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 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://gmatclub.com/forum/hunters-and-warriors-bison-were-hunted-by-the-native-134657.html?sort_by_oldest=true | 1,511,342,992,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00771.warc.gz | 609,454,094 | 50,672 | It is currently 22 Nov 2017, 02:29
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Hunters and warriors, bison were hunted by the native
Author Message
TAGS:
### Hide Tags
Intern
Joined: 06 Mar 2012
Posts: 29
Kudos [?]: 31 [0], given: 31
Hunters and warriors, bison were hunted by the native [#permalink]
### Show Tags
18 Jun 2012, 21:39
00:00
Difficulty:
15% (low)
Question Stats:
70% (00:42) correct 30% (00:42) wrong based on 46 sessions
### HideShow timer Statistics
Hunters and warriors, bison were hunted by the native Navajo, and they used the hides for shelters and meat for food.
A) bison were hunted by the native Navajo, and they used the hides for shelters and meat for food
B) the native Navajo hunted bison, using the hides for shelters and meat for food.
C) bison were hunted by the native...
D) the native Navajo had hunted bison, and they used the hides for shelters and meat for food
E) hunting by the native Navajo...
It's easy to throw out answers A, C and E because the introductory phase is modifying incorrectly.
My question:
[Reveal] Spoiler:
Why is B the correct answer? Isn't it unclear whether the Navajo or bison is "using"? I thought that using modifies back to the noun and in this case the direct object, bison.
[Reveal] Spoiler: OA
Kudos [?]: 31 [0], given: 31
BSchool Forum Moderator
Status: Flying over the cloud!
Joined: 17 Aug 2011
Posts: 888
Kudos [?]: 735 [0], given: 44
Location: Viet Nam
GMAT Date: 06-06-2014
GPA: 3.07
Re: Modifier ? via PowerScore [#permalink]
### Show Tags
18 Jun 2012, 23:42
Quote:
Why is B the correct answer? Isn't it unclear whether the Navajo or bison is "using"? I thought that using modifies back to the noun and in this case the direct object, bison.
Nope, the adjective clause after bison will modified the Subject of Main clause or all of Main clause. In the above case, the adjective clause modified the Subject of main clause.
_________________
Rules for posting in verbal gmat forum, read it before posting anything in verbal forum
Giving me + 1 kudos if my post is valuable with you
The more you like my post, the more you share to other's need
CR: Focus of the Week: Must be True Question
Kudos [?]: 735 [0], given: 44
Intern
Joined: 03 Jan 2012
Posts: 22
Kudos [?]: 8 [3], given: 5
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.5
WE: Engineering (Computer Software)
Re: Hunters and warriors, bison were hunted by the native Navajo [#permalink]
### Show Tags
19 Jun 2012, 04:16
3
KUDOS
damham17 wrote:
Hunters and warriors, bison were hunted by the native Navajo, and they used the hides for shelters and meat for food.
A) bison were hunted by the native Navajo, and they used the hides for shelters and meat for food
B) the native Navajo hunted bison, using the hides for shelters and meat for food.
C) bison were hunted by the native...
D) the native Navajo had hunted bison, and they used the hides for shelters and meat for food
E) hunting by the native Navajo...
It's easy to throw out answers A, C and E because the introductory phase is modifying incorrectly.
My question:
[Reveal] Spoiler:
Why is B the correct answer? Isn't it unclear whether the Navajo or bison is "using"? I thought that using modifies back to the noun and in this case the direct object, bison.
As per mGmat sc guide The verb "using" modifies the main subject but not the noun just before it. In d, they is ambiguous.
Kudos [?]: 8 [3], given: 5
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 4312
Kudos [?]: 8185 [0], given: 364
Location: India
WE: Education (Education)
Re: Hunters and warriors, bison were hunted by the native Navajo [#permalink]
### Show Tags
24 Jun 2012, 11:25
First thing: Can a bison use hides for shelters and meat for food? Herbivorous as it is, does a bison eat meat? Unless you want to surrender meaning to literal structure, bison cannot be the modified noun, just because of its proximity to the modifier.
_________________
Can you solve at least some SC questions without delving into the initial statement?
Narendran 98845 44509
Kudos [?]: 8185 [0], given: 364
e-GMAT Representative
Joined: 02 Nov 2011
Posts: 2355
Kudos [?]: 9290 [0], given: 341
Re: Hunters and warriors, bison were hunted by the native Navajo [#permalink]
### Show Tags
27 Jun 2012, 21:35
Hi @danham17,
When a verb-ing modifier is separated from a clause by using comma, the verb-ing modifier modifies the entire clause preceding it. In these structures, the verb-ing modifiers present either additional information about the preceding clause or the result of the preceding clause. Also, the verb-ing associates with the subject of the preceding clause. This is so because the verb-ing modifiers are constituted from verbs and denote an action. Hence, they need to associate with the subject of the preceding clause.
Now let's take a look at the sentence in question with correct answer (Choice B):
Hunters and warriors, the native Navajo hunted bison, using the hides for shelters and meat for food.
In this choice, the verb-ing modifier "using" is preceded by a comma. Hence, it is modifying the entire preceding clause. Here, "using" is correctly presenting additional information about the preceding clause. The native Navajo hunted bison and used the hides for shelters and meat for food. Also, "using" correctly associates with the subject "the native Navajo" because they used the skin and the meat. Hence, this is the correct answer.
Hope this helps.
Thanks.
_________________
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Kudos [?]: 9290 [0], given: 341
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 4312
Kudos [?]: 8185 [1], given: 364
Location: India
WE: Education (Education)
Re: Hunters and warriors, bison were hunted by the native Navajo [#permalink]
### Show Tags
28 Jun 2012, 01:58
1
KUDOS
An important point that has been ignored by many is the wrong use of past perfect tense in D; The antecedence of – they - is not under question at all here in D, because logically – they - cannot refer to bison or hides or shelters. If you don’t consider logic, when you have some doubt a pronoun’s reference, then you will be off the mark.
D is wrong because it says - had hunted -; Hunting is a general way of life and here is no timeline for that except that the tribes did that for their living. It is simultaneous with the habit of using hides for shelters and meat for food. Therefore, you need a simple past tense – hunted - , if you want to describe the avocation of the tribes.
_________________
Can you solve at least some SC questions without delving into the initial statement?
Narendran 98845 44509
Kudos [?]: 8185 [1], given: 364
Re: Hunters and warriors, bison were hunted by the native Navajo [#permalink] 28 Jun 2012, 01:58
Display posts from previous: Sort by | 1,953 | 7,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-47 | latest | en | 0.914485 |
http://staging.physicsclassroom.com/reasoning/momentum | 1,685,495,715,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00341.warc.gz | 37,822,748 | 23,400 | # Science Reasoning Center - Momentum
You have reached the Construction Zone. As of April 1, 2023 the redo of our Science Reasoning Center was elevated to our top priority. We should have a large amount of content ready by August 1, 2023. We release content once it is ready, allowing teachers to preview and to test our latest creations for potential adoption into their curriculum and unit planning. You can stop by periodically to view our progress. It will be like watching a skyscraper being built ... only more exciting. You can also keep abreast of our developments by a periodic visit to the What's New at TPC? page
The Legacy version of the Science Reasoning Center can be found here.
The following activities are planned for our Grand Opening. Once construction is completed, we will include a link to the activity:
#### Marshmallow Launcher
This activity describes an experiment in which students alter and test variables that impact the effectiveness of a marshmallow launcher. Data is provided in the form of a table and two diagrams. Questions target a student's ability to identify the effect of changes in a variable upon an outcome, to identify dependent and independent variables, to interpolate and extrapolate from provided data, to understand the design of an experiment, to draw conclusions that are consistent with the provided data, and to extend the findings of an experiment to a new situation.
#### Air Bag Inflation and Passenger Safety
This activity describes the factors that affect the safe functioning of air bags. Data regarding delay time before inflation, inflation rates and inflation pressures are presented in the form of a table and a graph. Questions target a student's ability to find basic information in a body of text, to recognize patterns within a table of data, to draw conclusions that are consistent with data, to translate information from a table to a graph, to interpret the slope of a graph, and to recognize the implications of experimental findings.
#### Collisions
This activity describes two experiments in which the inelastic and elastic collisions of two carts on a track are investigated. Mass, pre-collision velocities, and post-collision velocities of the carts are described using two data tables. Questions target a student's ability to understand the experimental design, to identify the effect of one variable on another, to draw conclusions consistent with the data, and to interpolate and extrapolate from the given data in order to predict the result of additional trials. | 486 | 2,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.929436 |
https://www.bogleheads.org/wiki/Importance_of_saving_rate | 1,722,987,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00519.warc.gz | 522,966,941 | 15,312 | # Importance of saving rate
Importance of saving rate compares a saving rate with Return on Investment (ROI), to understand its impact on your financial goals.
Investors often focus primarily on asset allocation, expenses associated with investments (such as transaction fees and expense ratios), tax efficiency (types of account), sector allocation, active vs passive, and so on, to maximize their return on investment to reach a financial goal. The importance of saving early is also well documented.
Beyond these, though, achieving your financial goals relies on a combination of market returns and the amount you save over time (that is, your saving rate).
## Saving rate
Your saving rate is the percentage of your income that you save towards a financial goal.[1]
To put the saving rate into perspective, suppose two investors Alice and Bob each have \$100,000 in income, and each saves a percentage of their income consistently for 30 years.
• Alice saves 4% of her income (\$4,000), and her return on investment on saving is 6%. After 30 years, she has \$316,233.
• Bob saves 6% of his income (\$6,000), but his return on investment on saving is only 4%. After 30 years, he has \$336,510. His higher saving rate has more than overcome his lower return on investments.
Increasing your saving rate is particularly important when the return on investments is low. However if return on investments is very high, then the impact of saving rate becomes relatively less important. Alternately, if your saving rate is less than 4%, the saving rate you use has a huge impact on the end balance, when compared the rate of return.
Table 1 below helps to show the relationship between saving rate and return on investment for a given saving rate. For example, with a saving rate of 1%, you need a 12% return on investment to save \$241,334. In contrast, for a saving rate of 2%, a return slightly over 8% is enough to achieve the same end balance of \$241,334.
Table 1: Saving rate vs Return on Investment (ROI) for savings.
Saving rate
ROI
1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 15% 20% 25%
1% \$34,785 \$69,570 \$104,355 \$139,140 \$173,924 \$208,709 \$243,494 \$278,279 \$313,064 \$347,849 \$521,773 \$695,698 \$869,622
2% \$40,568 \$81,136 \$121,704 \$162,272 \$202,840 \$243,408 \$283,977 \$324,545 \$365,113 \$405,681 \$608,521 \$811,362 \$1,014,202
3% \$47,575 \$95,151 \$142,726 \$190,302 \$237,877 \$285,452 \$333,028 \$380,603 \$428,179 \$475,754 \$713,631 \$951,508 \$1,189,385
4% \$56,085 \$112,170 \$168,255 \$224,340 \$280,425 \$336,510 \$392,595 \$448,680 \$504,764 \$560,849 \$841,274 \$1,121,699 \$1,402,123
5% \$66,439 \$132,878 \$199,317 \$265,755 \$332,194 \$398,633 \$465,072 \$531,511 \$597,950 \$664,388 \$996,583 \$1,328,777 \$1,660,971
6% \$79,058 \$158,116 \$237,175 \$316,233 \$395,291 \$474,349 \$553,407 \$632,465 \$711,524 \$790,582 \$1,185,873 \$1,581,164 \$1,976,455
7% \$94,461 \$188,922 \$283,382 \$377,843 \$472,304 \$566,765 \$661,226 \$755,686 \$850,147 \$944,608 \$1,416,912 \$1,889,216 \$2,361,520
8% \$113,283 \$226,566 \$339,850 \$453,133 \$566,416 \$679,699 \$792,982 \$906,266 \$1,019,549 \$1,132,832 \$1,699,248 \$2,265,664 \$2,832,080
9% \$136,308 \$272,615 \$408,923 \$545,230 \$681,538 \$817,845 \$954,153 \$1,090,460 \$1,226,768 \$1,363,075 \$2,044,613 \$2,726,151 \$3,407,688
10% \$164,494 \$328,988 \$493,482 \$657,976 \$822,470 \$986,964 \$1,151,458 \$1,315,952 \$1,480,446 \$1,644,940 \$2,467,410 \$3,289,880 \$4,112,351
11% \$199,021 \$398,042 \$597,063 \$796,084 \$995,104 \$1,194,125 \$1,393,146 \$1,592,167 \$1,791,188 \$1,990,209 \$2,985,313 \$3,980,418 \$4,975,522
12% \$241,333 \$482,665 \$723,998 \$965,331 \$1,206,663 \$1,447,996 \$1,689,329 \$1,930,661 \$2,171,994 \$2,413,327 \$3,619,990 \$4,826,654 \$6,033,317
Assumptions: Annual income: \$100,000. Future value is calculated assuming money is invested at the end of year for 30 years. Values are rounded to nearest dollar.
When starting out, your saving rate has the most impact on increasing your portfolio value. The return on investment becomes important only after your portfolio has grown significantly, or after it achieves a critical mass.[2]
To achieve a financial goal, you only have to control your savings rate rather than depend on market returns.[3]
Said another way, to achieve a particular financial goal, your saving rate is more important than returns.[4] Morningstar had recommended some suggestions to increase the saving rate.[5] | 1,377 | 4,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.954623 |
https://detmagazine.ru/physics-problem-solving-strategies-515.html | 1,620,447,198,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988837.67/warc/CC-MAIN-20210508031423-20210508061423-00322.warc.gz | 227,878,468 | 8,874 | # Physics Problem Solving Strategies
Physics problems begin as word problems and terminate as mathematical exercises.Before the mathematics portion of a problem begins, a student must translate the written information into mathematical variables.It is the habit of a good problem-solver to carefully read the verbal statement and to combine the attention to units (meters, kilograms, Joules, etc.) with their understanding of the meaning of physical quantities in order to accurately extract the numerical information and equate it with the appropriate symbol.
Physics problems begin as word problems and terminate as mathematical exercises.Before the mathematics portion of a problem begins, a student must translate the written information into mathematical variables.It is the habit of a good problem-solver to carefully read the verbal statement and to combine the attention to units (meters, kilograms, Joules, etc.) with their understanding of the meaning of physical quantities in order to accurately extract the numerical information and equate it with the appropriate symbol.
Tags: Moving Schools Essay8th Grade Persuasive Writing TopicsThesis Rapid Scorm Elearning 3.5Ways To Problem SolveWrite On A PaperWhat Is The Value Of A College Education EssayEssay Newspaper ReadingEssay On Educationist
In such cases, the time taken to plot out a strategy will pay huge dividends, preventing the loss of several frustrating minutes of impulsive attempts at solving the problem.
Good problem solvers use their background knowledge of physics and physics formulae to think about how the known information is related to each other and how it is related to the final unknown quantity.
Most good problem-solvers have unique little practices which make them different from other good problem-solvers.
Nonetheless, there are several habits which they all share in common.
One of the instructional goals of the Audio Help files is to assist students in becoming better and more confident problem-solvers.
If all students who are good problem-solvers could be observed doing problems, then one would not necessarily observe that they use the same approaches to solving problems.As mentioned earlier, physics problems begin as word problems and terminate as mathematical exercises.During the algebraic/mathematical part of the problem, the student must make substitution of known numerical information into a mathematical formula (and hopefully into the correct formula ).The path from known information to the unknown quantity is often not immediately obvious.The problem becomes like a jigsaw puzzle; the assembly of all the pieces into the whole can only occur after careful inspection, thought, analysis, and perhaps some wrong turns.Nonetheless, anyone who puts effort into disciplining themselves to be successful at solving problems can learn how to be proficient at the task.A student who devotes some time and attention to the list below and makes an effort to personalize it into their own approach to problems will improve their problem-solving ability.Many errors (and perhaps even most) can be traced back to this translation process.These errors are usually the result of a failure to visualize the physical situation described in the verbal statement of the problem or of a failure in missing some strategic information during the reading process.While a good problem-solver may not religiously adhere to these habitual practices, they become more reliant upon them as the problems become more difficult.The list below describes some of the habits which good problem-solvers share in common. | 648 | 3,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.922402 |
https://www.homeworklib.com/question/related/181321/x42-x15-2x510-another-is-3b7a-a2b-3a22b214ab-another-is-2xy3xy-5x2y-this-is-an-rational-algebraic-expression-my-topic-is-all-about-sums-and-differences-of-rational-algebraic-expression-please-answer-this-please-help-me | 1,563,413,051,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00233.warc.gz | 687,378,115 | 7,747 | x+4/2 - x-1/5 + 2x+5/10 another is 3b/7a - a/2b + 3a^2-2b^2/14ab another is 2x-y/3x+y + 5x/2y this is an rational algebraic expression my topic is all about sums and differences of rational algebraic expression please answer this please HELP ME related homework questions
• x+4/2 - x-1/5 + 2x+5/10 another is 3b/7a - a/2b + 3a^2-2b^2/14ab another is 2x-y/3x+y + 5x/2y this is an rational algebraic expression my topic is all about sums and differences of rational algebraic expression please answer this please HELP ME
x+4/2 - x-1/5 + 2x+5/10 another is 3b/7a - a/2b + 3a^2-2b^2/14ab another is 2x-y/3x+y + 5x/2y this is an rational algebraic expression my topic is all about sums and differences of rational algebraic expression please answer this please HELP ME!
• please answer this my topic is all about sums and differences of rational algebraic expression perform the indicated operation reduce answer to lowest term 3/5 + 4/5 another is x/3 - y/3 another is 7/a + 2/a please answer it thanks
please answer this my topic is all about sums and differences of rational algebraic expression perform the indicated operation reduce answer to lowest term 3/5 + 4/5 another is x/3 - y/3 another is 7/a + 2/a please answer it thanks!!!!!
• please answer this my topic is all about sums and differences of rational algebraic expression 3/5 + 4/5 another is x/3 - y/3 another is 7/a + 2/a please answer it thanks
please answer this my topic is all about sums and differences of rational algebraic expression 3/5 + 4/5 another is x/3 - y/3 another is 7/a + 2/a please answer it thanks!!!!!
• sums and differences of rational algebraic expression please answer the following 9ab/cd^2 - 7ab/cd^2 another 3/x-1 - 2/x-1 another is 4x/4x-y - 2x/4x-y another is 3x/2y - 3x-2y/3x+2y another is 8/3x - 3/4x + 4/12x another is x-3/3x+1 + x+5/3x+1
sums and differences of rational algebraic expression please answer the following 9ab/cd^2 - 7ab/cd^2 another 3/x-1 - 2/x-1 another is 4x/4x-y - 2x/4x-y another is 3x/2y - 3x-2y/3x+2y another is 8/3x - 3/4x + 4/12x another is x-3/3x+1 + x+5/3x+1
• my topic is all about sums and differences of a rational algebraic expression this is the problem 3a/ax+ay + 2a/ax+ay another problem is a/4a+1 + 2/4a^2-7a-2 another is 3/a-b + 4/a+2b please anyone else who can answer my assignmen
my topic is all about sums and differences of a rational algebraic expression this is the problem 3a/ax+ay + 2a/ax+ay another problem is a/4a+1 + 2/4a^2-7a-2 another is 3/a-b + 4/a+2b please anyone else who can answer my assignment please HELP ME!
• sums and differences of a rational algebraic expression 3/a + 2/ab + 1/abc another is 2x-1/6 + x+3/12 another is 2/a-b + 3/a+b another is 2a+b/a+b - 2a-b/a-b - 3a^2-b^2/a^2-b^2 another is 2x/2x^2+5x-3 + 5/x^2-9 another is x-3/12x^3-18x^2 - x+1/16x^2-24
sums and differences of a rational algebraic expression 3/a + 2/ab + 1/abc another is 2x-1/6 + x+3/12 another is 2/a-b + 3/a+b another is 2a+b/a+b - 2a-b/a-b - 3a^2-b^2/a^2-b^2 another is 2x/2x^2+5x-3 + 5/x^2-9 another is x-3/12x^3-18x^2 - x+1/16x^2-24x
• sums and differences of rational algebraic expression answer the following 2x/x^2-2x-15 - x-2/3x^2+9 another is t/2rs^2 - 2r/3rst + 3s/30r^2t another is 3x/2x-3 - 2x/2x+3 + 3/4x^2-9 please answer this i need it tomorrow morning 3:00 please answer it i wi
sums and differences of rational algebraic expression answer the following 2x/x^2-2x-15 - x-2/3x^2+9 another is t/2rs^2 - 2r/3rst + 3s/30r^2t another is 3x/2x-3 - 2x/2x+3 + 3/4x^2-9 please answer this i need it tomorrow morning 3:00 please answer it i will WAIT FOR IT!
• sums and differences of rational algebraic expression answer the following perform the indicated operation reduce answers to lowest term 2x/x^2-2x-15 - x-2/3x^2+9 another is t/2rs^2 - 2r/3rst + 3s/30r^2t another is 3x/2x-3 - 2x/2x+3 + 3/4x^2-9 please a
sums and differences of rational algebraic expression answer the following perform the indicated operation reduce answers to lowest term 2x/x^2-2x-15 - x-2/3x^2+9 another is t/2rs^2 - 2r/3rst + 3s/30r^2t another is 3x/2x-3 - 2x/2x+3 + 3/4x^2-9 please answer this i need it tomo...
• can you answer this please about sums and differences of a rational algebraic expression this is the problem 5/x-7 - x-1/x-7 - 2+x/x-7 another is x+1/2 - x-2/3 + 2x-1/9 another is 2x+3/3 - 3x-1/2 + x-4/12
can you answer this please about sums and differences of a rational algebraic expression this is the problem 5/x-7 - x-1/x-7 - 2+x/x-7 another is x+1/2 - x-2/3 + 2x-1/9 another is 2x+3/3 - 3x-1/2 + x-4/12
please can you answer this is all about sums and differences of rational algebraic expression...please i need it please ?this is the problem....3/5+4/5=? another x/3-y/3=? another is 7/a+2/a
• find the lcd of this rational algebraic expression 4/x+2,2/2-x,3/4-x^2 another rational algebraic expression you will also find the lcd 5/y+5,3y/2y+10,2/y+10
find the lcd of this rational algebraic expression 4/x+2,2/2-x,3/4-x^2 another rational algebraic expression you will also find the lcd 5/y+5,3y/2y+10,2/y+10
• Running head: DIFFERENCES AND SIMILARITIES Differences and Similarities in the Film and Play A Raisin in the Sun Jay Dias Tulsa Community College Differences and Similarities Differences and Similarities in the Film and Play A Raisin in
Running head: DIFFERENCES AND SIMILARITIESDifferences and Similarities in the Film and PlayA Raisin in the SunJay DiasTulsa Community CollegeDifferences and Similarities Differences and Similarities in the Film and Play A Raisin in the Sun A Raisin in the Sun by Lorraine Hansber...
Need Online Homework Help?
Get FREE HOMEWORK EXPERT | 2,027 | 5,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-30 | latest | en | 0.907704 |
https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_12_Problems/Problem_24&oldid=21883 | 1,624,105,514,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00105.warc.gz | 116,610,325 | 10,059 | # 2000 AMC 12 Problems/Problem 24
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
If circular arcs $AC$ and $BC$ have centers at $B$ and $A$, respectively, then there exists a circle tangent to both $AC$ and $BC$, and to $\overline{AB}$. If the length of $BC$ is $12$, then the circumference of the circle is
$\text {(A)}\ 24 \qquad \text {(B)}\ 25 \qquad \text {(C)}\ 26 \qquad \text {(D)}\ 27 \qquad \text {(E)}\ 28$
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks. | 177 | 557 | {"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": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.850359 |
https://in.mathworks.com/matlabcentral/cody/problems/120-radius-of-a-spherical-planet/solutions/2025329 | 1,606,573,860,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195656.78/warc/CC-MAIN-20201128125557-20201128155557-00321.warc.gz | 339,333,406 | 17,013 | Cody
# Problem 120. radius of a spherical planet
Solution 2025329
Submitted on 17 Nov 2019 by Le Huu Hai
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 4*pi; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = 400*pi; y_correct = 10; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = 40000*pi; y_correct = 100; assert(isequal(your_fcn_name(x),y_correct))
4 Pass
x = -4*pi; y_correct = 1i; assert(isequal(your_fcn_name(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 214 | 720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.686453 |
https://foreach.id/EN/common/length/dm-to-kiloparsec.html | 1,627,880,221,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154304.34/warc/CC-MAIN-20210802043814-20210802073814-00479.warc.gz | 262,569,803 | 9,369 | # Convert decimeter to kiloparsec (dm to kpc)
Batch Convert
• decimeter [dm]
• kiloparsec [kpc]
Copy
_
Copy
• decimeter [dm]
• kiloparsec [kpc]
## Decimeter to Kiloparsec (dm to kpc)
### Decimeter (Symbol or Abbreviation: dm)
Decimeter is one of length units. Decimeter abbreviated or symbolized by dm. The value of 1 decimeter is equal to 0.1 meter. In its relation with kiloparsec, 1 decimeter is equal to 3.2408e-21 kiloparsec.
Decimeter is one of the many units that represent length. This unit use the SI-recognized desi prefix where the value of one decimeter is equivalent to 0.1 meter or 1 x 10-1 meter.
#### Usage
One Decimeter is used to state a tenth of a meter. For volume measurement, one cubic decimeter is equivalent to one liter
#### Relation with other units
1 decimeter equals to 100,000,000,000 picometer
1 decimeter equals to 1,000,000,000 angstrom
1 decimeter equals to 100,000,000 nanometer
1 decimeter equals to 100,000 micrometer
1 decimeter equals to 100 millimeter
1 decimeter equals to 10 centimeter
1 decimeter equals to 0.1 meter
1 decimeter equals to 0.01 decameter
1 decimeter equals to 0.001 hectometer
1 decimeter equals to 0.0001 kilometer
1 decimeter equals to 0.000062137 mile
1 decimeter equals to 0.000053996 nautical mile
1 decimeter equals to 0.000017999 league
1 decimeter equals to 0.0004971 furlong
1 decimeter equals to 0.004971 chain
1 decimeter equals to 0.019884 rod
1 decimeter equals to 0.019884 pole
1 decimeter equals to 0.019884 perch
1 decimeter equals to 0.054681 fathom
1 decimeter equals to 0.10936 yard
1 decimeter equals to 0.32808 feet
1 decimeter equals to 3.937 inch
1 decimeter equals to 1 hand
1 decimeter equals to 0.14556 elo lama
1 decimeter equals to 0.00006636 pal jawa
1 decimeter equals to 0.000054 pal sumatra
1 decimeter equals to 20.787 cicero
1 decimeter equals to 1.7322 punt
1 decimeter equals to 6.6846e-13 astronomical unit
1 decimeter equals to 1.057e-17 light year
1 decimeter equals to 3.2408e-18 parsec
1 decimeter equals to 3.2408e-21 kiloparsec
1 decimeter equals to 3.2408e-24 megaparsec
1 decimeter equals to 3.2408e-27 gigaparsec
### Kiloparsec (Symbol or Abbreviation: kpc)
Kiloparsec is one of length units. Kiloparsec abbreviated or symbolized by kpc. The value of 1 kiloparsec is equal to 30857000000000000000 meter. In its relation with decimeter, 1 kiloparsec is equal to 308570000000000000000 decimeter.
#### Relation with other units
1 kiloparsec equals to 3.0857e+31 picometer
1 kiloparsec equals to 3.0857e+29 angstrom
1 kiloparsec equals to 3.0857e+28 nanometer
1 kiloparsec equals to 3.0857e+25 micrometer
1 kiloparsec equals to 3.0857e+22 millimeter
1 kiloparsec equals to 3.0857e+21 centimeter
1 kiloparsec equals to 308,570,000,000,000,000,000 decimeter
1 kiloparsec equals to 30,857,000,000,000,000,000 meter
1 kiloparsec equals to 3,085,700,000,000,000,000 decameter
1 kiloparsec equals to 308,570,000,000,000,000 hectometer
1 kiloparsec equals to 30,857,000,000,000,000 kilometer
1 kiloparsec equals to 19,173,000,000,000,000 mile
1 kiloparsec equals to 16,661,000,000,000,000 nautical mile
1 kiloparsec equals to 5,553,800,000,000,000 league
1 kiloparsec equals to 153,390,000,000,000,000 furlong
1 kiloparsec equals to 1,533,900,000,000,000,000 chain
1 kiloparsec equals to 6,135,500,000,000,000,000 rod
1 kiloparsec equals to 6,135,500,000,000,000,000 pole
1 kiloparsec equals to 6,135,500,000,000,000,000 perch
1 kiloparsec equals to 16,873,000,000,000,000,000 fathom
1 kiloparsec equals to 33,745,000,000,000,000,000 yard
1 kiloparsec equals to 101,240,000,000,000,000,000 feet
1 kiloparsec equals to 1.2148e+21 inch
1 kiloparsec equals to 308,570,000,000,000,000,000 hand
1 kiloparsec equals to 44,915,000,000,000,000,000 elo lama
1 kiloparsec equals to 20,476,000,000,000,000 pal jawa
1 kiloparsec equals to 16,663,000,000,000,000 pal sumatra
1 kiloparsec equals to 6.4141e+21 cicero
1 kiloparsec equals to 534,510,000,000,000,000,000 punt
1 kiloparsec equals to 206,260,000 astronomical unit
1 kiloparsec equals to 3,261.6 light year
1 kiloparsec equals to 1,000 parsec
1 kiloparsec equals to 0.001 megaparsec
1 kiloparsec equals to 0.000001 gigaparsec
### How to convert Decimeter to Kiloparsec (dm to kpc):
#### Conversion Table for Decimeter to Kiloparsec (dm to kpc)
decimeter (dm) kiloparsec (kpc)
0.01 dm 3.2408e-23 kpc
0.1 dm 3.2408e-22 kpc
1 dm 3.2408e-21 kpc
2 dm 6.4816e-21 kpc
3 dm 9.7223e-21 kpc
4 dm 1.2963e-20 kpc
5 dm 1.6204e-20 kpc
6 dm 1.9445e-20 kpc
7 dm 2.2685e-20 kpc
8 dm 2.5926e-20 kpc
9 dm 2.9167e-20 kpc
10 dm 3.2408e-20 kpc
20 dm 6.4816e-20 kpc
25 dm 8.1019e-20 kpc
50 dm 1.6204e-19 kpc
75 dm 2.4306e-19 kpc
100 dm 3.2408e-19 kpc
250 dm 8.1019e-19 kpc
500 dm 1.6204e-18 kpc
750 dm 2.4306e-18 kpc
1,000 dm 3.2408e-18 kpc
100,000 dm 3.2408e-16 kpc
1,000,000,000 dm 3.2408e-12 kpc
1,000,000,000,000 dm 3.2408e-9 kpc
#### Conversion Table for Kiloparsec to Decimeter (kpc to dm)
kiloparsec (kpc) decimeter (dm)
0.01 kpc 3,085,700,000,000,000,000 dm
0.1 kpc 30,857,000,000,000,000,000 dm
1 kpc 308,570,000,000,000,000,000 dm
2 kpc 617,140,000,000,000,000,000 dm
3 kpc 925,700,000,000,000,000,000 dm
4 kpc 1.2343e+21 dm
5 kpc 1.5428e+21 dm
6 kpc 1.8514e+21 dm
7 kpc 2.16e+21 dm
8 kpc 2.4685e+21 dm
9 kpc 2.7771e+21 dm
10 kpc 3.0857e+21 dm
20 kpc 6.1714e+21 dm
25 kpc 7.7142e+21 dm
50 kpc 1.5428e+22 dm
75 kpc 2.3143e+22 dm
100 kpc 3.0857e+22 dm
250 kpc 7.7142e+22 dm
500 kpc 1.5428e+23 dm
750 kpc 2.3143e+23 dm
1,000 kpc 3.0857e+23 dm
100,000 kpc 3.0857e+25 dm
1,000,000,000 kpc 3.0857e+29 dm
1,000,000,000,000 kpc 3.0857e+32 dm
#### Steps to Convert Decimeter to Kiloparsec (dm to kpc)
1. Example: Convert 20 decimeter to kiloparsec (20 dm to kpc). Then:
Where x is the value in kiloparsec (kpc) to find
2. 1 decimeter is equivalent to 3.2408e-21 kiloparsec (1 dm is equivalent to 3.2408e-21 kpc). Or:
3. 20 decimeter (dm) is equivalent to 20 times 3.2408e-21 kiloparsec (kpc). Or:
4. Retrieved 20 decimeter is equivalent to 6.4816e-20 kiloparsec (20 dm is equivalent to 6.4816e-20 kpc). Or:
▸▸
▸▸ | 2,432 | 6,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.674087 |
http://www.teachersdomain.org/resource/ket09.math.geo.pla.origin/ | 1,369,335,473,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703682988/warc/CC-MAIN-20130516112802-00065-ip-10-60-113-184.ec2.internal.warc.gz | 728,095,192 | 13,720 | Teachers' Domain is moving soon to its new and improved home — PBS LearningMedia! Learn More
# Points of Origin
Media Type:
Video
Running Time: 1m 52s
Size: 5.5 MB
or
Source: Quantus
### Collection Developed by:
Collection Credits
### Collection Funded by:
This animation from KET illustrates how an origin is used for positive and negative measurement along a straight line and on a flat plane. It also shows how an origin, latitude, and longitude identify locations on Earth and explores how measuring temperature differs from measuring height or weight.
Background Essay
Origin is a mathematical construct used to find an exact location or measurement along a line, on a flat plane, or in three-dimensional space. Identifying an origin, the point from which measurements originate, gives us a common frame of reference to describe a location or measurement.
Temperature scales provide a practical application of the concept of origins. The Fahrenheit and Celsius scales assign arbitrary numbers to the temperatures of certain natural phenomena such as the freezing or boiling point of water. On the Fahrenheit scale, the temperature at which water freezes is +32º. So 0º, the origin of this measurement, is very cold. On the Celsius scale, the freezing point of water is 0º, and +32º is quite warm. Both Fahrenheit and Celsius are relative scales, but because we understand their designated origins, they offer useful and reliable information.
On the Kelvin scale used by scientists, temperature measurement has a set origin, not a designated one. Absolute zero (0ºK) describes the point at which objects have the least amount of energy possible. Absolute zero is -273.15º Celsius and -459.67º Fahrenheit.
The system used to describe locations on Earth provides another example of how origins provide a common frame of reference. Think of the Earth as an enormous orange with 360 equal sections. The dividing lines between the sections are the meridians, which run between the North and South Poles. You can also think of meridians as halves of great circles encompassing the globe. These lines were arbitrarily established so we can determine longitude, a point’s east-west measurement.
The "prime meridian," which passes through Greenwich, England, has been designated as 0º longitude. Longitudes are designated as E or W, that is, east or west of the prime meridian. The end line for measuring east and west longitudes is the other half of the prime meridian, the 180º line on the opposite side of the Earth.
If you drew a circle around your orange halfway between its top and bottom, this circle would correspond to Earth’s equator. Unlike the prime meridian, the equator has a fixed location. For measurement purposes, the equator has been designated 0º latitude.
Above and below the equator are the latitude lines, called parallels because these concentric circles are parallel to the equator. Parallels help determine a point’s latitude, or location along a north-south line. Latitudes north of the equator are labeled N; latitudes south of the equator are labeled S.
This complex system for finding locations on the Earth works because we all have agreed to accept the equator and the prime meridian as the origins for our measurements.
To learn more about numerical values, check out Importance of the Origin, Two Points Determine a Line, and Finding the Intersection.
Discussion Questions
• Define what is meant by the term "origin."
• Think of using a stopwatch to time a race: explain what we mean by "origin" in this situation.
• The Kelvin temperature scale has a set origin. What are some other measurements whose origins are set rather than designated?
• Look up the longitude and latitude of a particular location (your home town, a favorite place, a place that interests you).
• What do these measurements reveal about the location’s relationship to the equator and the prime meridian?
Standards
to: | 813 | 3,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2013-20 | latest | en | 0.894397 |
http://mathforum.org/kb/message.jspa?messageID=9121129 | 1,513,212,857,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948532873.43/warc/CC-MAIN-20171214000736-20171214020736-00527.warc.gz | 188,638,137 | 6,639 | Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Topic: complicated equation including bessel functions
Replies: 22 Last Post: May 8, 2013 2:43 PM
Messages: [ Previous | Next ]
Steven Lord Posts: 18,038 Registered: 12/7/04
Re: complicated equation including bessel functions
Posted: May 8, 2013 1:15 PM
"ghasem " <shaban_sadeghi@yahoo.com> wrote in message
news:kmduru\$nov\$1@newscl01ah.mathworks.com...
>> Yes. Instead of trying to solve for a root of f(z) = besselj(1, z) [for
>> example] solve for solutions of f([x; y]) = [real(besselj(1, x+1i*y));
>> imag(besselj(1, x+1i*y))]
>>
>> function output = mysystem(inputs)
>> % The inputs are all purely real
>> realPartOfInput = inputs(1);
>> imagPartOfInput = inputs(2);
>>
>> % Build a complex number from the inputs
>> z = complex(realPartOfInput, imagPartOfInput);
>>
>> % Now you can work with the complex value z internally
>> complexOutput = besselj(1, z);
>>
>> % In order to return a value to FSOLVE, the outputs must be purely real
>> % Break the output into its parts
>> output = [real(complexOutput); imag(complexOutput)];
>>
>> Instead of having a system of one equation in one unknown you have a
>> system of two equations (the real and imaginary parts of the BESSEL
>> function) in two unknowns (the real and imaginary parts of z.)
>>
>> --
>> Steve Lord
>> slord@mathworks.com
>> To contact Technical Support use the Contact Us link on
>> http://www.mathworks.com
> ====================================
> Hi steve Lord.
> thank you for your attention.I know that.But when complex unknowns are
> within argument of bessel function,do you think by using of real and imag
> commands,we have a real and imag expression,really?
Yes.
> for example,suppose we have:
> syms a b % a,b are unknown
IF you want to operate symbolically you want to declare these as real.
Otherwise Symbolic Math Toolbox will assume that they could be complex, and
so:
> c = a+1j*b;
real(c) will be real(a)-imag(b).
If instead you told Symbolic Math Toolbox that a and b were real:
syms a b real % Use SYM instead inside a function
c = a+1j*b;
real(c) % returns a since b cannot contribute
But if you're trying to solve this symbolically, you should NOT be using
FSOLVE or FZERO. Those are for finding NUMERIC solutions to equations. The
SOLVE function solves equations SYMBOLICALLY. Don't mix the two. In this
case, trying to solve the system symbolically will probably take an
extremely long time and generate an incredibly large and complicated
expression (if it can be solved symbolically at all.) I recommend solving
the system numerically using the technique I described above.
> d = besseli(1,c);
> % now write real(d),we have:
> real(d) =
> conj(besseli(1, a + b*i))/2 + besseli(1, a + b*i)/2
> % now my question is:
> do you think that "conj(besseli(1, a + b*i))/2 + besseli(1, a + b*i)/2" is
> a purely real expression?
> I think that is not.
> So,we can not use from real and imag command to obtain purely real and
> imaginary expression,when we have complex argument in bessel functions...
> when use from real and imag commands in this case,again our equation is
> complex!
> are you agree with me?
See above about FSOLVE versus SOLVE. Either keep all your calculations
symbolic or keep all your calculations numeric. Mixing the two requires
--
Steve Lord
slord@mathworks.com
To contact Technical Support use the Contact Us link on
http://www.mathworks.com
Date Subject Author
5/7/13 ghasem
5/8/13 ghasem
5/8/13 Torsten
5/8/13 Bruno Luong
5/8/13 Torsten
5/8/13 ghasem
5/8/13 Bruno Luong
5/8/13 ghasem
5/8/13 Nasser Abbasi
5/8/13 Bruno Luong
5/8/13 ghasem
5/8/13 Torsten
5/8/13 ghasem
5/8/13 Torsten
5/8/13 Bruno Luong
5/8/13 ghasem
5/8/13 Bruno Luong
5/8/13 ghasem
5/8/13 Steven Lord
5/8/13 ghasem
5/8/13 Steven Lord
5/8/13 ghasem
5/8/13 ghasem | 1,141 | 3,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-51 | longest | en | 0.83693 |
https://algo.monster/liteproblems/2392 | 1,701,544,241,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00527.warc.gz | 114,885,012 | 108,161 | # 2392. Build a Matrix With Conditions
## Problem Description
In this problem, we are tasked with constructing a `k x k` matrix where each number from `1` to `k` appears exactly once and the rest of the cells are filled with `0`. The placement of numbers must follow certain conditions given by two 2D arrays: `rowConditions` and `colConditions`. `rowConditions[i] = [above_i, below_i]` indicates that the number `above_i` must be placed in a row above the number `below_i`. Similarly, `colConditions[i] = [left_i, right_i]` indicates that the number `left_i` must be placed in a column to the left of the number `right_i`.
The challenge lies in placing all numbers from `1` to `k` such that these above and left-below and right-side conditions are met simultaneously. If it's impossible to satisfy all conditions, we must return an empty matrix.
## Intuition
The given problem has a natural correspondence to topological sorting, which is a linear ordering of vertices of a directed graph such that for every directed edge `(u, v)`, vertex `u` comes before `v` in the ordering. Here, each vertex can represent a number from `1` to `k`, and the directed edges can represent the rows and columns ordering constraints provided by `rowConditions` and `colConditions`.
To arrive at the solution, we apply topological sort separately for the row and column conditions. This will give us a valid order for the rows and columns if a valid solution exists. If topological sort isnโt possible (which would indicate that there is a cycle in the graph of conditions), that implies there's no way to satisfy the conditions, and hence we return an empty matrix.
For the implementation of topological sort, we define a function `f` that does the following:
1. Build a directed graph from the conditions given. We also maintain an in-degree count array to keep track of how many edges are coming into each node.
2. Use a queue to maintain the nodes with an in-degree of `0`. These nodes can immediately be added to the sorted result as there are no preceding nodes according to the conditions.
3. When adding a node to the result, we also decrease the in-degrees of its connected nodes. If a nodeโs in-degree reaches `0`, we add it to the queue as itโs ready to be part of the sorted result.
4. If the resulting sorted array doesn't contain exactly `k` elements, we return `None`, signaling that a topological sort is not possible.
After carrying out the topological sort for both `rowConditions` and `colConditions`, we receive two ordered arrays, one for the rows and one for the columns. If either sort fails, we return an empty matrix. Otherwise, we use a mapping to translate column orders and populate the final answer matrix according to the sorted rows and columns.
## Solution Approach
The solution for building the matrix adheres to a two-step approach using topological sorting. Letโs break down each step with its corresponding algorithms, data structures, and patterns applied:
1. Topological Sorting via BFS (Breadth-First Search):
Topological sort requires us to arrange nodes (in this case, numbers from `1` to `k`) such that for every directed edge `(u, v)`, `u` is before `v`. We implement topological sorting through a modified BFS algorithm.
• Graph Representation: We use a `defaultdict` of lists to create an adjacency list representation of a graph. The keys are our nodes (numbers) and the values are lists of adjacent nodes that must appear after the key node.
• In-Degree Calculation: An array `indeg` is utilized to count how many edges point to each node. It essentially tells us if there are any conditions that need to be met before placing this node into our matrix.
• Queue for Nodes with Zero In-Degree: A `deque` is utilized to keep track of the nodes that have no incoming edges (in-degree of `0`). These are the nodes that can be placed immediately since there are no constraints on them.
• BFS and Updating In-Degree: From the nodes in the queue, we perform a BFS. When a node is placed into the result, we iterate over its adjacent nodes in the graph, decrease their in-degree by one, and if any of these adjacent nodes now have an in-degree of `0`, they are added to the queue.
• Checking for a Successful Sort: If we cannot visit all `k` nodes during the topological sort (i.e., the resulting list is shorter than `k`), this indicates that there are cyclic dependencies and we cannot build a valid matrix. Therefore, we return `None` to indicate failure.
2. Building the Result Matrix:
Once we have a valid topological sort for both rows (`rowConditions`) and columns (`colConditions`), we proceed to populate our `k x k` matrix.
• Mapping Columns: We create an array `m` that maps each number to its location in the column order. This allows us to know where to place each number in the matrix columns.
• Populating the Matrix: With `row` and `m` giving us the row and column indices respectively, we iterate through the sorted row indices, and for each number, we use the column index from the mapping `m`. This is where we place the value `v` using `ans[i][m[v]] = v`.
• Resulting Matrix: The result `ans` is the final `k x k` matrix that satisfies all the given conditions, filled with numbers from `1` to `k`, respecting the topological order, and with all remaining unfilled cells populated with zeroes.
In the case where either the row topological sort or column topological sort fails, we know that there is no valid solution and we return an empty matrix `[]`. Otherwise, the filled `ans` matrix is returned as the solution.
### Example Walkthrough
Let's go through an example to illustrate the solution approach described above. Assume we have a 3x3 matrix (k=3) and we need to place numbers 1, 2, and 3 in the matrix according to the given conditions.
Suppose `rowConditions` and `colConditions` are given as follows:
``````1rowConditions = [[2, 3], [1, 2]]
2colConditions = [[1, 3], [2, 3]]``````
Step 1: Topological Sorting
We start with topological sorting for the rows and columns separately. We'll use a Breadth-First Search (BFS) algorithm for this sorting.
• For rows:
1. Build a graph from the `rowConditions` and calculate the in-degrees:
• Node 2 should come before 3, and Node 1 should come before 2.
• Graph: 1 -> 2 -> 3 (1 points to 2, and 2 points to 3)
• In-degree: `[0, 1, 1]` since Node 1 has 0 in-degree (no conditions), Node 2 has 1 in-degree (from 1), and Node 3 has 1 in-degree (from 2).
2. Perform BFS:
• Add Node 1 to the result, queue next is Node 2 (decrement in-degrees and check).
• Add Node 2 to the result, queue next is Node 3 (decrement in-degrees and check).
• For columns:
1. Build a graph from the `colConditions` and calculate the in-degrees:
• Graph similar to row graph: 1 -> 3, 2 -> 3
• In-degree: `[0, 0, 2]` (Node 3 has 2 in-degrees, Node 1 and Node 2 have 0).
2. Perform BFS:
• Queue: Start with Node 1 and Node 2 (both have in-degree 0).
• Add Node 1 and Node 2 to the result, queue next is Node 3.
Step 2: Building the Result Matrix
Now, we use the obtained topological sorts to fill in our matrix.
• Row order from topological sort: `[1, 2, 3]`
• Column order from topological sort: `[1, 2, 3]`
Create a mapping of column indices: `m = [0, 1, 2]`
We now populate the matrix `ans`, using the order from row and column topological sorts and the mapping `m`:
• Place 1 at `ans[0][m[0]]`, which is `ans[0][0]`
• Place 2 at `ans[1][m[1]]`, which is `ans[1][1]`
• Place 3 at `ans[2][m[2]]`, which is `ans[2][2]`
The remaining cells are filled with zeros. Our final matrix looks like this:
``````1ans = [
2 [1, 0, 0],
3 [0, 2, 0],
4 [0, 0, 3]
5]``````
This matrix satisfies the given row and column conditions:
• Number 2 is below number 1 (row condition).
• Number 3 is below number 2 (row condition).
• Number 3 is to the right of number 1 (column condition).
• Number 3 is to the right of number 2 (column condition).
In conclusion, the matrix `ans` is the valid configuration we were looking to construct based on the given `rowConditions` and `colConditions`. If at any point during the topological sorting step we found a cycle or could not visit all nodes, that would have indicated there is no valid matrix configuration, and the result would have been an empty matrix. But in this example, everything checks out and we successfully built the matrix.
## Python Solution
``````1from collections import defaultdict, deque # for using defaultdict and deque
2
3class Solution:
4 def buildMatrix(self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]) -> List[List[int]]:
5
6 # Helper function to find the topological order based on the given conditions
7 def find_topological_order(conditions):
8 graph = defaultdict(list)
9 indegree = [0] * (k + 1)
10
11 # Create graph and indegree list
12 for u, v in conditions:
13 graph[u].append(v)
14 indegree[v] += 1
15
16 # Initialize queue with nodes having indegree 0
17 queue = deque([node for node in range(1, k+1) if indegree[node] == 0])
18 order = []
19
20 # Perform BFS to validate and get the topological order
21 while queue:
22 node = queue.popleft()
23 order.append(node)
24 for neighbor in graph[node]:
25 indegree[neighbor] -= 1
26 if indegree[neighbor] == 0:
27 queue.append(neighbor)
28
29 # If we can't order exactly `k` elements, return None
30 return order if len(order) == k else None
31
32 # Find topological order for both rows and columns
33 row_order = find_topological_order(rowConditions)
34 column_order = find_topological_order(colConditions)
35
36 # If either row or column topological order is impossible, return an empty matrix
37 if row_order is None or column_order is None:
38 return []
39
40 # Initialize the matrix to be returned
41 matrix = [[0] * k for _ in range(k)]
42 column_position = [0] * (k + 1)
43
44 # Fill in the column position for each element based on its order
45 for position, element in enumerate(column_order):
46 column_position[element] = position
47
48 # Place each element in the matrix based on the row and column orders
49 for i, elem_row in enumerate(row_order):
50 col = column_position[elem_row]
51 matrix[i][col] = elem_row
52
53 return matrix
54``````
## Java Solution
``````1class Solution {
2 private int size;
3
4 // Function to construct the matrix based on given row and column conditions
5 public int[][] buildMatrix(int k, int[][] rowConditions, int[][] colConditions) {
6 this.size = k;
7 // Determine the order of elements based on row conditions
8 List<Integer> rowOrder = getOrder(rowConditions);
9 // Determine the order of elements based on column conditions
10 List<Integer> colOrder = getOrder(colConditions);
11
12 // If we cannot satisfy either row or column conditions, return an empty matrix
13 if (rowOrder == null || colOrder == null) {
14 return new int[0][0];
15 }
16
17 // Initialize the answer matrix with zeroes
18 int[][] matrix = new int[size][size];
19 // Array to store the column indices for the elements in the matrix
20 int[] columnMapping = new int[size + 1];
21 // Map the value to its corresponding column index
22 for (int i = 0; i < size; ++i) {
23 columnMapping[colOrder.get(i)] = i;
24 }
25 // Fill the matrix with the correct values at the right positions
26 for (int i = 0; i < size; ++i) {
27 matrix[i][columnMapping[rowOrder.get(i)]] = rowOrder.get(i);
28 }
29 // Return the fully constructed matrix
30 return matrix;
31 }
32
33 // Function to determine the order of elements based on conditions (edges of a directed graph)
34 private List<Integer> getOrder(int[][] conditions) {
35 // Graph to represent conditions
36 List<Integer>[] graph = new List[size + 1];
37 // Initialize lists for all vertices in the graph
38 Arrays.setAll(graph, element -> new ArrayList<>());
39 // Array to store the number of incoming edges (in-degree) for each vertex
40 int[] incomingEdges = new int[size + 1];
41
42 // Build the graph based on the conditions
43 for (var edge : conditions) {
44 int from = edge[0], to = edge[1];
46 ++incomingEdges[to];
47 }
48
49 // Queue to store the vertices with no incoming edges
50 Deque<Integer> queue = new ArrayDeque<>();
51 for (int i = 1; i <= size; ++i) {
52 if (incomingEdges[i] == 0) {
53 queue.offer(i);
54 }
55 }
56
57 List<Integer> order = new ArrayList<>();
58 // Process vertices in topological order
59 while (!queue.isEmpty()) {
60 int vertex = queue.pollFirst();
62 // Decrease the in-degree of neighboring vertices and add to queue if they have no incoming edges left
63 for (int neighbour : graph[vertex]) {
64 if (--incomingEdges[neighbour] == 0) {
65 queue.offer(neighbour);
66 }
67 }
68 }
69
70 // If the size of the order list equals the size k, we successfully found an order, otherwise return null
71 return order.size() == size ? order : null;
72 }
73}
74``````
## C++ Solution
``````1#include <vector>
2#include <queue>
3
4class Solution {
5public:
6 int dimension;
7
8 // Helper function to process the topological sort on the given conditions
9 vector<int> topologicalSort(vector<vector<int>>& conditions) {
10 // Graph representation and array to track incoming edges
11 vector<vector<int>> graph(dimension + 1);
12 vector<int> inDegree(dimension + 1);
13
14 // Create graph and update in-degrees for each node
15 for (auto& edge : conditions) {
16 int from = edge[0], to = edge[1];
17 graph[from].push_back(to);
18 ++inDegree[to];
19 }
20
21 queue<int> queue;
22
23 // Initialize queue with nodes having no incoming edges
24 for (int i = 1; i <= dimension; ++i) {
25 if (inDegree[i] == 0) {
26 queue.push(i);
27 }
28 }
29
30 vector<int> sortedOrder;
31 // Perform topological sort
32 while (!queue.empty()) {
33 int count = queue.size();
34 while (count--) {
35 int node = queue.front();
36 sortedOrder.push_back(node);
37 queue.pop();
38 // Reduce in-degree for neighboring nodes and
39 // add them to the queue if their in-degree becomes zero
40 for (int neighbor : graph[node]) {
41 if (--inDegree[neighbor] == 0) {
42 queue.push(neighbor);
43 }
44 }
45 }
46 }
47 // If the sorted order does not include all nodes, return an empty array
48 return sortedOrder.size() == dimension ? sortedOrder : vector<int>();
49 }
50
51 // Public method to build the desired matrix based on row and column conditions
52 vector<vector<int>> buildMatrix(int dimension, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
53 this->dimension = dimension;
54 // Perform topological sort on both row and column conditions
55 vector<int> rowOrder = topologicalSort(rowConditions);
56 vector<int> colOrder = topologicalSort(colConditions);
57
58 // If either sort fails, return an empty matrix
59 if (rowOrder.empty() || colOrder.empty()) return {};
60
61 // Initialize the answer matrix with the correct dimensions
63 vector<int> colIndexMap(dimension + 1);
64
65 // Map the column order numbers to indices
66 for (int i = 0; i < dimension; ++i) {
67 colIndexMap[colOrder[i]] = i;
68 }
69
70 // Fill matrix using the sorted row and column orders
71 for (int i = 0; i < dimension; ++i) {
72 int rowIndex = i;
73 int colIndex = colIndexMap[rowOrder[i]];
75 }
77 }
78};
79``````
## Typescript Solution
``````1function buildMatrix(k: number, rowConditions: number[][], colConditions: number[][]): number[][] {
2 // Helper function to process conditions and perform topological sorting
3 function processConditions(conditions: number[][]): number[] {
4 // Create an adjacency list for the graph representing conditions
5 const graph = Array.from({ length: k + 1 }, () => []);
6 // Array to store in-degrees of each node
7 const inDegrees = new Array(k + 1).fill(0);
8
9 // Populate the adjacency list and update in-degrees
10 for (const [from, to] of conditions) {
11 graph[from].push(to);
12 ++inDegrees[to];
13 }
14
15 // Queue to manage nodes with in-degree of 0
16 const queue: number[] = [];
17 for (let i = 1; i < inDegrees.length; ++i) {
18 if (inDegrees[i] === 0) {
19 queue.push(i);
20 }
21 }
22
23 // List to store the order of nodes after topological sorting
24 const order: number[] = [];
25 while (queue.length) {
26 for (let i = queue.length; i > 0; --i) {
27 const node = queue.shift();
28 order.push(node);
29 // Decrease in-degrees of adjacent nodes and enqueue if in-degree becomes 0
30 for (const adjacent of graph[node]) {
31 if (--inDegrees[adjacent] === 0) {
33 }
34 }
35 }
36 }
37
38 // If the topologically sorted list has k elements, return it; otherwise, return empty array
39 return order.length === k ? order : [];
40 }
41
42 // Process the row conditions and column conditions through the helper function
43 const rowOrder = processConditions(rowConditions);
44 const colOrder = processConditions(colConditions);
45
46 // If we cannot find a valid order for either rows or columns, return an empty array
47 if (rowOrder.length === 0 || colOrder.length === 0) return [];
48
49 // Initialize the matrix with zeros
50 const matrix = Array.from({ length: k }, () => new Array(k).fill(0));
51
52 // Mapping to store the index of each value in colOrder
53 const colIndexMap = new Array(k + 1).fill(0);
54 for (let i = 0; i < k; ++i) {
55 colIndexMap[colOrder[i]] = i;
56 }
57
58 // Fill the matrix using the order of rows and mapped column indices
59 for (let i = 0; i < k; ++i) {
60 matrix[i][colIndexMap[rowOrder[i]]] = rowOrder[i];
61 }
62
63 // Return the constructed matrix
64 return matrix;
65}
66``````
## Time and Space Complexity
### Time Complexity
The given code consists primarily of two parts: constructing the order of elements according to the input conditions (`rowConditions` and `colConditions`) and then building the matrix based on these orders.
1. For constructing the order with the function `f(cond)`:
• It creates a graph where each element `i` points to the elements it must precede in sequence. The graph creation takes `O(E)` time where `E` is the total number of conditions in `rowConditions` or `colConditions`.
• It initializes an indegree array `indeg` which takes `O(k)` time.
• It processes each node exactly once using a queue to perform topological sort, which will be `O(V + E)`, where `V` is the number of vertices (which is `k`) and `E` is the number of edges.
• As `f(cond)` is called twice (once for rows and once for columns), the time complexity for this part is `2 * O(V + E) = O(2k + 2E) = O(k + E)`.
2. For the construction of the answer matrix:
• The code uses a map `m` to keep track of the index at which each value should be placed in the matrix, which takes `O(k)` time.
• It then iterates over each element of `row` and places it in the correct position in the `ans` matrix, which takes `O(k)` time.
Overall, the time complexity of the code is determined by the topological sort and the construction of the result matrix, which would be `O(k + E) + O(k) = O(k + E)`.
### Space Complexity
The space complexity of the code is determined by:
• The graph `g` and indegree array `indeg` which collectively take `O(k + E)` space.
• The queue `q` used for storing intermediate values while doing topological sort, which in the worst case could store all `k` vertices, taking `O(k)` space.
• The result array `res` that will contain `k` elements when the order is successfully constructed, taking `O(k)` space.
• The answer matrix `ans` which is a `k x k` matrix, so it takes `O(k^2)` space.
• Auxiliary space for the map `m` and variables used to iterate, which take `O(k)`.
When added together, the space complexity is `O(k + E) + O(k) + O(k) + O(k^2) + O(k) = O(k^2 + k + E)`.
Got a question?ย Ask the Teaching Assistantย anything you don't understand.
Still not clear? Ask in the Forum, ย Discordย orย Submitย the part you don't understand to our editors.
โ
โTA ๐จโ๐ซ | 5,292 | 21,538 | {"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} | 3.28125 | 3 | CC-MAIN-2023-50 | longest | en | 0.906186 |
https://gmatclub.com/forum/if-we-have-the-below-if-a-b-not-b-which-can-111293.html?fl=similar | 1,508,535,931,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00515.warc.gz | 695,035,702 | 43,656 | It is currently 20 Oct 2017, 14:45
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Events & Promotions
Events & Promotions in June
Open Detailed Calendar
If we have the below If A -----> ~B ( not B ) which can
Author Message
Director
Joined: 07 Jun 2004
Posts: 610
Kudos [?]: 922 [0], given: 22
Location: PA
If we have the below If A -----> ~B ( not B ) which can [#permalink]
Show Tags
22 Mar 2011, 09:04
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
If we have the below
If A -----> ~B ( not B )
which can we infer
~A ------> B
OR
~B ---------> A
can some one explain the contrapositive or negation which is right and valid in GMAT CR
_________________
If the Q jogged your mind do Kudos me : )
Kudos [?]: 922 [0], given: 22
Manager
Joined: 10 Nov 2010
Posts: 158
Kudos [?]: 334 [0], given: 6
Show Tags
22 Mar 2011, 09:31
In my opinion
B-------->~A
Kudos [?]: 334 [0], given: 6
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 871
Kudos [?]: 396 [0], given: 123
Show Tags
22 Mar 2011, 10:20
'When A is true, B is true' - original
'converse' says 'when B is true, A is true'.
Contra positive is 'when B is not true, A is not true'.
I tried using this approach didn't work for me especially when it comes to super hard questions. You have to dissect the fluff immediately to see the skyline. Too formalistic approach being "algebraic" pays off after a lot of hard work. The other way is using critical reasoning in day to day life. Just use CR in any damn conversation - what is the assumption my friend made? What is the "must" be true in my friend's argument? What will I say to completely destroy the argument / to strengthen the argument? Is my friend making a realistic assumption. And so on Hope that helps.
Kudos [?]: 396 [0], given: 123
Re: Clarification on Syllogism [#permalink] 22 Mar 2011, 10:20
Display posts from previous: Sort by | 708 | 2,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-43 | latest | en | 0.867275 |
http://mathhelpforum.com/calculus/21868-find-stationary-points-curve-y-3x-1-x-2-4-a.html | 1,480,985,961,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.44/warc/CC-MAIN-20161202170901-00464-ip-10-31-129-80.ec2.internal.warc.gz | 157,510,484 | 10,772 | # Thread: Find the stationary points on the curve y=(3x-1)(x-2)^4
1. ## Find the stationary points on the curve y=(3x-1)(x-2)^4
Find the stationary points on the curve y=(3x-1)(x-2)^4
I differentiated it using the udv + vdu, then i tried to work it out and i got no were close to the answer i cant figure how to do my working out properly.
back of book says it (2,0) and (2/3, 3/13/81)
can sum1 plz show me how they would do working out on this question.
thx
2. $f(x) = (3x-1)(x-2)^4$
$f'(x) = (3x-1)\cdot \frac{d}{\,dx}((x-2)^4) + (x-2)^4 \cdot \frac{d}{\,dx}(3x-1)$
The derivatives are:
$\frac{d}{\,dx} ((x-2)^4) = 4(x-2)^3$
And:
$\frac{d}{\,dx} (3x-1) = 3$
So:
$\Rightarrow f'(x) = 4(3x-1)(x-2)^3 + 3(x-2)^4$
Setting f'(x) = 0, we get:
$4(3x-1)(x-2)^3 + 3(x-2)^4 =0$
Now we can't simply divide by (x-2), or we will be missing out on a solution, so instead we factor out:
$(x-2)^3(4(3x-1)+3(x-2))=0$
$(x-2)^3(12x-4+3x-6)=0$
$(x-2)^3(15x-10)=0$
$\mbox{\Rightarrow x = 2 or x=\frac{2}{3}}\\$
Then, after substituting these values back into the original equation, we get the stationary points as:
$\lbrace{(x,y)2,0),\left(\frac{2}{3},\frac{256}{81}\right)\rbrac e}" alt="\lbrace{(x,y)2,0),\left(\frac{2}{3},\frac{256}{81}\right)\rbrac e}" />
3. thanx a lot
4. Hello, KavX!
Find the stationary points on the curve: . $y\:=\:(3x-1)(x-2)^4$
Advice: After doing a Product Rule, think of factoring.
We have: . $y' \;=\;(3x-1)\cdot4(x-2)^3\cdot 1 \:+\: 3\cdot(x-2)^4$
Simplify: . $y' \;=\;4(x-2)^3(3x-1) + 3(x-2)^4$
Factor: . . $y' \;=\;(x-2)^3\left[4(3x-1) + 3(x-2)\right]$
Simplify: . $y' \;=\;(x-2)^3(15x-10) \;=\;5(x-2)^3(3x-2)$
To solve $y' = 0$, we have: . $5(x-2)^3(3x-2) \;=\;0$
Got it?
5. aahh yer thanx for that | 773 | 1,747 | {"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": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-50 | longest | en | 0.607251 |
http://www1.maths.leeds.ac.uk/~kisilv/courses/math255.html | 1,539,866,318,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511806.8/warc/CC-MAIN-20181018105742-20181018131242-00303.warc.gz | 578,240,402 | 18,943 | Geometry
# Chapter 1 General Information
This is an online manual designed for students. The manual is available at the moment in HTML with frames (for easier navigation), HTML without frames and PDF formats. Each from these formats has its own advantages. Please select one better suit your needs.
There is on-line information on the following courses:
## 1.1 Web page
There is a Web page which contains this course description as well as other information related to this course. Point your Web browser to
## 1.2 Warnings and Disclaimers
Before proceeding with this interactive manual we stress the following:
• These Web pages are designed in order to help students as a source of additional information. They are NOT an obligatory part of the course.
• The main material introduced during lectures and is contained in Textbook. This interactive manual is NOT a substitution for any part of those primary sources of information.
• It is NOT required to be familiar with these pages in order to pass the examination.
• The entire contents of these pages is continuously improved and updated. Even for material of lectures took place weeks or months ago changes are made.
# Course Outline
1 General Information
1.1 Web page
1.2 Warnings and Disclaimers
0 Seven Top Reasons to Enjoy Geometry
1 Points and Lines Connected with a Triangle
1.1 The extended Law of Sines
1.2 Ceva's theorem
1.3 Points of intersect
1.4 The incircle and excircles
1.6 The orthic triangle
1.7 The medial triangle and Euler line
2 Some properties of Circles
2.1 The power of point with respect to a circle
2.2 The radical axis of two circles
2.3 Coaxal circles
2.5 Simpson line
2.6 Ptolemy's theorem and its extension
3 Collinearity and Concurrence
3.4 Menelaus's theorem
3.5 Pappus's theorem
3.6 Perspective triangles; Desargues's theorem
3.7 Hexagons
3.8 Pascal's theorem
4 Transformations
4.1 Translations
4.2 Rotations
4.3 Half-turn
4.4 Reflections
4.7 Dilation
4.8 Spiral symmetry
4.9 A genealogy of transformations
5 An Introduction to Inversive Geometry
5.1 Separation
5.2 Cross Ratio
5.3 Inversion
5.4 The inversive plane
6 An Introduction to Projective Geometry
6.1 Reciprocation
6.3 Conics
6.5 The projective plane
6.7 Stereographic and gnomonic projection
A Some Useful Theorems
B Some Useful Tricks
B.1 Look for a triangle
B.2 Investigate a particular case
Index
# Chapter 0 Seven Top Reasons to Enjoy Geometry
There are many reasons to enjoy Geometry. No, ``I need to pass the exam'' is not among them. These reasons are of a much pleasant nature:
• Geometry is elementary. To understand even most advanced results you need only to know simple notions like lines, circles, triangles, etc.
• Geometry is beautiful. The inner harmony of geometrical constructions is explicit.
• Geometry is real. It describes the property of the world around us.
• Geometry is principal. It was the first field of mathematical knowledge which set up a model for all other branches of mathematics.
• Geometry is reach. You may meet all variety of mathematical tools employed in geometry.
• Geometry is modern. Geometric results are used in all contemporary fields of mathematics and are source of inspiration for many new theories.
• Geometry is surprising. One is greatly impressed by the unexpected deep and beauty of geometrical results.
Exercise 1 Illustrate each of the above statements by at least one geometrical construction or theorem.
Exercise* 2 Give at least one more good reason to enjoy geometry.
# Chapter 1 Points and Lines Connected with a Triangle
## 1.1 The extended Law of Sines
Theorem 1 [Law of Sines] For a triangle ABC with circumradius R
a sinA = b sinB = c sinC = 2R
PROOF. The proof is based on the Theorem A.0.1 from the Useful Theorem Chapter . [¯]
Exercise 2 For any triangle ABC, even if B and C is an obtuse angle, a=bcosC+c cosB. Use the Law of Sines to deduce the additional formula
sin(B+C)=sinB cosC + cosB sinC.
Exercise 3 In any triangle ABC,
a(sinB −sinC) + b(sinC − sinA) + c (sinA −sinB) = 0.
Exercise 4 In any triangle ABC, (ABC)=abc/4R1 .
## 1.2 Ceva's theorem
The line segment joining a vertex of a triangle to any given point on the opposite side is called a cevian.
Theorem 1 [Ceva's theorem (1678)] If three cevians AX, BY, CZ, one through each vertex of a triangle ABC, are concurrent, then
BX XC CY YA AZ ZB =1
PROOF. Three cevians are concurrent so they pass through one point, say P. The proof follows from a consideration of areas of triangles ABP, BPC, CPA. The key point is Lemma on the area of two triangle with a common altitude . [¯]
Exercise 2 If X, Y, Z are midpoints of the sides, the three cevians are concurrent.
Exercise 3 Let XB/XC=p, YC/YA=q, and AX, BY, CZ are concurrent. Find AZ/ZB.
Exercise 4 Cevians perpendicular to the opposite sides are concurrent.
Exercise 5 Let ABC and A′B′C′ be two non-congruent triangles whose sides are respectively parallel. Then the three lines AA′, BB′, and CC′ (extended) are concurrent.
