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13800	https://journals.lww.com/ssjm/fulltext/2014/01020/evaluation_of_some_laboratory_parameters_of.3.aspx	Sub-Saharan African Journal of Medicine Log in or Register Get new issue alerts Get alerts;;) Submit a Manuscript Subscribe to eTOC;;) ### Secondary Logo Enter your Email address: Privacy Policy ### Journal Logo Articles Advanced Search Toggle navigation RegisterLogin Browsing History Home Current Issue Previous Issues For Authors Information for Authors Submit a Manuscript Published Ahead-of-Print Journal Info About the Journal Editorial Board Affiliated Society Advertising Subscriptions Reprints Rights and Permissions Articles Advanced Search Apr-Jun 2014 - Volume 1 - Issue 2 Previous Article Next Article Outline INTRODUCTION MATERIALS AND METHODS RESULTS DISCUSSION CONCLUSION Acknowledgement References Images Slideshow Gallery Export PowerPoint file Download PDF EPUB Cite Copy Export to RIS Export to EndNote Share Email Facebook X LinkedIn Favorites Permissions More Cite Permissions Image Gallery Article as EPUB Export All Images to PowerPoint FileAdd to My Favorites Email to Colleague Colleague's E-mail is Invalid Your Name: Colleague's Email: Separate multiple e-mails with a (;). Message: Your message has been successfully sent to your colleague. Some error has occurred while processing your request. Please try after some time. Export to End Note Procite Reference Manager [x] Save my selection Original Article Evaluation of some Laboratory Parameters of Malnourished Children in Magaria District, Zinder, Niger Republic Mahaman, YA; Akuyam, SA 1,; Danborno, B; Galadima, OM 2; Belemsigri, M 2; Moussa, SM 3 Author Information Department of Human Anatomy, Ahmadu Bello Univeresity, Zaria, Nigeria 1 Department of Chemical Pathology, Ahmadu Bello Univeresity Teaching Hospital, Zaria, Nigeria 2 Department of Paediatric Unit, Magaria Distric Hospital, Nigeria, Republic of Niger, Nigeria 3 Department of Paediatrics, Zinder National Hospital, Republic of Niger, Nigeria Address for correspondence: Dr. SA Akuyam, Department of Chemical Pathology, Ahmadu Bello Univeresity Teaching Hospital, Zaria, Nigeria. E-mail: shehuakuyam@gmail.com Received November 01, 2013 Accepted February 23, 2014 This is an open access journal, and articles are distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as appropriate credit is given and the new creations are licensed under the identical terms. Sub-Saharan African Journal of Medicine 1(2):p 77-81, Apr–Jun 2014. \| DOI: 10.4103/2384-5147.136816 Open Abstract Background and Aim: Protein energy malnutrition (PEM) is the most common nutritional disorder affecting children in developing countries including Niger Republic and continues to be a major health burden in these countries. There is paucity of data on the pattern of laboratory parameters among malnourished children in Niger Republic. The aim of the present study was to evaluate the pattern of some laboratory parameters among under-five malnourished children in Magaria District, Zinder, Niger Republic. Materials and Methods: A total of 220 malnourished children and 220 age- and sex-matched well-fed apparently healthy (controls) children were recruited for the study. Albumin, glucose, calcium, phosphorus and hemoglobin were estimated in the subjects and the data were analysed using Sigma-Stat 2.0 for Windows (San Rafael, CA). Results: Serum albumin, calcium, phosphorus and hemoglobin were significantly lower (P< 0.001) in PEM children compared to the controls. The serum levels of albumin and hemoglobin were also found to be significantly lower (P< 0.03) in children with edematous PEM than in children with non-edematous PEM. Conclusion: PEM children had reduced serum levels of the measured laboratory parameters when compared to apparently healthy well-fed children with resultant hypoalbuminaemia and anaemia, with more marked reduction in edematous than in non-edematous PEM children. This suggests that routine measurement of these laboratory parameters and their subsequent supplementation in PEM children could improve the management of this group of patients. INTRODUCTION Malnutrition is a complex problem which results