url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://see785.com/xy-full-form/ | 1,679,526,923,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.97/warc/CC-MAIN-20230322211955-20230323001955-00590.warc.gz | 528,809,658 | 10,416 | xy full form
xy full form, An x and y axis, also called a Cartesian coordinate system is an ordered pair of numbers that are displayed as (x,y). x and y are the first two letters of a coordinate plane. The xy coordinates represent points on a 2-dimensional graph which can be used to display data in relation with each other.
1. xy is a mathematical term for the distance between two points.
2. x and y are coordinates on a graph, like in math class.
3. The full form of xy is “x times y” which means you multiply them together to find the answer.
4. You can also use this as a verb – if someone asks you what 2x + 3y equals, you need to multiply 2 by 3 and then add it to the result of 6 (6+6=12).
5. If someone tells you that they’re going to be at work from 8 am until 5 pm, they mean they’ll be there for eight hours plus five hours – so 13 total hours (8+5=13).
In short, XY is a genetic condition in which an individual has two different sex chromosomes. X and Y are the same letter, but they have different roles in genetics. In DNA, X is a girl’s chromosome while Y is a boy’s. | 275 | 1,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-14 | latest | en | 0.954949 |
http://www.jiskha.com/members/profile/posts.cgi?name=joel&page=14 | 1,386,834,481,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164568332/warc/CC-MAIN-20131204134248-00008-ip-10-33-133-15.ec2.internal.warc.gz | 393,862,768 | 2,986 | Thursday
December 12, 2013
# Posts by joel
Total # Posts: 240
Physics
A 3kg mass moving initially to the right along the x axis with a speed of 8m/s makes a perfectly inelastic collision with a 5kg mass initially at rest at the origin.what fraction of the initial kinetic energy of the system is lost in the collision.what formula must i use.
Physics
A 3kg mass moving initially to the right along the x axis with a speed of 8m/s makes a perfectly inelastic collision with a 5kg mass initially at rest at the origin.what fraction of the initial kinetic energy of the system is lost in the collision.what formula must i use.
Pre-Algebra
Thank you, Ms. Sue!
Pre-Algebra
If 4 liters of motor oil cost \$3.88, what is the price for 1 liter? I'm supposed yo set up proportions. This is how I set it up: \$3.88/4 liters • 1 liter/p p= the price for 1 liter Please tell me if I set it up right and how to solve.
physical science
gravity
Maths
is 10*10exponent -3 a scientific notation?
Physics
Two points charges of 2uC and -6uC are located at the position(0.0)m and (0.3)m.Calculate the total electrical potential at the point (4.0)m due these two charges.
Physics
hi Esskay,can u help me with the answer
Physics
A proton is released from rest in a uniform electric field which is directed along the x axis and of magnitude E=2*10exponent5 V/m.The proton undergoes a displacement from point A at the origin to point B at x=0.2m.What is the change in the potential energy of the proton betwe...
Physics
if a ball is thrown straight upward and is caught 5s after being thrown,how fast must it have been going when it left the person s hand?help
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Members | 481 | 1,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.919769 |
https://www.mankier.com/3/zget01.f | 1,726,575,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651773.64/warc/CC-MAIN-20240917104423-20240917134423-00699.warc.gz | 800,065,651 | 3,074 | # zget01.f - Man Page
TESTING/LIN/zget01.f
## Synopsis
### Functions/Subroutines
subroutine zget01 (m, n, a, lda, afac, ldafac, ipiv, rwork, resid)
ZGET01
## Function/Subroutine Documentation
### subroutine zget01 (integer m, integer n, complex*16, dimension( lda, * ) a, integer lda, complex*16, dimension( ldafac, * ) afac, integer ldafac, integer, dimension( * ) ipiv, double precision, dimension( * ) rwork, double precision resid)
ZGET01
Purpose:
ZGET01 reconstructs a matrix A from its L*U factorization and
computes the residual
norm(L*U - A) / ( N * norm(A) * EPS ),
where EPS is the machine epsilon.
Parameters
M
M is INTEGER
The number of rows of the matrix A. M >= 0.
N
N is INTEGER
The number of columns of the matrix A. N >= 0.
A
A is COMPLEX*16 array, dimension (LDA,N)
The original M x N matrix A.
LDA
LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,M).
AFAC
AFAC is COMPLEX*16 array, dimension (LDAFAC,N)
The factored form of the matrix A. AFAC contains the factors
L and U from the L*U factorization as computed by ZGETRF.
Overwritten with the reconstructed matrix, and then with the
difference L*U - A.
LDAFAC
LDAFAC is INTEGER
The leading dimension of the array AFAC. LDAFAC >= max(1,M).
IPIV
IPIV is INTEGER array, dimension (N)
The pivot indices from ZGETRF.
RWORK
RWORK is DOUBLE PRECISION array, dimension (M)
RESID
RESID is DOUBLE PRECISION
norm(L*U - A) / ( N * norm(A) * EPS )
Author
Univ. of Tennessee
Univ. of California Berkeley
NAG Ltd.
Definition at line 106 of file zget01.f.
## Author
Generated automatically by Doxygen for LAPACK from the source code.
## Referenced By
The man page zget01(3) is an alias of zget01.f(3).
Tue Nov 28 2023 12:08:43 Version 3.12.0 LAPACK | 545 | 1,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.698265 |
https://bio.libretexts.org/Bookshelves/Biotechnology/Quality_Assurance_and_Regulatory_Affairs_for_the_Biosciences/10%3A_FDA_Enforcement/10.04%3A_Section_4- | 1,713,244,858,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00746.warc.gz | 121,798,367 | 30,852 | # 10.4: Recalls
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
## Recalls - Drugs, Biologics, Devices, & Food
A recall is the voluntary removal of a product by a manufacturer or at the request of the FDA. Recalls are almost always voluntary. The FDA can only issue a recall when they have the mandated power to do so. The FDA cannot recall a drug, or biologic, but can recall a medical device, some cosmetics, and food. When they do have recall authority, they can only do so when there is a substantial public health and safety risk. https://www.fda.gov/consumers/consumer-updates/fda-101-product-recalls
Recall Resources: https://www.fda.gov/safety/recalls-m...call-resources
### Food
The FDA can issue a food recall. In 2011, the FDA gained increased authority in regulating and responding to food product contamination via the new Food Safety Modernization Act (FSMA). The FSMA allows the FDA to suspend the services and production of food distributors if contamination is suspected. There need not be any proof of the source of the contamination. For more information on the FSMA: http://www.foodsafety.gov/news/fsma.html
### Medical Device
The FDA can issue a device recall. In 2012, the US Food and Drug Administration (FDA) announced that it was seeking to implement medical device recall authority under § 518(e) of the FD&C Act and Chapter 21, Section 810 of the CFR. Recall authority for medical devices would permit the FDA to order manufacturers to cease the distribution of a device and notify health professionals if FDA finds a reasonable probability that the device would cause serious adverse health reactions or death (fda.gov). A recall does not necessarily mean the product must be returned, sometimes it just needs to be adjusted, or clarification safety instructions provided. 21 CFR 7 provides Guidance for conducting an efficient voluntary recall.
Examples of the types of actions that may be considered device recalls:
1. Inspecting the device for problems
2. Repairing the device
3. Adjusting settings on the device
4. Re-labeling the device
5. Destroying device
6. Notifying patients of a problem
7. Monitoring patients for health issues
A medical device recall is either a correction or removal. A Correction addresses a problem with a medical device in the place where it is sold. A Removal approaches the problem by removing the device from where it is sold. In most cases, a company voluntarily recalls a device on its own. When the company has violated an FDA law, the company must recall the device (correction or removal) and notify the FDA. Legally, the FDA can require a company to recall a device if an organization refuses to do so under 21 CFR 810, Medical Device Recall Authority. 21 CFR 810 describes the procedures the FDA follows in exercising its medical device recall authority under section 518(e) of the FD&C Act.
A recall does not include a market withdrawal or a stock recovery. A market withdrawal is a firm's removal or correction of a distributed product which involves a minor violation that would not be subject to legal action by the FDA or which involves no violation, e.g., normal stock rotation practices, routine equipment adjustments, and repairs (fda.gov). In the end, almost all recalls are conducted voluntarily by the manufacturer. https://www.fda.gov/safety/recalls-market-withdrawals-safety-alerts
Explore!
A comprehensive, searchable recall medical device recall database: www.accessdata.fda.gov/scripts/cdrh/cfdocs/cfRES/res.cfm
To learn more about Device Recalls, visit the FDA device recall web page https://www.fda.gov/medicaldevices/safety/listofrecalls/ and also, watch the FDA Video here. http://fda.yorkcast.com/webcast/Play/1b95461f64be40ecbe3415195cb394911d
Device recalls following the same general recall procedure, as previously discussed for drugs. This includes classification of recall (I, II, or III) developing a recall strategy and providing the FDA with recall status reports.
Medical device safety alert: issued in situations where a medical device may present an unreasonable risk of substantial harm. In some cases, these situations also are considered recalls (fda.gov)
Market withdrawal occurs when a product has a minor violation that would not be subject to FDA legal action. The firm removes the product from the market or corrects the violation. For example, a product removed from the market due to tampering, without evidence of manufacturing or distribution problems would be a market withdrawal. (fda.gov)
## Drug Safety and Availability
The FDA offers many websites that address specifically drug safety and availability to communicate with customers. Here are a few you may want to explore. https://www.fda.gov/drugs/drug-safety-and-availability
1. Drug Safety Communications: https://www.fda.gov/drugs/drug-safety-and-availability/drug-safety-communications
2. Post-Market Drug Safety: https://www.fda.gov/drugs/drug-safety-and-availability/postmarket-drug-safety-information-patients-and-providers
3. What’s New – Human Drugs: https://www.fda.gov/drugs/news-events-human-drugs
## Drug Recall
A drug recall is a voluntary action taken by a company. www.fda.gov/drugs/drug-recalls/fdas-roledrug-recalls. Not all recalls are announced on FDA.gov or in the news media. Public notification is generally issued when a product that has been widely distributed or poses a serious health hazard is recalled. However, if a company does not issue public notification of a recall, FDA may do so if the agency determines it is necessary to protect patients. FDA evaluates the effectiveness of a recall by evaluating a company’s efforts to properly notify customers and remove the defective product from the market. If a recall is determined to be ineffective FDA will request the company take additional actions.
## Class I, II & III Recalls
• Class I Recall: "A reasonable probability that the use of or exposure to a violative product will cause serious adverse health consequences or death." (fda.gov)
• Class II Recall: "use of or exposure to a violative product may cause temporary or medically reversible adverse health consequences or where the probability of serious adverse health consequences is remote." (fda.gov)
• Class III Recall: "use of or exposure to a violative product is not likely to cause adverse health consequences." (fda.gov)
Read the article and watch the video “FDA 101: Product Recalls - From First Alert to Effectiveness Checks” and read FDA FAQs:
1. How does the FDA first learn about a problem with a product?
2. How does the FDA alert the product about a product recall?
3. Discuss an example of a recall, and state what class of recall it is and why.
Explore!
The FDA Drug Safety Podcasts are produced by FDA's CDER and provide emerging safety information about drugs in conjunction with the release of Public Health Advisories and other drug safety issues. (fda.gov). | 1,853 | 7,898 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-18 | latest | en | 0.590445 |
https://socratic.org/questions/how-do-you-convert-from-300-degrees-to-radians | 1,721,735,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00669.warc.gz | 477,076,499 | 6,439 | # How do you convert from 300 degrees to radians?
Feb 2, 2015
To do this conversion you have to think at what is a radian.
A radian is the angle that describes an arc of length equal to the radius.
Figure 1
To make our life easier let us make $r = 1$.
But what is the connection with degrees?
Consider an entire circle. We know that it spans 360° but how many radians?
If you try to draw them on top of your circle you'll find that you need a little bit more than 6 of the slices of figure 1 to cover your entire circle, i.e. 6 radians and a bit.
To get the exact number consider that an entire circle is a closed arc of length $2 \pi r$, the perimeter of the circle. If $r = 1$ you see that in an entire circle (that we know corresponds to 360° angle) we'll have $2 \pi = 6.28 \ldots$ radians!!!!
Now we have the key for our conversion:
360°=2pi
So:
if 360° is $2 \pi$
if I have 300° I'll have $x$ radians.
As a proportion:
360°:2pi=300°:x
and: x=2pi*300°/360°=5pi/3
If you want you can multiply the value of $\pi = 3.141 . .$ but I suggest to leave it as a fraction of pi.
Hope it helps | 323 | 1,098 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.917989 |
https://www.br.freelancer.com/projects/software-architecture-cplusplus-programming/algorithm-analysis/ | 1,519,107,348,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812880.33/warc/CC-MAIN-20180220050606-20180220070606-00738.warc.gz | 852,248,352 | 33,728 | # Algorithm analysis
Given an analysis of the running time (Big-Oh notation) for each of the following 4 program fragments. Note that the running time corresponds here to the number of times the operation sum++ is executed. Sqrt is the function that returns the square root of a given number.
(a) sum =0;
for(i=0;<sqrt(n)/2;i++)
sum++;
for(j=0;<sqrt(n)/4;j++)
sum++;
for(k=0;k<8+j;k++)
sum++;
(b) sum = 0;
for(i=0;i<sqrt(n)/2;i++)
for(j=I;8+I;j++)
for(k=j;k<8+j;k++)
sum++;
(c) sum = 0;
for(i=1;i<2*n;i++)
for(j=1;j<i*I;j++)
for(k=1;k<j;k++)
if (j%i==1)
sum++;
(d) sum=0;
for(i=1;i<2*n;i++)
for(j=1;j<i*I;j++)
for(k=1;k<j;k++)
if(j%i)
sum++;
2. if it takes 10ms to run program (b) for n = 100, how long will it take to run for n=400?
[url removed, login to view] it takes 10ms to run program(a) for n = 100, how large a problem can be solved in 40ms?
Habilidades: Programação C++ , Arquitetura de software
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4.3 | 477 | 1,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-09 | longest | en | 0.624532 |
https://www.univerkov.com/in-triangle-abc-sides-ab-and-bc-are-equal-and-the-outer-head-at-apex-c-123-degrees-find-the-value-of-angle-b/ | 1,721,359,415,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00265.warc.gz | 883,889,907 | 6,289 | # In triangle ABC, sides AB and BC are equal, and the outer head at apex C = 123 degrees. Find the value of angle B.
Since in this triangle the sides AB and BC are equal, then this triangle is isosceles, and in an isosceles triangle the angles at the base are equal:
∠А = ∠С.
Since the sum of the outer and inner angles of the triangle at one vertex is 180º, and the outer angle at the vertex C is 123º, then:
∠С = 180º – φ;
∠С = 180º – 123º = 57º;
∠А = ∠С = 57º.
Since the sum of all angles of a triangle is 180º, then:
∠В = 180º – ∠А – ∠С;
∠В = 180º – 57º – 57º = 66º.
Answer: the angle ∠B is equal to 66º.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 265 | 929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.87351 |
http://www.evi.com/q/90_feet_equals_how_many_yards | 1,419,132,004,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770633.72/warc/CC-MAIN-20141217075250-00133-ip-10-231-17-201.ec2.internal.warc.gz | 491,262,196 | 5,218 | # 90 feet equals how many yards
• tk10publ tk10canl
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• 90feet equals how many yards | 379 | 1,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2014-52 | latest | en | 0.981285 |
https://crosswordlabs.com/view/multiplication-295 | 1,718,877,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00653.warc.gz | 162,406,325 | 8,557 | # Multiplication
##### Across
1. 1. an operation used to separate a number of items into equal size groups
2. 2. an operation used to find the total number of items in a given number of equal size groups
3. 5. the number that is divided by another number
4. 6. the value you need to find to solve a problem
5. 10. changing the order of the factors does not change the product
6. 13. changing the grouping of three or more factors does not change the product
7. 14. a series of numbers or shapes that follow a rule to repeat or change
8. 16. the number by which another number is divided
##### Down
1. 1. when one of the factors of a product is written as a sum, multiplying each addend by the other factor before adding does not change the product
2. 3. the result of multiplication
3. 4. a number that is multiplied
4. 7. a mathematical statement that uses an equal sign to show that two expressions have the same value
5. 8. two numbers that are multiplied together to give a product.
6. 9. the amount left over when one number does not divide another number a whole number of times
7. 11. a procedure that is followed to go from one number or shape to the next in a pattern
8. 12. the product of a given number and any other whole number.
9. 15. the result of division | 322 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.890614 |
http://ciceroeuropa.eu/scholastic-close-reading-passages-for-2nd-grade-pdf/ | 1,534,750,541,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215858.81/warc/CC-MAIN-20180820062343-20180820082343-00541.warc.gz | 80,456,268 | 8,537 | # Scholastic close reading passages for 2nd grade pdf
Which representation helped you see a clear demonstration of the mean, this is still a bit confusing to me. The concordance rate in monozygotic twins while greater than that in dizygotes; they work with the described components of geometry described above, also differ in their level of complexity. Department of Community and Family Health College of Public Health, recorded Tegrity lectures allowed the instructor to reduce class teaching preparation time. According to Murray and Herrnstein; the reluctance of students to practise this compromises the value of education. I really had to think about the activity that goes with it to fully understand.
For the second portion of Problem B1, a structure that would continue through 2004. A student in the video as well as a student in Delores’s case study was able to justify that angles are present in shapes other than just triangles, minute writing section of the test for damaging standards of writing teaching in the classroom. Correlative study of hydrochloric acid, but I have attempted to elicit more information from those students volunteering them. While measurement is part of my life daily; i will cook dinner for my family tonight. African Americans are less likely to develop in their socialization, what is the sum of the measures of the exterior angles of each of these polygons?
Admission to undergraduate programs of universities or colleges. Intended for high school students. The test is intended to assess students’ readiness for college. On March 5, 2014, the College Board announced that a redesigned version of the SAT would be administered for the first time in 2016.
SAT is taken outside the United States. Scores on the SAT range from 400 to 1600, combining test results from two 800-point sections: mathematics, and critical reading and writing. SAT prep, free of charge. Flag of the United States.
The optional writing section measures writing skills taught in high school English classes and in entry, the article also goes on to share a website where students can create a virtual kaleidoscope. President of the ERB stated “It is a lesson we all learn at some point, i decided on this because the values where centered around five. Even after reviewing them slightly and gaining a bit of understanding about them, students then recorded data for the items they measured which included both the diameter and circumference of each object. These students’ standards are focused on gathering data, each individual data point presents some form of deviation from the mean, new York: College Entrance Examination Board. Even in high school, or by telephone, it reminded me of the Facebook questions that go around asking how many different squares or different shapes can you find in this image.
When working with the illuminations cube activity – many of the students in each example were so focused on the inside angle that many of them had not considered the outside angle. Babies that vomit, lexile measures help personalize instruction, the new exam was administered for the first time in March 2016. I found that the sum values could be 2, with this I found that the two smaller angles of the rhombus are 60 degrees. The module shared a great example of this put to use in the real world by show casing the use of mean when interpreting bacteria levels in water. Addison’s disease inadequately treated with 5 mg of deoxycorticosterone acetate who responded to an oral water load of 500 ml with a fall in plasma osmolality; and then to check those interpretations by calculating the balance of the deviations.
Template talk:Education in the U. They state that the SAT assesses how well the test takers analyze and solve problems—skills they learned in school that they will need in college. A large independent validity study on the SAT’s ability to predict college freshman GPA was performed by the University of California. The results of this study found how well various predictor variables could explain the variance in college freshman GPA.
It found that independently high school GPA could explain 15. When high school GPA and the SAT I were combined, they explained 20. When high school GPA and the SAT II were combined, they explained 22. When SAT I was added to the combination of high school GPA and SAT II, it added a . | 870 | 4,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-34 | latest | en | 0.967141 |
https://www.chegg.com/homework-help/questions-and-answers/partial-pressure-calculate-partial-pressure-gas-container-work-moles-ch4-8-g-16-g-05-moles-q3417956 | 1,529,613,016,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00318.warc.gz | 762,421,043 | 11,710 | Partial Pressure - Calculate the partial pressure of each gas in the container.
Here is my work:
moles CH4 = 8 g / 16 g) = 0.5 moles
moles C2H6 = 18 g / 30 g) = 0.6 moles
Total moles:
PV = nRT
n = PV/RT = (5.4 atm) x (10) / (0.0821 * 296K = 2.2234 moles
moles of propane = 2.2234- 0.5 - 0.6 = 1.1234 moles
partial pressure methane = (0.5 / 2.2234) x 5.4 atm = 1.2144 atm
partial pressure ethane = (0.6 / 2.2234) x 5.4 atm = 1.457 atm
partial pressure propane = 2.2234 - .5 - .6 = 1.1234 atm
I enter: "1.21,1.46,2.67" into mastering chemistry and it tells me I am close on methane and ehtane but a little high.
I have not even started B as I can't get A. | 268 | 660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-26 | latest | en | 0.819742 |
https://assignmentutor.com/elementary-data-analysis-dai-xie-math135/ | 1,712,978,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816535.76/warc/CC-MAIN-20240413021024-20240413051024-00236.warc.gz | 98,132,089 | 23,696 | assignmentutor-lab™ 为您的留学生涯保驾护航 在代写基础数据分析Elementary data Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写基础数据分析Elementary data Analysis代写方面经验极为丰富，各种基础数据分析Elementary data Analysis相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
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## 数学代写|基础数据分析代写Elementary data Analysis代考|Some General Theory for Linear Smoothers
Some key parts of the theory you are familiar with for linear regression models carries over more generally to linear smoothers. They are not quite so important any more, but they do have their uses, and they can serve as security objects during the transition to non-parametric regression.
Throughout this sub-section, we will temporarily assume that $Y=\mu(X)+\epsilon$, with the noise terms $\epsilon$ have constant variance $\sigma^2$, and being uncorrelated with each other at different observations. Also, we will define the smoothing, influence or hat matrix $\hat{\mathbf{w}}$ by $\hat{w}_{i j}=\hat{w}\left(x_i, x_j\right)$. This records how much influence observation $y_j$ had on the smoother’s fitted value for $\mu\left(x_i\right)$, which (remember) is $\widehat{\mu}\left(x_i\right)$ or $\widehat{\mu}_i$ for short ${ }^{14}$, hence the name “hat matrix” for $\hat{w}$.
It is easy to get the standard error of any predicted mean value $\widehat{\mu}(x)$, by first working out its variance:
\begin{aligned} \mathbb{V}[\widehat{\mu}(x)] &=\mathbb{V}\left[\sum_{j=1}^n w\left(x_j, x\right) Y_j\right] \ &=\sum_{j=1}^n \mathbb{V}\left[w\left(x_j, x\right) Y_j\right] \ &=\sum_{j=1}^n w^2\left(x_j, x\right) \mathbb{V}\left[Y_j\right] \ &=\sigma^2 \sum_{j=1}^n w^2\left(x_j, x\right) \end{aligned}
The second line uses the assumption that the noise is uncorrelated, and the last the assumption that the noise variance is constant. In particular, for a point $x_i$ which appeared in the training data, $\mathbb{V}\left[\widehat{\mu}\left(x_i\right)\right]=\sigma^2 \sum_j w_{i j}^2$.
Notice that this is the variance in the predicted mean value, $\widehat{\mu}(x)$. It is not an estimate of $\mathbb{V}[Y \mid X=x]$, though we will see how conditional variances can be estimated using nonparametric regression in Chapter 7.
Notice also that we have not had to assume that the noise is Gaussian. If we did add that assumption, this formula would also give us a confidence interval for the fitted value (though we would still have to worry about estimating $\sigma$ ).
## 数学代写|基础数据分析代写Elementary data Analysis代考|(Effective) Degrees of Freedom
For linear regression models, you will recall that the number of “degrees of freedom” was just defined as the number of coefficients (including the intercept). While degrees of freedom are less important for other sorts of regression than for linear models, they’re still worth knowing about, so I’ll explain here how they are calculated.
The first thing to realize is that we can’t use the number of parameters to define degrees of freedom in general, since most linear smoothers don’t have parameters. Instead, we have to go back to the reasons why the number of parameters matters in ordinary linear models ${ }^{15}$. We’ll start with an $n \times p$ data matrix of predictor variables $\mathbf{x}$ (possibly including an all- 1 column for an intercept), and an $n \times 1$ column matrix of response values $\mathbf{y}$. The ordinary least squares estimate of the $p$-dimensional coefficient vector $\beta$ is
$$\hat{\beta}=\left(\mathbf{x}^T \mathbf{x}\right)^{-1} \mathbf{x}^T \mathbf{y}$$
This lets us write the fitted values in terms of $\mathbf{x}$ and $\mathbf{y}$ :
\begin{aligned} \widehat{\mu} &=\mathbf{x} \hat{\beta} \ &=\left(\mathbf{x}\left(\mathbf{x}^T \mathbf{x}\right)^{-1} \mathbf{x}^T\right) \mathbf{y} \ &=\mathbf{w y} \end{aligned}
where w is the $n \times n$ matrix, with $w_{i j}$ saying how much of each observed $y_j$ contributes to each fitted $\hat{\mu}_i$. This is what, a little while ago, I called the influence or hat matrix, in the special case of ordinary least squares.
Notice that $\mathbf{w}$ depends only on the predictor variables in $\mathbf{x}$; the observed response values in $\mathbf{y}$ don’t matter. If we change around $\mathbf{y}$, the fitted values $\widehat{\mu}$ will also change, but only within the limits allowed by $\mathbf{w}$. There are $n$ independent coordinates along which y can change, so we say the data have $n$ degrees of freedom. Once $\mathbf{x}$ (and thus $\mathbf{w})$ are fixed, however, $\widehat{\mu}$ has to lie in a $p$-dimensional linear subspace in this $n$-dimensional space, and the residuals have to lie in the $(n-p)$-dimensional space orthogonal to it.
# 基础数据分析代考
## 数学代写|基础数据分析代写基本数据分析代考|Some General Theory for Linear – Smoothers
\begin{aligned} \mathbb{V}[\widehat{\mu}(x)] &=\mathbb{V}\left[\sum_{j=1}^n w\left(x_j, x\right) Y_j\right] \ &=\sum_{j=1}^n \mathbb{V}\left[w\left(x_j, x\right) Y_j\right] \ &=\sum_{j=1}^n w^2\left(x_j, x\right) \mathbb{V}\left[Y_j\right] \ &=\sigma^2 \sum_{j=1}^n w^2\left(x_j, x\right) \end{aligned}
## 数学代写|基础数据分析代写Elementary data Analysis代考|(Effective) Degrees of Freedom
$$\hat{\beta}=\left(\mathbf{x}^T \mathbf{x}\right)^{-1} \mathbf{x}^T \mathbf{y}$$这让我们可以将拟合值写成 $\mathbf{x}$ 和 $\mathbf{y}$ :
\begin{aligned} \widehat{\mu} &=\mathbf{x} \hat{\beta} \ &=\left(\mathbf{x}\left(\mathbf{x}^T \mathbf{x}\right)^{-1} \mathbf{x}^T\right) \mathbf{y} \ &=\mathbf{w y} \end{aligned}
## 有限元方法代写
assignmentutor™作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 2,344 | 6,510 | {"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} | 3.671875 | 4 | CC-MAIN-2024-18 | latest | en | 0.681711 |
https://www.hpmuseum.org/forum/showthread.php?mode=threaded&tid=5508&pid=49010 | 1,652,985,864,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00531.warc.gz | 943,909,368 | 6,368 | Calculating e^x-1 on classic HPs
01-11-2016, 10:20 PM (This post was last modified: 01-11-2016 10:22 PM by Dieter.)
Post: #1
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
Calculating e^x-1 on classic HPs
There is no doubt that the ln(1+x) and ex–1 functions that appeared with the 41C are extremely useful as they allow more precise calculations for arguments close to zero, which otherwise would round to zero. The 15C Advanced Functions Handbook included a short routine for emulating ln(1+x) on calculators that no not feature this function, and another (IMHO slightly better) method has been posted in this forum.
On the other hand, the ex–1 case seemed more challenging. Thomas Klemm posted a solution that included the hyperbolic sine for a short and elegant implementation of this function. However, not all HPs offer hyperbolic functions.
Here is a method that should work on most classic HPs. I think it works fine but I did not do any extensive tests. So try it and see what you get. Coded in VBA, the algorithm is as follows:
Code:
Function expm1(x) u = Exp(x) expm1 = (u - 1) + (x - Log(u)) * u End Function
On a classic HP it could look like this:
Code:
ENTER e^x ENTER ENTER LN CHS R^ + x x<>y 1 - + RTN
This returns ex in Y and ex–1 in X.
What do you think?
Dieter
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Messages In This Thread Calculating e^x-1 on classic HPs - Dieter - 01-11-2016 10:20 PM RE: Calculating e^x-1 on classic HPs - Thomas Klemm - 01-12-2016, 01:41 AM RE: Calculating e^x-1 on classic HPs - Dieter - 01-12-2016, 06:12 PM RE: Calculating e^x-1 on classic HPs - Claudio L. - 01-12-2016, 05:13 AM RE: Calculating e^x-1 on classic HPs - Dieter - 01-12-2016, 02:02 PM RE: Calculating e^x-1 on classic HPs - ElectroDuende - 01-12-2016, 03:34 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-12-2016, 06:30 PM RE: Calculating e^x-1 on classic HPs - Claudio L. - 01-12-2016, 09:44 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-12-2016, 08:16 PM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-13-2016, 03:46 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-13-2016, 06:43 PM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-14-2016, 02:47 AM RE: Calculating e^x-1 on classic HPs - Dieter - 01-14-2016, 01:44 PM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-14-2016, 03:30 PM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-14-2016, 06:50 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-14-2016, 07:38 PM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-14-2016, 07:50 PM RE: Calculating e^x-1 on classic HPs - Albert Chan - 02-02-2019, 07:32 PM RE: Calculating e^x-1 on classic HPs - Dieter - 02-03-2019, 06:46 PM RE: Calculating e^x-1 on classic HPs - Paul Dale - 01-14-2016, 07:20 AM RE: Calculating e^x-1 on classic HPs - Dieter - 01-14-2016, 01:48 PM RE: Calculating e^x-1 on classic HPs - emece67 - 01-14-2016, 03:55 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-14-2016, 06:42 PM RE: Calculating e^x-1 on classic HPs - emece67 - 01-14-2016, 09:14 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-14-2016, 09:55 PM RE: Calculating e^x-1 on classic HPs - Paul Dale - 01-15-2016, 01:19 AM RE: Calculating e^x-1 on classic HPs - Dieter - 01-15-2016, 02:00 PM RE: Calculating e^x-1 on classic HPs - matthiaspaul - 01-15-2016, 10:10 AM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-15-2016, 06:07 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-15-2016, 06:46 PM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-15-2016, 10:45 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-16-2016, 08:19 AM RE: Calculating e^x-1 on classic HPs - Gerson W. Barbosa - 01-16-2016, 12:40 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-16-2016, 01:32 PM RE: Calculating e^x-1 on classic HPs - Albert Chan - 02-02-2019, 07:22 PM RE: Calculating e^x-1 on classic HPs - Paul Dale - 01-14-2016, 09:22 PM RE: Calculating e^x-1 on classic HPs - Dieter - 01-14-2016, 09:49 PM RE: Calculating e^x-1 on classic HPs - Albert Chan - 02-26-2021, 03:00 PM RE: Calculating e^x-1 on classic HPs - Albert Chan - 03-01-2021, 02:23 PM RE: Calculating e^x-1 on classic HPs - Paul Dale - 01-15-2016, 01:15 AM RE: Calculating e^x-1 on classic HPs - Albert Chan - 02-03-2019, 08:28 PM RE: Calculating e^x-1 on classic HPs - Albert Chan - 02-04-2019, 07:28 PM RE: Calculating e^x-1 on classic HPs - Dieter - 02-04-2019, 07:59 PM
User(s) browsing this thread: 1 Guest(s) | 1,731 | 4,518 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2022-21 | latest | en | 0.828165 |
https://www.stockpricetrends.com/company/OTIS | 1,708,467,376,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00253.warc.gz | 1,091,071,065 | 39,760 | # Otis Worldwide Corporation (OTIS)
at otis, we are dedicated to connecting you to the people and places that matter. for over 160 years, we’ve continually reinvented the way our passengers move through the urban world with industry-leading elevators, escalators and moving walkways.
## Stock Price Trends
Stock price trends estimated using linear regression.
## Paying users area
The data is hidden behind and trends are not shown in the charts.
Unhide data and trends.
This is a one-time payment. There is no automatic renewal.
#### Key facts
• The primary trend is decreasing.
• The decline rate of the primary trend is 13.24% per annum.
• OTIS price at the close of February 16, 2024 was \$91.24 and was higher than the top border of the primary price channel by \$25.03 (37.81%). This indicates a possible reversal in the primary trend direction.
• The secondary trend is increasing.
• The growth rate of the secondary trend is 11.12% per annum.
• OTIS price at the close of February 16, 2024 was inside the secondary price channel.
• The direction of the secondary trend is opposite to the direction of the primary trend. This indicates a possible reversal in the direction of the primary trend.
### Linear Regression Model
Model equation:
Yi = α + β × Xi + εi
Top border of price channel:
Exp(Yi) = Exp(a + b × Xi + 2 × s)
Bottom border of price channel:
Exp(Yi) = Exp(a + b × Xi – 2 × s)
where:
i - observation number
Yi - natural logarithm of OTIS price
Xi - time index, 1 day interval
σ - standard deviation of εi
a - estimator of α
b - estimator of β
s - estimator of σ
Exp() - calculates the exponent of e
### Primary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### April 26, 2021 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### November 10, 2022 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The primary trend is decreasing.
• The decline rate of the primary trend is 13.24% per annum.
• OTIS price at the close of February 16, 2024 was \$91.24 and was higher than the top border of the primary price channel by \$25.03 (37.81%). This indicates a possible reversal in the primary trend direction.
### Secondary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### February 17, 2022 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### February 16, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The secondary trend is increasing.
• The growth rate of the secondary trend is 11.12% per annum.
• OTIS price at the close of February 16, 2024 was inside the secondary price channel. | 1,092 | 3,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | longest | en | 0.848709 |
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# Martin
### TU Wien
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https://svn.geocomp.uq.edu.au/escript/trunk/esys2/finley/test/python/AssembleTest.py?r1=82&sortdir=down&pathrev=97&r2=97&sortby=author | 1,657,038,069,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00764.warc.gz | 576,724,378 | 3,405 | # Diff of /trunk/esys2/finley/test/python/AssembleTest.py
revision 82 by jgs, Tue Oct 26 06:53:54 2004 UTC revision 97 by jgs, Tue Dec 14 05:39:33 2004 UTC
# Line 13 sys.path.append(esys_root+'/escript/py_s Line 13 sys.path.append(esys_root+'/escript/py_s
13
14 from escript import * from escript import *
15 from util import * from util import *
16 from linearPDE import * from linearPDEs import *
17
18 import finley import finley
19 from math import * from math import *
# Line 202 def TestSystem(numEqu,numComp,mydomain,r Line 202 def TestSystem(numEqu,numComp,mydomain,r
202 c_A2=c_A c_A2=c_A
203 text="A[%d,%d,%d,%d]"%(p,i,q,j) text="A[%d,%d,%d,%d]"%(p,i,q,j)
205 mypde1=linearPDE(domain=mydomain,A=c_A2,X=eval(x,elem)) mypde1=LinearPDE(domain=mydomain,A=c_A2,X=eval(x,elem))
206 mypde1.setReducedOrderForSolutionsTo(reduce) mypde1.setReducedOrderForSolutionsTo(reduce)
207 checkSystem(text+" const with X",mypde1.getOperator(),U,mypde1.getRightHandSide()) checkSystem(text+" const with X",mypde1.getOperator(),U,mypde1.getRightHandSide())
208 # check div( A grad(u) ) = Y # check div( A grad(u) ) = Y
209 y=-algebraicDiv(x) y=-algebraicDiv(x)
210 mypde2=linearPDE(domain=mydomain,Y=eval(y,elem),y=matmult(eval(x,face_elem),nrml)) mypde2=LinearPDE(domain=mydomain,Y=eval(y,elem),y=matmult(eval(x,face_elem),nrml))
211 mypde2.setReducedOrderForSolutionsTo(reduce) mypde2.setReducedOrderForSolutionsTo(reduce)
212 checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde2.getRightHandSide()) checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde2.getRightHandSide())
213
# Line 221 def TestSystem(numEqu,numComp,mydomain,r Line 221 def TestSystem(numEqu,numComp,mydomain,r
221 c_B2=c_B c_B2=c_B
222 text="B[%d,%d,%d]"%(p,i,q) text="B[%d,%d,%d]"%(p,i,q)
223 x=mult3_2(c_B,u) x=mult3_2(c_B,u)
224 mypde1=linearPDE(domain=mydomain,B=c_B2,X=eval(x,elem)) mypde1=LinearPDE(domain=mydomain,B=c_B2,X=eval(x,elem))
225 mypde1.setReducedOrderForSolutionsTo(reduce) mypde1.setReducedOrderForSolutionsTo(reduce)
226 checkSystem(text+" const with X",mypde1.getOperator(),U,mypde1.getRightHandSide()) checkSystem(text+" const with X",mypde1.getOperator(),U,mypde1.getRightHandSide())
227 # check div( B u ) = Y # check div( B u ) = Y
228 y=-algebraicDiv(x) y=-algebraicDiv(x)
229 mypde2=linearPDE(domain=mydomain,Y=eval(y,elem),y=matmult(eval(x,face_elem),nrml)) mypde2=LinearPDE(domain=mydomain,Y=eval(y,elem),y=matmult(eval(x,face_elem),nrml))
230 mypde2.setReducedOrderForSolutionsTo(reduce) mypde2.setReducedOrderForSolutionsTo(reduce)
231 checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde2.getRightHandSide()) checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde2.getRightHandSide())
232
# Line 240 def TestSystem(numEqu,numComp,mydomain,r Line 240 def TestSystem(numEqu,numComp,mydomain,r
240 c_C2=c_C c_C2=c_C
241 text="C[%d,%d,%d]"%(p,q,i) text="C[%d,%d,%d]"%(p,q,i)
243 mypde1=linearPDE(domain=mydomain,C=c_C2,Y=eval(y,elem)) mypde1=LinearPDE(domain=mydomain,C=c_C2,Y=eval(y,elem))
244 mypde1.setReducedOrderForSolutionsTo(reduce) mypde1.setReducedOrderForSolutionsTo(reduce)
245 checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde1.getRightHandSide()) checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde1.getRightHandSide())
246
# Line 255 def TestSystem(numEqu,numComp,mydomain,r Line 255 def TestSystem(numEqu,numComp,mydomain,r
255 c_D2=c_D c_D2=c_D
256 text="D[%d,%d]"%(p,q) text="D[%d,%d]"%(p,q)
257 y=mult2(c_D,u) y=mult2(c_D,u)
258 mypde1=linearPDE(domain=mydomain,D=c_D2,Y=eval(y,elem)) mypde1=LinearPDE(domain=mydomain,D=c_D2,Y=eval(y,elem))
259 mypde1.setReducedOrderForSolutionsTo(reduce) mypde1.setReducedOrderForSolutionsTo(reduce)
260 checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde1.getRightHandSide()) checkSystem(text+" const with Y",mypde1.getOperator(),U,mypde1.getRightHandSide())
261
Legend:
Removed from v.82 changed lines Added in v.97 | 1,503 | 4,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-27 | latest | en | 0.135791 |
https://www.newwaveautosales.com/council-of-experts/faq-water-pipe-questions/ | 1,632,860,061,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00533.warc.gz | 934,469,458 | 19,478 | FAQ: Water Pipe Questions?
How do you solve a tap question?
Tap C empties out 1/9 th part of the tank in 1 hour. Thus, in 1 hour (1/8 + 1/10 – 1/9) part of the tank is filled. (45 + 36 – 40)/360 = 41/360 th part of the tank is filled. Thus, tank will be filled completely in 360/41 hours, when all the three taps A, B and C are opened together.
How much time a tank would take to be filled by 3 pipes?
So when all 3 pipes work together, total inflow of water into the cistern per minute is 2+(8/9)+(1/2)=61/18 liters. Time needed to fill 122 liters = 122/(61/18) = (122*18)/61 = 2*18 = 36 minutes.
How do you solve a pipe and cistern problem?
Steps to Solve Pipes and Cistern with Trick Take the LCM of a given Number and that LCM will be the total capacity of cistern or. tank. Add or Subtract According to a particular question. The LCM from Step 1. will be the total work. Divide the total capacity of cistern or tank with the outcome of Step 2.
How long would it take for both pipes to fill the swimming pool together?
Answer: 10.5 hours/ 10 hours and 30 mins.
How long does it take to fill up a tank?
= (Volume of tank ) / (Volume of water flow in 1 sec) = 0.48/0.0025 = 192 sec. Time taken to fill the tank = 0.0025 / 0.48 = 0.005 sec.
How do you calculate tank capacity?
Find the volume of your tank. To determine the volume of a rectangular tank, multiply the length (l) times the width (w) times the height (h). The width is the horizontal distance from side to side.
Can fill a tank in 6 hours after half the tank is filled three more similar taps are opened what is the total time taken to fill the tank completely?
Explanation: Time taken by one tap to fill half of the tank = 3 hrs. So, total time taken = 3 hrs. 45 mins.
How do you calculate tank capacity in pipes and cisterns?
Let the capacity of the tank = LCM of (12,15, 6) = 60 litres. If all the pipes are opened simultaneously, (5+4-10) litres will be filled in a minute.
How much time will the leak take to empty the full cistern?
Leak will empty the full cistern in 90 hours. Clearly, both I and II are necessary to answer the question. Correct answer is (E).
What is the formula of time and work?
Important Time and Work Formula Work Done = Time Taken × Rate of Work. Rate of Work = 1 / Time Taken. Time Taken = 1 / Rate of Work. If a piece of work is done in x number of days, then the work done in one day = 1/x. | 677 | 2,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-39 | latest | en | 0.913489 |
http://www.manhattangmat.com/blog/index.php/2012/10/01/breaking-down-gmatprep-weighted-average-problems/comment-page-1/ | 1,406,813,440,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510273350.41/warc/CC-MAIN-20140728011753-00175-ip-10-146-231-18.ec2.internal.warc.gz | 646,878,703 | 16,404 | This week, we’re going to tackle a GMATPrep question from the quant side of things. We’ll tackle a medium-level question this week in order to learn how to master weighted average questions in general, and in the next article, we’ll try a very hard one “ just to see whether you learned the concept as well as you thought you did. : )
Before we begin, I want to mention that every weighted average problem I’ve seen on GMATPrep is a Data Sufficiency question. This doesn’t mean that they’ll never give us a Problem Solving weighted average problem, but it does seem to be the case that the test-writers are more concerned with whether we understand how weighted averages work than with whether we can actually do the calculations. So we’re going to work on that conceptual understanding today and then we’ll discuss a neat calculation shortcut next week (built on the same principles!), just in case we do need to solve.
* At a certain company, the average (arithmetic mean) number of years of experience is 9.8 years for the male employees and 9.1 years for the female employees. What is the ratio of the number of the company’s male employees to the number of the company’s female employees?
(1) There are 52 male employees at the company.
(2) The average number of years of experience for the company’s male and female employees combined is 9.3 years.
Given a certain company, we’re asked to determine the ratio (not a real number, just a ratio “ key point!) of two subgroups that together make up all employees: males to females.
So, what do we know? We know that male employees have an average of 9.8 years of experience. We also know that female employees have an average of 9.1 years of experience. What would be useful to solve? It would be useful to know about the actual number of male and female employees. Alternatively, it would be useful to know about the relationship between the number of male employees and the number of female employees. (For example, if they told me 60% of the employees were female, then I would know the ratio of males to females was 40:60, or 2:3, even though I wouldn’t know the actual number of employees.)
Most of what we’re going to do next is just to explain how weighted averages work. Once you understand how this works, you will not actually have to do these calculations on DS questions (this will take way longer than 2 minutes!); you’ll be able to determine conceptually whether enough info was provided to solve.
In the given problem, could there be equal numbers of male and female employees? Go take a look at the problem again and see what you think.
Let’s say that there are, in fact, 50 male employees and 50 female employees. If the male employees’ average experience is 9.8 years and the female employees’ average experience is 9.1 years, then what is the average experience for the whole group? That would just be the average of 9.8 and 9.1. Is the average of those two numbers 9.3 (the total group average given in statement 2)? No. So now we know we’ve got a weighted average problem; in other words, the number of male employees is not equal to the number of female employees. (Bonus question: can you tell, just based on what we’ve discussed so far, whether there are more male or female employees?)
In order to understand how weighted averages work, let’s calculate a few things and let’s start by using the weighted average formula to see what happens in a case where we have equal numbers of employees (which, again, is not true for this problem “ we’re just examining the concept).
We know the two sub-group averages, 9.8 and 9.1, and we’re also assuming an equal weighting of the two averages, 50:50, which simplifies to 1:1. Put that ratio, 1:1, in a form where the two parts add up to 1: ½:½. Each average gets paired with its adds up to 1 weighting:
[(9.8)(1/2) + (9.1)(1/2)] = 9.45
Because the weightings are equal, this can be simplified to the standard average formula (below); this is why we don’t bother calculating the adds up to 1 weighting when the weightings are equal.
(9.8 + 9.1) / 2 = 9.45
What if the weightings are not equal, though? Let’s say that there were 40 male employees and 60 female employees. Then, the ratio would be 40:60, or 2:3, and the adds up to 1 weighting would be 2/5: 3/5. (The easiest way to determine the adds up to 1 weighting is to first add the two parts of the given ratio, 2 and 3, to get 5. 5 becomes the denominator of both fractions and the original numbers, 2 and 3, become the numerators of each respective fraction: 2/5 and 3/5.)
The weighted average formula would become:
[(9.8)(2/5) + (9.1)(3/5)] = 9.38
Here’s the abstract version of this formula:
[(average #1)(a) + (average #2)(b)] = weighted average, where:
a + b = 1, and
a and b represent the relative weightings of the two sub-groups
In the given problem, we don’t know a and b, but we do know the two sub-group averages, 9.8 and 9.1, so we can write these two formulas:
9.8a + 9.1b = c
a+b = 1
We need to see whether we have enough information in the statements such that a and b could be calculated.
Statement 1 says There are 52 male employees at the company. That gives us an actual number for the male employees; that might be good. We want the male to female ratio, though; does this statement tell us anything about the other group, female employees? No. Not sufficient. Eliminate answers A and D.
Statement 2 says The average number of years of experience for the company’s male and female employees combined is 9.3 years.
That’s something we can add to one of our formulas: c, the weighted average, is 9.3:
9.8a + 9.1b = 9.3
a+b = 1
What do we have? We have two distinct, linear equations with two variables, a and b. Can we solve for a and b? Yes! Sufficient!
We can simplify this further (for future data sufficiency questions) by saying: if we know the two sub-group averages and we know the overall weighted average, then we know we can solve for a and b, the relative weightings of the two sub-groups. (Don’t bother to write the equations, of course “ it’s data sufficiency!) In this case, a:b represents the requested ratio (male:female).
Key Takeaways for Data Sufficiency Weighted Average Problems:
(1) Determine that you have a weighted average problem: this occurs when an average is discussed or could be calculated, but that average is not a standard 1:1 or equally weighted average.
(2) Carefully write down what you were asked to solve, then determine what you know, what you don’t know, and what you would need to know in order to solve (before you look at the statements). Remember that, if you have two sub-group averages and the overall average, then you can determine the relative weightings of the sub-groups.
(3) Check the given statements to see whether you can find a match (that is, a statement tells you what you had already decided you would need to know in order to solve).
Answer to bonus question: there are more female employees at the company because the weighted average, 9.3, is closer to 9.1 (the female employee figure) than to 9.8 (the male employee figure). Click here to read the second article in this series, where we’ll elaborate on this concept.
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
#### Stacey Koprince
Stacey Koprince is an Instructor and Trainer as well as the Director of Online Community for Manhattan Prep. She's also a management consultant who specializes in corporate strategy. She has been teaching various standardized tests for more than fifteen years and her entire teaching philosophy can be summed up in five words: teaching students how to think.
### 14 responses to Breaking Down GMATPrep Weighted Average Problems
1. Hi Stacey,
This really helped me refresh the weighted average concept. Thank you.
By the way, I think you meant to say “9.8a + 9.1b = 9.3″ instead of “9.8x + 9.1y = 9.3″.
SS
2. You’re welcome. And thanks! You’re right – I’ll alert the editing team!
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13. Russian ha ha ha Doll | 2,313 | 9,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2014-23 | longest | en | 0.963983 |
http://csbnews.org/bols-bridge-tip-fear-the-worst/?lang=en | 1,487,637,527,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170614.88/warc/CC-MAIN-20170219104610-00204-ip-10-171-10-108.ec2.internal.warc.gz | 54,040,283 | 21,329 | # BOLS Bridge Tip: Fear the Worst
When opponents bid unexpectedly high, ask yourself if your hand contains any nasty surprises. Photo: Marcelo Lerner, Terence Reese, Eduardo Berisso y Boris Schapiro en el Campeonato Mundial de Bridge, Buenos Aires 1965
By
On 10 October, 2012 At 13:19
Responses : Comments are off for this post
Terence Reese was probably the all-time great name in British bridge, both as a player and a writer. Born in 1913 he won the Bermuda Bowl in 1955 and the World Par Contest in 1961. He also won four European Teams Championships. Tragedy struck in the Bermuda Bowl of 1965 in Buenos Aires when he and his partner Boris Schapiro were accused of cheating. The British inquiry found them not guilty but Reese’s enthusiasm for playing bridge never really recovered. However, his writing thrived and he was the author of more than twenty books on the game, Reese on Play and The
Expert Game being considered classics. He died in his home in Hove, Sussex in 1996.
WHEN opponents bid unexpectedly high you have to ask yourself: does my hand contain any nasty surprises?
I take as my text (as the preachers say) a problem set by Howard Schenken in a 1971 Bridge World:
Both sides are vulnerable at IMP scoring and the dealer is North. South holds:
North Deals Both Vul. South helds:
♠ 9 6 2
♥ 6 3
♦ 9 7 5
♣ A K 9 4 2
West North East South
1♥ 1♠ Pass
2♠ 3♦ 4♠ ?
What call do you make? The great majority of the American panel were ready to double, some with confidence, such as:
Begin: Double. Throwing out the possibility of slam (which, I agree, could be on) I think I might get this for
800. Partner might even sneak in a club ruff.
Clarke: Double. If partner has the tops in his suits it is difficult to see how the opponents can escape for less than 800. If partner has two long red suits lacking some of the tops, I think he will pull my double.
Howard was not impressed by this argument. When opponents bid unexpectedly high you have to ask yourself: does my hand contain any nasty surprises for them? Obviously not. East knows he hasn’t got the ace and king of clubs.
There was a big majority for the double, though in many cases there was not much confidence. Thus:
Wolff: Double. I’d lead a trump. It looks as if East has one of partner’s suits stacked. To not double is too much like tiptoeing through the tulips.
That’s an entrancing picture of the rotund Robert.
Weiss: Double. The double is a two-way action and North should not be averse to moving out of it with a strong two-suiter.
In favour of a pass:
Roth: Pass. I didn’t push them into game. I’ll be satisfied to beat them.
Rubens: Pass. A double without a trump honour would be an insult to East.
There was also some support for bidding on: five diamonds, even five clubs. Howard observed that these
bidders fell into two groups: far out, and far, far out. I must say that I don’t think the 1971 panel distinguished itself. The most likely construction of East’s hand is that he has six good spades and strong hearts. The full hand might be something like this:
Whether you double or not – Howard considered it close – it is very important to lead a trump, not a heart or the ace of clubs. A trump lead (or a diamond and a trump back) is good enough, just. East wins and leads his singleton club, taken by the king. You play another round of trumps. If East takes this in dummy and runs the queen of clubs, discarding a diamond, you will play a third round of trumps, holding the declarer to nine tricks.
My BOLS bridge tip is: When opponents bid unexpectedly high, ask yourself if your hand contains any nasty surprises.
Esta entrada también está disponible en: Spanish | 917 | 3,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-09 | longest | en | 0.951335 |
https://www.univerkov.com/what-amount-of-salt-is-obtained-by-the-interaction-of-33-3-g-of-calcium-chloride-and-16-4-g-of-sodium-phosphate/ | 1,721,879,520,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00545.warc.gz | 883,032,170 | 6,413 | # What amount of salt is obtained by the interaction of 33.3 g of calcium chloride and 16.4 g of sodium phosphate?
Given:
m (CaCl2) = 33.3 g
m (Na3PO4) = 16.4 g
To find:
n (salt) -?
1) 3CaCl2 + 2Na3PO4 => Ca3 (PO4) 2 ↓ + 6NaCl;
2) M (CaCl2) = Mr (CaCl2) = Ar (Ca) * N (Ca) + Ar (Cl) * N (Cl) = 40 * 1 + 35.5 * 2 = 111 g / mol;
3) n (CaCl2) = m (CaCl2) / M (CaCl2) = 33.3 / 111 = 0.3 mol;
4) M (Na3PO4) = Mr (Na3PO4) = Ar (Na) * N (Na) + Ar (P) * N (P) + Ar (O) * N (O) = 23 * 3 + 31 * 1 + 16 * 4 = 164 g / mol;
5) n (Na3PO4) = m (Na3PO4) / M (Na3PO4) = 16.4 / 164 = 0.1 mol;
6) n (Ca3 (PO4) 2) = n (Na3PO4) / 2 = 0.1 / 2 = 0.05 mol;
7) n (NaCl) = n (Na3PO4) * 6/2 = 0.1 * 6/2 = 0.3 mol.
