams or one billion milligrams.
The tonne is equal to about 2204.6 avoirdupois pounds or 1.1023 short tons. The tonne is also equal to about 0.9842 long tons or 0.9072 tonnes-force.
## How to Convert Tonnes
Tonnes can be converted to other units of weight by using conversion factors or formulas. Here are some examples of how to convert tonnes to other units of weight in the US customary system and the SI system:
• To convert tonnes to avoirdupois pounds, multiply by 2204.6226. For example, 10 t = 10 x 2204.6226 = 22046.226 lb.
• To convert tonnes to short tons, multiply by 1.1023. For example, 5 t = 5 x 1.1023 = 5.5115 st.
• To convert tonnes to long tons, multiply by 0.9842. For example, 20 t = 20 x 0.9842 = 19.684 lt.
• To convert tonnes to kilograms, multiply by 1000. For example, 15 t = 15 x 1000 = 15000 kg.
• To convert tonnes to grams, multiply by 1000000. For example, 25 t = 25 x 1000000 = 25000000 g.
• To convert tonnes to milligrams, multiply by 1000000000. For example, 30 t = 30 x 1000000000 = 30000000000 mg.
## Where Tonnes are Used
Tonnes are used in different countries and regions for different applications and purposes. Here are some examples of where tonnes are used:
• In most countries that use the SI system, tonnes are used for measuring weight, especially for large masses, such as vehicles, ships, and buildings.
• In the United States, tonnes are sometimes used for measuring weight, especially for international trade and commerce.
• In Canada, Australia and New Zealand, tonnes are sometimes used for measuring weight, especially for international trade and commerce.
• In the United Kingdom, tonnes are sometimes used for measuring weight, especially for international trade and commerce.
• In India, tonnes are sometimes used for measuring weight, especially for international trade and commerce.
## History of Tonnes
Tonnes have a relatively short history that dates back to the late 18th century. Here are some highlights of the history of tonnes:
• The tonne was originally based on the weight of a cubic meter of water at its maximum density of four degrees Celsius, which was approximately equal to 1000 kilograms.
• The tonne was part of the original metric system that was introduced in France in 1795 as a decimal-based system of measurement that was designed to replace the traditional units that varied from region to region.
• The tonne was adopted by many countries around the world as part of the International System of Units (SI) that was established in 1960 as a universal system of measurement that was based on seven base units and several derived units.
• The tonne was also referred to as the metric ton or the megagram in some countries and contexts to distinguish it from the non-metric units of the short ton and the long ton.
## Example Conversions of Tonnes to Other Units
Here are some examples of conversions of tonnes to other units of weight:
• 1 t = 2204.6226 lb
• 1 t = 1.1023 st
• 1 t = 0.9842 lt
• 1 t = 1000 kg
• 1 t = 1000000 g
• 1 t = 1000000000 mg
• 1 t = 32.1507 oz t
• 1 t = 643.0149 dwt
• 1 t = 15432.3584 gr
• 1 t = 0.001 kN
Tonne also can be marked as Metric Ton.
Español Russian Français | 2,298 | 8,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | longest | en | 0.816056 |
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# Dividing Fractions and Other Memory Tips
I found this clever video to help students remember the order of operations for dividing fractions. It uses the “Keep, Change, Flip” mnemonic to teach students how to divide fractions. Keep the first fraction the same. Change the division sign to multiplication. Flip the second fraction over, then solve it the same way as a multiplication problem by multiplying the numerators and denominators. The song shows students this process with an example problem and explains the math behind it. I like this mnemonic because it works for all learning styles.
Keep, Change, Flip! by Flocabulary – a great song for remembering how to divide fractions!
• A picture of a concept helps the Visual Learner remember
• The rap dialog and repetitive pattern appeals to the Auditory Learner
• The Kinesthetic Learner moves his or her body to the sounds and remembers by the action
• It’s fun! When learning is enjoyable, it always enhances memory
When the phrase, “Keep, Change and Flip” is repeated multiple times, it is stored in long term memory. I would probably add an acronym for the phrase; “K C Flip” and call the dragon K.C. (Casey) and remember the pancake turner in his hand flipping the numbers.
While we are talking about division, take a look at this memory tip for remembering the meaning of “divisor,” “dividend” and “quotient.”
Use a rhyme for remembering “quotient”:
Quotient – “Kwo-shunt”
What a funny word,
Isn’t that absurd?
“Divisor” and “Dividend” are similar and often mixed up. Analyze the words. How are they different? One has a short word “or” at the end. The other has a longer word, “end”. The shorter word, “divisor” is always on the outside of the bracket trying to get in but the word is too small. The larger number, “dividend” is inside waiting to be divided.
Another memory tip for remembering the order of operations for long division:
Step 1 – Dad reminder to Divide
Step 2 – Mother reminder to Multiply
Step 3 – Sister reminder to Subtract
Step 4 – Brother reminder to Bring down
Step 5 – Rover (the dog) reminder to Repeat or notice the Remainder | 497 | 2,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.912483 |
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## #1 2006-08-13 12:23:39
MathsIsFun
Administrator
Offline
### Power Sets
Power Sets and Power Set Maker
Who would have thought that sets and binary numbers would have so much in common?
Have a read and a play. Tell me bugs. Etc.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #2 2012-03-16 23:49:21
anonimnystefy
Real Member
Offline
### Re: Power Sets
hi MIF
I couldn't resist making this the input:
{},{{}},{{{}}},{{};{{}}}
but then i saw it doesn't take sets as an input.But all in all it works very nicely
The page is nice as well,although it made me crave for ice-cream
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #3 2012-03-17 00:17:10
bob bundy
Moderator
Offline
### Re: Power Sets
Nice!
works for me ... actually I didn't try {m,e} but I did try {b,o,b}
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #4 2012-03-17 00:18:37
anonimnystefy
Real Member
Offline
### Re: Power Sets
hi bob
it works for me too.i just thought that maybe it would work for {},{{}},{{{}}},{{};{{}}} as well.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## Board footer
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https://www.tutoreye.com/homework-help/science/chemistry/coplanar?page=2 | 1,685,825,160,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00188.warc.gz | 1,154,852,969 | 30,357 | Coplanar homework Help at TutorEye
## Coplanar
When the atoms present in a compound are lying in the same plane they are called coplanar. For example- Ethyne molecule.
## Coplanar Sample Questions
Question 1: What is coplanarity in chemistry?
a) The atoms of molecules are the same. b) The atoms of a molecule are single bonded.
c) The atoms lie in the same plane. d) All of the above.
Answer: Option c) All the atoms lie in the same plane.
Explanation: When things are coplanar it means it lies in the same plane. There is an x, y, and z plane. So, if atoms in molecule are coplanar, they would be in the same plane
Question 2: Atoms in which of the following molecules are coplanar?
a) b)
c) d)
Answer: Option c)
Explanation: Biphenyl is hybridised. Molecules with Hybridisation are planar in nature. therefore, is coplanar.
Question 3: In which geometry, all the atoms are coplanar?
a) Tetrahedral b) Trigonal bipyramidal
c) Pyramidal. d) Trigonal planar
Answer: Option d) Trigonal planar
Explanation: In trigonal planar geometry, all four atoms are present in the same plane and hence, they are coplanar. In the rest of the mentioned geometries, there are some atoms lying perpendicular to other atoms.
Question 4: Which of the following molecules has all the atoms coplanar?
a) b)
c) d)
Answer: Option c)
Explanation: is hybridised and has 3 bonds and 0 lone pairs. The molecule has trigonal planar geometry with all the atoms lying in one plane and thus , all the atoms are coplanar.
Question 5: Atoms are coplanar in hybridization of -
a) b)
c) d)
Answer: Option b)
Explanation: hybridized atoms are coplanar, as the geometry of such molecules is planar and atoms lie in the same plane and are coplanar.
Question 6: In which of the following geometry, all the atoms are not coplanar?
a) Tetrahedral b) Linear
c) Square planar d) Trigonal planar
Answer: Option a) Tetrahedral
Explanation: When the atoms present in a compound are lying in the same plane they are called coplanar. In tetrahedral geometry there are a total of five atoms but only three are coplanar and the other two are at angle with them.
Question 7: Which of the following statement is true-
a) All atoms are coplanar in pyramidal geometry b) Hybridizationhave all atoms coplanar.
c) All hydrogens and carbons are coplanar in methane. d) Aromatic compounds are coplanar.
Answer: Option d) Aromatic compounds are coplanar.
Explanation: The atoms involved in the ring are in the same plane and therefore the structure of aromatic compounds are coplanar.
Question 8: Which of the molecules is nonplanar?
a) b)
c) d)
Answer: Option b)
Explanation: Diboranes has non planar structure because two of the hydrogen are present above and below the boron and hydrogen bond.
Question 9: Which of the following statements is incorrect?
a) Coplanar atoms lie in one plane. b) Square pyramidal geometry has all the coplanar atoms.
c) Molecules with hybridisation has coplanar atoms. d) Carbon Dioxide is coplanar
Answer: Option b) Square pyramidal geometry has all the coplanar atoms.
Explanation: In square pyramidal geometry there are a total of 6 atoms present. Five of the atoms are present in one plane and the sixth one is perpendicular to them. Thus, not all the atoms are coplanar in square pyramidal geometry.
Question 10: is a planar molecule
a) True b) False
Answer: Option b) TFalse
Explanation: Geometry of is trigonal bipyramidal, P and the four Cl atoms lie in one plane and one of the Cl atom lie above the plane and one lies below the plane. Therefore is nonplanar.
## Frequently Asked Questions
### What is coplanar?
When all the atoms present in a compound are lying in the same plane they are called coplanar. For example- Ethyne molecule.
The aromatic compounds have atoms in the ring lying on one plane and that’s why aromatic compounds are coplanar.
### What are coplanar points?
Coplanar points are two or more points which are lying in one plane. Coplanar molecules include atoms in the same plane. For example, In a Biphenyl ring, all the carbon and hydrogen atoms lie in the same plane.
### What is the difference between coplanar and collinear?
Collinear points or atoms are atoms lying in one line while coplanar atoms means atoms lying in one plane. A molecule can or can’t be both collinear and coplanar at same time. | 1,171 | 4,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | longest | en | 0.757224 |
https://groups.yahoo.com/neo/groups/ai-geostats/conversations/topics/80?l=1 | 1,501,113,521,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00059.warc.gz | 662,152,942 | 20,360 | ## Re: GEOSTATS: weighted interpolation method
Expand Messages
• Gregoire, Defining alternative metrics seems like a good idea on the face of it although a great deal of subjective input is needed and a sensible metric for
Message 1 of 2 , Feb 9, 1997
Gregoire,
Defining alternative metrics seems like a good idea on the face of it
although a great deal of subjective input is needed and a sensible metric
for one variable may not be sensible for other measured or mapped variables
available. This poses an interesting problem in which different metrics are
integrated into a single cost function for interpolation. An alternative I
try to employ if possible is to investigate linear/nonlinear regression /
ancova relationships between the variable of interest and any other
available covariates such as elevation, slope aspect, satelite image data,
and on and on depending on the particular application. After partialling
out large scale trends with categorical and continuous variables, the
residual process is often independent or only weakly spatially correlated
and predictions are made in a universal kriging type analysis where the
design matrix contains more than just geographic coordinates or a single
variable external drift. I do this in a MATLAB toolbox which I have been
developing over the last year or so. However, this could be conducted in
ARC by fitting regression models in S+, SAS, SPSS or your favorite
statistical software package followed by ordinary kriging the residuals and
adding. This can probably be done in GSLIB also without a great deal of
trouble. Especially in the case of pollutants, I have found this approach
especially useful for pollutants such as PCB or heavy metals.
>Greetings to all Arc/Users and or AI-geostatisticians,
>
>The IDW is an inverse distance weighted interpolation method.
>The problem here is the definition of the distance.
>2 measurements of a variable X close together can be separated
>geographically by a hill which won't be taken into account during
>the interpolation,
>
>Therefore, if I use a DEM (for exemple) to add another weight to
>the interpolation function ( with cost distance functions like
>those proposed by ESRI for ARc/Info) I wouldn't be limited
>anymore to euclidian distances.
>
>
>The use of a "cost distance weight", a measure of the "effort" to reach
>2 points, could also be usefull for the variogram calculation.
>
>Well, these are things I'm trying out for the moment with a pollutant.
>
>
>Any suggestions on how to do it with/without Grid (Arc/Info) ?
>
>
>Thanks for any help,
>
>
>Best regards,
>
>
>Gregoire
>--
>Gregoire Dubois (PhD student) Tel. 39-332-78.99.44
>Joint Research Centre Fax. 39-332-78.54.66
>Environment Inst. TP 321 Email: gregoire.dubois@...
>I-21020 Ispra (Va), ITALY URL: http://java.ei.jrc.it/rem/gregoire/
>--
>*To post a message to the list, send it to ai-geostats@....
>*As a general service to list users, please remember to post a summary
>of any useful responses to your questions.
>
--
*To post a message to the list, send it to ai-geostats@....
*As a general service to list users, please remember to post a summary
of any useful responses to your questions.
Your message has been successfully submitted and would be delivered to recipients shortly. | 765 | 3,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-30 | latest | en | 0.923067 |
https://metanumbers.com/125144 | 1,638,802,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00452.warc.gz | 458,327,375 | 7,365 | # 125144 (number)
125,144 (one hundred twenty-five thousand one hundred forty-four) is an even six-digits composite number following 125143 and preceding 125145. In scientific notation, it is written as 1.25144 × 105. The sum of its digits is 17. It has a total of 4 prime factors and 8 positive divisors. There are 62,568 positive integers (up to 125144) that are relatively prime to 125144.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 17
• Digital Root 8
## Name
Short name 125 thousand 144 one hundred twenty-five thousand one hundred forty-four
## Notation
Scientific notation 1.25144 × 105 125.144 × 103
## Prime Factorization of 125144
Prime Factorization 23 × 15643
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 31286 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 125,144 is 23 × 15643. Since it has a total of 4 prime factors, 125,144 is a composite number.
## Divisors of 125144
8 divisors
Even divisors 6 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 234660 Sum of all the positive divisors of n s(n) 109516 Sum of the proper positive divisors of n A(n) 29332.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 353.757 Returns the nth root of the product of n divisors H(n) 4.26639 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 125,144 can be divided by 8 positive divisors (out of which 6 are even, and 2 are odd). The sum of these divisors (counting 125,144) is 234,660, the average is 2,933,2.5.
## Other Arithmetic Functions (n = 125144)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 62568 Total number of positive integers not greater than n that are coprime to n λ(n) 31284 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 11717 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 62,568 positive integers (less than 125,144) that are coprime with 125,144. And there are approximately 11,717 prime numbers less than or equal to 125,144.
## Divisibility of 125144
m n mod m 2 3 4 5 6 7 8 9 0 2 0 4 2 5 0 8
The number 125,144 is divisible by 2, 4 and 8.
## Classification of 125144
• Refactorable
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (125144)
Base System Value
2 Binary 11110100011011000
3 Ternary 20100122222
4 Quaternary 132203120
5 Quinary 13001034
6 Senary 2403212
8 Octal 364330
10 Decimal 125144
12 Duodecimal 60508
20 Vigesimal fch4
36 Base36 2ok8
## Basic calculations (n = 125144)
### Multiplication
n×y
n×2 250288 375432 500576 625720
### Division
n÷y
n÷2 62572 41714.7 31286 25028.8
### Exponentiation
ny
n2 15661020736 1959882778985984 245267570493421981696 30693764841828800477364224
### Nth Root
y√n
2√n 353.757 50.0192 18.8084 10.4588
## 125144 as geometric shapes
### Circle
Diameter 250288 786303 4.92005e+10
### Sphere
Volume 8.20954e+15 1.96802e+11 786303
### Square
Length = n
Perimeter 500576 1.5661e+10 176980
### Cube
Length = n
Surface area 9.39661e+10 1.95988e+15 216756
### Equilateral Triangle
Length = n
Perimeter 375432 6.78142e+09 108378
### Triangular Pyramid
Length = n
Surface area 2.71257e+10 2.30974e+14 102180
## Cryptographic Hash Functions
md5 25034d2094f6312bd0e49f713efb5e45 a68c555fa2379f3c0c26d51f20761d1d19693503 25548c0fbbd165f31d4183fccf70ea95c551c29ee3618cb86f1f7f891d81aa9b 2406fb7c8dcf9a94cf9eb27e727155710a8502fc38bde399779fa15a096ddefa782c075897b769e3d045925c046f51f425bcbc3d7b4b70d3bca4bb3b37bd1d04 50b024eb2e0c2f00fe1bc7b8717b6056334817ee | 1,451 | 4,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.814103 |
https://www.physicsforums.com/threads/5th-order-equation-unsolvable-by-approximation-methods.862232/ | 1,512,964,675,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00496.warc.gz | 772,753,831 | 18,305 | # I 5th order equation unsolvable by approximation methods?
1. Mar 15, 2016
### arupel
I am accross this equation with no known analytical solutions:
x^5 - x + 1 = 0.
I asked this before and the answer was that you can get as good an approximation you want by approximation methods.
It is possible that when applying approximation methods, you will get singularities. These methods won't work. The equation is bullet proof.
The question is can we be cetain that there are approximation methods that will work or is this equation intrincally unvolvable even by approximatiuon methods?
My second question is messy. It is a stray thought.
Is this equaton one that satisified Godel's criteria for an unsolveable equation is tue but can be shown to be not provable and not unproveable? If it fits Godel's criteria it would be nice to show studends.
PS: Anyway, In teaching algebra, it is nice to show this unsolveable simple equation.
2. Mar 15, 2016
### jbriggs444
Binary search always works [on continuous functions defined over a contiguous range, given a pair of starting points]. If you can find a point where the graph of the function is positive and a point where the graph is negative then a binary search will converge toward a point where it is zero. By inspection, this polynomial is negative at x=-1000 and positive at x=+1000. Like all polynomials it is continuous and is defined over a contiguous range.
No, this has nothing to do with Godel.
Edit: Added some caveats to dot the i's and cross the t's
3. Mar 15, 2016
### Ssnow
Starting with the interval $[-2,-1]$, the solution is inside and you can approximate it with bisection method (for example) ... it seems to work ...
4. Mar 15, 2016
### pwsnafu
If you want any example for Godel see Goodstein's[/PLAIN] [Broken] theorem. Note that Goodstein can be expressed in PA, but not provable in PA. It is provable in ZFC.
Last edited by a moderator: May 7, 2017
5. Mar 15, 2016
### mathman
Galois proved that polynomial equations (in general) higher than 4th degree cannot be solved analytically.
6. Mar 15, 2016
### micromass
Staff Emeritus
That's a really inaccurate statement of what has actually been proved.
First, it's Abel and Ruffini who proved it first, not Galois. Although Galois' proof is superior.
Second, they are unsolvable using radicals, not with other analytic techniques. There are in fact some analytic techniques which can solve all 5 degree equations.
Third, this thread is about approximation, for which Galois theory is completely irrelevant.
7. Mar 15, 2016
### Thecla
A similar equation is discussed by Julian Havil in his book ,The Irrationals- A Story of the Numbers You Can't Count On". The equation is
x^5- x- 1=0. Julian says "no finite expression formed of the composition of radicals will provide an explicit expression for its roots, with its only real root
equal to 1.1673039782614186843...". p.133
Also the innocent looking equation: x^5- 5x+12=0 requires about 600 symbols for its exact expression in radicals for the only one real solution -1.842085966... p.133
8. Mar 15, 2016
### arupel
Thanks. I was hoping for a mystery; a continuous function which is absolutely unsolveble either analyticaly or by approximation methods, a platonic equation.
Sadly the mundane fact is that if it is a continious function it can be plotted. If it can be plotted it can be solved by the graph or analytically or by approximation methods, an obvious fact.
signing off | 868 | 3,482 | {"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} | 3.6875 | 4 | CC-MAIN-2017-51 | longest | en | 0.941328 |
https://www.physicsforums.com/threads/how-does-a-dielectric-interact-with-a-laser-beam.961685/ | 1,623,582,127,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00455.warc.gz | 862,143,657 | 16,964 | # How does a dielectric interact with a laser beam?
## Homework Statement
A laser tweezer is a laboratory instrument, which uses highly focused laser beams to ‘trap’, hold or move small sized objects. The principle of the operation is that in the focal spot, the light intensity is inhomogeneous, and acts on the particle with a force that points from the low intensity region towards the high intensity region.
In this problem below, the object to be trapped is a nano-ball made of latex, which is insulating, has no net electrical charge, and is much smaller than the wavelength of the light. The ball is compact, homogeneous with mass m, radius R and of relative dielectric constant of ##ε_r##.The nano-ball is placed into a well focused, polarized laser beam (see figure). Let us approximate the laser light
in all points of the focal region as a plane wave moving in the x direction, with angular frequency of ω and with an amplitude which varies point by point. The time averaged intensity of the laser light can be approximated as $$I= I_0 \left( 1- \frac {x^2} {a^2 } - \frac {y^2} {b^2} - \frac {z^2} {b^2} \right)$$ in the region of |x| ≪ a; |y| ≪ b; |z| ≪ b (here a, b > 0).
a) Let us determine the coordinates of the equilibrium position of the ball, that is, the point where the trapping
force and the force from radiation pressure are equal. We can assume that the distance of the equilibrium position
from the origin of the coordinate system is much smaller than the parameters a and b, but much larger than the ball radius R. Let us use the laws of Maxwellian electrodynamics.
## Homework Equations
Intensity
##I= \frac {P} {A}##
Where P and A denotes the power and cross sectional area respectively .
And the force of radiation is defined as
##F=\frac {Ic} {A}##
## The Attempt at a Solution
Due to symmetry of the problem,we can choose z=0 for equilibrium position.
Now the force of radiation is $$F_{rad} =\left( \frac {I_0c} {\pi R^2} \right)\left( 1- \frac {x^2} {a^2} - \frac {y^2} {b^2} \right)$$
But I can't figure out what will be the expression for trapping force. And how this ball will interact with the laser beam.
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https://123dok.net/document/1y9v5jlq-a-novel-monolithic-lagrangian-approach-for-modelling-crack-propagation-using-anisotropic-mesh-adaptation.html | 1,623,772,432,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621450.29/warc/CC-MAIN-20210615145601-20210615175601-00427.warc.gz | 89,654,715 | 35,323 | # A novel monolithic Lagrangian approach for modelling crack propagation using anisotropic mesh adaptation
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## crack propagation using anisotropic mesh adaptation
### of Mathematics Sarvajanik College of Engineering & Technology 2018, 5 (3), pp.53-65. �hal-01731909�
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Journal homepage:www.ijaamm.com
### propagation using anisotropic mesh adaptation
Research Article
Karim Hittia,, Stéphanie Feghalib, Faten Rafehc, Marc Bernackid, Pierre-Olivier Bouchardd
aDepartment of Mathematics, Faculty of Sciences and Issam Fares Faculty of Technology, University of Balamand, Lebanon bLebanese University, Faculty of sciences II, Mathematics Department, P.O. Box 90656 Fanar-Matn, Lebanon
cDepartment of Civil Engineering and Construction, Issam Fares Faculty of Technology, University of Balamand, Lebanon
dMines-ParisTech, PSL-Research University, CEMEF - Centre de mise en forme des matériaux, CNRS UMR 7635, CS 10207 rue Claude Daunesse,
06904 Sophia Antipolis Cedex, France
Received 17 January 2018; accepted (in revised version) 23 February 2018
Abstract: A monolithic Lagrangian finite element method is explored to model crack propagation. This approach is based on the level-set method coupled with anisotropic remeshing to define the crack faces and tip. Moreover, the Gθ method is used for computing the strain energy release rate and the propagation direction. Furthermore, a new technique for computing the convection velocity, which is a mixture between the propagation speed and the mechanical velocity, is introduced.
MSC: 74R99 • 74S05
Keywords: Crack propagation • Mesh adaptation • Gθ method • Monolithic approach • Lagrangian framework
### 1. Introduction
Since the last century, a large number of theories and applications have been developed to deal with fracture me-chanics and crack propagation [1–8]. The latter gives rise to many difficulties since it leads to an important displace-ment discontinuity. The Finite Eledisplace-ment Method (FEM) [4,5,9] has proved its efficiency in dealing with crack propa-gation by modifying the mesh’s topology and performing automatic remeshing [4,5,10–14].
Nevertheless, several remeshing-free methods have been proposed: Belytschko has suggested the Element free Galerkin method [3] where the discretisation is achieved by nodal and surface description of the model. The Strong Discontinuity Approach (SDA) [15–17] in which displacement jumps due to the presence of the crack are embedded locally in each cracked finite element without affecting neighbouring elements. The Extended Finite Element Method (XFEM) [18–20] in which the displacement-based approximation is enriched near a crack by incorporating both dis-continuous fields and the near tip asymptotic fields through a Partition of Unity method [21]. The arbitrary local mesh replacement method [22] based on two distinct meshes: one that surrounds the propagating crack front and moves with it, and the other one that fills the rest of the domain. However, these methods need improvement in order to
Corresponding author.
E-mail address(es): karim.hitti@balamand.edu.lb (Karim Hitti), stephanie.elfeghali@gmail.com (Stéphanie Feghali), faten.rafeh@balamand.edu.lb(Faten Rafeh),marc.bernacki@mines-paristech.fr(Marc Bernacki), pierre-olivier.bouchard@mines-paristech.fr(Pierre-Olivier Bouchard).
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properly deal with long crack propagation and crack coalescence such as the XFEM. Or also complex configurations such as the Element free method, and multiple cracks such as the arbitrary local mesh replacement method. And some of them, like the XFEM, would eventually need some type of remeshing.
Hence, remeshing techniques are important when dealing with crack propagation. Chiaruttini et al. [10] used remeshing along with the Cohesive Zone Model (CZM) to simulate crack propagation in structures subjected to mixed mode loadings. Furthermore, 3D remeshing techniques were used to model properly crack propagation us-ing the computation of a damage indicator [11] or also based on mesh intersection algorithms [12]. Two and three-dimensional remeshing, via the fracture numerical code FRANC [13,14], were also used to simulate crack growth. Remeshing, combined with nodal relaxation, was developed and used by Bouchard et al. [4,5] and it has proven its efficiency when dealing with complex and multiple crack propagations.
Furthermore, monolithic approaches, which consist in using a unique mesh for the different components, are widely used in various domains. Notably in fluid-structure interaction [23], heat transfer [24], incompressible fluid flows [25], microstructures’ behaviour [26,27]. In all the above-cited applications, the monolithic approach proved its efficiency. Hence, studying its feasibility in crack propagation is proposed in this paper. In this context, two frame-works, the Eulerian and Lagrangian can be compared. The Eulerian framework consists in describing the crack im-plicity using a level-set function and thus the air inside the crack is taken into account in the governing equations. Also this level-set function must be convected at each time step in order to simulate its propagation. On the other hand, in the Lagrangian framework the crack is explicitly described by nodes of the mesh and its propagation is performed by moving these nodes, and remeshing.
In this paper, a method for modelling quasi-static crack propagation using level-set functions, anisotropic remesh-ing and a novel method for choosremesh-ing the convection velocity is introduced. In the second section, the level-set method and the anisotropic remeshing technique are explained. Also the Gθ method used for computing the energy release rate along with the kinking angle is detailed. Section 3 introduces our method for computing the convection velocity of the crack. A comparison between results in an Eulerian and a Lagrangian framework is performed. Then the better one is compared with Bouchard et al.’s [5] technique. The fourth section concludes this paper.
All numerical calculations mentioned in this paper were performed with CimLib, a finite element C++ library [28].
### 2. Numerical modelling
In the perspective of modelling crack propagation, a monolithic approach is used. The monolithic method consists of using a unique mesh in which the different domains are considered using a level-set function. Moreover a level-set function is a signed distance functionα, defined over the domain Ω, that gives at any node X of coordinates x of the FE mesh the signed distance to the boundaryΓ. In turn, the interface Γ is given by the level 0 of the function α:
(α(x) = ±d(x,Γ),x ∈ Ω
Γ = {x ∈ Ω,α(x) = 0} (1)
Fig. 1illustrates the level-set function of a crack.
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Consequently, different structures are immersed in a larger domain of different material properties so that bound-ary conditions at the interface can be replaced naturally. Furthermore, the level-set function, defining the interface, is used to mix the different material properties using appropriate mixing laws [26,29]. Moreover, this function is used for meshing and remeshing operations in order to describe properly the considered interface and also to account for the discontinuities of the physical properties. Mesh adaptation is also needed at the crack tip to insure the computational accuracy of the singular fields. It has to be noted that, contrary to XFEM, these singular fields are not imposed in the shape functions of the elements containing the crack tip.
To this end, it is crucial to pre-adapt anisotropically the mesh at the materials’ interface, the crack faces and tip. By doing so, the mesh becomes locally refined, elements are stretched, which enables to sharply define the interface and to save a great number of elements compared to classical local isotropic refinement. This anisotropic adaptation is performed by constructing a metric map. This map is controlled by a directional error estimator based on local inter-polation error of the wanted fields which are in this case the level-set functions defining the interface, the crack faces and the pressure field. Furthermore, in our case, different metrics are obtained, one for each field. The intersection of these metrics [30] leads to the one used for mesh adaptation.
In the aim of constructing these metrics, an a posteriori error estimator [31] is used. This error estimator uses the eigenvalues and eigenvectors of the recovered Hessian matrix of a given function and a wanted number of elements in order to construct the metric field. In practice, the mesh is generated in several steps using the MTC mesher and remesher developed in [32]. The proposed mesh generation algorithm works well for 2D or 3D complex shapes. It allows the creation of meshes with extremely anisotropic elements stretched along the interfaces. The mesh size is then only refined in the direction of the high physical and mechanical properties gradients. This allows conserving a high accuracy for describing properly mechanical fields and geometry description. During propagation, the mesh is only modified in the vicinity of the interface and near the crack tip which keeps the computational work devoted to the mesh generation low.
Since we will begin by comparing our method in Eulerian and Lagrangian frameworks and the better approach with the one in [5],Fig. 2shows the result of the mesh adaptation technique on the construction of the initial mesh of the comparison case, which is a pre-cracked rectangular region with an off-center inclusion. Here anisotropic remeshing was applied using the level-set functions of the crack and the disc along with the pressure which is maximal at the crack tip.
2.2. Computing the strain energy release rate
The strain energy release rate G represents the energy required for a unitary crack increase. The criterion states that among all virtual and kinematically admissible crack propagation directions, the real increase is the one which maximizes the strain energy release rate. The kinking angle,θ0, is then determined by:
³ dG ´ θ=θ0= 0 ³ d2G 2 ´ θ=θ0≤ 0 (2)
Numerous numerical techniques can be used to compute G [5]. In this study, the Gθ method [33] is used. It consists in defining two contours C1and C2around the crack tip. These contours divide the domain near the crack tip into
three sub-domains: Cext, Cr i ngand Ci nt(seeFig. 3).
Afterwards, the strain energy release rate computation is performed by solving:
G = Z Cr i ng · Tr (σ∇u∇Θ) −1 2Tr (σ∇u)∇.Θ ¸ dCr i ng. (3)
where Tr (.) is the trace operator, σ is the stress tensor, u is the displacement field, and Θ(θ1,θ2) is a virtual unit
displacement field that may be expressed as:
θ1= (1 −I MI J )cos(θ) θ2= (1 −I MI J )si n(θ) (4)
Therefore, this virtual fieldΘ modifies only the shape of elements belonging to Cr i ng. The computation of G is
performed in the Cr i ng area and a strain energy release rate G can be computed for each possible direction of crack
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Fig. 2.Anisotropic meshing adaptation based on an a posteriori error estimation: a) in the Lagrangian framework, b) in the Eulerian framework and c) zoom at the crack tip
Fig. 3.Contours and domains used to compute G with the Gθ method [5]
2.2.1. Comparison with analytical values
The Gθ method has been implemented in CimLib [28]. At each crack increment, G is computed forθ varying from −70oto 70owith 1osteps. G’s curve is increasing and then decreasing (seeFig. 4) so that the determination of the angleθ0corresponding to the maximum strain energy release rate is straightforward.
In order to validate our implementation and use the Gθ method for computing strain energy release rates, two comparisons with known analytical results were performed. The first with a single edge crack (seeFig. 5for details and dimensions) and the second for a slant crack (seeFig. 6for details and dimensions)
In the case of the single edge crack (Fig. 5), the strain energy release rate is G = KI2/E where E is the Young modulus and KI, the stress intensity factor in mode I, is given by [34]:
KI= σnF
p
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Fig. 4.Curve of G as a function ofθ in case of a single edge crack
Fig. 5.Domain and dimensions used in the single edge crack case
Fig. 6.Domain and dimensions used in the slant crack case
with F = 1.0174/λ3/2whereλ = 1 − aW and σn, the normal stress, taken to be 50M P a.
Analytically, G = 5.31 × 10−3N /m and the obtained numerical result was G = 5.4 × 10−3N /m In the case of the slant crack, G =(K
2 I+K
2 I I)
E where KIand KI I(the stress intensity factor in mode II) are given by [34]:
KI ,I I= σnFI ,I I
p
πc (6)
with
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FI I' 0.4741si n(1.4675φ) (8)
whereφ = 45oandσnalso taken to be 50M P a.
Analytically, G = 6.6 × 10−3N /m and the obtained numerical result was 6.45 × 10−3N /m
These results validate our implementation and use of the Gθ method for computing strain energy release rate and hence the kinking angle.
### 3. Convection velocity
3.1. Mechanical equations
After mixing the different material properties, the incompressible Stokes equations modelling the mechanical problem are solved using a stabilized mixed FE method and a P1+/P1 interpolation. These equations can be writ-ten: µ∆v − ∇p = −g on Ω ∆.v = 0 v = vimp on Γu v = −vimp on Γl (9)
where v is the velocity, p the pressure, g the gravity,Ω is the computational domain, and µ = 1+νE is the first Lamé coefficient where E is Young’s Modulus andν is Poisson’s ratio. Also vimpis the imposed velocity onΓuandΓlwhich
are the upper and lower boundaries of the domain respectively. 3.2. Computation of the convection velocity
After obtaining the velocity and the pressure fields from the mechanical problem, the computation of G infers the calculations of the propagation direction,θ0and hence the speed at the crack tip, ˙a which can be written [35]:
˙ a =( 0 i f G < Gc cR q 1 −KIc KI ot her w i se (10)
This speed is greater than zero only if G ≥ Gcwhich is the fracture energy in function of KIc, the fracture toughness; cRis the Rayleigh wave speed. Hence, a velocity around the crack tip, vc t, can be given by:
vc t= ˙a(cosθ0, si nθ0). (11)
This velocity insures the proper propagation of the crack but not the crack opening. The latter can be insured by the mechanical velocity, v. These facts lead us to use a Lagrangian convection velocity which is the combination of vc t
and v. This convection velocity, denoted vconvec, is given by:
vconvec=
(v i f ˙
a = 0 ⇐⇒ G < Gc
vc t(H (β)) + v(1 − H(β)) i f a 6= 0 ⇐⇒ G ≥ G˙ c
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whereβ is the level-set function of a small circle centered at the crack tip and H(β) a smoothed Heaviside function used for mixing the different velocities on a certain predefined thickness.
3.3. Results
In all the below-studied cases, anisotropic mesh adaptation is performed, at each time step, to maintain proper propagation and computation of the singular fields at the crack tip. Hence each time step can be illustrated by the scheme inFig. 7.
Moreover, a rectangular domain with an off-center circular inclusion described inFig. 8will be used in all the comparative cases.
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Fig. 7.The scheme of a time step
Fig. 8.The domain of the comparative case
3.3.1. Comparison between Eulerian and Lagrangian
Before comparing our method with another one, a choice between the Eulerian and Lagrangian frameworks has to be maid. Hence, two crack propagation cases (one in each framework) in which the circular inclusion in the pre-described domain is less rigid than the matrix, were performed.Fig. 9shows the comparison between the two
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frame-Fig. 9.Comparison between the Lagrangian and Eulerian framework during crack propagation: a) the Lagrangian one and b) the Eulerian one
works at different time steps.
In this case, according to Bouchard et al. [5], the crack has to be drawn by the inclusion but not to the point of intersection. It can be seen that using a Lagrangian framework, the crack moves away from the inclusion before hitting it which is not the case in the Eulerian framework. This is due to the fact that in an Eulerian framework a level-set function defines the crack and this function needs to be convected or transported at each time step in order to simulate the propagation.
This type of convection is very delicate because it is largely affected by the mesh size and the time step and since the level-set function would, in general, loose its characteristic as a distance function and has to be reinitialized. This is problematic when the remeshing technique depending on the distance property is used at the interface. Further details can be found in [36]. A solution to this problem might lie in the use of the direct reinitialization technique [37]. It consists in discretizing the interface (isozero of the level-set function) into a collection of simple elements and, for every node, computing the distance to all elements of the collection and storing the smallest one which becomes the updated value of the distance function. This method was considered very time consuming [38] but efficient parallel implementation and computation reduced drastically its computational times [37].
Also capturing accurately the crack tip via a level-set function is very hard since it may lie inside an element of the mesh (seeFig. 10) unless extreme refinement near the tip which, on the other hand, would lead to expensive computational times.
Body-fitted meshing [39] used for analysis of void growth and coalescence could be a solution if used near the crack tip. It consists on defining the interface (the isozero of the level-set) by nodes of the mesh and redefining this interface after convection using body-fitted remeshing. In our case, this might solve the problem of capturing the crack tip since it would be a node of the finite element mesh.
Furthermore, in our case, in addition to the level-set defining the crack, a level-set function defining the circle used for mixing the velocities near the crack tip is used. This circle needs also to be displaced along with the crack tip which make things more complicated because any small difference in the velocity at neighboring nodes might lead to misplacement of this circle.Fig. 11shows the case where the circle is convected faster than the crack leading to an unusual propagation of the crack.
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Fig. 10.(a) The fine mesh near the tip (b) the inaccurate definition of the level-set function defining the crack at the tip
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Since working in a Lagrangian framework was less complicated and lead to better results than an Eulerian one, the following case will be based on the Lagrangian one. It is important to mention that working in the Eulerian framework could eventually give good results if explored furthermore but surely needs extreme remeshing especially near the crack tip or needs the use of one of the above-mentioned possible solutions. On the other hand, working in the Eulerian framework might be necessary for some applications notably the hydraulic fracking [40,41] where a fluid is present inside the crack and its pressure on the crack’s faces needs to be computed.
3.3.2. Comparison with Bouchard et al. [5]
Our method, in the Lagrangian framework, is compared with a case presented in [5] (seeFig. 8). The influence of this inclusion on the crack path is studied. The rectangular domain was submitted to a tensile test. The circular inclusion may be more or less rigid than the matrix. R is defined as the ratio of the matrix and inclusion’s Young modulus: R = Emat r i x/Ei ncl.
Figs. 12and13illustrate the comparison between our method and Bouchard et al.’s [5]. Fig. 12shows that for an inclusion less rigid than the matrix (R = 10), the crack is attracted to the inclusion. Conversely, if the inclusion is more rigid than the matrix (R = 0.1,Fig. 13), the crack is moving away from the inclusion.
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Fig. 13.Comparison between our method and the one in [5] for R = 0.1: a) Results of [5] and b) our results
Both figures show a very good agreement between the two methods. This validates that our technique computes the strain energy release rate and the kinking angle in a proper manner. Furthermore, our choice of Lagrangian con-vection velocity models correctly crack propagation.
### 4. Conclusion
A monolithic method was introduced to model crack propagation. This process is based on the use of level-set functions for interface description, on anisotropic mesh adaptation and on the choice of an appropriate convection velocity. The latter is a combination between the mechanical velocity and the speed at the crack tip. This speed is linked to the strain energy release rate which was computed by the Gθ method along with the kinking angle.
Furthermore, in the monolithic method, two frameworks, the Eulerian and Lagrangian, were compared. In our case, working in the Lagrangian framework was less complicated since it needed less remeshing and since the crack was convected naturally with the mesh. Also since in the Eulerian framework, extremely fine meshes are needed to properly describe the crack faces and tip. And the crack, described implicitly by a level-set function, is hard to convect. Moreover, our monolithic Lagrangian method proved its efficiency in correctly modelling the propagation of the crack after comparing it with the method in [5]. Nevertheless, the Eulerian framework might be essential in some applications like the the hydraulic fracking in order to take the fluid inside the crack into consideration. A solution to the weak points of the Eulerian framework could be in using a body-fitted immersed volume method at the crack tip as described in [39] or also in use of direct reinitialization [37] to solve the problem of level-set convection.
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### References
[1] A.A. Griffith, The phenomena of rupture and flow in solid, Phil. Trans. Roy. Soc. London A 221 (1920) 163-197. [2] G.R. Irwin, D.C. Washington, Analysis of stresses and strains near the end of a crack traversing a plate, J. Appl.
Mech. (1957) 361-364.
[3] T. Belytschko, Y.Y. Lu, L. Gu, Element free Galerkin methods, Int. J. Numer. Methods Engrg 37 (1994) 229-256. [4] P.-O. Bouchard, F. Bay, Y. Chastel, I. Tovena, Crack propagation modelling using an advanced remeshing
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[7] A. Sur, M. Kanoria, Analysis of thermoelastic response in a functionally graded infinite space subjected to a Mode-I crack, Mode-Int. J. Adv. Appl. Math. and Mech. 3(2) (2015) 33-44.
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[11] H. Proudhon, J. Li, F. Wang, A. Roos, V. Chiaruttini, S. Forest, 3D simulation of short fatigue crack propagation by finite element crystal plasticity and remeshing, Int. J. of Fatigue 82 (2016) 238-246.
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[13] T. Bittencourt, P. Wawrzynek, J. Sousa, A. Ingraffea, Quasi-automatic simulation of crack propagation for 2D LEFM problems, Engrg. Fract. Mech. 55 (2) (1996) 321-334.
[14] B.J. Carter, P.A. Wawrzynek, A.R. Ingraffea, Automated 3D crack growth simulation, Gallagher Special Issue of Int. J. Numer. Methods Engrg. 47 (2000) 229-253.
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[16] K. Garikipati, On strong discontinuities in inelastic solids and their numerical simulations, Ph.D. Thesis, Stanford University, 1996.
[17] J. Oliver, Continuum material failure in strong discontinuity settings, in: E. Oñate, D.R. Owen (Eds.), COMPLAS 2003, VII International Conference on Computational Plasticity, CIMNE, Barcelona, 2003.
[18] T. Belytschko, T. Black, Elastic crack growth in finite elements with minimal remeshing, Int. J. Numer. Methods Engrg. 45 (1999) 601-620.
[19] N. Moës, J. Dolbow, T. Belytschko, A finite element method for crack growth without remeshing, Int. J. Numer. Methods Engrg. 46 (1999) 131-150.
[20] N. Moës, E. Bechet, Modeling stationary and evolving discontinuities with finite elements, in: E. Oñate, D.R. Owen (Eds.), COMPLAS 2003, VII International Conference on Computational Plasticity, CIMNE, Barcelona, 2003. [21] I. Babuska, J.M. Melenk, The partition of unity method, Int. J. Numer. Methods Engrg. 40 (1997) 727-758. [22] M.M. Rashid, he arbitrary local mesh replacement method: an alternative to remeshing for crack propagation
analysis, Comput. Methods Appl. Mech. Engrg. 154 (1998) 133-150.
[23] E. Hachem, S. Feghali, R. Codina, T. Coupez, Anisotropic adaptive meshing and monolithic Variational Multiscale method for fluid-structure interaction, Comp. and Struc. 122 (2013) 88-100.
[24] E. Hachem, G. Jannoun, J. Veysset, M. Henri, R. Pierrot, I. Poitrault, E. Massoni, T. Coupez, Modeling of heat transfer and turbulent flows inside industrial furnaces, Sim. Mod. Practice and Theory 30 (2013) 35-53.
[25] T. Coupez, G. Jannoun, N. Nassif, H.C. Nguyen, H. Digonnet, E. Hachem, Adaptive time-step with anisotropic meshing for incompressible flows, J. of Comp. Phys. 241 (2013) 195-211.
[26] K. Hitti, T. Coupez, M. Bernacki, L. Silva, Elastic foam compression in a finite element (FE) context, European Journal of Computational Mechanics (2012) 1-29.
[27] K. Hitti, S. Feghali, M. Bernacki, Permeability computation on a Representative Volume Element (RVE) of unidi-rectional disordered fiber arrays, Journal of Computational Mathematics 34 (3) (2016) 233-239.
[28] H. Digonnet, L. Silva, T. Coupez, Cimlib: a fully parallel application for numerical simulations based on compo-nents assembly, Proceedings NUMIFORM 07, American Institute of Physics, p. 269, 2007.
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[30] F. Alauzet, P.J. Frey, Estimateur d’erreur géométrique et métriques anisotropes pour l’adaptation de maillage. Partie I : aspects théoriques, Internal Report, INRIA, 2003.
[31] Y. Mesri, W. Zerguine, H. Digonnet, L. Silva, T. Coupez, Dynamic parallel adaption for three dimensional un-structured meshes: application to interface tracking, Proceedings of the 17th International Meshing Roundtable, 195-212, 2008.
[32] T. Coupez, Génération de maillage et adaptation de maillage par optimisation locale, Revue européenne des éléments finis 9 (2000) 403-423.
[33] P. Destuynder, M. Djaoua, S. Lescure, Quelques remarques sur la mécanique de la rupture élastique, Journal de Mécanique Théorique et Appliqué 2(1) (1983) 113-135.
[34] T. Fett, Stress Intensity Factors - T-Stresses - WeightFunctions. Supplement Volume, Universitatsverlag Karlsruhe, Karlsruhe, 2010.
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in unstructured grids, Int. J. Num. Met. in Eng. 72 (9) (2007) 1095-1110.
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I High visibility within the field | 7,520 | 29,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-25 | latest | en | 0.831435 |
https://schoollearningcommons.info/question/length-of-a-rectangle-eceeds-its-breadth-by-4-m-if-the-perimeter-of-the-rectangle-is-84-m-find-i-24979593-78/ | 1,632,487,661,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00315.warc.gz | 530,785,005 | 13,604 | ## Length of a rectangle exceeds its breadth by 4 m. If the perimeter of the rectangle is 84 m, find its length and breadth.
Question
Length of a rectangle exceeds its breadth by 4 m. If the perimeter of the rectangle is 84 m,
in progress 0
1 month 2021-08-12T20:39:13+00:00 2 Answers 0 views 0
1. Step-by-step explanation:
So,length = x + 4
Perimeter of rectangle = 2(l + b)
60m = 2(x + x + 4)
60m = 2 X 2x + 4
60m ÷ 2 = 2x + 4
30m = 2x + 4
30m- 4 = 2x
26m = 2x
26m ÷ 2 = x
13m = x
So, | 197 | 501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.849486 |
https://serverlogic3.com/what-is-diagonal-matrix-in-data-structure/ | 1,701,245,168,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00633.warc.gz | 593,481,627 | 15,602 | # What Is Diagonal Matrix in Data Structure?
//
Heather Bennett
A diagonal matrix is a special type of square matrix where all the elements outside the main diagonal are zero. The main diagonal of a matrix consists of elements that have the same row and column index. In other words, a diagonal matrix has non-zero elements only on its main diagonal, and all other elements are zero.
## Properties of Diagonal Matrix
A diagonal matrix has several interesting properties:
• 1. Scalar Multiplication: When you multiply a scalar value by a diagonal matrix, each element on the main diagonal is multiplied by the scalar, while all other elements remain zero.
• 2.
Addition and Subtraction: Adding or subtracting two diagonal matrices is as simple as adding or subtracting their corresponding elements on the main diagonals.
• 3. Matrix Multiplication: Multiplying two diagonal matrices results in another diagonal matrix, where each element on the main diagonal is obtained by multiplying the corresponding elements from both matrices.
## Example
To understand better, let’s consider an example of a 3×3 diagonal matrix:
```1 0 0
0 4 0
0 0 -2
```
In this example, the first element (1) corresponds to row 1 and column 1, second element (4) corresponds to row 2 and column 2, and so on. All other elements outside the main diagonal are zero.
### Scalar Multiplication Example
If we multiply this matrix by a scalar value of 3:
```3 * (1 0 0)
(0 4 0)
(0 0 -2)
```
The resulting matrix will be:
```3 0 0
0 12 0
0 0 -6
```
Let’s consider another example where we add two diagonal matrices:
```(1 0 0) + (2 0 0) = (3 0 0)
(0 4 0) + (0 -1 0) = (3 -1 0)
(0 2 -2) + (1 -1 -2) = (1 -1 -4)
```
### Matrix Multiplication Example
Lastly, let’s multiply two diagonal matrices:
```(1 0 ) * (4 5 ) = (4 5 )
( ) * ( -2) = ( )
( ) * (-6 ) = (-6 )
```
The resulting matrix will also be a diagonal matrix:
```4
What Is Matrix in Data Structure?