from a long chain of interrelated events. It continues to be a major health burden in developing countries and affects mostly infants, young children and lactating mothers. Protein-energy malnutrition (PEM) previously referred to as protein-calorie malnutrition (PCM) describes the severe form of malnutrition seen in childhood (kwashiorkor, marasmic-kwashiorkor, marasmus and underweight) and the nutritionally determined growth retardation that precedes this clinical syndrome. It is the most common nutritional disorder affecting children in developing countries and the third most common disease of childhood in such countries. It manifests primarily by inadequate dietary intake of protein and energy, and always accompanied by deficiencies of other nutrients. Worldwide, an estimated 852 million people are undernourished with most (815 million) living in developing countries[5,6] and about 150 million under-five children are still malnourished although considerable progress has been achieved in reducing child malnutrition in developing world as a whole. Despite improved medical care, it has been reported that mortality of children with severe PEM remains high at 10-20%. Several reports across the world showed that PEM is associated with a significant reduction of most of the body's important nutrients including albumin, calcium and phosphorus and in the blood and tissues.[1,2,3,4,5,6,7,8,9,10,11,12,13] Serum protein concentrations are decreased in PEM and this is mainly due to hypoalbuminaemia. Life-threatening hypoglycaemia has been reported in severe PEM. It was also reported that the mean serum calcium concentrations in PEM was significantly low and the levels correlated significantly with weight-for-height.[11,14] The low serum calcium and phosphorus explains the radiological findings of poor mineralization of long bones in PEM patients as reported by Soliman et al. PEM has a great impact on the health care system, resulting in reduced quality of life of the affected patients and added financial costs to the institutions where the patients are receiving care. In view of its public health related problem, emphasis should be placed on the timely assessment of PEM. This involves the need to search and update the clinical and laboratory approaches for the assessment of malnourished children in Nigerien hospitals. These should include a comprehensive laboratory evaluation of serum laboratory parameters with a view to including them as part of patient's management. There is paucity of data on the pattern of serum concentrations of most of the important laboratory parameters including glucose, calcium, phosphorus and hemoglobin among others in Niger Republic. Estimation of serum albumin is usually the most routinely requested biochemical test for the assessment of PEM patients in most of the Nigerien hospitals. This has some limitations including being non-specific and non-sensitive for malnutrition. Therefore, there is the need to consider the estimation of other parameters which are readily available and more reliable for the assessment of PEM in our hospitals. This could improve the management of this group of subjects and hence reduce the morbidity and mortality from PEM. The aim of the present study was therefore to evaluate the pattern of serum concentrations of albumin, glucose, calcium, phosphorus and hemoglobin among children with PEM in Magaria District hospital, Niger Republic, with a view to knowing the pattern of these parameters in this part of the world, and this could improve the laboratory investigations of PEM. MATERIALS AND METHODS This is a descriptive study conducted in Magaria District Hospital, Zinder, Niger Republic. A total of 440 subjects were recruited for the study. These consisted of 220 malnourished and 220 age- and sex-matched children. The two hundred and twenty (220) malnourished children aged 6 to 36 months (mean age 16.67 ± 7.94) recruited for the study was made up of 129 males (mean age 16.35 ± 7.88) and 111 females (mean age 17.04 ± 8.04). These consisted of 80 patients (mean age 15.44 ± 7.16) with marasmus, 51 patients (mean age 24.09 ± 11.33) with marasmic-kwashiorkor, 41 patients (mean age 16.83 ± 8.35) with underweight