## 1.3 Points of intersect
The most important points and lines of intersect in a triangle are:
1. Circumcenter -the center of the circle circumscribed about a triangle, denoted by O. The circle is called circumcircle. Its radius ( circumradius) denoted by R.
2. The cevians that joint the vertices of a triangle to the midpoints of the opposite sides are called medians. They are concurrent (see Exercise 1.2.2) and the common point is centroid G, which is ``center of gravity'' of the triangle.
Theorem 1 A triangle is dissected by its medians into six smaller triangles of equal area.
3. The medians of a triangle divide one another in the ratio 2:1; in other words, the medians of a triangle ``trisect'' one another.
4. Cevians perpendicular to corresponding lines are altitudes. It follows from Exercise 1.2.4 that altitudes are concurrent, they intersect in orthocenter H. Feets of altitudes form the orthic triangle.
5. Bisectors are cevians which divide angles to two equal parts.
Theorem 2 Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides.
PROOF. There at least two ways to make a proof:
1. Applying Law of Sines to two resulting triangles.
2. Considering the ratio of areas of those two triangles.
[¯]
Theorem 3 The internal bisectors of the three angles of a triangle are concurrent.
PROOF. The proof follows from observation that points of bisectors are equidistant from the sides of the triangle. The point of concurrence I is incenter, that is the center of inscribed circle, which has all three sides for tangents. Its radius is inradius. [¯]
Exercise 4 The circumcenter and orthocenter of an obtuse-angled triangle lie outside the triangle.
Exercise 5 Find the ratio of the area of a given triangle to that of triangle whose sides have the same lengths as medians of the original triangle.
Exercise 6 Any triangle having two equal medians is isosceles.
Exercise 7 Any triangle having two equal altitudes is isosceles.
Exercise 8 Use Cevas Theorem to obtain another proof of Theorem 1.3.3.
Exercise 9 The product of two sides of a triangle is equal to the product of the circumdiameter and the altitude on the third side.
## 1.4 The incircle and excircles
Let the incircle touch sides BC, CA, AB at X, Y, Z. If x=AZ=AY, y=BX=BZ, and z=CX=CY. Let s = [ 1/2] (x+y+z) be semiperimeter.
Theorem 1 x=s−a, y=s−b, z=s−c.
PROOF. It follows from the Theorem on two tangents . [¯]
Theorem 2 (ABC)=sr.
PROOF. It follows from the Theorem on areas . [¯]
Theorem 3 The external bisectors of any two angles of a triangle are concurrent with the internal bisector of the third angle.
The circles with with above centers ( excenters) are excircles escribed to the triangle. Their radii ( exradii) denoted ra, rb, rc. Incircle together with excircles are four tritangent circles.
Exercise 4 The cevians AX, BY, CZ are concurrent. Their common point is the Gergonne point.
Exercise 5 ABC is the orthic triangle of the triangle formed by excenters.
Exercise 6 (ABC)=(s−a)ra=(s−b)rb=(s−c)rc
Exercise 7
1 ra + 1 rb + 1 rc = 1 r .
## 1.6 The orthic triangle
Theorem 1 The orthocenter H of an acute-angled triangle is the incenter of its orthic triangle.
PROOF. Let AD, CF, and BE are altitudes of ABC, O be the circumcenter and α = 90° −A. Then the following angles are α: OBC, OCB, ABE, ACF. Because CDFA is inscribe to a circle then FDA=FCA=α and by the same reason ADE=ABE=α. Thus DA bisect FDE. [¯]
We also see that OB⊥FD, OC⊥DE, OA⊥FE.
Exercise 2 AEF ∼ DBF ∼ DEC ∼ ABC.
Exercise 3 HAO=| B−C |.
## 1.7 The medial triangle and Euler line
The triangle A′B′C′ formed by joining the midpoints A′, B′, C′ of the sides of a given triangle ABC will be called the medial triangle. We have the following set of conclusions:
1. The medial triangle A′B′C′ is similar to the given ABC with the ration 1:2. In fact we have four equal triangles!
2. ABC and A′B′C′ have the same centroid G.
3. The circumcenter O of ABC is the orthocenter of A′B′C′.
4. ABC and A′B′C′ are homothetic with a center G and ratio 2:1.
As a consequence we obtain that
Theorem 1 The orthocenter, centroid and circumcenter of any triangle are collinear. The centroid divides the distance from the orthocenter to the circumcenter in the ratio 2:1.
The line on which these three points lie is called the Euler line of the triangle.
Theorem 2 The circumcenter of the medial triangle lies at the midpoint of segment HO of the Euler line of the parent triangle. The circumradius of the medial triangle equals half the circumradius of the parent triangle.
Exercise 3 OH2=9R2−a2−b2−c2.
Exercise 4 DA′=| b2−c2 |/2a.
Exercise 5 If ABC has the special property that its Euler line is parallel to its side BC, then tanB tanC = 3.
# Chapter 2 Some properties of Circles
## 2.1 The power of point with respect to a circle
Theorem 1 If two lines through a point P meet a circle at points A, A′ (possibly coincident) and B, B′ (possibly coincident), respectively, then PA ×PA′ = PB ×PB′.
PROOF. The proof follows from the similarity of triangles PAB′ and PBA′ in both cases if P inside or outside of the circle. Notably in the second case PA ×PA′ = PT2 where PT is tangent to circle and T belong to it. [¯]
Let R is the radius of the circle and d is the distance to its center. If P is inside then PA ×PA′ = R2 −d2 and if it is outside then PA ×PA′ = d2 − R2.
Theorem 2 Let O and I be the circumcenter and incenter, respectively, of a triangle with circumradius R and r; let d be the distance OI. Then
d2=R2−2rR.
PROOF. Let bisector AL meet circumcircle at L, ML be diameter of circumcircle, then LM ⊥BC. BLI is isosceles, thus BL=IL. Then
R2−d2
=
LI ×IA = BL ×IA
=
LM LB/LM IY/IA IY = LM sinA/2 sinA/2 IY
=
LM ×IY = 2rR.
This is an example of synthetic proof, compare with proof of Radial Axis Theorem . [¯]
If we adopt the Newton convention:
PA = −AP
(2.1)
then identity
d2−R2=PA ×PA′
became universally true for any secant or chord. Its value is power of P with respect to the circle .
Exercise 3 What is smallest possible value of the power of a point with respect to a circle of radius R? Which point has this critical power?
Exercise 4 What is the locus of points of constant power?
Exercise 5 If PT and PU are tangents from P to two concentric circles, with T on the smaller, and if the segment PT meets the larger circle at Q, then PT2 − PU2=QT2.
## 2.2 The radical axis of two circles
Theorem 1 The locus of all points whose powers with respect to two nonconcentric circles are equal is a line perpendicular to the line of centers of the two circles.
PROOF. The proof could be done by means of analytic geometry, or analytic proof. Namely we express the problem by means of equations and solve them afterwards. The key ingredient is the equation of circle known from the Calculus I course. [¯]
The locus of points of equal power with respect to two non-concentric circles is called their radical axis.
Exercise 2 Give a simple indication of radical axis when two circles intersect or are tangent.
Exercise 3 Let PAB, AQB, ABR, P′AB, AQ′B, ABR′ be six similar triangles all on the same side of their common side AB. Then points P, Q, R, P′, Q′, R′ all lie on one circle.
## 2.3 Coaxal circles
A pencil of coaxial circle is infinite family of circles, represented by the equation
x2 + y2 − 2ax + c = 0
for a fixed c and
1. a any if c < 0.
2. a ∈ [−√c, √c] if c > 0.
Any two circles from the pencil have the same radical axis-the y-axis.
Theorem 1 If the centers of three circles form a triangle, there is just one point whose powers with respect to the three circles are equal. Its name is radical center.
Exercise 2 Two circles are in contact internally at a point T. Let the chord AB of the largest circle be tangent to the smaller circle at point P. Then the line TP bisect ATB.
## 2.5 Simpson line
Theorem 1 The feet of the perpendiculars from a point to the sides of a triangle are collinear iff the point lies on the circumcircle.
PROOF. Let A′, B′, C′ be feets of perpediculars from point P on the circumcircle. Observations:
1. Quadragles PB′A′C and C′PB′A are inscribed to a circles.
2. From the above: A′PC=A′B′C and AB′C′=APC′.
3. C′PA′=APC;
4. From the above C′PA=A′PC.
5. From 2 and 4: AB′C′=CB′A′, thus A′, B′, and C′ are collinear.
[¯]
Exercise 2 What point on the circumcirle has CA as its Simpson line?
Exercise 3 The tangent at two points B and C on a circle meet at A. Let A1B1C1 be the pedal triangle of the isosceles triangle ABC for an arbitrary point P on the circle. Then PA12=PB1× PC1.
## 2.6 Ptolemy's theorem and its extension
The orthic triangle and medial triangle are two instances of a more general type of associated triangle. Let P be any point inside a given tringle ABC, and PA1, PB1, PC1 are three perpendicular to its sides. Then A1B1C1 is pedal triangle for pedal point P.
It is easy to see that AB1C1 inscribed to a circle with the diameter AP. Then from the Theorem of sines follows that:
Theorem 1 If the pedal point is distant x, y, z from the vertices of ABC, the pedal triangle has sides
ax 2R , by 2R , cz 2R .
(2.2)
The Simpson line is degenerate case of the pedal triangle, nevertheless the above formulas are true and moreover A1B1+B1C1=A1C1 we deduce c CP + a AP = b BP, thus
AB×CP + BC ×AP = AC ×BP.
Theorem 2 If a quadralaterial is inscribed in a circle, the sum of the product of the two pairs of opposite sides is equal to the product of the diagonals.
The inverse theorem could be modified accordingly to the triangle inequality A1B1 +B1C1 > A1C1.
Theorem 3 If ABC is a triangle and P is not on the arc CA of the circumcircle, then
AB×CP + BC ×AP > AC ×BP.
Exercise 4 If a point P lies on the arc CD of the circumcircle of a square ABCD, then PA(PA+PC)=PB(PB+PD).
# Chapter 3 Collinearity and Concurrence
A polygon is a cyclically ordered set of points in a plane, with no three successive points collinear, together with the line segments joining consecutive pairs of the points. First few names are triangle, quadrangle, pentagon, hexagon, and so on.
Two sides of a quadrangle are said to be adjacent or opposite according as they do or do not have a vertex in common. The lines joining pairs of opposite vertices are called diagonals.
There three different types of the quadrangles:
1. convex -both diagonals are inside;
2. re-entrant -one diagonal is in, another is out;
3. crossed -both diagonals are outside.
We agree to count the area of triangle positive or negative if its vertices are named in counterclockwise or clockwise order. For example
(ABC)=−(BAC).
(3.1)
For all convex and re-entrant quadrangles area is:
(ABCD)=(ABC)+(CDA).
Remark 1 Combined the idea of signed area with directed segments could extend the proof of the Ceva's theorem to the case, then points divides sides externally.
Theorem 2 [Varignon 1731] The figure formed when the midpoints of the sides of a quadrangle are joined in order is a parallelogram, and its area is half that of the quadrangle.
Theorem 3 The segments joining the midpoints of pairs of the opposite sides of the a quadrangle and segment joining the midpoints of the diagonals are concurrent and bisect one another.
Theorem 4 If one diagonal divides a quadrangle into two triangles of equal area, it bisect the other diagonal. Conversely, if one diagonal bisect the other, it bisect the area of the quadrangle.
Theorem 5 If a quadrangle ABCD has its opposite sides AD and BC (extended) meeting at W, while X and Y are the midpoints of the diagonals AC and BD, then (WXY)=1/4(ABCD).
## 3.2 Cyclic quadrangles; Brahmagupta's formula
Theorem 1 [Brahmagupta] If a cyclic quadrangle has sides a, b, c, d and semiperimeter s, its area K is given by
K2 = (s−a)(s−b)(s−c)(s−d).
Corollary 2 [Heron] Area of a triangle is given by
(ABC)2=s(s−a)(s−b)(s−c).
## 3.4 Menelaus's theorem
Theorem 1 [Menelaus] If points X, Y, Z on sides BC, CA, AB (suitable extended) of ∆ABC are collinear, then
BX CX CY AY AZ BZ =1.
Conversely, if this equation holds for points X, Y, Z on the three sides, then these three points are collinear.
For the directed segments it could be rewritten as follows:
BX XC CY YA AZ ZB = −1.
## 3.5 Pappus's theorem
The following theorem is the first belonging to projective geometry. It is formulated entirely in terms of collinearity.
Theorem 1 [Pappus, 300 A.D.] If A, C, E are three points on one line, B, D, F on another, and if the three lines AB, CD, EF meet DE, FA, BC, respectively, then three points of intersection L, M, N are collinear.
PROOF. Let lines AB, CD, EF form triangle UVW. Apply the Menelaus's Theorem to the five triads of points
LDE, AMF, BCN, ACE, BDF
on the sides of this triangle UVW. Then the product of first three identities divided by the last two ones gives
VL LM WM MU UN NV = −1.
Thus by Menelaus's Theorem L, M, N are collinear. [¯]
## 3.6 Perspective triangles; Desargues's theorem
If two specimens of a figure, composed of points and lines, can be put into correspondence in such a way that pairs of corresponding points are joined by concurrent lines, we say that two specimens are perspective from a point. If the correspondence is such that pairs of corresponding lines meet at collinear points, we say that two specimens are perspective from a line.
Theorem 1 [Desargues, 1650] If two triangles are perspective from a point then they are perspective from a line.
In other words, If two triangles are perspective from a point, and if their pairs of corresponding sides meet, then three points of intersection are collinear.
PROOF. Let PQR and P′Q′R′ are the triangles perspective from point O, and let D=RQ·R′Q′, E=PR·P′R′, F=PQ·P′Q′. Apply Menelaus's Theorem to triads DR′Q′, EP′R′, FQ′P′ and triangles OQR, ORP, OPQ. [¯]
The converse theorem is also true.
Theorem 2 If two triangles are perspective from a line, they are perspective from a point.
If two triangles are perspective from a line, and if two pairs of corresponding vertices are joined by intersecting lines, the triangles are perspective from the point of intersection of these lines.
## 3.7 Hexagons
Two vertices of a hexagon are said to be adjacent , alternate , opposite according as they are separated by one sides, two sides, or three sides. The join of two opposite vertices is called a diagonal .
Exercise 1 Count the number of ways a given hexagon could be labelled as ABCDEF (Answer: 12).
Exercise 2 Count number of different hexagons defined by given 6 point, no three collinear. (Answer: 60).
In term of hexagon we could reformulate Pappus's Theorem as follows:
If each set of three alternate vertices of a hexagon is a set of three collinear points, and the three pairs of opposite sides intersect, then the three points of intersection are collinear.
## 3.8 Pascal's theorem
Theorem 1 [Pascal's Theorem] If all six vertices of a hexagon lie on a circle and the three pairs of opposite sides intersect, then the three points of intersection are collinear.
PROOF. The proof consists of application Menelaus's Theorem four times. [¯]
This theorem of a projective nature and hexagon could be in fact inscribed in any conic . Under such a formulation it has an inverse:
Theorem 2 If the three pairs of opposite sides of a hexagon meet at three collinear points, then the six vertices lie on a conic.
Some degenerated cases of the Pascal's Theorem are of interest
Corollary 3 Let ABDE be a cyclic crossed quadrangle. Tangents to the circle in points B and E meet in a point N which is collinear with points L=AB·DE and M=BD·EA.
# Chapter 4 Transformations
The groups of transformations are very important in geometry. In fact they are could characterize different geometries as was stated by Felix Klein in his famous Erlangen program. We will consider most fundamental groups of transformations.
For Eucleadean geometry the important transformations are isometries. There are several of them: translations , rotations (particularly half-turn ), reflections .
## 4.1 Translations
We refer for properties of translations or vectors in the Calculus course . As geometrical application of vectors we could consider the deducing formula of parallelogramm area. Another example is
Exercise 1 Inscribe in a given circle a rectangle with two opposite sides equal and parallel to a given line segment a.
The characteristic property of translation among isometries is: each ray come to a parllel ray (prove it!).
## 4.2 Rotations
Other important isometries are rotations around a point O by an angle α.
The characteristic property of rotations among isometries is: each ray come to ray rotated by the α.
## 4.3 Half-turn
The half-turn is rotation by the angle 180° and is completely defined by its center. The characteristic property of half-turn among isometries is: each ray come to the opposite ray. Thus
Theorem 1 Composition of two half-turn is a traslation by the vector 2O1O2.
Using half-turns we could easily prove that if digonals of a quadrangle bisect each other then it is a parallelogramm.
## 4.4 Reflections
The third type of isometries is reflections in a mirror. It interesting that they give a geometrical solution for the following extremal problem : find the shortes path (which is physically the path of a light ray) between two points via a point of the mirror.
## 4.7 Dilation
Isometries transfor a figure into a congruent figure. Another important class is transformations which change each figure to into a similar figure, i.e. all distances increased in the same ratio, ration of magnification.
Exercise* 1 Prove that suchtransformations preserve collinearity and angles.
A simplest kind is dilation, which transforms each line into a parallel line. If a dilation is not a translation then its central dilation . Translations and half-turn are partucular cases of dilations with ratio 1 and −1 correspondingly.
## 4.8 Spiral symmetry
It is possible to see that the composition of a translation and a dilation or composition of two tarnslations are again a dilation (sinse parallel lines come to parallel lines). But composition of a dilation and rotation around the same point is something different- spiral similarity, which is a kind of direct similarity (preserves angles in magnitude and sign). They are completely determined by their center O, ratio k, and angle θ, we will denote it by O(k,θ).
Theorem 1 If squares, with centers O1, O2, O3, are erected externally on the sides of ∆ABC, then line segments O1O2 and CO3. are equal and perpendicular.
PROOF. It is follows from consideration of A(√2,45°) and C(√2,−45°). [¯]
It is interesting that there are no other direct similarities besides spiral ones:
Theorem 2 Any two directly similar figures are related either by a translation or by a spiral similarity.
Corollary 3 If ABC and A′B′C′ are two directly similar triangles, while AA′A′′, BB′B′′, CC′C′′ are three directly similar triangles, then ∆A′′B′′C′′ and ∆ABC are directly similar.
## 4.9 A genealogy of transformations
We could put the following transformation in a genealogical tree:
Transformation
Continuous transormation
Linear transformation
Similarity Procrustean stretch
Isometry Dilation Spiral similarity
Reflection Translation Rotation Central dilation
Half-turn
# Chapter 5 An Introduction to Inversive Geometry
## 5.1 Separation
Theorem 1 If four points A, B, C, D do not all lie on the circle or line, there exist two non-intercecting circles, one through A and C, the other through B and D.
Two distinct point pairs, AC and BD are said to sepatrate each other if A, B, C, D lie on a circle (or a line) in such an order that either of the arcs AC contains one but not both of the remaining points B and D. It is denoted by AC//BD. Another characterizations are
Theorem 2 Two distinct point pairs, AC and BD are said to sepatrate each other if every circle through A and C intersects (or coinsides with) every circle through B and D.
Alternatively
Theorem 3 The mutual distances of four distinct points A, B, C, D satisfy
AB ×CD + BC ×AD ≥ AC×BD,
with the equals sign only then AC//BD.
PROOF. It is follows directly from consideration of directed line segments if the points are collinear and is a consequence of the Ptolemy's theorem if points lie on a circle or are not collinear. [¯]
## 5.2 Cross Ratio
We introduce cross ratio as follows
Then we obtainfrom the Separation Theorem
Theorem 1 The cross ratios of four distinct points A, B, C, D satisfy
{AD, BC} + {AB, DC} = 1
iff AC // BD.
Now instead of defining separation in the term of circles we could define circles in the term of separation:
Definition 2 The circle determined by three points A, B, C is set of points consisiting of the three points themselves along with all the points X such that
BC //AX or CA //BX or AB //CX.
## 5.3 Inversion
For a given circle ω with the center O and radius k we define a point P′=i(P) being inverse to P if P′ ∈ OP and
OP ×OP′=k2.
It is obvious from this conditions that P=i(i(P)) for any point P (different from O). The inverse for O is not defined. There is a simple geometrical constraction.
Theorem 1 The inverse of any line a, not through 0, is a circle through O, and the diametre through O of the circle is perpendicular to a.
The inverse of any circle through O is a perpendicular to the diametr through O.
We could construct inverse points using Peaucellier's cell.
Considering images under inversion of three points we could observe
Theorem 2 For a suitable circel inversion, any three distinct points A, B, C can be inverted into the vertices of a triangle A′B′C′ congruent to a given triangle.
## 5.4 The inversive plane
Theorem 1 If a circle with center O and radius k invert point pair AB into A′B′, the distance are related by the equation
A′B′= k2 AB OA×OB .
Theorem 2 If A, B, C, D invert into A′, B′, C′, D′, then
{A′B′,C′D′}={AB,CD}.
Theorem 3 If A, B, C, D invert into A′, B′, C′, D′ and Ac//BD then A′C′//B′D′.
If we will think on lines as circles with infinite radius then we could state
Theorem 4 The inverse of any circle is a circle.
To define inversion for all points we may add to a plane a one special point: point at infinity p. Then inverse of O is p and vise verse. A plane together with pq form the inversive plane.
# Chapter 6 An Introduction to Projective Geometry
## 6.1 Reciprocation
Let ω be a circle with center O and radius k. Each point P (different from O) determine a corresponding line p, called the polar of P; it is the line perpendicular to OP through the inverse of P. Conversely, each line p determine a point P, the pole of p; it is the inverse of the foot of the perpendicular from O to p.
Theorem 1 If B lies on a, then b passes through A.
We say that A and B are conjugate points; a and b are conjugate line. Any point on a tangent a is conjugate to the point of contact A, which is self-conjugate point, and any line through A (on ω) is conjugate to the tangent a, which is a self-conjugate line.
Reciprocation allows us to introduce a vocabulary for projective duality
point line lie on pass through line joining two points intersection of two lines concurrent collinear quadrangle quadrilateral pole polar locus envelope tangent point of contact
Theorem 2 The pole of any secant AB (except a diameter) is the common point of the tangents at A and B. The polar of any exterior point is the line joining the points of contact of two tangents from this point. The pole of any line p (except a diameter) is the common point of the polars of two exterior points on p. The polar of any point P (except the center) is the line joining the poles of two secants through P.
## 6.3 Conics
We meet already conics (or conic sections) in the course of Calculus I. Their they was defined by means of equations in Cartesian coordinates or as sections of cones. Now we could give a projective definition. Let ω be a circle with center O.
Definition 1 A conic is the reciprocal of a circle with a center A and radius r. Let ε = OA/r be the eccentricity of the conic.
1. If ε < 1 then it is ellipce, particularly ε = 0 is the circle.
2. If ε = 1 then it is parabola.
3. If ε > 1 then it is hyperbola.
## 6.5 The projective plane
Similarly for definition of the inversive plane we could make an extension of Euclidean plane for the projective case. To define reciprocation for all points we need to introduce a one additional line: line at infinity l. This line is polar for O and its points ( points at infinity) are poles for lines through O. Those points are common points for pencil of parallel lines. Thus any two distinct lines a and b determine a unique point a·b.
Theorem 1 If P is not on the conic, its polar joins the points of intersection AB ·DE and AE·BD, where AD and BE are any two secant through P.
Theorem 2 With respect to any conic except a circle, a directrix is the polar of the corresponding focus.
## 6.7 Stereographic and gnomonic projection
In the same way as we introduce the inversion we could introduce in R3 with respect to a sphere Σ with a center O and radius K by relation OA×OA′=k2. As a corollary from the plane we see that the image of any sphere (including a plane as a limit case) is a sphere again.
If a plane is tangent to the sphere of inversion at A then its image is the sphere σ with a diameter OA. And the image of any point P in the plane is just another point of intersection of line OP with σ. Sphere σ is a model for inversive plane . The mapping between plane and sphere is stereographic projection. Its preserve angles between directions in any points.
If we take a sphere Σ and construct the map from a tangent plane to pairs of antipodal points as intersections of line OP and Σ then we obtain gnomonic map. It maps big circles (shortest distances on the sphere) to straight lines (shortest distances on the plane). Identifying pairs antipodal points on the sphere we obtain a model of projective plane .
# Appendix A Some Useful Theorems
Theorem 1 An angle inscribed in an arc of a circle has a measure which is a half of angular measure of the complementary arc.
Corollary 2 An angle inscribed in semicircle is is a right angle.
Theorem 3 Two tangemts to a circle from any external point are equal.
Theorem 4 The following geometric objects have indicated areas S:
1. Rectangle with sides a and b: S=ab.
2. Parallelogram with a base a and altitude h: S=ah.
3. Triangle with a side a and corresponding altitude ha: S=aha/2.
Corollary 5
1. Two triangles with a common altitude h have areas proportional to their sides: [(S1)/(S2)]=[(a1)/(a2)].
2. Two triangles with a common side a have areas proportional to their altitudes: [(S1)/(S2)]=[(h1)/(h2)].
# Appendix B Some Useful Tricks
## B.1 Look for a triangle-the golden rule of geometry
If the unknown element is a line segment or an angle-try to find a triangle, which contains this element and such that other parameters of the triangle are given or could be found from the given conditions.
If you are questioned about two elements (like a ratio of two line segments, for example) try to find two triangles with some common elements and each containing one of the unknown line segments.
## B.2 Investigate a particular case
If you meet a problem-try to investigate a particular case. If you study an angle inscribed in a circle-consider first a case when the angle goes through the center of circle. If this particular investigation was successful try to use this particular case for solution the general one.
## Bibliography
[1]
H.S.M. Coxeter and S.L. Greitzer. Geometry Revisited. This famous book was reprinted many times and translated to many foreign languages.
## Index (showing section)
adjacent, 3.1, 3.7 alternate, 3.7 altitudes, 1.3 feets of, 1.3 analytic geometry, 2.2 analytic proof, 2.2 Area, 3.2 area, 3.1, 3.2 cyclic quadrangle, 3.2 notation, 1.1 triangle, 3.2 Bisectors, 1.3 bisectors, 1.3 central dilation, 4.7 centroid, 1.3 cevian, 1.2 characteristic property of rotations, 4.2 characteristic property of translation, 4.1 Circumcenter, 1.3 circumcenter, 1.3 circumcircle, 1.3 circumradius, 1.3 congruent, 4.7 conic, 6.3 conjugate line, 6.1 conjugate points, 6.1 convex, 3.1 cross ratio, 5.2 crossed, 3.1 diagonal, 3.7 dilation, 4.7 central, 4.7 direct similarity, 4.8 eccentricity, 6.3 ellipce, 6.3 Euler line, 1.7 excenters, 1.4 excircles, 1.4 exradii, 1.4 Feets, 1.3 Gergonne point, 1.4 gnomonic map, 6.7 groups of transformations, 4.0 half-turn, 4.3 hexagon, 3.1 diagonal, 3.7 sides adjacent, 3.7 alternate, 3.7 opposite, 3.7 hyperbola, 6.3 incenter, 1.3 inradius, 1.3 inverse, 5.3 inversive plane, 5.4 isometries, 4.0 law of sines, 1.1 line at infinity, 6.5 medial, 1.7 medians, 1.3 mirror, 4.4 negative, 3.1 notation area, 1.1 opposite, 3.1, 3.7 orthic triangle, 1.3 orthocenter, 1.3 parabola, 6.3 Peaucellier's cell, 5.3 pedal point, 2.6 pedal triangle, 2.6 pencil of coaxial circle, 2.3 pencil of parallel lines, 6.5 pentagon, 3.1 perspective from a line, 3.6 perspective from a point, 3.6 point at infinity, 5.4 points at infinity, 6.5 polar, 6.1 pole, 6.1 polygon, 3.1 positive, 3.1 power of P with respect to the circle, 2.1 power of point, 2.1 projective geometry, 3.5 quadrangle, 3.1 area, 3.1 convex, 3.1 crossed, 3.1 re-entrant, 3.1 sides adjacent, 3.1 opposite, 3.1 radical axis, 2.2 radical center, 2.3 ration of magnification, 4.7 re-entrant, 3.1 reflections, 4.4 rotations, 4.2 self-conjugate line, 6.1 self-conjugate point, 6.1 semiperimeter, 1.4 sepatrate, 5.1 similar, 4.7 similarity direct, 4.8 spiral similarity, 4.8 stereographic projection, 6.7 synthetic proof, 2.1 theorem Ceva, 1.2 Pascal's, 3.8 Varignon's, 3.1 transformations, 4.0 translations, 4.1 triangle, 3.1 area negative, 3.1 positive, 3.1 centroid, 1.3 medial, 1.7 tritangent, 1.4 vectors, 4.1
### Footnotes:
1We alway denote area of a figure by its name enclosed in parentheses.
File translated from TEX by TTH, version 3.13.
On 22 May 2003, 12:56. | 9,519 | 35,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.850789 |
https://www.cscodehelp.com/%E7%A7%91%E7%A0%94%E4%BB%A3%E7%A0%81%E4%BB%A3%E5%86%99/%E7%A8%8B%E5%BA%8F%E4%BB%A3%E5%86%99%E4%BB%A3%E5%81%9A%E4%BB%A3%E8%80%83-computer-architecture-solutions8-4/ | 1,726,851,398,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00373.warc.gz | 663,994,149 | 13,001 | 程序代写代做代考 computer architecture Solutions8
Solutions8
Computer Architecture
Tutorial 4 – Floating Point Numbers – Answers
1) Binary fractions are:
a) 5.5 is 101.1
b) 8.25 is 1000.01
c) 9 is 1001
0.3 0.6, 1.2, 0.4, 0.8, 1.6, 1.2 01001 1001 1001 etc.
9.3 is 1001. 01001 1001 1001 repeating etc.
d) 11.46875 is 1011.01111
2) Convert the binary number 1001.1010101 to decimal.
1001 binary is 9 decimal
. 1 0 1 0 1 0 1
128 64 32 16 8 4 2 1 Sum=85
Fraction = 85 /128 = 0.6640625
Number = 9.6640625
3) a) 101.1 = 1.011 x 2
2
b) 1000.01 = 1.00001 x 2
3
c) 0.00010101 = 1.0101 x 2
–4
4) Convert –31.3 to IEEE Single Precision format.
First convert to a binary number -31.3 = -11111.01001 1001 1001
Next Normalise
1.11110 1001 1001 1001 1001 1001 x 2
4
Significand field is 1111 0100 1100 1100 1100 110 (23 bits with 1. omitted)
Exponent field is 4+127 = 131 = 1000 0011
Number is -ve therefore Sign field is 1
Sign Exponent Significand
1 1000 0011 1111 0100 1100 1100 1100 110
5) Convert the IEEE Single Precision format hex value C154 0000 to decimal.
C154 0000 = 1100 0001 0101 0100 0000 0000 0000 00000
Sign Exponent Significand
1 1000 0010 1010 1000 0000 0000 0000 000
Exponent field = 1000 0010 = 130 => Exponent = 130 – 127 = 3
Significand field = 10101 Adding Hidden Bit => 1.10101
Therefore number is 1.10101 x 2
3
= 1101.01 = Decimal 13.25
Sign is 1 therefore number is -13.25
6) Carry out the operation 31.3 + 13.25 in IEEE single precision arithmetic
Number Sign Exponent Significand
31.3 0 1000 0011 1111 0100 1100 1100 1100 110
13.25 0 1000 0010 1010 1000 0000 0000 0000 000
Significand of Larger Number = 1.1111 0100 1100 1100 1100 110
Significand of Smaller Number= 1.1010 1000 0000 0000 0000 000
Exponents differ by 1. Therefore shift binary point of Smaller Number 1 place.
Significand of Larger Number = 1.1111 0100 1100 1100 1100 1100
Significand of Smaller Number= 0.1101 0100 0000 0000 0000 0000
Significand of Sum = 10.1100 1000 1100 1100 1100 1100
Sum = 10.1100 1000 1100 1100 1100 1100 x 2
4
Normalise 1.01100 1000 1100 1100 1100 1100 x 2
5
Sign Exponent Significand
0 1000 0100 0110 0100 0110 0110 0110 011
7)
Fraction Binary Decimal
1/4 0.01 0.25
3/8 0.011 0.375
23/16 1.0111 1.4375
45/16 10.1101 2.8125
11/8 1.011 1.375
45/8 101.101 5.625
49/16 11.0001 3.0625 | 1,046 | 2,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-38 | latest | en | 0.451364 |
https://www.kylesconverter.com/area/dunams-to-square-rods | 1,702,124,750,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00560.warc.gz | 942,624,495 | 5,707 | # Convert Dunams to Square Rods
### Kyle's Converter > Area > Dunams > Dunams to Square Rods
Dunams Square Rods (sq rd) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Square Rods to Dunams
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Dunam:
One dunam is 1 000 square meters. Originated as an area of land that could be worked by an ox team, eventually redefined to a thousand square meters. 1 dunam = 1 000 m2
1 Square Rod:
A square rod is equal to the area of a square with sides of one rod each. A square rod is 2721/4 square feet. In SI base a square rod (sq rd) of 16.5 international feet is 25.29285264 square meters (m2).
Conversions Table
1 Dunams to Square Rods = 39.536970 Dunams to Square Rods = 2767.5803
2 Dunams to Square Rods = 79.073780 Dunams to Square Rods = 3162.9489
3 Dunams to Square Rods = 118.610690 Dunams to Square Rods = 3558.3175
4 Dunams to Square Rods = 158.1474100 Dunams to Square Rods = 3953.6861
5 Dunams to Square Rods = 197.6843200 Dunams to Square Rods = 7907.3722
6 Dunams to Square Rods = 237.2212300 Dunams to Square Rods = 11861.0583
7 Dunams to Square Rods = 276.758400 Dunams to Square Rods = 15814.7444
8 Dunams to Square Rods = 316.2949500 Dunams to Square Rods = 19768.4305
9 Dunams to Square Rods = 355.8317600 Dunams to Square Rods = 23722.1166
10 Dunams to Square Rods = 395.3686800 Dunams to Square Rods = 31629.4888
20 Dunams to Square Rods = 790.7372900 Dunams to Square Rods = 35583.1749
30 Dunams to Square Rods = 1186.10581,000 Dunams to Square Rods = 39536.861
40 Dunams to Square Rods = 1581.474410,000 Dunams to Square Rods = 395368.6103
50 Dunams to Square Rods = 1976.8431100,000 Dunams to Square Rods = 3953686.1035
60 Dunams to Square Rods = 2372.21171,000,000 Dunams to Square Rods = 39536861.0347 | 648 | 1,844 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.811161 |
https://6.flygentlebreezes.net/off-switch-schematic-symbol/ | 1,566,193,948,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00172.warc.gz | 350,432,357 | 35,424 | # Off Switch Schematic Symbol
In Wiring Diagram222 views
4.03 / 5 ( 216votes )
Top Suggestions Off Switch Schematic Symbol :
Off Switch Schematic Symbol The following symbols show the different components that can be found in an electrical circuit some of the more common components are switch lamp a switch is used to turn a circuit on closed switch This two article series presents an overview of circuit symbols and also provides some information on real circuits the body of a fet is often shorted to the source if you switch back and forth For instance a basic on off light switch is an example of a single pole i e one circuit single throw i e a switch contact in only one position device that is normally open no or spstno for.
Off Switch Schematic Symbol The transistor can be used as a switch if biased in the saturation and cut off regions electron flow going against the transistor symbol s arrow going back to the npn transistor in our example To understand a series circuit visualize it example of wiring up a three way light switch system suppose you have two light switches if you wire them in series opening either switch shuts the Rotating copper arms are designed to look like the switch symbols found on the diagrams and turn the lights on and off by completing points along the circuit resembling the circles of light.
Off Switch Schematic Symbol Here s how to determine if the problem is with its switch first turn off the power to the switch prong protruding from the male end a continuous circuit measured from these two ends will have a Taking a look back at seven days of news and headlines across the world of android this week s android circuit includes with a charging symbol and number the number is binary and adds up The lamp called node was created by odd matter studio using loops of copper to represent the wire in a circuit diagram reports dezeen to switch symbols advertisement so far the studio s built.
The schematic diagram represents the components using schematic symbols by comparing the pictorial diagram the thermostat supply the power to the compressor motor s start relay switch to check.
It's possible to get or download caterpillar-wiring diagram from several websites. If you take a close look at the diagram you will observe the circuit includes the battery, relay, temperature sensor, wire, and a control, normally the engine control module. With an extensive collection of electronic symbols and components, it's been used among the most completed, easy and useful wiring diagram drawing program. Off Switch Schematic Symbol. The wiring diagram on the opposite hand is particularly beneficial to an outside electrician. Sometimes wiring diagram may also refer to the architectural wiring program. The simplest approach to read a home wiring diagram is to begin at the source, or the major power supply. Basically, the home wiring diagram is simply utilized to reveal the DIYer where the wires are.
If you can't locate the information, get in touch with the manufacturer. The info in the diagram doesn't indicate a power or ground supply. The intention of the fuse is to safeguard the wiring and electrical components on its circuit. A typical watch's basic objective is to tell you the good time of day. When selecting the best type of computer cable to fulfill your requirements, it is very important to consider your upcoming technology plans.
Installing a tachometer on your Vehicles can assist in preventing critical repair problems, however. You might have a weak ground issue. The way the brain learns is a subject that still requires a good deal of study. How it learns can be associated by how it is able to create memories.
In a parallel circuit, each unit is directly linked to the power supply, so each system gets the exact voltage. There are 3 basic sorts of standard light switches. The circuit needs to be checked with a volt tester whatsoever points. Off Switch Schematic Symbol. Each circuit displays a distinctive voltage condition. You are able to easily step up the voltage to the necessary level utilizing an inexpensive buck-boost transformer and steer clear of such issues. The voltage is the sum of electrical power produced by the battery. Be sure that the new fuse isn't blown, and carries the very same amperage.
Each fuse is going to have a suitable amp rating for those devices it's protecting. The wiring is merely a bit complicated. Our automotive wiring diagrams permit you to relish your new mobile electronics in place of spend countless hours attempting to work out which wires goes to which Ford part or component. Overall the wiring is really straight forward. There's a lot wiring that you've got to tie into your truck's wiring harness, but it's much easier to do than it seems. A ground wire offers short circuit protection and there's no neutral wire used. There's one particular wire leading from the distributor which may be used for the tachometer.
When you have just a single cable going into the box, you're at the close of the run, and you've got the simplest scenario possible. All trailer plugs and sockets are extremely easy to wire. The adapter has the essential crosslinks between the signals. Wiring a 7-pin plug on your truck can be a bit intimidating when you're looking at it from beyond the box.
The control box may have over three terminals. After you have the correct size box and have fed the cable to it, you're almost prepared to permit the wiring begin. Then there's also a fuse box that's for the body controls that is situated under the dash. Off Switch Schematic Symbol. You will find that every circuit has to have a load and every load has to have a power side and a ground side. Make certain that the transformer nameplate power is enough to supply the load that you're connecting.
The bulb has to be in its socket. Your light can be wired to the receiver and don't require supply additional capacity to light as it can get power from receiver. In the event the brake lights aren't working, a police officer may block the vehicle and issue a warning to create the repair within a particular time limit. Even though you would still must power the relay with a power source or battery. Verify the power is off before trying to attach wires. In case it needs full capacity to begin, it won't operate in any way.
Replacing thermostat on your own without a Denver HVAC technician can be quite harrowing if you don't hook up the wiring correctly. After the plumbing was cut out, now you can get rid of the old pool pump. It's highly recommended to use a volt meter to make sure there is no voltage visiting the motor, sometimes breakers do not get the job done properly, also you might have turned off the incorrect breaker. Remote distance is left up to 500m. You may use a superior engine ground. The second, that's the most frequently encountered problem, is a weak ground in the computer system. Diagnosing an electrical short can be extremely tough and costly.
Don't ask me why I have such of an obsession with wires, but I do. My mother always said that ever since I've been able to walk, I would find things with wires and play with them and tear them apart, figure out how they worked and would be totally fascinated.
Top | 1,437 | 7,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.905354 |
http://betterlesson.com/lesson/resource/3019052/14-1-hvn150-introduction-to-quadratic-functions-mov | 1,488,121,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172017.60/warc/CC-MAIN-20170219104612-00647-ip-10-171-10-108.ec2.internal.warc.gz | 27,968,502 | 19,812 | ## 14 - 1 HVN150 Introduction to quadratic functions.MOV - Section 3: Huddle
14 - 1 HVN150 Introduction to quadratic functions.MOV
Lesson 1 of 9
## Big Idea: Basketball shots and parabolas - quadratic functions! Hook students into quadratic functions with this introductory lesson.
Print Lesson
11 teachers like this lesson
Standards:
Subject(s):
75 minutes
### Jeff Li MTP
##### Similar Lessons
###### Quadratic Equations, Day 1 of 2
Algebra II » The Complex Number System
Big Idea: Knowledge of quadratic equations is an important prerequisite for studying complex numbers.
Favorites(3)
Resources(13)
Fort Collins, CO
Environment: Suburban
###### Where are the Functions Farthest Apart? - Day 1 of 2
12th Grade Math » Functioning with Functions
Big Idea: Function combinations and maximization problems collide to create a challenging and mathematically rich task.
Favorites(3)
Resources(13)
Troy, MI
Environment: Suburban
###### Leap of Faith!
Algebra I » Bridge to 10th Grade
Big Idea: Students will find a linear relationship between the number of rubber bands and height.
Favorites(4)
Resources(14)
Washington, DC
Environment: Urban | 268 | 1,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-09 | latest | en | 0.765086 |
https://goprep.co/q28-a-manufacturing-company-makes-two-types-of-teaching-aids-i-1nlbpg | 1,606,703,271,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141204453.65/warc/CC-MAIN-20201130004748-20201130034748-00626.warc.gz | 305,118,022 | 37,428 | # A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs80 on each piece of type A, and Rs120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Let pieces of type A are manufactured per week and b pieces per week for type B
Profit on type A is 80rs per piece hence weekly profit is 80a. Similarly, profit on type B is 120rs per piece hence weekly profit is 120b
Total profit be z = 80a + 120b
Now we have to maximize profit which means we have to maximize z based on some constraints
Let us identify the constraints now
Constraint 1: labour hours for fabricating per week
Each piece of type A requires 9 labour hours of fabricating hence weekly there would be 9a hours of fabricating for type A
Now for type B 12 labour hours for each piece hence weekly for b pieces 12b labour hours of fabricating
The maximum labour hours for fabricating is given as 180hours weekly
Which means the total labour hours for fabricating of both type A and type B should not exceed 180
9a + 12b ≤ 180
3a + 4b ≤ 60 …(i)
Constraint 2: labour hours for finishing per week
Each piece of type A requires 1 labour hours of finishing hence weekly there would be 1a hours of finishing for type A
Now for type B 3 labour hours for each piece hence weekly for b pieces 3b labour hours of finishing
The maximum labour hours for finishing is given as a 30hours weekly
Which means the total labour hours for fabricating of both type A and type B should not exceed 30
a + 3b ≤ 30
a + 3b ≤ 30 …(ii)
Also, a, b represents the number of pieces manufactured per week hence it cannot be negative hence a ≥ 0 and b ≥ 0
Plot equations (i) and (ii)
Scale
On X-axis
1 cm = 2 pieces
On Y-axis
1 cm = 2 pieces
Now the corner points are O, A, B and C
Let us find values of z at these points
Hence the maximum value of z is 1680 at B(12, 6)
Hence for maximum profit, a = 12 pieces should be manufactured for type A and b = 6 pieces for type B per week
And the maximum profit is 1680rs
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos | 655 | 2,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-50 | longest | en | 0.896595 |
http://cn.comsol.com/support/learning-center/article/Terminology-and-Requirements-for-Swept-Meshing-51351/152 | 1,716,107,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00292.warc.gz | 6,909,519 | 14,767 | # 学习中心
Course:
Using Swept Meshes for Model Geometries
## Terminology and Requirements for Swept Meshing
In the first part of this course, we learned the circumstances in which a swept mesh can be useful. In Part 2, we dig into the details of a Swept operation: the terminology used for it and when exactly it can be used.
### Terminology of a Swept Mesh
Let's begin by presenting some terminology used in COMSOL Multiphysics® regarding swept meshing. Each face about a domain that is to be operated on by the swept mesher is classified as either a source face, a destination face, or a linking face. The Swept operation starts from the source faces and terminates at the destination faces, hence defining a sweep direction. The linking faces connect the source and destination faces.
The linking edges are the edges, or the chains of edges, that are approximately tangent to the sweep direction. The section edges are the edges, or chains of edges, that are approximately perpendicular to the sweep direction.
A section is a face component bounded by two loops of section edges.
Classification of the boundaries and edges about a domain used for swept meshing. The sections are depicted by the three shades of blue. Two faces are hidden to show the inside of the domain.
### Requirements for a Swept Mesh
The very first step before even trying to generate a swept mesh is to ensure that the domains are following several criteria, as described in the documentation. Since each domain within the geometry selection that will be swept follows the same logic, we will focus on one domain in this part. The different criteria from the documentation can be loosely summarized with four major rules:
1. No holes, isolated faces or voids in the domain
2. The source and destination must be connected face components
3. Each destination face matches one or more source faces, while each source face matches exactly one destination face or its subset
4. For each section, there must be at least two chains of linking edges
We will explore each of these requirements in detail as some technicalities were eluded deliberately in this list. If any of these requirements are not met, an error message will appear. Techniques on how to solve these and any other potential issues are covered in the next part of this swept meshing course.
### Requirements on the Shape of the Domain
#### Handling of Holes, Isolated Faces and Voids
No holes, isolated faces or voids are allowed in the domain. The only exception to this rule is if the holes or isolated faces penetrate both source and destination, as shown in the next picture.
Models with domains that include a hole and face that penetrate both the source and destination of the swept mesh.
A model with domains that include holes and isolated faces that do not penetrate the source and destination of the swept mesh.
Domains containing holes and isolated faces. Left: the domains are sweepable because the hole and face penetrate both source and destination. Note that the domain with the isolated face can be swept meshed because the interior boundary can be mapped meshed and follows the general direction of the sweep. Right: the geometries do not fulfill the criteria.
####
This domain has a hole that penetrates both the source and destination, so a swept mesh can be generated. Note that the hole is allowed to move and change size along the path and that the mesh is deformed accordingly.
#### How the Source and Destination Are Connected to the Domain
The source and destination must be connected face components.
Another way to interpret this rule on a domain level is to say that the domain cannot "branch off", it cannot start or end at different disconnected locations.
A domain with disconnected face components.
A domain with disconnected face components in more than one direction.
None of the two domains are sweepable because their top faces are disconnected face components (the right hand geometry has disconnected face components in more than one direction). Another way to interpret it is that the domain is "branching off", which is not supported.
On a side note, this rule does not mean that the source and destination must be be forming different connected face components, coincident source and destination faces are also supported. These are allowed as long as the source mesh matches the destination (i.e., including the potential twist of the face).
Following the square cross section of a meshed twisted torus, the surface mesh of the source needs to have a 180 degrees symmetry due to the half turn.
#### Visualizing the Sweep Path
It is often helpful to mentally visualize the sweep by following the cross section of the mesh. An interpretation of the first two rules is that the cross section must not have any abrupt changes in its shape. For example, a hole will create a void in the cross section, an isolated face will introduce a new edge edge, and a branching will either split, combine, add, or remove faces in the cross section. While this technique is not the most rigorous, 'cross section' and 'abrupt change' need to be properly defined, it will in most cases help to locate where a domain becomes unsweepable and will give a clue where it might help to partition the domain.
A visual interpretation on the reason why the previous domains could not be swept meshed. On the left, the cross section is splitting into two parts. On the right, the cross section's shape changes instantaneously when the first destination face is reached. A partial mesh is shown for visualization purposes in the portion where the domain could have been swept meshed if it was partitioned by the cross section, but in reality no mesh would be generated.
A way to help with this visualization is to add a Clip Plane that is parallel to the cross section of the sweep, and then drag the frame of the clip plane in the sweep direction. Make sure to select the option Show Cross Section.
This recording shows how to manually visualize the sweep path from the graphics window. This domain cannot be swept mesh because it encloses a void region and is therefore not in accordance with the first rule. This is especially apparent because the cross section becomes hollow in the middle of the sweep path.
#### The Special Case of Isolated Entities
Isolated edges and isolated vertices are edges and vertices that are positioned inside a domain or face without being part of any loop of edges delimiting a face. When isolated edges and vertices are present, the mesh is first generated without considering isolated entities. Then, in the second step, the entities are included and essentially "glued" to the existing mesh by moving some nearby mesh nodes. The mesh conforms to the isolated entities during the sweep.
The exterior of a mesh with and without isolated entities on the linking faces. Visualization of the mesh, without (left) or with (right) isolated entities located on the linking faces. The mesh on the right has been distorted in order to account for the additional entities during the sweep. Note that the total number of mesh elements remains constant because the isolated entities are snapped afterward; this means that they need to be placed reasonably so that the mesh is not distorted excessively.
The interior of a mesh with and without isolated entities on the linking faces. Visualization of the interior mesh, without (left) or with (right) isolated entities located within the domain. The mesh on the right has been distorted in order to account for the additional entities during the sweep.
### Source and Destination Faces Requirements
Each destination face matches one or more source faces, while each source face matches exactly one destination face or its subset. In other words, the source faces can contain more faces than the destination, as long as there is a way of imprinting the mesh onto the destination (up to some distortion, like we saw in this animation)
A geometry with more source faces than destination faces.
A meshed geometry with more source faces than destination faces.
This geometry can be swept meshed even though there are more source faces (magenta) than destination faces (yellow).
For the majority of cases there is no deformation between the source and destination, one can simply translate and rotate the source faces to overlap on top of the destination faces. In that case, it can help to mentally visualize copying the source mesh to the destination, and see if the copy operation is possible. This mental exercise is depicted in the next animations.
Only the source and destination faces are shown, following the same color convention as the previous 3D figure. Note that the source has more faces than the destination and that each destination face corresponds to one or more source faces.
An example where this rule is not adhered to: Since the destination has more faces, the mesh cannot be copied. The edges where the mesh copy failed are highlighted in red. Consequently, the direction of the sweep cannot be inverted for the previous 3D geometry.
Note that destination faces cannot contain any isolated edges, even if it would match perfectly isolated edges on the source faces. For isolated vertices on the destination faces (and their boundaries), they will be handled the same way as described in this section: the mesh will be locally distorted in order to snap those vertices to mesh nodes. For the source faces, they can both contain isolated edges and vertices.
For each section, there must be at least two chains of linking edges. Another way to interpret this rule is that there should at least be two linking faces per section.
A domain missing linking edges that can't be swept meshed.
A domain with linking edges that can be swept meshed.
The domain on the left cannot be swept meshed, as there is no linking edge on the second section (the transparent section in lighter blue is composed of one face only), but the domain on the right can be swept meshed because there are at least two linking edges in both section.
### Concluding Remarks
In this article we discussed how to distinguish domains that can or cannot be meshed with a swept mesh and four simple rules to check for. To see how the swept mesh was generated for several of the model examples included in this article, you can see the respective MPH-files attached. In the next part of the swept mesh course, we present the most important aspects to take into account so that a mesh of good quality can be generated. | 2,073 | 10,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | longest | en | 0.906891 |
https://www.weegy.com/?ConversationId=220E9219 | 1,603,782,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00368.warc.gz | 930,474,025 | 10,511 | What is the smallest number of states a presidential candidate can win to get the 270 electoral votes he or she needs? Which states are they? Hehe my SS teacher didn't even know the answer ;D
Maine’s four electoral votes and Nebraska’s five electoral votes are small compared to the 270 majority needed to win the presidency.
s
Question
Rating
Questions asked by the same visitor
What is 37.7 million divided by 55?
Weegy: 37.7 million divided by 55 is 685,454.55 (More)
Question
32,529,509
*
Get answers from Weegy and a team of really smart live experts.
Popular Conversations
Write the following equation in the general form Ax + By + C = 0. 2x ...
Weegy: 2x + y = 6 in general form is 2x + y - 6 = 0 User: Write the following equation in the general form Ax + By + ...
What is wailing?
There are twelve months in year.
Weegy: 12+12 = 24 User: Approximately how many hours does she play soccer in a year?
An ocean wave is an example of a(n) _____ wave form.
Weegy: An ocean waves are examples of surface wave. User: primary wave Weegy: In physics, a wave is disturbance or ...
Amendment two
Weegy: An amendment is a formal or official change made to a law, contract, constitution, or other legal document. ...
An athlete's arousel level refers to
S
L
L
Points 1951 [Total 6685] Ratings 0 Comments 1941 Invitations 1 Online
S
L
L
1
Points 1672 [Total 8434] Ratings 13 Comments 1542 Invitations 0 Offline
S
L
P
R
P
R
L
P
P
C
R
P
R
L
P
R
P
R
P
R
Points 1206 [Total 16880] Ratings 4 Comments 986 Invitations 18 Online
S
L
R
Points 1122 [Total 1456] Ratings 0 Comments 1122 Invitations 0 Online
S
L
Points 893 [Total 1061] Ratings 1 Comments 883 Invitations 0 Offline
S
L
P
P
P
1
P
L
1
Points 432 [Total 8583] Ratings 6 Comments 372 Invitations 0 Offline
S
L
P
1
L
Points 373 [Total 5909] Ratings 4 Comments 333 Invitations 0 Offline
S
L
Points 338 [Total 369] Ratings 0 Comments 338 Invitations 0 Offline
S
L
Points 203 [Total 203] Ratings 0 Comments 203 Invitations 0 Offline
S
L
Points 184 [Total 3011] Ratings 5 Comments 134 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 677 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.898343 |
https://cstheory.stackexchange.com/questions/39043/lower-bound-for-yaos-algorithm-on-general-addition-chains | 1,719,195,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00270.warc.gz | 161,908,031 | 38,193 | # Lower bound for Yao's algorithm on general addition chains?
An addition chain of size $$n$$ for given integers $$n_1,n_2\dots ,n_p$$ is a sequence of integers $$k_1,k_2\dots ,k_n$$ such that
• $$k_1=1$$,
• for all $$i$$ $$(2\le i\le m)$$ we have $$k_i=k_j+k_m$$ for some $$1\le j,m
• and $$\{n_1,n_2\dots ,n_p\}\subseteq \{k_1,k_2\dots k_n\}$$.
The general addition chain problem is the following:
Given: Numbers $$n_1,n_2,\dots,n_p$$
Problem: Compute a smallest addition chain for $$n_1,n_2,\dots,n_p$$.
This problem is known to be $$\mathsf{NP}$$-hard by [Dow81]. However, Yao [Yao76] presented an algorithm which on any input produces an addition chain of size $$\log N +c\cdot\sum_{i=1}^p\frac{\log n_i}{\log \log (n_i+2)}\le \log N +\frac{cp\log N}{\log \log (N+2)},$$ where $$N=\max_i(n_i)$$ and $$c$$ is a constant. In [Cha05] it is now claimed that this is an approximation algorithm for the general addition chain problem of ratio $$O\left(\frac{\log\left(\sum_i n_i\right)}{\log\log\left(\sum_i n_i\right)}\right),$$ i.e., Yao's algorithm produces for each input a chain which is at most by this factor larger than a smallest addition chain.
Now my question: Is there a lower bound on the approximation ratio of Yao's algorithm, e.g. is there a family of inputs such that Yao's algorithm produces addition chains which are by the described factor larger than a shortest addition chain?
[Dow81]: P. Downey, B. Leony, and P. Sethi, "Computing Sequences with Addition Chains", SIAM J. Computing, vol. 11, pp. 638-696, 1981
[Yao76]: A. C.-C. Yao, “On the evaluation of powers”, SIAM J. Comput., vol. 5, no. 1, pp. 100–103, 1976
[Cha05]: M. Charikar, E. Lehman, D. Liu, R. Panigrahy, M. Prabhakaran, A. Sahai, A. Shelat, "The Smallest Grammar Problem", IEEE Transactions on Information Theory. 51 (7): 2554–2576, 2005 | 604 | 1,838 | {"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-26 | latest | en | 0.796384 |
http://www7.economist.com/node/21561111/comments | 1,406,180,915,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997886087.7/warc/CC-MAIN-20140722025806-00016-ip-10-33-131-23.ec2.internal.warc.gz | 1,327,177,112 | 28,849 | ## The arrow of time
Backward ran sentences…
# To the relief of physicists, time really does have a preferred direction
See article
### Readers' comments
Reader comments are listed below. Comments are currently closed and new comments are no longer being accepted.
Sort:
Peter Sellers
Beautifully described! Hats off to your correspondent for making a difficult subject comprehensible.
CA-Oxonian
So then the question becomes: why does time have a preferred direction? If particles (in general) get mass from the Higg's field, do they get time from a time field? What happens with things that seem not to age, e.g. photons (which, interestingly, don't interact with the Higg's field either)?
QuietPro
Nothing new here. I figured all of this out next week.
Moran ya Simba
Interesting. I somehow have a slight suspicion that the fallacy of mistaking definitions for facts is part of the problem here. How would we recognize it if we ever encountered time going backwards? Even with the physical definitions of time as part of the space-time continuum, time is defined to go forward and if you travel faster than c (the speed of light) time does not go backwards as is often assumed but becomes complex mathematically (anyone's guess how to interpret that). Decreasing (local!!!) entropy is not the answer either because your standard fridge does that and if you think that makes it a time machine, then I have a 'time machine' I'd like to sell to you.
Time is *defined* to move forward (at uneven pace, btw) and in asking why we may be making a mistake similar to saying "I define this to be an apple; could it be an orange??" Or put differently, HOW would we *recognize* backward time if we ever encountered it???
WMcDill
About the elephant in the room. 1)Time goes only one direction, 2)there is an inexplicable preponderence of matter vs anti-matter. Observation: on the face of it, at the moment of the big bang, matter went one time direction, anti-matter the other direction. That is; positrons are just electrons going backward in time and thus, the missing anti-matter is 27.4 billion years in the past.
If time reversed, we wouldn't relive our past. This is a common misunderstanding. Due to the statistical nature of events it'd be slightly different and in the case of non-linear events potentially extremely different. The analogy between time and a film is superficially a good one but it really obscures and misleads when trying to understand the details.
But, you are correct, time always moves forward. It's the statistical nature of events that assure it. Entropy itself is just a result of the statistical nature of events, not the other way around. And entropy is certainly not the origin of the arrow of time.
DerMaulwurf in reply to CA-Oxonian
Well there is no separate time field, but it is lumped together with space into four-dimensional space-time.
And yes, finding a quantum field theory for that bugger has been driving physicists mad for a few decades by now.
And even if they solve that, there is still the issue of mass distorting space time (usually referred to as gravity).
boRmkKYLzU
Quantum mechanics is such an arcane field that I deeply respect the author for making this understandable to readers with no background in the field. Though quantum mechanics is rarely reported on in other publications, I noticed that The Economist's coverage of the Higgs boson was far more complete than other newspapers'.
salkagga in reply to method_acting
particles can and do decay, breaking into parts that constitute them (plus or minus some external remainder).
but particles can also be combined, as in nuclear fusion. if you watch a single event of fusion, and a single event of decay, how can you tell one from the other? this is the point, that the physical event on the smallest scale *works both ways*. there's no necessary direction of time (unless the physicists show otherwise through these sorts of painstaking experiments).
things really do work both ways. you have to look at the larger world to realize the difference (e.g. fusion just occurs in stars in particular conditions, decay occurs randomly everywhere).
method_acting
"A particle cannot, by itself, become disordered, so when you examine its behaviour in isolation the past and the future are hard to distinguish."
Well, there is particle decay. Is it a step forward in time, or backward?
Of course, it is alwasy forward in time! No matter what happens, that what was before, was before that what was after that.
Is that too difficult to understand? Or is it that scientists do want to put a notation of advancement into the notion of time?? Why do things advance, and what is advancement, to whom? These are very interesting questions, but they are different from the one about time. The whole arcicle is a big confusion, after all.
ashbird
Very glad to hear! This isn't just to the relief of physicists!
Anakha82
That's all very nice, but I think what we really want to know is this: can T violation be used to make a vortex manipulator?
TheSmartGuy in reply to method_acting
A good point!
Lets assume that we are first moving forwards in time. The past is backwards and future is forwards.
Now, lets assume we would suddenly start moving backwards in time. Our future would become our past and our past would become our future! That does not compute! Our future can't be our past, since we have not lived it yet!
A better way to describe this is that for every unit of time our past is the previous time unit. It doesn't matter if it was in the past or in the future. Actually that time unit doesn't even have that property. Time is measured compared to the previous unit of time, which always exists in the past by definition, since it was "previous". Alas, time always moves forwards.
persimagus
instead of adding a time feild to quantum mechanics it would be better to add the one we have it right now, gravity is its name.or is'nt it better to call it the finger print of entropy.Newtons legacy still is beyond standard model. How about a grand design to include the beatifull gravity within that, but wait a moment ,don,t we need super geometry the new realm for math far beyond revolutionary than euclids ? these are not basic objective questions of physics but they are more related to the subjective realm of reality , the ones which show how our brains through revolutionary ideas and innovations throhghout centuries enabled man to solve the much complicated questions in history of scinece .According to my humble idea,entroy,super-geometry(as some ppl like chinese eminent geometrician shing shen chern prefered to call it)and gravity stands at the same crossroad to the future of science . it would be better to focus on normal matter and try to grasp it fully before talking about anti matter or any thing metaphysicall like that.
Zambino in reply to CA-Oxonian
Indeed... or is the arrow of time an artifact of the expansion of the universe? Perhaps it varies with the rate of expansion?
marc75015
"there are innumerably fewer ways to arrange particles in an orderly fashion than in a jumbled mess. Any change in an existing arrangement is therefore likely to increase its disorder" finally after the disaster of supporting Obama The Economist is presenting its' excuses. Thank you. We accept. The Republicans of America. Now get back to reporting and stop your silly social engineering!
Giant Tortoise iawmaij
A major reason to investigate CP-violation is the apparent abudance of matter over anti-matter in the universe. That is a good thing, if matter and anti-matter are simple opposites, then the universe will be full of nothing but photons (light energy) as matter and anti-matter collide with each other to become photons, which means we humans will not even exist.
Many interesting things to work on physics :-)
this_Martin
uh oh... "nugatory" ?
See also, backward ran sentences... ;-)
Advertisement
###### Latest blog posts - All times are GMT
Food safety: Not yum!
Analects July 23rd, 23:41
Same-sex marriage and religion: When a pastor is re-frocked
Democracy in America July 23rd, 18:07
Rockaway!: New York's festival by the sea
Prospero July 23rd, 16:11
Egypt and Gaza: No longer a true mediator
Pomegranate July 23rd, 15:05
Travel mobile apps: Get with the programme
Gulliver July 23rd, 14:53
Most popular
Advertisement
Products and events
The Economist Radio is an on-demand social listening platform that allows you to listen, share and recommend The Economist audio content
Take our weekly news quiz to stay on top of the headlines
Try our new audio app and website, providing reporting and analysis from our correspondents around the world every weekday
Visit The Economist e-store and you’ll find a range of carefully selected products for business and pleasure, Economist books and diaries, and much more
Advertisement | 1,944 | 8,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2014-23 | latest | en | 0.950305 |
https://iris.uniroma3.it/handle/11590/311214 | 1,721,345,632,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00393.warc.gz | 272,550,713 | 12,076 | The eccentricity of a vertex vv in a graph GG is the maximum distance of vv from any other vertex of GG and vv is a contour vertex of GG if each vertex adjacent to vv has eccentricity not greater than the eccentricity of vv. The set of contour vertices of GG is geodetic if every vertex of GG lies on a shortest path between a pair of contour vertices. An induced connected subgraph HH of GG is isometric if, every two vertices of HH, have in HH the same distance as in GG. A graph is bridged if it does not contains an isometric cycle with length greater than 3. In this note, we show that the contour of a bridged graph is geodetic.
Mezzini, M., M., M. (2015). The contour of a bridged graph is geodetic. DISCRETE APPLIED MATHEMATICS, 204, 213-215 [10.1016/j.dam.2015.10.007].
### The contour of a bridged graph is geodetic
#### Abstract
The eccentricity of a vertex vv in a graph GG is the maximum distance of vv from any other vertex of GG and vv is a contour vertex of GG if each vertex adjacent to vv has eccentricity not greater than the eccentricity of vv. The set of contour vertices of GG is geodetic if every vertex of GG lies on a shortest path between a pair of contour vertices. An induced connected subgraph HH of GG is isometric if, every two vertices of HH, have in HH the same distance as in GG. A graph is bridged if it does not contains an isometric cycle with length greater than 3. In this note, we show that the contour of a bridged graph is geodetic.
##### Scheda breve Scheda completa Scheda completa (DC)
2015
Mezzini, M., M., M. (2015). The contour of a bridged graph is geodetic. DISCRETE APPLIED MATHEMATICS, 204, 213-215 [10.1016/j.dam.2015.10.007].
File in questo prodotto:
Non ci sono file associati a questo prodotto.
I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.
Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11590/311214`
• ND
• 7
• 5 | 532 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-30 | latest | en | 0.84666 |
http://stackoverflow.com/questions/19734934/binary-search-tree-remove-node | 1,419,788,843,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447558371.2/warc/CC-MAIN-20141224185918-00093-ip-10-231-17-201.ec2.internal.warc.gz | 89,600,867 | 16,358 | # Binary Search Tree Remove Node
I adapted my code from from this book: Data Structures and Algorithms by Mark Allen Weiss, 3rd edition.
Every time I run it, it crashes. By request I'll add the entire Binary Tree code(its long). Whenever I try to run it in debug mode, I end up cycling between the first three if else statements in the remove() function, and then I end up getting this output:
``````"Unhandled exception at 0x0007300d in Project 4Draft.exe: 0xC0000005: Access violation reading location 0x003c0000."
``````
I'm pretty sure this is a segfault, just trying to find the source. Also, when I run it, it doesn't step into findMin(), but i included it because its within remove, and It's not fully tested yet. Can anyone help me derive the source?
Here is the remove function:
``````void remove(const T& x, TreeNode * & tn) const {
if(tn == NULL)
return;
else if(x < tn->data)
remove(x, tn->left);
else if(tn->data < x)
remove(x, tn->right);
else if(tn->left != NULL && tn->right != NULL) {//Two Children(Swap the min of the right subtree, and swap
tn->data = findMin(tn->right)->data;
remove(tn->data,tn->right);
}
else{
TreeNode *oldNode = tn;
tn = (tn->left != NULL ) ? tn->left : tn->right;
delete oldNode;
}
}
``````
here is findMin():
``````TreeNode * findMin(TreeNode *x) const {
if(debugMode==true){
cout << "\nWERE IN FINDMIN\n";
}
if(x==NULL){
return NULL;
}
if(x->left==NULL){
if(debugMode==true){
cout << x;
}
return x;
}
return findMin(x->left);
};
``````
And here is what I call in my test file:
``````cout << "Checking remove()\n";
for(int i =SIZE; i>0;i++){
z.remove(x[i]);
}
cout << "DONE Checking remove()\n";
``````
-
Are you sure that your loop condition is correct?
``````for(int i =SIZE; i>0;i++){
z.remove(x[i]);
}
cout << "DONE Checking remove()\n";
``````
Maybe you should write something like:
``````for(int i = 0; i < SIZE; i++){
z.remove(x[i]);
}
``````
or
``````for(int i = SIZE - 1; i >= 0; i--){
z.remove(x[i]);
}
``````
-
Exactly right; the description makes it look like an infinite loop or stack smash, and this shows that it is. – Zach Stark Nov 1 '13 at 20:42
I see it now, I spent all of my time looking at the class functions. Thanks guys! – TaylorTheDeveloper Nov 1 '13 at 20:54 | 655 | 2,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2014-52 | latest | en | 0.768891 |
https://www.justintools.com/unit-conversion/fuel-consumption.php?k1=kilometers-per-liter&k2=feet-per-gallon-UK&q=8670000000000 | 1,713,576,359,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00223.warc.gz | 766,385,982 | 26,523 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Convert [Kilometers Per Liter] to [Feet Per Gallon UK], (km/L to ft/gal[UK])
## FUEL CONSUMPTION
8670000000000 Kilometers Per Liter
= 1.2931299311024E+17 Feet Per Gallon UK
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: fuel consumption
Conversion: Kilometers Per Liter to Feet Per Gallon UK
The base unit for fuel consumption is kilometers per liter (Non-SI/Derived Unit)
[Kilometers Per Liter] symbol/abbrevation: (km/L)
[Feet Per Gallon UK] symbol/abbrevation: (ft/gal[UK])
How to convert Kilometers Per Liter to Feet Per Gallon UK (km/L to ft/gal[UK])?
1 km/L = 14914.99343832 ft/gal[UK].