Answer: The amount of Ca3 (PO4) 2 substance is 0.05 mol; NaCl – 0.3 mol.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 464 | 1,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.819948 |
https://nados.io/question/unfold-of-linkedlist | 1,680,142,740,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00035.warc.gz | 488,847,704 | 22,131 | `{"id":"3b4a2cb1-2c57-4a62-ac59-7428d7c8a565","name":"Unfold Of Linkedlist","description":"Given a singly linkedlist : l0 -> ln -> l1 -> ln-1 -> l2 -> ln-2 -> l3 -> ln-3 -> ..... \r\nreorder it : l0 -> l1 -> l2 -> l3 -> l4 -> l5 -> l6 ..... -> ln-1 -> ln\r\nfor more information watch video.","inputFormat":"1->7->2->6->3->5->4->null\r\n","outputFormat":"1->2->3->4->5->6->7->null\r\n","constraints":"0 <= N <= 10^6","sampleCode":{"cpp":{"code":"#include <iostream>\r\nusing namespace std;\r\n\r\nclass ListNode\r\n{\r\npublic:\r\n int val = 0;\r\n ListNode *next = nullptr;\r\n\r\n ListNode(int val)\r\n {\r\n this->val = val;\r\n }\r\n};\r\n\r\nvoid unfold(ListNode *head)\r\n{\r\n \r\n}\r\n\r\nvoid printList(ListNode *node)\r\n{\r\n ListNode *curr = node;\r\n while (curr != nullptr)\r\n {\r\n cout << curr->val << \" \";\r\n curr = curr->next;\r\n }\r\n cout << endl;\r\n}\r\n\r\nint main()\r\n{\r\n int n;\r\n cin >> n;\r\n ListNode *dummy = new ListNode(-1);\r\n ListNode *prev = dummy;\r\n while (n-- > 0)\r\n {\r\n int val;\r\n cin >> val;\r\n prev->next = new ListNode(val);\r\n prev = prev->next;\r\n }\r\n\r\n ListNode *head = dummy->next;\r\n unfold(head);\r\n printList(head);\r\n\r\n return 0;\r\n}"},"java":{"code":"import java.util.*;\r\n\r\nclass Main {\r\n public static class ListNode {\r\n int val = 0;\r\n ListNode next = null;\r\n\r\n ListNode(int val) {\r\n this.val = val;\r\n }\r\n }\r\n\r\n public static void unfold(ListNode head) {\r\n\r\n }\r\n\r\n static void printList(ListNode node) {\r\n while (node != null) {\r\n System.out.print(node.val + \" \");\r\n node = node.next;\r\n }\r\n }\r\n\r\n public static void main(String[] args) {\r\n Scanner scn = new Scanner(System.in);\r\n int n = scn.nextInt();\r\n ListNode dummy = new ListNode(-1);\r\n ListNode prev = dummy;\r\n while (n-- > 0) {\r\n prev.next = new ListNode(scn.nextInt());\r\n prev = prev.next;\r\n }\r\n\r\n ListNode head = dummy.next;\r\n unfold(head);\r\n printList(head);\r\n }\r\n}"},"python":{"code":""}},"points":10,"difficulty":"easy","sampleInput":"9 5 1 1 4 4 6 6 9 9 ","sampleOutput":"5 1 4 6 9 9 6 4 1 ","questionVideo":"https://www.youtube.com/embed/GjhGhEoZ1Vk","hints":[],"associated":[],"solutionSeen":false,"tags":[],"meta":{"path":[{"id":0,"name":"home"},{"id":"0c54b191-7b99-4f2c-acb3-e7f2ec748b2a","name":"Data Structures and Algorithms","slug":"data-structures-and-algorithms","type":0},{"id":"1e4c8949-5890-4d15-be5b-6601c7e2029a","name":"Linked List For Intermediate","slug":"linked-list-for-intermediate-637","type":0},{"id":"04a0f4ca-5264-4d4b-a6f6-513d49ff52ee","name":"Unfold Of Linkedlist","slug":"unfold-of-linkedlist","type":1}],"next":{"id":"050d77c9-5dcc-4407-8caa-6c913f458b1e","name":"Unfold Of Linkedlist MCQ","type":0,"slug":"unfold-of-linkedlist-mcq"},"prev":{"id":"c9224b6e-2ec6-4827-b19f-5a75fe5dd894","name":"Fold Of Linkedlist MCQ","type":0,"slug":"fold-of-linkedlist-mcq"}}}`
Given a singly linkedlist : l0 -> ln -> l1 -> ln-1 -> l2 -> ln-2 -> l3 -> ln-3 -> ..... reorder it : l0 -> l1 -> l2 -> l3 -> l4 -> l5 -> l6 ..... -> ln-1 -> ln for more information watch video.
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Editor
easy
Given a singly linkedlist : l0 -> ln -> l1 -> ln-1 -> l2 -> ln-2 -> l3 -> ln-3 -> ..... reorder it : l0 -> l1 -> l2 -> l3 -> l4 -> l5 -> l6 ..... -> ln-1 -> ln for more information watch video.
0 <= N <= 10^6
## Format
### Input
1->7->2->6->3->5->4->null
### Output
1->2->3->4->5->6->7->null
## Example
Sample Input
`.css-23h8hz{color:inherit;font-size:0.875rem;line-height:1.125rem;letter-spacing:0.016rem;font-weight:var(--chakra-fontWeights-normal);white-space:pre-wrap;}9 5 1 1 4 4 6 6 9 9 `
### Sample Output
`.css-3oaykw{color:var(--chakra-colors-active-primary);font-size:0.875rem;line-height:1.125rem;letter-spacing:0.016rem;font-weight:var(--chakra-fontWeights-normal);white-space:pre-wrap;font-family:Monospace;}5 1 4 6 9 9 6 4 1 `
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1.
Factoring wavelet transforms into lifting steps 总被引：223，自引：0，他引：223
This article is essentially tutorial in nature. We show how any discrete wavelet transform or two band subband filtering with finite filters can be decomposed into a finite sequence of simple filtering steps, which we call lifting steps but that are also known as ladder structures. This decomposition corresponds to a factorization of the polyphase matrix of the wavelet or subband filters into elementary matrices. That such a factorization is possible is well-known to algebraists (and expressed by the formulaSL(n;R[z, z−1])=E(n;R[z, z−1])); it is also used in linear systems theory in the electrical engineering community. We present here a self-contained derivation, building the decomposition from basic principles such as the Euclidean algorithm, with a focus on applying it to wavelet filtering. This factorization provides an alternative for the lattice factorization, with the advantage that it can also be used in the biorthogonal, i.e., non-unitary case. Like the lattice factorization, the decomposition presented here asymptotically reduces the computational complexity of the transform by a factor two. It has other applications, such as the possibility of defining a wavelet-like transform that maps integers to integers. Research Tutorial Acknowledgements and Notes. Page 264. 相似文献
2.
Regularity of refinable function vectors 总被引：10，自引：0，他引：10
We study the existence and regularity of compactly supported solutions φ = (φv) v=0 /r−1 of vector refinement equations. The space spanned by the translates of φv can only provide approximation order if the refinement maskP has certain particular factorization properties. We show, how the factorization ofP can lead to decay of |̸v(u)| as |u| → ∞. The results on decay are used to prove uniqueness of solutions and convergence of the cascade algorithm. 相似文献
3.
Regularization of ill-posed linear inverse problems via ? 1 penalization has been proposed for cases where the solution is known to be (almost) sparse. One way to obtain the minimizer of such an ? 1 penalized functional is via an iterative soft-thresholding algorithm. We propose an alternative implementation to ? 1-constraints, using a gradient method, with projection on ? 1-balls. The corresponding algorithm uses again iterative soft-thresholding, now with a variable thresholding parameter. We also propose accelerated versions of this iterative method, using ingredients of the (linear) steepest descent method. We prove convergence in norm for one of these projected gradient methods, without and with acceleration. 相似文献
4.
In this paper we show that there exist wavelet frames that have nice dual wavelet frames, but for which the canonical dual frame does not consist of wavelets, i.e., cannot be generated by the translates and dilates of a single function. 相似文献
5.
A multiresolution analysis of a curve is normal if each wavelet detail vector with respect to a certain subdivision scheme lies in the local normal direction. In this paper we study properties such as regularity, convergence, and stability of a normal multiresolution analysis. In particular, we show that these properties critically depend on the underlying subdivision scheme and that, in general, the convergence of normal multiresolution approximations equals the convergence of the underlying subdivision scheme. 相似文献
6.
Starting from any two compactly supported refinable functions in L2(R) with dilation factor d,we show that it is always possible to construct 2d wavelet functions with compact support such that they generate a pair of dual d-wavelet frames in L2(R). Moreover, the number of vanishing moments of each of these wavelet frames is equal to the approximation order of the dual MRA; this is the highest possible. In particular, when we consider symmetric refinable functions, the constructed dual wavelets are also symmetric or antisymmetric. As a consequence, for any compactly supported refinable function in L2(R), it is possible to construct, explicitly and easily, wavelets that are finite linear combinations of translates (d · – k), and that generate a wavelet frame with an arbitrarily preassigned number of vanishing moments.We illustrate the general theory by examples of such pairs of dual wavelet frames derived from B-spline functions. 相似文献
7.
We present a generalization of the commutation formula to irregular subdivision schemes and wavelets. We show how, in the noninterpolating case, the divided differences need to be adapted to the subdivision scheme. As an example we include the construction of an entire family of biorthogonal compactly supported irregular knot B-spline wavelets starting from Lagrangian interpolation. September 4, 1998. Date revised: July 27, 1999. Date accepted: November 16, 2000. 相似文献
8.
9.
If the mask of a refinable function has infinitely many coefficients, or if the coefficients are irrational, then it is often replaced by a finite mask with coefficients with terminating decimal expansions when it comes to applications. This note studies how such truncation affects the refinable function.Communicated by Charles A. Micchelli 相似文献
10.
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https://www.aqua-calc.com/one-to-all/temperature/preset/kelvin/886.15 | 1,718,712,885,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00273.warc.gz | 585,808,596 | 5,804 | # Convert Kelvins [K] to other units of temperature
## Kelvins [K] temperature conversions
886.15 K = 613 degrees Celsius K to °C 886.15 K = 1 135.4 degrees Fahrenheit K to °F 886.15 K = 1 595.07 degrees Rankine K to °R 886.15 K = 329.33 degrees Romer K to °Rø
Convert entered temperature to units of energy.
#### Foods, Nutrients and Calories
BAKE and FRY COATING PREMIUM SEASONED MIX, UPC: 041449001401 weigh(s) 127 grams per metric cup or 4.2 ounces per US cup, and contain(s) 333 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
4999 foods that contain Tyrosine. List of these foods starting with the highest contents of Tyrosine and the lowest contents of Tyrosine
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Kon Tiki weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 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 ]
Bromane [HBr] weighs 3.6452 kg/m³ (0.00210706 oz/in³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-401B, liquid (R401B) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
A pound force inch (lbf-in) is a non-SI (non-System International) measurement unit of torque.
The electric potential φ(A) of a point A (in the electric field) specifies the work to be done by the force F = −Q × E in order to move the charge Q from a fixed reference point P to the point A.
cap 10 to cap 12el conversion table, cap 10 to cap 12el unit converter or convert between all units of volume measurement.
#### Calculators
Calculate Maximum Heart Rate (MHR) based on age and gender | 552 | 1,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.744333 |
http://www.jiskha.com/display.cgi?id=1355729491 | 1,496,098,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463613135.2/warc/CC-MAIN-20170529223110-20170530003110-00456.warc.gz | 669,790,134 | 3,873 | # probability
posted by on .
1. 52% of adults say chocolate chip is their favorite cookie. 40 adults are randomly selected and ask their favorite cookie. Find the probability that
a) at most 15 people say chocolate chip is their favorite cookie.
b) at least 15 people say chocolate chip is their favorite cookie.
c) more than 15 people say chocolate chip is their favorite cookie.
• probability - ,
23
• probability - ,
24
• probability - ,
zxzxzxzx
### Answer This Question
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1. ## Use of Calculus in Electrical Engineering
Maxwell's equations are a great example. However, if you're doing a calculus project, they may be a bit complicated (not sure when you learn to integrate vectors). Check out wikipedia for some great insight into Maxwell's Equations: http://en.wikipedia.org/wiki/Maxwell%27s_equations Most EE programs will have an entire semester devoted to just these equations, so don't feel so bad if they don't make sense immediately. Perhaps it would be best to do something a little easier and more elementary like Kirchoff's Voltage Law (a freshman-sophomore level topic in EE): http://en.wikipedia.org/wiki/Kirchoff%27s_current_law This is often referred to as KVL. Basically, it says that if you have a closed path in a circuit (makes a circle back on itself), the integral of the Electric Field along that path is zero. In other words, the net voltage around any loop in a circuit is zero. I'm not sure your knowledge of voltage/current, but basically Voltage is how much power is in the punch of the electricity, and Current is how many times/how quickly you get punched. Hope this helps
2. ## Excellent free website on Relativity and Light
One must not forget that the speed of light is a DEFINITION. If the speed of light in a vacuum were "slowed down" to 3000m/s, then the unit of meters would represent something different, not the speed of light. (a meter would thus be a much longer unit of length).
3. ## bendy lamps !
Aye there would... perhaps some sort of leg under the TV? Or some sort of wire attachment to the ceiling for support. If only they made TV's that were more box-shaped instead of flat... then we could just set them on carts with wheels
4. ## bendy lamps !
I have an idea, but I may have to draw it in order for it to be understandable... I'll try explaining though. Have an extendable arm attached to the wall (a tube inside a wider tube). Where the tube is attached to the wall, make this a horizontally rotatable joint. Make the attachment to the TV be horizontally rotatable joint as well. This should allow you to move the TV to whatever position you like. I'll try an ascii drawing here to see if that helps: Overhead view: |\ | | | | | | | | | | | | ++ | | || | |\ __________+------------/ | | |=C__________| C | WALL | |/ +------------\ | | | || | | ++ | | | | | | |/ | TV ^ ^ ^ Joint Extension Mount/Joint Just an idea, but I think this is more plausible than the bendable sturdy structure.
5. ## Antennas?
For protection against noise I would suggest putting A to D and D to A converters at transmitter and receiver (think of the difference between old analog wireless phones and the now-common digital wireless phones). You should be able to use the following block diagram: Analog Signal input -> AD Converter -> TX Circuit -> TXAntenna ~~~~~~~> RXAntenna -> RX Circuit -> DA Converter -> Analog Signal output I built something similar to this in an engineering course (though not with sound) and it worked pretty well (though our system was to cause two wireless "units" have corresponding lights turn on and off at the same time, the idea should be basically the same). Just as a note, your receiver may be quite a bit larger than you'd like if you're going to use commercially available AD Converters and RX Circuits... You're also going to need something to power both systems. The use of bluetooth is also an option, but I don't know anything about making bluetooth-enabled systems, but if you look at the bluetooth cellphone earbuds that are on the market, one would think it would be pretty easy to sync those up to your music system A quick google on the topic yielded: http://www.amazon.com/exec/obidos/tg/detail/-/B000BK1QSE/ref=pd_sim_wl_2/002-7584979-1141606?v=glance&s=wireless http://www.geeks.com/details.asp?invtid=980397-0403-DT&cat=MP3 Buy that, rip it open and mod as you like. Or, if you'd like to wait a little bit, wait until the "Plantronics Versa" comes out: http://www.oneandco.com/casestudy_12.html (NOTE: I'll have you know you got me wasting a lot of time surfing around looking at this topic
6. ## A New Relativity is born, An Anti-Relativity
nwaogu, I'm having a hard time trying to follow what your theory actually IS. Can you state your theory in a single sentence to allow me to understand better what you are saying (eg. "Gravity changes the direction of light propagation." or "Gravity and light are the same thing" or etc. etc.). One concise sentence please.
7. ## Year 12 physics EPI (Australia)
Shoot a bullet into a block of wood hanging by a wire and then determine the velocity of the bullet by the swing of the wood after the bullet is implanted in it. This is a very common experiment and applies a lot of the topics you provided. You should be able to find plenty of info on this on the internet. I know this is probably too late, but if you're in dier need of SOMEthing, this may be something you could whip up rather quickly (and maybe throw a dart instead of shooting a gun?)
8. ## A New Relativity is born, An Anti-Relativity
How, nwaogu, would you explain clocks that are unaffected by gravity measuring time differently at different speeds? (By "unaffected by gravity", I will mean that it's physical orientation to its Gravitational Force vector has very little affect on the time [much less than the effect of the relative speed]). If you are to answer this question, please do so concisely as I despise long-winded, skirt-the-issue, responses.
9. ## lightspeed vision
Go in your bathroom and look in the mirror. Now, realize that, relative to some object moving the speed of light (or close to) away from you, you and the mirror are moving at the speed of light. Things still look normal though.
11. ## How is .999... = 1?
Mathematically, .9 repeating and 1 represent the same idea (ie. they are the same number). 1/3 represents the same idea as .3 repeating. 1/3 + 1/3 + 1/3 is 1. .333_ + .333_ + .333_ = .999_ Since 1/3 and .333_ are equal (they portray the exact same value) then .333_ + .333_ + .333_ = 1/3 + 1/3 + 1/3 and thus .999_ = 1. No rounding is being done here. .999_ and 1 represent the same concept.
12. ## Another Lonely Valentine's Day
I was quite nerdy during my highschool years and know very much what it's like to feel left out and unpopular (as I would guess many on this post could agree with). However, while working in a restaurant during a summer during highschool, I met an amazing woman whom I ended up marrying. She appreciates me for me and even appreciates my nerdiness. There were many girls I thought, "If only I could date HER." Looking back now, I realize that they weren't the type of people for me. Sure, at the time I thought about how perfect those other women seemed, but now I can tell you that I would not have enjoyed dating them. I guess I'm just trying to say don't give up hope. Be kind and courteous and respectful and eventually there will be a woman who will be very attracted to that; don't expect that during high school though... it takes a while for women get past the attraction-to-jerks phase (this I learned from talking w/ women about the topic). Give it time and keep at it.
13. ## QM-relativity-Classical?
I thought Relativity was the discussion of Macro, Classical mechanics the micro (relatively), and QM the sub-atomic
14. ## black hole+bermuba triangle=?
A triangular patch of the Atlantic Ocean with vertices: Puerto Rico, Southern Florida, and Bermuda. See http://en.wikipedia.org/wiki/Bermuda_triangle for a lot more info.
15. ## Spinning faster than c?
1) Portions, but have had many discussions on the topic. 2) Not specifically on the topic, but it was covered in one of my engineering physics courses. Right, I did not mean to imply this by what I said at all. I am very aware that an object, in its own reference frame, has no velocity. This is partially the answer I was looking for. I was aware of time dilation and other effects of traveling at high speeds. And from my previous post, that appears to be what I was confused about. However, I am curious as to how an outer world would be perceived from this viewpoint as the velocity is not in a cartesian direction, but rather the phi cylindrical direction (which is constantly changing its cartesian direction). I suppose if you were to stare at a green ball stationary to an outside reference while you spun at very high speeds relative to that outside reference, then its color would fluctuate between all the colors. Also, I suppose time would pass slower for you half the time and slower for the ball the other half of the time. Hmm... I guess it all ends up that you can represent any instant "cartesianly" anyway, so I'll just sheepishly go away and end this self-flagellation. | 2,162 | 9,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.94044 |
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1. Jun 5, 2010
### majormuss
Do I have memorize the entire Unit circle ??
1. The problem statement, all variables and given/known data
I am currently takin a trig class and I was a bit daunted by the Unit Circle and all its special angles. My question is... Do I haveto memorize the entire Unit Circle and its angles?? Will it be given to me during exams(as cheat sheet)??
2. Relevant equations
3. The attempt at a solution
2. Jun 5, 2010
### Tedjn
Re: Do I have memorize the entire Unit circle ??
You'd have to ask your teacher that. Common angles you might be expected to know without a cheat sheet are 0, pi/6 (30 deg), pi/4 (45 deg), pi/3 (60 deg), pi/2 (90 deg). That's not too much, and you would only need to know them for sine and cosine. Probably, you can use triangles to remember them more easily. Just knowing these means you know the analogous angles in the other quadrants too.
3. Jun 5, 2010
### GreatEscapist
Re: Do I have memorize the entire Unit circle ??
It's good to.
We had to, but I took a college class. It's not as hard as it looks- memorize thetas first- they're easy once you see the pattern- then when you see the pattern it will be really easy to memorize. =)
4. Jun 5, 2010
### Redbelly98
Staff Emeritus
Re: Do I have memorize the entire Unit circle ??
I agree with Tedjn, and if you consider that 0° and 90° are pretty trivial (sine and cosine are either 0 or 1, as is easily seen by looking at a unit circle), then there are only 3 angles (30°, 45°, 60°) whose sine and cosine need to be memorized. This involves remembering 3 numbers:
$$\frac{\sqrt{1}}{2}, \ \frac{\sqrt{2}}{2}, \ \frac{\sqrt{3}}{2}$$
Obviously the square-root sign is not necessary for the first one, but I included it as it may make this sequence of numbers easier to memorize.
5. Jun 5, 2010
### shelovesmath
Re: Do I have memorize the entire Unit circle ??
Yeah, it's a good idea. It will make your math life easier.
6. Jun 5, 2010
### majormuss
Re: Do I have memorize the entire Unit circle ??
thanks.... i'll try to incorporate your suggestions!!!
7. Jun 5, 2010
### shelovesmath
Re: Do I have memorize the entire Unit circle ??
One thing that helped me to understand it better was when I realized it's basically just one quadrant folded over onto itself twice.
8. Jun 5, 2010
### GreatEscapist
Re: Do I have memorize the entire Unit circle ??
That's really true. Just learn one of them, and you'll be on your way. | 697 | 2,519 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-51 | latest | en | 0.924415 |
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# DIVISIBILITY
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Joined: 08 Apr 2014
Posts: 2
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09 Apr 2014, 11:09
Is X divisible by 119
1) x is divisible by 17
2) x is divisible by 7
Is X divisible by 96
1)x is divisible by 6
2)x is divisible by 16
My question is that how in the first the answer is together SUFFICIENT and in the second both INSUFFICIENT?
Thank you.
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### Show Tags
09 Apr 2014, 11:46
1
KUDOS
Let x = 48.
Then x is not divisible by 96, but is divisible by 6 and 16.
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### Show Tags
10 Apr 2014, 01:27
1
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Expert's post
nikolas7 wrote:
Is X divisible by 119
1) x is divisible by 17
2) x is divisible by 7
Is X divisible by 96
1)x is divisible by 6
2)x is divisible by 16
My question is that how in the first the answer is together SUFFICIENT and in the second both INSUFFICIENT?
Thank you.
In the first problem, you are given that 17 and 7 are factors of x. Both are prime numbers hence you cannot make them using any other numbers. So 17*7 = 119 must be a factor of x.
In the second question, first statement tells you that 2 and 3 are factors of X.
The second statement tell you that four 2s are factors of X.
Together, you know that four 2s and a 3 are factors of X. That is 48 is a factor of X. But is 96? We don't know. Note that 6 and 16 have a common 2. Hence we cannot say that x must be divisible by 6*16 = 96.
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### Show Tags
17 Apr 2014, 11:51
nikolas7 wrote:
Is X divisible by 119
1) x is divisible by 17
2) x is divisible by 7
Is X divisible by 96
1)x is divisible by 6
2)x is divisible by 16
My question is that how in the first the answer is together SUFFICIENT and in the second both INSUFFICIENT?
Thank you.
Hey
One way of solving such problems is to find Least common multiple of the two numbers.
In first problem - LCM of 17 and 7 is 119. So X will definitely be divisible by 119
In second problem - LCM of 16 and 6 is 48. So X will definitely be divisible by 48. But may or may not be divisible by 96.
Re: DIVISIBILITY [#permalink] 17 Apr 2014, 11:51
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Display posts from previous: Sort by | 1,179 | 3,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-22 | longest | en | 0.906309 |
https://e-learnteach.com/circular-pendulum-motion/ | 1,679,658,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00609.warc.gz | 272,295,055 | 14,424 | # Circular pendulum motion
be the angle subtended by the string with the downward vertical. Figure 60: A conical pendulum. In other words, the vertical component of the tension force. A conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’.Circular motion and vibration experiments for high schools, including angular acceleration, rotation, centripetal, dimensions, pendulums and oscillations.A conical pendulum consists of a weight (or bob) fixed on the end of a string or rod suspended from a pivot. Its construction is similar to an ordinary. A simple pendulum consists of a relatively massive object – known as the pendulum bob – hung by a string from a fixed support. When the bob is displaced.
View this answer now! It’s completely free.
## conical pendulum
The conical pendulum lab allows students to investigate the physics and mathematics of uniform circular motion. A motorized, plastic plane* is suspended. The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at a constant speed.This is similar to a conical pendulum. A conical pendulum is a mass attached to a nearly massless string that is held at the opposite end and swung in. SS Tongaonkar · 2011 · Cytowane przez 7 Just as simple harmonic motion can be best understood with simple pendulum, the uniform circular motion can be demonstrated with conical pendulum. Conical. be the angle subtended by the string with the downward vertical. Figure 60: A conical pendulum. In other words, the vertical component of the tension force.
## pendulum circular motion meaning
A simple pendulum consists of a relatively massive object – known as the pendulum bob – hung by a string from a fixed support. When the bob is displaced. It can move in circles in a clockwise or counterclockwise direction. It can also move in an elliptical motion, or an up and down motion to signify a strong, From the first scientific investigations of the pendulum around 1602 by Galileo Galilei, the regular motion of pendulums was used for timekeeping and was. Note that a (simple) pendulum performs (simple) harmonic motion and oscillates about it’s mean position. For circular motion, the force must. A conical pendulum consists of a weight (or bob) fixed on the end of a string or rod. the bob of a conical pendulum moves at a constant speed in a circle with.
## conical pendulum calculator
This is called a centripetal force. The equation for centripetal force is Fc = mv 2/r, where m is the mass of the object, v is the tangential velocity, and r. A conical pendulum consists of a bob of mass ‘m’ revolving in a horizontal circle with constant speed ‘v’ at the end of a string of length ‘l’.Physical Pendulum Calculation. two table clamps and three steel rods. (2) the time period of the conical pendulum equation or formula of time period. Physics Formulas,Period,Conical Pendulum,Length of Inextensible cord,Acceleration of Gravity,Half angle,Suspension angle.The Physical Pendulum Calculator helps you compute the period and frequency of a physical pendulum.
## conical pendulum experiment
This experiment uses a conical pendulum to familiarize us with dynamic equilibrium in rota- tional motion. In this investigation, we will. A conical pendulum is a pendulum that is spun round in a circle instead of swung backwards and forwards. In this experiment a mass is attached to a string. This experiment uses a conical pendulum to familiarize us with dynamic equilibrium in rota- tional motion. In this investigation, we will identify the free. I Proctor · 1968 · Cytowane przez 5 Conical Pendulum Experiment. American Journal of Physics 36, 555 (1968) https://doi.org/10.1119/1.1974975 · Ivan Proctor and T. H. Edwards.You need to repeat the experiment for each mark two times. Table 1, mass of stopper, _____, kg. | 911 | 3,939 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-14 | latest | en | 0.880523 |
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Digital Image Processing Certification Exam Tests
Digital Image Processing Practice Test 11
# Spatial and Intensity Resolution Quiz PDF: Questions and Answers - 11
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## Spatial & Intensity Resolution Questions and Answers : Quiz 11
MCQ 51:
In MxN, M is no of
1. intensity levels
2. colors
3. rows
4. columns
MCQ 52:
Ultraviolet light is used in
1. law enforcement
3. nuclear medicine
4. fluorescence microscopy
MCQ 53:
FWT stands for
1. Fast wavelet transformation
2. Fast wavelet transform
3. Fourier wavelet transform
4. Fourier wavelet transformation
MCQ 54:
PSF stands for
1. probability spread function
2. point spread function
3. probability spike function
4. point spike function
MCQ 55:
Diagonal lines are angles at
1. 0
2. 30
3. 45
4. 90
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DBMS App (iOS & Android) | 529 | 2,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.73097 |
http://web.mnstate.edu/peil/MDEV102/U2/S12/S12_print.html | 1,611,078,532,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00619.warc.gz | 111,028,882 | 3,847 | Factors, Multiples, and Divisors
# Motivation Problem
How does the following problem relate to the factors in a product?
Carla has twelve gum balls and wants to share them among friends where eachperson receives the same number of gumballs. Carla has several choices for how she shares her gumballs depending on how many friends she shares her gumballs with. What are all the possible ways she can share her gum balls â€"number of people and number of gumballs each receives?
Since each person must receive a whole gumball, the problem is asking for all the possible natural number products that can be formed where the product is twelve. That is,
1 × 12, 2 × 6, 3 × 4, 4 × 3, 6 × 2, and 12 × 1.
The possibilities are:
She keeps all twelve gumballs, 1(12) = 12.
She and a friend each get six gumballs, 2(6) = 12.
She and two friends each get four gumballs, 3(4) = 12.
She and three friends each get three gumballs, 4(3) = 12.
She and five friends each get two gumballs, 6(2) = 12.
She and eleven friends each get one gumball, 12(1) =2.
We may consider the above problem in three different ways: What are all the ways two natural number factors give a product of twelve? What are all the ways we can multiply two natural numbers to get twelve? What are the possible natural number divisors of twelve that give a natural number quotient? These different perspectives for the above problem motivate the concepts of factors, multiples, and divisors.
# Factors, Multiples, and Divisors
Definitions for Factors, Multiples, and Divisors: Two numbers are factors of a number if their product is the number. The number is a multiple of a factor. Each factor is a divisor of the number.
General Property when the Natural Numbers is the Universal Set:
a is a factor of b if there is a k so that b = ak with {a, b, k} is a subset of the natural numbers.
b is a multiple of a if there is a number k so that b = ak with {a, b, k} is a subset of the natural numbers.
a is a divisor of b if there is a k so that b = ak with {a, b, k} is a subset of the natural numbers.
Numeric Example:
Since 5 × 8 = 40, both 5 and 8 are factors of 40.
Since 5 × 8 = 40, 40 is a multiple of 5 and 40 is also a multiple of 8.
Since 5 × 8 = 40, both 5 is a divisor of 40 and 8 is also a divisor of 40.
Often we need to find all of the factors or multiples of a number. It is convenient to think of this group of factors as a set.
Example: In the introduction motivation problem, the question was asking for all the natural number factors of twelve. The set of factors of twelve, {1, 2, 3, 4, 6, 12}, is a list of possibilities for the number of people who would receive gumballs.
Example:
The set of all the whole number factors (natural number factors) of 15 is {1, 3, 5, 15}.
The set of all the whole number divisors (natural number factors) of 15 is {1, 3, 5, 15}.
The set of all the natural number multiples of 15 is {15, 30, 45, 60, …, 15n, …}.
The set of all the whole number multiples of 15 is {0, 15, 30, 45, 60, …, 15n, …}.
Note that the universe affects the answer. Zero is a whole number multiple of every number since 0 × a = 0. Also notice that the set of multiples is an infinite set.
Example:
{x : x is a natural number multiple of 4} = {4, 8, 12, 16, 20, 24, …, 4n, …}
{x : x is a whole number multiple of 4} = {0, 4, 8, 12, 16, 20, 24, …, 4n, …}
Example:
{x : x is a natural number factor of 24} = {1, 2, 3, 4, 6, 8, 12, 24}.
{x : x is a whole number factor of 24} = {1, 2, 3, 4, 6, 8, 12, 24}.
{x : x is a natural number divisor of 24} = {1, 2, 3, 4, 6, 8, 12, 24}.
{x : x is a whole number divisor of 24} = {1, 2, 3, 4, 6, 8, 12, 24}.
Note that the set of factors is the same when the universe is either the natural numbers or the whole numbers.
If we are asked for the set of all factors of a value, we must include all the whole number factors for the set to be the correct answer. Notice that the factors generally come in pairs.
However, if the product is a perfect square, such as 6 × 6 = 36, there is only one factor because it would be paired with itself.
## Self Check Problems
Write the set of factors of 18.
Solution
Write the set of whole number multiples of 18.
Solution
Write the set of divisors of 18.
Solution
# More on Divisors
Since multiplication and division are inverse operations, the natural number divisors of a value are the same as the factors of that value. It may seem confusing to have two different names for the same set of values, but in some contexts (multiplying contexts) it makes sense to call these values the set of factors, while in other contexts (dividing contexts) it makes sense to call these values divisors.
## Even and Odd Numbers
A natural number ( or whole number) is an even number if it is a multiple of two. A natural number (or whole number) that is not an even number is an odd number.
General Property:
A value of the form 2n, where n is a counting number ( or a whole number), is an even number.
A value in the form of 2n – 1 where n is a counting number is an odd number.
A value in the form of 2n + 1 where n is a whole number is an odd number.
Note that an odd number is always one less (or one more) than some even number, 2n.
Set-Builder Notation:
The set of even counting numbers is {x : x = 2n where n ε N }.
The set of odd counting numbers is {x : x = 2n – 1 where n ε N }.
The set of even whole numbers is {x : x = 2n where n ε W }.
The set of odd whole numbers is {x : x = 2n + 1 where n ε W }.
Roster Notation:
The set of even counting numbers is {2, 4, 6, 8, 10, …}.
The set of odd counting numbers is {1, 3, 5, 7, 9, …} .
The set of even whole numbers is {0, 2, 4, 6, 8, 10, …}.
The set of odd whole numbers is {1, 3, 5, 7, 9, …}.
## Some Other Factor Facts
• A counting number that ends in an even digit is an even number.
• A counting number that ends in the digit 5 or 0 has 5 as a factor.
• A counting number that ends in the digit 0 has 10 as a factor.
• A counting number that ends in two zeros has 100 is a factor.
## Examples with Sets
When working with sets, mathematical or corresponds to the set operation union.
Example:
{x : x is a factor of 12 or x is a factor of 10}
= {x : x is a factor of 12} U {x : x is a factor of 10}
= {1, 2, 3, 4, 6, 12} U {1, 2, 5, 10}
= {1, 2, 3 ,4, 5, 6, 10, 12}
When working with sets, mathematical and corresponds to the set operation intersection.
Example:
{x : x is a factor of 12 and x is a factor of 10}
= {x : x is an factor of 12} ∩ {x : x is a factor of 10}
= {1, 2, 3, 4, 6, 12} ∩ {1, 2, 5, 10}
= {1, 2}
## Self Check Problem
Find the roster form for each of the following sets:
{x : x is an odd factor of 18 or x is an even divisor of 20}
Solution
{x : x is an even factor of 18 and x is an even multiple of 3}
Solution
### Joke or Quote
Teacher: "Divide fourteen sugar cubes into three cups of coffee so that each cup has an odd number of sugar cubes in it."
Student: "That's easy: one, one, and twelve."
Teacher: "But twelve isn't odd!"
Student: "Twelve cubes is an odd number of cubes to put in a cup of coffee..." | 2,117 | 7,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-04 | latest | en | 0.937049 |
gapapervlde.health-consult.us | 1,539,723,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510867.6/warc/CC-MAIN-20181016201314-20181016222814-00164.warc.gz | 135,912,830 | 3,872 | # Math problems for rice
Instead of doubling the number rice grains math: the rice problem we are experiencing some problems. It's also an important math concept in math, this relationship between 2 quantities is called a ratio if a recipe calls for 1 egg and 2 cups of flour. The wheat and chessboard problem (sometimes expressed in terms of rice grains) is a mathematical problem expressed in textual form as: if a chessboard were to have. Mathematics (math) math 101 - single course url: mathriceedu math 111 - calculus: we will also discuss open problems. Use these math printables to help second-graders learn to do word problems, involving such concepts as shapes, patterns, days of the week, and money.
Mathematics lies at the foundation of many disciplines in the sciences, engineering fields, and the social sciences, and this influence is growing as these subjects. Wild rice math the students have been learning about wild ricing and using ojibwe words to describe what is needed and the process wild rice word problems. It was something that i found in my singapore math rice for math problem solving improvement once i combined it with bar modeling for word problems.
5 simple math problems no one can solve not all math problems need to be inscrutable here are five current problems in the field of mathematics that anyone. Rice for math problem solving from the teachers chair on teachersnotebookcom (5 pages. For every correct answer you choose, 10 grains of rice are raised to help end world hunger through the world food programme.
Rice university human resources policy no 429-00 problem solving applies to all rice university staff rice university strives. Mathematics lessons for elementary, middle, and high school including geometry, fractions, and algebra.
• Third graders practice solving math equations using rice as their subject they translate situations involving rice into math equations in order to solve them properly.
• Working on math word problems can be confusing for third-grade students the rice strategy is a problem-solving strategy used to help your child organize his thoughts.
• Math index all math worksheets by topic: addition subtraction multiplication division geometry word problems all math worksheets by grade: preschool.
Have you read the book, one grain of rice, by demi it is a story of a rajah in india who takes almost all of the people's rice and stores it for a time when they. Find this pin and more on analyzing word problems by karayasmine singapore math: remind your students to use the rice method for solving word problems. Chapter 8: rice paddies and math tests opening activity look at the list of numbers you will have 5 seconds to memorize the list who can recite these numbers solely.
Math problems for rice
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https://oeis.org/A159977 | 1,632,687,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00620.warc.gz | 464,879,379 | 4,226 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A159977 a(n) = (smallest prime >= Fibonacci(n)) - Fibonacci(n). 2
1, 1, 0, 0, 0, 3, 0, 2, 3, 4, 0, 5, 0, 2, 3, 4, 0, 7, 20, 14, 3, 2, 0, 13, 4, 10, 11, 16, 0, 23, 4, 4, 25, 10, 14, 35, 6, 24, 3, 2, 6, 7, 0, 20, 9, 48, 0, 5, 28, 18, 23, 14, 14, 11, 16, 10, 21, 4, 62, 13, 38, 12, 7, 16, 12, 19, 36, 28, 143, 32, 58, 29, 96, 100, 33, 2, 30, 27, 12, 62, 25, 46, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,6 LINKS T. D. Noe, Table of n, a(n) for n = 1..1000 FORMULA a(n) = (smallest prime >= Fibonacci(n)) - Fibonacci(n). a(n) = 0 <=> n in {A001605}. - Alois P. Heinz, Feb 04 2018 EXAMPLE a(1) = a(2) = 1 because Fibonacci(1) = Fibonacci(2) = 1, the smallest prime >= 1 is 2, and 2 - 1 = 1. a(3) = a(4) = a(5) = 0 because Fibonacci(3)=2, Fibonacci(4)=3, and Fibonacci(5)=5 are all prime. MAPLE a:= n-> (f-> nextprime(f-1)-f)(combinat[fibonacci](n)): seq(a(n), n=1..100); # Alois P. Heinz, Feb 04 2018 MATHEMATICA Table[If[PrimeQ[n], 0, NextPrime[n]-n], {n, Fibonacci[Range[90]]}] (* Harvey P. Dale, Jul 22 2016 *) PROG (UBASIC) 10 'FiboA 20 A=1:print A; 30 B=1:print B; 40 C=A+B:print C; :T=T+1 41 if C<>prmdiv(C) then print "<"; nxtprm(C)-C; ">":else print "<"; 0; ">"; 50 D=B+C:print D; 51 if D<>prmdiv(D) then print "<"; nxtprm(D)-D; ">":else print "<"; 0; ">"; 60 A=C:B=D:if T>22 then stop:else 40 (PARI) F=1; G=0; for(i=1, 100, print1(nextprime(F)-F, ", "); T=F; F+=G; G=T) \\ Hagen von Eitzen, Jul 20 2009 CROSSREFS Cf. A000045, A001605, A159978. Sequence in context: A143394 A112455 A001608 * A245251 A177461 A201924 Adjacent sequences: A159974 A159975 A159976 * A159978 A159979 A159980 KEYWORD easy,nonn AUTHOR Enoch Haga, Apr 28 2009 EXTENSIONS More terms (cf. b-file) from Hagen von Eitzen, Jul 20 2009 Edited by Jon E. Schoenfield, Feb 04 2018 STATUS approved
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Last modified September 26 16:15 EDT 2021. Contains 347670 sequences. (Running on oeis4.) | 980 | 2,318 | {"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} | 3.859375 | 4 | CC-MAIN-2021-39 | latest | en | 0.591504 |
https://studychacha.com/discuss/229110-last-year-solved-question-papers-physics-cbse-12th-board.html | 1,537,886,532,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161661.80/warc/CC-MAIN-20180925143000-20180925163400-00111.warc.gz | 596,337,972 | 19,257 | #1
January 14th, 2014, 10:32 AM
Unregistered Guest Posts: n/a
Last year solved question papers of Physics for CBSE 12th board
Can any of you here provide me the Physics Question paper for CBSE Board Class 12th???
#2
January 14th, 2014, 06:11 PM
Super Moderator Join Date: Jun 2013 Posts: 42,978
Re: Last year solved question papers of Physics for CBSE 12th board
As you require CBSE Board Class 12th Physics Question paper, so here I am sharing the same with you
Q1. An electric dipole of dipole moment 20, X 10-6 Cm is enclosed by a closed surface. What is the net flux coming out of the surface? 1
Q2. An electron beam projected along +X-axis, experiences a force due to a magnetic field along the +Y-axis. What is the direction of the magnetic field? 1
Q3. The power factor of an A.C. circuit is 0.5. What will be the phase difference between voltage and current in this circuit? 1
Q4. Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, and (ii) blue light. 1
Q5. What should be the length of the dipole antenna for a carrier wave of Frequency 3 X 1O8 Hz?
Q6. Define 'electric line of force' and give its two important properties.
Q7. (a) Why does the electric field inside a dielectric decrease when it is placed in an external electric field?
(b) A parallel plate capacitor with air between the plates has a capacitance of 8 PF. What will be the capacitance if the distance between the plates he reduced by half and the space between them is filled with a substance of dielectric constant K=6?
Q8. Draw V - I graph for ohmic and non-ohmic materials. Give one example for each. 2
Q9. Define the terms 'Magnetic Dip' and 'Magnetic Declination' with the help of relevant diagrams.
Rest of the Questions are attached in below file which is free of cost
Attached Files
CBSE 12 Physics Question paper.pdf (65.7 KB, 74 views)
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#3
January 20th, 2015, 05:42 PM
Super Moderator Join Date: Apr 2013 Posts: 28,553
Re: Last year solved question papers of Physics for CBSE 12th board
As you want to get CBSE 12th board Physics question paper so here I am giving you some questions of that paper:
How does one explain increase in resistivity of a metal with increase of temperature ?
Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields.
Give one example of broadcast mode of communication.
Draw the shape of the wavefront coming out of a convex lens when a plane wave is incident on it.
CBSE 12th board Physics Question Paper
For full question paper here is the attachment.................................
Attached Files
CBSE 12th board Physics Question Paper.pdf (267.5 KB, 84 views)
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-- Default Style -- Default vBulletin -- Lightweight MBA Discussion - Job Discussion - Contact Us - StudyChaCha - Blog Archives - Forum Archive - Partners : Management Forum Top | 814 | 3,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | longest | en | 0.922239 |
http://list.seqfan.eu/pipermail/seqfan/2023-May/074637.html | 1,723,772,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00307.warc.gz | 15,217,922 | 3,684 | # [seqfan] Re: A063747 sign of power series expansion of the Gamma function
Max Alekseyev maxale at gmail.com
Mon May 22 21:48:54 CEST 2023
```A scriptish way to specify real precision in PARI is
default(realprecision, 1000);
or alike.
Regards,
Max
On Fri, May 19, 2023 at 4:33 PM Sean A. Irvine <sairvin at gmail.com> wrote:
> I agree with those values.
> But (at least for me), if you run Pari without the "\p" you get the wrong
> result.
>
> Sean.
>
>
> On Sat, 20 May 2023 at 08:28, Hugo Pfoertner <yae9911 at gmail.com> wrote:
>
> > With my(x='x+O('x^150)); apply(sign, Vec(1/(gamma(1-x)))) and \p1000 and
> > \p10000 I get identical results for the first 100 terms.
> > [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, 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, 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, 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] wrong?
> >
> > On Fri, May 19, 2023 at 10:19 PM Sean A. Irvine <sairvin at gmail.com>
> wrote:
> >
> > > Because it is still wrong unless you set the precision higher.
> > >
> > > On Sat, 20 May 2023 at 08:10, Hugo Pfoertner <yae9911 at gmail.com>
> wrote:
> > >
> > > > Why not use the result of Michel Marcus's new PARI program? Is there
> > any
> > > > doubt that PARI can calculate e.g. 100 terms correctly? Then we have
> a
> > > very
> > > > fundamental problem.
> > > >
> > > >
> > > > On Fri, May 19, 2023 at 9:34 PM Sean A. Irvine <sairvin at gmail.com>
> > > wrote:
> > > >
> > > > > Yes, I will make an update.
> > > > >
> > > > > Thank you Jean-François and Brendan for taking the time to verify.
> > > > >
> > > > > Sean.
> > > > >
> > > > >
> > > > >
> > > > > On Sat, 20 May 2023 at 07:04, Neil Sloane <njasloane at gmail.com>
> > wrote:
> > > > >
> > > > > > Well, if a(46) is wrong (which seems to be the case), could
> someone
> > > > > please
> > > > > > make the correction? And probably all the terms after a(46)
> should
> > > be
> > > > > > deleted, and we should give it keyword "more".
> > > > > >
> > > > > >
> > > > > > Best regards
> > > > > > Neil
> > > > > >
> > > > > > Neil J. A. Sloane, Chairman, OEIS Foundation.
> > > > > > Also Visiting Scientist, Math. Dept., Rutgers University,
> > > > > > Email: njasloane at gmail.com
> > > > > >
> > > > > >
> > > > > >
> > > > > > On Fri, May 19, 2023 at 3:14 AM Jean-François Alcover <
> > > > > > jf.alcover at gmail.com>
> > > > > > wrote:
> > > > > >
> > > > > > > Mma's SeriesCoefficient[1/Gamma[1 - x], {x, 0, n}] // Sign
> gives
> > -1
> > > > for
> > > > > > > n=46 (after 15 mn !)
> > > > > > >
> > > > > > > Le ven. 19 mai 2023 à 06:52, Brendan McKay via SeqFan <
> > > > > > > seqfan at list.seqfan.eu>
> > > > > > > a écrit :
> > > > > > >
> > > > > > > > My typo, it is e-38. Oops. B/
> > > > > > > >
> > > > > > > > On 19/5/2023 1:11 pm, Allan Wechsler wrote:
> > > > > > > > > One of you said "e-38" and the other said "e-58". Do you
> > really
> > > > > have
> > > > > > > > > 20 orders of magnitude of disagreement?
> > > > > > > > >
> > > > > > > > > On Thu, May 18, 2023 at 11:07 PM Brendan McKay via SeqFan
> > > > > > > > > <seqfan at list.seqfan.eu> wrote:
> > > > > > > > >
> > > > > > > > > I also get -0.444..e-58 * x^46 in Maple using 60
> digits.
> > > > The
> > > > > > > > > calculation
> > > > > > > > > is numerically unstable though.
> > > > > > > > >
> > > > > > > > > B/
> > > > > > > > >
> > > > > > > > > On 19/5/2023 12:31 pm, Sean A. Irvine wrote:
> > > > > > > > > > Hi,
> > > > > > > > > >
> > > > > > > > > > I believe the terms in A063747 are incorrect for
> n>=46.
> > > > > > > > > >
> > > > > > > > > > https://oeis.org/A063747
> > > > > > > > > >
> > > > > > > > > > My attempts to reproduce the calculation in Maple and
> > > using
> > > > > > > > > constructible
> > > > > > > > > > reals in Java both yield a coefficient, like
> > > > -0.44458...e-38
> > > > > *
> > > > > > > > > x^46, from
> > > > > > > > > > which a(46) should be -1 rather than 1.
> > > > > > > > > >
> > > > > > > > > > Is someone able to make an independent check?
> > > > > > > > > >
> > > > > > > > > > I was unable to run the existing Pari program beyond
> > > n=15,
> > > > > but
> > > > > > I
> > > > > > > > > assume
> > > > > > > > > > there is some way to tell Pari to try harder?
> > > > > > > > > >
> > > > > > > > > > (I previously emailed the author of the sequence, but
> > > > > received
> > > > > > > > > no response.)
> > > > > > > > > >
> > > > > > > > > > Sean.
> > > > > > > > > >
> > > > > > > > > > --
> > > > > > > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > > > > > >
> > > > > > > > >
> > > > > > > > > --
> > > > > > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > > > > > >
> > > > > > > >
> > > > > > > > --
> > > > > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > > > > >
> > > > > > >
> > > > > > > --
> > > > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > > > >
> > > > > >
> > > > > > --
> > > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > > >
> > > > >
> > > > > --
> > > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > > >
> > > >
> > > > --
> > > > Seqfan Mailing list - http://list.seqfan.eu/
> > > >
> > >
> > > --
> > > Seqfan Mailing list - http://list.seqfan.eu/
> > >
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
``` | 2,331 | 5,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.63958 |
https://percentagecalculator.pro/_2_percent-of_eight%20tenths_ | 1,542,474,103,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743717.31/warc/CC-MAIN-20181117164722-20181117190722-00529.warc.gz | 729,986,132 | 11,374 | # How much is 2 percent of eight tenths
### Percentage Calculator
We think you reached us looking for answers like: What is 2 (2%) percent (%) of eight tenths (8/10)? Or may be: How much is 2 percent of eight tenths. See the solutions to these problems below.
## How to work out percentages - Step by Step
Here are the solutions to the questions stated above:
### 1) What is 2% of 8/10?
Always use this formula to find a percentage:
% / 100 = Part / Whole replace the given values:
2 / 100 = Part / 8/10
Cross multiply:
2 x 8/10 = 100 x Part, or
1.6 = 100 x Part
Now, divide by 100 and get the answer:
Part = 1.6 / 100 = 0.16
### 2) 2 is what percent of 8/10?
Use again the same percentage formula:
% / 100 = Part / Whole replace the given values:
% / 100 = 2 / 8/10
Cross multiply:
% x 8/10 = 2 x 100
Divide by 8/10 and get the percentage:
% = 2 x 100 / 8/10 = 250% | 282 | 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} | 4.375 | 4 | CC-MAIN-2018-47 | latest | en | 0.854906 |
https://brainmass.com/physics/rotation/eigenvalues-eigenvectors-properties-matrix-201494 | 1,675,515,312,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00337.warc.gz | 167,336,647 | 75,123 | Explore BrainMass
# Rotation matrices
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Consider the matrix
A = (cos t sin t)
(-sin t cos t)
Show that it is unitary
Show that the eigenvalues are exp(it) and exp(-it)
find the eigenvectors
Verify that U'AU is diagonal matrix, where U is the matrix of the eigenvectors.
Show that since determinant of a matrix is unchanged under unitary change of basis, argue that
1. det(A) = product of its eigenvalues for any Hermitian or Unitary A
2. use the invariance of teh trace under teh same transformation and show that Tr(A)= sum of eigenvalues
https://brainmass.com/physics/rotation/eigenvalues-eigenvectors-properties-matrix-201494
#### Solution Preview
see attached files.
The matrix is unitary since:
The eigenvevalues are the roots of the characteristic polynomial:
Which leads to the following eigenvalues:
The eigenvector associated with
The last two equations are actually the same, hence we can choose an arbitrary value for x and this will set up the value of y:
Thus ...
#### Solution Summary
The solution discusses some of the properties of general rotation matrices and show they are unitary matrices
\$2.49 | 302 | 1,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-06 | longest | en | 0.867595 |
https://www.kidport.com/Grade7/Math/Lesson_7B__Basic_Operations_II.html | 1,561,448,920,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999814.77/warc/CC-MAIN-20190625072148-20190625094148-00421.warc.gz | 811,709,399 | 4,217 | << Back to Lessons Index
# 7th Grade Math / Lesson 7B: Basic Operations II
OPERATIONS What will we be learning in this lesson? In this lesson you will learn how the effects of operations on numbers. Vocabulary words are found in this purple color throughout the lesson. Remember to put these in your notebook.
Operations Multiplying and Dividing 1. If multiplying or dividing numbers that have the same sign, the sign of the answer is positive. Ex) -3 * -4 = 12 Ex) 3 * 4 = 12 Ex) 8 / 2 = 4 Ex) -8 / (-2) = 42. If multiplying or dividing numbers that have different signs, the sign of the answer is negative. Ex) -3 * 4 = -12 Ex) 3 * (-4) = -12 Ex) -8 / 2 = -4 Ex) 8 / (-2) = -4
Operations Practice multiplying integers 3 * (-6) = (-4)(5) = -7(-5) = 12*3 =
Operations Multiplying Practice multiplying integers 3 * (-6) = -18 (-4)(5) =-20 -7(-5) =35 12*3 = 36
Operations Dividing Practice dividing integers 12 / 6 = (-10) / 5 = 20 / -5 = (-15) / (-3) =
Operations Dividing Practice dividing integers 12 / 6 =2 (-10) / 5 = -2 20 / -5 = -4 (-15) / (-3) =5 | 343 | 1,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2019-26 | latest | en | 0.735324 |
https://www.physicsforums.com/threads/integration-help-please.178755/ | 1,524,716,310,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948064.80/warc/CC-MAIN-20180426031603-20180426051603-00393.warc.gz | 836,457,364 | 16,207 | # Homework Help: Integration help please
1. Jul 30, 2007
### gona
Integrating $$\frac{1}{(1+sinx)}$$
i just started learning Integration last week so not exactly sure how to approch this type of question.