and 48 patients (mean age 23.39 ± 6.19) with kwashiorkor. Similarly, the two hundred and twenty (220) apparently healthy well-fed children who were age- and sex-matched with the patients were recruited as control. The study was approved by the Ethical Committee of Magaria District Hospital, Zindar, Niger Republic. All children with apparent deformities that affect anthropometric measures like limb deformity were excluded. All children whose parents/guardians declined to give consent for inclusion were also excluded from the study. Informed consent for inclusion into the study was sought from the parents/guardians of the selected children and was granted. At the clinic, children who satisfied the study inclusion criteria were consecutively selected using simple random technique. A full medical history was obtained from the parents/guardians of the selected children. This was followed by a detailed physical examination, anthropometric measurements and collection of blood samples. The main findings were documented in the study proforma. For each child with PEM consecutively recruited, an age- and sex-matched control was consecutively selected. At the point of first presentation or subsequent presentation during treatment of the children, blood specimens (5 ml each) were collected from peripheral vein by veinepuncture into a plain container using a sterile technique. These were centrifuged for 5 min at 1000 g. The sera were separated from the cells and were transferred to sample bottles and then analysed for serum albumin, glucose, calcium, phosphorus and hemoglobin on the same day of blood collection. However, in a situation when it is not possible to carry out the analysis on the day of specimens' collection due to logistic problem, they used to be kept frozen at -20° C until the following day. The chemicals for the measurements of serum albumin, glucose, calcium, phosphorus and hemoglobin were procured from RANDOX laboratories Ltd. (United Kingdom). All the chemicals were of analytical grade or higher. Hettich Universal 32 Centrifuge (Germany) was used to spin the blood specimens. ESSE-3-2007 FAST chemistry analyzer (Italy) was used for the measurements of serum albumin, calcium, phosphorus and glucose concentrations, while HemoCue ® B-hemoglobin (HB 201 +) photometer (Sweden) was used for measuring hemoglobin concentrations. Serum albumin (ALB) concentration was measured using method of Doumas et al., while glucose (GLU) was measured by GLU oxidase method. Serum calcium (CAL) was measured by the method of Ray and Chauhan, while phosphorus (PHOS) was measured by the method of Gomorri, as modified by Henry. The hemoglobin concentrations were measured by the method of Vanzetti. The data were expressed as mean±standard deviation (Mean ± SD) and were analysed using Sigma-Stat 2.0 for Windows (San Rafael, CA). Student's t-test was used to assess differences between data from malnourished and apparently healthy well-fed children, as well as edematous from non-edematous PEM children. One way analysis of variance (ANOVA) was applied to assess differences between data from different classes of malnutrition. A P-value of equal to or less than 0.05 (P ≤ 0.05) was considered as statistically significant. RESULTS The results of laboratory parameters of the malnourished and well-fed children were shown in Table 1. Most of parameters were significantly lower (P< 0.001) in PEM children compared to the well-fed ones except glucose which was slightly lower in PEM than in well-fed children. Table 1: Biochemical parameters (Mean ± SD) of malnourished and well-fed children The results of laboratory parameters according to different classes of malnutrition are presented in Table 2. The results in this table showed that there were significant differences (P< 0.03) and P< 0.00) in albumin and hemoglobin levels, respectively, between the four different classes of PEM. Post hoc analysis has shown that the values were significantly lower in kwashiorkor compared with marasmus and underweight, and also lower in marasmic-kwashiorkor compared with underweight. The values were not significantly different between kwashiorkor and marasmic-kwashiorkor, as well as between marasmus and