8670000000000 x 14914.99343832 ft/gal[UK] = 1.2931299311024E+17 Feet Per Gallon UK.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [fuel consumption] => (kilometers per liter), 1 Kilometers Per Liter (km/L) is equal to 1 kilometers-per-liter, while 1 Feet Per Gallon UK (ft/gal[UK]) = 6.7046626881562E-5 kilometers-per-liter.
8670000000000 Kilometers Per Liter to common fuel-consumption units
8670000000000 km/L = 20393090341155 miles per gallon US (MPG[US])
8670000000000 km/L = 24491110320164 miles per gallon UK (MPG[UK])
8670000000000 km/L = 39414644790857 kilometers per gallon US (km/gal)
8670000000000 km/L = 8670000000000 kilometers per liter (km/L)
8670000000000 km/L = 8.67E+15 meters per liter (m/L)
8670000000000 km/L = 5387288236697.7 miles per liter (mi/L)
8670000000000 km/L = 2.4550705801643E+17 meters per cubic foot (m/ft3)
8670000000000 km/L = 1.4207585441338E+14 meters per cubic inch (m/in3)
8670000000000 km/L = 8.67E+18 meters per cubic meter (m/m3)
8670000000000 km/L = 1.0767511194575E+17 feet per gallon US (ft/gal[US])
(Kilometers Per Liter) to (Feet Per Gallon UK) conversions | 677 | 2,172 | {"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} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.722292 |
https://community.powerbi.com/t5/Desktop/Total-average-in-PBI-table-is-not-correct/m-p/635143/highlight/true | 1,585,398,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491857.4/warc/CC-MAIN-20200328104722-20200328134722-00263.warc.gz | 440,411,886 | 129,011 | cancel
Showing results for
Did you mean:
Helper I
## Total average in PBI table is not correct
Hi,
I don't understand why average in total is not correct, using standard table in powerbi.
In rows values are group by customers correctly (as average):
Customer A | 17,44
Customer B | 4,09
Customer C | 3,30
Customer D | 2,93
Customer E | 2,91
Customer F | 2,58
Customer G | 2,47
TOTAL: 4,07 -> not correct, should be average = 5,10
1 ACCEPTED SOLUTION
Accepted Solutions
Anonymous
Not applicable
## Re: Total average in PBI table is not correct
This is the expected behaviour. The total average will not take an average of the averages per customer. The total average will take an average from the original data.
4 REPLIES 4
Anonymous
Not applicable
## Re: Total average in PBI table is not correct
This is the expected behaviour. The total average will not take an average of the averages per customer. The total average will take an average from the original data.
Highlighted
Helper I
## Re: Total average in PBI table is not correct
Of course!!! I was trying to calculate average from average 🙂
And its a difference like in example:
select (1+2) / 2. = 1.5
select (1+2+3) / 3. = 2
select (1.5 + 2) / 2 = 1.75 -> AVG from AVG
select (1+2+1+2+3) / 5. = 1.8 -> correct AVG
Thank's for help in showing the way 🙂
Regular Visitor
## Re: Total average in PBI table is not correct
Hello,
do you find the solution?
I have the same problem.
I don't know how to calculate average of averages.
Helper I
## Re: Total average in PBI table is not correct
Calculating average from average its not a good idea because:
select (1+2) / 2. = 1.5 -> this is average
select (1+2+3) / 3. = 2 -> this is average
then if you calculate avg from avge you have: select (1.5 + 2) / 2 = 1.75 -> AVG from AVG
but proper calculate is:
select (1+2+1+2+3) / 5. = 1.8 -> correct AVG
So it's not an error of PBI but mistake in calculating average from average (1.75 vs 1.8 on above example).
Announcements
#### New Ranks Launched March 24th!
The time has come: We are finally able to share more details on the brand-new ranks coming to the Power BI Community!
#### ‘Better Together’ Contest Finalists Announced!
Congrats to the finalists of our ‘Better Together’-themed T-shirt design contest! Click for the top entries.
#### Arun 'Triple A' Event Video, Q&A, and Slides
Missed the Arun 'Triple A' event or want to revisit it? We've got you covered! Check out the video, Q&A, and slides now.
#### Join THE global Power Platform event series.
Attend for two days of expert-led learning and innovation on topics like AI and Analytics, powered by Dynamic Communities.
#### Community Summit North America
Innovate, Collaborate, Grow. The top training and networking event across the globe for Microsoft Business Applications
Top Solution Authors
Top Kudoed Authors | 780 | 2,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-16 | latest | en | 0.86681 |
https://www.studypool.com/discuss/1023179/find-and-simplify-the-different-quotient-f-x-h-f-x-h-f-x-x-2-9x-2-f-x-1?free | 1,481,206,714,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541529.0/warc/CC-MAIN-20161202170901-00066-ip-10-31-129-80.ec2.internal.warc.gz | 993,711,001 | 14,186 | ##### ​Find and simplify the different quotient f(x+h)-f(x) / h f(x)=x^2-9x+2 f(x+
Algebra Tutor: None Selected Time limit: 1 Day
Find and simplify the different quotient
f(x+h)-f(x) / h
f(x)=x^2-9x+2
f(x+h)-f(x) / h =???? simplify
Jun 6th, 2015
f(x)=x^2-9x+2
f(x+h)= (x+h)^2-9(x+h)+2= x^2+2xh+h^2-9x-9h+2
f(x+h)-f(x) / h = (x^2+2xh+h^2-9x-9h+2-x^2+9x-2)/h
=(2xh+h^2-9h)/h= h(h+2x-9)/h = h+2x-9
Jun 7th, 2015
...
Jun 6th, 2015
...
Jun 6th, 2015
Dec 8th, 2016
check_circle | 256 | 486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2016-50 | longest | en | 0.652387 |
https://gambiter.com/chess/Checkmate_pattern.html | 1,718,785,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00791.warc.gz | 231,743,754 | 10,709 | Home :: Chess
# Checkmate pattern
Checkmate
In chess several checkmate patterns occur sufficiently frequently, or are otherwise of such interest to scholars, so as to have acquired specific names in chess commentary. The diagrams that follow show these checkmates with White checkmating Black.
## Anastasia's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Anastasia's mate
In Anastasia's mate, a knight and rook team up to trap the opposing king between the side of the board on one side and a friendly piece on the other. This checkmate got its name from the novel Anastasia und das Schachspiel by Johann Jakob Wilhelm Heinse.
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
In Anderssen's mate (named for Adolf Anderssen), the rook or queen is supported by a diagonally-attacking piece such as a pawn or bishop as it checkmates the opposing king along the eighth rank.
Sometimes a distinction is drawn between Anderssen's mate, where the rook is supported by a pawn (which itself is supported by another piece, as in the diagram), and Mayet's mate, where the rook is supported by a distant bishop.
## Arabian mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Arabian mate
In the Arabian mate, the knight and the rook team up to trap the opposing king on a corner of the board. The rook sits on a square adjacent to the king both to prevent escape along the diagonal and to deliver checkmate while the knight sits two squares away diagonally from the king to prevent escape on the other two squares and to protect the rook.
## Back-rank mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal back-rank mate
The back-rank mate occurs when a rook or queen checkmates a king that is blocked in by friendly pieces (usually pawns) on his first rank.
## Bishop and knight mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Bishop and knight mate
The Bishop and knight checkmate occurs when the king teams up with a bishop and knight to force the opponent king to the corner of the board. The bishop and knight endgame can be difficult to master: some positions may require up to 34 moves of perfect play before checkmate can be delivered.
## Blackburne's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Blackburne's mate
Blackburne's mate is named for Joseph Henry Blackburne and is a rare method of checkmating. The checkmate utilizes the black rook (it could be a bishop or queen instead) to confine the black king's escape to the f8 square. One of the bishops confines the black king's movement by operating at a distance, while the knight and the other bishop operate within close range. Threatening Blackburne's mate can be used to weaken Black's position.
## Boden's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Boden's mate
In Boden's mate two attacking bishops on criss-crossing diagonals deliver mate to a king obstructed by friendly pieces, usually a rook and a pawn.
## Box mate (Rook mate)
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
A typical Box mate
The Box mate is one of the four basic checkmates along with Queen mate, king and two bishops checkmate, and bishop and knight checkmate. It occurs when the side with the king and rook box in the bare king to the corner or edge of the board. The mate is delivered by the rook along the edge rank or file, and escape towards the centre of the board is blocked by the king.
## Corner mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Corner mate
The Corner mate is a common method of checkmating. It works by confining the king to the corner using a rook or queen and using a knight to engage the checkmate.
## Cozio's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Cozio's mate. After 1.Qh6+, Black is forced to play 1...Kg3
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
White then checkmates with 2.Qh2
Cozio's mate is a common method of checkmating. The checkmate is an upside down version of the Dovetail mate. It was named after a study Carlo Cozio that was published in 1766.
## Damiano's bishop mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Damiano's bishop mate
Damiano's bishop mate is a classic method of checkmating. The checkmate utilizes a queen and bishop, where the bishop is used to support the queen and the queen is used to engage the checkmate. The checkmate is named after Pedro Damiano.
## Damiano's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Damiano's mate
Damiano's mate is a classic method of checkmating and one of the oldest. It works by confining the king with a pawn and using a queen to initiate the final blow. Damiano's mate is often arrived at by first sacrificing a rook on the h-file, then checking the king with the queen on the h-file, and then moving in for the mate. The checkmate was first published by Pedro Damiano in 1512. In Damiano's publication he failed to place the white king on the board which resulted in it not being entered into many chess databases due to their rejection of illegal positions.
## David and Goliath mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal David and Goliath mate
The David and Goliath mate is a common method of checkmating. Although the David and Goliath mate can take many forms, it is characterized generally as a mate in which a pawn is the final attacking piece and where enemy pawns are nearby. Its name is taken from the biblical account of David and Goliath.
## Double bishop mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Double bishop mate
The Double bishop mate is a classic method of checkmating. It is a similar to Boden's mate, but a bit simpler. The checkmate involves attacking the king using two bishops, resulting in the king being trapped behind a pawn that has not been moved.
## Dovetail mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Dovetail mate
The Dovetail mate is a common method of checkmating. It involves trapping the black king in a pattern shown to the right. It does not matter how the queen is supported and it does not matter which type Black's other two pieces are so long as neither is an unpinned knight.
## Epaulette mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Epaulette mate
Epaulette or epaulet mate is, in its broadest definition, a checkmate where two parallel retreat squares for a checked king are occupied by its own pieces, preventing its escape. The most common Epaulette mate involves the king on its back rank, trapped between two rooks. The perceived visual similarity between the rooks and epaulettes, ornamental shoulder pieces worn on military uniforms, gives the checkmate its name.
### Example games
• Van Wely-Morozevich, Wijk aan Zee 2001; Loek van Wely is forced to resign after blundering into an unavoidable Epaulette mate against Alexander Morozevich.
• Carlsen-Ernst, Wijk aan Zee 2004; a thirteen-year-old Magnus Carlsen achieves an unusual "sideways" Epaulette mate against Sipke Ernst on his way to winning the C Group at the Corus chess tournament in 2004.
• Anand-Carlsen, Tal Memorial Blitz World Championship 2009; when Carlsen essentially clinched the blitz world champion title.
## Greco's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Greco's mate
Greco's mate is a common method of checkmating. The checkmate is named after the famous Italian checkmate cataloguer Gioachino Greco. It works by using the bishop to contain the black king by use of the black g-pawn and subsequently using the queen or a rook to checkmate the king by moving it to the edge of the board.
## h-file mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal h-file mate
The h-file mate involves the use of a rook attacking the black king supported by a bishop. It often comes about after the black king castles on its kingside in a fianchetto position. White usually arrives at this position after a series of sacrifices on the h-file.
Although it is called the h-file mate, it can also occur on other files, so for example with the uncastled black king on e8 and a white rook on d8 protected by a white bishop on g5.
## Hook mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Hook mate
The Hook mate involves the use of a white rook, knight, and pawn along with one black pawn to limit the black king's escape. The rook is protected by the knight and the knight is protected by the pawn.
## King and two bishops mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
A typical king and two bishops checkmate
King and two bishops checkmate is one of the four basic checkmates along with Queen mate, Rook mate, and the bishop and knight checkmate. It occurs when the king with two bishops force the bare king to the corner of the board to force a possible mate.
## King and two knights mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Checkmate with a king and two knights, but it cannot be forced
In a two knights endgame, the side with the king and two knights cannot checkmate a bare king by force. This endgame should be a draw if the bare king plays correctly. A mate only occurs if the player with the bare king blunders. In some circumstances, if the side with the king also has a pawn, it is possible to set up this type of checkmate.
## Légal mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Légal mate
In the Légal mate, two knights and a bishop coordinate to administer checkmate.
## Lolli's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Lolli's mate
Lolli's mate is a common method of checkmating. The checkmate involves infiltrating Black's fianchetto position using both a pawn and queen. The queen often gets to the h6 square by means of sacrifices on the h-file. It is named after Giambattista Lolli.
## Max Lange's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Max Lange's mate
Max Lange's mate is a common method of checkmating. The checkmate is named after Max Lange. It works by using the bishop and queen to checkmate the king.
## Morphy's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Morphy's mate
Morphy's mate is a common method of checkmating. It was named after Paul Morphy. It works by using the bishop to attack the black king and a rook and Black's own pawn to confine it. In many respects it is very similar to the Corner mate.
## Opera mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Opera mate
The Opera mate is a common method of checkmating. It works by attacking the king on the back rank with a rook using a bishop to protect it. A pawn or other piece other than a knight of the enemy king's is used to restrict its movement. The checkmate was named after its implementation by Paul Morphy in 1858 at a game at the Paris opera against Duke Karl of Brunswick and Count Isouard, see Opera game.
## Pillsbury's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
Pillsbury's mate example. 1.Rxg7+ Kh8 2.Rg1+ Rf6 3.Bxf6#
Pillsbury's mate is a common method of checkmating and is named for Harry Nelson Pillsbury. It works by attacking the king with either the rook or bishop as shown to the right. The king can be either on the g8 or h8 square during the checkmate.
## Queen mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
A typical Queen mate
Queen mate is one of the four major checkmates along with Box mate, king and two bishops checkmate, and bishop and knight checkmate. It occurs when the side with the king and queen force the bare king to the edge or corner of the board. The queen checkmates the bare king with the support of the allied king.
## Réti's mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Réti's mate
Réti's mate is a famous method of checkmating. The checkmate is named after Richard Réti, who delivered it in an 11-move game against Savielly Tartakower in 1910 in Vienna. It works by trapping the enemy king with four of its own pieces that are situated on flight squares and then attacking it with a bishop that is protected by a rook or queen.
## Smothered mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
A typical Smothered mate
Smothered mate is a common method of checkmating. It occurs when a knight checkmates a king that is smothered (surrounded) by his friendly pieces and he has nowhere to move nor is there any way to capture the knight. It is also known as Philidor's Legacy after François-André Danican Philidor, though its documentation predates Philidor by several hundred years.
## Suffocation mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Suffocation mate
The Suffocation mate is a common method of checkmating. It works by using the knight to attack the enemy king and the bishop to confine the king's escape routes.
## Swallow's tail mate
a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
An archetypal Swallow's tail mate
Swallow's tail mate also known as the Guéridon mate is a common method of checkmating. It works by attacking the enemy king with a queen that is protected by a rook or other piece. The enemy king's own pieces (in this example, rooks) block its means of escape. It resembles the Epaulette mate. | 4,443 | 13,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.89343 |
https://coderforevers.com/dsa/greedy-algorithms/ | 1,550,807,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247513222.88/warc/CC-MAIN-20190222033812-20190222055812-00310.warc.gz | 516,131,025 | 19,445 | # Data Structures – Greedy Algorithms
29
views
(Last Updated On: May 27, 2018)
## In this DSA tutorial, you will learn what is Greedy Algorithms and how to use it.
An algorithm is designed to achieve the optimum solution for a given problem. In Data Structures Greedy Algorithms approach, decisions are made from the given solution domain. As being greedy, the closest solution that seems to provide an optimum solution is chosen.
Data structures Greedy algorithms try to find a localized optimum solution, which may eventually lead to globally optimized solutions. However, generally greedy algorithms do not provide globally optimized solutions.
#### Counting Coins
This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. If we are provided coins of ₹ 1, 2, 5 and 10 and we are asked to count ₹ 18 then the greedy procedure will be −
• Select one ₹ 10 coins, the remaining count is 8
• Then select one ₹ 5 coin, the remaining count is 3
• Then select one ₹ 2 coin, the remaining count is 1
• And finally, the selection of one ₹ 1 coins solves the problem
Though, it seems to be working fine, for this count we need to pick only 4 coins. But if we slightly change the problem then the same approach may not be able to produce the same optimum result.
For the currency system, where we have coins of 1, 7, 10 value, counting coins for value 18 will be absolutely optimum but for count like 15, it may use more coins than necessary. For example, the greedy approach will use 10 + 1 + 1 + 1 + 1 + 1, total 6 coins. Whereas the same problem could be solved by using only 3 coins (7 + 7 + 1)
Hence, we may conclude that the greedy approach picks an immediate optimized solution and may fail where global optimization is a major concern.
#### Examples
Most networking algorithms use the greedy approach. Here is a list of few of them −
• Travelling Salesman Problem
• Prim’s Minimal Spanning Tree Algorithm
• Kruskal’s Minimal Spanning Tree Algorithm
• Dijkstra’s Minimal Spanning Tree Algorithm
• Graph – Map Coloring
• Graph – Vertex Cover
• Knapsack Problem
• Job Scheduling Problem
There are lots of similar problems that uses the greedy approach to find an optimum solution.
Data Structures Algorithms Books: Algorithms Plus Data Structures
Ask your questions and clarify your/others doubts on Greedy Algorithms by commenting.
PREVIOUS ASYMPTOTIC ANALYSIS NEXT DIVIDE AND CONQUER | 571 | 2,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-09 | longest | en | 0.899775 |
https://www.physicsforums.com/threads/lagrangian-of-two-masses-connected-by-string-on-inclined-pln.893370/ | 1,511,130,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00397.warc.gz | 859,302,566 | 16,945 | # Lagrangian of two masses connected by string on inclined pln
Tags:
1. Nov 14, 2016
### Elvis 123456789
Two masses, m1 and m2, are attached by a light string of length D. Mass m1 starts at rest on an inclined plane and mass m2 hangs as shown. The pulley is frictionless but has a moment of inertia I and radius R. Find the Lagrangian of the system and determine the acceleration of the masses using the Lagrangian. Though there are three coordinates of interest (along the plane for mass m1, down for mass m2, and an angle for the rotation of the pulley), there are two constraints.
2. Relevant equations
∂L/∂q - d/dt(∂L/∂(q-dot)2) = 0
L = T - V
3. The attempt at a solution
If I define the x-direction to be in the direction of the inclined plane then
L = 0.5*m1*(x-dot)2 + 0.5*m2*(y-dot)2 + 0.5*I*(phi-dot)2 + m1*g*x*sin(θ) + m*g*y
where φ is the angle that the pulley is rotating through
The length of the string is constant so the length of string on the plane plus the bit on the pulley plus the rest that is hanging holding up m2 is equal to D
so x + R(π/2 + θ) + y = D -----> x-dot = -(y-dot) ≡ q-dot
and the other constraint involves the string moving on the pulley without slipping
y = Rφ ----> y-dot = R(φ-dot) ---> q-dot = R(φ-dot)
then L = 0.5*m1*(q-dot)2 + 0.5*m2*(q-dot)2 + 0.5*I*(q-dot/R)2 + m1*g*q*sin(θ) + m*g*(D - R(π/2 + θ) - q)
i dont know if what i have so far is correct. Anybody care to give me a hand? The setup is shown in the attachment.
#### Attached Files:
• ###### hw 7.png
File size:
12 KB
Views:
38
2. Nov 14, 2016
### haruspex
That all looks right, except that you appear to have renamed m2 as m, and it would be simpler to redefine the zero potential of that so that the term becomes simply -m2gq. | 550 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-47 | longest | en | 0.857807 |
https://www.quanswer.com/en/abody-of-mass-5kgs-is-raised-to-a-height-1-8metres-what-is-the-kinetic-energy-when-it-is-half-way-down | 1,721,926,931,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00539.warc.gz | 798,127,752 | 12,600 | > #potential-energy , #energy , #physics > Abody of Mass 5kgs is raised to a height 1.8metres . what is the kinetic energy when it is half way down?
Mar 14
# Abody of Mass 5kgs is raised to a height 1.8metres . what is the kinetic energy when it is half way down?
1 view
Sort By
ChatGPT
Mar 14
The potential energy of an object at a height is given by the formula:
Potential energy = mass x gravity x height
Given:
Mass = 5 kg
Height = 1.8 meters
Gravity = 9.81 m/s^2 (acceleration due to gravity)
Potential energy = 5 kg x 9.81 m/s^2 x 1.8 m
Potential energy = 88.29 Joules
Therefore, the potential energy of the object at a height of 1.8 meters is 88.29 Joules.
0
× | 212 | 674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.87515 |
http://nrich.maths.org/public/leg.php?code=-68&cl=3&cldcmpid=1051 | 1,475,076,637,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661555.40/warc/CC-MAIN-20160924173741-00192-ip-10-143-35-109.ec2.internal.warc.gz | 201,174,571 | 10,660 | # Search by Topic
#### Resources tagged with Visualising similar to Is it Magic or Is it Maths?:
Filter by: Content type:
Stage:
Challenge level:
### Concrete Wheel
##### Stage: 3 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Convex Polygons
##### Stage: 3 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Cube Paths
##### Stage: 3 Challenge Level:
Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B?
### Tetra Square
##### Stage: 3 Challenge Level:
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.
### A Tilted Square
##### Stage: 4 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Seven Squares - Group-worthy Task
##### Stage: 3 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Bands and Bridges: Bringing Topology Back
##### Stage: 2 and 3
Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.
### Trice
##### Stage: 3 Challenge Level:
ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
### Euromaths
##### Stage: 3 Challenge Level:
How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Tessellating Hexagons
##### Stage: 3 Challenge Level:
Which hexagons tessellate?
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Weighty Problem
##### Stage: 3 Challenge Level:
The diagram shows a very heavy kitchen cabinet. It cannot be lifted but it can be pivoted around a corner. The task is to move it, without sliding, in a series of turns about the corners so that it. . . .
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Triangle Inequality
##### Stage: 3 Challenge Level:
ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm.
### The Old Goats
##### Stage: 3 Challenge Level:
A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . .
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Mystic Rose
##### Stage: 3 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Chess
##### Stage: 3 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### Christmas Chocolates
##### Stage: 3 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Partially Painted Cube
##### Stage: 4 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Cutting a Cube
##### Stage: 3 Challenge Level:
A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical?
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Painting Cubes
##### Stage: 3 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Threesomes
##### Stage: 3 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Königsberg
##### Stage: 3 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Zooming in on the Squares
##### Stage: 2 and 3
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
### 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?
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Rolling Triangle
##### Stage: 3 Challenge Level:
The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks.
### Playground Snapshot
##### Stage: 2 and 3 Challenge Level:
The image in this problem is part of a piece of equipment found in the playground of a school. How would you describe it to someone over the phone?
### Icosagram
##### Stage: 3 Challenge Level:
Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . .
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Flight of the Flibbins
##### Stage: 3 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### 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. . . .
### Hidden Squares
##### Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
##### Stage: 3 Challenge Level:
Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite?
### Reflecting Squarely
##### Stage: 3 Challenge Level:
In how many ways can you fit all three pieces together to make shapes with line symmetry?
### Take Ten
##### Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
### One and Three
##### Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . . | 2,479 | 10,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-40 | longest | en | 0.878031 |
http://dopovidi-nanu.org.ua/en/archive/2020/1/6 | 1,603,623,942,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888931.67/warc/CC-MAIN-20201025100059-20201025130059-00617.warc.gz | 30,334,402 | 7,160 | # Joint finite elements with vector approximation of unknowns for the calculation of thin shells of complex geometry
1Storozhuk, EA1S.P. Timoshenko Institute of Mechanics of the NAS of Ukraine, Kyiv Dopov. Nac. akad. nauk Ukr. 2020, 1:39-48 https://doi.org/10.15407/dopovidi2020.01.039 Section: Mechanics Language: Ukrainian Abstract: The formulation of boundaryvalue problems for thin shells of complex shape under the action of a static load is given. The basic equations are given on the basis of the theory of shells, in which the Kirchhoff—Love hypotheses hold. The geometric relationships are written in the vector form, and the physical ones are based on Hooke’s law for isotropic materials. Using the finiteelement method, a technique has been developed for numerically solving twodimensional static problems for thin shells of complex geometry. The resolving equations in displacements are obtained from the stationary conditions of a discrete analog of the Lagrange functional. Two variants of joint finite elements with 36 and 20 degrees of freedom are proposed. A feature of the developed modifications of the finiteelement method is the vector form of approximation of the sought quantities and the discrete execution of the geometric part of the Kirchhoff—Love hypotheses. The finite elements of thin shells of complex shape constructed in this way satisfy the continuity conditions for the displacement vectors and rotation angles and accurately describe the translational part of the movements of the finite elements as rigid bodies. Keywords: complexshape shells, finiteelement method, Kirchhoff—Love discrete hypotheses, static load, vector approximation
References:
1. Golovanov, A. I. & Kornishin, M. S. (1989). Introduction to the finite element method of the statics of thin shells. Kazan: Publ. Kazan. branch of physical and technical Institute (in Russian).
2. Zienkiewicz, O. C. (1975). The finite element method in technology. Moscow: Mir (in Russian).
3. Storozhuk, E. A. (2009). Variational vectordifference method in nonlinear problems of the theory of thin shells with curved holes. System technologies. Math. problems of techn. mechanics, No. 3 (62), pp. 149156 (In Ukrainian).
4. Kiseleva, T. A., Klochkov, Yu. V. & Nikolaev, A. P. (2015). Comparison of scalar and vector FEM forms in the case of an elliptic cylinder. J. Comput. Math. Math. Phys, 55, No. 3, pp. 422431. Doi: https://doi.org/10.1134/S0965542515030094
5. Areias, P.M.A., Song, J.H. & Belytschko, T. (2005). A finitestrain quadrilateral shell element based on discrete Kirchhoff–Love constraints. Int. J. Numer. Meth. Eng., 64, No. 9, pp. 11661206. Doi: https://doi.org/10.1002/nme.1389
6. Guz, A. N., Storozhuk, E. A. & Chernyshenko, I. S. (2003). Physically and Geometrically Nonlinear Static Problems for ThinWalled Multiply Connected Shells. Int. Appl. Mech, 39, No. 6, pp. 679687. Doi: https://doi.org/10.1023/A:1025793808397
7. Novozhilov, V. V., Chernykh, K. F. & Mikhailovsky, E. I. (1991). Linear theory of thin shells. Leningrad: Politekhnika (in Russian). | 817 | 3,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-45 | latest | en | 0.84099 |
https://bzoj.llf0703.com/p/2578.html | 1,571,415,624,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684226.55/warc/CC-MAIN-20191018154409-20191018181909-00543.warc.gz | 406,278,095 | 2,788 | # pku3985 Knight
### 题目描述
You must have heard of the Knight's Tour problem. In that problem, a knight is placed on an empty chess board and you are to determine whether it can visit each square on the board exactly once.
Let's consider a variation of the knight's tour problem. In this problem, a knight is place on an infinite plane and it's restricted to make certain moves. For example, it may be placed at (0, 0) and can make two kinds of moves: Denote its current place as (x,y), it can only move to (x+1,y+2) or (x+2,y+1). The goal of this problem is to make the knight reach a destination position as quickly as possible (i.e. make as less moves as possible).
### 输入格式
The first line contains an integer T ( T < 20) indicating the number of test cases.
Each test case begins with a line containing four integer: fx fy tx ty(-5000<=fx,fy,tx,ty<=5000). The knight is originally placed at (fx, fy) and (tx, ty) is its destination.
The following line contains an integer m(0 < m <= 10), indicating how many kinds of moves the knight can make.
Each of the following m lines contains two integer mx my(-10<=mx,my<=10; |mx|+|my|>0), which means that if a knight stands at (x,y), it can move to (x+mx,y+my).
### 输出格式
Output one line for each test case. It contains an integer indicating the least number of moves needed for the knight to reach its destination. Output "IMPOSSIBLE" if the knight may never gets to its target position.
```2
0 0 6 6
5
1 2
2 1
2 2
1 3
3 1
0 0 5 5
2
1 2
2 1
```
```3
IMPOSSIBLE
``` | 437 | 1,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-43 | latest | en | 0.905441 |
https://mirrorinfo.online/knowledge-base/how-to-transfer-amperes | 1,718,899,842,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00410.warc.gz | 334,840,967 | 7,890 | # How to transfer amperes
Amperes – standard system unit of measure of the current (SI). Quite large to household measures therefore multiple units (kiloamper) in practice are used seldom. But in characteristics of the electronic equipment (especially tiny) submultiple unit – milliampere often meets. The household electrical accessories are usually described by such parameter as power (it is measured in watts). It is necessary to connect electrical household appliances to the power supply network having restriction on current. To avoid continuous operation of safety locks, it is necessary to represent how in practice to transfer amperes to other units of measure.
## It is required to you
• - tester;
• - calculator;
• - technical documentation on electric devices.
## Instruction
1. If to translate amperes in other units of measure of current is required (submultiple or multiple), then just increase the number of amperes by the corresponding coefficient. So, for example, to transfer amperes to milliamperes increase number of amperes by 1000, and for the translation of amperes in kiloamper – increase by 0.001. Respectively, at transfer of amperes to megaamperes increase current by 0.000001, and at transfer to microamperes – increase by 1000000. Other submultiple and multiple units of measure of current in life and at the solution of standard tasks are practically not used.
2. To estimate the maximum total power of consumers of the electric power which your power supply network is capable to sustain increase current (in amperes) on which safety locks, on tension in network (220 volts) are calculated. The received value will equal the admissible power of at the same time connected electric devices expressed in watts. Thus (it is clean practically) amperes can be transferred in watts.
3. It is similarly possible to calculate the maximum power of electrical accessories at its connection to autonomous power sources. As a rule, on accumulators and batteries tension and the maximum current on which the power source is calculated is specified. At connection of too powerful consumer, the source of current can fail very quickly or, even to ignite.
4. For determination of power consumption study technical documentation of the electrodevice or look for information on the case of the device. Power of electrical accessories is specified in watts (W, W), kilowatts (kW, kW) or milliwatts (MW, mW).
5. Example. The household electrical network is calculated on the maximum current of 20 amperes. Question. How many hundred-watt electrobulbs can be included at the same time? Decision.1. Estimate the maximum power of loading of the power supply network: 20 (A) * 220 (C) = 4400 (W).2. Divide the general admissible power of network into the power of one bulb: 4400 (W) / 100 (W) = 44 (pieces). Answer. It is at the same time possible to connect 44 bulbs.
Author: «MirrorInfo» Dream Team | 641 | 2,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-26 | latest | en | 0.916179 |
http://www.physicsforums.com/showthread.php?t=248149 | 1,369,246,168,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702185502/warc/CC-MAIN-20130516110305-00080-ip-10-60-113-184.ec2.internal.warc.gz | 655,976,279 | 10,012 | ## sitting n married couples on a round table
Question: A total of 2n people, consisting of n married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let C_i denote the event that couple i are seated next to each other, i = 1, 2, ... n.
(a) Find P(C_i).
There are two different ways to seat C_i together on a round table and the possible orderings of the rest of the couples is (2n - 2)!, hence we have:
P(C_i) = (2 * (2n - 2)!)/(2n - 1)!) = 2/(2n - 1)
The the (2n - 1)! being the total number of orderings (C_i included) on a round table. The book gives an answer of 2/(2n + 1).
What am I doing wrong?
(b) For j <> i, find P(C_j | C_i)
Instead of using the P(C_j | C_i) = P(C_j C_i) / P(C_i) formula, we may cut to the chase and say that since a couple has already been selected, we can view the round table as a straight line. So the number of ways to arrange C_j are:
i) 2 choices on which of the couple to sit
ii) Pick a place among (2n - 3) seats
iii) Place the other couple on the other side (1 choice)
2 (2n - 3)
So P(C_j | C_i) = ( 2 (2n - 3) (2n - 4)! )/ (2n - 2)! = 2 / (2n - 2)
Which is in the back of the book
(c) When n is large, approximate the probability that there are no married couples who are seated next to each other.
I imagine we're supposed to use the Poisson distribution?
Mentor
Quote by lizzyb What am I doing wrong?
In short, you are ignoring that the table is round, and this changes things. There is a very easy way to solve this problem: What is the probability that the couple is not seated next to one another?
Quote by D H In short, you are ignoring that the table is round, and this changes things. There is a very easy way to solve this problem: What is the probability that the couple is not seated next to one another?
There are 2n places for the first member of the couple, then 2n-2 places for the other member.
So P(E_i) = ( (2n)(2n - 2)(2n-2)! )/(2n-1)!??
Mentor
## sitting n married couples on a round table
You did not answer my question. I asked you to compute the probability that the couple are not seated next to one another. Call this probability q. The probability that the couple are seated next to one another is p=1-q.
Quote by D H You did not answer my question. I asked you to compute the probability that the couple are not seated next to one another. Call this probability q. The probability that the couple are seated next to one another is p=1-q.
That's what I meant by ( (2n)(2n - 2)(2n-2)! )/(2n-1) but that's probably not right.
Since its a round table, we can say that the location of the first person placed upon it doesn't matter, but for the other member of that couple, there are 2n-2 places to put him or her (since there are two places on either side of the first seat where his/her partner is). After these two have been placed, there are still the other people, of which there are (2n-4)! possible seating arrangements. This all goes over (2n-1)! since that is the total number of possible permutations of 2n people on a round table:
q = (2n - 2)(2n - 4)!/(2n - 1)! = 1 / (2n - 1)(2n - 3)
does that look right to you?
For part (a) you did it correctly! The book is wrong :) There are $$(2n-1)!$$ ways to arrange 2n people. Since the couple with index i are sitting together, you can think of them as one person thus you get $$(2n-2)!$$. But, there are 2 ways to arrange couple with index i. Thus $$\boxed{P(C_{i})=\frac{2(2n-2)!}{(2n-1)!}=\frac{2}{2n-1}}$$ (b) Well, you can further simplify it into $$\frac{1}{n-1}$$ (c) Yes, you use the Poisson r.v. distribution with parameter $$\lambda = pn$$. Now what is p? Then just find the probability that no couple are sitting together i.e. $$\boxed{P[X=0] = e^{-\lambda}}$$ by definition of Poisson
Mentor
I took a second look, sorry.
Quote by lizzyb hence we have: P(C_i) = 2/(2n - 1). The book gives an answer of 2/(2n + 1). What am I doing wrong?
You are doing nothing wrong. The book is what is wrong. The answer is 2/(2n-1) (n>1).
Consider one member of the couple in question. If the couple is not seated adjacently, both seats next to this one member of the couple must be filled someone other than the other member of the couple. These are the only two seats one need be concerned with, and the probability neither is the other member of the couple is (2n-2)/(2n-1)*(2n-3)/(2n-2)=(2n-3)/(2n-1)=1-2/(2n-1). The probability the couple *are* seated adjacently is thus 2/(2n-1).
As a sanity check, look at the case n=2. The only way a couple is not seated adjacently at a table of four is when they are seated across from one another. There are 4*2 such seatings out of a total of 24, so the probability they are seated adjacently is 16/24=2/3.
Great! Thank you for verifying part a as well as the other way of understanding it as well. As for part (c), I suppose I'm to guess a decent value of p based on the answers of (a) and (b)? (a) P(C_i) = 2/(2n - 1) (b) P(C_j | C_i) = 1/(n-1) (i <> j) And this p should be a general guesstimate of the probability that a couple sits together? Both (a) and (b) are similar to 1/n ... that gives the right answer in the back of the book but I can't say I fully understand why it works. Thanks for your help! [Added later] No, 1/n makes sense but is that the right question to ask myself in answering "what value of p" - that is, "what is the general probability of a single couple sitting together"?
Yes, which you have already found namely $$P(C_{i})$$ | 1,594 | 5,492 | {"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} | 4.1875 | 4 | CC-MAIN-2013-20 | longest | en | 0.947808 |
https://questionpaper4exam.com/solved-ssc-previous-year-question-with-answer/ | 1,656,870,262,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00452.warc.gz | 522,065,199 | 27,853 | # SOLVED SSC PREVIOUS YEAR QUESTION WITH ANSWER
Q 1. Two pipes A and B can fill a tank in 6 hours and 8 hours respectively. If both the pipes are opened together, then after how many hours should B be closed so that the tank is full in 4 hours?
a. 2/3
b. 1
c. 2
d. 8/3
Q 2.Two numbers are in the ratio of 3:5. If 9 be subtracted from each, then they are in the ratio of 12:23. Find the number?
a. 15:28
b. 36:115
c. 33:55
d. 60:69
Q 3. A dishonest shopkeeper, using a faulty balance make a profit of 5% while buying as well as while selling his goods. His actual gain percent in the whole process amounts to?
a. 11
b. 10
c. 10.25
d. 10.5
Q 4. The price of an article has been reduced by 25%. In order to restore the original price, the reduced price must be increased by?
a. 25%
b. 30%
c. 33%
d. 33.1/3%
Q 5. A cube of edge 6 cm is painted on all sides and then cut into unit cubes. The number of unit cubes with no sides painted is?
a. 0
b. 64
c. 186
d. 108
Q 6. Internal bisector of angles B and angle C of a triangle ABC meet at O. If angle BAC = 80 degree, then the value of angle BOC is?
a. 120 degree
b. 140 degree
c. 110 degree
d. 130 degree
Q 7. If sec Ɵ+ tan Ɵ= 2 then the value of sec Ɵ is ?
a. 4/5
b. 5
c. 5/4
d. 2
Q 8. If each interior angle is double of each exterior angle of a regular polygon with n sides, then the value of n is?
a. 8
b. 10
c. 5
d. 6
Q 9. The side BC of a triangle ABC is produced to D. If angle ACD =108 degree and angle B = ½ of angle A , then angle A in degree?
a. 36
b. 72
c. 108
d. 59
Q 10. In triangle ABC, AD is the median and AD = ½ BC. If angle BAD = 30 degree, then measure of angle ACB is?
a. 90 degree
b. 45 degree
c. 30 degree
d. 60 degree
Q 11. If a+b+1 = 0, then the value of (a3 + b3 + 1 – 3ab) is?
a. 3
b. 0
c. -1
d. 1
Q 12. A discount of 30% on the marked price of toy reduces its selling price by rs. 30. What is the new selling price ?
a. 70
b. 21
c. 130
d. 100
Q 13. Pipe A alone can fill a tank in 8 hours. Pipe B alone can fill it in 6 hours. If both the pipes are opened and after 2 hours pipe A is closed, then the other pipe will fill the tank?
a. 6 hours
b. 3.1/2 hours
c. 4 hours
d. 2.1/2 hours
Q 14. A sells an article to B at a profit of 20% and B sell it to C at a profit of 25%. If C pays rs. 1200, then the cost price of the article originally is?
a. 700
b. 600
c. 1000
d. 800
Q 15. If the ratio of cost price and selling price be 9:10, then the percentage in profit is?
a. 11%
b. 10%
c. 11.1/9%
d. 9%
Sharing is caring:- | 996 | 2,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-27 | longest | en | 0.621399 |
https://www.astronomy.ohio-state.edu/pogge.1/Ast161/Au06/Quizzes/q2study.txt | 1,709,569,005,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-10/segments/1707947476452.25/warc/CC-MAIN-20240304133241-20240304163241-00073.warc.gz | 653,909,067 | 1,854 | Astronomy 161 -- Autumn 2006 In-Class Quiz 2 Study Guide Unit 3: The Revolutions of the Heavenly Orbs ------------------------------------ Aristotelian World View Assumptions (uniform circular motion, fixed unmoving earth) Early Geocentric Systems Eudoxus, Pythagoras, Aristotle Epicyclic Systems Hipparchus & Ptolemy Early Heliocentric System Aristarchus of Samos Ptolemaic Geocentric System Epicycles Equants Preserving Appearances - esp. retrograde motion & change in brightness of superior planets at opposition. Problems: complex, no way to measure planetary distances Copernicus Motivations & Assumptions (disliked equant, wanted to restore uniform circular motion) Copernican Heliocentric System Sun at the center Earth rotates on its axis every 24 hours Earth orbits (revolves) around the sun once a year His use of epicycles and why he used them. Successes: a) explains superior & inferior planets b) explains retrograde motion c) gives a geometric way to measure planetary distances Problems: (a) moving earth (b) stellar parallaxes Tycho Brahe: his observations & their significance Johannes Kepler: his theoretical work & its significance Kepler's Three Laws of Planetary Motion First Law Second (Equal Areas) Law Third (Harmonic) Law Galileo's telescope observations & their significance The Moon Sunspots Phases of Venus Moons of Jupiter Isaac Newton: work and its significance Laws of Motion First Law (Law of Inertia) Second Law (F=ma) Third Law (Action & Reaction) Unit 4: Gravitation, Light, & Matter ---------------------------- Newtonian Gravity Inverse-Square Law Force Dependence of the gravitational force on masses and distance of the two bodies. Newton's Generalized forms of Kepler's Laws Shapes of Orbits Orbit about the Center of Mass Circular and Escape Velocity Measuring Masses with Newton's form of Kepler's 3rd Law Tests of Newton's Laws: Return of Halley's Comet Discovery of Neptune Tides Basic causes of gravitational tides Earth Tides caused by the Sun and Moon Tidal Locking Tidal Evolution of the Moon's Orbit and Earth's Rotation Lunar Recession Increasing Length of the Day Rotation and Revolution of the Earth: Observational demonstrations of the Earth's motion: Demonstrations of the Rotation of the Earth: The Coriolis Effect The Foucault Pendulum Demonstration of the Earth's Orbit around the Sun: Stellar Parallaxes | 519 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-10 | latest | en | 0.790294 |
https://www.lotterypost.com/thread/199392 | 1,513,611,227,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948617816.91/warc/CC-MAIN-20171218141805-20171218163805-00582.warc.gz | 774,414,888 | 20,766 | Welcome Guest
You last visited December 18, 2017, 9:44 am
All times shown are
Eastern Time (GMT-5:00)
Randomness?
Topic closed. 18 replies. Last post 8 years ago by KnuckleHead.
Page 1 of 2
United States
Member #73037
April 3, 2009
147 Posts
Offline
Posted: August 25, 2009, 5:46 pm - IP Logged
Afternoon all,
Can anyone tell me if there is a formula to detect the degree of randomness in a specific game?
Also,
Is there a formula to model a mechanical or air blown draw machine?
I ask the 1st question because I'm wondering if the randomness is preset @ .50% or with histories of draws and data to show what each number frequency is along with the number skips, can these figures help to discover to what degree a lotteries randomness is? Could knowing the degree of randomness in a specific game help in predicting the next draw?
I'm just going through data trying to figure out a new way to put the data to work.
Any ideas, thanks ahead of time.
The only DUMB question is the one question you DID NOT ask...
mid-Ohio
United States
Member #9
March 24, 2001
20147 Posts
Offline
Posted: August 25, 2009, 5:59 pm - IP Logged
There are formulas for figuring the odds of certain events happening like a match 2,3,4,5 and 6. If certain events happen a lot more or a lot less often than the odds would suggest you probably could say the events happening aren't totally random.
* you don't need to buy more tickets, just buy a winning one *
NY
United States
Member #23835
October 16, 2005
3655 Posts
Offline
Posted: August 27, 2009, 12:56 pm - IP Logged
I don't think there's any way to directly calculate the "randomness" of a game. What you can do is calculate the difference between the actual results and what you would expect from a perfectly random outcome. That goes beyond what I learned about probability and statistics, nevermind how much I still remember. Here's what I can tell you about it, though. First, probability is all about the relative chances of different outcomes, and can't do anything to predict specific outcomes, and it can't tell you with certainty if a result differs from the expected results because of a factor other than randomness. Rolling a die 100 times and getting a single 6 and forty 5's is unlikely, but can still be the result of perfectly random probability with a perfectly balanced die. The other is that the vast majority of people won't recognize random results as random. Very few people could list 100 results for a coin they didn't really flip and fool somebody who understands probability, because they think that equal probability means equal results.
Even if you could acurately measure the "randomness" of a game, that information would be completely useless, except for the possibility that it would encourage you to keep looking for an accurate prediction. If you could accurately calculate the deviations from perfectly random, that info might be useful, but it would have to be a substantial deviation and you'd have to know which way the deviations occur.
As for a formula for modeling the motion of the balls in the machine, I'd think it comes down to fairly simple mechanics, but the mechanics are interacting in very complex ways. There's not a chance that you could accurately account for all of the variables.
United States
Member #73037
April 3, 2009
147 Posts
Offline
Posted: August 28, 2009, 8:52 am - IP Logged
Morning RJOH and KY Floyd,
In the Pick3 game that I follow, my very rudamentary formulas are picking 2 of 3 about 30% of the time. About every 13th to 19th draw, my rudamentary formulas actually pick the correct 3 numbers. I keep my winning numbers in a color coded pattern file based on which line the winning number shows up in. It's looking like a checker board of colors and patterns.
I have calculated the number frequencies, skips/gaps, the -+ between the past drawn numbers in a column only basis (I call "distance") and a column to column basis (I call "difference").
I'm looking at the number results and trying to figure out if theres a way to assign a degree of randomness[?] (or would it be probability?) based on these figures.
I may be off, but, I'm thinking that if a formula could take all of these figures, it may be able to choose the next numbers more accurately.
OK. Here's where I get more confused. Why do the formulas that I've read through all calculate the probability first (usually based at 50%) instead of analyzing the history and then calculating each numbers own probability based on the history? Wouldn't that be a more accurate calculation?
None of the other games that I watch are producing any results like the Pick3 game. My formulas are kept in several different formula formats for each game.
Anyway, I'm looking at this data and trying to figure out a different way to utillize it. Trying to increase the 30% figure and lower the 13th to 19th figure.
I figure if I post my strange/weird ideas, it may trigger something in someone else. As I've said before, I'm still fairly new at all this so my ideas may not make sense to "old hands".
But, I'm still trolling for ideas...
Thanks for indulging my warped mind.
The only DUMB question is the one question you DID NOT ask...
United States
Member #13130
March 30, 2005
2171 Posts
Offline
Posted: August 28, 2009, 10:32 am - IP Logged
Afternoon all,
Can anyone tell me if there is a formula to detect the degree of randomness in a specific game?
Also,
Is there a formula to model a mechanical or air blown draw machine?
I ask the 1st question because I'm wondering if the randomness is preset @ .50% or with histories of draws and data to show what each number frequency is along with the number skips, can these figures help to discover to what degree a lotteries randomness is? Could knowing the degree of randomness in a specific game help in predicting the next draw?
I'm just going through data trying to figure out a new way to put the data to work.
Any ideas, thanks ahead of time.
You may be thinking of the Chi-Square test.
In neo-conned Amerika, bank robs you.
Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.
United States
Member #73037
April 3, 2009
147 Posts
Offline
Posted: August 28, 2009, 11:16 am - IP Logged
You may be thinking of the Chi-Square test.
Morning time-treat,
Please don't take this incorrectly...my posted name isn't a joke of any kind. When it comes to math, I am deficient.
I have extreme trouble attempting to understand/comprehend what I'm reading in discovered PDF files that I think may be helpful. (unfortunately, that's why it appears to all that I ask dumb questions in my posts). I'm reading everything that I can discover, but without the understanding or comprehending what is needed for implementing any of them. I appoligize for my knuckheadedness (just very bad math teachers when I was in school). Without teachers that take an interest in their students understanding the concepts properly, the students get passed along without learning. I'm one of them...
Anyway, what is Chi-Square test and (please) how could it apply to my idea or mis-idea?
Thank you.
The only DUMB question is the one question you DID NOT ask...
NASHVILLE, TENN
United States
Member #33372
February 20, 2006
1044 Posts
Offline
Posted: August 28, 2009, 11:20 am - IP Logged
From what I am gathering from my reading about probability, there is little if any attempt on the part of mathematicians to look beyond (how can I put this?) "the odds", "the possibilities", "what might happen". The reason, I feel, is because there is no definite formula, algorithym, or whatever, which will result in a correct answer each time.
We all know 4 + 4 = 8. That is certain. In the area of probability, we can not go beyond the "certain" with any degree of accuracy. So mathematicians stay away, preferring just to state the odds. Us lotto players, on the other hand, are looking past the odds. We want to know the degree of uncertainty.
I don't think anyone will develop a formula which will predict the lottory each and every time. That would be impossible. What I think we will do is develop a strategy that results in winning more money than we risk over any period of time. Members are working on it. I am working on it. A few inhabitants of this world have claimed to have found it.
As far as "degree of uncertianty" goes, there is not a definition for it within the lottory or rather I should say I haven't read of one. With the many systems posted here, I would think one and only one defintion would be impossible. Uncertainty has different meanings for different systems.
The idea of being able to calculate a degree of uncertainty is extremely appealing, tho.
COLUMBUS,GA.
United States
Member #4924
June 3, 2004
6161 Posts
Offline
Posted: August 28, 2009, 12:37 pm - IP Logged
From what I am gathering from my reading about probability, there is little if any attempt on the part of mathematicians to look beyond (how can I put this?) "the odds", "the possibilities", "what might happen". The reason, I feel, is because there is no definite formula, algorithym, or whatever, which will result in a correct answer each time.
We all know 4 + 4 = 8. That is certain. In the area of probability, we can not go beyond the "certain" with any degree of accuracy. So mathematicians stay away, preferring just to state the odds. Us lotto players, on the other hand, are looking past the odds. We want to know the degree of uncertainty.
I don't think anyone will develop a formula which will predict the lottory each and every time. That would be impossible. What I think we will do is develop a strategy that results in winning more money than we risk over any period of time. Members are working on it. I am working on it. A few inhabitants of this world have claimed to have found it.
As far as "degree of uncertianty" goes, there is not a definition for it within the lottory or rather I should say I haven't read of one. With the many systems posted here, I would think one and only one defintion would be impossible. Uncertainty has different meanings for different systems.
The idea of being able to calculate a degree of uncertainty is extremely appealing, tho.
There is a formula for "Degree of Certainty". Do a search for Thoth, he has some charts listed.
United States
Member #13130
March 30, 2005
2171 Posts
Offline
Posted: August 28, 2009, 2:08 pm - IP Logged
Morning time-treat,
Please don't take this incorrectly...my posted name isn't a joke of any kind. When it comes to math, I am deficient.
I have extreme trouble attempting to understand/comprehend what I'm reading in discovered PDF files that I think may be helpful. (unfortunately, that's why it appears to all that I ask dumb questions in my posts). I'm reading everything that I can discover, but without the understanding or comprehending what is needed for implementing any of them. I appoligize for my knuckheadedness (just very bad math teachers when I was in school). Without teachers that take an interest in their students understanding the concepts properly, the students get passed along without learning. I'm one of them...
Anyway, what is Chi-Square test and (please) how could it apply to my idea or mis-idea?
Thank you.
Your orig post asked about degree of randomness; the Chi-Square test takes the data you have and compares it to what is "supposed to happen". The difference tells you how far from "random" your real-world data is.
Here is another formula some people like -- Fundamental Formula of Gambling (FFG) -- http://www.lotterypost.com/thread/78519/160313 -- it let's you develop ideas based on how often something should or should not happen, given X number of trials. Raven62 made a chart, here. -- http://www.lotterypost.com/thread/141288/677883
In neo-conned Amerika, bank robs you.
Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.
United States
Member #73037
April 3, 2009
147 Posts
Offline
Posted: August 28, 2009, 5:10 pm - IP Logged
Thank you both, I'm searching...
The only DUMB question is the one question you DID NOT ask...
mid-Ohio
United States
Member #9
March 24, 2001
20147 Posts
Offline
Posted: August 29, 2009, 11:08 am - IP Logged
Your orig post asked about degree of randomness; the Chi-Square test takes the data you have and compares it to what is "supposed to happen". The difference tells you how far from "random" your real-world data is.
Here is another formula some people like -- Fundamental Formula of Gambling (FFG) -- http://www.lotterypost.com/thread/78519/160313 -- it let's you develop ideas based on how often something should or should not happen, given X number of trials. Raven62 made a chart, here. -- http://www.lotterypost.com/thread/141288/677883
the Chi-Square test takes the data you have and compares it to what is "supposed to happen". The difference tells you how far from "random" your real-world data is.
What is "supposed to happen" must never happens or else one could always play what is "suppose to happen" and hit a jackpot once in a while.
* you don't need to buy more tickets, just buy a winning one *
United States
Member #13130
March 30, 2005
2171 Posts
Offline
Posted: August 29, 2009, 2:32 pm - IP Logged
the Chi-Square test takes the data you have and compares it to what is "supposed to happen". The difference tells you how far from "random" your real-world data is.
What is "supposed to happen" must never happens or else one could always play what is "suppose to happen" and hit a jackpot once in a while.
Well, we know heads is "supposed" to come up half the time, and tails the other half. What we don't know is what will happen on the next flip of the coin.
I didn't want KH to get confused by me giving too exact of an explanation.
In neo-conned Amerika, bank robs you.
Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.
mid-Ohio
United States
Member #9
March 24, 2001
20147 Posts
Offline
Posted: August 29, 2009, 2:45 pm - IP Logged
Well, we know heads is "supposed" to come up half the time, and tails the other half. What we don't know is what will happen on the next flip of the coin.
I didn't want KH to get confused by me giving too exact of an explanation.
In that case if you stucked with heads, you would be right half the time. With most lotteries you only have to be right once, maybe twice is you're real lucky.
* you don't need to buy more tickets, just buy a winning one *
United States
Member #13130
March 30, 2005
2171 Posts
Offline
Posted: August 30, 2009, 9:08 am - IP Logged
'Reproduced part of the chart with some popular fractions
degree of certainty probability # of trials 1/2 1/4 1/6 1/8 1/10 1/16 1/32 5.0% 0.07 0.18 0.28 0.38 0.49 0.79 1.62 10.0% 0.15 0.37 0.58 0.79 1.00 1.63 3.32 15.0% 0.23 0.56 0.89 1.22 1.54 2.52 5.12 20.0% 0.32 0.78 1.22 1.67 2.12 3.46 7.03 25.0% 0.42 1.00 1.58 2.15 2.73 4.46 9.06 30.0% 0.51 1.24 1.96 2.67 3.39 5.53 11.23 35.0% 0.62 1.50 2.36 3.23 4.09 6.67 13.57 40.0% 0.74 1.78 2.80 3.83 4.85 7.92 16.09 45.0% 0.86 2.08 3.28 4.48 5.67 9.26 18.83 50.0% 1.00 2.41 3.80 5.19 6.58 10.74 21.83 55.0% 1.15 2.78 4.38 5.98 7.58 12.37 25.15 60.0% 1.32 3.19 5.03 6.86 8.70 14.20 28.86 65.0% 1.51 3.65 5.76 7.86 9.96 16.27 33.07 70.0% 1.74 4.19 6.60 9.02 11.43 18.66 37.92 75.0% 2.00 4.82 7.60 10.38 13.16 21.48 43.66 80.0% 2.32 5.59 8.83 12.05 15.28 24.94 50.69 85.0% 2.74 6.59 10.41 14.21 18.01 29.40 59.75 90.0% 3.32 8.00 12.63 17.24 21.85 35.68 72.53 95.0% 4.32 10.41 16.43 22.43 28.43 46.42 94.36 99.0% 6.64 16.01 25.26 34.49 43.71 71.36 145.05 99.5% 7.64 18.42 29.06 39.68 50.29 82.10 166.88
'Have to truncate at decimal, since you can't have a partial trial.
degree of certainty - real-world expectation that the event will happen within X number of trials
probability - raw mathematical odds
# of trials - where yellow and blue meet ... usually.
In neo-conned Amerika, bank robs you.
Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.
United States
Member #73037
April 3, 2009
147 Posts
Offline
Posted: August 30, 2009, 11:07 am - IP Logged
Morning time-treat,
I've read several files that discuss "# of trials". I'm looking at your chart. I comprehended "# of trials" to mean "# of draws to check", is this correct or incorrect?, because how do you check 1/2 or 1/4 , etc. of a games data?
I'm missing (not comprehending) somethimg here...
Also, what is "truncate at decimal" mean?
Thanks to all you for the examples and explanations. I'm trying to get up to speed...
The only DUMB question is the one question you DID NOT ask...
Page 1 of 2 | 4,640 | 16,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-51 | latest | en | 0.917113 |
http://www.qacollections.com/How-much-does-an-average-15-month-old-baby-weigh | 1,539,633,074,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509690.35/warc/CC-MAIN-20181015184452-20181015205952-00369.warc.gz | 526,478,704 | 5,079 | How much does an average 15 month old baby weigh?
20-26 pounds Hope it helped
Top Q&A For: How much does an average 15 month old baby weigh
How much should a four-month-old baby weigh?
A four-month-old baby will weigh, on average, between 12 and 15 pounds, though this obviously will vary. A good rule of thumb is that the child will weigh slightly less than double its birth weight... Read More »
How much does a 1 to 2 month old baby weigh?
wieght=age in months+9 divided by 2.like for 1 month babyweight=(1+9)/5=10/2=5kg
Related Questions | 145 | 548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-43 | latest | en | 0.953011 |
https://proofwiki.org/wiki/Closed_Ball_contains_Smaller_Closed_Ball | 1,580,058,752,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690095.81/warc/CC-MAIN-20200126165718-20200126195718-00446.warc.gz | 620,988,561 | 9,650 | # Closed Ball contains Smaller Closed Ball
## Theorem
Let $M = \struct{A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{\gt 0}$ such that $\epsilon \le \delta$.
Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-ball on $a$.
Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.
Then:
$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
## Proof
$\displaystyle x \in \map {B^-_\epsilon} a$ $\leadstoandfrom$ $\displaystyle \map d {x, a} \le \epsilon$ Definition of closed ball $\displaystyle$ $\leadsto$ $\displaystyle \map d {x, a} \le \delta$ As $\epsilon \le \delta$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in \map {B^-_\delta} a$ Definition of closed ball
By definition of subset:
$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
$\blacksquare$ | 274 | 813 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-05 | longest | en | 0.531395 |
https://essaymartials.com/telecommunications-engineering-homework-help-32/ | 1,717,047,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00439.warc.gz | 205,220,218 | 15,777 | # Telecommunications Engineering homework help
After you read and made yourself familiar with the following topics, please answer /discuss the following 15 questions. Your answers should have detailed explanations. Topics: Time & frequency domain, AM, FM and PM modulation, Noise in FM, Transmitters, Topologies, SSAM transmitter, Power measurements, Receiver topology, Demodulation and Receiver variations.
Questions:
1) A sine wave carrier has a frequency of 75KHz. A modulation signal has 2 sine waves at 2KHz and 4KHZ. If a signal is applied to a full carrier AM (Amplitude Modulation) transmitter, what are the output frequencies?
2) A sine wave carrier has a frequency of 75KHz. A modulation signal has 2 sine waves at 2KHz and 4KHZ. If a signal is applied to a full carrier DSBSC (Double Side Band Suppressed Carrier), AM (Amplitude Modulation) transmitter, what are the output frequencies?
3) A sine wave carrier has a frequency of 75KHz. A modulation signal has 2 sine waves at 2KHz and 4KHZ. If a signal is applied to a full carrier SSBSC (Single Side Band Suppressed Carrier), USB (Upper Side Band) AM (Amplitude Modulation) transmitter, what are the output frequencies?
4) A sine wave carrier has a frequency of 75KHz. A modulation signal has 2 sine waves at 2KHz and 4KHZ. If a signal is applied to a full carrier SSBSC (Single Side Band Suppressed Carrier), LSB (Lower Side Band) AM (Amplitude Modulation) transmitter, what are the output frequencies?
5) A modulation signal has 3 sine waves at 4KHz, 8KHZ and 12KHz. The carrier frequency is 185KHz. If a signal is applied to a SSBSC (LSB) AM transmitter, what are the output frequencies?
6) How is modulation done in a full carrier AM (time domain) circuit? Do we just add the carrier signal to the modulating signals?
7) Explain the problems caused by overmodulation of an AM signal.
8) It is well known that AM (Amplitude Modulation) is highly effected by noise since noise has effect on signal amplitude and not signal frequency. So, for an AM system, can we say the “envelope” is due to the amplitude of un-modulated carrier signal?
9) Explain what Quadrature AM is and its main problem.
10) What do AM, FM, PM, DSB, DSBSC, SSBSC, NBFM, WBFM stand for?
11) It is known that there are 3 parameters in a sine wave that may be varied for signal modulation. Those are: Amplitude, Frequency and Phase parameters. Explain the relationship between Phase and Frequency of a signal.
12) FM (Frequency Modulation) was originally designed with the idea of having noise immunity compared with AM (Amplitude Modulation). Explain why.
13) An SSB transmitter radiates 500 watts at 100% modulation. What will it radiate with no modulation?
14) Is this a correct statement: The FM modulation index: increases with both deviation and modulation frequency? Why?
15) Explain “Threshold” and “Capture” effects.
## Why US?
##### 100% Confidentiality
Information about customers is confidential and never disclosed to third parties.
##### Timely Delivery
No missed deadlines – 97% of assignments are completed in time.
##### Original Writing
We complete all papers from scratch. You can get a plagiarism report.
##### Money Back
If you are convinced that our writer has not followed your requirements, feel free to ask for a refund. | 784 | 3,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-22 | latest | en | 0.890707 |
https://techcommunity.microsoft.com/t5/excel/breach-or-not-breach-result-using-if-function/m-p/780787/highlight/true | 1,579,619,390,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00297.warc.gz | 694,380,856 | 52,586 | • 507K Members
• 8,301 Online
• 604K Conversations
## Breach or not breach result using IF Function
Occasional Visitor
# Breach or not breach result using IF Function
Hi,
Can anyone help to get result breach or not breach for the below condition
Trader A limit = 600 for USD/AED and 800 for Other CCY
Trader B limit = 700 for USD/AED and 600 for Other CCY
Trader C limit = 500 for USD/AED and 900 for Other CCY
Type of contract = USD/AED or Other CCY
Column A trader name
Column B amount
Column C Type of contract
Condition
Line no 1 for Trader A
If column C is USD/AED and amount >=600 then Breach
If column C is Other CCY and amount >= 800 then Breach, if not then Not Breach
Line 2, Line 3 and so on apply the limit condition as per Trader B and C
Please advice, I tried apply Nested if(and and result arrived successfully but few are incorrect not sure why? even though it match with other successful rows.
2 Replies
Highlighted
# Re: Breach or not breach result using IF Function
It would greatly help if you upload a sample file with realistic data, the solution will be close to what you are looking for. please provide expected results too for few rows so that formula can be tested. thank you !!
# Re: Breach or not breach result using IF Function
Perhaps, you are looking for something like that in the attached file. There are drop-down lists in Columns F and H for TraderName and ContractType, respectively. The formula in I2, copied down rows, is:
=VLOOKUP(F2,
A\$2:D\$4,
MATCH(H2,A\$1:D\$1,0),0)
Conversly, the formula in J2, copied down rows, is:
=IF(G2>=I2,
"Yes",
"No")
Related Conversations
Help with an IF AND formula
aanaya6 in Excel on
3 Replies
IF FUNCTION ISN'T WORKING NO MATTER HOW SIMPLE THE COMMAND IS
thomasea in Excel on
6 Replies
Which formula to use
Ramon Haagen in Excel on
2 Replies
Excel If Functions
Mfouad2255 in Excel on
10 Replies
function talking to table storage
donquijote in Compute on
0 Replies
Calculated column help
gopalaraoa in SharePoint on
1 Replies | 549 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-05 | latest | en | 0.808986 |
https://www.coursehero.com/file/8656499/By-28-B-y-B-y-1-B-y-B-z-0-B-z-B-z/ | 1,500,753,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424090.53/warc/CC-MAIN-20170722182647-20170722202647-00432.warc.gz | 765,165,992 | 22,750 | By 28 b y b y 1 b y b z 0 b z b z
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: ral Theorem, there is a 3 × 3 orthogonal matrix B of determinant one such that B T AB is diagonal. We introduce new coordinates y1 by setting x = B y, y = y2 y3 and equation (2.7) becomes yT (B T AB )y = 1. Thus after a suitable linear change of coordinates, the equation (2.7) can be put in the form λ1 0 y1 0 y1 y2 y3 0 λ2 0 y2 = 1, y3 0 0 λ3 39 4 1 0.5 0 -0.5 -1 2 0 -2 -2 -2 0 2 -4 Figure 2.2: An ellipsoid. or 2 2 2 λ1 y1 + λ2 y2 + λ3 y3 = 1, where λ1 , λ2 , and λ3 are the eigenvalues of A. It is relatively easy to sketch the quadric surface in the coordinates (y1 , y2 , y3 ). If the eigenvalues are all nonzero, we find that there are four cases: • If λ1 , λ2 , and λ3 are all positive, the quadric surface is an ellipsoid . • If two of the λ’s are positive and one is negative, the quadric surface is an hyperboloid of one sheet . • If two of the λ’s are negative and one is positive, the quadric surface is an hyperboloid of two sheets . • If λ1 , λ2 , and λ3 are all negative, the equation represents the empty set . Just as in the cas...
View Full Document
Ask a homework question - tutors are online | 426 | 1,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2017-30 | latest | en | 0.813689 |
https://www.polymathlove.com/special-polynomials/fractional-exponents/college-algebra-help.html | 1,566,117,402,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00117.warc.gz | 935,660,509 | 12,294 | Algebra Tutorials!
Sunday 18th of August
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
### Our users:
Thank you! I was having problems understanding exponential expressions, when my friend told me about the Algebrator Software. I can't believe how easy the software made it for me to understand how each step is preformed.
Catherine, IL
I am very much relieved to see my son doing well in Algebra. He was always making mistakes, as his concepts of arithmetic were not at all clear. Then I decided to buy Algebrator. I was amazed to see the change in him. Now he enjoys his mathematics and the mistakes are considerably reduced.
Dr. Stephen Wordell, KA
As a teacher I praise Algebrator because the students love it and find it most stimulating and interesting. What they see is what they learn and moreover it is relevant to the curriculum.
C.P., Massachusetts
### Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?
#### Search phrases used on 2012-04-21:
• Montana Education
• ks3 mental math test worksheet
• Books on permutation and combination
• Loans for College Students
• variable expressions worksheets
• +trigonometry word problems
• subtract algebraic equations
• Hosted Microsoft CRM
• permutation and combination
• best algebra 2 books
• Cards
• mixed numbers equations
• Visa Platinum
• sample codes in vb 6 f or calculating grade points of students
• 8th grade math practice worksheet
• boolean algebra expression simplification
• Lawyers Anchorage AK
• Indianapolis Lawyers
• online ti 84 calculator
• online non linear simultaneous equation solver
• free tutorial learning percentages
• Calcium Vitamins
• Mosby's Medical
• Laptop
• how to solve algebra story problems
• Fitness Books
• completing the square cubes difference
• english and math practice sheets for 6th grade
• solving simultaneous equations
• Homeowners Policy
• Indiana Gary Cemetery
• Manchester Injury Lawyers
• simplifying variable expressions with parenthesis
• free ged video lectures
• 6th grade free homework worksheets
• Journal of Pediatric Health Care
• Flight to Oregon Portland
• CRM Software
• practice workbook prentice hall pre-algebra chapter 1 answers
• algebra substitution
• maple example "linear algebra"
• calculate integral in ti 84 plus
• Funding of Education
• How to Handle Divorce
• pre algebra worksheets
• formula for square root in programming
• Convert a Fraction to a Decimal Point
• college algebra probabilities
• free pdf accounting books
• Find Credit Score
• Flat Vergleich
• Solving Systems of Linear Equations in Excel
• FREE MATHS EXAMINATION PAPERS
• Two Step Equations
• Trafalgar Travel
• factoring Kumon
• Group Calls
• WHERE CAN I BUY ONTARIO HIGH SCHOOL TEXT BOOK
• whats the basic steps invovled problem solving in addition- algebra
• layla richards
• Hotel in Downers Grove
• math taks for algebra
• printable matrix worksheets
• ks3 coordinates pictures maths
• holt algebra 2 workbook
• topics for algebraic expression
• surds solver
• boolean algebra-samples of logical circuits
• solving fraction equations
• divide m2 into lineal metre
• ti 84 plus emulator
• solve linear equations in ti84
• grade 3 math text, alberta
• greatest common divisor calculator
• examples of math trivia
• fun worksheet for 7th graders
• java solving equations
• Low Cost Secured UK Loans | 955 | 3,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.925901 |
https://alberta-davidson.com/flow-chart-coding/ | 1,600,531,412,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00521.warc.gz | 244,874,492 | 15,706 | # Flow Chart Coding
Flow Chart Coding has a variety pictures that similar to find out the most recent pictures of Flow Chart Coding here, and after that you can get the pictures through our best flow chart coding collection. Flow Chart Coding pictures in here are posted and uploaded by centralvalleybaptist.net for your flow chart coding images collection. The images that existed in Flow Chart Coding are consisting of best images and high environment pictures.