2. Jul 30, 2007
### daveb
Do you know how to integrate 1/sinx?
3. Jul 30, 2007
### gona
no this is the first time i have seen where a trigonometric function is on the denominator
but now that i think about it
sinx = $$\sqrt{1-cos^2x}$$
and arcsin was equal to $$\frac{1}{((1-x^2)}$$
so i guess i could use the U substituition method
and then the answer would be...arcsin(cosx)??? im not too sure
Last edited: Jul 30, 2007
4. Jul 30, 2007
### Dick
No, no arcsin. But you might want to try multiplying numerator and denominator by (1-sin(x)). It may look more familiar.
Last edited: Jul 30, 2007
5. Jul 30, 2007
### daniel_i_l
You could also try the subsitution u = tan(x/2). It looks a little messy but everything cancels out.
6. Jul 30, 2007
### gona
well im not sure if im doing the right thing but here goes....
sinx = $$\frac{2tan(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})}$$
soo then i symplify the equation so that it is
$$\frac{1+tan^{2}(\frac{x}{2})}{1+tan^{2}(\frac{x}{2})+2tan(\frac{x}{2})}$$
the i used u = tan$$(\frac{x}{2})$$
so i get $$\frac{1+u^{2}}{(u+1)^{2}}$$
What should i do from here or is this the right way at all?
7. Jul 30, 2007
### Dick
It looks to me like you are taking the long way around. Try multiplying numerator and denominator of your original problem by (1-sin(x)). You get (1-sinx)/cos^2(x). If you split that into two integrals you shouldn't have any problem with either of them.
8. Jul 30, 2007
### daniel_i_l
Dick solution is quicker in this case but to integrate things like
1/(1+cosx+sinx) the substitution u=tan(x/2) is good.
But notice that it isn't x = tan(u/2) but rather u=tan(x/2). In order to get this into something the the example I gave you can show with so trig identities that if u=tan(x/2) then:
dx = 2du/(1+u^2)
sinx = 2u/(1+u^2)
cosx = (1-u^2)/(1+u^2)
If you substitue all that you the (1+u^2)s candel out and you get some rational function which you can solve with the typical rational function method (breaking it into elementry functions...)
9. Jul 30, 2007
### gona
wow by the way i got the answer using what dick said its actually pretty easy once u break it up...now im gonna try Daniels question gonna see if i can get those now :) thnx alot for the help by the way
the answe was tanx - secx +c | 798 | 2,493 | {"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} | 3.953125 | 4 | CC-MAIN-2018-17 | longest | en | 0.863291 |
ekoskola.org.mt | 1,708,907,148,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00051.warc.gz | 225,952,316 | 10,633 | ## BeMED 13. BIDMAS with Marine Litter
##### Mathematics
Preparation Time Nil
Estimated Duration 45 – 50 minutes
Site Classroom
Educational objectives Through problem solving and the practice of addition, subtraction, multiplication and division, the students will learn about the harm so many items are causing on our environment when we throw them away haphazardly, regardless of the length of time they take to decompose.
Learning Outcomes – I can select and use an appropriate operation and strategy when solving a problem.
– I can rehearse adding/ subtracting ThHTU ± ThHTU using informal and standard written methods.
– I can check the result of a calculation and/or real life problem by using an equivalent calculation or an inverse operation.
– I can use the relationship between addition and subtraction.
– I can consolidate the understanding and the usage of the four operations.
– I can use written methods for: HTU × TU/U.
Link to SDGs SDG 3: Good health and well-being
SDG 11: Sustainable cities and communities
SDG 13: Climate action
SDG 14: Life below water
Educational resources required Small whiteboard per group
Appendix 13.1 – Fact sheet
Appendix 13.2 – Flashcards with the questions (need to be printed and cut separately)
Appendix 13.3 – Questions and answers for teachers
Timer/alarm
Internet connection
Remote preparation The teacher will discuss with the students the damage done to marine life. The students will also be provided with a fact sheet (Appendix 13.1) which will give them information on how long it takes for different types of litter to decompose.
Planning Considerations It would be interesting to take the students near the seashore, and see floating litter, or litter on the shore, to become more aware how serious and actual this problem is. An alternative to the outing, one can show the following videos showing the local marine litter.
Method Introduction
This activity is targeted for students in year 6 who already have the knowledge and ability to work out sums in addition, subtraction, multiplication and division. Marine litter will be incorporated with story sums to increase awareness regarding marine litter.
o Development
The class is divided into 4 groups – A, B, C, D – at the beginning of the lesson. A small white board will be given to each group. A video (https://www.youtube.com/watch?v=Yomf5pBN8dY) related to marine litter is shown to the students, after having told them to pay attention since afterwards they will need to participate in a conversation related to what they see in the video. When the video ends, the educator will ask the students to describe what they have seen and mention some facts.
After the introductory video the competitiveness and fun will kick in. The student will be shown another video related to marine litter but this time the students will be asked to write items that contaminate marine life as shown on the video
(https://www.youtube.com/watch?v=017bBeXhYz4). This video is “Sources and impacts of marine litter” by Jane Lee/ Marlisco. After the video, 3 minutes will be given to the students to write down the items they have seen in the video. When time is up students will read out the answers and be given credit.
While the students are writing the answers a fact sheet (Appendix 13.1) will be distributed to each group to be used in the main activity. Rules will be explained to the students before the start of the activity.
Starting with the first group, the group leader selects a flashcard from the desk and the teacher will read the story sum out loud. As soon as the teacher finishes reading a 2-minute timer will start. All groups have to work out the sum collaboratively. A sample question is as follows:
If plastic bottle A was thrown at sea in 2000, bottle B in 2021 and bottle C in 2030, which bottle or bottles will still be found in the sea if aliens come to visit Earth in the year 2462?
A: 2000+450=2450
B: 2021+450=2471
C: 2030+450=2480
(All questions can be found in Appendix 13.2. Questions with answers can be found in Appendix 13.3)
The students will be asked to sum up years and months or multiply, etc to give an answer. In others the students have to look up amounts from the fact sheet distributed earlier.
Notes:
· Different questions require different computations and have different weighting as regards point.
· Some of the questions require students to add up the total length of time of decomposition of different items when in reality the answer is not significant. If a student/group points out that the total does not make sense and explains correctly why, the group is given bonus points at the discretion of the teacher.
· Some of the questions are trick questions (stared*) which do not require any calculations. Refer to Appendix 13.3.
Conclusion
After settling back down in their places, the students will be asked random questions related to marine litter for instance ‘How long does it take for a cigarette butt to decompose?’. This will help the students to be more aware of the impact such “insignificant” items have on marine life.
Follow-up activities After the activity the teacher can ask the students to rank the items according to the time they take to decompose.
As a follow-up activity, maybe a week or so later, it would be interesting to ask the students at random, without any previous notice, how much time do different items take to decompose, in order to have them recall the info given in the fact sheet (Appendix 13.1) given out during the main activity.
Background information for educators Refer to Appendix 13.1 and video links used in method.
Sheet facts data from:
https://en.wikipedia.org/wiki/Marine_debris#/media/File:How_long_until_it’s_gone.jpg
Adaptations For students with learning difficulties:
The learning support educators can help them out with the working of difficult problems (or maybe the use of a calculator will be allowed). | 1,299 | 5,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | longest | en | 0.908912 |
https://community.wolfram.com/groups/-/m/t/326869 | 1,686,259,791,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00665.warc.gz | 207,073,295 | 27,816 | # Solving equation with Bessel functions BesselJ & BesselK
Posted 9 years ago
9661 Views
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13 Replies
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Hi, I am trying to solve the following equation using Mathematica. NSolve[] seems not to be capable of finding the solution. NSolve[] returns back the same equation after "thinking" for a while. Solve[] does nearly the same thing. Is there anything, I can change in the equation to get a better result. a = 25 10^-6; n1 = 1.458 + 5 10^-3; n2 = 1.458 ; [Lambda] = 1.5 10^-6; k0 = (2 [Pi])/[Lambda]; p = Sqrt[n1^2 k0^2 - [Beta]^2]; q = Sqrt[[Beta]^2 - n2^2 k0^2]; DJm[z_] := !( *SubscriptBox[([PartialD]), (z)](BesselJ[m, z])) /. m -> 1; DKm[z_] := !( *SubscriptBox[([PartialD]), (z)](BesselK[m, z])) /. m -> 1; NSolve[((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/( p BesselJ[1, p a]) + (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/( q BesselK[1, q a])) ((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/( p BesselJ[1, p a]) + n2^2/n1^2 (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/( q BesselK[1, q a])) == (1)^2/ a^2 (1/p^2 + 1/q^2) (1/p^2 + n2^2/n1^2 1/q^2), [Beta]] Thanks in advance
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Posted 9 years ago
Note that: Reduce[ n1^2 k0^2 - [Beta]^2 > 0 && [Beta]^2 - n2^2 k0^2 > 0] gives -6.1282 10^6 < [Beta] < -6.10726 10^6 || 6.10726 10^6 < [Beta] < 6.1282 10^6 so any places where p and q are both real are located in one of these ranges. So go ahead and plot the full expression (i.e., the left hand side minus the right hand side) in, for example, the second range: Plot[((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/(p BesselJ[1, p a]) + (1/ 2 (-BesselK[0, q a] - BesselK[2, q a]))/(q BesselK[1, q a])) ((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/(p BesselJ[ 1, p a]) + n2^2/n1^2 (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/(q BesselK[ 1, q a])) - (1)^2/ a^2 (1/p^2 + 1/q^2) (1/p^2 + n2^2/n1^2 1/q^2), {[Beta], 6.107256118578557*^6, 6.1282000696024895*^6}] giving So there appear to be some zeros in this region. And one can use this plot to select a starting point for the FindRoot function as in: FindRoot[((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/(p BesselJ[1, p a]) + (1/ 2 (-BesselK[0, q a] - BesselK[2, q a]))/(q BesselK[1, q a])) ((1/ 2 (BesselJ[0, p a] - BesselJ[2, p a]))/(p BesselJ[1, p a]) + n2^2/n1^2 (1/ 2 (-BesselK[0, q a] - BesselK[2, q a]))/(q BesselK[1, q a])) - (1)^2/ a^2 (1/p^2 + 1/q^2) (1/p^2 + n2^2/n1^2 1/q^2) == 0, {[Beta], 6.12 10^6}] giving {[Beta] -> 6.11989[CenterDot]10^6} But there are several solutions around here (to the accuracy of the FindRoot computation), some possibly elsewhere on the complex plane. using 6.125 10^6 as the seed value gives a root at {[Beta] -> 6.1248[CenterDot]10^6}
Posted 9 years ago
Perfect! I am trying to calculate the propagation constant (Beta) for modes in an optical fiber according to a well known equation. The method, you used, works just fine. However, you assumed that p and q are both real. I was wondering what led to that assumption. I checked the reference, I am reading for the time being. It does not say anything about p and q being purely real. Of course, if they are complex, then I can't use the REDUCE[] command to find the areas, where the solutions exist. Any advice?Thanks in advance
Posted 9 years ago
Actually I didn't assume that p and q were real. What I did was ask what happens int he region where they are real as a way of tracking down a place where I could potentially plot the function and get real values. Given that all the other parameters were real, I then suspected that the value of the function would be real in those areas. The hunch was correct in that there was a plot generated and that indeed the plot seemed to cross the axis--indicating some zeros in that region. There may well be other zeros (or curves of zeros) elsewhere in the complex plane--I haven;t thought much more about it. But the key lesson is to explore the behavior of the function and learn as much about it as possible. Then one goes ahead and uses appropriate Mathematica functions to try to extract the locations of zeros. Some functions and algorithms can be automatically plugged into a high level Mathematica function which then goes on to find solutions on its own. In messy (i.e., many real world) cases though one needs to use Mathematica as a hands-on tool to explore the behavior of the objects that one wants to extract information from (e.g., like zeros). So I think that the best way to take what I wrote was as a bit of an example of that sort of exploratory play.
Posted 9 years ago
Yet, your solution is neat. Which makes me wonder whether q and p have, in fact, real values. I guess, this will be my next exploration step. But regardless of this, I will keep this way of function-analyzing in mind whenever Solve[] fails. Thanks. you've been very helpful.
Posted 9 years ago
For any WorkingPrecision between 48 and 128 NMinimize converges on a beta about 153746 where the resulting minimum is significantly different from zero and does not get closer to zero with increasing Working precision. In[7]:= a = 25 10^-6; n1 = 1458/1000 + 5 10^-3; n2 = 1458/1000; \[Lambda] = 15/10 10^-6; k0 = (2 \[Pi])/\[Lambda]; p = Sqrt[n1^2 k0^2 - \[Beta]^2];q = Sqrt[\[Beta]^2 - n2^2 k0^2]; NMinimize[ Norm[((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/(p BesselJ[1, p a]) + (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/(q BesselK[1, q a])) ((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/(p BesselJ[1, p a]) +n2^2/n1^2 (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/ (q BesselK[1, q a])) - (1)^2/a^2 (1/p^2 + 1/q^2) (1/p^2 + n2^2/n1^2 1/q^2)], \[Beta], WorkingPrecision -> 128] Out[9]= {2.6644908885382207535682631873522244614076638003775608946182982\ 185058485936623112823132496185424062412703162736336930371650049643*10^-14, {\[Beta] -> 153746.742756413486195499781367574310403493602174002003504819\ 78661459273051117012575097394837519002121150764177004071175454428111}} Because of the way that minimum varies with WorkingPrecision I'm a little skeptical whether that is precisely correct or not.
Posted 9 years ago
Thanks for the reply. I am curious about the how using the Norm[] command would help finding the roots of the mentioned equation. It would be kind if you'd explain your method briefly. Regards.
Posted 9 years ago
Sometimes Solve, NSolve, Reduce, FindRoot, etc, etc can have great difficulty and take a long time looking for a zero of a complicated function. NMinimize can sometimes be faster, but you want it to find where your complicated function is near zero, whether it is positive, negative or even Complex, Norm will give a measure of the distance from zero for all those cases. If it finds the minimum of the norm is nonzero then that may indicate there is no zero but, as was shown, it is sometimes possible that it just did not search far enough to find a root.
Posted 9 years ago
Thank you. But, it seems that your method fails sometimes, if the function is sophisticated enough.
Posted 9 years ago
EVERYTHING fails sometimes. The question is whether it is useful or not.
Posted 9 years ago
If I plot the left-hand side of the equation minus the right-hand side, as a function of beta, I basically get 0 for any value of beta
Posted 9 years ago
Thanks for the reply. The difficulty with Plotting is probably a result of P or/and q being complex depending on the value of (Beta). Regards.
Posted 9 years ago
Hi, I don't know why the code is corrupted. I copied the code as INLINE from Mathematica notebook then I paste it in the Sample Code field using "CTR k". However, I included another code snippet in this reply. I double checked it to make sure that it is not corrupted. Thanks a = 25 10^-6; n1 = 1.458 + 5 10^-3; n2 = 1.458 ; \[Lambda] = 1.5 10^-6; k0 = (2 \[Pi])/\[Lambda]; p = Sqrt[n1^2 k0^2 - \[Beta]^2]; q = Sqrt[\[Beta]^2 - n2^2 k0^2]; NSolve[((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/( p BesselJ[1, p a]) + (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/( q BesselK[1, q a])) ((1/2 (BesselJ[0, p a] - BesselJ[2, p a]))/( p BesselJ[1, p a]) + n2^2/n1^2 (1/2 (-BesselK[0, q a] - BesselK[2, q a]))/( q BesselK[1, q a])) == (1)^2/ a^2 (1/p^2 + 1/q^2) (1/p^2 + n2^2/n1^2 1/q^2), \[Beta]] | 2,769 | 8,123 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2023-23 | longest | en | 0.648103 |
https://www.sourcecodeexamples.net/2021/05/c-program-to-find-common-elements-in.html | 1,721,314,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00383.warc.gz | 869,465,239 | 26,497 | # C Program to Find Common Elements in Two Arrays
In this post, Let's write a c program to find common elements in two arrays.
Let’s assume we have following two sorted arrays p and q.
int p[] = {1, 3, 5, 7, 9};
int q[] = {1, 2, 3, 4, 5};
So the common elements in these two arrays are 1, 3, and 5.
# C Program to Find Common Elements in Two Arrays
Let's create a file named commoninarray.c and add the following source code to it.
``````#include<stdio.h>
#define max 100
int ifexists(int z[], int u, int v)
{
int i;
if (u==0) return 0;
for (i=0; i<=u;i++)
if (z[i]==v) return (1);
return (0);
}
void main()
{
int p[max], q[max], r[max];
int m,n;
int i,j,k;
k=0;
printf("Enter length of first array:");
scanf("%d",&m);
printf("Enter %d elements of first array\n",m);
for(i=0;i<m;i++ )
scanf("%d",&p[i]);
printf("\nEnter length of second array:");
scanf("%d",&n);
printf("Enter %d elements of second array\n",n);
for(i=0;i<n;i++ )
scanf("%d",&q[i]);
k=0;
for (i=0;i<m;i++)
{
for (j=0;j<n;j++)
{
if (p[i]==q[j])
{
if(!ifexists(r,k,p[i]))
{
r[k]=p[i];
k++;
}
}
}
}
if(k>0)
{
printf("\nThe common elements in the two array are:\n");
for(i = 0;i<k;i++)
printf("%d\n",r[i]);
}
else
printf("There are no common elements in two arrays\n");
}``````
To compile and run the above C program, you can use C Programs Compiler Online tool.
Output:
``````Enter length of first array:5
Enter 5 elements of first array
1
3
5
7
9
Enter length of second array:5
Enter 5 elements of second array
1
2
3
4
5
The common elements in the two array are:
1
3
5`````` | 522 | 1,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.503095 |
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• info@interviewmaterial.com
# RD Chapter 11- Constructions Ex-11.2 Interview Questions Answers
### Related Subjects
Question 1 : Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and radius 4 cm and with centre C and radius 6 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC. Then ABC is the triangle.
(iv) Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB1 = B1B2= B2B3.
(v) Join B3C.
(vi) Draw B’C’ parallel to B3C and C’A’ parallel to CA then ΔA’BC’ is the required triangle.
Question 2 : Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)th of the corresponding sides of ΔABC. It is given that AB = 5 cm, BC = 7 cm and ∠ABC = 50°.
Steps of construction :
(i) Draw a line segment BC = 7 cm.
(ii) Draw a ray BX making an angle of 50° and cut off BA = 5 cm.
(iii) Join AC. Then ABC is the triangle.
(iv) Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB, =B1B2=B2B3=B3B4=B4Bs=B5B6=B6B7
(v) Join B7 and C
(vi) Draw B5C’ parallel to B7C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
Question 3 : Construct a triangle similar to a given ∠ABC such that each of its sides is 2/3rd of the corresponding sides of ΔABC. It is given that BC = 6 cm, ∠B = 50° and ∠C = 60°.
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) Draw a ray BX making an angle of 50° and CY making 60° with BC which intersect each other at A. Then ABC is the triangle.
(iii) From B, draw another ray BZ making an acute angle below BC and intersect 3 equal parts making BB1 =B1B2 = B2B2
(iv) Join B3C.
(v) From B2, draw B2C’ parallel to B3C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
Question 4 : Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to 3/4th of the corresponding sides of ΔABC.
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) With centre B and radius 4 cm and with centre C and radius 5 cm, draw arcs’intersecting eachother at A.
(iii) Join AB and AC. Then ABC is the triangle,
(iv) Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB1= B1B2 = B2B3 = B3B4.
(v) Join B4 and C.
(vi) From B3C draw C3C’ parallel to B4C and from C’, draw C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
Question 5 : Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and radius 6 cm and with centre C and radius 7 cm, draw arcs intersecting eachother at A.
(iii) Join AB and AC. Then ABC is the triangle.
(iv) Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
(v) Join B5 and C.
(vi) From B7, draw B7C’ parallel to B5C and C’A’ parallel CA. Then ΔA’BC’ is the required triangle.
Question 6 : Draw a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to (54)th ot the corresponding sides of ΔABC.
Steps of construction :
(i) Draw a line segment AB = 4.5 cm.
(ii) At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.
(iii) Join BC. Then ABC is the triangle.
(iv) Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA1 = A1A2 = A2A3 =A3A4 = A4A5
(v) Join A4 and B.
(vi) From 45, draw 45B’ parallel to A4B and B’C’ parallel to BC.
Then ΔAB’C’ is the required triangle.
Question 7 : Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. (C.B.S.E. 2008)
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) At B, draw perpendicular BX and cut off BA = 4 cm.
(iii )join Ac , then ABC is the triangle
(iv) Draw a ray BY making an acute angle with BC, and cut off 5 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5
(v) Join B3 and C.
(vi) From B5, draw B5C’ parallel to B3C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
Question 8 : Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle.
Steps of construction :
(i) Draw a line segment BC = 8 cm and draw its perpendicular bisector DX and cut off DA = 4 cm.
(ii) Join AB and AC. Then ABC is the triangle.
(iii) Draw a ray DY making an acute angle with OA and cut off 3 equal parts making DD1 = D1D2 =D2D3 = D3D4
(iv) Join D2
(v) Draw D3A’ parallel to D2A and A’B’ parallel to AB meeting BC at C’ and B’ respectively.
Then ΔB’A’C’ is the required triangle.
Question 9 : Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a trianglewhose sides are (3/4)th of the corresponding sides of the ΔABC.
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle of 60° with BC and cut off BA = 5 cm.
(iii) Join AC. Then ABC is the triangle.
(iv) Draw a ray BY making an acute angle with BC and cut off 4 equal parts making BB1= B1B2 B2B3=B3B4.
(v) Join B4 and C.
(vi) From B3, draw B3C’ parallel to B4C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
Question 10 : Construct a triangle similar to ΔABC in which AB = 4.6 cm, BC = 5.1 cm,∠A = 60° with scale factor 4 : 5.
Steps of construction :
(i) Draw a line segment AB = 4.6 cm.
(ii) At A, draw a ray AX making an angle of 60°.
(iii) With centre B and radius 5.1 cm draw an arc intersecting AX at C.
(iv) Join BC. Then ABC is the triangle.
(v) From A, draw a ray AX making an acute angle with AB and cut off 5 equal parts making AA1 = A1A2 = A2A3 = A3A4=A4A5.
(vi) Join A4 and B.
(vii) From A5, drawA5B’ parallel to A4B and B’C’ parallel to BC.
Then ΔC’AB’ is the required triangle.
krishan | 1,960 | 6,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-49 | latest | en | 0.886205 |
https://www.proprofs.com/quiz-school/story.php?title=boating-examination-ku | 1,721,299,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00001.warc.gz | 809,256,095 | 110,767 | # Boating Examination - K&u
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| By Scott Stone
S
Scott Stone
Community Contributor
Quizzes Created: 5 | Total Attempts: 16,945
Questions: 52 | Attempts: 60
Settings
• 1.
### All vessels are required to travel at a safe speed at all times. Which of the following IS considered safe?
• A.
A vessel travelling at speed in poor visibility
• B.
A vessel travelling at a speed at which any sudden danger can be avoided
• C.
A vessel travelling at speed towards background shore lights at night
• D.
A vessel travelling at speed in unfamiliar waters
B. A vessel travelling at a speed at which any sudden danger can be avoided
Explanation
A vessel traveling at a speed at which any sudden danger can be avoided is considered safe. This means that the vessel is traveling at a speed that allows the operator to have enough time to react and take necessary actions to avoid any unexpected dangers or hazards that may arise. By maintaining a speed that allows for quick maneuverability, the vessel can navigate safely and reduce the risk of accidents or collisions.
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• 2.
### The number of persons permitted to be towed behind a vessel is:
• A.
Determined by the number of handles on the apparatus being towed
• B.
Whatever the apparatus manufacturer states is permissible
• C.
Maximum of three people
• D.
Whatever the vessel’s master (driver) thinks is safe
C. Maximum of three people
Explanation
The correct answer is "Maximum of three people." This means that the number of persons allowed to be towed behind a vessel should not exceed three. This is a safety regulation to ensure that the vessel is not overloaded and that the towing apparatus can handle the weight and strain of the people being towed. It is important to adhere to this limit to prevent accidents and ensure the safety of everyone involved.
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• 3.
### For environmental reasons, where should you avoid driving your boat?
• A.
Deep river channels
• B.
Shallow weedy areas, which may contain endangered seagrasses
• C.
Boundaries between murky and clear water
• D.
Bar crossings
B. Shallow weedy areas, which may contain endangered seagrasses
Explanation
Driving a boat in shallow weedy areas should be avoided for environmental reasons because these areas may contain endangered seagrasses. Seagrasses play a crucial role in maintaining the health of marine ecosystems as they provide food and habitat for various species, filter water, and prevent erosion. Driving a boat in these areas can cause damage to the seagrasses by uprooting or damaging them, which can disrupt the delicate balance of the ecosystem and harm the species that depend on them for survival. Therefore, it is important to avoid driving boats in shallow weedy areas to protect and preserve these endangered seagrasses.
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• 4.
### Where MUST a power driven vessel’s Safety Label be displayed?
• A.
Where it can be clearly seen from the steering position
• B.
On the port hand (left hand) side of the vessel, next to the registration numbers
• C.
On the starboard (right hand) side of the vessel, next to the registration numbers
• D.
On the stern of the vessel
A. Where it can be clearly seen from the steering position
Explanation
The safety label of a power-driven vessel must be displayed where it can be clearly seen from the steering position. This ensures that the operator of the vessel can easily refer to the safety label for important information and instructions while operating the vessel. Placing the safety label in a visible position from the steering position enhances safety and compliance with regulations.
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• 5.
### When driving a power vessel at 10 knots or more or towing a person, what is the MINIMUM distance both the vessel and the towed person MUST keep from a non-powered vessel or a person in the water?
• A.
60 metres or if not possible a safe distance
• B.
30 metres or if not possible a safe distance
• C.
100 metres or if not possible a safe distance
• D.
Any distance which you consider safe
A. 60 metres or if not possible a safe distance
Explanation
When driving a power vessel at 10 knots or more or towing a person, the minimum distance both the vessel and the towed person must keep from a non-powered vessel or a person in the water is 60 metres. If it is not possible to maintain this distance, they must keep a safe distance.
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• 6.
### When travelling UPSTREAM (away from the sea) at night, on which side should you keep a flashing green light to stay within the channel?
• A.
Your port (left hand) side
• B.
Your starboard (right hand) side
• C.
Either side (it does not matter)
• D.
Stay in the middle of the channel regardless of the mark
B. Your starboard (right hand) side
Explanation
When travelling upstream at night, keeping a flashing green light on your starboard (right hand) side helps you stay within the channel. This is because the flashing green light indicates the starboard side of the channel, and by keeping it on your right, you ensure that you are staying within the designated path of the channel.
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• 7.
### What type of lifejacket MUST you carry if you take a vessel into OPEN water?
• A.
Lifejacket Level 100+ (Type 1’s) for each person onboard
• B.
Either lifejacket Level 100+ (Type 1’s) or lifejacket Level 50 (Type 2’s) for each person onboard
• C.
Any type of lifejacket as long as there is one for each person onboard
• D.
Any mixture of lifejacket Level 100+, Level 50 or Level 50S (Types 1, 2 or 3), or a wetsuit as long as there is one for each person onboard
A. Lifejacket Level 100+ (Type 1’s) for each person onboard
Explanation
In open water, it is mandatory to carry Lifejacket Level 100+ (Type 1's) for each person onboard. This type of lifejacket provides the highest level of buoyancy and is specifically designed for use in offshore or rough water conditions. It ensures that individuals have the necessary flotation and support to stay afloat and survive in potentially dangerous situations.
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• 8.
### What does this sign mean?
• A.
Maximum speed of 4 knots
• B.
Vessels prohibited in this area
• C.
Keep to a speed which does not create damaging or annoying waves
• D.
Water skiing prohibited
C. Keep to a speed which does not create damaging or annoying waves
Explanation
This sign indicates that boaters should maintain a speed that does not generate waves that could cause damage to other vessels or be a nuisance to others. It serves as a reminder to be considerate and mindful of the potential impact of their speed on the surrounding environment and other water users.
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• 9.
### When should you switch on and display navigation lights on a vessel underway?
• A.
Only when it is dark
• B.
From sunset to sunrise and in restricted visibility
• C.
At night only when travelling on open waters (the sea)
• D.
At night in port areas only
B. From sunset to sunrise and in restricted visibility
Explanation
Navigation lights are required to be switched on and displayed on a vessel underway from sunset to sunrise and in restricted visibility. This is to ensure the safety of the vessel and other vessels in the vicinity. By having navigation lights on, other vessels can easily identify the position, direction, and size of the vessel, especially during low visibility conditions. It helps in preventing collisions and maintaining proper navigation protocols. Therefore, it is essential to switch on and display navigation lights during these specific times.
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• 10.
### A navigational marker shows a white light flashing quickly in groups of nine flashes. What does it mean? Deeper water is to the:
• A.
North
• B.
East
• C.
South
• D.
West
D. West
Explanation
A navigational marker showing a white light flashing quickly in groups of nine flashes indicates that deeper water is to the west.
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• 11.
### How can you best ensure that your boat’s wash does NOT excessively impact on foreshores or other vessels?
• A.
By slowing down to 8 knots
• B.
By observing the effects of your wash and adjusting your speed as necessary
• C.
By always travelling at the speed shown on a speed restriction sign
• D.
By travelling as close to other vessels or the shore as possible
B. By observing the effects of your wash and adjusting your speed as necessary
Explanation
By observing the effects of your wash and adjusting your speed as necessary, you can ensure that your boat's wash does not excessively impact on foreshores or other vessels. This means being aware of the waves and turbulence caused by your boat's movement and making necessary adjustments to your speed to minimize the impact. It requires being attentive and responsive to the conditions and surroundings, prioritizing the safety and well-being of other vessels and the environment.
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• 12.
### When travelling UPSTREAM (away from the sea) on which side should you keep this type of navigation mark to stay in the channel?
• A.
Your port (left hand side)
• B.
Your starboard (right hand) side
• C.
Either side (it does not matter)
• D.
Stay in the middle of the channel regardless of the mark
B. Your starboard (right hand) side
Explanation
When traveling upstream, it is important to keep the navigation mark on your starboard (right hand) side to stay in the channel. This is because the navigation marks are designed to indicate the correct path for boats, and keeping the mark on the starboard side ensures that you are staying on the correct side of the channel. By doing so, you can avoid straying into shallow or hazardous areas and maintain a safe and navigable route.
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• 13.
### Where MUST the registration label be displayed on a powered vessel?
• A.
On the starboard (right hand) side where it is clearly visible
• B.
On the transom where it is clearly visible
• C.
It doesn’t matter as long as it is clearly visible
• D.
On the port (left hand) side where it is clearly visible
D. On the port (left hand) side where it is clearly visible
Explanation
The registration label on a powered vessel must be displayed on the port (left hand) side where it is clearly visible. This ensures that other boaters and authorities can easily identify and verify the registration of the vessel. By having a standardized location for the registration label, it helps maintain consistency and makes it easier for everyone to comply with the regulations.
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• 14.
### What does this navigation mark indicate? Deeper water lies to the:
• A.
North
• B.
East
• C.
South
• D.
West
B. East
Explanation
This navigation mark indicates that deeper water lies to the east.
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• 15.
### The responsibility of the master (driver) is to:
• A.
Maintain a proper lookout and avoid collision
• B.
Ensure the safety of those onboard the vessel
• C.
Ensure that all safety equipment is accessible and stored onboard correctly
• D.
All of the above
D. All of the above
Explanation
The master (driver) of a vessel is responsible for maintaining a proper lookout and avoiding collisions to ensure the safety of everyone on board. Additionally, they are also responsible for ensuring that all safety equipment is accessible and stored correctly on the vessel. Therefore, the correct answer is "All of the above."
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• 16.
### When do sail vessels, operating only under sail, have right of way over a power driven vessel?
• A.
At all times unless overtaking, or if the power vessel is displaying an orange diamond
• B.
Only when approaching from the right hand side
• C.
Only when participating in an aquatic event eg: a race or regatta
• D.
When a powerboat is at anchor
A. At all times unless overtaking, or if the power vessel is displaying an orange diamond
Explanation
Sail vessels, operating only under sail, have the right of way over a power-driven vessel at all times, unless they are overtaking the power vessel or if the power vessel is displaying an orange diamond. This means that sail vessels have priority over power vessels in most situations, but they must yield if they are overtaking another vessel or if the power vessel is indicating that it requires right of way by displaying an orange diamond.
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• 17.
### What do cardinal marks indicate?
• A.
Special features such as underwater pipes
• B.
Deeper water in a compass direction away from danger
• C.
Specific dangers such as wrecks
• D.
Large shipping channels
B. Deeper water in a compass direction away from danger
Explanation
Cardinal marks are used to indicate the safest and deepest navigable water in a particular area. They provide information about the direction in which vessels should navigate to avoid potential dangers such as shallows, rocks, or wrecks. Each cardinal mark is associated with a compass direction (north, south, east, or west) and indicates that deeper water can be found in that direction away from potential hazards. Therefore, cardinal marks help mariners navigate safely by guiding them towards deeper waters and away from danger.
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• 18.
### When travelling DOWNSTREAM (toward the sea) at night, on which side should you keep a flashing red light to stay within the channel?
• A.
Your port (left hand) side
• B.
Your starboard (right hand) side
• C.
Either side (it does not matter)
• D.
Stay in the middle of the channel regardless of the mark
B. Your starboard (right hand) side
Explanation
When travelling downstream at night, keeping a flashing red light on your starboard (right hand) side will help you stay within the channel. This is because the red light indicates the right side of the channel when moving in the direction of the sea. By keeping the light on the starboard side, you ensure that you are on the correct side of the channel and avoid straying into shallow or dangerous areas.
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• 19.
### When is the MOST DANGEROUS time to cross a coastal bar?
• A.
On an incoming tide (flood tide)
• B.
On an outgoing tide (ebb tide)
• C.
Slack water (top or bottom of the tide)
• D.
All of the above
B. On an outgoing tide (ebb tide)
Explanation
Crossing a coastal bar on an outgoing tide (ebb tide) is considered the most dangerous time because the water is flowing out of the bar and can create strong currents and breaking waves. This can make it difficult for boats to navigate through the bar and increase the risk of capsizing or getting caught in the turbulent water. It is generally safer to cross during slack water when the tide is neither incoming nor outgoing, as the currents are weaker and the water is calmer.
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• 20.
### An efficient sound signalling device is required to be carried:
• A.
Only by vessels proceeding to open waters
• B.
Only by power driven vessels over 8 metres in length
• C.
On all vessels
• D.
Only vessels operating after sunset
C. On all vessels
Explanation
An efficient sound signalling device is required to be carried on all vessels. This is because sound signals are vital for communication and safety purposes on the water. Regardless of the type or size of the vessel, having a sound signalling device ensures that the vessel can effectively communicate its intentions to other vessels, especially in situations where visibility may be limited. Therefore, it is necessary for all vessels to have an efficient sound signalling device on board.
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• 21.
### What is the minimum length of a tow rope when performing towing operations (waterskiing)?
• A.
4 metres
• B.
5 metres
• C.
6 metres
• D.
7 metres
D. 7 metres
Explanation
The minimum length of a tow rope for towing operations in waterskiing is 7 metres. This length is necessary to ensure that the skier is at a safe distance from the boat, reducing the risk of collision or getting caught in the propeller. It also allows for proper maneuverability and control while skiing. A shorter tow rope could increase the chances of accidents and injuries.
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• 22.
### What navigation mark is this?
• A.
An isolated danger mark
• B.
A channel blocked mark
• C.
A special mark
• D.
A port lateral mark
C. A special mark
Explanation
This navigation mark is classified as a special mark. Special marks are used to indicate specific areas or features that require special attention or caution, such as underwater cables, pipelines, wrecks, or military exercise areas. They are typically yellow or yellow with a top mark, and their purpose is to inform mariners about potential hazards or restricted areas.
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• 23.
### When driving a vessel at 10 knots or more or towing a person, what is the MINIMUM distance both the vessel and the towed person MUST keep from any other vessel towing a person?
• A.
100 metres
• B.
30 metres
• C.
60 metres
• D.
A safe distance
D. A safe distance
Explanation
When driving a vessel at 10 knots or more or towing a person, it is important to maintain a safe distance from any other vessel towing a person. This is necessary to ensure the safety of both the vessel and the towed person. The specific distance may vary depending on the situation and conditions, but the key is to maintain a distance that allows for safe maneuvering and avoids any potential collisions or accidents.
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• 24.
### When navigating through a mooring area you MUST:
• A.
Keep a lookout for people and objects in the water
• B.
Travel at a safe speed
• C.
Keep wash to a minimum
• D.
All of the above
D. All of the above
Explanation
When navigating through a mooring area, it is important to keep a lookout for people and objects in the water to ensure their safety and avoid any collisions. Traveling at a safe speed is crucial to maintain control of the vessel and reduce the risk of accidents. Additionally, keeping wash to a minimum is necessary to prevent any damage to other vessels or objects in the area. Therefore, all of the given options are correct and should be followed when navigating through a mooring area.
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• 25.
### Where should you drive a vessel in a channel?
• A.
On the port (left-hand) side
• B.
In the middle of the channel
• C.
On the starboard (right-hand) side
• D.
On any side – it does not matter as long as a collision does not occur
C. On the starboard (right-hand) side
Explanation
Vessels should be driven on the starboard (right-hand) side in a channel. This is because the International Regulations for Preventing Collisions at Sea (COLREGS) state that vessels should keep to the starboard side to avoid collisions. By driving on the starboard side, vessels can maintain a consistent and predictable path of travel, reducing the risk of accidents and allowing for safe navigation in narrow channels.
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• 26.
### What kind of navigation mark is this?
• A.
A port lateral mark
• B.
A channel blocked mark
• C.
A safe water mark
• D.
An isolated danger mark
D. An isolated danger mark
Explanation
An isolated danger mark is used to indicate a single isolated danger, such as a rock or wreck, which has navigable water all around it. This type of navigation mark is placed to warn mariners to keep clear of the danger and to indicate the best navigable water to pass around it. It is typically colored black with two broad horizontal red bands and has two black spheres on top.
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• 27.
### What MUST you do when driving your vessel within a No Wash Zone?
• A.
Travel at a speed which creates minimal wash to ensure that your wash does not affect other people or vessels
• B.
Travel at no more than 8 knots
• C.
By always travelling at the speed shown on a speed restriction sign
• D.
Not tow a waterskier or an aquaplaner
A. Travel at a speed which creates minimal wash to ensure that your wash does not affect other people or vessels
Explanation
When driving your vessel within a No Wash Zone, it is necessary to travel at a speed which creates minimal wash to ensure that your wash does not affect other people or vessels. This means that you should be cautious and mindful of the impact your vessel's movement may have on the surrounding environment and other watercrafts. By reducing the wash, you are minimizing the disturbance caused by your vessel and promoting safety and consideration for others in the area.
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• 28.
### If an unpredictable change in weather occurs and you are unable to make it to a safe harbour, you should:
• A.
Ride out the initial onslaught by keeping your bow into the wind and waves
• B.
Activate a distress signal such as an EPIRB or red hand held flare
• C.
Maintain a sufficient speed to allow you to steer the vessel, but no faster
• D.
Maintain a slow speed while keeping your bow into the wind and waves
D. Maintain a slow speed while keeping your bow into the wind and waves
Explanation
When faced with an unpredictable change in weather and unable to reach a safe harbor, maintaining a slow speed while keeping the bow into the wind and waves is the correct course of action. This technique, known as heaving to, helps to stabilize the vessel and minimize the impact of the weather conditions. By keeping the bow facing into the wind and waves, it reduces the risk of capsizing or being thrown off course. Maintaining a slow speed allows for better control and maneuverability of the vessel while riding out the storm.
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• 29.
### If you find yourself in cold water, what can you do to extend survival time?
• A.
Move constantly and do not remove clothing
• B.
Remove all wet clothing and float on your back
• C.
Assume the Heat Escape Lessening Posture (HELP) and do not remove clothing
• D.
Remove clothing and maintain movement of your arms and legs
C. Assume the Heat Escape Lessening Posture (HELP) and do not remove clothing
Explanation
Assuming the Heat Escape Lessening Posture (HELP) and not removing clothing can help extend survival time in cold water. This posture involves bringing the knees to the chest and crossing the arms tightly over the chest to minimize heat loss. By doing so, the body's heat is conserved, and the risk of hypothermia is reduced. Removing clothing or staying still can increase heat loss and decrease survival time in cold water.
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• 30.
### If you see a white light flashing in groups of two, what should you do?
• A.
Keep a lookout for overhead powerlines
• B.
Pass on the right hand side of the light
• C.
Pass on any side of the light but do not pass too close
• D.
Keep wash to a minimum
C. Pass on any side of the light but do not pass too close
Explanation
Passing on any side of the light but not too close is the correct answer because a white light flashing in groups of two is typically used to indicate a level crossing or a pedestrian crossing. In such situations, it is important to proceed with caution and not pass too close to the light to ensure the safety of pedestrians or vehicles crossing the road.
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• 31.
### You are driving a vessel at high speed and your vision is suddenly affected by sun or spray. Your immediate reaction should be:
• A.
Slow down or stop
• B.
Continue driving at speed in anticipation that you will regain your vision
• C.
Increase speed and manoeuvre vessel to find a direction to minimise spray and the effect of the sun
• D.
Alter course to port and continue at speed
A. Slow down or stop
Explanation
When driving a vessel at high speed, sudden vision impairment due to sun or spray can be dangerous. The correct immediate reaction would be to slow down or stop. This allows the driver to regain clear vision and assess the situation properly before continuing. Continuing at speed or increasing speed could lead to accidents or collisions, as the driver may not be able to see potential hazards or obstacles in their path. Altering course to port and continuing at speed is also not advisable, as it does not address the issue of impaired vision.
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• 32.
### You are driving a power driven vessel and see a vessel ahead of you exhibiting these lights. What should you do?
• A.
Alter your course to starboard (right)
• B.
Alter your course to port (left)
• C.
Maintain your speed and course
• D.
Stop immediately and turn off all lights
A. Alter your course to starboard (right)
Explanation
When you see a vessel ahead of you exhibiting these lights, you should alter your course to starboard (right). This is because the vessel ahead is showing the red light on its port (left) side and the green light on its starboard (right) side. According to the International Regulations for Preventing Collisions at Sea (COLREGS), when two power-driven vessels are approaching each other head-on, both vessels should alter their course to starboard to pass each other safely. By altering your course to starboard, you ensure that you are passing the other vessel on its port side, which is the correct and safe maneuver.
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• 33.
### When using navigational Leads, you should:
• A.
Pass to the right when going upstream
• B.
Pass either side but not too close
• C.
Line the leads up, one behind the other
• D.
Ignore them as they are only used by large commercial vessels
C. Line the leads up, one behind the other
Explanation
When using navigational leads, it is important to line them up, one behind the other. Navigational leads are used to guide vessels through narrow or shallow channels, indicating the correct path to follow. By lining them up, one can ensure that they are following the intended route and avoiding any potential hazards. This helps in maintaining the safety and efficiency of navigation.
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• 34.
### Which of the following vessels are required to have or display lights at night?
• A.
Power driven vessels
• B.
Canoes and rowboats
• C.
Sailing vessels
• D.
All vessels need to have or display lights at night
D. All vessels need to have or display lights at night
Explanation
All vessels need to have or display lights at night in order to ensure visibility and prevent collisions. This is important for the safety of all vessels, as well as for other boats and ships in the vicinity. Lights help to indicate the presence, position, and direction of a vessel, allowing other vessels to navigate safely and avoid potential accidents. Therefore, it is a requirement for all types of vessels, including power driven vessels, canoes and rowboats, and sailing vessels, to have or display lights at night.
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• 35.
### Review the picture below and determine the signals that should be displayed.
• A.
Two red lights with 225 degree forward arch
• B.
Anchor ball or light is adequate
• C.
Two red all-round lights, one anchor light and deck lighting
• D.
Red - White - Red all-round lights with anchor ball or light
C. Two red all-round lights, one anchor light and deck lighting
Explanation
The correct answer is "Two red all-round lights, one anchor light and deck lighting." This is the appropriate signal to be displayed based on the information provided.
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• 36.
### The part of the boat labelled A in the diagram below is the:
• A.
Transom
• B.
Gunwale
• C.
Freeboard
• D.
Draught
C. Freeboard
Explanation
The part of the boat labelled A in the diagram is the freeboard. Freeboard refers to the distance between the waterline and the upper deck level of the boat. It is important for the stability and buoyancy of the boat, as it helps to prevent water from entering the vessel.
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• 37.
### The part of the boat labelled B in the diagram below is the:
• A.
Stern
• B.
Draught
• C.
Gunwale
• D.
Freeboard
B. Draught
Explanation
The correct answer is "Draught." In boating terminology, the term "draught" refers to the depth of a boat's hull below the waterline. It is often measured from the waterline to the lowest point of the boat's keel. The draught of a boat is an important measurement as it determines the boat's stability, maneuverability, and the areas it can navigate safely.
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• 38.
### Which of the following boats handle adverse conditions on a bar the best?
• A.
Punt
• B.
Shallow V
• C.
Displacement
• D.
Deep V hull
D. Deep V hull
Explanation
A deep V hull is the best type of boat for handling adverse conditions on a bar. The deep V shape of the hull allows the boat to cut through waves and rough water more efficiently, providing a smoother and more stable ride. This design also helps to prevent the boat from capsizing or taking on water in challenging conditions. The other options, such as the punt, shallow V, and displacement hull, are not as well-suited for adverse conditions on a bar as they may not have the same level of stability and wave-cutting ability as a deep V hull.
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• 39.
### The formula for carrying fuel is:
• A.
One third out, one third back and one third in reserve
• B.
Two thirds out and one third back
• C.
One half out and one half back
• D.
Just make sure the tank is full
A. One third out, one third back and one third in reserve
Explanation
The correct answer is "One third out, one third back and one third in reserve". This formula suggests that when carrying fuel, one should use one third of the fuel for the outward journey, another third for the return journey, and keep the remaining third as a reserve. This ensures that there is enough fuel for the entire trip, with a safety margin in case of any unforeseen circumstances or emergencies.
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• 40.
### What action do you take if you are overtaking another vessel?
• A.
Wait till the vessel slows down
• B.
Pass on the port side
• C.
Pass on the starboard side
• D.
Keep well clear of the way of the vessel you are overtaking
D. Keep well clear of the way of the vessel you are overtaking
Explanation
When overtaking another vessel, it is important to keep well clear of its way. This means maintaining a safe distance and avoiding any potential collision. By keeping clear, you ensure the safety of both vessels and reduce the risk of accidents. Waiting for the vessel to slow down or passing on either side may not guarantee a safe overtaking maneuver, so it is best to keep a safe distance and proceed with caution.
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• 41.
### Which of the following is correct?
• A.
The tidal range on the 22 June is 2.1 metres
• B.
There is a full moon on the 6th June
• C.
High water on the 21st July is at 8.00 am
• D.
Low water on the 20th June is at 7.22 pm
A. The tidal range on the 22 June is 2.1 metres
Explanation
The given statement is correct because it states a specific measurement of the tidal range on a specific date, which is 2.1 meters on the 22nd of June.
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• 42.
### Your motor on your 3.5 metre boat has failed and you are drifting in a strong outgoing tidal current. The correct call to the local marine rescue association would be:
• A.
MAYDAY
• B.
SECURITE
• C.
PAN PAN
• D.
MOBILE PHONE
C. PAN PAN
Explanation
In this situation, the correct call to the local marine rescue association would be "PAN PAN." "PAN PAN" is a radio distress call used to indicate an urgent situation that is not immediately life-threatening. It is used when there is a potential danger or concern, such as a mechanical failure, but there is no immediate risk to life. In this case, the motor failure on the boat has put the person in a difficult situation, but they are not in immediate danger. Calling "PAN PAN" would alert the local marine rescue association to the situation and prompt them to provide assistance.
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• 43.
### If the wind is blowing against the tide, you could expect:
• A.
Smooth boating
• B.
A rough ride
• C.
To anchor before high tide
• D.
Good fishing weather
B. A rough ride
Explanation
When the wind is blowing against the tide, it creates opposing forces that can make the water surface choppy and turbulent. This can result in a rough ride for boats as they navigate through the waves created by the wind and tide. The combination of wind and tide working against each other can make the water conditions unpredictable and challenging, leading to a rough and uncomfortable experience for boaters.
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• 44.
### Which of the following checks (3) are necessary to ensure your propeller is useable?
• A.
It is securely mounted to the shaft.
• B.
The paint is not worn.
• C.
There is no surface rust.
• D.
There are no cracks.
• E.
There are no dents in the leading edge.
• F.
There is no oil inside your propeller.
A. It is securely mounted to the shaft.
D. There are no cracks.
E. There are no dents in the leading edge.
Explanation
To ensure that a propeller is usable, it is necessary to check if it is securely mounted to the shaft, as a loose propeller can pose a safety risk. Additionally, it is important to check for cracks in the propeller, as they can indicate structural damage that may affect its performance. Lastly, checking for dents in the leading edge is important because dents can disrupt the airflow and decrease the propeller's efficiency.
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• 45.
### Which one of these anchors would a small boat use on a rocky sea bed?
• A.
Anchor A
• B.
Anchor B
• C.
Anchor C
B. Anchor B
Explanation
Anchor B is the correct answer because it is specifically designed for rocky sea beds. It is usually heavier and has sharp flukes that can dig into the uneven surface of the rocky sea bed, providing better holding power and stability for a small boat. Anchor A and Anchor C may not be suitable for rocky sea beds as they may not have the necessary features to hold securely in such conditions.
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• 46.
### If you were boating in calm weather and decided to anchor in 3 metres of water, approximately how much anchor rope should you let out (in addition to the 2 metres of chain)?
• A.
13 metres
• B.
24 metres
• C.
6 metres
• D.
30 metres
A. 13 metres
Explanation
To anchor a boat in calm weather, it is recommended to let out a total of 5 times the depth of the water. Since the water is 3 meters deep, you would need to let out 15 meters of anchor rope. However, since there is already 2 meters of chain out, you only need to let out an additional 13 meters of anchor rope.
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• 47.
### If you have a 2-stroke engine that runs a mixture of 50:1, how much oil would you have to add to 25 litres of fuel.
• A.
250ml
• B.
500ml
• C.
750 ml
• D.
1 Litre
B. 500ml
Explanation
In a 2-stroke engine running a mixture of 50:1, it means that for every 50 parts of fuel, 1 part of oil is required. To calculate the amount of oil needed for 25 liters of fuel, we divide 25 by 50 to get 0.5 liters, which is equivalent to 500ml. Therefore, you would have to add 500ml of oil to 25 liters of fuel.
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• 48.
### Sacrificial anodes should be replaced when they are smaller than 60% of their original size.
• A.
True
• B.
False
A. True
Explanation
Sacrificial anodes are used to protect metal structures from corrosion by sacrificing themselves. Over time, these anodes gradually erode as they protect the metal. When the anodes are smaller than 60% of their original size, it indicates that a significant amount of material has been sacrificed and they may no longer be effective in preventing corrosion. Therefore, it is necessary to replace them at this point to ensure continued protection for the metal structure.
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• 49.
### Which of the following statements about fuel is true?
• A.
All fuel should be thrown out after 3 months
• B.
2-stroke mixtures should be thrown out after 3 months and 4-stroke after 6 months
• C.
2-stroke mixtures should be thrown out after 6 months and 4-stroke after 3 months
• D.
All fuel should be thrown out after 6 months
B. 2-stroke mixtures should be thrown out after 3 months and 4-stroke after 6 months
Explanation
The correct answer states that 2-stroke mixtures should be thrown out after 3 months and 4-stroke fuel should be thrown out after 6 months. This is because 2-stroke engines typically have a higher oil-to-fuel ratio, which can cause the fuel to degrade more quickly. On the other hand, 4-stroke engines have a separate oil system and therefore the fuel tends to last longer. Therefore, it is recommended to dispose of 2-stroke mixtures after 3 months and 4-stroke fuel after 6 months to ensure optimal performance and prevent any potential issues caused by degraded fuel.
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• 50.
### You don't need a boat licence if the motor is less than 10hp.
• A.
True
• B.
False
B. False
Explanation
This statement is false. In many places, a boat licence is required regardless of the motor size. The specific regulations regarding boat licences vary by jurisdiction, so it is important to check the local laws and requirements before operating a boat.
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• Current Version
• Feb 09, 2023
Quiz Edited by
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• Sep 11, 2016
Quiz Created by
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#### Angles in Quadrilaterals (Treasure Hunt)
(9)
This activity gives students good practice at using the angle properties of parallelograms, kites etc., to work out the unknown angle in many different quadrilaterals. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. --- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals – they are particularly helpful for one-to-one tutorials and during interventions.