underweight. The finding was confirmed when the two groups comparison was carried out by combining the data from kwashiorkor and marasmic-kwashiorkor (edematous PEM) children and marasmus and underweight (non-edematous PEM) children as shown in Table 3. The results in this table showed that serum albumin and hemoglobin levels were significantly lower (P< 0.03, P< 0.00, respectively) in edematous PEM children compared to the non-edematous ones. Serum calcium and phosphorus levels were not significantly different between edematous and non-edematous PEM children. Table 2: Biochemical parameters (Mean ± SD) of malnourished children by class of PEM Table 3: Biochemical parameters (Mean ± SD) of malnourished children by edema status DISCUSSION The present study has evaluated the pattern of some laboratory parameters among children with protein-energy malnutrition (PEM) in Magaria District, Zinder, Niger Republic. The results of this study showed that all laboratory parameters studied were lower in PEM patients than in the well-fed control group. However, with exception of hemoglobin, all the other results were low-normal that is close to the lower limit of the reference ranges. This finding agrees with previous reports by Soliman et al., Rahman and Begum, Adegbusi and Sule, Kalra et al., and Waterlow who found that serum total calcium and phosphorus concentrations were markedly decreased according to the type of PEM. Nassar et al., reported that calcium and phosphorus levels were within reference ranges in non-edematous and edematous groups of PEM patients but lower in the edematous than non-edematous patients. The occurrence of reduced levels of serum calcium and phosphorus in malnourished children could be due to dietary deficiency of these nutrients or infection owing to increased metabolic loses. The dietary deficiency of calcium could be as a result of intake of calcium-free diet such as gruel which is often prepared from phytates-containing cereals (whole-grain cereals), as it is known that most weaning diets in Zinder community are gruel. It was observed that diets made from products containing phytates which limit calcium availability can lead to calcium deficiency. An overall prevalence of hypophosphataemia of 86% at presentation and of severe hypophophataemia of 6% was observed among PEM children by Kimutai et al. The occurrence of hypophosphataemia in PEM children could be due to long term low dietary phosphate intake or increased lost of the phosphate iron in the urine owing to excessive catabolic release from intracellular compartment. This is because during starvation, in order to maintain normal phosphate levels there is catabolic release of phosphate from intracellular stores which is subsequently lost in urine and therefore significant total body phosphate depletion can occur despite normal serum phosphate levels. Again, severely malnourished children could have a long term relatively low phosphate intake and this predisposes them to significant total phosphate depletion. The slightly reduced serum glucose levels among PEM children observed in the present study is in agreement with previous reports by Reddy who reported that the fasting blood glucose levels were lower in malnourished than in normal children, but Mishra et al., reported that the mean glucose concentrations in PEM and control subjects were not significantly different. In Indian children, blood glucose levels were reported to be normal but glucose tolerance was impaired, suggesting impaired utilization in PEM patients. The slight reduction of serum glucose in PEM children agrees with previous report by Buchanan et al., who observed impaired glycogenolysis in severe kwashiorkor leading to hypoglycaemia. In different studies, hypoglycaemia has also been reported in severe PEM. The decreased serum glucose could be due to reduced intake, infection or some metabolic abnormalities such as impaired glycogenolysis and gluconeogenesis. Hemoglobin concentrations were found to be significantly decreased in the malnourished compared to the well-fed children. Anaemia was observed in all classes of PEM in this study which is more pronounced in edematous than non-edematous PEM children. This is in