Planning Flowcharts
A flowchart is a type of diagram that represents an algorithm, workflow or process. Flowchart can moreover be defined as a diagramatic representation of an algorithm (step by step way in to solve a task).
The flowchart shows the steps as boxes of various kinds, and their order by connecting the boxes past arrows. This diagrammatic representation illustrates a answer model to a pure problem. Flowcharts are used in analyzing, designing, documenting or managing a process or program in various fields.
Flowchart of inter frame coding
Teaching Kids to Code with Lucidchart Lucidchart Blog
How to Use Code to FlowChart Converter to Let Code Visual
You Might Also Like :
Flowcharts are used in designing and documenting easy processes or programs. once new types of diagrams, they help visualize what is going upon and thereby put up to understand a process, and perhaps after that find less-obvious features within the process, following flaws and bottlenecks. There are substitute types of flowcharts: each type has its own set of boxes and notations. The two most common types of boxes in a flowchart are:
a running step, usually called activity, and denoted as a rectangular box.
a decision, usually denoted as a diamond.
A flowchart is described as “cross-functional” bearing in mind the chart is divided into exchange vertical or horizontal parts, to characterize the direct of swing organizational units. A parable appearing in a particular allowance is within the direct of that organizational unit. A cross-functional flowchart allows the author to correctly find the responsibility for performing an do something or making a decision, and to play the liability of each organizational unit for interchange parts of a single process.
Flowcharts depict certain aspects of processes and are usually complemented by further types of diagram. For instance, Kaoru Ishikawa, defined the flowchart as one of the seven basic tools of setting control, bordering to the histogram, Pareto chart, check sheet, run chart, cause-and-effect diagram, and the scatter diagram. Similarly, in UML, a satisfactory concept-modeling notation used in software development, the protest diagram, which is a type of flowchart, is just one of many exchange diagram types.
Nassi-Shneiderman diagrams and Drakon-charts are an oscillate notation for process flow.
Planning Flowcharts
Converiting a flowchart into code
Convert PHP Code to Flowchart Easily
Common stand-in names include: flow chart, process flowchart, on the go flowchart, process map, process chart, working process chart, business process model, process model, process flow diagram, operate flow diagram, situation flow diagram. The terms “flowchart” and “flow chart” are used interchangeably.
The underlying graph structure of a flowchart is a flow graph, which abstracts away node types, their contents and further ancillary information.
design flowchart in programming with examples programiz draw a flowchart to find all the roots of a quadratic equation ax 2 bx c=0 draw a flowchart to find the fibonacci series till term≤1000 though flowchart are useful in efficient coding debugging and analysis of a program drawing flowchart in very plicated in case of plex programs and often ignored flow chart coding agroclasi free stock charts and broker charts do not have scans or if they do there is not the ability to create custom criteria scans are a necessity and the better the scan the faster your chart analysis and stock pick selection code2flow online interactive code to flowchart converter line automatic pseudo code to flowchart generator with plugins for jira and confluence svg png and pdf export unlimited free trial free download flow chart coding software supershareware flow chart coding in title code to flow chart generator developer tools mercial \$99 95 1 2 mb code visual to flowchart developer tools ware \$99 95 1 2 mb flow chart medical coding medical billing classes medical billing flow chart by karna – docstoc medical billing flow chartc download legal documents this flowchart was created in india healthcare solutions when i worked for that pany and i prepared it from a template which was already used for training but was too plex to use flowcharts for coding flowcharts are to coding what logic is to math see how a simple flowchart makes sense of the coding process links edutopia benefits of coding for your st 6 useful flowchart tips to create better flowcharts flowchart color coding you can use a color scheme in your flowchart to identify various things you can use it to highlight processes that belong to different parties to highlight risky processes decisions to highlight a specific path in a process and for many other things medical coding flowchart medical billing programs automated coding workflow and cac practice guidance workflow and automation assessment in order to properly prepare for cac an organization should conduct a thorough review of its current coding workflow … medical billing and coding flowchart medical billing medical chart auditing – medical coding audits – healthcare … auditing medical charts is the first step and a critical part of any successful medical office operation coding process workflow tasneef ba 2 coding process flow chart example the coding flow chart is a chart and or text policy which describes the process by which coding is done in the facility
puter Programming
Flowgorithm Flowchart Programming Language
Flowcharts in Programming Visualizing Logic and Flow of | 1,165 | 6,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.92892 |
https://stats.stackexchange.com/questions/625925/can-you-do-a-log-transformation-for-excess-kurtosis-or-is-that-mainly-used-for/625927 | 1,701,276,388,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00516.warc.gz | 645,762,714 | 43,247 | # Can you do a log transformation for excess kurtosis, or is that mainly used for skewness?
I am planning on doing a regression analysis on STATA on the financial performance of private equity funds. On my descriptive statistics, I saw higher levels of kurtosis and skewness. I decreased these by using the winsorization method for outliers. However, I still have high levels, especially on the kurtosis.
1. Would it be correct to do log transformations on all of the variables (dependent and control) in order to decrease kurtosis, or is this method usually only used for skewness?
2. If yes, is it okay that I do the log transformation for the variables that have excess kurtosis, even if they have normal ranges of skewness?
3. I have researched online and have seen that -2 to +2 are the "normal" ranges for kurtosis and skewness...is this correct for both?
If you cant see the picture, to summarize, I have kurtosis levels ranging for 2.50-22 and skewness levels ranging from 1-4 on the dependent and control variables.
• Welcome to Cross Validated! What do you want to “fix” by altering the kurtosis?
– Dave
Sep 7 at 11:42
• Thank you for your answer, happy to be here! I am a beginner in statistics and I assumed its better to lower the kurtosis to address the non-normality and thus do log transformations on the dependent variables and control variables.
– Lucy
Sep 7 at 12:16
• It is a mathematical fact that taking the logarithm will decrease (make more negative) the skewness of the distribution. It might as a byproduct decrease the kurtosis -- but that's a secondary effect and should not be a consideration in your analysis. // There is no such thing as "normal" ranges of skewness and kurtosis. I have seen reasonable, realistic datasets with kurtosis into the hundreds and thousands. // For a principled approach to identifying transformations in regression, see stats.stackexchange.com/a/35717/919. Also see stats.stackexchange.com/a/60455/919.
– whuber
Sep 7 at 12:20
• Thank you for your answer whuber! In that case, would you say that I could just leave the kurtosis and skewness as is and perform an OLS or robust regression? If it is possible, I would rather not do other transformations to the data if not needed, as I am only a beginner in statistics currently!
– Lucy
Sep 7 at 12:54
• The entire discussion becomes moot when you Winsorize first, because now you are trying to model some sort of truncated distribution that has no connection to to processes you are trying to model. Sep 8 at 11:45
First, if you are doing OLS regression, it does not assume anything about the distribution of the variables. It makes assumptions about the errors, which are approximated by the residuals.
Second, in my opinion, you should not transform variables in order to make the data fit a model; rather, you should use a model that fits the data. Two such models are robust regression (actually a collection of methods) and quantile regression. You should transform for substantive reasons. Taking logs of money variables often makes substantive sense.
Third, I'm not even sure logs will reduce kurtosis. Kurtosis behaves in counterintuitive ways.
• Many thanks for your message, Peter! All the dependent variables and one control variable ("final close size USD") are monetary variables. Would it makes sense then to soley log transform these ones then? And then do for instance, a robust regression? Or, as a second option, is the degree of kurtosis and skewness okay to leave as it is, and to just continue on with a robust regression instead of an OLS regression?
– Lucy
Sep 7 at 12:29
• The reason that I think logs make sense for monetary variables is that we often think of money in multiplicative terms rather than additive ones. If you make $30,000 a year, a$5,000 raise is huge. If you make \$300,000 a year, it's not even cost of living. Sep 7 at 13:16
• Right, I see what you are saying. Then would you recommend that I do the logs for the monetary variables, as they are monetary, not because of kurtosis or skewness, and then carry out a OLS or robust regression?
– Lucy
Sep 7 at 13:20
• Well, that's up to you, as it is a substantive question. Probably. But, in your application, would you think about the variables multiplicatively or additively? Sep 7 at 13:42
• Yes, they would be in multiplicative terms!
– Lucy
Sep 7 at 13:54 | 1,044 | 4,368 | {"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-2023-50 | longest | en | 0.915278 |
https://gmatclub.com/forum/unlike-auto-insurance-the-frequency-of-claims-158190.html | 1,563,455,907,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00473.warc.gz | 403,810,353 | 153,773 | Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.
It is currently 18 Jul 2019, 06:18
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Unlike auto insurance, the frequency of claims
Author Message
TAGS:
Hide Tags
Intern
Joined: 07 Mar 2012
Posts: 32
Location: India
Concentration: General Management, Nonprofit
GMAT Date: 09-13-2013
GPA: 4
WE: Engineering (Energy and Utilities)
Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
19 Aug 2013, 02:38
7
27
00:00
Difficulty:
5% (low)
Question Stats:
90% (00:56) correct 10% (01:13) wrong based on 2921 sessions
HideShow timer Statistics
The Official Guide for GMAT Verbal Review 2017
Practice Question
Question No.: SC 293
Page: 273
Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
(A) Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage
(B) Unlike with auto insurance, the frequency of claims do not affect the premiums for personal property coverage
(C) Unlike the frequency of claims for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(E) Unlike with the premiums for auto insurance, the premiums for personal property coverage is not affected by the frequency of claims
_________________
Remember that potential unused turns into pain. So dedicate yourself to expressing your best.
Manager
Joined: 21 Aug 2012
Posts: 105
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
19 Aug 2013, 02:47
8
gpsao wrote:
Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
(A) Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage
Cannot compare auto insurance with frequency of claims
(B) Unlike with auto insurance, the frequency of claims do not affect the premiums for personal property coverage
Same as A..
(C) Unlike the frequency of claims for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
Frequency of claims cannot be compared with premiums for personal property coverage
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
CORRECT... Proper comparision..!!!
(E) Unlike with the premiums for auto insurance, the premiums for personal property coverage is not affected by the frequency of claims
"with" shows that we comparing premiums for auto insurance + something else to premiums for personal property
RULE: Like/Unlike always followed by NOUN.
LIKE/UNLIKE never followed by a CLAUSE.
Thanks,
Jai
KUDOS if it HELPED...!!!!
_________________
MODULUS Concept ---> http://gmatclub.com/forum/inequalities-158054.html#p1257636
HEXAGON Theory ---> http://gmatclub.com/forum/hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308
General Discussion
Manager
Joined: 31 May 2012
Posts: 109
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
27 Aug 2013, 17:31
"Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims...
To make it more concise, is it correct to skip repeated usage of premiums?
Unlike the premiums for auto insurance, these for personal property coverage are not affected by the frequency of claims...
Retired Moderator
Joined: 16 Jun 2012
Posts: 1003
Location: United States
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
27 Aug 2013, 22:54
2
1
umeshpatil wrote:
"Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims...
To make it more concise, is it correct to skip repeated usage of premiums?
Unlike the premiums for auto insurance, these for personal property coverage are not affected by the frequency of claims...
Hi umesh.
Good question! But you can not use "these" here.
(1) the premiums for auto insurance and the premiums for personal property are different. On the other hand, if you want to indicate a "New copy" of the antecedent, you must use "that" or "those". You use "these" or "this" if you indicate same antecedent.
(2) "That" or "Those" indicating a New copy must be modified. On the other hand, you have to add a description for the antecedent before using pronouns
Let see the first part: Unlike the premiums for auto insurance" <== It does not provide enough descriptions of the antecedent, so the pronoun "those of" is not be modified enough ==> the usage of pronoun "those for" is not correct.
If you want to use pronoun, the correct sentence should be: The premiums for auto insurance are affected by the frequency of claims, but those for personal property are not.
I would say the question from OG is good.
Hope it helps.
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Chris Bangle - Former BMW Chief of Design.
Manager
Joined: 21 Aug 2012
Posts: 105
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
27 Aug 2013, 23:04
1
pqhai wrote:
umeshpatil wrote:
"Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims...
To make it more concise, is it correct to skip repeated usage of premiums?
Unlike the premiums for auto insurance, these for personal property coverage are not affected by the frequency of claims...
Hi umesh.
Good question! But you can not use "these" here.
(1) the premiums for auto insurance and the premiums for personal property are different. On the other hand, if you want to indicate a "New copy" of the antecedent, you must use "that" or "those". You use "these" or "this" if you indicate same antecedent.
(2) "That" or "Those" indicating a New copy must be modified. On the other hand, you have to add a description for the antecedent before using pronouns
Let see the first part: Unlike the premiums for auto insurance" <== It's not a complete sentence, so the usage of pronoun "those for" is not correct.
If you want to use pronoun, the correct sentence should be: The premiums for auto insurance are affected by the frequency of claims, but those for personal property are not.
I would say the question from OG is good.
Hope it helps.
Hi pqhai,
Let see the first part: Unlike the premiums for auto insurance" <== It's not a complete sentence, so the usage of pronoun "those for" is not correct.
During comparison, Like/Unlike is always followed by a noun. If we make it a complete sentence, then for clause we use "As" instead of "like/unlike" because we are comparing actions in that case.
If i say:
Like that of John, Mary's car is white.
As John has a white car, Mary has a white car.
Like John's car, that of Mary is white.
Here, "Unlike the John's Car" is not a clause, but still usage of "that" is correct.
Thanks,
Jai
KUDOS if it HELPED..!!!
_________________
MODULUS Concept ---> http://gmatclub.com/forum/inequalities-158054.html#p1257636
HEXAGON Theory ---> http://gmatclub.com/forum/hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308
Retired Moderator
Joined: 16 Jun 2012
Posts: 1003
Location: United States
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
28 Aug 2013, 00:05
jaituteja wrote:
Hi pqhai,
Let see the first part: Unlike the premiums for auto insurance" <== It's not a complete sentence, so the usage of pronoun "those for" is not correct.
During comparison, Like/Unlike is always followed by a noun. If we make it a complete sentence, then for clause we use "As" instead of "like/unlike" because we are comparing actions in that case.
If i say:
Like that of John, Mary's car is white.
As John has a white car, Mary has a white car.
Like John's car, that of Mary is white.
Here, "Unlike the John's Car" is not a clause, but still usage of "that" is correct.
Thanks,
Jai
KUDOS if it HELPED..!!!
Thanks Jai
Sorry for confusing. I say "It's not a complete sentence", I want to convey the idea the phrase "unlike....." does not provide enough description to hold a pronoun "those" valid (or the pronoun is not be modified enough). I will change my post a bit to avoid confusing.
Regarding your example, I, in turn, politely disagree with you.
You use pronoun "those" or "that to indicate "new copy" ==> A "new copy" must be modified. In other word, you have to add a description to indicate how the new copy is different from the previous version.
I think your example should be: Marry's car, unlike that of John's, is white <== "that" is pronoun which is a "new copy" of Marry's car. "that of John's" is modified by "Marry's car is white". Thus, the sentence is correct.
However, in your example, the clause "that of...." is a main clause and is not be modified by anything.
Any discussion is welcome.
Cheer
PS: I highly recommend MGMAT - Sentence correction - Chapter Pronoun which describes very well how to use pronoun properly. (Just my recommendation).
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Chris Bangle - Former BMW Chief of Design.
Manager
Joined: 21 Aug 2012
Posts: 105
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
28 Aug 2013, 01:34
pqhai wrote:
jaituteja wrote:
Hi pqhai,
Let see the first part: Unlike the premiums for auto insurance" <== It's not a complete sentence, so the usage of pronoun "those for" is not correct.
During comparison, Like/Unlike is always followed by a noun. If we make it a complete sentence, then for clause we use "As" instead of "like/unlike" because we are comparing actions in that case.
If i say:
Like that of John, Mary's car is white.
As John has a white car, Mary has a white car.
Like John's car, that of Mary is white.
Here, "Unlike the John's Car" is not a clause, but still usage of "that" is correct.
Thanks,
Jai
KUDOS if it HELPED..!!!
Thanks Jai
Sorry for confusing. I say "It's not a complete sentence", I want to convey the idea the phrase "unlike....." does not provide enough description to hold a pronoun "those" valid (or the pronoun is not be modified enough). I will change my post a bit to avoid confusing.
Regarding your example, I, in turn, politely disagree with you.
You use pronoun "those" or "that to indicate "new copy" ==> A "new copy" must be modified. In other word, you have to add a description to indicate how the new copy is different from the previous version.
I think your example should be: Marry's car, unlike that of John's, is white <== "that" is pronoun which is a "new copy" of Marry's car. "that of John's" is modified by "Marry's car is white". Thus, the sentence is correct.
However, in your example, the clause "that of...." is a main clause and is not be modified by anything.
Any discussion is welcome.
Cheer
PS: I highly recommend MGMAT - Sentence correction - Chapter Pronoun which describes very well how to use pronoun properly. (Just my recommendation).
Hi pqhai,
I agree with your explanation for the "new copy" thing.
I think Marry's car, unlike that of John's, is white is not correct.
It should be unlike that of John... since we are using "that" we don't need to use the possessive form of John.
the sentence should be
Marry's car, unlike that of John, is white
OR
Marry's car, unlike John's, is white
We are comparing the car of Marry to that of John,not mary's car to John.
However, in your example, the clause "that of...." is a main clause and is not be modified by anything.
Thanks,
Jai
KUDOS if it HELPED..!!!
_________________
MODULUS Concept ---> http://gmatclub.com/forum/inequalities-158054.html#p1257636
HEXAGON Theory ---> http://gmatclub.com/forum/hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308
Retired Moderator
Joined: 16 Jun 2012
Posts: 1003
Location: United States
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
28 Aug 2013, 11:39
jaituteja wrote:
Hi pqhai,
I agree with your explanation for the "new copy" thing.
I think Marry's car, unlike that of John's, is white is not correct.
It should be unlike that of John... since we are using "that" we don't need to use the possessive form of John.
the sentence should be
Marry's car, unlike that of John, is white
OR
Marry's car, unlike John's, is white
We are comparing the car of Marry to that of John,not mary's car to John.
However, in your example, the clause "that of...." is a main clause and is not be modified by anything.
Thanks,
Jai
KUDOS if it HELPED..!!!
Hi Jai
Yes, we're comparing Marry's car with John's car. We do not compare Marry's car with John. You can see the possessive form - John's. If you repeat the same thing, we can omit the noun.==> John's means John's car.
Pronoun topic is huge. If you use pronoun, you will use a representative rather than repeat an antecedent. You just use "this, that, these, those" to indicate the antecedent. Logically, the readers need to understand the nature of the antecedent first. Therefore, pronouns must be modified.
For example:
Incorrect: Unlike John's car, that of Marry is white.
Because you're comparing X to Y. X is antecedent and Y is the pronoun. Logically, Y is just a new copy of X, so X must show its characteristics first.
In your sentence, "that" is the pronoun ==> it indicates something appears before it in the sentence ==> The "thing" appears before the pronoun must be understood by readers first. However, without a description about the pronoun, readers do not understand the nature of the antecedent. You cannot understand the phrase "Unlike John's car" if it stands alone.
Correct: Marry's car, unlike that of John, is white ==> The antecedent is Marry's car. Even without the pronoun, readers still understand the nature of Marry's car completely (Marry's car is white). Then the usage of pronoun (a new copy) will be correct.
Hope it helps.
_________________
Please +1 KUDO if my post helps. Thank you.
"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."
Chris Bangle - Former BMW Chief of Design.
SVP
Joined: 14 Apr 2009
Posts: 2281
Location: New York, NY
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
30 Aug 2013, 10:24
Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
(A) Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage
(B) Unlike with auto insurance, the frequency of claims do not affect the premiums for personal property coverage
(C) Unlike the frequency of claims for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(E) Unlike with the premiums for auto insurance, the premiums for personal property coverage is not affected by the frequency of claims
Note the keyword "unlike" appears in all answer choices.
So because of that, all we need to do is make sure that the left side and right side of the comma are consistent.
In (A) - "auto insurance" is not consistent with "frequency of claims" - so don't even bother reading the rest of the answer choice.
Just glance down - and it is not until we get to (D) that we see "premiums" being compared with "being" - only then should we spend the time to actually read through (D).
This is a speed technique we implement in our fast solutions to SC that allows you to arrive at the correct answer without necessarily reading every answer choice.
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4512
Location: India
GPA: 3.5
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
23 Jun 2016, 10:24
Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
(A) Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage
(B) Unlike with auto insurance, the frequency of claims do not affect the premiums for personal property coverage
(C) Unlike the frequency of claims for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(E) Unlike with the premiums for auto insurance, the premiums for personal property coverage is not affected by the frequency of claims
Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
Correct idiomatic usage of unlike is : Unlike X, Y
All except (D) are incorrect...
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2351
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
05 Jan 2017, 11:39
Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
(A) Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage - Illogical comparison- illogically compares auto insurance and the frequency of claims
(B) Unlike with auto insurance, the frequency of claims do not affect the premiums for personal property coverage - Unlike with is unidiomatic ; Subject-verb agreement issue - Frequency and do not affect
(C) Unlike the frequency of claims for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims - illogically compares the frequency of claims and premiums
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims - Correct
(E) Unlike with the premiums for auto insurance, the premiums for personal property coverage is not affected by the frequency of claims - Unlike with is unidiomatic ; Subject-verb agreement issue - Premiums and is
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
+1 Kudos if you find this post helpful
Manager
Joined: 28 Jan 2018
Posts: 154
Location: Taiwan
GMAT 1: 690 Q47 V38
GPA: 3.34
Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
11 Feb 2018, 19:44
Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
(A) Unlike auto insurance, the frequency of claims does not affect the premiums for personal property coverage
(B) Unlike with auto insurance, the frequency of claims do not affect the premiums for personal property coverage
(C) Unlike the frequency of claims for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims
(E) Unlike with the premiums for auto insurance, the premiums for personal property coverage is not affected by the frequency of claims
"Unlike" suggests a comparison between two subjects. The comparison has to be logical.
Also, I'm certain the use of "unlike with" is unidiomatic.
Manager
Joined: 28 Jan 2018
Posts: 154
Location: Taiwan
GMAT 1: 690 Q47 V38
GPA: 3.34
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
11 Feb 2018, 19:56
Somehow the meaning of entire sentence is not clear to me.
(D) Unlike the premiums for auto insurance, the premiums for personal property coverage are not affected by the frequency of claims, but if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
What does it refer to? Ideally should refer to frequency because it cannot refer to
a) insurance company - the company cannot decline
frequency is only thing left. However, how can frequency decline to renew to the policy. Doesn't make sense to me.
Pls help!
if the insurance company is able to prove excessive loss due to owner negligence, it may decline to renew the policy.
Why can't the company decline?
You can rewrite this:
if the insurance company is able to prove excessive loss due to owner negligence, insurance company may decline to renew the policy.
You only read the word "decline" and forgot about the "to renew" part.
How can the frequency decline to renew?
"Insurance company" is the only logical reference.
EMPOWERgmat Instructor
Joined: 23 Feb 2015
Posts: 497
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
16 Feb 2018, 16:03
Hi lary301254M7,
Thank you for your question. I think the misunderstanding with grad2020 is the meaning of the word "decline." There are two general meanings for "decline":
1. To reduce in size
2. To refuse someone
In this case, the phrase "it may decline to renew the policy" means that the insurance company (what "it" is referring to) can refuse to renew the policy - not that they will somehow lose value or size.
I hope this helps clear it up!
_________________
Director
Joined: 02 Oct 2017
Posts: 727
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
10 May 2018, 02:29
Unlike is used to compare nouns of same nature
A and B - wrong as auto insurance is compared with frequency
E- with after unlike create meaning error
Posted from my mobile device
_________________
Give kudos if you like the post
Intern
Joined: 06 Feb 2017
Posts: 37
Location: India
Schools: HBS '22, HEC '22
GMAT 1: 570 Q39 V28
GMAT 2: 620 Q49 V26
GPA: 4
Re: Unlike auto insurance, the frequency of claims [#permalink]
Show Tags
25 Sep 2018, 17:42
Very basic Parallelism question.
only in option D, premiums for auto insurance are compared with the premiums for personal property coverage.
_________________
I hope this helped. If this was indeed helpful, then you may say Thank You by giving a Kudos!
Re: Unlike auto insurance, the frequency of claims [#permalink] 25 Sep 2018, 17:42
Display posts from previous: Sort by | 5,834 | 24,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-30 | latest | en | 0.896479 |
http://math.stackexchange.com/questions/150627/conductor-of-abc-frey-hellegouarch-curves-and-twists | 1,466,910,529,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783394605.61/warc/CC-MAIN-20160624154954-00171-ip-10-164-35-72.ec2.internal.warc.gz | 204,436,141 | 17,212 | # Conductor of $ABC$, Frey-Hellegouarch curves, and twists
In page 109 of de Weger's paper, he says that for coprime $A, B, C$ the conductor $N$ of the Frey-Hellegouarch curve $$E: y^2 = x(x - A)(x + B)$$ equals $N(A,B,C)$ (product of primes dividing $ABC$ without multiplicity, and where $A + B = C$) times an absolutely bounded power of $2$. Why is this the case?
Also, on page 114, he says that the conductor $N_q$ of the twisted curve
$$E_q : qy^2 = x(x - A)(x+B)$$
where $q$ is a squarefree integer (ie the quadratic twist of $E$), is $lcm(N,q^2)$ and the difference in the power of $2$ is at most $2^8$. Why is this the case?
Thanks
-
• The highest power of $2$ that can divide the conductor of an elliptic curve is $8$, if I remember correctly.
• The formula for the conductor $N_q$ (which is valid provided that $q$ is coprime to $N$) can be checked directly from the definition of the conductor in terms of $\ell$-adic Tate modules. You can also think of it in terms of how the conductor of a newform changes when you make a twist. This is discussed in classical language in the article of Atkin and Lehner. It is also easily verified using representation-theoretic language. | 343 | 1,191 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2016-26 | latest | en | 0.922419 |
http://examcrazy.com/air-conditioning-system-outdoor-and-indoor-design-dry-bulb-temperatures | 1,537,900,411,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162385.83/warc/CC-MAIN-20180925182856-20180925203256-00111.warc.gz | 80,295,970 | 28,455 | Question:-
For an air-conditioning system, the outdoor and indoor design dry bulb temperatures are 45°C and 25°C respectively. The space to be airconditioned is 20 m x 30 m x 5 m and infiltration is estimated to be one air change. If the density and specific heat of air are 1.2 (kg of dry air)/m3 and 1.02 kJ/k (kg of dry air)°C, then the sensible heat load due to infiltration is, nearly
Option (A)
122.4 kW
Option (B)
61.2 kW
Option(C)
12.24 kW
Option(D)
20.4 kW
Correct Option:
(D)
Question Solution:
SHL = 0.0204 x (cmm) x DT
= 0.0204 x [1 x {(20 x 30 x 5) / 60}] x (45-25)
= 0.0204 x 50 x 20 = 20.4 kW | 213 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-39 | latest | en | 0.796511 |
http://mathoverflow.net/questions/6450?sort=newest | 1,371,675,942,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708808740/warc/CC-MAIN-20130516125328-00018-ip-10-60-113-184.ec2.internal.warc.gz | 159,631,117 | 9,648 | MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
## Shifted Dirichlet series
If $\sum_{n=1}^\infty \frac{a_n}{n^s}$ converges, does $\sum_{n=1}^\infty \frac{a_n}{(n+1)^s}$ also converge?
-
Yes, because $(n+1)^{-s} = n^{-s} + sn^{-s-1} + O(|s|^2n^{-\sigma - 2})$. The first series necessarily converges in the open half-plane strictly to the right of s, and converges absolutely in the half-plane strictly to the right of s + 1. I hope I am not doing homework from a course in analytic number theory here. | 189 | 612 | {"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} | 2.8125 | 3 | CC-MAIN-2013-20 | latest | en | 0.776158 |
https://en.zdam.xyz/problem/13119/ | 1,670,576,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00737.warc.gz | 271,625,234 | 5,625 | #### Problem 7E
7. The total cost of repaying a student loan at an interest rate of $r \%$ per year is $C=f(r)$.
(a) What is the meaning of the derivative $f^{\prime}(r) ?$ What are its units?
(b) What does the statement $f^{\prime}(10)=1200$ mean?
(c) Is $f^{\prime}(r)$ always positive or does it change sign? | 98 | 315 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-49 | latest | en | 0.924689 |
https://brilliant.org/practice/stability/?subtopic=random-variables&chapter=markov-chains | 1,656,541,768,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00633.warc.gz | 209,815,479 | 13,091 | Probability
# Markov Chains - Stationary Distributions
Which states, if any, are periodic in the Markov chain pictured?
The only states of weather are sun and rain. If it's sunny today, there is an $80\%$ chance it will be sunny tomorrow, and if it's rainy today, there is a $60\%$ chance it's rainy tomorrow. In the long run, on what fraction of days will it be sunny?
What is the stationary distribution of the Markov chain pictured below?
In a particular (very large) deck of cards, the suit of each card is dependent upon the suit of the previous card. If the dependencies are given by the Markov chain pictured, determine the stationary distribution of suits using the order (Spades, Hearts, Diamonds, Clubs).
A player wins a game of tennis by being the first to win $2$ games more than the other player, provided they have won at least $4$ games. If the game is tied at $3$ to $3$ and the probabilities are given by the Markov chain pictured, determine the probability that player A wins.
× | 233 | 1,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-27 | latest | en | 0.942953 |
https://community.qlik.com/t5/QlikView-App-Development/Can-AGGR-dimension-be-dynamic/m-p/71916 | 1,569,250,349,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576965.71/warc/CC-MAIN-20190923125729-20190923151729-00202.warc.gz | 425,428,484 | 39,747 | # QlikView App Development
Discussion Board for collaboration related to QlikView App Development.
Highlighted
Contributor III
## Can AGGR dimension be dynamic?
I have a pivot table with a linear guage (bar) for the measure.
I have three dimensions: region, manager, salesrep
MAX(AGGR(sum(sales),salesrep)) yields the max metric across all employees but I want the max for salesrep under the manager under the region.
Same for the other dimensions. If the pivot only shows region, the max should be for the region, not the sales rep.
I've tried a few variations with dimensionality
MAX(AGGR(sum(sales),PICK(DIMENSIONALITY(),region,manager,salesrep)))
but this often resulted in Max Value appearing on all the rows.
If there is only one dimension its simple but what needs to be done for multiple dimensions ?
Tags (3)
5 Replies
MVP
## Re: Can AGGR dimension be dynamic?
How do you decide your dimension? is it a cycle or drill-down group?
Contributor III
## Re: Can AGGR dimension be dynamic?
Neither - its a pivot table with 3 dimensions. The pivot table does the drill-down.
Valued Contributor III
## Re: Can AGGR dimension be dynamic?
I had the same problem but didn't find a solution...
For cyclic groups you can use getcurrentfield(), but for dynamically expanded / collapsed pivots this is not possible I think.
New Contributor III
## Re: Can AGGR dimension be dynamic?
Did you try to move Pick with Dimentionality out of aggr function?
PICK( DIMENSIONALITY(), MAX(AGGR(sum(sales),region)), MAX(AGGR(sum(sales),manager)), MAX(AGGR(sum(sales),salesrep)) ).
I know that it`s not best solution from repeating code point of view, but maybe it`s solve your problem.
MVP
## Re: Can AGGR dimension be dynamic?
I would think that this might be want you want to get calculated:
Pick(Dimensionality()
, Max(TOTAL Aggr(Sum(Amount),Region,Manager,SalesRep))
, Max(TOTAL <Region> Aggr(Sum(Amount),Region,Manager,SalesRep))
, Max(TOTAL <Region,Manager> Aggr(Sum(Amount),Region,Manager,SalesRep))
)
Or if you want the total sum per dimension and the max amount for the dimension :
Pick(Dimensionality()
, Max(TOTAL Aggr(Sum(Amount),Region))
, Max(TOTAL <Region> Aggr(Sum(Amount),Region,Manager))
, Max(TOTAL <Region,Manager> Aggr(Sum(Amount),Region,Manager,SalesRep))
) | 575 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-39 | latest | en | 0.806081 |
https://www.bellandcomusic.com/major-9th-chords.html | 1,721,398,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00858.warc.gz | 602,585,618 | 6,071 | # Major 9th Chords... maj9
• Major 9th chords are 5-note chords spanning 14 semitones - octave plus a major second or 2 semitones.
• Written as maj9 or △9 with maj9 most commonly used.
• They are called compound intervals as they span more than an octave.
• Intervals: root, major third, perfect fifth, major 7th, major 9th.
• Chord formula: 1 - 3 - 5 - 7 - 9
Example: Calculating a Gmaj9 chord
• Chords can be calculated from the major scale of the chord's root note, therefore Gmaj9 is calculated from the G major scale which consists of 1 sharp - F#.
• Because a 9th chord exceeds one octave, we need to add an extra note to the existing one-octave scale (9 notes in total).
The scale reads as follows: G - A - B - C - D - E - F# - G - A
• Formula for maj9 chords = 1 - 3 - 5 - 7 - 9
• Substitute notes from the scale into the formula
• Gmaj9 = G - B - D - F# - A
Major 9th Chords Table (maj9)
(maj9) A B♭ B C C# D E♭ E F F# G A♭
Root A B♭ B C C# D E♭ E F F# G A♭
Major 3rd C# D D# E E#(F) F# G G# A A# B C
Perfect 5th E F F# G G# A B♭ B C C# D E♭
Major 7th G# A A# B B#(C) C# D D# E E#(F) F# G
Major 9th B C C# D D# E F F# G G# A B♭
## Major 9th Chords - RH / LH
Chord Chart Legend
Right Handers
↓ major 9th chords - Part 1
Left Handers
Right Handers
↓ maj9 Chords - Part 2
Left Handers
Right Handers
Left Handers
#### Take Note:
Don't confuse a 9 chord with a maj9 chord. A ninth chord is built on a dominant or flattened 7th chord, whereas a major 9th chord is built on a major 7th chord.
• Formula for a 9 chord: 1 - 3 - 5 - ♭7 - 9 (dom 7th chord with an added 9th)
• Formula for a maj9 chord: 1 - 3 - 5 - 7 - 9 (maj 7th chord with an added 9th) | 609 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-30 | latest | en | 0.860581 |
http://gpuzzles.com/mind-teasers/police-investigating-murder-riddle/ | 1,477,266,432,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719453.9/warc/CC-MAIN-20161020183839-00238-ip-10-171-6-4.ec2.internal.warc.gz | 110,172,714 | 12,013 | • Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Police Investigating Murder Riddle
Difficulty Popularity
A couple went to river rafting on a hill station. Two days after their departure, the husband returned alone. He informed the police that her wife was swept away by the waves and died.
On the next day, police arrived to his doorstep and when he opened the door, they arrested him for murdering his wife. They told him that his travel agent had called them.
He was shocked. How did the travel agent know about the murder? Can you suggest how did he know ?
Discussion
Suggestions
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Maths Equation Brain Teaser
Difficulty Popularity
Arrange four eights(9) and a one(1) and only one mathematical symbol to make it equal to 100
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Secret Code Riddle
Difficulty Popularity
You are giving an intelligence test. In that, you are provided by the code - MOD OAT AIM DUE TIE
You know that only one word from this code is true and the rest ones are only put in to make the task difficult for you. To understand, you are provided with a clue – If you are told any one of the characters of the code word, you can find out the word easily.
Can you deduce the actual code word from the entire code?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Akbar Birbal Puzzle
Difficulty Popularity
Akbar summoned Birbal out of anger. He told him that he will have to face death. He asked him to make a statement and if the statement is true he will be buried alive and if the statement is false, he will be thrown at lions. After hearing Birbal’s statement, Akbar could do nothing but smile. He gave him 5 gold bars and let him go.
What did Birbal say?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : No Death Penalty For Kill Riddle
Difficulty Popularity
Ramsay was killed by Sansa in front of everyone but the judge was not sure how to punish Sansa. Why is the judge confused?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Math Number Riddle
Difficulty Popularity
A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were:
The fifth number plus the third number equals fourteen.
The fourth number is one more than the second number.
The first number is one less than twice the second number.
The second number plus the third number equals ten.
The sum of all five numbers is 30.
What were the five numbers and in what order?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Hard Logic Brain Teaser
Difficulty Popularity
There are 100 doors. 100 strangers have been gathered in the adjacent room. The first one goes and opens every door. The second one goes and shuts down all the even numbered doors – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third door (third, sixth, ninth… and so on.) i.e. if the door is open, he shuts it and if the door is shut, he switches opens it. All the 100 strangers progresses in the similar fashion.
After the last person has done what he wanted, which doors will be left open and which ones will be shut at the end?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Crossing The River In Minimum Time Puzzle
Difficulty Popularity
Four people need to cross a dark river at night.
* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Logical Bank Brain Teaser
Difficulty Popularity
A bank customer had Rs.100 in his account. He then made 6 withdrawals, totaling Rs.. He kept a record of these withdrawals, and the balance remaining in the account, as follows:
Withdrawals Balance left
Rs.50 Rs.50
Rs.25 Rs.25
Rs.10 Rs.15
Rs.8 Rs.7
Rs.5 Rs.2
Rs.2 Rs.0
Rs.100 Rs.99
Why are the Totals not exactly right ?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Dirt Hole Tricky Riddle
Difficulty Popularity
How many cubic meters of dirt are in a hole of 20 meter long and 30 meter wide.
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Bridge Breaks Brain Teaser
Difficulty Popularity
There is a 30km long bridge. The bridge can only support up to a weight of 2000 kg. A car that weighs 2000 kg needs to cross that bridge. When the car has reached midway of the bridge, a bird comes and sits on top of the car. The bird weighs 300 grams.
Now, does the bridge breaks down at this point of time or not?
### Latest Puzzles
24 October
##### Count The Queue Riddle
Rohini is standing in a queue in which h...
23 October
##### Classic What Am I
I got many teeth, but I don't bite.Who A...
22 October
##### Frequent Maths Digit Brain Teaser
Can you identify the most common digit b...
21 October
##### Color Of Stairs Brain Teaser
I live in a one story house made up of r...
20 October
##### Make Equation True MatchStick Puzzle
Can you move one matchstick to make belo...
19 October
##### Murder Suspect Riddle
It was the first day of the college when...
18 October
##### Batman Forever Movie Riddle | What Am I
This riddle was asked in movie "Batman F... | 1,376 | 5,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2016-44 | longest | en | 0.971439 |
http://masterexcel.net/tag/vlookup/ | 1,582,632,534,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146066.89/warc/CC-MAIN-20200225110721-20200225140721-00240.warc.gz | 89,135,986 | 13,130 | ### Excel Magic Trick 1493 Part 2
DESCRIPTION In this video learn how to take Invoice Shipping and Discount Amounts from an Invoice Level (Header) Transaction or Fact Table and allocate those amounts to the Invoice Line Item Level Fact Table so [Read More >>>]
### Excel Magic Trick 1489
DESCRIPTION In this video learn how to create a single Excel formula that will lookup a product price and lookup the volume / quantity discount and deliver the correct discounted Product Price. See how to [Read More >>>]
### Excel Magic Trick 1488
DESCRIPTION Vote For Favourite Formula to Lookup Correct Price Based on Effective Date WORKBOOKS
### Excel Magic Trick 1486
DESCRIPTION In this video learn how to Lookup the correct product price based on the latest Effective Date. The Lookup table has multiple listings of the product, each with a different effective date and price. [Read More >>>]
### Excel Magic Trick 1484
DESCRIPTION In this video learn how to Lookup the correct product price based on the latest Effective Date. The Lookup table has multiple listings of the product, each with a different effective date and price. [Read More >>>]
### Excel Magic Trick 1481
DESCRIPTION In this video learn how to how to perform lookup from multiple tables. Learn two methods: 1) VLOOKUP and SWITCH Function or 2) VLOOKUP and INDIRECT Function with Defined Names. (00:01) Introduction (01:30) VLOOKUP [Read More >>>]
### Excel Magic Trick 1480
DESCRIPTION In this video learn how to do a two-way lookup to get the correct Price based on Quantity Discount & Product Name. Learn how to use the VLOOKUP and MATCH Functions. WORKBOOKS
### Excel Magic Trick 1479
DESCRIPTION In this video learn the Basics of VLOOKUP to looks up a price based on a Bulk Quantity Discount. Also learn about the MATCH Function to automatically pick out the correct Column Number for [Read More >>>]
### Office 2016 33
DESCRIPTION In this video learn about how to use Excel Tables, One-To-Many Relationships and the Data Model as a substitute for a VLOOKUP Function Helper Column for a Large Data Set. Discuss pros and cons. [Read More >>>]
### Excel Magic Trick 1458
DESCRIPTION Learn how to lookup the price for PVC Pipe in three different tables using VLOOKUP and INDIRECT Function and Defines Names. Defined Names allow us to assign a Named Reference to a lookup table. [Read More >>>] | 509 | 2,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-10 | latest | en | 0.806392 |
https://www.irif.fr/~buccia/COURS/PLC/prolog/sol4.txt | 1,680,093,607,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00233.warc.gz | 906,413,492 | 1,093 | concat([],L,L). concat([A|G],L,[A|R]):-concat(G,L,R). genere(0,[]):-!. genere(N,[N|L]):-M is N-1, genere(M,L). perm([],[]). perm([X|L],Z):-perm(L,W),insertion(X,W,Z). insertion(X,L,[X|L]). insertion(X,[Y|L],[Y|G]):-insertion(X,L,G). nombremagique(N,M):-M is (N*(N*N+1))//2. nth([H|_],1,H). nth([_|L],N,R):-N>0,P is N-1, nth(L,P,R). accumuler(_,_,_,0,0). accumuler(L,N,M,P,R):-P>0, S is P-1, nth(L,N,X), Next is N+M, accumuler(L,Next,M,S,T), R is T+X. sommeligne(L,N,I,R):- Init is N*(I-1)+1,accumuler(L, Init, 1, N, R). sommecolonne(L,N,I,R):- accumuler(L, I, N, N, R). testlignes(L,N,V):-auxlignes(L,N,V,1). auxlignes(_,N,_,K):-K=:=N+1,!. auxlignes(L,N,V,K):-sommeligne(L,N,K,V), S is K+1,auxlignes(L,N,V,S). testcolonnes(L,N,V):-auxcolonnes(L,N,V,1). auxcolonnes(_,N,_,K):-K=:=N+1,!. auxcolonnes(L,N,V,K):-sommecolonne(L,N,K,V), S is K+1,auxcolonnes(L,N,V,S). diago(L,N,R):-S is N+1, accumuler(L,1,S,N,R). adiago(L,N,R):-S is N-1, accumuler(L,N,S,N,R). carremagique(N,L1):-P is N*N, nombremagique(N,M),genere(P,L),perm(L,L1), testlignes(L1,N,M),testcolonnes(L1,N,M),diago(L1,N,M),adiago(L1,N,M). | 495 | 1,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.542383 |
https://pt.slideshare.net/MilkieGochkon/clil-pillar1-task1-61458717 | 1,618,477,683,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084601.32/warc/CC-MAIN-20210415065312-20210415095312-00616.warc.gz | 579,205,668 | 27,779 | O slideshow foi denunciado.
Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.
Próximos SlideShares
Carregando em…5
×
# Clil pillar1 task1
Clil pillar1 task1
• Full Name
Comment goes here.
Are you sure you want to Yes No
Your message goes here
• Entre para ver os comentários
• Seja a primeira pessoa a gostar disto
### Clil pillar1 task1
1. 1. Unit Mathematics Topic: Pythagoras’ theorem Matthayomsuksa 2
2. 2. b a c
3. 3. b c b > a!
4. 4. b c b-a
5. 5. b c b-a c*
6. 6. c c c c b-a a a a a
7. 7. 𝑐2 = 4 × 𝑎𝑏 2 + (𝑏 − 𝑎)2
8. 8. 𝑐2 = 4 × 𝑎𝑏 2 + (𝑏 − 𝑎)2 𝑐2 = 2𝑎𝑏 + (𝑏 − 𝑎)2 (𝑏 − 𝑎)2 = 𝑏2 − 2𝑎𝑏 + 𝑎2
9. 9. 𝑐2 = 2𝑎𝑏 + (𝑏2 −2𝑎𝑏 + 𝑎2 ) 𝑐2 = 2𝑎𝑏 + (𝑏2−2𝑎𝑏 + 𝑎2) 𝒄 𝟐 = 𝒂 𝟐 + 𝒃 𝟐
10. 10. Key terms hypotenuse (n.) the longest side of right triangle that has one angle of 90° EG. The longest side of the triangle is called the "hypotenuse".
11. 11. Content words triangle (n.) a flat shape with three straight sides EG. Pythagoras discovered an amazing fact about triangle.
12. 12. Pythagoras' Theorem Over 2000 years ago Pythagoras discovered an amazing fact about triangle. “When the triangle has a right angle (90°) and squares are made on each of the three sides, then the biggest square has the exact same area as the other two squares put together.” It is called "Pythagoras' Theorem" and can be written in one short equation. a2 + b2 = c2
13. 13. The longest side of the triangle is called the "hypotenuse", so the formal definition is If we know the lengths of two sides of a right angled triangle, we can find the length of the third side. (But remember it only works on right angled triangles!) In a right triangle: the square of the hypotenuse is equal to the sum of the squares of the other two sides.
14. 14. Present simple tense We use simple present to talk about something that is true in the present or something that is always true. Past Present Future
15. 15. Language Patterns symbols speak = equals or equal to + plus or positive - minus, subtract by or negative x or . multiplies or times ÷ or / divided by x2 x squared or x to the power of two x square root of x
16. 16. True or False Direction: Write "T" if the statement is true or "F" if the statement is false. Then rewrite to correct the information if the statement is false.
17. 17. 1. Pythagoras' equation can be written in 𝒂 𝟐 + 𝒃 𝟐 = 𝒄 𝟑. 𝒂 𝟐 + 𝒃 𝟐 = 𝒄 𝟐
18. 18. 2. If we know the lengths of 2 sides of a right angled triangle, we can find the length of the remaining side. | 898 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-17 | latest | en | 0.696591 |
http://www.stata.com/statalist/archive/2008-09/msg00620.html | 1,432,657,062,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928864.73/warc/CC-MAIN-20150521113208-00216-ip-10-180-206-219.ec2.internal.warc.gz | 722,180,064 | 3,012 | # Re: st: treatment of missing values in a matrix dissimilarity score
From "Eva Poen" To statalist@hsphsun2.harvard.edu Subject Re: st: treatment of missing values in a matrix dissimilarity score Date Mon, 15 Sep 2008 20:49:49 +0100
```Zach,
Stata seems to do listwise deletion, which appears fair in this
situation; it is not clear to me how comparable similarity measures
would be between observations when they are calculated based on
different sets of variables.
You want to calculate similarities only across non-missing elements. I
don't think there is an easy way to do this. One approach is to enlist
the help of -mvpatterns- which you can locate using -findit-.
-mvpatterns x1 x2 x3 x4- will tell you which patterns of missing
values exist in your data. You can then run the -matrix dissimilarity-
command separately for each pattern:
matrix dissimilarity m1=x1 x3 if x2 >=. & x4 >=., match proportion
for example, for all those observations that have x1 and x3
non-missing but x2 and x4 missing.
Eva
2008/9/15 Elkins, Zachary S <zelkins@austin.utexas.edu>:
> I'd like to calculate the matches between observations across a set of binary variables. I suspect that there are multiple ways to do this. I've specified:
>
> matrix dissimilarity m=x1 x2 x3 x4, match proportion
>
> However, some values in x1-x4 are missing. Based on the results, it appears that Stata treats missing values as if they were 1 and I don't see how to modify that. I'd like to calculate to calculate similarities across only non-missing elements (the number of which will be different for each pair, of course).
>
> Thanks for any help.
>
> Zach Elkins
>
>
> ______________________________
> Zachary Elkins
> Assistant Professor
> Department of Government
> University of Texas at Austin
> 1 University Station A1800
> Austin, TX 78712-0119
>
> p: (512) 232-7250
> f: (512) 471-1061
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 610 | 2,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-22 | longest | en | 0.899269 |
https://brokereybosh.netlify.app/shaner55521veca/calculate-interest-rate-on-mortgage-loan-xom | 1,726,307,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00572.warc.gz | 123,667,553 | 12,272 | ## Calculate interest rate on mortgage loan
Updated daily Mon-Fri, see mortgage rates from Chase broken down by term and The interest rate table below is updated daily, Monday through Friday, to give mortgage calculator to get a customized estimate of your mortgage rate and Jul 24, 2018 When calculating your DTI, all debt counts, including your new mortgage payment plus car loans, student loans, medical debt, credit card debt,
Bankrate’s mortgage calculator gives you a monthly payment estimate after you input the home price, your down payment, the interest rate and length of the loan term. Use the calculator to price Determine the monthly payments for any fixed-rate loan. Just enter the amount and terms, and our mortgage calculator does the rest. Click on “Show Amortization” Table to see how much interest you’ll pay each month and over the lifetime of the loan. The mortgage calculator will also show how Use our free mortgage calculator to quickly estimate what your new home will cost. Includes taxes, insurance, PMI and the latest mortgage rates. Mortgage Calculator | Zillow Equation for mortgage payments. M = the total monthly mortgage payment. P = the principal loan amount. r = your monthly interest rate. Lenders provide you an annual rate so you’ll need to divide that figure by 12 (the number of months in a year n = number of payments over the loan’s lifetime. To calculate mortgage interest, start by multiplying your monthly payment by the total number of payments you'll make. Then, subtract the principal amount from that number to get your mortgage interest. For example, if you're paying \$1,250 dollars a month on a 15-year, \$180,000 loan, you would start by multiplying \$1,250 by 15 to get \$225,000.
## Normally, the shorter the loan term, the lower the interest rate. Interest rate—the rate of interest charged by a mortgage lender. It can be fixed (otherwise known as a fixed-rate mortgage, or FRM), or adjustable (otherwise known as an adjustable rate mortgage, or ARM). The calculator above is only usable for fixed rates.
Equation for mortgage payments. M = the total monthly mortgage payment. P = the principal loan amount. r = your monthly interest rate. Lenders provide you an annual rate so you’ll need to divide that figure by 12 (the number of months in a year n = number of payments over the loan’s lifetime. To calculate mortgage interest, start by multiplying your monthly payment by the total number of payments you'll make. Then, subtract the principal amount from that number to get your mortgage interest. For example, if you're paying \$1,250 dollars a month on a 15-year, \$180,000 loan, you would start by multiplying \$1,250 by 15 to get \$225,000. For a \$300,000, 30-year mortgage with a 10-year, interest-only period at a 5 percent interest rate, your interest-only monthly payment would be \$1,250.00. A traditional loan payment at the same interest rate (with principal and interest factored in) would be \$1,870 per month. Another option is an adjustable-rate mortgage, or ARM, which has an initial, fixed-rate interest period of three, five, seven or 10 years. After the initial time frame, an ARM resets and interest rates can go up or down for the remaining life of the loan.
### Mortgage interest rates may be at an all time low, but there's still a big difference We take a look at the factors that determine your mortgage rate and calculate how Not only your down payment, but your loan length determines your rate,
Dec 23, 2016 Finally, using the mortgage calculator, input the original loan amount and find the interest rate that corresponds with the adjusted monthly Rocket Mortgage by Quicken Loans — Best for online loan applications. Mortgage Type, Interest Rate, APR. 30-year fixed, 4.125%, 4.395%. 15-year fixed The most common mortgage terms are 15 years and 30 years. Interest rate: Annual fixed interest rate for this mortgage. Monthly payment (PI): Monthly principal Dec 5, 2017 Generally, interest on student loans is calculated daily. Use this calculator to Loan Principal (Balance Owed)* Annual Interest Rate (%)*. Mortgage Calculator. email. Use this calculator to figure out Interest rate ? Enter the interest rate, or the Mortgage Loan Type? Choose the mortgage term.
### Jul 24, 2018 When calculating your DTI, all debt counts, including your new mortgage payment plus car loans, student loans, medical debt, credit card debt,
Mortgage calculator. The loan amount, the interest rate, and the term of the mortgage can have a dramatic effect on the total amount you will eventually pay for View your estimated monthly mortgage payment and get preapproved today. Figure out a monthly payment that works for you and your budget. -- Call 800- 531-0341 to speak with a loan specialist. -- Unlike interest rates, APR factors in the amount borrowed, the interest rate, one-time fees and discounts to determine a Use the Loan Comparison Calculator from Investors Bank to determine which mortgage Interest charges, origination fees, fees paid for a specific interest rate Updated daily Mon-Fri, see mortgage rates from Chase broken down by term and The interest rate table below is updated daily, Monday through Friday, to give mortgage calculator to get a customized estimate of your mortgage rate and Jul 24, 2018 When calculating your DTI, all debt counts, including your new mortgage payment plus car loans, student loans, medical debt, credit card debt, Interest rates are a big deal. They can also be a little confusing. M&T Bank's free, easy-to-use home loan calculator eliminates the confusion. Try it now!
## Interest rates are a big deal. They can also be a little confusing. M&T Bank's free, easy-to-use home loan calculator eliminates the confusion. Try it now!
Use the Loan Comparison Calculator from Investors Bank to determine which mortgage Interest charges, origination fees, fees paid for a specific interest rate Updated daily Mon-Fri, see mortgage rates from Chase broken down by term and The interest rate table below is updated daily, Monday through Friday, to give mortgage calculator to get a customized estimate of your mortgage rate and
The mortgage calculator with taxes and insurance estimates your monthly home mortgage payment and shows amortization table. The loan calculator estimates your car, auto, moto or student loan payments, shows amortization schedule and charts. Compare mortgage rates from multiple lenders in one place. It's fast, free, and anonymous. Balloon Mortgage Calculator. These loans are usually 5 to 10 years long and require borrowers to repay only a fraction of the loan during that time. Although balloon loans are often easier to qualify for than a traditional 30 year mortgage loan, and charge lower interest rates, there is a catch. Mortgage payment calculator to calculate your home loan payments. Use our mortgage interest rate calculator to view loan amortization tables and see how quickly you can pay off your home loan. | 1,461 | 6,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.932105 |
http://slideplayer.com/slide/228699/ | 1,544,482,580,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823445.39/warc/CC-MAIN-20181210212544-20181210234044-00184.warc.gz | 261,728,524 | 22,135 | # Lecture 44 Numerical Analysis. Solution of Non-Linear Equations.
## Presentation on theme: "Lecture 44 Numerical Analysis. Solution of Non-Linear Equations."— Presentation transcript:
Lecture 44 Numerical Analysis
Solution of Non-Linear Equations
Bisection Method Regula-Falsi Method Method of iteration Newton - Raphson Method Mullers Method Graeffes Root Squaring Method
Newton - Raphson Method
An approximation to the root is given by
Better and successive approximations x 2, x 3, …, x n to the root are obtained from N-R Formula
Newtons algorithm To find a solution to f(x)=0 given an initial approximation p 0
INPUT initial approximation p 0 ; tolerance TOL; maximum number of iterations N 0 OUTPUT approximate solution p or message of failure
Step 1 Set I = 1 Step 2 While i < N0 do Steps 3-6
Step 3 Setp = p 0 – f ( p 0 ) / f ( p 0 ) (compute p i ). Step 4 If Abs (p – p 0 ) < TOL OUTPUT ( p ); If Abs (p – p 0 ) < TOL OUTPUT ( p ); (The procedure was successful.) STOP
Step 5 Set i = i + 1 Step 6 Set p 0 = p (Update p 0 ) Step 7 OUTPUT (The method failed after N 0 iterations, N 0 =,N 0 ) The procedure was unsuccessful STOP
Example Using Maple to solve a non-linear equation.
Solution The Maple command will be as follows, Fsolve ( cos (x) -x);
alg023(); alg023(); This is Newton's Method This is Newton's Method Input the function F(x) in terms of x For example: > cos(x)-x Input initial approximation > 0.7853981635 Input tolerance > 0.00005
Input maximum number of iterations - no decimal point > 25 Select output destination 1. Screen 2. Text file Enter 1 or 2 > 1
Select amount of output 1. Answer only 2. All intermediate approximations Enter 1 or 2 > 2
Newton's Method I P F(P) I P F(P) 1 0.739536134 -7.5487470e-04 1 0.739536134 -7.5487470e-04 2 0.739085178 -7.5100000e-08 2 0.739085178 -7.5100000e-08 3 0.739085133 0.0000000e-01 3 0.739085133 0.0000000e-01 Approximate solution = 0.73908513 with F(P) = 0.0000000000 Number of iterations = 3 Tolerance = 5.0000000000e-05
Another Example > alg023(); Input the function F(x) in terms of x, > sin(x)-1 Input initial approximation > 0.17853 Input tolerance > 0.00005
Input maximum number of iterations – no decimal point > 25 Select output destination 1. Screen 2. Text file Enter 1 or 2 >2
Select amount of output 1. Answer only 2. All intermediate approximations Enter 1 or 2 > 2
Newton's Method I P F(P) I P F(P) 1 1.01422964e+00 -1.5092616e-01 1 1.01422964e+00 -1.5092616e-01 2 1.29992628e+00 -3.6461537e-02 2 1.29992628e+00 -3.6461537e-02 3 1.43619550e+00 -9.0450225e-03 3 1.43619550e+00 -9.0450225e-03 4 1.50359771e+00 -2.2569777e-03 4 1.50359771e+00 -2.2569777e-03 5 1.53720967e+00 -5.6397880e-04 5 1.53720967e+00 -5.6397880e-04 6 1.55400458e+00 -1.4097820e-04 6 1.55400458e+00 -1.4097820e-04 7 1.56240065e+00 -3.5243500e-05 7 1.56240065e+00 -3.5243500e-05More…
8 1.56659852e+00 -8.8108000e-06 8 1.56659852e+00 -8.8108000e-06 9 1.56869743e+00 -2.2027000e-06 9 1.56869743e+00 -2.2027000e-06 10 1.56974688e+00 -5.5070000e-07 10 1.56974688e+00 -5.5070000e-07 11 1.57027163e+00 -1.3770000e-07 11 1.57027163e+00 -1.3770000e-07 12 1.57053407e+00 -3.4400000e-08 12 1.57053407e+00 -3.4400000e-08 13 1.57066524e+00 -8.6000000e-09 13 1.57066524e+00 -8.6000000e-09 14 1.57073085e+00 -2.1000000e-09 14 1.57073085e+00 -2.1000000e-09 15 1.57076292e+00 -6.0000000e-10 15 1.57076292e+00 -6.0000000e-10
Approximate solution = 1.57076292 with F(P) =6.0000000000e-10 Number of iterations = 15 Tolerance = 5.0000000000e-05
Bisection Method > alg021(); This is the Bisection Method. Input the function F(x) in terms of x For example: > x^3+4*x^2-10
Input endpoints A < B separated by blank > 1 2 Input tolerance > 0.0005 Input maximum number of iterations - no decimal point > 25
Select output destination 1.Screen, 2. Text file Enter 1 or 2 > 1 Select amount of output 1. Answer only 2. All intermediate approximations Enter 1 or 2 > 2
Bisection Method I P F(P) I P F(P) 1 1.50000000e+00 2.3750000e+00 1 1.50000000e+00 2.3750000e+00 2 1.25000000e+00 -1.7968750e+00 2 1.25000000e+00 -1.7968750e+00 3 1.37500000e+00 1.6210938e-01 3 1.37500000e+00 1.6210938e-01
4 1.31250000e+00 -8.4838867e-01 5 1.34375000e+00 -3.5098267e-01 5 1.34375000e+00 -3.5098267e-01 6 1.35937500e+00 -9.6408842e-02 6 1.35937500e+00 -9.6408842e-02 7 1.36718750e+00 3.2355780e-02 7 1.36718750e+00 3.2355780e-02 8 1.36328125e+00 -3.2149969e-02 8 1.36328125e+00 -3.2149969e-02 9 1.36523438e+00 7.2030000e-05 9 1.36523438e+00 7.2030000e-05 10 1.36425781e+00 -1.6046697e-02 10 1.36425781e+00 -1.6046697e-02 11 1.36474609e+00 -7.9892590e-03 11 1.36474609e+00 -7.9892590e-03 Approximate solution P = 1.36474609 with F(P) = -.00798926 Number of iterations = 11 Tolerance = 5.00000000e-04
alg021(); Another example of the Bisection Method. Input the function F(x) in terms of x, > cos(x) Input endpoints A < B separated by blank > 1 2 Input tolerance > 0.0005
Input maximum number of iterations - no decimal point > 25 Select output destination 1. Screen, 2. Text file Enter 1 or 2 > 1 Select amount of output 1. Answer only 2. All intermediate approximations Enter 1 or 2 > 2
Bisection Method 1 P F(P) 1 P F(P) 1 1.50000000e+00 7.0737202e-02 1 1.50000000e+00 7.0737202e-02 2 1.75000000e+00 -1.7824606e-01 2 1.75000000e+00 -1.7824606e-01 3 1.62500000e+00 -5.4177135e-02 3 1.62500000e+00 -5.4177135e-02 4 1.56250000e+00 8.2962316e-03 4 1.56250000e+00 8.2962316e-03 5 1.59375000e+00 -2.2951658e-02 5 1.59375000e+00 -2.2951658e-02
6 1.57812500e+00 -7.3286076e-03 7 1.57031250e+00 4.8382678e-04 7 1.57031250e+00 4.8382678e-04 8 1.57421875e+00 -3.4224165e-03 8 1.57421875e+00 -3.4224165e-03 9 1.57226563e+00 -1.4692977e-03 9 1.57226563e+00 -1.4692977e-03 10 1.57128906e+00 -4.9273519e-04 10 1.57128906e+00 -4.9273519e-04 11 1.57080078e+00 -4.4542051e-06 11 1.57080078e+00 -4.4542051e-06 Approximate solution P = 1.57080078 with F(P) = -.00000445 Number of iterations = 11 Tolerance = 5.00000000e-04
alg025(); This is the Method of False alg025(); This is the Method of FalsePosition Input the function F(x) in terms of x > cos(x)-x Input endpoints P0 < P1 separated by a blank space 0.5 0.7853981635 0.5 0.7853981635 Input tolerance >0.0005
Input maximum number of iterations - no decimal point > 25 Select output destination 1. Screen 2. Text file Enter 1 or 2 >1
Select amount of output 1. Answer only 2. All intermediate approximations Enter 1 or 2 > 2
METHOD OF FALSE POSITION I P F(P) 2 7.36384139e-01 4.51771860e-03 2 7.36384139e-01 4.51771860e-03 3 7.39058139e-01 4.51772000e-05 3 7.39058139e-01 4.51772000e-05 4 7.39084864e-01 4.50900000e-07 4 7.39084864e-01 4.50900000e-07 Approximate solution P =.73908486 with F(P) =.00000045 Number of iterations = 4 Tolerance = 5.00000000e-04
System of Linear Equations
Gaussian Elimination Gauss-Jordon Elimination Crouts Reduction Jacobis Gauss- Seidal Iteration Relaxation Matrix Inversion
> alg061(); This is Gaussian Elimination to solve a linear system. The array will be input from a text file in the order: A(1,1), A(1,2),..., A(1,N+1), A(2,1), A(2,2),..., A(2,N+1),..., A(N,1), A(N,2),..., A(N,N+1) Place as many entries as desired on each line, but separate entries with at least one blank.
Has the input file been created? - enter Y or N. > y Input the file name in the form - drive:\name.ext for example: A:\DATA.DTA > d:\maple00\dta\alg061.dta Input the number of equations - an integer. > 4 Choice of output method: 1. Output to screen2. Output to text file Please enter 1 or 2. > 1
GAUSSIAN ELIMINATION The reduced system - output by rows: 1.00000000 -1.00000000 2.00000000 -1.00000000 -8.00000000 1.00000000 -1.00000000 2.00000000 -1.00000000 -8.00000000 0.00000000 2.00000000 -1.00000000 1.00000000 6.00000000 0.00000000 2.00000000 -1.00000000 1.00000000 6.00000000 0.00000000 0.00000000 -1.00000000 -1.00000000 -4.00000000 0.00000000 0.00000000 -1.00000000 -1.00000000 -4.00000000 0.00000000 0.00000000 0.00000000 2.00000000 4.00000000 0.00000000 0.00000000 0.00000000 2.00000000 4.00000000 Has solution vector: -7.00000000 3.00000000 2.00000000 2.00000000 -7.00000000 3.00000000 2.00000000 2.00000000 with 1 row interchange (s)
> alg071(); This is the Jacobi Method for Linear Systems. The array will be input from a text file in the order A(1,1), A(1,2),..., A(1,n+1), A(2,1), A(2,2),..., A(2,n+1),..., A(n,1), A(n,2),..., A(n,n+1) Place as many entries as desired on each line, but separate entries with at least one blank.
The initial approximation should follow in the same format has the input file been created? - enter Y or N. > y Input the file name in the form - drive:\name.ext for example: A:\DATA.DTA > d:\maple00\alg071.dta Input the number of equations - an integer. > 4
Input the tolerance. > 0.001 Input maximum number of iterations. > 15 Choice of output method: 1. Output to screen 2. Output to text file Please enter 1 or 2. > 1
JACOBI ITERATIVE METHOD FOR LINEAR SYSTEMS The solution vector is : 1.00011860 1.99976795 -.99982814 0.99978598 using 10 iterations with Tolerance 1.0000000000e-03
Lecture 44 Numerical Analysis
Similar presentations | 3,691 | 9,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.697782 |
https://club.tidalcycles.org/t/week-2-lesson-3-combining-patterns-with-arithmetic-plus-the-hurry-function/489?page=2 | 1,726,005,559,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00554.warc.gz | 152,891,022 | 4,417 | Week 2 lesson 3 - Combining patterns with arithmetic, plus the 'hurry' function
Hello. I'm not sure if I should be "waking up" this thread, but I have a question about the examples on the worksheet. There's one example that includes this rhythm:
d1 \$ n "0(5,8) [4 1]" # sound "drum"
I don't get how "0(5,8)" works, how this combination of parenthesis and comma is changing the rhythm and I couldn't find any tip in previous worksheet.
Btw, lesson 5 from week 1 is missing on the "Course I (> 1.6)" page.
Thank you!
EDIT: just remembered this has to do with Euclidean Patterns! So it is applying the pattern to the 0:drum sound fitting it in the first half of the cycle, right?
Waking up threads is encouraged Yes as you say, it's a Euclidean pattern squashed into the first half.
Here are all the lessons, to make sure you're not missing any.
1 Like | 227 | 859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.941954 |
https://www.physicsforums.com/threads/simple-complex-numbers-integral.254682/ | 1,531,905,067,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590074.12/warc/CC-MAIN-20180718080513-20180718100513-00281.warc.gz | 952,701,796 | 12,924 | # Homework Help: Simple complex numbers integral
1. Sep 8, 2008
### elcotufa
1. The problem statement, all variables and given/known data
Integrate using complex numbers
$$\int\limits_0^{2\pi} cos^4(\theta)$$
2. Relevant equations
$$cos^4(\theta)= (\frac{e^{j\theta} + e^{-j\theta}}2)^4$$
3. The attempt at a solution
$$\frac 1{2^4} (e^{j\theta} + e^{-j\theta})^4$$
I got
$$\frac 1{2^4} \int^{2\pi}_0 (e^{4j\theta}+4e^{2j\theta}+4e^{-2j\theta}+e^{-4j\theta}+6)$$
After this I am not sure what to do
The integral of $$\int e^{4j}$$ would be $$\frac{e^{4j\theta}}{4j}$$?
How do I cancel them?
Input appreciated
2. Sep 8, 2008
### Dick
Sure, that's the integral of e^(4j*theta). You'll notice if you evaluate it from 0 to 2*pi the result is 0. The same for all the other exponentials. The only term that contributes is the 6.
3. Sep 8, 2008
thanks man | 323 | 862 | {"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} | 3.390625 | 3 | CC-MAIN-2018-30 | latest | en | 0.682409 |
https://www.numbersaplenty.com/2724 | 1,720,785,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514387.30/warc/CC-MAIN-20240712094214-20240712124214-00283.warc.gz | 572,574,912 | 3,404 | Search a number
2724 = 223227
BaseRepresentation
bin101010100100
310201220
4222210
541344
620340
710641
oct5244
93656
102724
112057
1216b0
131317
14dc8
15c19
hexaa4
2724 has 12 divisors (see below), whose sum is σ = 6384. Its totient is φ = 904.