What Is Matrix in Data Structure? A matrix is a two-dimensional data structure that consists of a collection of elements arranged in rows and columns. It is often used to represent mathematical or tabular data.
What Is Matrix in Data Structure With Example?
What Is Matrix in Data Structure With Example? A matrix is a two-dimensional data structure that represents a collection of elements arranged in rows and columns. It is an essential concept in computer science and is widely used to solve various computational problems.
What Is Matrices in Data Structure?
What Is Matrices in Data Structure? In the field of data structure, a matrix is a rectangular array of elements arranged in rows and columns. It is a fundamental concept used in various applications, such as mathematics, computer science, physics, and engineering.
What Is Band Matrix in Data Structure?
What Is Band Matrix in Data Structure? A band matrix is a special type of matrix that has non-zero elements only in a specific band around its diagonal. In other words, most of the elements in a band matrix are zero, except for those within a certain number of diagonals from the main diagonal.
What Is Tri Diagonal Matrix in Data Structure?
A tri diagonal matrix is a special type of square matrix where the elements outside the main diagonal and the two diagonals adjacent to it are all zero. In other words, a tri diagonal matrix has non-zero elements only in the main diagonal, the diagonal above it, and the diagonal below it. Tri diagonal matrices find application in various fields, including numerical analysis, computational mathematics, and computer science.
Which Data Structure Is Best for Matrix?
The choice of the right data structure is crucial when working with matrices. A matrix is a two-dimensional array that consists of rows and columns, and it is commonly used in various fields such as mathematics, computer graphics, and machine learning. In this article, we will explore different data structures and determine which one is best for handling matrices.
What Is Matrix Data Structure?
The Matrix Data Structure: A Comprehensive Overview
Matrices are an essential part of computer science and mathematics. They provide a structured way to organize and store data in a two-dimensional grid format. In this article, we will explore the concept of a matrix data structure, its properties, and its applications.
What Is Matrix Transpose in Data Structure?
Matrix transpose, also known as matrix transposition, is an important concept in the field of data structures. It involves rearranging the elements of a matrix such that the rows become columns and vice versa. This operation is denoted by the superscript ‘T’ or by writing the matrix with its rows and columns interchanged.
What Is Transpose Matrix in Data Structure?
What Is Transpose Matrix in Data Structure? A transpose matrix is a fundamental concept in data structures that involves rearranging the rows and columns of a given matrix. This operation essentially flips the matrix over its main diagonal, resulting in a new matrix where the rows become columns and the columns become rows.
What Is a Matrix Data Structure?
A matrix data structure is a two-dimensional array that represents a collection of elements arranged in rows and columns. It is commonly used to store and manipulate data in various applications, such as mathematical operations, computer graphics, and scientific simulations. In this article, we will explore the concept of a matrix data structure and its significance in programming.
``` | 1,241 | 5,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2023-50 | longest | en | 0.843924 |
https://www.convertunits.com/from/mol/to/micromole | 1,669,636,132,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00524.warc.gz | 754,488,957 | 23,886 | ## ››Convert mole to micromole
mol micromole
How many mol in 1 micromole? The answer is 1.0E-6.
We assume you are converting between mole and micromole.
You can view more details on each measurement unit:
mol or micromole
The SI base unit for amount of substance is the mole.
1 mole is equal to 1000000 micromole.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between moles and micromoles.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of mol to micromole
1 mol to micromole = 1000000 micromole
2 mol to micromole = 2000000 micromole
3 mol to micromole = 3000000 micromole
4 mol to micromole = 4000000 micromole
5 mol to micromole = 5000000 micromole
6 mol to micromole = 6000000 micromole
7 mol to micromole = 7000000 micromole
8 mol to micromole = 8000000 micromole
9 mol to micromole = 9000000 micromole
10 mol to micromole = 10000000 micromole
## ››Want other units?
You can do the reverse unit conversion from micromole to mol, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Mole
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is "mol."
## ››Definition: Micromole
The SI prefix "micro" represents a factor of 10-6, or in exponential notation, 1E-6.
So 1 micromole = 10-6 moles.
The definition of a mole is as follows:
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is "mol."
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 591 | 2,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-49 | latest | en | 0.880009 |
https://www.coursehero.com/tutors-problems/Economics/8217376-1-Explain-why-voter-turnouts-are-generally-low-using-the-idea-of-rati/ | 1,548,177,675,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583857993.67/warc/CC-MAIN-20190122161345-20190122183345-00484.warc.gz | 752,059,676 | 51,589 | View the step-by-step solution to:
# TO VOTE OR NOT TO VOTE --that is the question-By Dewey D. Heinsma There are numerous reasons why voter turnout has steadily decreased over the last...
1. Explain why voter turnouts are generally low using the idea of rational ignorance. Does the Risker and Ordeshook voting model account for the idea of rational ignorance? Why or why not?
TO VOTE OR NOT TO VOTE --that is the question-- By Dewey D. Heinsma There are numerous reasons why voter turnout has steadily decreased over the last 40 years. One cited reason is the free time factor and the perception that voters are too busy. In contrast, John Samples presented three myths regarding declining voter turnout. Nonetheless, during the 2008 presidential election , there were 132,645,504 total voters out of an eligible voting age population of 212,702,354, which yields a 62.4% participation rate. Risker and Ordeshook (1968) presented a model for determining whether a person will vote. It is based on rational choice theory . The formula is presented below. The likelihood a rational person will vote = (pB + U) - C > 0. p = the probability that a person's vote will influence the outcome of an election. According to your text, the expected probability ( p ) of any one vote determining an election outcome is low. B = perceived benefit received from the voter's candidate winning the election or vote on a proposition U = the level of utility received from voting (e.g., "I voted today" stickers; researching an issue; allegiance to a political party) C = Explicit (financial cost) and implicit (opportunity cost) costs associated with casting a vote For a person to vote, the benefits minus the costs must be greater than zero (Risker & Ordeshook, 1968). However, these listed variables are intrinsically inexact, making it difficult to isolate why voters choose to vote. In contrast, public choice theory helps us explain the behavior of voters and why voting is the most common decision making method in the public sector
Answer: The voter turnouts are generally low because of the free time factor and the perception
that voters are too busy. Also Samples presented three myths regarding declining voter
turnouts....
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Browse Documents | 554 | 2,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-04 | latest | en | 0.942008 |
https://www.physicsforums.com/threads/how-much-torque-is-required-to-rotate-a-1500lb-steel-drum.897435/ | 1,718,946,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00325.warc.gz | 840,838,569 | 17,863 | # How Much Torque is Required to Rotate a 1500lb Steel Drum?
• SevenToFive
In summary, the conversation discusses the need to rotate a 1500lb steel drum with an 8 foot diameter to wrap up air filter material. The drum is supported by bearings and needs to turn at 0.02rpm. The torque required is calculated using the moment of inertia equation for a hollow cylinder and is found to be 8.81lb-in. The group discusses if this value seems low and considers the bearing friction and efficiency, as well as the tension of the filter material, as potential factors affecting the required torque. They also review the correct formula for calculating moment of inertia and consider the slow rotational speed of 0.02rpm. Finally, they discuss the possibility of the wrapping mechanism adding additional load to the drum.
SevenToFive
We have a 1500lb steel drum that has a 8 foot diameter that we need to rotate to wrap up air filter material. The drum is supported by bearings on both sides and needs to turn at 0.02rpm.
I calculated the torque, using T=Moment of Inertia * Angular Acceleration. For the moment of inertia I used the equation for a hollow cylinder.
Gravity =386.088in/sec^2
For the weight I figured 2500lbs to factor in losses through friction(0.5). Outside diameter with the roll of air filter material would be 109 inches, inside diameter =100 inches. Final rpm = 0.02, Acceleration time is 5 seconds.
The angular acceleration =0.02*2*3.14/60/5 = 0.00042Rad/sec^2
Moment of inertia = 21041.835lb-in-sec^2
T=8.81lb-in
However that value seems very low to me. Any help is greatly appreciated.
SevenToFive said:
T=8.81lb-in
... and, in "ft-lbs?" Can you turn it by hand? With a "cheater?"
Bystander said:
... and, in "ft-lbs?"
Went the wrong way, sorry. The "by hand" inquiry stands.
Calculate the torque required to turn the drum based strictly upon the bearing friction as a comparison for judging if your calculated starting torque is greater than the basic required rotating torque.
JBA said:
Calculate the torque required to turn the drum based strictly upon the bearing friction as a comparison for judging if your calculated starting torque is greater than the basic required rotating torque.
I would have 1500lbs, 0.50 for bearing efficiency, and 48" from the center of the drum to the outside, which would give me 36000lb-in, which seems a bit on the high side.
I am not sure to what the term "bearing efficiency" refers, since the rolling resistance of an anti-friction bearing is generally based upon the "bearing friction factor" and 0.50 is an extremely high "bearing friction factor" for any type of anti-friction (rolling type) bearing.
See the below website for a typical anti-friction (rolling bearing) resistance torque calculation.
http://www.skf.com/us/products/bear...ction/estimating-frictional-moment/index.html
Also, I believe you used the wrong Moment of Inertia equation for your calculation; and, should have used the "Polar Moment of Inertia" equation for a cylinder rotating about its axis.
.02 RPM is VERY VERY slow (once an hour?).. I think in this case the amount of energy needed to accelerate it can pretty much be ignored, you're BIG one is friction which will be orders of magnitude higher.
Looking at this http://www.skf.com/ca/en/products/b...ction/estimating-frictional-moment/index.html and getting some guidelines, the formula is
M = 0,5 μ P d
It seems to me you've mistaken 0.5 as the friction coefficient μ, it is NOT.
From their table, I'm going to choose "Self aligning ball bearings" since most pillow block bearings in industrial applications would fall into that category.. μ for them is 0.0010. You haven't mentioned the shaft size, for that kind of load I'm going to take a stab at guessing a 3" bearing (the d component in the formula)
So that brings me to the following
M = 0.5 * .0010 * (1500 lb * .454kg/lb) * (3in * 25.4 mm/in)
M = 0.5 * .0010 * 681 * 76.2
M = 25.9N*mm (This is far less than I had estimated in my mind)
However, is the wrapping mechanism going to put any load on the drum? that may be the largest force working against you.
Rx7man said:
However, is the wrapping mechanism going to put any load on the drum? that may be the largest force working against you.
+1.
What is the tension in the filter material?
## 1. What is torque and how does it relate to rotating a drum?
Torque is a measure of the turning force on an object. In the case of rotating a drum, torque is the force applied to the drum in order to make it spin.
## 2. How is the torque needed to rotate a drum calculated?
The torque needed to rotate a drum depends on several factors such as the mass of the drum, the inertia of the drum, and the speed at which it needs to rotate. It can be calculated using the formula: torque = mass x acceleration x radius.
## 3. What factors can affect the torque required to rotate a drum?
The torque required to rotate a drum can be affected by the size and weight of the drum, the friction between the drum and its bearings, and the speed at which it needs to rotate. Additionally, the shape and distribution of the load inside the drum can also affect the torque needed.
## 4. How can torque be applied to rotate a drum?
Torque can be applied to rotate a drum through various methods such as using a motor, a gear system, or manual force. The amount of torque needed will depend on the method used and the factors mentioned in the previous questions.
## 5. How can torque be controlled to rotate a drum at a specific speed?
To control the speed at which a drum rotates, the torque can be adjusted accordingly. This can be done by changing the motor speed, adjusting the gear ratio, or applying more or less force manually. Additionally, using sensors and controllers can also help regulate the torque and maintain a specific speed for the drum rotation.
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4K | 1,525 | 6,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.928522 |
https://space.stackexchange.com/questions/18367/how-far-could-material-from-earth-or-mars-be-transported-due-to-asteroid-impact | 1,702,030,796,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00280.warc.gz | 592,957,919 | 42,943 | # How far could material from Earth or Mars be transported due to asteroid impact?
NASA has issued reports to the effect that small bits of Mars have come to Earth due to asteroid impacts - asteroid hits Mars, chunks are ejected into space far enough to eventually fall into Earth's gravity well. The reverse is also plausible - material from Earth could have been transported to Mars the same way.
That said, my question is: how far could such transport reach through the solar system? Could material ejected from an early Earth have made it as far out as Jupiter? What about material from Mars?
## 2 Answers
It is possible for dust and debris to travel basically anywhere in the solar system, given lots and lots of time.
The first major hurdle to this travel is that the impact needs to accelerate the debris to a speed higher than the planets escape velocity. In the case of Mars, that's a speed of about 5km/s. Any debris traveling under this speed will eventually fall back down and land on Mars. Anything higher will drift off into interplanetary space.
Once in interplanetary space, there is a set of pathways known as The Interplanetary Transport Network, which are very low-energy routes between bodies in the solar system. With a small bit more speed than the escape velocity it is possible for the debris to enter a path in this network, and eventually arrive at some other planet in the solar system. The only cost is time; it can take millions or billions of years on some routes.
In an even bigger collision, and in some gravitational slingshot scenarios, the debris can reach the sun escape velocity, and head off into interstellar space.
There is no limit in either direction.
Kim has shown that matter can leave the solar system.
In the opposite direction, towards inner planets, it's a matter of angular momentum with respect to the Sun. Since objects move slower the farther out they are, it's easier for them to lose all their momentum in a collision. Then the Sun's gravity pulls them inwards nearly radially. It's slow, but given enough time, such an object crosses all inner planet's orbits and hits the Sun. If it didn't collide with another body. (This assumes the original body was close to the ecliptic plane and the collision didn't change that too drastically.)
In other words, there are no conservation laws that prohibit rocks from leaving or entering the Solar System as far as they like. Those laws only restrict the time frame and the velocity.
This opens up the (slim) possibility that we might find some Pluto/Sedna/Quaoar debris on Antarctica :-)
• Collisions of this nature are very likely to change the orbital plane. Why wouldn't it? There are no forces to constrain collision debris to a particular direction. Sep 28, 2016 at 10:14
• Also, it is incorrect to say that it is easier for objects to lose all their momemtum in a collision. It just depends on which angle they are hit at. If they are hit from behind, they will accelerate and move to a further out orbit, if they are hit from the prograde direction they will slow down, and move to a smaller orbit closer to the sun. In fact, it is only the FAR side of the orbit which will be lowered towards the sun, making the orbit more elliptical. It is not conceivable that the object can be STOPPED, so it falls straight down and hits the sun. Sep 28, 2016 at 10:36
• @Innovine My description is imprecise, sorry. What I meant to convey was that in a collison not the whole body would be stopped, but that it is likely that some of the zillions of debris rocks, each with its own random momentum, there could be some with near zero momentum.
– Jens
Sep 28, 2016 at 11:35 | 802 | 3,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | longest | en | 0.95572 |
https://metricsystem.net/derived-units/special-names/lumen/ | 1,686,161,692,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654012.67/warc/CC-MAIN-20230607175304-20230607205304-00190.warc.gz | 422,229,201 | 33,391 | lumen
SI coherent derived unit with special name and symbol
Name Symbol Derived quantity Expressed in terms of SI base units
lumen lm luminous flux cd
Definition
The lumen, symbol lm, is the SI coherent derived unit of luminous flux,
One lumen is equal to the luminous flux emitted in a solid angle of one steradian by a uniform point source having a luminous intensity of one candela.
The lumen is defined in relation to the candela as:
$1 \ \text{lm} = 1 \ \text{cd} \ \text{sr}$
Luminous flux
Luminous intensity is a measure of the wavelength-weighted power emitted by a light source in a particular direction per unit solid angle, based on the luminosity function, a standardised model of the sensitivity of the human eye across the visible range of the electromagnetic spectrum.
Luminous flux is a measure of the luminous energy emitted by a source, in all directions, per unit time. It can be thought of as the power, or brightness of a light source.
If a source radiating a power of one watt of monochromatic light, in the colour for which the eye is most efficient (a wavelength of 555 nm, in the green region of the optical spectrum), it will have a luminous flux of 683 lumens.
One lumen represents at least 1683 watts of visible light power, depending on the spectral distribution.
Luminous flux is analogous to radiant flux, the corresponding objective physical quantity used in the measurement science of radiometry.
Luminous flux differs from radiant flux in that radiant flux includes all frequencies of emitted electromagnetic waves, whereas luminous flux is weighted according to the human eye’s sensitivity to different electromagnetic wavelengths. Essentially, luminous flux is a measure of the visible component of a radiant flux.
Lumen vs candela
The difference between the units lumen and candela is that the candela describes the luminous flux in a particular direction, per unit solid angle, whereas the lumen describes the total luminous flux in all directions.
$1 \ \text{cd} = 1 \ \dfrac{\text{lm}}{\text{sr}}$
one candela = one lumen per steradian one candela in all directions = a total luminous flux of 4π lumens
A complete sphere has a solid angle of 4π steradians. Therefore, a light source that uniformly radiates one candela (one lumen per steradian), in all directions has a total luminous flux of 4π lumens.
$1 \ \text{cd} \times 4\pi \ \text{sr} \ = \ 4\pi \ \text{cd} \ \text{sr} \ \approx \ 12.57 \ \text{lm}$
Lumen vs lux
The difference between the units lumen and lux is that the lux takes into account the area over which the luminous flux is spread.
A luminous flux of one lumen, concentrated onto an area of one square metre, lights that square metre with an illuminance of one lux. The same one lumen, spread out over ten square metres, produces a dimmer illuminance of 0.1 lux.
$1 \ \text{lx} = 1 \ \dfrac{\text{lm}} {\text{m}^2}$
Light bulbs
The sale of incandescent light bulbs has largely been replaced by energy-efficient light bulbs, including light bulbs using LED technologies. This has rendered meaningless the old method of comparing bulb brightnesses by their power consumption, measured in watts.
The lumen is now the standard unit used to describe the brightness of all domestic light bulbs, and is used alongside the power consumption in watts.
Photometry units
Name Symbol Quantity Expressed in terms of SI base units Expressed in terms of other SI units candela cd luminous intensity cd lm/sr lumen lm luminous flux cd cd sr lux lx illuminance m-2 cd lm/m2 candela per square metre m-2 cd luminance m-2 cd cd/m2 lumen second lm s luminous energy s cd lm s lumen per watt lm W‑1 luminous efficacy kg‑1 m‑2 s3 cd lm/W | 904 | 3,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-23 | latest | en | 0.870951 |
https://www.thestudentroom.co.uk/showthread.php?t=2558777 | 1,604,015,393,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00499.warc.gz | 919,246,031 | 34,716 | # C3 differentiation help please
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#1
Given that y=secx+tanx, find dy/dx and show that it can be written in the form 1/1-sinx
My working out:
dy/dx = secxtanx+sec2x
then I converted them
= 1/cosx (√ sec2x -1 ) + 1/cos2x
Am I on the right track? If not please help mee!
0
6 years ago
#2
(Original post by queenfatso)
Given that y=secx+tanx, find dy/dx and show that it can be written in the form 1/1-sinx
My working out:
dy/dx = secxtanx+sec2x
then I converted them
= 1/cosx (√ sec2x -1 ) + 1/cos2x
Am I on the right track? If not please help mee!
Write the whole thing as one fraction: . Rewrite the denominator in terms of a familiar identity and then think about difference of two squares.
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48.69% | 405 | 1,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.897144 |
https://mb.com.ph/2021/03/05/where-are-you-in-the-vaccine-queue/ | 1,652,949,447,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00766.warc.gz | 459,857,856 | 60,083 | ## Where are you in the vaccine queue?
Published March 5, 2021, 4:07 PM
• There’s an online tool that predicts when you are likely to get the jab
• It’s called the Vaccine Queue Calculator and it estimates how many people are ahead of you in the queue
• It will also predict how long you might have to wait to get the vaccine jab
The much-awaited vaccination drive against the coronavirus disease (COVID-19) finally started in the Philippines this week, after over a year of facing the pandemic which caused more than 12,000 casualties and plunged the country’s economic growth.
With around 110 million people in the Philippines, who will be the first in the queue and how long will it be until you get vaccinated? Kenneth Alambra, a civil engineer from the University of the Philippines Los Baños, and Reina Sagnip, a researcher from the De La Salle University in Manila, teamed up to create a vaccine queue calculator to answer these questions.
“The Vaccine Queue Calculator for the Philippines will estimate how many people are ahead of you in the queue to get a COVID vaccine in the Philippines. It also predicts how long you might have to wait to get your vaccine doses,” Kenneth and Reina said.
Through the Philippine Vaccine Queue Calculator, one will have an idea of when they can get vaccinated.
Vaccine queue calculator
How to use the vaccine queue calculator? First, several questions will be asked from the user such as age, whether one is a frontline health worker, frontline personnel in essential sectors, or just employed but not a frontliner.
It will also ask if the user is a person with comorbidities; an indigent, or a person with disability (as determined by the Department of Social Welfare and Development); or is included in the socio-demographic groups at significant higher risk.
The calculator will then guess how many people are ahead of you in the queue and estimate when you are likely to be vaccinated using the target vaccination rate of the government.
Given a vaccination rate of 1,274,980 a week and an uptake of 56 percent, the calculator estimates that a 25-year-old who is not a frontline worker, without comorbidities, and is not indigent, could receive the two doses of vaccine between August 4 and Oct. 24, 2021.
“We’ve based our vaccine queue calculator on the data provided by The Philippine National Deployment and Vaccination Plan for COVID-19 Vaccines for the priority list, and the likely vaccination rate to achieve the government’s target to vaccinate 50 to 70 percent of the adult population for the rest of 2021,” the pair explained.
The uptake rate, meanwhile, is based on a February 2021 survey which shows that only 56 percent are willing to take the COVID-19 vaccine.
Who gets the vaccine first?
The government has published a priority list which is divided into the following groups for the vaccination plan: A1: Frontline workers in health facilities both national and local, private and public, health professionals and non-professionals like students, nursing aides, janitors, barangay health workers, etc.
A2: Senior citizens aged 60 years old and above
A3: Persons with comorbidities not otherwise included in the preceding categories
A4: Frontline personnel in essential sectors including uniformed personnel and those in working sectors identified by the IATF as essential during ECQ
A5: Indigent population not otherwise included in the preceding categories
B1: Teachers, social workers
B2: Other government workers
B3: Other essential workers
B4: Socio-demographic groups at significantly higher risk other than senior citizens and indigenous people
B5: Overseas Filipino workers
B6: Other remaining workforce
C: Rest of the Filipino population not otherwise included in the above groups
Be patient
While waiting to get the vaccine might be frustrating, prioritizing those who are most at risk will save more lives and speed up the country’s recovery to move forward to the new normal. | 835 | 3,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.942552 |
http://ordnur.com/tag/business-risk/ | 1,539,963,472,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512411.13/warc/CC-MAIN-20181019145850-20181019171350-00169.warc.gz | 270,447,612 | 16,503 | How to Determine Optimal Capital Structure
Many people get confused about the optimal capital structure for a company because they observed that different company holding different capital structure. Capital can be formed by using only equity or combination of debt and equity, but cannot form only using debt capital. For a company debt and equity can be 50/50, for other company it can be 40/60, or 60/40, or 65/35 or any other proportion. Here question may arise that what proportion of debt and equity should use to form an optimal capital structure?
The optimal capital structure mainly depends on the two important things. These are:
• The financial risk of that company
Business risk arises when a company unable to generate a sufficient amount of cash flow to pay off its operating expenditures. Operating expenditure can be rent expense, salaries expense, wages, depreciation etc. The main thing is if the business unable to cover its business operation related expenditure then the company will exercise business risk. Lower the cash inflow, the higher the business risk and higher cash inflow from the operation will reduce the business risk.
So when forming an optimal capital structure it is required to keep in mind that the number of business risks which may arise. If there is a possibility of higher business risk then it will be a wise decision to use higher equity capital rather using debt capital. On the other hand, the company which generates a huge amount of cash flow from its business operation can use a larger amount of debt to maximize its return.
Business risk can be divided into two risk categories. One is systematic (non-diversifiable) and other is unsystematic (diversifiable) risk. As we cannot reduce the systematic we must diversify our investment so that we can reduce the risk of the company.
Financial Risk
Financial risk directly related to the interest amount payable to the debt provider. That is when a company is unable to generate sufficient cash flow to pay off the financial obligations properly then that company experience financial risk. A company with a sound financial position ensures less financial risk and the cost of debt financing is lower as the investment in this company is less risky.
Although business risk and financial risk are two broad things we have to keep in mind that some other factors like availability of equity capital, cost of equity, cost of debt, cash conversion cycle of the business, the profitability of the firm and corporate tax rate may involve with determining the optimal capital structure. Also, optimal capital structure (debt and equity ratio) depends on the type of business and economic condition of the company.
What are the Basic Types of Risk?
We know that future is uncertain, because of uncertainty; involvement of risk can be traced to our every part of life. When we talk about any investment we have to think about risk and return, higher the risk higher the rates of return and lower the risk lower the rates of return. Our life is directly related with economic activities where risk is the considerable element that cannot be overlooked.
To minimize the risk people go for savings and some people take the help of insurance companies/ agencies by paying insurance premium. Risk can be categories into different perspective but here we only discuss about the business risk.
Before discussing the types of risk, let’s have some idea of risk. Risk is the deviation between the actual outcome and expected outcome. Some risk can be minimized and some risk cannot be minimized. Some risk arisen from the micro economic factor and some from macro economic factors.
Basic types of risk that we may found are:
Basic Types of Risk
Unsystematic Risk
Unsystematic risk is that portion of risk which can be minimize through diversification of the investment by forming portfolio. If we form a portfolio using the negatively correlated investment securities then it would be possible to minimize the risk at lower level. This types of risk is known as diversiable risk Theoretically it is possible to eliminate the portion of unsystematic risk but in real sense it is not possible to eliminate the risk through diversification.
Systematic Risk
Systematic risk is that portion of risk which cannot minimize through diversification of the investments. Systematic risk is mainly arisen from the macro economic variables which are beyond our control. Beta is the measure of the systematic risk. Sometimes this risk is also known as systematic market risk. Sources of systematic risk are given below with short explanation.
Business risk is the risk which mainly arise when a firm or business organization unable to generate sufficient revenue to maintain its operating expenditure through providing service or selling products, that is risk is directly related with the operation of the firm.
Financial Risk
When a firm is unable to pay off its fixed financial obligation then this type of risk may arise. This type of risk is involved with the levered firm which uses debt capital for business. In some cases this risk can lead a lead a company to bankruptcy.
Liquidity Risk
This risk is involved with the marketability of a security or investment that is the capacity to generate asset into cash as much quicker as possible. If an investment is takes less time to convert into cash then it is liquid asset or investment.
Country Risk
Unstable political condition of a country is responsible for this type of risk. If this risk is more than an economy definitely fall, so does business. In our Bangladesh this type of risk is higher.
Exchange Rate Risk
Exchange of currency is required when a country is involved with import and export. For importing product or services foreign currency basically dollar is used. So if there is more fluctuation of the exchange rate frequently then a business may incur loss. This probable loss is the risk for the business.
Although every economic activity is involved with risk, we need to be more cautious to minimize the risk. If we can minimize the risk of doing business then it will be possible to generate profit for the company/ business organization.
Written by
Md. Nahian Mahmud Shaikat
Student of MBA
Jahangirnagar University
Email: [email protected] | 1,201 | 6,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-43 | latest | en | 0.91517 |
http://yakasee.com/xgU6Q154/eV5649gJ/ | 1,566,492,687,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317274.5/warc/CC-MAIN-20190822151657-20190822173657-00029.warc.gz | 336,321,529 | 9,754 | # Ccss Nbt Worksheets Two Digit Addition And Subtraction Within Year
Published at Friday, 10 May 2019. Color by Number. By .
So many positive things have been said about coloring sheets and coloring apps in general. Lately, a new trend has emerged on the Internet and it is absolutely raging: color by number and paint by number coloring sheets and pages. Let’s put all these online apps aside for a moment, and focus on the benefits of color by number coloring books alone, especially for young users. Okay, you are given a picture, the numbers are there within it, and you are already told what the color for a corresponding number is. How can it be beneficial? If nothing, it seems a bit simple and stupid, don’t you think? Well, not really!.
Color by number worksheets are a popular way to introduce your child to both numbers and colors. Your child will learn to recognize and identify numbers and match them to the numbers used to label the crayons, paints, or markers to determine which color to use in each area. While this can help with basic color by number addition and math skills, it can also be useful in exposing your child to different colors. While your child will learn to tell the difference between red, blue, pink, and green, he may also have the opportunity to learn about lesser known shades. Color by number activities for older children can serve as a great educational tool, especially when it comes to math. Advanced coloring activities can teach kids about addition, subtraction, multiplication, and division, while also providing skills in how to properly use a legend. Not only can these skills help to improve logic and mathematical ability, but they can also be transferred to other crucial life lessons, such as reading and understanding a map.
Color by number printable build fine motor skills, For younger students, coloring “between the lines” helps to build fine motor skills in the hands and fingers. It helps with children’s dexterity, hand-eye coordination, and skills with manipulating tools. This is not as much of a benefit for older kids, but it is definitely something the younger kids should practice. This will eventually lead to future skills with writing letters and writing in cursive. Color by number printable introduce students to colors, Color by number pages are a good way to teach color theory to children. Initial color by number pages can introduce colors like red, yellow, blue, green – the primary colors and secondary colors. Later color by number pages can introduce more complicated colors like magenta, cerulean, and amber. Color by number pages can help introduce kids to fine arts.
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https://gonitzoggo.com/problem/291 | 1,723,628,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00191.warc.gz | 220,258,720 | 5,503 | ### লম্ব-সমান্তরালের হযবরল
In the figure given below, $P$ is a point on $AB$ such that $AP:PB=5:4$. $PQ$ is parallel to $AC$ and $QD$ is parallel to $CP$. $AR$ and $QS$ are perpendicular to $CP$. Length of $QS=6$ then ratio of $AP:PD=a:b$ where $a,b$ are relatively co-prime and positive number. Then $a+b=?$
Geometry BdMO
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#### Statistics
Attempt 234
Solve 183
First Solve MSI23 | 163 | 426 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.698998 |
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[quote="Trojan Horse"]Okay, so one of the lights is controlled by exactly 2 switches. Question: do we know exactly how those 2 switches affect that light? Do we know for sure that the 2 switches are connected in an "XOR" fashion; that is, flipping either switch always flips the light? Or is it possible that the switches control the light in some other way? Maybe the light turns on only if both switches are in the up position, or if at least one switch is in the down position, or whatever. (I once lived in an apartment where the only way to turn the hall light on was to flip BOTH of two switches to the down position. So I'm not assuming anything here.)[/quote]
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ChickenMarengo
Posted: Sun Mar 07, 2010 12:12 am Post subject: 1
Zag wrote:
if n=3, this is actually the same as n=1.
You can actually do slightly better with n=3 (but still worse on average than your answer of n=2):
You can ignore any flips that change all 3 bulbs. You're only interested in the ones that flip a single bulb.
Once you've found them label the switches and bulbs so that ABC=1 and ABD=2. Then A and B = 3, C = 1, D = 2.
The probability of finding the 2 interesting flips in the first 2 flips is 1/6, of needing 3 flips is 1/3, and of needing 4 flips is 1/2, for an average of 3 1/3 flips.
Zag
Posted: Sat Mar 06, 2010 11:25 pm Post subject: 0
Then the answer is clearly n=2, which can always be solved in 3 turns.
obviously, n=0 and n=4 are useless.
if n=1, then your first three tries might each change a different bulb, and you will need a fourth try to know which bulb is double-switched.
if n=3, this is actually the same as n=1.
with n=2, you switch AB, BC, CD.
One of these situations must be the case (once you number the bulbs to fit).
AB=no change, BC=12, CD=23 (A and B = 1, C=2, D=3)
AB=12, BC=12, CD=13 (A and C = 1, B=2, D=3)
AB=12, BC=23, CD=13 (A and D = 1, B=2, C=3)
AB=12, BC=no change, CD=23 (A=1, B and C=2, D=3)
AB=12, BC=23, CD=23 (A=1, B and D = 2, C=3)
AB=12, BC=23, CD=no change (A=1, B=2, C and D = 3)
bonanova
Posted: Sat Mar 06, 2010 9:03 pm Post subject: -1
In all cases if you flip a switch, one and only one bulb will change its state.
In my entrance hall, I have the case you talked about.
I have not been able to fix those connectons.
Anyone have a clue on that issue, I'd be glad to hear it.
Trojan Horse
Posted: Sat Mar 06, 2010 8:20 pm Post subject: -2
Okay, so one of the lights is controlled by exactly 2 switches.
Question: do we know exactly how those 2 switches affect that light? Do we know for sure that the 2 switches are connected in an "XOR" fashion; that is, flipping either switch always flips the light? Or is it possible that the switches control the light in some other way? Maybe the light turns on only if both switches are in the up position, or if at least one switch is in the down position, or whatever.
(I once lived in an apartment where the only way to turn the hall light on was to flip BOTH of two switches to the down position. So I'm not assuming anything here.)
bonanova
Posted: Fri Mar 05, 2010 9:31 am Post subject: -3
Here's a piece of fluff for the back of a napkin at lunch time.
Might not even need the napkin...
In my kitchen there are 3 ceiling lights and 4 wall switches.
Here's the deal:
1. Each switch controls 1 and only 1 light.
2. Each light is controlled by at least one switch.
3. When you enter the kitchen [you might not want to open the fridge] you find the switches positioned randomly up or down.
4. You are to determine how the lights and switches are connected.
5. You may [repeatedly, but using the same value of n each time] simultaneously change any n [0 < n < 4] of the switches and note what happens.
Here's the question:
If you want to minimize the number of times you execute Step 5 above, is there a preferred value of n? | 1,299 | 4,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2013-20 | latest | en | 0.873993 |
https://plainmath.org/precalculus/84388-in-a-and-b-cosine-is-given-find-sine-a | 1,701,812,301,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00184.warc.gz | 529,606,080 | 21,355 | Violet Woodward
2022-07-28
In A and B, cosine is given..find sine and tangent if x liesin the specific interval.
A.) $\mathrm{cos}\left(x\right)=1/3x\in \left[\pi /2,0\right]$
B.) $\mathrm{cos}\left(x\right)=-5/13x\in \left[\pi /2,\pi \right]$
C.) $s=\mathrm{sec}\left(\pi x/2\right)$...Find the period. What symmetries does the function have?
D.) $\mathrm{cos}\left(x+\pi /4\right)-1$. What is the period?
eyiliweyouc
(A) since $\left[0,\pi /2\right]$ comes in first interval so sine and tan are positive
so;$\mathrm{sin}\left(x\right)=\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}=\left(\sqrt{9-{x}^{2}}\right)/3$
so;$\mathrm{tan}\left(x\right)=\mathrm{sin}\left(x\right)/\mathrm{cos}\left(x\right)=\left(\sqrt{9-{x}^{2}}\right)/x$
(B) since$\left[\pi /2,\pi \right]$ comes in second interval sosine is positive and tan is negative.
so; $\mathrm{sin}\left(x\right)=\sqrt{1-{\mathrm{cos}}^{2}\left(x\right)}=\left(\sqrt{169-25{x}^{2}}\right)/13$
so;$\mathrm{tan}\left(x\right)=\mathrm{sin}\left(x\right)/\mathrm{cos}\left(x\right)=\left(\sqrt{169-25{x}^{2}}\right)/-5x$
(c)the period of $\mathrm{sec}\left(\pi x/2\right)=2$
(d)the period of $\mathrm{tan}\left(x+\pi /4\right)=\pi /2$
Do you have a similar question? | 455 | 1,217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 25, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.399448 |
https://www.chittorgarh.com/ipo-hni-funding-cost-calculator/metro-brands-ipo/1193/ | 1,679,708,302,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945292.83/warc/CC-MAIN-20230325002113-20230325032113-00441.warc.gz | 819,474,031 | 20,031 | # Metro Brands IPO HNI Funding Cost Calculator
## Metro Brands IPO Shares to Apply By Amount Calculator
Enter the amount you would like to invest in Metro Brands IPO and find how many shares you should apply. In an IPO, shares are applied in lots. This calculator helps you to find an exact number of shares to apply for the given amount.
Amount to Invest (Rs) Lot Size (Shares) 30 Shares Issue Price (Rs) 500 Lots to Apply 0 Shares to Apply 0 Total Amount (Rs) 0
## Metro Brands IPO Funding Interest Rates Calculator
Metro Brands IPO funding cost is derived based on the NII (HNI) over subscription and interest rate considering the 7-day loan period. The table provides an estimated interest cost per share for rate of interest (ROI) ranging from 7 to 10% over a range of 10 different over subscription intervals.
IPO Price ₹500 NII IPO Quota ₹205 Cr NII Over subscription (Estimated in times) NII IPO Collection (Incremental in ₹ Cr) Number of Days Interest Rate (%)
### Metro Brands IPO HNI Cost for 7 days
NII IPO Collection (₹ Cr) 0 0 0 0 0 0 0 0 0 0 NII Over subscription (times) 0 0 0 0 0 0 0 0 0 0 Interest Rate ↓ ↓ HNI Funding Cost (Rs per share) ↓ 0% 0 0 0 0 0 0 0 0 0 0 7% 0 0 0 0 0 0 0 0 0 0 7.5% 0 0 0 0 0 0 0 0 0 0 8% 0 0 0 0 0 0 0 0 0 0 8.5% 0 0 0 0 0 0 0 0 0 0 9% 0 0 0 0 0 0 0 0 0 0 9.5% 0 0 0 0 0 0 0 0 0 0 10% 0 0 0 0 0 0 0 0 0 0
IPO Funding Cost = IPO Price * NII Over Subscription * (Interest Rate/100) * (7/365 days)
where,
• IPO Price is the upper price band in case of a book-built issue.
• NII Over subscription is the number of times over subscription in NII category. The table reflects data considering 2 time over subscription as a starting point with the flexibility to change as desired.
• ROI is the interest rate at which the loan gets granted.
• 7-days is the loan funding period.
Note: The over subscription intervals are derived based on the set incremental HNI IPO collection value. This value is currently set to 1500 which can be changed as desired based on the IPO issue size and price.
## FAQs
You can apply for Metro Brands IPO online through the Netbanking ASBA facility provided by your bank or through offline mode by filling and submitting the physical IPO application form.
Steps to apply online in Metro Brands IPO as HNI
2. Go to Investment/IPO section.
3. Select the desired IPO.
4. Enter the Investor category/Status as HNI.
5. Fill in the bid details viz. Quantity/lots and price.
6. Enter the personal information viz., Applicant name, PAN No., Email id, Bank account no., depository details.
7. Apply.
Steps to apply offline in Metro Brands IPO as HNI
1. Procure the IPO application form from the nearest broker office, syndicate member office, or download the application form from NSE/BSE website.
2. Fill in the below details in IPO Form:
• Applicant Name
• PAN Number
• Bid Details - Quantity, Price
• ASBA Bank Account Number
• Tick the investor category as HNI.
3. Submit the form to the nearest syndicate member/broker office.
Points to Note
• An HNI cannot place a bid at the cut-off price.
• An HNI cannot apply through UPI considering the UPI IPO Transaction limit of ₹2 lakhs.
You can invest in Metro Brands IPO as HNI through ASBA Netbanking or offline mode.
You need to enter all the required details viz. the bid quantity, amount, depository details, bank account no., applicant name, PAN No. to apply for IPO online or offline. You need to ensure to select the investor category as HNI.
It is important to note that as an HNI, you cannot bid in IPO at the cut-off price.
You cannot apply in Metro Brands IPO through the Zerodha IPO application(Console) in the HNI category as Zerodha offers a UPI-based IPO application that has a limit of a maximum of ₹2 lakh per transaction.
You can apply using the ASBA facility offered by the banks to invest in the HNI category.
ICICI Direct offers a UPI-based IPO application with an upper limit of ₹2 lakh per transaction. Thus, an HNI cannot apply for Metro Brands IPO through ICICI Direct.
To apply as an HNI in Metro Brands IPO, you can apply online through Netbanking using the ASBA facility offered by banks or submit a physical IPO ASBA application form.
The three main factors that decide the feasibility to invest in Metro Brands IPO using the IPO funding route.
1. IPO Subscription Rate - This is one of the most critical factors that help decide whether an HNI should opt for a loan or not. The higher the subscription, the lower is the allotment. In the case of lesser allotment in IPO, selling that small chunk may not cover the entire funding cost and thus reduce the chances of earning profits.
2. IPO Listing Price - If the IPO listing is expected at a good premium, taking an IPO loan should not be a burden as selling the shares on listing would earn handsome profit to cover the funding interest cost and the IPO price.
3. IPO Funding Rate - Generally, IPO funding is available at around 7%-10%. The above IPO funding cost table shows an estimated cost that can help you arrive at the break-even point to assess the probability of earning profit based on the expected listing price and over subscription. | 1,336 | 5,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.886176 |
https://www.biocodekb.com/2020/10/01/discrete-mathematics-in-action-in-r/ | 1,624,460,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00421.warc.gz | 609,233,123 | 10,508 | Discrete Mathematics in Action in R
An operator in R is a symbol that tells the compiler to perform specific mathematical or logical manipulations. R language is rich in built-in operators and has many operators to carry out different mathematical and logical operations.
Operators in R can mainly be classified into the following categories.
R Arithmetic Operators
These operators are used to carry out mathematical operations like addition and multiplication.
R Relational Operators
Relational operators are used to compare between values. Relational Operators are also known as comparators which help us to see how one R-Object relates to another R-object. For example, we can check whether two objects are equal or not, which can be accomplished with the help of == (double equal) sign. A logical operator which is TRUE on both sides, will return a logical value as TRUE since TRUE equals TRUE.
Precedence of relational operators is lower than arithmetic operators. To avoid confusion, it is good practice to use brackets to control ordering.
R Logical Operators
Logical operators are used to carry out Boolean operations like AND, OR etc. Operators & and | perform element-wise operation producing result having length of the longer operand.
But && and || examines only the first element of the operands resulting into a single length logical vector. Zero is considered FALSE and non-zero numbers are taken as TRUE. The basic way to define a logical variable is by relational operators by comparing two expressions. For example, we often ask if a variable x is bigger than a certain number?
A more sophisticated way is to combine two simple logical statements using logical operators.
It compares two numerical expressions and returns a Boolean variable: either 0 (FALSE) or 1 (TRUE). In practice, we often use two or more conditions to decide what to do. For example, scholarship is often given if the candidate has done well in both academic and extracurricular activities.
To combine different conditions, there are three logical operators:
1. &&
2. ||
3. !
First, && is similar to AND in English, that is, A && B is true only if both A and B are true. Second, is similar to OR in English, that is, A || B is false only if both A and B are false. Third, ! is similar to NOT in English, that is, !A is true only if A is false.
Different relational operators, the precedence of logical operators can be high. While and and or operators are lower in precedence than relational operators, not has a very high precedence and almost always evaluated first. – Hence, it is always a good practice to use brackets to control operation ordering.
R Assignment Operators
These operators are used to assign values to variables. The operators <- and = can be used, almost interchangeably, to assign to variable in the same environment.
The <<- operator is used for assigning to variables in the parent environments (more like global assignments). The rightward assignments, although available, are rarely used. Similarly, the inequality operator can also be used for numerical and other R-Objects like matrices, vectors, and lists.
• What if you want to check whether given two R-Objects which one of them is greater or less than the other. Well, of course, we can, using simple < and > sign (less-than and greater-than signs).
Conditional Statements
R provides us a way to combine all of these operators and use the result to change the behavior of our R scripts.
The syntax of the if condition is defined below:
if(condition){ // expression }
The if statement takes an input a condition ( inside parentheses ) which we can relate to a relational or some logical operator and if that condition is TRUE then the R code associated with the condition ( inside curly brackets ) is executed.
Operation on Vectors
The above mentioned operators work on vectors. We can use the function c() (as in concatenate) to make vectors in R.
All operations are carried out in element-wise fashion. When there is a mismatch in length (number of elements) of operand vectors, the elements in the shorter one is recycled in a cyclic manner to match the length of the longer one. R will issue a warning if the length of the longer vector is not an integral multiple of the shorter vector. | 902 | 4,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-25 | latest | en | 0.908259 |
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# Bazne funkcije i IF funkcija - 13 The average age of all...
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Tree Height Age Yield Profit Height =Apple >10 <16 =Pear Tree Height Age Yield Profit Apple 18 20 14 105 Pear 12 12 10 96 Cherry 13 14 9 105 Apple 14 15 10 75 Pear 9 8 8 76.8 Apple 8 9 6 45 BAZNE FUNKCIJE 1 This function looks at the records of apple trees between a height of 10 and 16 and counts how many of the Age fields in those records contain numbers. (1) 1 This function looks at the records of apple trees between a height of 10 and 16 and counts how many of the Profit fields in those records are not blank. (1) 105 The maximum profit of apple and pear trees. (105) 75 The minimum profit of apple trees over 10 in height. (75) 225 The total profit from apple trees. (225) 35 The total profit from apple trees with a height between 10 and 16. (75) 140 The product of the yields from apple trees with a height greater than 10. (140) 12 The average yield of apple trees over 10 feet in height. (12)
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Unformatted text preview: 13 The average age of all trees in the database. (13) 2.9664793948 The estimated standard deviation in the yield of apple and pear trees if the data in the database is only a sample of the total orchard population. (2.97) 2.6532998323 The true standard deviation in the yield of apple and pear trees if the data in the database is the entire population. (2.65) 8.8 The estimated variance in the yield of apple and pear trees if the data in the database is only a sample of the total orchard population. (8.8) 7.04 The true variance in the yield of apple and pear trees if the data in the database is the entire orchard population. ( 7.04) Returns the #NUM! error value because more than one record meets the criteria. IF FUNKCIJA 3 3 3 5 4 nemam sve ocjene!...
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## Bazne funkcije i IF funkcija - 13 The average age of all...
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Ask a homework question - tutors are online | 604 | 2,257 | {"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} | 3.34375 | 3 | CC-MAIN-2016-50 | longest | en | 0.817007 |
https://www.mahanmk.com/question-papers/s/amvi-main-2017-question-paper/ | 1,550,548,099,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489304.34/warc/CC-MAIN-20190219020906-20190219042906-00584.warc.gz | 908,112,292 | 39,301 | # AMVI Main 2017
1 The neutral axis of the cross-section of a beam is that axis at which the bending stress is : A. zero B. minimum C. maximum D. infinite
2 Moment of inertia of quarter circle of radius 'r' about 'x' axis passing through centroid is : A. Ix=0.055 r4 B. Ix=0.11 r4 C. Ix= 0.4 r4 D. None of these
3 Thin cylindrical shell of length 'L' diameter 'd' and thickness 't', subjected to internal pressure P. What is the change in length if it is made up of material having modulus of elasticity E and poisson's ratio 'μ' ? A. B. C. D. None of these
4 Angle between major principal plane and minor principal plane for strained body is : A. 45° B. 30° C. 60° D. 90°
5 A steel plate is bent into a circular arc of radius 10 meters. If the plate section be 120 mm wide and 20 mm thick, the maximum stress induced will be : (Take E=2 x 105 N/mm2) A. 400 N/mm2 B. 200 N/mm2 C. 100 N/mm2 D. 150 N/mm2
6 The ratio of critical load of columns of same dimensions and same material. One is fixed at both ends and other is fixed at one end and hinged at other end : A. B. 2 C. 0.5 D. 4
7 Moment of inertia of hollow rectangular section having outer depth 'D' and breadth ‘B' and dimensions of inner rectangle are depth'd' and width 'b' about horizontal axis passing through centroid is : A. B. C. D.
8 The ratio of the largest load in a test to the original cross-sectional area of the test piece is : A. elastic unit B. yield stress C. ultimate stress D. breaking stress
9 The shear force diagram for a simply supported beam carrying a uniformly distributed toad of w per unit length, consists of: A. one right angled triangle B. two right angled triangles C. one equilateral triangle D. two equilateral triangles
10 The stress at which the extension of the material takes place considerably as compared to the increase in load, is called ; A. elastic limit B. yield point C. ultimate point D. breaking point
11 The shear stress distribution over a beam cross-section is shown in figure. The beam is of : A. equal flange I-section B. unequal flange I-section C. circular cross-section D. T - section
12 The relation between Young's modulus (E), shear mod alus (G) & bulk modulus (K) is given by : A. B. C. D.
13 What is the shape of shearing stress distribution across w rectangular cross-section beam ? A. Triangular B. Parabolic only C. Rectangular only D. A combination of rectangular and parabolic shape
14 A beam has triangular cross-section having base b & altitude h. If the section of beam is subjected to a shear force F, the shear stress at the level of neutral axis in the cross-section is given by : A. B. C. D.