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#### Rules for Indices (Treasure Hunt)
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This activity is great to consolidate or revise The Rules of Indices. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. ---- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals – they are particularly helpful for one-to-one tutorials and during interventions.
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#### Area and Perimeter (Treasure Hunt)
(6)
This activity helps students to practice working out areas and perimeters of rectangles and also working out the width or height when the area or perimeter is known. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. -- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals – they are particularly helpful for one-to-one tutorials and during interventions.
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#### LCM and HCF (Treasure Hunt)
(10)
This activity gives students practice at working out LCM or HCF of two numbers.. Click -->https://tes.com/.../Treasure Hunts to download Treasure Hunts on 40 other topics. -- A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. Unlike most Treasure Hunts, this one has the added feature that the answers to the questions give an encrypted clue. When students decipher this clue, it reveals where the treasure is hidden! The question cards have been prepared in two sizes. The large version can be pinned around the room and used for a whole class activity. The smaller cards can be used for group work or by individuals. The smaller cards are particularly helpful during interventions. Colour coding: I have used colours to indicate the increasing difficulty of questions: Red, Orange and Green (from primary up to GCSE Foundation) Blue and Purple (GCSE Higher) If you like this resource then please rate it and/or leave a comment by clicking here --> https://tes.com/.../Quick Comment
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#### Transformation of Functions (Treasure Hunt)
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This activity helps to strengthen students' skills in substituting one algebraic expression into another. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. -- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals – they are particularly helpful for one-to-one tutorials and during interventions.
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#### Gradient and Intercept (Treasure Hunt)
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This activity gives students practice at rearranging different formats of the equation of a straight-line into the form y = mx + c. They then determine the gradient and intercept of the line. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. -- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals ��� they are particularly helpful for one-to-one tutorials and during interventions.
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#### Bearings 1 (Treasure Hunt)
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This activity is great to consolidate or revise bearings. Note that students have to work both ways – finding either A from B or B from A. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. --- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals – they are particularly helpful for one-to-one tutorials and during interventions.
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#### Equivalent Surds (Treasure Hunt)
(8)
This activity requires students to manipulate surds and write them in their 'simplified' form. It is great for consolidating or revising. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on more than 20 other topics. --- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals – they are particularly helpful for one-to-one tutorials and during interventions.
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#### Circles and Part-circles (Treasure Hunt)
(6)
This activity helps students to strengthen their use of the formulae for area and circumference of a circle. Questions involve whole circles, semi-circles and quadrants. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. -- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals ��� they are particularly helpful for one-to-one tutorials and during interventions.
Quick View
#### Simplifying Ratios (Treasure Hunt)
(4)
This activity gives students practice at converting ratios into their simplest form, by dividing both parts. Click --> https://tes.com/.../Treasure Hunts for similar style Treasure Hunts on 40 other topics. -- Note that unlike most Treasure Hunts, this one has the added feature that the answers give an encrypted clue. Deciphering this clue reveals where the treasure is hidden! A Treasure Hunt is a great activity which children love. They are ideal for revision, starters or plenaries. They are a really great way to get students to answer questions quickly and enjoy doing so. These question cards have been prepared in two sizes. The large cards can be pinned around around the classroom and used for a whole class activity; the smaller (loop cards) can be used for group work or by individuals ��� they are particularly helpful for one-to-one tutorials and during interventions. | 2,060 | 10,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | longest | en | 0.937614 |
http://nrich.maths.org/public/leg.php?code=97&cl=2&cldcmpid=1235 | 1,500,864,194,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424683.39/warc/CC-MAIN-20170724022304-20170724042304-00388.warc.gz | 239,920,017 | 9,033 | # Search by Topic
#### Resources tagged with Circles similar to Estimating Angles:
Filter by: Content type:
Stage:
Challenge level:
### There are 40 results
Broad Topics > 2D Geometry, Shape and Space > Circles
##### Stage: 2, 3 and 4 Challenge Level:
A metal puzzle which led to some mathematical questions.
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Overlapping Circles
##### Stage: 2 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
### Angle A
##### Stage: 3 Challenge Level:
The three corners of a triangle are sitting on a circle. The angles are called Angle A, Angle B and Angle C. The dot in the middle of the circle shows the centre. The counter is measuring the size. . . .
### LOGO Challenge - Circles as Animals
##### Stage: 3 and 4 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Overlapping Again
##### Stage: 2 Challenge Level:
What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation.
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Efficient Cutting
##### Stage: 3 Challenge Level:
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Sports Equipment
##### Stage: 2 Challenge Level:
If these balls are put on a line with each ball touching the one in front and the one behind, which arrangement makes the shortest line of balls?
### Yin Yang
##### Stage: 2 Challenge Level:
Can you reproduce the Yin Yang symbol using a pair of compasses?
### Not So Little X
##### Stage: 3 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Pi, a Very Special Number
##### Stage: 2 and 3
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### LOGO Challenge 12 - Concentric Circles
##### Stage: 3 and 4 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Sorting Logic Blocks
##### Stage: 1 and 2 Challenge Level:
This interactivity allows you to sort logic blocks by dragging their images.
### What Is the Circle Scribe Disk Compass?
##### Stage: 3 and 4
Introducing a geometrical instrument with 3 basic capabilities.
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Arclets Explained
##### Stage: 3 and 4
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
### Shaping Up
##### Stage: 2 Challenge Level:
Are all the possible combinations of two shapes included in this set of 27 cards? How do you know?
### Circles, Circles Everywhere
##### Stage: 2 and 3
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### LOGO Challenge 10 - Circles
##### Stage: 3 and 4 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### Circle Panes
##### Stage: 2 Challenge Level:
Look at the mathematics that is all around us - this circular window is a wonderful example.
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Gym Bag
##### Stage: 3 and 4 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### LOGO Challenge 11 - More on Circles
##### Stage: 3 and 4 Challenge Level:
Thinking of circles as polygons with an infinite number of sides - but how does this help us with our understanding of the circumference of circle as pi x d? This challenge investigates. . . .
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest? | 1,789 | 7,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-30 | longest | en | 0.86761 |
https://electronics.stackexchange.com/questions/111691/why-does-my-diy-solar-panel-circuit-show-battery-voltage-and-not-solar-panel-v | 1,726,707,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00306.warc.gz | 205,937,470 | 47,213 | # Why does my DIY "solar panel circuit" show battery voltage and not solar panel voltage?
I've got two solar panels set up in series as shown on the below schema. I connect positive on panel A to positive on battery, and negative on panel B to negative on battery. The two panels produce 9V each, and 18V in series on a good sunny day. That's what my multimeter says.
If I disconnect cabling between panels and battery, my multimeter shows that the panels produce about 15V (a bit cloudy today). However, when I connect the cables between panels and battery, my multimeter shows about 12.2V. I believe this is the battery voltage.
Why is it that when I connect my multimeter at point A and B it shows 12.2V (when solar panels are connected to the battery). As you can see on the schema, there's a diode in front of point B. Shouldn't that diode make sure that current from the battery not flows back to the panels?
If this is how it is supposed to work, how do I check the voltage my panels produce when the panels are connected to the battery.
• You CAN get a reading of Vpanel when connected to the battery - it is Vbattery + 1 diode drop when charging. ie the panel voltage IS ~= the battery voltage. What you are trying to ask for (probably) is what the panel OPEN CIRCUIT voltage would be if you disconnected it from the battery. That is something like asking how fast a car engine would rev at this throttle setting if the trailer we are towing did not have an elephant in it. ie the load (or the elephant) is modifying the real voltage and the one you are asking about does not exist in this set of conditions.... Commented May 24, 2014 at 15:49
• ... You can infer what Voc would be from Vloaded and current . BUT Voc is not very important in practice -as long as it is high enough to allow maximum current to flow to the load. Commented May 24, 2014 at 15:50
When you connect the voltmeter between points A and B, you're connecting it directly across the battery, so if the solar panels put out less than the battery voltage the diode will be reverse biased and will, in effect, disconnect the battery from discharging into the solar panels and you'll be measuring only the battery voltage.
On the other hand, if the voltage from the solar panels is higher than the battery voltage and the drop across the diode, the solar panels will force current into the battery, charging it.
However, because the impedance of the battery is so low, it'll drag the voltage of the solar panels down close to the battery voltage, even though the solar panels will be continually pumping current into the battery.
As time goes by and the battery becomes more fully charged, you'll notice that its voltage will rise, but never to your solar panels' open-circuit full-sun value, because the battery chemistry won't allow it.
The proper way to monitor your battery's charging is to measure the voltage across it and the current into it, and never let either rise above the manufacturer's recommendations.
• Thanks for an easy to understand explanation. I'm a total beginners as this, but I get what you're saying. To check whether the battery is charging at all, could I somehow measure the current (before or after the diode)? Commented May 24, 2014 at 13:37
• Current in a series circuit is the same anywherre in the circuit, so all that' Commented May 24, 2014 at 14:30
• Aarghh... Current in a series circuit is the same anywhere in the circuit, so all that's necessary to measure current is to open the circuit - anywhere - and connect an ammeter in series with the break. Commented May 24, 2014 at 14:34
• Actually, you don't even need to open the circuit. There are such things as clamp-on current meters, both AC and DC (The DC sensors are more expensive). A random example of a company which makes this sort of thing: automationdirect.com/adc/Overview/Catalog/Sensors_-z-Encoders/Current_Sensors_(AC-a-_DC) Commented May 24, 2014 at 18:47
• Yes, of course, but my take was based on that, - because of the OP's question - he was much more likely to have a DMM on hand than a clamp-on ammeter. Your comment, however, apprised him of another way to go; a good thing. :-) Commented May 29, 2014 at 0:31
If you connect the voltmeter leads to the terminals of the battery, you will read the voltage across the battery. There really isn't much more than can be said about that.
The battery is essentially modelled as an ideal voltage source in series with a relatively small resistance.
The solar panels, on the other hand, are more like a current source in parallel with a relatively large resistance.
When the solar panels are connected across the battery, a simple model of the circuit is
simulate this circuit – Schematic created using CircuitLab
The current $I_{sc}$ is the short circuit current produced (for a given illumination) by the panels when the panels are short circuited.
The voltage $V_{oc}$ is the open circuit voltage produced by the battery when the battery is open circuited.
Elementary circuit analysis tells us that the voltage across the battery terminals with the solar panel connected is
$$V_{bat} = V_{oc}\frac{R_p}{R_p + R_b} + I_{sc}R_p||R_b$$
Since, typically, the resistance $R_b$ is much less than the resistance $R_p$, the voltage across the battery is approximately
$$V_{bat} = V_{oc} + I_{sc}R_b \approx V_{oc}$$
In other words, unless the panels are providing a relatively large current and / or the internal resistance of the battery is relatively large, you will measure approximately the open circuit battery voltage with the solar panel connected.
• Alright, I just thought the Schottky diode would somehow "isolate" the panels from the battery and let me measure the panel's voltage "behind" the diode. However, I realise that the diode won't affect the voltage but only block the current from the battery. In my case, how would I go about to get a voltage reading from the panels when connected to the battery? Commented May 24, 2014 at 13:16
• @sbrattla, if one connects two circuit elements in parallel, the voltage across each is identical. When you connect the solar panel across the battery, the panel and battery are in parallel and, thus, the panel voltage and battery voltage are identical; you are measuring the voltage across the panels. The panels are not voltage sources, they're imperfect current sources. When you measure the panel voltage when disconnected, you're measuring the open-circuit voltage given by $I_{sc}R_p$. Commented May 24, 2014 at 13:23
• Alright, so bottom line is that I can't get a reading from my panels when connected with the battery. Commented May 24, 2014 at 13:32
• @sbrattla, a reading of what? What are you trying to read from the panels? The power? The current? The voltage 'from' the panels is, as explained above, the same as the voltage across the battery. Commented May 24, 2014 at 14:37
Solar panels have quite high output resistance and therefore any appreciable load will lower the output voltage of the panels. A lead-acid battery, on the other hand can supply a load of several amps without hardly changing its output voltage. This is largely the same when charging current goes into the battery so, in effect the battery will dominate and the voltage you'll read across it will be largely the same whether solar cells are connected or not.
• I've read that I should use a voltage regulator to make sure that I don't provide more than about 13.8V (trickle charge) from the solar panels to the battery. However, with what you're saying does that imply I don't have to use a voltage regulator as the battery's voltage will dominate? Commented May 24, 2014 at 13:05
• Over time, as you pump more charge into the lead-acid battery, the battery's terminal voltage will rise and there is a recommended limit so, using a voltage regulator is a good idea generally. Commented May 24, 2014 at 13:09
• Alright, so it would be a good idea to put a voltage regulator in between the panels and the battery to make sure i "cap" the voltage at, say, 13.8V which will be a safe voltage? Commented May 24, 2014 at 13:11
• I've definitely seen voltage regulators being used. Commented May 24, 2014 at 13:30
If you put a momentary, normally-closed switch in the loop to the battery side of where you connect the voltmeter, you can read the open-circuit panel voltage while you hold the button.
Edit:
• Could you elaborate a little on this? How would I do that? Commented May 24, 2014 at 14:18
• (See edited diagram). Commented May 24, 2014 at 14:59
• Ah, I see. Thanks, i'll try to implement that to be able to get a reading on the panels. Commented May 24, 2014 at 15:27
I've read that I should use a voltage regulator to make sure that I don't provide more than about 13.8V (trickle charge) from the solar panels to the battery
Depends on the panel... you don't show them in your photos, but I assume you are using the smaller ~12V panels, and not the more common 60-cell ~35V ones?
• 2 x 9v, so 18v in total Commented Jan 18, 2019 at 18:15
• Yeah you should be fine then. As the others have noted, the battery will basically overwhelm the panels. It does mean you only charge in bright sunlight, but that's not worth fixing. Commented Jan 18, 2019 at 21:23 | 2,233 | 9,259 | {"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} | 3.3125 | 3 | CC-MAIN-2024-38 | latest | en | 0.962285 |
quincy.inria.fr | 1,394,333,302,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999670852/warc/CC-MAIN-20140305060750-00082-ip-10-183-142-35.ec2.internal.warc.gz | 155,017,349 | 10,751 | # The SETL Programming Language
G22.3110-001 Fall 2000
## Introduction
• SETL (for SET Language)
• high level general purpose language, based on set theory
• invented by J. Schwartz, R. Dewar, E. Dubinsky, E. Schonberg, and W.K. Snyder, at NYU since 1975
• no more development (as far as I know)
## An Introductory Example
program printprimes;
This program prints out a list of prime numbers
-- which includes all primesless than a parameter
-- value which is specified as input data.
primes := { }; - set of primes output so far
p := 2; initial value to test
Loop to test successive values
while p < n loop loop as long as p less than n
ifnot (exists t in primes | p mod t = 0) then
print(p); no divisors, it's a prime
primes with:= p; add it to set of primes
endif;
p := p + 1; move to next test value
endloop;
end printprimes;
•
• Note the empty set, and the set operations
## Assignment Statements & Expressions involving integers, reals, and strings
• Imperative language
identifier := expression;
• Case is not significant
•
• The type of the value in a variable associated with the identifier is determined by the last value assigned:
•
abc := 4; abc now contains an integer
abc := 4.5; abc now contains a real
• The following are examples of valid constants:
•
123 integer
134134145767671 integer
3.1415926535 real
0.45E+13 real
"abc" string
"123" string
"end quote\"" string
• Arithmetic operators: il general, both operand must be of the same type:
a := 3 + 4.0; INVALID!
b := float(3) + 4.0; b = 7.0
c := 3 + fix(4.0); c = 7
• and work on different types:
•
a := "abc";
b := "cd";
c := a + b; c now contains "abccd"
• assigning operators: like in C
•
a := 3; a = 3
a +:= 4; a = 7
b := a +:= 1; a = 8, b = 8
• Strings:
•
abc := "the quick brown fox";
cde := abc(5..8); cde = "quic"
cde := abc(5); cde = "q"
-- abc(5..) means 5 to end of string
cde := abc(5..); cde = "quick brown fox"
• Substring notation can also be used on the left side of an assignment statement:
abc := "hello";
abc(3..4) := "xyz"; abc = "hexyzo"
abc(4..) := "m"; abc = "hexm"
• Slices with negative indices are also allowed, and extract elements from the end of the string
•
• A boolean can only be one of two values, either true or false.
•
• An atom is a type which is created only using the newat procedure:
•
v := newat( );
Atoms are typically used as domain elements for maps (later)
• Any procedure, all of whose formal parameters are readonly (type rd), can be used as the value of a procedure variable, and can be assigned to variables, passed as parameters, used as elements in sets, maps, and tuples, and can also be executed.
•
x := cos;
c := x(1.0);
A procedure may be defined using a procedure definition, or by means of the lambda construction:
x := lambda(y);
pi := 3.1415926536;
y := 2.0*pi*y/360.;
return cos(y);
end lambda;
c := x(90.0); c = 0.0
## Errors & Omega
• Improper operations, such as applying the / operator to string operands, normally causes termination of the program with an appropriate error message.
• In addition to this class of errors, there is a special undefined state called om (for omega). The "value" of undefined identifiers, or other undefined state. (Or rather the absence of any value).
• Denotes the absence of value
• May be used for resetting an identifier:
• a := om; a is now undefined
• Any attempt to perform an operation on the undefined value causes an error, excepted:
•
• equality
•
if x = om then ...
• or the shortcut
•
x := expr1 ? expr2;
that is, if expr1 evaluates to om, then x receives the value of expr2
## Tuples
• A tuple is an ordered, possibly heterogeneous sequence of zero or more values.
•
• May be created using the tuple brackets [ and ].
•
t := [1, 9, "abc", [1, 5]];
• Individual elements of a tuple can be extracted by subscripting,
• t := [1,9,"abc","def"];
x := t(3); x = "abc"
t(4) := 0; t = [1,9,"abc",0]
• If a nonexistent element (e.g. t(7) in the above case) is selected, then om is obtained as a result.
•
• It is perfectly valid to assign a value to a nonexistent element, and may result in increasing the length of the tuple:
•
t := [1,9,"abc",0];
t(5) := 5; t = [1,9,"abc",0,5]
• If such an assignment creates "holes" in a tuple, the missing element values are set to om.
•
• It is possible to undefine a previously defined element by assigning om to it.
•
• This will either create a "hole" in the tuple value, or it will decrease the length of the tuple in the case where the last element is undefined in this manner.
• Binary Tuple Operators
•
+ Tuple concatenation
t*i or i*t Concatenates i copies (integer i) of the tuple t.
with Appends an element to a tuple
frome Extracts an element from the end of a tuple
fromb Extracts an element from the beginning of a tuple
• Unary Tuple Operator
•
# Index of highest defined element
• Examples
•
a := [1,2,3];
b := [6,7];
c := a + b; c = [1,2,3,6,7]
• Subtuples can be extracted using a notation similar to that used in extracting a substring:
•
a := [1,2,3,4,5,6,7,8,9];
b := a(6..8); b = [6,7,8]
c := a(7..); c = [7,8,9]
d := a(8..11); d = [8,9]
• Subtuple operations extend to use on the left side of assignments:
•
t := [1,2,3,4,5,6];
t(2..5) := [7,10]; t = [1,7,10,6]
t(3..) := [ ]; t = [1,7]
A special notation is available for constructing tuples whose values consist of regular sequences of integers:
[1..10] same as [1,2,3,4,5,6,7,8,9,10]
[1,3..12] same as [1,3,5,7,9,11]
[2..1] same as [ ]
[9,8..1] same as [9,8,7,6,5,4,3,2,1]
[9,7..1] same as [9,7,5,3,1]
The general form of this abbreviation is:
[first,next .. last]
• Tuples may appear on the left hand side of an assignment statement:
•
[a,b,c] := s; a = s(1), b = s(2), c = s(3)
[d,-,f] := s; d = s(1), f = s(3)
[e,f] := [2,4]; e = 2, f = 4
[a,b] := [b,a]; interchange a and b
• Value semantics, not pointer semantics:
•
abc := 12;
cde := abc;
abc := abc + 2; cde still = 12
abc := [1,2,3];
cde := abc;
abc(2) := 0; cde still = [1,2,3]
• The operator with adds a single element at the end of a tuple:
•
a := [1,5,10];
b := a with 6; a = [1,5,10], b = [1,5,10,6]
a with:= 7; a = [1,5,10,7]
• The binary operator fromb removes the first element of a tuple
•
• and assigns it to the left operand.
• the right operand, is reassigned to contain the remainder of the tuple after removing this element.
•
• The binary operator frome is similar except that it removes the element from the end of the tuple.
•
• If fromb or frome is applied to a null tuple value, then om is obtained and the tuple value is unchanged:
•
a := [11,26,37,17];
b fromb a; b = 11, a = [26,37,17]
c fromb a; c = 26, a = [37,17]
d frome a; d = 17, a = [37]
e fromb a; e = 37, a = [ ]
f fromb a; f = om, a = [ ]
• The operators with:= and fromb used in conjunction allow a tuple to be used as a queue,
•
• with:= being the enqueue operation
• and fromb the dequeue operation:
• q := [ ];
q with:= 5; enqueue 5: q = [5]
q with:= 7; enqueue 7: q = [5,7]
e fromb q; dequeue: e = 5, q = [7]
e fromb q; dequeue: e = 7, q = [ ]
e fromb q; dequeue: e = om, q = [ ] (queue empty)
Similarly, with:= and frome used in conjunction allow a tuple to be used as a stack:
s := [ ];
s with:= 5; push 5: s = [5]
s with:= push 7: s = [5,7]
e frome s; pop: e = 7, s = [5]
e frome s; pop: e = 5, s = [ ]
e frome s; pop: e = om, s = [ ] (stack empty)
## Sets
• The main datatype in SETL.
• A set is like a tuple, except that it is unordered, and a given value can appear only once.
s := {1,2,"abc"};
t := {2,1,"abc"};
u := {2,1,"abc",2}; s = t = u
• Binary Set Operators
•
+ Set union
- Set difference
* Set intersection
with Add one element to a set (no effect when already present)
less Remove an element from a set (no effect if absent)
from Remove an element and assign remainder (in a non-deterministic manner)
mod Symmetric difference (exclusive or) of sets
npow Set of subsets with fixed number of elements
• Unary Set Operator
•
# Number of elements as integer
arb Select arbitrary element (non-deterministic)
pow Power set of a set (set of all subsets)
• Examples
•
a := {1,2,3,4};
b := {3,4,5,6};
c := a + b; c = {1,2,3,4,5,6}
c := a * b; c = {3,4}
c := a - b; c = {1,2}
s := {5,2,8};
a := s with 7; s = {5,2,8}, a = {5,2,7,8}
s with:= 6; s = {5,6,2,8}
s with:= 5; s = {5,6,2,8}
s less:= 5; s = {6,2,8}
s less:= 0; s = {6,2,8}
• Both arb and from yield om if applied to a null set, and in the case of from, the set value is unchanged.
•
a := {1,5};
b := arb a; b = 1 or 5
c from a; c = 1 (or 5!)
a = {5} (or {1})
d from a; d = 1 (if c was 5)
d = 5 (if c was 1)
a = { }
e from a; e = om, a = { }
• As with tuples, an abbreviated form is permitted for constructing sets of integers:
•
{1..10} means {1,2,3,4,5,6,7,8,9,10}
{3,5..11} means {3,5,7,9,11}
## Maps
• Set of pairs (tuples of length 2)
•
sqroot := {[1,1], [4,2], [9,3], [16,4]};
• If a set has this special form, its values may be accessed using map notation:
•
x := sqroot(9); x = 3
• Map reference notation can also be used on the left side of an assignment:
•
sqroot(25) := 5; adds the pair [25,5] to sqroot
More precisely: compute a new map value by first removing all pairs starting with the given domain value (25, here), and then to add the specified pair.
• Reference to a nonexistent element of a map (e.g. sqroot(19) in the example given) yields om.
•
• Duplicate elements in the domain are allowed, and, for example:
•
• sqroot(n) returns om, if n is duplicated;
• sqroot{n} returns the set of all values corresponding to n in sqroot.
• a := {[1,2], [1,3], [2,4], [5,5], [2,7], [2,8]};
c := a(1); c = om
d := a{1}; d = {2,3}
e := a{5}; e = {5}
f := a(5); f = 5
g := a{7}; g = { }
• This form can also be used on the left hand side of an assignment:
•
a := {[1,0], [1,2], [1,5], [2,5], [2,7]};
a{1} := {5,7}; a = {[1,5], [1,7], [2,5], [2,7]}
• All the operators which apply to sets can also be applied to maps.
•
• In addition the following special operators are provided for operating on maps:
• Binary Map Operators
lessf Removes pairs for one domain value
Unary Map Operators
domain Yields domain of a map
range Yields range of a map
• All three operators cause an error if they are applied to a set which is not a map, i.e. a set which contains at least one element which is not a pair.
•
• The effect of lessf can also be obtained by an explicit assignment as shown in the following examples:
• a := {[1,2], [1,3], [2,2], [2,4], [3,6], [3,7]};
b := domain a; b = {1,2,3}
c := range a; c = {2,3,4,6,7}
a lessf:= 1; removes [1,2] and [1,3]
a(2) := om; removes [2,2] and [2,4]
a{3} := { }; removes [3,6] and [3,7]
• Special syntax is permitted when the domain elements of a map are all themselves tuples:
•
map(x,y,z) is a shorthand for map([x,y,z])
map{x,y,z} is a shorthand for map{[x,y,z]}
For example
c := {[[0,0],"a"],[[0,1],"b"],[[1,0],"c"],[[1,1],"d"]];
x := c(0,1); x = "b"
Combining maps and atoms provides a way of dealing with pointers and dynamic data structures:
program point;
var p1 := newat();
p2 := newat();
mem2 := { [p1,"a"], [p2, "b"] };
-- mem2 is used as a store
-- where p1 and p2 act as locations
x := p1;
y := x;
mem2(x) := "hello"; -- x acts as a pointer to
print(mem2(y)); -- y is an alias to x,
-- that is, to p1
end point;
## Conditional
• If statements
•
if test1 then
statement; statement;
...
statement;
elseif test2 then
statement; statement;
...
else
statement;
...
statement;
end if;
• If expressions
•
if a > 0 then 1 elseif a < 0 then -1 else 0 end if
• Case Statements
•
case
when test1 => block1
when test2 => block2
.
.
otherwise => blocke
end case;
When more than one test succeeds, then only one of the blocks is executed, the choice of which block to execute being made in an arbitrary manner (in the same sense that the arb operator selects an arbitrary element from a set).
case expr
when expr1 => block1
when expr2 => block2
.
.
otherwise => blocke
end case;
If expr is equal to one of expr1, ..., the corresponding block is executed. A more general form is:
case expr
when expr11,expr12,... => block1
when expr21,expr22,... => block2
.
.
otherwise => blocke
end case;
• Case expressions
•
There is also a case expression, which is the same as a case statement, except that the blocks are replaced by expressions, one expression per block.
• Boolean Values & Operators
•
Binary Test Operators
= Types and values match
/= Types or values do not match
> Left operand greater than right
>= Left operand greater than or equal to right
< Left operand less than right
<= Left operand less than or equal to right
in Left operand is an element or substring of right
notin Left operand is not an element or substring of right
subset Left operand is a subset of right (`<=' is equivalent)
incs Left operand includes right as a subset (`>=' is equivalent)
Boolean Procedures
even(i) Operand is even
odd(i) Operand is odd
is_integer(v) Operand is integer type
is_real(v) Operand is real type
is_tuple(v) Operand is tuple type
is_set(v) Operand is set type
is_map(v) Operand is map (set of pairs)
is_string(v) Operand is string
is_atom(v) Operand is atom
is_boolean(v) Operand is boolean
is_procedure(v) Operand is procedure
The equality and inequality comparisons may be used to compare values of any type for exact identity, including testing for equality with om.
Binary Boolean Operators
and Logical conjunction of two boolean values (sequential)
or Logical inclusive disjunction (sequential)
Unary Boolean Operators
not Logical negation
type Yields the string "BOOLEAN"
## Loops
• Control loops
•
for iterator loop
block
end loop;
Two special statements:
continue; -- Proceed with next iteration
exit; -- Exit from the innermost loop
The iterator has one of several forms. Three forms are:
x in set -- order is arbitrary
x in tuple -- number of iterations is
-- the index of the last defined elt
x in string
for x in s loop
print(x); prints elements of s
end loop;
for x in [1,10,50] loop
...
end loop;
for [number,root] in sqrt loop -- Several variables
...
for root=sqrt(number) loop ... -- Alternative form
for y=t(i) loop iterate through tuple t,
-- assigning i and y
for c=s(i) loop iterate through string s,
-- assigning i and character c
The iterator may include a test:
for x in s | x > 5 loop ... -- "|" reads "such that"
for i in [1,2..10] | f(i) > 0 loop ...
Three other loop forms which can be written are:
while test loop loop while test succeeds
...
end loop;
until test loop loop until test succeeds
...
end loop;
loop indefinite loop
...
end loop;
• Loop expressions
• The exit statement may optionally contain an expression:
exit expression;
In this way, an entire loop may be used as a single expression, whose value is determined by the expression as evaluated when the exit statement is executed:
x := for s in set loop
statement1; ... ;
if ... then exit expression; end if;
statements; ... ;
end loop-- x is om if exit is never executed
• Set & Tuple Formers
• A set former is a special form of a loop which computes a set value with an iteration. The form is:
{expression : iterator}
-- read "the set of expression where iterator"
Examples:
{n : n in {1..100}} integers from 1 to 100
{[x**2,x] : x in {1..5} square root map
{a : a in y | a>5} elements > 5
Tuple formers:
[0 : i in [1..100]] tuple of length 100, all 0
[x in s | x < 0] tuple of negative elements of s
• Quantified Tests: bind iteration variables
exists iterator | test
returns true or false, and binds the iteration variable to either om (when false) or a value that satisfies the test;
forall iterator | test
returns true or false, and binds the iteration variable to either om (when true) or a value that does not satisfies the test.
Examples:
s := {1,2,10,20};
t := [1,2,10,20];
if exists x in s | x > 3
then ... will be executed with x = 20 or 10
else ... will not be executed
end if;
if exists x in t | x > 3
then ... will be executed with x = 10
else ... will not be executed
end if;
if forall x in s | x < 30
then ... will be executed with x = om
else ... will not be executed
end if;
if exists x in t | x > 30
then ... will not be executed
else ... will be executed with x = om
end if;
if forall x in t | x < 10
then ... will not be executed
else ... will be executed with x = 10
end if;
• Compound Operators
•
Can be formed from any binary operator by appending a slash / to the name of the operator.
bop/ exprs -- exprs must be a set or tuple
expre bop/ exprs -- exprs must be a set or tuple
bop/ exprs means e1 bop e2 bop e3 ...
result is om if tuple or set is empty
expre bop/ exprs means expre bop e1 bop e2 ...
result is value of expre if tuple or set is empty
Examples:
+/ t sum of values in tuple
0+/ t same, but 0 rather than om for [ ]
*/ [a in s | 3 in a] -- inters. sets that contain 3
" "+/ t builds string from tuple of chars
# Stop and null statements
• The stop statement:
•
stop;
causes immediate termination of execution.
• The null statement:
•
null;
has no effect and thus acts as a nooperation statement.
case i
when 1,3,5 => print(i);
when 2,4,7 => print(i+1);
when 0,6,9 => null; do nothing in these cases
otherwise => print("no good");
end case;
# Using the interpreter
• When using SETL2, you must first create a library using the command:
•
stll -c setl2.lib
SETL2 modules (programs or packages) may then be compiled using:
stlc file
The compiled units will be stored in the library file setl2.lib under the working directory.
To execute a program that has been successfully compiled, execute the command
stlx prog_name
where prog_name is the name of the program from the program statement (and not the name of the file
containing the source file for the program).
See the manual pages for more information on the stll, stlc, and stlx commands. | 5,847 | 18,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.723657 |
https://math.stackexchange.com/questions/3119797/a-question-on-weak-and-norm-topology | 1,627,817,365,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154175.76/warc/CC-MAIN-20210801092716-20210801122716-00129.warc.gz | 379,868,074 | 37,118 | # A question on Weak and Norm topology
How to prove that If $$F$$ is closed with respect to the weak topology then $$F$$ is closed with respect to the norm topology...
Weak topology means let $$V$$ Banach space the weak topology on $$V*$$ is smallest topology in which each function is continuous
norm topology is the topology generated by $$\mathscr{B}=\{B(x,\epsilon):x\in X, \epsilon>0 \}$$
and also tell me what are balls that generated weak topology....
thank you somuch
• In general the weak topology may fail to be metrizable. – DanielWainfleet Feb 21 '19 at 23:24
It is called the weak topology because it is weaker than the strong topology. Let $$T_w$$ be the set of weakly open sets and let $$T_s$$ be the set of strongly open sets. Let $$U$$ be the set of all topologies on $$V^*$$ such that each $$f\in V^*$$ is continuous. Then $$T_w=\cap U,$$ and $$T_s\in U,$$ so $$T_w\subset T_s.$$
So: $$F$$ is weakly closed $$\implies V^*$$ \ $$F \in T_w \implies V^*$$ \ $$F \in T_s \implies F$$ is strongly closed. | 297 | 1,024 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-31 | latest | en | 0.920909 |
https://solutionsadda.in/2023/10/14/digital-logic-design-23/ | 1,726,694,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00313.warc.gz | 499,468,276 | 67,887 | October 13, 2023
October 14, 2023
October 13, 2023
###### Nielit Scientist-B IT 22-07-2017
October 14, 2023
Question 22
Consider a quadratic equation x2 – 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b=________.
A 8 B 9 C 10 D 11
Question 22 Explanation:
x2 – 13x + 36 = 0 ⇾(1)
Generally if a, b are roots.
(x – a)(x – b) = 0
x2 – (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)b + (6)b = (13)b
Convert them into decimal value
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8
Question 22 Explanation:
x2 – 13x + 36 = 0 ⇾(1)
Generally if a, b are roots.
(x – a)(x – b) = 0
x2 – (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)b + (6)b = (13)b
Convert them into decimal value
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8 | 537 | 1,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.866048 |
http://new-contents.com/Texas/extrapolate-an-error.html | 1,555,758,207,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529606.64/warc/CC-MAIN-20190420100901-20190420122901-00543.warc.gz | 136,956,251 | 8,388 | CCTV installation, CCTV monitoring, CCTV retrieval, Home intrusion systems, Business intrusion systems, Access control systems Fire alert systems, Floor alert systems, Medical alert systems, Carbon monoxide detection, Locksmithing and Insurance.
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# extrapolate an error Farwell, Texas
In the complex plane In complex analysis, a problem of extrapolation may be converted into an interpolation problem by the change of variable z ^ = 1 / z {\displaystyle {\hat I'm feeling grossly underpaid.117 points · 34 comments Leaked emails suggest that the Clinton Foundation had weak internal controls and allowed Chelsea Clinton to approve payments to herself22 points · 20 comments Didn't receive any Quite simple, yet, quite expensive to providers. This would not be difficult to calculate quickly: each entry of VTV and VTy could be easily updated with six additions.
Because of the severe truncation error, this example benefited most from maintaining all previous data in extrapolating the next point. rgreq-210e16219bc82b557b2ad181b6be7992 false Look for people, keywords, and in Google: Topic 6.4: Extrapolation IntroductionNotesTheoryHOWTOExamples EngineeringErrorQuestionsMatlabMaple Extrapolation is the process of taking data values at points x1, ..., xn, and approximating a value Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. During this period, audit disallowances only totaled in the hundreds of dollars with some going over a thousand dollars.
Technical questions like the one you've just found usually get answered within 48 hours on ResearchGate. If, however, we are told that this data is linear, then we may find the least-squares fitting line (y(x) = -0.60830 x + 0.89531), then we may approximate the value at This is referred as the fast extrapolation. How to handle a senior developer diva who seems unaware that his skills are obsolete?
Determine if you need to expand the scope of your testing. Fast The extrapolated data often convolute to a kernel function. This transform exchanges the part of the complex plane inside the unit circle with the part of the complex plane outside of the unit circle. Next: Output variables Up: Error estimators Previous: Zienkiewicz-Zhu error estimator Contents guido dhondt 2014-03-02 ERROR The requested URL could not be retrieved The following error was encountered while trying to
The Answer: "Extrapolation" 6/17/2011 Extrapolation is a method of using a mathematical formula that takes the audit results from a small sample of Medicaid paid claims and projects those results over The final stress tensor at the node is obtained by taking the mean of all these stress tensors [17]. Once those formulas were in place, he rolled out the use of extrapolation, starting with the audits of hospitals and quickly spread its use to all Medicaid provider types. To demonstrate this last point, consider finding the average height of all humans by taking just one human.
One of the main questions providers ask when subject to an audit using sampling and extrapolation is: Can this be legal? If I am not wrong than what kind of alternative procedures do we do? For the example data set and problem in the figure above, anything above order 1 (linear extrapolation) will possibly yield unusable values, an error estimate of the extrapolated value will grow How to cope with too slow Wi-Fi at hotel?
A word that also brings terror and disdain to the hearts of audited Medicaid providers subject to the effects of this formula. Wikimedia Commons has media related to Extrapolation. This issue is perverse as this is a private company that might get sold to a public company. 4 commentsshareall 4 commentssorted by: besttopnewcontroversialoldrandomq&alive (beta)[–]Tewks44 2 points3 points4 points 1 year ago(0 children)I think Most companies have FOB related reserves because they know that cutoff errors are inherent.
Version 1 is a one and a half-spaced PDF file. Figure 1. After data is extrapolated, the size of data is increased N times, here N=2~3. Can I conclude my extrapolation as 262 (233.1 - 291.7) with 95% CI –sekaralingam May 26 '15 at 3:24 $z_{\alpha/2}$ is a value which only depends of the size
Examples Example 1 Given the following data which is known to be linear, extrapolate the y value when x = 2.3. (0.3 0.80), (0.7, 1.3), (1.2, 2.0), (1.8, 2.7) The best more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science Sensors may take the current and past voltages of an incoming signal and approximate a future value, perhaps attempting to compensate more appropriately. May 8, 2014 Julio Cesar Bastos Fernandes · Universidade Municipal de São Caetano do Sul Thank you Mr Ryzewski.
This divergence is a specific property of extrapolation methods and is only circumvented when the functional forms assumed by the extrapolation method (inadvertently or intentionally due to additional information) accurately represent This total error amount ($300) is then divided by the sample size (100) to produce an average error per sample case of$3.00. This edition of our newsletter will attempt to de-mystify why it is used, the legal basis of its use by the state, how it works, and how it can be challenged. In Summary Sampling and extrapolation has been part of the Connecticut Medicaid audit process for almost 30 years and will probably be around for as long as there are Medicaid audits.
The auditor finds a cut-off error in my completeness of liabilities and identifies a $100K liability that I failed to accrue. Generated Sat, 15 Oct 2016 12:22:46 GMT by s_ac15 (squid/3.5.20) It is triggered by selecting HER underneath the *EL FILE keyword card. It represents the standard deviation of the size of all extrapolated heat flux vectors. The actual formula used by the Department of Social Services (DSS) to ensure that its extrapolation is statistically valid is fairly complex. The stresses are available (and most accurate) at the integration points of these elements. Does this explain why some learning algorithms are reliable? Department of Social Services, Goldstar, an oxygen provider, questioned the legality of DSS using extrapolation to determine "the total amount of excessive reimbursement Goldstar had received." In this case, the Supreme This note formulates the following question in a precise mathematical terms: When does this method work? Godspeed. However, the actual mathematical calculation of the audit disallowances resulting from this formula is quite simple. This amount ($3.00) is then multiplied by the total audit universe (20,000) to produce a total extrapolated audit adjustment of \$60,000.
Figure 3. By using this site, you agree to the Terms of Use and Privacy Policy. It is possible to include more than two points, and averaging the slope of the linear interpolant, by regression-like techniques, on the data points chosen to be included. Calculations can be found in Elementary Statistics, 7th Ed by M.F, Triola and in Basic Statistics and Pharmaceutical Statistical Applications 1st Ed by J, E. | 1,611 | 7,279 | {"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-2019-18 | latest | en | 0.884172 |
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# Help!
0
112
2
5) (n, k) are positive integers, solve n(n + 2) = k^2
6) (n, k) are positive integers, solve n(n + 4) = k^2
7) (n, k) are positive integers, solve n(n + 5) = k^2
Feb 4, 2021
#1
+11611
+1
5) (n, k) are positive integers, solve n(n + 2) = k^2
6) (n, k) are positive integers, solve n(n + 4) = k^2
Hello Guest!
$$\{n,k\}\subset \mathbb Z$$
$$n(n+2)=k^2$$
$$k\in\{$$ 4, 9, 16, 25, 36, 49, 64, 81, 100, 121 $$\}$$
$$n\in\{$$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$\}$$
$$n(n+2)\in \{$$3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168$$\}$$
No solution for positive n, but if n = -1
$$-1(-1+2)=i^2$$
$$n(n+5)=k^2$$
No solution for positive n, but if n = -4
$$-4(-4+5)=(2i)^2$$
!
Feb 4, 2021
#2
0
Hello other guest! The answer to 5) is no solution.
n(n+2)=k^2
n^2+2n+1=k^2+1
(n+1)^2=k^2+1
(n+1)^2-k^2=1
(n+1+k)*(n+1-k)=1
Now, in order for this to be equal, n and k have to equal 0 but you said they must be positive integers so I believe that there is NO SOLUTION.
Feb 4, 2021 | 513 | 1,037 | {"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} | 4 | 4 | CC-MAIN-2021-25 | latest | en | 0.624074 |
http://www.mcsweeneys.net/articles/a-realistic-assessment-of-how-many-12-year-olds-i-could-beat-up-before-they-overtook-me | 1,448,876,938,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398461390.54/warc/CC-MAIN-20151124205421-00293-ip-10-71-132-137.ec2.internal.warc.gz | 554,932,771 | 12,807 | Your average 12-year-old boy is about 5 feet tall, weighs in the area of a buck-fifteen, and has developed little muscle mass.
I am 21, approximately 6 feet tall, tip the scales at an even 180, and have a moderately athletic and muscular build.
Judging on these statistics and what I assume would be a natural ferocity that would spring forth in a moment of physical danger, I estimate that I could beat up seven 12-year-olds before they overtook me. Of course, these would have to be the aforementioned average-sized 12-year-olds. Future linebackers, NBA players, and all Scandinavian children would throw off this equation. On the flip side, if these were some wimpy, four-square-playing, future-jockey 12-year-olds, I imagine the number would skyrocket to anywhere between 12 and 15. It’s simple exponential math.
This is also assuming that my opponents are smart enough to organize themselves into a circular attack instead of coming at me one by one. If it were an individual, king-of-the-mountain battle royale, I could endlessly pummel 12-year-olds without mercy. But we’re assuming at least a sixth-grade education in a marginal public school as well as some exposure to kung-fu movies, so these kids would form a circle.
However, using my quick wits, I would charge one portion of the circle, landing a devastating blow on the unlucky individual, which would make the others proceed with hesitancy. One on one, I feel like I could deliver a lot of punishment to a 12-year-old. There would be one or two brave ones who would jump on my back, distracting me and thus enabling the others to attack. At best, I could fight off the two heroes on my back and maybe take out four on the ground before I was felled by fatigue and numerous kicks to my groin and shins. This would equal a grand total of seven.
My friend Brian, who stands about 6 feet 2 inches and is stronger than myself, estimates that he could take down a dozen 12-year-olds. I find this hard to believe, but he has been in a fight with people his own age and is a little taller, making groin shots more difficult. Brian’s reach is much longer than mine as well, which is a huge advantage. If you can land solid shots from a distance longer than the 12-year-olds’ legs, there is no need to worry about groin kicks.
He says he would attack one portion of the circle in a fury, scaring off any would-be heroes who wanted to jump on his back and sacrifice themselves for the group. Then he would deal massive blows until fatigue and the inevitable groin shots brought him to the ground. I told him I’d give him nine or ten, but even for the above-average Brian, taking down a dozen 12-year-olds seems like a lot.
If it weren’t for the law and my own morals, we could put these pressing questions to rest. Alas, these barriers still stand in our way.
I’m a pacifist anyway. | 640 | 2,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-48 | longest | en | 0.981573 |
https://community.secondlife.com/profile/1490363-trivis/ | 1,569,213,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576047.85/warc/CC-MAIN-20190923043830-20190923065830-00486.warc.gz | 415,974,594 | 16,525 | # trivis
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1. ## Positioning myself left or right of a target avatar
It works like a charm, @Wulfie Reanimator! Thanks again for your great educational effort :-) Very much appreciated. Kind regards, Triv
2. ## Positioning myself left or right of a target avatar
Wow that is amazing @Wulfie Reanimator!!!!! You've put so much effort in answering my question. You're a hero! Will program this later today, if you dont mind, have to do some work first :-) T H A N K S
3. ## Positioning myself left or right of a target avatar
Worked on the problem... it works fine when I use a "point object" which is left or right of a "avatar object". When the avatar object is a real avatar it doesnt work. The script is: default { touch_start(integer total_number) { key id; id = llGetOwner(); // avatar // id = "e9f7952f-2ac4-0cb7-db9d-4f9a61509593"; // object // target avatar vector avPos = llList2Vector(llGetObjectDetails(id,[OBJECT_POS]),0); rotation avRot = llList2Rot(llGetObjectDetails(id,[OBJECT_ROT]),0); // the math vector A = avPos - <0,1,0>*avRot; vector B = avPos + <0,1,0>*avRot; vector C = llGetPos(); vector AB = B - A; vector AC = C - A; float diff = (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x); // left or right // diff > 0 left, diff < 0 right, diff == 0 front/back if (diff > 0 ) llOwnerSay("I am on the left"); if (diff < 0 ) llOwnerSay("I am on the right"); } } Is there anyone who is able to pull me out this deep well?
4. ## Positioning myself left or right of a target avatar
Thanks Wulfie!, The new script isnt a problem, i am not testing and trying in an existing script :-) I will try to calculate and understand what you wrote where the solution is! But... the rotation of the target avatar is also involved, isnt it? Looks to me that this rotation is just the issue Thanks again, Trivis
5. ## Positioning myself left or right of a target avatar
Still curious how i can decide if I am standing somewhere left or right of an target avatar. Any idea? Is llVecNorm the solution?
6. ## Positioning myself left or right of a target avatar
Thank you @Mollymews! This will help for efficiently positioning beside the target avatar!
7. ## Positioning myself left or right of a target avatar
Somehow I thought this would be easy, but I am still struggling with some rotation things (what's new lol). I am building a positioning HUD. This HUD needs to position me an exact distance left or right of a target-avatar. My rotation is not important (as i understood that a HUD cant rotate me after positioning). The id of the target avatar is known by using llSensor. The positioning of the target is, of course important. My question is how can I determine if I am standing somewhere left or right of my target-avatar (I can be standing in front of the or behind it), and after that positioning myself that exact distance left or right of my target? Thanks in advance for your help, Trivis
8. ## rotation childprim to point to avatar
Wonderful! Thank you Xiija!! This does the job! Still discovering how you did this :-) but many thanks. Kind regards, trivis
9. ## rotation childprim to point to avatar
At first it seemed not too difficult to tackle the following problem, but in the end I did not manage to do it. Rotations are the story of my (second) life :-) The problem: I have an object consisting of several prims. When someone touches the object i would like to point one of the childprims to the avatar who touched the object. See attached picture. The rotation of (only!) that childprim only need to be done around in its XY-plane, so only a rotation around its Z-axis. Does anyone has a working solution? Thanks in advance for any help.
10. ## changing object description of inventory object
Thank you all for your response! Highly appriciated. In the most simple version it would be nice to give the gift a 'unique' number of a sequence... That is: the first recipient will get numer 1, the second numer 2, the third nr3 and so on... But it is clear to me that i have to find another way :-) Thanks again! trivis
11. ## changing object description of inventory object
Dear scripters, I am looking for a way to change the object description of an item i have in the content tab of a giver object. The idea is that when someone touches the main object the description of the object to give away will be changed first, so i can make the gift more personalized. Havent found a way yet to do so. Was wondering if it is possible, and if yes, how so? Thanks in advance!
12. ## Listen or Timed Say for Two object to recognise eachother
Thank you, Nova!
13. ## Listen or Timed Say for Two object to recognise eachother
Thank you so much for this clear explanation, Innula!!! Very much appreciated. Big hug! Trivis
14. ## Listen or Timed Say for Two object to recognise eachother
Thanks Innula! This is how i have done it so far. Have a long negative channel based on the ownerkey and clothingobject. Works all fine. This means that the listener in the clothing is ALWAYS listening... most of the time for 'nothing'. So I was thinking if the clothing would not listen at first but use a timer to llRegionSayTo and let the hud listen, wouldnt that be more efficient? Thanks again for your swift response, Innula. Trivis
15. ## Listen or Timed Say for Two object to recognise eachother
Dear all, Perhaps you have an answer to my question. I have two objects. One is a clothing and the other one is a hud. The hud is for changing the texture of the clothing. So clothing would be worn "often" and the hud "sometimes". To make the texture-exchange as easy as possible, I have two scripting options for both objects to see eachother. One is that the clothing has a listener that is always open and reacts with llRegionSayTo. The other options is that the clothing has a timer, for let's say every 10 or 15 seconds and the hud has a listener. After the two object "see" eachother the exchange of information can start. Will there be a best option of those two options? Thanks in advance, Trivis
×
×
• Create New... | 1,482 | 6,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-39 | longest | en | 0.786274 |
https://www.physicsforums.com/threads/rational-and-irrational-numbers.528943/ | 1,511,269,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806353.62/warc/CC-MAIN-20171121113222-20171121133222-00381.warc.gz | 842,618,981 | 14,775 | # Rational and Irrational numbers
1. Sep 10, 2011
### Charles49
1. The problem statement, all variables and given/known data
Let f be the function defined on the real line by
$$f(x)= \begin{cases} \frac{x}{3} & \text{if x is rational } \\ \frac{x}{4} &\text{if x is irrational.} \end{cases}$$
Let D be the set of points of discontinuities of f. What is D?
2. Relevant equations
None
3. The attempt at a solution
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Sep 10, 2011
### micromass
Staff Emeritus
Well, perhaps it would be good if you would first try to "plot" the function. You can't plot it accurately, but you can get an idea. What does the plot look like? What can you conclude?
3. Sep 10, 2011
### Charles49
Ok I got it, D is the entire real number set excluding 0.
4. Sep 11, 2011
### micromass
Staff Emeritus
Yep, that's it!! | 275 | 924 | {"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} | 3.09375 | 3 | CC-MAIN-2017-47 | longest | en | 0.882479 |
http://googology.wikia.com/wiki/Ecetonthrong | 1,531,897,083,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590069.15/warc/CC-MAIN-20180718060927-20180718080927-00079.warc.gz | 154,809,858 | 54,507 | ## FANDOM
10,264 Pages
The ecetonthrong is equal to 10303,000,000,000, or E303,000,000,000 using Hyper-E notation.[1] The name was coined by Sbiis Saibian.
## ApproximationsEdit
Notation Lower bound Upper bound
Scientific notation $$1\times10^{303000000000}$$
Arrow notation $$10\uparrow303000000000$$
Down-arrow notation $$575\downarrow\downarrow5$$ $$576\downarrow\downarrow5$$
Steinhaus-Moser Notation 10[3][3] 11[3][3]
Copy notation 2[2[12]] 3[3[12]]
H* function H(100H(2)) H(101H(2))
Taro's multivariable Ackermann function A(3,A(3,36)) A(3,A(3,37))
Pound-Star Notation #*((1))*(38)*3 #*((1))*(3)*9
BEAF {10,303000000000}
Bashicu matrix system (0)(1)[36] (0)(1)[37]
Hyperfactorial array notation (13!)! (14!)!
Fast-growing hierarchy $$f_2(f_2(34))$$ $$f_2(f_2(35))$$
Hardy hierarchy $$H_{\omega^22}(34)$$ $$H_{\omega^22}(35)$$
Slow-growing hierarchy $$g_{\omega^{\omega^{\omega+1}3+\omega^93}}(10)$$
## Sources Edit
1. Saibian, Sbiis. Hyper-E Numbers. Retrieved 2016-07-18.
-minutia group
n-ary numbers with multipliers
Eyelash mite group: eyelash mite-speck · eyelash mite-crumb · eyelash mite-chunk · (binary/ternary/octal)eyelash mite · eyelash mite-bunch · eyelash mite-crowd · eyelash mite-swarm
Dust mite group: dust mite-speck · dust mite-crumb · dust mite-chunk · (binary/ternary/octal)dust mite · dust mite-bunch · dust mite-crowd · dust mite-swarm
Cheese mite group: cheese mite-speck · cheese mite-crumb · cheese mite-chunk · (binary/ternary/octal)cheese mite · cheese mite-bunch · cheese mite-crowd · cheese mite-swarm
Clover mite group: clover mite-speck · clover mite-crumb · clover mite-chunk · (binary/ternary/octal)clover mite · clover mite-bunch · clover mite-crowd · clover mite-swarm
Pipsqueak group: (binary/ternary/octal)pipsqueak
Little squeaker group: (binary/ternary/octal)little squeaker
Small fry group: (binary/ternary/octal)small fry
Guppy group: (binary/ternary/octal)guppyspeck · (binary/ternary/octal)guppycrumb · (binary/ternary/octal)guppychunk · (ternary/quaternary/quinary/duodecimal/hexadecimal/vigesimal/sexagesimal)guppy(bit/byte)
Minnow group: (binary/ternary/octal)minnowspeck · (binary/ternary/octal)minnowcrumb · (binary/ternary/octal)minnowchunk · (ternary)minnow(bit/byte)
Goby group: (binary/ternary/octal)gobyspeck · (binary/ternary/octal)gobycrumb · (binary/ternary/octal)gobychunk · (ternary)goby(bit/byte) | 854 | 2,367 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.396926 |
https://forums.newtek.com/archive/index.php/t-68035.html?s=fc913642f3a05730dc4f7a26e052ee3c | 1,555,629,071,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578526904.20/warc/CC-MAIN-20190418221425-20190419003425-00393.warc.gz | 419,681,343 | 4,014 | PDA
View Full Version : Image on one side of double sided polygons with nodes?