The previous prime is 2719. The next prime is 2729. The reversal of 2724 is 4272.
2724 = T22 + T23 + ... + T29.
It is an interprime number because it is at equal distance from previous prime (2719) and next prime (2729).
It is a tau number, because it is divible by the number of its divisors (12).
It is a nialpdrome in base 4, base 14 and base 16.
It is not an unprimeable number, because it can be changed into a prime (2729) by changing a digit.
2724 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a pernicious number, because its binary representation contains a prime number (5) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 102 + ... + 125.
It is an arithmetic number, because the mean of its divisors is an integer number (532).
22724 is an apocalyptic number.
It is an amenable number.
2724 is an abundant number, since it is smaller than the sum of its proper divisors (3660).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
It is a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (3192).
2724 is a wasteful number, since it uses less digits than its factorization.
2724 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 234 (or 232 counting only the distinct ones).
The product of its digits is 112, while the sum is 15.
The square root of 2724 is about 52.1919534028. The cubic root of 2724 is about 13.9659034208.
Adding to 2724 its reverse (4272), we get a palindrome (6996).
It can be divided in two parts, 272 and 4, that added together give a triangular number (276 = T23).
The spelling of 2724 in words is "two thousand, seven hundred twenty-four", and thus it is an iban number.
Divisors: 1 2 3 4 6 12 227 454 681 908 1362 2724 | 629 | 2,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.911337 |
http://math.stackexchange.com/questions/39467/omega-saturation-of-mathbbr | 1,469,724,057,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828286.80/warc/CC-MAIN-20160723071028-00160-ip-10-185-27-174.ec2.internal.warc.gz | 155,739,991 | 18,093 | # $\omega$-saturation of $(\mathbb{R},<)$
Could anyone of you explain me why $(\mathbb{R},<)$ is $\omega$-saturated?
EDIT: do you know also why the theory of Boole algebras without atoms is $\omega$-categoric?
-
en.wikipedia.org/wiki/Saturated_model – user9413 May 16 '11 at 18:50
I think you should ask your second question as a separate question. – JDH May 16 '11 at 19:31
Your structure is an endless dense linear ordering, and this is a theory that admits elimination of quantifiers. Thus, every assertion in this language is equivalent to a quantifier-free assertion. If one has finitely many parameters, then the only possible consistent types are the assertions about how those parameters are ordered, and how the new variable $x$ fits into the resulting intervals. That is, the variable $x$ is either equal to one of them, or between two successive ones, or above all or below all of them. All such types are already realized in $\mathbb{R}$, since it has such points already for each of these possible patterns, and so the structure is $\omega$-saturated.
A different approach: the theory of dense linear orderings is $\omega$-categorical.
From that it is easy to see that if a theory is $\kappa$-categorical, then every model of the theory of cardinality at least $\kappa$ is $\kappa$-saturated (just extend the set of parameters to an elementary submodel of cardinality $\kappa$). | 349 | 1,395 | {"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} | 2.671875 | 3 | CC-MAIN-2016-30 | latest | en | 0.899582 |
https://betterthisworld.com/latest-news/como-estara-el-tiempo-hoy-unpacking-today-s-weather-forecast/ | 1,720,957,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00236.warc.gz | 123,830,245 | 40,569 | Connect with us
Latest News
# ¿Cómo Estará El Tiempo Hoy – Unpacking Today’s Weather Forecast
Published
on
## ¿Cómo Estará El Tiempo Hoy
Ever wondered, “How’s the weather going to be today?” Knowing your day’s weather can indeed make a world of difference. From planning your outfits to scheduling outdoor activities, it’s a question that influences our everyday life. And in this digital age where information is just a click away, getting an accurate weather forecast has never been easier.
Nowadays, I find myself checking my smartphone first thing in the morning for the day’s weather forecast. Weather apps and websites have made predicting atmospheric conditions as simple as pie. But how accurate are these forecasts? Can they truly predict what Mother Nature has up her sleeve?
The science behind weather forecasting is fascinating – it involves complex computer models and tons of data from satellites, radars, and sensors all over the globe. Through this article, I’ll delve into how meteorologists predict daily weather patterns and discuss how reliable these predictions really are. By understanding the process behind these forecasts, you’ll be able to better interpret and utilize them in your daily routine.
### Understanding Weather Forecasts
We’ve all been there. You’re planning an outdoor event, and you need to know what the weather’s going to be like. That’s when we turn to weather forecasts. But how exactly do they work?
Weather forecasts are predictions about the state of the atmosphere at a specific time in a particular location. They’re created using data collected from around the world by satellites, aircraft, ships, weather stations and buoys.
The raw data that’s collected is processed by supercomputers running complex mathematical models called atmospheric models. These models simulate the Earth’s atmosphere and oceans to predict future conditions.
Now, you might wonder just how accurate these forecasts are? Well, here’s where it gets interesting:
• Short-term forecasts (1-3 days) have an accuracy rate of over 80%.
• Medium-range forecasts (4-7 days) drop slightly with a 70% accuracy rate.
• Long-term forecasts (8+ days) aren’t as reliable with only a 50% chance of being correct.
Forecasts can sometimes seem vague or ambiguous (“chance of rain”, “partly cloudy”), but that’s because weather is inherently unpredictable! Meteorologists use terms such as “probability” or “percent chance” to convey this uncertainty.
So next time you check your local forecast remember it’s not just guesswork – there’s some seriously high-tech science behind those predictions!
## Different Ways to Check the Weather
I often get asked, “How do I check the weather?” Honestly, there’s no one-size-fits-all answer. It all depends on your personal preferences and what resources you have available. Let’s delve into some of the most common methods.
Your smartphone is probably your best friend when it comes to checking the weather. There are a plethora of weather apps out there, such as The Weather Channel, AccuWeather, and Weather Underground. These apps give detailed forecasts for your exact location and provide hourly updates. They’ll even send severe weather alerts straight to your phone so you’re always in the know.
If you’re more of an old-school type, newspapers and local TV news channels are still reliable sources for daily forecasts. They might not provide hour-by-hour updates like an app would, but they’ve got the basics covered: temperature highs and lows, chances of rain or snow -you get the picture.
Then there’s my personal favorite: nature itself! Cloud patterns can tell us quite a bit about upcoming storms while wind direction can hint at changes in temperature or humidity levels.
Lastly remember that websites like NOAA (National Oceanic and Atmospheric Administration) offer comprehensive forecasts too. Not only do they cover basic weather predictions but also dive deep into climatology data if you’re interested.
So there you go – from tech-savvy solutions to traditional methods and freebie nature clues – multiple ways exist for staying ahead of Mother Nature’s curveball!
In today’s world where information is readily available at our fingertips, there’s no excuse for being caught off-guard by sudden shifts in the weather. By staying informed through reliable mediums, we can ensure that whether it rains or shines, we’re prepared to face whatever Mother Nature throws our way!
To sum up, understanding “¿cómo estará el tiempo hoy?” isn’t just about knowing if you’ll need an umbrella or sunscreen before stepping out. It’s about making smart decisions that affect your everyday life based on accurate and timely information. So next time when you wonder what the weather will be like today, remember: you’ve got plenty resources to find out! | 953 | 4,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.904181 |
https://iq.opengenus.org/select-random-node-from-binary-tree/ | 1,670,500,317,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711336.41/warc/CC-MAIN-20221208114402-20221208144402-00433.warc.gz | 342,076,952 | 23,433 | ×
Search anything:
# Select Random Node from Binary Tree
#### Problems on Binary Tree Algorithms Randomized Algorithm
Get this book -> Problems on Array: For Interviews and Competitive Programming
In this article, we have presented two algorithms to select a random node from Binary Tree while maintaining the uniform randomness.
1. Introduction to Randomized Algorithms
2. Efficient Approach
3. Second approach : Using ArrayList
# Introduction to Randomized Algorithms
In Mathematics , we all have studied about the concept of random events,experiments and the probabilities govering them. For solving the above problem we generally use the concept of Probability and Randomised Algorithm.
Now before jumping at the problem, lets first take a quick glance at the concept of Randomised algorithms in programming.
What is Randomised Algorithm?
An algorithm that takes help of random numbers to decide the upcoming/next action in its logic is called randomised algorithm. The fate of the entire algorithm depends on the random number generated.
There are basically two types of Randomises algorithms :
1. Las Vegas algorithm : Allowed to return only true values
2. Monte Carlo algorithm : Allowed to return wrong values but with small probability.
What is Probability?
The part of Mathematics that deals with finding the chance of the occurence of a random event is called probability. For example , we have a coin , the probability of finding heads on tossing the coin is 0.5. Its general formula is :
P(X) = (No. of favorable outcomes) / (No. of possible outcomes)
## 1. Efficient Approach :
Let's jump first to the Algorithm :
Step 1: START
Step 2: The first task is to create a Binary tree
(i) Declare a static class(TreeNode in this case) that will basically contain the data and pointers for the left and right child of every root.
(ii) Then create a class with binary tree class(TreeNode as return type that will create new node for our binary tree
Step 3: Declare a method with Binary Tree class as return type that will basically count the number of children for a given root
Step 4: Then we will define a random number generator
(i) randomNode()
index = int index = (int) (Math.random() * (root.child + 1));
Step 5: Having got the random index, we now have to just return the corresponding node of the binary tree
Step 5: STOP
Java Implementation of the Algorithm:
``````// Program to select a random node from a Binary tree
import java.lang.*;
class Opengenus_random_treeNode
{
static class TreeNode
{
int val;
int child;
TreeNode left, right;
}
static TreeNode Node(int data)
{
TreeNode temp = new TreeNode();
temp.val = data;
temp.left = temp.right = null; //No child initially for the Binary tree
temp.child = 0;
return temp;
}
static int getChild(TreeNode root)
{
if (root == null)
return 0;
return getChild(root.left) +
getChild(root.right) + 1;
}
// the following function just counts the number of children for a given root
static TreeNode ChildCount(TreeNode root)
{
if (root == null)
return null;
root.child = getChild(root) - 1;
root.left = ChildCount(root.left);
root.right = ChildCount(root.right);
return root;
}
// returns number of children for a particular root of the binary tree
static int children(TreeNode root)
{
if (root == null)
return 0;
return root.child + 1;
}
//The following function is basically used to return the required random node of the binary tree
static int randomNodeUtil(TreeNode root, int count)
{
if (root == null)
return 0;
if (count == children(root.left))
return root.val;
if (count < children(root.left))
return randomNodeUtil(root.left, count);
return randomNodeUtil(root.right,
count - children(root.left) - 1);
}
// Returns Random node
static int randomNode(TreeNode root)
{
int index = (int) (Math.random() * (root.child + 1));
return randomNodeUtil(root, index);
}
public static void main(String[] asdf)
{
//Now let's create the binary Tree
TreeNode root = Node(1);
root.left = Node(2);
root.right = Node(3);
root.left.right = Node(12);
root.left.right = Node(89);
root.right.left = Node(68);
root.right.right = Node(99);
root.right.left = Node(68);
ChildCount(root);
System.out.println( "Random Node : " +
randomNode(root));
}
}
``````
Output :
1. First Execution :
``````Random Node : 89
``````
2. Second Execution :
``````Random Node : 1
``````
3. Third Execution :
``````Random Node : 89
``````
Explanation of the above approach :
So in this particular solution we have tried to basically modify the structure of the tree. Let's take a ride through an example :
(a) Store the count of children for each node
The value on the left is node and on the right is the count of children
(b) Consider the inorder traversal of the tree and generate a number smaller than or equal to the number of nodes in the tree i.e. random number
(c) Now traverse the tree and visit the desired node using the counts.
(d) On reaching a particular node, we go either to the left subtree or right subtree using the random number, if the random number is less than count then go to the left else to the right, remember that we are moving either to the left or right at a time.
In the above implementation, we use getChild() method to get the count of children for the root, randomNode() method to return the random node of the tree.
Note that the child attribute stores the count of children for each node of the tree.
Time Complexity : O(h), where h is the height of the tree
Auxiliary Space Complexity : O(1)
## 2. Second Approach : Using ArrayList
Having gone through the above mentioned tedious method, let's jump on to a simple method that will basically make use of a inorder traversal of tree and then we will generate a random number between 0 to n-1, then use the number as array index and display the number at that particular index.
Let's take a look at the algorithm :
Step 1: START
Step 2: First create a binary tree using the steps mentioned in the first approach
Step 3: Now use a method inOrder() that takes a node as input parameter to traverse through the binary tree in inorder fashion as also store the values in a ArrayList simultaneously.
Step 4: Now define a method getrandom() that takes a node as input parameter, in this first call the inOrder() method to store the values in the arraylist, then find the size of the binary tree and now just generate a random number between 0 to n-1.
Step 5: After generating the number display the value of the ArrayList at the generated index
Step 6: STOP
Java implementation of the Algorithm :
``````import java.util.ArrayList;
// Using auxillary array to find the random node in a given binary tree
class Node {
int item;
Node left, right;
public Node(int key) {
item = key;
left = right = null;
}
}
class Tree {
// Using a arraylist to store the inorder traversal of the given binary tree
static ArrayList<Integer> list = new ArrayList<Integer>();
// root of Tree
Node root;
Tree() {
root = null;
}
// Now lets find the inorder traversal of the given binary tree
static void inOrder(Node node) {
if (node == null)
return;
// traverse the left child
inOrder(node.left);
// traverse the right child
inOrder(node.right);
}
public void getrandom(Node val)
{
inOrder(val);
// getting the count of node of the binary tree
int n = list.size();
int min = 0;
int max = n - 1;
//Generate random int value from 0 to n-1
int b = (int)(Math.random()*(max-min+1)+min);
// displaying the value at the generated index
int random = list.get(b);
System.out.println("Random Node : " + random);
}
public static void main(String[] args) {
Tree tree = new Tree();
tree.root = new Node(1);
tree.root.left = new Node(12);
tree.root.right = new Node(9);
tree.root.left.left = new Node(5);
tree.root.left.right = new Node(6);
tree.getrandom(tree.root);
}
}
``````
Output :
1. First Execution :
``````Random Node : 9
``````
2. Second Execution :
``````Random Node : 1
``````
3. Third Execution :
``````Random Node : 12
``````
Explanation of the approach:
(a) Form the required binary tree
(b) Now use the inOrder() method to get the nodes in inOrder fashion and also store them in the given arraylist list
(c) Using the getRandom() method generate a random number between 0 to n-1, then get the value at the generated random number from the arraylist using get() method and finally display the result.
Time Complexity : O(n), where n is the number of nodes in the tree
Auxiliary Space Complexity : Theta(1)
## MCQs on Randomised algorithms
1. Unix sort command uses?
(A) Quick sort
(B) Bucket sort
(C) Insertion sort
(D) Merge sort
Ans : A
2. Which of the following is a type of randomized algorithm?
(A) New York algorithm
(B) California algorithm
(C) Monte Carlo algorithm
Ans : C
3. Which among the follwing can be solved in Computer Science?
(A) NP = co-NP problem
(B) P = NP problem
(C) P = BPP problem
(D) None of these
Ans : A, B, and C
4. Which one of the following algorithm is fast in execution?
(A) Las Vegas algorithm
(B) Duke algorithm
(C) Atlantic City algorithm
(D) None of these
Ans : C
5. Which of the following the application of randomized algorithm?
(A) Quick sort
(B) Min cut
(C) All of the mentioned
(D) Verifying Matrix Multiplication
Ans : C
Thank You!
#### Suraj Kumar
Android Developer, Content Writer and Open Source Enthusiast!
Improved & Reviewed by:
Select Random Node from Binary Tree | 2,199 | 9,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-49 | longest | en | 0.884757 |
https://askfilo.com/math-question-answers/two-balls-are-drawn-at-random-with-replacement-frotf8 | 1,719,247,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00094.warc.gz | 88,667,147 | 43,384 | World's only instant tutoring platform
Question
Medium
Solving time: 3 mins
# Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that(i) both balls are red.(ii) First ball is black and second is red.(iii) One of them is black and other is red.
## Text solutionVerified
Given A box containing 10 black and 8 red balls.
Total number of balls in box =18
(i) Both balls are red.
Probability of getting a red ball in first draw =8 / 18=4 / 9
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw =8 / 18=4 / 9
Now, Probability of getting both balls red
(ii) First ball is black and second is red.
Probability of getting a black ball in first draw =10 / 18=5 / 9
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw =8 / 18=4 / 9
Now, Probability of getting first ball is black and second is red
(iii) One of them is black and other is red.
Probability of getting a black ball in first draw =10 / 18=5 / 9
As the ball is replaced after first throw,
Hence, Probability of getting a red ball in second draw =8 / 18=4 / 9
Now, Probability of getting first ball is black and second is red Probability of getting a red ball in first draw =8 / 18=4 / 9
As the ball is replaced after first throw,
Hence, Probability of getting a black ball in second draw =10 / 18=5 / 9
Now, Probability of getting first ball is red and second is black Therefore, Probability of getting one of them is black and other is red:
= Probability of getting first ball is black and second is red + Probability of getting first ball is red and second is black
=20 / 81+20 / 81=40 / 81
38
Share
Report
## Video solutions (4)
Learn from their 1-to-1 discussion with Filo tutors.
4 mins
77
Share
Report
4 mins
96
Share
Report
3 mins
63
Share
Report
Found 5 tutors discussing this question
Discuss this question LIVE
13 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Trusted by 4 million+ students
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that(i) both balls are red.(ii) First ball is black and second is red.(iii) One of them is black and other is red. Updated On Jan 6, 2024 Topic Probability Subject Mathematics Class Class 12 Answer Type Text solution:1 Video solution: 4 Upvotes 350 Avg. Video Duration 6 min | 727 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.928535 |
https://www.houseofmath.com/study-tips/problem-solving/8 | 1,679,532,185,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00136.warc.gz | 878,979,647 | 26,789 | # Guess
Guess a solution! It can sometimes pay off to guess a solution and see how it fits. This is especially smart when you can’t find a systematic approach.
Testing a solution shows you how you need to proceed for the next guess: Should you go up or down in value?
Jasmine, Martin and Qiang are 37 years in total. Martin is four years older than Jasmine. Qiang is three years younger than Jasmine. How old are they?
Here, you can try to guess Jasmin’s age. Then you also know how old Martin and Qiang are. If the sum is greater than 37, then Jasmine is younger than what you guessed. If the sum is less than 37, then Jasmine is older than what you guessed. By adjusting, you will reach the right solution. | 166 | 712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-14 | latest | en | 0.948538 |
https://www.contextgarden.net/index.php?title=Math/Display&diff=22146&oldid=22145 | 1,590,714,114,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347401004.26/warc/CC-MAIN-20200528232803-20200529022803-00444.warc.gz | 678,254,523 | 12,102 | # Difference between revisions of "Math/Display"
< Math|
Warning!
If you consider using ConTeXt for a paper with a proper formula numbering please wait until it is implemented in mkiv. At the moment only a very limited formula numbering is possilbe:
• number formula(s) with a plain number
• number formula(s) with a subnumber
Not possible is at the moment:
• Table of formulas
• Named subformulas
• Mixure of formulas and subformulas in one block
• Reference on a (sub-)formula without having it listed in the Table of formulas
For more details and actual state see mailing list and Summary of formula numbering problem.
# Display Math
Display math is enclosed in a \startformula / \stopformula pair. Thus
The famous result (once more) is given by \startformula c^2 = a^2 + b^2. \stopformula This, when typeset, produces the following:
# Numbering Formulae
ConTeXt provides an easy way to number the display maths equations. Simply, put \placeformula before \startformula and you will get numbered equations. Thus,
The famous result (once more) is given by \placeformula \startformula c^2 = a^2 + b^2. \stopformula This, when typeset, produces the following:
The \placeformula command is optional, and produces the equation number; leaving it off produces an unnumbered equation.
## Changing format of numbers
You can use \setupformulas to change the format of numbers. For example to get bold numbers inside square brackets use
\setupformulas[left={[},right={]},numberstyle=bold] which gives
To get equations also numbered by section, add the command:
\setupnumber[formula][way=bysection]
To the start of your document.
To get alphabets instead of numbers, use
\setupformulas[conversion=Character] which gives
## Changing Formula alignment
Normally a formula is centered, but in case you want to align it left or right, you can set up formulas to behave that way. Normally a formula will adapt its left indentation to the environment:
In the next examples we explicitly align formulas to the left (\raggedleft), center and right (\raggedright):
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
\setupformulas[align=left]
\startformula c^2 = a^2 + b^2 \stopformula
\setupformulas[align=middle]
\startformula c^2 = a^2 + b^2 \stopformula
\setupformulas[align=right]
\startformula c^2 = a^2 + b^2 \stopformula
Or in print:
With formula numbers the code is:
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
\setupformulas[align=left]
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
\setupformulas[align=middle]
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
\setupformulas[align=right]
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
And the formulas look like:
When tracing is turned on (\tracemathtrue) you can visualize the bounding box of the formula,
As you can see, the dimensions are the natural ones, but if needed you can force a normalized line:
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
\setupformulas[align=middle,strut=yes]
\tracemathtrue
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
This time we get a more spacy result. [Ed. Note: For this example equation, there appears to be no visible change.]
We will now show a couple of more settings and combinations of settings. In centered formulas, the number takes no space
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
\tracemathtrue
\setupformulas[align=middle]
\startformula c^2 = a^2 + b^2 \stopformula
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
You can influence the placement of the whole box with the parameters leftmargin and rightmargin.
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
Some example text, again, to show where the right and left margins of the text block are.
\tracemathtrue
\setupformulas[align=right,leftmargin=3em]
\startformula c^2 = a^2 + b^2 \stopformula
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
\setupformulas[align=left,rightmargin=1em]
\startformula c^2 = a^2 + b^2 \stopformula
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
You can also inherit the margin from the environment.
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
Some example text, again, to show where the right and left margins of the text block are.
\tracemathtrue
\setupformulas[align=right,margin=standard]
\startformula c^2 = a^2 + b^2 \stopformula
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
The distance between the formula and the number is only applied when the formula is left or right aligned.
\setuppapersize[A5]
\setuplayout[textwidth=8cm]
\tracemathtrue
\setupformulas[align=left,distance=2em]
\startformula c^2 = a^2 + b^2 \stopformula
\placeformula \startformula c^2 = a^2 + b^2 \stopformula
# Referencing formulae
Equations can be referred to by simply adding a label to \placeformula and using \ref to create the reference:
The famous result (and again) is given by \placeformula[formulalabel] \startformula c^2 = a^2 + b^2. \stopformula And now we can refer to formula \ref[][formulalabel]. This, when typeset, produces the following:
Note, that \ref expects two arguments, therefore you need the brackets twice.
By default, only the formula number appears as a reference. This can be changed by using \definereferenceformat. For example, to create a command \eqref which shows the formula number in brackets, use
\definereferenceformat[eqref][left=(,right=)]
See References for more examples of \definereferenceformat.
# Sub-Formula Numbering
## Automatic Sub-Formula Numbering
To use subformula numbering, you can use \startsubformulas/\stopsubformulas. For example
Examples: \startsubformulas[eq:1] \placeformula[eq:first] \startformula c^2 = a^2 + b^2 \stopformula \placeformula[eq:second] \startformula c^2 = a^2 + b^2 \stopformula \stopsubformulas Formula (\in[eq:1]) states the Pythagora's Theorem twice, once in (\in[eq:first]) and again in (\in[eq:second]).
## The Manual Method
Sometimes, you need more fine grained control over numbering of subformulas. In that case one can make use of the optional agument of \placeformula command and the related \placesubformula commands which can be used to produce sub-formula numbering. For example:
Examples: \placeformula{a} \startformula c^2 = a^2 + b^2 \stopformula \placesubformula{b} \startformula c^2 = a^2 + b^2 \stopformula
What's going on here is simpler than it might appear at first glance. Both \placeformula and \placesubformula produce equation numbers with the optional tag added at the end; the sole difference is that the former increments the equation number first, while the latter does not (and thus can be used for the second and subsequent formulas that use the same formula number but presumably have different tags).
This is sufficient for cases where the standard ConTeXt equation numbers suffice, and where only one equation number is needed per formula. However, there are many cases where this is insufficient, and \placeformula defines \formulanumber and \subformulanumber commands, which provide hooks to allow the use of ConTeXt-managed formula numbers with plain TeX equation numbering. These, when used within a formula, simply return the formula number in properly formatted form, as can be seen in this simple example with plain TeX's \eqno. Note that the optional tag is inherited from \placeformula.
More examples: \placeformula{c} \startformula \let\doplaceformulanumber\empty c^2 = a^2 + b^2 \eqno{\formulanumber} \stopformula
In order for this to work properly, we need to turn off ConTeXt's automatic formula number placement; thus the \let command to empty \doplaceformulanumber, which must be placed after the start of the formula. In many practical examples, however, this is not necessary; ConTeXt redefines \displaylines and \eqalignno to do this automatically.
For more control over sub-formula numbering, \formulanumber and \subformulanumber have an optional argument parallel to that of \placeformula, as demonstrated in this use of plain TeX's \eqalignno, which places multiple equation numbers within one formula.
\placeformula \startformula \eqalignno{ c^2 &= a^2 + b^2 &\formulanumber{a} \cr c &= \left(a^2 + b^2\right)^{\vfrac{1}{2}} &\subformulanumber{b}\cr a^2 + b^2 &= c^2 &\subformulanumber{c} \cr d^2 &= e^2 &\formulanumber\cr} \stopformula
Note that both \formulanumber and \subformulanumber can be used within the same formula, and the formula number is incremented as expected. Also, if an optional argument is specified in both \placefigure and \formulanumber, the latter takes precedence.
More examples for left-located equation number: \setupformulas[location=left] \placeformula{d} \startformula \let\doplaceformulanumber\empty c^2 = a^2 + b^2 \leqno{\formulanumber} \stopformula and \placeformula \startformula \leqalignno{c^2 &= a^2 + b^2 &\formulanumber{a} \cr a^2 + b^2 &= c^2 &\subformulanumber{b} \cr d^2 &= e^2 &\formulanumber\cr} \stopformula
-- 23:46, 15 Aug 2005 (CEST) Prinse Wang
# List of Formulas
You can have a list of the formulas contained in a document by using \placenamedformula instead of \placeformula. Only the formulas written with \placenamedformula are not put in the list, so that you can control precisely the content of the list.
\placenamedformula takes as first parameter the name of the formula put in the list. The other \placeformula features are still available. The list can be formatted like any other list.
Example:
\subsubject{List of Formulas}
\placelist[formula][criterium=text,alternative=c]
\subsubject{Formulas}
\placenamedformula[one]{First listed Formula}
\startformula a = 1 \stopformula \endgraf
\placeformula
\startformula a = 2 \stopformula \endgraf
\placenamedformula{Second listed Formula}{b}
\startformula a = 3 \stopformula \endgraf
Gives:
# Shaded background for part of a displayed equation
To highlight part of a formula, you can give it a gray background using \mframed: the following is the code you can use in mkii (see below what one has to do in mkiv):
\setuppapersize[A5]
\setupcolors[state=start]
\def\graymath{\mframed[frame=off,
background=color,
backgroundcolor=gray,
backgroundoffset=3pt]}
\startformula
\ln (1+x) =\, \graymath{x - {x^2\over2}} \,+ {x^3\over3}-\cdots.
\stopformula
In mkiv the code is slightly different: one may define \graymath directly using \definemathframed
\setuppapersize[A5]
\definemathframed[graymath]
[
frame=off,
location=mathematics,
background=color,
backgroundcolor=lightgray,
backgroundoffset=2pt
]
\starttext
Since for $|x| < 1$ we have
\startformula
\log(1+x) = \graymath{x- \displaystyle{x^2\over2}} + {x^3 \over 3} + \cdots
\stopformula
we may write $\log(1+x) = x + O(x^2)$.
\stoptext
The result is shown below (possibly the framed part of the formula is not aligned correctly with the remainder of the formula because the mkiv engine on Context Garden is not up to date…). | 2,953 | 10,888 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-24 | latest | en | 0.772424 |
https://www.physicsforums.com/threads/bayes-theory-confusion.607965/ | 1,540,187,434,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514708.24/warc/CC-MAIN-20181022050544-20181022072044-00281.warc.gz | 1,049,150,395 | 15,173 | # Bayes' Theory Confusion
1. May 22, 2012
### WCMU101
I'm struggling with Bayes' theory. Please consider the following:
Example 1:
Submarine sinks if one missile hits it. Two ships aim at submarine and fire one missile each. Ship 1 shoots missile X1, ship 2 shoots missile X2.
P(X1 hitting = 0.8)
P(X2 hitting = 0.5)
P(X1 and X2 hitting | submarine seen sinking) = P(submarine seen sinking | X1 and X2 hitting)*P(X1 and X2 hitting)/P(submarine seen sinking)
P(X1 and X2 hitting | submarine seen sinking) = 1*(.4)/(1-.5*.2) = .4/.9 = .444444...
Example 2:
Same as example one except:
P(X1 hitting = 0.5)
P(X2 hitting = 0.5)
Thus,
P(X1 and X2 hitting | submarine seen sinking) = P(submarine seen sinking | X1 and X2 hitting)*P(X1 and X2 hitting)/P(submarine seen sinking)
P(X1 and X2 hitting | submarine seen sinking) = 1*(.25)/(.75) = .3333....
However, some would argue that in example two, P(X1 and X2 hitting | submarine seen sinking) = 0.5 - since you could say we know for sure that 1 hit, thus the probability that the other one hit is still 0.5 (independent events). Who is correct?
The reason I showed the first example, is because I don't see how you could come up with an answer without Bayes' in the first example - you don't know which one hit???
Let me show you another example which I posted in an earlier thread (https://www.physicsforums.com/showthread.php?t=607946):
Example 3:
2 fair dice. What is the probability of both showing six if I have observed
at least one six.
Thus, the way I did it was:
P(2 6's | observing at least 1 6) = P(observing at least 1 6 | observing 2 6's)*P(2 6's)/P(observing at least 1 6)
P(2 6's | observing at least 1 6) = 1*(1/36)/(1-25/36) = 1/11
However, as the person who replied to my post pointed out, an answer of 1/6 here would appear to be the logical answer - knowing that the first dice came up 6, the probability that the second dice came up 6 is still 1/6 - independent events.
So, as you can see, I really am struggling with Bayes' theory here. Could somebody please help me out of the darkness!
Thanks.
Nick.
Last edited: May 22, 2012
2. May 22, 2012
### haruspex
No, that doesn't work. Knowing that at least one hit is not the same as knowing a particular one hit. This is a classic stumbling block in probability. It probably has a name.
Let's make it as simple as possible. I toss a coin twice. It came up heads at least once. What's the probability it was heads both times?
A priori, we have HH, HT, TH, TT, equally likely.
The given information rules out only TT, leaving two head & tail combinations to one HH.
But if instead you'd been told the first was a head that would rule out TT and TH, leaving HH and HT.
3. May 22, 2012
### viraltux
precisely because you don't know which one hit you have these results, you can calculate both "without" Bayes dividing all favorable outcomes between all possible outcomes, for instance:
0.8*0.5 / ( 0.8*0.5 + 0.8*0.5 + 0.2*0.5) = 4/9
0.5*0.5 / (0.5*0.5 + 0.5*(1-0.5) + (1-0.5)*0.5) = 1/3
Your problem with the theorem of Bayes is in how you interpret it, when you wonder who is correct, well, both, because both measure different things, 1/3 is the probability that both missiles hit if the submarine is sinking and you don't know if it was hit by both missiles or just one, if you know one missile hit, let's say X1, then 1/2 is the probability X2 hit the submarine too.
4. May 22, 2012
### WCMU101
Thank you both very much - those answers were excellent! My headache has disappeared. Also, thank you haruspex for replying to my other posts - I should get you to do my phone interview for me haha
Thanks again!
Nick.
5. May 22, 2012
### haruspex
That doesn't seem right. If you know exactly one hit then the probability the other hit as well is zero.
The key issue is whether you know that a particular missile hit (the first one, the red one, the one that had further to go, whatever) or merely that at least one did. This seems paradoxical because the puzzle is usually worded as for volunteered information. If Joe tells me the red missile hit, why did he choose to word it that way? The interpretation, for the purposes of the puzzle, is that Joe decided to tell me the fate of the red missile, whatever that was. But I could instead assume that he was really just telling me that at least one had hit, and if both hit then he would have mentally tossed a coin to decide which one to mention. In that view, specifying the red missile changes nothing.
The puzzle is more robust if structured as question and answer:
Scenario 1: Joe, did the red missile hit? A: Yes.
Scenario 2: Joe, did either missile hit? A: Yes.
6. May 22, 2012
### viraltux
I think haruspex wants to go philosophy on my ***
Be careful WCMU101, don't do that in your interview! | 1,345 | 4,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.954961 |
https://library.curriki.org/oer/?rid=87896 | 1,660,096,607,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00385.warc.gz | 336,887,687 | 19,841 | Function graphs
Supports the following Standards of Mathematical Practice
• 2 Reason abstractly and quantitatively.
• 4 Model with mathematics.
• 5 Use appropriate tools strategically.
• 6 Attend to precision.
## Collection Contents
### Understanding Functions
by Allen Wolmer
A YouTube video from Robert WirtzSupports the following Standards of Mathematical Practice• 2 Reason abstractly and quantitatively.• 4 Model with mathematics.• 5 Use appropriate tools strategically.• 6 Attend to precision.
Member Rating
Curriki RatingC
'C' - Curriki rating
### Understanding and comparing functions? Lesson 1
by Allen Wolmer
A YouTube video from LearnZillionSupports the following Standards of Mathematical Practice• 2 Reason abstractly and quantitatively.• 4 Model with mathematics.• 5 Use appropriate tools strategically.• 6 Attend to precision.
Member Rating
Curriki RatingC
'C' - Curriki rating
### Understanding and comparing functions? Lesson 2
by Allen Wolmer
A YouTube video from LearnZillionSupports the following Standards of Mathematical Practice• 2 Reason abstractly and quantitatively.• 4 Model with mathematics.• 5 Use appropriate tools strategically.• 6 Attend to precision.
Member Rating
Curriki RatingC
'C' - Curriki rating
### Understanding and comparing functions Lesson 3
by Allen Wolmer
A YouTube video from LearnZillionSupports the following Standards of Mathematical Practice• 2 Reason abstractly and quantitatively.• 4 Model with mathematics.• 5 Use appropriate tools strategically.• 6 Attend to precision.
Member Rating
Curriki RatingC
'C' - Curriki rating
### Introduction to Functions
by Sal Khan
This video provides an introduction to the notion of a function.
Member Rating
Curriki RatingC
'C' - Curriki rating
### Functions Part 2
by Sal Khan
This video continues to explore the properties of a function.
Member Rating
Curriki RatingC
'C' - Curriki rating
Non-profit Tax ID # 203478467 | 437 | 1,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-33 | latest | en | 0.777799 |
https://www.exceltip.com/summing/how-to-sum-column-in-a-excel-by-matching-heading.html | 1,679,619,801,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945218.30/warc/CC-MAIN-20230323225049-20230324015049-00667.warc.gz | 856,993,462 | 19,355 | How to Sum Column in a Excel by Matching Heading
If you want to get the sum of a column by just using the column name, you can do this in 3 easy ways in Excel. Let's explore these ways.
Unlike other articles, let's see the scenario first.
Here I have a table of sales done by different salesmen in different months.
Now the task is to get the sum of given month's sales in Cell C10. If we change the month in B10, the sum should change and returns that month's sum, without changing anything in the formula.
Method 1: Sum Whole Column in Table Using SUMPRODUCT function.
The syntax of the SUMPRODUCT method to sum matching column is:
Columns: It is the 2-dimensional range of the columns that you want to sum. It should not contain headers. In the table above it is C3:N7.
Headers: It is the header range of columns that you want to sum. In the above data, it is C2:N2.
Heading: It is the heading that you want to match. In the example above, it is in B10.
Without further delay let's use the formula.
=SUMPRODUCT((C3:N7)*(C2:N2=B10))
and this will return:
How does it work?
It is simple. In the formula, the statement C2:N2=B10 returns an array that contains all FALSE values except one that matches B10. Now the formula is
=SUMPRODUCT((C3:N7)*{FALSE,FALSE,FALSE,FALSE,TRUE,FALSE,FALSE,FALSE,FALSE,FALSE,FALSE,FALSE})
Now C3:N7 is multiplied to each value of this array. Every column becomes zero except the column that is multiplied by TRUE. Now the formula becomes:
=SUMPRODUCT({0,0,0,0,6,0,0,0,0,0,0,0;0,0,0,0,12,0,0,0,0,0,0,0;0,0,0,0,15,0,0,0,0,0,0,0;0,0,0,0,15,0,0,0,0,0,0,0;0,0,0,0,8,0,0,0,0,0,0,0})
Now this array is summed up, and we get the sum of the column that matches the column in the cell B10.
Method 2: Sum Whole Column in Table Using INDEX-MATCH function.
The syntax of the method to sum the matching column heading in excel is:
All the variables in this method are the same as the SUMPRODUCT method. Let's just implement it to solve the problem. Write this formula in C10.
=SUM(INDEX(C3:N7,,MATCH(B10,C2:N2,0)))
This returns:
How does it work?
The formula is solved inside out. First, the MATCH function returns the index of the matching month from the range C2:N2. Since we have May in B1o, we get 5. Now the formula becomes
=SUM(INDEX(C3:N7,,5))
Next, the INDEX function returns values from the 5th column of C3:N7. Now the formula becomes:
=SUM({6;12;15;15;8})
And finally, we get the sum of these values.
Method 3: Sum Whole Column in Table Using Named Range and INDIRECT function
Everything get's simple if you name your ranges as column headings. In this method, we first need to name the columns as their heading names.
Select the table including the headings and press CTRL+SHIFT+F3. It will open a dialog to create a name from the ranges. Check the top row and hit the OK button.
It will name all the data columns as their headings.
Now the generic formula to sum the matching column will be:
Heading: It is the name of the column that you want to sum. In this example, it is B10 that contains may as of now.
To implement this generic formula, write this formula in cell C10.
=SUM(INDIRECT(B10))
This returns the sum of May month:
Another method is similar to this. In this method, we use excel tables and it's structured naming. Let's say if you have named the above table as table1. Then this formula will work the same as the above formula.
=SUM(INDIRECT("Table1["&B10&"]"))
How does it work?
In this formula, the INDIRECT function takes the reference of the name and converts it into actual name reference. The procedure onwards is simple. The SUM function sums up the named range.
So yeah guys, this how you can sum the matching column in excel. I hope it is helpful and explanatory to you. If you have any doubts regarding this article or any other excel/VBA related topic, ask in the comments section below.
Related Articles:
How to Sum by Matching Row and Column in ExcelThe SUMPRODUCT is the most versatile function when it comes to sum and count values with tricky criteria. The generic function to sum by matching column and row is...
SUMIF with 3D Reference in Excel | The fun fact is that the normal Excel 3D referencing does not work with conditional functions, like SUMIF function. In this article, we will learn how to get 3D referencing working with SUMIF function.
Relative and Absolute Reference in Excel | Referencing in excel is an important topic for every beginner. Even experienced excel users do mistakes in referencing.
Dynamic Worksheet Reference | Give reference sheets dynamically using the INDIRECT function of excel. This is simple...
Expanding References in Excel | The expanding reference expands when copied down or rightwards. We use the \$ sign before the column and row number to do so. Here is one example...
All About Absolute Reference | The default reference type in excel is relative but if you want the reference of cells and ranges to be absolute use the \$ sign. Here are all the aspects of absolute referencing in Excel.
Popular Articles:
50 Excel Shortcuts to Increase Your Productivity | Get faster at your task. These 50 shortcuts will make your work even faster on Excel.
The VLOOKUP Function in Excel | This is one of the most used and popular functions of excel that is used to lookup value from different ranges and sheets.
COUNTIF in Excel 2016 | Count values with conditions using this amazing function. You don't need filter your data to count specific value. Countif function is essential to prepare your dashboard.
How to Use SUMIF Function in Excel | This is another dashboard essential function. This helps you sum up values on specific conditions.
Terms and Conditions of use
The applications/code on this site are distributed as is and without warranties or liability. In no event shall the owner of the copyrights, or the authors of the applications/code be liable for any loss of profit, any problems or any damage resulting from the use or evaluation of the applications/code. | 1,457 | 6,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-14 | latest | en | 0.871797 |
https://www.mathcelebrity.com/community/threads/let-f-x-3x-6-and-g-x-2x-5-which-of-the-following-is-f-x-g-x.3173/ | 1,686,277,399,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00263.warc.gz | 935,774,089 | 9,061 | # Let f(x) = 3x - 6 and g(x)= -2x + 5. Which of the following is f(x) - g(x)?
Discussion in 'Calculator Requests' started by math_celebrity, Sep 9, 2020.
Tags:
Let f(x) = 3x - 6 and g(x)= -2x + 5. Which of the following is f(x) - g(x)?
f(x) - g(x) = 3x - 6 - (-2x + 5)
Distribute the negative sign where double negative equals a plus:
f(x) - g(x) = 3x - 6 + 2x - 5
Combine like terms:
f(x) - g(x) = (3 + 2)x - 6 - 5
f(x) - g(x) = 5x - 11 | 189 | 443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-23 | longest | en | 0.782877 |
https://gitlab.inria.fr/why3/why3/-/commit/de7a797b8393a8a666b92fffe0c5e155b2554e19 | 1,624,334,209,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488507640.82/warc/CC-MAIN-20210622033023-20210622063023-00437.warc.gz | 259,479,050 | 54,486 | Commit de7a797b by MARCHE Claude
### Updated sessions for easier replay
parent 414a7182
This diff is collapsed.
... ... @@ -239,7 +239,7 @@ ... ... @@ -256,247 +256,247 @@ ... ... @@ -505,6 +505,9 @@ ... ... @@ -512,7 +515,7 @@ ... ... @@ -521,133 +524,130 @@ ... ... @@ -665,7 +665,7 @@ ... ... @@ -854,8 +854,8 @@ ... ...
... ... @@ -72,8 +72,8 @@ Definition distance (d:Z) (i:Z): Prop := (path d i) /\ forall (d':Z), (path d' i) -> (d <= d')%Z. Require Import Why3. Ltac ae := why3 "Alt-Ergo,0.95.2," timelimit 5. Ltac z3 := why3 "Z3,4.3.1," timelimit 5. Ltac ae := why3 "Alt-Ergo,0.95.2," timelimit 30. Ltac z3 := why3 "Z3,4.3.1," timelimit 30. (* Why3 goal *) Theorem WP_parameter_distance : let o := n in ((0%Z <= o)%Z -> forall (g:Z) ... ... @@ -102,6 +102,9 @@ Theorem WP_parameter_distance : let o := n in ((0%Z <= o)%Z -> forall (g:Z) ((count < n)%Z -> forall (k:Z), (((0%Z < k)%Z \/ (0%Z = k)) /\ (k < n)%Z) -> forall (d':Z), (path d' k) -> ((map.Map.get d2 k) <= d')%Z))))). (* Why3 intros o h1 g g1 (h2,(h3,h4)) h5 g2 (h6,h7) o1 h8 d d1 (h9,(h10,h11)) o2 h12 count d2 g3 ((h13,(h14,h15)),(h16,h17)) h18 k (h19,h20) d' h21. *) (* intros o _ _ h3 g _ o1 h4 h5 o2 h6 count d g1 ((h7,(h8,h9)),(h10,h11)) h12 k (h13,h14) d' h15. ... ...
... ... @@ -4,10 +4,10 @@ ... ... @@ -97,7 +97,7 @@ ... ...
... ... @@ -114,7 +114,7 @@ ... ... @@ -642,8 +642,13 @@ ... ...
No preview for this file type
... ... @@ -2,12 +2,13 @@ | 613 | 1,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-25 | latest | en | 0.165296 |
http://lutanho.net/diagram/logarithmic_scale.html | 1,638,942,418,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00061.warc.gz | 57,230,403 | 2,060 | ### JavaScript Diagram Builder - Logarithmic Scale Example
To draw a coordinate diagram with a logarithmic scale, define a function-scale (like shown in the example below) and use GridDelta=1 (increasing values) or GridDelta=-1 (decreasing values) and use SubGrids=-1 to get a logatithmic sub-grid. Use the logarithmic values in the functions SetBorder and ScreenX or ScreenY, as it is shown below.
This diagram was generated by
<SCRIPT Language="JavaScript">
Hosts=new Array(28174, 80000, 290000, 500000, 727000, 1200000, 2217000, 4852000, 9472000,
16146000, 29670000, 43230000, 72398092, 109574429, 147344723);
function LogScale(vv)
{ if ((vv>3)||(vv<-3)) return("10<sup>"+vv+"</sup>");
if (vv>=0) return(Math.round(Math.exp(vv*Math.LN10)));
else return(1/Math.round(Math.exp(-vv*Math.LN10)));
}
document.open();
var D=new Diagram();
D.SetFrame(100, 140, 580, 460);
D.SetBorder(1988, 2002, Math.log(Hosts[0])/Math.LN10, Math.log(Hosts[14])/Math.LN10);
D.SetText("Year", "Hosts", "<B>Internet growth</B>");
D.XGridDelta=2;
D.XSubGrids=2;
D.YGridDelta=1;
D.YSubGrids=-1;
D.YScale="function LogScale";
D.SetGridColor("#FFFFFF", "#EEEEEE");
D.Draw("#DDDDDD", "#000000", true);
for (var n=1; n<Hosts.length; n++)
{ new Line(D.ScreenX(1987+n), D.ScreenY(Math.log(Hosts[n-1])/Math.LN10),
D.ScreenX(1988+n), D.ScreenY(Math.log(Hosts[n])/Math.LN10), "#0000ff", 2, "internet hosts");
}
for (n=0; n<Hosts.length; n++)
{ new Dot(D.ScreenX(1988+n), D.ScreenY(Math.log(Hosts[n])/Math.LN10), 10, 1, "#ff0000",
eval(1988+n)+": "+Hosts[n]+" hosts");
}
document.close();
</SCRIPT>
« Relative Position Browser Support » | 538 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.368225 |
http://lists.electorama.com/pipermail/election-methods-electorama.com/2002-October/074072.html | 1,686,204,577,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654097.42/warc/CC-MAIN-20230608035801-20230608065801-00396.warc.gz | 24,944,685 | 2,988 | # [EM] Matrix Representation of Election Methods
Alex Small asmall at physics.ucsb.edu
Sun Oct 20 16:26:20 PDT 2002
```I'm homing in on a strategy for figuring out if strong FBC can ever be
satisfied (aside from Random Ballot). At least initially, it's easiest if
I restrict the class of methods under consideration, and also restrict the
number of candidates to 3.
All methods will be ranked methods. The electorate can be specified by a
vector
E = [N(A>B>C), N(A>C>B), N(B>A>C), N(B>C>A), N(C>A>B), N(C>B>A)]
where each element specifies the number of voters with a particular
preference order. It is convenient, but not necessary, to impose the
condition
E*[1,1,1,1,1,1] = 1
All ranked election methods that I'm aware of can be reduced to a set of
linear inequalities involving the elements of E. For instance, in
plurality voting the criteria for A's victory are
[1,1,-1,-1,0,0]*E > 0 (A gets more first-place votes than B)
[1,1,0,0,-1,-1]*E > 0 (A gets more first-place votes than C)
This can be expressed in matrix form.
| 1 1 -1 -1 0 0 | | x |
E* | | = | |
| 1 1 0 0 -1 -1 | | y |
where x>0, y>0. More generally, we can always require that E*(specified
matrix) gives a vector with all elements greater than zero.
The victory conditions for B and C can be obtained by multiplying the 6x2
matrix above by a 6x6 matrix (which I won't write out) that "swaps" A and
B or A and C. Some methods might require more matrices. For instance,
Condorcet methods would have two possible victory conditions for A: A is
the CW, or A satisfies some condition for breaking cyclic ambiguities (we
might need two conditions for ambiguities, one for the case A>B>C>A and
the other for C>B>A>C, but that's a minor point).
Restricting my attention to such "linear" methods, which can be expressed
in matrix form, will make life much easier. It will allow me to use
linear algebra to explore strong FBC, and physics grad students know
linear algebra like the back of our teeth (not willingly, but it's beaten
into our brains during Quantum Mechanics). Obviously this simplification
comes at the expense of generality, but it's a starting point. In
particular, it gives an easy formalism for discussing favorite betrayals:
Insincere voting can be represented by adding to E some vector d with only
two non-zero elements, both elements having equal magnitude and opposite
sign. For instance, if some people in the A>B>C faction switch to the
B>A>C faction for strategic advantage, we can express this as
E(new) = E + [-n,0,n,0,0,0]
Not all such representations of insincere voting are favorite betrayals,
of course.
I want to end this brain-dump session with a question:
Can anybody think of ranked election methods that can't be expressed in
matrix form, i.e. with linear inequalities? I'd think that such methods
are rather absurd, but I'm not sure that they can be ruled out.
Alex
---- | 792 | 2,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-23 | latest | en | 0.901487 |
http://gmatclub.com/forum/each-december-31-in-country-q-a-tally-is-made-of-the-16336.html?fl=similar | 1,484,632,676,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279468.17/warc/CC-MAIN-20170116095119-00243-ip-10-171-10-70.ec2.internal.warc.gz | 115,673,014 | 59,893 | Each December 31 in Country Q, a tally is made of the : GMAT Critical Reasoning (CR)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 16 Jan 2017, 21:57
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Each December 31 in Country Q, a tally is made of the
Author Message
TAGS:
### Hide Tags
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7
Kudos [?]: 98 [0], given: 0
Each December 31 in Country Q, a tally is made of the [#permalink]
### Show Tags
11 May 2005, 21:51
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
Each December 31 in Country Q, a tally is made of the country’s total available coal supplies—that is, the total amount of coal that has been mined throughout the country but not consumed. In 1991 that amount was considerably lower than it had been in 1990. Furthermore, Country Q has not imported or exported coal since 1970.
If the statements above are true, which one of the following must also be true on the basis of them?
(A) In Country Q, more coal was mined in 1990 than was mined in 1991.
(B) In Country Q, the amount of coal consumed in 1991 was greater than the amount of coal mined in 1991.
(C) In Country Q, the amount of coal consumed in 1990 was greater than the amount of coal consumed in 1991.
(D) In Country Q, the amount of coal consumed in 1991 was greater than the amount of coal consumed in 1990.
(E) In Country Q, more coal was consumed during the first half of 1991 than was consumed during the first half of 1990.
If you have any questions
New!
Manager
Joined: 22 Apr 2005
Posts: 129
Location: Los Angeles
Followers: 1
Kudos [?]: 5 [0], given: 0
### Show Tags
11 May 2005, 23:45
Ambiguous question - choosing the obvious.
D
Manager
Joined: 08 Mar 2005
Posts: 101
Followers: 1
Kudos [?]: 1 [0], given: 0
### Show Tags
12 May 2005, 00:48
Assuming available coal in 1990 was 100 kg. according to the statement available coal in 1991 must be less than 100 i.e 95.
So that means that not only the coal mined in 1991 was consumed but also some of the coal available from previous year 1990 was consumed.
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7
Kudos [?]: 98 [0], given: 0
### Show Tags
12 May 2005, 05:16
meenu wrote:
The answer should be 'B'. Assuming available coal in 1990 was 100 kg. according to the statement available coal in 1991 must be less than 100 i.e 95. So that means that not only the coal mined in 1991 was consumed but also some of the coal available from previous year 1990 was consumed.
does the passage support for this reasoning?
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 30
Kudos [?]: 355 [0], given: 0
### Show Tags
12 May 2005, 05:27
Only B fits the statements given in the argument. If more coal was consumed in 1991 than the amt of coal mined in that year, then the total available coal supply would be less than 1990. (since the definition of the total amt is the am to coal mined but not consumed)
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 30
Kudos [?]: 355 [0], given: 0
### Show Tags
12 May 2005, 05:28
Tyr wrote:
Ambiguous question - choosing the obvious.
D
(D) In Country Q, the amount of coal consumed in 1991 was greater than the amount of coal consumed in 1990.
Could well be that amt of coal mined in 1991 was less to begin with, so D is not strong.
Director
Joined: 27 Dec 2004
Posts: 905
Followers: 1
Kudos [?]: 43 [0], given: 0
### Show Tags
12 May 2005, 05:43
It is B.
Total available supplies = mined not consumed
If the Total available in 1991 is lower than the Total available 1990, then it makes sense that the total consumed in 1991 is greater than the total consumed in 1990.
"cause and effect"
VP
Joined: 18 Nov 2004
Posts: 1440
Followers: 2
Kudos [?]: 37 [0], given: 0
### Show Tags
12 May 2005, 06:55
This is an ambiguous question i.e. poorly worded one. If u go by what is written, none of the answers has to be correct. This has been discussed here before, let's see if I can find that thread.
Senior Manager
Joined: 07 Nov 2004
Posts: 458
Followers: 2
Kudos [?]: 87 [0], given: 0
### Show Tags
12 May 2005, 11:14
I still remember Chunjunwu posted this question before and the answer is B. Anyhow, the stem sounds ambiguous clear to me.
VP
Joined: 25 Nov 2004
Posts: 1493
Followers: 7
Kudos [?]: 98 [0], given: 0
### Show Tags
12 May 2005, 16:58
banerjeea_98 wrote:
This is an ambiguous question i.e. poorly worded one. If u go by what is written, none of the answers has to be correct. This has been discussed here before, let's see if I can find that thread.
yes. i also think so and this the reason i posted it. however, OA is B.
12 May 2005, 16:58
Similar topics Replies Last post
Similar
Topics:
3 OG12 Q31 5 30 Sep 2013, 20:03
18 Each December 31 in Country Q, a tally is made of the 25 22 Jan 2011, 02:18
Each December 31 in Country Q, a tally is made of the 17 12 Feb 2010, 04:14
Seven countries signed a treaty binding each of them to 7 03 Jul 2007, 12:20
Seven countries signed a treaty binding each of them to 4 04 Apr 2007, 15:38
Display posts from previous: Sort by | 1,695 | 5,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-04 | latest | en | 0.947986 |
http://docplayer.net/22262745-Teaching-basic-facts-considerations-for-instruction.html | 1,539,795,364,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511203.16/warc/CC-MAIN-20181017153433-20181017174933-00075.warc.gz | 103,729,232 | 24,518 | # Teaching Basic Facts: Considerations for Instruction
Save this PDF as:
Size: px
Start display at page:
## Transcription
4 Common Areas of Difficulty Prerequisite skills not mastered: Knowledge of numbers and what number represent. Counting. Specific Basic Facts skills: Understanding symbols. Fluency with facts. Strategies to calculate answers. About the Basic Facts Addition: 100 facts. Single-digit addend + single-digit addend = sum = 7; = 17 Subtraction: 100 facts. Minuend single-digit subtrahend = single-digit difference. 8 2 = 6; 13 9 = 4 Multiplication: 100 facts. Single-digit factor single-digit factor = product. 2 7 = 14; 8 6 = 48 Division: 90 facts. Dividend single-digit divisor = single-digit quotient. 9 3 = 3; 56 7 = 8 Developing Conceptual Understanding Manipulatives can be used to help students understand the concepts behind the basic facts. Examples of manipulatives include: Unifix cubes, plastic clips, chips, or dominoes. 4 Basic Facts: Considerations for Instruction
5 Activities and Strategies Related to Specific Standards Represent addition and subtraction. (K) Use objects (e.g., chips, fingers) to show two groups. Put the two groups together. Use objects (e.g., chips, fingers) to show one group. Take an amount away from the group. Tell stories to show addition (i.e., putting together, adding on). Tell stories to show subtraction (i.e., taking away, comparing). Use a number line. Move forward for addition. Move backwards for subtraction. (Also, compare the difference between two numbers for subtraction.) Decompose numbers less than or equal to 10 into pairs in more than one way. (K) Use two different colors of the same manipulative to show all possible combinations of a specific number = = = = = = = = = = = = 5 Demonstrate commutative property of addition; the order of the addends does not matter. For example, is the same as Basic Facts: Considerations for Instruction 5
6 For any number from 1 to 9, find the number that makes 10 when added to the given number. (K) Use a Tens Frame and manipulatives to teach all possible combinations that make 10. Use a set of 10 clips or 10 cubes to teach all combinations that make 10. Use fingers. Start with some (e.g., 3) fingers held up. Count up to 10. Learn operation symbols (plus, minus, and equal signs) and how to use equation notation. Solve addition and subtraction word problems within 10. (K) Use addition and subtraction within 20 to solve word problems. (1) Use addition and subtraction within 100 to solve one- and two-step word problems involving situations. (1) Present word problems written and orally. Provide situations that involve: Adding to Maisie has 4 buttons. She buys 3 buttons at the store. How many buttons does Maisie have now? Taking from Maisie had 9 buttons, and then her brother took 2 of them. How many buttons does Maisie have left? Putting together Maisie has 5 buttons. Jamey has 4 buttons. How many buttons do they have altogether? Comparing Maisie has 8 buttons and Jamey has 2 buttons. How many fewer buttons does Jamey have than Maisie? Use stories where the unknown is in all positions. For the comparing problem from above: Maisie has 8 buttons and Jamey has 2 buttons. How many fewer buttons does Jamey have than Maisie? (unknown: difference) Maisie has 8 buttons. She has 6 more buttons than Jamey. How many buttons does Jamey have? (unknown: subtrahend) 6 Basic Facts: Considerations for Instruction
7 Jamey has 2 buttons. Maisie has 6 more buttons than Jamey. How many buttons does Maisie have? (unknown: minuend) Act problems out with objects or students. Determine whether a group of objects (up to 20) has an odd or even number of members. (2) Practice skip counting by twos. Start at 0 to skip count even numbers. Start at 1 to skip count odd numbers. Write an equation to express an even number as a sum of two equal addends. (2) Teach the addition doubles with rhymes or chants. Look at doubles patterns on an addition chart Basic Facts: Considerations for Instruction 7
8 Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns. (2) Write an equation to express the total as a sum of equal addends. (2) Arrange objects in arrays and skip count rows or columns. This helps with learning multiplication as repeated addition. Learn operation symbols (multiplication, division, and equal signs) and how to use equation notation. Interpret products of whole numbers. (3) Use objects (e.g., chips and plates) to show a number of groups and the number within each group. Interpret whole-number quotients of whole numbers. (3) Use objects (e.g., chips and plates) to show an amount divided evenly into groups. 8 Basic Facts: Considerations for Instruction
11 Determine the unknown whole number in a multiplication or division equation relating three whole numbers. (3) Introduce unknowns within stories and using manipulatives. Apply properties of operations as strategies to multiply and divide. (3) Teach the commutative property of multiplication. Teach how the commutative property can help students solve multiplication and division facts. Say, If you know 3 8 = 24, you also know that 8 3 = 24. Understand division as an unknown-factor problem. (3) Teach the reciprocal property of multiplication and division. Say, If you know 7 5 = 35, you know that 35 7 = 5 and 35 5 = 7. Teach of thinking of division as, What number do I multiply to get to? Let s see. 21 divided by 7. What number can I multiply to 7 to get to 21? I can count by 7. 7, 14, 21. I counted 7 3 times. 21 divided by 7 is c_07/15 Basic Facts: Considerations for Instruction 11
### Basic Facts INTENSIVE INTERVENTION. National Center on. at American Institutes for Research
National Center on INTENSIVE INTERVENTION at American Institutes for Research Basic Facts 1000 Thomas Jefferson Street, NW Washington, DC 20007 E-mail: NCII@air.org While permission to reprint this publication
### Common Core State Standards for Math Grades K - 7 2012
correlated to the Grades K - 7 The Common Core State Standards recommend more focused and coherent content that will provide the time for students to discuss, reason with, reflect upon, and practice more
### ALIGNMENT OF MATH PERSPECTIVES RESOURCES WITH COMMON CORE STATE STANDARDS IN MATHEMATICS
ALIGNMENT OF MATH PERSPECTIVES RESOURCES WITH COMMON CORE STATE STANDARDS IN MATHEMATICS KINDERGARTEN COUNTING AND CARDINALITY Count to tell the number of objects Understand the relationship between numbers
### Sample Computation Concept:
Sample Computation Concept: Use Place-Value Understanding and Properties of Addition and Subtraction to Add and Subtract College and Career Ready Standards Addressed 2.NBT.7 Add and subtract within 1,000,
### Unit Plan Components. Appropriate Length of Time. Unit Name: Addition and Subtraction to 20. Common Core Standards
Unit Plan Components Big Goal Standards Big Ideas Unpacked Standards Scaffolded Learning Resources Appropriate Length of Time Aligned Assessment Comments This unit plan identifies a big goal for the overall
### Computation Strategies for Basic Number Facts +, -, x,
Computation Strategies for Basic Number Facts +, -, x, Addition Subtraction Multiplication Division Proficiency with basic facts aids estimation and computation of multi-digit numbers. The enclosed strategies
### Computation of Fractions
National Center on INTENSIVE INTERVENTION at American Institutes for Research Computation of Fractions 000 Thomas Jefferson Street, NW Washington, DC 0007 E-mail: NCII@air.org While permission to reprint
### K 3 Mathematics Core Course Objectives
K 3 Mathematics Core Course Objectives The Massachusetts Department of Elementary and Secondary Education (ESE) partnered with WestEd to convene panels of expert educators to review and develop statements
### Common Core State Standards DECONSTRUCTED. Booklet I: Kindergarten to Second Grade, Math FOR INTERNAL USE ONLY
Common Core State Standards DECONSTRUCTED Booklet I: Kindergarten to Second Grade, Math How to use this booklet You cannot teach a Common Core Standard you must teach the skills inside of each standard.
### Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve word problems that call for addition of three whole numbers
### ~ Gateway Regional School District~ Unit Plan
~ Gateway Regional School District~ Unit Plan (revised 2/15/13) Content Area: Go Math! Unit: Multiplication Facts and Strategies Grade(s): 3 Time Line: 13 days Date: December Domain/Content Standard(s):
### Sample Fraction Addition and Subtraction Concepts Activities 1 3
Sample Fraction Addition and Subtraction Concepts Activities 1 3 College- and Career-Ready Standard Addressed: Build fractions from unit fractions by applying and extending previous understandings of operations
### Kindergarten Common Core Standards & Learning Targets
Kindergarten Common Core Standards & Learning Targets CCS Standards: Counting and Cardinality K.CC.1. Count to 100 by ones and by tens. K.CC.2. Count forward beginning from a given number within the known
### Math Content by Strand 1
Math Content by Strand 1 Number and Operations with Whole Numbers Multiplication and Division Grade 3 In Grade 3, students investigate the properties of multiplication and division, including the inverse
### 3: Q I know all of the Simple and Complex Learning Goals and my understanding goes beyond the grade level target.
Topic Proficiency Scale Domain: Operations and Algebraic Thinking Critical Area: Multiplication & Division within 100 EXPECTED MASTERY: 7 s, 8 s, and 9 s Facts 4.0 I know all of the Simple and Complex
### Fractions as Numbers INTENSIVE INTERVENTION. National Center on. at American Institutes for Research
National Center on INTENSIVE INTERVENTION at American Institutes for Research Fractions as Numbers 000 Thomas Jefferson Street, NW Washington, DC 0007 E-mail: NCII@air.org While permission to reprint this
### Math Journal HMH Mega Math. itools Number
Lesson 1.1 Algebra Number Patterns CC.3.OA.9 Identify arithmetic patterns (including patterns in the addition table or multiplication table), and explain them using properties of operations. Identify and
### Math Content by Strand 1
Math Content by Strand 1 Number and Operations: Whole Numbers Addition and Subtraction and the Number System Kindergarten Young students develop their understanding of the operations of addition and subtraction
### Nursery Math Assessment
Nursery Math Assessment Name: Age: Class: The Counting Board Building the Stair Puts away blocks 1-10 when chosen randomly (by looking at the size of the block) Puts away blocks 1-10 sequentially (+1 sequence)
### Content Elaborations. Standards
Grade Two Mathematics Domain Operations and Algebraic Thinking Cluster Represent and solve problems involving addition and subtraction Pacing Quarter 1: Stepping Stones Modules 1, 2, 3 Quarter 2: Stepping
### Grade 3: Module 1 Parent Letter. What s It All About?
Grade 3: Module 1 Parent Letter What s It All About? In this first module, students review addition and arrays and then begin to work with multiplication facts. The focus is on solving problems with the
### Year 1. Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on.
Year 1 Add with numbers up to 20 Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on. +1 +1 +1 Children should: Have access to a wide
### 3 rd Grade Mathematics Unpacked Content
3 rd Grade Mathematics Unpacked Content This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these
### AERO/Common Core Alignment 3-5
AERO/Common Core Alignment 3-5 Note: In yellow are the AERO Standards and inconsistencies between AERO and Common Core are noted by the strikethrough ( eeeeee) notation. AERO Common Core Mapping 1 Table
### Sue. Rubric for the Critical Area of Mathematics Grades K - 2
Rubric for the Critical Area of Mathematics Grades K - 2 The intent of this document is to provide support as teachers transition to Common Core s. It draws attention to the most critical skills for their
### Huntsville City Schools Second Grade Math Pacing Guide
Huntsville City Schools Second Grade Math Pacing Guide 2016-2017 Thoughtful and effective planning throughout the school year is crucial for student mastery of standards. Once a standard is introduced,
### Math Content by Strand 1
Math Content by Strand 1 The Base-Ten Number System: Place Value Introduction 2 Learning about whole number computation must be closely connected to learning about the baseten number system. The base-ten
### Explanation of the Standard
Jenny Kerola CCLM Project 2 July 15, 2011 Grade level: 2 Domain: Operations and Algebraic Thinking (OA) Cluster: Add and Subtract within 20 Standard: 2.OA.2 2. Fluently add and subtract within 20 using
### Common Core State Standard I Can Statements 1 st Grade Mathematics
CCSS Key: Operations and Algebraic Thinking (OA) Number and Operations in Base Ten (NBT) Measurement and Data (MD) Geometry (G) Common Core State Standard 1 st Grade Mathematics Common Core State Standards
### Montessori Mathematics Curriculum
1 Montessori Mathematics Curriculum Presentations on Preparation for Math One-to-One Correspondence Books Counting Songs 3 Little Speckled Frogs Sorting/Classification Sequence/Patterns Do As I Do Sequence/Seriation
### 1) Make Sense and Persevere in Solving Problems.
Standards for Mathematical Practice in Second Grade The Common Core State Standards for Mathematical Practice are practices expected to be integrated into every mathematics lesson for all students Grades
### First Grade Mathematics Curriculum Guide Plainwell Community Schools. Topic Pacing EnVision Math Lessons Common Core State Standards Topic 1
First Grade Mathematics Curriculum Guide Plainwell Community Schools Topic Pacing EnVision Math Lessons Common Core State Standards Topic 1 Understanding Addition @ 10 Days September 1-1: Spatial Patterns
### Progressing toward the standard
Report Card Language: The student can fluently multiply and divide within 100. CCSS: 3.OA.7 Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and
### Correlation to the Common Core State Standards. Math in Focus
Correlation to the Common Core State Standards Math in Focus Correlation to the Common Core State Standards Table of Contents Explanation of Correlation.......................... 1 Grade 1...........................................