15 What is the ratio of maximum shear stress to average shear stress in a beam of circular section ? A. 1.5 B. 2 C. 1.33 D. 2.5
16 Defect such as slight eccentricity, a wavy surface, or a slight taper caused by previous operations can be corrected by : A. Grinding operation B. Lapping operation C. Honing operation D. Turning operation
17 Chipping of tool may occur due to : (a) tool material is too brittle (b) As a result of crack that is already in the tool (c) Excessive static loading of the tool (d) Weak design of tool, such as high positive rake angle Answer Options : A. Only (a) and (b) B. Only (b) and (c) C. Only (a) and (c) D. All (a), (b), (c) and (d)
18 'Microstoning is the following type of operation : A. Roughing B. Finishing C. Super-finishing D. Polishing
19 In orthogonal cutting system the cutting edge is : A. In line with direction of tool travel B. Perpendicular to direction of tool travel C. Perpendicular to shear plane D. Perpendicular to direction of depth of cut
20 Colour code for part to be machined for wooden patterns is represented by following colour : A. Red or orange B. Yellow C. Black on core prints D. Green
рдЖрддрд╛ рдорд┐рд│рд╡рд╛ рд╕рд░реНрд╡ рдЬрд╛рд╣рд┐рд░рд╛рддреА Telegram Channel рд╡рд░ (рдЕрдЧрджреА рдореЛрдлрдд) - рдпреЗрдереЗ рдХреНрд▓рд┐рдХ рдХрд░рд╛
Maha NMK Here You will get the list of all district from Maharashtra. You Can Select Any Disctrict From The List And Get Latest Recruitment News For The District. Keep Visiting MahaNMK Daily For The Latest Recruitment new and Free Job Alert 2018 | 1,187 | 4,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.852873 |
https://stackoverflow.com/questions/16943860/a-basic-confusion-on-quicksort | 1,723,391,151,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00359.warc.gz | 436,744,116 | 41,640 | # A basic confusion on quicksort
Suppose we choose a pivot as the first element of an array in case of a quicksort. Now the best/worst case complexity is `O(n^2)` whereas in average case it is `O(nlogn)`. Is not it weird (best case complexity is greater than worst case complexity)?
• The best case is `O(nlogn)`, instead of `O(n^2)`. Commented Jun 5, 2013 at 17:29
The best case complexity is `O(nlogn)`, as the average case. The worst case is `O(n^2)`. Check http://en.wikipedia.org/wiki/Quick_sort.
While other algorithms like Merge Sort and Heap Sort have a better worst case complexity (`O(nlogn)`), usually Quick Sort is faster - this is why it's the most common used sorting algorithm. An interesting answer about this can be found at Why is quicksort better than mergesort?.
The best-case of quicksort `0(nlogn)` is when the chosen pivot splits the subarray in two +- equally sized parts in every iteration.
So choosing the first element as the pivot in an already sorted array, will get you `0(n^2)`. ;) | 261 | 1,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.858352 |
https://stacks.math.columbia.edu/tag/05QN | 1,716,747,105,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00445.warc.gz | 465,076,954 | 14,588 | 13.4 Elementary results on triangulated categories
Most of the results in this section are proved for pre-triangulated categories and a fortiori hold in any triangulated category.
Lemma 13.4.1. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. Then $g \circ f = 0$, $h \circ g = 0$ and $f[1] \circ h = 0$.
Proof. By TR1 we know $(X, X, 0, 1, 0, 0)$ is a distinguished triangle. Apply TR3 to
$\xymatrix{ X \ar[r] \ar[d]^1 & X \ar[r] \ar[d]^ f & 0 \ar[r] \ar@{-->}[d] & X[1] \ar[d]^{1[1]} \\ X \ar[r]^ f & Y \ar[r]^ g & Z \ar[r]^ h & X[1] }$
Of course the dotted arrow is the zero map. Hence the commutativity of the diagram implies that $g \circ f = 0$. For the other cases rotate the triangle, i.e., apply TR2. $\square$
Lemma 13.4.2. Let $\mathcal{D}$ be a pre-triangulated category. For any object $W$ of $\mathcal{D}$ the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, -)$ is homological, and the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, W)$ is cohomological.
Proof. Consider a distinguished triangle $(X, Y, Z, f, g, h)$. We have already seen that $g \circ f = 0$, see Lemma 13.4.1. Suppose $a : W \to Y$ is a morphism such that $g \circ a = 0$. Then we get a commutative diagram
$\xymatrix{ W \ar[r]_1 \ar@{..>}[d]^ b & W \ar[r] \ar[d]^ a & 0 \ar[r] \ar[d]^0 & W[1] \ar@{..>}[d]^{b[1]} \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }$
Both rows are distinguished triangles (use TR1 for the top row). Hence we can fill the dotted arrow $b$ (first rotate using TR2, then apply TR3, and then rotate back). This proves the lemma. $\square$
Lemma 13.4.3. Let $\mathcal{D}$ be a pre-triangulated category. Let
$(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$
be a morphism of distinguished triangles. If two among $a, b, c$ are isomorphisms so is the third.
Proof. Assume that $a$ and $c$ are isomorphisms. For any object $W$ of $\mathcal{D}$ write $H_ W( - ) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, -)$. Then we get a commutative diagram of abelian groups
$\xymatrix{ H_ W(Z[-1]) \ar[r] \ar[d] & H_ W(X) \ar[r] \ar[d] & H_ W(Y) \ar[r] \ar[d] & H_ W(Z) \ar[r] \ar[d] & H_ W(X[1]) \ar[d] \\ H_ W(Z'[-1]) \ar[r] & H_ W(X') \ar[r] & H_ W(Y') \ar[r] & H_ W(Z') \ar[r] & H_ W(X'[1]) }$
By assumption the right two and left two vertical arrows are bijective. As $H_ W$ is homological by Lemma 13.4.2 and the five lemma (Homology, Lemma 12.5.20) it follows that the middle vertical arrow is an isomorphism. Hence by Yoneda's lemma, see Categories, Lemma 4.3.5 we see that $b$ is an isomorphism. This implies the other cases by rotating (using TR2). $\square$
Remark 13.4.4. Let $\mathcal{D}$ be an additive category with translation functors $[n]$ as in Definition 13.3.1. Let us call a triangle $(X, Y, Z, f, g, h)$ special1 if for every object $W$ of $\mathcal{D}$ the long sequence of abelian groups
$\ldots \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, Z) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, X[1]) \to \ldots$
is exact. The proof of Lemma 13.4.3 shows that if
$(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$
is a morphism of special triangles and if two among $a, b, c$ are isomorphisms so is the third. There is a dual statement for co-special triangles, i.e., triangles which turn into long exact sequences on applying the functor $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(-, W)$. Thus distinguished triangles are special and co-special, but in general there are many more (co-)special triangles, than there are distinguished triangles.
Lemma 13.4.5. Let $\mathcal{D}$ be a pre-triangulated category. Let
$(0, b, 0), (0, b', 0) : (X, Y, Z, f, g, h) \to (X, Y, Z, f, g, h)$
be endomorphisms of a distinguished triangle. Then $bb' = 0$.
Proof. Picture
$\xymatrix{ X \ar[r] \ar[d]^0 & Y \ar[r] \ar[d]^{b, b'} \ar@{..>}[ld]^\alpha & Z \ar[r] \ar[d]^0 \ar@{..>}[ld]^\beta & X[1] \ar[d]^0 \\ X \ar[r] & Y \ar[r] & Z \ar[r] & X[1] }$
Applying Lemma 13.4.2 we find dotted arrows $\alpha$ and $\beta$ such that $b' = f \circ \alpha$ and $b = \beta \circ g$. Then $bb' = \beta \circ g \circ f \circ \alpha = 0$ as $g \circ f = 0$ by Lemma 13.4.1. $\square$
Lemma 13.4.6. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. If
$\xymatrix{ Z \ar[r]_ h \ar[d]_ c & X[1] \ar[d]^{a[1]} \\ Z \ar[r]^ h & X[1] }$
is commutative and $a^2 = a$, $c^2 = c$, then there exists a morphism $b : Y \to Y$ with $b^2 = b$ such that $(a, b, c)$ is an endomorphism of the triangle $(X, Y, Z, f, g, h)$.
Proof. By TR3 there exists a morphism $b'$ such that $(a, b', c)$ is an endomorphism of $(X, Y, Z, f, g, h)$. Then $(0, (b')^2 - b', 0)$ is also an endomorphism. By Lemma 13.4.5 we see that $(b')^2 - b'$ has square zero. Set $b = b' - (2b' - 1)((b')^2 - b') = 3(b')^2 - 2(b')^3$. A computation shows that $(a, b, c)$ is an endomorphism and that $b^2 - b = (4(b')^2 - 4b' - 3)((b')^2 - b')^2 = 0$. $\square$
Lemma 13.4.7. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. There exists a distinguished triangle $(X, Y, Z, f, g, h)$ which is unique up to (nonunique) isomorphism of triangles. More precisely, given a second such distinguished triangle $(X, Y, Z', f, g', h')$ there exists an isomorphism
$(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Z', f, g', h')$
Proof. Existence by TR1. Uniqueness up to isomorphism by TR3 and Lemma 13.4.3. $\square$
Lemma 13.4.8. Let $\mathcal{D}$ be a pre-triangulated category. Let
$(a, b, c) : (X, Y, Z, f, g, h) \to (X', Y', Z', f', g', h')$
be a morphism of distinguished triangles. If one of the following conditions holds
1. $\mathop{\mathrm{Hom}}\nolimits (Y, X') = 0$,
2. $\mathop{\mathrm{Hom}}\nolimits (Z, Y') = 0$,
3. $\mathop{\mathrm{Hom}}\nolimits (X, X') = \mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$,
4. $\mathop{\mathrm{Hom}}\nolimits (Z, X') = \mathop{\mathrm{Hom}}\nolimits (Z, Z') = 0$, or
5. $\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = \mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$
then $b$ is the unique morphism from $Y \to Y'$ such that $(a, b, c)$ is a morphism of triangles.
Proof. If we have a second morphism of triangles $(a, b', c)$ then $(0, b - b', 0)$ is a morphism of triangles. Hence we have to show: the only morphism $b : Y \to Y'$ such that $X \to Y \to Y'$ and $Y \to Y' \to Z'$ are zero is $0$. We will use Lemma 13.4.2 without further mention. In particular, condition (3) implies (1). Given condition (1) if the composition $g' \circ b : Y \to Y' \to Z'$ is zero, then $b$ lifts to a morphism $Y \to X'$ which has to be zero. This proves (1).
The proof of (2) and (4) are dual to this argument.
Assume (5). Consider the diagram
$\xymatrix{ X \ar[r]_ f \ar[d]^0 & Y \ar[r]_ g \ar[d]^ b & Z \ar[r]_ h \ar[d]^0 \ar@{..>}[ld]^\epsilon & X[1] \ar[d]^0 \\ X' \ar[r]^{f'} & Y' \ar[r]^{g'} & Z' \ar[r]^{h'} & X'[1] }$
We may choose $\epsilon$ such that $b = \epsilon \circ g$. Then $g' \circ \epsilon \circ g = 0$ which implies that $g' \circ \epsilon = \delta \circ h$ for some $\delta \in \mathop{\mathrm{Hom}}\nolimits (X[1], Z')$. Since $\mathop{\mathrm{Hom}}\nolimits (X[1], Z') = 0$ we conclude that $g' \circ \epsilon = 0$. Hence $\epsilon = f' \circ \gamma$ for some $\gamma \in \mathop{\mathrm{Hom}}\nolimits (Z, X')$. Since $\mathop{\mathrm{Hom}}\nolimits (Z, X') = 0$ we conclude that $\epsilon = 0$ and hence $b = 0$ as desired. $\square$
Lemma 13.4.9. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent
1. $f$ is an isomorphism,
2. $(X, Y, 0, f, 0, 0)$ is a distinguished triangle, and
3. for any distinguished triangle $(X, Y, Z, f, g, h)$ we have $Z = 0$.
Proof. By TR1 the triangle $(X, X, 0, 1, 0, 0)$ is distinguished. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. By TR3 there is a map of distinguished triangles $(1, f, 0) : (X, X, 0) \to (X, Y, Z)$. If $f$ is an isomorphism, then $(1, f, 0)$ is an isomorphism of triangles by Lemma 13.4.3 and $Z = 0$. Conversely, if $Z = 0$, then $(1, f, 0)$ is an isomorphism of triangles as well, hence $f$ is an isomorphism. $\square$
Lemma 13.4.10. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ be triangles. The following are equivalent
1. $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$ is a distinguished triangle,
2. both $(X, Y, Z, f, g, h)$ and $(X', Y', Z', f', g', h')$ are distinguished triangles.
Proof. Assume (2). By TR1 we may choose a distinguished triangle $(X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. By TR3 we can find morphisms of distinguished triangles $(X, Y, Z, f, g, h) \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$ and $(X', Y', Z', f', g', h') \to (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h'')$. Taking the direct sum of these morphisms we obtain a morphism of triangles
$\xymatrix{ (X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \ar[d]^{(1, 1, c)} \\ (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h''). }$
In the terminology of Remark 13.4.4 this is a map of special triangles (because a direct sum of special triangles is special) and we conclude that $c$ is an isomorphism. Thus (1) holds.
Assume (1). We will show that $(X, Y, Z, f, g, h)$ is a distinguished triangle. First observe that $(X, Y, Z, f, g, h)$ is a special triangle (terminology from Remark 13.4.4) as a direct summand of the distinguished hence special triangle $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h')$. Using TR1 let $(X, Y, Q, f, g'', h'')$ be a distinguished triangle. By TR3 there exists a morphism of distinguished triangles $(X \oplus X', Y \oplus Y', Z \oplus Z', f \oplus f', g \oplus g', h \oplus h') \to (X, Y, Q, f, g'', h'')$. Composing this with the inclusion map we get a morphism of triangles
$(1, 1, c) : (X, Y, Z, f, g, h) \longrightarrow (X, Y, Q, f, g'', h'')$
By Remark 13.4.4 we find that $c$ is an isomorphism and we conclude that (2) holds. $\square$
Lemma 13.4.11. Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle.
1. If $h = 0$, then there exists a right inverse $s : Z \to Y$ to $g$.
2. For any right inverse $s : Z \to Y$ of $g$ the map $f \oplus s : X \oplus Z \to Y$ is an isomorphism.
3. For any objects $X', Z'$ of $\mathcal{D}$ the triangle $(X', X' \oplus Z', Z', (1, 0), (0, 1), 0)$ is distinguished.
Proof. To see (1) use that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, Z) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(Z, X[1])$ is exact by Lemma 13.4.2. By the same token, if $s$ is as in (2), then $h = 0$ and the sequence
$0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, X) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(W, Z) \to 0$
is split exact (split by $s : Z \to Y$). Hence by Yoneda's lemma we see that $X \oplus Z \to Y$ is an isomorphism. The last assertion follows from TR1 and Lemma 13.4.10. $\square$
Lemma 13.4.12. Let $\mathcal{D}$ be a pre-triangulated category. Let $f : X \to Y$ be a morphism of $\mathcal{D}$. The following are equivalent
1. $f$ has a kernel,
2. $f$ has a cokernel,
3. $f$ is the isomorphic to a composition $K \oplus Z \to Z \to Z \oplus Q$ of a projection and coprojection for some objects $K, Z, Q$ of $\mathcal{D}$.
Proof. Any morphism isomorphic to a map of the form $X' \oplus Z \to Z \oplus Y'$ has both a kernel and a cokernel. Hence (3) $\Rightarrow$ (1), (2). Next we prove (1) $\Rightarrow$ (3). Suppose first that $f : X \to Y$ is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$. By Lemma 13.4.1 the composition $f \circ h[-1] = 0$. As $f$ is a monomorphism we see that $h[-1] = 0$ and hence $h = 0$. Then Lemma 13.4.11 implies that $Y = X \oplus Z$, i.e., we see that (3) holds. Next, assume $f$ has a kernel $K$. As $K \to X$ is a monomorphism we conclude $X = K \oplus X'$ and $f|_{X'} : X' \to Y$ is a monomorphism. Hence $Y = X' \oplus Y'$ and we win. The implication (2) $\Rightarrow$ (3) is dual to this. $\square$
Lemma 13.4.13. Let $\mathcal{D}$ be a pre-triangulated category. Let $I$ be a set.
1. Let $X_ i$, $i \in I$ be a family of objects of $\mathcal{D}$.
1. If $\prod X_ i$ exists, then $(\prod X_ i)[1] = \prod X_ i[1]$.
2. If $\bigoplus X_ i$ exists, then $(\bigoplus X_ i)[1] = \bigoplus X_ i[1]$.
2. Let $X_ i \to Y_ i \to Z_ i \to X_ i[1]$ be a family of distinguished triangles of $\mathcal{D}$.
1. If $\prod X_ i$, $\prod Y_ i$, $\prod Z_ i$ exist, then $\prod X_ i \to \prod Y_ i \to \prod Z_ i \to \prod X_ i[1]$ is a distinguished triangle.
2. If $\bigoplus X_ i$, $\bigoplus Y_ i$, $\bigoplus Z_ i$ exist, then $\bigoplus X_ i \to \bigoplus Y_ i \to \bigoplus Z_ i \to \bigoplus X_ i[1]$ is a distinguished triangle.
Proof. Part (1) is true because $[1]$ is an autoequivalence of $\mathcal{D}$ and because direct sums and products are defined in terms of the category structure. Let us prove (2)(a). Choose a distinguished triangle $\prod X_ i \to \prod Y_ i \to Z \to \prod X_ i[1]$. For each $j$ we can use TR3 to choose a morphism $p_ j : Z \to Z_ j$ fitting into a morphism of distinguished triangles with the projection maps $\prod X_ i \to X_ j$ and $\prod Y_ i \to Y_ j$. Using the definition of products we obtain a map $\prod p_ i : Z \to \prod Z_ i$ fitting into a morphism of triangles from the distinguished triangle to the triangle made out of the products. Observe that the “product” triangle $\prod X_ i \to \prod Y_ i \to \prod Z_ i \to \prod X_ i[1]$ is special in the terminology of Remark 13.4.4 because products of exact sequences of abelian groups are exact. Hence Remark 13.4.4 shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual. $\square$
Lemma 13.4.14. Let $\mathcal{D}$ be a pre-triangulated category. If $\mathcal{D}$ has countable products, then $\mathcal{D}$ is Karoubian. If $\mathcal{D}$ has countable coproducts, then $\mathcal{D}$ is Karoubian.
Proof. Assume $\mathcal{D}$ has countable products. By Homology, Lemma 12.4.3 it suffices to check that morphisms which have a right inverse have kernels. Any morphism which has a right inverse is an epimorphism, hence has a kernel by Lemma 13.4.12. The second statement is dual to the first. $\square$
The following lemma makes it slightly easier to prove that a pre-triangulated category is triangulated.
Lemma 13.4.15. Let $\mathcal{D}$ be a pre-triangulated category. In order to prove TR4 it suffices to show that given any pair of composable morphisms $f : X \to Y$ and $g : Y \to Z$ there exist
1. isomorphisms $i : X' \to X$, $j : Y' \to Y$ and $k : Z' \to Z$, and then setting $f' = j^{-1}fi : X' \to Y'$ and $g' = k^{-1}gj : Y' \to Z'$ there exist
2. distinguished triangles $(X', Y', Q_1, f', p_1, d_1)$, $(X', Z', Q_2, g' \circ f', p_2, d_2)$ and $(Y', Z', Q_3, g', p_3, d_3)$, such that the assertion of TR4 holds.
Proof. The replacement of $X, Y, Z$ by $X', Y', Z'$ is harmless by our definition of distinguished triangles and their isomorphisms. The lemma follows from the fact that the distinguished triangles $(X', Y', Q_1, f', p_1, d_1)$, $(X', Z', Q_2, g' \circ f', p_2, d_2)$ and $(Y', Z', Q_3, g', p_3, d_3)$ are unique up to isomorphism by Lemma 13.4.7. $\square$
Lemma 13.4.16. Let $\mathcal{D}$ be a pre-triangulated category. Assume that $\mathcal{D}'$ is an additive full subcategory of $\mathcal{D}$. The following are equivalent
1. there exists a set of triangles $\mathcal{T}'$ such that $(\mathcal{D}', \mathcal{T}')$ is a pre-triangulated subcategory of $\mathcal{D}$,
2. $\mathcal{D}'$ is preserved under $[1]$ and $[1] : \mathcal{D}' \to \mathcal{D}'$ is an auto-equivalence and given any morphism $f : X \to Y$ in $\mathcal{D}'$ there exists a distinguished triangle $(X, Y, Z, f, g, h)$ in $\mathcal{D}$ such that $Z$ is isomorphic to an object of $\mathcal{D}'$.
In this case $\mathcal{T}'$ as in (1) is the set of distinguished triangles $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ such that $X, Y, Z \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$. Finally, if $\mathcal{D}$ is a triangulated category, then (1) and (2) are also equivalent to
1. $\mathcal{D}'$ is a triangulated subcategory.
Proof. Omitted. $\square$
Lemma 13.4.17. An exact functor of pre-triangulated categories is additive.
Proof. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Since $(0, 0, 0, 1_0, 1_0, 0)$ is a distinguished triangle of $\mathcal{D}$ the triangle
$(F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, F(0))$
is distinguished in $\mathcal{D}'$. This implies that $1_{F(0)} \circ 1_{F(0)}$ is zero, see Lemma 13.4.1. Hence $F(0)$ is the zero object of $\mathcal{D}'$. This also implies that $F$ applied to any zero morphism is zero (since a morphism in an additive category is zero if and only if it factors through the zero object). Next, using that $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle by Lemma 13.4.11 part (3), we see that $(F(X), F(X \oplus Y), F(Y), F(1, 0), F(0, 1), 0)$ is one too. This implies that the map $F(X) \oplus F(Y) \to F(X \oplus Y)$ is an isomorphism by Lemma 13.4.11 part (2). To finish we apply Homology, Lemma 12.7.1. $\square$
Lemma 13.4.18. Let $F : \mathcal{D} \to \mathcal{D}'$ be a fully faithful exact functor of pre-triangulated categories. Then a triangle $(X, Y, Z, f, g, h)$ of $\mathcal{D}$ is distinguished if and only if $(F(X), F(Y), F(Z), F(f), F(g), F(h))$ is distinguished in $\mathcal{D}'$.
Proof. The “only if” part is clear. Assume $(F(X), F(Y), F(Z))$ is distinguished in $\mathcal{D}'$. Pick a distinguished triangle $(X, Y, Z', f, g', h')$ in $\mathcal{D}$. By Lemma 13.4.7 there exists an isomorphism of triangles
$(1, 1, c') : (F(X), F(Y), F(Z)) \longrightarrow (F(X), F(Y), F(Z')).$
Since $F$ is fully faithful, there exists a morphism $c : Z \to Z'$ such that $F(c) = c'$. Then $(1, 1, c)$ is an isomorphism between $(X, Y, Z)$ and $(X, Y, Z')$. Hence $(X, Y, Z)$ is distinguished by TR1. $\square$
Lemma 13.4.19. Let $\mathcal{D}, \mathcal{D}', \mathcal{D}''$ be pre-triangulated categories. Let $F : \mathcal{D} \to \mathcal{D}'$ and $F' : \mathcal{D}' \to \mathcal{D}''$ be exact functors. Then $F' \circ F$ is an exact functor.
Proof. Omitted. $\square$
Lemma 13.4.20. Let $\mathcal{D}$ be a pre-triangulated category. Let $\mathcal{A}$ be an abelian category. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor.
1. Let $\mathcal{D}'$ be a pre-triangulated category. Let $F : \mathcal{D}' \to \mathcal{D}$ be an exact functor. Then the composition $H \circ F$ is a homological functor as well.
2. Let $\mathcal{A}'$ be an abelian category. Let $G : \mathcal{A} \to \mathcal{A}'$ be an exact functor. Then $G \circ H$ is a homological functor as well.
Proof. Omitted. $\square$
Lemma 13.4.21. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ be an abelian category. Let $G : \mathcal{A} \to \mathcal{D}$ be a $\delta$-functor.
1. Let $\mathcal{D}'$ be a triangulated category. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor. Then the composition $F \circ G$ is a $\delta$-functor as well.
2. Let $\mathcal{A}'$ be an abelian category. Let $H : \mathcal{A}' \to \mathcal{A}$ be an exact functor. Then $G \circ H$ is a $\delta$-functor as well.
Proof. Omitted. $\square$
Lemma 13.4.22. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $G : \mathcal{A} \to \mathcal{D}$ be a $\delta$-functor. Let $H : \mathcal{D} \to \mathcal{B}$ be a homological functor. Assume that $H^{-1}(G(A)) = 0$ for all $A$ in $\mathcal{A}$. Then the collection
$\{ H^ n \circ G, H^ n(\delta _{A \to B \to C})\} _{n \geq 0}$
is a $\delta$-functor from $\mathcal{A} \to \mathcal{B}$, see Homology, Definition 12.12.1.
Proof. The notation signifies the following. If $0 \to A \xrightarrow {a} B \xrightarrow {b} C \to 0$ is a short exact sequence in $\mathcal{A}$, then
$\delta = \delta _{A \to B \to C} : G(C) \to G(A)[1]$
is a morphism in $\mathcal{D}$ such that $(G(A), G(B), G(C), a, b, \delta )$ is a distinguished triangle, see Definition 13.3.6. Then $H^ n(\delta ) : H^ n(G(C)) \to H^ n(G(A)[1]) = H^{n + 1}(G(A))$ is clearly functorial in the short exact sequence. Finally, the long exact cohomology sequence (13.3.5.1) combined with the vanishing of $H^{-1}(G(C))$ gives a long exact sequence
$0 \to H^0(G(A)) \to H^0(G(B)) \to H^0(G(C)) \xrightarrow {H^0(\delta )} H^1(G(A)) \to \ldots$
in $\mathcal{B}$ as desired. $\square$
The proof of the following result uses TR4.
Proposition 13.4.23. Let $\mathcal{D}$ be a triangulated category. Any commutative diagram
$\xymatrix{ X \ar[r] \ar[d] & Y \ar[d] \\ X' \ar[r] & Y' }$
can be extended to a diagram
$\xymatrix{ X \ar[r] \ar[d] & Y \ar[r] \ar[d] & Z \ar[r] \ar[d] & X[1] \ar[d] \\ X' \ar[r] \ar[d] & Y' \ar[r] \ar[d] & Z' \ar[r] \ar[d] & X'[1] \ar[d] \\ X'' \ar[r] \ar[d] & Y'' \ar[r] \ar[d] & Z'' \ar[r] \ar[d] & X''[1] \ar[d] \\ X[1] \ar[r] & Y[1] \ar[r] & Z[1] \ar[r] & X[2] }$
where all the squares are commutative, except for the lower right square which is anticommutative. Moreover, each of the rows and columns are distinguished triangles. Finally, the morphisms on the bottom row (resp. right column) are obtained from the morphisms of the top row (resp. left column) by applying $[1]$.
Proof. During this proof we avoid writing the arrows in order to make the proof legible. Choose distinguished triangles $(X, Y, Z)$, $(X', Y', Z')$, $(X, X', X'')$, $(Y, Y', Y'')$, and $(X, Y', A)$. Note that the morphism $X \to Y'$ is both equal to the composition $X \to Y \to Y'$ and equal to the composition $X \to X' \to Y'$. Hence, we can find morphisms
1. $a : Z \to A$ and $b : A \to Y''$, and
2. $a' : X'' \to A$ and $b' : A \to Z'$
as in TR4. Denote $c : Y'' \to Z[1]$ the composition $Y'' \to Y[1] \to Z[1]$ and denote $c' : Z' \to X''[1]$ the composition $Z' \to X'[1] \to X''[1]$. The conclusion of our application TR4 are that
1. $(Z, A, Y'', a, b, c)$, $(X'', A, Z', a', b', c')$ are distinguished triangles,
2. $(X, Y, Z) \to (X, Y', A)$, $(X, Y', A) \to (Y, Y', Y'')$, $(X, X', X'') \to (X, Y', A)$, $(X, Y', A) \to (X', Y', Z')$ are morphisms of triangles.
First using that $(X, X', X'') \to (X, Y', A)$ and $(X, Y', A) \to (Y, Y', Y'')$. are morphisms of triangles we see the first of the diagrams
$\vcenter { \xymatrix{ X' \ar[r] \ar[d] & Y' \ar[d] \\ X'' \ar[r]^{b \circ a'} \ar[d] & Y'' \ar[d] \\ X[1] \ar[r] & Y[1] } } \quad \text{and}\quad \vcenter { \xymatrix{ Y \ar[r] \ar[d] & Z \ar[d]^{b' \circ a} \ar[r] & X[1] \ar[d] \\ Y' \ar[r] & Z' \ar[r] & X'[1] } }$
is commutative. The second is commutative too using that $(X, Y, Z) \to (X, Y', A)$ and $(X, Y', A) \to (X', Y', Z')$ are morphisms of triangles. At this point we choose a distinguished triangle $(X'', Y'' , Z'')$ starting with the map $b \circ a' : X'' \to Y''$.
Next we apply TR4 one more time to the morphisms $X'' \to A \to Y''$ and the triangles $(X'', A, Z', a', b', c')$, $(X'', Y'', Z'')$, and $(A, Y'', Z[1], b, c , -a[1])$ to get morphisms $a'' : Z' \to Z''$ and $b'' : Z'' \to Z[1]$. Then $(Z', Z'', Z[1], a'', b'', - b'[1] \circ a[1])$ is a distinguished triangle, hence also $(Z, Z', Z'', -b' \circ a, a'', -b'')$ and hence also $(Z, Z', Z'', b' \circ a, a'', b'')$. Moreover, $(X'', A, Z') \to (X'', Y'', Z'')$ and $(X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])$ are morphisms of triangles. At this point we have defined all the distinguished triangles and all the morphisms, and all that's left is to verify some commutativity relations.
To see that the middle square in the diagram commutes, note that the arrow $Y' \to Z'$ factors as $Y' \to A \to Z'$ because $(X, Y', A) \to (X', Y', Z')$ is a morphism of triangles. Similarly, the morphism $Y' \to Y''$ factors as $Y' \to A \to Y''$ because $(X, Y', A) \to (Y, Y', Y'')$ is a morphism of triangles. Hence the middle square commutes because the square with sides $(A, Z', Z'', Y'')$ commutes as $(X'', A, Z') \to (X'', Y'', Z'')$ is a morphism of triangles (by TR4). The square with sides $(Y'', Z'', Y[1], Z[1])$ commutes because $(X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])$ is a morphism of triangles and $c : Y'' \to Z[1]$ is the composition $Y'' \to Y[1] \to Z[1]$. The square with sides $(Z', X'[1], X''[1], Z'')$ is commutative because $(X'', A, Z') \to (X'', Y'', Z'')$ is a morphism of triangles and $c' : Z' \to X''[1]$ is the composition $Z' \to X'[1] \to X''[1]$. Finally, we have to show that the square with sides $(Z'', X''[1], Z[1], X[2])$ anticommutes. This holds because $(X'', Y'', Z'') \to (A, Y'', Z[1], b, c , -a[1])$ is a morphism of triangles and we're done. $\square$
[1] This is nonstandard notation.
Comments (13)
Comment #360 by Fan on
In Lemma 13.4.9 (1), shouldn't $s$ be a right inverse? ($g \cdot s = 1_Z$)
Comment #361 by Fan on
In Lemma 13.4.10, (1) implies (3), the implication from the distinguished triangle $(X, Y, Z, f, g, h)$ to $h = 0$ (assuming $\ker f = 0$) seems to have only tangential relation to Lemma 4.2. I think the implication works as follows: Since $[-1] \cdot [1] = [0] = id$, $[1]$ is an equivalence of categories. Since $f$ is a monomorphism, so is $f[1]$. By Lemma 4.1, $f[1] \cdot h = 0$, so $h = 0$. I don't see how to show $h = 0$ directly without involving $f[1]$, as the text literally suggests.
Comment #362 by Fan on
In Lemma 13.4.7, the implication (3) to (1) seems to use Lemma 4.3 rather than Lemma 4.6.
Comment #363 by Fan on
Sorry for being a little picky about this section, but I'm just learning derived category and the lecturer tells us to read the stack project!
In the implication (2) to (1) in Lemma 4.8, the argument following the diagram is presumably a repetition of deriving Lemma 4.3 from Lemma 4.2, so can be replaced by a direct application of Lemma 4.3.
Comment #364 by Fan on
By the same token in the implication (1) to (2) in Lemma 4.8, one can quote Lemma 4.3 without explicitly mentioning the exactness of the Hom functor,
Comment #365 by Fan on
Lemma 13.4.16, the first sentence should say "the only if part is clear," because what is proved next is the "if" part.
Comment #368 by Fan on
Please ignore the first comment. I realized that the definition of left and right inverse is the other way round than Wikipedia.
Comment #381 by on
@#368: Actually I agree with the suggestion of #360 and I changed it.
Comment #3050 by Matthieu Romagny on
Remark 13.4.4, very last sentence: ... than there are distinguished triangles.
Comment #3051 by Matthieu Romagny on
Remark 13.4.11, condition (3): I don't understand what is 'a map $K \oplus Z \to Z \oplus Q$ induced by $id_Z$'. Maybe could it be good to be more explicit.
Comment #5367 by Frid on
In lemma 13.4.8, (2), is it $\text{Hom}(Z, Y') = 0$?
Comment #5604 by on
Thanks very much Frid! Sorry for the mistake. Fixed here.
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Division
Subtrction
Mixed
100
4 x 5
What is 20
100
6 + 2
What is 8
100
2 / 6
3
100
7 - 3
What is 10
100
4 x 1 + 2
What is 6
200
7 x 9
What is 63
200
4 + 42
What 46
200
8 / 4
What is 2
200
8 - 6
What 2
200
9 x 2 - 4
What is 14
300
25x5
125
300
8 + 8 + 2
18
300
2 / 10
What is 5
300
200 - 4 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 - 1 -1 -1 -1 -1 -1 -1 -1 - 1 - 1 - 1 - 1 - 1 - 1 - 1
What is 171
300
4 + 3 - 2
What is 5
400
4 x 5 x 2
What is 40
400
7 + 2
9
400
10 / 2
5
400
1 - 1
What is 0
400
5 + 2 x 2
What is 14
500
9 x 9 x 9 x 9 x 9 x 9 x 9
4782969
500
8 + 9 + 9 + 9 + 9
What is 44
500
5 / 25=
5
500
1000 - 7
What is 993
500
6 plus 6 plus 6
What is 18 | 404 | 673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.685413 |
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# Load carrying capacity and life
Operating life
## Axial load carrying capacity of cylindrical roller bearings
Top
Schaeffler KG introduced the Expanded calculation of the adjusted rating life
in 1997. This method was standardised for the first time in
DIN ISO 281 Appendix 1 and has been a constituent part of the international
standard ISO 281 since 2007.
As part of the international standardisation work, the life adjustment factor aDIN
was renamed as aISO but without any change to the calculation method.
Top
## Fatigue theory as a principle
The basis of the rating life calculation in accordance with ISO 281 is
Lundberg and Palmgren's fatigue theory which always gives a final rating
life.
However, modern, high quality bearings can exceed by a considerable margin
the values calculated for the basic rating life under favourable operating
conditions. Ioannides and Harris have developed a further model of fatigue in
rolling contact that expands on the Lundberg/Palmgren theory and gives a
better description of the performance capability of modern bearings.
The method Expanded calculation of the adjusted rating life takes account of
the following influences:
## The influencing factors, especially those relating to contamination, are
extremely complex. A great deal of experience is essential for an accurate
assessment. As a result the Engineering Service of Schaeffler Group
Industrial should be consulted for further advice.
The tables and diagrams can only give guide values.
Top
Dimensioning of
rolling bearings
## The required size of a rolling bearing is dependent on the demands made
on its:
rating life
operational reliability.
Top
carrying capacity and life
## The dynamic load carrying capacity is described in terms of the basic
DIN ISO 281.
The basic dynamic load ratings for rolling bearings are matched to empirically
proven performance standards and those published in previous FAG and INA
catalogues.
The fatigue behaviour of the material determines the dynamic load carrying
capacity of the rolling bearing.
The dynamic load carrying capacity is described in terms of the basic
dynamic load rating and the basic rating life.
The fatigue life is dependent on:
## the statistical probability of the first appearance of failure.
The basic dynamic load rating C applies to rotating rolling bearings. It is:
## a constant, concentrically acting axial load Ca for axial bearings.
The basic dynamic load rating C is that load of constant magnitude and
direction which a sufficiently large number of apparently identical bearings can
endure for a basic rating life of one million revolutions.
Top
## Calculation of the rating life
The methods for calculating the rating life are:
basic rating life L10 and L10h to ISO 281, see link
adjusted rating life Lna to DIN ISO 281:1990 (no longer a constituent
part of ISO 281), see link
## Basic rating life
The basic rating life L10 and L10h is calculated as follows:
L10
106 revolutions
The basic rating life in millions of revolutions is the life reached or exceeded by 90% of a
sufficiently large group of apparently identical bearings before the first evidence of
material fatigue develops
L10h
h
The basic rating life as defined for L10 but expressed in operating hours
C
N
P
N
p
Life exponent; for roller bearings: p = 10/3 for ball bearings: p = 3
n
min1
Operating speed.
## The equivalent dynamic load P is a calculated value. This value is constant
for axial bearings.
The load value P gives the same rating life as the combined load occurring in
practice.
P
N
Fr
N
Fa
N
X
Radial factor given in the dimension tables or product description
Y
Axial factor given in the dimension tables or product description.
## This calculation cannot be applied to radial needle roller bearings, axial
needle roller bearings and axial cylindrical roller bearings. Combined loads
are not permissible with these bearings.
Top
speed, other influences are known such as:
lubrication
or
## This calculation method was replaced in ISO 281:2007 by the calculation of
Lna
106 revolutions
Adjusted rating life for special material characteristics and operating conditions with a
requisite reliability of (100 n) %
L10
106 revolutions
Basic rating life
a1
Life adjustment factor for a requisite reliability other than 90%. In ISO 281:2007, the
values for the life adjustment factor a1 have been redefined, see table
a2
Life adjustment factor for special material characteristics. For standard rolling bearing
steels: a2 = 1
a3
Life adjustment factor for special operating conditions; in particular lubrication, Figure 1.
## Good cleanliness and suitable
Very high cleanliness and low
Contamination in the lubricant
= viscosity ratio
Figure 1
Viscosity ratio
The viscosity ratio is an indication of the quality of lubricant film formation:
mm2s1
Kinematic viscosity of the lubricant at operating temperature
1
mm2s1
Reference viscosity of the lubricant at operating temperature.
## The reference viscosity 1 is determined from the mean bearing diameter
dM = (D + d)/2 and the operating speed n, Figure 2.
The nominal viscosity of the oil at +40 C is determined from the required
operating viscosity and the operating temperature , Figure 3. In the case of
greases, is the operating viscosity of the base oil.
In the case of heavily loaded bearings with a high proportion of sliding
contact, the temperature in the contact area of the rolling elements may be up
to 20 K higher than the temperature measured on the stationary ring (without
## the influence of any external heat).
Taking account of EP additives in calculation of the expanded adjusted rating
1 = reference viscosity
dM = mean bearing diameter
n = speed
Figure 2
Reference viscosity 1
= operating viscosity
= operating temperature
40 = viscosity at +40 C
Figure 3
V/T diagram for mineral oils
Top
The calculation of the expanded adjusted rating life Lnm was standardised in
DIN ISO 281 Appendix 1. Since 2007, it has been standardised in the
worldwide standard ISO 281. Computer-aided calculation in accordance with
DIN ISO 281 Appendix 4 has been specified since 2008 in ISO/TS 16 281.
Lnm is calculated as follows:
Lnm
106 revolutions
Expanded adjusted rating life to ISO 281
a1
Life adjustment factor for a requisite reliability other than 90%, see table
aISO
Life adjustment factor for operating conditions
L10
106 revolutions
The values for the life adjustment factor a1 were redefined in ISO 281:2007
and differ from the previous data.
Requisite reliability
%
90
95
96
life
Lnm
L10m
L5m
L4m
a1
1
0,64
0,55
Requisite reliability
%
97
98
99
99,2
99,4
99,6
99,8
99,9
99,92
99,94
99,95
life
Lnm
L3m
L2m
L1m
L0,8m
L0,6m
L0,4m
L0,2m
L0,1m
L0,08m
L0,06m
L0,05m
a1
0,47
0,37
0,25
0,22
0,19
0,16
0,12
0,093
0,087
0,08
0,077
The standardised method for calculating the life adjustment factor aISO
essentially takes account of:
## contamination in the lubricant.
aISO
Life adjustment factor for operating conditions,
Figure 4 to Figure 7
eC
Life adjustment factor for contamination, see table
Cu
N
P
N
For 4, calculation should be carried out using = 4.
For 0,1, this calculation method cannot be used.
lubricant
## In accordance with ISO 281, EP additives can be taken into consideration in
the following way:
## For a viscosity ratio 1 and a contamination factor eC 0,2,
calculation can be carried out using the value = 1 for lubricants with
EP additives that have proven effective. With severe contamination
(contamination factor eC 0,2), the effectiveness of the additives
under these contamination conditions must be proven. The
effectiveness of the EP additives can be demonstrated in the actual
application or on a rolling bearing test rig FE 8 to DIN 51 819-1.If the
EP additives are proven effective and calculation is carried out using
the value = 1, the life adjustment factor must be restricted to
aISO 3. If the value aISO calculated for the actual is greater than 3,
this value can be used in calculation.
Figure 4
Figure 5
for axial roller bearings
Figure 6
Figure 7
for axial ball bearings
Top
The fatigue limit load Cu in accordance with ISO 281 is defined as the load
below which, under laboratory conditions, no fatigue occurs in the material.
Top
for contamination
The life adjustment factor for contamination ec takes into consideration the
influence of contamination in the lubrication gap on the rating life, see table.
The rating life is reduced by solid particles in the lubrication gap and is
dependent on:
## Due to the complex nature of the interaction between these influencing
factors, only an approximate guide value can be attained. The values in the
tables are valid for contamination by solid particles (factor eC). They do not
take account of other contamination such as that caused by water or other
fluids.
Under severe contamination (eC 0) the bearings may fail due to wear. In
this case, the operating life is substantially less than the calculated life.
Factor eC
Contamination
Extreme cleanliness
Factor eC
dM
dM
1)
100 mm
1
100 mm1)
1
0,8 to 0,6
0,9 to 0,8
0,6 to 0,5
0,8 to 0,6
0,5 to 0,3
0,6 to 0,4
0,3 to 0,1
0,4 to 0,2
0,1 to 0
0,1 to 0
## Particle size within lubricant film
thickness
Laboratory conditions
High cleanliness
Oil filtered through extremely fine filter
Sealed, greased bearings
Standard cleanliness
## Oil filtered through fine filter
Slight contamination
## Slight contamination of oil
Typical contamination
material from other machine elements
Heavy contamination
contaminated
sealed
## Very heavy contamination
______
1
dM = mean bearing diameter (d + D)/2.
Top
## Equivalent operating values
The rating life formulae assume a constant bearing load P and constant
bearing speed n. If the load and speed are not constant, equivalent
operating values can be determined that induce the same fatigue as the
actual conditions.
The equivalent operating values calculated here already take account of the
life adjustment factors a3 or aISO. They must not be applied again when
If the load and speed vary over a time period T, the speed n and equivalent
bearing load P are calculated as follows:
Variation in steps
If the load and speed vary in steps over a time period T, n and P are
calculated as follows:
Top
If the function F describes the variation in the load over a time period T and
the speed is constant, P is calculated as follows:
Top
constant speed
If the load varies in steps over a time period T and the speed is constant, P
is calculated as follows:
Top
If the speed varies but the load remains constant, the following applies:
Top
speed varying in steps
If the speed varies in steps, the following applies:
Top
Under oscillating bearing motion
The equivalent speed is calculated as follows:
The formula is valid only if the angle of oscillation is greater than twice the
angular pitch of the rolling elements. If the angle of oscillation is smaller,
there is a risk of false brinelling.
Figure 8
Angle of oscillation
Top
Symbols, units and definitions
n
min1
Mean speed
T
min
Time period under consideration
P
N
p
Life exponent;
for roller bearings: p = 10/3
for ball bearings: p = 3
ai, a(t)
Life adjustment factor aISO for current operating condition,
ni, n(t)
min1
Bearing speed for current operating condition
qi
%
Duration of operating condition as a proportion of the total operating period;
qi = (ti/T) 100
Fi, F(t)
N
Bearing load during the current operating condition
nosc
min1
Frequency of oscillating motion
Top
## Required rating life
If no information is available on the rating life, the guide values from the
following tables may be used.
Do not overspecify the bearing. If the calculated life is 60 000 h, this
normally means that the bearing arrangement is overspecified. Pay attention
to the minimum load for the bearings; see the design and safety guidelines in
the product sections.
Motor vehicles
Motorcycles
Passenger car powertrains
Passenger car gearboxes protected
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
400
2 000
400
2 400
500
1 100
500
1 200
200
500
200
500
against contamination
Passenger car wheel bearings
Light commercial vehicles
Medium commercial vehicles
Heavy commercial vehicles
Buses
Internal combustion engines
1 400
2 000
2 900
4 000
2 900
900
Mounting location
Rail vehicles
Mounting location
Wheelset bearings for freight wagons
Tram carriages
Passenger carriages
Goods wagons
Tipper wagons
Powered units
Locomotives, external bearings
Locomotives, internal bearings
Gearboxes for rail vehicles
Shipbuilding
Mounting location
Marine thrust blocks
Marine shaft bearings
Large marine gearboxes
Small marine gearboxes
Boat propulsion systems
Agricultural machinery
Mounting location
Tractors
Self-propelled machinery
Seasonal machinery
Construction machinery
Mounting location
5 300
4 000
5 300
8 800
11 000
4 000
1 500
2 400
3 600
5 000
3 600
900
7 000
5 000
7 000
12 000
16 000
5 000
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
7 800
21 000 35 000 50 000
20 000 35 000
20 000 35 000
20 000 35 000
35 000 50 000
35 000 50 000
75 000 110 000
14 000 46 000 20 000 75 000
Ball bearings
Roller bearings
from
to
from
to
20 000
50 000
50 000
200 000
14 000
46 000
20 000
75 000
4 000
14 000
5 000
20 000
1 700
7 800
2 000
10 000
Ball bearings
Roller bearings
from
to
from
to
1 700
4 000
2 000
5 000
1 700
4 000
2 000
5 000
500
1 700
500
2 000
Ball bearings
Roller bearings
from
to
from
to
## Excavators, travelling gear
Excavators, slewing gear
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
500
1 700
500
2 000
1 700
4 000
2 000
5 000
1 700
4 000
2 000
5 000
generators
Vibrator bodies
500
Mounting location
Electric motors
Mounting location
Electric motors for household
appliances
Series motors
Large motors
Electric traction motors
## Rolling mills, steelworks equipment
Mounting location
Roll stands
Rolling mill gearboxes
Roller tables
Centrifugal casting machines
Machine tools
500
2 000
Ball bearings
Roller bearings
from
to
from
to
1 700
4 000
21 000
32 000
14 000
32 000
63 000
21 000
35 000
50 000
20 000
50 000
110 000
35 000
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
500
14 000 500
20 000
14 000 32 000 20 000
50 000
7 800
21 000 10 000
35 000
21 000 46 000 35 000
75 000
Drilling spindles
Grinding spindles
Workpiece spindles in grinding
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
14 000 46 000 20 000 75 000
14 000 32 000 20 000 50 000
7 800
21 000 10 000 35 000
21 000 63 000 35 000 110 000
machines
Machine tool gearboxes
Presses, flywheels
Presses, eccentric shafts
Electric tools and compressed air tools
14 000
21 000
14 000
4 000
Mounting location
Woodworking machinery
1 700
32 000
32 000
21 000
14 000
20 000
35 000
20 000
5 000
50 000
50 000
35 000
20 000
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
Milling spindles and cutter blocks
14 000 32 000 20 000
50 000
Saw frames, main bearings
35 000
50 000
Saw frames, connecting rod bearings 10 000
20 000
Mounting location
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
4 000
14 000 5 000
20 000
Mounting location
Circular saws
Gearboxes in
general machine building
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
4 000
14 000
5 000
20 000
4 000
14 000
5 000
20 000
14 000
46 000
20 000
75 000
Mounting location
Universal gearboxes
Geared motors
Large gearboxes, stationary
Conveying equipment
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
Belt drives, mining
75 000 150 000
Conveyor belt rollers, mining
46 000 63 000 75 000 110 000
Conveyor belt rollers, general
7 800
21 000 10 000 35 000
Belt drums
50 000 75 000
Bucket wheel excavators, travel drive 7 800
21 000 10 000 35 000
Bucket wheel excavators, bucket wheel 75 000 200 000
Bucket wheel excavators, bucket wheel 46 000 83 000 75 000 150 000
Mounting location
drive
Winding cable sheaves
Sheaves
32 000
7 800
## Pumps, fans, compressors
46 000
21 000
50 000
10 000
75 000
35 000
Ventilators, fans
Large fans
Piston pumps
Centrifugal pumps
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
21 000 46 000 35 000 75 000
32 000 63 000 50 000 110 000
21 000 46 000 35 000 75 000
14 000 46 000 20 000 75 000
500
7 800
500
10 000
engines
Gear pumps
Compressors
500
4 000
Mounting location
Centrifuges, stirrers
Mounting location
Centrifuges
Large stirrers
7 800
21 000
500
5 000
10 000
35 000
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
7 800
14 000
10 000
20 000
21 000
32 000
35 000
50 000
Textile machinery
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
Spinning machines, spinning spindles 21 000 46 000 35 000
75 000
Weaving and knitting machines
14 000 32 000 20 000
50 000
Mounting location
Plastics processing
Mounting location
Plastics worm extruders
Rubber and plastics calenders
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
14 000 21 000 20 000
35 000
21 000 46 000 35 000
75 000
Jaw crushers
Gyratory crushers, roll crushers
Rigid hammer mills, hammer mills,
## Recommended rating life in h
Ball bearings Roller bearings
from
to from
to
20 000
35 000
20 000
35 000
50 000
110 000
impact crushers
Tube mills
Vibration grinding mills
Grinding track mills
Vibrating screens
Briquette presses
Rotary furnace track rollers
Mounting location
## Paper and printing machinery
Mounting location
Paper machinery, wet section
Paper machinery, dry section
Paper machinery, refiners
Paper machinery, calenders
Printing machinery
50 000
5 000
50 000
10 000
35 000
50 000
100 000
20 000
110 000
20 000
50 000
110 000
## Recommended rating life in h
Ball bearings
Roller bearings
from
to
from
to
110 000
150 000
150 000 250 000
80 000
120 000
80 000
110 000
32 000 46 000 50 000
75 000
Top
Operating life
The operating life is defined as the life actually achieved by the bearing. It
may differ significantly from the calculated value.