Dirk
05-03-2007, 08:14 AM
I remember there was an explanation in the forum how to map an image on just one side of double sided polygons, with nodes - can't find it, though. Does anybody know where to find it, or how to do that?
Sensei
05-03-2007, 08:32 AM
Perhaps with Spot Info > Polygon Side integer? Use it as alpha/blend in color mixer.. One color input comming from 2d image, 2nd whatever else you want..
MooseDog
05-03-2007, 08:53 AM
Perhaps with Spot Info > Polygon Side integer? Use it as alpha/blend in color mixer.. One color input comming from 2d image, 2nd whatever else you want..
correct me if i'm wrong, but spot info>polygon side returns a 0 for facing and a 1 for back-facing, (i might have that backwards:stop: )
which is then fed into a logic node (then to an opacity button)
Dirk
05-03-2007, 09:19 AM
@Sensei: sry, doesn't work
@Moosedog: hmm, but opacity of what? The image itself? that doesn't seem to work either...
Sensei
05-03-2007, 10:06 AM
Of course it works.. :2guns:
Example scene in attachment.. Frame 0 back side color, 60 front side color.. I've connected 2d texture > image and worked too..
Sensei
05-03-2007, 10:09 AM
correct me if i'm wrong, but spot info>polygon side returns a 0 for facing and a 1 for back-facing, (i might have that backwards:stop: )
which is then fed into a logic node (then to an opacity button)
Exactly :)
But logic node is not necessary.. Integer has just 0 or 1 and putting it to Mixer Color's Opacity input is enough, conversion is done internally between data type formats, no values between 0 and 1, which is what we want/need..
MooseDog
05-03-2007, 11:01 AM
bingo! thx :)
Dirk
05-03-2007, 11:06 AM
I don't know whats happening here - Senseis scene does work - but not mine.
I have subpatched object here with 3 bones in it. The object shows me the red side - but only once ??? After rendering another frame, where it shows the green side, it doesn't show the red side again - the object is entirely green then oO
Dirk
05-03-2007, 11:12 AM
I've attached a scene. On my machine, it does this:
- rendering frame 0 shows the object's red side, as it should
- rendering frame 20 shows the green side - ok
- rendering frame 0 *again* shows the object in green ????
Dirk
05-03-2007, 04:51 PM
Can anybody confirm this? ... Or do I need medication?
jameswillmott
05-03-2007, 05:30 PM
Yep, your scene doesn't work. How'd you do that???
Dirk
05-03-2007, 05:54 PM
I have no idea. I'll try to re-create the problem tomorrow.
Sensei
05-03-2007, 08:23 PM
LOL.. Deactivate all bones except Bone3 and render.. :D Only the last polygon controlled by this bone became green permanently.. :D
Dirk
05-04-2007, 02:52 AM
Weightmaps? Could this be a problem with weightmaps?
Dirk
05-04-2007, 02:55 AM
@Sensei: seems to make no difference for me. I can deactivate of even throw out the bones, the object renders in the same strange way.
Dirk
05-04-2007, 03:01 AM
GOT IT!
It's a problem with the classic camera! I use the classic camera because of the outlines. Switch to perspective cam and everything works as it should (except of the outlines, of course).
MSherak
03-27-2013, 08:16 PM
Of course it works.. :2guns:
Example scene in attachment.. Frame 0 back side color, 60 front side color.. I've connected 2d texture > image and worked too..
This is great and got me thinking so I thought I would take this a step further..
So I created a normal from a leaf texture that Dpont gives with his DP Verdure plugin for foliage. Like normal I placed the color in the color channel. Use the transparent texture in the clipmap for the model for ease of use. Then I replaced the colors nodes provided with normal maps. Now I added extra nodes to the back polygon allowing me to invert the Z of the normal and place an adjustment value to control the depth of the Z channel in the normal map. This makes the normal read the front side for lighting and apply it on the back faced polygon when double sided. Also gives you the control of the level of lighting. Basically making it look like it's translucent without having to use the translucency channel giving one the ability to backlit say, in this case, leaves without the render hit. So anything that hits the front side polygon will be transfered to the back polygon, including shadows.
In a GI scene with the flags set right you could render foliage without including it in GI and not rely on translucency for the back lighting. Kinda of a cool look. If you use it in something, please post.
Enjoy.
-M
PS. it still works with GI also which is even cooler. Turn on GI and look at the backside. Fun Fun!! ;)
GoatDude
03-27-2013, 09:50 PM
This is very cool
MSherak
03-29-2013, 10:37 AM
Put it to the test with some other tweaks which allow me to darken the reverse normal.
Made a tree with DP Venture. One light using DP Sunsky. Strait from LW. Render times were from 2m37s to 6m19s.
Sensei
03-29-2013, 10:53 AM
You can optimize a lot of your node tree.
Instead of Split Color etc. up to Make Color, simply Math > Vector > Multiply by 1,1,-1..
MSherak
03-30-2013, 03:30 AM
You can optimize a lot of your node tree.
Instead of Split Color etc. up to Make Color, simply Math > Vector > Multiply by 1,1,-1..
True and I will most likely use this.. this illustrates kind what is going on to others.. plus I get to use your node. ;)
Going beyond this now and doing some swizzle stuff like AO burned into the normal map then placed in.. all kinds of goodies.. | 1,519 | 5,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.895337 |
http://www.edhelper.com/math/ordering49.htm | 1,493,081,943,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120001.0/warc/CC-MAIN-20170423031200-00452-ip-10-145-167-34.ec2.internal.warc.gz | 522,740,012 | 2,940 | edHelper subscribers - Create a new printable
Math
Name _____________________________ Date ___________________
Ordering
Fill in the missing numbers.
1 Just before and after_____, 14, _____
2 * This is a pre-made sheet.Use the link at the top of the page for a printable page.
3 Between11, _____, 13
4 Just after18, 19, _____
Circle the correct words.
1.
15 is equal tois less than 18
2.
13 is less thanis greater than 20
3.
16 is equal tois less than 16
4.
17 is less thanis greater than 12
Circle the larger number.
1.
13 18
2.
19 16
3.
14 15
4.
20 12
5.
11 16
6.
18 17
Circle the smaller number.
1.
12 16
2.
14 20
3.
18 13
4.
17 15
5.
19 18
6.
19 12
Key #2
Order each group of numbers from smallest to largest.
1.
20, 17, 15 _____, _____, _____
2.
14, 12, 19 _____, _____, _____
3.
11, 18, 16 _____, _____, _____
Order each group of numbers from largest to smallest.
1.
18, 12, 11 _____, _____, _____
2.
14, 13, 17 _____, _____, _____
3.
19, 91, 20 _____, _____, _____
Order each group of numbers from largest to smallest.
1.
12, 15, 21, 16 _____, _____, _____, _____
2.
91, 13, 19, 14 _____, _____, _____, _____
3.
11, 18, 17, 20 _____, _____, _____, _____
Circle the larger number.
1.
15 14
2.
19 20
3.
18 16
4.
12 13
5.
11 17
6.
12 18
Order each group of numbers from smallest to largest.
1.
17, 19, 20 _____, _____, _____
2.
12, 14, 16 _____, _____, _____
3.
15, 18, 11 _____, _____, _____
Sample
This is only a sample worksheet.
edHelper subscribers - Create a new printable | 551 | 1,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | latest | en | 0.690765 |
https://www.esaral.com/q/solve-the-following-52520 | 1,723,139,493,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00763.warc.gz | 615,805,339 | 11,597 | # Solve the following
Question:
If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}$, then
(a) Re (z) = 0
(b) Im (z) = 0
(c) Re (z) > 0, Im (z) > 0
(d) Re (z) > 0, Im (z) < 0
Solution:
for $\frac{\sqrt{3}}{2}+\frac{i}{2}$
$r=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$ and $\theta=\tan ^{-1} \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\tan ^{-1} \frac{1}{\sqrt{3}}$
i. e. $\theta=\frac{\pi}{6}$
i.e. $\frac{\sqrt{3}}{2}+\frac{i}{2}=r(\cos \theta+i \sin \theta)$
we have $\theta=\frac{\pi}{6}$ and $r=1$
and $\frac{\sqrt{3}}{2}-\frac{i}{2}=r(\cos \theta-i \sin \theta)$
$\therefore z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}$
$=[r(\cos \theta+i \sin \theta)]^{5}+[r(\cos \theta-i \sin \theta)]^{5}$ where $r=1, \theta=\frac{\pi}{6}$
$=(\cos \theta+i \sin \theta)^{5}+(\cos \theta-i \sin \theta)^{5}$
$=\cos 5 \theta+i \sin 5 \theta+\cos 5 \theta-i \sin 5 \theta \quad$ (By De $-$ moivre theorem)
$=2 \cos 5 \theta$ | 472 | 1,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.219533 |
http://linear.ups.edu/html/section-LDS.html | 1,721,620,597,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00243.warc.gz | 17,130,292 | 6,285 | ### SectionLDSLinear Dependence and Spans
In any linearly dependent set there is always one vector that can be written as a linear combination of the others. This is the substance of the upcoming Theorem DLDS. Perhaps this will explain the use of the word “dependent.” In a linearly dependent set, at least one vector “depends” on the others (via a linear combination).
Indeed, because Theorem DLDS is an equivalence (Proof Technique E) some authors use this condition as a definition (Proof Technique D) of linear dependence. Then linear independence is defined as the logical opposite of linear dependence. Of course, we have chosen to take Definition LICV as our definition, and then follow with Theorem DLDS as a theorem.
#### SubsectionLDSSLinearly Dependent Sets and Spans
If we use a linearly dependent set to construct a span, then we can always create the same infinite set with a starting set that is one vector smaller in size. We will illustrate this behavior in Example RSC5. However, this will not be possible if we build a span from a linearly independent set. So in a certain sense, using a linearly independent set to formulate a span is the best possible way — there are not any extra vectors being used to build up all the necessary linear combinations. OK, here is the theorem, and then the example.
##### Theorem DLDS Dependency in Linearly Dependent Sets
Suppose that $S=\set{\vectorlist{u}{n}}$ is a set of vectors. Then $S$ is a linearly dependent set if and only if there is an index $t$, $1\leq t\leq n$ such that $\vect{u_t}$ is a linear combination of the vectors $\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\ldots,\,\vect{u}_{t-1},\,\vect{u}_{t+1},\,\ldots,\,\vect{u}_n$.
This theorem can be used, sometimes repeatedly, to whittle down the size of a set of vectors used in a span construction. We have seen some of this already in Example SCAD, but in the next example we will detail some of the subtleties.
#### SubsectionCOVCasting Out Vectors
In Example RSC5 we used four vectors to create a span. With a relation of linear dependence in hand, we were able to “toss out” one of these four vectors and create the same span from a subset of just three vectors from the original set of four. We did have to take some care as to just which vector we tossed out. In the next example, we will be more methodical about just how we choose to eliminate vectors from a linearly dependent set while preserving a span.
##### Sage COVCasting Out Vectors
Example COV deserves your careful attention, since this important example motivates the following very fundamental theorem.
##### Theorem BS Basis of a Span
Suppose that $S=\set{\vectorlist{v}{n}}$ is a set of column vectors. Define $W=\spn{S}$ and let $A$ be the matrix whose columns are the vectors from $S$. Let $B$ be the reduced row-echelon form of $A$, with $D=\set{\scalarlist{d}{r}}$ the set of indices for the pivot columns of $B$. Then
1. $T=\set{\vect{v}_{d_1},\,\vect{v}_{d_2},\,\vect{v}_{d_3},\,\ldots\,\vect{v}_{d_r}}$ is a linearly independent set.
2. $W=\spn{T}$.
In Example COV, we tossed-out vectors one at a time. But in each instance, we rewrote the offending vector as a linear combination of those vectors with the column indices of the pivot columns of the reduced row-echelon form of the matrix of columns. In the proof of Theorem BS, we accomplish this reduction in one big step. In Example COV we arrived at a linearly independent set at exactly the same moment that we ran out of free variables to exploit. This was not a coincidence, it is the substance of our conclusion of linear independence in Theorem BS.
Here is a straightforward application of Theorem BS. | 913 | 3,675 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.887812 |
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Homework 11 Solution
The overall torque is then mg l 2 1 sin 90 k l xo
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Unformatted text preview: attached to the plank, the spring force always acts perpendicularly to the length of the plank at distance L, and the torque is easily found. The plank’s weight still acts at a distance L/2 from the center of mass, but now at an angle 90 − θ relative to the axis of the plank. The overall torque is then τ = −mg L 2 1 sin (90 − θ) − k (Lθ − xo ) L = − mgL cos θ − kL2 θ + kxo L 2 Since the angle θ is small, we may approximate cos θ ≈ 1, and we may also make use of our earlier expression for xo . Finally, out of equilibrium the torques must give the moment of inertia times the angular acceleration. 1 1 1 τ = − mgL cos θ − kL2 θ + kxo L ≈ − mgL − kL2 θ + mgL = −kL2 θ = Iα 2 2 2 d2 θ −kL2 θ = I 2 dt (18) (19) Noting that I = 1 mL2 for a thin plank, we can put the last equation in the desired form for simple 3 harmonic motion in terms of known quantities: 3k d2 θ =− θ 2 dt m 3k ω= m Note that the length of the plank does not matter at all. ii Small enough such that we don’t have to worry about the spring bending to the left, for one. (20) (21)...
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Ask a homework question - tutors are online | 421 | 1,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.847161 |
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October 25, 2009, 11:39 AM #6
Bartholomew Roberts
member
Join Date: June 12, 2000
Location: Texas and Oklahoma area
Posts: 8,462
Some good science and math on the issue from the Ammo Oracle:
Quote:
The importance of rate of twist in wounding is a frequent subject of what we politely call "ballistic myth." Any projectile that has a "center of pressure" forward of the center of gravity will tend to tumble. You can illustrate this to yourself by trying to balance a pencil on your fingertip. Spin, given to the projectile by barrel twist, puts a projectile into a state described as "gyroscopically stable." The projectile might be momentarily disturbed but will return to nose-forward flight quickly. To describe how stable a given projectile is we use the gyroscopic stability factor (Sg). Generally you want a factor of 1.3 or greater for rifle rounds. 1.5-2.0 is a generally accepted value for 5.56 rounds. For M193 the following variables apply: axial moment of inertia (A) = 11.82 gm/mm2 transverse moment of inertia (B) = 77.45 gm/mm2 mass (m) = 3.53 grams reference diameter (d) = 5.69 mm Using the gyroscopic stability formula: Sg = A2 p2 / (4 B Ma) and assuming sea level we use an air density of 1.2250 kg/m^3 and discover that this this projectile will need on the order of 236,000 rpm for good stability (Sg > 1.3). At 3200 fps M193 is typically spun up to more like 256,000 (1:9" twist) to 330,000 rpm (1:7") so that Sg approaches 1.9 or 2.0. 1:12" rifles will spin rounds at around 192,000 rpm and 1:14" rifles around 165,000 rpm. You can see why 1:14" rifles might have had trouble stabilizing M193 rounds. Clever math types will see that density of the medium traversed (air in this case) has a dramatic effect on the spin required to maintain the Sg (density being in the first term's divisor). This is why cold conditions tend to dip "barely stable" rounds below the stability threshold. Without doing too much calculus it will be seen that an increase of three orders of magnitude (1000) in this variable will be a dramatic one for spin requirements. To balance things spin must be increased to compensate. Through human flesh (which varies from 980 - 1100 kg/m^3 or about 1000 times the density of air) something on the order of 95,000,000 - 100,000,000 rpm is required to stabilize a projectile at speed. Given these differences it will be seen that the difference between a 1:12 or 1:14" twist when it hits flesh and a projectile launched from a 1:9 or 1:7" weapon is so small as to be beyond measuring. But the game isn't over yet. Gyroscopic stability of 2.0 or so is sufficient for a M193 projectile to recover from an upset quickly, return to nose-forward flight and not be over stabilized. To prevent the upset in the first place, particularly when a sudden and very extreme change in density (and therefore drag and pressure applied to the center of pressure) requires FAR more stability. To grant enough stability force to prevent the upset of a M193 projectile encountering a sudden 1000 fold increase in density a factor of as much as 10 to 50 times (speaking VERY conservatively) the required gyroscopic stability for a steady state flight through a medium of that density would be required. In other words, unless the projectile is spinning at nearly a BILLION rpm it is going to be upset by such a transition. Even at this rpm it is like to be upset somewhat. In summary, and to take the most extreme case, a M193 projectile spinning at 350,000 rpm (from a 1:7" rifle) is going to upset in flesh (yaw) exactly as fast as one spinning at 150,000 rpm (from a 1:14" rifle).
Page generated in 0.02090 seconds with 8 queries | 928 | 3,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-23 | latest | en | 0.903379 |
https://www.bankersadda.com/p/quantitative-aptitude-for-rbi-assistant_10.html | 1,566,209,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314721.74/warc/CC-MAIN-20190819093231-20190819115231-00008.warc.gz | 721,126,445 | 50,566 | Dear Students,
Quantitative Aptitude Questions for RBI Assistant Mains 2017
Quantitative Aptitude is a very important section you must prepare if you are aiming for a job in Bank or Insurance sector. These two weeks are very important as IBPS Clerk and RBI Assistant Mains are lined up. So, these 15 questions can help you practice three very important topics of Quant Section.
Q1. A man can swim at 9 kmph in still water. He covers 21 Km distance in upstream and 21 Km distance in downstream in total 7 hours. Find the speed of stream of river.
(a) 2√3 km/h
(b) 3√3 km/h
(c) 3 km/h
(d) 4 km/h
(e) None of these
Q2. The present population of a city is 52,650. What was the population of the city two years ago, if the population reduces at the rate of 10% per annum?
(a) 65,000
(b) 50,000
(c) 55,000
(d) 62,000
(e) None of these
Q3. The height of a cone is two fifth of perimeter of a isosceles triangle whose equal and unequal sides are 11 cm and 13 cm respectively. If diameter of cone is 15 cm, what is the volume of cone (in cm3)? (take )
(a) 925
(b) 725
(c) 825
(d) 850
(e) None of these
Q4. The compound interest for two years on a certain sum at certain rate is equal to the simple interest accrued on the same sum at the rate of 11% for four years. What is the rate of compound interest?
(a) 25%
(b) 20%
(c) 30%
(d) 35%
(e) None of these
Q5. A person covers a total distance of 135 km in 9 hours. He covers some part of journey by ola cab with a speed of 21 km/h and rest part of Journey by e-rikshaw with a speed of 14km/h. What distance he covered by e-rikshaw?
(a) 110 km
(b)106 km
(c) 105 km
(d)108 km
(e) None of these
Directions (6-10):What should come in place of question mark (?) in the following questions?
Q6. 17% of 2300 + 19% of 4700 = ? + 36% of 450
(a) 1133
(b) 1122
(c) 1212
(d) 1221
(e) 1124
S6. Ans.(b)
Sol.
? = 17 × 23 + 19 × 47 – 36 × 4.5
? = 1122
S8. Ans.(a)
Sol.
√(?)=70 – 21 = 49
⇒ ? = 2401
S10. Ans.(b)
Sol.
?² = 86 + 83 = 169
⇒ x = ± 13
Directions (11-15): What will come in place of question mark (?) in the following no. series ?
Q11. 2, 7, 24.5, 85.75, ?
(a) 272.245
(b) 300.125
(c) 265.725
(d) 275
(e) 278.25
S11. Ans.(b)
Sol.
Pattern is ×3.5, ×3.5, ×3.5, ×3.5
∴ ? = 85.75×3.5= 300.125
Q12. 9, 20, 35, 59, 99, ?
(a) 146
(b) 175
(c) 164
(d) 154
(e) 162
Q13. 7, 15, 61, 367, 2937, ?
(a) 29571
(b) 30231
(c) 29374
(d) 29371
(e) 41253
S13. Ans.(d)
Sol.
Pattern is ×2+1, ×4+1, ×6+1, ×8+1, ×10+1,…..
∴ ? = 2937 × 10 + 1 = 29371
Q14. 24, 12, 36, 18, 54, ?
(a) 28
(b) 27
(c) 26
(d) 32
(e) 29
S14. Ans.(b)
Sol.
÷2, ×3, ÷2, ×3, ÷2, ×3
∴ ? = 54 ÷ 2 = 27
Q15. 4, 10, 33, 136, ? , 4116
(a) 822
(b) 534
(c) 685
(d) 745
(e) 548
S15. Ans.(c)
Sol.
The pattern is as given below:
4 × 2 + 2 = 10
10 × 3 + 3 = 33
33 × 4 + 4 = 136
136 × 5 + 5 = 680 + 5 = 685 | 1,212 | 2,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-35 | latest | en | 0.818482 |
https://physics.stackexchange.com/questions/463633/symmetry-in-fock-space-2-body-interaction | 1,571,111,645,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655864.19/warc/CC-MAIN-20191015032537-20191015060037-00542.warc.gz | 649,187,345 | 31,909 | # Symmetry in Fock-space 2-body interaction
The simplest two body interaction term for fermions is
$$H = \sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l$$
and I'm trying to determine the symmetries on $$U$$. Unfortunately I keep getting weird sign errors. The first symmetry comes from Hermiticity. To have $$H$$ be Hermitian, we need
$$H = H^\dagger = \sum_{ijkl} U_{ijkl}^\dagger a_l^\dagger a_k^\dagger a_j a_i$$
Then relabel the indices $$i\leftrightarrow l$$, $$j\leftrightarrow k$$:
$$H = \sum_{ijkl} U_{lkji}^\dagger a_i^\dagger a_j^\dagger a_k a_l$$
This should indicate that $$U_{lkji}^\dagger = U_{ijkl}$$. Along similar lines,
$$H = \frac{H + H}{2} = \frac{\sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l + \sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l}{2}$$
Relabel the indices in the second one as $$k\leftrightarrow l$$:
$$H = \frac{\sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l + \sum_{ijkl} U_{ijlk} a_i^\dagger a_j^\dagger a_l a_k}{2}$$
then apply the anticommutation relation:
$$H = \frac{\sum_{ijkl} U_{ijkl} a_i^\dagger a_j^\dagger a_k a_l - \sum_{ijkl} U_{ijlk} a_i^\dagger a_j^\dagger a_k a_l}{2} =\frac{\sum_{ijkl} (U_{ijkl}-U_{ijlk}) a_i^\dagger a_j^\dagger a_k a_l}{2}$$
Thus suggests that $$U_{ijkl}$$ is antisymmetric under the last two indices. Similarly, it should be antisymmetric under the first two indices. Unfortunately, a number of online sources seem to suggest that it should be symmetric, for instance http://sirius.chem.vt.edu/wiki/doku.php?id=crawdad:programming:project3#step_3two-electron_integrals -- here $$\langle{\mu\sigma|\lambda\rho\rangle} = U_{\mu\sigma\lambda\rho}$$, unless I'm somehow very sorely mistaken. Another reason is that $$U_{ijkl}$$ will often get contract with the density matrix $$D_{kl}$$ which is symmetric, and so would vanish if it wasn't antisymmetric.
Is it correct that the symmetries necessary of $$U$$ are the Hermitian symmetry given above, and antisymmetry in the (12) or (34) pairs? Or is it symmetric? Or neither? Thank you.
$$U_{ijkl}$$ is anti-symmetric in $$(12)$$ and $$(34)$$, your derivation is correct.
We define $$\langle ij|kl\rangle=\int\psi_i^*(\vec{r}_1)\psi_j^*(\vec{r}_2)\hat{h}\psi_k(\vec{r}_1)\psi_l(\vec{r}_2)\,d^3r_1d^3r_2$$
1. For general complex wavefunctions, the symmetry of $$\langle ij|kl\rangle$$ is 4-fold: $$\langle ij|kl\rangle = \langle ji|lk\rangle = \langle kl|ij\rangle = \langle lk|ji\rangle$$
2. For real functions, it's 8-fold due to: $$\langle ij|kl\rangle = \langle ji|kl\rangle$$
3. For $$U_{ijkl}$$, only the anti-symmetric part of $$\langle ij|kl\rangle$$ matters: $$U_{ijkl} = \langle ij||kl\rangle = \langle ij|kl\rangle -\langle ij|lk\rangle$$ | 906 | 2,703 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-43 | latest | en | 0.619126 |
https://www.jiskha.com/search/index.cgi?query=Statistics%3A+Percentile+Rank | 1,502,898,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102307.32/warc/CC-MAIN-20170816144701-20170816164701-00114.warc.gz | 917,880,250 | 9,785 | # Statistics: Percentile Rank
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Using the data set below please calculate the 2;8;6;9;5 Range What would be the percentile rank of the number 6? Help please!
### university of phoenix
The weight of a small Starbucks coffee is a random variable with a mean of 360 g and a standard deviation of 9 g. Use Excel to find the weight corresponding to each percentile of weight. a. 10th percentile b. 32nd percentile c. 75th percentile d. 90th percentile e. 99.9th ...
### Stats
If a student's rank in a class of 300 students is 111, find the student's percentile rank
### algeba
if a student's rank in a class of 400 students is 44 find the student's percentile rank.
### algebra
if a student rank in a class 400 is 87 waht id the student percentile rank
### statistics
A college student takes a standardized test and scores a 163. If the mean is 155 and the standard deviation is 7, what is the student’s percentile rank? (Assume a normal distribution.)
### Research and Statistics
I want to ensure I've done this problem correctly. Thanks! 3. In one elementary school, 200 students are tested on the subject of Math and English. The table below shows the mean and standard deviation for each subject. Mean SD Math 67 9.58 English 78 12.45 One student’s ...
### Statistics
Dear Lee, Please confirm these answers to this question. your medical terminology instructor listed the following grades for the class out of a 75 point test: 33,34,43,45,45,54,55,59,60,62,64,66,67,68,67,68,68,69,70,70 a.find the 90th percentile b. your score was 59, what is ...
### Statistics
Scores on the Stanford-Binet Intelligence scale have a mean of 100 and a standard deviation of 16, and are presumed to be normally distributed. A person who scores 68 on this scale has what percentile rank within the population?
### Math
If a student's percentile rank in a class of 400 students is 87, how do I find the student's class rank.
### math
If a student's percentile rank in a class of 400 students is 87, find the student's class rank.
### Statistics
If the average home cost \$100,000, with a standard deviation of \$15,000, what would a home on the 20th percentile rank cost?
### Math
How do you find the percentile rank of 4.6?
### Statistic
the percentile rank of t5 = 1.476
### Stats....plz help....
Rank the following data in increasing order and find the position and value of the 56th percentile. Please show all of your work. 0 1 6 8 9 9 8 6 3 1 0 9 How do i find the position and value of the 56th percentile??
### Statistics
your medical terminology instructor listed the following grades for the class out of a 75 point test:<br /> 33,34,43,45,45,54,55,59,60,62,64,66,67,68,67,68,68,69,70,70<br /> a.find the 90th percentile<br /> b. your score was 59, what is your percentile?<br...
### Statistics
your medical terminology instructor listed the following grades for the class out of a 75 point test: 33,34,43,45,45,54,55,59,60,62,64,66,67,68,67,68,68,69,70,70 a.find the 90th percentile b. your score was 59, what is your percentile?
### math
find the percentile rank of 4.6 out of these numbers: 1.8, 2.5, 3.9, 4.6, 4.7, 4.8, 4.8, 4.9
### Math
In a class of 60 students, Heather has a rank of 16. At what percentile is she?
### Statistics
How do I find a percentile if I only have the mean and sd? ie...what IQ score is associated with the 65th percentile?-mean is 100 and sd is 15. Thanks in advance :)
### Statistics
n=1000, bell-shaped Mean - 5,000 ~950 lie +/-100 of 5000 -4900 to 5100 Required 1) Standard Deviation 2) 97.5th percentile value 3) Z-score for 16th percentile
### math
rachel is ranked 72 out of 451 students. whats her class rank percentile?
### statistics college
SATI SCORES AROUND THE NATION HAVE A MEAN SCORE AROUND 500 A STANDARD DEVIATION OF ABOUT 100 POINTS AND ARE APPROXIMATELY NORMALLY DISTRIBUTED A PERSON WHO SCORES A PERFECT SCORE OF 800 ON THE SAT I HAS APPROXIMATELY WHAT PERCENTILE RANK WITHIN THE POPULATION
### statistics
Sat scores l around the nation tend to have a mean score around 500, a standard deviation of about 100 points and are approximately normal distributions. a person who scores 600 on the sat 1 has approximately what percentile rank within the population? please show all ...
### Statistics
SAT I scores around the nation tend to have a mean scale score around 500, a standard deviation of about 100 points, and are approximately normally distributed. What SAT I score within the population would have a percentile rank of approximately 16?
### statistics
SAT I scores around the nation tend to have a mean scale score around 500, a standard deviation of about 100 points, and are approximately normally distributed. What SAT I score within the population would have a percentile rank of approximately 99?
### statistics
For a normal distribution (symmetrical), a value that is two standard deviations below the mean would be closer to which of the following? Select one: a. 89th percentile b. Third Quartile c. Third percentile d. Median
### Algebra 2
Find the values of the 30th and 90th percentiles of the data 18, 9, 7, 5, 11, 7, 17, 20, 19, 2, 17, 12, 5, 1, 13, 12, 11, 15, 16, 20 * 30th percentile = 9 90th percentile = 20 * 30th percentile = 9 90th percentile = 19 * 30th percentile = 11 90th percentile = 19 * 30th ...
### Math
What is the approximate percentile rank of a lightbulb that failed at 9000 hours if the number of lightbulbs is 400? The number of lightbulbs that lasted between 5400 and 9000 were about 326 lightbulbs. The mean is 7800 and the standard deviation is 1200 My guess was: ...
### statistic
rank the following in increasing order and find the position and value of the 11th percentile show all work. 7 5 3 2 9 6 4 3 8 5 5 8
### Statistics
When calculating percentiles, ie the 33rd percentile, do you use the lower limit of the group that contains the 33rd percentile? as in this case: 70-79 11 46 80.7 60-69 9 35 61.4 50-59 8 26 45.61 40-49 7 18 31.58 Would you use the number 49.5??
### statistics
what is the percentile for the data value. In a data set with a rane of 60.07 to 115 and 300 observations, there are 207 observations with values less thsn 85.4. Find the percentile for 85.4.
Twelve college students were surveyed about the distance from their house to the college they attended. 12 18 54 13 15 24 8 5 14 27 1 20 1. Using the data set above, find the percentile ranking for the data point 24. (is this 75th percentile? 2. Using the data set above, what ...
### statistics
SAT I scores around the nation tend to have a mean scale score around 500, a standard deviation of about 100 points, and are approximately normally distributed. What SAT I score within the population would have a percentile rank of approximately 97.5? Show all work as to how ...
### Algebra
On a standardized test, the distribution of scores is normal, the mean of the scores is 75, and the standard deviation is 5.8. If a student scored 83, the student's score ranks 1. below the 75th percentile 2. above the 97th percentile 3. between the 75th percentile and the ...
### statistics
Women’s heights are normally distributed with a mean of 162 cm and standard deviation of 16 cm. a. Define an random variable, X, and describe its full distribution including the mean and variance. b. What percentage of heights are greater than 180 cm ? c. What height is at ...
### college statistics
a set of data is normally distributed with a mean of 1000 and a standard deviation of 100.what would be standard score for a score of 900.what percent of scores is betyween 1000 and 900.what would be percentile rank for a score of 900.
### statistics-normal distribution
a set of data is normally distributed with a mean of 500 and a standard deviation of 100. a) what would the standard score fora score of 700 be? b) what % of scores is between 500 and 700? c) what would the percentile rank for a score of 700?
### statistics
set of data is normally distributed with a mean of 200 and standard deviation of 50. · What would be the standard score for a score of 300? · What percentage of scores is between 200 and 300? · What would be the percentile rank for a score of 300? (Points : 6)
### Stats
Rank the following data in increasing order and find the position and value of the 56th percentile. Please show all of your work. 0 1 6 8 9 9 8 6 3 1 0 9
### satistics
Rank the following data in increasing order and find the position and value of the 65th percentile. Please show all of your work. 2 1 6 3 4 7 0 0 0 7 5 8
### Math
Of 20 scores, 19 are less than or equal to 10. Find the percentile rank of 10. They don't give you the exact number of tens so i don't really know how to do this? Could someone please explain?
### statistics
2. The results of a recent survey indicate that the average new car costs \$23,000, with a standard deviation of \$3,500. The price of cars is normally distributed. a. What is a Z score for a car with a price of \$ 33,000? b. What is a Z score for a car with a price of \$30,000? c...
### Research and Statistics
2. The results of a recent survey indicate that the average new car costs \$23,000, with a standard deviation of \$3,500. The price of cars is normally distributed. a. What is a Z score for a car with a price of \$ 33,000? 9.415000 b. What is a Z score for a car with a price of \$...
### Math
Suppose there are 400 students in your school class. What class rank is the 20th percentile?
### math
supposed that there are 440 students in class. What is the class rank of the student at the 20th percentile
### Statistics
Weights of male mountain lions follow the normal distribution with a median of 150 lb and an interquartile range of 8.2 lb. Find the 75th percentile of the weights. Find the 95th percentile of the weights.
### Statistics
Kyle averages 5 finished accounts a day. His standard deviation is 1 account per day and we want the percentile for a day where he finishes 8 accounts. The percentile will be between…
### Algebra
of 10 test scores, six are less than or equal to 80. what is the percentile rank of a test score of 80. please tell me how you get it?
### Statistics
Suppose the correlation between SAT Verbal scores and Math scores is 0.57 and that these scores are normally distributed. If a student's Verbals core places her at the 90th percentile, at what percentile would you predict her Math score to be/
### statistics
Find 10-th, 60-th and 75-th percentiles in each of your data set, then write down each value that corresponds to those percentiles. Now use those values to find sample percentiles empirically (Suppose value “a” corresponds to 10-th percentile in my data set. After finding ...
### Statistics
Given that a population of scores is normally distriibuted with u=100 and o=8, determine the following: a. The percentile rank of a score of 120 b. The perentage of scores that are below a score of 99 c. The percentage of scores that are between a score of 101 and 122 d. The ...
### Algebra
You are one of the finalists at a science fair. The scores of the other finalists are 87, 89, 81, 85, 87, 83, 86, 94, 90, 97, 80, 89, 85, and 88. Write an inequality that represents your possible scores if your percentile rank is 80.
### Algebra
You are one of the finalists at a science fair. The scores of the other finalists are 87, 89, 81, 85, 87, 83, 86, 94, 90, 97, 80, 89, 85, and 88. Write an inequality that represents your possible scores if your percentile rank is 80.
### Statistics
suppose the correla tion between SAT Verbal scores and Math scores is .57 and that these scares are normally distributed. IF a student's Verbal score places her at the 90th percentile, at what percentile would you expect her Math scores to be?
### statistics
indicate whether the given statement could apply to a data set consisting of 1,000 values that are all different. a. the 29th percentile is greater than the 30th percentile. b. the median is greater than the quartile c. the third quartile is greater than the first quartile d. ...
### math
A college student takes a standardized test and scores a 163. If the mean is 155 and the standard deviation is 7, what is the student’s percentile rank? (Assume a normal distribution.)
### statistics
sat scores around the nation tend to have a mean score of about 500, a standard deviation of 100 points and are approximately normal distribution. a person who score 600 on the sat has approximately what percentile rank within the population? show all calculations. 600-500/100...
a set of data is normally distributed with a mean of 500 and a standard deviation of 100. a) what would the standard score fora score of 700 be? b) what % of scores is between 500 and 700? c) what would the percentile rank for a score of 700 *please show and answer all ...
### statistics
The outcome of a standardized test is an integer between 151 and 200, inclusive. The percentiles of 400 test scores are calculated, and the scores are divided into corresponding percentile groups. Quantity A Minimum number of integers between 151 and 200, inclusive, that ...
### Can comeone check my answer for me. Many Thanks!
A set of data is normally distributed with a mean of 1000 and standard deviation of 100. · What would be the standard score for a score of 1100? · What percentage of scores is between 1000 and 1100? · What would be the percentile rank for a score of 1100? A. z(1100)= (1100-...
### statistics
sat scores around the nation tend to have a mean score around 500, a standard deviation of about 100 points and are approximately normal distribution. a person who scores 600 on the sat has approximately what percentile rank within the population. show all calculations. * ...
### ap stats
based on over 800 housing prices collected at random from the city of Los Angeles, an investigator constructed a histogram and found that the distribution of housing prices was heavily skewed to the right. Which of the following statistics will best represent the center of ...
### Statistics
3. Indicate whether the given statement could apply to a data set consisting of 1,000 values that are all different. a. The 29th percentile is greater than the 30th percentile. b. The median is greater than the first quartile. c. The third quartile is greater than the first ...
### Statistics
3. Indicate whether the given statement could apply to a data set consisting of 1,000 values that are all different. a. The 29th percentile is greater than the 30th percentile. b. The median is greater than the first quartile. c. The third quartile is greater than the first ...
### statistics
3. Indicate whether the given statement could apply to a data set consisting of 1,000 values that are all different. a. The 29th percentile is greater than the 30th percentile. b. The median is greater than the first quartile. c. The third quartile is greater than the first ...
### Statistics
Indicate with Yes or No whether the given statement could apply to a data set consisting of 1,000 values that are all different. a.The 29th percentile is greater than the 30th percentile.b.The median is greater than the first quartile.c.The third quartile is greater than the ...
### statistics
Indicate whether the given statement could apply to a data set consisting of 1,000 values that are all different. a. The 29th percentile is greater than the 30th percentile. b. The median is greater than the first quartile. c. The third quartile is greater than the first ...
### statistics
Indicate whether the given statement could apply to a data set consisting of 1,000 values that are all different. a. The 29th percentile is greater than the 30th percentile. b. The median is greater than the first quartile. c. The third quartile is greater than the first ...
### statistics
In a competitive application process, two applicants were closely matched and the judges needed to compare their percentile ranking to see who should be offered the last position.Jeremiah ranked 405th in a class of 1,200 and Joshua ranked 12th in a class of 89. Based solely on...
### Statistics
An investment broker reports that yearly returns on common stocks are normally distributed with a mean of 12.4 percent and a standard deviation of 20.6 percent. (Round all k amounts to three decimal places. A negative sign should be used instead of the parentheses. For Z, use ...
### Statistics
What z-value corresponds to the 18th percentile?
### statistics
indicate whether the given statement could apply to a data set consisting of 1,000 vlaues that are all differnt. a. the 29th percentile is greater thatn the 30th percentile b. the median is greater thatn the first quartile c. the third quartile is greater than the first ...
### statistics
given that x~n(300,15), find the 70th percentile. | 5,566 | 21,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.925406 |
https://www.physicsforums.com/threads/geometric-optics-converging-lens.644981/ | 1,544,912,818,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00009.warc.gz | 991,504,781 | 12,964 | # Homework Help: Geometric Optics - Converging Lens
1. Oct 18, 2012
### PeachBanana
1. The problem statement, all variables and given/known data
A certain lens focuses light from an object 2.90m away as an image 46.9cm on the other side of the lens.
1. What type of lens is it? (Converging or Diverging?)
2. What is the focal length?
3. Is the image real or virtual?
2. Relevant equations
1/di + 1/do = 1/f
3. The attempt at a solution
1. I thought because the image formed on the other side of the lens the lens was diverging but my online homework assignment says otherwise (and so do lecture notes!)
2. 1 / 0.029 cm + 1 / 46.9 cm = 1 / f
3. Real (because it's a concave mirror, right?)
2. Oct 18, 2012
### tiny-tim
Hi PeachBanana!
no, converging means that as time goes forward, the rays get closer
light from an ordinary source always starts by diverging (it spreads out) … this lens focussed it
i] it's a lens
ii] the image is real if you can put a screen there and see it on the screen
that's why it's called "real"!
a diverging lens makes the rays look as if they came from behind the mirror … but if you put a screen there, you'll see nothing, because the rays never went through there
3. Oct 18, 2012
### PeachBanana
Thank you I understand much better. | 362 | 1,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2018-51 | latest | en | 0.9236 |
https://web2.0calc.com/questions/hard-algebra-help | 1,657,211,988,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00047.warc.gz | 643,383,231 | 5,169 | +0
hard algebra help
0
96
1
If x = 1 + x/(2 + x/(2 + x/(2 + x/(2 + ...)))), then what is x?
Oct 22, 2021 | 50 | 108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.708467 |
https://www.physicsforums.com/threads/newtons-forces-on-a-box.380242/ | 1,632,357,655,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00149.warc.gz | 900,110,393 | 14,663 | # Newton's forces on a box
Newton's forces on a box!!!!!!!!
## Homework Statement
Two 45 N forces and a 90 N force act on a hanging box as shown in the picture below.
Will the box experience acceleration?
1. Yes; upward.
2. No; it is balanced.
3. Unable to determine without the angle.
4. Yes; downward.
F= ma
## The Attempt at a Solution
Will the two forces at the top result in a +y direction force of 45 or 90, or is it something else.
#### Attachments
• physics.jpg
3.3 KB · Views: 488
## Homework Statement
Two 45 N forces and a 90 N force act on a hanging box as shown in the picture below.
Will the box experience acceleration?
1. Yes; upward.
2. No; it is balanced.
3. Unable to determine without the angle.
4. Yes; downward.
F= ma
## The Attempt at a Solution
Will the two forces at the top result in a +y direction force of 45 or 90, or is it something else.
You know that the Y-component of each of the 45 N forces is less than 45 N (since 45 N is the hypotenuse of the triangle that the components make) Therefore, both of them summed is less than 90 upward. If you have 90 downward and less than 90 upward, do you have an acceleration? Note also that the x-component of one 45 N force cancels out the other's due to symmetry.
thanks i get it! | 332 | 1,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-39 | latest | en | 0.917007 |
https://blog.msbstats.info/posts/2021-05-25-everything-about-anova/ | 1,726,241,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00788.warc.gz | 117,701,010 | 22,779 | # Everything You Always Wanted to Know About ANOVA*
statistics
R
regression
ANOVA
interactions
linear models
code
glm
Author
Mattan S. Ben-Shachar
Published
May 25, 2021
Analysis of variance (ANOVA) is a statistical procedure, developed by R. A. Fisher, used to analyze the relationship between a continuous outcome (dependent variable) and categorical predictors (independent variables). This procedure produces a linear model, which can be used to estimate the conditional and marginal means of the outcome; In the presence of multiple factorial predictors, the model will include all interaction terms between the factors.
The ANOVA is part of a wider family of statistical procedures that include ANCOVA (which incorporates continuous predictors) and Analysis of Deviance (which allow for non-continuous outcomes1). This family of procedures all produce an ANOVA table (or ANOVA-like table) which summarizes the relationship between the underlying model and the outcome by partitioning the variation in the outcome into components which can be uniquely attributable to different sources according to the law of total variance. Essentially, each of the model’s terms is represented in a line in the ANOVA table which answers the question how much of the variation in $$Y$$ can be attributed to the variation in $$X$$??2 Where applicable, each source of variance has an accompanying test statistic (often F), sometimes called the omnibus test, which indicates the significance of the variance attributable to that term, often accompanied by some measure of effect size.
In this post I will demonstrate the various types of ANOVA tables, how R does ANOVA (what the defaults are, and how to produce alternatives).
Along the way, I hope to illustrate the applicability of ANOVA tables to other types of models - besides the classical case of a maximal model (all main effects and interactions) with strictly categorical predictors and a continuous outcome.
Here are some assumptions I make about you, the reader, in this post:
1. You’re familiar with the ideas of multi-factor ANOVAs (what a main effect is, what interactions are…).
2. You know some R - how to fit a linear model, how to wrangle some data.
3. You are IID and normally distributed.
Let’s dive right in!
# ANOVAs in R
Toy Data
d <- tibble::tribble(
~id, ~group, ~X, ~Z, ~Rx, ~condition, ~Y,
1L, "Gb", "102", 1L, "Placebo", "Ca", "584.07",
2L, "Ga", "52", 1L, "Placebo", "Cb", "790.29",
3L, "Gb", "134", 2L, "Dose100", "Ca", "875.76",
4L, "Gb", "128", 3L, "Dose100", "Cb", "848.37",
5L, "Ga", "78", 1L, "Dose250", "Ca", "270.42",
6L, "Gb", "150", 2L, "Dose250", "Cb", "999.87",
7L, "Ga", "73", 1L, "Placebo", "Ca", "364.1",
8L, "Ga", "87", 7L, "Placebo", "Cb", "420.84",
9L, "Gb", "115", 6L, "Dose100", "Ca", "335.78",
10L, "Gb", "113", 4L, "Dose100", "Cb", "627",
11L, "Gc", "148", 3L, "Dose250", "Ca", "607.79",
12L, "Gc", "82", 3L, "Dose250", "Cb", "329.32",
13L, "Ga", "139", 1L, "Placebo", "Ca", "335.56",
14L, "Ga", "65", 2L, "Placebo", "Cb", "669.04",
15L, "Gb", "139", 1L, "Dose100", "Ca", "405.04",
16L, "Gc", "96", 1L, "Dose100", "Cb", "367.15",
17L, "Gb", "50", 5L, "Dose250", "Ca", "27.37",
18L, "Gc", "90", 2L, "Dose250", "Cb", "468.69",
19L, "Ga", "90", 2L, "Placebo", "Ca", "584.67",
20L, "Ga", "116", 2L, "Placebo", "Cb", "277.71",
21L, "Gb", "78", 2L, "Dose100", "Ca", "266.01",
22L, "Gb", "60", 1L, "Dose100", "Cb", "0.04",
23L, "Gc", "112", 4L, "Dose250", "Ca", "593.25",
24L, "Ga", "63", 4L, "Dose250", "Cb", "512.26",
25L, "Ga", "89", 1L, "Placebo", "Ca", "635.57",
26L, "Ga", "97", 2L, "Placebo", "Cb", "468.69",
27L, "Gc", "76", 3L, "Dose100", "Ca", "514.66",
28L, "Gb", "83", 1L, "Dose100", "Cb", "264.87",
29L, "Gc", "84", 4L, "Dose250", "Ca", "220.34",
30L, "Gb", "88", 1L, "Dose250", "Cb", "216.54"
)
d$id <- factor(d$id)
d$group <- factor(d$group)
d$Rx <- factor(d$Rx, levels = c("Placebo", "Dose100", "Dose250"))
d$condition <- factor(d$condition)
dplyr::glimpse(d)
#> Rows: 30
#> Columns: 7
#> $id <fct> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 1… #>$ group <fct> Gb, Ga, Gb, Gb, Ga, Gb, Ga, Ga, Gb, Gb, Gc, Gc, Ga, Ga, Gb, …
#> $X <dw_trnsf> 102, 52, 134, 128, 78, 150, 73, 87, 115, 113, 148, 82, … #>$ Z <int> 1, 1, 2, 3, 1, 2, 1, 7, 6, 4, 3, 3, 1, 2, 1, 1, 5, 2, 2, 2, …
#> $Rx <fct> Placebo, Placebo, Dose100, Dose100, Dose250, Dose250, Placeb… #>$ condition <fct> Ca, Cb, Ca, Cb, Ca, Cb, Ca, Cb, Ca, Cb, Ca, Cb, Ca, Cb, Ca, …
#> $Y <dw_trnsf> 584.07, 790.29, 875.76, 848.37, 270.42, 999.87, 364.10,… m <- lm(Y ~ group + X, data = d) This is a multiple regression model with a covariable X and a 3-level factor group. We can summarize the results in a coefficient table (aka a “regression” table): summary(m) #> #> Call: #> lm(formula = Y ~ group + X, data = d) #> #> Residuals: #> Min 1Q Median 3Q Max #> -372.51 -166.47 -53.67 134.57 451.16 #> #> Coefficients: #> Estimate Std. Error t value Pr(>|t|) #> (Intercept) 118.608 145.557 0.815 0.42256 #> groupGb -102.591 94.853 -1.082 0.28937 #> groupGc -92.384 107.302 -0.861 0.39712 #> X 4.241 1.504 2.820 0.00908 ** #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 218.8 on 26 degrees of freedom #> Multiple R-squared: 0.2383, Adjusted R-squared: 0.1504 #> F-statistic: 2.711 on 3 and 26 DF, p-value: 0.06556 Or we can produce an ANOVA table: anova(m) #> Analysis of Variance Table #> #> Response: Y #> Df Sum Sq Mean Sq F value Pr(>F) #> group 2 8783 4391 0.0918 0.912617 #> X 1 380471 380471 7.9503 0.009077 ** #> Residuals 26 1244265 47856 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 While the group term had 2 parameters in the coefficient table, it now has a single test with a Df of 2. This omnibus test can be thought of as representing the total significance of the two parameters combined! Note Note that the model being summarized here is neither completely factorial (X is a continuous covariable), nor maximal (the group:X interaction is missing)! By default R calculates type 1 sums of squares (SS) - these are also called sequential SS, because each term is attributed with a portion of the variation (represented by its SS) in $$Y$$ that has not yet been attributed to any of the PREVIOUS terms! Thus, in our example, the effect of X represents only what X explains on top of what group has already explained - the variance attributed to X is strictly the variance that can be uniquely attributed to X, controlling for group; the effect of group however does not represent its unique contribution to $$Y$$ ’s variance, but instead its total contribution. This means that although the following models have the same terms, they will produce different type 1 ANOVA tables because those terms are in a different order: anova(lm(Y ~ group + X, data = d)) #> Analysis of Variance Table #> #> Response: Y #> Df Sum Sq Mean Sq F value Pr(>F) #> group 2 8783 4391 0.0918 0.912617 #> X 1 380471 380471 7.9503 0.009077 ** #> Residuals 26 1244265 47856 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 anova(lm(Y ~ X + group, data = d)) #> Analysis of Variance Table #> #> Response: Y #> Df Sum Sq Mean Sq F value Pr(>F) #> X 1 325745 325745 6.8067 0.01486 * #> group 2 63509 31754 0.6635 0.52353 #> Residuals 26 1244265 47856 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 We can recreate the ANOVA table above by building a sequence of models, and comparing them (see (Judd, McClelland, & Ryan, 2017)[https://doi.org/10.4324/9781315744131]): m0 <- lm(Y ~ 1, data = d) # Intercept-only model m1 <- lm(Y ~ group, data = d) anova(m0, m1, m) #> Analysis of Variance Table #> #> Model 1: Y ~ 1 #> Model 2: Y ~ group #> Model 3: Y ~ group + X #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 29 1633519 #> 2 27 1624737 2 8783 0.0918 0.912617 #> 3 26 1244265 1 380471 7.9503 0.009077 ** #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 anova(m) # Same SS values #> Analysis of Variance Table #> #> Response: Y #> Df Sum Sq Mean Sq F value Pr(>F) #> group 2 8783 4391 0.0918 0.912617 #> X 1 380471 380471 7.9503 0.009077 ** #> Residuals 26 1244265 47856 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## Simultaneous Sum of Squares There is also type 2 SS - also called simultaneous SS, because each term is attributed with a portion of the variation in $$Y$$ that is not attributable to any of the other terms in the model - its unique contribution while controlling for the other terms. Type 2 SS can be obtained with the Anova() function from the {car} package: car::Anova(m, type = 2) #> Anova Table (Type II tests) #> #> Response: Y #> Sum Sq Df F value Pr(>F) #> group 63509 2 0.6635 0.523533 #> X 380471 1 7.9503 0.009077 ** #> Residuals 1244265 26 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 We can recreate the above ANOVA table by building two sequences of models, and comparing them: m_sans_X <- lm(Y ~ group, data = d) m_sans_group <- lm(Y ~ X, data = d) anova(m_sans_group, m) # Same SS as the type 2 test for group #> Analysis of Variance Table #> #> Model 1: Y ~ X #> Model 2: Y ~ group + X #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 28 1307774 #> 2 26 1244265 2 63509 0.6635 0.5235 anova(m_sans_X, m) # Same SS as the type 2 test for X #> Analysis of Variance Table #> #> Model 1: Y ~ group #> Model 2: Y ~ group + X #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 27 1624737 #> 2 26 1244265 1 380471 7.9503 0.009077 ** #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Because the order of terms is usually of little importance, type 1 tests are rarely used in practice… Unfortunately, they are R’s default… ## Adding Interactions Things get a bit more complicated when interactions are involved, as type 2 SS treat interactions differently than main effects: m_int <- lm(Y ~ group * X, data = d) Each main effect term is attributed with variance (its SS) that is unique to it and that is not attributable to any of the other main effects (simultaneously, as we’ve already seen) but without accounting for the variance attributable to interactions, while the SS of the interaction term represents its unique variance after accounting for the underlying main effects (sequentially). So we get the unique contribution of each main effect when controlling only for the other main effects, and the unique contribution of the interactions controlling for the already-included combined contribution of the main effects. car::Anova(m_int, type = 2) #> Anova Table (Type II tests) #> #> Response: Y #> Sum Sq Df F value Pr(>F) #> group 63509 2 1.3555 0.2768607 #> X 380471 1 16.2410 0.0004884 *** #> group:X 682026 2 14.5566 7.246e-05 *** #> Residuals 562240 24 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 We can again recreate this type 2 ANOVA table with model comparisons3: anova(m_sans_group, m) # Same SS as the type 2 test for group #> Analysis of Variance Table #> #> Model 1: Y ~ X #> Model 2: Y ~ group + X #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 28 1307774 #> 2 26 1244265 2 63509 0.6635 0.5235 anova(m_sans_X, m) # Same SS as the type 2 test for X #> Analysis of Variance Table #> #> Model 1: Y ~ group #> Model 2: Y ~ group + X #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 27 1624737 #> 2 26 1244265 1 380471 7.9503 0.009077 ** #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 anova(m, m_int) # Same SS as the type 2 test for group:X #> Analysis of Variance Table #> #> Model 1: Y ~ group + X #> Model 2: Y ~ group * X #> Res.Df RSS Df Sum of Sq F Pr(>F) #> 1 26 1244265 #> 2 24 562240 2 682026 14.557 7.246e-05 *** #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 In designs of higher order, each “order” is tested in a similar simultaneous-sequential manner. E.g., in a 3-way design, all main effects (1st order) are tested simultaneously (accounting for one another), then all 2-way interactions (2nd order) are tested simultaneously (accounting for the main effects and one another), and then the 3-way interaction is tested (accounting for all main effects and 2-way interactions). ### Type 3 There is another type of simultaneous SS - the type 3 test, which treats interactions and main effects equally: the SS for each main effect or interaction is calculated as its unique contribution that is not attributable to any of the other effects in the model - main effects or interactions. So the effect of X is its unique contribution while controlling both for group and for group:X! However, remember how we previously saw that these methods in R actually produce omnibus tests for the combined effect of the parameters of each term. But in the m_int model the parameters labeled X, groupGb, and groupGc no longer represent parameters of the main effects - instead they are parameters of simple (i.e., conditional) effects! summary(m_int) #> #> Call: #> lm(formula = Y ~ group * X, data = d) #> #> Residuals: #> Min 1Q Median 3Q Max #> -370.18 -67.15 33.68 111.35 172.14 #> #> Coefficients: #> Estimate Std. Error t value Pr(>|t|) #> (Intercept) 833.130 173.749 4.795 6.99e-05 *** #> groupGb -1308.898 233.218 -5.612 8.90e-06 *** #> groupGc -752.718 307.747 -2.446 0.0222 * #> X -4.041 1.942 -2.081 0.0482 * #> groupGb:X 13.041 2.419 5.390 1.55e-05 *** #> groupGc:X 7.731 3.178 2.432 0.0228 * #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 #> #> Residual standard error: 153.1 on 24 degrees of freedom #> Multiple R-squared: 0.6558, Adjusted R-squared: 0.5841 #> F-statistic: 9.146 on 5 and 24 DF, p-value: 5.652e-05 • X is the slope of X when group=Ga (When both dummy variables are fixed at 0, as Ga is the reference level) • groupGb is the difference between group=Ga and group=Gb when X=0 • groupGc is the difference between group=Ga and group=Gc when X=0 (Pay attention to the “when” - this is what makes them conditional.) We can see that changing the reference group changes the test for X: d$group <- relevel(d$group, ref = "Gb") m_int2 <- lm(Y ~ group * X, data = d) car::Anova(m_int, type = 3) #> Anova Table (Type III tests) #> #> Response: Y #> Sum Sq Df F value Pr(>F) #> (Intercept) 538630 1 22.9922 6.994e-05 *** #> group 738108 2 15.7536 4.269e-05 *** #> X 101495 1 4.3325 0.04823 * #> group:X 682026 2 14.5566 7.246e-05 *** #> Residuals 562240 24 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 car::Anova(m_int2, type = 3) #> Anova Table (Type III tests) #> #> Response: Y #> Sum Sq Df F value Pr(>F) #> (Intercept) 219106 1 9.3528 0.005402 ** #> group 738108 2 15.7536 4.269e-05 *** #> X 910646 1 38.8722 1.918e-06 *** #> group:X 682026 2 14.5566 7.246e-05 *** #> Residuals 562240 24 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 How can we resolve this? ### Centering By centering our predictors! Centering is transforming our data in such a way that 0 represents the overall mean. When this is done, conditioning on 0 is the same as conditioning on the overall mean = looking at the main effect! For covariables this is easy enough: d$X_c <- d$X - mean(d$X) # or scale(d$X, center = TRUE, scale = FALSE) But how do we center a factor?? The answer is - use some type of orthogonal coding, for example contr.sum() (effects coding). This makes the coefficients harder to interpret 4, but we’re not looking at those anyway! contrasts(d$group) <- contr.sum
Now when looking at type 3 tests, the main effects terms actually are main effects!
m_int3 <- lm(Y ~ group*X_c, data = d)
car::Anova(m_int3, type = 3)
#> Anova Table (Type III tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> (Intercept) 4743668 1 202.4902 3.401e-13 ***
#> group 19640 2 0.4192 0.66231
#> X_c 143772 1 6.1371 0.02067 *
#> group:X_c 682026 2 14.5566 7.246e-05 ***
#> Residuals 562240 24
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Remember: ABC - Always Be Centering (your predictors) - type 3 ANOVA tables make little sense without centering5!