### Mathematics K-8 Critical Areas of Focus
Mathematics K-8 Critical Areas of Focus The Common Core State Standards (CCSS) for Mathematics include critical areas for instruction in the introduction for each grade, K-8. The critical areas are designed
### Composing and Decomposing Whole Numbers
Grade 2 Mathematics, Quarter 1, Unit 1.1 Composing and Decomposing Whole Numbers Overview Number of instructional days: 10 (1 day = 45 60 minutes) Content to be learned Demonstrate understanding of mathematical
### Supporting Rigorous Mathematics Teaching and Learning
Supporting Rigorous Mathematics Teaching and Learning Mathematical Structures: Addition and Subtraction Word Problem Types PARTICIPANT HANDOUT Tennessee Department of Education Elementary School Mathematics,
Grade 1 and 2: Basic Addition and Subtraction Facts Series 2: First and Second Grade Texas Essential Knowledge and Skills (TEKS): (1.3) Number, operation and quantitative reasoning. The student recognizes
### Grades Three though Five Number Talks Based on Number Talks by Sherry Parrish, Math Solutions 2010
Grades Three though Five Number Talks Based on Number Talks by Sherry Parrish, Math Solutions 2010 Number Talks is a ten-minute classroom routine included in this year s Scope and Sequence. Kindergarten
### Unit Descriptions USER GUIDE
LEARNING Unit Descriptions USER GUIDE Pre-K - KINDERGARTEN UNITS Counting Build 1 to 10 Optimally. Students build and identify numbers from static and flashed sets of 1 to 10 objects using the least number
### Progression in written calculations in response to the New Maths Curriculum. September 2014
Progression in written calculations in response to the New Maths Curriculum This policy has been written in response to the New National Curriculum, and aims to ensure consistency in the mathematical written
### Standard Content I Can Vocabulary Assessment Time Frame (Marking Period) 1.OA.1
1.OA.1 Use addition and subtraction within I can add and Chapter Tests 20 to solve word problems subtract to 20 so 1 1.OA.3 involving situations of adding to, I can solve word 2 1 st marking period taking
### Mathematics K-8 Critical Areas of Focus
Mathematics K-8 Critical Areas of Focus The Common Core State Standards (CCSS) for Mathematics include critical areas for instruction in the introduction for each grade, K-8. The critical areas are designed
### Topic: 1 - Understanding Addition and Subtraction
8 days / September Topic: 1 - Understanding Addition and Subtraction Represent and solve problems involving addition and subtraction. 2.OA.1. Use addition and subtraction within 100 to solve one- and two-step
### Second Grade Math Standards and I Can Statements
Second Grade Math Standards and I Can Statements Standard CC.2.OA.1 Use addition and subtraction within 100 to solve one and two-step word problems involving situations of adding to, taking from, putting
### Decision One: Curriculum Map
Understand and apply the properties of operations and the relationship between addition and subtraction while representing and solving problems involving addition and subtraction equations. Operations
### Third Grade Common Core Standards & Learning Targets
Third Grade Common Core Standards & Learning Targets CCS Standards: Operations and Algebraic Thinking 3.OA.1. Interpret products of whole numbers, e.g., interpret 5 7 as the total number of objects in
### Part 3. Number and Operations in Base Ten
Part 3 Number and Operations in Base Ten Number and Operations in Base Ten Domain Overview KINDERGARTEN The work in kindergarten forms the foundation for students to develop an understanding of the base
### Sample Fraction Equivalence Activities (1 4)
Sample Fraction Equivalence Activities ( 4) Common Core State Standard(s) Addressed in these activities: 3.NF.3. Explain equivalence of fractions in special cases and compare fractions by reasoning about
### 1 ST GRADE COMMON CORE STANDARDS FOR SAXON MATH
1 ST GRADE COMMON CORE STANDARDS FOR SAXON MATH Calendar The following tables show the CCSS focus of The Meeting activities, which appear at the beginning of each numbered lesson and are taught daily,
### Overview. Essential Questions. Grade 4 Mathematics, Quarter 2, Unit 2.1 Multiplying Multi-Digit Whole Numbers
Multiplying Multi-Digit Whole Numbers Overview Number of instruction days: 5 7 (1 day = 90 minutes) Content to Be Learned Use strategies based on place value and properties of operations to multiply a
### Grade 4 Multi-Digit Multiplication and Division with One Divisor Unit of Instruction
Grade 4 Multi-Digit Multiplication and Division with One Divisor Unit of Instruction This is a progressive unit of instruction beginning with students investigating the concrete area model of fraction
### Next Generation Standards and Objectives for Mathematics in West Virginia Schools
Next Generation Standards and Objectives for Mathematics in West Virginia Schools Descriptive Analysis of Second Grade Objectives Descriptive Analysis of the Objective a narrative of what the child knows,
### SCOPE & SEQUENCE. Kindergarten, First Grade, and Second Grade. Skills and Activities
SCOPE & SEQUENCE Kindergarten, First Grade, and Second Grade Skills and Activities INNOVATIVE LEARNING CONCEPTS INC. creators of TOUCHMATH TouchMath materials were first published in 1975. Innovative Learning
### Math News! Focus Area Topic A. 3 rd Grade Math. Grade 3, Module 3, Topic A. Properties of Multiplication and Division
Grade, Module, Topic A rd Grade Math Module : Multiplication and Division with Units of 0,, 6-9, covers Multiplication and Division with Units of 0,, 6-9 Topic A. Topic A. Properties of Multiplication
### GRADE 2 MODULE 5 - Addition and Subtraction Within 1,000 with Word Problems to 100
GRADE 2 MODULE 5 - Addition and Subtraction Within 1,000 with Word Problems to 100 New or Recently Introduced Terms Algorithm (a step-by-step procedure to solve a particular type of problem) Compensation
### UNIT TITLE: UNIT TITLE: Addition and Subtraction within 20 GRADE: 1st TIME: 9 weeks
Domain: Operations and Algebraic Thinking Cluster 1: Represent and solve problems involving addition and subtraction (OA.1.1, 1.2). Cluster 2: Understand and apply properties of operations and the relationship
### 5 th Grade Remediation Guide
5 th Grade Remediation Guide Focused remediation helps target the skills students need to more quickly access and practice on-grade level content. This chart is a reference guide for teachers to help them
### I have 15 apples to share between four friends and myself. We each get 3. So I have 5 groups of 3 apples. Dina Mendola CCLM^2 Project Summer 2012
Dina Mendola CCLM^2 Project Summer 2012 DRAFT DOCUMENT. This material was developed as part of the Leadership for the Common Core in Mathematics (CCLM^2) project at the University of Wisconsin-Milwaukee.
### BREATHITT COUNTY SCHOOLS KINDERGARTEN MATH CURRICULUM
Weeks 1-2 Instructional Emphasis: Counting and Cardinality Topic 1: One to Five K.CC.3: Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count
### Calculations Year 2 What We Do and How We Do It
Calculation Policy Calculations Year 2 What We Do and How We Do It The following calculation policy has been devised to meet requirements of the calculation strand of the National Curriculum 2014 for the
### 2013 Texas Education Agency. All Rights Reserved 2013 Introduction to the Revised Mathematics TEKS: Vertical Alignment Chart Kindergarten Algebra I 1
2013 Texas Education Agency. All Rights Reserved 2013 Introduction to the Revised Mathematics TEKS: Vertical Alignment Chart Kindergarten Algebra I 1 The materials are copyrighted (c) and trademarked (tm)
### MATHEMATICS TUTORING TIPS
DRAFT MATHEMATICS TUTORING TIPS Targeted Intervention: Practice and Strategies UTAH STATE OFFICE OF EDUCATION 250 East 500 South P.O. Box 144200 Salt Lake City, UT 84114-4200 Larry K. Shumway, Ed.D. State
### 3.OA.B Understand properties of multiplication and the relationship between multiplication and division. Related Grade 2 Standards
Claim 1: Concepts and Procedures Students can explain and apply mathematical concepts and carry out mathematical procedures with precision and fluency. Content Domain: Operations and Algebraic Thinking
### GO Math! is built. for the COMMON CORE. Includes Complete Common Core Correlation Grades K 6
GO Math! is built for the COMMON CORE Includes Complete Common Core Correlation Grades K 6 It s New! NEW Write-in Student Edition Children record, represent, solve, and explain as they discover and build
### First Grade Math Standards and Learning Targets (I Can Statements)
First Grade Math Standards and Learning Targets (I Can Statements) Operations and Algebraic Thinking 1.OA.1 Use addition and subtraction within 20 to solve word problems involving situations of adding
### Mathematical Goals. Common Core State Standards
Mathematical Goals Common Core State Standards Lesson 2.2: Story Problems and Centers Introduction and Background Information By the end of the lesson students will: Represent the two numbers in a 2-digit
### 3rd Grade Texas Mathematics: Unpacked Content
3rd Grade Texas Mathematics: Unpacked Content What is the purpose of this document? To increase student achievement by ensuring educators understand specifically what the new standards mean a student must
### Just want the standards alone? You can find the standards alone at
5 th Grade Mathematics Unpacked Content This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these
### COMMON CORE STATE STANDARDS FOR MATHEMATICS 3-5 DOMAIN PROGRESSIONS
COMMON CORE STATE STANDARDS FOR MATHEMATICS 3-5 DOMAIN PROGRESSIONS Compiled by Dewey Gottlieb, Hawaii Department of Education June 2010 Operations and Algebraic Thinking Represent and solve problems involving
### Mathematics Calculation and Number Fluency Policy. Curriculum MMXIV. Chacewater School. + - x
Mathematics Calculation and Number Fluency Policy Curriculum MMXIV Chacewater School + - x Autumn 2014 Introduction The purpose of this document is to build on the successes of the Calculation Policy which
### Bonneygrove Primary School Calculations Policy
Bonneygrove Primary School Calculations Policy Rationale At Bonneygrove, we strongly encourage children to independently use a variety of practical resources to support their learning for each stage of
### Mini-Posters. LES 1 Shopping at the Co-op
Mini-Posters LES 1 Shopping at the Co-op Application Situation 1 p. 1 Application Situation 2 p. 48 Application Situation 3 p. 53 Application Situation 4 p. 54 Table of Contents APPLICATION SITUATION 1
### Overview. Essential Questions. Grade 4 Mathematics, Quarter 4, Unit 4.1 Dividing Whole Numbers With Remainders
Dividing Whole Numbers With Remainders Overview Number of instruction days: 7 9 (1 day = 90 minutes) Content to Be Learned Solve for whole-number quotients with remainders of up to four-digit dividends
### Algorithm set of steps used to solve a mathematical computation. Area The number of square units that covers a shape or figure
Fifth Grade CCSS Math Vocabulary Word List *Terms with an asterisk are meant for teacher knowledge only students need to learn the concept but not necessarily the term. Addend Any number being added Algorithm
### Next Generation Standards and Objectives for Mathematics in West Virginia Schools
Next Generation Standards and Objectives for Mathematics in West Virginia Schools Descriptive Analysis of Third Grade Objectives Descriptive Analysis of the Objective a narrative of what the child knows,
### Third Grade Math Standards and I Can Statements
Third Grade Math Standards and I Can Statements Standard - CC.3.OA.1 Interpret products of whole numbers, e.g., interpret 5 7 as the total number of objects in 5 groups of 7 objects each. For example,
### Every Day Counts: Partner Games. and Math in Focus Alignment Guide. Grades K 5
Every Day Counts: s and Math in Focus Alignment Guide Grades K 5 7171_Prtnrs_AlgnmtChrt.indd 1 9/22/10 6:04:49 PM Every Day Counts : s s offers 20 simple yet effective games to help students learn, review,
Ohio Standards Connection Patterns, Functions and Algebra Benchmark E Solve open sentences and explain strategies. Indicator 4 Solve open sentences by representing an expression in more than one way using
### Multiply Using the Distributive Property
Multiply Using the Distributive Property Common Core Standard: Use place value understanding and properties of operations to perform multi-digit arithmetic. 4.NBT.5 Multiply a whole number of up to four
### Swavesey Primary School Calculation Policy. Addition and Subtraction
Addition and Subtraction Key Objectives KS1 Foundation Stage Say and use number names in order in familiar contexts Know that a number identifies how many objects in a set Count reliably up to 10 everyday
### Georgia Department of Education THIRD GRADE MATHEMATICS UNIT 3 STANDARDS
Dear Parents, THIRD GRADE MATHEMATICS UNIT 3 STANDARDS We want to make sure that you have an understanding of the mathematics your child will be learning this year. Below you will find the standards we
### Objectives. Key Skills Addition. Subtraction
Objectives read, write and interpret mathematical statements involving addition (+), subtraction (-) and equals (=) signs represent and use number bonds and related subtraction facts within 20 add and
### Objective: Model the distributive property with arrays to decompose units as a strategy to multiply.
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 10 3 1 Lesson 10 Objective: Model the distributive property with arrays to decompose units as a Suggested Lesson Structure Fluency Practice Application Problem
### Math News! 5 th Grade Math
Math News! Grade 5, Module 1, Topic A 5 th Grade Math Module 1: Place Value and Decimal Fractions Math Parent Letter This document is created to give parents and students a better understanding of the
### Multiplication. Year 1 multiply with concrete objects, arrays and pictorial representations
Year 1 multiply with concrete objects, arrays and pictorial representations Children will experience equal groups of objects and will count in 2s and 10s and begin to count in 5s. They will work on practical
### Sequenced Units for the Common Core State Standards in Mathematics Grade 1
Sequenced Units for the Common Core State Standards in Mathematics In Kindergarten, students learned to count in order, count to find out "how many", and model addition and subtraction situations with
### Objectives. Key Skills Addition. Subtraction
Objectives read, write and interpret mathematical statements involving addition (+), subtraction (-) and equals (=) signs represent and use number bonds and related subtraction facts within 20 add and
### Grade 4 Unit 1 Whole Number Computation and Application. Assessment Plan
Standards Addressed: Grade 4 Unit 1 Whole Number Computation and Application Assessment Plan ISBE 4 th Grade UNIT 1 MAP 3.OA.8 Solve two-step word problems using the four operations. Represent these problems
### Standards for Mathematical Practice in Kindergarten
Standards for Mathematical Practice in Kindergarten The Common Core State Standards for Mathematical Practice are expected to be integrated into every mathematics lesson for all students Grades K-12. Below
### Addition and Subtraction Within 1,000 with Word Problems to 100
Grade 2 Module 5 Addition and Subtraction Within 1,000 with Word Problems to 100 OVERVIEW In Module 4, students developed addition and subtraction fluency within 100 and began developing conceptual understanding
### Georgia s Early Intervention Program (EIP) Mathematics K- 5 Rubrics
2016-2017 Georgia s Early Intervention Program (EIP) Mathematics K- 5 Rubrics NOTE: The EIP eligibility criteria for student placement and exit decisions must be supported by and consistent with multiple
### Reteaching. Properties of Operations
- Properties of Operations The commutative properties state that changing the order of addends or factors in a multiplication or addition expression does not change the sum or the product. Examples: 5
### Grade 2. M4: and M:5 Addition and Subtraction of Numbers to 1000. M3: Place Value, Counting, and Comparison of Numbers to 1000
Grade 2 Key Areas of Focus for Grades K-2: Addition and subtraction-concepts, skills and problem solving Expected Fluency: Add and Subtract within 20 Add and Subtract within 100 Module M1: Mastery of Sums
### Pocantico Hills School District Grade 1 Math Curriculum Draft
Pocantico Hills School District Grade 1 Math Curriculum Draft Patterns /Number Sense/Statistics Content Strands: Performance Indicators 1.A.1 Determine and discuss patterns in arithmetic (what comes next
### Written methods for addition of whole numbers
Written methods for addition of whole numbers The aim is that children use mental methods when appropriate, but for calculations that they cannot do in their heads they use an efficient written method
### The Crescent Primary School Calculation Policy
The Crescent Primary School Calculation Policy Examples of calculation methods for each year group and the progression between each method. January 2015 Our Calculation Policy This calculation policy has | 7,010 | 32,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | longest | en | 0.868009 |
https://dokumen.tips/documents/mrsm-trial-spm-2013-physicspdf.html | 1,716,423,518,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00138.warc.gz | 166,817,059 | 51,978 | 93
# MRSM Trial SPM 2013 Physics.pdf
Embed Size (px)
DESCRIPTION
MRSM TRIAL SPM 2013 PHYSICS
Citation preview
CONFIDBNTIAL
PHYSICSPaper ISeptember2013
1 % hours
453u1
MAKTAB RENDAH SAINS MARA
SIJIL PELAJARAN MALAYSIATRIAL EXAMINATION 2OI3
f----.r
@tstptusEnLuuBtRSry
PHYSICS
Paper 1
One hour and fifteen minutes
1.
2.
3.
DO NOT OPEN THIS QUESTTON BOOKLET UNTrL TOLD TO DO SO
This paper is written in English and bahasa MelayuKertas soalan ini adalah dolam dwibahasa.
The question in English is written on top while the bahasa Melayu version is below.Soalan di atas adalah dqlam bahasa Inggeris dan soalan dalam bahasa Melayu terdapat di bm,ahnya.
Candidates are required to read the information at the back of the booklet.Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.
www. mysctroolchildren. com
This question booklet consists of 43 printed pages and 1 btank page
[Turn page over453ltl @ rtat Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
The following information maybe usefuI. The symbols have their usual meaning.
Maklumatberikutmungkinberfaedah.Simbol-simbolmempunyaimakna yang biasa.
v-u1. u: ,
2. ,2:y2 + 2as
3. s:ut+%af
4. Momentum: mv
5. F:ma
6. Kinetic energy I Tenaga kinetik: % mv2
7. Gravitational potential energy I Tenaga knupayaan graviti - mghI
8. Elastic potential energy I Tenaga keupayaan kenyal - ) Fx2
s. p -YV
10. Pressure / Tekanan, P : hPS
1 1. Pressure / Tekanan, p : !A
12. Heat I Haba, Q: mcT
13. Heat / Heat, Q: ml
PV14. * =constant /pemalarT
15. E - mc2
16. v:fl17. Power, P: Energy
time
Kuasa, , :tenagamasa
11118. -+-J uv
4s3ul
453lll @ ffat< Cipta Bahagian Pendidikan Menengah MARA
&
CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
ax19. )" _
D
sin i20. n: .
s1n r
1
n: .
sln c
.r r real dePth21. n _
4s3t/l
[Turn page overCONFIDENTIAL
apparent depth
dalam nyatan:dalam ketara
22. O:Itn. V:EO
24. V:IR
25- PowerlKuasa,P:IV
N. V,26.--Np vP
27. Efficien cy I Kecekapan : 'r'Y: xfi[YoI ovo
28. g-10ms-'
29. c: 3.0 x 108 m s-'
453tll @ gak Cipta Batragian Pendidikan Menengah MARA
i --- . 1., -!-i:.. :?i*:r*::i*.i
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFTDENTIAL
I Which of the following is a derived quantity?
Antara berilrut yang manakah kuantiti terbitan?
A MassJisim
B TimeMasa
C LengthPanjang
D WeightBerat
Which of the following is most suitable to measure the depth of a test tube?
Antara berikut yang manakah paling sesuai untuk mengukur kedalaman sebuah tabung uii?
453ut
C
r\
rs)-\ t}\\:
-s t.t'
=tb-
453lll @ Hak Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
Which of the following frequencies is the same as 185.6 MHz?
Manakah antarafrekuensi berikut sama dengan 185.6 MHz?
1.856 x 10-r Hz
1.856 x !02 Hz
1.856 x 106 Hz
1.856 x 108 Hz
Diagram I shows a pattern on a ticker tape.
A
B
C
D
453ut
[Turn page oyerCONFIDENTIAL
Direction of motionArah gerakan
Diagram 1
Rajah I
Which statement describes the motion of the trolley?
Perryataan manakah yang menerangkan gerakan troli tersebut?
It moves with constant velocityIa bergerak dengan halaju seragqrn
It moves with an accelerationIa bergerak dengan pecutan
It moves at constant speed and then deceleratesIa bergerak dengan halaju seragam dan kemudian nyahpecut
It moves at constant speed and then acceleratesIa bergerak dengan halaju seragam dan kemudian memecut
A
B
C
D
at a a t a I I {} a t {
453llt @ Hak Cipta Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4531t1
Diagram 2 shows a displacement-time graph for the motion of a car.
Rajah 2 menunjukkan graf sesqran-masa bagi gerakan sebuah kereta.
DisplacementSesaran
TimeMasa
Diagram 2Rajah 2
Based on the graph, which statement explains the motion of the car from A to B?
Berdasarkan graf, pernyataan manakah yang menerangkan pergerakan kereta tersebut dari A ke B?
A StationaryTidak bergerak
B DeceleratingNyahpecut
C AcceleratingMemecut
D Constant velocityHalaju seragam
Diagram 3 shows a velocity-time graph for the motion of an object.
Rajah j menunjukkan graf halaju-masa bagi satu objek bergerak
u/ m s-L
t/s
Diagram 3Rajah j
What is the displacement of the object in the first 5 s?
Berapakah sesaran objek tersebut dalam masa 5 saat pertama?
20m
50m
70m
100 m
Hak Cipta Bahagian Pendidikan Menengah MARA CONX'IDENTIAL
A
C
D
4531/r @
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFTDENTTAL
7 Which of the following vehicles takes the longest time to start moving?
Kenderaan yang manakah mengambil masa yang paling lama untuk mula bergerak?
453u1
A
Mass : 750 kgJisim
Mass : 250 kgJisim
DC
Mass :Jisim
12 000 kg Mass: 5 000 kgJisim
[Turn page over
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
453LllCONFIDENTIAL
; t,lifu t**i***et;ruv iSri''i',
Diagram 4 shows an aeroplane flying at consffi4[,ffi And'velocity'
Rajah 4 menunjukkan sebuah kapar terbang bergerak pada rcetrnggran dan kelaiudn yang tetap'
UpthrustDcya tuiah atas
ThrustDaya tuiahan
Diagram 4Rajah 4
Which Physics concept explains the above situation?
ApakahkonsepFizikyangmenerongkansituasidiatas?
Resolution of forcesLerqian daYa
Forces in equilibriumKeseimbangan daYa
DragDaya seret
A
C
4531tL @ I{ak Cipta Bahagian Pendidikan Menengah I'IARA r..;;\$ibf
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
Diagram 5 shows an air bag being released during collision.
Rajah 5 menunjukkan sebuah beg udara berfungsi semosa pelanggaran.
Diagram 5Rajah 5
The released air bag is to reduce
inertiainersia
the change of momentum of the driverperubahan momentum pado pemandu
the rate of change of momentum acting on the driverkadar perubahan momentum yang bertindak ke atas pemandu
the collision time between the driver and the steeringmasa perlanggaran antara pemandu dan stering kereta
9 453u1
[Turn page oyerCONFIDENTIAL
A
B
C
D
453u1 @ Hak Cipta Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 453Llt
10 Diagram 6 shows a ticker tape representing the motion of a 1.0 kg trolley before,during and after a collision.
Rajah I menunjukkan pita detik yang mewakili gerakan sebuah troli berjisism 1.0 kg sebelum, semasadan selepas perlanggaran.
ttltlltll
{<- 1.6qn } l<- 4.0 cm
After collisionSelepos perlanggaran
Diagram 6
Rajah 6
If the ticker timer produces 50 dots every second, what is the impulsive force duringthe collision?
Jika jongka masa detik menghasilkan 50 titik setiap saat, berapakah drya impuls semasa perlanggaran ?
A 225N
B 180 N
c 2.25 N
D 2.00 N
10
I
II
7.6 cm
Before collisionSebelum perlanggaran
aI
a a a a' a t l a I ?3.e11....*...trl
453lll @ uat Cipta Bahagian Pendidikan Menengah MARA i.': \$@6)III\$ffiENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 453uL
Diagram 7 shows a roller coaster sliding down its track.
Raj ah 7 m enunjukkan " r o ll er co ast er " menggel on s or m enuruni I andas anrry a.
Diagram 7Rajah 7
Which graph shows the relationship between the gravitational potential energy (PE)and kinetic energy (KE) of the roller coaster as it moves down the track?
Grof manakah menunjul*an hubungan antara tenaga kanpayaan graviti (PE) dan tenago kinetik (KE)"roller coaster" ketika menuruni landasan?
[Turn page overCONFIDENTIAL
11
11
BA
DC
4s3ut o Hak Cipta Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 12 453TIL
t2 Diagram 8 shows a load M supported by three different arrangements of identicalsprings.
Rajah 8 menunjukkan beban M yang digantung mengikut tiga susunan berbeza, menggunakan springyang serupa.
Diagram 8Rajah 8
Which comparison is correct about the extension of arrangements P, Q and R?
Perbandingan yang manakah betul tentang pemanjangan susunan P, Q dan R?
A Q<R<P
B P<Q<R
c R<P<Q
D R<Q<P
453LlL @ ftat Cipta Bahagian Pendidikan Menengah MARA ' CONtr'IDEI.ITIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
www. m vschoolc.h ildren .com
4s3u1
Diagram 9 shows three different shoes.
Rajah 9 menunjukkan tiga kasut berlainan.
Diagram 9Rajah 9
Arrange the shoes based on the pressure exerted on the ground in ascending order ifworn by the same person.
Susunkan kasut di atas berdasarkan tekanan yang dikenakan pado permukaan tanah dalam urutanmenaikjika kasut-kasut tersebut dipakai oleh orongyang sama.
X,Y,ZY,Z,XZ,Y,XZ,X,Y
L4 Diagram 10 shows a container with water spurting out from a hole, P.
Rajah I0 menunjukkan satu bekas di mana air memancut keluar dari lubang, P.
13
13
A
B
C
D
ContainerBekas
WaterAir
Diagram 10Rajah I0
Distance Y can be reduced by
Jarak Y boleh dihtrangkan dengan
A replacing water with cooking oilmenggantikan air dengan minyak masak
B using a taller containermenggunakan bekas yang lebih tinggi
C lowering the level of hole Pmerendahkan kedudukan lubang P
D increasing the water depthmenambahkan kedalaman air
453lll @ tlat Cipta Bahagian Pendidikan Menengah MARA
"\
Y->
[Turn page overCONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL t4 453111
15 Diagram 11 shows a manometer connected to ia gas container.
Rajah I I menunjukkan sebuah manometer yang disambung ke sebuah bekas berisi gas.
Gas in containerGas di dalam bekas
LiquidCecair
t6
Diagram 11
Rajah 1I
What happens to h if the temperature of the gas in the container increases?
A DecreasesBerlcurang
B IncreasesBertambah
C Remains constantTidak berubah
Which of the applications below uses Pascal's principle?
Yang manakah aplikasi di banah menggunakan prinsip Pascal?
BA
DC
453L11 @ Uat Cipta Batragian Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4s3ut
17 Diagraml2 shows a hydraulic system. The total weight of load X and piston K is 80 N.
Rajah l2 menunjukkan satu sistem hidraulik.Jumlah berat beban X dan piston K ialah 80 N.
15
Piston KOmboh K
Area/Luas2
40 cm
Piston LOmboh L
Area/Luas2
20 cm
Diagram 12Rajah 12
What is the magnitude of the force to be exerted on piston L in order to lift load X?
Berapakah magnitud drya yang perlu dikenakan pada piston L untuk mengangkat bebanfr
A
B
C
D
10N
40N
80N
160 N
18 Diagrams 13 shows a floating hot air balloon.
Rajah l3 menunjukkan sebuah belon udara panas terapung.
Diagram 13Rajah I j
The situation can be explained using
Situasi ini boleh diterangkan menggunakan
A Pascal's principlePrinsip Pascal
B Bernoulli's principlePrinsip Bernoulli
C Archimedes'principlePrinsip Archimedes
453111 @ Hak Cipta Bahagian Pendidikan Menengah MARA[Turn page over
CONFIDENTIAL
tltrrtrar::r-,-3
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 16 453T/1
19 Diagram 14 shows a submarine with volume "\$200'\$d 5ubmerg ed 52 m below thewatgr surface.
r! :i *. , ; ;,Rajah l4 menunjukkan sebuah kapal selqm dengan isipadu SZ00 mt berada 52 m di bawah permukaanair.
(Density of water : 1000 kg *-r)(Kenmpatan air = 1000 kS *-)
K: G
* k
-
E
*
* -
G 52m:
I r_Fi
E
Diagram 14Rajah t4
what is the buoyant force acting on the submarine?Berapakah nilai daya apungan yang bertindak keatas kapal selam tersebut?
A S.2x tOs N
B 5.2 xl06 N
C 5.2x107N
D 5.2x108N
4531n @ nar ciptaBahagian pendidikan Menengah MARA "KHffiffiilIhttp://edu.joshuatly.com/
http://fb.me/edu.joshuatly
20
CONFIDENTIAL L7
Diagram 15 shows a glass of iced tea.
Rajah 15 menunjukkan segelas teh ais.
Diagram l5Rajah l5
What happens when the iced tea and glass are in thermal equilibrium?
Apakahyang berlaku apabila teh ais dan gelas dalam keseimbangan terma?
The temperature of the glass decreasesSuhu gelas berhtrangan
The temperature of the iced tea decreasesSuhu teh ais meningkat
There is heat flow from the glass to the iced teaAda pengaliran haba dari gelas ke teh ais
There is no net heat flow from the glass to the iced teaTiada pengaliran haba bersih dari gelas lcepada teh ais
453u1 @
4s31ll
[Turn page overCONFIDENTIAL
A
C
D
Hak Cipta Batragian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONF'IDENTIAL 453UL
21 Diagram 16 shows three different pots with the,saine rnass and heated with the sameamount of heat for 5 minutes.
Raiah 16 menunjukkan tiga jenis periuk dengan jisim yang sama dan dipanaskan dengan jumlah habayang sama dalam masa 5 minit.
Aluminum potPeriuk aluminium
Copper potPeriuk kuprum
Glass potPeriuk kaca
Specific heat capacityMuatan haba tentu
: 9oo J kg-t og-t
Specific heat capacityMuatan haba tentu
:387 J kg't og-t
Specific heat capacityMuatan haba tentu
: 840 3'Pn'l og-t
Diagram 16Rajah 16
Which statement is correct after 5 minutes of heating?
Pernyataan manakah benar selepas 5 minit dipanaskan?
A The copper pot has the lowest temperaturePeriukkuprum mempunyai suhu paling rendah
B The aluminium pot has the highest temperaturePeriuk aluminium mempunyai suhu paling tinggi
C The temperature of the copper pot is equal to glass potSuhu periuk htprum samo dengan periuk kaca
D The temperature of the copper pot is higher than glass potSuhu periuk kuprum lebih tinggi dari periuk kaca
18
EIOIqIel(I)ILI:ol-clololol-clC)l.nl>lEI
=l
=t
453LlL @ Uat Cipta Bahagian Pendidikan Menengah MARA COhIFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL t9 453u1
22 Diagram 17 shows the temperature-time graph for a substance X when heated by a100 W electric heater.
Rajah 17 menunjukkan graf suhu-masa bagi suatu bahan X bila dipanaskan oleh pemanas elelarikIOO W.
80
70
60
50
40
30
Diagrarn 17
RajahlT
Determine the mass of the substance.
Tentukan jisim bahan tersebut.
[Specific heat capacity of the substance is 500 J kg-'oC-']
[Muatan haba tentu bahan itu ialah 500 J kg-t oC'J
A 0.03 kg
B 0.75 kg
C l.20kg
D 2.00 kg
[Turn page overCONFIDENTIAL
e/ 0c
453111 O ftak Cipta Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
23 Diagram 18 shows ice melting.
20 45311L
meltingpeleburan
Diagram 18Rajah 18
The heat absorbed during the process is
Tenaga haba yang diserap semasa proses adalah
A latent heat of fusionhaba pendam pelalatran
B latent heat of vaporizationhab a pendam pengeuv ap an
C heat capacrty of solidmuatan haba pepeial
D heat capacity of liquidmuatan haba cecair
453LlL @ gat Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTTAL 453ut
24 Diagram 19 shows the same balloon before and after it was placed in a refrigerator foran hour.
2t
Rajah 19 menunjukkan sebiji belon yang sqma sebelum dan selepasselama satu jam.
diletakkan di dalam peti sejuk
www. mvgchoolchildren.com
BeforeSebelum
Diagram l9Rajah 19
The situation above can be explained bySituasi di atas boleh dijelaskan oteh
A Gas lawHukum gas
B Boyle's lawHulrum Boyle
C Pressure lawHulrumTekanan
D Charles'lawHulam Charles
AfterSelepas
[Turn page overCONFIDENTIAL453lll @ uat< cipta Bahagian pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 22 453u1
25 Diagram 20 shows the path of light rays reflected by a concave mirror and points A,B, C and D in front of the mirror.
Raiah 20 menunjukkan lintasan sinar cahaya dipantulkan oleh sebuah cermin celamg dan titik-titik A,B, C dan D di hadapan cermin.
Concave mirrorCermin celamg
Diagram 20Rajah 20
Which point is the centre of curvature of the concave mirror?
Titikmanakah pusat kelengkungan cermin cekung itu?
453LlL @ Hak Cipta Batragian Pendidikan Menengah MARA CONF'IDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
t
l,'il
CONFIDENTIAL 23
26 on Diagram 21, which light ruy undergoes the phenomenon of total
reflection?pada Rajah 2 I sinar cahaya manakah yang menuniukkan fenomena pantulan dalam penuh?
Diagrarr2lRaiah 21
4s3fl1
internal
453[tl @ Uat< Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
27 Which of the followingray diagrams is correct?
Manakoh di antara rajah sinar yang berilafi adalah betul?
453u1
KeyKelrunci
F : Focal pointTrtikfolax
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 25
28 Diagram 22 shows an object which is placed 25 cm fromThe focal length of the lens is 20 cm.
fokus kanta itu ialah 20 cm.
4s31n
the centre of a convex lens.
sebuah kanta cembung. Panjang
Diagram22Rajah 22
Calculate the image distance.
Kirokan jarak imej.
A 25cm
B 45cm
C 90cm
D 100 cm
What does wave carry when propagating from one point to another?
Apalrah yang dibawa oleh gelombang apabila merambat dari satu titik ke titikyang lain?
EnergyTenaga
PowerKuasa
ParticleZarah
MediumMedium
29
A
B
C
D
453lll @ Uat Cipta Batragian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 26 4s31lt
30 Diagram 23 shows a graph of displacement against time of a simple pendulum
Rajah 23 menunjukkan graf sesaran lawan masa bagi sebuah bandul ringkas.
DisplacementSesaran
Diagram23Rajah 23
What is the frequency of oscillations?
Ap akah fr ekuens i get ar an t ers ebut?
A 0.4H2
B OSHZ
C 2.0H2
D 2,5 HZ
31 Which of the following diagrams shows the correct pattern of the reflection of waterwaves?
Manakah di antara rajah berikut yang menunjukkan corak pantulan gelombang air yang betul?
[Turn page overCONFIDENTIAL
A
D
Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 27
32 Diagram 24 shows a spoon in a glass of water.
Diagram24. Rajah 24
What phenomenon explains the situation above?
Apakahfenomena yang menjelaskan situosi di atas?
Reflection of lightPantulan cahaya
Refraction of lightPembiasan cahaya
Diffraction of lightPembelauan cahaya
Interference of lightInterferens cahaya
A
C
453u1
[Turn page oYerCONF'IDENTIAL,453tll @ uat Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 28
33 Diagram 25 shows diffraction of water waves.
Rajah 25 menunjukkan pembelauan gelombang air.
453ut
Diagram25Rajah 25
Which quantity decreases when the water waves diffract?
Manakah di antara htantiti berilafi yang berlarang apabila gelombang air mengalami pembelauan?.
A SpeedLaju
B PeriodTempoh
C AmplitudeAmplitud
WavelengthPanjang gelombang
45311L @ rut cipta Bttffitn Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 29
34 Diagram 26 shows an astronaut in outer space'
Rajah 26 menuniukan seorang angkasawan di ruang angkasa lepas.
4s3ul
Diagran26Raiah 26
An astronaut could not hear direct sound in outer space because
Angkasanan tidak dapat mendengar sebarang bunyi secara terus di angkasa lepas kerana
sound cannot travel in vacuumbunyi tidak dapat bergerak melalui valatm
the density of outer space is lowketumpatan angkasa lepas rendah
the speed of sound is greater than the speed of lightkelajuan bunyi lebih tinggi daripada kelaiuan cahaya
the speed of light is greater than the speed of soundkelajuan cahaya lebih tinggi daripada keloiuan bunyi
A
B
C
D
,453llL @ Uat Cipta Bahagian Pendidikan Menengah MARA
r
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
35
CONFIDENTIAL 30
Diagram 27 shows a Global Positioning Sytem (GPS) device.
Rajah 27 menunjukan sebuah peranti Sistem Kedudukan Sejagat.
Diagram2TRajah 27
Which part of electromagnetic spectrum is used by the device?
Bahagian manakah daripada spektrum eleldromagnet yang digunakan oleh alat tersebut?
X raysSinar X
MicrowaveGelombang milvo
InfraredInfra merah
4s3Llt
A
C
't Bahagian Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
4s31lL
CONFIDENTIAL31
36 which diagram shows the correct electric field pattern?
Rajah manakah yang menuniukkan corak medan elektrik yang betul?
A
B
C
D
[Turn Page over
CONFIDENTIAL4531/1@rukCiptaBahagianPendidikanMenengahMARAhttp://edu.joshuatly.com/
http://fb.me/edu.joshuatly
CONFIDENTIAL
37 Diagram 28 shows an electric circuit.
Rajah 28 menunjukkan satu litar elektrik.
32 453u1
Diagram 28Rajah 28
Which gaph shows the relationship between potential difference, V and current, .Iwhen the rheostat is adjusted?
Graf manakah yang menunjukkan hubungon antara beza keupcryaan Y dan qrus, I apabila reostatdilaraskan?
BA
Dc
4s3llt o Hak Cipa Bahagian Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
.:#
CONFIDENTHL
38 Diagram 29 shptrry,an'electric circuit.
33 453uL
39
Diagram29Rajah 29
At which point the current is the least?
Diagram 30 shows a voltage{urrent graph.
Raj ah j 0 menunjukkan gr af vo ltan- arus.
VIY
3.02.8
2.0
0.5
DiagramRajah30
Determine the value of electromotive force.
Tentukan nilai bagi doya gerak elektrik
A 1.87 V
B 2.00 v
c 2.80 v
D 3.00 v
fl\$SUf @ Hak Cipta Bahagian Pendidikan Menengah MARA
1.51.0 IIA
[Tp',ff
3V
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 34 453IIL
40 Diagram 31 shows a right burb connected to a resistor and a battery.
Rajah 3 r menunjukkan sebuah mentor yang disambungkon kepada perintang dan sebuah bateri'
Diagram 31Raiah 31
Calculate the power used by the light bulb'
Hitungkan lausa yang digunakan oleh mentol'
A 0.90 w
B 1.92 w
c 2.25 W
D 3.00 v/
3.0 v
Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIA'http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 35 453u1
4t Diagram 32 shows the iurangement of apparatus to investigate the magnetic fieldproduced by the current in a skaight wire
Rajoh 32 menunjukkan susunan rodas untuk mengkaji medan magnet yang dihasilkan oleh arus dalamwayar lurus
Compasskompas
Diagam32Rajah j2
What are the directions shown by the pointers of compass X and Y when the circuit isswitched on?
Apakoh arahyang ditunjuk oleh jarum kompasX danY apabila suis ditutup?
[Turn page overCONF'IDENTIAL
A
B
C
D
Compass XKompas X
Compass YKompas Y
@ @
@ @
@ @
@ @
-6tA @ uat Cipta Bahagian Pendidikan Menengah L{ARAhttp://edu.joshuatly.com/
http://fb.me/edu.joshuatly
CONFIDENTIAL 36 4s3ul
42 Diagram 33 shows a. solenoid connected to a galvanometer.
<_N S
Diagram 33Rajah 33
Which of the following factors will increase the deflection of galvanometer pointer?
Antara berikut yang manakah akan meningkatkan pesongan pada penuniuk galvanometer?
Use thinner wireGunakanwayar yctng lebih nipis
Increase the currentTambahkan arus
Move the magnet away from the solenoidMengger okkan magnet meni auhi s ol enoid
Increase the number of turns of the solenoidT amb ahkan bilangan lilitan s olenoid
A
B
C
D
453llt @ Uak Cipta Bahagian Pendidikan Menengah MARA CONFTDENTI/http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 37
43 Diagram 34 shows a circuit where the bulb lights up atnormal brightness.
Raiah34 menunjukkan litar di mana mentol menyala dengan kecerahan normal.
n{iL-
Which of the following is the transformer X?Antara berilail adalah transformer -fr
Primary coilGegelung primer
Secondary coilGegelung sekunder
Primarycoil mGegetung nrtmer
! *lTcr_/
Secondary coilGegelung sekunder
Primary coilGegelung primer
Secondary coilGegelung sekunder
[Turn page overt @ Uat< Cipta Bahagian Pendidikan Menengah MARA , ;,;.,r':dl ft.,i.u:pr,; .1f CON.I{DEfrfIAl
453U1
240 VTransformer
x
Diagram 34Rajoh 34
8S12V,24W
A
B
C
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 38 4531/7
44 Which characteristics are the most suitable for transmission of electricity through the
National Grid Network?
Ciri manakah yang paling sesuai untuk penghantaran elektrik melalui Rangkaian Grid Nasional?
Type of currentJenis arus
Magnitude of voltageMasnitud voltan
Direct currentArus terus
HighTinggi
Direct currentArus terus
LowKecil
Alternating currentArus ulangalik
LowRendah
Alternating currentArus ulangalik
HighTinggi
45 Which of the following information cannot be obtained directly from a cathode rayoscilloscope (C.R.O) display?
Antara maHumat berikut, yang manakah tidak boleh didapati secara terus dari paparan osiloskop
sinar katod (O. S. K.)?
A FrequencyFrekuensi
Waveforms displayP ap ar an b entuk gel omb ang
Short time interval measurementUkuran sela masa
Potential difference measurementUhtran beza keupayaan
A
B
C
D
B
C
D
453ltt @" Hak Cipta Bahagian Perrdidikan Menengah MARA
-].
CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 39
46 Diagram 3 5 shows an automatic lamp circuit.
Rajah 35 menunjukkan sebuah litar lampu automatik
Dry cellsSel kering
Diagram 35Rajah 35
Which of the following is correct about the circuit?
Antara berihi, yang manakah betul mengenai litar tersebut?
A
B
C
D
S urrounding bri ghtnes sKecer ah an p ers ekit ar an
Resistance of LDRRintangan LDR
LarnpLampu
DarkGelap
LowRendah
Does not light upTidak menyala
DarkGelap
HighTinggi
Lights upMenyala
BrightCerah
HighTinggi
Does not light upTidak menyala
BrightCerqh
LowRendah
Lights upMenyala
*'ttl, @ gat Cipta Bahagian Pendidikan Menengah MARA[Turn pn
CONFIThttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
0011
CONFIDENTIAI, 40
47 Diagram 36 shows a.logic gate circuit.
Rajah 36 menunjukkan litar get logik.
48
4s31/1
*a-ffilr-; Jtffi+iiiiIIT'IttrtItlr a3
aa,ItE!tt.t.xr @
BA
P
a
Diagram 36Rajah j6
Which of the following is the correct output, R?
Manakah di antara berilcut adalah betul untuk output, R?
Diagram 37 shows a nuclide notation for cobalt-60
Rajah 37 menunjukkan notasi nuklid kobalt-60.
60
27
Diagram 37Rajah 37
Which of the following combination is correct?
A
B
C
D
DC
Co
Proton numberNombor proton
Nucleon numberNombor nucleon
Number of neutronsBilangan neutron
27 60 33
60 27 33
27 33 60
33 60 27
A Hak Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4t 4s3ul
49 The diagram 3 8 shows how the thickness of steel sheet is monitored using aradioisotope and a detector.
RollerPenggelek
Detector1 Pengesan
Steel sheet
Diagram 38Rajah 38
Which is the most suitable radiation?
S inaran manakah yang paling sesuai?
A X-raySinar-X
Gamma ray,' Sinar gammal
Beta particlesZarah beta
Alpha particlesZarah alfa
[Turq page oYefCONT'IDENTIAL
I
1. (D
"t o
:' : '()E,o)(rE.o-tU';(Do, CT'-o)Jro,
"::CIt.+
l=,t{'ttl3s<lutlot=lotolot=l=:lo-l-t(Dtilc).tol3
7-I. @ uat< Cipta Bahagian Pendidikanhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4531/2
. Index Number : ..............................................................
Identity Card No.: ……………………………………..…
Name : …………………………………………………...
College No. : ………………. Class : ………..
MAKTAB RENDAH SAINS MARA
PHYSICS
Paper 2
Two hours and thirty minutes
DO NOT OPEN THIS BOOKLET UNTIL
BEING TOLD TO DO SO.
1. Write down your name and class in the
space provided.
2. The questions are written in English and
bahasa Melayu 3 Candidates are required to read the
information at the back of the booklet
_________________________________________________________________________
Examiner’s Code
Section Question Marks Score
1 4
2 5
4 7
6 8
7 10
A
8 12
9 20
B 10 20
11 20
C 12 20
Total
4531/2
PH
YS
ICS
2
3 6
5 8
[Turn over
SIJIL PELAJARAN MALAYSIA
TRIAL EXAMINATION 2013
4531/2
PHYSICS
Paper 2
September
2013
2 ½ hours
This booklet consists of 36 printed pages and 1 unprinted pages
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
(Maklumat berikut mungkin berfaedah. Simbol-simbol mempunyai makna yang biasa.)
1. t
sv =
2. t
uva
−=
3. asuv 222 +=
4. 2
2
1atuts +=
5. Momentum = mv
6. maF =
7. Kinetic energy (Tenaga kinetik ) = ½ mv 2
8. Potential energy (Tenaga keupayaan) = mgh
9. Density (Ketumpatan) , V
m=ρ
10. Pressure (Tekanan ), A
FP =
11. Pressure (Tekanan), ghP ρ=
12. Heat (Haba), θmcQ=
13. Heat (Haba), mlQ =
14. =T
PV constant (pemalar )
15. λfv =
16. Wavelength (panjang gelombang), λ = D
ax
17. Power (Kuasa), P = t
E
18. vuf
111 +=
19. Linear magnification (Pembesaran linear), u
vM =
20. Refractive index (indeks biasan), r
in
sin
sin=
21. Refractive index (indeks biasan), n = real depth (dalam nyata) apparent depth (dalam ketara)
22. ItQ =
23. IRV =
CONFIDENTIAL 1 4531/2 The following information may be useful. The symbols have their usual meaning.
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
24. Power (Kuasa), IVP =
25. p
s
p
s
V
V
N
N =
26. E = mc2
27. Efficiency (Kecekapan) = %100×pp
ss
VI
VI
28. g = 10 m s -2
29. Atmospheric pressure at sea level (Tekanan atmosfera pada aras laut) = 1 x 10 5 Pa
[Turn over 4531/2 ©2013 Copyright Bahagian Pendidikan dan Latihan (Menengah) MARA CONFIDENTIAL
CONFIDENTIAL 2 4531/2
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 3
Section ABahagion A
[ 60 marks ][60 markah)
Answer all questions in this section
Jswab semud soalan dalam bahagian ini
Diagram 1 shows plane waves travelling towards a barier.
4s3U2
Incident waveGelombang tuju
Normal lineGaris normal
BarrierPenghalang
Diagram 1
Rajah I
(a) What is the incident angle in Diagram I ?
Berapakah sudut tuju dalam Rajah l?
ForExaminer's
Use
L (a)
I mark]fl markahl
[Turn page overCONFIDENTIAL453112 @ Hak Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
1 (b)
1 (c)
Total A1
CONFIDENTIAL
(D) Complete the following sentenc€*
bracket.
Lengkapkan ayat berikut dengan men
The angle of incident isSudut tuju adalah
(c) Draw the pattern and direction of the reflected waves on Diagram 1.
Lukiskan corak dan arah gelombang pantulan pada Rajah I.
the same assdmct
453L12
the
ddiiaf rrflectionsudut pantulan
[1 mark]\$ markahl
[2 rnarks][2 markahl
col{m\$rtALA\$Ln o Hak cipta Bahagian MARA
i :a-n: "..,.-*;* L*
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4s3t/2
2 Diagram 2.1 shows the reading of a measuring instrument when the switch is open.Diagram 2.2 shows the reading of the measuring instrument when the switch isclosed.
Raiah 2.1 menunjukkan bacaan satu alat pengukur apabila suis dibuka.Raioh 2.2 menunjukkan bacaan alat penguhr tersebut apabila suis ditutup.
Diagram 2.1
Rajah 2.1
"a,
Diagram2.2Rajah 2.2
[Turn page overCONFIDENTIAL453t12 @ Uat< Cipta Bahagian Pendidikan Menengah MARA
ForExaminer's
Use
(D(D
o'(,o)E(D
.U'
(D(Dct:r(D3o)o)-IEl=t=I=klutIC,l-lololc)l5t:lo-I(Dl=Itrlol3
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
2 (a)
2 (b)(i)
2 (b)(ii)
2 (c)(i)
2 (c)(ii)
Total A2
CONT'IDENTIAL 453u2
2 (a) What is the physical quantity measure{Apakah kuantiti fizik yang diukur oleh alat ini?
" " ":!!r.lqf il'"!.]..t-';1"..'". t " " " j ; ;j; lfl markah]
(b) (0 Name the type of zerc error found on this instrument.Namakan jenis ralat sifur yang terdapat pada alat ini.
I I mark]I I markahl
(ii) What is the value of the zero error in Diagran2.l?Berapakah nilai ralat sifar dalam Rajah 2.1?
I l mark][l markahl
(c) (D What is the reading shown in Diagran2.Z?Berapakah bacaan yang dilunjukkan dalam Rajah 2.2?
What is the actual reading of the physical quantity measured?
B er ap akah b ac aan s eb enar ku antiti fizik y ang diukur?
I l mark]ll markahl
(iD
I l mark]ll markahl
CONtr.TDENTIAL453112 @ IIat< Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 7 453112
3 Diagram 3 shows a copper rod being moved downwards between two oppositemagnetic poles. I
The rod is connected to the galvanometer and the pointer of the galvanometer
. deflects.
Raiah 3 menunjukkan rod kuprum digerakkan ke bowah di antara dua kutub magnet yqngbertentangan. Rod disambung ke galvanometer dan jarum penunjuk galvanometer terpesong.
Copper rodRod laryrum
Diagram 3
Rajah 3
(a) (i) Name the physical quantity shown by the deflection of galvanometer.N am a kan ku ant it i fi z ik y an g dit unj ukkan o I eh p e s o n gan ga lv an o m et er.
I l mark]ll markah)
How is this physical quantity produced?Bagaimanakah kuantiti fizik ini dihasilkan?
I l mark][l markahl
[Turn page overCONF'IDENTIAL
(ii)
3i
!':!
II
Direction of motion
453112 @ Uat Cipta Bahagian Pendidikan Menengah MARA
ForExaminer's
Use
3 (a)(D
3 (aXii)
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
3 (b)
3 (c)(i)
m
3 (c)(ii)
3(A
Total A3
CONFIDENTIAL
(b) Name the physics principle involved in Diagram 3.Namakan prinsipfizikyang terlibat dalam Rajah j.
4s3112
(c) (i)
I l mark]ll markahl
What happens to the galvanometer pointer when the rod is movedupwards?
Apakahyang berlaku kepada penunjuk galvanometer apabila rod digerakkan ke atas?
@
I l mark][l markahJ
(ii) Name the physics rule used to determine the direction of the physicalquantity measured.
Namakan hukum fizik yang digunakan untuk menentukan arah kuantiti fizikal yangdiukur.
I I mark][l markahl
State one way to increase the degree of deflection of the pointer in Diagram 3.
Nyatakan satu kaedah untukmeningkatkan darjah pesongan jarum penunjuk dalam Rajah 3.
I l mark]ll markah)
453112 @ FIat< Cipta Bahagian Pendidikan Menengah MARA CONF'IDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
4 Diagram 4 shows an alrangement of logic gates in an electronic device.Rajah 4 menunjulckan susunan bagi get-get logik dalam satu peranti elektronik.
Diagrarn 4Rajoh 4
Based on Diagram 4,Berdasarkan pada Rajah 4,
(a) Name the logic gate P.
Namakan get logik P.
4s3ll2
ts{aiF(a
Ht-vDrUIHt"lnF>(
rHF
II
4 (a)
I l mark]fl rnarkahl
(b) (i) Write a Boolean algebra for gate P.
Tuliskan algebra Booleqn untuk get P.
4 (6)S)
I l markJ[l markah]
[Turn page overCONIFIDENTIAL
*,,
*wB.
%:
#,?,€.*E:
s:
453112 @ Hat< Cipta Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
4 (DXii)
CONFIDENTIAL t0 453L12
(ii) Complete the truth table below,fgf,the,s,utput C, D and X.Lengkapkan jadual kebenaran di bawalt bagi=gutpat G D dan X.
[ 3 marks]13 markah)
Using two switches, one dry cell and a bulb, draw an electric circuit which
represents a logic gate where the output is the same as X.
Menggunakan dua suis, satu sel kering dan mentol, lukiskan satu litar eleldrik yang mattakili
get logik yang outputnya sama dengan X.
J--/-
aSwitch A
Suis A
Swirch BSuis B
Dry celISel kerins
BulbMentol
[ 2 marks ][2 markah]
(c)
4 (c)
Total A4
453112 @ gat Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 453u2
5 Diagram 5.1 shows an audio generator connected to a speaker.Raiah 5.1 menunjukkan sebuah penjana audio disambungkan kepada pembesar suara.
Diagram 5.1Rajah 5.1
Diagrams 5.2(a) and 5.2(b) show the diffraction of sound waves from the samespeaker when sounds of low and high frequency are produced respectively.
Rajah 5.2(a) and 5.2(b) masing-masing menunjukkan pembelauan gelombang bunyi dari pembesarsuara yang samq opabila bunyi denganfrelarcnsi rendah danfrekuensi tinggi dihasilkan.
www. myschoolchildren.com
11
iI\$!FI
F
Low frequency soundBunyi b erfr ekuensi r endah
Diagram 5.2 (a)Rajah 5.2 (a)
High frequency soundBunyi b erfrekuens i t inggi
SpeakerPembesar suara
Diagram 5.2 (b)Rajah s.2 (b)
ia-
*is,rr-l
\$,
;.
[Turn page overCONFIDENTIAL
Audio generatorPeniana audio
453112 @ ffat Cipta Bahagian Pendidikan Menengah MARA
ForExaminer's
Use
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
s (b)
s (c)(i)
s (c)(ii)
s (4(i)
s (4(ii)
s (e)
Total A5
CONF'IDENTIAL 12,
(") What is the meaning of diffraciion? r -r: i., l
Apa yang dimat<sudkan dengan pembelauan?,;
4s3u2
I l mark]ll markahl
(b) In Diagram 5.2(a) mark and label the wavelength with t.Dalam rajah 5.2(a) tanda dan labelkan jarak gelombang dengan )..
Observe Diagram 5.2(a) and Diagram 5.2(b).Perhatikan Rajah 5.2(a) dan Rajah 5.2@.
(c) (i) Compare the wavelengths.B andingkan j or ak gelomb ang.
I l mark]ll markahl
(ii) Compare the shape of the diffracted sound waves.B andingkan b entuk p em b el auan gel omb ang bunyi.
Ilmark]IL markah]
(i) What happens to the amplitude of the diffracted sound waves?Apa akan terjadi kepado amplitud gelombang bunyi yang terbelau?
Ilmark]ll markahl
[ 2 marks ]12 markahl
(e) State the relatiorship between wavelenglh and the amplitude ofthe difupted waves.Nyatakan hubungan antara panjang gelombang dan amplitud gelombangyang terbelau.
Ilmark]ll markahl
CONFIDENTIAT
I l mark]fl markahl
@
!l
dI
IiI{T
sITila
f;
*
453112 @ ftaf Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 13 \$\$i I.*"HI
6 Diagram 6.1 and Diagram 6.2 show trapped gas being heated in an enclosed space.
Rajah 6.1 dan Rajah 6.2 menunjuklcan gqs yqng terperangkap sedang dipanaskan di dalam ruqngtertutup.
Atmospheric PressureTekanan Atmosfera
Gas moleculeMolekul gas
40 0c
Atmospheric PressureTekanan Atmosfera
70 0c
Heat source -Sumber haba
Diagram 6.1Rajah 6.1
(a) State the SI unit for temperature.Nyatakan unit SI bagi suhu.
Diagram 6.2Rajah 6.2
ti!r:trttttt:
Gas moleculeMolekul gas
6 (a)
6 (DXi)
I l mark]ll markahl
6 (bxii)
[Turn page oYerCONFIDENTIAL
(b) Based on Diagr€rms 6.1 and 6.2,Berdasarkan Rajah 6.1 dan 6.2,
(i) Compare the volume of the trapped gas.
(ii) Compare the temperature of the trapped gas.
Bandingkan suhu gas yarug terperangkap.
453112 @ Uat< Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL L4
Compare the pressure exerted on the piston.Bandingkan tekananyang dikenakan ke atas omboh.
(iii)
(c) State the relationship between volume and temperature of the gas.
Nyatakan hubungan antara isipadu dan suhu gas.
453u2For
Examiner'sUse
6 (bxiii)
6 (c)
m1
6 (e)
[ETotal A6
[E
I I mark]ll marlcahl
@ Name the Law associated with the relationship above.
Namakan Hukum yang dikaitkan dengan hubungkait di atas.
I l mark]fl markah]
(e) A gas of volume 35 m3 at temperature 40oC is heated at a fixed p_ressure.
Calculate the volume of the gas when its temperature reaches 70oC,
Kirakan isipadu gas opabila sultu mencapai 7fC.
[ 2 marks ]12 markahl
453L12 @ fUk Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 15 453u2
Diagram 7.1 shows dinosaur fossil unearthed at an archaeological site.Raiah 7.1 menunjukkanfosil dinasour yang dijumpai di satu tapak arkeologi.
Dinasour fossilFosil dinasour
DiagramT.lRajah 7.1
Scientist can estimate the age of the dinosaur fossil by determining the amount ofundecayed radioisotope carbon-l4. Carbon-l4 ,'tQ decay emits beta particle. The
half-life of carbon-L4 is 5730 years.
Saintis boleh menganggarkan usia fosil tersebut dengan menentukan jumlah radioisotop karbon-L4yang masih belum mereput. Pereputan karbon-|4 memancar zarah beta. Separuh hayat karbon-l4adalah 5730 tahun.
(") What is the meaning of half-life?Apokah maksud separuh hayat?
I l mark][t markah]
(i) Based on the information above, write the equation for the decay ofcarbon- 14.
Berdasarkan maklumat di atas, tuliskan persamaan bagi reputan karbon-|4.
I l mark][l markah\
[1mark]fl markahl
[Turn page overCONFIDENTIAL
(b)
(i0
453112 @ ttak Cipta Bahagian Pendidikan Menengah MARA
7 (bxii)
H'S-ii.,,rl;
ForExaminer's
Use
7 (a)
7 (b)(i)
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use,
7 (cXi)
7 (cXii)
7 (c[iii),
CONFIDENTIAL 453u2
(c) (i) Sketch the graph of activity against time to show the decay of carbon-l4.(The activity of carbon-l4 is 16 counts per minute in living organisms)
Lakarkan graf aktiviti melawon mqsa untuk reputan karbon-!4.(Aktiviti karbon-|4 dalam benda hidup adalah 16 bilangan per minit)
ActivityAHiviti
TimeMasa
Diagtam7.2Rajah 7.2
[2 marks ]12 markahl
(ir) Show how you determine the halFlife of carbon-L4.Tunjul;kan bagaimana anda menentukan nilai separuh hayat karbon-|4.
[ 1 mark]f I markah)
if the current decay rate of
[ 2 marks ][2 markahl
4\$ln @ ftat Cipta Bahagiao Pendidikan Menengah MARA
t6
%.B
E
!I
1,6
(iii) Deterrnine the age of the dinosaur fossilcarbon-14 is 2 cormts p.er minute.
Tentukan umur fosil dinasour tersebut sekiranyaadalahZ bilangan per minit.
CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
BFr
FEI
F
B
I
F
E
r
F
E
rI
rE
t
tE
t
r
CONFIDENTIAL 17
@ The carbon-I4 decay releases 2.56 x l0-r5 J of energy.Calculate the mass defect.
Pereputan karbon-|4 membebaskan 2.56 x l0 t5 J tenaga.Hitungkan kecacatan j is im.
453u2
[ 2 marks ][2 markahl
[Turn page overCONFIDENTIAL453112 @ ftat Cipta Bahagian Pendidikan Menengah IvIARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
8 (a)
CONF'IDENTIAL
8 Diagram 8 shows a 150 kg billboard hung
I8
by
453112
two identical cables and where all the
\\:-S---
--s-//
forces are in equilibrium.
CableKabel
//I
IIIII
\\\
(a)
Diagram 8Rajah 8
What is the meaning of forces in equilibrium?Apakah maksud keseimbangan daya?
I l mark]I L marlmhl
453t12 @ ftak Cipta Batragian Pendidikan Menengah MARA CONF'IDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
(b) (i)
(c) (i)
(ii)
t9 4s3u2
In Diagram 8, mark and label the weight, W of the billboard, the tensionsof the cable, T1 and Tz and their directions.
Pada Rajah 8, tanda dan labelkan berat papan iklan, W, tegangan kabel T1 dan T2 sertaarahrrya.
[ 2 marks ]12 markahl
Calculate the weight of the billboard.Hitung berat papqn iHan tersebut.
I l mark]ll markahl
In the space below, sketch a diagram of triangle of forces that act on thebillboard in Diagram 8.
Pada ruang di bawah, lakarkan rajah segitiga keseimbangan dayo yang bertindak keatas papan iHan dalam Rajah 8.
,
\
[ 2 marks ]12 markahl
(ii) Based on the diagram of the triangle of forces, write the relationshipbetween W, Tt and Tz.
Berdasarkan lukisan rajah segitiga keseimbangan dryo, tuliskan hubungan antara W,
T dan Tz.
I l mark]f I markahl
[Turn page oYerCOI\FIDENTIAL453112 @ Uat< Cipta Bahagian Pendidikan Menengah MARA
ForExaminer's
Use
8 (D)(i)
I (bxii)
8 (cXi)
8 (cXii)
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
8 (dxi)
CONF'IDENTIAL 20 453Lt2
@ Table 8 shows the characteristics and a:rangement of cables for hanging theheavy billboard.
Jadual 8 menunjukkan ciri-ciri dan susunon kabel untuk menggantung papan iklan yang berat.
CableKabel
Morimum tension can be supported by the cable' Tesanpan maksimum vane daoat ditamouns oleh kabel
Angle / 0Sudut / 0
PHighTinssi
600
a LowRendah
600
R LowRendah
300
Based on Table 8, state the suitable characteristics of the cable to be used forhanging the heavy billboard.
Give one reason for the suitability of each characteristic.Berdasarkan Jadual 8, nyatalcan ciri-ciri kabel yang sesuai digunakan untuk menggantungpapan iHan yang berat.
Berikan satu sebab untuk kesesuaian setiap ciri tersebut.
(i) Morimum tension which can be supported by the cableTegangan malcsimum yang dapat ditampung oleh kabel
ReasonSebab
[ 2 marks ][2 markahl
EIqtctEIEl-clololot-Elolal>{EI
'I=t=l
453112 @ ftak Cipta Batragian Pendidikan Menengah MARA CONBIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4531t2
(ii) The angle of dSudut 0
ReasonSebab
[ 2 marks ][2 markah]
(e) Based on your answer in8(d), determine the most suitable cable for hanging thebillboard.
Berdasarkon iawapan di 8(d), tentukan kabel yang paling sesuai untuk digunakan bagimenggantung p apan ikl an.
IL markah]
[Turn page oYerCOIIFIDENTIAL
21
8 (e)
8 (4Gi)
Total A8
ffiLl2,Q ffat Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 22
Section BBahagian B
[ 20 marks ][20 markahl
Answer any one question from this sectionJawab mana-mana satu soalan daripada bahagian ini
4s3ll2
n: 1.66p - 2.8 kg rn-'
Diagrams 9.1 afi9.2 show a light ray travelling through two gemstones withdifferent critical angle, c.
Rajah 9.1 dan 9.2 menunjukkan sinar cahaya melalui 2 batu permata yang berlainan sudut genting, c.
n:2.42p:3.5kgtn-'
453L12 @ ftat Cipta Bahagian Pendidikan Menengah MARA
Diagram 9.1Rajah 9.1
(a) What is the meaning of critical angle?Apakah maksud sudut genting?
I I mark]ll markahl
(b) Based on Diagrams 9.1 and 9.2, compare *re critical angla, c, density, p ffiLd
, refractive index, zl.
Berdasarkan kepada Rajah 9.1 dan 9.2, bandingkan sudut genting, c, ketumpatan, p dan indeks
biasan' n' t 3 marks ]13 markahl
(c) State the relationship between refractive index andNyatakan hubungan antara indeks biasan dan
(i) DensityKetumpatan
(ii) Critical angleSudut genting
[ 2 markah ]12 markahl
CONFIDENTIAL
Diagram 9.2Rajah 9.2
c :24.4o c: 37 .0"
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTTAL 2g 4531/2
@ Diagram 9.3 shows a submarine whieh uses a glass prism periscope to seeobjects above the water surface.
Rajah 9,3 menunjukkan sebuah kapal selam yqng menggunakan periskop prisma kaca untukmelihat objek di atas permukaan air.
Diagram 9.3Rajah 9.3
With the help of a diagram, explain how the observer in Diagran 9.3 can see
objects above the water surface.
Dengan bantuan gambarajoh, terangkan bagaimana pemerhati dalam Rajah 9.3objekyang berada di atas permukaan air.
boleh melihat
[ 4 marks ]14 markah)
[Turn page overCONFIDENTIAL453112 @ Hat< Cipta Bahagian Pendidikan Menengah MARA4
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
(e) Diagram 9.4 shows
Inner coreTeras dalam
9.4.4
You are required to give some suggestions in designingendoscope.Explain your suggestions based on the following aspects:
Anda dikehendaki memberikan beberapa cadangan untuk mereka bentukcekap.
T er an gkan c a dan gan and a b er d as ar kan as p ek- as p.ek b er i lafi :
24 4s3u2
an optical fibre in an endoscope which is used in medicine.
gentian optik dalam sebuah endoskop yang digunakan dalam bidang
.ffiDiagram
Rajah 9
(i) Comparison between the refractive index of the inner core and the outercladding.
Perbandingan antara indela biasan teros dalam dan pembalut luar
(ii) Purity of the inner core
Ketulenan teras dalam.
(iii) Thickness
Ketebalan
(iv) Strength
Kekuatan
(v) Density
Ketumpatan
[ 10 marks ]110 markahf
a more efficient
endoskop yang lebih
453112 @ gat< Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 25
10 Diagram 10.1(a) shows a non-rechargeable dry cell.Diagram 10.1(b) shows a rechargeable dry cell.
Rajah 10.1(a) menunjukkan sel kering boleh dicas semula.Rajah 10.1(6) menunjukkan sel kering tidak boleh dicas semula.
Diagram 10 (a)Rajah 10 (a)
4531/2
Diagram 10 (b)Rajah t 0 (b)
Table 10 shows the characteristics of the rechargeable and non-chargeable dry cells.
Jadual l0 menunjukkan ciri-ciri sel kering yang boleh dicas semula dan sel kering yang tidak bolehdicas semula.
(a) What is the meaning of electromotive force?Apakahyang dimaksudkan dengan daya gerak eleWrik?
I l mark]ll markahl
(b) Using Table 10, compare theMenggunakan Jadual 10, bandingkan
(i) electromotive force between the rechargeable and non-rechargeable drycells.
daya gerak elek*ik antara sel kering baleh dicas semula dan sel kering tidak boleh dicassemula.
(ii) current between the rechargeable and non-rechargeable dry cells.arus antara sel kering boleh dicas semula dan sel kering tidak boleh dicas semula.
(iii) internal resistance between the rechargeable and non-rechargeable dry
cells.rintangan dalam qntara sel kering boleh dicas sernula dan sel kering tidak boleh dicas
semula.
[Turn page overCOSI-FIDEf.{TIAL
CharacteristicsCiri-ciri
Rechargeable dry cellSel kering boleh dicos semula
Dry cellSel kering tidak boleh dicas
semula
Electromotive forceDaya gerak eleWrik
1.5 v 1.5 V
CurrentArus
0.6 A 0.4 A
Internal resistanceRintangan dalam
0.50 c) 1.75 {t
453112 @ nat Cipta Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
]CONFIDENTIAL 453112
www. myschoolchildren. com(c) (i)
(ii) Deduce the relationship between internal resistance and voltage drop.Rumuskan hubungan antara rintangan dalam dan kejatuhanvoltan.
[ 5 marks ]15 markah)
@ Explain why batteries connected in parallel and in series will affect theeffectiveness of the electromotive force of the batteries differently.
Terangkan mengapa bateri yang disambungkan secara selari dan sesiri memberi kesan berbezapada kecekapan dayo gerak elektrik bateri.
[ 4 marks ][4 markah]
(e) Diagram 10.3 shows a hand dryer.This hand dryer needs to be switched on to use it. It also takes a long time todry hands.
Rajah 10.3 menunjukkan sebuah pengering tangan.Pengering tangan perlu dihidupkan suis untuk menggunakannya. Ia juga mengambil masa yanglama untuk mengeringkan tangan.
SwitchSzls
Diagram 10.3Rajah I0.i
26
Relate internal resistance and current.Hubungkaitkan rintangan dalam dan arus
453112 @ Uat Cipta Bahagian Pendidikan Menengah MARA CONF'IDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 27 4531/2
You are required to give some suggestions to design a hand dryer which is moreefficient.
Using the knowledge on electricity and electronics, explain your suggestionsbased on the following aspects:
Anda dikehendaki memberi beberapa cadangan untuk merekabentuk sebuah pengering tanganyang lebih cekap.