This may be due to wear or fatigue as a result of:
## insufficient or excessive operating clearance
contamination
insufficient lubrication
brinelling)
## Due to the wide variety of possible installation and operating conditions, it is
not possible to precisely predetermine the operating life. The most reliable
way of arriving at a close estimate is by comparison with similar applications.
Top
of cylindrical roller bearings
## Radial cylindrical roller bearings used as semi-locating and locating
bearings can support axial forces in one or both directions in addition to
The axial load carrying capacity is dependent on:
the size of the sliding surfaces between the ribs and the end faces of
the rolling elements
## Ribs subjected to load must be supported across their entire height.
The permissible axial load Fa per must not be exceeded, in order to avoid
unacceptably high temperatures.
The limiting load Fa max must not be exceeded, in order to avoid unacceptable
## pressure at the contact surfaces.
The ratio Fa/Fr must not exceed a value of 0,4. For bearings of TB design, the
value 0,6 is permissible.
permissible.
Bearings of TB design
In the case of bearings of TB design, the axial load carrying capacity has
been significantly improved through the use of new calculation and
manufacturing methods.
Optimum contact conditions between the roller and rib are ensured by means
of a special curvature of the roller end faces. As a result, axial surface
pressures on the rib are significantly reduced and a lubricant film with
improved load-carrying capabilities is achieved. Under normal operating
conditions, wear and fatigue at the rib contact running and roller end faces is
completely eliminated. The axial frictional torque is reduced by up to 50%. The
bearing temperature during operation is therefore significantly lower.
Top
Permissible and
Fa per and Fa max are calculated as follows:
Bearings of standard design
Bearings of TB design
## Bearings of standard and TB design
Fa per
N
Fa max
N
kS
Factor dependent on the lubrication method,
see table
kB
Factor dependent on the bearing series,
see table
dM
mm
Mean bearing diameter (d + D)/2
n
min1
Operating speed.
## Misalignment caused by shaft deflection for example, may lead to
alternating stresses on the inner ring ribs. In this instance, the axial load
must be restricted to Fas in accordance with the formula where the bearing is
tilted up to a maximum of 2 angular minutes.
## For more severe tilting, a separate strength analysis is required.
Factor kS
for lubrication method
Factor
Lubrication methods1)
Minimal heat dissipation, drip feed oil lubrication, oil mist
kS
7,5 to 10
## lubrication, low operating viscosity ( 0,5 1)
Little heat dissipation, oil sump lubrication, oil spray lubrication,
10
to 15
## low oil flow
Good heat dissipation, recirculating oil lubrication (pressure oil
12
to 18
lubrication)
Very good heat dissipation,
16
to 24
## recirculating oil lubrication with oil cooling,
high operating viscosity ( 2 1)
______
1
The precondition for these kS values is the reference viscosity 1 according
to the section Oil lubrication. Doped oils should be used such as
CLP (DIN 51 517) and HLP (DIN 51 524) of ISO-VG classes 32 to 460 and
ATF oils (DIN 51 502) and gearbox oils (DIN 51 512) of SAE viscosity
classes 75 W to 140 W.
Bearing factor kB
Series
SL1818, SL0148
SL1829, SL0149
SL1830, SL1850
SL1822
LSL1923, ZSL1923
SL1923
NJ2..-E, NJ22..-E, NUP2..-E, NUP22..-E
NJ3..-E, NJ23..-E, NUP3..-E, NUP23..-E
NJ4
Factor
kB
4,5
11
17
20
28
30
15
20
22
Top
Very high static loads or shock loads can cause plastic deformation on the
raceways and rolling elements. This deformation limits the static load
carrying capacity of the rolling bearing with respect to the permissible noise
## level during operation of the bearing.
If a rolling bearing operates with only infrequent rotary motion or completely
without rotary motion, its size is determined in accordance with the basic
According to DIN ISO 76, this is:
a constant, concentrically acting axial load C0a for axial bearings.
The basic static load rating C0 is that load under which the Hertzian pressure
at the most heavily loaded point between the rolling elements and raceways
reaches the following values:
for roller bearings, 4 000 N/mm2
for ball bearings, 4 200 N/mm2
for self-aligning ball bearings, 4 600 N/mm2.
Under normal contact conditions, this load causes a permanent deformation at
the contact points of approx. 1/10 000 of the rolling element diameter.
In addition to dimensioning on the basis of the fatigue limit life, it is advisable
to check the static load safety factor. The guide values and shock loads
occurring in operation to table must be taken into consideration, see table,
The static load safety factor S0 is the ratio between the basic static load rating
C0 and the equivalent static load P0:
S0
C0 (C0r, C0a)
N
P0 (P0r, P0a)
N
Guide values for axial spherical roller bearings and high precision bearings:
see corresponding product description.
For drawn cup needle roller bearings, S0 3 is necessary.
Guide values
Operating conditions
S0
for roller
bearings
Smooth, low-vibration, normal operation with 1
for ball
bearings
0,5
## minimal demands for smooth running;
bearings with slight rotary motion
Normal operation with higher requirements
## The equivalent static load P0 is a calculated value. It corresponds to a radial
P0 induces the same load at the centre point of the most heavily loaded
contact point between the rolling element and raceway as the combined load
occurring in practice.
P0
N
F0r
N
F0a
N | 6,442 | 23,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.90058 |
http://www.numbersaplenty.com/2600015 | 1,586,546,663,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00413.warc.gz | 253,828,902 | 3,440 | Search a number
2600015 = 511411153
BaseRepresentation
bin1001111010110001001111
311220002112212
421322301033
51131200030
6131421035
731046135
oct11726117
94802485
102600015
111516480
12a5477b
13700592
144b9755
15365595
hex27ac4f
2600015 has 16 divisors (see below), whose sum is σ = 3489696. Its totient is φ = 1843200.
The previous prime is 2600011. The next prime is 2600033. The reversal of 2600015 is 5100062.
Adding to 2600015 its reverse (5100062), we get a palindrome (7700077).
It is not a de Polignac number, because 2600015 - 22 = 2600011 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (2600011) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1679 + ... + 2831.
It is an arithmetic number, because the mean of its divisors is an integer number (218106).
22600015 is an apocalyptic number.
2600015 is a deficient number, since it is larger than the sum of its proper divisors (889681).
2600015 is a wasteful number, since it uses less digits than its factorization.
2600015 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1210.
The product of its (nonzero) digits is 60, while the sum is 14.
The square root of 2600015 is about 1612.4562009555. The cubic root of 2600015 is about 137.5071511432.
The spelling of 2600015 in words is "two million, six hundred thousand, fifteen". | 488 | 1,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.867113 |
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# Beams and Slabs - PowerPoint PPT Presentation
Beams and Slabs. October 30, 2007. Beams and Slabs. A beam is a linear structural member with loading applied perpendicular to its long axis The load on a beam is a bending load Bending is the tendency of a member to bow A lintel is a term sometimes used for a short beam
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### Beams and Slabs
October 30, 2007
Beams and Slabs
• A beam is a linear structural member with loading applied perpendicular to its long axis
• Bending is the tendency of a member to bow
• A lintel is a term sometimes used for a short beam
• A slab is a 2 dimensional piece with loaded the same way – it is an extension of the beam just as the bearing wall is an extension of the column.
Beams
• Deflection or sag is important
• With columns, we want to know how much weight they can take before failure
• With beams, the maximum weight is usually less important than the deflection. People will become uncomfortable walking on a sagging floor well before it actually comes close to the breaking point.
Simply supported beam
One end fixed – no horizontal translation or vertical movement
• Other end on a roller to allow for the shortening or lengthening of the beam
Beam stresses
• Compression at the top of the beam
• Tension at the bottom
• Because compression and tension are occurring in parallel, shear is also present
Shear
• Shear effects are important for beams because their two faces are moving opposite to one another
Shear
• Stack of planks on top of one another will sag a lot more than a single piece of wood
Shear
• Glue or a small piece called a KEY may be used prevent this slippage
Materials
• The best materials for beams are those which have similar strength in tension and compression
• Wood and steel are good materials
• Concrete and masonry are poor choices for long beams
• Lintels are short beams and are often made from concrete
Reinforcement
• Concrete beams are reinforced with steal to prevent tensile cracking
• Steel reinforcing bars are commonly on the lower half of the beam
Prestressed and posttensioned concrete beams
• Steel must begin to stretch before it can offer any bending resistance.
• This little bit of stretch can result in cracks in the bottom surface of a concrete beam
• Steel is often stretched and installed before the concrete is poured
• Once the concrete hardens, the steel is released putting the concrete in compression - camber
• Alternately, the steel is placed in hollow sheathes in the concrete and there is no bond between the two
Beam deflection
• Factors that affect beam deflection include
• Span
• Depth
• Width
• Material
• Cross-sectional shape
• Longitudinal shape
Span
• The deflection of a beam increases as the cube of the span
• Doubling the length, increases the sag by a factor of 8
Width
• Deflection varies with its cross-sectional dimensions
• Deflection is inversely proportional to the horizontal dimension
• Doubling the width reduced the deflection by one half
0.5d
Depth
• Deflection is inversely proportional to vertical depth
• Doubling the depth, reduces the deflection by a factor of 8
• A beam has more stiffness by increasing the depth, as compared to the width
8 d
Material strength
• Deflection is inversely proportional to the modulus of elasticity of the material
• Aluminum will deflect 3 times as much as a comparable steel beam
• Deflection at midspan is effected by the location of the load
• Increases as the load moves toward the middle of the beam
Cross sectional shape
• A problem with beams is the understressing of the material near the center of the cross section
• Distributing the material away from the central axis increases bending resistance
• Box and I-shapes are examples
Deflection
• If the span or length is increased by a factor 2, the deflection increases by a factor 8 (23)
• If the depth increases by a factor of 2, the deflection decreases by a factor of 8 (23)
• If the width increases by a factor of 2, the deflection decreases by a factor of 2
width
depth
length
Multiple Beams
• To span a space (eg. bedroom on 2nd floor of a house), we would want multiple, closely spaced beams, known as Joists.
Joists
• Joists are closely spaced beams
• Rectangular space
• Which way should the beams be laid, N-S or E-W?
• Shorter direction results in less sag
• Works well for smaller spaces
• DVD beam slide 24
Beam Grid
• Increase the structural stability by using a Beam Grid
• Interlocking 2 dimensional beams
• better support than the 1D joists
• the continuity of the beams is assumed
• steel can be welded
• reinforced concrete with bars that bypass each other
• wood is not suitable
Slabs
• 2D horizontal supports
• Just as a continuous bearing wall is more than just a bunch of columns, a slab is more than just a bunch of beams
• the load is distributed over a wider area
• How the slab behaves (how it deflects under load) depends on how it is supported
Types of Slabs
• A one-way slab has supports on two parallel sides, but not the other two
• A two-way slab has supports in both directions (either interlocking beams, or dropped panels)
• stronger and more efficient than a one-way slab if the distances in each direction are comparable
• Flat plates just rest on random columns
• the slab must be reinforced at the points of support, since there is so much stress there
Ribbing
• The slab need not be a solid piece
• removes a lot of needless weight if the middle portion of the slab (which experiences little stress) can be removed.
• most of the concrete is at the top of the slab,
• more “ribs” with reinforcing steel are placed at the bottom to take the tension.
• Thus one has a slab on top of joists (i.e. slab on top of beams or a beam grid).
Isostatic Ribs
• Usually the ribs are laid out in convenient, rectangular forms
• In principle it is more efficient to use isostatic ribs that follow the lines of the bending stresses and so prevent shear
• Expensive!
Cantilever and Continuous Beams
• So far we’ve considered simply supported beams (or slabs) that just rest on one support at each end
• We get slightly different conditions if a beam is actually clamped at each end, or if it rests on multiple supports
• A beam that is clamped at each end is a Cantilever
• A beam that rests on multiple supports is a Continuous Beam
Cantilever
• A a cantilever which is fixed at one end and loaded perpendicular to its axis
• A cantilever can be vertical or horizontal
• A flagpole is an example of a cantilever
• Assuming the latter, how does this change in attachment affect the beam?
LOWER portion is in compression
• UPPER portion is in tension
• exactly the opposite of the simply supported beam
• DVD beams slide 18
Cantilever deflection
• The sag of a cantilever again depends on span, width, depth, and material
• just treat it as half a simply-supported beam for these cases
• more material away from the axis is good
• I-beam or hollow tube are good choices for a cantilever
• DVD Beams slide 5
Cantilever deflection – con’t
• The sag also depends on the shape of the cantilever
• Unlike a simply-supported beam, the cantilever has the greatest stress near the clamped end, so we want the most material at that end and little material at the free end
• A tapered shape is most efficient
• need depth near support, but not out near load
Falling Water
• The most famous cantilevered structure
• a house built for the Kauffman family
• built by the famous architect Frank Lloyd Wright
Video!
Overhanging Beam
• In contrast to a cantilever is an overhanging beam
• As the name implies, the overhanging beam is a beam that extends out beyond its last support
• Unlike a cantilever the overhanging beam is allowed to rotate instead of being held rigidly
• DVD beam slide 5
• Which case would lead to more deflection?
Continuous Beams
• Some projects require a large open area, unbroken by supporting columns
• But for many projects it is acceptable (and much more efficient) to space columns across a space
Continuous beams
• In this case, we could place individual short beams that extend from one column to the next
• Or we could use one long beam that stretched across all the columns
Continuous beams
• NO, the continuous beam will continue to affect portions beyond the closest support
• Instead of only a convex shape between each column, we now get a concave shape over the supports
• DVD beam slide 12
• Just as the load bends the beam down, the support bends it up
• That means that the length of stressed beam between each set of supports is shortened, and hence the deflection is less!
• Or, put another way, for the same amount of deflection, we could use a narrower beam, saving material and hence cost
Gerber Beam
• Alternately, instead of a single long piece, we CAN break the beam down into shorter segments and STILL get the reduced deflection of a continuous beam. How?
• The key is the inflection points of the continuous beam
• We can’t sever the beam at the supports because the stress is high there.
• But at the ¼ and ¾ points where the beam changes from concave up to concave down, there is NO stress.
• Thus at THOSE points it is OK to break the longer segment and replace it with shorter beams joined by pinned connections
• Such an arrangement is called a Gerber beam | 2,326 | 9,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-30 | latest | en | 0.931362 |
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• ### How to calculate critical speed of ball mill
Critical Speed Of Ball Mill Calculation Ecssr. calculating critical speed in a ball mill calculate the critical speed of the ball mill [ 48 4562 Ratings ] The Gulin product line, consisting of more than 30 machines, sets the standard for our industry We plan to help you meet your needs with our equipment, with our distribution and product support system, and,
• ### Ball Mills - an overview | ScienceDirect Topics
The critical speed n (rpm) when the balls remain attached to the wall with the aid of centrifugal force is: n = 42.3 D m. where D m is the mill diameter in meters. The optimum rotational speed is usually set at 65%–80% of the critical speed. These data are approximate and may not be valid for metal particles that tend to agglomerate by welding. The minimal magnitude of the ball size is
• ### How to calculate critical speed of ball mill
Critical Speed Of Ball Mill Calculation Ecssr. calculating critical speed in a ball mill calculate the critical speed of the ball mill [ 48 4562 Ratings ] The Gulin product line, consisting of more than 30 machines, sets the standard for our industry We plan to help you meet your needs with our equipment, with our distribution and product support system, and,
• ### Ball Mill Critical Speed & Working Principle -
20.06.2015· Learn about Ball Mill Critical Speed and its effect on inner charge movements. The effect of Ball Mill RPM speed going from sub-critical to super-critical helps understand the Ball Mill Working
• ### Ball Mill Critical Speed Mineral Processing
Ball mills have been successfully run at speeds between 60 and 90 percent of critical speed, but most mills operate at speeds between 65 and 79 percent of critical speed. Rod mills speed should be limited to a maximum of 70% of critical speed and preferably should be in the 60 to 68 percent critical speed
• ### critical speed of ball mill depends on
This formula calculates the critical speed of any ball mill. Most ball mills operate most efficiently between 65% and 75% of their critical speed. Whatever your requirements, you 'll find the perfect service-oriented solution to match your specific needs with our help.We are here for your questions anytime 24/7, welcome your consultation. Chat Online. Ball Mill. Feeding Granularity: ≤20
• ### Why Critical Speed Maintain In Ball Mill
why critical speed maintain in ball mill. why critical speed maintain in ball mill Understanding Journal Bearings EDGE understanding journal bearings and is integral to bearing design and application Since they have significant damping fluid film journal bearings have a strong impact on the vibration characteristics of machinery The types of machinery we are concerned with range
• ### Critical Speed Crusher- SPECIAL Mining machine
Critical Parts Of Grinding Mill Worldcrushers. May 06 2013 speed and ball mills usually operate at 65 to 75 of the critical speed critical speed of mill clinker grinding mill a grinding mills critical speed is the lowest rpm which cause an infinitely small particle on the
• ### Ball Mill Operating Speed - YouTube
18.10.2016· Ball mill, Grinding machine, Wet ball mill, Dry ball mill, Milling machine, - Duration: 3:57. Eric Zhang 40,669 views
• ### how to calculate critical speed of ball mill
Normal mill speeds are 70 — 80% of the critical speed. Ball mill Read More. Ball mill . A ball mill is a type of grinder used to grind and blend materials for use in mineral dressing The grinding works on the principle of critical speed. The critical Read More. Critical rotation speed for ball-milling - ScienceDirect. Critical rotation speed of dry ball-mill was studied by
• ### Ball Mill Parameter Selection – Power, Rotate
2.2 Rotation Speed Calculation of Ball Mill Critical Speed_ When the ball mill cylinder is rotated, there is no relative slip between the grinding medium and the cylinder wall, and it just starts to run in a state of rotation with the cylinder of the mill. This instantaneous speed of the mill is as follows: N0 —- mill working speed, r/min; K'b — speed ratio, %. There are many layers of
• ### Ball Mill Critical Speed Mineral Processing
Ball mills have been successfully run at speeds between 60 and 90 percent of critical speed, but most mills operate at speeds between 65 and 79 percent of critical speed. Rod mills speed should be limited to a maximum of 70% of critical speed and preferably should be in the 60 to 68 percent critical speed
• ### Critical Speed of Ball Mill
Critical Speed of Ball Mill has a direct impact on the movement of steel balls and coal and coal grinding process. If the rotational speed is very low, the steel ball and coal block is hard to be taken up, can be only in the lower part, and output of grinding coal is very small. If the speed is very high, centrifugal force on the ball and coal is greater than its gravity, ball and coal will
• ### Why Critical Speed Maintain In Ball Mill
why critical speed maintain in ball mill. why critical speed maintain in ball mill Understanding Journal Bearings EDGE understanding journal bearings and is integral to bearing design and application Since they have significant damping fluid film journal bearings have a strong impact on the vibration characteristics of machinery The types of machinery we are concerned with range
• ### critical speed calculation ball mill 42 29
Mill Critical Speed Formula Derivation (2 replies) In practice Ball Mills are driven at a speed of 50-90% of the critical speed, the factor being influenced by economic consideration. Mill capacity can be increased by increasing speed but there is very little increase in efficiency (i.e kWht -1) when the mill is operated above about 40-50% of the critical speed.
• ### Why Critical Speed Maintain In Ball Mill
why critical speed maintain in ball mill. why critical speed maintain in ball mill Understanding Journal Bearings EDGE understanding journal bearings and is integral to bearing design and application Since they have significant damping fluid film journal bearings have a strong impact on the vibration characteristics of machinery The types of machinery we are concerned with range
• ### How to calculate critical speed of ball mill
Critical Speed Of Ball Mill Calculation Ecssr. calculating critical speed in a ball mill calculate the critical speed of the ball mill [ 48 4562 Ratings ] The Gulin product line, consisting of more than 30 machines, sets the standard for our industry We plan to help you meet your needs with our equipment, with our distribution and product support system, and, | 2,307 | 10,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.860716 |
https://ltwork.net/which-event-played-an-important-part-in-the-civil-rights--10052035 | 1,669,563,411,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00426.warc.gz | 410,107,983 | 11,344 | # Which event played an important part in the civil rights movement? freedom rides to washington, d.
###### Question:
Which event played an important part in the civil rights movement? freedom rides to washington, d. c. a march on washington, d. c.
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-- 0.050922-- | 1,008 | 4,343 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.930637 |
https://www.easycalculation.com/physics/fluid-mechanics/hazen-williams-velocity.php | 1,653,815,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00410.warc.gz | 830,817,671 | 7,469 | English
# Hazen Williams - Mean Fluid Velocity Calculator English Español
The Hazen Williams formula is an empirical equation that can be used to calculate the pressure loss per one foot of pipe of a known diameter due to friction dependent on the flow. Here we can calculate for Mean Flow Velocity, Friction Coefficient, Hydraulic Radius, Hydraulic Grade Line Slope.
## Hazen Williams - Mean Fluid Velocity Calculation
I want to calculate Mean Flow Velocity(v) Friction Coefficient(C) Hydraulic Radius(r) Hydraulic Grade Line Slope(s) Friction Coefficient(C) = Hydraulic Radius(r) = m Hydraulic Grade Line Slope(s) = Mean Flow Velocity(v) = m/s
The Hazen Williams formula is an empirical equation that can be used to calculate the pressure loss per one foot of pipe of a known diameter due to friction dependent on the flow. Here we can calculate for Mean Flow Velocity, Friction Coefficient, Hydraulic Radius, Hydraulic Grade Line Slope. | 198 | 945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | longest | en | 0.796444 |
https://us.metamath.org/mpeuni/pcohtpylem.html | 1,725,921,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00025.warc.gz | 551,195,358 | 17,106 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > pcohtpylem Structured version Visualization version GIF version
Theorem pcohtpylem 23713
Description: Lemma for pcohtpy 23714. (Contributed by Jeff Madsen, 15-Jun-2010.) (Revised by Mario Carneiro, 24-Feb-2015.)
Hypotheses
Ref Expression
pcohtpy.4 (𝜑 → (𝐹‘1) = (𝐺‘0))
pcohtpy.5 (𝜑𝐹( ≃ph𝐽)𝐻)
pcohtpy.6 (𝜑𝐺( ≃ph𝐽)𝐾)
pcohtpylem.7 𝑃 = (𝑥 ∈ (0[,]1), 𝑦 ∈ (0[,]1) ↦ if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)))
pcohtpylem.8 (𝜑𝑀 ∈ (𝐹(PHtpy‘𝐽)𝐻))
pcohtpylem.9 (𝜑𝑁 ∈ (𝐺(PHtpy‘𝐽)𝐾))
Assertion
Ref Expression
pcohtpylem (𝜑𝑃 ∈ ((𝐹(*𝑝𝐽)𝐺)(PHtpy‘𝐽)(𝐻(*𝑝𝐽)𝐾)))
Distinct variable groups: 𝑥,𝑦,𝐹 𝑥,𝑀,𝑦 𝑥,𝑁,𝑦 𝜑,𝑥,𝑦 𝑥,𝐺,𝑦 𝑥,𝐻,𝑦 𝑥,𝐽,𝑦 𝑥,𝐾,𝑦
Allowed substitution hints: 𝑃(𝑥,𝑦)
Proof of Theorem pcohtpylem
Dummy variables 𝑠 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 pcohtpy.5 . . . . 5 (𝜑𝐹( ≃ph𝐽)𝐻)
2 isphtpc 23688 . . . . 5 (𝐹( ≃ph𝐽)𝐻 ↔ (𝐹 ∈ (II Cn 𝐽) ∧ 𝐻 ∈ (II Cn 𝐽) ∧ (𝐹(PHtpy‘𝐽)𝐻) ≠ ∅))
31, 2sylib 221 . . . 4 (𝜑 → (𝐹 ∈ (II Cn 𝐽) ∧ 𝐻 ∈ (II Cn 𝐽) ∧ (𝐹(PHtpy‘𝐽)𝐻) ≠ ∅))
43simp1d 1140 . . 3 (𝜑𝐹 ∈ (II Cn 𝐽))
5 pcohtpy.6 . . . . 5 (𝜑𝐺( ≃ph𝐽)𝐾)
6 isphtpc 23688 . . . . 5 (𝐺( ≃ph𝐽)𝐾 ↔ (𝐺 ∈ (II Cn 𝐽) ∧ 𝐾 ∈ (II Cn 𝐽) ∧ (𝐺(PHtpy‘𝐽)𝐾) ≠ ∅))
75, 6sylib 221 . . . 4 (𝜑 → (𝐺 ∈ (II Cn 𝐽) ∧ 𝐾 ∈ (II Cn 𝐽) ∧ (𝐺(PHtpy‘𝐽)𝐾) ≠ ∅))
87simp1d 1140 . . 3 (𝜑𝐺 ∈ (II Cn 𝐽))
9 pcohtpy.4 . . 3 (𝜑 → (𝐹‘1) = (𝐺‘0))
104, 8, 9pcocn 23711 . 2 (𝜑 → (𝐹(*𝑝𝐽)𝐺) ∈ (II Cn 𝐽))
113simp2d 1141 . . 3 (𝜑𝐻 ∈ (II Cn 𝐽))
127simp2d 1141 . . 3 (𝜑𝐾 ∈ (II Cn 𝐽))
13 pcohtpylem.8 . . . . . 6 (𝜑𝑀 ∈ (𝐹(PHtpy‘𝐽)𝐻))
144, 11, 13phtpy01 23679 . . . . 5 (𝜑 → ((𝐹‘0) = (𝐻‘0) ∧ (𝐹‘1) = (𝐻‘1)))
1514simprd 500 . . . 4 (𝜑 → (𝐹‘1) = (𝐻‘1))
16 pcohtpylem.9 . . . . . 6 (𝜑𝑁 ∈ (𝐺(PHtpy‘𝐽)𝐾))
178, 12, 16phtpy01 23679 . . . . 5 (𝜑 → ((𝐺‘0) = (𝐾‘0) ∧ (𝐺‘1) = (𝐾‘1)))
1817simpld 499 . . . 4 (𝜑 → (𝐺‘0) = (𝐾‘0))
199, 15, 183eqtr3d 2802 . . 3 (𝜑 → (𝐻‘1) = (𝐾‘0))
2011, 12, 19pcocn 23711 . 2 (𝜑 → (𝐻(*𝑝𝐽)𝐾) ∈ (II Cn 𝐽))
21 pcohtpylem.7 . . 3 𝑃 = (𝑥 ∈ (0[,]1), 𝑦 ∈ (0[,]1) ↦ if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)))
22 eqid 2759 . . . 4 (topGen‘ran (,)) = (topGen‘ran (,))
23 eqid 2759 . . . 4 ((topGen‘ran (,)) ↾t (0[,](1 / 2))) = ((topGen‘ran (,)) ↾t (0[,](1 / 2)))
24 eqid 2759 . . . 4 ((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) = ((topGen‘ran (,)) ↾t ((1 / 2)[,]1))
25 dfii2 23576 . . . 4 II = ((topGen‘ran (,)) ↾t (0[,]1))
26 0red 10675 . . . 4 (𝜑 → 0 ∈ ℝ)
27 1red 10673 . . . 4 (𝜑 → 1 ∈ ℝ)
28 halfre 11881 . . . . . 6 (1 / 2) ∈ ℝ
29 halfge0 11884 . . . . . 6 0 ≤ (1 / 2)
30 1re 10672 . . . . . . 7 1 ∈ ℝ
31 halflt1 11885 . . . . . . 7 (1 / 2) < 1
3228, 30, 31ltleii 10794 . . . . . 6 (1 / 2) ≤ 1
33 elicc01 12891 . . . . . 6 ((1 / 2) ∈ (0[,]1) ↔ ((1 / 2) ∈ ℝ ∧ 0 ≤ (1 / 2) ∧ (1 / 2) ≤ 1))
3428, 29, 32, 33mpbir3an 1339 . . . . 5 (1 / 2) ∈ (0[,]1)
3534a1i 11 . . . 4 (𝜑 → (1 / 2) ∈ (0[,]1))
36 iitopon 23573 . . . . 5 II ∈ (TopOn‘(0[,]1))
3736a1i 11 . . . 4 (𝜑 → II ∈ (TopOn‘(0[,]1)))
389adantr 485 . . . . . 6 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (𝐹‘1) = (𝐺‘0))
394, 11, 13phtpyi 23678 . . . . . . . 8 ((𝜑𝑦 ∈ (0[,]1)) → ((0𝑀𝑦) = (𝐹‘0) ∧ (1𝑀𝑦) = (𝐹‘1)))
4039simprd 500 . . . . . . 7 ((𝜑𝑦 ∈ (0[,]1)) → (1𝑀𝑦) = (𝐹‘1))
4140adantrl 716 . . . . . 6 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (1𝑀𝑦) = (𝐹‘1))
428, 12, 16phtpyi 23678 . . . . . . . 8 ((𝜑𝑦 ∈ (0[,]1)) → ((0𝑁𝑦) = (𝐺‘0) ∧ (1𝑁𝑦) = (𝐺‘1)))
4342simpld 499 . . . . . . 7 ((𝜑𝑦 ∈ (0[,]1)) → (0𝑁𝑦) = (𝐺‘0))
4443adantrl 716 . . . . . 6 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (0𝑁𝑦) = (𝐺‘0))
4538, 41, 443eqtr4d 2804 . . . . 5 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (1𝑀𝑦) = (0𝑁𝑦))
46 simprl 771 . . . . . . . 8 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → 𝑥 = (1 / 2))
4746oveq2d 7167 . . . . . . 7 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (2 · 𝑥) = (2 · (1 / 2)))
48 2cn 11742 . . . . . . . 8 2 ∈ ℂ
49 2ne0 11771 . . . . . . . 8 2 ≠ 0
5048, 49recidi 11402 . . . . . . 7 (2 · (1 / 2)) = 1
5147, 50eqtrdi 2810 . . . . . 6 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (2 · 𝑥) = 1)
5251oveq1d 7166 . . . . 5 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → ((2 · 𝑥)𝑀𝑦) = (1𝑀𝑦))
5351oveq1d 7166 . . . . . . 7 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → ((2 · 𝑥) − 1) = (1 − 1))
54 1m1e0 11739 . . . . . . 7 (1 − 1) = 0
5553, 54eqtrdi 2810 . . . . . 6 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → ((2 · 𝑥) − 1) = 0)
5655oveq1d 7166 . . . . 5 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → (((2 · 𝑥) − 1)𝑁𝑦) = (0𝑁𝑦))
5745, 52, 563eqtr4d 2804 . . . 4 ((𝜑 ∧ (𝑥 = (1 / 2) ∧ 𝑦 ∈ (0[,]1))) → ((2 · 𝑥)𝑀𝑦) = (((2 · 𝑥) − 1)𝑁𝑦))
58 retopon 23458 . . . . . . 7 (topGen‘ran (,)) ∈ (TopOn‘ℝ)
59 0re 10674 . . . . . . . 8 0 ∈ ℝ
60 iccssre 12854 . . . . . . . 8 ((0 ∈ ℝ ∧ (1 / 2) ∈ ℝ) → (0[,](1 / 2)) ⊆ ℝ)
6159, 28, 60mp2an 692 . . . . . . 7 (0[,](1 / 2)) ⊆ ℝ
62 resttopon 21854 . . . . . . 7 (((topGen‘ran (,)) ∈ (TopOn‘ℝ) ∧ (0[,](1 / 2)) ⊆ ℝ) → ((topGen‘ran (,)) ↾t (0[,](1 / 2))) ∈ (TopOn‘(0[,](1 / 2))))
6358, 61, 62mp2an 692 . . . . . 6 ((topGen‘ran (,)) ↾t (0[,](1 / 2))) ∈ (TopOn‘(0[,](1 / 2)))
6463a1i 11 . . . . 5 (𝜑 → ((topGen‘ran (,)) ↾t (0[,](1 / 2))) ∈ (TopOn‘(0[,](1 / 2))))
6564, 37cnmpt1st 22361 . . . . . 6 (𝜑 → (𝑥 ∈ (0[,](1 / 2)), 𝑦 ∈ (0[,]1) ↦ 𝑥) ∈ ((((topGen‘ran (,)) ↾t (0[,](1 / 2))) ×t II) Cn ((topGen‘ran (,)) ↾t (0[,](1 / 2)))))
6623iihalf1cn 23626 . . . . . . 7 (𝑧 ∈ (0[,](1 / 2)) ↦ (2 · 𝑧)) ∈ (((topGen‘ran (,)) ↾t (0[,](1 / 2))) Cn II)
6766a1i 11 . . . . . 6 (𝜑 → (𝑧 ∈ (0[,](1 / 2)) ↦ (2 · 𝑧)) ∈ (((topGen‘ran (,)) ↾t (0[,](1 / 2))) Cn II))
68 oveq2 7159 . . . . . 6 (𝑧 = 𝑥 → (2 · 𝑧) = (2 · 𝑥))
6964, 37, 65, 64, 67, 68cnmpt21 22364 . . . . 5 (𝜑 → (𝑥 ∈ (0[,](1 / 2)), 𝑦 ∈ (0[,]1) ↦ (2 · 𝑥)) ∈ ((((topGen‘ran (,)) ↾t (0[,](1 / 2))) ×t II) Cn II))
7064, 37cnmpt2nd 22362 . . . . 5 (𝜑 → (𝑥 ∈ (0[,](1 / 2)), 𝑦 ∈ (0[,]1) ↦ 𝑦) ∈ ((((topGen‘ran (,)) ↾t (0[,](1 / 2))) ×t II) Cn II))
714, 11phtpycn 23677 . . . . . 6 (𝜑 → (𝐹(PHtpy‘𝐽)𝐻) ⊆ ((II ×t II) Cn 𝐽))
7271, 13sseldd 3894 . . . . 5 (𝜑𝑀 ∈ ((II ×t II) Cn 𝐽))
7364, 37, 69, 70, 72cnmpt22f 22368 . . . 4 (𝜑 → (𝑥 ∈ (0[,](1 / 2)), 𝑦 ∈ (0[,]1) ↦ ((2 · 𝑥)𝑀𝑦)) ∈ ((((topGen‘ran (,)) ↾t (0[,](1 / 2))) ×t II) Cn 𝐽))
74 iccssre 12854 . . . . . . . 8 (((1 / 2) ∈ ℝ ∧ 1 ∈ ℝ) → ((1 / 2)[,]1) ⊆ ℝ)
7528, 30, 74mp2an 692 . . . . . . 7 ((1 / 2)[,]1) ⊆ ℝ
76 resttopon 21854 . . . . . . 7 (((topGen‘ran (,)) ∈ (TopOn‘ℝ) ∧ ((1 / 2)[,]1) ⊆ ℝ) → ((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ∈ (TopOn‘((1 / 2)[,]1)))
7758, 75, 76mp2an 692 . . . . . 6 ((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ∈ (TopOn‘((1 / 2)[,]1))
7877a1i 11 . . . . 5 (𝜑 → ((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ∈ (TopOn‘((1 / 2)[,]1)))
7978, 37cnmpt1st 22361 . . . . . 6 (𝜑 → (𝑥 ∈ ((1 / 2)[,]1), 𝑦 ∈ (0[,]1) ↦ 𝑥) ∈ ((((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ×t II) Cn ((topGen‘ran (,)) ↾t ((1 / 2)[,]1))))
8024iihalf2cn 23628 . . . . . . 7 (𝑧 ∈ ((1 / 2)[,]1) ↦ ((2 · 𝑧) − 1)) ∈ (((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) Cn II)
8180a1i 11 . . . . . 6 (𝜑 → (𝑧 ∈ ((1 / 2)[,]1) ↦ ((2 · 𝑧) − 1)) ∈ (((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) Cn II))
8268oveq1d 7166 . . . . . 6 (𝑧 = 𝑥 → ((2 · 𝑧) − 1) = ((2 · 𝑥) − 1))
8378, 37, 79, 78, 81, 82cnmpt21 22364 . . . . 5 (𝜑 → (𝑥 ∈ ((1 / 2)[,]1), 𝑦 ∈ (0[,]1) ↦ ((2 · 𝑥) − 1)) ∈ ((((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ×t II) Cn II))
8478, 37cnmpt2nd 22362 . . . . 5 (𝜑 → (𝑥 ∈ ((1 / 2)[,]1), 𝑦 ∈ (0[,]1) ↦ 𝑦) ∈ ((((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ×t II) Cn II))
858, 12phtpycn 23677 . . . . . 6 (𝜑 → (𝐺(PHtpy‘𝐽)𝐾) ⊆ ((II ×t II) Cn 𝐽))
8685, 16sseldd 3894 . . . . 5 (𝜑𝑁 ∈ ((II ×t II) Cn 𝐽))
8778, 37, 83, 84, 86cnmpt22f 22368 . . . 4 (𝜑 → (𝑥 ∈ ((1 / 2)[,]1), 𝑦 ∈ (0[,]1) ↦ (((2 · 𝑥) − 1)𝑁𝑦)) ∈ ((((topGen‘ran (,)) ↾t ((1 / 2)[,]1)) ×t II) Cn 𝐽))
8822, 23, 24, 25, 26, 27, 35, 37, 57, 73, 87cnmpopc 23622 . . 3 (𝜑 → (𝑥 ∈ (0[,]1), 𝑦 ∈ (0[,]1) ↦ if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦))) ∈ ((II ×t II) Cn 𝐽))
8921, 88eqeltrid 2857 . 2 (𝜑𝑃 ∈ ((II ×t II) Cn 𝐽))
90 simpll 767 . . . . . 6 (((𝜑𝑠 ∈ (0[,]1)) ∧ 𝑠 ≤ (1 / 2)) → 𝜑)
91 elii1 23629 . . . . . . . 8 (𝑠 ∈ (0[,](1 / 2)) ↔ (𝑠 ∈ (0[,]1) ∧ 𝑠 ≤ (1 / 2)))
92 iihalf1 23625 . . . . . . . 8 (𝑠 ∈ (0[,](1 / 2)) → (2 · 𝑠) ∈ (0[,]1))
9391, 92sylbir 238 . . . . . . 7 ((𝑠 ∈ (0[,]1) ∧ 𝑠 ≤ (1 / 2)) → (2 · 𝑠) ∈ (0[,]1))
9493adantll 714 . . . . . 6 (((𝜑𝑠 ∈ (0[,]1)) ∧ 𝑠 ≤ (1 / 2)) → (2 · 𝑠) ∈ (0[,]1))
954, 11phtpyhtpy 23676 . . . . . . . 8 (𝜑 → (𝐹(PHtpy‘𝐽)𝐻) ⊆ (𝐹(II Htpy 𝐽)𝐻))
9695, 13sseldd 3894 . . . . . . 7 (𝜑𝑀 ∈ (𝐹(II Htpy 𝐽)𝐻))
9737, 4, 11, 96htpyi 23668 . . . . . 6 ((𝜑 ∧ (2 · 𝑠) ∈ (0[,]1)) → (((2 · 𝑠)𝑀0) = (𝐹‘(2 · 𝑠)) ∧ ((2 · 𝑠)𝑀1) = (𝐻‘(2 · 𝑠))))
9890, 94, 97syl2anc 588 . . . . 5 (((𝜑𝑠 ∈ (0[,]1)) ∧ 𝑠 ≤ (1 / 2)) → (((2 · 𝑠)𝑀0) = (𝐹‘(2 · 𝑠)) ∧ ((2 · 𝑠)𝑀1) = (𝐻‘(2 · 𝑠))))
9998simpld 499 . . . 4 (((𝜑𝑠 ∈ (0[,]1)) ∧ 𝑠 ≤ (1 / 2)) → ((2 · 𝑠)𝑀0) = (𝐹‘(2 · 𝑠)))
100 simpll 767 . . . . . 6 (((𝜑𝑠 ∈ (0[,]1)) ∧ ¬ 𝑠 ≤ (1 / 2)) → 𝜑)
101 elii2 23630 . . . . . . . 8 ((𝑠 ∈ (0[,]1) ∧ ¬ 𝑠 ≤ (1 / 2)) → 𝑠 ∈ ((1 / 2)[,]1))
102101adantll 714 . . . . . . 7 (((𝜑𝑠 ∈ (0[,]1)) ∧ ¬ 𝑠 ≤ (1 / 2)) → 𝑠 ∈ ((1 / 2)[,]1))
103 iihalf2 23627 . . . . . . 7 (𝑠 ∈ ((1 / 2)[,]1) → ((2 · 𝑠) − 1) ∈ (0[,]1))
104102, 103syl 17 . . . . . 6 (((𝜑𝑠 ∈ (0[,]1)) ∧ ¬ 𝑠 ≤ (1 / 2)) → ((2 · 𝑠) − 1) ∈ (0[,]1))
1058, 12phtpyhtpy 23676 . . . . . . . 8 (𝜑 → (𝐺(PHtpy‘𝐽)𝐾) ⊆ (𝐺(II Htpy 𝐽)𝐾))
106105, 16sseldd 3894 . . . . . . 7 (𝜑𝑁 ∈ (𝐺(II Htpy 𝐽)𝐾))
10737, 8, 12, 106htpyi 23668 . . . . . 6 ((𝜑 ∧ ((2 · 𝑠) − 1) ∈ (0[,]1)) → ((((2 · 𝑠) − 1)𝑁0) = (𝐺‘((2 · 𝑠) − 1)) ∧ (((2 · 𝑠) − 1)𝑁1) = (𝐾‘((2 · 𝑠) − 1))))
108100, 104, 107syl2anc 588 . . . . 5 (((𝜑𝑠 ∈ (0[,]1)) ∧ ¬ 𝑠 ≤ (1 / 2)) → ((((2 · 𝑠) − 1)𝑁0) = (𝐺‘((2 · 𝑠) − 1)) ∧ (((2 · 𝑠) − 1)𝑁1) = (𝐾‘((2 · 𝑠) − 1))))
109108simpld 499 . . . 4 (((𝜑𝑠 ∈ (0[,]1)) ∧ ¬ 𝑠 ≤ (1 / 2)) → (((2 · 𝑠) − 1)𝑁0) = (𝐺‘((2 · 𝑠) − 1)))
11099, 109ifeq12da 4454 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀0), (((2 · 𝑠) − 1)𝑁0)) = if(𝑠 ≤ (1 / 2), (𝐹‘(2 · 𝑠)), (𝐺‘((2 · 𝑠) − 1))))
111 simpr 489 . . . 4 ((𝜑𝑠 ∈ (0[,]1)) → 𝑠 ∈ (0[,]1))
112 0elunit 12894 . . . 4 0 ∈ (0[,]1)
113 simpl 487 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 0) → 𝑥 = 𝑠)
114113breq1d 5043 . . . . . 6 ((𝑥 = 𝑠𝑦 = 0) → (𝑥 ≤ (1 / 2) ↔ 𝑠 ≤ (1 / 2)))
115113oveq2d 7167 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 0) → (2 · 𝑥) = (2 · 𝑠))
116 simpr 489 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 0) → 𝑦 = 0)
117115, 116oveq12d 7169 . . . . . 6 ((𝑥 = 𝑠𝑦 = 0) → ((2 · 𝑥)𝑀𝑦) = ((2 · 𝑠)𝑀0))
118115oveq1d 7166 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 0) → ((2 · 𝑥) − 1) = ((2 · 𝑠) − 1))
119118, 116oveq12d 7169 . . . . . 6 ((𝑥 = 𝑠𝑦 = 0) → (((2 · 𝑥) − 1)𝑁𝑦) = (((2 · 𝑠) − 1)𝑁0))
120114, 117, 119ifbieq12d 4449 . . . . 5 ((𝑥 = 𝑠𝑦 = 0) → if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)) = if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀0), (((2 · 𝑠) − 1)𝑁0)))
121 ovex 7184 . . . . . 6 ((2 · 𝑠)𝑀0) ∈ V
122 ovex 7184 . . . . . 6 (((2 · 𝑠) − 1)𝑁0) ∈ V
123121, 122ifex 4471 . . . . 5 if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀0), (((2 · 𝑠) − 1)𝑁0)) ∈ V
124120, 21, 123ovmpoa 7301 . . . 4 ((𝑠 ∈ (0[,]1) ∧ 0 ∈ (0[,]1)) → (𝑠𝑃0) = if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀0), (((2 · 𝑠) − 1)𝑁0)))
125111, 112, 124sylancl 590 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → (𝑠𝑃0) = if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀0), (((2 · 𝑠) − 1)𝑁0)))
1264, 8pcovalg 23706 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → ((𝐹(*𝑝𝐽)𝐺)‘𝑠) = if(𝑠 ≤ (1 / 2), (𝐹‘(2 · 𝑠)), (𝐺‘((2 · 𝑠) − 1))))
127110, 125, 1263eqtr4d 2804 . 2 ((𝜑𝑠 ∈ (0[,]1)) → (𝑠𝑃0) = ((𝐹(*𝑝𝐽)𝐺)‘𝑠))
12898simprd 500 . . . 4 (((𝜑𝑠 ∈ (0[,]1)) ∧ 𝑠 ≤ (1 / 2)) → ((2 · 𝑠)𝑀1) = (𝐻‘(2 · 𝑠)))
129108simprd 500 . . . 4 (((𝜑𝑠 ∈ (0[,]1)) ∧ ¬ 𝑠 ≤ (1 / 2)) → (((2 · 𝑠) − 1)𝑁1) = (𝐾‘((2 · 𝑠) − 1)))
130128, 129ifeq12da 4454 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀1), (((2 · 𝑠) − 1)𝑁1)) = if(𝑠 ≤ (1 / 2), (𝐻‘(2 · 𝑠)), (𝐾‘((2 · 𝑠) − 1))))
131 1elunit 12895 . . . 4 1 ∈ (0[,]1)
132 simpl 487 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 1) → 𝑥 = 𝑠)
133132breq1d 5043 . . . . . 6 ((𝑥 = 𝑠𝑦 = 1) → (𝑥 ≤ (1 / 2) ↔ 𝑠 ≤ (1 / 2)))
134132oveq2d 7167 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 1) → (2 · 𝑥) = (2 · 𝑠))
135 simpr 489 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 1) → 𝑦 = 1)
136134, 135oveq12d 7169 . . . . . 6 ((𝑥 = 𝑠𝑦 = 1) → ((2 · 𝑥)𝑀𝑦) = ((2 · 𝑠)𝑀1))
137134oveq1d 7166 . . . . . . 7 ((𝑥 = 𝑠𝑦 = 1) → ((2 · 𝑥) − 1) = ((2 · 𝑠) − 1))
138137, 135oveq12d 7169 . . . . . 6 ((𝑥 = 𝑠𝑦 = 1) → (((2 · 𝑥) − 1)𝑁𝑦) = (((2 · 𝑠) − 1)𝑁1))
139133, 136, 138ifbieq12d 4449 . . . . 5 ((𝑥 = 𝑠𝑦 = 1) → if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)) = if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀1), (((2 · 𝑠) − 1)𝑁1)))
140 ovex 7184 . . . . . 6 ((2 · 𝑠)𝑀1) ∈ V
141 ovex 7184 . . . . . 6 (((2 · 𝑠) − 1)𝑁1) ∈ V
142140, 141ifex 4471 . . . . 5 if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀1), (((2 · 𝑠) − 1)𝑁1)) ∈ V
143139, 21, 142ovmpoa 7301 . . . 4 ((𝑠 ∈ (0[,]1) ∧ 1 ∈ (0[,]1)) → (𝑠𝑃1) = if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀1), (((2 · 𝑠) − 1)𝑁1)))
144111, 131, 143sylancl 590 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → (𝑠𝑃1) = if(𝑠 ≤ (1 / 2), ((2 · 𝑠)𝑀1), (((2 · 𝑠) − 1)𝑁1)))
14511, 12pcovalg 23706 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → ((𝐻(*𝑝𝐽)𝐾)‘𝑠) = if(𝑠 ≤ (1 / 2), (𝐻‘(2 · 𝑠)), (𝐾‘((2 · 𝑠) − 1))))
146130, 144, 1453eqtr4d 2804 . 2 ((𝜑𝑠 ∈ (0[,]1)) → (𝑠𝑃1) = ((𝐻(*𝑝𝐽)𝐾)‘𝑠))
1474, 11, 13phtpyi 23678 . . . 4 ((𝜑𝑠 ∈ (0[,]1)) → ((0𝑀𝑠) = (𝐹‘0) ∧ (1𝑀𝑠) = (𝐹‘1)))
148147simpld 499 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → (0𝑀𝑠) = (𝐹‘0))
149 simpl 487 . . . . . . . 8 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → 𝑥 = 0)
150149, 29eqbrtrdi 5072 . . . . . . 7 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → 𝑥 ≤ (1 / 2))
151150iftrued 4429 . . . . . 6 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)) = ((2 · 𝑥)𝑀𝑦))
152149oveq2d 7167 . . . . . . . 8 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → (2 · 𝑥) = (2 · 0))
153 2t0e0 11836 . . . . . . . 8 (2 · 0) = 0
154152, 153eqtrdi 2810 . . . . . . 7 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → (2 · 𝑥) = 0)
155 simpr 489 . . . . . . 7 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → 𝑦 = 𝑠)
156154, 155oveq12d 7169 . . . . . 6 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → ((2 · 𝑥)𝑀𝑦) = (0𝑀𝑠))
157151, 156eqtrd 2794 . . . . 5 ((𝑥 = 0 ∧ 𝑦 = 𝑠) → if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)) = (0𝑀𝑠))
158 ovex 7184 . . . . 5 (0𝑀𝑠) ∈ V
159157, 21, 158ovmpoa 7301 . . . 4 ((0 ∈ (0[,]1) ∧ 𝑠 ∈ (0[,]1)) → (0𝑃𝑠) = (0𝑀𝑠))
160112, 111, 159sylancr 591 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → (0𝑃𝑠) = (0𝑀𝑠))
1614, 8pco0 23708 . . . 4 (𝜑 → ((𝐹(*𝑝𝐽)𝐺)‘0) = (𝐹‘0))
162161adantr 485 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → ((𝐹(*𝑝𝐽)𝐺)‘0) = (𝐹‘0))
163148, 160, 1623eqtr4d 2804 . 2 ((𝜑𝑠 ∈ (0[,]1)) → (0𝑃𝑠) = ((𝐹(*𝑝𝐽)𝐺)‘0))
1648, 12, 16phtpyi 23678 . . . 4 ((𝜑𝑠 ∈ (0[,]1)) → ((0𝑁𝑠) = (𝐺‘0) ∧ (1𝑁𝑠) = (𝐺‘1)))
165164simprd 500 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → (1𝑁𝑠) = (𝐺‘1))
16628, 30ltnlei 10792 . . . . . . . . 9 ((1 / 2) < 1 ↔ ¬ 1 ≤ (1 / 2))
16731, 166mpbi 233 . . . . . . . 8 ¬ 1 ≤ (1 / 2)
168 simpl 487 . . . . . . . . 9 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → 𝑥 = 1)
169168breq1d 5043 . . . . . . . 8 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → (𝑥 ≤ (1 / 2) ↔ 1 ≤ (1 / 2)))
170167, 169mtbiri 331 . . . . . . 7 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → ¬ 𝑥 ≤ (1 / 2))
171170iffalsed 4432 . . . . . 6 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)) = (((2 · 𝑥) − 1)𝑁𝑦))
172168oveq2d 7167 . . . . . . . . . 10 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → (2 · 𝑥) = (2 · 1))
173 2t1e2 11830 . . . . . . . . . 10 (2 · 1) = 2
174172, 173eqtrdi 2810 . . . . . . . . 9 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → (2 · 𝑥) = 2)
175174oveq1d 7166 . . . . . . . 8 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → ((2 · 𝑥) − 1) = (2 − 1))
176 2m1e1 11793 . . . . . . . 8 (2 − 1) = 1
177175, 176eqtrdi 2810 . . . . . . 7 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → ((2 · 𝑥) − 1) = 1)
178 simpr 489 . . . . . . 7 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → 𝑦 = 𝑠)
179177, 178oveq12d 7169 . . . . . 6 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → (((2 · 𝑥) − 1)𝑁𝑦) = (1𝑁𝑠))
180171, 179eqtrd 2794 . . . . 5 ((𝑥 = 1 ∧ 𝑦 = 𝑠) → if(𝑥 ≤ (1 / 2), ((2 · 𝑥)𝑀𝑦), (((2 · 𝑥) − 1)𝑁𝑦)) = (1𝑁𝑠))
181 ovex 7184 . . . . 5 (1𝑁𝑠) ∈ V
182180, 21, 181ovmpoa 7301 . . . 4 ((1 ∈ (0[,]1) ∧ 𝑠 ∈ (0[,]1)) → (1𝑃𝑠) = (1𝑁𝑠))
183131, 111, 182sylancr 591 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → (1𝑃𝑠) = (1𝑁𝑠))
1844, 8pco1 23709 . . . 4 (𝜑 → ((𝐹(*𝑝𝐽)𝐺)‘1) = (𝐺‘1))
185184adantr 485 . . 3 ((𝜑𝑠 ∈ (0[,]1)) → ((𝐹(*𝑝𝐽)𝐺)‘1) = (𝐺‘1))
186165, 183, 1853eqtr4d 2804 . 2 ((𝜑𝑠 ∈ (0[,]1)) → (1𝑃𝑠) = ((𝐹(*𝑝𝐽)𝐺)‘1))
18710, 20, 89, 127, 146, 163, 186isphtpy2d 23681 1 (𝜑𝑃 ∈ ((𝐹(*𝑝𝐽)𝐺)(PHtpy‘𝐽)(𝐻(*𝑝𝐽)𝐾)))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 400 ∧ w3a 1085 = wceq 1539 ∈ wcel 2112 ≠ wne 2952 ⊆ wss 3859 ∅c0 4226 ifcif 4421 class class class wbr 5033 ↦ cmpt 5113 ran crn 5526 ‘cfv 6336 (class class class)co 7151 ∈ cmpo 7153 ℝcr 10567 0cc0 10568 1c1 10569 · cmul 10573 < clt 10706 ≤ cle 10707 − cmin 10901 / cdiv 11328 2c2 11722 (,)cioo 12772 [,]cicc 12775 ↾t crest 16745 topGenctg 16762 TopOnctopon 21603 Cn ccn 21917 ×t ctx 22253 IIcii 23569 Htpy chtpy 23661 PHtpycphtpy 23662 ≃phcphtpc 23663 *𝑝cpco 23694 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2730 ax-rep 5157 ax-sep 5170 ax-nul 5177 ax-pow 5235 ax-pr 5299 ax-un 7460 ax-cnex 10624 ax-resscn 10625 ax-1cn 10626 ax-icn 10627 ax-addcl 10628 ax-addrcl 10629 ax-mulcl 10630 ax-mulrcl 10631 ax-mulcom 10632 ax-addass 10633 ax-mulass 10634 ax-distr 10635 ax-i2m1 10636 ax-1ne0 10637 ax-1rid 10638 ax-rnegex 10639 ax-rrecex 10640 ax-cnre 10641 ax-pre-lttri 10642 ax-pre-lttrn 10643 ax-pre-ltadd 10644 ax-pre-mulgt0 10645 ax-pre-sup 10646 ax-mulf 10648 This theorem depends on definitions: df-bi 210 df-an 401 df-or 846 df-3or 1086 df-3an 1087 df-tru 1542 df-fal 1552 df-ex 1783 df-nf 1787 df-sb 2071 df-mo 2558 df-eu 2589 df-clab 2737 df-cleq 2751 df-clel 2831 df-nfc 2902 df-ne 2953 df-nel 3057 df-ral 3076 df-rex 3077 df-reu 3078 df-rmo 3079 df-rab 3080 df-v 3412 df-sbc 3698 df-csb 3807 df-dif 3862 df-un 3864 df-in 3866 df-ss 3876 df-pss 3878 df-nul 4227 df-if 4422 df-pw 4497 df-sn 4524 df-pr 4526 df-tp 4528 df-op 4530 df-uni 4800 df-int 4840 df-iun 4886 df-iin 4887 df-br 5034 df-opab 5096 df-mpt 5114 df-tr 5140 df-id 5431 df-eprel 5436 df-po 5444 df-so 5445 df-fr 5484 df-se 5485 df-we 5486 df-xp 5531 df-rel 5532 df-cnv 5533 df-co 5534 df-dm 5535 df-rn 5536 df-res 5537 df-ima 5538 df-pred 6127 df-ord 6173 df-on 6174 df-lim 6175 df-suc 6176 df-iota 6295 df-fun 6338 df-fn 6339 df-f 6340 df-f1 6341 df-fo 6342 df-f1o 6343 df-fv 6344 df-isom 6345 df-riota 7109 df-ov 7154 df-oprab 7155 df-mpo 7156 df-of 7406 df-om 7581 df-1st 7694 df-2nd 7695 df-supp 7837 df-wrecs 7958 df-recs 8019 df-rdg 8057 df-1o 8113 df-2o 8114 df-oadd 8117 df-er 8300 df-map 8419 df-ixp 8481 df-en 8529 df-dom 8530 df-sdom 8531 df-fin 8532 df-fsupp 8860 df-fi 8901 df-sup 8932 df-inf 8933 df-oi 9000 df-card 9394 df-pnf 10708 df-mnf 10709 df-xr 10710 df-ltxr 10711 df-le 10712 df-sub 10903 df-neg 10904 df-div 11329 df-nn 11668 df-2 11730 df-3 11731 df-4 11732 df-5 11733 df-6 11734 df-7 11735 df-8 11736 df-9 11737 df-n0 11928 df-z 12014 df-dec 12131 df-uz 12276 df-q 12382 df-rp 12424 df-xneg 12541 df-xadd 12542 df-xmul 12543 df-ioo 12776 df-icc 12779 df-fz 12933 df-fzo 13076 df-seq 13412 df-exp 13473 df-hash 13734 df-cj 14499 df-re 14500 df-im 14501 df-sqrt 14635 df-abs 14636 df-struct 16536 df-ndx 16537 df-slot 16538 df-base 16540 df-sets 16541 df-ress 16542 df-plusg 16629 df-mulr 16630 df-starv 16631 df-sca 16632 df-vsca 16633 df-ip 16634 df-tset 16635 df-ple 16636 df-ds 16638 df-unif 16639 df-hom 16640 df-cco 16641 df-rest 16747 df-topn 16748 df-0g 16766 df-gsum 16767 df-topgen 16768 df-pt 16769 df-prds 16772 df-xrs 16826 df-qtop 16831 df-imas 16832 df-xps 16834 df-mre 16908 df-mrc 16909 df-acs 16911 df-mgm 17911 df-sgrp 17960 df-mnd 17971 df-submnd 18016 df-mulg 18285 df-cntz 18507 df-cmn 18968 df-psmet 20151 df-xmet 20152 df-met 20153 df-bl 20154 df-mopn 20155 df-cnfld 20160 df-top 21587 df-topon 21604 df-topsp 21626 df-bases 21639 df-cld 21712 df-cn 21920 df-cnp 21921 df-tx 22255 df-hmeo 22448 df-xms 23015 df-ms 23016 df-tms 23017 df-ii 23571 df-htpy 23664 df-phtpy 23665 df-phtpc 23686 df-pco 23699 This theorem is referenced by: pcohtpy 23714
Copyright terms: Public domain W3C validator | 13,193 | 19,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-38 | latest | en | 0.215204 |
https://small--loans.com/blog/how-do-i-apply-for-a-personal-loan | 1,713,487,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00794.warc.gz | 476,504,223 | 44,201 | Category