agreement with the findings of El-Nawawy et al., who reported anaemia in PEM and suggested that it is a result of ineffective erythropoiesis. Some degrees of anaemia which is more pronounced in the marasmic-kwashiorkor patients than in other forms of PEM were also observed in the previous studies. The presence of anaemia as found in the present and previous studies could also be related to the nutritional deficiency of iron and other essential minerals and vitamins, as well as infections. CONCLUSION Concentrations of albumin, calcium, phosphorus, glucose and hemoglobin were significantly reduced in PEM children compared with apparently healthy well-fed controls with resultant hypoalbuminaemia and anaemia, and the reductions were more pronounced in edematous than non-edematous PEM children. This suggests that routine measurement of these laboratory parameters and their subsequent supplementation in PEM children could improve the management of this group of patients. Moreover, the community should be educated on the use of locally available rich sources of the above mentioned nutrients. Adhering to these recommendations could reduce morbidity and mortality from PEM in Niger Republic and probably elsewhere in the world. Acknowledgement The authors acknowledge and appreciate the support given by Mrs Mariama Aboubacar, Mrs Hadiza Mahaman Yacoubou, Mr Mahaman Yacoubou, Mr Mahaman Yacoubou Abdoul Kader, Mr Nassirou Nazongo and Mr Abdou Salam of the Magaria district Hospital in the area of blood collection and anthropometric measurement. The authors also thank Mr Lawan Ousman and Mr Issoufou Abdou of Zinder National Hospital for their assistance in the area of laboratory sample analysis for the biochemical parameters. References Muller O, Krawinkel M. Malnutrition and health in developing countries. Can Med Ass J 2005;171:279–93. Cited Here Wellcome Trust Working Party. Classification of protein-energy malnutrition. Lancet 1970;2:302–3. Cited Here Jancen AA, Mannet WT. Assessment of nutritional status: A comparison of methods. J Trop Pediatr 1982;28:38–40. Cited Here Nassar MF, Dina AA, Salwa RE, Soad MG. Markers of bone metabolism in Protein Energy Malnutrition. Intern J Food Nutr Public Health 2010;3:59–70. Cited Here World Health Organisation. Reducing risks and producing healthy life. World Health Report, 2002, 1211 Geneva 27, Switzerland. Cited Here Food and Agricultural Organization (FAO). The state of food insecurity in the world, Monitoring Progress towards the World Summit and Millennium Development Goals. Food and Agricultural Organization of the United Nations, Viale delle Terme di Caracalla, 00100 Rome, Italy, 2004. Cited Here World Health Organisation. United Nations High Commissioner for Refugees, International Federation of Red Cross and Red Crescent Societies and World Food Programme. The Management of Nutrition in Major Emergencies. World Health Organization, Geneva, Switzerland, 2000. Cited Here Central Bureau of Statistics (CBS), Ministry of Health (MOH), Kenya. Kenya Demographic and Health Survey Data, Nairobi, 2003. Calverton, Maryland: CBS, MOH and ORC Macro. Cited Here Johnson AM. Low levels of plasma protein: malnutrition or inflammation? Clin Chem Lab Med 1999;37:91–6. Cited Here Reddy V. Protein-energy malnutrition. In: Standfield P, Brueton M, Chan M, Waterston T, editors. Diseases of children in the subtropics and tropics. 4th ed. London: ELBS with Edward Arnold; 1993. p. 335–57. Cited Here Mishra SK, Bastola SP, Jha B. Biochemical nutritional indicators in children with protein energy malnutrition attending Kanti Children Hospital, Kathmandu, Nepal. Kathmandu Univ Med J 2009;7:129–34. Cited Here Soliman AT, Madina EH, Morsi MR. Radiological, biochemical and hormonal changes in malnourished children with rachitic manifestations. J Trop Paediatr 1996;42:34–7. Cited Here Reddy V. Protein-energy malnutrition. In: Stanfield P, Brueton M, Chan M, Waterston T, editors. Diseases of children in the subtropics and tropics. 