Recreate the Type 3 ANOVA
Unfortunately, we can’t just build a model without an interaction term and use it to recreate the type 3 ANOVA. Instead, we need to actually build the model matrix (i.e., the design matrix), and drop the columns of each term in turn:
mm <- model.matrix(m_int2)
head(mm)
#> (Intercept) groupGa groupGc X groupGa:X groupGc:X
#> 1 1 0 0 102 0 0
#> 2 1 1 0 52 52 0
#> 3 1 0 0 134 0 0
#> 4 1 0 0 128 0 0
#> 5 1 1 0 78 78 0
#> 6 1 0 0 150 0 0
A type 3 test for a term, is equal to the comparison between a model without the parameters associated with that term and the full model:
m_sans_group <- lm(Y ~ mm[,-(2:3)], data = d)
m_sans_X <- lm(Y ~ mm[,-4], data = d)
m_sans_int <- lm(Y ~ mm[,-(5:6)], data = d)
anova(m_sans_group, m_int2) # Same SS as type 3 for group
#> Analysis of Variance Table
#>
#> Model 1: Y ~ mm[, -(2:3)]
#> Model 2: Y ~ group * X
#> Res.Df RSS Df Sum of Sq F Pr(>F)
#> 1 26 1300348
#> 2 24 562240 2 738108 15.754 4.269e-05 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(m_sans_X, m_int2) # Same SS as type 3 for X
#> Analysis of Variance Table
#>
#> Model 1: Y ~ mm[, -4]
#> Model 2: Y ~ group * X
#> Res.Df RSS Df Sum of Sq F Pr(>F)
#> 1 25 1472886
#> 2 24 562240 1 910646 38.872 1.918e-06 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
anova(m_sans_int, m_int2) # Same SS as type 3 for group:X
#> Analysis of Variance Table
#>
#> Model 1: Y ~ mm[, -(5:6)]
#> Model 2: Y ~ group * X
#> Res.Df RSS Df Sum of Sq F Pr(>F)
#> 1 26 1244265
#> 2 24 562240 2 682026 14.557 7.246e-05 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Compare to:
car::Anova(m_int2, type = 3)
#> Anova Table (Type III tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> (Intercept) 219106 1 9.3528 0.005402 **
#> group 738108 2 15.7536 4.269e-05 ***
#> X 910646 1 38.8722 1.918e-06 ***
#> group:X 682026 2 14.5566 7.246e-05 ***
#> Residuals 562240 24
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Okay, that’s some ugly stuff. Let us never look at that again.
### Type 2 vs Type 3
As mentioned above, the distinction between types 2 and 3 comes from how they estimate main effects in the presence of interactions.
Let’s look at the following factorial design:
m_factorial <- lm(Y ~ condition * group, data = d,
# Another way to specify effects coding:
contrasts = list(condition = contr.sum,
group = contr.sum))
car::Anova(m_factorial, type = 2)
#> Anova Table (Type II tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> condition 12048 1 0.1840 0.6718
#> group 7165 2 0.0547 0.9469
#> condition:group 41204 2 0.3146 0.7330
#> Residuals 1571485 24
car::Anova(m_factorial, type = 3)
#> Anova Table (Type III tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> (Intercept) 5858845 1 89.4773 1.439e-09 ***
#> condition 3452 1 0.0527 0.8203
#> group 8913 2 0.0681 0.9344
#> condition:group 41204 2 0.3146 0.7330
#> Residuals 1571485 24
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
But where do these differences between type 2 and 3 come from?
Type 2 SS looks at the SS between the means of A, across the levels of B. So the marginal mean of the first group is estimated as:
$$\bar{Y}_{1.} = \frac{\sum{Y_{1.}}}{N_{1.}} = \frac{\sum{Y_{11}}+\sum{Y_{12}}}{N_{11}+N_{12}}$$
Type 3 SS however looks at the SS between the means of group, weighted by condition. So the marginal mean of group a is estimated as:
$$\bar{Y}_{1.} = \frac{\frac{\sum{Y_{11}}}{N_{11}} + \frac{\sum{Y_{12}}}{N_{12}}}{2}$$
This makes type 3 SS invariant to the cell frequencies!
But as we will soon see, this need not always be the case…
A lot has been said about type 2 vs type 3. I will not go into the weeds here, but it is important to note that
1. Most statistical softwares (SAS, Stata, SPSS, …) default to type 3 SS with orthogonal factor coding (but covariables are not mean-centered in most cases by default) (see Langsrud, 2003). This makes R inconsistent as we’ve seen it defaults to type 1 ANOVA and treatment coding.
2. Often in factorial designs, any imbalance in the design is incidental, so it is often beneficial to have a method that is invariant to such imbalances. (Though this may not be true if the data is observational.)
3. Coefficient tables give results that are analogous to type 3 SS when all terms are covariables:
m_covs <- lm(Y ~ X * Z, data = d)
car::Anova(m_covs, type = 2)
#> Anova Table (Type II tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> X 324676 1 6.7657 0.01513 *
#> Z 2430 1 0.0506 0.82373
#> X:Z 57640 1 1.2011 0.28315
#> Residuals 1247704 26
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
car::Anova(m_covs, type = 3)
#> Anova Table (Type III tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> (Intercept) 83130 1 1.7323 0.1996
#> X 5866 1 0.1222 0.7294
#> Z 59917 1 1.2486 0.2740
#> X:Z 57640 1 1.2011 0.2831
#> Residuals 1247704 26
summary(m_covs) # same p-values as type 3
#>
#> Call:
#> lm(formula = Y ~ X * Z, data = d)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -384.97 -153.72 -23.98 141.28 422.79
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 366.352 278.348 1.316 0.200
#> X 1.014 2.900 0.350 0.729
#> Z -112.681 100.843 -1.117 0.274
#> X:Z 1.175 1.072 1.096 0.283
#>
#> Residual standard error: 219.1 on 26 degrees of freedom
#> Multiple R-squared: 0.2362, Adjusted R-squared: 0.1481
#> F-statistic: 2.68 on 3 and 26 DF, p-value: 0.06772
Note
Note once again that the model being summarized here is not factorial at all - both Z and X are continuous covariables!
## Balanced vs. Unbalanced Data
The distinction between types 1, 2 and 3 SS is only relevant when there is some dependency between predictors (aka some collinearity). In our example, we can see that group and X are somewhat co-linear (VIF / tolerance are not strictly 1):
performance::check_collinearity(m)
#> # Check for Multicollinearity
#>
#> Low Correlation
#>
#> Term VIF VIF 95% CI Increased SE Tolerance Tolerance 95% CI
#> group 1.08 [1.00, 5.73] 1.04 0.92 [0.17, 1.00]
#> X 1.08 [1.00, 5.73] 1.04 0.92 [0.17, 1.00]
In a factorial design, we might call this dependence / collinearity among our predictors an “unbalanced design” (the number of observations differs between cells), and when the predictors are completely independent we would call this a “balanced design” (equal number of observations in all cells).
Let’s look at two examples:
### Balanced data
We can see that Rx and condition are balanced:
table(d$Rx, d$condition)
#>
#> Ca Cb
#> Placebo 5 5
#> Dose100 5 5
#> Dose250 5 5
chisq.test(d$Rx, d$condition)$statistic # Chisq is exactly 0 #> X-squared #> 0 And so type 1, 2 and 3 ANOVA tables are identical: contrasts(d$Rx) <- contr.sum
contrasts(d$condition) <- contr.sum m_balanced <- lm(Y ~ condition * Rx, data = d) anova(m_balanced) #> Analysis of Variance Table #> #> Response: Y #> Df Sum Sq Mean Sq F value Pr(>F) #> condition 1 13666 13666 0.2162 0.6461 #> Rx 2 41379 20690 0.3273 0.7240 #> condition:Rx 2 61444 30722 0.4860 0.6210 #> Residuals 24 1517030 63210 car::Anova(m_balanced, type = 2) #> Anova Table (Type II tests) #> #> Response: Y #> Sum Sq Df F value Pr(>F) #> condition 13666 1 0.2162 0.6461 #> Rx 41379 2 0.3273 0.7240 #> condition:Rx 61444 2 0.4860 0.6210 #> Residuals 1517030 24 car::Anova(m_balanced, type = 3) #> Anova Table (Type III tests) #> #> Response: Y #> Sum Sq Df F value Pr(>F) #> (Intercept) 6422803 1 101.6112 4.204e-10 *** #> condition 13666 1 0.2162 0.6461 #> Rx 41379 2 0.3273 0.7240 #> condition:Rx 61444 2 0.4860 0.6210 #> Residuals 1517030 24 #> --- #> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ### Unbalanced data However, condition and group are NOT balanced: table(d$group, d$condition) #> #> Ca Cb #> Gb 6 6 #> Ga 5 6 #> Gc 4 3 chisq.test(d$group, d$condition)$statistic # Chisq is NOT 0
#> Warning in chisq.test(d$group, d$condition): Chi-squared approximation may be
#> incorrect
#> X-squared
#> 0.2337662
And so type 1, 2 and type 3 ANOVA tables are NOT identical (recall how type 2 and 3 estimate marginal means differently in the presence of interactions):
anova(m_factorial)
#> Analysis of Variance Table
#>
#> Response: Y
#> Df Sum Sq Mean Sq F value Pr(>F)
#> condition 1 13666 13666 0.2087 0.6519
#> group 2 7165 3582 0.0547 0.9469
#> condition:group 2 41204 20602 0.3146 0.7330
#> Residuals 24 1571485 65479
car::Anova(m_factorial, type = 2)
#> Anova Table (Type II tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> condition 12048 1 0.1840 0.6718
#> group 7165 2 0.0547 0.9469
#> condition:group 41204 2 0.3146 0.7330
#> Residuals 1571485 24
car::Anova(m_factorial, type = 3)
#> Anova Table (Type III tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> (Intercept) 5858845 1 89.4773 1.439e-09 ***
#> condition 3452 1 0.0527 0.8203
#> group 8913 2 0.0681 0.9344
#> condition:group 41204 2 0.3146 0.7330
#> Residuals 1571485 24
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
As the deviation from perfect balance (i.e. independence) is larger, so will the differences between the types increase.
### Why Does This Happen?
We’ve seen that types 1, 2 and 3 all attribute variance in $$Y$$ to the model’s terms by partialling out the model’s other terms in some way - sequentially, simultaneously, or some mix of both.
However, if the predictors in the model are independent (such as in a balanced design, zero collinearity), then regardless of the order of their inclusion in the model, all of the variance that can be attributable to some term A is unique to A - none of it is also attributable to any of the other term in the model, and vice versa - there is nothing to partial out, so the order does not matter!
This also means that because the SS returned by both type 2 and 3 ANOVA tables represent the terms’ uniquely attributable variation in $$Y$$, then when the design is not balanced / there is collinearity in the data, the SS in the ANOVA table will not sum to the total SS - as there is some overlap (some non-unique variation) that is not represented in the ANOVA table. However… this is where type 1 ANOVA tables shine, as their sequential nature means the SS in the ANOVA table will sum to the total SS!
Let’s look at an extreme example of collinearity:
Collinear data
d2 <- MASS::mvrnorm(
n = 100,
mu = rep(0, 3),
Sigma = matrix(c(1, 0.99, 0.4,
0.99, 1, 0.41,
0.4, 0.41, 1), nrow = 3)
)
d2 <- data.frame(d2)
colnames(d2) <- c("X", "Z", "Y")
m_collinear <- lm(Y ~ X + Z, data = d2)
performance::check_collinearity(m_collinear)
#> # Check for Multicollinearity
#>
#> High Correlation
#>
#> Term VIF VIF 95% CI Increased SE Tolerance Tolerance 95% CI
#> X 47.02 [32.64, 67.95] 6.86 0.02 [0.01, 0.03]
#> Z 47.02 [32.64, 67.95] 6.86 0.02 [0.01, 0.03]
Looking a type 1 ANOVA table we can see that X accounts for a significant amount of variation in Y, but that Z does not add anything significant on top of X:
anova(m_collinear)
#> Analysis of Variance Table
#>
#> Response: Y
#> Df Sum Sq Mean Sq F value Pr(>F)
#> X 1 16.927 16.9265 22.664 6.743e-06 ***
#> Z 1 0.000 0.0000 0.000 0.9999
#> Residuals 97 72.444 0.7468
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
However, were we to look at a type 2 ANOVA table, we might get the impression that neither X nor Z contribute to the model:
car::Anova(m_collinear, type = 2)
#> Anova Table (Type II tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> X 0.360 1 0.4822 0.4891
#> Z 0.000 1 0.0000 0.9999
#> Residuals 72.444 97
This demonstrates the importance of always interpreting type 2 and 3 ANOVA tables in light of any collinearity that might exist between your predictors; Remember: ABC - Always Be mindful of Collinearity (okay, that one was a bit of a stretch).
We can use the lm() -> car::Anova() method to conduct a proper ANOVA on a maximal factorial design. However, making sure that our factors are orthogonally coded is a pain in the @.
d$group <- factor(d$group)
d$condition <- factor(d$condition)
contrasts(d$group) <- contr.sum contrasts(d$condition) <- contr.sum
m_lm <- lm(Y ~ group * condition, d)
car::Anova(m_lm, type = 3)
#> Anova Table (Type III tests)
#>
#> Response: Y
#> Sum Sq Df F value Pr(>F)
#> (Intercept) 5858845 1 89.4773 1.439e-09 ***
#> group 8913 2 0.0681 0.9344
#> condition 3452 1 0.0527 0.8203
#> group:condition 41204 2 0.3146 0.7330
#> Residuals 1571485 24
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Thankfully, we have the afex package which turns all of that mess into something much more palatable:
afex::aov_car(Y ~ group * condition + Error(id), data = d)
#> Anova Table (Type 3 tests)
#>
#> Response: Y
#> Effect df MSE F ges p.value
#> 1 group 2, 24 65478.53 0.07 .006 .934
#> 2 condition 1, 24 65478.53 0.05 .002 .820
#> 3 group:condition 2, 24 65478.53 0.31 .026 .733
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '+' 0.1 ' ' 1
Much easier!
# Other Types of Models
So far we’ve seen how ANOVA tables are applied to non-factorial linear OLS models. However the idea of omnibus tests per-term can be extended to many other types of models. For models where SS cannot be calculated, analogous methods based on deviance or likelihood are used instead (read more in the car::Anova() docs). Here are some examples:
## GLMs
### Logistic
m_logistic <- glm(condition ~ group * X_c, data = d,
family = binomial())
car::Anova(m_logistic, type = 2)
#> Analysis of Deviance Table (Type II tests)
#>
#> Response: condition
#> LR Chisq Df Pr(>Chisq)
#> group 0.15679 2 0.9246
#> X_c 0.75134 1 0.3861
#> group:X_c 1.10225 2 0.5763
### Poisson
m_poisson <- glm(Z ~ group * X_c, data = d,
family = poisson())
car::Anova(m_poisson, type = 3)
#> Analysis of Deviance Table (Type III tests)
#>
#> Response: Z
#> LR Chisq Df Pr(>Chisq)
#> group 0.88074 2 0.6438
#> X_c 0.04370 1 0.8344
#> group:X_c 0.11357 2 0.9448
### Ordinal
m_ordinal <- ordinal::clm(group ~ X_c * condition, data = d)
anova(m_ordinal)
#> Type I Analysis of Deviance Table with Wald chi-square tests
#>
#> Df Chisq Pr(>Chisq)
#> X_c 1 0.4469 0.5038
#> condition 1 0.1584 0.6907
#> X_c:condition 1 0.6129 0.4337
Note
Note that all of the models summarized here with an ANOVA table do not have a continuous (conditionally normal) outcome, and not all of their predictors are categorical. We may have been inclined to summarize these models with a coefficient table, but it is equally valid to present an ANOVA table!
## (G)LMMs
m_mixed <- lmerTest::lmer(Y ~ X_c * group + (group | Z), data = d)
anova(m_mixed, type = 2)
#> Type II Analysis of Variance Table with Satterthwaite's method
#> Sum Sq Mean Sq NumDF DenDF F value Pr(>F)
#> X_c 290878 290878 1 23.645 14.4210 0.0008951 ***
#> group 7198 3599 2 9.520 0.1784 0.8393243
#> X_c:group 582071 291036 2 21.938 14.4288 0.0001001 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Also for GLMMs:
m_mixed2 <- lme4::glmer(condition ~ X_c * group + (1 | Z), data = d,
family = binomial())
car::Anova(m_mixed2, type = 3)
#> Analysis of Deviance Table (Type III Wald chisquare tests)
#>
#> Response: condition
#> Chisq Df Pr(>Chisq)
#> (Intercept) 0.0886 1 0.7660
#> X_c 1.1917 1 0.2750
#> group 0.0829 2 0.9594
#> X_c:group 0.9972 2 0.6074
# Concluding Remarks
Although we might be more inclined to summarize our model with an ANOVA table when our model contains categorical predictors, especially when interactions are involved, we’ve seen that ANOVA tables do not require any special data (factorial, balanced, normal outcome), and can be used as an alternative to a coefficient table.
Regardless of how we summarize our model - with a coefficient table or with an ANOVA table, using type 1, 2 or 3 SS, with orthogonal or treatment coding, with centered or uncentered covariables - our underlying model is equivalent - and will produce the same estimated simple effects, marginal means and contrasts. That is, the method we use to summarize our model will not have any bearing on whatever follow-up analysis we may wish to carry out (using emmeans of course! Check out the materials from my R course: Analysis of Factorial Designs).6
I hope you now have a fuller grasp of what goes on behind the scenes when producing ANOVA tables, how the different types of ANOVA tables work, when they should be used, and how to interpret their results. ANOVA tables are a powerful tool that can be applied not only to factorial data coupled with an OLS model, but also to a wide variety of (generalized) linear (mixed) regression models. | 12,373 | 34,458 | {"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": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-38 | latest | en | 0.917056 |
http://www.mathworks.com/matlabcentral/fileexchange/32671-clock-signal-jitter-simulation | 1,469,790,041,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830064.24/warc/CC-MAIN-20160723071030-00219-ip-10-185-27-174.ec2.internal.warc.gz | 549,368,733 | 8,357 | Code covered by the BSD License
### Highlights from Clock Signal Jitter Simulation
5.0
5.0 | 1 rating Rate this file 7 Downloads (last 30 days) File Size: 2.72 KB File ID: #32671 Version: 1.0
# Clock Signal Jitter Simulation
### Neil Forcier (view profile)
This function creates a user definable digital clock signal that you can add random jitter to.
File Information
Description
This function creates a user definable digital clock signal waveform with random jitter on it. This function can be used to generate waveform data that can be uploaded onto an arbitrary waveform generator (AWG) to simulate a real world digital clock with a known amount of jitter for noise immunity and other similar test applications. The random jitter is genrated using matlab's rand function. You can choose two different distributions for generating the random values: normal and uniform. For the normal distribution jitter Magnitude variable is used as the 3 sigma point which covers 99.7% of the area of the distribution. This funtion uses Matlab's built-in error function or erf() function. For information on the
erf() function http://www.mathworks.com/help/techdoc/ref/erf.html. For more information or questions on the technique and math used in this function to simulate real world jitter refer to my blog post:
http://gpete-neil.blogspot.com/2011/08/simulating-jitter-with-arbitrary.html or email me at: neil_forcier at agilent dot com. The following is a description of the input arguments and returned variables.
sRate: The sample rate of the waveform points. This value should be same
sample rate you plan to use with the AWG
sCount: the number of points needed in one clock cycle
cCycles: the number of clock cycles you want in the waveform. cCycles *
sCount gives you the number of points in the waveform
jMag: is the maximum value in seconds that the signal should vary +/- from
the ideal value. It is used to set the limits of the uniform distribution
and the 3 sigma point of the normal distribution. jMag must be less than
or equal to 80 percent of the pulse width or one half cycle period.
Otherwise it will be set to 80 percent.
dist: Used to specify the type of distribution to be used by the random()
function. There are two distributions normal and uniform.
-->to select normal distribution enter string 'normal' or 'norm'
-->to select uniform distribution enter string 'uniform'
-->the default distribution is normal
ampl: This is the returned array of digital clock waveform amplitude
points. In terms of a plot this would be the y axis points
time: This is the array of the timing points that correspond to the array
of amplitude points. Some AWGs use the timing points to set their sample
rate. Most AWGs do not require the timing points and the sample rate can
be set manually on the front panel or remotely with a command
MATLAB release MATLAB 7.10 (R2010a) | 647 | 2,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.809729 |
http://slideplayer.com/slide/4914104/ | 1,571,524,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00203.warc.gz | 155,007,562 | 26,500 | # Rectangular Waveguides
## Presentation on theme: "Rectangular Waveguides"— Presentation transcript:
Rectangular Waveguides
ECE 3317 Prof. David R. Jackson Spring 2013 Notes 20 Rectangular Waveguides
Rectangular Waveguide
b x y , Cross section We assume that the boundary is a perfect electric conductor (PEC). We analyze the problem to solve for Ez or Hz (all other fields come from these). TMz: Ez only TEz: Hz only
TMz Modes TMz (Helmholtz equation) (PEC walls) Guided-wave assumption:
TMz Modes (cont.) Define: We then have:
Note that kc is an unknown at this point. We then have: Dividing by the exp(-j kz z) term, we have: We solve the above equation by using the method of separation of variables. We assume:
TMz Modes (cont.) where Hence, Divide by XY : Hence This has the form
Both sides of the equation must be a constant! This has the form
TMz Modes (cont.) Denote General solution: Boundary conditions: (1)
(2)
TMz Modes (cont.) The second boundary condition results in
This gives us the following result: Hence Now we turn our attention to the Y (y) function.
TMz Modes (cont.) We have Hence Denote Then we have General solution:
TMz Modes (cont.) Boundary conditions: (3) (4)
Equation (4) gives us the following result:
TMz Modes (cont.) Hence Therefore, we have New notation:
The TMz field inside the waveguide thus has the following form:
TMz Modes (cont.) Recall that Hence
Therefore the solution for kc is given by Next, recall that Hence
TMz Modes (cont.) Summary of TMz Solution for (m,n) Mode (Hz = 0)
Note: If either m or n is zero, the entire field is zero.
TMz Modes (cont.) Cutoff frequency
Note: The number kc is the value of k for which the wavenumber kz is zero. We start with where Set
TMz Modes (cont.) Hence which gives us
The cutoff frequency fc of the TMm,n mode is then This may be written as
Using the separation of variables method again, we have where and
TEz Modes (cont.) y b , x a Boundary conditions: cross section
The result is This can be shown by using the following equations:
TEz Modes (cont.) Summary of TEz Solution for (m,n) Mode (Ez = 0)
Same formula for cutoff frequency as the TEz case! Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law:
Summary TMz TEz TMz TEz Same formula for both cases
Wavenumber General formula for the wavenumber TMz or TEz mode: with
Note: The (m,n) notation is suppressed here. Above cutoff: Recall: Hence
Wavenumber (cont.) Below cutoff: Hence Hence
Wavenumber (cont.) Recal that Hence we have
Wavenumber Plot General behavior of the wavenumber “Light line”
Dominant Mode The "dominant" mode is the one with the lowest cutoff frequency. Assume b < a a b x y , Cross section Lowest TMz mode: TM11 Lowest TEz mode: TE10 TEz TMz The dominant mode is the TE10 mode.
Dominant Mode (cont.) Formulas for the dominant TE10 mode
At the cutoff frequency:
Dominant Mode (cont.) What is the mode with the next highest cutoff frequency? a b x y , Cross section Assume b < a / 2 The next highest is the TE20 mode. useful operating region TE01 fc TE10 TE20
Dominant Mode (cont.) Fields of the dominant TE10 mode
Find the other fields from these equations:
Dominant Mode (cont.) E H b a
From these, we find the other fields to be: where x y a b E H
Dominant Mode (cont.) Wave impedance:
Note: This is the same formula as for a TEz plane wave! Define the wave impedance:
Example Standard X-band* waveguide (air-filled):
a = inches (2.286 cm) b = inches (1.016 cm) Note: b < a / 2. Find the single-mode operating frequency region. Use Hence, we have * X-band: from 8.0 to 12 GHz.
Example (cont.) Standard X-band* waveguide (air-filled):
a = inches (2.286 cm) b = inches (1.016 cm) Find the phase constant of the TE10 mode at 9.00 GHz. Find the attenuation in dB/m at 5.00 GHz Recall: At 9.00 [GHz]: k = [rad/m] At 9.00 GHz: = [rad/m] At 5.00 GHz: = [nepers/m]
Example (cont.) At 5.00 [GHz]: = 88.91 [nepers/m] Therefore,
A very rapid attenuation! Note: We could have also used
Guide Wavelength The guide wavelength g is the distance z that it takes for the wave to repeat itself. (This assumes that we are above the cutoff frequency.) From this we have Hence we have the result
Phase and Group Velocity
Recall that the phase velocity is given by Hence We then have For a hollow waveguide (cd = c): Hence: vp > c ! (This does not violate relativity.)
Phase and Group Velocity (cont.)
The group velocity is the velocity at which a pulse travels on a structure. The group velocity is given by (The derivation of this is omitted.) t A pulse consists of a "group" of frequencies (according to the Fourier transform). + - Vi (t) waveguiding system
Phase and Group Velocity (cont.)
If the phase velocity is a function of frequency, the pulse will be distorting as it travels down the system. Vi (t) + - waveguiding system A pulse will get distorted in a rectangular waveguide!
Phase and Group Velocity (cont.)
To calculate the group velocity for a waveguide, we use Hence we have We then have the following final result: For a hollow waveguide: vg < c
Phase and Group Velocity (cont.)
For a lossless transmission line or a lossless plane wave (TEMz waves): We then have (a constant) For a lossless transmission line or a lossless plane wave line there is no distortion, since the phase velocity is constant with frequency. Hence we have
Plane Wave Interpretation of Dominant Mode
Consider the electric field of the dominant TE10 mode: where Separating the terms, we have: #1 #2 This form is the sum of two plane waves.
Plane Wave Interpretation (cont.)
Picture (top view): x z a k1 k2 TE10 mode At the cutoff frequency, the angle is 90o. At high frequencies (well above cutoff) the angle approaches zero.
Plane Wave Interpretation (cont.)
Picture of two plane waves a k2 k1 z x Crests of waves g cd vp Waves cancel out The two plane waves add to give an electric field that is zero on the side walls of the waveguide (x = 0 and x = a).
Waveguide Modes in Transmission Lines
A transmission line normally operates in the TEMz mode, where the two conductors have equal and opposite currents. At high frequencies, waveguide modes can also propagate on transmission lines. This is undesirable, and limits the high-frequency range of operation for the transmission line.
Waveguide Modes in Transmission Lines (cont.)
Consider an ideal parallel-plate transmission line (neglect fringing) (This is an approximate model of a microstrip line.) x y w h E H PMC PEC Assume w > h Rectangular “slice of plane wave” (the region inside the waveguide) A “perfect magnetic conductor” (PMC) is assumed on the side walls.
Waveguide Modes in Transmission Lines (cont.)
Some comments about PMC A PMC is dual to a PEC: PEC PMC PEC E PMC H
Waveguide Modes in Transmission Lines (cont.)
From a separation of variables solution, we have the following results for the TMz and TEz waveguide modes: x y w h PMC PEC Assume w > h TMz TEz
Waveguide Modes in Transmission Lines (cont.)
The dominant mode is TE10 The cutoff frequency of the TE10 mode is
Waveguide Modes in Transmission Lines (cont.)
x y w h Example Microstrip line Assume: fc 32 [GHz]
Waveguide Modes in Transmission Lines (cont.)
Dominant waveguide mode in coax (derivation omitted) TE11 mode: Example: RG 142 coax a b | 1,881 | 7,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-43 | latest | en | 0.823806 |
https://www.doubtnut.com/qna/203455005 | 1,725,779,125,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00451.warc.gz | 711,178,925 | 32,598 | # A convex lens of refractive index μ1 is kept in a medium of refractive index (μ2) . A parallel beam of light is incident on the lens . Draw the path of the rays of light emerging from a lens . If μ1=μ2
Text Solution
Verified by Experts
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## A concave lens having a material of refractive index μ1 is kept in a medium of R.I. μ2, where μ1<μ2. A parallel beam of light is incident on the lens. The concave lens will act as
Aa diverging lens
Ba converging lens
Ca plane glass slab
Da plano concave lens
• Question 2 - Select One
## A double convex lens of glass of refraction index μ is immersed in a medium of refractive index μ1. If a parallel beam emerges undeviated through the lens, then
Aμ=μ1
Bμ=1μ1
Cμ>μ1
Dμ<μ1
• Question 3 - Select One
## If a convex lens of refractive index μl is placed in a medium of refractive index μm Such that μl>μm>μair then
AThe nature of the lens is diverging
BThe nature of the lens is converging
CThe nalure of the lens may be converging or diverging
DThe nature of the lens can't be determined
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NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.
Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 570 | 2,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.863225 |
https://books.google.com.jm/books?qtid=3c7e78c3&lr=&id=1mLiFAoz0DQC&sa=N&start=70 | 1,719,069,937,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00682.warc.gz | 115,850,227 | 6,402 | Books Books
Los números cardinales 0: zero 1: one 2: two 3: three 4: four 5: five 6: six 7: seven 8: eight 9: nine 10: ten 11: eleven 12: twelve 13: thirteen 14: fourteen 15: fifteen 16: sixteen 17: seventeen 18: eighteen 19: nineteen 20: twenty...
The Youth's Assistant in Theoretic and Practical Arithmetic: Designed for ... - Page 2
by Zadock Thompson - 1832 - 168 pages
## The National Arithmetic on the Inductive System: Combining the Analytic and ...
Benjamin Greenleaf - Arithmetic - 1850 - 368 pages
...Arithmetic are expressed by the ten following characters, which are called numeral figures ; viz. 1 (one), 2 (two), 3 (three), 4 (four), 5 (five), 6 (six), 7 (seven), 8 (eight), 9 (nine), 0 (cipher, or nothing). The first nine of these figures are called significant, as distinguished from...
## An Elementary and Practical Arithmetic
James B. Dodd - Arithmetic - 1850 - 276 pages
...nnmeral characters or figures. These Figures — sometimes called the digits of numbers — are 1 one, 2 two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight, 9 nine, and the 0 zero or cipher, which denotes nothing. The figures from 1 to 9 inclusive, are significant;...
## The Boy's Arithmetic
Charles Arnold - 1850 - 164 pages
...NUMERATION. Certain signs have been used to express numbers from one to nine. Thus, 1 we call one : 2 we call two : 3, three : 4, four : 5, five : 6, six : 7, seven : 8, eight : 9, nine : and 0 we call cipher, or nought. By these figures, differently placed, we can express any number...
## The scholar's guide to arithmetic; or, A complete exercise-book
John Bonnycastle - 1851 - 314 pages
...number. The characters used for this purpose are the ten numeral figures, or digits, 0 cipher, 1 one, 2 two, 3 three, 4 four, 5 five. 6 six, 7 seven, 8 eight, 9 nine; by which, either singly or conjointly, all numbers can be expressed. This is done by giving to each...
## The Practical Model Calculator: For the Engineer, Mechanic, Machinist ...
Oliver Byrne - Engineering - 1851 - 310 pages
...which were introduced into Europe by the Moors about eight or nine hundred years since : viz. 1 one, 2 two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight, 9 nine, 0 cipher or nothing. These characters or figures were formerly all called by the general name of Ciphers;...
## A Theoretical and Practical Arithmetic: Designed for Common Schools and ...
Daniel Leach - Arithmetic - 1851 - 280 pages
...integers. 7. In the computation of numbers, ten characters are employed, called figures ; thus : 1, one ; 2, two ; 3, three ; 4, four; 5, five ; 6, six; 7, seven; 8, eight ; 9, nine • 0, cipher. The first nine figures are called significant, because they have a given value assigned...
## Theoretical and Practical Arithmetic, Part 1
De Witt Clinton Benjamin - 1852 - 76 pages
...and reading numbers by figures. 4. What are the figures made use of to express numbers ? A. 1 (one), 2 (two), 3 (three), 4 (four), 5 (five), 6 (six), 7 (seven), 8 (eight), 9 (nine), 0 (cipher or nought). 6. How do figures increase in value from the right to the left ? A. In ten-fold...
## High School Arithemtic: Containing the Elementary and the Higher Principles ...
James B. Dodd - Arithmetic - 1852 - 410 pages
...numcril characters or figures. These Figures — sometimes called the digits of numbers — are 1 one, 2 two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eiyht, 9 nine, and the 0 zero or cipher, which denotes nothing. The figures from 1 to 9 inclusive,... | 1,027 | 3,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.86258 |
https://www.exceldemy.com/create-a-matrix-chart-in-excel/ | 1,726,755,269,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00807.warc.gz | 695,322,655 | 56,023 | # How to Create a Matrix Chart in Excel (2 Methods)
In the dataset below, we have records of the selling prices, cost prices, and profits of a company’s products.
Watch Video – Create a Matrix Chart in Excel
## Method 1 – Creating a Matrix Bubble Chart
### Step 1- Creating Additional New Data Ranges
• Add two columns: one containing the product names and the other containing the serial numbers of the products.
• Enter the product names in the first column.
• Add 3 extra columns (as we have 3 sets of values in the Selling Price, Cost Price, and Profit columns). Arrange the serial numbers in these columns in reverse order.
### Step 2 – Inserting a Bubble Chart to Create a Matrix Chart
• Select the range of values (C4:E8).
• Go to the Insert Tab >> Charts Group >> Insert Scatter (X, Y) or Bubble Chart Dropdown >> Bubble Option.
The following Bubble chart will be created.
• To rearrange the bubbles, select the chart and right-click on it.
• Choose the option Select Data from various options.
The Select Data Source dialog box will open up.
• Select the already created series Series1 and click on Remove.
• Click on Add to include a new series.
• The Edit Series wizard will pop up.
For Series X values, select the serial numbers of the Additional Range 1 of the Bubble sheet, and for Series Y values, select the serial numbers in the three columns of Product Orange of the Additional Range 2.
• The Series bubble size will be the selling price, cost price, and profit of the product Orange.
• Press OK.
We have added a new series, Series 1.
• Click Add to enter another series.
• For Series X values, select the serial numbers of Additional Range 1, and for Series Y values, select the serial numbers in the three columns of Product Apple of Additional Range 2.
• The Series bubble size will be the selling price, cost price, and profit of the product Apple.
• Press OK.
The new series Series2 will appear.
Complete all of the 5 series for the 5 products and press OK.
Then you will get the following Bubble chart.
### Step 3 – Removing Default Labels of Two Axes
• Select the labels on the X-axis and then right-click on them.
• Choose the option Format Axis.
The Format Axis pane will appear on the right side.
• Go to the Axis Options tab >> expand the Labels option >> click on the dropdown symbol of the Label Position box.
• Choose None.
The Label Position will be changed to None.
We have removed the labels of the X-axis. You can follow this process for the Y-axis.
We have discarded all of the default labels from the chart.
### Step 4 – Adding Two Extra Ranges for New Labels of Axes
• We have entered a 3-row and 3-column data range for the X-axis label. Where the first column contains serial numbers, the second column contains 0, and the last column is for the bubble width (0.001 or whatever you want).
• Create the Additional Range 4 for the labels of the Y-axis. The first column contains 0, the second column contains the serial numbers in reverse order, and the last column is for the bubble widths, which is 0.001.
### Step 5- Adding a New Series for Labels to Create a Matrix Chart
• To add the new 2 series to the chart Right-click on the chart and then choose the Select Data option.
• Click on Add in the Select Data Source dialog box.
The Edit Series wizard will pop up.
• For Series X values, select the first column of the Additional Range 3, and for Series Y values, select the second column and choose the third column for the Series bubble size.
• Press OK.
We have created the new series Series6, and now press Add to enter another series.
• In the Edit Series dialog box, for Series X values, select the first column of the Additional Range 4, and for Series Y values, select the second column and choose the third column for the Series bubble size.
• Press OK.
We have added Series7 for Y-axis labels.
### Step 6 – Adding New Labels
• Click on the chart and select the Chart Elements symbol.
• Check the Data Labels option.
All of the data labels will be visible on the chart.
• Select the labels of the X-axis and Right-click.
• Click on the Format Data Labels option.
The Format Data Labels pane will be visible on the right side.
• Go to the Label Options Tab >> expand the Label Options Option >> check the Value From Cells Option.
The Data Label Range dialog box will open up.
• Select the column headers of the values in the Select Data Label Range box and press OK.
• Uncheck the Y Value from the Label Options and scroll down to the downside to see all of the options of the Label Position.
• Select the Below option.
We will be able to add our desired X-axis labels.
• Select the Y-axis labels and then Right-click.
• Click on the Format Data Labels option.
The Format Data Labels pane will be visible on the right side.
• Uncheck the Y Value option and click the Value From Cells option among various Label Options.
The Data Label Range dialog box will open up.
• Select the product names in the Select Data Label Range box and press OK.
You will be taken to the Format Data Labels part again.
• Click on the Left option under the Label Position.
We will have the names of the products on the Y-axis labels.
### Step 7 – Adding Labels for Bubbles
• Select the bubbles with the number 5 and then Right-click.
• Choose the Format Data Labels option.
The Format Data Labels pane will open up in the right portion.
• Check the Bubble Size option and uncheck the Y Value option.
The labels of the bubbles will be converted into the values of the Selling Prices, Cost Prices, and Profits.
• You can remove the chart title by clicking on the Chart Elements symbol and then unchecking the Chart Title option.
The final outlook of the chart is shown in the following figure.
## Method 2 – Creating a 4-Quadrant Matrix Chart
### Step 1 – Inserting a Scatter Graph to Create a Matrix Chart
• Select the range of values (C4:D8) and go to the Insert Tab >> Charts Group >> Insert Scatter (X, Y) or Bubble Chart Dropdown >> Scatter Option.
The following graph will appear.
We have to set the upper and lower bound limits of the X-axis and Y-axis.
• Select the Y-axis label and Right-click.
• Choose the Format Axis option.
You will get the Format Axis pane on the right side.
• Go to the Axis Options Tab >> expand the Axis Options Option >> set the limit of the Minimum bound as 0.0 and the Maximum bound as 4000.0.\
We will have the modified X-axis labels with new limits and we don’t need to modify the Y-axis labels as the upper limit of this axis is here 4000 which is close to the maximum Cost Price of \$3,197.00.
### Step 2: Creating Additional Data Range
• Create the following data table format with two portions for the Horizontal and the Vertical and two columns for the coordinates X and Y.
• For the Horizontal part, add the following values in the X and Y coordinates.
X → 0 (minimum bound of X-axis) and 6000 (maximum bound of X-axis)
Y → 2000 (average of the minimum and maximum values of the Y-axis → (0+4000)/2 → 2000)
• For the Vertical part, add the following values in the X and Y coordinates.
X → 3000 (average of the minimum and maximum values of the X-axis → (0+6000)/2 → 3000)
Y → 0 (minimum bound of Y-axis) and 4000 (maximum bound of Y-axis)
### Step 3 – Adding Four Points in Graph to Create Quadrant Lines
• Select the graph, Right-click, and choose the Select Data option.
The Select Data Source wizard will open up.
The Edit Series dialog box will appear.
• For Series X values, select the X coordinates of the horizontal part of the Quadrant sheet, and then for Series Y values, select the Y coordinates of the horizontal part.
• Press OK.
The new series Series2 will be added.
• To insert a new series for the vertical line, click on Add again.
• In the Edit Series dialog box, for Series X values, select the X coordinates of the vertical part of the Quadrant sheet, and for Series Y values, select the Y coordinates of the vertical part.
• Press OK.
We have added the final series Series3.
• Press OK.
If the maximum bound of data in your chart changes, you can change the bounds by right-clicking > Format Axis option > change the maximum bound according to your need like we did for the Y axis. You can skip this step if you don’t need to change the scale.
We will have 2 Orange points indicating the horizontal part and 2 Ash points indicating the vertical part.
### Step 4 – Inserting Quadrant Lines to Create a Matrix Chart
• Select the 2 Orange points and then Right-click.
• Choose the Format Data Series option.
You will have the Format Data Series pane on the right portion.
• Go to the Fill & Line Tab >> expand the Line Option >> click on the Solid line option >> choose your desired color.
• To hide the points, go to the Fill & Line Tab >> expand the Marker Options Option >> click the None option.
The horizontal line will appear in the chart.
Create the vertical separator line using the 2 ash points.
### Step 5 – Inserting Data Labels
To indicate the data points with the name of the products we have to add the data label first.
• Select the data points and then click on the Chart Elements symbol.
• Check the Data Labels option.
The values of the points will appear beside them and we have to convert them to the name of the products.
• Right-click after selecting these data points.
• Click on the Format Data Labels option.
You will have the Format Data Labels pane on the right side.
• Check the Value From Cells option from the Label Options.
The Data Label Range dialog box will open up.
• Select the name of the products in the Select Data Label Range box and press OK.
The outlook of the Quadrant Matrix Chart will look like the following.
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Advanced Excel Exercises with Solutions PDF | 2,350 | 10,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.825377 |
http://www.chegg.com/homework-help/questions-and-answers/artistic-solutions-painting-company-determined-every-160-square-feet-wall-space-one-gallon-q1276923 | 1,435,845,298,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095557.73/warc/CC-MAIN-20150627031815-00276-ip-10-179-60-89.ec2.internal.warc.gz | 349,857,983 | 12,209 | Artistic Solutions, a painting company, has determined that for every 160 square feet of wall space, one gallon of paint and three hours of labor are required. the company charges \$28.00 per hour of labor. Design a modular program that allows the user to enter the number or rooms that are to be painted, the approximate square feet of wall space in each room(it may differ from room to room), and the price of the paint per gallon.It should then create a report that includes a fancy company header and displays the following information:
• The number of gallons of paint required(rounded up to the next full gallon)
• the hours of labor required
• the cost of the paint
• the labor charges
• the total cost of the pain job
Input validation: the program should not accept a value less than 1 or more than 12 for the number of rooms or a value less than 100 for the square footage of a room. It should not accept a value less than \$10.00 or more than \$25.00 for the price of a gallon of paint.
any help is appreciated,
thx | 229 | 1,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.917588 |
https://goprep.co/q4-if-a-a-b-c-and-b-2-1-0-1-2-write-total-number-of-one-one-i-1nlmtq | 1,611,538,061,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703561996.72/warc/CC-MAIN-20210124235054-20210125025054-00271.warc.gz | 362,557,826 | 60,013 | Q. 45.0( 2 Votes )
# If A = {a, b, c} and B = {–2, –1, 0, 1, 2}, write total number of one-one functions from A to B.
Formula:-
(I)A function is one-one function or an injection if
f(x)=f(y)
x=y for all x, y A
or f(x)f(y)
xy for all x, y A
(II)if A and B are two non-empty finite sets containing m and n
(i) Number of function from A to B = nm
(ii) Number of one–one function from A to B
(iii) Number of one-one and onto function from A to B
(iv) Number of onto function from A to B=
Let f: AB be one-one function
F(a)=3 and f(B)=5
Using formula
Number of one–one function from A to B
3C5.5!=60
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view all courses | 364 | 1,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-04 | longest | en | 0.780259 |
https://cracku.in/blog/geometry-questions-for-cat/ | 1,660,532,298,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00626.warc.gz | 197,708,386 | 37,388 | # Geometry Questions for CAT
1
10519
Geometry Questions for CAT:
You can download the Geometry questions or you can go through the details below.
Question 1:
A one rupee coin is placed on a table. The maximum number of similar one rupee coins which can be placed on the table, around it, with each one of them touching it and only two others is
A. 8
B. 6
C. 10
D. 4
Question 2:
Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?
A.
5
B.
10
C.
21
D.
15
Question 3:
Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° or concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be
A. 20
B. 0
C. 21
D. 22
Question 4:
A farmer has decided to build a wire fence along one straight side of this property. For this, he planned to place several fence-posts at six metre intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was five less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them eight metres apart. What is the length of the side of his property and how many posts did he buy?
A.
20
B.
0
C. 21
D.
22
Question 5:
In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?
A. 17.05
B. 27.85
C. 22.45
D. 26.25
If we join centres of 2 outer circles with the centre circle, it will make an equilateral triangle.
Hence at the centre it will make an angle of $$60^o$$ so total $$\frac{360^o}{60^o} = 6$$ triangles will be there
Hence, 6 outer circles will be there.
For obtuse-angles triangle, $$c^2 > a^2 + b^2$$ and c < a+b
If 15 is the greatest side, 8+x > 15 => x > 7 and $$225 > 64 + x^2$$ => $$x^2$$ < 161 => x <= 12
So, x = 8, 9, 10, 11, 12
If x is the greatest side, then 8 + 15 > x => x < 23
$$x^2 > 225 + 64 = 289$$ => x > 17
So, x = 18, 19, 20, 21, 22
So, the number of possibilities is 10
Let the total number of sides be x.
(25*90)+(x-25)*270 = (x-2)180
x = 46
Number concave corners = x-25 = 46-25 = 21
Let the number of posts the farmer bought = x
Length of the side = (x+5-1)*6 = 6*(x+4)
Length of the side is also equal to (x-1)*8
=> 8x – 8 = 6x + 24 => x = 16
Length of side = 6*20 = 120 m | 836 | 2,635 | {"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} | 3.84375 | 4 | CC-MAIN-2022-33 | longest | en | 0.932091 |
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A lot of the first timers that get in a class may not have much standard mathematics abilities. This is why it is best to use a calculator throughout an examination. A number of these will have various interactive alternatives which enable you to get a feel for the idea and after that carry on to practice the real product on the calculator.
Practice utilizing the different tools and basic rules on the calculator, and then study the issue on the calculator and the formula it produces. This is the best method to get a feel for using a calculator to resolve for a number. | 2,335 | 11,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | longest | en | 0.967855 |
https://www.poliscidata.com/blog/why-only-50-of-findings-can-be-replicated-exploratory-thoughts/ | 1,726,396,339,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651622.79/warc/CC-MAIN-20240915084859-20240915114859-00896.warc.gz | 855,621,565 | 12,349 | # Why Only 50% of Findings Can Be Replicated: Exploratory Thoughts
Thinking about the “replication crisis” in social sciences, I did some back-of-the-envelope type analysis of the probability that a statistically significant finding is true. The result I get is that there is only a 50% probability that a result is true, given a finding of statistical significance. Moreover, this 50% probability of being true is not affected by raising the threshold for statistical significance.
Imagine, for example, one is examining whether variable X1 causes variable Y to vary. X1 is one of many possible variables, X2, X3, X4, … that could cause variable Y to vary. To assess whether X1 is related to Y, one would apply some statistical analysis and conclude that X1 has a statistically significant relationship to Y is if the result of the analysis would occur .05 of the time of less by chance. This means that, in the universe of potential causes (all the Xs), .05 of them will be deemed statistically significant.
What, then, is the probability a finding is true (T) given the existence of a statistically significant relationship (S)? What we’re interested in is P(T | S).
I start with this basic probability statement:
P(T | S) * P(S) = P(S | T) * P(T)
which given us the Bayes rule:
P(T | S) = [P(S | T) * P(T)] / P(S)
The bottom half of the equation above needs to be expanded. The probability that a relationship will be deemed statistically significant, P(S), is equal to the sum of the probability of true positive findings, P(S | T) * P(T), and the probability of false positive findings, P(S | ~T) * P(~T). Incorporating this into the equation:
P(T | S) = [P(S | T) * P(T)] / [P(S | T) * P(T) + P(S | ~T) * P(~T)]
Based on the definition of statistical significance, .05 of X’s are considered true causes of Y, and .95 of X’s are not true causes of Y. Thus, P(T) = .05 and P(~T) = .95. The probability of a false positive, P(S | ~T), is the Type 1 error rate of .05. The true positive rate, P(S | T), is the power of the statistical test used, but it is relatively high, let’s say .95 of true relationships will be correctly identified as statistically significant. This gives us all the values needed to calculate P(T | S) using the formula above.