# How Do I Apply For A Personal Loan?
When applying for a personal loan, the first step is to research different lenders and compare their interest rates, fees, and terms. Once you have chosen a lender, you will need to gather all necessary documents such as proof of income, identification, and credit history.
Next, you will need to fill out the loan application form provided by the lender. This form will ask for personal information such as your name, address, employment status, and the amount you wish to borrow.
After submitting your application, the lender will review your information and make a decision on whether to approve or deny your loan request. If approved, you will receive the loan funds in your bank account.
It is important to carefully review the terms and conditions of the loan agreement before signing it. Make sure you understand the interest rate, repayment schedule, and any fees associated with the loan.
Once you have received the funds, it is important to make timely payments on the loan to avoid any additional fees or damage to your credit score.
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## How do I calculate my monthly payments for a personal loan?
To calculate your monthly payments for a personal loan, you will need to know the loan amount, interest rate, and loan term (in months).
1. Determine the loan amount: This is the total amount of money you are borrowing.
2. Determine the interest rate: This is the annual percentage rate (APR) that you will be charged for borrowing the money.
3. Determine the loan term: This is the length of time you have to repay the loan, typically in months.
Once you have this information, you can use the following formula to calculate your monthly payments:
M = P[r(1+r)^n]/[(1+r)^n-1]
Where: M = Monthly payment P = Loan amount r = Monthly interest rate (annual interest rate divided by 12) n = Loan term in months
For example, if you borrow \$10,000 with an annual interest rate of 5% and a loan term of 36 months, the calculation would be as follows:
P = \$10,000 r = 0.05/12 = 0.0041667 n = 36
M = \$10,000[(0.0041667*(1+0.0041667)^36]/[(1+0.0041667)^36-1] M = \$299.71
Therefore, your monthly payment for this personal loan would be approximately \$299.71. You can use an online loan calculator to easily calculate your monthly payments based on your specific loan details.
## How long does it take to get approved for a personal loan?
The approval process for a personal loan can vary depending on the lender and the individual's financial situation. In general, it can take anywhere from a few minutes to a few business days to get approved for a personal loan. Some online lenders may offer instant approval decisions, while traditional banks and credit unions may take longer to review and approve the application. It's also important to note that some lenders may require additional documentation or information before making a final decision on the loan application.
## What factors do lenders consider when approving a personal loan?
1. Credit score: Lenders typically consider the borrower's credit score as a primary factor when deciding whether to approve a personal loan. A higher credit score indicates a lower credit risk, making it more likely that the loan application will be approved.
2. Income and employment history: Lenders will also look at the borrower's income and employment history to assess their ability to repay the loan. A stable job and steady income can increase the likelihood of loan approval.
3. Debt-to-income ratio: Lenders will analyze the borrower's debt-to-income ratio, which is the ratio of their monthly debt payments to their monthly income. A lower debt-to-income ratio indicates that the borrower has more disposable income available to make loan payments.
4. Loan amount and purpose: The amount of the loan requested and the purpose for which it will be used can also impact the approval decision. Lenders may be more willing to approve a loan for a specific purpose, such as debt consolidation or home improvement, than for a discretionary expense.
5. Credit history and payment history: Lenders will review the borrower's credit history and payment history to assess their past behavior with credit and determine if they are likely to repay the loan on time.
6. Collateral: Some personal loans may be secured by collateral, such as a car or a home. Lenders will consider the value of the collateral and the borrower's equity in it when deciding whether to approve the loan.
7. Loan term and interest rate: The loan term and interest rate offered by the lender can also impact the approval decision. Some lenders may be more selective in approving loans with longer terms or higher interest rates.
8. Other financial obligations: Lenders will also consider any other financial obligations the borrower may have, such as outstanding loans or credit card debt, when assessing their ability to repay the personal loan.
## How do I know if a personal loan offer is a scam?
There are several red flags to watch out for when determining if a personal loan offer is a scam:
1. Unsolicited offers: If you receive an email, text, or phone call out of the blue offering you a loan without you having applied for one, it could be a scam.
2. Upfront fees: Legitimate lenders do not require you to pay any upfront fees before receiving a loan. If a lender asks for payment before providing the loan, it is likely a scam.
3. No credit check required: Reputable lenders will always conduct a credit check before approving a loan. If a lender claims they can provide a loan without checking your credit history, it is likely a scam.
4. Pressure to act quickly: Scammers often create a sense of urgency to pressure you into making a decision quickly. Legitimate lenders will give you time to review the terms and conditions before agreeing to a loan.
5. Lack of physical address or contact information: If a lender does not provide a physical address, phone number, or other contact information, it could be a scam. Make sure to research the lender and verify their legitimacy before providing any personal information.
6. Unrealistic loan terms: If a lender is offering a loan with extremely low interest rates, no repayment schedule, or other unrealistic terms, it is likely a scam. Always be wary of offers that seem too good to be true.
If you have any doubts about a personal loan offer, it is best to do thorough research, check the lender's credibility, and trust your instincts. It is always safer to work with reputable financial institutions or lenders who have a proven track record of providing legitimate loans.
## How do I request a personal loan deferment or forbearance?
To request a personal loan deferment or forbearance, you should contact your loan servicer directly. They will be able to provide you with the necessary forms and information on how to apply for a deferment or forbearance. You may be required to provide documentation of your current financial situation or hardship in order to qualify for the deferment or forbearance. It is important to reach out to your loan servicer as soon as possible if you are experiencing financial difficulties and need assistance with your loan payments.
## What is the impact of a personal loan on my credit score?
Taking out a personal loan can have both positive and negative impacts on your credit score, depending on how you manage the loan.
Positive impacts:
1. Establishing a positive payment history: Making timely payments on your personal loan can demonstrate to credit bureaus that you are a responsible borrower, which can help improve your credit score.
2. Diversifying your credit mix: Having different types of credit, such as installment loans like personal loans, can help improve your credit score by showing that you can manage different types of debts.
Negative impacts:
1. Increased debt-to-income ratio: Taking on a personal loan increases your overall debt, which can negatively impact your credit score if you have a high debt-to-income ratio.
2. Hard credit inquiry: When you apply for a personal loan, the lender will perform a hard credit inquiry, which can temporarily lower your credit score.
Overall, if you make timely payments and manage your personal loan responsibly, it can have a positive impact on your credit score in the long run. However, if you miss payments or accumulate too much debt, it can have a negative impact on your credit score.
## Related Posts:
A personal loan is a type of loan that is borrowed from a bank, credit union, or online lender for personal use. It is typically an unsecured loan, meaning it does not require collateral, such as a house or a car, to secure the loan. Here's how a personal ...
If you are a nurse looking to apply for a loan, there are several options available to you. Here are some places where you can apply for a loan:Banks: Traditional banks are a common choice for individuals seeking a loan, including nurses. You can visit your lo...
If you are looking to apply for a personal loan for a 6-month term, there are various options available to you. Most commonly, you can apply for such loans through banks, credit unions, online lenders, and even peer-to-peer lending platforms. These institution... | 1,998 | 9,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-18 | longest | en | 0.956431 |
https://tutelman.com/golf/shafts/allAboutSpines2.php | 1,669,795,684,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710733.87/warc/CC-MAIN-20221130060525-20221130090525-00507.warc.gz | 628,107,560 | 10,369 | # Finding spines: the instruments
## Feel Finders
For years, there have been clubmaking tools on the market called "spine finders". I refer to them as "bearing-based spine finders", and have seen them referred to as "feel finders". I like the latter term, and have adopted it. (Sorry, but I don't know to whom to attribute the term.) This is the technology that Bill Day was using when he wrote his "terminology" paper.
Here is how feel finders work:
A shaft is fit into three freely-rotating ball bearing assemblies, so it can rotate as it pleases. The bearings are not aligned (or can be forced out of line -- see the red force arrows on the picture), so the shaft must bend as it rests in the bearings. Now the shaft is rotated, and the fingers feel the rotation. There are two kinds of features that you feel:
• A "bump", a position where it takes some pressure to get the shaft there from either direction. This is generally considered to be a spine, the direction of maximum stiffness of the shaft.
• A "stable" direction, where the shaft rotates to naturally and rests there if not held. This is how the NBP got the name "natural bending position". It is the direction the shaft "wants" to rest if bent.
The principle is that the shaft will naturally settle into the position that takes the least force to produce the bend imposed by the bearings. That's a good principle. All other things being equal, it will find the direction of least stiffness. Where that is the case, spine finders do find the spine and NBP.
But sometimes -- often, in fact -- all things are not equal. If the shaft has any residual bend, it can throw off the measurement. Let's see why.
Consider a shaft that has no spine at all (that is, the stiffness is the same in all directions), but 3mm of residual bend as shown in the picture. You would never notice 3mm of bend unless you looked for it. Now let's put the shaft in a feel finder that deflects the shaft by 20mm (almost an inch).
The shaft will just snap into position in the feel finder, indicating a very strong spine. Remembering that we started out by saying there is no spine, let's understand why this happens.
• If you turn it so the residual bend is in the same direction as the spine-finder-induced bend, then the shaft is only deflected by 17mm (20mm minus the 3mm of built-in bend).
• If you turn it so the residual bend is in the opposite direction to the spine-finder-induced bend, then the shaft is deflected by 23mm (20mm plus the 3mm of built-in bend).
That's a difference of 6mm, or 30% of the total 20mm bend. So the shaft is deflected more in one direction than the other direction. 30% more, in fact. The result: it will take 30% more force to deflect it. So there's a lot more force in one direction than the other. And that difference in force is how a feel finder locates the spine.
Because of the forces as you turn the shaft, you will come to the conclusion that there is a spine on one side and an NBP directly opposite. There are two things wrong with this conclusion:
1. We started out by saying the shaft has no spine, so the conclusion has to be wrong.
2. We know from physics that, if there is a spine in one direction, the opposite direction will have a spine, not an NBP. So the feel finder is telling us something that is physically impossible.
This explains what Bill Day calls a Type 1 shaft, a shaft with a single spine on one side and a single NBP opposite. It also explains why he says this is common in steel shafts. The vast majority of steel shafts have very little spine, so any residual bend will overwhelm what spine there is. The residual bend will be all that the feel finder can measure.
What happens when there is both spine and residual bend? Unless the spine effects are large enough to overwhelm the residual bend, you will get spine and NBP directions that do not conform to the laws of phyics -- and therefore are wrong. We will see typical examples a little later. Really, the only time you can trust the output of a feel finder is when it tells you that you have a well-behaved Type 2 shaft, with 180º between the spines and 180º between the NBPs. If it says anything else about the shaft, it is paying too much attention to the residual bend, and coming up with a wrong answer.
Feel finders come in a wide variety of shapes, some of which use quantitative measures and not just the sense of feel. Let's look at a few variations:
It is not necessary for the bearings to be equally spaced over the shaft. Many of the commercial feel-finders have two bearings only 6"-10" apart, for the butt of the shaft to be inserted. This allows the device to be packaged as two bearings in a tube. The JB spine finder at the right is a prime example of this very common construction.. When the two bearings for the shaft butt are packaged together, the third bearing is often hand-held. Or it may be omitted altogether, leaving the fingers acting as the tip bearing. Here the JB finder is demonstrated by Jerry Ballard himself. There are quite a few instruments designed primarily as deflection analyzers, but which also advertise themselves as spine finders. One might be led by their advertising to believe that they are not just feel finders, because they measure the load on the shaft due to a deflection -- the definition of shaft stiffness. But one look at their configuration shows them to be identical to the "principles" diagram above. And, if we tested the same steel shaft (with no spine and 3mm residual bend), the instrument would tell us -- very precisely, in fact -- that it is a Type 1 shaft with a 30% spine. So, no matter how precise the output reading is, you still can't tell how much of the indicated load is the spine and how much is an artifact of residual bend. The GolfMechanix Auditor (shown in the photo) is a good example of this type of instrument. Others that fit the description are the Flexmaster, the NeuFinder 2, and the MCC Multi-Match. I have heard such gadgets defended as being more precise than feel finders. And they are. But -- much more important -- they are not a bit more accurate than feel finders, which is not very accurate at all. (If you are confused about the difference between precision and accuracy, see my article on the subject.)
## FLO
The "gold standard" for finding spine is FLO, raising two questions:
1. What is FLO?
2. Why does it find the spine better than other methods?
FLO stands for "Flat Line Oscillation". It refers to the direction in which the shaft vibrates back and forth in a flat line, as opposed to wobbling out of the line.
To test for FLO, place the shaft in a frequency meter clamp and weight the tip. (You don't need a frequency meter to find FLO, but you do need a very secure clamp and a tip weight to get the shaft to vibrate. And the frequency meter is handy to distinguish spine from NBP.) In the picture, David Dugally of The Golf Coast has attached a laser pointer to the tip of the shaft, so it traces out the path of the shaft on the cardboard in the background. The path is a nice, straight line. That's what you want to see; it indicates FLO. The plane in which the shaft is vibrating is either the spine plane or the NBP plane. If the shaft has a signrificant spine, then the path of the tip will wobble from its original straight line into an oval shape -- unless the shaft is oscillating in the plane of the spine or the NBP. So, to find the spine and NBP, test the vibration of the shaft in different orientations. Where you get Flat Line Oscillation with no tendency to wobble into an oval, then the plane of vibration is either the spine or the NBP. You can tell which FLO plane is the spine and which the NBP by using a frequency meter. The higher of the two frequencies is the spine. (In a pinch, a feel-finder may be useful to tell which FLO plane is which.)
And here is an actual video of a shaft in non-FLO motion. (Click on the snapshot to watch the video.) It starts in a good up-down motion, but "ovals" out of it in a few vibrations. After ovaling a while, it briefly flattens again in a different direction from the original. But it ovals again before returning to the original up-down motion. This shaft has a very prominent spine; most will take longer to go out of flat.
FLO finds the real spine and NBP, unpolluted by non-spine properties like residual bend. How does it do that? If a shaft is bent in any direction but a spine or NBP, the resulting spring force is not exactly in line with the bend. That causes the beginning of wobble. Let it vibrate outside the FLO plane for enough cycles, and the wobble becomes very visible. If you want all the details, they are available in another article.
In finding FLO, you can use any convenient tip weight, as long as it is secured solidly to the shaft tip. The FLO planes will not change with the amount of weight. They won't even change if the weight is off-center -- which means there's no advantage to testing for FLO with the actual clubhead. John Kaufman did an interesting experiment to confirm this. From the CSFA Tech Notes:
Flat Line Oscillation (FLO) vs Center of Gravity
TECH NOTE 29: In the spine alignment process we try to determine the two planes in a shaft that produce flat line oscillation, i.e. no wobble. This is very often done with a tip weight rather than the actual clubhead. The questions arises," will the FLO change when I install the head because its center of gravity is offset from the centerline of the shaft?" Golfsmith claims it will not change. I was curious. If a change in FLO does occur is it due to the change in cg or because it's hard to twang the club straight up and down when your finger is twanging somewhere out on the clubhead rather than on the centerline of the shaft?
To test this I built a tip weight with an arm sticking out of the side of the piece that attaches to the shaft. This piece weighed about 100 grams. I placed another weight on this arm and held it in place with a setscrew. This second weight also weighs about 100 grams. This arrangement allowed me to vary the cg of the test weight by more than an inch in a direction at right angles to the shaft. I attached a very small key chain laser to the tip weight. I built a trigger release system to eliminate any variations due to finger plucking. I placed a 10" disc on the butt end of the shaft. With degree lines on this disc it was easy to repeat alignments as I searched for the FLO. I determined the FLO plane as accurately as I could with the laser projected about 20 feet across the room. I then adjusted the side weight to move the cg about an inch. I did not find any variation in the FLO plane. I guess Golfsmith was right.
You may have heard of PURED shafts. They are shafts whose FLO has been found using a proprietary machine from SST PURE (Dick Weiss' company). The machine uses a computer-controlled process to home in on the FLO plane by examining the wobble quantitatively and deducing the FLO plane from the details of the wobble. To do that, you need a mathematical model of the wobble. Fortunately, the vibration of a clamped shaft with a spine is remarkably simple as mathematical models go. You can see the model and a derivation of it in my article on FLO.
Let's finish the section on FLO by repeating: One does not find FLO because FLO is important in characterizing the motion of the shaft during the swing. One finds FLO because FLO is a reliable way to find the stiffest and softest flex directions of a shaft.
## Differential deflection
Is FLO the only way to find true spine? Does measuring the force needed to deflect the shaft always result in unreliable answers?
Not necessarily. Measuring the force needed to deflect the shaft is actually a very sound way to find the spine; it goes right back to the definitions of spine and NBP. But you have to be sure you are deflecting the shaft by the same amount in all directions. The reason feel finders are unreliable is that a shaft with residual bend is pre-deflected, and bearing-based spine finders don't allow for that fact.
But there are instruments around that can do it right. In order to do it right, you have to deflect the shaft by the same amount for every load reading in every direction. Since the shaft may have residual bend, this is tricky. The way to do it is:
• Pre-load the shaft by some small deflection, and note the load.
• Increase the deflection by a carefully controlled amount (call it D), and note the load.
• The difference between the two loads is the force due to a deflection of D.
This is called "differential deflection" because you use the difference between the loads at two deflections. The important things in this measurement are (a) the shaft must be pre-loaded and "read" in every position and (b) the deflection D is very precisely repeatable. You can't use bearings for this, because the shaft must hold its position stably, even when it doesn't "want" to stay there. (We know from using feel finders that there are positions the shaft will rotate away from when deflected.)
Here are a few instruments that can find spine and NBP via differential deflection:
So differential deflection has a problem in that it needs precise instrumentation and data analysis to find the spine and NBP.
It has another weakenss as well. It tends to take longer than FLO-finding to find the spine and NBP. You have to do a full reading -- that is, rotate to next position, pre-load, tare, load, record the reading -- for each angle of the shaft. 10* increments is probably right. That means 18 full readings for each shaft, before you begin the data analysis.
Before you ask -- and everyone asks -- yes, you have to do a pre-load and tare for every reading. If you don't, you get the numbers that a feel finder would give.
## Bottom line
• If you have a frequency meter (or even just a proper clamp), FLO is probably the best and quickest way to find spine.
• Differential detection takes longer, and isn't very precise if the spine size is small. But, if you don't have a frequency meter and do have a differential deflection instrument, it gives correct answers for your efforts.
• Feel finding -- to paraphrase H.L. Mencken -- is quick, simple, and wrong. It gives wrong directions if the shaft has any residual bend. And, for shafts with small spines, it confidently gives very clear spine directions -- unfortunately based on little but the residual bend itself. Its major value is to give a starting point to look for FLO and, if you don't have a frequency meter, a hint which FLO is spine and which is NBP. | 3,307 | 14,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-49 | longest | en | 0.960135 |
https://study.com/academy/topic/evaluating-calculating-probability.html | 1,580,296,899,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251796127.92/warc/CC-MAIN-20200129102701-20200129132701-00559.warc.gz | 664,905,340 | 24,260 | # Ch 11: Evaluating & Calculating Probability
If you need some help calculating and evaluating probability in math, take a look at our helpful chapter. These video and text lessons allow you to work at your own pace as you review all the topics you need to know in order to catch up on missed class time or study for an upcoming quiz.
## Evaluating & Calculating Probability - Chapter Summary
This fun chapter covers how probability is calculated and evaluated. Our lessons allow you to review the formula for theoretical probability, the uses for conditional probability and how simple conditional probabilities are calculated. These mobile-friendly video and text lessons can be reviewed as many times as you need. Rather than watch a whole video lesson, you can use the video tabs to skip ahead. Once you complete this chapter, you should be able to:
• Define probability as it's used in math
• Explain how predictions are used in experimental probability
• Describe the probability of compound, complementary and simple events
• Give an example of a probability of a compound event
• Calculate the probability of permutations
• Differentiate between the probabilities of dependent and independent events
• Understand how to calculate the probability of combinations
• Use manipulatives to solve probability problems
11 Lessons in Chapter 11: Evaluating & Calculating Probability
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
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Test your knowledge of the entire course with a 50 question practice final exam.
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### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 391 | 1,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-05 | longest | en | 0.917037 |
https://quizzerweb.com.ng/learn/discuss/physics/28089/ | 1,606,247,721,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00371.warc.gz | 465,521,057 | 26,526 | # Two spanners X and Y of lengths 15cm and 20cm respectively are used in...
Question 328
A
1 : 50
B
3 : 20
C
3:4
D
4:3
E
20:3
##### Explanation
Rx = 15/0.2 = 75
Ry = 20/0.2 = 100
Rx : Ry = 75 : 100 = 3 : 4
Try this quiz in in E-test/CBT Mode
switch to
Question Map | 118 | 267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.666136 |
http://docplayer.net/21081549-Which-shapes-make-floor-tilings.html | 1,540,001,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512501.27/warc/CC-MAIN-20181020013721-20181020035221-00219.warc.gz | 106,572,892 | 25,224 | # Which shapes make floor tilings?
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1 Which shapes make floor tilings? Suppose you are trying to tile your bathroom floor. You are allowed to pick only one shape and size of tile. The tile has to be a regular polygon (meaning all the same sides and all the same interior angles, such as a square or an equilateral triangle). Our goal is to figure out which shapes will work. Vertices from one polygon (the corners where edges meet) have to touch vertices from other polygons. For a tiling to work, it means you would be able to cover the floor with only same-size copies of that tile without having gaps. Don t worry about what happens at the walls or with fixtures such as sinks or bathtubs we ll get a tile cutter later. 1. Sketch several regular polygons: an equilateral triangle, a square, a pentagon, a hexagon, an octagon. Can you draw a seven sided regular polygon? Make a guess as to which of these shapes will work to make a tiling. Are there reasons for your guesses? 2. Make cutouts using the separate sheet. There are regular triangles, squares, pentagons, hexagons, and octagons. From your experimentation, which ones work as tiles and which ones don t work? Do you think that tilings might be possible with other regular polygons that you don t have on the next sheet, such as regular heptagons (seven-sided polygons) or regular decagons (ten-sided polygons)? Why or why not? 3. As mentioned before, if you can make a tiling, vertices must touch other vertices at certain points. Around those points, there are a number of angles from each of the polygons sharing that vertex. What is the sum of the degrees in all of those angles? Why is that the sum? 1
2 4. In item 3, you should have found that all the tiles meeting at a point have to have angles adding up to 360. We can use this to see which regular polygons can possibly be used in a tiling pattern. The key is to find out the size of each angle in a regular polygon. We would like to be able to answer questions like this one: what is the size of each angle in a regular heptagon? A regular three sided polygon is also called an equilateral triangle. How large is each angle? Why? A regular 4 sided polygon is a square. How large is each angle of the square? A regular hexagon can be divided into 6 equilateral triangles, as pictured at right. Using the angles in the equilateral triangles, determine the size of the angle marked in the first hexagon. 5. Use the results you have found so far to fill in the table below. Leave the second column blank for 5 and 7 sided polygons. Number of Sides Size of Each Angle (in degrees) Do you see a pattern in the table? Can you guess the angle size for regular pentagons and heptagons? 2
3 6. The dissection method used for the hexagon can be applied to other polygons. Consider the figure at right of a regular octagon, subdivided into triangles. One of the eight identical triangles is shown beside the octagon. Work in your group to determine the three angles of this triangle. Here are two hints: (i) the top angle is one of 8 identical angles that fit together around the center point of the octagon; (ii) the two base angles of the triangle are equal. Find the three angles below, showing all your work and explaining your logic. Based on your answers, how large are the angles in the original octagon? Why? 7. Repeat the steps of item 6 to figure out the size of the angles in a regular pentagon, as pictured at right. 8. Based on the prior two exercises, can you find a formula for the size of each angle in a regular polygon with n sides? If so, do that now, and then skip step 9. [Hints: Imagine that the polygon is divided into n triangles. One angle of the triangle is at the center of the polygon; how big is that angle? How big are the other two angles in the triangle?]. If you are not sure how to do this step, go on to step 9. 3
4 9. Repeat the steps of item 6 to figure out the size of the angles in a regular pentagon, as pictured at right. Then try step 8 again. 10. The correct answer to step 8 is given as follows: The size of each angle in a regular polygon with n sides is 180. Verify that this gives the correct answers for the polygons you have already considered: triangle, square, hexagon, octagon, pentagon. 360 n 4
5 11. Now that you know how to determine the size of angles in regular polygons, decide which polygons can tile a floor. Your goal is a complete list of the possible tiles, and an explanation of why no other tiles can work. (Hint: you need at least 3 tiles meeting at each corner point of the tiling why?) 5
6 Discussion. If you make a tiling pattern using regular polygons, there must be at least tiles at each corner. With exactly 3 tiles, each one must have an angle of 120 to completely fill up the space where the tiles fit together. We know that the hexagon has a 120 and in fact hexagons do make a tiling. What about 4 tiles at each corner? For that to work, each tile has to have an angle of 90 to fill up the space where the 4 tiles meet. We know that each angle of a square is 90 and in fact squares do make a tiling. Could 5 tiles meet at each corner? If so, each angle would have to be 360/5 = 72. That is not possible because no regular polygon has a 72 angle. To see why this is certain, notice that the angle of a square (the 4 sided regular polygon) is 90, which is too big, and the angle of an equilateral triangle (the 3 sided regular polygon) is 60, which is too small. Since there is no polygon in between these two cases, no regular polygon can have an angle of 72. So pentagons cannot make a tiling pattern. Could 6 tiles meet at each corner? That would require 360/6 = 60, which is the angle in an equilateral triangle. And we saw earlier that you can make a tiling pattern with equilateral triangles. Could 7 or more tiles meet at each corner? For seven tiles the angle would have to be 360/7 = That is smaller than the angles in an equilateral triangle. But notice that in our formula , as we increase the number of sides that also increases the angle. In fact, if you increase n the number of sides, n, that results in a smaller center angle 360/n. Then, we subtract a smaller amount from 180, and that produces a larger result. So if we want an angle that is smaller than 60 we have to decrease the number of sides below 3. But that is not possible. The same logic works for any number of tiles greater than 7: if there are t tiles then the angle of each tile would have to be 360/t, and with t > 6 that gives an angle less than 60. But as we just saw, all regular polygons have angles greater than 60. To summarize, the above remarks show that there are just three regular polygon tiling patterns. With hexagons you can fit three together at each corner, with squares you can fit 4 together, and with equilateral triangles you can fit 6 together. No other arrangement is possible. 6
7 7
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https://www.jiskha.com/questions/1113399/present-value-of-an-annuity-sandra-and-regina-earn-the-same-salary-however-regina-has | 1,586,355,359,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00076.warc.gz | 965,210,325 | 5,194 | # Finance
Present Value of an Annuity Sandra and Regina earn the same salary. However, Regina has been far more financially responsible. She pays her bills on time and pays off her credit card debt quickly. Sandra had been less financially responsible. She often buys too many shoes and has allowed her credit card balance to balloon. If she is short on cash for a month, she simply decides to not even pay the minimum balance due on her credit card. Now they both are looking to buy apartments. Regina decides she can afford to make \$3,500 payments, but Sandra can only make \$1,500 payments and pay off her credit card debt, too. Regina qualifies for a 6 percent, 30-year mortgage, but because of her bad credit rating Sandra will be charged 7.5 percent on a 30-year mortgage. Both will put 20 percent down. How is Sandra's bad credit going to impact her apartment search?
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posted by .
we have a problem that my parents cant help with.
846*2=(...+40+...)*2=(800 over ...+...over2+6 over...)=(400+...+3)...=...
...means blank space for a number. Can you help with this problem, so that I can figure out the rest of the problems.Thanks
• math -
from what I can tell you are using * for division (over).
Usually / is used for division and * is used for multiplication.
So I will interpret as follows ...
846/2=(...+40+...)/2=(800 over ...+...over2+6 over...)=(400+...+3)...=...
846/2=(800+40+6)/2
=(800 over 2+40 over 2+6 over 2)
=(400+20+3)
=423
• math -
846*2=(800+40+6)*2
=(800*2 +40*2 +6*2
= 1600 + 80 + 12 = 1692
I don't know why the word "over" appears.
It shouldn't.
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# The city council will certainly vote to approve the new
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The city council will certainly vote to approve the new [#permalink]
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10 Sep 2005, 05:59
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The city council will certainly vote to approve the new downtown redevelopment plan, despite the objections of environmentalists. After all, most of the campaign contributions received by members of the city council come from real estate development firms, which stand to benefit from the plan.
Which of the following statements, if true, would most weaken the argument above?
(A) Several members of the city council receive sizable campaign contributions from environmental lobbying groups.
(B) Members of the city council are required to report the size and source of each campaign contribution they receive.
(C) Not every real estate development firm in the city will be able to participate in, and profit from, the new downtown redevelopment plan.
(D) The members of the city council have often voted in ways that are opposed to the interests of their campaign contributors.
(E) Some environmentalists have stated that the new downtown redevelopment plan might be environmentally sound if certain minor modifications are made.
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10 Sep 2005, 13:14
D is my choice.
C is out of scope.
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hey ya......
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10 Sep 2005, 16:32
I think it is a pretty clear D.
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10 Sep 2005, 16:51
D.
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11 Sep 2005, 02:33
"D" as well
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15 Sep 2005, 19:12
D for me.
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16 Sep 2005, 03:05
The two choices I boiled down to were : A/D...
I finally choose A. Can somebody explain why A is incorrect?
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16 Sep 2005, 09:23
I picked D.
I don't think that A is right for a few reasons. It says that SEVERAL members recieved the donations. We have no idea how many members are on the city council. There could be thousands.
Last edited by popew626 on 16 Sep 2005, 16:40, edited 1 time in total.
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16 Sep 2005, 10:29
I wud go for D.
A is not the most weakening sentense.
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16 Sep 2005, 15:04
Same....D
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16 Sep 2005, 15:36
My choice is D.
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16 Sep 2005, 18:26
I chose A between A and D! I dont understand D at all! Pls someone explain the connection!
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Cheers, Rahul.
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17 Sep 2005, 11:03
rahulraao wrote:
:wall
I chose A between A and D! I dont understand D at all! Pls someone explain the connection!
The assumption in the argument is that the city council members will vote according to the wishes or goals of who donated the most money to each council member's campaign.
D pretty much negates this assumption, thus weakening the argument.
The argument says the plan will CERTAINLY get approved, i.e. without a doubt. BUT, as D says, the council members have OFTEN voted in ways OPPOSED to the interests of those that contribute to the member's campaigns.
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17 Sep 2005, 11:51
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hey ya......
17 Sep 2005, 11:51
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Display posts from previous: Sort by | 1,674 | 5,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-22 | latest | en | 0.935393 |
https://www.physicsforums.com/threads/expansion-of-space-earth-moon.629553/ | 1,670,142,238,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00750.warc.gz | 1,011,837,590 | 16,448 | # Expansion of space: Earth - Moon
Michel_vdg
Hi,
I made a (rough) calculation of how far the Moon would be moving away from Earth according to Hubble's law = 72.6 (km/s)/Mpc
--
1 megaParsec = 3.08 × 1022 m ≈ 4 × 1022 m
Distance between Moon and Earth ≈ 4 x 108 m
Expansion rate ≈ 7.2 104 (m/s)/4 × 1022 m = 7.2 (m/s)/4 × 1018 m
Thus expansion for Earth - Moon ≈ 7.2 x 10-10 m/s
--
Is this calculation correct? If so, than that it would be about one picometer per second, or is Earth's gravity force simply keeping the Moon in it place?
Thanks,
m.
Homework Helper
Hubble's law applies to intergalactic distances. It does not apply to stars within our own galaxy, much less distance between planets in a single solar system not distance between a planet and it moons.
Michel_vdg
ok, thanks.
Than how can we calculate the expansion of the Universe on a local scale, or is there no expansion locally because gravity keeps it all together?
Gold Member
Correct. Gravitational bound objects are essentially immune to the effects of expansion. The force of expansion is overwhelmed by gravity over 'short' distances. By 'short' distances we are talking the level of galactic clusters.
Nik_2213
FWIW, the Moon *is* receding, but due tidal friction, currently ~ 38 mm / yr.
http://en.wikipedia.org/wiki/Orbit_of_the_Moon
IIRC, this is measured by laser ranging using the retro-reflector panels positioned by Apollo astronauts or mounted on Russian rover.
Lsos
Why do I keep hearing that the expansion of space will eventually rip even atoms apart, if it's too weak to affect anything below intergalactic distances?
Gold Member
2021 Award
Why do I keep hearing that the expansion of space will eventually rip even atoms apart, if it's too weak to affect anything below intergalactic distances?
The so-called "Big Rip" scenario that you describe is hypothetical and not widely believed to be likely. There is no evidence for it, just speculation.
I SEEM to recall (sorry I can't give a link, so this is hearsay) that it has even been disproven, but that may be a stretch. Call it unlikely.
Staff Emeritus
Gold Member
The influence of the cosmological expansion on local systems
F. I. Cooperstock, V. Faraoni, D. N. Vollick
http://arxiv.org/abs/astro-ph/9803097v1
They calculate this kind of thing. The effect is extremely small. The predicted general-relativistic effect on the radius of the Earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus.
Michel_vdg
The so-called "Big Rip" scenario that you describe is hypothetical and not widely believed to be likely. There is no evidence for it, just speculation.
I SEEM to recall (sorry I can't give a link, so this is hearsay) that it has even been disproven, but that may be a stretch. Call it unlikely.
Here's an interesting blogpost by a theoretical astrophysicist on this topic, and somewhere on it there is also a link to that paper:
http://astrokatie.blogspot.co.uk/2012/07/you-dont-have-to-blow-up-universe-to-be.html
Michel_vdg
The influence of the cosmological expansion on local systems
F. I. Cooperstock, V. Faraoni, D. N. Vollick
http://arxiv.org/abs/astro-ph/9803097v1
They calculate this kind of thing. The effect is extremely small. The predicted general-relativistic effect on the radius of the Earth's orbit since the time of the dinosaurs is calculated to be about as big as the diameter of an atomic nucleus.
Thanks, this was the kind of information I was curious about. | 878 | 3,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-49 | latest | en | 0.918014 |
https://stats.libretexts.org/Bookshelves/Applied_Statistics/Business_Statistics_(OpenStax)/04%3A_Discrete_Random_Variables/4.05%3A_Poisson_Distribution | 1,725,781,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00176.warc.gz | 519,484,568 | 33,957 | # 4.5: Poisson Distribution
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Another useful probability distribution is the Poisson distribution, or waiting time distribution. This distribution is used to determine how many checkout clerks are needed to keep the waiting time in line to specified levels, how may telephone lines are needed to keep the system from overloading, and many other practical applications. A modification of the Poisson, the Pascal, invented nearly four centuries ago, is used today by telecommunications companies worldwide for load factors, satellite hookup levels and Internet capacity problems. The distribution gets its name from Simeon Poisson who presented it in 1837 as an extension of the binomial distribution which we will see can be estimated with the Poisson.
There are two main characteristics of a Poisson experiment.
1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate.
2. The events are independent of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages and it is assumed that there is no relationship between when misspellings occur.
3. The random variable $$X$$ = the number of occurrences in the interval of interest.
##### Example $$\PageIndex{1}$$
A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let $$X$$ = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.
$$P (X < 5)$$
##### Example $$\PageIndex{2}$$
You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast.
This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast.
a. What is the interval of interest?
a. one broadcast measured in minutes
b. What is the average number of times the news reporter says "uh" during one broadcast?
b. 2
c. Let $$X$$ = ____________. What values does $$X$$ take on?
c. Let $$X$$ = the number of times the news reporter says "uh" during one broadcast.
$$x = 0, 1, 2, 3$$, ...
d. The probability question is $$P$$ (______).
d. $$P (X > 2)$$
## Poisson Probability Distribution Function
$$X \sim \text{Poisson} (\mu)$$
Read this as "$$X$$ is a random variable with a Poisson distribution." The parameter is $$\mu$$; $$\mu$$ = the mean for the interval of interest. The mean is the number of occurrences that occur on average during the interval period.
The formula for computing probabilities that are from a Poisson process is:
$p(x)=\frac{\mu^{x} e^{-\mu}}{x !}\nonumber$
where $$p(x)=P(X=x)$$ is the probability of $$x$$ occurrences, $$\mu$$ is the expected number of occurrences based upon historical data, $$e$$ is the natural logarithm approximately equal to 2.718.