Menggunakan pengetahuan tentang elektrik dan elektronik, terangkan cadangan andab er d as ar kan asp ek- asp e k b er ikut :
(i) The power of the fanKuasa kipos
(ii) The diameter of the heating elementDiameter unsur pemanas
(iii) Material for the heating elementBahanunsur pemanas
(iv) Safety featuresCiri keselamatan
(v) Device to replace the switchAlat bagi menggantikan suis
[ 10 marks ]ll0 markahl
[Turn page oYerCONFIDENTIAL453112 @ nak Cipta Bahagian Pendidikan Menengah IvIA!,d
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 28
Section CBahagiun C
[ 20 marks ][20 markah]
Answer any one question from this sectionJawab mana-mana satu soalan daripada bahagian ini
11 Diagram 11.1 shows a boy looking at a lorry moving in front of him.Rajah I1.1 menunjukkan seorang budak lelaki melihat sebuah lori bergerak dihadapannya.
(a) (i)
Diagram 11.IRajah I 1.1
What is the meaning of pressure?Apakah yang dimalcsudkan dengan tekanan?
453Lt2
I l mark]ll markahl
(ii) When the speeding lorry moves in front of the boy, he feels he is pulledtowards the lorty.
By using appropriate physics concept(s), explain the above situation.
Apabila lori yang sedang memecut melintas di hadapan budak itu, dia berasa tertarik kearah lori.
Dengan menggunakan konsepfizikyang sesuai, terangkan situasi di atas.
[ 4 marks ][4 markah)
..\ t/ /
453L12 @ gat Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 29 4531/2
movement of a ball after being kicked from a(b) Diagram Ll.z shows the curvedcorner.
Raiah I 1.2 menuniukkan pergerakan melengkung sebiji bola selepas ditendang dari satu sudut.
Region AKanasan A
Region BKawasan B@.
Diagram 11.2Rajah 11.2
Based on Diagram 11.2,Berdasarkan Raj ah I 1.2,
(i) Which region experienced low pressure?Kawasan manakah yong mengalami tekanan rendah?
(ii) Explain how the curving effect is produced.Ter an gkan b a g aim an aka h kes an m e I en ghtn g t erj a d i.
I I mark]I I markah)
[ 2 marks ]12 markah)
reduced when a football match takes(iii) Explain why the curving effect isplace during heavy rain.
Terangkan mengapakah kesan lengkungan berkurang apabila perlawanan berlangsungketika hujan.
[ 2 marks ][2 markah]
[Turn page overCONFIDENTIALtfi\$31,12 @ gar Cipta Bahagian pendidikan Meneqlgah !vIARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 30 4531t2
(c) Table I 1.1 shows the characteristics of four different track helmets used for atrack cycling competition in a velodrome.
You are required to determine the most suitable helmet.
Jadual II.1 menunjukkan ciri-ciri bagi empat topi keledar trek berlainan yang digunakandalam pertandingan trek berbasikal di dalam velodrom.
Anda dikehendaki menentukan topi keledar trekyang paling sesuai.
Study the specifications of all the four helmets based on the following aspects:Kaji spesifikasi keempat-empat topi keledar trek berdasarkan aspek aspek berikut:
(i) Front end of helmetB ahagian hadapan helmet
(ii) Material for inner shellBahan untuk bahagian dalam
(iii) V/idth of strapKelebaran tali
(iv) Presence of air holeKehadiran lubangudara
Explain the suitabihty of each characteristic of the track helmets and determinethe most suitable track helmet to be used for the indoor cycling competition.
Terangkan kesesuian setiap ciri topi keledar trek dan tentukan topi keledar trek yang palingsesuai untuk digunakan dalam pertandingan di dalam danan.
Beri sebab-sebab bagi pilihan anda.
[ 10 marks ]lLO markahl
#
\$54*4II
ij
l
453112 @ Uat< Cipta Bahagian Pendidikan Menengah MARA COI\-FIDENTIAI
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4537/2
Table I I .1Jadual I LI
[Turn page overq{tSUZ @ nat cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
3I
HelmetTopi
Keledar
Characteristics of Track HelmetCtrl-clri Topl Keledar Trek
K
Material for interior shell:Rubber
Bahan bahagian dalam:Getah
Narrow strapTali halus
Front end: Short & roundBahagian hadapan: Pendek & butat
L
Material for interior shell:Rubber
Bahan bahagian dalam:Getah
Na:row strap Front end: Long & pointedBahagian hadapan: Panjang& tirusTali halus
M
Material for interior shell:Polystyrene
Bahan untuk bahagian dalam:Polisterina
Wide strapTali lebar
Front end: Long & pointedBahagian hadapan: Ponjang & tirus
N
Front end: Short & roundBahagian hadapan: Bulat
Material for interior shell:
Polystyrene
Bahan untuk bahagian dalam:Polisterin
Wide strapTali lebar
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 33
12 Figure l2.l(a) shows an n-type semiconductor produced by a doping process.Raiah lZ.l(a) menunjukkan semikonduktor jenis n yang dihasilkan melalui proses pendopan.
oo OO
.Ooo oo
o
.;)\-/
oo
oo
O
I
o
ooo
oo
a
oc
a
.;)VC
o
oo
.;)Vo
453U2
Extra free electronLebihan elehron bebas
ao oa
Diagram l2.lRajah 12.1
What is the meaning of doping?Apakah yang dimalcsudkan dengan pendopan?
I I mark]I I markahl
With the help of a diagram(s), explain how a semiconductor diode functions.Dengan bantuan gambaraJah, terangkan bagaimana diod semikonduhor berfungsi.
[ 4 marks ]l4 markahl
Draw a circuit to produ ce ahalf-wave rectifier and show the waveformsfor the input voltago and the output voltage.
Lukiskan lttar yang dt *t*@ untuk menghasilkan relaifikasiseparuh gelombang dantunjufi:kan benfi* galo bagl voltan input dan yoltan output.
(iv) Explain how *"
[ 3 marks ][3 markahJ
in a rectifier circuit smoothens thecr[Tent output. i
Teranglan yang disambungkan dalam litar rektifikasi
t 2 marks l12 markahl
melictnlun oan
[Turn page overCONT.IDENTIAL
o o
a
a
oo
o
(a) (i)
(iii)
(ii)
453112 @ ttat< Cipta Batragian
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
&oNFIDENTIAL 34 4s3ln I
(b) Diagram 12.2 shows the characleristics of four transistor circuits, P, Q, R ard S Iproposed for switching on lights automatically when it is dark. The lights !switch on when the base voltage is at least 5V. I
i you are required to determine the most suitable circuit II',, Jadual 12.2 merunjuk*an ciri-ciri bagi empat litar transistor, P, Q, R dan S yang dicadangkan I'. untuk memthlp suis sectra a*omatik apabila gelap.Lampu aakan menyala apabila voltan tapak 1
sehrang-htrangtya 5 Y,
Anda dikehendaki menentukan litar yang paling sesuai.
Studythe specifications of all the four circuits based on the following aspects:Kaji spesifikasi keempat-empat litar berdasarkan aspek aspek berilafi:
(i) The type of connection of the transistor at B-EJenis sambungan transistor di B-E
(ii) The electric component connecting the two circuitsAlat penyambung kedua-dua litar
(iii) The magnitude of the base voltage of the transistorMagnitud u o lt an t ap ak tr ans is t or
(rv) Resistance of thebase resistorRintangan bagi perintang tapak
Explain the suitability of each aspect and then determine the most suitablecircuit.Give reasons for your choice.
Terangkan kesesuian setiap aspek dan tentukan litar yang paling sesuai.
Berikan sebab untuk pilihan anda.
I l0 marks ]I l0 rnarkah]
453L12 @ Hat Cipta Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
TypeJenis
Connection of Base-EmitterS ambungan Tap ak- P en geluar
DiagramRajah
P
Reverse biasPincang songsang
240 V
t5V
+sY .,i-
aForward bias
RReverse bias
Pincang songsang
240Y
S
".' ;i .i.
,,\$lo iu,1,;11,:i.\$.-
:. :L.,f ,q,,
b
,3f:;; 3It-!
. ":.!-&.,.''-i-
.di. ri#,s '.l:";mi'-,ijr
IIIIIt-
CONFIDENTIAL
END OT QUESTION PAPERKERTAS SOAL/IN TAMAT
4s3u2
[Turn page ovofCONF'IDENTI''
35
(D(D
D1'o)'ro(D
a;(D(D
<t,x(D3q)a)
l=l=t<1U,l()I=lolot6-l=I::lo-l-t(DIPlc)lol3
'.g+
*frEhfgian Pendidikan Menengah MARAB.r"t
ffim .'*:ry]:S "'trT:E* T
ffifiF"t. ,rffiS-- rre&E-:<:. - .ffisrtlir'
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
453u3PhysicsPaper 3Sept 2013I % hours
4s31/3
NAME:
CLASS:
tizFr(t)
t(r)
Iah{
Fr
MAKTAB RENDAH SAINS MARA
SIJIL PELAJARAN MALAYSIATRIAL EXAMINATION 2013
PHYSICS
Paper 3
One hour and thirty minutes
DO NOT OPEN TIrrS QUESTION BOOKLET trNTrL TOLD TO DO SO
1.
2.
3.
Write down your n€Lme, college no.and your class in the space provided.Tulis nqmq, no. maktab dan kelas anda padaruangyqng disediakan.
The questions are written in Englishand bahasa Melayu.Kertas soalan ini adalah dalam dtyibahasa.
Candidates' are required to read theinformation at the back of thebooklet.Calon dikehendaki membaca maklumat dihalaman belakang buku soalan ini.
www. mvschoolchildren.com
For Examiner's Use
Section Question Marks Score
A1 t6
2 l2
B1 12
2 1.2
Total
This booklet consists of printed 19 pages and
453113a- 2013 Copyright Bahagian Pendidikan Menengah MARA
I blank page
[Turn overCONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 2
Section ABahagian A
128 marksJ[28 markah]
Answer all questions in this sectionJawab semuo soalan dalam bohagian ini
4531t3
A student ca:ries out an experiment to investigate the relationship between the timetaken for ice to melt, r and the mass of meltedice, m.
The arrangement of the apparatus is shown in Diagram I .l .
Seorang murid menjalankan satu elrsperimen untuk mengkaji hubungan antara masa untuk aismelebur, t dan jisim ais yangtelah melebur, m.
Heater -----------+Pemanas
To power supplyKe bekalan kuasa
IceAis
Retort standKaki retot
BeakerBikar
oo
o
Lever balanceNeraca tuas
Diagram 1.1Rajah l.l
453ll3a 201,3 Copyright Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 4s3u3
The mass of the empty beaker, m, is 100 g.
The heater is switched on and the stopwatch started simultaneously. At time,t : 20.0 s, the mass of the melted ice in the beakat, //t; on the lever balance isrecorded. Diagram 1.2 shows the reading of the lever balance.
The experiment is continued and the reading of the lever balance is recorded at
| - 40.0 s, 60.0 s, 80.0 s and 100.0 s. The coffesponding readings of the leverbalance are shown in Diagrams 1.3, 1 .4, L5 and 1.6.
Jisim bikar kosong, mo adalah 100 g.
Pemanas dan jam randik dihidupkan serentak Selepas masa, t bersamaan 20.0 s, jisim ais yang telahmelebur di dalam bikar, mi di atas neraca tuas dicatatkan. Rajah 1.2 menunjukkan bacaan padaneraca tuas.
Elrsperimen diteruskan dan bacaan pada neraca tuas dicatatkan untuk masa, t : 40.0 s, 60.0 s, 80.0 s
dan 1.6.
Diagram 1.2Rajah 1.2
4531/3@ 2Ol3 Copyright Bahagian Pendidikan Menengah MARA
t
ftli
m
_ 20.0 s
o6
g
[Turn overCONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
Diagram 1.3Rajah l.j
4s3Lt3
| - 40.0s
flli :
m: o6
4531/3O 2013 Copyright Bahagian Pendidikan Menengah MARA CONFIDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAI,
Diagram 1.4
Rajah 1.4
4S3tR
t - 60.0s
lTli :
m: o6
[Turn overCONFIDENTTAL
5
453113c^ 2013 Copyright Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
Diagram 1.5
Rajah 1.5
453113
t - 80.0s
tlli :
m: o6
453I-l3@ 2013 Copyright Bahagian Pendidikan Menengah MARA CONFIDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
Diagram 1.6
Rajah 1.6
4s3U3
t - 100.0 s
t/7i :
m: o6
[Turn oYerCONFIDENTIAL4531/3@ 2013 Copyright Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
(a)
www.mvschoolchildren.com 4\$Ll3
For the experiment described on pages 2 and 3, identify:Bagi el<sperimen yang diterangkan di halaman 2 dan 3, kenal pasti:
The manipulated variableP em b o I ehub ah dim anipul as ikan
| 1. marklll markahl
(ii) The responding variableP em b o I ehub ah dim al ar kan
| | markllI markahl
(iii) The constant variableP em b o lehub ah dim al ar kan
I I marklf I markahl
(b) For this part of the question, write your answers in the spaces
provided in the corresponding diagrams.Untuk bahagian soalan ini, tulis jawapan anda dalam ruang yang disediakan
dal am r aj ah-r aj ah yqng s epadan.
(i) Based on Diagram 1.2, 1.3, 1.4, 1.5 and 1.6 on pages
7, record the readings of mi.
Berdasarkan Rajah 1.2, 1.3, 1.4, 1.5 dan 1.6 di halaman 3,4, 5,
bacaan mi.
(i)
3, 4,5, 6 and
6 dan 7, catatkan
[2 marks ]12 markah'l
4531/30 2013 Copyright Bahagian Pendidikan Menengah MARA CONF'IDENTIAL
1(aXi)
1(aXii)
1(aXiii)
1(6Xi)
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use|!:raF{a
taI(nl.rlrlA{
1(c)
[trr(d)
[trTotalA1m
1(bxiil
1(bxiii)
CONFIDENTIAL
(c)
(d)
9 4\$Lt3
(ii) Calculate m for each value of mi in l(b)(i) using equation:
m : (m,- mo)
Record the value of m onpages 3,4,5,6 and7.
Hitungkan nilai m bagi setiap nilai mi di l(b)(i) dengan menggunakan
persqmaan:
m: (mi- ms)
Catatkan niloi m pada halaman 3, 4, 5, 6 dan 7.
[2 marlcs J
[2 markah)
(iiD Tabulate your results for t, mi and m inthe space below.Jadualkan keputusan anda bagi semua nilai t, mi dan m dalam ruangan dibm,yah.
13 marks l13 markahl
On the graph paper on page 10, plot a graph of m against f.Pada sehelai kertas graf di halaman 10, lukiskan graf m melawan t.
| 5 marks ll5
.markahl
Based on your graph in 1(c), state the relationship between m an.d t.
Berdasarkan grafanda di 1(c), nyatakan hubungan antara m dan t.
| | mark)[l markoh]
[Turn overCONFIDENTIAL4531/3@ 2013 Copyright Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
ForExaminer's
Use
2(a)(i)
2(a\(ii)
4531/3@ 2013 Copyright Bahagian Pendidikan Menengah MARA
CONFIDENTIAL 11 4537t3
A student carries out an experiment to investigate the relationship betweenthe length of a simple pendulum, {, and the perio d, T.
In this experiment, a slotted weight is hung on a string that is attached to aretort stand. The slotted weight is pulled to the side slightly and released.The time for 20 complete oscillations is recorded. The results of theexperiment are shown in the graph f against { inDiagram 2.! onpage 13.
Seorang murid menjalankan elcperimen bandul ringkas untuk menyiasat hubungan
antara panjang bandul, t dengan tempoh, T.
Dalam elrsperimen ini, pemberat digantung pada seutas tali yang disambung pada kakiretot. Pemberat drtarik ke tepi dan dilepaskan. Masa untuk 20 ayttnan lengkap
direkodkan. Keputusan eksperimen tersebut ditunjukknn oleh graf t' mela,nan I padaRajah 2.1 di halaman 13.
(a) Based on the graph in DiagranZ.L:Berdasarkan grafpada Rajah 2.1:
(i) State the relationship between t' and t.Nyatakan hubungan antara f dengan t.
| | marklI I markahJ
(ii) Determine the value of T when f,:0.35 m.
Show on the graph, how you detennine the value of T.
Tentukan nilai T apabilal : 0.35 m.
Tunjulckan pada graf bagaimana anda menentukan nilai T.
rrt -t- 13 marlcs l
13 markahl
[Turn overCONF'IDENTIAL
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL12
(b) calculate the gradient, m, of thegraph t' againstt.show on the graph how you determine the varue of m.
Hitung kecerunan, m, bagi graf t' melawan &.Tunjukkan pada graf itu bagaimana anda menentukan nilai m.
m:[3 marks
[ 3 markah
453113@ zor3 copvright Bahagian pendidikan Menengah MARA COI\TFIDENTTAL
4531/3
t
i
l
{
2(b\m
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
0.4
Diagram 2.1
Rajah 2.1
13
Graph of. t' against IGraf T melawan {
453u3
0.7 / (m)
[Turn overCONFIDENTIAL4531/30 2013 Copyright Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL t4
(c) The gravitational acceleration, g, is calculated using the equation:
r: 39'49
m
Use your graph to determine the gravitational acceleration.
Pecatan graviti, g, dikira menggunakan persamaan:
r:39'49m
Gunakan graf anda untukmenentukan pecutan groviti.
o: * S-26
12 marks ff 2 markahl
@ (i) Predict what happens to the gradient of the graph, m if theexperiment is carried out on the moon.Ramalkan apakahyang berlalru kepada kecerunan graf, m sekiranyaeksperimen ini dijalankan di bulan.
I L marklll markahl
(ii) Give one reason for the answer inz (d)(i)Berikan satu sebab bagi jawapan di 2(d)(i)
[ | marklf I markahl
(e) State one precaution that should be taken to improve the result of thisexperiment.Nyatakan satu langkah berjaga-jago yang perlu diambit untuk memperbaikikeputus an elcsp erimen ini.
I L marklf I markahl
4531/3O 2013 Copyright Bahagian Pendidikan Menengah MARA CONFIDENTIAL
453u3
2(c'lE2(AG\m2(d)(ii\[r
Z(e)
[trTotalA2E
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL
Observer's eye at position AMata pemerhati pada kedudukan A
Diagram 3.1
Rajah 3.1
Observer's eye at position BMata pemerhati pado kedudukan B
Diagran3.2Rajah 3.2
4531t3
ll marklfl markahf
l1 markJll markah)
[Turn overCONFIDENTIAL
15
Section BBaltagian B
t 12 marlrs l[ 12 markahl
Answer any one question from this sectionJawab manq-mana satu soalan daripada bahagian ini.
Diagram 3.1 and Diagram 3.2 show the path of a light ray from a fish to an
observer's eye. It is observed thatthe bending of light is different when theposition of the observer changes.
Rajah 3.1 dan Rajah 3.2 menunjukkan rajah sinar dari ikan ke mata pemerhati. Didapatipembengkokan cahaya adalah berbeza apabila kedudukan mata pemerhati berubah.
Observe the magnitude of the angle, d in both diagrams.Perhatikan nilai sudut, 0 pada kedua-dua rajah.
Based on the information and observation ;
B er d as ar kan m aklum at dan p em er h at i an t ers ebut ;
(a) State one suitable inference,Nyatakan satu inferens yang sesuai.
(b) State one suitable hypothesisNyatakan satu hipotesis yang sesuai
-lr-t r-v l-
: \$bld[r* r- ---
13 Copyright Bahagian Pendidikan Menengah MARAhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL t6 453u3
(c) With the use of apparatus such as a glass block, a ruy box and otherapparutus, describe an experiment to investigate the hypothesis statedin 3 (b).Dengan menggunakan radas seperti blok kaca, kotak sinar dan radas lain,terangkan satueksperimen untuk menyiasat hipotesis yang dinyatakan di 3(b).
In your description, state clearly the following :
Dalam penerangan anda nyatakan dengan jelas perkara berikut:
(i) The aim of the experimentTujuan eksperimen
(ii) The variable in the experimentP embolehubah dalam elcsperimen
(iii) The list of apparatus and materialsSenarai radas dan bahan
(iv) The arrangement of the apparatusSusunan radas
(v) The procedure of the experiment, which includes one methodof controlling the manipulated variable and one method ofmeasuring the responding variableProsedur eksperimen yang mesti termasuk satu kaedah mengawal pembolehubah dimanipulasikan dan sata knedah mengukur pemboleh ubah bergerakbalas.
(vi) The way to tabulate the dataCara menjadualkan data
(vii) The way to analyse the dataCara menganalisis data
ll0 marks J
I l0 marlcs ]
453113a 201,3 Copyright Bahagian Pendidikan Menengah MARA CONF'IDENTIALhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL t7 453u3
Diagrams 4.1 and 4,2 show a moving coil ammeter. The magnitude ofcurrent in Diagram 4.1 is lower than in Diagran 4.2.
Concave magnetMagnet cekung '
CoilGegelung
Diagram 4.1Rajah 4.1
Observe the deflection of the pointer in both diagrams.P er hat ikan p es o n gan j arum p enuniik p a da ke du a- du a, oj oh.
Based on the information and observation ;B er d as ar k an m aHum at d an p em er h at i an t er s ebut ;
(a) State one suitable inference,Nyatakan satu inferens yang sesuai.
(b) State one suitable hypothesisNyatakan satu hipotesis yang sesuai
Diagram 4.2Rajah 4.2
f I marklll markahl
| | marklll markahl
.Aqffi inapyright Bahagian Pendidikan Menengah MARA
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly
CONFIDENTIAL 18 453u3
(c) With the use of apparatus such as a d.c power supply, ammeter,C-shaped steel yoke, bare copper rod and other apparatus, desqibe anexperiment to investigate the hypothesis stated in a @).
Dengan menggunakan radas seperti bekalan a.t, ammeter, dening besi berbentuk-C, dawai kuprum tak berpenebat dan rqdas lain, terangkan satueksperimen untukmenyiasat hipotesis yang dinyatakan di 4(b).
In your description, state clearly the following :
Dalam penerangan anda nyatakan dengan jelas perkara berikut:
(i) The aim of the experimentTujuan eksperimen
(ii) The variable in the experimentP embo\ehubah dalam eksperimen
(iii) The list of apparatus and materialsSenarai radas dan bahan
(iv) The arrangement of the apparatusSusunan radas
(v) The procedure of the experiment, which includes one methodof controlling the manipulated variable and one method ofmeasuring the responding variableProsedur eksperimen yang mesti termasuk satu kaedah mengawal pembolehubah dimanipulasikan don satu kaedah mengukur pemboleh ubah bergerakbalas.
(vi) The way to tabulate the dataCara menjadualkan data
(vii) The way to analyse the dataCara menganalisis data
ll0 marks )I l0 marks ]
END OF QUESTION PAPERKERTAS SOALAN TAMAT
free Q papers, free skema at: www.myschoolchildren.com
453113@ 2A\$ Copynght Bahagian Pendidikan Menengah MARA CONF'IDET
http://edu.joshuatly.com/ http://fb.me/edu.joshuatly | 23,373 | 71,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.247749 |
https://www.ademcetinkaya.com/2023/05/maai-mid-america-apartment-communities.html | 1,714,020,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00551.warc.gz | 556,197,925 | 60,541 | Outlook: Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 22 May 2023 for (n+16 weeks)
Methodology : Transductive Learning (ML)
## Abstract
Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock prediction model is evaluated with Transductive Learning (ML) and Lasso Regression1,2,3,4 and it is concluded that the MAA^I stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Sell
## Key Points
1. Market Outlook
2. Can statistics predict the future?
3. Is it better to buy and sell or hold?
## MAA^I Target Price Prediction Modeling Methodology
We consider Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock Decision Process with Transductive Learning (ML) where A is the set of discrete actions of MAA^I stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Lasso Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Transductive Learning (ML)) X S(n):→ (n+16 weeks) $∑ i = 1 n s i$
n:Time series to forecast
p:Price signals of MAA^I stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## MAA^I Stock Forecast (Buy or Sell) for (n+16 weeks)
Sample Set: Neural Network
Stock/Index: MAA^I Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock
Time series to forecast n: 22 May 2023 for (n+16 weeks)
According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Sell
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
## IFRS Reconciliation Adjustments for Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock
1. An entity shall assess whether contractual cash flows are solely payments of principal and interest on the principal amount outstanding for the currency in which the financial asset is denominated.
2. When an entity designates a financial liability as at fair value through profit or loss, it must determine whether presenting in other comprehensive income the effects of changes in the liability's credit risk would create or enlarge an accounting mismatch in profit or loss. An accounting mismatch would be created or enlarged if presenting the effects of changes in the liability's credit risk in other comprehensive income would result in a greater mismatch in profit or loss than if those amounts were presented in profit or loss
3. If there are changes in circumstances that affect hedge effectiveness, an entity may have to change the method for assessing whether a hedging relationship meets the hedge effectiveness requirements in order to ensure that the relevant characteristics of the hedging relationship, including the sources of hedge ineffectiveness, are still captured.
4. For the purpose of applying paragraph 6.5.11, at the point when an entity amends the description of a hedged item as required in paragraph 6.9.1(b), the amount accumulated in the cash flow hedge reserve shall be deemed to be based on the alternative benchmark rate on which the hedged future cash flows are determined.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
## Conclusions
Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock prediction model is evaluated with Transductive Learning (ML) and Lasso Regression1,2,3,4 and it is concluded that the MAA^I stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Sell
### MAA^I Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2C
Balance SheetBa3Caa2
Leverage RatiosCaa2Caa2
Cash FlowBaa2Ba2
Rates of Return and ProfitabilityBa2B3
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
### Prediction Confidence Score
Trust metric by Neural Network: 78 out of 100 with 665 signals.
## References
1. Bera, A. M. L. Higgins (1997), "ARCH and bilinearity as competing models for nonlinear dependence," Journal of Business Economic Statistics, 15, 43–50.
2. S. J. Russell and P. Norvig. Artificial Intelligence: A Modern Approach. Prentice Hall, Englewood Cliffs, NJ, 3nd edition, 2010
3. Efron B, Hastie T. 2016. Computer Age Statistical Inference, Vol. 5. Cambridge, UK: Cambridge Univ. Press
4. M. Babes, E. M. de Cote, and M. L. Littman. Social reward shaping in the prisoner's dilemma. In 7th International Joint Conference on Autonomous Agents and Multiagent Systems (AAMAS 2008), Estoril, Portugal, May 12-16, 2008, Volume 3, pages 1389–1392, 2008.
5. S. Devlin, L. Yliniemi, D. Kudenko, and K. Tumer. Potential-based difference rewards for multiagent reinforcement learning. In Proceedings of the Thirteenth International Joint Conference on Autonomous Agents and Multiagent Systems, May 2014
6. Bell RM, Koren Y. 2007. Lessons from the Netflix prize challenge. ACM SIGKDD Explor. Newsl. 9:75–79
7. Wan M, Wang D, Goldman M, Taddy M, Rao J, et al. 2017. Modeling consumer preferences and price sensitiv- ities from large-scale grocery shopping transaction logs. In Proceedings of the 26th International Conference on the World Wide Web, pp. 1103–12. New York: ACM
Frequently Asked QuestionsQ: What is the prediction methodology for MAA^I stock?
A: MAA^I stock prediction methodology: We evaluate the prediction models Transductive Learning (ML) and Lasso Regression
Q: Is MAA^I stock a buy or sell?
A: The dominant strategy among neural network is to Sell MAA^I Stock.
Q: Is Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock stock a good investment?
A: The consensus rating for Mid-America Apartment Communities Inc. 8.50% Series I Cumulative Redeemable Preferred Stock is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of MAA^I stock?
A: The consensus rating for MAA^I is Sell.
Q: What is the prediction period for MAA^I stock?
A: The prediction period for MAA^I is (n+16 weeks)
• Real-time stock market analysis | 1,934 | 7,868 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-18 | latest | en | 0.834525 |
http://www.instructorweb.com/lesson/candyfractions.asp | 1,521,803,279,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648207.96/warc/CC-MAIN-20180323102828-20180323122828-00614.warc.gz | 397,179,135 | 5,509 | CANDY FRACTIONS LESSON PLAN MATH ACTIVITY ELEMENTARY TEACHING AND LEARNING LANGUAGE ARTS WORKSHEET STUDENTS PRIMARY EDUCATION CURRICULUM KIDS THEME UNIT RESOURCES ACTIVITY
A fun elementary fractions lesson for your class.
CANDY FRACTIONS
Content Area
Elementary Math - Fractions
Content Targets
Identifying parts of a whole, percentages, working with fractions as discrete models, graphing data using bar graphs, recording data in a table, classifying and sorting objects, number sense
Learning Objective
After learning about the different types of fraction models, students will be able to analyze a discrete model in order to identify fractions (as fractions, decimals and percentages) by working with a candy sample to fill out a table and draw a bar graph.
CANDY FRACTIONS LESSON PLAN
• How to distinguish between an area model, discrete model and number line
• How to sort and categorize candies by color
• What it means to record data
• What it means to show the same quantity as a fraction, decimal and percentage
Key Vocabulary
Area model, discrete model, numerator, denominator, percentage, fraction, decimal
Lesson Materials
Candy Fractions worksheets
• A cookie, some coins and a ruler
• Small paper cups
• A bag of small, unwrapped candies
• Pencils, crayons, colored tape.
Introduction
Write the fraction ¼ on the board. Have students identify the numerator (part) and denominator (whole). Explain to students that there are other ways to write this fraction. Show students how to show ¼ as a decimal and as a percentage. Write one or two more fractions and have students convert or tell you how to convert them in to decimals and percentages.
Erase everything on the board except ¼. Ask a student to show the class how to draw this as a model. Most likely, the student will make a pie chart and shade in one fourth. Show the students a cookie and with colored tape, mark off ¼. Tell students that this is an area model. Remind students that an area model can be a square, rectangle or any other plane. Explain that there are two other models that we could use to show ¼. Pull out a ruler and tell students that this is an example of a number line.
Explain that 12 inches is the whole and ask students what they think ¼ of the 12 in. is. Mark off three inches with the tape to show ¼. Next, tell students there is yet another way to show ¼. Set out some coins (pennies and nickels) and ask a student volunteer to show that ¼ of a group of coins are pennies. A student can show 1 penny and 3 nickels, 2 pennies and 6 nickels, etc. Tell students that this is a discrete model, since the parts of the whole are separate objects.
Display the three models of ¼ in the classroom for students to refer to.
Body
Ask 3 students with fair hair to stand up in front of the class. Ask 2 students with dark hair to stand up in front of the class.
• Ask students what is the whole (5).
• Ask student students what fraction of the students has fair hair (3/5).
• What fraction has brown hair? (2/5)
• What type of model is this? (discrete)
Pass out small cups of candy to students (10-20 pieces per cup) and the Candy Fraction packet (see printables). Tell students that today we are going to be working with a discrete model- pieces of candy. We are going to divide our candy samples by color to find out what colors we have as fractions of our whole sample.
Read through the whole packet with students and do steps 1 and 2 together as a class. Chose a color to do together as an example and answer any questions that come up (fill in the table and bar graph for that one color as a class). Remind students that they will have different answers since they all have different samples. Students can then fill in their packet with the rest of the colors.
Closure
To review the major concepts, have students choose one fraction they found and draw sketches of it as an area model, discrete model and number line.
Have students share their findings with a partner and post their tables and bar graphs on a bulletin board.
Print these worksheets for this lesson.
Elementary Fractions Lesson Plan - Children - Primary Education - Home School Child Teachers Free - Fourth Grade - Third Grade - Fifth Grade - Homeschool Kids - Fraction Lesson Plan Worksheet | 952 | 4,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-13 | latest | en | 0.872004 |
http://docplayer.net/22132033-Connect-areas-perimeters.html | 1,544,727,185,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825029.40/warc/CC-MAIN-20181213171808-20181213193308-00164.warc.gz | 75,362,797 | 25,200 | CONNECT: Areas, Perimeters
Save this PDF as:
Size: px
Start display at page:
Transcription
1 CONNECT: Areas, Perimeters 1. AREAS OF PLANE SHAPES A plane figure or shape is a two-dimensional, flat shape. Here are 3 plane shapes: All of them have two dimensions that we usually call length and width (or sometimes height). Plane shapes do not have thickness, which means we can draw them on paper. The amount of space inside each shape or the amount of space each figure occupies is called the area of that shape. Units to measure area When you hear the word area, you may think of school maths when you had to find the area of different shapes, mostly rectangles and squares. But we can actually find the area of ANY shape, even the strange one above. To do this we firstly need to think about the basic unit of area. Try to think of a small shape that, if it was repeated over and over again, perhaps turned upside down or back-to-front, would cover each of the shapes above. Using a shape like this is called tessellating and examples can be found in patchwork, tiling, mosaics and mosques. Designs range from simple to ornate and can be very attractive. I have copied some examples over the page. 1
2 Retrieved January 22, 2013 from and Retrieved January 22, 2013 from and But what if we need to compare the areas of our shapes on the first page? Say we want to cover them all perhaps or, even worse these are the shapes of your late great aunt s farm paddocks and she has left them to the family who is going to get the biggest share?! The small shape that has been accepted throughout the world for tessellating to determine areas is the square. In the Metric System of measurement, the unit of length is the metre and so for area, the unit is the square metre written as m 2. Smaller measures are given in square centimetres (cm 2 ) or even smaller square millimetres (mm 2 ) and large measures are given in hectares (ha). Note that ha doesn t have the superscript 2 as it is already a square unit 1ha measures an area 100m by 100m, that is, 1ha = m 2. 2
3 100m 100m 1ha To be able to work out the areas of the shapes above, we could draw a grid of squares on each shape and count them. (We would need to approximate for the irregular shapes.) The area of any rectangle With regular shapes, the area is more straightforward. To find the area of this rectangle, we can cover it with a grid of squares with length 1cm and width 1cm, that is they are each square centimetres (1cm 2 ), and count them. In this rectangle, there are 12 square centimetres, so its area is 12 cm 2. Is there a faster way to find the area of this rectangle? Yes! We know that the rectangle is 6cm long and 2cm wide, so we can fit two rows each of 6 centimetre squares in the complete rectangle. So, its area is 12cm 2. We can also calculate this as 6cm x 2cm = 12 cm 2. We can generalise this with a formula. If we let the length of the rectangle be l units and the width of the rectangle be w units, then to calculate the area (A units 2 ) we multiply the length by the width and so we have: A = l x w units 2 3
4 We can apply this formula to any rectangle as long as we know its dimensions (the size of its length and width). You may have seen this formula before, using b (represents breadth) instead of w width and breadth are the same and in this case, the area would be A = l b units 2. Both formulas give the same result. To find the area of the rectangle given above, we know that l = 6cm and w = 2cm, so A = 6cm x 2cm = 12 cm 2. To find the area of any rectangle, the dimensions must be measured using the same unit such as centimetres, kilometres and so on. We cannot find the area of a rectangle where the length is given in kilometres and the width is in metres, for example, we must convert one of those measures to the other. The following diagrams are not drawn to scale. See if you can find the area of each rectangle. You can check your results with the solutions at the end of this resource cm 11mm 1.5cm 4.2cm Areas of other shapes What about triangles? Can you see that the area of one of the triangles formed is half the area of the whole rectangle? So we can say its area is 6cm 2 straight away. 4
5 But what about this one? We could do this: Now we have 2 triangles that are both half their respective rectangles in area. So the area of the complete triangle is 2 cm 2 (for the one on the left) + 4 cm 2 (for the one on the right) which makes 6 cm 2. Here is the formula for the area of a triangle you might remember it from school. With a triangle, we ll refer to the length (of the rectangle) as the base of the triangle and the width (of the rectangle) as the height of the triangle. height base The area of a triangle is given by half the area of a rectangle, so the area is given by A = ½ x b x h units 2, where A units 2 represents the area of the triangle, b units represents the length of the base of the triangle and h units represents the height of the triangle. The height of the triangle must be at right angles to the base of the triangle and the units of measure must be the same. 5
6 Example: Find the area of this triangle. 8.5cm 15cm The area is ½ x 15cm x 8.5cm = 63.75cm 2. Find the area of this triangle: 17mm 6.9mm Notice that in this triangle, the base has to be extended so that we can draw a height at right angles to it. The area can still be calculated in the same way, though. So the area = ½ x 17mm x 6.9mm = 58.65mm 2 Here are some for you to try. You can check your results with the solutions at the end. Calculate the areas of these triangles (the diagrams are not drawn to scale): km 65km 6
7 m 10.1m What about circles? If we know the size of the radius of a circle, we can work out the area. Diagram retrieved 22 January 2013 from The radius is any line joining the centre of the circle to the circumference and is half the diameter. If we let r units be the length of the radius, then the area (given by A units 2 ) is A = πr 2 units 2 π is a Greek letter (pronounced pie in Australia) and is always the same number. It is the answer when you divide the length of the circumference of a circle by the length of the diameter of that circle. It is always and can be calculated correctly to many decimal places. If you use a calculator when you work with π, your answer will always be accurate and you will need to round it as it will contain many decimal places. 7
8 When you want to calculate using A = πr 2, you type π in your calculator (often you need to use the Shift key) followed by x, then the radius followed by x 2, and =. Example: Calculate the area of this circle, correct to 3 decimal places. Area = πr 2 In this circle, the radius is 5cm long, so r = 5cm, and the area is πr 2 = π 5 2 cm 2 = cm cm 2 (rounded to 3 decimal places) Here are two for you to try. Remember, if you are given a diameter instead of a radius, you need to calculate half of it (to work out the radius) before calculating the area. (The diagrams are not to scale.) Calculate the area of each of the following circles:
9 Converting between square units 11mm 4.2cm This rectangle was one of the questions from page 4, where you were required to find its area. In the solutions, I changed the centimetre measurements into mm and calculated the area in mm 2. The area is 462mm 2. Now I m going to calculate the area using cm 2 answers. and we will compare our First, change 11mm to cm, by dividing by 10, so 11mm = cm, = 1.1cm. Area = 4.2cm x 1.1cm = 4.62cm 2 Is this what you expected? The area in mm 2 is 100 times as big as the area in cm 2. This is because we are no longer dealing with one dimension (length) where 1cm = 10mm, but with 2 dimensions (length and width): 10mm 10mm 1cm 2 So, 1cm 2 = 100mm 2 In the same way, 1m 2 = 100cm x 100cm = 10000cm 2 and 1km 2 = 1000m x 1000m = m 2 9
10 Here is an example: I have measured a space for a small window in cm 2, but the builder needs the measurement in mm 2. My measurement was 3 600cm 2. What should I tell the builder? 1cm 2 = 100mm 2, so 3 600cm 2 = x 100mm 2 = mm 2 A second example: I have measured a block of land in m 2 but would like to know how many hectares this is. My measurement is m 2. I have put the solution at the end. If you need help with any of the Maths covered in this resource (or any other Maths topics), you can make an appointment with Learning Development through Reception: phone (02) , or Level 3 (top floor), Building 11, or through your campus. 10
11 Solutions Areas of rectangles, (page 4) cm 1.5cm Area = 3.7cm x 1.5cm = 5.55cm mm 4.2cm To deal with this rectangle, we must make sure both measurements are in the same unit. It is probably easier to make them both mm, so 4.2cm = 4.2 x 10mm, that is 42mm. Now the area is found using the formula. Area = 42mm x 11mm = 462mm 2 (If you calculated the area in cm 2, please see page 9.) 11
12 Areas of triangles (page 6, 7) km 65km Area = ½ x 65km x 8.8km = 286km m 10.1m Area = ½ x 14.1m x 10.1m = m 2 Areas of circles (page 8) 1. Area = π 4 2 cm 2 = cm cm 2 (rounded to 3 decimal places) 12
13 2. Firstly, we are given the diameter instead of the radius, so we must find half of 38cm to get the length of the radius. (38 2 = 19). So the area is Area = π 19 2 cm 2 = cm cm 2 (rounded to 3 decimal places) Converting between square units (page 10) 1ha = m 2 The measurement is m 2. How many lots of m 2 are in m 2? We need to divide by to find the number of ha, and obtain 4.25ha. 13
CONNECT: Volume, Surface Area
CONNECT: Volume, Surface Area 1. VOLUMES OF SOLIDS A solid is a three-dimensional (3D) object, that is, it has length, width and height. One of these dimensions is sometimes called thickness or depth.
Calculating Area, Perimeter and Volume
Calculating Area, Perimeter and Volume You will be given a formula table to complete your math assessment; however, we strongly recommend that you memorize the following formulae which will be used regularly
FROM THE SPECIFIC TO THE GENERAL
CONNECT: Algebra FROM THE SPECIFIC TO THE GENERAL How do you react when you see the word Algebra? Many people find the concept of Algebra difficult, so if you are one of them, please relax, as you have
Granby Primary School Year 5 & 6 Supporting your child with maths A handbook for year 5 & 6 parents H M Hopps 2016 G r a n b y P r i m a r y S c h o o l 1 P a g e Many parents want to help their children
Convert between units of area and determine the scale factor of two similar figures.
CHAPTER 5 Units of Area c GOAL Convert between units of area and determine the scale factor of two. You will need a ruler centimetre grid paper a protractor a calculator Learn about the Math The area of
Dŵr y Felin Comprehensive School. Perimeter, Area and Volume Methodology Booklet
Dŵr y Felin Comprehensive School Perimeter, Area and Volume Methodology Booklet Perimeter, Area & Volume Perimeters, Area & Volume are key concepts within the Shape & Space aspect of Mathematics. Pupils
Functional Skills Mathematics
Functional Skills Mathematics Level Learning Resource Perimeter and Area MSS1/L.7 Contents Perimeter and Circumference MSS1/L.7 Pages 3-6 Finding the Area of Regular Shapes MSS1/L.7 Page 7-10 Finding the
CONNECT: Volume, Surface Area
CONNECT: Volume, Surface Area 2. SURFACE AREAS OF SOLIDS If you need to know more about plane shapes, areas, perimeters, solids or volumes of solids, please refer to CONNECT: Areas, Perimeters 1. AREAS
CALCULATING THE AREA OF A FLOWER BED AND CALCULATING NUMBER OF PLANTS NEEDED
This resource has been produced as a result of a grant awarded by LSIS. The grant was made available through the Skills for Life Support Programme in 2010. The resource has been developed by (managers
AREA. AREA is the amount of surface inside a flat shape. (flat means 2 dimensional)
AREA AREA is the amount of surface inside a flat shape. (flat means 2 dimensional) Area is always measured in units 2 The most basic questions that you will see will involve calculating the area of a square
CONNECT: Algebra. 3x = 20 5 REARRANGING FORMULAE
CONNECT: Algebra REARRANGING FORMULAE Before you read this resource, you need to be familiar with how to solve equations. If you are not sure of the techniques involved in that topic, please refer to CONNECT:
Basic Math for the Small Public Water Systems Operator
Basic Math for the Small Public Water Systems Operator Small Public Water Systems Technology Assistance Center Penn State Harrisburg Introduction Area In this module we will learn how to calculate the
Working in 2 & 3 dimensions Revision Guide
Tips for Revising Working in 2 & 3 dimensions Make sure you know what you will be tested on. The main topics are listed below. The examples show you what to do. List the topics and plan a revision timetable.
Circumference and area of a circle
c Circumference and area of a circle 22 CHAPTER 22.1 Circumference of a circle The circumference is the special name of the perimeter of a circle, that is, the distance all around it. Measure the circumference
Show that when a circle is inscribed inside a square the diameter of the circle is the same length as the side of the square.
Week & Day Week 6 Day 1 Concept/Skill Perimeter of a square when given the radius of an inscribed circle Standard 7.MG:2.1 Use formulas routinely for finding the perimeter and area of basic twodimensional
Area Long-Term Memory Review Review 1
Review 1 1. To find the perimeter of any shape you all sides of the shape.. To find the area of a square, you the length and width. 4. What best identifies the following shape. Find the area and perimeter
EDEXCEL FUNCTIONAL SKILLS PILOT. Maths Level 1. Chapter 5. Working with shape and space
EDEXCEL FUNCTIONAL SKILLS PILOT Maths Level 1 Chapter 5 Working with shape and space SECTION H 1 Calculating perimeter 86 2 Calculating area 87 3 Calculating volume 89 4 Angles 91 5 Line symmetry 92 6
Perimeter is the length of the boundary of a two dimensional figure.
Section 2.2: Perimeter and Area Perimeter is the length of the boundary of a two dimensional figure. The perimeter of a circle is called the circumference. The perimeter of any two dimensional figure whose
Tallahassee Community College PERIMETER
Tallahassee Community College 47 PERIMETER The perimeter of a plane figure is the distance around it. Perimeter is measured in linear units because we are finding the total of the lengths of the sides
The Area is the width times the height: Area = w h
Geometry Handout Rectangle and Square Area of a Rectangle and Square (square has all sides equal) The Area is the width times the height: Area = w h Example: A rectangle is 6 m wide and 3 m high; what
Developing Conceptual Understanding of Number. Set J: Perimeter and Area
Developing Conceptual Understanding of Number Set J: Perimeter and Area Carole Bilyk cbilyk@gov.mb.ca Wayne Watt wwatt@mts.net Perimeter and Area Vocabulary perimeter area centimetres right angle Notes
Characteristics of the Four Main Geometrical Figures
Math 40 9.7 & 9.8: The Big Four Square, Rectangle, Triangle, Circle Pre Algebra We will be focusing our attention on the formulas for the area and perimeter of a square, rectangle, triangle, and a circle.
Grade 9 Mathematics Unit 3: Shape and Space Sub Unit #1: Surface Area. Determine the area of various shapes Circumference
1 P a g e Grade 9 Mathematics Unit 3: Shape and Space Sub Unit #1: Surface Area Lesson Topic I Can 1 Area, Perimeter, and Determine the area of various shapes Circumference Determine the perimeter of various
Section 7.2 Area. The Area of Rectangles and Triangles
Section 7. Area The Area of Rectangles and Triangles We encounter two dimensional objects all the time. We see objects that take on the shapes similar to squares, rectangle, trapezoids, triangles, and
EDEXCEL FUNCTIONAL SKILLS PILOT. Maths Level 2. Chapter 5. Working with shape and space
EDEXCEL FUNCTIONAL SKILLS PILOT Maths Level 2 Chapter 5 Working with shape and space SECTION H 1 Perimeter 75 2 Area 77 3 Volume 79 4 2-D Representations of 3-D Objects 81 5 Remember what you have learned
Number & Place Value. Addition & Subtraction. Digit Value: determine the value of each digit. determine the value of each digit
Number & Place Value Addition & Subtraction UKS2 The principal focus of mathematics teaching in upper key stage 2 is to ensure that pupils extend their understanding of the number system and place value
Geometry - Calculating Area and Perimeter
Geometry - Calculating Area and Perimeter In order to complete any of mechanical trades assessments, you will need to memorize certain formulas. These are listed below: (The formulas for circle geometry
The GED math test gives you a page of math formulas that
Math Smart 643 The GED Math Formulas The GED math test gives you a page of math formulas that you can use on the test, but just seeing the formulas doesn t do you any good. The important thing is understanding
PERIMETERS AND AREAS
PERIMETERS AND AREAS 1. PERIMETER OF POLYGONS The Perimeter of a polygon is the distance around the outside of the polygon. It is the sum of the lengths of all the sides. Examples: The perimeter of this
Areas of Polygons. Goal. At-Home Help. 1. A hockey team chose this logo for their uniforms.
-NEM-WBAns-CH // : PM Page Areas of Polygons Estimate and measure the area of polygons.. A hockey team chose this logo for their uniforms. A grid is like an area ruler. Each full square on the grid has
Q1. The grid below is made of right-angled triangles like this: Shade triangles on the grid to make a quadrilateral.
Q1. The grid below is made of right-angled triangles like this: Shade triangles on the grid to make a quadrilateral. Your quadrilateral must have an area of 24 cm 2 and a perimeter of 26 cm. Page 1 of
Calculating Perimeter
Calculating Perimeter and Area Formulas are equations used to make specific calculations. Common formulas (equations) include: P = 2l + 2w perimeter of a rectangle A = l + w area of a square or rectangle
Geometry Notes VOLUME AND SURFACE AREA
Volume and Surface Area Page 1 of 19 VOLUME AND SURFACE AREA Objectives: After completing this section, you should be able to do the following: Calculate the volume of given geometric figures. Calculate
Perimeter of Triangle = Sum of all Sides Perimeter of Triangle = Side + Side + Side
Chapter 11 Perimeter As a present, your parents have bought you a pet, a small puppy for you to play with and take care of. For the first few weeks it is quite content living inside the house with the
Charlesworth School Year Group Maths Targets
Charlesworth School Year Group Maths Targets Year One Maths Target Sheet Key Statement KS1 Maths Targets (Expected) These skills must be secure to move beyond expected. I can compare, describe and solve
Geometry. Geometry is the study of shapes and sizes. The next few pages will review some basic geometry facts. Enjoy the short lesson on geometry.
Geometry Introduction: We live in a world of shapes and figures. Objects around us have length, width and height. They also occupy space. On the job, many times people make decision about what they know
Quick Reference ebook
This file is distributed FREE OF CHARGE by the publisher Quick Reference Handbooks and the author. Quick Reference ebook Click on Contents or Index in the left panel to locate a topic. The math facts listed
Fractions Associate a fraction with division to calculate decimal fraction equivalents (e.g ) for a simple fraction (e.g. 3/8).
: Autumn 1 Numeracy Curriculum Objectives Number, Place Value and Rounding Read, write, order and compare numbers up to 10,000,000 and determine the value of each digit. Round any whole number to a required
LEFT HAND SIDE = RIGHT HAND SIDE
SIPLE FORULA What is a formula? When you do a calculation, you might add numbers together, subtract numbers, multiply or divide them. Take addition as an example: 7 + 45 = 82 Two given numbers added together
16 Circles and Cylinders
16 Circles and Cylinders 16.1 Introduction to Circles In this section we consider the circle, looking at drawing circles and at the lines that split circles into different parts. A chord joins any two
Circumference and Area of a Circle
Overview Math Concepts Materials Students explore how to derive pi (π) as a ratio. Students also study the circumference and area of a circle using formulas. numbers and operations TI-30XS MultiView two-dimensional
Calculating the Surface Area of a Cylinder
Calculating the Measurement Calculating The Surface Area of a Cylinder PRESENTED BY CANADA GOOSE Mathematics, Grade 8 Introduction Welcome to today s topic Parts of Presentation, questions, Q&A Housekeeping
LESSON 10 GEOMETRY I: PERIMETER & AREA
LESSON 10 GEOMETRY I: PERIMETER & AREA INTRODUCTION Geometry is the study of shapes and space. In this lesson, we will focus on shapes and measures of one-dimension and two-dimensions. In the next lesson,
9 Area, Perimeter and Volume
9 Area, Perimeter and Volume 9.1 2-D Shapes The following table gives the names of some 2-D shapes. In this section we will consider the properties of some of these shapes. Rectangle All angles are right
Q1. Here is a flag. Calculate the area of the shaded cross. Q2. The diagram shows a right-angled triangle inside a circle.
Q1. Here is a flag. Calculate the area of the shaded cross. 2 marks Q2. The diagram shows a right-angled triangle inside a circle. The circle has a radius of 5 centimetres. Calculate the area of the triangle.
Geometry Chapter 9 Extending Perimeter, Circumference, and Area
Geometry Chapter 9 Extending Perimeter, Circumference, and Area Lesson 1 Developing Formulas for Triangles and Quadrilaterals Learning Target (LT-1) Solve problems involving the perimeter and area of triangles
The area of a figure is the measure of the size of the region enclosed by the figure. Formulas for the area of common figures: square: A = s 2
The area of a figure is the measure of the size of the region enclosed by the figure. Formulas for the area of common figures: square: A = s 2 s s rectangle: A = l w parallelogram: A = b h h b triangle:
Trades Math Practice Test and Review
Trades Math Practice Test and Review This material is intended as a review only. To help prepare for the assessment, the following resources are also available:. online review material (free of charge)
Perimeter, Area, and Volume
Perimeter, Area, and Volume Perimeter of Common Geometric Figures The perimeter of a geometric figure is defined as the distance around the outside of the figure. Perimeter is calculated by adding all
EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES. Maths Level 1. Chapter 5. Working with shape and space
Shape and space 5 EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES Maths Level 1 Chapter 5 Working with shape and space SECTION H 1 Calculating perimeter 2 Calculating area 3 Calculating volume 4 Angles
Geometry Chapter 9 Extending Perimeter, Circumference, and Area
Geometry Chapter 9 Extending Perimeter, Circumference, and Area Lesson 1 Developing Formulas for Triangles and Quadrilaterals Learning Targets LT9-1: Solve problems involving the perimeter and area of
GAP CLOSING. 2D Measurement. Intermediate / Senior Student Book
GAP CLOSING 2D Measurement Intermediate / Senior Student Book 2-D Measurement Diagnostic...3 Areas of Parallelograms, Triangles, and Trapezoids...6 Areas of Composite Shapes...14 Circumferences and Areas
Finding Volume of Rectangular Prisms
MA.FL.7.G.2.1 Justify and apply formulas for surface area and volume of pyramids, prisms, cylinders, and cones. MA.7.G.2.2 Use formulas to find surface areas and volume of three-dimensional composite shapes.
Year 4 (Entry into Year 5) 25 Hour Revision Course Mathematics
Year 4 (Entry into Year 5) 25 Hour Revision Course Mathematics Section 1 Geometry 4 hours ~2~ Shape Properties Any two-dimensional shape made up of straight lines is called a polygon. Although circles
Area of a triangle: The area of a triangle can be found with the following formula: 1. 2. 3. 12in
Area Review Area of a triangle: The area of a triangle can be found with the following formula: 1 A 2 bh or A bh 2 Solve: Find the area of each triangle. 1. 2. 3. 5in4in 11in 12in 9in 21in 14in 19in 13in
Perimeter and Area. Chapter 11 11.1 INTRODUCTION 11.2 SQUARES AND RECTANGLES TRY THESE
PERIMETER AND AREA 205 Perimeter and Area Chapter 11 11.1 INTRODUCTION In Class VI, you have already learnt perimeters of plane figures and areas of squares and rectangles. Perimeter is the distance around
Build your skills: Perimeter and area Part 1. Working out the perimeter and area of different shapes
Working out the perimeter and area of different shapes This task has two parts. Part 1 In this part, you can brush up your skills and find out about perimeter and area. Part 2 In the second part, you can
Within each area, these outcomes are broken down into more detailed step-by-step learning stages for each of the three terms.
MATHEMATICS PROGRAMME OF STUDY COVERAGE all topics are revisited several times during each academic year. Existing learning is consolidated and then built upon and extended. Listed below are the end of
Curriculum overview for Year 1 Mathematics
Curriculum overview for Year 1 Counting forward and back from any number to 100 in ones, twos, fives and tens identifying one more and less using objects and pictures (inc number lines) using the language
MODULE FRAMEWORK EN ASSESSMENT SHEET
MODULE FRAMEWORK EN ASSESSMENT SHEET LEARNING OUTCOMES (LOS) ASSESSMENT STANDARDS (ASS) FORMATIVE ASSESSMENT ASs Pages and (mark out of 4) LOs (ave. out of 4) SUMMATIVE ASSESSMENT Tasks or tests Ave for
Junior Math Circles November 18, D Geometry II
1 University of Waterloo Faculty of Mathematics Junior Math Circles November 18, 009 D Geometry II Centre for Education in Mathematics and Computing Two-dimensional shapes have a perimeter and an area.
CALCULATING PERIMETER. WHAT IS PERIMETER? Perimeter is the total length or distance around a figure.
CALCULATING PERIMETER WHAT IS PERIMETER? Perimeter is the total length or distance around a figure. HOW DO WE CALCULATE PERIMETER? The formula one can use to calculate perimeter depends on the type of
Revision Notes Adult Numeracy Level 1
Revision Notes Adult Numeracy Level 1 Numbers The number 5 703 428 has been entered into a table. It shows the value of each column. The 7 is in the hundred thousands column The 0 cannot be missed out
Year 3 End of year expectations
Number and Place Value Count in 4s, 8s, 50s and 100s from any number Read and write numbers up to 1000 in numbers and words Compare and order numbers up to 1000 Recognise the place value of each digit
AUTUMN UNIT 3. first half. Perimeter. Centimetres and millimetres. Metres and centimetres. Area. 3D shapes PART 3 MEASURES AND PROPERTIES OF SHAPES
PART AUTUMN first half MEASURES AND PROPERTIES OF SHAPES SECTION Perimeter SECTION Centimetres and millimetres SECTION Metres and centimetres SECTION Key Stage National Strategy CROWN COPYRIGHT 00 Area
LESSON SUMMARY. Measuring Shapes
LESSON SUMMARY CXC CSEC MATHEMATICS UNIT SIX: Measurement Lesson 11 Measuring Shapes Textbook: Mathematics, A Complete Course by Raymond Toolsie, Volume 1 (Some helpful exercises and page numbers are given
Year 6 Maths Objectives
Year 6 Maths Objectives Place Value COUNTING COMPARING NUMBERS IDENTIFYING, REPRESENTING & ESTIMATING NUMBERS READING & WRITING NUMBERS UNDERSTANDING PLACE VALUE ROUNDING PROBLEM SOLVING use negative numbers
Dyffryn School Ysgol Y Dyffryn Mathematics Faculty
Dyffryn School Ysgol Y Dyffryn Mathematics Faculty Formulae and Facts Booklet Higher Tier Number Facts Sum This means add. Difference This means take away. Product This means multiply. Share This means
Marie has a winter hat made from a circle, a rectangular strip and eight trapezoid shaped pieces. y inches. 3 inches. 24 inches
Winter Hat This problem gives you the chance to: calculate the dimensions of material needed for a hat use circle, circumference and area, trapezoid and rectangle Marie has a winter hat made from a circle,
Paper 2. Year 9 mathematics test. Calculator allowed. Remember: First name. Last name. Class. Date
Ma KEY STAGE 3 Year 9 mathematics test Tier 5 7 Paper 2 Calculator allowed First name Last name Class Date Please read this page, but do not open your booklet until your teacher tells you to start. Write
Basic Garden Math. This document is organized into the following sections:
Basic Garden Math Gardening is an activity which occasionally requires the use of math, such as when you are computing how much fertilizer to use or how much compost to buy. Luckily, the math involved
Unit #10 Volume and Surface Area
10.0 Unit Preview Unit #10 Volume and Surface Area By the end of this unit I should be able to: Determine the volume of any right angled prism Determine the surface area of any right angled prism Determine
Shape, space and measures 1
Shape, space and measures 1 contents There are three lessons in this unit, Shape, space and measures 1. S1.1 Lines, length and perimeter 3 S1.2 Area of a rectangle 6 S1.3 Solving problems 9 Resource sheets
I Perimeter, Area, Learning Goals 304
U N I T Perimeter, Area, Greeting cards come in a variety of shapes and sizes. You can buy a greeting card for just about any occasion! Learning Goals measure and calculate perimeter estimate, measure,
EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES. Maths Level 2. Chapter 5. Shape and space
Shape and space 5 EDEXCEL FUNCTIONAL SKILLS PILOT TEACHER S NOTES Maths Level 2 Chapter 5 Shape and space SECTION H 1 Perimeter 2 Area 3 Volume 4 2-D Representations of 3-D Objects 5 Remember what you
Circumference of a Circle
Circumference of a Circle A circle is a shape with all points the same distance from the center. It is named by the center. The circle to the left is called circle A since the center is at point A. If
DATE PERIOD. Estimate the product of a decimal and a whole number by rounding the Estimation
A Multiplying Decimals by Whole Numbers (pages 135 138) When you multiply a decimal by a whole number, you can estimate to find where to put the decimal point in the product. You can also place the decimal
Paper 2. Year 9 mathematics test. Calculator allowed. Remember: First name. Last name. Class. Date
Ma KEY STAGE 3 Year 9 mathematics test Tier 6 8 Paper 2 Calculator allowed First name Last name Class Date Please read this page, but do not open your booklet until your teacher tells you to start. Write
Chapter 1 Measurement
Chapter 1 Measurement Math 1201 1 Chapter 1 Measurement Sections 1.1-1.3: Goals: Converting between imperial units by unit analysis Converting between SI units Converting between SI and imperial units
Name Revision Sheet 1
Name Revision Sheet 1 1 What is 8? Show your working 11 Solve the equation y 1 Round 79 to the nearest 10. 1 Expand ( x 1 0 ) Use BIDMAS to work out 5 1 How many lines of symmetry does a square have? 1
Student Outcomes. Lesson Notes. Classwork. Exercises 1 3 (4 minutes)
Student Outcomes Students give an informal derivation of the relationship between the circumference and area of a circle. Students know the formula for the area of a circle and use it to solve problems.
Oral and Mental calculation
Oral and Mental calculation Read and write any integer and know what each digit represents. Read and write decimal notation for tenths and hundredths and know what each digit represents. Order and compare
Geometry Notes PERIMETER AND AREA
Perimeter and Area Page 1 of 57 PERIMETER AND AREA Objectives: After completing this section, you should be able to do the following: Calculate the area of given geometric figures. Calculate the perimeter
Name: Class: Date: Geometry Chapter 3 Review
Name: Class: Date: ID: A Geometry Chapter 3 Review. 1. The area of a rectangular field is 6800 square meters. If the width of the field is 80 meters, what is the perimeter of the field? Draw a diagram
Lesson 21. Circles. Objectives
Student Name: Date: Contact Person Name: Phone Number: Lesson 1 Circles Objectives Understand the concepts of radius and diameter Determine the circumference of a circle, given the diameter or radius Determine
Add and subtract 1-digit and 2-digit numbers to 20, including zero. Measure and begin to record length, mass, volume and time
Year 1 Maths - Key Objectives Count to and across 100 from any number Count, read and write numbers to 100 in numerals Read and write mathematical symbols: +, - and = Identify "one more" and "one less"
CARPENTRY MATH ASSESSMENT REVIEW
CARPENTRY MATH ASSESSMENT REVIEW This material is intended as a review. The following Learning Centres have more resources available to help you prepare for your assessment Nanaimo ABE Learning Centre:
Surface Area Quick Review: CH 5
I hope you had an exceptional Christmas Break.. Now it's time to learn some more math!! :) Surface Area Quick Review: CH 5 Find the surface area of each of these shapes: 8 cm 12 cm 4cm 11 cm 7 cm Find
Lesson 6. Unit 3. Building with Blocks. Area
Math 5 Lesson 6 Area Building with Blocks Lian is making a fort for her action figures out of building blocks. She has 42 cm blocks. She wants to have the largest area possible. How can she arrange her
The Measurement of Area
The Measurement of Area a h Area = ½(a+b) h b Area = ½b h b h 2000 Andrew Harris Area 1 2000 Andrew Harris Contents Defining Area 3 Progression in Learning about Area 3 1. Pre-measurement Experiences 3
Pizza! Pizza! Assessment
Pizza! Pizza! Assessment 1. A local pizza restaurant sends pizzas to the high school twelve to a carton. If the pizzas are one inch thick, what is the volume of the cylindrical shipping carton for the
Measurements 1. BIRKBECK MATHS SUPPORT www.mathsupport.wordpress.com. In this section we will look at. Helping you practice. Online Quizzes and Videos
BIRKBECK MATHS SUPPORT www.mathsupport.wordpress.com Measurements 1 In this section we will look at - Examples of everyday measurement - Some units we use to take measurements - Symbols for units and converting
Imperial Length Measurements
Unit I Measuring Length 1 Section 2.1 Imperial Length Measurements Goals Reading Fractions Reading Halves on a Measuring Tape Reading Quarters on a Measuring Tape Reading Eights on a Measuring Tape Reading
Area and Perimeter. Practice: Find the perimeter of each. Square with side length of 6 cm. Rectangle with side lengths of 4 cm and 7 cm
Area and Perimeter Perimeter: add up all the sides (the outside of the polygon) Practice: Find the perimeter of each Square with side length of 6 cm Rectangle with side lengths of 4 cm and 7 cm Parallelogram
Lesson 22. Circumference and Area of a Circle. Circumference. Chapter 2: Perimeter, Area & Volume. Radius and Diameter. Name of Lecturer: Mr. J.
Lesson 22 Chapter 2: Perimeter, Area & Volume Circumference and Area of a Circle Circumference The distance around the edge of a circle (or any curvy shape). It is a kind of perimeter. Radius and Diameter
Q1. Lindy has 4 triangles, all the same size. She uses them to make a star. Calculate the perimeter of the star. 2 marks.
Q1. Lindy has 4 triangles, all the same size. She uses them to make a star. Calculate the perimeter of the star. Page 1 of 16 Q2. Liam has two rectangular tiles like this. He makes this L shape. What is
CHAPTER 27 AREAS OF COMMON SHAPES
EXERCISE 113 Page 65 CHAPTER 7 AREAS OF COMMON SHAPES 1. Find the angles p and q in the diagram below: p = 180 75 = 105 (interior opposite angles of a parallelogram are equal) q = 180 105 0 = 35. Find
Integrated Algebra: Geometry
Integrated Algebra: Geometry Topics of Study: o Perimeter and Circumference o Area Shaded Area Composite Area o Volume o Surface Area o Relative Error Links to Useful Websites & Videos: o Perimeter and | 8,848 | 35,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.968691 |
https://docs.trifacta.com/pages/viewpage.action?pageId=148810024 | 1,652,963,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00486.warc.gz | 264,310,299 | 25,833 | Page tree
Returns `true` if the argument is an odd value. Argument can be an Integer, a function returning Integers, or a column reference.
Since the function returns a Boolean value, it can be used as a function or a conditional.
Basic Usage
Integer literal value:
isodd('3')
Output: Returns the value `true`.
Column reference value:
isodd(countStudents)
Output: If the value in the `countStudents` column is an odd number, then return `true`
Syntax and Arguments
isodd(int_value)
ArgumentRequired?Data TypeDescription
int_valueYintegerThis value can be an Integer, a function returning an Integer, or a column reference.
int_value
Name of the columns, expressions, or literals to compare.
• Missing values generate missing string results.
Usage Notes:
Required?Data TypeExample Value
YesColumn reference, function, or Integer literal value`myColumn`
Examples
Example - Basic Equal and Notequal Functions
This example demonstrate the following comparison functions.
In this example, the dataset contains current measurements of the sides of rectangular areas next to the size of those areas as previously reported. Using these functions, you can perform some light analysis of the data.
Source:
sideAsideBreportedArea
41456
6635
8432
1515200
4728
12670
9981
Transformation:
In the first test, you are determining if the four-sided area is a square, based on a comparison of the measured values for `sideA` and `sideB`:
Transformation Name `New formula` `Single row formula` `EQUAL(sideA, sideB)` `'isSquare'`
Next, you can use the reported sides to calculate the area of the shape and compare it to the area previously reported:
Transformation Name `New formula` `Single row formula` `NOTEQUAL(sideA * sideB, reportedArea)` `'isValidData'`
You can also compute if the reportedArea can be divided into even square units:
Transformation Name `New formula` `Single row formula` `ISEVEN(reportedArea)` `'isReportedAreaEven'`
You can test if either measured side is an odd number of units:
Transformation Name `New formula` `Single row formula` `IF((ISODD(sideA) == true) OR (ISODD(sideB) == true),TRUE,FALSE)` `'isSideOdd'`
Results:
sideAsideBreportedAreaisSquareisValidDataisReportedAreaEvenisSideOdd
41456FALSEFALSETRUEFALSE
6635TRUETRUETRUEFALSE
8432FALSEFALSETRUEFALSE
1515200TRUETRUETRUETRUE
4728FALSEFALSETRUETRUE
12670FALSETRUETRUEFALSE
9981TRUEFALSEFALSEFALSE
• Page:
• Page:
• Page:
• Page:
• Page:
• Page:
• Page:
• Page:
• Page:
• Page: | 614 | 2,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-21 | latest | en | 0.605491 |
https://ptcouncil.net/how-many-diagonals-does-a-dodecagon-have/ | 1,659,955,494,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570793.14/warc/CC-MAIN-20220808092125-20220808122125-00163.warc.gz | 444,564,553 | 6,211 | A dodecagon is a polygon v 12 sides, 12 angles, and also 12 vertices. Words dodecagon comes from the Greek native "dōdeka" which means 12 and also "gōnon" which method angle. This polygon can be regular, irregular, concave, or convex, depending on its properties.
You are watching: How many diagonals does a dodecagon have
1 What is a Dodecagon? 2 Types that Dodecagons 3 Properties that a Dodecagon 4 Perimeter of a Dodecagon 5 Area the a Dodecagon 6 FAQs on Dodecagon
A dodecagon is a 12-sided polygon that encloses space. Dodecagons deserve to be continuous in which all interior angles and also sides room equal in measure. Lock can additionally be irregular, with different angles and sides of different measurements. The following number shows a regular and also an irregular dodecagon.