4th ed. London: Edward Arnold; 1991. p. 335–8. Cited Here Ighogboja IS, Okonjo MC, Onyeocha BK. Serum calcium, inorganic phosphorus and magnesium levels in malnourished pre-school children. Niger J Paeditr 1996;23:336. Cited Here Doumas BT, Watson WA, Biggs HG. Albumin standard and the measurement of serum albumin with bromocresol green. Clin Chim Acta 1971;31:87–96. Cited Here Trinder P. Determination of glucose in blood using glucose oxidase with an alternative oxygen acceptor. Ann Clin Biochem 1969;6:24–7. Cited Here Ray SB, Chauhan C. Estimation of calcium in plasma. Ann Biochem 1967;20:155–9. Cited Here Henry RJ. Clinical Chemistry, Principles and Techniques. 2 nd ed. Harper and Row; London: Lang Medical Books/McGraw-Hill, Medical Publishing Division; 1974. p. 525. Cited Here Vanzetti G. Hemoglobin estimation. J Lab Clin Med 1966;67:116. Cited Here Rahman MZ, Begum BA. Serum total protein, albumin and A/G ratio in different grades of protein energy malnutrition. Mymensingh Med J 2005;l: 38–40. Cited Here Adegbusi HS, Sule MS. Anthropometric and biochemical assessment among under five children in Kusada Local Government Area, Katsina State, Nigeria. Bayero J Pure Appl Sci 2011;4:137–40. Cited Here Kalra K, Mital VP, Pal R, Goyal RK, Dayal RS. Serum electrolyte studies in malnutrition. Indian Paediatr 1975;9:1135–9. Cited Here Waterlow JC. Summary of causes and mechanisms of linear growth retardation. Eur J Clin Nutr 1994;48:S210. Cited Here de Vizia B. Calcium and phosphorus metabolism in infants and growing child. In: Annales Nestle. Vol. 45. Switzerland: Vevey/Roven Press Ltd.; 1987. p. 36–44. Cited Here Kimutai D, Maleche-Obimbo E, Kamenwa R, Murila F. Hypo-phosphataemia in children under five years with kwashiorkor and marasmic kwashiorkwor. East Afr Med J 2009;86:330–6. Cited Here Berkelhammer C, Bear RA. A clinical approach to common electrolyte problems: Hypophosphataemia. Can Med Assoc 1984;130:17–23. Cited Here Buchanan N, Moodley G, Eyberg-Bloom SR, Hansen JD. Hypoglycaemia associated with severe kwashiorkor. South Afr Med J 2003;50:1442–6. Cited Here El-Nawawy A, Barakat S, Elwalily T, Abdel-Moneim DA, Hussein M. Evaluation of erythropoiesis in protein energy malnutrition. East Mediterr Health J 2002;8:281–9. Cited Here Etukudo MH, Agbedana EO, Akinyinka OO, Osifo BO. Plasma electrolytes, total cholesterol, liver enzymes and selected anti-oxidant status in protein-energy malnutrition. Afr J Med Sci 1999;28:81–5. Cited Here View full references list Keywords: Biochemical parameters; malnourished children; PEM children; protein-energy malnutrition; well-fed children © 2014 Sub-Saharan African Journal of Medicine \| Published by Wolters Kluwer – Medknow View full article text Source Evaluation of some Laboratory Parameters of Malnourished Children in Magaria District, Zinder, Niger Republic Sub-Saharan African Journal of Medicine1(2):77-81, Apr-Jun 2014. 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13801	https://brainly.com/question/31641648	[FREE] Predict the molecular geometry of each interior atom in acetic acid [ \textrm{CH}_3\textrm{COOH} ]. Drag - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +31,2k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +33,8k Ace exams faster, with practice that adapts to you Practice Worksheets +6,3k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Predict the molecular geometry of each interior atom in acetic acid [CH 3COOH]. Drag the appropriate labels to their respective targets: Linear Trigonal planar Bent Tetrahedral Atoms: Carbon (left) Carbon (right) Oxygen Number of Electron Groups: Number of Lone Pairs: Geometry: Molecular: SubmitMy AnswersGive Up Help 2 See answers Explain with Learning Companion NEW Asked by AddieBffMaddie3010 • 04/18/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 778147 people 778K 0.0 0 Upload your school material for a more relevant answer Here is the molecular geometry of each interior atom in acetic acid (CH₃COOH): Carbon (left): tetrahedral Carbon (right): trigonal planar Oxygen (center): bent The structure of acetic acid has two possible resonance structures. The geometry of the atoms in each resonance structure is the same, but the location of the double bond between the carbon and oxygen atoms alternates between the two carbons. Molecular geometry refers to the three-dimensional arrangement of atoms and the spatial relationships among them in a molecule. It