In order to use the Poisson distribution, certain assumptions must hold. These are: the probability of an occurrence is unchanged within the interval, there cannot be simultaneous occurrences within the interval, and finally, that the probability of a occurrence among intervals is independent, the same assumption of the binomial distribution.
In a way, the Poisson distribution can be thought of as a clever way to convert a continuous random variable, usually time, into a discrete random variable by breaking up time into discrete independent intervals. This way of thinking about the Poisson helps us understand why it can be used to estimate the probability for the discrete random variable from the binomial distribution. The Poisson is asking for the probability of a number of successes during a period of time while the binomial is asking for the probability of a certain number of successes for a given number of trials.
##### Example $$\PageIndex{3}$$
Leah's answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $$\frac{1}{4}$$ hour.)
$$x = 0, 1, 2, 3$$, ...
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives
$$\left(\frac{1}{8}\right)$$(6)= 0.75 calls in 15 minutes, on average. So, $$\mu = 0.75$$ for this problem.
$$X \sim \text{Poisson} (0.75)$$
Find $$P (X > 1)$$:
$P(X > 1) = 1 - P(X=0) = 1 - \frac{0.75^{0} e^{-0.75}}{0 !} = 0.1734\nonumber$
Probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734.
The graph of $$X \sim \text{Poisson} (0.75)$$ is:
The $$y$$-axis contains the probability of $$X$$ = the number of calls in 15 minutes.
##### Example $$\PageIndex{4}$$
According to a survey a university professor gets, on average, 7 emails per day. Let X = the number of emails a professor receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: $$X\sim\text{Poisson}(7)$$. The mean is 7 emails.
1. What is the probability that an email user receives exactly 2 emails per day?
2. What is the probability that an email user receives at most 2 emails per day?
3. What is the standard deviation?
a. $$P(X=2)=p(2)=\frac{\mu^{x_{e}-\mu}}{x !}=\frac{7^{2} e^{-7}}{2 !}=0.022$$
b. $$P(X \leq 2)=p(0)+p(1)+p(2)=\frac{7^{0} e^{-7}}{0 !}+\frac{7^{1} e^{-7}}{1 !}+\frac{7^{2} e^{-7}}{2 !}=0.029$$
c. Standard Deviation = $$\sigma=\sqrt{\mu}=\sqrt{7} \approx 2.65$$
##### Example $$\PageIndex{5}$$
Text message users receive or send an average of 41.5 text messages per day.
1. How many text messages does a text message user receive or send per hour?
2. What is the probability that a text message user receives or sends two messages per hour?
3. What is the probability that a text message user receives or sends more than two messages per hour?
a.Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is $$\frac{41.5}{24}$$ ≈ 1.7292.
b.$$P(X=2)=p(2)=\frac{\mu^{x} e^{-\mu}}{x !}=\frac{1.729^{2} e^{-1.729}}{2 !}=0.265$$
c.$$P(>x2)=1-P(X \leq 2) = 1 - \left(p(0) + p(1) + p(2)\right)=1-\left[\frac{7^{0} e^{-7}}{0 !}+\frac{7^{1} e^{7}}{1 !}+\frac{7^{2} e^{-7}}{2 !}\right]=0.250$$
## Estimating the Binomial Distribution with the Poisson Distribution
The Poisson distribution can provide an approximation for the binomial. We say that the binomial distribution approaches the Poisson. The binomial distribution approaches the Poisson distribution is as n gets larger and p is small such that np becomes a constant value. There are several rules of thumb for when one can say they will use a Poisson to estimate a binomial. One suggests that np, the mean of the binomial, should be less than 25. Another author suggests that it should be less than 7. And another, noting that the mean and variance of the Poisson are both the same, suggests that np and npq, the mean and variance of the binomial, should be greater than 5. There is no one broadly accepted rule of thumb for when one can use the Poisson to estimate the binomial.
As we move through these probability distributions we are getting to more sophisticated distributions that, in a sense, contain the less sophisticated distributions within them. This proposition has been proven by mathematicians. This gets us to the highest level of sophistication in the next probability distribution which can be used as an approximation to all of those that we have discussed so far. This is the normal distribution.
##### Example $$\PageIndex{6}$$
On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
Let X = the number of days with low seismic activity.
Using the binomial distribution:
$P\left(X=10\right)=p(10)=\frac{200 !}{10 !(200-10) !} \times .0102^{10} \times .9898^{190}=0.000039\nonumber$
Using the Poisson distribution:
Calculate $$\mu = np = 200(0.0102) \approx 2.04$$
$P\left(X=10\right)=p(10)=\frac{\mu^{x} e^{-\mu}}{x !}=\frac{2.04^{10} e^{-2.04}}{10 !}=0.000045\nonumber$
We expect the approximation to be good because $$n$$ is large (greater than 20) and $$p$$ is small (less than 0.05). The results are close—both probabilities reported are almost 0.
##### Example $$\PageIndex{7}$$
A survey of 500 seniors in the Price Business School yields the following information. 75% go straight to work after graduation. 15% go on to work on their MBA. 9% stay to get a minor in another program. 1% go on to get a Master's in Finance.
What is the probability that more than 2 seniors go to graduate school for their Master's in finance?
This is clearly a binomial probability distribution problem. The choices are binary when we define the results as "Graduate School in Finance" versus "all other options." The random variable is discrete, and the events are, we could assume, independent. Solving as a binomial problem, we have:
Binomial Solution
$n\cdot p=500\cdot 0.01=5=\mu\nonumber$
$p(0)=\frac{500 !}{0 !(500-0) !} 0.01^{0}(1-0.01)^{500^{-0}}=0.00657\nonumber$
$p(1)=\frac{500 !}{1 !(500-1) !} 0.01^{1}(1-0.01)^{500}=0.03318\nonumber$
$p(2)=\frac{500 !}{2 !(500-2) !} 0.01^{2}(1-0.01)^{500^{2}}=0.08363\nonumber$
Adding all 3 together = 0.12339
$1−0.12339=0.87661\nonumber$
Poisson approximation
$n\cdot p=500\cdot 0.01=5=\mu\nonumber$
$n \cdot p \cdot(1-p)=500 \cdot 0.01 \cdot(0.99) \approx 5=\sigma^{2}=\mu\nonumber$
$p(x)=\frac{e^{-n p}(n p)^{x}}{x !}=\left\{p(0)=\frac{e^{-5} \cdot 5^{0}}{0 !}\right\}+\left\{p(1)=\frac{e^{-5} \cdot 5^{1}}{1 !}\right\}+\left\{p(2)=\frac{e^{-5} \cdot 5^{2}}{2 !}\right\}\nonumber$
$0.0067+0.0337+0.0842=0.1247\nonumber$
$1−0.1247=0.8753\nonumber$
An approximation that is off by 1 one thousandth is certainly an acceptable approximation.
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Two buildings face each other. The height of one building is greater than that of the other. The smaller building is 80 m high. The angles of depression of the top and the bottom of the smaller building from the top of the taller building are 45º and 60º respectively. Find the height of the taller building.
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Your query on height and distance is solved here.
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# Conservative Fields
## Prerequisites
Students should be able to:
• Find $d\vec{r}$ for a straight or curved path.
• Evaluate the dot product between two vectors qualitatively.
• Use chop-multiply-add ideas to construct an integral.
• Estimate integrals by approximation.
Reading: GVC § Independence of PathFinding potential Functions
## Homework for Static Fields
1. (PathIndependence) Students explicitly compute the work required to bring a charge from infinity using two different paths. Include part (e) as an additional path that cannot be solved by integration. (You must invoke path-independence of conservative fields.)
The gravitational field due to a spherical shell of mass is given by: %/* $\Vec g =\begin{cases} 0&r<b\\ -\frac{4}{3}\pi\rho\,G\left({r}-{b^3\over r^2}\right)\hat{r}&b<r<a\\ -\frac{4}{3}\pi\rho\, G\left({a^3-b^3\over r^2}\right)\hat{r}&a<r\\ \end{cases}$
where $b$ is the inside radius of the shell, $a$ is the outside radius of the shell, and $\rho$ is the constant mass density.
1. Using an explicit line integral, calculate the work required to bring a test mass, of mass $m_0$, from infinity to a point $P$, which is a distance $c$ (where $c>a$) from the center of the shell.
2. Using an explicit line integral, calculate the work required to bring the test mass along the same path, from infinity to the point $Q$ a distance $d$ (where $b<d<a$) from the center of the shell.
3. Using an explicit line integral, calculate the work required to bring the test mass along the same radial path from infinity all the way to the center of the shell.
4. Using an explicit line integral, calculate the work required to bring in the test mass along the path drawn below, to the point $P$ of question a. Compare the work to your answer from question a.
5. What is the work required to bring the test mass from infinity along the path drawn below to the point $P$ of question a. Explain your reasoning.
2. (MurderMystery) This question is adapted from the in-class activity
(Throughout this problem, assume all of the constants-including invisible factors of 1-carry the necessary dimensions so that the fields in this problem are dimensionally correct.) Consider the vector field in rectangular coordinates: $$\vec{E} = \frac{q}{4 \pi \epsilon_{0}} [(2 x y^3z+z)\hat{x} + (3x^2 y^2 z) \hat{y}+(x^2 y^3+x)\hat{z}]$$
1. Using only the $x$-component of $\vec{E}$, find as much information as possible about the potential from which this electric field might have come.
2. Repeat this exercise for the $y$- and $z$-components of $\vec{E}$. Does this field come from a potential?
3. Consider the different vector field: $$\vec{E} = \frac{q}{4 \pi \epsilon_{0}} (-y \hat{x} + x \hat{y})$$ Does this field come from a potential?
4. Consider the different vector field: $$\vec{E}=\frac{q}{4 \pi \epsilon_{0}} \left(s \hat{\phi}\right)$$ Does this field come from a potential?
1. I am going to turn the murder mystery method activity into a homework problem.
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2 초 512 MB120272128.767%
문제
Ilya loves dating! He is planning his dates for the next t days (these days are numbered from 1 to t). For each day x, he knows that he can plan at most ax dates.
Ilya has n known girls numbered from 1 to n. Girl i is willing to go on a date with him at most once, on any day in [li, ri], and Ilya will get pi pleasure for dating this girl. For each i < n, it is true that li ≤ li+1 and ri ≤ ri+1.
Ilya’s goal is to plan dates with some girls to maximize total pleasure. Help him! Find the maximum total pleasure he can get if he will properly choose the girls and the days of dates with them.
입력
The first line contains two integers n and t: the number of girls and the number of days (1 ≤ n, t ≤ 300 000).
The next line contains t integers a1, a2, . . . , at: here, ai is the maximum number of dates Ilya can plan on day i (0 ≤ ai ≤ n).
The next n lines contain the descriptions of the girls. The i-th of these lines contains three integers, li, ri, and pi: the borders of the i-th girl’s days segment and the pleasure Ilya will get for dating her (1 ≤ li ≤ ri ≤ t, 1 ≤ pi ≤ 109).
It is guaranteed that, for each i < n, it is true that li ≤ li+1 and ri ≤ ri+1.
출력
Print one integer: the maximum total pleasure Ilya can get if he will properly choose the girls and the days of dates with them.
예제 입력 1
3 5
0 1 0 1 0
1 2 2
2 4 1
3 5 5
예제 출력 1
7
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``` CHAPTER 2 & 6 – EXPONENTS, LOGS, GEOMETRIC SEQUENCES & SERIES
DAYS 1-2
Review Exponent Laws Examples of Each Law
1. xm xn = xm + n a3 a 7 a10
xm a3 1
2. n
x mn 7
a 37 a 4 4 (avoid negative exponents!)
x a a
3. a. (xm)n = xmn
b. (xmyn)c = xmcync a 3 4
a34 a12
c
xm x mc
c. n nc
y y
1 1
4. a. xa a 3
xa a3
1
b. a
xa
x
a
x
a
y ya
c. or a
y x x
5. a. x0 = 1
b. x = x1
6. a. (x)2 = (x)(x)
b. x2 = (1)(x)(x)
a
m 3
m 3
7. x n n x m or n
x a 4 4 a3 4
Page 1 M30P LP CH 2 & 6 Logs Exp
Examples: Simplify. (Positive Exponents!!)
27x3 (2 xy 3 ) 2 (3 x 3 y )3 3y 7
1. (3xy2)3 ANS: 2. ANS:
y6 (6 x 7 y ) 2 x3
3
6 x 5 y y3
3. 2
ANS: 4. 70 22 + 21 + 3x0 + (8x)0 ANS: 1.5
2 xy 27 x12
2 5
4 1
5. 31 27 3
ANS: 6. 81 4
ANS:
9 243
82 x 3 2 x 5
7. ANS: 23 x20
16 x 4
Assignment
Simplify (positive exponents only)
(3 xy )3 (4 x 3 y 5 ) 3y14
1. (2x3y)4 ANS: 16x12y4 2. ANS:
(6 x 4 y 3 ) 2 x2
3
3 2 x6 10 x 6 y 7 125x18
3. (2x y) ANS: 4. ANS:
4 y2 2y
9
y6
1
5. (8 x12 y 6 ) 3 ANS: 2x4y2 6. 80 + (7x)0 6x0 + (2)4 22 ANS: 8
3 5
1
7. 16 4 (8) 3 ANS: 24 8. ( 3 8) 5 ANS:
32
9 x 3 27 2 x 7
9. 25x 7 87 x ANS: 22x + 14 10. ANS: 34 x35
81x 2
3
20 x 2 y 6 1 ( xyz 2 )3 (2 x 3 z ) 4 2z 7
11. ANS: 12. ANS:
5 xy
4
64x3 y 6 ( 2 x 5 y 3 z ) 3 y6
Extra:
pg. 333 # 1 – 12
pg. 71 # 1 – 36
Page 2 M30P LP CH 2 & 6 Logs Exp
DAYS 1-2
A. Solving Exponential Equations
B. Solving Equations with Exponents Other Than 1
When the bases are the same, (or can be manipulated to be the same) and the exponent is in the variable,
we can solve by comparing exponents. Remember if ax = ay then x = y.
Examples:
43
1. 32x 7 = 3x + 2 ANS: 9 2. 93x 4 = 27x 17 ANS:
3
Bases are the same, exponents equal Get bases to be the same
2x 7 x 2 93 x4 27 x17
3
2 3 x 4
33
x 17
x9
3 3
6 x 8 3 x 51
6 x 8 3 x 51
3x 43
43
x
3
43 x 2 10 1
3. 1 ANS: 4. 5x 4 ANS: –7
8x2 3 125
43 x 2 8 x 2
1
2
2 3 x2
23 5x4
x2
53
6 x 4 3x 6 5 x 4 5 3
3x 10 x 4 3
10 x 7
x
3
2
5. x3 = 8 ANS: 2 6. x 5 4 ans : 32
2
What number to the power of 3 What number to the power of
5
equals 8? equals 4?
1 5
Or, x 8 3
Or, x 4 2
Page 3 M30P LP CH 2 & 6 Logs Exp
5
25 2
3 2
26
7. 2( x 1) 2
250 ANS: 8. 3x 11 14
5
ANS: or 200.47
25 3
3 2
( x 1) 2
125 divide by 2 3 x 5 25
2
2
x 1 125 3 25
x5
1 3
x 1 5
25 25 2
26 x 200.47
x 3
25
9. 8x+ 3 = 1 ANS: –3
8x+ 3 = 1
8x+ 3 = 80 Anything to the power of zero is 1
x+ 3= 0
x= - 3
Assignment
pg. 89 # 1, 3, 6, 9, 10, 11, 12
plus:
1. Solve for x: (* nearest hundredth)
4 1
1
a. x 16
3
ANS: 8 b. 2 x 2
10 ANS:
25
2 5
c. (2 x 3) 3 9 ANS: 15 *d. 7 x 6 1 57 ANS: 12.13
3 2
1
*e. (7 x 2) 5 ANS: 2.375
f. 3x 3
7 19 ANS: or 0.125
8
Page 4 M30P LP CH 2 & 6 Logs Exp
DAYS 1-2
Review
1. Simplify
a a 2 x 3 y
=
a a
2 x 3 y
a a
2x 3y
a 2 x 3 y
a3 y
a a a a a a
x y x y x y x y x y x y
a2 x
m a b mb c m a b mb c m a 2b c
2. Simplify a c
= a c a c m 2 a 2 b
m m m
2 x 5 2
4x 5 x
2
3. Solve for x
2
x 5 2
4x 5 x
2
22 x 10 22
x 2 5 x
22 x 10 22 x 10 x
2
2 x 10 2 x 2 10 x
0 2 x 2 12 x 10
0 x2 6x 5
0 ( x 5)( x 1)
x 5,1
2
27 3
64
9
4. Solve 1
In calculator (27 / 64) ^ (2 / 3) /(125 ^ (1/ 3))
3
80
125
Exponential Function – this is a short lesson. Feel free to start the day with some review type
questions.
y = ax, a > 0
2 Basic Situations
1. Start low and the shoot up. 2. Start high and shoot down.
Page 5 M30P LP CH 2 & 6 Logs Exp
Examples:
I. Graph and find:
1. x–int
2. y–int
3. horizontal asymptote (what is non–permissible value for y?) What y value is never reached?
4. domain (always x R ) An exponent can be positive or negative.
5. range
a. y = 2x b. y = 5x
1. none 1. none
2. (0, 1) 2. (0, 1)
3. y=0 3. y=0
4. xR 4. xR
5. y>0 5. y>0
c. y = 2x 3 d. y = 2x + 3 Need x 3 in brackets!
1. (1.58, 0) 1. none
2. (0, 2) 2. (0, 8)
3. y = 3 3. y=0
4. xR 4. xR
5. y > 3 5. y<0
x
1
e. y = 5x + 1 or y = + 1 f. y = 3x 1 27
5
1. none 1. (4, 0)
80
2. (0, 2) 2. (0, )
3
3. y = 1 3. y = 27
4. x R 4. x R
5. y > 1 5. y > 27
g. y = 3 2x Note: is the same as
1. none
2. (0, 3)
3. y=0
4. xR
5. y>0
Page 6 M30P LP CH 2 & 6 Logs Exp
Assignment
Graph and find:
1. x–int
2. y–int
3. horizontal asymptote
4. domain
5. range
1. 2. 3. 4. 5.
A. y = 2x + 5
B. y = 3x 2 9
C. y = 5x + 1
x
1
D. y 27
3
E. y = 2x 2 2
F. y = 5 2x
y = ax yY bx y = cx y = dx
1
Plus: = We know the values of a, b, c, and d to be 2, 3, ,
10 3
9 and 6 (but not in this order). Which is which?
8
7
6
5
4
3
2
1
X
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-1 1 2 3 4 5 6 7 8 9 10
1
-2 ANS: a = , b = 6, c = 3, d = 2
-3 3
-4
Plus pg. 76 # 2 – 6 -5
-6
-7
-8
-9
1.
-10 2. 3. 4. 5.
x
A. y = 2 + 5 none (0, 6) y=5 xR y>5
x2
B. y = 3 9 8 y = 9
Created with an unregistered version of Advanced Grapher - http:/ / www.serpik.com/ agrapher/
(4, 0) xR y > 9
(0, 8 )
9
C. y = 5 + 1x
(0, 0) (0, 0) y=1 xR y<1
1
x
(3, 0) (0, 26) y = 27 xR y > 27
D. y 27
3
E. y = 2x 2 2 (3, 0) 3 y = 2 xR y > –2
(0, 1 )
4
F. y = 5 2 x
none (0, 5) y=0 xR y<0
Page 7 M30P LP CH 2 & 6 Logs Exp
DAY 3 Short Lesson (Start Day 4 as Well)
Determining Equation of Exponential Function
(Algebraically & Using the Regression Function on Calculator)
Regression Function
curve of best fit
We will find the exponential function equation in the form y = abx. Most problems can be solved using
this form and the calculator uses this form only.
Example:
1. Exponential function of the form y = abx contains the points (1, 10) and (3, 250). Find a & b.
250 ab3
10 ab
25 b 2
5 b (b 0) 10 a(5), 2 a
y = 2(5)x
* This can be done using the graphing calculator: Hit Stat then hit Edit on your calculator
1. clear lists L1 and L2
2. enter x values in L1 and y values in L2
3. you can graph the points using stat plot (you may have to change the window)
4. go to stat, then calc, then exponential regression (you do not need to enter L1 and L2 as they are
defaulted)
5. if r2 and r do not appear, you have to turn diagnostic on by hitting 2nd 0 (catalog) and selecting
Diagnostic On (enter must be hit twice)
* The r value gives a coefficient correlation which shows how well the equation represents the data (1 is
a perfect correlation)
Page 8 M30P LP CH 2 & 6 Logs Exp
1.
The population of an ant colony was studied over time.
Day Population of
Colony
0 98
1 268
2 681
3 1411
4 2622
a. What is the exponential regression equation that best fits the data?
ANS: y 111.92 2.28x
b. What would you expect the ant population to be on day 15?
ANS: y 111.92 2.2815 26 177 635
2. Use an exponential regression equation for the data below.
Wolf What is the exponential regression equation?
Year
Population
0 500
1 548
2 800 ANS: y 458.79 1.34 x
3 1 168
How many wolves would you expect in year 5?
ANS: y 458.79 1.345 1982
Page 9 M30P LP CH 2 & 6 Logs Exp
Assignment
1. Determine the exponential function given points (1, 6) and (4, 48) on the function. Do both ways.
ANS: y = 3 2x
2. Tuition Fees. Use the data below.
a. Determine the exponential function for each using regression function.
ANS: Educ. y = 873.80 1.12x
Eng. y = 1077.34 1.20x
b. Estimate tuition fees in 2005 for Education. ANS: \$4783
Education Engineering
1991 (1) \$990 \$1300
1992 (2) \$1094 \$1539
1993 (3) \$1223 \$1865
1994 (4) \$1358 \$2251
1995 (5) \$1574 \$2681
3.
Joe invested \$5000 in an account and tracked the growth.
Year (L1) 0 1 2 3 4 5
Value (L2) \$5000 \$5350 \$5725 \$6125 \$6554 \$7013
a. Write the exponential regression equation that best models this data.
ANS: y 5000 1.07 x
b. How much would he have after 10 years?
ANS: y 5000 1.0710 \$9835.76
Page 10 M30P LP CH 2 & 6 Logs Exp
DAY 4
Logarithms (Critical Day = Must Know Cold)
Log Form Exponential Form
y log b x by x
* b > 0, b 1 (will be explained when we get to functions)
Common Logs – base of 10 (log 5; base 10 is understood)
We write log 5 instead of writing log10 5 . When no base is written, 10 is assumed.
Natural Logs – base e (ln 5; e 2.71828 )
We write ln 5 instead of writing ln e 5 . When no base is written, e is assumed.
Examples:
1. Change to exponential form.
a. logmp = r ANS: mr = p b. log232 = 5 ANS: 25 = 32
c. logm10 + 2 = x ANS: mx 2 = 10 d. 7 = log3(4x 5) ANS: 37 = 4x 5
Rewrite as
log m 10 x 2
2. Change to log form.
a. 34 = 81 ANS: log 3 81 4 b. mc = d ANS: log m d c
c. (x + m)3 = a + b ANS: logx + m(a + b) = 3
3. Evaluating logarithms.
5
a. log381 = x ANS: 4 b. log432 ANS:
2
3x = 81 4 x 32
3x = 34 x=4 2
2 x
25
5
2x 5 x
2
Page 11 M30P LP CH 2 & 6 Logs Exp
1
c. log 2 ANS: –6 d. log 5 25 ANS: 4
64
x
2x
1 5 25
64 x4
1
2x
26
2 x 26 x 6
5
e. log3 9 3 ANS: f. log1111 ANS: 1
2
11x 11
3x 9 3
x 1
1
3x 32 32
5
5
3 3
x 2
x
2
g. log51 ANS: 0 h. log m m n ANS: n
5x 1 mx mn
x0 anything to power 0 xn
equals 1
5 7
10 7
i. log a7 ANS: j. log 4 m10 ANS:
a3 21 8
m5
x 5
a3 a 7
4x 7
3x 5
m m10
5
a a
2 7
4x 7
3x 5
5 10
2 7 40 x 35
21x 10
35 7
10 x
x 40 8
21
Page 12 M30P LP CH 2 & 6 Logs Exp
Assignment for Day 4
pg. 98 – 99 # 1, 4, 7, 10, 13, 16, 19, 21, 23, 24, 25, 27, 28, 29, 31, 32
plus:
11 10
33 50
1. log 5 a 2 ANS: 2. log 5 a a 7 ANS:
a3 10 7
10
x
5
a a7
5x 11
a a23 1x 10
a a 5 7
5 x 11
x 10
3 2
10 x 33 5 7
7 x 50
33
x 50
10 x
7
Worksheet B 1- 5 is good for review as well.
Page 13 M30P LP CH 2 & 6 Logs Exp
DAY 5 (Shorter Lesson) Can Start Laws of Logs Day 6
Solving Basic Log Equations
Determine Exact:
3 2
1. log168 = x ANS: 2. log x 9 ANS: 27
4 3
2
16 8
x x3 9
3
2 4 x
23 x 92 If you get 364.5, you typed
2 4 x 23 x 27
4x 3
3 9 ^ 3/ 2 not 9 ^ (3/ 2)
x
4
3 1 4
3. log 25 x ANS: 4. log 8
x ANS: 4
2 125 3
4
3
83 x
25 2
x
4
x 0.008
1 1 3 4
8 x 8 4
2 6
125
Don’t type 25 ^ 3 / 2
Solve for x 0.01 :
5
1. log7x = 1.73 ANS: 28.97 2. log x 294 ANS: 30.27
3
5
71.73
x x 294
3
3
x = 28.97 x 294 30.27 5
Page 14 M30P LP CH 2 & 6 Logs Exp
2 2
3. log (2 x 7) 29 ANS: 74.58 4. 3 log 2x 3 5 ANS: 11 264.30
3
2
2 103 2 x 3 5
2x 7 3 29 2
3
1000 2 x 3 5
2 x 7 29 2
2
2 x 7 156.17 1005 2 x 3
2 x 149.17 2
502.5 x 3
x 74.58 3
x 502.5 2 11264.30
Page 15 M30P LP CH 2 & 6 Logs Exp
Assignment for day 5
pg. 99 #33 – 40
plus: Solve for x 0.01
3
1. log11x = 0.43 ANS: 2.80 2. log x 4 ANS: 25.40
7
3
11 0.43
x x 4
7
x 2.80 7
4 x 25.40
3
5
3. log (2 x3) 16 ANS: 4.14 4. log 3.4 (7 x 11) 1.9 ANS: 3.03
3
5
2 x 3 3 16
3.41.9 7 x 11
3
2 x 3 16 5
10.23 7 x 11
2 x 3 5.28
21.23 7 x
2 x 8.28
x 3.03
x 4.14
2
5. log 6 (3 x 5
3) 1.3 ANS: 0.11
2
61.3 3 x 5 3
2
10.27 3 x 5 3
2
7.27 3 x 5
2
7.27
x5
3
2
2.42 x 5
5
2.42 2 x
x 0.11
Page 16 M30P LP CH 2 & 6 Logs Exp
DAY 6
Laws of Logarithms Formulas 1–4 are on formula sheet.
M
1. loga M N loga M loga N 2. log a log a M log a N
N
log a c
3. log a M n n log a M 4. log b c
log a b
5. log x x 1 * not on formula sheet
6. log x x k k * not on formula sheet (This is based on #3 and #6)
7. M logM x x ( M log M x y, log M y log M x, y x) * not on formula sheet
Examples:
1. Express as a single logarithm and evaluate whenever possible.
a. log 250 + log 4 ANS: 3 b. log354 log32 ANS: 3
log 250 4 log 3 54 2
log1000 log10 1000 log 3 27
10 1000
x
3x 27
x3 x3
ab
c. logx5x3 + logx3x7 logx15x ANS: 9 d. log a + log b log c ANS: log
c
5 x3 3x 7
log x No work needed. Rules of logs.
15 x
15 x10
log x
15 x
log x x 9
x a x9
a9
ab
e. log a log c + log b ANS: log
c
No work needed. Rules of logs.
Page 17 M30P LP CH 2 & 6 Logs Exp
1 p
f. log p 2 log m 3log r ANS: log 2 3
2 m r
1
log p log m 2 log r 3
2
log p log m 2 log r 3
p
log 2 3
m r
g. 2log 3x + 3log 2x2 ANS: log (72x8)
log(3 x) 2 log(2 x 2 )3
log(9 x 2 ) log(8 x 6 )
log 72 x8
2. Expand the following.
mx 2
a. log 3 ANS: log3 m 2 log3 x log3 p
p
3
x2 y5 2 5
b. log p ANS: log p x log p y 3log p m
m3 3 3
2 5
3
x2 y5 x3 y3 2 5
log p 3
log p 3
log p x log p y 3log p m
m m 3 3
3. Given log3 x 5 and log3 y 8 , evaluate:
9x2
a. log 3 xy ANS: 13 b. log 3 ANS: 4
y
log 3 x log 3 y log3 9 x 2 log3 y
5 8 13 log 3 (3 x) 2 log 3 y
2 log 3 (3 x) log 3 y
2 log 3 3 log 3 x log 3 y
2 1 5 8 2 6 8 4
Page 18 M30P LP CH 2 & 6 Logs Exp
y
c. log 3 ANS: 3
3
1
log 3 y 2 log 3 3
1
log 3 y log 3 3
2
1
8 1 4 1 3
2
4. Given log54 = x and log53 = y, express in terms of x and/or y.
a. log 5 12 ANS: x + y b. log 5 45 ANS: 2y + 1
log 5 4 log 5 3 log 5 3 log 5 3 log 5 5
x y y y 1 2 y 1
4 2
c. log 5 ANS: x 2y d. log5 3 32 ANS: y
9 3
2
log 5 4 log 5 9 log 5 3 3
log 5 4 log 5 3 log 5 3 2
log 5 3
x y y 3
x 2y 2
y
3
1
e. log52 ANS: x
2
log 5 4
1
log 5 4 2
1
log 5 4
2
1
x
2
Page 19 M30P LP CH 2 & 6 Logs Exp
5. Simplify.
a. 3log3 7 ANS: 7 b. mlogm y ANS: y
Rule #5 Rule #5
c. 4log2 5 ANS: 25 d. x3log x 2 y ANS: 8y3
22
log x 2 y
3
log 2 5
x
log x 8 y 3
22log 2 5 x
log 2 52 8 y3
2
2 log 2 25
25
e. 7log7 3log7 y ANS: 3y
7 log7 (3 y )
7 log7 (3 y )
3y
Assignment
pg. 106 # 1, 3, 7, 11, 15, 18, 19, 22, 25, 26, 29, 30, 32, 33, 53 a – c
plus:
# I of Worksheet A
1. m2logm 5 ANS: 25 2. 9log3 2 x ANS: 4x2
m log m (52 ) 3
2 log 3 2 x
log 3 2 x
2
m log m 25 32log3 2 x 3
25
log 3 4 x 2
3 4x2
3. 5log5 x 2log5 y ANS: xy2
log5 x log5 y 2
5
log5 xy 2
5
xy 2
Page 20 M30P LP CH 2 & 6 Logs Exp
DAY 7
Solving Complex Log Equations Using Properties
* Either Log = Number (or algebraic expression) or Log = Log
Examples:
1. log3 x log3 (2 x 10) log 3 3 ANS: 6
log 3 x log 3 6 x 30
x 6 x 30
30 5 x
x6
2. 1 log3 5x 6 log3 x 2 ANS: 6
1 log 3 5 x 6 log 3 x 2
log 3 3 log 3 5 x 6 log 3 x 2
5x 6 5x 6
1 log 3
3 x2
x2
3x 6 5 x 6 5x 6
or 31
12 2 x x2
3x 6 5 x 6
x6
12 2 x
x6
13
3. log2 3x 1 log2 x 1 6 ANS: ,5
3
log 2 3x 1 log 2 x 1 6 log 2 2
log 2 3x 1 log 2 x 1 log 2 26
log 2 3x 1 log 2 x 1 log 2 64
3x 1 x 1 64
3x 2 2 x 1 64
b b 2 4ac
3x 2 x 65 0
2
use quadratic formula x
2a
13
x ,5 can ' t have negative answer in logs
3
Different Technique below
Page 21 M30P LP CH 2 & 6 Logs Exp
log 2 3 x 1 log 2 x 1 6
log 2 3x 1 x 1 6
26 3 x 1 x 1
64 3 x 2 2 x 1
b b 2 4ac
3x 2 2 x 65 0 use quadratic formula x
2a
13
x ,5 can ' t have negative answer in logs
3
4. 3log 2 x 4 log 2 x 21 ANS: 8
log 2 x3 log 2 x 4 21log 2 x
log 2 x log 2 x
3 4
log 2
2
21
7 log 2 x = 21
or
x 7 2097152 log 2 x = 3 divide by 7
3
1 2 = x= 8
x 2097152 7 8
5. 3 logx 2x 3 logx (x 6)
3log x x log x 2 x 3 log x x 6 3 log x 2 x 3 log x x 6
log x x3 log x 2 x 3 x 6
3 log x 2 x 3 x 6
x 3 2 x 2 9 x 18 or x3 2 x 2 9 x 18
x 3 2 x 2 9 x 18 0 x3 2 x 2 9 x 18 0
x 3 , 2,3 x 3 , 2,3
log3 x 5log3 x 6 0
2
6. (use substitution) ANS: 9 and 27
Let y log3 x
y2 5 y 6 0
y 3 y 2 0
log 3 x 3 0 log 3 x 2 0
log 3 x 3 log 3 x 2
3 x
3
32 x
27 x 9x
Assignment
pg. 113 – 114 # 21, 23, 25, 40 a, c, e, f, h
Could also use Worksheet A (assign all)
Page 22 M30P LP CH 2 & 6 Logs Exp
DAY 8
Solving Exponential Equations Using Logs
Method I: Take log of each side using any base (we always use base 10 because it is on the
calculator)
Method II: Change from exponential form to log form. You can use this method when only one
base has a variable for an exponent.
i.e. only one base has a variable exponent. Questions 1, 4, 5
Examples:
1. 2x = 19 Method I log 2x = log 19 Method II 2x = 19
xlog 2 = log 19 log219 = x
log19 log19
x= =x
log 2 log 2
x = 4.25 4.25 = x
4 log 7 2 log 5 log 5 2 log 7
2. 53x 2 = 7x + 4 ANS: 3. 5 3x = 7x 2 ANS:
3log 5 log 7 log 7 log 3
or 3.82 or 6.49
log 53 x 2 log 7 x 4 log 5 log 3x log 7 x 2
3x 2 log 5 x 4 log 7 log 5 x log 3 x 2 log 7
3x log 5 2 log 5 x log 7 4 log 7 log 5 x log 3 x log 7 2 log 7
3x log 5 x log 7 4 log 7 2 log 5 log 5 2 log 7 x log 7 x log 3
x 3log 5 log 7 4 log 7 2 log 5 log 5 2 log 7 x log 7 log 3
4 log 7 2 log 5 log 5 2 log 7
x 3.82 x 6.49
3log 5 log 7 log 7 log 3
5x
2 log 3 29
4. 3 2 29 ANS: x = or 1.23
5
5x
log 3 2 log 29
5x
log 3 log 29
2
5 x log 29
2 log 3
2 log 29 2 log 3 29
x 1.23
5log 3 5
Page 23 M30P LP CH 2 & 6 Logs Exp
5
5
5. 13 7 2x
ANS: x = or 1.90
2 log 7 13
5
log13 log 7 2 x
5
log13 log 7
2x
log13 5
log 7 2 x
2 x log13 5log 7
5log 7 5log13 7
x 1.90
2 log13 2
Assignment
pg. 113 # 1, 2, 4, 9, 11, 13, 17, 20
plus: questions from worksheet A or B
Page 24 M30P LP CH 2 & 6 Logs Exp
DAY 9 Can do questions a, b and c in 30 minutes. Great quiz day.
Graphing Log Functions
Consider the equation y log b x where b > 0 and b 1
It could also be written as b y x
It is the inverse of b x y , reflected over line y = x
Examples:
1. Find domain, range, x & y–intercepts, equation of asymptote and sketch for:
a. y = log2x (2y = x) b. y = log2(x + 4) + 2
D: x > 0 D: x > 4
R: y R R: y R
x–int: (1, 0) x–int: (–3.75, 0)
y–int: none y–int: (0, 4)
asymptote: x = 0 asymptote: x = 4
Graph in calc as Graph in calc as
y1 log( x) / log(2) y1 log( x 4) / log(2) 2
Domain: although the graph appears to start at (0, –2), it actually Domain: although the graph appears to start at (–4, –1), it
shoots down the y axis and allows y to get incredibly small. actually shoots down x = –4 and allows y to get incredibly
Thus D: x > 0. small. Thus D: x > –4.
Range: The graph shoots down for every y and thus R: y R . Range: The graph shoots down for every y and thus R: y R .
x – intercept: solve for y = 0 x – intercept: solve for y = 0
y log 2 x 4 2
y log 2 x 0 log 2 x 4 2
0 log 2 x 2 log 2 x 4
20 x 1 1
22 x 4 x 4 3.75
4
Asymptote: since x can not be zero, the vertical y–intercept: solve for x = 0
asymptote is given by x = 0 y log2 (0 4) 2
y log2 4 2 3
Asymptote: since x can not be negative 4, the vertical
asymptote is given by x = –4
Page 25 M30P LP CH 2 & 6 Logs Exp
c. y log3 x 9 d. y log4 8 x 1
D: x> 9 D: x < 8
R: y R R. y R
x–int: (8, 0) x–int: (4, 0)
y–int: (0, 2) y–int: (0.5, 0)
asymptote: x = 9 asymptote: x = 8
Graph in calc as Graph in calc as
y1 log( x 9) / log(3) y1 log(8 x) / log(4) 1
Domain: although the graph appears to start at (–9, 1), it actually Domain: although the graph appears to start at (8, –2), it
shoots up x = –9 and allows y to get incredibly big. Thus actually shoots down x = 8 and allows y to get incredibly
D: x > –9. small. Thus D: x < 8.
Range: The graph shoots down for every y and thus R: y R . Range: The graph shoots down for every y and thus R: y R .
x – intercept: solve for y = 0 x – intercept: solve for y = 0
y log 3 x 9
y log 4 8 x 1
0 log 3 x 9
0 log 4 8 x 1
0 log 3 x 9
1 log 4 8 x
30 x 9
41 8 x x 84 4
1 x9 x 8
y–intercept: solve for x = 0 y–intercept: solve for x = 0
y log3 0 9 y log 4 8 0 1
y log3 9 2 y log 4 8 1 0.5
Asymptote: since x can not be –9, the vertical Asymptote: since x can not be 8, the vertical
asymptote is given by x = –9 asymptote is given by x = 8
Page 26 M30P LP CH 2 & 6 Logs Exp
Day 9 Assignment
Domain Range x–int y–int asymptote
y = log3(x 2)
y = log5x 1
y = log2(x + 8) 1
y = log3(9 x)
also known as
y = log3((x 9)
Domain Range x–int y–int asymptote
y = log3(x 2) x>2 yR (3, 0) none x=2
y = log5x 1 x>0 yR (5, 0) none x=0
y = log2(x + 8) 1 x > 8 yR (6, 0) (0, 2) x = 8
y = log3(9 x)
also known as x<9 yR (8, 0) (0, 2) x=9
y = log3((x 9)
Page 27 M30P LP CH 2 & 6 Logs Exp
DAY 10
Application of Logarithms
A. Log Functions
1. Richter Scale a. The Richter scale reading of earthquake A is 6.8
and of earthquake B is 7.9. How many times as
I
R = log10I or (R = log10 , use I o 1) intense is B compared to A? ANS: 12.59
Io
R = Richter scale reading 6.8 log I
I = actual intensity 106.8 I 6309573.445
7.9 log I
107.9 I 79432823.47
107.9
6.8
101.1 12.59
10
or
79432823.47
12.59
6309573.445
b. Earthquake A has a Richter scale rating of 6.8. Earthquake C was 50 times as intense. What was
earthquakes C Richter scale rating?
6.8 log I 106.8 50 IC 315478672.2
106.8 I A
R log10 I
R log10 106.8 or R log10 315478672.2
R 8.5 R 8.5
Page 28 M30P LP CH 2 & 6 Logs Exp
2. Sound a. The decibel reading of sound A is 79.8. Sound B
is 250 times as intense. What is its decibel
dB = 10log L
dB = decibel reading 79.8 10log L
L = actual loudness 7.98 log L
107.98 L 95499258.6
3. The sound of the final explosion was heard
over 4500km away and covered 1/13th of the
Earth's surface. 107.98 250 2.37 1010
5. The final death toll from pyroclastic flows,
dB 10 log 107.98 250
volcanic bombs, and tsunamis was calculated
to be a devastating 36,417.
dB 103.78
KRAKATOA VOLCANO ERUPTION-1883 A.D., CRACKED ONE FOOT THICK CONCRETE
AT 300 MILES, CREATED A 3000 FOOT TIDAL WAVE, SOUND PRESSURE CAUSED BAROMETERS TO
FLUCTUATE WILDLY AT 100 MILES
INDICATING LEVELS OF AT LEAST 170-190 DB (P) AT THIS DISTANCE OF 100 MILES
EVEN WHEN SHOUTING IN SOMEONES EAR, COULD NOT BE HEARD AT 100 MILES
CAUSED FOG TO APPEAR AND DISAPPEAR INSTANTLY AT HUNDREDS OF MILES
ROCKS WERE THROWN TO A HIEGHT OF 34 MILES. DUST AND DEBRIS FELL CONTINOUSLY FOR 10
DAYS AFTER BLAST. PRODUCED VERY COLORFUL SUNSETS FOR ONE YEAR, EJECTED 4 CUBIC MILES OF
THE EARTH. CREATED ANTI-NODE OF NEGATIVE PRESSURE AT THE EXACT OPPOSITE SIDE OF THE
EARTH.SOUND COVERED 1 / 10 OF THE WORLDS SURFACE, SHOCK (SOUND) WAVES “ECHOED” AROUND THE
EARTH 36 TIMES AND LASTED FOR ABOUT A MONTH!
3. Welding Glasses a. What shade number would be required so that
approximately 25% of the light would pass
through the glasses? ANS: 6
20 log T
Sn = 2
3
20log 0.25
Sn = shade number Sn 2
3
T = fraction of light transmitted through Sn 6
glass, if 20%, then T = 0.2
b. If you have shade number 9, what percent of
of the light is transmitted through the glasses?
ANS: 8.9%
20 log T
9 2
3
20 log T
7
3
21 20 log T
21
log T
20
21
10 20
T 0.089 8.9%
Page 29 M30P LP CH 2 & 6 Logs Exp
B. Using Logs to Solve Exponential Functions
1. Exponential Growth
t
A(t ) A0 (2) d
A(t) = resulting number
A0 = original number
t = total time
d = doubling time
Example:
a. Bacteria double every 4 hours. How long before you have 3000 bacteria if you start with 20
bacteria?
t
3000 20(2) 4
t
150 2 4
t
log150 log 2
4
4 log150 t log 2
4 log150
t
log 2
t = 28.92
2. Exponential Decay
t
1 h
A(t ) A0
2
A(t) = resulting mass
A0 = original mass
t = total time
h = half life
Page 30 M30P LP CH 2 & 6 Logs Exp
Example:
a. 60g of radioactive material decays to 20g in 40 years. Find its half life.
40
1h
20 60
2
40
1 1h
3 2
1 40 1
log log
3 h 2
1 1
h log 40 log
3 2
1
40 log
h 2
1
log
3
h 25.24
Assignment
pg. 99 # 46 a, d (do not use trial and error), 50 a
pg. 107 # 48 a, b
pg. 114 # 43, 44 a, b
plus:
20 log T
1. Sn = 2
3
a. Find the shade number if 3% of light passes through. (nearest whole number) ANS: 12
20log 0.03
Sn 2 12
3
b. Find the percent of light that passes through if the shade number is 15. ( 0.01)
ANS: 1.12%
20 log T
15 2
3
20 log T
13
3
39 20 log T
39
log T
20
39
10 20 T 0.012 1.12%
Page 31 M30P LP CH 2 & 6 Logs Exp
DAY 11
Geometric Sequence
Notation
t5 = 5th term
t1 = a = 1st term
tn = general term
S5 = sum of 1st 5 terms
S1 = t1 = a
Sn = sum of 1st n terms
Geometric Sequence
ratio of consecutive terms is a constant (r = common ratio)
each term multiplied by a common ratio to get the next term
o 5, 10, 20, 40… each term is multiplied by 2.
General form
tn = arn 1
Develop the Formula:
2, 6, 18, 54, ...
2, 2 3, 2 3 3, 2333
t1 t2 t3 t4
therefore, tn ar n 1 tn 2(3) n 1
Examples:
1. 4, 8, 16, ..., Find:
a. t9 ANS: 1024 b. tn ANS: tn = 2n + 1
t9 4(2)91 1024 tn 4(2) n 1
tn 22 2n 1
tn 2n 1
c. n for 8 388 608 ANS: 22
8388608 2n 1
log 8388608 log 2n 1
8388608 2n 1
log 8388608 (n 1) log 2
log 2 8388608 n 1
or log 8388608
23 n 1 n 1
log 2
n 22
23 n 1
n 22
Page 32 M30P LP CH 2 & 6 Logs Exp
5 2
2. , ,... is a geometric sequence. Find:
8 3
16
a. r (exact) ANS: b. t25 (0.0001) ANS: 2.9415
15
2
251
t2 16 5 16
r 3 t25 2.9415
t1 5 15 8 15
8
3. 0.6, 0.72, ... is a geometric sequence. Find:
6
a. r ANS: or 1.2 b. smallest value of n so that the term is 57
5
ANS: 25.977, so n = 26
t2 0.72 6
r 1.2 tn ar n 1
t1 0.6 5
57 0.6(1.2)n 1
95 1.2n 1
log1.2 95 n 1
24.977 n 1
25.977 n 26
4. t3 = 12, t7 = 192. Find a & r. (substitution or elimination) ANS: 3 and 2 or 3 and 2
12 ar 31 192 ar 7 1
12 ar 2 192 ar 6
12 192
a a
r2 r6
t3 12
t2 6
12 192 r 2
6
r2 r t 6
a 2 3
12r 192r 2
6
r 2
r 6 192
r 2 12
r 4 16
r 2
Page 33 M30P LP CH 2 & 6 Logs Exp
Geometric Means
term(s) between 2 terms of a geometric sequence so that together they form a geometric sequence
Examples:
1. Find 2 geometric means between 5 & 135. ANS: 15, 45
5, ______, _______, -135
135 5r 41
135 5r 3
5 3 15 153 45
27 r 3
r 3
2. Find the missing terms: ____, 4, ____, 8 ANS: 2 2, 4 2 or 2 2, 4 2
tn ar n 1
4 2 4 2, 4 2
8 4r 31
4 4 4 2 4 2
2 r2 4 2 , , 2 2, 2 2
2 2 2 2
r 2
Example:
1. Find x so that x 1, x + 1, and 2x 1 form a geometric sequence. ANS: 0, 5
t 2 t3
t1 t2
x 1 2x 1
x 1 x 1
2 x 1 x 1 x 1 x 1
2 x 2 3x 1 x 2 2 x 1
x2 5x 0
x( x 5) 0
x 0,5
Page 34 M30P LP CH 2 & 6 Logs Exp
Assignment for day 11
pg. 300 – 301 # 1, 3, 7, 8, 12, 17, 20, 23 a, c, 30 a
plus:
1. 3, 3.6, ... is a geometric sequence. Find:
a. t52 (0.01) ANS: 32 761.58 b. smallest value of n so that the term is at least 200
ANS: n = 25
t 3.6 200 3(1.2) n 1
r 2 1.2
t1 3 200
1.2n 1
tn ar n 1 3
200
t52 3(1.2)51 log1.2 n 1
3
t52 32761.58
n 24.03 25
1
2. x 2, x + 3, 3x 1 form a geometric sequence. Find x. ANS: , 7
2
t 2 t3
t1 t2
x 3 3x 1
x2 x3
3x 1 x 2 x 3 x 3
3x 2 7 x 2 x 2 6 x 9
2 x 2 13x 7 0
1
x ,7
2
Page 35 M30P LP CH 2 & 6 Logs Exp
DAY 12 (Less than a day)
Application of Geometric Sequence
Compound Interest
A P 1 i
n
i = interest rate per compounding period
n = # of compounding periods
Example:
A P r compounding time
period
1. ? \$5000 10%/a quarterly 3 years
2. \$9000 \$3000 9.3%/a monthly ?
3. 5x x ? semi–annually 18 years
A P r compounding time
period
1. (\$6724.44)
2. (143 months)
3. (9.1%)
Work
1.