Dodecagons deserve to be of different types depending top top the measure up of their sides, angles, and also many such properties. Let us go through the various varieties of dodecagons.
Regular Dodecagon
A consistent dodecagon has actually all the 12 sides of equal length, all angle of same measure, and the vertices space equidistant native the center. It is a 12-sided polygon that is symmetrical. Watch the first dodecagon displayed in the figure given over which reflects a consistent dodecagon.
Irregular Dodecagon
Irregular dodecagons have sides of various shapes and also angles.There have the right to be an infinite amount of variations. Hence, they all look quite various from every other, yet they all have 12 sides. Observe the second dodecagon presented in the figure given over which shows an irregular dodecagon.
Concave Dodecagon
A concave dodecagon contends least one line segment that have the right to be drawn in between the point out on that is boundary but lies external of it. It contends least one of its interior angles higher than 180°.
Convex Dodecagon
A dodecagon whereby no heat segment between any kind of two clues on its boundary lies exterior of that is dubbed a convex dodecagon. Nobody of its interior angles is better than 180°.
## Properties that a Dodecagon
The nature of a dodecagon are noted below i m sorry explain around its angles, triangles and also its diagonals.
Interior angles of a Dodecagon
Each inner angle the a continuous dodecagon is same to 150°. This deserve to be calculation by making use of the formula:
(frac180n–360 n), wherein n = the variety of sides the the polygon. In a dodecagon, n = 12. Currently substituting this value in the formula.
(eginalign frac180(12)–360 12 = 150^circ endalign)
The sum of the internal angles of a dodecagon have the right to be calculated v the help of the formula: (n - 2 ) × 180° = (12 – 2) × 180° = 1800°.
Exterior angles of a Dodecagon
Each exterior edge of a consistent dodecagon is equal to 30°. If we observe the figure given above, we can see that the exterior angle and interior angle type a directly angle. Therefore, 180° - 150° = 30°. Thus, each exterior angle has a measure of 30°. The amount of the exterior angles of a constant dodecagon is 360°.
Diagonals of a Dodecagon
The number of distinct diagonals that have the right to be drawn in a dodecagon from all its vertices can be calculate by utilizing the formula: 1/2 × n × (n-3), wherein n = variety of sides. In this case, n = 12. Substituting the worths in the formula: 1/2 × n × (n-3) = 1/2 × 12 × (12-3) = 54
Therefore, there are 54 diagonals in a dodecagon.
Triangles in a Dodecagon
A dodecagon can be broken into a collection of triangles by the diagonals which are drawn from the vertices. The number of triangles which are created by these diagonals, deserve to be calculated through the formula: (n - 2), where n = the number of sides. In this case, n = 12. So, 12 - 2 = 10. Therefore, 10 triangles can be developed in a dodecagon.
The adhering to table recollects and also lists every the crucial properties the a dodecagon debated above.
Properties Values Interior angle 150° Exterior angle 30° Number that diagonals 54 Number the triangles 10 Sum that the inner angles 1800°
## Perimeter of a Dodecagon
The perimeter that a regular dodecagon can be discovered by recognize the amount of all its sides, or, by multiply the size of one side of the dodecagon through the total variety of sides. This have the right to be stood for by the formula: ns = s × 12; whereby s = length of the side. Let us assume that the next of a constant dodecagon steps 10 units. Thus, the perimeter will certainly be: 10 × 12 = 120 units.
## Area of a Dodecagon
The formula for finding the area that a regular dodecagon is: A = 3 × ( 2 + √3 ) × s2 , where A = the area that the dodecagon, s = the length of that side. Because that example, if the side of a continuous dodecagon procedures 8 units, the area the this dodecagon will certainly be: A = 3 × ( 2 + √3 ) × s2 . Substituting the worth of its side, A = 3 × ( 2 + √3 ) × 82 . Therefore, the area = 716.554 square units.
Important Notes
The adhering to points should be retained in mental while solving troubles related come a dodecagon.
Dodecagon is a 12-sided polygon v 12 angles and also 12 vertices.The amount of the internal angles that a dodecagon is 1800°.The area of a dodecagon is calculated with the formula: A = 3 × ( 2 + √3 ) × s2The perimeter of a dodecagon is calculated v the formula: s × 12.
## Related posts on Dodecagon
Check the end the complying with pages pertained to a dodecagon.
Example 1: Identify the dodecagon from the complying with polygons.
Solution:
A polygon through 12 political parties is recognized as a dodecagon. Therefore, number (a) is a dodecagon.
Example 2: There is an open park in the shape of a continual dodecagon. The neighborhood wants come buy a fencing cable to place it about the boundary of the park. If the size of one next of the park is 100 meters, calculation the size of the fencing wire forced to location all follow me the park's borders.
Solution:
Given, the size of one side of the park = 100 meters. The perimeter the the park deserve to be calculated using the formula: Perimeter of a dodecagon = s × 12, whereby s = the length of the side. Substituting the value in the formula: 100 × 12 = 1200 meters.
Therefore, the size of the required wire is 1200 meters.
See more: How Long Is 144 Inches Is How Many Feet ? 144 In To Ft 144 Inches To Feet
Example 3: If each side the a dodecagon is 5 units, uncover the area that the dodecagon. | 1,728 | 6,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-33 | latest | en | 0.939493 |
http://www.javaprogrammingforums.com/%20whats-wrong-my-code/29401-simulation-grid-game-printingthethread.html | 1,495,848,547,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608726.4/warc/CC-MAIN-20170527001952-20170527021952-00071.warc.gz | 679,391,814 | 4,666 | # Simulation for Grid Game?
• May 11th, 2013, 10:03 AM
lah2015
Simulation for Grid Game?
I am trying to simulate a grid game using java (for loops, if/else). The game is played as follows: The player draws a card that has one of four colors (Red, green, blue, yellow). Then, the player rolls two dice and finds the product of the two numbers. The player tries to find the number on a grid with the multiples of numbers one through 6. Each number correlates with a color, and if the number the player gets matches the color you win. If not, you lose.
Here is the code I have attempted so far but I am getting errors I don't know how to fix, please help!
Code :
```public class ColorGrid { public static void main(String args[]) { //Pick a Color int color=(int)(Math.random()*4)+1; //Product of 2 Dice int a=(int)(Math.random()*6)+1; int b=(int)(Math.random()*6)+1; int product=(a*b); int trial=1; int win=0; int lose=0; //Red for(color=1,trial<=500,trial++); { if(product==2,6,15,24) { win++; } else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) { lose++; } }```
Errors:
Code :
```ColorGrid.java:16: error: not a statement for(color=1,trial<=500,trial++); ^ ColorGrid.java:16: error: ';' expected for(color=1,trial<=500,trial++); ^ ColorGrid.java:17: error: illegal start of expression { ^ ColorGrid.java:18: error: ')' expected if(product==2,6,15,24) ^ ColorGrid.java:18: error: ';' expected if(product==2,6,15,24) ^ ColorGrid.java:18: error: illegal start of expression if(product==2,6,15,24) ^ ColorGrid.java:18: error: ';' expected if(product==2,6,15,24) ^ ColorGrid.java:18: error: illegal start of expression if(product==2,6,15,24) ^ ColorGrid.java:18: error: ';' expected if(product==2,6,15,24) ^ ColorGrid.java:18: error: illegal start of expression if(product==2,6,15,24) ^ ColorGrid.java:18: error: ';' expected if(product==2,6,15,24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: ';' expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: ';' expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: ';' expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: ';' expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: ';' expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: ';' expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: illegal start of type else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:22: error: <identifier> expected else(product==!2)&&(product==!6)&&(product==!15)&&(product==!24) ^ ColorGrid.java:23: error: ';' expected { ^ ColorGrid.java:24: error: illegal start of type lose++; ^ ColorGrid.java:24: error: <identifier> expected lose++; ^ ColorGrid.java:24: error: ';' expected lose++; ^ ColorGrid.java:30: error: class, interface, or enum expected } ^```
• May 11th, 2013, 10:37 AM
Norm
Re: Simulation for Grid Game?
Quote:
I am getting errors
Please copy the full text of the error messages and paste it here.
Take a look at the tutorial for the correct syntax:
http://docs.oracle.com/javase/tutori...dbolts/if.html
http://docs.oracle.com/javase/tutori...bolts/for.html
http://docs.oracle.com/javase/tutori...bolts/op2.html
• May 11th, 2013, 10:40 AM
lah2015
Re: Simulation for Grid Game?
• May 11th, 2013, 10:44 AM
Norm
Re: Simulation for Grid Game?
There are several syntax errors in the code. You need to read the tutorials at the links I posted to see the correct way to write the code.
• May 11th, 2013, 10:46 AM
theoriginalanomaly
Re: Simulation for Grid Game?
in the for statement you need to separate with semicolons. for (color = 1; trial <=500; trial++) and not put a semicolon after the parentheses. In your if statements, you want to write != not ==!(int). Also in your compound else statement. Else is used to be done as long as the if statement is not used. It doesn't require any logic checks. If you want a logic check, you would need to use else if. And in your if statement, you cannot have commas like that to check multiple values.
Code java:
```for (color = 1; trial <= 500; trial++) if ((product == 2) && (product == ... etc) // notice the double opening parentheses. This makes the compound if statement else lose++ // or else if ((product != 2) && (product != 4)) //```
• May 11th, 2013, 10:48 AM
lah2015
Re: Simulation for Grid Game?
So, is it okay to have an if-else statement nested within a for loop?
• May 11th, 2013, 10:51 AM
Norm
Re: Simulation for Grid Game?
Yes, a for loop can contain if statements and many other types of statements.
• May 11th, 2013, 11:05 AM
lah2015
Re: Simulation for Grid Game?
Ok, now it is compiling but it is not completing 500 trials like I want it to, instead it is only completing one.
Code :
```public class ColorGrid { public static void main(String args[]) { //Pick a Color int color=(int)(Math.random()*4)+1; //Product of 2 Dice int a=(int)(Math.random()*6)+1; int b=(int)(Math.random()*6)+1; int product=(a*b); int trial=1; int win=0; int lose=0; //Red for(trial=1; trial<=500; trial++); { if(color==1) if((product==2)||(product==6)||(product==15)||(product==24)) { win++; } else lose++; //Blue else if(color==2) if((product==4)||(product==12)||(product==20)||(product==36)) { win++; } else lose++; //Green else if(color==3) if((product==3)||(product==10)||(product==16)||(product==30)) { win++; } else lose++; //Yellow else if(color==4) if((product==1)||(product==5)||(product==9)||(product==9)||(product==18)||(product==25)) { win++; } else lose++; } System.out.println("Number of Wins: "+win); System.out.println("Number of Losses: "+lose); } }```
Is it something with the for loop that is incorrect?
• May 11th, 2013, 11:14 AM
theoriginalanomaly
Re: Simulation for Grid Game?
You didn't remove the semicolon from the end of the for loop. for (trial = 1; trial<=500;trial++) no semicolon. Other then that, you are sending the same numbers through the for loop, so even if it ran 500 times you would get either 500 wins or 500 losses.
• May 11th, 2013, 11:15 AM
Norm
Re: Simulation for Grid Game?
Quote:
it is only completing one.
Why do you think it is only completing one loop?
Add a call to the println() method inside the loop that prints out the values of trial and the value of color.
The print out will show you the number of loops the program is making.
Please copy the program's output that shows what you are talking about.
• May 11th, 2013, 11:16 AM
lah2015
Re: Simulation for Grid Game?
That should not be the case, but I don't really know how to fix that. I should be getting a combination of wins and losses?
Norm, at first the output only showed:
Number of Wins: 0
Number of Losses: 1
Now, after removing the semi-colon, it is:
Number of Wins: 0
Number of Losses: 500
or vice versa.
But I need it to be a combination of wins and losses. What am I doing wrong?
• May 11th, 2013, 11:21 AM
theoriginalanomaly
Re: Simulation for Grid Game?
You are assigning the values before the loop. Then going through the loop, with those assigned values.
• May 11th, 2013, 11:25 AM
lah2015
Re: Simulation for Grid Game?
How can I make it reassign the values each time it goes through the loop?
• May 11th, 2013, 11:37 AM
theoriginalanomaly
Re: Simulation for Grid Game?
By removing the "How can" and the "?" from your last sentence.
• May 11th, 2013, 11:42 AM
lah2015
Re: Simulation for Grid Game?
Sorry hehe silly question! Thank you so much. | 2,611 | 8,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.465675 |
https://www.routledge.com/Linear-and-Non-Linear-Deformations-of-Elastic-Solids-1st-Edition/Roy-Bera/p/book/9780367333652 | 1,571,401,020,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00150.warc.gz | 1,058,093,985 | 25,042 | # Linear and Non-Linear Deformations of Elastic Solids
## 1st Edition
CRC Press
536 pages | 87 B/W Illus.
Hardback: 9780367333652
pub: 2019-12-20
SAVE ~\$33.99
Available for pre-order. Item will ship after 20th December 2019
\$169.95
\$135.96
x
FREE Standard Shipping!
### Description
Linear and Non-Linear Deformations of Elastic Solids aims to compile the advances in the field of linear and non-linear elasticity through discussion of advanced topics. Broadly classified into two parts, it includes crack, contact, scattering and wave propagation in linear elastic solids and bending vibration, stability in non-linear elastic solids supported by MATLAB examples. This book is aimed at graduate students and researchers in applied mathematics, solid mechanics, applied mechanics, structural mechanics and includes comprehensive discussion of related analytical/numerical methods.
PART ONE
1. Basic fundamentals and an overview
1. Introduction
2. Basic stress system
3. Equation of motion and various potentials
4. Various transforms used
5. General form of the elastic wave equation
6. Reciprocity principle and representation theorem
7. General solution of the equation of motion for an arbitrary force system
8. Green’s function in an infinite medium
9. Principle of fracture mechanics
2. One or two-dimensional singular integral equation in contact and crack and method of solution
1. Introduction
2. Crack boundary condition
3. Boundary condition for punch or indentation problem
4. Basic form of singular integral equation for crack and punch problems
5. Method of solution of one-dimensional singular integral equation
6. Basic integral equation in crack and punch problem in planar surface
7. Direct method of solution for two-dimensional singular integral solution in elliptic region
8. Potential method for two-dimensional singular integral solution
9. Derivation in terms of Jacobi’s polynomial
10. Applications
3. Two-dimensional contact and crack problem in isotropic elastic media:Complex variable technique
1. Introduction
2. Complex representation of the plane elasticity problem
3. Complex potentials in semi-infinite medium
4. First fundamental problem for the semi-infinite medium
5. Green’s function in infinite and semi-infinite media
6. Contact problem for the half plane
7. Flat punch
8. Hertz indentation
9. Stress in the medium for Hertz’ indentation
10. Formulation of the crack problem
11. Line crack at the interface of two elastic media
12. Stress intensity factor in interface medium
13. Stress intensity factor
14. Crack tip singularity: stress intensity factor determination in wedge
15. General Observation
4. Two-dimensional contact and crack problem in anisotropic media
1. Introduction
2. Green’s function in an anisotropic medium
3. Line source and dislocation in an infinite medium
4. Green’s function in a half space
5. Green’s function of two-dimensional anisotropic plates containing an elliptic hole
6. Contact problem under a punch
7. Hertzian Contact solution in bonded dissimilar materials in presence of a loading
8. Fully open crack between dissimilar anisotropic composites
9. Formulation of the integral equation
10. The Comninou interface crack
11. Method of Solution
5. Complete solution to three-dimensional indentation and crack problems in isotropic elastic media
1. Introduction
2. Circular crack and Punch problem
3. Point dislocation in front of a crack
4. Dislocation outside a circular punch
5. Elastic field around a circular crack and punch: Fabrikant’s method
7. Basic solutions in three-dimensional Contact problem in isotropic elastic media
8. Formulation of the integral equation and its solution for the contact problem
9. Alternative method of solution
10. Complete solutions of the elastic field inside the elastic half space
12. Stresses on the axis of symmetry
13. Surface displacement for elliptic contact
14. Circular contact: A Particular case
15. Line contact
16. Tangential indentation
17. Elliptic crack in an isotropic elastic medium
18. Indentation stress field for Hertzian contact
19. Features of Hertzian fracture
6. Three-dimensional interface crack in isotropic and anisotropic elastic media
1. Introduction
2. Formulation of the problem
3. Analytical Solution of the pair of integral equations
4. Energy release rate
5. Interface crack in anisotropic medium
6. Constant normal pressure
7. General observation
7. Three-dimensional elliptic indentation and crack problem in piezoelectric media
1. Introduction
2. Basic solution in piezoelectric medium
4. Integral equations for contact and crack problem
5. Formulation of the integral equations
6. Method of solution for contact problem
7. Total mechanical load and electric charge
8. Limiting case of transversely isotropic media
10. Complete solution in the medium
11. Complete field
12. Crack tip field
13. Crack in piezoelectric medium
15. General observation and discussion
8. Crack-microcrack interaction and crack and punch in plate and layered media
1. Introduction
2. Two-dimensional Crack-microcrack interaction
3. Kachanov’s method for two-dimensional crack interaction problem
4. Three-dimensional crack interaction
5. Interaction between circular cracks: Kachanov’s method
7. Summary of numerical results: Interaction between circular cracks
8. Interaction between elliptic cracks
10. Interaction between circular and elliptic cracks
9. Weight function theory
1. Introduction
2. Basic theory
3. Application
4. Axisymmetric weight function for a circular crack
5. Crack face weight functions for circular crack
6. Crack face weight functions for half plane crack
7. Weight function theory for an elliptic crack in an infinite medium
8. Determination of the potentials
9. Approximate method for the determination of the weight function
10. The Petroski Achenbach method
11. Discussion and some applications of the weight function theory
10. Surface displacement in an elastic half space due to an earthquake source on an inclined fault plane
1. Introduction
2. Statement of the problem
3. Reduction by Cagniard’s technique
4. Reduction in case of S - wave
5. Complete form of surface displacement
6. Discussion
11. Earth response to uniform self similar crack motion
1. Introduction
2. Formulation
3. Formulation of the problem
4. Method of homogeneous solution
5. Body force equivalents and surface displacement
6. Discussion
12. Growth of a semi-infinite crack at a varying velocity
1. Introduction
2. Growth of a half plane infinite crack at a varying velocity
3. Wiener -Hopf Method
4. Reduction of the integral equation
12.5 Discussion
13. Dynamic response of elliptical footings
1. Introduction
2. Basic solutions for forced vibration of elliptic disc
14. Two-dimensional low frequency scattering of acoustic wave by a rough surface
1. Introduction
2. Statement of the scattering problem
3. Scattering cross section
4. Examples
15. Scattering and impact response of a half plane crack in transversely isotropic and isotropic media
1. Introduction
2. Formulation of the problem
3. Limiting case: Isotropic medium
4. Diffraction by a line crack in a transversely isotropic medium
5. Line crack in an isotropic medium
6. Stopping of a line crack
16. Scattering from an elliptic crack
1. Introduction
2. Formulation of the problem
3. Low frequency case
4. Mid frequency case
5. Effective elastic moduli and attenuation coefficient
6. Numerical results and general discussion
7. Dynamic crack opening displacement
8. Dynamic stress intensity factor
9. Scattering Cross-Section and Back-Scattered Displacement
17. Two-dimensional crack and contact problems – Transform method
1. Introduction
2. Formulation
3. Anti plane fracture analysis of a functionally graded piezoelectric layer on a substrate
4. Discussions
18. Effective moduli of elastic inclusion and inhomogeneity
1. Introduction
2. Ellipsodal inclusion
3. Eshelby tensor
4. Equivalent inclusion method - Ellipsoidal inhomogeneity
5. Wu’s result
6. Self consistent scheme - Energy equivalent method
7. Effective medium theory of composites
8. Self consistent theory : Various approximate schemes
9. Mori Tanaka’s method and Kuster Toks model
10. Kuster Toks
11. Differential effective medium theory
12. Effective dynamic elastic moduli of a random distribution of inclusion
13. Propagation of elastic waves in composites with random set of spherical inclusions (EMM Version)
14. General Remark
19. Numerical method in elasto-static and elasto-dynamic crack problem
1. Introduction
2. Three-dimensional elasto-static case
3. Derivation of singular integral equation from body force method
4. CPV and hypersingular integral equation
5. Numerical implementation
6. Discussion of the results obtained by various workers
7. Boundary integral method in elastic wave scattering problem
8. Formulation of BIM
9. Discretisation and regularisation technique for BIE
10. Alternate method
11. Zhang and Achenbach’s method for two-dimensional BI
12. Alternate BIM for anisotropic piezoelectric media
13. Two dimensional BIE for anisotropic media
14. Details of numerical scheme
15. Stress intensity factor evaluation
16. Element free BIM
17. Discussion
PART TWO
20. Large Amplitude Free Vibration of Rotating Non-Homogeneous Beams with Non- Linear Spring and Mass System
1. Introduction
2. Formulation of the Problem
3. Solution Methodology
4. Linear solution
5. Non-linear solution
6. Results and discussions
7. Conclusion
8. Appendix
21. Stability of an Anisotropic Right-Angled Isosceles Triangular Plate under Large Deflection
1. Introduction
2. Constitutive Equations
3. Governing Equations for an Anisotropic Right-Angled Triangular Plate
4. Stability Analysis of an Anisotropic Right-Angled Isosceles Triangular Plate underLarge Deflection
22. Large Amplitude Free Vibrations of Irregular Plates using Complex Variable Technique
1. Introduction
2. Governing Equation
3. Applications of the Method
4. Experimental Verification
5. Discussion on Numerical and Experimental Results
6. Conclusion
23. Large Amplitude Vibrations of Thin Elastic Plates using Conformal Transformation
1. Introduction
2. Governing Equations
3. Applications of the Method
4. Results and Conclusions
24. Large Deflection of a Circular Plate on Elastic Foundation
1. Introduction
2. Governing Equations
3. Solution for a Circular Plate under Transverse Load
4. Numerical Results and Discussions
25. A Modified Approach to the Nonlinear Analysis of Thin Elastic Plates
1. Introduction
5. Large Deflection of Elastic Plates under Uniform Load
6. Large Deflection of Circular Elastic Plates under a Concentrated Load at the Centre
26. Large Amplitude Free Vibration of Parabolic Plates
1. Introduction
2. Governing Equations
3. Transverse Vibration of Parabolic Plates
4. Solution of the Problem
5. Numerical Results
6. Observations and Conclusions
27. Large Amplitude Free Vibration of Sandwich Parabolic Plates
1. Introduction
2. Governing Equations
3. Equation for Sandwich Parabolic Plate
4. Solution of the Problem
5. Numerical Results and Discussions
28. Large Amplitude Vibration of Orthotropic Sandwich Elliptic Plates
1. Introduction
2. Governing Equations
3. Stress-Strain Relations for each Face-Sheet of the Sandwich Plate
4. Strain and Displacement Relations of the Sandwich Elliptic Plate
5. Derivation of Strain Energy of the Sandwich Plate
6. Vibration of an Orthotropic Sandwich Elliptic Plate
7. Solution of the Problem
8. Numerical Results and Discussions
9. Conclusion
29. Large Amplitude Vibration of Heated Orthotropic Sandwich Elliptic Plates
1. Introduction
2. Governing Equations
3. Stress-Strain-Temperature Relations for each Face-Sheet of the Heated Sandwich Plate
4. Strain and Displacement Relations of the Sandwich Plate
5. Strain Energy of a Heated Sandwich Plate
6. Governing Equation for the Heated Sandwich Elliptic Plate
7. Solution of the Problem
8. Numerical Results and Discussions
9. Conclusion
30. Stability Analysis of Thermal Bending and Buckling of Plates due toLargeDeflection
1. Introduction
2. Governing Equations
3. Solution for Simply Supported Rectangular Plate
4. Solution for Clamped Circular Plate
5. Solution for Clamped Elliptic Plate
31. Stability of Thin Plates under Edge Thrust due to Large Deflections, Buckling being Resisted by a Force Proportional to the Displacement
1. Introduction
2. Governing Equations
3. Solution for Simply Supported Rectangular Plate
4. Solution for Clamped Circular Plate
5. Conclusion
32. Large Deflection of Clamped Cylindrical Shell
1. Introduction
2. Nonlinear Analysis of Clamped Cylindrical Shells under Static Load
3. Large Amplitude Free Vibration of Clamped Cylindrical Shells
4. Discussion
33. Large Deflection of Heated Orthotropic Thin Cylindrical Shell
1. Introduction
2. Governing Equations
3. Solution of the Problem
4. Numerical Computations and Discussion
5. Observation and Conclusion
34. Nonlinear Vibration and Stability of an Orthotropic Sandwich Shell of Double Curvature with Orthotropic Core
1. Introduction
2. Governing Equations
3. Stability of a Shallow Sandwich Shell
4. Solution for movable edge
6.
7. Numerical Results and Discussions
35. Nonlinear Vibrations of a Heated Orthotropic Sandwich Shell of Double Curvature with Orthotropic Core
1. Introduction
4. Appendix
36. Nonlinear Vibration of Spherical Shells of Variable Thickness
1. Introduction
2. Governing Equations
3. Solution for Spherical Shell of Variable Thickness
4. Numerical Computations and Graphs
5. Conclusions
Prof. Arabinda Roy graduated from Presidency College, followed by M.Sc.degree from the University of Calcutta .He has two Doctoral degrees to his credit, one from Calcutta University (1969) and the other from the University of Cambridge (U.K) in 1972 under Dr. E.R. Lapwood. He was awarded Commonwealth Research Fellowship to work at Emmanuel College, Cambridge (1969-72).He started his professional career in 1974 at Geological Survey of India as Senior Geophysicist. He later joined his alma mater the Department of Applied Mathematics, University of Calcutta from where he retired as a Professor in 2007 after serving for more than three decades. He was a visiting Professor at I.I.M.A.S., UNAM for a year in 1981. He worked as Principal investigator in a UGC research project. Four students worked under him for Ph.D. degree. Roy’s research interests primarily include theoretical Seismology, wave propagation, vibration and scattering problems, contact and crack theory and associated fields. He has oft-quoted publications in journals of international repute.
Rasajit Kumar Bera
Prof. Rasajit Kumar Bera is a Gold Medalist in Applied Mathematics from University of Calcutta. He received his Ph.D. from Jadavpur University in 1968. Previously, he was a faculty in Presidency College, Kolkata, Bengal Engineering College, Shibpur and joined as Professor & Head of the Department of Science in NITTTR, Kolkata in 1993 from where he retired. He was then invited to act as Professor& Head of the Department of Mathematics in Heritage Institute of Technology, Kolkata where he taught for more than ten years. He guided ten students for Ph.D. Degrees in different topics of Applied Mathematics including Fractional Calculus. 130 Research Papers of him have been published in National and International Journals. He has also contributed in more than ten books. Mathematical Physics for Engineers published by New Age International Publishers and Encyclopaedia of Thermal Stresses edited by Prof. R.B. Hetnarski, published by Springer are among them. Prof. Bera is an associate editor of International Journal of Applied and Computational Mathematics – A Springer Scopus indexed journal. His research interest includes Mathematical Theory of Linear and Non-Linear Elasticity, Generalized Thermo-Elasticity, Thermo-Elasticity in Random Media, Numerical Methods and Computation, Fractional Calculus. Among many Research Projects completed by him, a Major Project from Bhaba Atomic Research Centre(BARC) on Fractional Calculus applied to describe Reactor Kinetics & Flux Matching is worth mentioning. | 3,624 | 16,110 | {"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} | 2.6875 | 3 | CC-MAIN-2019-43 | latest | en | 0.833163 |
https://ryanscharf.com/correlation-regression/ | 1,719,126,193,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00128.warc.gz | 431,453,594 | 26,280 | Select Page
1a. H0: A cookie will break at more than 70 pounds of force or greater; u >= 70
Ha: A cookie will break at less than 70 pounds of force on average; u < 70
1b. yes
1c. p = 0.0359 There’s a 3.59% chance of finding a sample mean of 69.1 if u >=70.
1d. p = 0.0002 since p < 0.5, there is enough evidence to dismiss H0
1e. p = 0.0227 since p < 0.5, there is enough evidence to dismiss H0
2a. cor(mod8\$Cost.per.serving, mod8\$Fibber.per.serving)
2b. .228
2c: | 192 | 485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.731821 |
https://www.mrexcel.com/board/threads/suimifs-only-unique-values.804671/ | 1,725,968,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00271.warc.gz | 857,624,121 | 18,144 | # SUIMIFS only unique values
#### quiqueperez
##### New Member
Hi,
I'm trying to add the values in column A based on certain criteria in other columns using a SUMIFS formula.
However, some of the rows can be classed as duplicates for my purpose and I need to exclude them.
IN the attached example, I need to add the length of the ducts in column A. However, some ducts are duplicated (duct A is listed twice) and they need to be added only once.
If I wanted to add the length of the ducts in region HH, I would be expecting 25+12=37m
Is there any SUMIFS formula I can use for this.
Something like SUMIFS(Column A, Region, HH, only when the values in column Duct are unique)
Column values to add Duct Cable Region 25 Duct A Cable 1 HH 25 Duct A Cable 2 HH 12 Duct B Cable 1 HH 8 Duct C Cable 2 HV
<tbody>
</tbody>
Many thanks!
Enrique
### Excel Facts
Does the VLOOKUP table have to be sorted?
No! when you are using an exact match, the VLOOKUP table can be in any order. Best-selling items at the top is actually the best.
maybe...
Excel 2012
ABCDE
225Duct ACable 1HH37
325Duct ACable 2HH
412Duct BCable 1HH
58Duct CCable 2HV
<tbody>
</tbody>
Sheet4
Array Formulas
CellFormula
E2{=SUM(IF(FREQUENCY(IF((B2:B5<>"")*(D2:D5="HH"),MATCH(A2:A5,A2:A5,0)),ROW(A2:A5)-ROW(A2)+1),A2:A5))}
<tbody>
</tbody>
Entered with Ctrl+Shift+Enter. If entered correctly, Excel will surround with curly braces {}.
Note: Do not try and enter the {} manually yourself
<tbody>
</tbody>
Replies
7
Views
202
Replies
1
Views
294
Replies
8
Views
741
Replies
13
Views
659
Replies
6
Views
298
### Forum statistics
1,221,053
Messages
6,157,640
Members
451,426
Latest member
VinnyDoesntKnowExcelCode
### We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
### Which adblocker are you using?
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
### Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
### Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back | 737 | 2,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.831734 |
https://stats.stackexchange.com/questions/545155/how-to-reconcile-these-two-versions-of-a-linear-model | 1,652,892,841,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00174.warc.gz | 608,909,681 | 66,746 | # How to reconcile these two versions of a "linear model"?
Outline
I am taking a course in which the professor, unless I'm badly misunderstanding something, is discussing two varieties of linear models.
Version 1
The general linear model is $$\boldsymbol Y = \boldsymbol X \boldsymbol \beta + \boldsymbol \epsilon$$. Written out, it is
\begin{align*} \begin{pmatrix} Y_1\\ \vdots\\ Y_n \end{pmatrix} = \begin{pmatrix} h_1(X_1) & \dots & h_p(X_1)\\ \vdots & & \vdots\\ h_1(X_n) & \dots & h_p(X_n) \end{pmatrix} \begin{pmatrix} \beta_1\\ \vdots\\ \beta_p \end{pmatrix} + \begin{pmatrix} \epsilon_1\\ \vdots\\ \epsilon_n \end{pmatrix} \end{align*}
Written out, this takes the form
\begin{align*} Y_i = \beta_1 h_1(X_i) + \dots + \beta_p h_p(X_i) + \epsilon_i \end{align*}
for $$i = 1, \dots, n$$.
Version 2
\begin{align*} \begin{pmatrix} Y_1\\ \vdots\\ Y_n \end{pmatrix} = \begin{pmatrix} X_{11} & \dots & X_{1p}\\ \vdots & & \vdots\\ X_{n1} & \dots & X_{np} \end{pmatrix} \begin{pmatrix} \beta_1\\ \vdots\\ \beta_p \end{pmatrix} + \begin{pmatrix} \epsilon_1\\ \vdots\\ \epsilon_n \end{pmatrix}. \end{align*}
Or written out,
\begin{align*} Y_i = X_{i1} + \dots + X_{ip} + \epsilon_i. \end{align*} for $$i = 1, \dots, n$$.
It took me a while to realize that these two are quite different, because in the first one the rows of $$\boldsymbol X$$ are functions of the same explanatory variable $$X_i$$. I guess you could say that Version 1 only involves one explanatory variable. Version 2, on the other hand, looks more like a regular setup for multiple regression, with each $$Y_i$$ being written as a linear combination of the $$i$$th observations of all $$p$$ predictors.
How do I reconcile these? I guess you could combine them by making $$\boldsymbol X$$ $$p^2$$ columns wide and $$\boldsymbol \beta$$ $$p^2$$ rows long, by applying the functions $$h_1, \dots, h_p$$ to all $$p$$ explanatory variables in each row of $$\boldsymbol X$$? (I suppose there is no reason why the number of functions $$h$$ needs to be equal to the number of explanatory variables $$X$$.)
In other words, if I wanted to combine these two paradigms, would I be looking at something like
\begin{align*} \begin{pmatrix} Y_1\\ \vdots\\ Y_n \end{pmatrix} = \begin{pmatrix} h_1(X_{11}) & h_2(X_{11}) & \dots & h_p(X_{11}) & \dots & h_1(X_{1p}) & h_2(X_{1p}) & \dots & h_p(X_{1p})\\ \vdots & \vdots & & \vdots & & \vdots & \vdots & & \vdots\\ h_1(X_{n1}) & h_2(X_{n1}) & \dots & h_p(X_{n1}) & \dots & h_1(X_{np}) & h_2(X_{np}) & \dots & h_p(X_{np}) \end{pmatrix} \begin{pmatrix} \beta_1\\ \vdots\\ \beta_{p^2} \end{pmatrix} + \begin{pmatrix} \epsilon_1\\ \vdots\\ \epsilon_n \end{pmatrix}? \end{align*}
I appreciate any help.
You have it exactly right.
For instance, you might have two predictors $$X_1$$ and $$X_2$$. In your model, you decide to use $$X_1$$ untransformed, "as-is": $$h_1(X_1)=X_1$$. For your second predictor, you decide to use $$X_2$$ both untransformed, $$h_1(X_2)=X_2$$ and squared, $$h_2(X_2)=X_2^2$$. Your model contains three predictors $$X_1, X_2, X_2^2$$, so your parameter vector is also of length three.
Unless you also add an intercept, that is. Which in this framework you could consider yet another function, namely the one that sends everything to $$1$$: $$h_0(X)=1$$.
Consider what happens if every $$h_j$$ is the identity function.
Spoiler alert: It's exactly the same as the other model.
What your professor is showing you is the idea of nonlinear basis functions that allow you to introduce curvature, not just lines and planes. The gist is that, once you do the transformation, you have some other features and then fit the linear regression to those new features. Let's go through an example.
EXAMPLE
We start out with a variable $$y$$ that we want to predict, given two features $$x_1$$ and $$x_2$$. The basic linear regression would be $$y = \beta_0 + \beta_1x_1 +\beta_2x_2 +\epsilon$$. However, you know from your domain knowledge (your understanding of the physics, biology, economics, etc) that $$y$$ should depend on $$x_1^2$$, not $$x_1$$. Enter the $$h$$ function $$h(x_1) = x_1^2$$. Now write your new linear regression equation, $$y = \beta_0 + \beta_1h(x_1) +\beta_2x_2 +\epsilon$$. You can think of this as $$h(x_1) = x_3$$. Then you wind up with a linear regression $$y = \beta_0 + \beta_1x_3 +\beta_2x_2 +\epsilon$$, which is the usual format.
The idea is that the matrix multipliction does not know or care how you got your $$x_3$$, only that you got it.
(In fact, your professor has not made this complicated enough. It is routine to use interaction terms, such as $$y = \beta_0 + \beta_1x_1 +\beta_2x_2 \beta_3 x_1x_2 +\epsilon$$. This involves some function $$h(x_1, x_2) = x_1x_2$$, yet your definition does not allow for that. Really, each of your $$h$$ functions should be functions of all of the features.) | 1,574 | 4,862 | {"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": 49, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-21 | longest | en | 0.703637 |
https://citizendium.org/wiki/Basel_problem | 1,656,991,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00576.warc.gz | 217,347,405 | 8,739 | # Basel problem
Main Article
Discussion
Related Articles [?]
Bibliography [?]
Citable Version [?]
This editable Main Article is under development and subject to a disclaimer.
The Basel problem is a question about infinite series, first formulated by Pietro Mengoli in 1650[1] Many of the most talented and influential mathematicians of that time attempted and failed to solve the problem. Its solution was one of the early triumphs of Leonard Euler, and immediately made him renown in European mathematical circles.
Mengoli knew that the series of positive numbers ${\displaystyle \scriptstyle \sum _{n=1}^{\infty }n^{-2}}$ is convergent. In the Basel problem, he asked for its sum. Euler showed that
${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}.}$
He actually showed considerable more than this, evaluating the sums of several related series which, from a modern viewpoint, are values of the Riemann zeta function at positive even integers. His argument was highly non-rigorous, assuming that results known to be true for finite sums and products also extend to infinite series and products. Even so, he had already calculated its value to 20 decimal places[2], and up to this precision, the value matched his claimed ${\displaystyle \pi ^{2}/6}$. Thus, mathematicians of the time who disbelieved the validity of his argument could not argue that the value he had found was incorrect.
## Notes
1. See Andre Weil, Basic Number Theory, p.184
2. See William Dunham, Journey Through Genius | 369 | 1,530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.9406 |
http://www.docstoc.com/docs/10726252/Leaving-Certificate-Maths-Syllabus-Higher-and-Ordinary-Level-(File | 1,430,966,820,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430460160413.1/warc/CC-MAIN-20150501060240-00004-ip-10-235-10-82.ec2.internal.warc.gz | 361,935,014 | 39,435 | # Leaving Certificate Maths Syllabus Higher and Ordinary Level (File by shelseaZvansky
VIEWS: 12 PAGES: 3
• pg 1
``` 23
Algebra
1. Manipulation of formulae including simple
algebraic fractions.
Laws of indices: xa xb = xa+b ; (xy)a = xa ya ; (xa )b =
xab.
Use of fractional and negative indices, e.g. (-8)2/3 ,
(1/4)-1/2 . Solution of equations such as 5x = 1/25.
Solution of quadratic equations with rational
coefficients.
The Factor Theorem for polynomials of degree two
or three.
Factorisation of such polynomials (the linear and
2. Unique solution of simultaneous linear equations
with two unknowns.
Solution of one linear and one quadratic equation
with two unknowns (e.g. 2x – y = 1, x2 + y2 = 9).
3. Inequalities: solution of inequalities of the form g(x)
< k, where g(x) = ax + b, and a, b, k ∈ Q.
4. Complex numbers: Argand diagram, modulus,
complex conjugate.
Geometry
1. Synthetic geometry: “Cuts” will not be asked
(See Option: Further
Theorems (to be proved): Gemetry).
I: The sum of the degree-measures of the angles of a
triangle is 180o .
24
Corollary I: The degree-measure of an exterior angle of a
triangle is equal to the degree-measure of the sum of the
two remote interior angles.
Corollary II: An exterior angle of a triangle is greater than
either remote interior angle.
II: Opposite sides of a parallelogram have equal lengths.
III: If three parallel lines make intercepts of equal length
on a transversal, then they will also make intercepts of
equal length on any other transversal.
IV: A line which is parallel to one side-line of a triangle,
and cuts a second side, will cut the third side in the same
proportion as the second.
V: If the three angles of one triangle have degree-
measures equal, respectively, to the degree-measures of
the angles of a second triangle, then the lengths of the
corresponding sides of the two triangles are proportional.
VI: (Pythagoras): In a right-angled triangle, the square of
the length of the side opposite to the right-angle is equal
to the sum of the squares of the lengths of the other two
sides.
VII: (Converse of Pythagoras’ Theorem): If the square of
the length of one side of a triangle is equal to the sum of
the squares of the lengths of the other two sides, then the
triangle has a right-angle and this is opposite the longest
side.
VIII: The products of the lengths of the sides of a triangle
by the corresponding altitudes are equal.
IX: If the lengths of two sides of a triangle are unequal,
then the degree-measures of the angles opposite to them
are unequal, with the greater angle opposite to the longer
side.
X: The sum of the lengths of any two sides of a triangle is
greater than that of the third side.
25
2. Coordinate Geometry:
Coordinates; distance between points; area of triangle;
midpoint of line segment; slope.
Line:
⎯ equation of line in the forms y = mx + c and y – y1 =
(x – x1);
⎯ line through two given points;
⎯ lines parallel to and lines perpendicular to a given line
and through a given point;
⎯ intersection of two lines.
Circle:
⎯ the equation x2 + y2 = a2 ;
⎯ intersection of a line and a circle; Restricted to a circle
centre the origin.
⎯ proving a line is a tangent to a circle;
⎯ equation of circle in the form (x – h)2 + (y – k)2
Given equation, obtain
= a2.
centre; and vice versa.
3. Enlargements:
Enlargement of a rectilinear figure by the ray method.
Centre of enlargement. Scale factor k. Two cases to be
considered:
⎯ k>1, k ∈ Q (enlargement);
⎯ 0<k<1, k ∈ Q (reduction).
A triangle abc with centre of enlargement a, enlarged by a
scale factor k, gives an image triangle ab′ c′ with bc
parallel to b′ c′.
Object length, image length, calculation of scale factor.
Finding the centre of enlargement.
A region when enlarged by a scale factor k has its area
multiplied by a factor k2 .
Trigonometry
Trigonometry of triangle; area of triangle; use of sine and Proofs not required.
cosine rules.
```
To top | 1,064 | 4,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2015-18 | longest | en | 0.875278 |
https://worddisk.com/wiki/Frequency/ | 1,718,734,412,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00181.warc.gz | 566,889,920 | 24,113 | Frequency
# Frequency
Number of occurrences or cycles per unit time
Frequency (symbol f), most often measured in hertz (symbol: Hz), is the number of occurrences of a repeating event per unit of time.[1] It is also occasionally referred to as temporal frequency for clarity and to distinguish it from spatial frequency. Ordinary frequency is related to angular frequency (symbol ω, with SI unit radian per second) by a factor of 2π. The period (symbol T) is the interval of time between events, so the period is the reciprocal of the frequency: T = 1/f.[2]
Quick Facts Common symbols, SI unit ...
Frequency is an important parameter used in science and engineering to specify the rate of oscillatory and vibratory phenomena, such as mechanical vibrations, audio signals (sound), radio waves, and light.
For example, if a heart beats at a frequency of 120 times per minute (2 hertz), the period—the interval between beats—is half a second (60 seconds divided by 120 beats).
## Definitions and units
For cyclical phenomena such as oscillations, waves, or for examples of simple harmonic motion, the term frequency is defined as the number of cycles or repetitions per unit of time. The conventional symbol for frequency is f or ν (the Greek letter nu) is also used.[3] The period T is the time taken to complete one cycle of an oscillation or rotation. The frequency and the period are related by the equation[4]
The term temporal frequency is used to emphasise that the frequency is characterised by the number of occurrences of a repeating event per unit time.
The SI unit of frequency is the hertz (Hz),[4] named after the German physicist Heinrich Hertz by the International Electrotechnical Commission in 1930. It was adopted by the CGPM (Conférence générale des poids et mesures) in 1960, officially replacing the previous name, cycle per second (cps). The SI unit for the period, as for all measurements of time, is the second.[5] A traditional unit of frequency used with rotating mechanical devices, where it is termed rotational frequency, is revolution per minute, abbreviated r/min or rpm.[6] 60 rpm is equivalent to one hertz.[7]
## Period versus frequency
As a matter of convenience, longer and slower waves, such as ocean surface waves, are more typically described by wave period rather than frequency.[8] Short and fast waves, like audio and radio, are usually described by their frequency. Some commonly used conversions are listed below:
More information Period ...
${\displaystyle y(t)=\sin \theta (t)=\sin(\omega t)=\sin(2\mathrm {\pi } ft)}$
The unit of angular frequency is the radian per second (rad/s) but, for discrete-time signals, can also be expressed as radians per sampling interval, which is a dimensionless quantity. Angular frequency is frequency multiplied by 2π.
## In wave propagation
For periodic waves in nondispersive media (that is, media in which the wave speed is independent of frequency), frequency has an inverse relationship to the wavelength, λ (lambda). Even in dispersive media, the frequency f of a sinusoidal wave is equal to the phase velocity v of the wave divided by the wavelength λ of the wave:
${\displaystyle f={\frac {v}{\lambda }}.}$
In the special case of electromagnetic waves in vacuum, then v = c, where c is the speed of light in vacuum, and this expression becomes
${\displaystyle f={\frac {c}{\lambda }}.}$
When monochromatic waves travel from one medium to another, their frequency remains the same—only their wavelength and speed change.
## Measurement
Measurement of frequency can be done in the following ways:
### Counting
Calculating the frequency of a repeating event is accomplished by counting the number of times that event occurs within a specific time period, then dividing the count by the period. For example, if 71 events occur within 15 seconds the frequency is:
${\displaystyle f={\frac {71}{15\,{\text{s}}}}\approx 4.73\,{\text{Hz}}.}$
If the number of counts is not very large, it is more accurate to measure the time interval for a predetermined number of occurrences, rather than the number of occurrences within a specified time.[citation needed] The latter method introduces a random error into the count of between zero and one count, so on average half a count. This is called gating error and causes an average error in the calculated frequency of ${\textstyle \Delta f={\frac {1}{2T_{\text{m}}}}}$, or a fractional error of ${\textstyle {\frac {\Delta f}{f}}={\frac {1}{2fT_{\text{m}}}}}$ where ${\displaystyle T_{\text{m}}}$ is the timing interval and ${\displaystyle f}$ is the measured frequency. This error decreases with frequency, so it is generally a problem at low frequencies where the number of counts N is small.
A resonant-reed frequency meter, an obsolete device used from about 1900 to the 1940s for measuring the frequency of alternating current. It consists of a strip of metal with reeds of graduated lengths, vibrated by an electromagnet. When the unknown frequency is applied to the electromagnet, the reed which is resonant at that frequency will vibrate with large amplitude, visible next to the scale.
### Stroboscope
An old method of measuring the frequency of rotating or vibrating objects is to use a stroboscope. This is an intense repetitively flashing light (strobe light) whose frequency can be adjusted with a calibrated timing circuit. The strobe light is pointed at the rotating object and the frequency adjusted up and down. When the frequency of the strobe equals the frequency of the rotating or vibrating object, the object completes one cycle of oscillation and returns to its original position between the flashes of light, so when illuminated by the strobe the object appears stationary. Then the frequency can be read from the calibrated readout on the stroboscope. A downside of this method is that an object rotating at an integer multiple of the strobing frequency will also appear stationary.
### Frequency counter
Higher frequencies are usually measured with a frequency counter. This is an electronic instrument which measures the frequency of an applied repetitive electronic signal and displays the result in hertz on a digital display. It uses digital logic to count the number of cycles during a time interval established by a precision quartz time base. Cyclic processes that are not electrical, such as the rotation rate of a shaft, mechanical vibrations, or sound waves, can be converted to a repetitive electronic signal by transducers and the signal applied to a frequency counter. As of 2018, frequency counters can cover the range up to about 100 GHz. This represents the limit of direct counting methods; frequencies above this must be measured by indirect methods.
### Heterodyne methods
Above the range of frequency counters, frequencies of electromagnetic signals are often measured indirectly utilizing heterodyning (frequency conversion). A reference signal of a known frequency near the unknown frequency is mixed with the unknown frequency in a nonlinear mixing device such as a diode. This creates a heterodyne or "beat" signal at the difference between the two frequencies. If the two signals are close together in frequency the heterodyne is low enough to be measured by a frequency counter. This process only measures the difference between the unknown frequency and the reference frequency. To convert higher frequencies, several stages of heterodyning can be used. Current research is extending this method to infrared and light frequencies (optical heterodyne detection).
## Examples
### Light
Visible light is an electromagnetic wave, consisting of oscillating electric and magnetic fields traveling through space. The frequency of the wave determines its color: 400 THz (4×1014 Hz) is red light, 800 THz (8×1014 Hz) is violet light, and between these (in the range 400–800 THz) are all the other colors of the visible spectrum. An electromagnetic wave with a frequency less than 4×1014 Hz will be invisible to the human eye; such waves are called infrared (IR) radiation. At even lower frequency, the wave is called a microwave, and at still lower frequencies it is called a radio wave. Likewise, an electromagnetic wave with a frequency higher than 8×1014 Hz will also be invisible to the human eye; such waves are called ultraviolet (UV) radiation. Even higher-frequency waves are called X-rays, and higher still are gamma rays.
All of these waves, from the lowest-frequency radio waves to the highest-frequency gamma rays, are fundamentally the same, and they are all called electromagnetic radiation. They all travel through vacuum at the same speed (the speed of light), giving them wavelengths inversely proportional to their frequencies.
where c is the speed of light (c in vacuum or less in other media), f is the frequency and λ is the wavelength.
In dispersive media, such as glass, the speed depends somewhat on frequency, so the wavelength is not quite inversely proportional to frequency.
### Sound
Sound propagates as mechanical vibration waves of pressure and displacement, in air or other substances.[10] In general, frequency components of a sound determine its "color", its timbre. When speaking about the frequency (in singular) of a sound, it means the property that most determines its pitch.[11]
The frequencies an ear can hear are limited to a specific range of frequencies. The audible frequency range for humans is typically given as being between about 20 Hz and 20,000 Hz (20 kHz), though the high frequency limit usually reduces with age. Other species have different hearing ranges. For example, some dog breeds can perceive vibrations up to 60,000 Hz.[12]
In many media, such as air, the speed of sound is approximately independent of frequency, so the wavelength of the sound waves (distance between repetitions) is approximately inversely proportional to frequency.
### Line current
In Europe, Africa, Australia, southern South America, most of Asia, and Russia, the frequency of the alternating current in household electrical outlets is 50 Hz (close to the tone G), whereas in North America and northern South America, the frequency of the alternating current in household electrical outlets is 60 Hz (between the tones B and B; that is, a minor third above the European frequency). The frequency of the 'hum' in an audio recording can show in which of these general regions the recording was made.
## Aperiodic frequency
Aperiodic frequency is the rate of incidence or occurrence of non-cyclic phenomena, including random processes such as radioactive decay. It is expressed with the unit of reciprocal second (s−1)[13] or, in the case of radioactivity, becquerels.[14]
It is defined as a rate, f = Nt, involving the number of entities counted or the number of events happened (N) during a given time durationt);[citation needed] it is a physical quantity of type temporal rate.
## Notes
1. The term spatial period, sometimes used in place of wavelength, analogously corresponds to the (temporal) period.[9]
## References
1. "Definition of FREQUENCY". Retrieved 3 October 2016.
2. "Definition of PERIOD". Retrieved 3 October 2016.
3. "Resolution 12 of the 11th CGPM (1960)". BIPM (International Bureau of Weights and Measures). Archived from the original on 8 April 2020. Retrieved 21 January 2021.
4. "Special Publication 811: NIST Guide to the SI, Chapter 8". NIST. 28 January 2016. Retrieved 2022-11-08.
5. Davies 1997, p. 275.
6. Boreman, Glenn D. "Spatial Frequency". SPIE. Retrieved 22 January 2021.
7. "Definition of SOUND". Retrieved 3 October 2016.
8. Pilhofer, Michael (2007). Music Theory for Dummies. For Dummies. p. 97. ISBN 978-0-470-16794-6.
9. Condon, Tim (2003). Elert, Glenn (ed.). "Frequency range of dog hearing". The Physics Factbook. Retrieved 2008-10-22.
10. Lombardi, Michael A. (2007). "Fundamentals of Time and Frequency". In Bishop, Robert H. (ed.). Mechatronic Systems, Sensors, and Actuators: Fundamentals and Modeling. Austin: CRC Press. ISBN 9781420009002.
11. Newell, David B; Tiesinga, Eite (2019). The international system of units (SI) (PDF) (Report). Gaithersburg, MD: National Institute of Standards and Technology. doi:10.6028/nist.sp.330-2019. sub§2.3.4, Table 4.
## Sources
• Giancoli, D.C. (1988). Physics for Scientists and Engineers (2nd ed.). Prentice Hall. ISBN 978-0-13-669201-0.
###### Share this article:
This article uses material from the Wikipedia article Frequency, and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses. | 2,834 | 12,668 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.917095 |
https://quantumcomputing.stackexchange.com/questions/1841/what-happens-if-two-separately-entangled-qubits-are-passed-through-a-c-not-gate | 1,722,803,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00199.warc.gz | 397,882,906 | 42,952 | # What happens if two separately entangled qubits are passed through a C-NOT gate?
Suppose I transform a state as follows:
1. I start with the state $\lvert 0\rangle \otimes \lvert0\rangle \otimes \lvert0\rangle \otimes \lvert 0 \rangle$.
2. I entangle the 1st and 2nd qubits (with an H gate and C-NOT).
3. I then then entangle the 3rd and 4th qubits the same way.
If I try to apply H gate and C-NOT to the 2nd and 3rd qubits afterwords, will the whole system become entangled? What happens to the 1st and 4th qubits in that case?
• Commented Apr 19, 2018 at 14:00
• I've focused your post down to the first question you asked, which is the more interesting of the two. You should try to avoid asking more than one question per post unless they are very closely related. Commented Apr 19, 2018 at 14:01
• It would also be nice if the question included an explicit quantum circuit to inequivocally visualize the gates that are being applied. Commented Apr 19, 2018 at 14:25
• Thanks for your questions! As others have said, it is better to have one question per post. If you repost the second question as separate question, I'm sure you'll get a detailed answer to that too. Though the DaftWullie's answer also does a good job. Commented Apr 19, 2018 at 14:51
• Thank you for your very quick response. I am a noobie to this quantum computing field. I recently watched 'quantum computing for the determined' [link] (youtu.be/X2q1PuI2RFI?list=PL1826E60FD05B44E4) playlist from youtube. Now, I am trying to create a programming library to emulate QC (i know there are already). Can anybody link me some source, that I can actually learn all the technical stuff? like, I didn't know the purpose of 'ρ' until the answer. (do I need to ask this as a new question?) Commented Apr 19, 2018 at 17:21
$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>} %$So, first you're entangling qubits 1 and 2, and qubits 3 and 4, so overall you have the quantum state $$\left(\ket{00}+\ket{11}\right)\otimes\left(\ket{00}+\ket{11}\right)/2$$ Then you apply a Hadamard on qubit 2, $$(\ket{0}(\ket{0}+\ket{1})+\ket{1}(\ket{0}-\ket{1}))\otimes(\ket{00}+\ket{11})/(2\sqrt{2})$$ before applying a controlled-NOT from qubit 2 (control) to qubit 3 (target), right? This gives you $$(\ket{0}\otimes(\ket{0}\otimes(\ket{00}+\ket{11})+\ket{1}\otimes(\ket{10}+\ket{01}))+\ket{1}\otimes(\ket{0}\otimes(\ket{00}+\ket{11})-\ket{1}\otimes(\ket{10}+\ket{01})))/(2\sqrt{2})$$ Let's rearrange this slightly as $$\ket{\Psi}=((\ket{0}-\ket{1})\ket{1}(\ket{10}+\ket{01})+(\ket{0}+\ket{1})\ket{0}(\ket{00}+\ket{11}))/(2\sqrt{2})$$ Note that we need the full state of the whole system. You cannot really talk about the states of qubits 1 and 4 separately due to the entanglement.
The question of "is it still entangled" is straightforwardly "yes", but that is a actually a triviality of a much more complex issue. It is entangled in the sense that it is not a product state $\ket{\psi_1}\otimes\ket{\psi_2}\otimes\ket{\psi_3}\otimes\ket{\psi_4}$.
One simple way to see that this state is entangled is to pick a bipartition, i.e. a split of the qubits into two parties. For instance, let's take qubit 1 as one party (A), and all the others as party B. If we work out the reduced state of party A, a product state (unentangled) would have to give a pure state. Meanwhile, if the reduced state is not pure, i.e. has a rank greater than 1, the state is definitely entangled. For example, in this case $$\rho_A=\text{Tr}(\ket{\Psi}\bra{\Psi})=\frac{\mathbb{I}}{2},$$ has rank 2. Actually, it doesn't matter what you did between qubits 2 and 3, as $\rho_A$ is independent of that unitary; it cannot remove the entanglement created with qubit 1 (just possibly spread it between qubits 2 and 3). The fact that you have to look at different bipartitions to see which qubits are entangled with which already starts to indicate some of the complexity. For pure states, it is sufficient to look at each of the bipartitions of 1 qubit with the rest. If each of these reduced density matrices is rank 1, your whole state is separable.
Related to your question, you might like to look up issues of "monogamy of entanglement" -- the more entangled qubit 1 is with qubit 2, the less entangled qubit 1 is with qubit 3 (for example), and that can be quantified in a number of different ways. Equally, you can ask questions about "what sort of entanglement is there?". One approach is to look at what types of entanglement can be converted into different types (often termed "SLOCC equivalence classes"). For example, with 3 qubits, people make the distinction between W-state entanglement, which looks like $\ket{001}+\ket{010}+\ket{100}$ and GHZ-entanglement that looks like $\ket{000}+\ket{111}$, as well as bipartite entanglement between different pairs of qubits, and a separable state on the other. It gets really messy for 4 qubits! | 1,513 | 5,006 | {"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.94202 |
http://mathhelpforum.com/trigonometry/155116-ooh-tricky-question.html | 1,481,224,138,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00046-ip-10-31-129-80.ec2.internal.warc.gz | 174,872,080 | 13,240 | # Thread: ooh a tricky question
1. ## ooh a tricky question
i don't know weather its grouped correctly(sorry)
for $0 < \theta < \frac{\pi}{2}$if
$x = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^{2n} \theta)$
$y = \displaystyle\Sigma_{n=0}^{\infty}(\sin^{2n} \theta)$
$z = \displaystyle\Sigma_{n=0}^{\infty}(\sin^{2n} \theta \cos^{2n} \theta)$
then which of the following are true(multiple choice)
(A) $\frac{1}{x} + \frac{1}{y} = 1$
(B) $x + y +xy =0$
(C) $xyz = xy + z$
(D) $xyz = x + y + z$
2. You may try using mathematical induction.
3. THE ACTUAL TIME GIVEN FOR PROBLEM IS 3 MINUTES using mathematical induction it takes 10 minutes at least
i think we should expand it and take the G.P sum of n terms and do something might work
this is a tricky sum for me
4. Originally Posted by grgrsanjay
i don't know weather its grouped correctly(sorry)
for $0 < \theta < \frac{\pi}{2}$if
$x = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta)$
$y = \displaystyle\Sigma_{n=0}^{\infty}(\sin^2n \theta)$
$z = \displaystyle\Sigma_{n=0}^{\infty}(\sin^2n \theta \cos^2n \theta)$
then which of the following are true(multiple choice)
(A) $\frac{1}{x} + \frac{1}{y} = 1$
(B) $x + y +xy =0$
(C) $xyz = xy + z$
(D) $xyz = x + y + z$
There is something wrong with this question, because $x+y = \displaystyle\Sigma_{n=0}^{\infty}(\cos ^2n \theta + \sin^2n \theta) = \Sigma_{n=0}^{\infty}1$, which is infinite. So at least one of the series for x and y fails to converge. If for example $\theta = \pi/4$ then the three series for x, y, z all diverge, so none of the conditions (A), (B), (C), (D) is true.
5. oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n
6. Originally Posted by grgrsanjay
oh sorry it was just a typing error i corrected the question actually it sin to the power of 2n
In that case, these are geometric series, and the sums are given by $x = \dfrac1{1-\cos^2\theta} = \dfrac1{\sin^2\theta}$ and $y = \dfrac1{1-\sin^2\theta} = \dfrac1{\cos^2\theta}$, from which it should be clear that (A) is correct.
7. yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
8. Originally Posted by grgrsanjay
yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
Is this a multiple choice question or isn't it?
9. Originally Posted by grgrsanjay
which of the following are true(multiple choice)says it is multiple
(A) $\frac{1}{x} + \frac{1}{y} = 1$
(B) $x + y +xy =0$
(C) $xyz = xy + z$
(D) $xyz = x + y + z$
me too getting c and d as answer with a(when substituting)
10. Originally Posted by ggn
says it is multiple
Yes thank you I can read, my point is that in a valid multiple choice question it's not possible to have more than one correct option, therefore if A is correct then B,C,D are automatically incorrect. If the OP wanted to call into question the validity of the problem, the OP could have specified that.
11. Originally Posted by grgrsanjay
yea i got (A) correct so any other options are correct??? but when substitute a value (C) and (D) also seem correct to me
That's right! I only looked at (A), because when I saw that was correct I assumed that it would be the only correct choice. But (C) and (D) are also correct.
12. yea but how to prove (C) and (D) are correct?? tell me please
13. Originally Posted by grgrsanjay
yea but how to prove (C) and (D) are correct?? tell me please
The formulas for x, y and z are $x = \dfrac1{\sin^2\theta},\quad y = \dfrac1{\cos^2\theta},\quad z = \dfrac1{1-\sin^2\theta\cos^2\theta}$. Substitute those into the equations (C) and (D), and check that both sides agree.
14. what have you used to get that formula??
15. Originally Posted by grgrsanjay
what have you used to get that formula??
Sum of a infinite geometric series, $\displaystyle\sum_{n=0}^\infty x^n = \frac1{1-x}$.
Page 1 of 2 12 Last | 1,263 | 3,939 | {"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": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.798883 |
https://effectiveinventory.com/using-residual-inventory-analysis-in-fine-tuning-your-safety-stock-quantities/ | 1,685,740,808,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648858.14/warc/CC-MAIN-20230602204755-20230602234755-00217.warc.gz | 282,549,551 | 27,926 | Select Page
#### Using Residual Inventory Analysis in Fine Tuning Your Safety Stock Quantities
Over the last several months we have been discussing various ways of calculating safety stock quantities. If safety stock quantities are too low, they will not provide adequate insurance to prevent stockouts in case of unusually high demand or delays in receiving a replenishment shipment. If safety stock quantities are too high, part of your inventory investment will be tied up in non-productive stock. This excess inventory will not contribute to your efforts in achieving the goal of effective inventory management.
There is a simple, best practice analysis that will help ensure that the safety stock quantity you maintain for each item is “just right”. To perform this analysis you need to record for each item, the forecast quantity, actual usage and the safety stock quantity for each of the previous three months. For each product in each month add the safety stock quantity to the forecast:
Forecast + Safety Stock = Planned Total Available Stock for the Month
The result is the planned total available stock for the month. This is the quantity you are intending to sell or use plus your insurance stock to cover unanticipated demand or delays in receiving replenishment shipments. Subtract from this quantity the actual usage quantity for the month. The result is called “residual inventory”.
Planned Total Available Stock for the Month – Actual Usage = Residual Inventory.
To convert the residual inventory quantity into a number of day’s supply, divide this quantity by daily demand (e.g., your monthly forecast divided by 30):
Residual Inventory ÷ (Forecast ÷ 30) = Residual Inventory Day’s Supply
In a specific month if the residual inventory is less than a minimum number of day’s supply (three or four days is a typical number) the forecast plus safety stock quantity was not adequate to meet actual usage and it is highly probable that a stock out occurred. You should consider increasing the safety stock quantity. Why not base this analysis on zero day’s supply? Because you might have experienced some lost sales because the on hand quantity was not adequate to meet a customer need (e.g., they wanted 5 pieces but you only had one piece of the item in stock) and the customer didn’t place an order.
Calculate how many potential stockouts occurred during the three month period and divide this quantity by the number of possible stockouts. That is the months with actual sales or usage during the previous three months. For example:
12,000 instances of usage for products in 3 months = 12,000 opportunities for a stock out
600 residual inventory values less than a three day supply
Potential Stockout percentage = 600 ÷ 12,000 = 5%
An estimated customer service level is the inverse of the stockout percentage or 95%. This means that 95% of the time you should have adequate stock to meet customers’ expectations of product availability.
If residual inventory analysis shows that a product consistently has a residual inventory quantity representing more than an “x” day (typical value is 21 days) supply, consider reducing the safety stock quantity for this item and invest the money saved it in additional safety stock for a critical product that has recently experienced one or more stockouts.
In today’s competitive environment it is critical to make sure that every dollar invested in inventory is contributing to achieving the goal of effective inventory management. Determining what items you should stock based on the number customer orders for the product and fine tuning safety stock quantities with residual inventory analysis are valuable tools in this effort. | 723 | 3,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.881256 |
http://mathhelpforum.com/discrete-math/102250-anyone-here-good-probability-i-can-t-get-answer-question.html | 1,481,168,428,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00296-ip-10-31-129-80.ec2.internal.warc.gz | 179,995,684 | 10,793 | # Thread: Anyone here is good at Probability? I can't get the answer for this question...
1. ## Anyone here is good at Probability? I can't get the answer for this question...
2. Originally Posted by daominhhai129
Let R be a right move: $P(R)=\frac{2}{6}$.
Let L be a left move: $P(L)=\frac{4}{6}$.
To end at B in three moves, we need RRL, RLR, or LRR.
Each of those has the same probability and $P(RRL)=\frac{2}{6}\cdot \frac{2}{6}\cdot\frac{4}{6}$
3. thank you very much..but my answer sheets say it's 2/9....i don't know whether my answer sheet is wrong or i do wrongly...
4. Originally Posted by daominhhai129
my answer sheets say it's 2/9..
No the answer sheet is correct.
$3P(RRL)=3\left(\frac{2}{6}\cdot \frac{2}{6}\cdot\frac{4}{6}\right)=\frac{2}{9}$
5. oh haha..thank you thank you..i'm so stupid..ya..now i can understand..thanks | 275 | 844 | {"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} | 3.78125 | 4 | CC-MAIN-2016-50 | longest | en | 0.8899 |
http://www.nzmaths.co.nz/resource/investigating-digits | 1,448,748,044,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398454160.51/warc/CC-MAIN-20151124205414-00320-ip-10-71-132-137.ec2.internal.warc.gz | 600,410,928 | 9,747 | Te Kete Ipurangi
Communities
Schools
### Te Kete Ipurangi user options:
Level Three > Number and Algebra
# Investigating Digits
Keywords:
Purpose:
This is a level 3 algebra link activity from the Figure It Out series.
Achievement Objectives:
Achievement Objective: NA3-7: Generalise the properties of addition and subtraction with whole numbers.
AO elaboration and other teaching resources
Specific Learning Outcomes:
investigate patterns involving subtraction
Required Resource Materials:
FIO, Link, Number, Book Five, Investigating Digits, pages 12-13
A classmate
Activity:
#### Activities One and Two
Both activities continue the theme of patterns in number. Some astute students may ask, “Why, with twodigit numbers, is the final result always 9?” This is a very good question. You could help the students to recognise that the difference between any reversed pairs of two-digit numbers is always a multiple of 9 (for example, 35 leads to 53 – 35, which is 18; 69 leads to 96 – 69, which is 27; 81 – 18 is 63, and so on), and that finding the difference between reversed pairs of two-digit multiples of 9 (excluding 99) eventually results in 9 itself.
When the students have tackled questions 2 and 3 of Activity One, ask them to share the strategies that they used to determine the number of one-subtraction pairs and to identify the seven two-subtraction pairs. Did they notice, for example, that the one-subtraction pairs all have a difference of 9 and that the difference between the two-subtraction pairs has to be either 45 or 54? A similar approach could be taken with the tasks in Activity Two and the Investigation.
Activity One
1. Four subtractions
2. There are seven more, apart from 21 and 12:
32 and 23
43 and 34
54 and 45
65 and 56
76 and 67
87 and 78
98 and 89
3. 61 and 16
72 and 27
83 and 38
94 and 49
71 and 17
82 and 28
93 and 39
4. 99
Activity Two
1. a. Five steps
b. Three steps
c. Four steps
2. Possible five-step numbers are 123, 234, 345, 456, 567, 678, 789 or different arrangements of these numbers (for example, 123, 132, 213, 231, 312, or 321). In five-step numbers, the arrangements of the digits to form the smallest possible number is always consecutive, with reverse consecutive numbers for the largest number. The result from the first step is always 198. (So, numbers that cannot be rearranged as consecutive whole numbers will not be five-step numbers.)
Investigation
a. You should find that all the numbers you try end up as 6 174.
b. When 6 174 is rearranged and subtracted, the result is 6 174. | 661 | 2,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | longest | en | 0.898574 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.