describes the shape and orientation of the molecule in space and is determined by the arrangement of the atoms and the lone pairs of electrons around the central atom(s) in the molecule. Molecular geometry is an important concept in chemistry as it determines many of the physical and chemical properties of a molecule, including its reactivity, polarity, and intermolecular forces. The geometry of a molecule can be determined experimentally using techniques such as X-ray crystallography or nuclear magnetic resonance spectroscopy, or it can be predicted using computational methods and models. Molecular geometry refers to the three-dimensional arrangement of atoms and the spatial relationships among them in a molecule. It describes the shape and orientation of the molecule in space and is determined by the arrangement of the atoms and the lone pairs of electrons around the central atom(s) in the molecule. To know more about molecular geometry here: brainly.com/question/31323866# SPJ11 Answered by Amul543 •3.1K answers•778.1K people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 778147 people 778K 0.0 0 Concept Development Studies in Chemistry - John S. Hutchinson Chemistry: Atoms First 2e - Paul Flowers, Edward J. Neth, William R. Robinson, Klaus Theopold, Richard Langley Chemistry 2e - Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson Upload your school material for a more relevant answer In acetic acid (CH₃COOH), the left carbon has tetrahedral geometry, the right carbon has trigonal planar geometry, and the oxygen has bent geometry. The left carbon has four electron groups and no lone pairs, while the right carbon has three electron groups and no lone pairs, and the oxygen has four electron groups with two lone pairs. Understanding these geometries helps predict the molecule's overall shape and properties. Explanation To determine the molecular geometry of each interior atom in acetic acid (CH₃COOH), we first need to look closely at its structure. Acetic acid consists of two carbon atoms, two oxygen atoms, and four hydrogen atoms. Here’s a breakdown of the geometries: Carbon (left): The left carbon is bonded to three hydrogen atoms and one carbon atom. It has four regions of electron density, which corresponds to a tetrahedral geometry. Therefore, this carbon atom's geometry is: Geometry: Tetrahedral Number of Electron Groups: 4 Number of Lone Pairs: 0 Carbon (right): The right carbon is bonded to one hydroxyl group (–OH) and one carbon atom. It has three regions of electron density around it due to these three bonds, corresponding to a trigonal planar geometry. Therefore, this carbon atom's geometry is: Geometry: Trigonal planar Number of Electron Groups: 3 Number of Lone Pairs: 0 Oxygen: The oxygen atom in the hydroxyl group (–OH) possesses two lone pairs and is bonded to one hydrogen and one carbon atom. Thus, it has four regions of electron density, resulting in a bent geometry. Therefore, this oxygen atom's geometry is: Geometry: Bent Number of Electron Groups: 4 Number of Lone Pairs: 2 So, in summary: Carbon (left): Tetrahedral Carbon (right): Trigonal planar Oxygen: Bent Examples & Evidence For example, the tetrahedral geometry of the left carbon allows for bond angles of approximately 109.5 degrees between the hydrogen atoms. The trigonal planar geometry of the right carbon allows for bond angles of approximately 120 degrees. The bent geometry of oxygen, due to its lone pairs, results in bond angles less than 109.5 degrees between the bonds. The geometries can be confirmed using VSEPR (Valence Shell Electron Pair Repulsion) theory, which predicts the shape of molecules based on the repulsion between electron pairs. Molecular models and Lewis structures can provide visual confirmation of the predicted shapes. Thanks 0 0.0 (0 votes) Advertisement Community Answer This answer helped 4110845 people 4M 0.0 0 Acetic acid consists of a methyl carbon with a tetrahedral geometry and an acid carbon with a trigonal planar geometry. The molecule acetic acid, CH₃COOH, consists of two carbon atoms with different molecular geometries. To predict the molecular geometry of each interior atom, we analyze the number of electron groups around each carbon. The