0.10
i 0.025 A P(1 i ) n
4
n 3 4 12 A 5000(1 0.025)12
A \$6724.44
2.
0.093
i 0.00775 A P(1 i ) n
12
9000 3000(1 0.00775) n
3 1.00775n
log1.00775 3 n
n 142.3 months 11.9 years
3.
n 18 2 36 A P(1 i ) n
5 x x(1 i)36
5 (1 i)36 4.57% semi-annual = 9.1% annually
1
5 1 i
36
1.0457 1 i 0.0457 4.57%
Page 36 M30P LP CH 2 & 6 Logs Exp
Appreciation & Depreciation
Examples:
1. A building increases in value by 7% per year. Find its value in 15 years if current value is
\$50 000. ANS: 137 951.58
t16 = 50 000(1 + 0.07)16 1 or use A = 50 000(1 + 0.07)15
2. A building decreases in value by 9% per year. Find its value in 12 years if current value is
\$38 000. ANS: \$12 254.07
t13 = 38 000(1 – 0.09)13 1 or use A = 38 000(1 – 0.09)12
3. Water purifier removes 5% of impurities every hour.
a. What % is removed after 6 hours? ANS: A = 1(1 0.05)6;
A = 0.735; 26.5% is removed
A P (1 i ) n
A 100(1 0.05) 6
A 73.51%
100 73.51 26.49%
b. How long to remove 75% of impurities? ANS: 27 hours
A P (1 i ) n
25 100(1 0.05) n
0.25 0.95n
log 0.95 0.25 n
n 27 hours
Assignment
pg. 304 # 1, 6, 13, 15, 20 or Logs Worksheet E Applications
plus: Solve for the unknown variables
A P r compounding time
period
1. \$20 000 \$8000 ? semi–annually 7y
2. 3x x 8.9%/a monthly ?
ANSWERS 1. r 13.5% 2. time 149 months
3. Purifier removes 3% every hour. How long to remove 75% of impurities (0.1)? ANS: 45.5 h
4. Land valued at \$43 000 is projected to increase by 9%/a. Find its estimated value in 11 years.
(nearest thousand) ANS: \$111 000
Page 37 M30P LP CH 2 & 6 Logs Exp
DAY 13
Geometric Series Adding of geometric sequences
a(r n 1)
Sn ,r 1
r 1
rt a
Sn n ,r 1
r 1
(use when you know last term or are finding last term)
Examples:
1. 3, 6, 12, ..., Find:
a. S15 ANS: 98 301 b. n if Sn = 25 165 821 ANS: 23
a (r n 1) a (r n 1)
Sn Sn
r 1 r 1
3 215 1 3 2n 1
Sn 98301 25165821
2 1 2 1
25165821 3 2n 1
8388607 2n 1
8388607 2n
log 2 8388607 n
n 23
2. 11 12 + ... is a geometric series. Find:
a. S25 (0.0001) ANS: 51.5812
t2 12
r
t1 11
12 25
11 1
S25 11 51.5812
12
1
11
Page 38 M30P LP CH 2 & 6 Logs Exp
3. 6, 7, ... is a geometric series. Find the smallest value of n so that the sum is at least 1000.
ANS: n = 22
t2 7 a (r n 1)
r Sn
t1 6 r 1
7n
6 1
1000
6
7
1
6
500 7n
6 1
3 6
n
250 7
1
9 6
n
259 7
9 6
259
log 7 n 21.8 22
6 9
4. 5 + 10 + 20 + ... + 1280. Find the sum. ANS: 2 555
We do not know the number of terms, so we need the second formula.
rt a
Sn n
r 1
2 1280 5
Sn 2555
2 1
Page 39 M30P LP CH 2 & 6 Logs Exp
2
5. Bouncing Ball. A ball bounces up of its height from which it falls. It is dropped from a height
3
of 81m. Find:
a. height after it hits the ground for 10th time ANS: 1.40m
(54, 36, 24, ...) 54 is first bounce
tn ar n 1
10 1
2
tn 54 1.4m
3
b. distance travelled until it hits the ground for the eighth time ANS: 386.04m
81 54 54 36 36 24 24 16 16 etc.
Bounce 1 2 3 4 5 etc.
Obviously there is a geometric sequence. The key is to notice how the 54, 36, 24 etc. are
all doubled. So, calculate the sum, double it and then add the extra 81, which occurs only
once.
a r n 1
Sn
r 1
2 7
54 1
3
S7 152.52m 152.52 152.52 81 386.04 m
2
1
3
6. You have 2 400 acres of treed land. 5% of the trees are removed each year. How many acres will
be removed in 11 years? ANS: 1034.88
(trees removed each year: 120, 114, 108.3, ... Find S11
OR # of trees each year: 2400, 2280 (after 1 year = 2nd term), 2166, ... Find t12 and take 2400 t12)
A P(1 i ) n
A 2400(1 0.05)11
A 2400(0.95)11 1365.12
2400 1365.12 1034.88 acres
Page 40 M30P LP CH 2 & 6 Logs Exp
a(r n 1)
Sn ,r 1
r 1
rt a
Sn n ,r 1
r 1
Assignment for day 13
pg. 309 # 1, 5, 6, 22 a, 24 (if you do ex. 6)
plus:
1. 7 + 14 + 28 + ... + 1 835 008. Find the sum. ANS: 3 670 009
rtn a 2 1835008 7
Sn 3670009
r 1 2 1
2. 11 + 13 + ... is a geometric series. Find:
a. S13 (0.1) ANS: 470.3 b. n (smallest) so that Sn 5 000 ANS: n = 27
a r n 1
13 13 Sn
11 1 r 1
a r n 1 11
Sn 470.3 13 n
11 1
r 1 13
1 11
11 5000
13
1
11
13 n
909.09 11 1
11
n
13
82.645 1
11
n
13
83.645
11
log 13 83.645 n 26.49 27
11
Page 41 M30P LP CH 2 & 6 Logs Exp
3. 5 + 11 ... is a geometric series. Find (0.1):
a. t23 ANS: 170 713 938.7 b. S24 ANS: 258 204 830.7
t2 11 11 t2 11 11
r r
t1 5 5 t1 5 5
tn ar n 1 a r n 1
Sn
11
231
r 1
t23 5
5 11 24
5 1
t23 170713938.7 5
S 24 258204830.7
11
1
5
4. A bouncing ball bounces up 0.7 of its height from which it fell. It is dropped from a height of
4 000cm. Find:
a. height after it hits the ground for 7th time (0.1) ANS: 329.4cm
tn ar n 1
tn 2800 0.7
7 1
tn 329.4cm
b. distance traveled until it hits ground for 7th time (whole number) ANS: 20 471cm
4000 2800 2800 1960 1960 1372 1372 960 960 etc.
Bounce 1 2 3 4 5 etc.
Obviously there is a geometric sequence. The key is to notice how the 2800, 1960, 1372,
960 etc. are all doubled. So, calculate the sum, double it and then add the extra 4000,
which occurs only once.
a r n 1
Sn
r 1
2800 0.76 1
S6 8235.276cm 8235.276 8235.276 4000 20471cm
0.7 1
Page 42 M30P LP CH 2 & 6 Logs Exp
DAY 14 (Short Day)
Sigma Notation ( )
Examples:
1. Expand and evaluate (Find sum).
7
a. 5k k
k 3
2
= 24 + 36 + 50 + 66 + 84 = 260
This means put 3, 4, 5, 6 and 7 into the formula 5k k 2 and add your answers.
25
b. 5(2
k 2
k 1
) ANS: 167 772 150 (10 + 20 + 30 + ...)
* a is not necessarily 5, but r is definitely 2 (if coefficient of k is 1),
# of terms = 25 2 + 1 = 24
38
c. 2(1.1)
k 5
k 3
ANS: 594.05 (2.42 + 2.662 + 2.9282 + ...)
* a = 2.42, r = 1.1, n = 38–5+1=34
2. Write using notation.
17
a. 4 + 8 + 16 + ... + 262 144 ANS: 4(2)
k 1
k 1
* Find n first (17) then use k = 1 to k = 17, a = 4, r = 2.
12
b. 5 15 45 ... 885 735 ANS: 5(3)
k 1
k 1
k 1
1 12
1 12
c. 27 + 9 + 3 + ... + ANS: 27 or 3 k4
6561 k 1 3 k 1
21
3. How many terms in the series:
k p 3
4(5) k 1 ANS: 19 p
Page 43 M30P LP CH 2 & 6 Logs Exp
Assignment
pg. 309 # 8, 16
plus:
1. Find sum:
16 19
a. 3(2)k 1
k 2
ANS: 196 602 b. 100(0.6)
k 5
k 3
ANS: 89.96
a 3 2
2 1
6 a 100(0.6)53 36
t2 100(0.6) 63 21.6
t2 3 2
31
12
21.6
r2 r 0.6
36
n 16 2 1 15
n 19 5 1 15
6 215 1
S15 196602 36 0.615 1
2 1 S15 89.96
0.6 1
2. Write using notation.
12
a. 2 + 10 + 50 + ... + 97 656 250 ANS: 2(5)
k 1
k 1
n 1
tn ar
97656250 2(5) n 1
48828125 5n 1
log 5 48828125 n 1
n 12
k 1
9
3
b. 64 + 96 + 144 + ... + 1 640.25 ANS: 64 2
k 1
tn ar n 1
1640.25 64(1.5) n 1
25.629 1.5n 1
log1.5 25.629 n 1
n9
Page 44 M30P LP CH 2 & 6 Logs Exp
Decibel records
1976
The Who were listed as the "record holder", at 126 dB, measured at a distance of 32 meters from the speakers at a
concert at The Valley (home ground of Charlton Athletic F.C.) on 31 May 1976.[2]
1984 and 1994
The heavy metal band Manowar is one claimant of the title of "loudest band in the world",[5] citing a measurement
of 129.5 dB in 1994 in Hanover.[6] However, Guinness Book of World Records listed Manowar as the record holder
for the loudest musical performance for an earlier performance in 1984. Guinness does not recognize Manowar's
later claim, because it no longer includes a category of loudest band, reportedly because it does not want to
encourage hearing damage. Manowar broke their own record again in 2008 at the Magic Circle Fest with 139dB.
1997
Metallica has styled itself the "loudest band in the world". However, after one concert on 11 November 1997,
which the band dubbed the "Million Decibel March", the Philadelphia Inquirer reported that "neighbors who [had]
feared the worst from the self-styled Loudest Band in the World complained more about the sound from the news
2007
British punk band Gallows allegedly broke Manowar's previous record for loudest band in the world, claiming to
have achieved 132.5 dB; however, this record was claimed in an isolated studio environment as opposed to live.
Other dates
Other previous record holders include Deep Purple (117 decibels), Iron Maiden (1985 Guinness World Book of
Records), The Rolling Stones, Motörhead, AC/DC, My Bloody Valentine (132 decibels), and KISS (120 decibels).
Although it has never been made official, some reports suggest that New York noise rock band, Swans reached 140
decibels at some gigs.
Environmental Noise
Whisper Quiet Library 30dB
Normal conversation (3-5') 60-70dB
Telephone dial tone 80dB
City Traffic (inside car) 85dB
Train whistle at 500', Truck Traffic 90dB
Subway train at 200' 95dB
Level at which sustained exposure may result in hearing loss 90 - 95dB
Power mower at 3' 107dB
Snowmobile, Motorcycle 100dB
Power saw at 3' 110dB
Sandblasting, Loud Rock Concert 115dB
Pain begins 125dB
Pneumatic riveter at 4' 125dB
Even short term exposure can cause permanent damage -
140dB
Loudest recommended exposure WITH hearing protection
Jet engine at 100', Gun Blast 140dB
Death of hearing tissue 180dB
Loudest sound possible 194dB
Page 45 M30P LP CH 2 & 6 Logs Exp
```
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How are you planning on using Docstoc? | 23,111 | 59,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.155889 |
http://mathhelpforum.com/algebra/201757-help-polynomial.html | 1,527,131,308,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034502-00548.warc.gz | 186,597,467 | 9,357 | # Thread: help with this polynomial
1. ## help with this polynomial
Hi;
got this question
Given that
x -2 is a factor of the polynomial x^3 - kx^2 -24x + 28, find k.
I get k=-3 they get k=3.
Pretty sure I'm right just need confirmation.
Thanks.
2. ## Re: help with this polynomial
confirm it yourself ...
$\displaystyle f(x) = x^3 + 3x^2 - 24x + 28$ if $\displaystyle k = -3$
or
$\displaystyle f(x) = x^3 - 3x^2 - 24x + 28$ if $\displaystyle k = 3$
which one yields $\displaystyle f(2) = 0$ ?
3. ## Re: help with this polynomial
Got it thanks Skeeter its the top one. | 192 | 581 | {"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} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.787909 |
https://www.numbersaplenty.com/200002 | 1,656,816,874,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00340.warc.gz | 1,003,875,510 | 3,363 | Search a number
200002 = 2119091
BaseRepresentation
bin110000110101000010
3101011100111
4300311002
522400002
64141534
71462045
oct606502
9334314
10200002
111272a0
12978aa
137005a
1452c5c
153e3d7
hex30d42
200002 has 8 divisors (see below), whose sum is σ = 327312. Its totient is φ = 90900.
The previous prime is 199999. The next prime is 200003.
200002 is nontrivially palindromic in base 10.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a junction number, because it is equal to n+sod(n) for n = 199964 and 200000.
It is not an unprimeable number, because it can be changed into a prime (200003) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 4524 + ... + 4567.
It is an arithmetic number, because the mean of its divisors is an integer number (40914).
2200002 is an apocalyptic number.
200002 is a gapful number since it is divisible by the number (22) formed by its first and last digit.
200002 is a deficient number, since it is larger than the sum of its proper divisors (127310).
200002 is a wasteful number, since it uses less digits than its factorization.
200002 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 9104.
The product of its (nonzero) digits is 4, while the sum is 4.
The square root of 200002 is about 447.2158315623. The cubic root of 200002 is about 58.4805496981.
It can be divided in two parts, 20000 and 2, that multiplied together give a square (40000 = 2002).
The spelling of 200002 in words is "two hundred thousand, two", and thus it is an iban number.
Divisors: 1 2 11 22 9091 18182 100001 200002 | 537 | 1,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.90468 |
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Topic: phone book probability
Replies: 1 Last Post: Nov 5, 2008 1:32 PM
Messages: [ Previous | Next ]
loren Posts: 1 Registered: 10/17/08
phone book probability
Posted: Oct 17, 2008 12:44 PM
Let's say you have 50 phone books, one for each state in the US. For each last name in each phone book, you compute the frequency with which that name appears. So "Smith" might comprise 1% of the phone book, while "Spears" might be 0.01%.
Now let's say I gave you a group of 10 last names, and asked you to determine which state these people were most likely from, based on the phone book counts.
How do we go about solving this problem?
Even if you can point me to the area of probability where this problem lies, I'd be happy to go look it up and teach myself, but I do not know where to begin.
Date Subject Author
10/17/08 loren
11/5/08 Austin | 278 | 1,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-17 | latest | en | 0.94944 |
http://mathhelpforum.com/advanced-algebra/55900-isomorphism.html | 1,481,425,737,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543782.28/warc/CC-MAIN-20161202170903-00458-ip-10-31-129-80.ec2.internal.warc.gz | 169,056,005 | 10,239 | 1. ## Isomorphism...
Let B be an n x n invertible matrix. Define Ф: Mnxn (F) -> Mnxn (F) by Ф (A) = B^-1 AB. Prove that Ф is an isomorphism
2. Originally Posted by leungsta
Let B be an n x n invertible matrix. Define Ф: Mnxn (F) -> Mnxn (F) by Ф (A) = B^-1 AB. Prove that Ф is an isomorphism
Just check three conditions: (i)it is one-to-one, (ii)it is onto, (iii) $\phi (AB) = \phi (A) \phi (B)$.
3. Originally Posted by ThePerfectHacker
Just check three conditions: (i)it is one-to-one, (ii)it is onto, (iii) $\phi (AB) = \phi (A) \phi (B)$.
Quick note: Make sure it is a different B from the one defined in the question. It is also a ring isomorphism.
4. Originally Posted by whipflip15
Quick note: Make sure it is a different B from the one defined in the question. It is also a ring isomorphism.
Thank you for correcting the notational mistake.
5. Originally Posted by ThePerfectHacker
Just check three conditions: (i)it is one-to-one, (ii)it is onto, (iii) $\phi (AB) = \phi (A) \phi (B)$.
This is correct. Another way for showing is to proof as the matrix is invertible. That is a constraint for isomorphism.
If you have a linear map which is bijective then this linear map always has a inverse map.
greetings
Herbststurm | 382 | 1,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.888344 |
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1 AHP Lecture Notes
# 1 AHP Lecture Notes - Analytic Hierarchy Process The...
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1 Slide Analytic Hierarchy Process The Analytic Hierarchy Process (AHP) , is a procedure designed to quantify managerial judgments of the relative importance of each of several conflicting criteria used in the decision making process.
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2 Slide Analytic Hierarchy Process Step 1: List the Overall Goal, Criteria, and Decision Alternatives Step 2: Develop a Pairwise Comparison Matrix Rate the relative importance between each pair of decision alternatives. The matrix lists the alternatives horizontally and vertically and has the numerical ratings comparing the horizontal (first) alternative with the vertical (second) alternative. Ratings are given as follows: . . . continued For each criterion, perform steps 2 through 5
3 Slide Analytic Hierarchy Process Step 2: Pairwise Comparison Matrix (continued) Compared to the second alternative, the first alternative is: Numerical rating extremely preferred 9 very strongly preferred 7 strongly preferred 5 moderately preferred 3 equally preferred 1 Intermediate numeric ratings of 8, 6, 4, 2 can be assigned. A reciprocal rating (i.e. 1/9, 1/8, etc.) is assigned when the second alternative is preferred to the first. The value of 1 is always assigned when comparing an alternative with itself.
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4 Slide Analytic Hierarchy Process Step 3: Develop a Normalized Matrix Divide each number in a column of the pairwise comparison matrix by its column sum. Step 4: Develop the Priority Vector Average each row of the normalized matrix. These row averages form the priority vector of alternative preferences with respect to the particular criterion. The values in this vector sum to 1.
5 Slide Analytic Hierarchy Process Step 5: Calculate a Consistency Ratio The consistency of the subjective input in the pairwise comparison matrix can be measured by calculating a consistency ratio. A consistency ratio of less than .1 is good. For ratios which are greater than .1, the subjective input should be re- evaluated. Step 6: Develop a Priority Matrix After steps 2 through 5 has been performed for all criteria, the results of step 4 are summarized in a priority matrix by listing the decision alternatives horizontally and the criteria vertically. The column entries are the priority vectors for each criterion.
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Ask a homework question - tutors are online | 652 | 3,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-13 | latest | en | 0.813687 |
http://www.chegg.com/homework-help/questions-and-answers/man-average-useful-power-output-400-w-minimumtime-would-take-lift-fifty150-kg-boxes-height-q790710 | 1,455,434,309,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701169261.41/warc/CC-MAIN-20160205193929-00004-ip-10-236-182-209.ec2.internal.warc.gz | 330,541,862 | 13,053 | If a man has an average useful power output of 40.0 W, what minimumtime would it take him to lift fifty15.0-kg boxes to a height of1.70 m? The value of g is9.80 m/s2. Please break this down into steps I canduplicate for other answers.
Thanks
Mike | 70 | 246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2016-07 | longest | en | 0.901028 |
https://playtaptales.com/numbers/novendecinongentilliseptuagintaoctingentillion/ | 1,675,758,263,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00543.warc.gz | 457,893,648 | 4,459 | ## Novendecinongentilliseptuagintaoctingentillion
A Novendecinongentilliseptuagintaoctingentillion (1 Novendecinongentilliseptuagintaoctingentillion) is 10 to the power of 2759613 (10^2759613). This is a truly astronomical number!
## How many zeros in a Novendecinongentilliseptuagintaoctingentillion?
There are 2,759,613 zeros in a Novendecinongentilliseptuagintaoctingentillion.
## What's before Novendecinongentilliseptuagintaoctingentillion?
A Novendecinongentillinovensexagintaoctingentillion is smaller than a Novendecinongentilliseptuagintaoctingentillion.
## What's after Novendecinongentilliseptuagintaoctingentillion?
A Novendecinongentilliunseptuagintaoctingentillion is larger than a Novendecinongentilliseptuagintaoctingentillion.
## Novendecinongentilliseptuagintaoctingentillionaire
A Novendecinongentilliseptuagintaoctingentillionaire is someone whos assets, net worth or wealth is 1 or more Novendecinongentilliseptuagintaoctingentillion. It is unlikely anyone will ever be a true Novendecinongentilliseptuagintaoctingentillionaire. If you want to be a Novendecinongentilliseptuagintaoctingentillionaire, play Tap Tales!
## Is Novendecinongentilliseptuagintaoctingentillion the largest number?
Novendecinongentilliseptuagintaoctingentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
## Novendecinongentilliseptuagintaoctingentillion written out
Novendecinongentilliseptuagintaoctingentillion is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00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https://homework.zookal.com/questions-and-answers/4-according-to-the-bureau-of-labor-statistics-the-average-748358584 | 1,618,983,367,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039508673.81/warc/CC-MAIN-20210421035139-20210421065139-00072.warc.gz | 392,767,842 | 26,253 | 1. Math
2. Statistics And Probability
3. 4 according to the bureau of labor statistics the average...
# Question: 4 according to the bureau of labor statistics the average...
###### Question details
4. According to the Bureau of Labor Statistics, the average weekly pay for a U.S. production worker was $441.84. Assume that available data indicate that production worker wages were normally distributed with a standard deviation of$90. What is the probability that a worker earned between $400 and$500? (Round to four decimal places) How much did a production worker have to earn to be in the top 20% of wage earners? (Round to two decimal places) For a randomly selected production worker, what is the probability the worker earned less than \$250 per week? (Round to four decimal places) | 169 | 792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.955303 |
https://www.freelancer.pt/job-search/bayes-probability-code/ | 1,537,818,989,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160641.81/warc/CC-MAIN-20180924185233-20180924205633-00117.warc.gz | 744,881,643 | 40,723 | # Bayes probability code jobs
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3,607 bayes probability code trabalhos encontrados, preços em EUR
...04202 0.4375 0.05999 0.5 0.09025 0.5625 0.10540 0.625 0.13685 0.6875 0.14058 0.75 0.13857 0.8125 0.10240 0.875 0.07448 0.9375 0.03702 1 0.01557 I need your help to find a probability distribution to find the best fit the data points above (by calculating the MSE and also shown with a plot). The data above is arbitrarily given and could change. Would
€22 (Avg Bid)
€22 Média
12 ofertas
Hey there, I need some help solving a few probabilities questions and draw the decision-making tree for 3 questions (8 sub-questions). You need to have a solid mathematical background, as well as the ability to articulate your answers. A full brief will be given. As the task is highly technical, please provide a short pitch, why you'd be qualified to take the task, when you are biding ...
€164 (Avg Bid)
€164 Média
16 ofertas
...___________________________________________________ Work Already Done I already have code of CogNos. here it is the link of code and how to install in ns2 : [login to view URL] Now this code is running perfectly fine in my system and it also showing GUI In this code there is a user called primary user (PU) and they have just implemented that how cognitive
€236 (Avg Bid)
€236 Média
2 ofertas
Need to work on mathematics and stats & probability .. If you are good at those subjects... then only apply, I'm looking out for a freelancer who can work for me on montly basis , Only from India
€13 (Avg Bid)
€13 Média
17 ofertas
Probability expertise/... 1 day left
I need probability expertise with [login to view URL] about weather forecasting and estimation
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12 ofertas
Data Analysis 1 day left
...mean)/sigma. A google search “z-score numpy array” will give you plenty of ideas about implementation and you could achieve an efficient implementation with just one line of code! So let me throw in an enhanced requirement: the user also inputs a scalar value (the second input) to indicate the desired column for transformation. Thus, only the specified
€4 / hr (Avg Bid)
€4 / hr Média
13 ofertas
Probability expertise/ 1 day left
I need probability expertise with good
€30 (Avg Bid)
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21 ofertas
If you are good in probability expert and having good skills you must bid on it
€20 (Avg Bid)
€20 Média
7 ofertas
Need to work on mathematics and stats & probability .. If you are good at those subjects... then only apply, I'm looking out for a freelancer who can work for me on montly basis
€17 (Avg Bid)
€17 Média
33 ofertas
There wi...have comparative plots in python especially, Influence VS seed set size, Influence VS threshold, Influence VS activation probability and running time VS seed set size, running time VS threshold, running time VS activation probability. I will provide algorithms and some partially solved modules (in python) so that you can solve it with ease.
€94 (Avg Bid)
€94 Média
8 ofertas
We need help in probability help.
€22 (Avg Bid)
€22 Média
6 ofertas
I need probability expertise with good
€20 (Avg Bid)
€20 Média
17 ofertas
Expert in probanility and along with good understanding .
€40 (Avg Bid)
€40 Média
31 ofertas
...Facebook, Gmail and Mobile number. After completion of the registration, user will be landed to dashboard/profile page, here if the user has arrived using the friend's referral code, then both the referrer and referral will be credited with some fixed amount of coins in their respective account as referral coins. Player can refer using any of social networking
€5048 (Avg Bid)
€5048 Média
13 ofertas
1. There will be a custom scenario to be built in SUMO which contains vehicles and RSUs 2. Each RSU generates a fire message at random probability which has to be sent to the Fire station and hospital's RSU. 3. For the message to travel cars act as the carrier. 4. The message should follow the GPSR(Greedy Perimeter Stateless Routing) 5. This should
€88 (Avg Bid)
€88 Média
2 ofertas
Create a probability formula relating to blockchain Proof-of-Work systems: Given: 1. a hash of N bits, 2. a 'difficulty' of L leading zero bits, 3. a matching hash (generic solution S) is found on average in M minutes. (by varying a part of the input data, nonce) Someone selects one predetermined hash fitting the above requirements and searches only
€434 (Avg Bid)
€434 Média
6 ofertas
...in xls format i want to compute the conditional probabilities using Bayes theorem. Compute 10 cases of your choice. Results in microsoft word 2003 or microsoft word 2007 .Use Equation Editor. look at the picture attached, to have an idea ! theory : [login to view URL] Attribute Information: Attribute Information: ------------------------
€16 (Avg Bid)
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6 ofertas
...last name, or only one name element) and it will output its guess as to what ethnicity group this person belongs to based on the training of the dataset AND a confidence or probability number which tells how sure the system is of this ethnicity being the correct answer. For example, should the program receive input "john smith", it should output ethnicity
€177 (Avg Bid)
€177 Média
20 ofertas
...of 50 chromosomes and each chromosome should be 20 bits long. Your crossover operator should be one point crossover with a probability of pcross(.8) and your mutation operation should randomly flip each bit with a probability of pmut(.o1). Use a roulette proportional selection algorithm. Have the GA stop when the optimal fitness is 20(size of the
€24 (Avg Bid)
€24 Média
5 ofertas
...question is in part the basis of our research, along with answering the following questions: 1. Do chosen lottery numbers from the lottery occur in drawings with equal probability? 2. Is there a game strategy that outperforms others? 3. Is the performance of a strategy effective in multiple states? This research aims to prove that strategies
€336 (Avg Bid)
€336 Média
30 ofertas
presentation of the project: The project is based on 2 parts the first one is to do the trading backtesti...project is based on 2 parts the first one is to do the trading backtesting based on tradingview. The second part is to create a bot of buy and sell of crypto based on a probability of yesterday of the best parameter to gain a maximum of profit.
€416 (Avg Bid)
€416 Média
15 ofertas
Budget is \$10 AUD. ONLY BID IF YOU'RE INTERESTED. Message me for details.
€14 (Avg Bid)
€14 Média
3 ofertas
Only apply if you can see the following book in your University library catalogue: "I. G. Petrowsky Selected Works: Differential Equations and Probability Theory" Author: Oleinik, O.A Published by CRC (1996) ISBN 10: 2881249795 ISBN 13: 9782881249792 I need scans or images of pages 352 to 399. Readable images, scanned or photographed (with phone)
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€23 Média
3 ofertas
...proponents of blood and organ donating in an effort to prolong and enhance the health of others in need, while many are reluctant to participate in this practice due to the probability of transmitting various diseases, as well as other medical complications, which can result from the donating process. Choose a position on this issue and write a 2-3 page
€28 (Avg Bid)
€28 Média
28 ofertas
I have a project relates to linear programming in python. The code is in python and uses gurobi solver. But you can use the solver that you prefer to make it. Since it is in python or julia language. for more details contact me
€34 (Avg Bid)
€34 Média
15 ofertas
FREELANCER SUCKS. I did not want to post a contest. I feel they are unfair to designers. I simply wanted to have design...ads). To prove you read this post in it's entirety and confirm you fit the job requirements, please put "Qualified Ad Designer" in the subject line. If you do not, the probability of me even opening your submission is very low.
€128 (Avg Bid)
€128
21 inscrições
...of 50 chromosomes and each chromosome should be 20 bits long. Your crossover operator should be one point crossover with a probability of pcross(.8) and your mutation operation should randomly flip each bit with a probability of pmut(.o1). Use a roulette proportional selection algorithm. Have the GA stop when the optimal fitness is 20(size of the
€20 (Avg Bid)
€20 Média
3 ofertas
Role :Application Designer Role Description :Assist in defining requirements and designing applications to meet business pro...Asset Creation 4 Skill building Professional Experience atleast 2yrs 1 Machine Learning and Deep Learning 2 Unix 3 Python 4 Data Visualization Knowledge 1 Statistics and Probability 2 Distributed Computing 3 ML/ DL toolkits
€8 / hr (Avg Bid)
€8 / hr Média
19 ofertas
I am looking for someone to analyze all of the most profitable items being sold on Amazon and current trends of top 5 items to sell. Goal is to establ...sell on FBA/ Amazon. I am seeking detailed information from a researched that will provide facts and figures on which are the top trending 10 products that have highest probability to sell. Thank you.
€23 (Avg Bid)
€23 Média
5 ofertas
...the concept of social media. Free maintenance for 6 months. Token system in surveys where a token number will be given to the participant which will decide his winning probability. Directory system where users can add their business or concerned credentials which they want to promote. Survey advertisement on front page. News will be shown when
€263 (Avg Bid)
€263 Média
19 ofertas
i have an article u need to find probability value coefficient of variability and sample size reqd and chances that the study will succeed.
€19 (Avg Bid)
€19 Média
17 ofertas
We need an expert who will be able to perform nested simulation in matlab. The work is very simple, where you need to generate random numbers and based ...matlab. The work is very simple, where you need to generate random numbers and based on each outer scenarios you need to perform another inner simulation and calculate the probability loss function.
€19 (Avg Bid)
€19 Média
16 ofertas
...valubet tool: -By comparing the live and pre-match odds of a bookmaker compared to others bookmakers, detect if there is a valubet through a small calculation. -Display the probability and the value. -Filter the valubet according to several criteria (see attachment) Contact me if you want to have an example of a tool / site. Waiting for your proposals
€2290 (Avg Bid)
€2290 Média
11 ofertas
Hi i need someone I want to add a new classifier to WEKA platform for classification task using Co-Training approach and Naive Bayes classifier.
€109 (Avg Bid)
€109 Média
7 ofertas
My project is a java code for semi-supervised learning technique. I want to add a new classifier to WEKA for classification task using co-training and naive bayes classifier.
€402 (Avg Bid)
€402 Média
6 ofertas
Hi i need someone I want to add a new classifier to WEKA platform for classification task using Co-Training approach and Naive Bayes classifier.
€119 (Avg Bid)
€119 Média
18 ofertas
...it pull for a specific country (as an example, the USA), and for a specific time (as an example, July 2018). Your program will scrape that data, create a database w/ the probability that a real Internet user will have that user agent, and will generate that user agent. Just like your other program. It is essential that the breakout by device, browser
€85 (Avg Bid)
€85 Média
1 ofertas
...Odds of 10 means 1/10= 10% chance of winning the bet as per the bookmaker estimation (and of course 90% chance of losing the bet). Always the inverse of the odds is the probability. Turnover: If you bet for example 10 x 10 EUR = 100 EUR is your betting turnover ROI = Return on Investment. If you have for example an ROI of 5%, you expect to win 50
€9 / hr (Avg Bid)
€9 / hr Média
9 ofertas
...machine learning models in R -- Dataset : "Statlog (Heart) Data Set" from [login to view URL] 4 different algorithms : 1. Multilayer Perceptron Network 2. The Naïve Bayes classifier 3. Classification and Regression Trees (CART). 4. k-Nearest Neighbors (kNN). A. For every algorithm use 10-fold crossvalidation to estimate accuracy. B. Visualize
€249 (Avg Bid)
€249 Média
5 ofertas
...the spreadsheet is to list all the resources, their current utilization, the current projects which is current and expected projects, project probability, project hours total, net project hours after probability, project start-date, project end-date. I have to be able to calculate and project out the Project Hours for the resources until they are at
€5 / hr (Avg Bid)
€5 / hr Média
9 ofertas
...the spreadsheet is to list all the resources, their current utilization, the current projects which is current and expected projects, project probability, project hours total, net project hours after probability, project start-date, project end-date. I have to be able to calculate and project out the Project Hours for the resources until they are at
€5 / hr (Avg Bid)
€5 / hr Média
23 ofertas
...then searches our Collection and comes back with the id = for example: Person is Jim Doe, probability 99% or so. YOUR ROLE would be to provide us with a sample apk, demonstrating the above functionality to our senior engineer. He will then take your demo code and integrate it with our App. That app is used to identify individuals, either through face
€479 (Avg Bid)
€479 Média
20 ofertas
...Each time I am looking for at least one red marble in the 5 that withdrew. So, on average, how may red marbles will I get in the 10000 trials? (In other words, what is the probability of getting at least one marble when N = 5?) Case 3: Same as case 2, except I make N a random variable, Poisson distributed with a mean value of 5? Provide mathematical
€21 (Avg Bid)
€21 Média
6 ofertas
Calculator App Logo Ended
...represents "Diabetes Complications Calculator" (DCC). I have attached sketches and directions and fonts as a starting point .The project is an algorithm to predict the probability for someone who has diabetes to get diabetes complications. I want it to be a distinctive asset in a way that when people see the logo they say "ah that’s the Diabetes complications
€13 (Avg Bid)
€13
34 inscrições
...xtrm2A_kMy13oUa9fD6p89COs4Gw1YQPEs/edit?usp=sharing you must: 2. Create a simple rest service API that given a text will respond one of the values in column C and it's probability of matching. I'm only interested in bids that can give a brief description of how they will go about developing this (i.e: Tensorflow with a Java servlet). Deliverables:
€154 (Avg Bid)
€154 Média
13 ofertas
...xtrm2A_kMy13oUa9fD6p89COs4Gw1YQPEs/edit?usp=sharing you must: 2. Create a simple rest service API that given a text will respond one of the values in column C and it's probability of matching. I'm only interested in bids that can give a brief description of how they will go about developing this (i.e: Tensorflow with a Java servlet). Deliverables:
€2032 (Avg Bid)
€2032 Média
2 ofertas
presentation of the project: The project is based on 2 parts the first one is to do the bac...on 2 parts the first one is to do the backtesting based on 1 day or 1 month .... ( so its a customized period to choose using tradingview) The second part is based on a probability of buying the cryptocurrencies based on parameters using the machine learning.
€256 (Avg Bid)
€256 Média
5 ofertas
...Odds of 10 means 1/10= 10% chance of winning the bet as per the bookmaker estimation (and of course 90% chance of losing the bet). Always the inverse of the odds is the probability. Turnover: If you bet for example 10 x 10 EUR = 100 EUR is your betting turnover ROI = Return on Investment. If you have for example an ROI of 5%, you expect to win 50
€9 / hr (Avg Bid)
€9 / hr Média
18 ofertas
Pablo – can you create a present value excel that would include a probability life calculation.
€128 (Avg Bid)
€128 Média
1 ofertas
...should be shown on chart • Prediction Distribution (should be for both Density and Distribution) • Y_label: Probability Density • X_label: Provability of Event (range: 0.0 – 1.0) • Ideal threshold marked (range: 0.0 – 1.0) • Y_label: Probability Frequency • X_label: Provability of Event (range: 0.0 – 1.0) • Ideal threshold marke...
€20 (Avg Bid)
€20 Média
6 ofertas | 4,015 | 16,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-39 | longest | en | 0.846463 |
http://www.barnesandnoble.com/w/3000-easy-shopping-math-puzzles-kalman-toth-ma-mphil/1118142189?ean=9781495253546&itm=1&usri=kalman+toth | 1,469,613,637,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00214-ip-10-185-27-174.ec2.internal.warc.gz | 316,807,707 | 23,202 | # 3000 Easy Shopping Math Puzzles
All puzzles are fun and entertaining. Sometimes we learn an unexpected appreciation for a particular area of learning when it is presented in an enjoyable manner. Most of us love to shop. So, Shopping Math Puzzles - word puzzles based upon shopping - will prove particularly enjoyable, even if word problems were not one of your favorite math assignments in
## Overview
All puzzles are fun and entertaining. Sometimes we learn an unexpected appreciation for a particular area of learning when it is presented in an enjoyable manner. Most of us love to shop. So, Shopping Math Puzzles - word puzzles based upon shopping - will prove particularly enjoyable, even if word problems were not one of your favorite math assignments in school.
Each page in this book contains twenty shopping, word-problem puzzles. Some are multiple-choice, with four possible answers provided, and some require a numerical answer. The correct answers are in reverse order at the bottom of each page.
Benefits of Solving Shopping Math Puzzles
The benefits of solving Shopping Math Puzzles are many. They help teach you how to solve problems in everyday life. You began to practice and solve mathematical word puzzles in elementary school, and you continue as an adult to encounter math word "puzzles" everyday, in real-life situations.
Shopping Math Puzzles is an excellent resource to use when teachers test the understanding of a new concept. The puzzles aid in not only understanding the application of math principles, but also improve reading comprehension.
Shopping Math Puzzles
Provide an Interesting and Enjoyable Activity
Shopping Math Puzzles are implicitly interesting, partly because solving them does not involve a sequence of very similar steps that are designed to practice the same skill. The novelty of the puzzles adds to their interest. Solving Shopping Math Puzzles involves detective work, which most people enjoy. We become deeply involved in the process required for solving the puzzle, and relish getting the answer after having struggled.
Create Greater Understanding of Mathematical Processes
Solving Shopping Math Puzzles requires an understanding of the mathematical process. As we struggle with a puzzle, we often immerse ourselves in obtaining the solution so much that we apply mathematical methods spontaneously. This application gives us a deep understanding of problem solving methods.
Promote Positive Attitudes
As we work through a shopping math puzzle, we become so involved we forget that the puzzle solving involves mathematics. Success when we find the answer helps us gain a positive attitude. This is especially effective when these Shopping Math Puzzles are used with students.
Introduce and Enhance Research Methods
The method used to solve Shopping Math Puzzles is similar to the Scientific Method used by a researcher. The solver moves step-by-step through the process. Hence, while solving the puzzle, you learn to apply the thinking processes of the scientific method.
Promote Flexibility and Creativity
Solving a shopping math puzzle provides an opportunity to explore ideas and to extend creativity. You need to be creative and come up with new ways of tackling problems. What works on one puzzles, may not work on the next. Flexibility in your approach is necessary.
## Product Details
ISBN-13:
9781495253546
Publisher:
CreateSpace Publishing
Publication date:
01/18/2014
Pages:
162
Product dimensions:
7.44(w) x 9.69(h) x 0.35(d)
## Customer Reviews
Average Review:
Write a Review
and post it to your social network | 710 | 3,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.938465 |
https://meilyngift.com/qa/how-many-is-a-gazillion.html | 1,638,968,264,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363510.40/warc/CC-MAIN-20211208114112-20211208144112-00235.warc.gz | 442,499,317 | 42,500 | # How Many Is A Gazillion?
## Is Jillion a real word?
A jillion is an enormous number of something.
The word is modeled on actual numbers like million and billion, so it almost sounds like a real quantity.
But like zillion, jillion is imprecise.
Its origin is vague too, described as an “arbitrary coinage” first used around 1940..
## What’s higher than a bazillion?
We can say that zillion and jillion are roughly in the same class in terms of vastness. Beyond these lie the more ginormous bazillion and bajillion. Beyond even these are the incomprehensible gazillion and gajillion [4] .
## How many is a sextillion?
NameShort scaleLong scalequintillion1 with 18 zeros1 with 30 zerossextillion1 with 21 zeros1 with 36 zerosseptillion1 with 24 zeros1 with 42 zerosoctillion1 with 27 zeros1 with 48 zeros6 more rows
## Is a gazillion a real number?
gazillion Add to list Share. If you have an enormous, indefinite number of things, you can say you have a gazillion. … Like zillion and jillion, gazillion is a made-up word meaning “a whole bunch” that’s modeled after actual numbers such as million and billion.
## Which is bigger gazillion or bazillion?
To answer your question, it’s a gazillion. B/C it goes mazillion, bazillion, and then eventually gazillion.
## How many zeros are in a bajillion?
six zeros1,000 has three zeros. That means that the next large number, ten thousand (10,000), has four zeros. The same goes when you get into the millions. One million has six zeros (1,000,000).
## How many zeros are in a gazillion?
Etymology of Gaz Therefore a Gazillion has (28819 x 3) zeros and a Gazillion is…
## What is the highest number?
The number googol is a one with a hundred zeros. It got its name from a nine-year old boy. A googol is more than all the hairs in the world.
## What comes next after billion?
After a billion, of course, is trillion. Then comes quadrillion, quintrillion, sextillion, septillion, octillion, nonillion, and decillion.
## WHAT IS A zillion?
A zillion is a huge but nonspecific number. … Zillion sounds like an actual number because of its similarity to billion, million, and trillion, and it is modeled on these real numerical values. However, like its cousin jillion, zillion is an informal way to talk about a number that’s enormous but indefinite.
## What is a bajillion?
US, informal. : a huge, unspecified number : bazillion Twister cost, as I understand it, 570 bajillion dollars.
## What is this number 1000000000000000000000000?
Some Very Big, and Very Small NumbersNameThe NumberSymbolquintillion1,000,000,000,000,000,000Equadrillion1,000,000,000,000,000PVery Small !quadrillionth0.000 000 000 000 001f6 more rows | 686 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-49 | longest | en | 0.914749 |
https://programming-articles.com/how-to-convert-a-string-to-an-integer-in-javascript/ | 1,652,748,899,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00307.warc.gz | 540,023,822 | 14,907 | # How to convert a string to an integer in JavaScript?
The simplest way would be to use the native `Number` function:
``````var x = Number("1000")
``````
If that doesn’t work for you, then there are the parseIntunary plusparseFloat with floor, and Math.round methods.
parseInt:
``````var x = parseInt("1000", 10); // you want to use radix 10
// so you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])
``````
unary plus if your string is already in the form of an integer:
``````var x = +"1000";
``````
if your string is or might be a float and you want an integer:
``````var x = Math.floor("1000.01"); //floor automatically converts string to number
``````
or, if you’re going to be using Math.floor several times:
``````var floor = Math.floor;
var x = floor("1000.01");
``````
If you’re the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.
``````var floor = Math.floor;
var x = floor(parseFloat("1000.01"));
``````
Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:
``````var round = Math.round;
var x = round("1000"); //equivalent to round("1000",0)``````
## How to convert a string to an integer in JavaScript?
To convert a string to an integer in JavaScript, use parseInt() function:
``````var number = parseInt("10");
``````
But there is a problem. If you try to convert “010” using parseInt function, it detects as octal number, and will return number 8. So, you need to specify a radix (from 2 to 36). In this case base 10.
``````parseInt(string, radix)
``````
Example:
``````var result = parseInt("010", 10) == 10; // Returns true
var result = parseInt("010") == 10; // Returns false
``````
Note that `parseInt` ignores bad data after parsing anything valid.
This guid will parse as 51:
``var result = parseInt('51e3daf6-b521-446a-9f5b-a1bb4d8bac36', 10) == 51; // Returns true``
There are two main ways to convert a string to a number in javascript. One way is to parse it and the other way is to change its type to a Number. All of the tricks in the other answers (e.g. unary plus) involve implicitly coercing the type of the string to a number. You can also do the same thing explicitly with the Number function.
Parsing
``````var parsed = parseInt("97", 10);
``````
parseInt and parseFloat are the two functions used for parsing strings to numbers. Parsing will stop silently if it hits a character it doesn’t recognise, which can be useful for parsing strings like “92px”, but it’s also somewhat dangerous, since it won’t give you any kind of error on bad input, instead you’ll get back NaN unless the string starts with a number. Whitespace at the beginning of the string is ignored. Here’s an example of it doing something different to what you want, and giving no indication that anything went wrong:
``````var widgetsSold = parseInt("97,800", 10); // widgetsSold is now 97
``````
It’s good practice to always specify the radix as the second argument. In older browsers, if the string started with a 0, it would be interpreted as octal if the radix wasn’t specified which took a lot of people by surprise. The behaviour for hexadecimal is triggered by having the string start with 0x if no radix is specified, e.g. `0xff`. The standard actually changed with ecmascript 5, so modern browsers no longer trigger octal when there’s a leading 0 if no radix has been specified. parseInt understands radixes up to base 36, in which case both upper and lower case letters are treated as equivalent.
Changing the Type of a String to a Number
All of the other tricks mentioned above that don’t use parseInt, involve implicitly coercing the string into a number. I prefer to do this explicitly,
``````var cast = Number("97");
``````
This has different behavior to the parse methods (although it still ignores whitespace). It’s more strict: if it doesn’t understand the whole of the string than it returns `NaN`, so you can’t use it for strings like `97px`. Since you want a primitive number rather than a Number wrapper object, make sure you don’t put `new` in front of the Number function.
Obviously, converting to a Number gives you a value that might be a float rather than an integer, so if you want an integer, you need to modify it. There are a few ways of doing this:
``````var rounded = Math.floor(Number("97.654")); // other options are Math.ceil, Math.round
var fixed = Number("97.654").toFixed(0); // rounded rather than truncated
var bitwised = Number("97.654")|0; // do not use for large numbers
``````
Any bitwise operator (here I’ve done a bitwise or, but you could also do double negation as in an earlier answer or a bitshift) will convert the value to a 32bit integer, and most of them will convert to a signed integer. Note that this will not do want you want for large integers. If the integer cannot be represented in 32bits, it will wrap.
``````~~"3000000000.654" === -1294967296
// This is the same as
Number("3000000000.654")|0
"3000000000.654" >>> 0 === 3000000000 // unsigned right shift gives you an extra bit
"300000000000.654" >>> 0 === 3647256576 // but still fails with larger numbers
``````
To work correctly with larger numbers, you should use the rounding methods
``````Math.floor("3000000000.654") === 3000000000
// This is the same as
Math.floor(Number("3000000000.654"))
``````
Bear in mind that coeercion understands exponential notation and Infinity, so `2e2` is `200` rather than NaN, while the parse methods don’t.
Custom
It’s unlikely that either of these methods do exactly what you want. For example, usually I would want an error thrown if parsing fails, and I don’t need support for Infinity, exponentials or leading whitespace. Depending on your usecase, sometimes it makes sense to write a custom conversion function.
Always check that the output of Number or one of the parse methods is the sort of number you expect. You will almost certainly want to use `isNaN` to make sure the number is not NaN (usually the only way you find out that the parse failed).
## Fastest method of converting a string to integer in JavaScript:
``````var x = "1000"*1;
``````
### Test
Here is a little comparison of speed (Mac Os only)… 🙂
For chrome ‘plus’ and ‘mul’ are fastest (>700,000,00 op/sec), ‘Math.floor’ is slowest. For Firefox ‘plus’ is slowest (!) ‘mul’ is fastest (>900,000,000 op/sec). In Safari ‘parseInt’ is fastes, ‘number’ is slowest (but resulats are quite similar, >13,000,000 <31,000,000). So Safari for cast string to int is more than 10x slower than other browsers. So the winner is ‘mul‘ 🙂
Update
I also test `var x = ~~"1000";` – on Chrome and Safari is a little bit slower than `var x = "1000"*1` (<1%), on Firefox is a little bit faster (<1%). I update above picture and test.