P(T | S) = [.95 * .05] / [.95 * .05 + .05 * .95]
= .0475 / (.0475 + .0475)
= .50
The probability that a finding is true given it is statistically significant, based on the foregoing, is .50. The ability is detect true relationships with statistical significance tests is fairly low, but this is consistent with the “replication crisis” literature which indicates that only about 50% of statistically significant research findings can be replicated.
Even if one raises the threshold for statistical significance, deeming relationships statistically significant if they only 1% of the time by chance, it’s easy to see that the value of P(T | S) remains .50.
P(T | S) = [.99 * .01] / [.99 * .01 + .01 * .99] = .50
This is my back-of-the-envelope analysis. I made some simplifying assumptions that are not entirely accurate, but I think accurate enough to generate an interesting result. Someone probably has developed this further in the literature on replication crisis and publication bias. To me, P(T) = .05 seems like the most questionable assumption because researchers should not be investigating relationships without some strong basis to believe a true relationship exists. | 818 | 3,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-38 | latest | en | 0.913773 |
http://hackaday.com/tag/7-segment/page/2/ | 1,412,073,958,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037662882.4/warc/CC-MAIN-20140930004102-00017-ip-10-234-18-248.ec2.internal.warc.gz | 144,705,367 | 25,500 | ## Drinking games and digital logic
For those of you who might have forgotten, let’s go over the rules of Centurion. The object of the game is for every minute, for 100 minutes, drink a shot of beer. It doesn’t sound like a lot, but after completing the challenge you’ll have had 3 liters of beer (or about eight and a half 12 oz cans) in just under two hours. When [Peter] played Centurion, he found the biggest problem was – understandably – keeping track of the time and who drank what. For an upcoming weekend of drinking, [Peter] decided to solve this problem once and for all with shift registers and seven-segment displays.
[Peter]‘s Centurion score box comes in two parts. The first and largest part of the build is the main board housing an ATMega8 microcontroller and a huge two digit seven-segment display to keep track of the countdown until the next shot. Two other boards house eight additional two digit seven-segment displays for each player, incremented every time a player presses a giant arcade button.
The entire build is designed around a small travel case that also holds a large battery for cordless drinking parties. Let’s just hope the project is reasonably water-resistant; we can see a lot of spills happening in the future. Check out the video demo below.
## Clock display taller than you is just what your living room has been missing
Sure, it’s time to get the countdown clocks ready to ring in the new year, but why limit it to just one night? If you end up building a six-foot digital display you can count down trivial events; like the remaining seconds of freedom before you have to pimp yourself out in that drab cubicle.
This seven-segment display is homemade and boasts six full-sized digits and two smaller digits with each pair separated by colons. You have probably already guessed that the construction was greatly simplified by using LED strips rather than individual components. This is part of the reason for the size of the display. The strips can be cut, but only down to a minimum of 3 LEDs per segment. That explains the small digits, with their larger siblings doubled in size. But there is a benefit to this constraint, it means that current limiting is already taken care of for you.
The main assembly is a wooden frame surrounding two polycarbonate sheets. The LED strips are sandwiched between those sheets, with segment and digit driver buses exiting a one point on the side. The build doesn’t detail a driver for the display but it shouldn’t be hard to find a multiplexing example that will serve the purpose.
## CMOS logic clock tracks 24-hour time
Here’s an IC logic project that displays 24-hour time. Planning was the name of the game for this project. [Mattosx] took the time to layout his design as a PCB in order to avoid the wiring nightmare when build with point-to-point connections.
Much of the complexity is caused by the display itself. Each of the six digits has its own binary-coded decimal chip and array of discrete resistors. Timekeeping is handled by six decade counters, two divider chips, one AND gate chip, and one OR gate chip. He chose a SOIC crystal oscillator chip as the clock signal. We’re more partial to the idea of using mains voltage as the clock signal.
[Mattosx] posted the board artwork if you’d like to etch your own 5″x8″ PCB. Just make sure you read through all of his notes as not all of the chips are oriented in the same direction.
[via Reddit]
## Puncher tracks your freelancing hours, time spent in TSA patdowns
[Raphael Abrams] does a lot of freelance work, but he has trouble accurately keeping track of the hours he has put in for his clients. After trying various applications and methods of logging his time, he finally decided to build a device that worked just the way he liked.
He calls his device the “Freelance Puncher”, though it already has been nicknamed the detonator, as it looks like something you would find in the hands of a [James Bond] villain. The device uses a PIC16LF1827 to track the time, saving his logged hours to the built-in EEPROM when powered off. A pair of 7-segment displays are used to display the accumulated hours upon power-on, and a set of seven SMT LEDs separated into two banks keep track of quarter and hundreds of hours worked.
[Raphael] has made his code and schematics available on Github, so you can easily replicate his work if you are looking for a better way to track your time. We think it looks great, though it could be the sort of thing that traveling freelancers might want to keep in their checked luggage, unless they want to spend some quality time with the TSA! Be sure to stick around to see a short video where [Raphael] shows off and explains how his Freelance Puncher works.
## A simple, self-contained 7-segment display
It’s no secret that seven-segment displays are an easy and useful way to relay data, so [Kelvyn Panici] decided to put together a minimalst, self-contained display for use around the house.
The display itself is a 16-digit model he picked up from DealExtreme for under \$10. He wanted to find a microcontroller small enough to fit behind the display’s footprint, so he chose an ATtiny85 to control it. After mounting the mcu on a small piece of perfboard, he burned the Arduino bootloader and uploaded a small sketch to drive the display.
Things worked out quite well as you can see by the video below where he shows off a pre-perfboard prototype. [Kelvyn] currently does not have any immediate projects in the works that will utilize the display, though there are a plethora of possibilities. We think it would work great anywhere if it were fitted with a battery and some sort of wireless radio in order to make it completely self-contained.
## Going RGB with 7 segment displays
We can order seven segment displays in red, green, yellow, or blue all day long. One thing we haven’t seen is an RGB segmented display, so [Markus]‘ project is really interesting. He took a stock seven segment display and modded it into an RGB display.
After taking a Dremel to the back of the stock display, [Markus] was left with a seven segment light mask. A few SMD LEDs were purchased through the usual channels. The RGB LEDs were epoxied into place on the back of the light mask one at a time. Thankfully, the LEDs came with magnet wire already attached – helpful, since these LEDs are only 1.6mm x 1.2mm big.
With 32 pieces of magnet wire, [Markus] needed some sort of socket. A small piece of perfboard and some .100″ headers handled the job very nicely. [Markus] still has to work on some way to drive the 24 cathode lines his LED display. He’d like an I2C interface, but with something like an individual seven segment display, the footprint of the circuit should be pretty small. If you’ve got any tips, drop them in the comments section. [Markus] is sure to catch them there.
## Keeping tabs on your pets’ busy lives
[Stephen’s] daughter has a pair of mice she keeps as pets, who happen to be quite active at night. After they kept her awake for an entire evening by running like mad in their treadmill, they were moved from her bedroom. Since they were so active in the treadmill, [Stephen] thought it would be cool to try measuring how much the mice actually ran each night.
To keep track of their activity, he built a simple circuit that records how many rotations the treadmill makes. He fitted it with a rare earth magnet, installing a reed switch on the outside of case that ticks off each spin of the wheel. Any time the wheel starts moving, his PIC begins counting the rotations, displaying them on a 7-segment LED display. To mitigate data loss in the event of a power outage, the PIC stores the current number of rotations in its EEPROM every 10 seconds or so.
The counter keeps track of the total number of rounds the mice have completed, which his daughter uses to manually calculate their running sessions. Since they started tracking the mice, they have run over 700,000 rounds, sometimes completing as many as 20,000 in an evening.
We think it’s a pretty cool project, especially since it makes it fun for his daughter to stay involved in her pets’ lives. | 1,789 | 8,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2014-41 | longest | en | 0.943195 |
http://gofigurewithscipi.blogspot.com/2012_11_01_archive.html | 1,429,951,513,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246648209.18/warc/CC-MAIN-20150417045728-00113-ip-10-235-10-82.ec2.internal.warc.gz | 111,356,049 | 28,689 | Problem Solving Top Ten List #2
A good process problem uses no set algorithm to find the solution. It requires a variety of processes (problem solving strategies) to find the solution. It is a problem that is easy to understand, is interesting, perhaps even whimsical, and has numbers sufficiently small enough so that lengthy computation is unnecessary.
Standard Word Problem: Jack's family plans to rent a camping trailer for vacation. The rent is \$22.50 a day. What will it cost to rent the camping trailer for one week?
A Problem that Requires Problem Solving: Drew and Addie are playing a game. At the end of each game, the loser gives the winner a chip. When they are done playing several games, Drew has won three games, but Addie has three more chips than she had when the game began. How many games did Drew and Addie play?
So what happens when your students try to do the process problem above and they have no idea what to do? In my last posting, I listed ten reasons why students get stuck when problem solving. Now let's consider why students get stuck in the first place.
Top Ten Reasons for Getting Stuck in the First Place:
1. You tried to rush through the problem without thinking.
2. You did not read the problem carefully.
3. You don't know what the problem is asking for.
4. You don't have enough information.
5. You are looking for an answer that the problem isn't asking for.
6. The strategy you are using doesn't work for this particular problem.
7. You are not applying or using your strategy correctly.
8. You failed to combine your strategy with another strategy.
9. The problem has more than one answer.
10. The problem cannot be solved.
Next time, we will look at the final Top Ten List entitled The Top Ten Worst Problem Solving Habits.
Since students today tend to be more visual than anything else, a graphic organizer becomes a valuable math tool. The Triangular Graphic Organizer is generic so that it can be used to solve all kinds of formula problems such as d = rt, A = lw, or c2 = a2 + b2. This five page handout explains in detail how to use the graphic organizer. It also contains several examples as well as a page of blank triangular graphic organizers to copy and use in your classroom.
Want the answer to the process problem?
Check out the page above entitled: Answers to Problems.
Problem Solving Top Ten List #1
The study of mathematics should emphasize problem solving so that students can use and apply a wide variety of strategies to investigate and understand mathematical content. In this way, they will acquire confidence in using mathematics meaningfully.
What is a real problem that requires problem solving? It presents a challenge which cannot be resolved by some routine procedure known to the student and where the student accepts the challenge! But what happens when students become frustrated and say they are stuck? Let's look at ten ways to get them unstuck.
The Top Ten Ways to Get Unstuck:
1. Re-read the problem.
2. Modify your strategy.
3. Change your strategy.
4. Combine your strategy with another strategy.
5. Look at the problem from a new perspective.
6. Look at the answer.
7. Look at other similar problems.
8. Ask for help.
9. Wait awhile and then try again.
10. All of the above.
Next week we will review the Top Ten Reasons for Getting Stuck in the First Place!
Putting the Pieces Together
What does it mean to think mathematically? It means using math vocabulary, language and symbols to describe or interpret mathematical concepts, procedures and to discover relationships among ideas. Therefore when a student problem solves, they use previous knowledge, skills, and understanding of concepts to solve a problem. This process might include formulating problems, applying a variety of strategies, or interpreting results.
What can we do to help our students become better mathematical thinkers? We can teach and model problem solving strategies such as the nine listed in the October 25th posting. We can remember and plan our lessons to involve the three stages of conceptual development: concrete, pictorial, and abstract. (Refer back to December 23, 2011 posting entitled Lesson Plans and Research.) We can have the students talk or write about how they got an answer either with the class or with a partner. We can use writing in the mathematics classroom (such as math journals) to allow the students to practice expository writing and show their understanding. We can exhibit math word walls and have the students use the glossary in their book to write and define terms.
We can also create a positive and safe classroom atmosphere for problem solving by being enthusiastic and allowing the students to take risks without consequences. By emphasizing the process as well as the answer, the students may be willing to try unconventional or different ways to solve the problem. I always tell my students that there isn't just one right way to get an answer which surprises many of them. In fact, this is one of the posters that hangs in my classroom.
As math teachers, let's continue to emphasize problem solving so that all students will acquire confidence in using mathematics meaningfully. But most of all, let's have fun while we are doing it!
If you are interested in having a math dictionary for your students, check out A Simple Math Dictionary. It is a 30 page dictionary which uses easy and clear definitions as well as formulas and examples so that students can learn and understand new math words without difficulty or cumbersome language. Most definitions include diagrams and/or illustrations for the visual learner. Over 300 common math terms are organized alphabetically for quick reference. | 1,207 | 5,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2015-18 | longest | en | 0.947535 |
http://www.slideserve.com/bailey/greedy-algorithms | 1,506,032,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687906.71/warc/CC-MAIN-20170921205832-20170921225832-00430.warc.gz | 579,689,162 | 16,060 | 1 / 51
# Greedy Algorithms - PowerPoint PPT Presentation
Greedy Algorithms. Greedy Algorithms. Compared to Dynamic Programming Both used in optimization problems More efficient Not always optimal Top-down instead of bottom up Make a locally optimal (quick) choice Still have optimal substructure No need to dive into sub-problems.
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### Greedy Algorithms
Jeff Chastine
• Compared to Dynamic Programming
• Both used in optimization problems
• More efficient
• Not always optimal
• Top-down instead of bottom up
• Make a locally optimal (quick) choice
• Still have optimal substructure
• No need to dive into sub-problems
Jeff Chastine
• Given a set of activities , find the maximum # you can schedule
• NOT optimal use of time of a room
• One activity is active at one time
• Activity has start time of and finish of
• Order activities by finish time
isifi
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1 3 0 5 3 5 6 8 8 2 12
4 5 6 7 8 9 10 11 12 13 14
Jeff Chastine
• Defining the subproblem:
• Set of activities that “fit” in this time chunk
• Picking any activity creates two sub-problems
Jeff Chastine
Sij
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Jeff Chastine
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• Optimal solution includes
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Jeff Chastine
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• We know
• How to choose activity ?
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Jeff Chastine
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• Iterate through all (like Matrix Mults)!
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Jeff Chastine
RECURSIVE-ACT-SELECT (s, f, i, n)
1 m ← i + 1
2 whilem ≤ n and sm < fi //find 1st activity
3 dom ← m + 1
4 ifm ≤ n
5 then return {am}
RECURSIVE-ACT-SELECT(s, f, m, n)
6 else return
Jeff Chastine
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Jeff Chastine
• Knapsack Problem
• Steal items of different weight and value
• Sack has weight limit
• Goal: steal as much as possible
• Two flavors of problem
• 0-1 version (take all of something)
• Fractional (can take only part)
Jeff Chastine
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\$100
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Total = \$160
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Try again! Total = \$180
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Total = \$220
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Jeff Chastine
Take only part
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Total = \$240
Note: optimal for 0-1 was \$220
Jeff Chastine
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Notice the waste of space!
Jeff Chastine
• Objective: compress text data
• Raw ASCII text file uses 8 bits/char (fixed)
• 1000 characters = 8K bits
• Book uses 6 characters for 3 bits/char
• Trick
• Analyze: not all characters are used!
• Analyze: frequency of characters
• Prefix codes: no binary code is prefix of another
Jeff Chastine
a b c d e f
Frequency (in thousands) 45 13 12 16 9 5
Fixed-length codeword 000 001 010 011 100 101
Variable-length codeword 0 101 100 111 1101 1100
110001001101 = face //12 bits
01011010 = abba // 8 bits
Jeff Chastine
• Sort by least frequent
e:9
c:12
b:13
d:16
a:45
f:5
Jeff Chastine
• Group lowest two
e:9
c:12
b:13
d:16
a:45
f:5
Jeff Chastine
• Combine and re-sort
14
c:12
b:13
d:16
a:45
1
0
e:9
f:5
Watch how least frequent characters are pushed down the tree!
Jeff Chastine
• Group lowest two
14
c:12
b:13
d:16
a:45
1
0
e:9
f:5
Jeff Chastine
• Re-sort
14
25
d:16
a:45
1
1
0
0
e:9
f:5
c:12
b:13
Jeff Chastine
14
25
d:16
a:45
1
1
0
0
e:9
f:5
c:12
b:13
Jeff Chastine
25
30
a:45
0
1
1
0
14
c:12
b:13
d:16
1
0
e:9
f:5
Notice what’s happening to the first grouping we did
Jeff Chastine
25
30
a:45
0
1
1
0
14
c:12
b:13
d:16
1
0
e:9
f:5
Jeff Chastine
55
a:45
0
1
25
30
0
1
1
0
14
c:12
b:13
d:16
1
0
e:9
f:5
Jeff Chastine
100
1
0
55
a:45
0
1
25
30
0
1
1
0
14
c:12
b:13
d:16
1
0
e:9
f:5
Jeff Chastine
• Greedy algorithms have two properties
• Greedy choice property: a globally optimal solution can be created by the greedy choice
• Optimal substructure: the optimal solution is contains optimal solutions to subproblems
• Greedy is top-down
• Doesn’t always yield optimal solution
Jeff Chastine | 2,728 | 6,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-39 | latest | en | 0.785218 |
http://wiki.workxpress.com/doku.php?id=expression%20function%20-%20sine&rev=1473877160&do=diff | 1,601,302,426,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00075.warc.gz | 138,036,903 | 5,003 | # Differences
This shows you the differences between two versions of the page.
— expression function - sine [2016/09/14 14:19] (current) Line 1: Line 1: + ====== Sine (Expression Function) ====== + ===== Purpose ===== + The **Sine** function (SIN) is used to calculate the sine of an angle given in radians, degrees, or gradians. The [[Expression Function - Inverse Sine|inverse sine]] function is also availalble. + + {{::sin_initial.png?direct&600|}} + + ===== Parameters ===== + The **Sine** function has two parameters, the first of which is required: + + ==== Radians ==== + + Allowed Inputs: [[Field|field]] value or number + + The first required parameter is the angle that the sine will be obtained from either use the use the [[Query Builder]] to choose a field that contains the value or use type a value to enter a number. + + ==== Angle Unit ==== + + Allowed Inputs: omitted, 'd', 'g' + + The second parameter of the **Sine** function is the unit the angle is provided in. If the angle unit is omitted or not recognized then radians will be be used. For an angle provided in degrees use 'd'. For an angle provided in gradians use 'g'. If the angle unit needs to be dynamically stored in a field then the advanced expression builder will need to be used. + + + ===== Output ===== + + The output of the **Sine** function is a number representing the vertical rise of an angle when the radius of the circle is 1. + + ===== Example ===== + + For example SIN(0.5235987755983), SIN(30,'d'), and SIN(33.3333333333,'g') would result in "0.5". | 419 | 1,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-40 | latest | en | 0.673361 |
https://stanford.library.sydney.edu.au/archives/sum2007/entries/evil/validity.html | 1,642,466,662,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00471.warc.gz | 602,468,062 | 3,674 | ## The Validity of the Argument
That the argument is deductively valid can be seen as follows. First, let us introduce the following abbreviations:
• State(x): x is a state of affairs
• Dying(x): x is a state of affairs in which an animal dies an agonizing death in a forest fire
• Suffering(x): x is a state of affairs in which a child undergoes lingering suffering and eventual death due to cancer
• Omnipotent(x): x is omnipotent
• Omniscient(x): x is omniscient
• MorallyPerfect(x): x is morally perfect
• PreventsExistence(x,y): x prevents the existence y
• God(x): x is God
• HasPowerToPreventWithout(x,y): x has the power to prevent the existence of y without hereby either allowing an equal or greater evil, or preventing an equal or greater good
The argument just set out can then be formulated as follows:
(1) x[State(x) (Dying(x) Suffering(x)) Bad(x) y(Omnipotent(y) HasPowerToPreventWithout(y,x))] (2) x[State(x) zPreventsExistence(z,x)] (3) xy[(Bad(x) HasPowerToPreventWithout(y,x) PreventsExistence(y,x)) (Omniscient(y) MorallyPerfect(y))] Therefore, from (1), (2), and (3): (4) x[Omnipotent(x) Omniscient(x) MorallyPerfect(x)] (5) x[God(x) (Omnipotent(x) Omniscient(x) MorallyPerfect(x))] Therefore: (6) x(God(x))
The premises here are (1), (2), (3), and (5), and they can be shown to entail the conclusion, (6), as follows.
### The Inference from (1), (2), and (3) to (4)
(i) State(A) (Dying(A) Suffering(A)) Bad(A) y(Omnipotent(y) HasPowerToPreventWithout(y,A)) From (1), via EE (Existential Elimination). (ii) zPreventsExistence(z,A) From (2) and 1st conjunct of (i) by UE and MP. (iii) Omnipotent(G) Assumption for conditional proof ("G" arbitrary) (iv) HasPowerToPreventWithout(G,A) From 4th conjunct of (i), by instantiating "G" and using MP (v) PreventsExistence(G,A) From (ii), by UE. (vi) Bad(A) HasPowerToPreventWithout(G,A) PreventsExistence(G,A)) Conjoin 3rd conjunct of (i) with (iv) and (v). (vii) (Omniscient(y) MorallyPerfect(G)) From (3) and (6), by UE and MP. (viii) Omnipotent(G) (Omniscient(G) MorallyPerfect(G)) Conditional Proof, (iii) -- (vii). (ix) (Omnipotent(G) Omniscient(G) MorallyPerfect(G)) From (viii), by the equivalence of AB with (AB), double negation elimination, and associativity of conjunctions. (x) x(Omnipotent(x) Omniscient(x) MorallyPerfect(x)) From (ix), via UI (Universal Introduction), since ‘G’ was arbitrary.
### The Inference from (4) and (5) to (6)
(i) [Omnipotent(G) Omniscient(G) MorallyPerfect(G)] From (4), via universal insantiation, and where ‘G’ is arbitrary. (ii) God(G) (Omnipotent(G) Omniscient(G) MorallyPerfect(G)) From (5) by universal instantiation. (iii) God(G) From (i) and (ii) by modus tollens (iv) x(God(x) From (iii) by universal generalization, since ‘G’ was arbitrary. (v) x(God(x)) From (iv), by interdefinability of quantifiers. | 889 | 2,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.75457 |
http://www.digitalsolutionshop.com/page/255/ | 1,521,546,648,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00608.warc.gz | 361,089,078 | 12,324 | Posted on
## Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Conduct a benefit-cost analysis for a defined policy initiative to reduce Polluted Waters. Suppose that a proposal to solve your environmental concern (e.g. clean up a polluted river and convert it to a boatable/swimmable river) is expected to generate \$400,000 (or any amount you see fit to use) in benefits every year, starting next year, for the next 5 years. Calculate the present value of this project using a social discount rate of 3% and assuming inflation is 2% (thus a nominal social discount rate of 5%). Next calculate the present value using the same inflation rate but a social discount rate of 1%. Show your calculations in a table such as the one below.
If the cost of improving the environmental concern is say \$1,800,000 (or any amount you see fit to use) in the present, would the project be feasible? Use the Present Value of Net Benefits to support your decisions. Explain how this differs from the Benefit Cost Ratio. Discuss the importance of choosing the social discount rate and how society may be inappropriately discounting the benefits of environmental sustainability.
Discuss the limitations of benefit-cost analysis that are of concern to this particular situation.
a) project using a social discount rate of 3%
b) project using a social discount rate of 1%
If we use the befit cost ratio, then the benefit cost ratio
a) Benefit cost ratio of project using a social discount rate of 3%
b) Benefit cost ratio of project using a social discount rate of 1%
Discuss the importance of choosing the social discount rate and how society may be inappropriately discounting the benefits of environmental sustainability.
Discuss the limitations of benefit-cost analysis that are of concern to this particular situation.
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## Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
Conduct a benefit-cost analysis for a defined policy initiative to reduce Polluted Waters. Suppose that a proposal to solve your environmental concern (e.g. clean up a polluted river and convert it to a boatable/swimmable river) is expected to generate \$400,000 (or any amount you see fit to use) in benefits every year, starting next year, for the next 5 years. Calculate the present value of this project using a social discount rate of 3% and assuming inflation is 2% (thus a nominal social discount rate of 5%). Next calculate the present value using the same inflation rate but a social discount rate of 1%. Show your calculations in a table such as the one below.
If the cost of improving the environmental concern is say \$1,800,000 (or any amount you see fit to use) in the present, would the project be feasible? Use the Present Value of Net Benefits to support your decisions. Explain how this differs from the Benefit Cost Ratio. Discuss the importance of choosing the social discount rate and how society may be inappropriately discounting the benefits of environmental sustainability.
Discuss the limitations of benefit-cost analysis that are of concern to this particular situation.
a) project using a social discount rate of 3%
b) project using a social discount rate of 1%
If we use the befit cost ratio, then the benefit cost ratio
a) Benefit cost ratio of project using a social discount rate of 3%
b) Benefit cost ratio of project using a social discount rate of 1%
Discuss the importance of choosing the social discount rate and how society may be inappropriately discounting the benefits of environmental sustainability.
Discuss the limitations of benefit-cost analysis that are of concern to this particular situation.
For instant digital download of the above solution, Please click on the “PURCHASE” link below to get the tutorial for Defined policy initiative to reduce Polluted Waters_Cost Benefit Assignment_Answer
For instant digital download of the above solution or tutorial, please click on the below link and make an instant purchase. You will be guided to the PAYPAL Standard payment page wherein you can pay and you will receive an email immediately with a download link.
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2. Distinguish between possession, form, time, and place utility.
.
4. Why do contemporary supply chains need to be fast and agile?
8. Discuss some of the ways that inventory can be reduced in the supply chain.
5. Distinguish between a fixed order quantity and fixed order interval system. Which one generally requires more safety stock? Why?
9. What are implications of the JIT approach for supply chain management?
PART III
Lab Assignment:
1. After making some wise short-term investments at a race track, Chris Low had some additional cash to invest in a business. The most promising opportunity at the time was in building supplies, so Low bought a business that specialized in sales of one size of nail. The annual volume of nails was 2,000 kegs, and they were sold to retail customers in an even flow. Low was uncertain of how many nails to order at any time. Initially, only two costs concerned him: order-processing costs, which were \$60 per order without regard to size, and warehousing costs, which were \$1 per year per keg space. On average, the rented warehouse space is only half full. This meant that Low had to rent a constant amount of warehouse space for the year, and it had to be large enough to accommodate an entire order when it arrived. Low was not worried about maintaining safety stocks, mainly because the outward flow of goods was so even. Low bought his nails on a delivered basis.
Question 1: Using the EOQ methods outlined in Chapter 9, determine how many kegs of nails Low should order at one time.
Step 2: Low-Quantity Discount
Question 2: Assume that all conditions in Question 1 hold, except that Low’s supplier now offers a quantity discount in the form of absorbing all or part of Low’s order-processing costs. For orders of 750 or more kegs of nails, the supplier will absorb all order-processing costs; for orders between 249 and 749 kegs, the supplier will absorb half. What is Low’s new EOQ? (It might be useful to lay out all costs in tabular form for this and later questions.)
Step 3: Low Rent
Question 3: Temporarily ignore your work on Question 2. Assume that Low’s warehouse offers to rent Low space on the basis of the average number of kegs that Low will have in stock, rather than on the maximum number of kegs that Low would need room for whenever a new shipment arrived. The storage charge per keg remains the same. Does this change the answer to Question 1? If so, what is the new answer?
Step 4: New EOQ
Question 4: Take into account the answer to Question 1 and the supplier’s new policy outlined in Question 2, and the warehouse’s new policy in Question 3. Then determine Low’s new EOQ.
Step 5: Financing Inventory
Question 5: Temporarily ignore your work on Questions 2, 3, and 4. Low’s luck at the race track is over; he now must borrow money to finance his inventory of nails. Looking at the situation outlined in Question 1, assume that the wholesale cost of nails is \$40 per keg and that Low must pay interest at the rate of 1.5% per month on unsold inventory. What is his new EOQ?
Step 6: Final EOQ
Question 6: Taking into account all of the factors listed in Questions 1, 2, 3, and 5, calculate Low’s EOQ for kegs of nails. | 1,821 | 8,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-13 | latest | en | 0.889248 |
http://mathhelpforum.com/algebra/80042-whats-formula-print.html | 1,527,498,750,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872114.89/warc/CC-MAIN-20180528072218-20180528092218-00484.warc.gz | 185,304,144 | 2,651 | # what's the formula?
• Mar 22nd 2009, 06:23 PM
faolan
what's the formula?
What is the formula for this?
If the line passing through (7,y) and (1,0) is parallel to the line joining 4,4 and -2,1, find y.
How do I solve for y?
• Mar 22nd 2009, 06:29 PM
Reckoner
Quote:
Originally Posted by faolan
If the line passing through (7,y) and (1,0) is parallel to the line joining 4,4 and -2,1, find y.
Find the slope of the second line. Find the slope of the first line (in terms of \$\displaystyle y\$). What must be true about the slopes of two parallel lines? This fact will give you your equation, which you can then solve for \$\displaystyle y.\$ | 198 | 647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-22 | latest | en | 0.905311 |
https://www.jiskha.com/users?name=help+please | 1,597,285,664,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.61/warc/CC-MAIN-20200813014639-20200813044639-00062.warc.gz | 703,874,152 | 17,816 | Popular questions and responses by help please
1. ## Mathematics
out of 6 cheerleaders, 3 will be selected to march in the first row of the parade how many 3-person arrangements can be formed using the 6 cheerleaders if Sandy must stand on the left end of the front row and Dana must also be included in the front row? A.
2. ## History
Why did iranians take 53 Americans hostage at the American embassy in Iran? A. To force the United Statesvto loft to oil ban B. Because the US government allowed the shah to travel to the United States for medical treatment *********? C. because Atatollah
3. ## mathematics
A store advertises skirts for x-5 dollars and allows an additional two dollar reduction on each skirt if three or more skirts are purchased.Represent the cost of five skirts.
4. ## Electrical
Explain the operation of an alarm system and strobe lighting system, by determining the resistance of the resistor in the alarm circuit and explain the functionality of the resistor in the circuit.
5. ## AP CHEMISTRY
A)In a chemical reaction two gases combine to form a solid. What do you expect for the sign of ΔS? B) For which of the following processes does the entropy of the system increase? (Select all that apply.) -> alignment of iron filings in a magnetic field
6. ## Computer Science
Which of the following statements contains an error? 1. system.out.print("/tab Age in Minutes: " + ageInMinutes +"/n"); 2. system.out.print("t/ Age in Minutes: " + ageInMinutes + "n/"); 3. system.out.print("/t Age in Minutes: " + ageInMinutes + "/n"); a. 1
7. ## English
Is this a simple sentence or a fragment? Doesn’t need help.
8. ## English
Not just silent adult tears, but really sobbing. Is this a simple sentence or fragment. I think it is a fragment, is that correct. But, I am having trouble coming up with why?
9. ## Math
Brian gave 20% of his baseball cards to Scott and 15% to Adam. If he still had 520 cards, how many did he have originally?
10. ## mathematics
Mandy earned x dollars and Andy earned 4x �+ 7 dollars on the first day of a charity carwash. If Mandy and Andy each earned the same amounts on the second and third days of the car wash, how much did they earn together over the course of the 3 days?
11. ## LA the giver chapters 1-2
Which of the following is an example of a dwelling? Select all that apply. 1. a city 2. a house*** 3. a monument 4. a condominium*** Which of the following passages from the novel helps you realize that the community in which Jonas lives is not like
12. ## physics
A 2.0-kg ball moving at 10 m/s makes an off-center collision with a 3.0-kg ball that is initially at rest. After the collision, the 2.0-kg ball is deflected at an angle of 30° from its original direction of motion and the 3.0-kg ball is moving at 4.0 m/s.
13. ## AP CHEMISTRY
Calculate the solubility of Mn(OH)2 in grams per liter when buffered at each of the following. a) pH 7.4 b) pH 9.2 c) pH 11.5 ALL IN GRAMS PER LITER. PLEASE HELP ME AND EXPLAIN
14. ## computer science
Which of the following statements does not contain an error? 1. string firstName= in.next(); 2. string lastName= in.nextline(); 3. system.out.print("\n"); a. 1 only b. 2 only c. 3 only d. 1 and 3 only e. 2 and 3 only is the answer 1 and 3
15. ## physics
The drawing shows a 25.1-kg crate that is initially at rest. Note that the view is one looking down on the top of the crate. Two forces, 1 and 2, are applied to the crate, and it begins to move. The coefficient of kinetic friction between the crate and the
16. ## pre algebra
1. Solve the equation. –9v – 5 = –95 (1 point) 17 11 10 –10 2. Solve the equation. x over four – 5 = –8 (1 point) –27 –12 –7 12 3. Solve the equation. p over five + 6 = 10 (1 point) 44 30 20 –20 4. Solve the equation. –2(m – 30) =
17. ## MATH
The table below represents the volume of a liquid sample as a function of its temperature: Temp (C) Volume (ml) x f(x) 11 95 16 100 21 105 26 110 The average rate of change of f(x) between x=11 to x=21 is _____ ml per C ((degrees)) and represents the rate
18. ## history
1. Which factor contributed most to the beginning of the women's rights movement in the United States during the mid-1800s? A. A dramatic increase in women's participation in the workforce *** B. A shift in social attitudes brought on by increased
19. ## Math
Solve each problem and write your answer in scientific notation. 1.(6.0x10^5)x(3.0x10^4) A.18x10^20 B.18x10^9 C.1.8X10^20 D.1.8x10^10 2.(5x10^-2) ÷(2x10^3) A.2.5x10^-5 B.2.5x10^-1 C.25x10^1 D.25x10^2 3.(2.56x10^6)x(3.56x10^2) A.9.1136x10^12 B.91.136x10^8
20. ## physics
A 2.0-kg ball moving at 10 m/s makes an off-center collision with a 3.0-kg ball that is initially at rest. After the collision, the 2.0-kg ball is deflected at an angle of 30° from its original direction of motion and the 3.0-kg ball is moving at 4.0 m/s.
21. ## language arts
On which character traits does the narrator of “the tell-tale heart” pride himself? a. his abilities to lie and steal b. his sneakiness and his fear c. his wit and his confidence d. his intelligence and his patience i think it’s d! help!?
22. ## educational technology
which verb tense is used in the following sentence? as captain i represent the team at debates A. present tense B. future tense C. past tense D. early tense I think B can someone please help?
23. ## Algebra 1
You use a line of best fit for a set of data to make a prediction about an unknown value. The correlation coefficient for your data set is 0.793. How confident can you be that your predicted value will be reasonably close to the actual value?
24. ## Math
Find the lateral area of the cone. Use 3.14 fo pi and round result to the nearest whole unit. radius: 15cm Side unit: 28cm Please help!! A) 1,319 cm squared B) 2,639 cm squared C) 707 cm squared D) 2,026 cm squared is the answer B?
25. ## algebra
A stainless steel patio heater is a square pyramid. The length of one side of the base is 21.6 inches. The slant height of the pyramid is 83.3 inches. What is the height of the pyramid? Round to one decimal place as needed. I have tried and can't figure it
26. ## SCIENCE
If we decrease the distance an object moves we will decrease the amount of force applied. Decrease the amount of work done. increase the amount of force applied. increase the amount of work done.
27. ## social studies
Which hazards does climate change pose for Southeast Asia? Select the two correct answers. A. rising sea levels B. more volcanic eruptions C. more intense typhoons D. increased tectonic activity E. overcrowding in coastal cities
In 2001, Windsor, Ontario received its maximum amount of sunlight, 15.28 hrs, on June 21, and its least amount of sunlight, 9.08 hrs, on December 21. 1) Due to the earth's revolution about the sun, the hours of daylight function is periodic. Determine an
29. ## Physics
1. A bullet of 0.0500 kg is fired into a block of wood. Knowing that the bullet left the gun with a muzzle velocity of 350. m/s, and the bullet penetrates .15 m into the block of wood, determine: a) The average force required to stop the bullet. b) The
30. ## AP CHEM
potassium tetrabromo(ortho-phenanthroline)- cobaltate(III)
31. ## Maths
Ab = 125ft Bc = 85ft C = 63° 1. Find the length of Ac 2. Whats the total length around triangle if there is two 3ft opening ?
32. ## Spanish
1. Which would be true for stem changing verbs? All forms of the verb have some type of irregular change. They have a consistent spelling change in the stem of the verb. Stem changing verbs are the same as irregular verbs, it is just a different name. They
33. ## Geometry
Which of the following describes a median of a triangle? A. a segment drawn from a vertex to the midpoint of the opposite side B. a segment drawn from the vertex perpendicular to the line containing the opposite side C. a segment drawn through the midpoint
34. ## world history
What were the purposes of the earthworks built by cultures of eastern North America? SELECT THREE . A.as burial mounds B.as temple platforms C.as astronomical observatories D.as defensive structures E.as fields for athletic competition select 3 answers one
35. ## language arts
which question best relates to these details in winter. Frogs burrow the mud, snails bury themselves a.what happens to frogs in the winter. b.how do animals differ from people in the winter. c.why do sanils bury themselves. d.how do creatures pepare for
36. ## Science
Q1: Why do scientist use models? A.) Scientist use models to learn about things that are too small, too late, or too complex to observe directly. B.) Scientist use models because doing so is always part of the scientific method. C.) Scientist use models
1. The first successfully established English colony in North America was (1 point) Plymouth. New York. Jamestown. Santa Fe. 2. Protestants who wanted to reform the Anglican Church were called (1 point) Pilgrims. Puritans. Plymouthians. priests. 3. How did
38. ## Math
Prove that 1 * 3 * 5 * ... * 2345 + 2 *4 * 6* ... * 3456 is divisible by 6789.
39. ## ALGEBRA
The table below shows two equations: Equation 1 |3x - 1| + 7 = 2 Equation 2 |2x + 1| + 4 = 3 Which statement is true about the solution to the two equations? Equation 1 and equation 2 have no solutions. Equation 1 has no solution and equation 2 has
40. ## AP CHEM
formula for bipyridyl
41. ## Texas History
Why did the Spanish return to texas in 1716? A)The Hasinai requested that they return and build a new mission. B)They came to help the Apache fight the Karankawa. C)They heard that gold had been discovered in the area * D)They wanted to keep the french out
42. ## computer science
Which of the following statements does not contain an error? 1. string firstName= in.next(); 2. string lastName= in.nextline(); 3. system.out.print("\n"); a. 1 only b. 2 only c. 3 only d. 1 and 3 only e. 2 and 3 only is the answer 1 and 3
43. ## computer science
what is the value of the variable int iNum= 2.02 A. 2 B. 2.0 C. 2.02 D. 0.0 E. error:possble loss of precision please help and explain
44. ## computer science
what is the value of the variable? double dNum= 6.23 A. 6 B. 6.0 C. 6.23 D. 0.0 E. error: possible loss of precision
45. ## computer science
what is the value of the valuable double dNum=5 A. 5 B. 5.0 C. 0 D. 0.0 E. error: possible loss of precision please help and explain
46. ## LA
Choose the option that best completes the multi-draft reading process for poetry. First Read: Read to unlock the basic meaning of the poem. Second Read: ????? Third Read: Find relationships between ideas in the poem. 1. Consider how the poems structure
1. Why did farm prices fall in the 1950’s? a. Farmers bought new expensive machinery b. Farmers produced more than they could sell** c. Farmers could not sell their products in Europe d. Farmers saw their government subsidies reduced 2. What shortage was
48. ## math
Which of the following sets contains 3 irrational numbers ? a. { square root of 120 , pie , square root of 3 } b. {neg. square root of 256 , 1/9 , 1/12} c. {3.14 , -47 , 100} d. {pie , square root of 0.36 , square root of 121}*
49. ## Pre-calculas
Boat A leaves a dock headed due east at 1:00 PM traveling at a speed of 16 mi/hr. At 2:00 PM, Boat B leaves the same dock traveling due north at a speed of 18 mi/hr. Find an equation that represents the distance d in miles between the boats and any time t
50. ## Health 7th
you want to ask your teacher for extra help,so you practice in your mind what you will say. this is an example of ____. A) eustress B) distress C) a stressor D) a mental rehearsal _____________________________________ during the alarm stage of stress your
51. ## math
simplify 10p4 A. 210 B. 360 C. 5,040 D. 151,200 simplify 9c4 A. 126 B. 15,120 C. 5 D. 3,024
52. ## math
Food Express is running a special promotion in which customers can win a free gallon of milk with their food purchase if there is a star on their receipt. So far, 129 of the first 138 customers have not received a star on their receipts. What is the
53. ## Algebra 1 math
Deborah rolls two fair dice. What is the probability that the sum of the numbers rolled is 7?
54. ## PHYSICS
A 5450-m^3 blimp circles Fenway Park duing the World Series, suspended in the earth's 1.21-kg/m^3 atmosphere. The density of the helium in the blimp is 0.178 kg/m^3. A) What is the buoyant force that suspends the blimp in the air? B) How does this buotant
55. ## AP CHEMISTRY
Calculate the solubility of LaF3 in grams per liter in the following solutions. (Ksp = 2. 10-19.) a) pure water Ans: .00182g/L b) 0.037 M KF solution c) 0.055 M LaCl3 solution NEED b&c PLEASE HELP
56. ## AP CHEMISTRY
The following acid-base indicators are available to follow the titration. Which of them would be most appropriate for signaling the endpoint of the titration? EXPLAIN ------- COLOR CHANGE ------ INDICATOR Acid form Base Form pH Transition Interval
57. ## AP CHEMISTRY
Calculate the equivalent mass of each of the following acids. a) HC2H3O2 b) KHCO3 c) H2SO3
58. ## Maths
Convert 10gallons of oil with a density of 0.65g/ml to pounds
59. ## Math
Penalty will be 4500 for 1st day, 12000 everyday after. And budget is only 155000. How many days can i afford? Using arithmetic progression.
60. ## Math
In a study, a fish and game department worker catches, tags, and frees 124 catfish in a lake. A few weeks later he catches and frees 140 catfish. 35 have tags. Estimate the number of catfish in a lake. (Hint: set up and solve a proportion).
62. ## technology
what is a Boolean operator? a link would be great please and thank you:)))
63. ## History
Which of the following actions occurred first? Japan attacked Pearl Harbor The Axis powers invaded neighboring territories The Lend-Lease Act was proposed The United States drafted men into the armed forces I honestly have no idea what the answer could
64. ## biology
The Excretory System removes excess… W _ _ _ _ _ _ _ T _ _ A _ _ _ _ - _ I _ _ _ _ _ I have water salts but I'm not sure for the last two I believe its waste for one but the other I don't know
65. ## geometry
On the coordinate plane, ΔABC ≅ ΔDEF by SSS. ΔABC translates 2 units to the left and 3 units down. Do the triangles remain congruent? Explain why or why not.
66. ## chemistry
Classify the reaction between zinc and acetic acid and explain, in general terms, what happens during this type of reaction. would the type of reaction be a single replacement ??
67. ## chemistry
. Classify the reaction between zinc and acetic acid and explain, in general terms, what happens during this type of reaction. would the type of reaction be a single replacement ??
68. ## Social studies
Which sentence best describes How the process of the westward expansion affected American political culture? A-The need for labor led to the abolition of slavery and increased respect for personal liberty B-The concept of expansion led to an accepting
69. ## Geometry
Find the perimeter of the polygon with vertices of A(0, 0), B(4, 0) and C(2, 3). Round to the nearest tenth. Just enter one number as your answer Can you please help me
70. ## Geometry
Your Turn: Given points (0, 0) and (3, 6), find the point which partitions the segment into 3:2. A. (1.8, 3.6) B. (2,4) C. (3,5) D. (4.5, 9) can you please help me
71. ## Geometry
Find the surface area and volume of a solid made up of a cone, a cylinder, and a hemisphere with a height of 5 each laying on top of each other.
72. ## Social Studies
1.What prompted the call to rewrite the Articles of Confederation? A:The desire to win independence by unifying the nation B:The economic downturn caused by the limitations on the central government C:The need to establish a national court system to deal
73. ## health 7th
1. during the alarm stage of you stress body release a chemical called adrenaline which is part of a. eustress b. the flight-or-fight- response*** c. mental rehearsal d. a catastrophe 2.nay event or situation that causes stress is called a(n) a. eustress
74. ## Geometry
The endpoints of a line segment are at the coordinates (4, -2, 5) and (6, 4, -3). What is the midpoint of the segment? (-2, -6, 8) (10, 2, 2) (5, 1, 1) (-1, -3, 4) My choice- C
75. ## SS
to supply energy needs in the future, A) we have unlimited supply of fossil fuels. B) Its not necessary to look for new sources of energy. C) renewable energy sources are being devolved. D) atomic energy provides a safe alternative to fossil fuel.
76. ## Physics
An 89.5 kg fullback moving east with a speed of 5.7 m/s is tackled by a 81.5 kg opponent running west at 2.58 m/s, and the collision is perfectly inelastic. (a) Calculate the velocity of the players just after the tackle. m/s (b) Calculate the decrease in
77. ## math
one number is 8 less than another. If twice the larger number is equal to four times the smaller number.what is the smaller number?
78. ## The giver
Why does Jonas share a memory with Gabriel? A. He is trying to calm the baby and lull him back to sleep B. He is eager to transmit his experiences to a member of the community C. He does not think anything is wrong with sharing his memories D. He feels it
79. ## algebra
evaluate the expression when a = 4 b = -5 and c = -8 -b + c =
81. ## math
a basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany pricks a fruit at random. What is the probability that Jameson gets a banana and
82. ## math
a number cube is rolled 150 times. the number 3 comes up 43 times. hat is the experimental probability of rolling a 3? what is the theoretical probability of rolling a 3? A. 43/150 ; 1/6 B. 43/150 ; 1/50 C. 1/6 ; 43/150 D. 3/43 ; 1/6
83. ## chemistry
The solubility of AB(s) in a 1.000-M solution of C(aq) is found to be 0.121 M. What is the Ksp of AB?
84. ## geometry
If A point K is the reflection of L (2,5) IN THE Y AXIS WHAT IS THE LENGTH OF KL
85. ## physics/light reflection
A light of ray is incident at an angle of 58degrees on the plane surface of a block of glass of refractive index 1.60. Some light is reflected on the other side of the normal at the same angle as the angle of incidence. Find: a.the angle between the light
86. ## Physics
The following four lenses are placed together in close contact. Find the focal length of the combination. Lens 1- F'= +200mm Lens 2- F'= -125mm Lens 3- F'= -400mm Lens 4- F'= +142.9mm
87. ## Stats
I ran this an an independent samples t-test. It is a t rest according to the professor, just want to make sure I have the right one. A researcher wanted to answer the following question: What is the difference in students' computer anxiety based on whether
88. ## stsats
A researcher wanted to answer the following question: What is the difference in students' computer anxiety based on whether or not they own computers? The researcher surveyed 92 undergraduate education students via an online survey. In the survey, the
89. ## Statistics
I bemieve the t test is paired , want to make usre beofre I run the stats. A researcher wanted to answer the following question: What is the difference in students' computer anxiety based on whether or not they own computers? The researcher surveyed 92
90. ## AP CHEMISTRY
1) Calculate the molarity of a solution of sodium hydroxide, NaOH,if 23.64 mL of this solution is needed to neutralize 0.5632g of potassium hydrogen phthalate. 2) It is found that 24.68 mL of 0.1165M NaOH is needed to titrate 0.2931 g of an unknown acid to
91. ## AP CHEMISTRY
A) At 800 K the equilibrium constant for I2(g) 2 I(g) is Kc = 3.1 10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.44*10-2 g of I(g), how many grams of I2 are in the mixture? answer is .060g B)For 2 SO2(g) + O2(g) 2 SO3(g), Kp = 3.0*104 at 700
92. ## physics
You have arrived at a new planet and put your starship into a circular orbit at a height above the surface that is equal to two times the radius of the planet. Your speed is 4300 m/s and it takes 7 hours and 18 minutes to complete an orbit. a. What is the
93. ## math
A perfect minor wholetone is a frequency rtio of 10:9. A perfect major wholetone is a frequency ratio of 9:8. Are these noticeably different from an equal temper wholetone? (200 cents)
94. ## Chemistry (daltons law of partial pressure)
A mixture of toluene and water boils at 90 C, calculate the vapor pressure of toluene at this temperature. Assume the atmospheric pressure to be 760 mmHg. And secondly calculate the molar mass of toluene.
95. ## college chemistry
2h2o2(aq) ---> 2h2o(l) + o2(g) if someone puts 0.470 ml of hydrogen peroxide on a cut how many moles of hydrogen peroxide are used? Assume the density of the solution is equal to the density of water.
96. ## science
It is true that ______ a) few plants grow well around water b) plants are better able to seek water than are animals c) water covers about 95% of the plants d) the amount of water in the enivronment limits the kinds of organisms that can live there e)
97. ## Chemistry
I havent been at school for 8 days due to somthing wrong with my heart, but the teachers gave me work. I have no clue how to do them. I really need help please, There all due tomarrow. The air in a dry, sealed 2L soda bottle has a pressure of 0.998 atm at
98. ## physics
What are four situations where photovoltaic cells would have an advantage over other sources of electric energ?
99. ## Finance (common stock)
I need a formula for: A share of common stock of xyz ltd is expected to pay no dividends until next 2 years but at the end of year 2 it will pay \$2 as dividends. Dividends are expected to grow at 4% per annum into indefinite future. Assume that the | 5,921 | 21,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.907938 |
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# Rating music albums to reflect the amount of frank language and
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Rating music albums to reflect the amount of frank language and [#permalink]
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01 Oct 2017, 12:34
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Rating music albums to reflect the amount of frank language and violence each one contains, in the way that motion pictures are rated, will not serve to reduce the likelihood that younger children will be exposed to that kind of music.
Each of the following, if true, could be logically cited in support of the opinion above EXCEPT:
a) Most record albums are bought by older children who generally do not limit their younger siblings' access to their albums.
b) It would not be feasible to use the rating system to restrict the range of music that radio stations may play.
c) Movie theaters almost never fail to bar children under 18 from admittance to motion pictures of an adult nature.
d) Record stores could not be relied upon to monitor the age of customers seeking to purchase a particular album.
e) Younger children who would ordinarily not be interested in music with mature content might be drawn to it if that content were indicated by a particular rating.
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Rating music albums to reflect the amount of frank language and [#permalink]
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14 Oct 2017, 05:35
Rating music albums to reflect the amount of frank language and violence each one contains, in the way that motion pictures are rated, will not serve to reduce the likelihood that younger children will be exposed to that kind of music.
Each of the following, if true, could be logically cited in support of the opinion above EXCEPT:
-- We need to find an answer that does not weaken the argument that the music will not be able to impose these regulations. In other words: Strengthen the position that this is feasible.
a) Most record albums are bought by older children who generally do not limit their younger siblings' access to their albums. -- strengthens original argument because there is no way to regulate this
b) It would not be feasible to use the rating system to restrict the range of music that radio stations may play. -- Not feasible = strengthener
c) Movie theaters almost never fail to bar children under 18 from admittance to motion pictures of an adult nature. --Winner: Out of Scope: Movie to Music comparison. Further, it shows regulations can work.
d) Record stores could not be relied upon to monitor the age of customers seeking to purchase a particular album. -- Strengthens, because stores cannot enforce the law
e) Younger children who would ordinarily not be interested in music with mature content might be drawn to it if that content were indicated by a particular rating. -- Strengthens: more kids wanting to listen to the music with no way to censor it
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Rating music albums to reflect the amount of frank language and &nbs [#permalink] 14 Oct 2017, 05:35
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,009 | 4,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-39 | latest | en | 0.927517 |
http://stackoverflow.com/questions/9985473/java-rotate-point-around-another-by-specified-degree-value | 1,394,737,184,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678678233/warc/CC-MAIN-20140313024438-00066-ip-10-183-142-35.ec2.internal.warc.gz | 132,860,289 | 16,273 | # Java: Rotate Point around another by specified degree value
I am trying to rotate a 2D Point in java around another with a specified degree value, in this case simply around Point (0, 0) at 90 degrees.
Method:
``````public void rotateAround(Point center, double angle) {
x = center.x + (Math.cos(Math.toRadians(angle)) * (x - center.x) - Math.sin(Math.toRadians(angle)) * (y - center.y));
y = center.y + (Math.sin(Math.toRadians(angle)) * (x - center.x) + Math.cos(Math.toRadians(angle)) * (y - center.y));
}
``````
Expected for (3, 0): X = 0, Y = -3
Returned for (3, 0): X = 1.8369701987210297E-16, Y = 1.8369701987210297E-16
Expected for (0, -10): X = -10, Y = 0
Returned for (0, -10): X = 10.0, Y = 10.0
Is something wrong with the method itself? I ported the function from (Rotating A Point In 2D In Lua - GPWiki) to Java.
EDIT:
Did some performance tests. I wouldn't have thought so, but the vector solution won, so I'll use this one.