## How to convert a string to an integer in JavaScript?- Answer #5:
To convert a String into Integer, I recommend using parseFloat and NOT parseInt. Here’s why:
Using parseFloat:
``````parseFloat('2.34cms') //Output: 2.34
parseFloat('12.5') //Output: 12.5
parseFloat('012.3') //Output: 12.3
``````
Using parseInt:
``````parseInt('2.34cms') //Output: 2
parseInt('12.5') //Output: 12
parseInt('012.3') //Output: 12
``````
So if you have noticed parseInt discards the values after the decimals, whereas parseFloat lets you work with floating point numbers and hence more suitable if you want to retain the values after decimals. Use parseInt if and only if you are sure that you want the integer value. | 2,009 | 7,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | longest | en | 0.700237 |
https://thetopsites.net/article/53290617.shtml | 1,620,908,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989916.34/warc/CC-MAIN-20210513111525-20210513141525-00034.warc.gz | 512,828,178 | 6,812 | ## Finding the Average in a sequence of number in c while ignoring the negative numbers in the sequence
program to find average of n numbers in python
c program to find positive or negative number in array
c program to count positive and negative numbers using while loop
average of numbers in c
c++ program to find average of n numbers using while loop
how to find the average of 3 numbers in python
c program to find average of 5 numbers
c program to find average of n numbers using functions
I have to write a C program that reads a sequence of integers (positive, negative, or zero) and calculate the average of the positive integers only. If there are no positive numbers, you should display the following statement followed by a new line No positive numbers!
Here is my code, I just need help with how to ignore the negative numbers in the input sequence.
```#include<stdio.h>
int main(){
int num; //number of elements
int i;
int sequence[100]; //numeber if sequence
int sum = 0.0; //sum if the sequence
float avg; // the average
printf("Enter the number of number in the sequence:\n");
scanf("%d", &num);
while (num < 0 || sequence[num] < 0) {
printf("No positine numbers!\n");
}
printf("Enter the sequence:\n");
for (i=0; i < num; i++) {
scanf("%d", &sequence[i]);
sum += sequence[i];
}
avg = (float) sum / num;
printf("Average is %.2f\n", avg);
return(0);
}
```
C Exercises: Counts the number of positive numbers and print the , C programming, exercises, solution: Write a C program that read 5 numbers and counts the number of positive numbers and print the average C Program to find Sum and Average of n Number using For Loop. This C program allows the user to enter the number (n) he wishes to calculate the average and sum. Next, it will ask the user to enter individual items up to a declared number.
If you want the average of the positive numbers, you should use a different variable than num for your final avg calculation.
I would do something like this:
``` int PositiveNumCount = 0;
float avg;
for (i=0; i < num; i++) {
scanf("%d", &sequence[i]);
if(sequence[i] > 0){ sum += sequence[i]; PositiveNumCount++;}
}
avg = (float) sum / PositiveNumCount;
```
Find negative numbers from array of integers using C# program , How do you find a negative number in an array? C/C++ :: Calculate Sum And Average Of A Sequence Of Numbers Feb 17, 2014. Write a C program that calculates the sum and average of a sequence of numbers. Assume the first number specifies the number of values remaining to be entered. If the first number is less than 1, then the program should display an "Invalid Entry Please enter a
Here:
```while (num < 0 || sequence[num] < 0) {
```
you are testing `sequence[num]` before you have written any values to the array, and the test of `num` should be `num <= 0` since if there are zero values, there are also no positive values. You can only decide that there are no positive values after you have entered them all.
The array `sequence` is unnecessary - you can simply accumulate the positive values to `sum` and discard them, maintaining a count of valid values. The avoids any buffer overrun issues also.
`sum` is an `int`, but you have unnecessarily initialised it with a `double`.
```avg = (float) sum / num ;
```
is incorrect since `num` includes negative values - you need a separate count of valid values contributing to the average (unless the intent is to treat all negative values as zero).
Another issue is that you have not validated the input matched the format specifier.
```int main()
{
printf("Enter the number of number in the sequence:\n");
int num = 0 ;
int check = scanf( "%d", &num ) ;
int positive_count = 0 ;
int sum = 0 ;
if( check != 0 && num >= 0 )
{
printf( "Enter the sequence:\n" ) ;
int i = 0 ;
while( i < num )
{
int value = 0 ;
check = scanf( "%d", &value ) ;
if( check != 0 )
{
i++ ;
if( value >= 0 )
{
positive_count++ ;
sum += value ;
}
}
}
}
if( positive_count != 0 )
{
float avg = (float) sum / positive_count ;
printf( "Average is %.2f\n", avg ) ;
}
else
{
printf( "No positive numbers!\n" ) ;
}
return 0 ;
}
```
Average of even numbers till a given even number?, How do you find the average of 5 numbers in C? How To Write a C program To Find The Average of Three Numbers in This Video. some related videos. 4. Write A Program In C To Find a Number of Positive and Negative
```printf("Enter the sequence:\n");
for (i=0; i < num; i++) {
scanf("%d", &sequence[i]);
if(sequeunce[i] > 0){ //This if statement is what your looking for, but there are several other problems in your code. GL!
sum += sequence[i];
}
}
```
The above if statement is what the OP was asking for: "Here is my code, I just need help with how to ignore the negative numbers in the input sequence."
There are many issues with his posted code, including using variables before they are initialized and infinite looping paths. This is the shortest answer to his question.
How do I add negative numbers in C? : C_Programming, status (or a status above 255). signed int x = 1 -3 +5; ? You need to give us some code if you really want help. Enter number: 45.6 Average = 27.69. This program calculates the average if the number of data are from 1 to 100. If user enters value of n above 100 or below 100 then, while loop is executed which asks user to enter value of n until it is between 1 and 100.
Solved: Write A Program That Calculates The Average Of A S , The User Can Enter As Many Non-negative Numbers As They Want, And They For This Program, Zero Counts As A Number That Goes Into The Average. the user if they want to calculate another average for a different set of numbers. Fill in the method with a loop that runs while the user enters non-negative numbers. Number of ways to obtain each numbers in range [1, b+c] by adding any two numbers in range [a, b] and [b, c] Count of subarrays of size K which is a permutation of numbers from 1 to K; Count of triplets of numbers 1 to N such that middle element is always largest; Minimum LCM of all pairs in a given array
Move all negative numbers to beginning and positive to end with , Segregating negative and positive maintaining order and O(1) space · Find ratio of zeroes, positive numbers and negative numbers in the Array · C program to I have to design a C++ program that reads a sequence of positive integer values that ends with zero and find the length of the longest increasing subsequence in the given sequence. For example, for the following sequence of integer numbers:
Average of even numbers till a given even number, We can calculate average by adding each even numbers till n and then dividing sum by count. C++. Please help out here. I want to create a program whereby a user inputs several numbers (let's say 6 numbers from his/ her head). The program should then go ahead and calculate the sum of all these numbers. I however have to use a loop statement, either For statement, While statement or do/while statement. This is what I have so far:
• Please explain each line of your coding attempt, by adding a comment. It is very unclear what your thinking is. What is the idea of `sequence[num] < 0`? Where in your code is anything related to "only positive numbers"? | 1,744 | 7,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-21 | latest | en | 0.829451 |
https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus_(OpenStax)/01%3A_Functions/1.0R%3A_1.R%3A_Functions_(Review) | 1,590,620,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00144.warc.gz | 437,215,606 | 26,491 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.R: Functions (Review)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
## 1.1: Functions and Function Notation
For the exercises 1-4, determine whether the relation is a function.
1) $$\{(a,b),(c,d),(e,d)\}$$
function
2) $$\{(5,2),(6,1),(6,2),(4,8)\}$$
3) $$y^2+4=x$$$,$for $$x$$ the independent variable and $$y$$ the dependent variable
not a function
4) Is the graph in the Figure below a function?
For the exercises 5-6, evaluate the function at the indicated values: $$f(-3); f(2); f(-a); -f(a); f(a+h)$$
5) $$f(x)=-2x^2+3x$$
$$f(-3)=-27; f(2)=-2;f(-a)=-2a^2-3a;-f(a)=2a^2-3a;f(a+h)=-2a^2+3a-4ah+3h-2h^2$$
6) $$f(x)=2|3x-1|$$
For the exercises 7-8, determine whether the functions are one-to-one.
7) $$f(x)=-3 x+5$$
one-to-one
8) $$f(x)=|x-3|$$
For the exercises 9-11, use the vertical line test to determine if the relation whose graph is provided is a function.
9)
function
10)
11)
function
For the exercises 12-13, graph the functions.
12) $$f(x)=|x+1|$$
13) $$f(x)=x^{2}-2$$
For the exercises 14-17, use the Figure below to approximate the values.
14) $$f(2)$$
15) $$f(-2)$$
$$2$$
16) If $$f(x)=-2$$, then solve for $$x$$
17) If $$f(x)=1$$, then solve for $$x$$
$$x=-1.8$$ or $$x=1.8$$
For the exercises 18-19, use the function $$h(t)=-16 t^{2}+80t$$ to find the values.
18) $$\dfrac{h(2)-h(1)}{2-1}$$
19) $$\dfrac{h(a)-h(1)}{a-1}$$
$$\dfrac{-64+80 a-16 a^{2}}{-1+a}=-16 a+64$$
## 1.2: Domain and Range
For the exercises 1-4, find the domain of each function, expressing answers using interval notation.
1) $$f(x)=\dfrac{2}{3 x+2}$$
2) $$f(x)=\frac{x-3}{x^{2}-4 x-12}$$
$$(-\infty,-2) \cup(-2,6) \cup(6, \infty)$$
3)
4) Graph this piecewise function: $$f(x)=\left\{\begin{array}{ll}{x+1} & {x<-2} \\ {-2 x-3} & {x \geq-2}\end{array}\right.$$
## 1.3: Rates of Change and Behavior of Graphs
For the exercises 1-3, find the average rate of change of the functions from $$x=1$$ to $$x=2$$
1) $$f(x)=4 x-3$$
2) $$f(x)=10 x^{2}+x$$
$$31$$
3) $$f(x)=-\dfrac{2}{x^{2}}$$
For the exercises 4-6, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant.
4)
increasing $$(2, \infty)$$; decreasing $$(-\infty, 2)$$
5)
6)
increasing $$(-3,1)$$; constant $$(-\infty,-3) \cup(1, \infty)$$
7) Find the local minimum of the function graphed in Exercise 4.
8) Find the local extrema for the function graphed in Exercise 5.
local minimum $$(-2,-3)$$; local maximum $$(1,3)$$
9) For the graph in the Figure in Exercise 10, the domain of the function is $$[-3,3]$$. The range is $$[-10,10]$$. Find the absolute minimum of the function on this interval.
10) Find the absolute maximum of the function graphed in the Figure below.
$$(-1.8,10)$$
## 1.4: Composition of Functions
For the exercises 1-5, find $$(f \circ g)(x)$$ and $$(g \circ f)(x)$$ for each pair of functions.
1) $$f(x)=4-x, g(x)=-4x$$
2) $$f(x)=3 x+2, g(x)=5-6x$$
$$(f \circ g)(x)=17-18 x ;(g \circ f)(x)=-7-18x$$
3) $$f(x)=x^{2}+2 x, g(x)=5 x+1$$
4) $$f(x)=\sqrt{x+2}, g(x)=\dfrac{1}{x}$$
$$(f \circ g)(x)=\sqrt{\dfrac{1}{x}+2} ;(g \circ f)(x)=\dfrac{1}{\sqrt{x+2}}$$
5) $$f(x)=\dfrac{x+3}{2}, g(x)=\sqrt{1-x}$$
For the exercises 6-9, find $$(f \circ g)$$ and the domain for $$(f \circ g)(x)$$ for each pair of functions.
6) $$f(x)=\frac{x+1}{x+4}, g(x)=\frac{1}{x}$$
$$(f \circ g)(x)=\dfrac{1+x}{1+4 x}, x \neq 0, x \neq-\dfrac{1}{4}$$
7) $$f(x)=\dfrac{1}{x+3}, g(x)=\dfrac{1}{x-9}$$
8) $$f(x)=\dfrac{1}{x}, g(x)=\sqrt{x}$$
$$(f \circ g)(x)=\frac{1}{\sqrt{x}}, x>0$$
9) $$f(x)=\frac{1}{x^{2}-1}, g(x)=\sqrt{x+1}$$
For the exercises 10-11, express each function $$H$$ as a composition of two functions $$f$$ and $$g$$ where $$H(x)=(f \circ g)(x)$$
10) $$H(x)=\sqrt{\frac{2 x-1}{3 x+4}}$$
sample: $$g(x)=\dfrac{2 x-1}{3 x+4}; f(x)=\sqrt{x}$$
11) $$H(x)=\dfrac{1}{\left(3 x^{2}-4\right)^{-3}}$$
## 1.5: Transformation of Functions
For the exercises 1-8, sketch a graph of the given function.
1) $$f(x)=(x-3)^{2}$$
2) $$f(x)=(x+4)^{3}$$
3) $$f(x)=\sqrt{x}+5$$
4) $$f(x)=-x^{3}$$
5) $$f(x)=\sqrt[3]{-x}$$
6) $$f(x)=5 \sqrt{-x}-4$$
7) $$f(x)=4[|x-2|-6]$$
8) $$f(x)=-(x+2)^{2}-1$$
For the exercises 9-10, sketch the graph of the function $$g$$ if the graph of the function $$f$$ is shown in the Figure below.
9) $$g(x)=f(x-1)$$
10) $$g(x)=3 f(x)$$
For the exercises 11-12, write the equation for the standard function represented by each of the graphs below.
11)
$$f(x)=|x-3|$$
12)
For the exercises 13-15, determine whether each function below is even, odd, or neither.
13) $$f(x)=3 x^{4}$$
even
14) $$g(x)=\sqrt{x}$$
15) $$h(x)=\frac{1}{x}+3 x$$
odd
For the exercises 16-18, analyze the graph and determine whether the graphed function is even, odd, or neither.
16)
17)
even
18)
## 1.6: Absolute Value Functions
For the exercises 1-3, write an equation for the transformation of $$f(x)=|x|$$.
1)
$$f(x)=\dfrac{1}{2}|x+2|+1$$
2)
3)
$$f(x)=-3|x-3|+3$$
For the exercises 4-6, graph the absolute value function.
4) $$f(x)=|x-5|$$
5) $$f(x)=-|x-3|$$
6) $$f(x)=|2 x-4|$$
For the exercises 7-8, solve the absolute value equation.
7) $$|x+4|=18$$
$$x=-22, x=14$$
8) $$\left|\dfrac{1}{3} x+5\right|=\left|\dfrac{3}{4} x-2\right|$$
For the exercises 9-10, solve the inequality and express the solution using interval notation.
9) $$|3 x-2|<7$$
$$\left(-\dfrac{5}{3}, 3\right)$$
10) $$\left|\dfrac{1}{3} x-2\right| \leq 7$$
## 1.7: Inverse Functions
For the exercises 1-2, find $$f^{-1}(x)$$ for each function.
1) $$f(x)=9+10 x$$
2) $$f(x)=\dfrac{x}{x+2}$$
$$f^{-1}(x)=\dfrac{-2 x}{x-1}$$
3) For the following exercise, find a domain on which the function $$f$$ is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of $$f$$ restricted to that domain. $f(x)=x^{2}+1$
4) Given $$f(x)=x^{3}-5$$ and $$g(x)=\sqrt[3]{x+5}$$ :
1. Find $$f(g(x))$$ and $$g(f(x))$$.
2. What does the answer tell us about the relationship between $$f(x)$$ and $$g(x) ?$$
1. $$f(g(x))=x$$ and $$g(f(x))=x$$
2. This tells us that $$f$$ and $$g$$ are inverse functions
For the exercises 5-8, use a graphing utility to determine whether each function is one-to-one.
5) $$f(x)=\dfrac{1}{x}$$
The function is one-to-one.
6) $$f(x)=-3 x^{2}+x$$
The function is not one-to-one.
7) If $$f(5)=2,$$ find $$f^{-1}(2)$$
$$5$$
8) If $$f(1)=4,$$ find $$f^{-1}(4)$$
## Practice Test
For the exercises 1-2, determine whether each of the following relations is a function.
1) $$y=2 x+8$$
The relation is a function.
2) $$\{(2,1),(3,2),(-1,1),(0,-2)\}$$
For the exercises 3-4, evaluate the function $$f(x)=-3 x^{2}+2 x$$ at the given input.
3) $$f(-2)$$
$$-16$$
4) $$f(a)$$
5) Show that the function $$f(x)=-2(x-1)^{2}+3$$ is not one-to-one.
The graph is a parabola and the graph fails the horizontal line test.
6) Write the domain of the function $$f(x)=\sqrt{3-x}$$ in interval notation.
7) Given $$f(x)=2 x^{2}-5 x,$$ find $$f(a+1)-f(1)$$
$$2 a^{2}-a$$
8) Graph the function $$f(x)=\left\{\begin{array}{ccc}{x+1} & {\text { if }} & {-2<x<3} \\ {-x} & {\text { if }} & {x \geq 3}\end{array}\right.$$
9) Find the average rate of change of the function $$f(x)=3-2 x^{2}+x$$ by finding $$\dfrac{f(b)-f(a)}{b-a}$$
$$-2(a+b)+1$$
For the exercises 10-11, use the functions $$f(x)=3-2 x^{2}+x$$ and $$g(x)=\sqrt{x}$$ to find the composite functions.
10) $$(g \circ f)(x)$$
11) $$(g \circ f)(1)$$
$$\sqrt{2}$$
12) Express $$H(x)=\sqrt[3]{5 x^{2}-3 x}$$ a composition of two functions, $$f$$ and $$g,$$ where $$(f \circ g)(x)=H(x)$$
For the exercises 13-14, graph the functions by translating, stretching, and/or compressing a toolkit function.
13) $$f(x)=\sqrt{x+6}-1$$
14) $$f(x)=\dfrac{1}{x+2}-1$$
For the exercises 15-17, determine whether the functions are even, odd, or neither.
15) $$f(x)=-\dfrac{5}{x^{2}}+9 x^{6}$$
even
16) $$f(x)=-\dfrac{5}{x^{3}}+9 x^{5}$$
17) $$f(x)=\dfrac{1}{x}$$
odd
18) Graph the absolute value function $$f(x)=-2|x-1|+3$$.
19) Solve $$|2 x-3|=17$$.
$$x=-7$$ and $$x=10$$
20) Solve $$-\left|\dfrac{1}{3} x-3\right| \geq 17$$ . Express the solution in interval notation.
For the exercises 21-22, find the inverse of the function.
21) $$f(x)=3 x-5$$
$$f^{-1}(x)=\dfrac{x+5}{3}$$
22) $$f(x)=\dfrac{4}{x+7}$$
For the exercises 23-26, use the graph of $$g$$ shown in the Figure below.
23) On what intervals is the function increasing?
$$(-\infty,-1.1)$$ and $$(1.1, \infty)$$
24) On what intervals is the function decreasing?
25) Approximate the local minimum of the function. Express the answer as an ordered pair.
$$(1.1,-0.9)$$
26) Approximate the local maximum of the function. Express the answer as an ordered pair.
For the exercises 27-29, use the graph of the piecewise function shown in the Figure below.
27) Find $$f(2)$$.
$$f(2)=2$$
28) Find $$f(-2)$$.
29) Write an equation for the piecewise function.
$$f(x)=\left\{\begin{array}{cl}{|x|} & {\text { if } x \leq 2} \\ {3} & {\text { if } x>2}\end{array}\right.$$
For the exercises 30-35, use the values listed in the Table below.
$$x$$ $$F(x)$$ 0 1 1 3 2 5 3 7 4 9 5 11 6 13 7 15 8 17
30) Find $$F(6)$$.
31) Solve the equation $$F(x)=5$$
$$x=2$$
32) Is the graph increasing or decreasing on its domain?
33) Is the function represented by the graph one-to-one?
34) Find $$F^{-1}(15)$$.
35) Given $$f(x)=-2 x+11,$$ find $$f^{-1}(x)$$.
$$f^{-1}(x)=-\dfrac{x-11}{2}$$ | 4,243 | 10,775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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} | 3.78125 | 4 | CC-MAIN-2020-24 | latest | en | 0.273877 |
https://www.rp-photonics.com/saturation_power.html | 1,642,941,233,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00298.warc.gz | 1,004,473,836 | 9,823 | Encyclopedia … combined with a great Buyer's Guide!
# Saturation Power
Definition: a measure of the incident optical power required for achieving significant saturation of an absorber or a gain medium
German: Sättigungsleistung
Formula symbol: Psat
Units: W
Author:
The saturation power of a laser gain medium is the optical power of an input signal which in the steady state leads to a reduction in the gain to half of its small-signal value. Similarly, the saturation power of a saturable absorber is defined, although saturable absorbers are usually not meant to be operated in the steady state.
The saturation intensity is the corresponding optical intensity, i.e., the saturation power per unit area.
Usually it is assumed that the gain is small, i.e. input and output powers are similar. For high gain, it is common to refer to the output power.
For a low-gain laser amplifier, saturation intensity and power can be calculated according to
where h ν is the photon energy at the signal wavelength, σem and σabs are the emission and absorption cross sections at the emission wavelength, τ is the upper-state lifetime, and A is the effective mode area. The quantity σabs is zero for four-level gain media but should not be forgotten for quasi-three-level gain media.
## Calculator for the Saturation Power
Enter input values with units, where appropriate. After you have modified some inputs, click the "calc" button to recalculate the output.
A comparison with the equations for the saturation energy shows that the saturation power can be calculated as the saturation energy divided by the upper-state lifetime.
## Importance of the Saturation Power
The saturation power plays an important role in various areas of laser physics and laser or amplifier design. Some examples are:
• It determines the amount of input power of an amplifier required for achieving most of the possible output power.
• The laser intensity in the gain medium of a four-level laser equals the saturation intensity if the pump power is twice the threshold pump power. This is remarkable, because the laser intensity in this situation is thus determined only by a property of the gain medium, not by resonator losses etc.
• For a saturable absorber, as used e.g. in a mode-locked laser, the ratio of continuous-wave intracavity power to saturation power is an important parameter for the initial pulse formation process.
The saturation power should not be confused with the saturated output power, which is usually not precisely defined but means the output power achieved for an input signal power which causes significant amplifier saturation. Obviously, the saturated output power (other than the saturation power) depends on the pump power.
## Pump Saturation
A subtle detail is that the saturation characteristics can be modified if the pump intensity of the gain medium is comparable to or higher than the pump saturation power (which is defined as above, but based on photon energy and transition cross sections at the pump wavelength). In this situation, the rule that the gain is reduced by a factor of 2 for an intensity equal to the saturation intensity (as defined above) no longer holds.
For two reasons, however, pump saturation effects are in most cases not very important for lasers and amplifiers:
• In many solid-state bulk lasers, the pump intensity is well below the pump saturation intensity, so that such effects do not occur. (On the other hand, in certain fiber lasers such effects can be very strong.)
• Even if pump saturation is significant at some location within a laser crystal or amplifier, it may overall not be important, provided that the pump absorption within the whole gain medium is complete. In such a situation, local pump saturation simply means that some of the pump power is absorbed at another location within the same medium, effectively not changing the overall saturation characteristics.
## Questions and Comments from Users
Here you can submit questions and comments. As far as they get accepted by the author, they will appear above this paragraph together with the author’s answer. The author will decide on acceptance based on certain criteria. Essentially, the issue must be of sufficiently broad interest. | 826 | 4,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-05 | longest | en | 0.895348 |
https://math.stackexchange.com/questions/2213338/finding-marginal-density-of-a-joint-density-function | 1,722,884,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00123.warc.gz | 321,810,167 | 37,229 | # Finding marginal density of a joint density function
I don't like/don't understand the solution to the SOA practice problem #171 so I'm trying to find the marginal density of x in another way without drawing a diamond.
Here is the joint density: $f(x,y)=\begin{cases} {{1\over2}}&0<|x|+|y|<1\\{0}&otherwise\end{cases}$
I thought I could do the following:
$f_x(x)= \int_{-1}^{1-x}{1\over2}dy \,+\,\int_0^{1+x}{1\over2}dy$
Then I tried
$f_x(x)= \int_{0}^{1-x}{1\over2}dy \,+\,\int_{1-x}^{1+x}{1\over2}dy$
but I'm not getting the same density as them with their drawing. If it comes down to it, I can try to learn their method but I would really like to just do it the way I know how. Can anyone tell me where my limits are going wrong?
• Did you sketch a drawing of the support region? Commented Apr 1, 2017 at 17:13
• @leonbloy it's a diamond, and I thought the 2nd one had the right limits for sure, but it doesn't work. Commented Apr 1, 2017 at 17:15
For a fixed $0<x<1$, the support region (check this) goes from $y=x-1$ to $y=1-x$
Hence, in that range, $$f_x(x)= \int_{x-1}^{1-x}\frac12 dy= 1 - x$$
The range $-1<x<0$ is similar.
• so for $-1<x<0\, f_x(x)=\int_{-x-1}^{x+1}{1\over2}dy=x+1$? Commented Apr 1, 2017 at 18:30
• yes. you should check that the answer makes sense, for example, that $f_x$ integrates to 1 Commented Apr 1, 2017 at 19:27 | 476 | 1,359 | {"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-2024-33 | latest | en | 0.920736 |
https://www.studysmarter.us/explanations/physics/electricity/resistivity/ | 1,656,298,904,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00060.warc.gz | 1,089,924,260 | 35,756 | Suggested languages for you:
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# Resistivity
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When we build an electric circuit, we want it to be as efficient as possible. This means that we want low resistance, so it's only logical that we use materials like copper and not wood or rubber in our circuits. But why? Because materials like wood and rubber have high resistivities compared to copper.
An informal definition of resistivity is ‘the characteristic obstruction materials offer to the flow of charges per unit of length and cross-section’, which is related to the concept of electrical resistance.
## What is electrical resistance?
We often explore electric phenomena in circuits where we can use materials to guide electric charges for different purposes. We use three basic quantities to characterise circuits: resistance, voltage, and current.
Electrical resistance (or simply resistance) is a measure of the opposition of a medium to the movement of charges inside of it. It is measured in ohms (Ω).
The voltage or potential difference is the amount of energy per unit of charge needed to move charges between two points of a circuit. It is usually supplied by batteries, and it is measured in volts (V).
The electric current, or simply the current, is the number of charges that pass a cross-section (transversal cut) of a conductor per unit of time. It is measured in amperes (A).
The role of resistance is most easily seen in Ohm’s law, which governs the behaviour of ohmic conductors and certain ranges of non-ohmic conductors. Its equation is the following:
Here R is the resistance, V is the voltage, and I is the electric current. If a circuit has high resistance, less current will be produced (and vice-versa). Since the current is the flow of charges, it is clear that the bigger the resistance, the bigger the opposition to the movement of charges.
The greater the resistance, the smaller the current. Check out our explanation on the Basics of Electricity and Circuits for more info.
See our explanation on Current-Voltage Characteristics. You’ll have more info on why Ohm’s law is not universal – only some conductors behave as this law predicts, and they are called ohmic conductors. The relationship between voltage, current, and resistance may be as complex as we desire (non-ohmic conductors), but if we limit ourselves to a small region of these quantities, we can always use Ohm’s law in that range.
Above, we have defined resistance, the role it fulfils in circuits, and the movement of charges. However, the definition we have given does not include information about its fundamental nature, i.e. how resistance is generated due to microscopic phenomena. To study these matters in-depth, let’s look at the concept of resistivity.
## The definition of resistivity
Studying the relationship between resistivity and resistance allows us to see why resistivity is a characteristic property of materials and resistance is not.
Resistivity is a quantity that measures the resistance of a conductor per unit of length and cross-section. It is different for each material and depends on certain physical conditions, such as temperature. It is measured in ohm-meters or Ωm and is denoted by the Greek letter ρ.
### Factors affecting resistivity
#### Temperature
Resistivity grows with temperature because the temperature is a measure of the average kinetic energy of the particles of a material. If the particles of the conductor move faster (on average), they are more likely to interfere with the movement of charges.
#### Metallic nature
Another factor that determines the resistivity of a material is its metallic nature. Metals are known to favour the movement of charges, which implies that their characteristic resistivity is lower than the resistivity of other materials like wood or rubber. When we consider metals, their atomic structure and microscopic spatial disposition will determine how easy it is for charges to move, which will ultimately determine the exact value of the resistivity.
Some examples of the characteristic resistivity values of materials are shown below:
Material Resistivity at 20ºC (Ω·m) Silver 1.59 · 10-8 Copper 1.68 · 10-8 Iron 9.71 · 10-8 Carbon 3 · 10-5 - 60 · 10-5 Mercury 98 · 10-8 Silicones 1 · 10-3 - 500 · 10-3 Glass 1 · 109 - 1 · 1013 Rubber 1 · 1013 - 1 · 1015 Air 1.3 · 1016 - 3.3 · 1016
Resistivity is a characteristic property of materials that does not depend on their length and their cross-section.
## The resistivity equation
If we know the resistivity of a material, we can calculate the resistance of a conductor made out of this material by multiplying it by the length and dividing it by the cross-section. Here is the equation that captures the relationship between resistance and resistivity:
Here, R is the resistance, ρ is the resistivity, L is the length of the conductor, and A is its cross-section.
To interpret this equation, we must remember that current is the number of charges that go through a cross-section of a conductor per unit of time.
No matter the shape of a conductor, we can always find the cross-section as the surface perpendicular to the direction of the current at each point.
Now, since we know that resistance measures the opposition of the material to the current flow, why should we consider the material's length? Because the length also directly affects the resistance: the longer a medium (or object) is, the greater the resistance. This means that resistance and length are directly proportional. On the other hand, the resistance is inversely proportional to the medium’s cross-sectional area.
Length
You are in a very crowded street. The street is the conductor, and you are a charge trying to reach the other end of the street by avoiding the people standing in the street. It will be less tiring to walk just one block instead of three blocks because you avoid fewer people by walking just one block (the shorter the distance, the fewer people you encounter, which means that the shorter the length of a conductor, the less resistance there is).
Cross-section
The role of the cross-section is much easier to explain. In the end, we know that resistance measures the opposition to the flow of a current, but the current depends on the cross-section. If we double the size of the cross-section, we also double the current. That means that the opposition (resistance) is still acting, but due to the characteristics of the medium, we get more current (which means less resistance).
Imagine you are at the end of a crowded street, and you have several friends uniformly separated from each other at the other end of the street. If you were to count how many of your friends reach your end of the street per unit of time, you would count double that amount if you were in a street that was two times wider (and, consequently, where you had double the number of friends).
You have a proportional growth in friends by enlarging the street because you are considering a uniform density of charges in a material (following the analogy).
• Resistance grows with the length of conductors since the moving charges find more particles that obstruct them.
• Resistance decreases with the cross-section since the bigger the cross-section, the bigger the number of charges crossing it per unit of time.
## How to calculate resistance using resistivity
Consider two materials, silver and carbon. Silver is really expensive and difficult to get while obtaining carbon is relatively easy. We want to make a cable to connect two parts of a circuit separated by 1 meter. Since silver is hard to get, we only have a wire with a cross-section of 1cm2 (0.0001m2).
How wide should the carbon wire be to transfer current as efficiently as silver?
By using the equation of the resistance in terms of the length, the resistivity (found in the table), and the cross-section, we can calculate the resistance of the silver wire:
We now solve the same equation for the cross-section of carbon and the same resistance:
If we were considering approximately cylindrical wires, this would imply using a cable with a diameter of roughly 0.5m, which is big compared to the silver cable.
If we had considered a copper cable, the diameter would need to be almost the same as for silver (around 1.1cm), which explains why we use copper instead of carbon to make the cables we use.
## Resistivity - key takeaways
• Resistance is a measure of the opposition of a medium to the flow of charges flowing through it.
• Resistivity is a measure of the opposition of a medium to the flow of charges inside of it per unit of length and cross-section. It is a more fundamental quantity than resistance because it does not depend on the size or width of the conductor, just on the properties of the material.
• Resistivity is characteristic for each material at certain external conditions since it is determined by the microscopic characteristics of the material. For instance, temperature affects the resistivity of a material.
• Resistance grows with the length of conductors since the moving charges find more particles that obstruct them.
• Resistance decreases with the cross-section since the bigger the cross-section, the bigger the number of charges crossing it per unit of time.
Resistivity is a quantity that measures the characteristic opposition of a material to the movement of charges inside of it per unit of length and cross-section.
If we know the resistivity of a material, we can calculate the resistance of a conductor made out of it by multiplying it by the length and dividing it by the cross-section.
Resistivity is a quantity that measures the resistance of a conductor per unit of length and cross-section. It is different for each material and depends on certain physical conditions, such as temperature.
The equation for resistivity is ρ=R·A/L
## Final Resistivity Quiz
Question
What is resistance? Choose the correct statement.
Resistance is a quantity that quantifies the obstruction of materials to the movement of charges.
Show question
Question
How is resistivity measured? Choose the correct statement.
Resistivity is measured in ohms-meter (Ω·m).
Show question
Question
Choose the correct statement about the relationship between voltage, current, and resistance.
Voltage, current, and resistance can be related by an arbitrary equation.
Show question
Question
Choose the correct statement.
Silver is a better conductor than copper.
Show question
Question
Choose the correct statement.
Resistivity is a characteristic property of materials that is related to their microscopic structure, and that depends on external factors like temperature.
Show question
Question
Does resistance grow with the length of the conductor?
Yes, resistance grows with the length of a conductor.
Show question
Question
Does resistance grow with the cross-section of the conductor?
No, resistance decreases with the cross-section of a conductor.
Show question
Question
Does resistivity grow with the length of the conductor?
Resistivity is a characteristic property of materials that does not depend on their length.
Show question
Question
Does resistivity decrease with the cross-section of conductors?
Resistivity is a characteristic property of materials that does not depend on their cross-section.
Show question
Question
Are materials with lower resistivity better or worse conductors?
Better conductors.
Show question
Question
Does temperature increase or decrease resistivity?
Temperature increases resistivity.
Show question
Question
What is the name of the conductors that obey Ohm’s law?
Ohmic materials.
Show question
Question
Which quantity better captures the microscopic properties of materials: resistance or resistivity?
Resistivity.
Show question
Question
For non-ohmic conductors, does an increase in resistance imply a decrease in electric current?
Yes, it is independent of the exact form of the mathematical relationship.
Show question
Question
What is the definition of the electric current?
Electric current is the number of charges that go through the cross-section of a material per unit of time.
Show question
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No, I'll do it now | 2,704 | 13,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.933834 |
https://im.kendallhunt.com/MS/teachers/3/2/4/practice.html | 1,719,286,565,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00547.warc.gz | 267,727,678 | 26,043 | # Lesson 4
Dilations on a Square Grid
### Problem 1
Triangle $$ABC$$ is dilated using $$D$$ as the center of dilation with scale factor 2.
The image is triangle $$A’B’C’$$. Clare says the two triangles are congruent, because their angle measures are the same. Do you agree? Explain how you know.
### Problem 2
On graph paper, sketch the image of quadrilateral PQRS under the following dilations:
1. The dilation centered at $$R$$ with scale factor 2.
2. The dilation centered at $$O$$ with scale factor $$\frac{1}{2}$$.
3. The dilation centered at $$S$$ with scale factor $$\frac{1}{2}$$.
### Problem 3
The diagram shows three lines with some marked angle measures.
Find the missing angle measures marked with question marks.
### Solution
(From Unit 1, Lesson 14.)
### Problem 4
Describe a sequence of translations, rotations, and reflections that takes Polygon P to Polygon Q.
### Solution
Point $$B$$ has coordinates $$(\text-2,\text-5)$$. After a translation 4 units down, a reflection across the $$y$$-axis, and a translation 6 units up, what are the coordinates of the image? | 280 | 1,096 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.858682 |
http://openstudy.com/updates/515b4f42e4b07077e0c17cd5 | 1,449,000,470,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398471436.90/warc/CC-MAIN-20151124205431-00088-ip-10-71-132-137.ec2.internal.warc.gz | 171,777,006 | 9,855 | ## NicholasWilliams 2 years ago If f(x) = x + 5, find f(-9).
1. Sophiaaaa
2. some_someone
\[f(x) = x + 5\] \[f(-9) = (-9) + 5\] \[f(-9) = -4\]
3. precal
f(-9)=-4 is function notation it also, represents (-9,-4) this is just coordinate notation | 99 | 247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | longest | en | 0.517592 |
http://inst.eecs.berkeley.edu/~cs61a/su13/hw/hw6.html | 1,550,923,154,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249500704.80/warc/CC-MAIN-20190223102155-20190223124155-00053.warc.gz | 127,179,317 | 5,278 | # 61A Homework 6
Due by 11:59pm on Monday, 7/15
Submission. See the online submission instructions. We have provided a starter file for the questions below.
Readings. Section 2.4 of the online lecture notes.
Q1. Read Albert's debugging guidelines. Then fix the problems with the following code. (Only fix the bugs; do not change the structure of the computation.).
```def divide_by_fact(dividend, n):
"""Recursively divide dividend by the factorial of n.
>>> divide_by_fact(120, 4)
5.0
"""
if n < 0:
return dividend
return divide_by_fact(dividend / n, n - 1)
```
Q2. Write a recursive function that divides an input sequence into a tuple of smaller sequences that each contain 4 or 5 elements from the original sequence. For example, an input sequence of 14 elements should be divided into sequences of 4, 5, and 5 elements. Use as few 5-element sequences as necessary in the result, and all 5-element sequences should be at the end. Finally, preserve the relative ordering of elements from the original sequence.
Hint: You may assume that the input sequence has length at least 12. Think carefully about how many base cases you need, and what they should be.
```def group(seq):
"""Divide a sequence of at least 12 elements into groups of 4 or
5. Groups of 5 will be at the end. Returns a tuple of sequences, each
corresponding to a group.
>>> group(range(14))
(range(0, 4), range(4, 9), range(9, 14))
>>> group(tuple(range(17)))
((0, 1, 2, 3), (4, 5, 6, 7), (8, 9, 10, 11), (12, 13, 14, 15, 16))
"""
num = len(seq)
assert num >= 12
```
Pizza Sort
Leonardo da Vinci is well know for his inventions, his paintings, and his numerous contributions to many other fields. It is less known that he only turned to art after failing miserably as an apprentice in a pizzeria. (Granted, modern pizza was invented centuries after his era, but its precursors have been around since Roman times.)
One day, little Leonardo was at the pizzeria, mindlessly flipping over partially cooked crusts with a spatula. At that moment, inspiration struck, and he realized that he could sort the crusts in a stack, so that the largest ended up at the bottom and the smallest at the top, with each crust smaller than the one below it. He named his procedure "pizza sort" and wrote it down in a journal.
Unfortunately, that journal was lost to time, so pizza sort, perhaps Leonardo's greatest invention, remained unknown. Until now. Last year, a previously unknown journal was found in the hands of a private collector, and it appears to be that elusive journal in which Leonardo recorded his pizza sort. Unfortunately, time has not been kind to the journal, so many details are missing. But archaeologists have managed to recover one key piece of the puzzle: the only allowed move in pizza sort is to place the spatula under a crust in the stack and flip the set of crusts above the spatula:
```"""
====
==
========== <--- spatula underneath this crust
========
||
||
\||/
\/
========== }
== } flipped
==== }
========
"""
```
Q3. Implement this flip operation in code, with the stack of crusts represented as a list starting from the bottom of the stack. Write a function partial_reverse to reverse the subset of a list starting from start until the end of the list. This reversal should be in-place, meaning that the original list is modified. Do not create a new list inside your function, even if you do not return it.
Note: As can be seen from the doctests, partial_reverse should return None. (Don't put in an explicit return None, however; either leave out any return statement or just use return with no expression.)
```def partial_reverse(lst, start):
"""Reverse part of a list in-place, starting with start up to the end of
the list.
>>> a = [1, 2, 3, 4, 5, 6, 7]
>>> partial_reverse(a, 2)
>>> a
[1, 2, 7, 6, 5, 4, 3]
>>> partial_reverse(a, 5)
>>> a
[1, 2, 7, 6, 5, 3, 4]
"""
```
The archaeologists also found some indecipherable symbols along with the description of the above procedure. As they could make neither heads nor tails of it, they called it "the da Vinci code." Hoping to crack it, they asked Eva Lu Ator, a leading computer scientist, for help. Upon closer examination, she realized that the da Vinci code was actual code, in a rudimentary programming language that Leonardo had invented. This code would find the position of the largest crust in a stack of pizzas.
Q4. Help Eva to translate the da Vinci code into Python. Write a function that takes in a sequence and returns the index of the largest element in the sequence:
```def index_largest(seq):
"""Return the index of the largest element in the sequence.
>>> index_largest([8, 5, 7, 3 ,1])
0
>>> index_largest((4, 3, 7, 2, 1))
2
"""
assert len(seq) > 0
```
Q5. Finally, help the world learn the secret of the da Vinci code by writing the pizza_sort function. This function takes in a list and sorts its elements in descending order, using only the previous two functions and the built-in len function. It is well-known that Leonardo hated iteration, so make sure to use recursion instead. (You may define a helper function if you want.) Lastly, you may use slicing; we are talking about pizza, after all.
(Keep in mind, however, that slicing returns a new list, but you need to sort the existing list in place.)
```def pizza_sort(lst):
"""Perform an in-place pizza sort on the given list, resulting in
elements in descending order.
>>> a = [8, 5, 7, 3, 1, 9, 2]
>>> pizza_sort(a)
>>> a
[9, 8, 7, 5, 3, 2, 1]
"""
```
Marvel in the genius that was Leonardo da Vinci!
Q6. Define a function make_accumulator that returns an accumulator function, which takes one numerical argument and returns the sum of all arguments ever passed to accumulator. Use a list and not a nonlocal statement:
```def make_accumulator():
"""Return an accumulator function that takes a single numeric argument and
accumulates that argument into total, then returns total.
>>> acc = make_accumulator()
>>> acc(15)
15
>>> acc(10)
25
>>> acc2 = make_accumulator()
>>> acc2(7)
7
>>> acc3 = acc2
>>> acc3(6)
13
>>> acc2(5)
18
>>> acc(4)
29
"""
```
Q7. Define a function make_accumulator_nonlocal that returns an accumulator function, which takes one numerical argument and returns the sum of all arguments ever passed to accumulator. Use a nonlocal statement, but no list or dict:
```def make_accumulator_nonlocal():
"""Return an accumulator function that takes a single numeric argument and
accumulates that argument into total, then returns total.
>>> acc = make_accumulator_nonlocal()
>>> acc(15)
15
>>> acc(10)
25
>>> acc2 = make_accumulator_nonlocal()
>>> acc2(7)
7
>>> acc3 = acc2
>>> acc3(6)
13
>>> acc2(5)
18
>>> acc(4)
29
"""
```
Q8. (Extra for Experts) Recall the count_change function from earlier in the semester. (Below, we have converted it to use a tuple rather than a function to represent a coin sequence, but it is otherwise the same.) It was quite slow on larger inputs, since it repeated the same computation many times. Instead, write a function make_count_change that produces a more efficient version of count_change. This more efficient version should only perform any computation if it is called with a new pair of arguments. If it is called with a pair of arguments previously seen, then it should just return the previously computed value. Once you are done, compare the two versions on a large number such as 500 to make sure that your code is faster than the original version.
```# Old version
def count_change(a, coins=(50, 25, 10, 5, 1)):
if a == 0:
return 1
elif a < 0 or len(coins) == 0:
return 0
return count_change(a, coins[1:]) + count_change(a - coins[0], coins)
# Version 2.0
def make_count_change():
"""Return a function to efficiently count the number of ways to make
change.
>>> cc = make_count_change()
>>> cc(500, (50, 25, 10, 5, 1))
59576
""" | 1,992 | 7,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.918734 |
https://de.mathworks.com/matlabcentral/answers/46480-how-to-setup-a-color-matrix-in-surf-x-y-z-c-that-has-two-components-one-fixed-one-related-to-data | 1,621,384,156,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00525.warc.gz | 218,333,571 | 24,871 | # how to setup a color matrix in surf(x,y,z,C) that has two components, one fixed, one related to data?
3 views (last 30 days)
Francisco on 22 Aug 2012
I am solving a wave like equation using the perfectly matched layer method, where an absorbing layer surrounds the domain of interest. I am able to compute the solution and want to plot it. So far what I have been doing is assigning the absorbing layer a fixed color, and the interior of the domain a checkerboard pattern using the code below and surf(X,Y,Z,C). However, I am wondering if it is possible to keep the color of the exterior boundary fixed, but have the interior of the domain have a color assigned related to the Z data.
C=zeros(N+2*Nl);
for n=1:N+2*Nl
for p=1:N+2*Nl
if n<=Nl || n>=(N+Nl) ||p<=Nl || p>=(N+Nl)
C(n,p)=1;
else
if mod(floor(n/10)+floor(p/10),2)==0
C(n,p)=4/3;
else
C(n,p)=5/3;
end
end
end
end
So far the plot looks like this one, but I would like to have instead of the checkerboard pattern the interior to have a colormap related to the z-data, and it would look as if you just used surf(X,Y,Z), but keeping the one color exterior layer.
José-Luis on 22 Aug 2012
Could you edit your question so the code is a bit clearer?
Patrick Kalita on 28 Aug 2012
This approach certain has its limitations, but it's also pretty easy to implement: If you set the CData values on the border to be -Inf, those locations will always get the color at the "bottom" of the colormap (which is actually the first row). For example:
%%Generate some X, Y, and ZData
[X, Y] = meshgrid(linspace(-3, 3, 60));
Z = zeros(60);
border = true(60);
border(6:55, 6:55) = false;
Z(~border) = peaks(X(~border), Y(~border));
%%Make CData and plot
C = Z;
C(border) = -Inf;
surf(X, Y, Z, C); | 504 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-21 | latest | en | 0.85843 |
https://www.coursehero.com/file/6334025/hw2-sol/ | 1,512,985,340,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513330.14/warc/CC-MAIN-20171211090353-20171211110353-00246.warc.gz | 737,277,966 | 24,004 | # hw2-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...
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Unformatted text preview: AMS 311 (Fall, 2010) Joe Mitchell PROBABILITY THEORY Homework Set # 2 – Solution Notes (1). (14 points) Consider an experiment in which three fair dice are rolled. What is the conditional probability that at least one lands on 6, given that the sum of the three dice is at least 15? We consider the sample space, S = { (1 , 1 , 1) , (1 , 1 , 2) ,... } , to be the set of all 6 3 equally likely outcomes (triples of integers between 1 and 6), where, e.g., (2,4,4) is the outcome in which the first die lands 2, the second lands 4, and the third lands 4. Let E = { (6 , 1 , 1) , (6 , 1 , 2) ,..., (6 , 6 , 6) , (1 , 6 , 1) , (1 , 6 , 2) ,... } be the event that at least one of the two dice lands on 6. Note that | E | = 1 + 3 · 5 + 3 · 25 = 91, so P ( E ) = 91 / 216. (Reasoning: there is 1 outcome with three 6’s, 3 · 5 outcomes of the form “66x” (3 choices for position of the “x”, and 5 choices for “x”), 3 · 5 2 outcomes of the form “6xy” (3 choices for position of the “6”, and 5 2 = 25 choices for the digits “x” and “y”, each between 1 and 5).) Let F = { (4 , 5 , 6) , (4 , 6 , 5) , (5 , 5 , 5) , (5 , 5 , 6) , (5 , 6 , 5) , (5 , 6 , 6) , (6 , 4 , 5) ,... , (6 , 6 , 6) } be the event that the sum of the three dice is at least 15. Note that | F | = 20, so P ( F ) = 20 / 216 = 5 / 54. (To do the counting, either list the entries explicitly, or note that there are 3 outcomes using the digits “366”, 3! outcomes using the digits “456”, 3 with digits “466”, 1 with “555”, 3 with “556”, 3 with “566”, and 1 with “666”.) Now, E ∩ F = F \ { (5 , 5 , 5) } , since all elements of F have at least one 6, except the one outcome (5,5,5). Thus, P ( E ∩ F ) = (20 − 1) / 216 = 19 / 216. We want to compute P ( E | F ) = P ( E ∩ F ) P ( F ) = P ( E ∩ F ) P ( F ) = 19 / 216 20 / 216 = 19 20 (2). (14 points) A poll shows that 41 percent of the women and 17 percent of the men that took a weight-loss class kept the weight off for at least one year after completing the class. These people then attended a success party at the end of a year. If 72 percent of the original class were male, (a) what percentage of those attending the party were women? (b) what percentage of the original class attended the party? Select a random member of the class. Let A be the event that this member attends the party (i.e., this member kept the weight off for at least a year). Let W be the event that this member is a woman. We are told that P ( A | W ) = 0 . 41 and P ( A | W c ) = 0 . 17, and that P ( W ) = 1 − . 72 = 0 . 28. (a). We want to compute P ( W | A ) = P ( W ∩ A ) P ( A ) = P ( A | W ) P ( W ) P ( A | W ) P ( W ) + P ( A | W c ) P ( W c ) = ( . 41)( . 28) ( . 41)( . 28) + ( . 17)( . 72) ≈ . 484 Thus, 48.4% of the attendees were women....
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hw2-sol - AMS 311(Fall 2010 Joe Mitchell PROBABILITY THEORY...
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Ask a homework question - tutors are online | 1,138 | 3,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-51 | latest | en | 0.859477 |
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