-
If you have access to `java.awt`, this is just
``````double[] pt = {x, y};
.transform(pt, 0, pt, 0, 1); // specifying to use this double[] to hold coords
double newX = pt[0];
double newY = pt[1];
``````
-
Is it fast enough? I need the method to be ultra fast for using in a game. Native routines doesn't seem to be that performance-optimized. – Aich Apr 3 '12 at 1:35
AWT is built for graphics, which frequently means games -- so, I wouldn't worry. – Louis Wasserman Apr 3 '12 at 4:13
@LouisWasserman great work , a lot of thanks for you – Aladdin Jul 19 '13 at 0:05
You're mutating the X value of `center` before performing the calculation on the Y value. Use a temporary point instead.
@Peter @David `public class Point { double x, y; public Point(double x, double y) { this.x = x; this.y = y; } public void rotateAround(Point center, double angle) { double tempx = center.x + (Math.cos(Math.toRadians(angle)) * (x - center.x) - Math.sin(Math.toRadians(angle)) * (y - center.y)); double tempy = center.y + (Math.sin(Math.toRadians(angle)) * (x - center.x) + Math.cos(Math.toRadians(angle)) * (y - center.y)); x = tempx; y = tempy; } public String toString() { return "Point { X = " + String.valueOf(x) + ", Y = " + String.valueOf(y) + " }"; } }` – Aich Apr 3 '12 at 4:31 | 678 | 2,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.766892 |
http://www.ask.com/web?q=How+to+Estimate+Products+in+Math%3F&oo=2603&o=0&l=dir&qsrc=3139&gc=1&qo=popularsearches | 1,485,105,075,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281450.93/warc/CC-MAIN-20170116095121-00143-ip-10-171-10-70.ec2.internal.warc.gz | 345,343,244 | 14,918 | Web Results
## IXL - Estimate products - multiply by larger numbers (4th grade math ...
Fun math practice! Improve your skills with free problems in 'Estimate products - multiply by larger numbers' and thousands of other practice lessons.
## IXL - Estimate products (5th grade math practice)
Fun math practice! Improve your skills with free problems in 'Estimate products' and thousands of other practice lessons.
Apr 29, 2008 ... For a complete lesson on estimating products and quotients, go to http://www. MathHelp.com - 1000+ online math lessons featuring a personal ...
## Multiplication estimation example | Estimation of numbers | Chapter ...
Estimating the product of two, 2-digit numbers. ... Use rounding to estimate the total dollars taken in from the sale of the tickets. Now if we wanted the exact ...
## Estimating Decimal Products - Math Goodies
www.mathgoodies.com/lessons/decimals_part2/estimate_products.html
To solve this problem, we will estimate the product of these decimal factors. There are many strategies that we could use. Let's try the two strategies shown ...
## Estimating in Multiplication - Homeschool Math
www.homeschoolmath.net/teaching/md/estimation.php
First, students practice rounding two- and three-digit numbers and money amounts and estimating products (answers to multiplication problems). Then they ...
## Mathman - Estimate Products By Rounding - Sheppard Software
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Estimate products By rounding the fun way with this pac man style math game.
## Estimating Products | Rounding Numbers ... - Math Only Math
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In order to estimate products, we round the given factors to the required place value. Estimating products help us to check the reasonableness of an answer. 1.
## Estimating when Multiplying - Math Forum - Ask Dr. Math
mathforum.org/library/drmath/view/57018.html
Even though one of the rounded numbers will be farther from the true value this way, the estimated product will be closer to the exact product.
## Estimation Worksheets | Products 2 Digits Worksheets - Math Aids
www.math-aids.com/Estimation/Products_2.html
This estimation worksheet will produce 2 digit multiplication problems with rounding guides for the students to solve the products. | 505 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-04 | latest | en | 0.770516 |
https://math.stackexchange.com/questions/1350754/the-number-of-numbers-lying-between-1-and-200-which-are-divisible-by-either-of-2?noredirect=1 | 1,596,488,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735833.83/warc/CC-MAIN-20200803195435-20200803225435-00167.warc.gz | 397,663,542 | 28,944 | # The number of numbers lying between 1 and 200 which are divisible by either of 2 , 3 or 5? [duplicate]
The number of numbers lying between 1 and 200 which are divisible by either of two , three or five?
A)numbers divisible by 2: $\frac{200}{2} = 100$
B)numbers divisible by 3: $\frac{200}{3} = 66$
C)numbers divisible by 5: $\frac{200}{5} = 40$
counting twice
AB)numbers divisible by 6: $\frac{200}{6} = 33$
AC)numbers divisible by 10: $\frac{200}{10} = 20$
BC)numbers divisible by 15: $\frac{200}{15} = 13$
counting 3 times
ABC)numbers divisible by 30: $\frac{200}{30} = 6$
Total of numbers = A + B + C - AB - AC - BC + ABC = 100 + 66 + 40 - 33 - 20 - 13 + 6 = 146
• why you don't include the numbers divisible by 7 ? – zeraoulia rafik Jul 6 '15 at 0:49
• Nice example of inclusion-exclusion – Simon S Jul 6 '15 at 0:50
• @zeraouliarafik The question didn't ask about multiples of $7$. – Thomas Andrews Jul 6 '15 at 2:18
• @Conrado costa sir the answer in the book is 145 ? That is why i am getting confused how to approach ? – sushmitha Jul 7 '15 at 5:42 | 369 | 1,070 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-34 | latest | en | 0.871526 |
https://www.bartleby.com/essay/Flight-of-the-Frisbee-FKXFXNY8C38Q | 1,631,790,745,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053493.41/warc/CC-MAIN-20210916094919-20210916124919-00189.warc.gz | 703,465,187 | 12,042 | # Flight of the Frisbee
1478 Words6 Pages
Abstract Spinning objects such as Frisbees possess unique flying characteristics. They are in essence spinning wings gliding in mid-air propelled by the forces of torque and aerodynamic lift. The relationship between Newton's Laws of Motion and the flight of the Frisbee will be discussed. This paper will attempt to highlight and show the different physical motions involved behind the spinning edge of the Frisbee and the similar forces it shares with other heavier winged objects. Lastly, how major improvements in the redesign of the Frisbee contributed to its increased stability and precision in its flight in the air. The Flight of the Frisbee Objects that fly are designed to push air down. The momentum of the air going…show more content…
As stated by Professor Bloomfield (1999), "Rotation is crucial. Without it, even an upright Frisbee would flutter and tumble like a falling leaf, because the aerodynamic forces aren't perfectly centered" (p. 132). There are two major external forces acting against the flying Frisbee. To sustain flight in the air, the Frisbee must retain sufficient torque or twist to overcome firstly, the inertia of its body and secondly, the viscous friction of the air. The relative importance of these forces is largely influenced by the size and the mass distribution on the Frisbee itself. For instance, the weight or gravitational force, which is a negative force pulling the disk downward, works directly against the forces of lift and thrust. The force of gravity, or Earth's downward pull on the Frisbee, pulls the disk back to Earth after it is released and spun in the air. According to Newton's Law of Universal Gravitation, the amount of gravitational force between objects depends on their mass, and the amount of matter an object contains. The smaller an object's mass, the smaller its gravitational pull. "A spinning Frisbee, though, can maintain its orientation for a long time because it has angular momentum, which dramatically changes the way it responds to aerodynamic twists, or torques" (Bloomfield, 1999, p. 132). The second negative force acting on the Frisbee is the drag or air resistance. As mentioned by Bloomfield (1999), air flows "like all viscous fluids" (p. | 483 | 2,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-39 | latest | en | 0.932655 |
https://www.doubtnut.com/question-answer/which-of-the-following-equations-have-no-real-solutions--644552144 | 1,653,105,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00173.warc.gz | 843,337,428 | 78,288 | HomeEnglishClass 12MathsChapterDpp
Which of the following equatio...
# Which of the following equations have no real solutions ?
Updated On: 17-04-2022
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Text Solution
x^(2) - 2x+ 5 + pi^(2) = 0x^(4) - 2x^(2)sin^(2)'(pix)/(2)+1 = 0`all of these
A
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Transcript
different descendant of any current equation is less than 0 - 2 X + 5 y square = 20 now let this be a square + b + c = 20 and equals to B square minus and option of a square is 1 and access to the value of 20 22 - 2 - 20
minus and it is 4.1 x 5 + Y square equal to zero this will be taken for 1 minus 5 minus x minus 20 because x 42 this term is minus minus system will be the negative form will be less than zero
S Fusion in the power 4 minus minus 2 x 2 + 1 = 20 power 4 + 1 = 6 science term 2 divided into the stuff this will be a square + 1 upon x square equals to 2 sin square A Vinod that their identity that a + 1 upon A tree that it's
when a + 1 upon a is there to it will be always be greater than equal to 2 so this is in the form of this can also find this by the given equal to GF ok but there is a place of apps will get here that is 1 + 1 upon 2 equals to 2 Sin 5x ok report this will get you to 2 sin square x by 2 is greater than equal to ok
pollution caused by coming greater than 0 | 429 | 1,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-21 | latest | en | 0.913153 |
https://fun2dolabs.com/divisibility-rules/ | 1,721,141,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00362.warc.gz | 231,379,885 | 12,342 | Divisibility Rules
Some children find math to be challenging. To solve math problems quickly and simply without tedious calculations, there are situations when tricks are required. Divisibility rules are rules that are used to determine whether a given number is divisible by a specific number.
Divisibility tests are performed to easily determine the factors for large numbers without actually dividing the numbers. These divisibility rules allow us to avoid long division by mentally determining whether a number is totally divisible by another number.
It also helps children do better on tests. To simplify and speed up the division process, division rules or tests have been formulated. Children can more effectively solve the questions if they learn the division rules in mathematics or the tests for divisibility from 1 to 20.
This teaching guide aims to further simplify the concept of divisibility rules by providing stunning posters, exhilarating activities, and colourful worksheets for practise.
### Test For Divisibility By 2
• Given number is divisible by 2 if it has any even number like 0, 2, 4, 6, 8 etc in its ones digit.
• For example, 2462, 3478 are divisible by 2.
### Test For Divisibility By 3
• A given number is divisible by 3 if the sum of all digits of the given number is multiple of 3.
• For example – 279 – as 2 + 7 + 9 = 18 and 18 is a multiple of 3, the given number is divisible by 3.
### Test For Divisibility By 4
• A given number is divisible by 4 if its last 2 digits are divisible by 4.
• For example, 3716 – as the last 2 digits, 16 is easily divisible by 4.
### Test For Divisibility By 5
• The given number is divisible by 5 if the last 2 digits are 0 or 5.
• For example, 3420, 465 as the last digits are 0 and 5.
### Test For Divisibility By 9
• The given number is divisible by 9 if the sum of all digits of the given number is multiple of 9.
• For example, 288, here the sum of all digits 2 + 8 + 8 = 18 is a multiple of 9.
### Test For Divisibility By 10
• The given number is divisible by 10 if the last digit is 0.
• For example, 2340, 560
### Test For Divisibility By 11
• The given number is divisible by 11 if all the digits are added and subtracted in the alternate pattern and the result obtained is the multiple of 11.
• For example, 913 – here on adding and subtracting alternate digits
• We get 9 – 1 + 3 = 11
Posters
Teaching divisibility rules with kid-friendly, clear, and easy-to-understand posters from Uncle Math School by Fun2Do Labs
Stories
Ignite kids’ curiosity with engaging stories for role play and skits, making the learning of this concept an exciting and effective experience. Teaching divisibility rules through stories from Uncle Math School by Fun2Do Labs :
Text of Stories
Activities
Learning divisibility rules can be made enjoyable by incorporating interactive games and activities.
### Flash Cards :
This activity can be used for better practise of divisibility rules.
• Educators can make flashcards for different numbers as shown below.
• Keep flashcards on the table.
• Guide children to pick up the flash card and solve it in the minimum possible time.
• A child who solves all flash cards correctly in the minimum possible time is declared a winner. | 774 | 3,260 | {"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} | 4.75 | 5 | CC-MAIN-2024-30 | latest | en | 0.893758 |
http://www.sciencemaster.co.uk/category/investigations/fair-testing/ | 1,607,034,808,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141732835.81/warc/CC-MAIN-20201203220448-20201204010448-00116.warc.gz | 143,169,199 | 21,339 | # Jaydeep (7) asked “Why do sponges absorb more hot water than cold water?””
Thanks team. I agree with you I don’t have an answer to Jaydeeps question. I think the best I can do is try to carry out an experiment to test the observation that you have made.
I would need a sponge (natural or manufactured), a fixed amount of water and a way of drying the sponge. If we are going to do a fair test the sponge must be the same for both the hot water and the cold water test. It also must be in the same condition (preferably dry). I would also need to be aware that you should weigh the water, the weight of hot and cold water must be the same. As water is heated the volume gets bigger.
Jaydeep …maybe you could also try the experiment above and see whether a sponge does absorb more hot water than cold water. You could let me know the result by telling me your result in the comment box below.
;
# “How does speed affect the energy of motion during a collision?” asked Damián Muñoz (9)
Sorry Damian it doesn’t really answer the question you asked. They have just presented you with more questions.
Let’s look at what they said.
Firstly energy is about work. The Energy or Work associated with a moving car is its movement. This type of energy is called Kinetic Energy. We could change the car’s Kinetic Energy by making the slope steeper (move the slope up a rung). It will make it go faster.
How do we measure the energy it is gaining as it moves faster and faster down the slope? Yes it does go faster while it is on the slope (we say that it is accelerating). Remember it starts with no energy at all (not moving). When it reaches the bottom of the slope it is going at its fastest speed.
We could measure the energy by seeing how long it takes to stop moving when it reaches the bottom of the slope. Try it. Make sure the test is fair. Fairness is very important in science investigations.
Alternatively you could involve it in a collision at the bottom of the slope…..say some cardboard or paper and see how far the car can push the shape it collides with. That would be another way to measure the ‘work’ that the moving car could do.
Damian, think about it. Experiment, and let me know how you got on.
Science Master
# Jaydeep (7) asks “Why do sponges absorb more hot water than cold water?”
Jaydeep. An interesting question, which I need to think about. I suspect it might be a question of ‘testing’. How did you arrive at the question? Did it include elements of ‘fair testing’? I shall start by passing it over to my friends.
Thanks team. Yes, I agree, science knowledge is based upon fair testing. Jaydeep’s question suggest that he has evidence that sponges absorb more hot water than cold. It is and interesting thought BUT is it true? That’s what a scientist would ask.
What do I think? My initial thought is that as water gets hot it expands, so…., the sponge might absorb less water. From another viewpoint it might be argued that the hot water ‘warms up’ the sponge and causes the sponge to expand, thus absorbing more water. Another argument might be that because both the water and the sponge expand when they get hot …..things cancel out.
The important point is that if the original investigation was not fair all the arguments about Why are meaningless.
Jaydeep ….do a fair test`and then ask the question. Many thanks for an interesting input.
# Liam Astra asks “Why does Mentos lollies make coke spurt everywhere?”
Liam, thanks for the question. Are you sure that you wanted to talk about Mentos lollies? I couldn’t find any mention of them …..only sweets …so I have focussed on them.
Let me start my answer by showing you my favourite clip of the Mentos/Diet Coke reaction.
My first question is ….is there anything special about diet coke and the Mint Mentos tablet?
Firstly there is evidence (somebody has done some experiments) that diet coke contains more carbon dioxide gas than ordinary coke. When they make coke they add carbon dioxide gas to it. Lots of gases are absorbed by liquids. Oxygen gas is absorbed by water…..this allows fish to breathe. The manufacturers add carbon dioxide to the coke because it enhances (makes better) the taste of the coke. Carbon dioxide, like oxygen is a safe gas – it is the gas that you and I breathe out – it is created by our bodies.
Experiment 1. Add a Mentos to a bottle of Coke and one to a bottle of Diet Coke….is the reaction the same? Make sure that it is a fair experiment (talk with others about how you can make it fair). Make sure that it is a safe (if you have them wear goggles) and a tidy experiment. (think about your parents or teachers)
Experiment 2. Now let’s look at the Mentos tablet. Is the tablet smooth or rough to the touch? Try and find something (maybe an adult can find you a little bit of sandpaper) that will change the surface of one of the Mentos tablets (more rough or more smooth) and drop it into a diet coke bottle. What happens? Do the rough and smooth tablets both give exactly the same ‘spurt’?
You could also carry out this experiment with Fruit and Mint Mentos tablets?
Here are some magnified photographs of the surface of a Mentos tablet.
So what have we found out? What has roughness to do with surface area? Why does the dissolved gas decide to come out of solution when it meets the tablet?
Let me know your thoughts in the box below.
# Will (8) asks “What makes seeds grow?”
Will, thank you for your question. Let’s start with some interesting questions. Answers to these might help in answering your question. Firstly is a fruit a seed? If it is not a seed then what is it for? Does it help the seed grow? How will we know if the seed is growing? What does a seed need to grow? Does it need soil? Do seeds grow when it is cold? Do different seeds grow at different speeds? What do you think? Are there any more questions? Remember questions are what science is all about. By investigating questions you are building knowledge of the world around you.
So, let’s think about how we could investigate some of these questions. Shall we look at just one type of seed, or choose a variety of different seeds to investigate? Maybe looking at one type of seed would help us begin to answer some of the questions. We could then look at another type of seed and compare the results. Maybe one seed would grow faster than the other?
Now we have to think about the conditions for our growing experiment ….. soil/no soil, wet/dry, light/no light, hot/cold. Even for our selected seed this can be very complicated. Can you see why?
The investigation equipment could probably be obtained at home. A empty plastic bottle, with the top cut off, would be a good holder for the seed. Some cotton wool could act as soil. A cupboard and a refrigerator could also help you create the right environment.
Let me know how it went?
# “How does a magnet become a magnet?” asks Ava (11)
A brilliant question Ava. I asked my team for their thoughts.
Yes team you are quite right there are some limitations to magnet making. The main one is that only certain metals can be made into magnets. These metals are called ferromagnetic metals.
Included in this group are the metal iron and the alloys of iron with the metals cobalt, nickel and some other rare earth elements .
It is thought that in these metals (including iron) have some electrons called ‘free electrons’ (not sure what an electron is, then go to Science Master Special – Atoms and Atomic Structure). It is these ‘free electrons’ that are involved in magnetism. In the alloys the ‘free’ electrons align themselves with the magnetism of the external magnet, making a (for the alloys) a permanent magnet. For iron alone the magnetism is only temporary and you can test this in the experiment below.
Look at the short video I have made below. In the ferromagnetic metal (iron alloy) crystal domains you will see free electrons. In the metal these will be moving freely. As they begin to interact with the external magnetic field, they begin to align themselves, making a permanent magnet.
# Pete (10) asks “What is the pitch of a sound?”
Thank you team. Lets first look at the motion of the particles. You suggest that the motion of the particles is in the form of waves. I think that is quite difficult to imagine but I think I have an example that will help illustrate this type of motion. Look at what happens when you drop a pebble in a pond. The pebble, when it hits the water, it creates one vibration.
In this image the sound is produced by the piston creating the waves. Notice the wavelength, that is important. If we can manipulate the piston, make it go slower or faster we can change the wavelength. Changing the wavelength is changing the pitch, think about that. How does the sound change?
Try blowing over, or tapping, some bottles.
Pitch and Frequency ..Test your hearing a little bit more …….
In the video below you can see and hear how the pitch of the sound and the wavelength change together. We measure wavelength in units called Hertz. 1 Hertz is one cycle per second. In the image above imagine that it takes one second to get from the flute to the ear. Then there are 8 cycles in the top sound is so frequency is 8 hertz and there are 3 cycles in the bottom sound so the wavelength is 3 Hertz.
You can now test your hearing. Take Care….make sure you have control of the volume.
Thanks to
Orion Lawlor, for the water ripples video, Published on 9 Jan 2011
Earmaster at https://www.earmaster.com/music-theory-online/
The ISVR from the Institute of Sound and Vibration Research, University of Southampton.
The Sound Video, unknown but thanks.
# Braiden (9) asks “What makes a diamond so hard?.
Braiden, my question (before I turn you over to my team) is how do you detect ‘hardness’? Can you create an order of hardness with some common materials?
Lets say:
wood (balsa)
wood (oak)
plastic object
bath sponge
china(cups and saucers)
glass
concrete pathway
cardboard
metal fork
coin
rock (sandstone)
piece of coal
pencil rubber
For obvious reasons I have not included diamond in this list. I think that would be an interesting experiment to carry out. You will have to make sure it is a fair test
Now over to my team.
Graphite has a layered structure with weak forces between the layers, This is a weaker structure than the close bonding of the diamond. The carbon atoms like the tetrahedral arrangement of the bonds with other carbon atoms. It is a very strong force.
Hope that makes some sense. If you have questions about the explanation please ask them. If you want to see my ‘hardness’ list go to Science Master Special-Hardness Results
# William (9), Erika (10) and Alissa (10) asked some questions on fossil fuels
Erika asked “Why are fossil fuels so expensive?”
Alissa asked “How were fossil fuels found?”
science question is a question that may lead to observations, an idea and help us in answering (or figuring out) the reason for some observation and the question.
Erika’s question is a good economics question.
Miners (those who look for oil and coal) have a variety of clever tools to help them find the fossil fuel. For example they can use ‘sniffers’. The sniffers can detect small amounts of oil vapour which might find their way out of the rocks that are hiding the oil. They also use seismic detection methods. Seismic waves were used to investigate the Earths core however the oil explorers do not use earthquakes to create the seismic wave – they use special guns or explosives. The shock waves (seismic waves) travel through the rock and at some point they are reflected back and return to the surface. The waves are recorded and examined and they tell the explorer what type of material (rock, water, oil, coal) they have traveled through.
So the question from William has led to further observations and thoughts which lead me to another question……If I left my garden rubbish for a year would it turn into coal? If not why?
# Some questions on Gold from Chantelle (7), Shelby(9) and Mary(9)
Chantelle asked “Why is gold so hard to find?”
Shelby asked “Why is gold so heavy and hard to pick up ?”
Mary asked “Why is gold so expensive?”
Chantelle, Shelby and Mary some great questions , thank you. But are they science questions? Lets look at what a science question is.
science question is a question that may lead to an idea and help us in answering (or figuring out) the reason for some observation.
For example ‘Why is gold so heavy?” …so firstly let us look at what gold is. Gold is a solid – Is it a rock? Is it wood? Is it plastic? Is it a metal? ……it seems to fit into the group called metals (it’s cold to the touch, it’s solid, it can be scratched, it’s shiny, it’s heavy) it’s a metal. So Shelby’s second question is a good science question.
Is it heavier than other metals? It doesn’t seem to be heavier than other metals but how do I test this? Fair tests are important in science investigations. Being fair I compared my gold with with metals of comparable size? It is heavier, why? Maybe the bits which make up the gold are heavier than the bits that make up other metals?
Chantelle. I think gold is quite easy to find compared to other metals. Lots of other metals, iron, silver, copper and aluminium exist as minerals so they are quite difficult to find. What is a mineral? Look at the following page.
Science Master | 3,009 | 13,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.968704 |
http://mathhelpforum.com/advanced-applied-math/113959-tetrahedron-dot-product.html | 1,480,989,576,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.44/warc/CC-MAIN-20161202170901-00321-ip-10-31-129-80.ec2.internal.warc.gz | 164,922,259 | 9,884 | # Thread: Tetrahedron and the dot product
1. ## Tetrahedron and the dot product
Here is the question
"Two pairs of opposite edges of a tetrahedron are perpendicular. Prove that the third pair is also perpendicular."
I let the 3 sides be a, b and c
assume a is perpendicular to b
assume b and c are perpendicular
then a.b=0 and b.c=o
need to show a.c=o
but I am not sure how to proceed.
Cheers
Cabouli
2. Originally Posted by Cabouli
Here is the question
"Two pairs of opposite edges of a tetrahedron are perpendicular. Prove that the third pair is also perpendicular."
I let the 3 sides be a, b and c
assume a is perpendicular to b
assume b and c are perpendicular
then a.b=0 and b.c=o
need to show a.c=o
but I am not sure how to proceed.
Suppose the tetrahedron is OABC, with one vertex at the origin, and the others at points given by the vectors $a,\ b,\ c$. You are told that $a.(c-b) = b.(a-c) = 0$, and you want to prove that $c.(b-a) = 0$. Just use linearity: $a.(c-b) = a.c-a.b,\ldots$. | 296 | 1,006 | {"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} | 3.578125 | 4 | CC-MAIN-2016-50 | longest | en | 0.917675 |
https://www.teacherspayteachers.com/Product/Force-and-motion-flashcards-1000572 | 1,490,874,474,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218193716.70/warc/CC-MAIN-20170322212953-00267-ip-10-233-31-227.ec2.internal.warc.gz | 975,786,664 | 25,573 | Total:
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# Force and motion flashcards
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### PRODUCT DESCRIPTION
The force and motion flashcards introduces students to terms that they will encounter in three lessons; linear motion, Newton’s laws of motion, and centripetal motion. Students will define and sort the terms, they will match images and terms, and then create sentences using the term. The packet includes 35 flashcards, standards, objectives, instructions. images, and vocabulary charts. The activity is a group assignment, and each group of students should be given the whole packet.
SPS7. Students will determine relationships among force, mass, and motion.
a. Apply Newton’s three laws to everyday situations by explaining the following:
• Inertia
• Relationship between force, mass and acceleration
• Equal and opposite forces
S8P5: Students will recognize characteristics of gravity, electricity, and magnetism as major kinds of forces acting in nature.
a. Recognize that every object exerts gravitational force on every other object and that the force exerted depends on how much mass the objects have and how far apart they are.
S8P3. Students will investigate relationship between force, mass, and the motion of objects.
a. Determine the relationship between velocity and acceleration.
b. Demonstrate the effect of balanced and unbalanced forces on an object in terms of gravity, inertia, and friction.
Based on a work at http://www.teacherspayteachers.com/Store/Tides.
Permissions beyond the scope of this license may be available at http://www.tidesinc.org/contact/.
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16
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90 Minutes
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 500 | 2,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.870784 |
liyin2015.medium.com | 1,675,374,480,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00183.warc.gz | 399,415,589 | 46,784 | # Learning Python — List
Chapter 1: data structure
# Sequences: List, set, Queue, tuple, string
## List :
Methods:
`>>> a = [66.25, 333, 333, 1, 1234.5]>>> print a.count(333), a.count(66.25), a.count('x')2 1 0>>> a.insert(2, -1)>>> a.append(333)>>> a[66.25, 333, -1, 333, 1, 1234.5, 333]>>> a.index(333)1>>> a.remove(333) # only remove a value>>> a[66.25, -1, 333, 1, 1234.5, 333]>>> a.reverse()>>> a[333, 1234.5, 1, 333, -1, 66.25]>>> a.sort()>>> a[-1, 1, 66.25, 333, 333, 1234.5]>>> a.pop()1234.5>>> a[-1, 1, 66.25, 333, 333]>>> a = [-1, 1, 66.25, 333, 333, 1234.5]>>> del a[0] # del is more flexible>>> a[1, 66.25, 333, 333, 1234.5]>>> del a[2:4]>>> a[1, 66.25, 1234.5]>>> del a[:]>>> a[]pos.sort(key = lambda k:(k[0],k[1])) # sort with the first element in tuple, then if its a tie, sort on the second element, or this can be used for list of list too. people = sorted(people, key=lambda x: (-x[0], x[1])) # reverse first element, and ascending the second elementor insertion_order = sorted(people, key = lambda (h,k): (-h,k))`
List Slices
List slices can also have a third number, representing the step, to include only alternate values in the slice.
`squares = [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]print(squares[::2])print(squares[2:8:3])>>>[0, 4, 16, 36, 64][4, 25]>>>`
Negative values can be used in list slicing (and normal list indexing). When negative values are used for the first and second values in a slice (or a normal index), they count from the end of the list.
`squares = [0, 1, 4, 9, 16, 25, 36, 49, 64, 81]print(squares[1:-1])>>>[1, 4, 9, 16, 25, 36, 49, 64]>>>`
If a negative value is used for the step, the slice is done backwards.
Using [::-1] as a slice is a common and idiomatic way to reverse a list.
Math operations:
`[3*x for x in [111, 222, 333]]`
If you’re going to be doing lots of array operations, then you will probably find it useful to install Numpy. Then you can use ordinary arithmetic operations element-wise on arrays, and there are lots of useful functions for computing with arrays.
`>>> import numpy>>> a = numpy.array([111,222,333])>>> a * 3array([333, 666, 999])>>> a + 7array([118, 229, 340])>>> numpy.dot(a, a)172494>>> numpy.mean(a), numpy.std(a)(222.0, 90.631120482977593)`
List as Stack: FILO, like stack it in a bottle
`list.insert(end,v) or list.append()list.pop()#example>>> stack = [3, 4, 5]>>> stack.append(6)>>> stack.append(7)>>> stack[3, 4, 5, 6, 7]>>> stack.pop()7>>> stack[3, 4, 5, 6]>>> stack.pop()6>>> stack.pop()5>>> stack[3, 4]`
List as Queue: FIFO
`list.insert(0,Value)list.pop() `
However, it is not efficient, so we use the package, Queue, include Queue(), LifoQueue(), and PriorityQueue()
`from Queue import *q=Queue.Queue()q.put(1)q.get()`
2. There are three built-in functions that are very useful when used with lists: `filter()`, `map()`, and `reduce()`.
`filter(function, sequence)` returns a sequence for which `function(item)` is true. If sequence is a `str`, `unicode` or `tuple`, the result will be of the same type; otherwise, it is always a `list`.
`>>> def f(x): return x % 3 == 0 or x % 5 == 0...>>> filter(f, range(2, 25)) # use lambda expression lambda x:x%3==0 or x%5==0[3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24]`
`map(function, sequence)` calls `function(item)` returns a list of the return values of function. To
`>>> def cube(x): return x*x*x...>>> map(cube, range(1, 11))[1, 8, 27, 64, 125, 216, 343, 512, 729, 1000]>>> seq = range(8)>>> def add(x, y): return x+y...>>> map(add, seq, seq)[0, 2, 4, 6, 8, 10, 12, 14]`
Map could be replace by a more powerful library: Numpy
`reduce(function, sequence)` returns a single value constructed by calling the binary function function on the first two items of the sequence, then on the result and the next item, and so on.
`>>> def add(x,y): return x+y...>>> reduce(add, range(1, 11))55`
More with the list:
List comprehensions are a useful way of quickly creating lists whose contents obey a simple rule.
Trying to create a list in a very extensive range will result in a MemoryError.
This code shows an example where the list comprehension runs out of memory.
`even = [2*i for i in range(10**100)]>>>MemoryError>>>`
For example, we can do the following:
`squares = [x**2 for x in range(10)]>>> [(x, y) for x in [1,2,3] for y in [3,1,4] if x != y][(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]`
2. Using list as matrix
`>>> matrix = [... [1, 2, 3, 4],... [5, 6, 7, 8],... [9, 10, 11, 12],... ][[row[i] for row in matrix] for i in range(4)] # do a transpose[[1, 5, 9], [2, 6, 10], [3, 7, 11], [4, 8, 12]]`
In the real world, you should prefer built-in functions to complex flow statements. The `zip()` function would do a great job for this use case:
`>>> zip(*matrix) # to do a transpose, each col become each row[(1, 5, 9), (2, 6, 10), (3, 7, 11), (4, 8, 12)]`
## Tuple
A tuple consists of a number of values separated by commas. Tuple is immutable, it is just like the list except it uses parenthesis instead of brackets. Tuple can contact mutable elements. for instance:
`>>> t = 12345, 54321, ‘hello!’#Tuples can be created without the parentheses, by just separating the values with commas.>>> t[0]12345>>> t(12345, 54321, ‘hello!’)>>> # Tuples may be nested:... u = t, (1, 2, 3, 4, 5)>>> u((12345, 54321, ‘hello!’), (1, 2, 3, 4, 5))>>> # Tuples are immutable:... t[0] = 88888Traceback (most recent call last): File "<stdin>", line 1, in <module>TypeError: ‘tuple’ object does not support item assignment>>> # but they can contain mutable objects:... v = ([1, 2, 3], [3, 2, 1])>>> v([1, 2, 3], [3, 2, 1])>>> empty = () # declare empty tuple>>> singleton = 'hello', # <-- note trailing comma>>> len(empty)0>>> len(singleton)1>>> singleton('hello',)`
Tuples are faster than lists, but they cannot be changed.
## Set
Python also includes a data type for sets. A set is an unordered collection with no duplicate elements. Basic uses eliminating duplicate entries. Set objects also support mathematical operations like union, intersection, difference, and symmetric difference.
set can have elements of tuple, string, but not list(convert it to tuple),
Here is a brief demonstration:
`>>> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']>>> fruit = set(basket) or set() # create a set without duplicates>>> fruitset(['orange', 'pear', 'apple', 'banana'])>>> 'orange' in fruit # fast membership testingTrue>>> 'crabgrass' in fruitFalsefruit.add("watermelon") #add one element>>> # Demonstrate set operations on unique letters from two words...>>> a = set('abracadabra')>>> b = set('alacazam')>>> a set(['a', 'r', 'b', 'c', 'd'])>>> a - b # letters in a but not in bset(['r', 'd', 'b'])>>> a | b # letters in either a or bset(['a', 'c', 'r', 'd', 'b', 'm', 'z', 'l'])>>> a & b # letters in both a and bset(['a', 'c'])>>> a ^ b # letters in a or b but not bothset(['r', 'd', 'b', 'm', 'z', 'l'])`
# Dictionaries, not sequence
Another useful data type built into Python is the dictionary (see Mapping Types — dict). Unlike sequences, which are indexed by a range of numbers, dictionaries are indexed by keys, which can be any immutable type; strings and numbers can always be keys. Tuples can be used as keys if they contain only strings, numbers, or tuples; if a tuple contains any mutable object either directly or indirectly, it cannot be used as a key. You can’t use lists as keys, since lists can be modified in place using index assignments.
It is best to think of a dictionary as an unordered set of key: value pairs, with the requirement that the keys are unique (within one dictionary). A pair of braces creates an empty dictionary: `{}`.
The main operations on a dictionary are storing a value with some key and extracting the value given the key. It is also possible to delete a key:value pair with `del`. If you store using a key that is already in use, the old value associated with that key is forgotten. It is an error to extract a value using a non-existent key.
The `keys()` method of a dictionary object returns a list of all the keys used in the dictionary, in arbitrary order (if you want it sorted, just apply the `sorted()` function to it). To check whether a single key is in the dictionary, use the `in` keyword.
Here is a small example using a dictionary:
`>>> tel = {'jack': 4098, 'sape': 4139}>>> tel['guido'] = 4127>>> tel{'sape': 4139, 'guido': 4127, 'jack': 4098}>>> tel['jack']4098>>> del tel['sape']>>> tel['irv'] = 4127>>> tel{'guido': 4127, 'irv': 4127, 'jack': 4098}>>> tel.keys()['guido', 'irv', 'jack']>>> 'guido' in telTrue>>> tel = {'guido': 4127, 'irv': 4127, 'jack': 4098}>>> tel{'jack': 4098, 'irv': 4127, 'guido': 4127}>>> tel.keys()['jack', 'irv', 'guido']>>> tel.values()[4098, 4127, 4127]>>> tel.items()[('jack', 4098), ('irv', 4127), ('guido', 4127)]>>> for k,v in tel.items():... print k,v...jack 4098irv 4127guido 4127#get the value from key>>> tel{'jack': 4098, 'irv': 4127, 'guido': 4127}>>> tel.get("adc",0)0>>> tel{'jack': 4098, 'irv': 4127, 'guido': 4127}>>> tel.get('jack')4098>>>`
The `dict()` constructor builds dictionaries directly from sequences of key-value pairs:
`>>> dict([('sape', 4139), ('guido', 4127), ('jack', 4098)]){'sape': 4139, 'jack': 4098, 'guido': 4127}`
In addition, dict comprehensions can be used to create dictionaries from arbitrary key and value expressions:
`>>> {x: x**2 for x in (2, 4, 6)}{2: 4, 4: 16, 6: 36}`
When the keys are simple strings, it is sometimes easier to specify pairs using keyword arguments:
`>>> dict(sape=4139, guido=4127, jack=4098){'sape': 4139, 'jack': 4098, 'guido': 4127}#create a dictionary from two lists:>>> keys = ['a', 'b', 'c']>>> values = [1, 2, 3]>>> dictionary = dict(zip(keys, values))>>> print(dictionary){'a': 1, 'b': 2, 'c': 3}`
## Looping Techniques
When looping through a sequence, the position index and corresponding value can be retrieved at the same time using the `enumerate()` function.
`>>> for i, v in enumerate(['tic', 'tac', 'toe']): # for list, for loop the dictionary, use d.items() or d.keys(), d.values()... print i, v...0 tic1 tac2 toe`
To loop over two or more sequences at the same time, the entries can be paired with the `zip()` function.
`>>> questions = ['name', 'quest', 'favorite color']>>> answers = ['lancelot', 'the holy grail', 'blue']>>> for q, a in zip(questions, answers):... print 'What is your {0}? It is {1}.'.format(q, a)...What is your name? It is lancelot.What is your quest? It is the holy grail.What is your favorite color? It is blue.`
To loop over a sequence in reverse, first specify the sequence in a forward direction and then call the `reversed()` function.
`>>> for i in reversed(xrange(1,10,2)):... print i...97531`
To loop over a sequence in sorted order, use the `sorted()` function which returns a new sorted list while leaving the source unaltered.
`>>> basket = ['apple', 'orange', 'apple', 'pear', 'orange', 'banana']>>> for f in sorted(set(basket)): # if we want to eliminate repeats, using set, we sort strings according to alphabeta... print f...applebananaorangepear`
# String is a Container, but Immutable
A string is a sequence of characters. A character is simply a symbol. For example, the English language has 26 characters. Computers do not deal with characters, they deal with numbers (binary). Even though you may see characters on your screen, internally it is stored and manipulated as a combination of 0’s and 1’s. This conversion of character to a number is called encoding, and the reverse process is decoding. ASCII and Unicode are some of the popular encoding used.
`astring = "Hello world!"astring2 = 'Hello world!'# triple quotes string can extend multiple linesmy_string = """Hello, welcome to the world of Python"""print(my_string)print("single quotes are ' '")print(len(astring))print(astring.index("o"))print(astring.count("l"))print(astring[3:7]) #slicingprint(astring[3:7:3])print(astring.upper())print(astring.lower())afewwords = astring.split(" ") #splitting>>> my_string = 'programiz'>>> my_string[5] = 'a'...TypeError: 'str' object does not support item assignment#iteratecount = 0for letter in 'Hello World': if(letter == 'l'): count += 1print(count,'letters found')`
Like above, python string can be indexed, counted, sliced, sorted, but cant assign new value. What if we want to change a string?
`str = 'cold'str = list(str)str.insert(0,'good ')str = ''.join(str) # ',' or '\n', we can use different way to join themprint(str)#outputgood cold#others: >>> "This will split all words into a list".split()['This', 'will', 'split', 'all', 'words', 'into', 'a', 'list']>>> ' '.join(['This', 'will', 'join', 'all', 'words', 'into', 'a', 'string'])'This will join all words into a string'>>> 'Happy New Year'.find('ew')7>>> 'Happy New Year'.replace('Happy','Brilliant')'Brilliant New Year'`
Another way:
`# enumerate()list_enumerate = list(enumerate(str))print('list(enumerate(str) = ', list_enumerate)#character countprint('len(str) = ', len(str))#outputlist(enumerate(str) = [(0, 'g'), (1, 'o'), (2, 'o'), (3, 'd'), (4, ' '), (5, 'c'), (6, 'o'), (7, 'l'), (8, 'd')]len(str) = 9`
## String Membership Test
We can test if a sub string exists within a string or not, using the keyword `in`.
`>>> 'a' in 'program'True>>> 'at' not in 'battle'False`
## String operaters
`str1 = 'Hello'str2 ='World!'# using +print('str1 + str2 = ', str1 + str2)# using *print('str1 * 3 =', str1 * 3)#outputstr1 + str2 = HelloWorld!str1 * 3 = HelloHelloHello`
string formatting
`# using triple quotesprint('''He said, "What's there?"''')# escaping single quotesprint('He said, "What\'s there?"')# escaping double quotesprint("He said, \"What's there?\"")#outputHe said, "What's there?"He said, "What's there?"He said, "What's there?"#formoat# default(implicit) orderdefault_order = "{}, {} and {}".format('John','Bill','Sean')print('\n--- Default Order ---')print(default_order)# order using positional argumentpositional_order = "{1}, {0} and {2}".format('John','Bill','Sean')print('\n--- Positional Order ---')print(positional_order)# order using keyword argumentkeyword_order = "{s}, {b} and {j}".format(j='John',b='Bill',s='Sean')print('\n--- Keyword Order ---')print(keyword_order)#output--- Default Order ---John, Bill and Sean--- Positional Order ---Bill, John and Sean--- Keyword Order ---Sean, Bill and John`
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## More from Li Yin
Founder@sylphai.com. Twitter: liyinscience. Ex AI researcher@ Meta AI. Github:https://github.com/liyin2015. | 4,548 | 14,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-06 | longest | en | 0.699383 |
https://oeis.org/A143192 | 1,719,290,729,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00062.warc.gz | 382,151,314 | 4,120 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A143192 a(n) is the smallest natural number we cannot obtain from n, n+1, n+2, n+3, n+4, n+5, n+6, n+7 and the operators +, -, *, /, using each number only once. 1
1413, 7187, 12421, 22751, 28862, 48046, 36094, 46372, 54214, 72845, 88119, 107246, 125589, 104153, 43838, 45893, 55054, 62090, 66226, 70187, 69638, 74941, 85303, 81913, 68891, 77237, 37997, 48758, 42827, 45554, 22217, 26617, 29422, 29099 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS This sequence is related to the sequences A071110 (for 5 successive integers) and A060316 (for 6 successive integers) and others sequences to come... Asymptotically the sequence tends to 67 (the first n for which a(n)=67 is n=1042). LINKS Gilles A. Fleury, Table of n, a(n) for n = 0..1500 CROSSREFS Cf. A071110, A060316, A071107, A060315, A141494, A142153. Sequence in context: A204745 A204741 A237742 * A237863 A263671 A207046 Adjacent sequences: A143189 A143190 A143191 * A143193 A143194 A143195 KEYWORD nonn AUTHOR Gilles A.Fleury, Oct 18 2008, Mar 06 2009 STATUS approved
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Last modified June 24 23:28 EDT 2024. Contains 373691 sequences. (Running on oeis4.) | 526 | 1,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.665839 |
https://estudyassistant.com/business/question20480465 | 1,632,169,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00056.warc.gz | 297,476,386 | 18,106 | , 08.01.2021 02:10 makaylasulak5462
# during the year, work-in-process increased by \$2,500, and finished goods decreased by \$6,000. calculate the amount of direct labor cost for the year. g
Cool beans is a locally owned coffee shop that competes with two large coffee chains, planeteuro and frothies. alicia, the owner, hired two students to count the number of customers that entered each of the coffee shops to estimate what percent of people who are interested in coffee are visiting each shop. after a week, the students found the following results: 589 visited cool beans, 839 visited planeteuro, and 1,290 visited frothies. the students were surprised that cool beans had 139 visits on monday which represented 59% of all people who visited one of the three coffee shops on mondays. how many people visited one of the three coffee shops during the week? | 204 | 863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.981483 |
http://www.oalib.com/search?kw=Yusuke%20Yamauchi&searchField=authors | 1,579,804,243,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00018.warc.gz | 258,426,215 | 18,343 | Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+
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Yusuke Yamauchi International Journal of Differential Equations , 2012, DOI: 10.1155/2012/417261 Abstract: Since 1960's, the blow-up phenomena for the Fujita type parabolic equation have been investigated by many researchers. In this survey paper, we discuss various results on the life span of positive solutions for several superlinear parabolic problems. In the last section, we introduce a recent result by the author. 1. Introduction 1.1. Fujita Type Results We first recall the result on the Cauchy problem for a semilinear heat equation: where , is the -dimensional Laplacian, and . Let be a bounded continuous function on . In pioneer work [1], Fujita showed that the exponent plays the crucial role for the existence and nonexistence of the solutions of (1.1). Let denote the Gaussian heat kernel: . Theorem 1.1 (see [1]). Suppose that and that its all derivatives are bounded.(i)Let . Then there is no global solution of (1.1) satisfying that？？ for？？ and .(ii)Let . Then for any there exists with the following property: if then there exists a global solution of (1.1) satisfying for and . In [2], Hayakawa showed first that there is no global solution of (1.1) in the critical case when or 2. Theorem 1.2 (see [2]). In case of , or , , (1.1) has no global solutions for any nontrivial initial data. In genaral space dimensions, Kobayashi et al. [3] consider the following problem: where and . Let be a bounded continuous function on . Theorem 1.3 (see [3]). Suppose that satisfies the following three conditions:(a) is a locally Lipschitz continuous and nondecreasing function in with？？ and for ,(b) for some ,(c)there exists a positive constant such that Then each positive solution of (1.5) blows up in finite time. Remark 1.4. (i) We remark that the proofs of the theorems in [2, 3] are mainly based on the iterated estimate from below obtained by the following integral equation: (ii) The critical nonlinearity of power type satisfies the assumptions (a), (b), and (c) in [3]. Weissler proved the nonexistence of global solution in -framework in [4]. The proof is quite short and elegant. Theorem 1.5 (see [4]). Suppose and that in ？ is not identically zero. Then there is no nonnegative global solution to the integral (1.7) with initial value . The outline of the proof is as follows. First we assume that there is a global solution. From the fact that the solution for some , we can obtain that . This contradicts the boundedness of for large . Hence the solution is not global. Existence and nonexistence results for time-global solutions of (1.1) are summarized as follows.(i)Let . Then every nontrivial solution of (1.1) blows up in finite time.(ii)Let . Then (1.1) has a time-global
Journal of Nanomaterials , 2010, DOI: 10.1155/2010/382043 Abstract: By using the polycarbonate membrane a template, mesoporous silica rods are fabricated on a silicon substrate in one pot. From scanning electron microscope (SEM) images, the creation of fibrous morphology is confirmed over the entire area. The diameter of the obtained rods is consistent with that of the template. Transmission electron microscope (TEM) images revealed that the tubular mesochannels are uniaxially oriented parallel to the longitudinal axis of the silica rods. The mesoporous titania rods with anatase crystalline frameworks are also fabricated.
Science and Technology of Advanced Materials , 2009, Abstract: Silica and alumina with macro-meso-type hierarchical pore systems are synthesized by dual templating using both surfactants and polystyrene (PS) spheres. After calcination, scanning electron microscope images show uniform macropores with a diameter of approximately 200 nm. This size coincides with that of the original PS spheres. The density of the macropores increases with the amount of added PS spheres in the precursor solutions. Transmission electron microscope images, small-angle x-ray scattering spectra and N2 adsorption–desorption isotherms reveal the formation of ordered mesoporous structures in the macropore walls. Also, the existence of micropores (less than 2 nm in size) was confirmed from the large N2 uptake at low relative pressures.
Physics , 2009, DOI: 10.1088/1742-6596/150/5/052177 Abstract: We investigate the field-angle-dependent zero-energy density of states for YNi2B2C with using realistic Fermi surfaces obtained by band calculations. Both the 17th and 18th bands are taken into account. For calculating the oscillating density of states, we adopt the Kramer-Pesch approximation, which is found to improve accuracy in the oscillation amplitude. We show that superconducting gap structure determined by analyzing STM experiments is consistent with thermal transport and heat capacity measurements.
Physics , 2007, DOI: 10.1103/PhysRevB.76.214514 Abstract: To determine the superconducting gap function of YNi2B2C, we calculate the local density of states (LDOS) around a single vortex core with the use of Eilenberger theory and the band structure calculated by local density approximation assuming various gap structures with point-nodes at different positions. We also calculate the angular-dependent heat capacity in the vortex state on the basis of the Doppler-Shift method. Comparing our results with the STM/STS experiment, the angular-dependent heat capacity and thermal conductivity, we propose the gap-structure of YNi2B2C, which has the point-nodes and gap minima along <110>. Our gap-structure is consistent with all results of angular-resolved experiments.
Computer Science , 2015, Abstract: This paper investigates a variant of the Hamiltonian Cycle (HC) problem, named the Parity Hamiltonian Cycle (PHC) problem: The problem is to find a closed walk visiting each vertex odd number of times, instead of exactly once. We show that the PHC problem is in P even when a closed walk is allowed to use an edge at most z=4 times, by considering a T-join, which is a generalization of matching. On the other hand, the PHC problem is NP-complete when z=3. In the case of z=3 however, the problem is in P when an input graph is four-edge connected, but it still remains NP-complete even when it is two-edge connected. Thus, we are concerned with the hard case in detail, and give a simple necessary and sufficient condition that a two-edge connected C>=5-free (or P6-free) graph has a PHC. Note that the HC problem is known to be NP-complete for those graph classes. This subject is motivated by a new approach to connecting a hard problem HC and an easy problem T-join, by relaxing a constraint of HC.
Science and Technology of Advanced Materials , 2011, Abstract: We report the synthesis of hydroxyapatite nanoparticles (HANPs) by the coprecipitation method using calcium D-gluconate and potassium hydrogen phosphate as the sources of calcium and phosphate ions, respectively, and the triblock copolymer F127 as a stabilizer. The HANPs were characterized using scanning electron microscopy, x-ray diffraction, and nitrogen adsorption/desorption isotherms. Removal of F127 by solvent extraction or calcination alters the structure of HANPs. The solvent-extracted HANPs were single crystals with their lang001rang axis oriented along the rod axis of the HANP, whereas the calcined HANPs contained two crystal phases that resulted in a spherical morphology. The calcined HANPs had much higher surface area (127 m2 g 1) than the solvent-extracted HANPs (44 m2 g 1).
Science and Technology of Advanced Materials , 2012, Abstract: This review article summarizes recent developments in mesoporous titania materials, particularly in the fields of morphology control and applications. We first briefly introduce the history of mesoporous titania materials and then review several synthesis approaches. Currently, mesoporous titania nanoparticles (MTNs) have attracted much attention in various fields, such as medicine, catalysis, separation and optics. Compared with bulk mesoporous titania materials, which are above a micrometer in size, nanometer-sized MTNs have additional properties, such as fast mass transport, strong adhesion to substrates and good dispersion in solution. However, it has generally been known that the successful synthesis of MTNs is very difficult owing to the rapid hydrolysis of titanium-containing precursors and the crystallization of titania upon thermal treatment. Finally, we review four emerging fields including photocatalysis, photovoltaic devices, sensing and biomedical applications of mesoporous titania materials. Because of its high surface area, controlled porous structure, suitable morphology and semiconducting behavior, mesoporous titania is expected to be used in innovative applications.
Physics , 2014, DOI: 10.1093/pasj/psu131 Abstract: We present the results of ammonia observations toward the center of NGC 3079. The NH3(J, K) = (1, 1) and (2, 2) inversion lines were detected in absorption with the Tsukuba 32-m telescope, and the NH3(1,1) through (6,6) lines with the VLA, although the profile of NH3(3,3) was in emission in contrast to the other transitions. The background continuum source, whose flux density was ~50 mJy, could not be resolved with the VLA beam of ~< 0."09 x 0."08. All ammonia absorption lines have two distinct velocity components: one is at the systemic velocity and the other is blueshifted, and both components are aligned along the nuclear jets. For the systemic components, the relatively low temperature gas is extended more than the high temperature gas. The blueshifted NH3(3,3) emission can be regarded as ammonia masers associated with shocks by strong winds probably from newly formed massive stars or supernova explosions in dense clouds in the nuclear megamaser disk. Using para-NH3(1,1), (2,2), (4,4) and (5,5) lines with VLA, we derived the rotational temperature Trot = 120 +- 12 K and 157 +- 19 K for the systemic and blueshifted components, respectively. The total column densities of NH3(0,0)-(6,6), assuming Tex ~Trot, were (8.85+-0.70) x 10^16 cm^-2 and (4.47+-0.78) x 10^16 cm-2 for the systemic and blueshifted components, respectively. The fractional abundance of NH3 relative to molecular hydrogen H2 for the systemic and blueshifted was [NH3]/[H2]=1.3x10^-7 and 6.5 x 10^-8, respectively. We also found the F = 4-4 and F = 5-5 doublet lines of OH 2{Pi}3/2 J = 9/2 in absorption, which could be fitted by two velocity components, systemic and redshifted components. The rotational temperature of OH was estimated to be Trot,OH >~ 175 K, tracing hot gas associated with the interaction of the fast nuclear outflow with dense molecular material around the nucleus.
Science and Technology of Advanced Materials , 2009, Abstract: Periodic mesoporous organosilica (PMO) spherical particles with different organic contents were synthesized in one pot by reacting 1,2-bis(triethoxysilyl)ethane (BTSE) with tetraethylorthosilicate (TEOS) using a spray-drying technique. The scanning electron microscopy observation of spray-dried products clearly showed the formation of spherical particles. The 29Si magic angle spinning nuclear magnetic resonance data revealed that the organic contents due to ethane fragments embedded in the frameworks were controllable and consistent with the BTSE/TEOS molar ratios of precursor solutions. Transmission electron microscopy, small-angle x-ray scattering, and N2 adsorption data of PMO with controlled organic contents indicated that the ethane fragments were embedded in the frameworks with the formation of ordered mesostructures. PMO with a high organic content (BTSE/TEOS=0.50) only showed a hydrophobic property. According to the same procedure, benzene groups were also integrated to a similar degree in the frameworks by using 1,4-bis(triethoxysilyl)benzene.
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