methyl carbon (left carbon) in acetic acid is bonded to three hydrogen atoms and one carbon atom, making it surrounded by four regions of electron density. In this case, the molecular geometry is tetrahedral. The acid carbon (right carbon) is double-bonded to one oxygen atom, single-bonded to the methyl carbon, and single-bonded to a hydroxyl group. This carbon atom is surrounded by three regions of electron density, giving it a trigonal planar molecular geometry. Therefore, the methyl carbon has a tetrahedral geometry, while the acidic carbon has a trigonal planar geometry due to its different bonding and lone pairs arrangement. Answered by ShabnamKoundal •10.3K answers•4.1M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Chemistry solutions and answers Community Answer 4.6 5 Predict the approximate molecular geometry around each carbon atom of acetonitrile. A central carbon atom is connected to a second carbon atom through a single bond and to a nitrogen atom through a triple bond. The outer carbon atom is connected to three hydrogen atoms. The nitrogen atom has one lone pair of electrons. The carbon atom in CH3 is: trigonal planar linear trigonal pyramidal tetrahedral bent The nitrile carbon atom ( C≡N ) is: trigonal pyramidal bent tetrahedral trigonal planar linear Community Answer Determine the molecular geometry around each carbon atom in maltose. a) linear b) trigonal pyramidal c) bent d) tetrahedral e) trigonal planar Community Answer Assign a molecular geometry to each interior atom in adenine. Drag the appropriate labels to their respective targets. Reset Help NH2 tetrahedral trigonal planar CH trigonal pyramidal НС bent Submit Request Answer Assign a molecular geometry to each interior atom in thymine. Drag the appropriate labels to their respective targets. Reset Help tetrahedral НС trigonal planar NH trigonal pyramidal HaC bent Submit Request Answer Assign a molecular geometry to each interior atom in guanine. Drag the appropriate labels to their respective targets Reset Help tetrahedral trigonal planar HN trigonal pyramidal H2N Community Answer 5.0 2 The electron geometry and the molecular geometry of ammonia (NH3) are, respectively: The electron geometry and the molecular geometry of ammonia (NH3) are, respectively: tetrahedral, trigonal pyramidal. tetrahedral, tetrahedral. trigonal planar, bent. tetrahedral, bent. none of the above g Community Answer Use VSEPR theory to predict the electron-pair arrangement and the molecular geometry of sulfur dioxide, SO2. - The electron-pair arrangement is trigonal-planar, the molecular geometry is trigonal-planar. - The electron-pair arrangement is trigonal-planar, the molecular geometry is bent. - The electron-pair arrangement is tetrahedral, the molecular geometry is bent. - The electron-pair arrangement is tetrahedral, the molecular geometry is linear. - The electron-pair arrangement is trigonal-bipyramidal, the molecular geometry is linear. Community Answer HELP PLEASE !!!!!!!!!!! Select the electron domain geometry that results in a bent molecular shape. Check all of the boxes that apply. linear with two atoms and zero lone pairs trigonal planar with three atoms and zero lone pairs trigonal planar with two atoms and one lone pair tetrahedral with three atoms and one lone pair tetrahedral with two atoms and two lone pairs Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta New questions in Chemistry Consider the chemical equations shown here. C H 4(g)+2 O 2(g)→C O 2(g)+2 H 2O(g)Δ H 1=−802 k J 2 H 2O(g)→2 H 2O(l)Δ H 2=−88 k J Which equation shows how to calculate Δ H r x n for the equation below? C H 4(g)+2 O 2(g)→C O 2(g)+2 H 2O(l) What is Δ H r x n for the overall reaction? Prepare a "Coal and Petroleum" fact file. What is a double bond? A. A covalent bond between 2 atoms where each atom contributes 2 electrons. B. A covalent bond between 2 atoms where each atom contributes 4 electrons. C. An ionic bond between 2 atoms where 2 electrons have been transferred. D. A covalent bond between 2 atoms where each atom contributes 1 electron. Why do noble gases rarely form bonds with other atoms? A. The noble gases are not reactive, so they don't need full valence octets. B. Gases do not form bonds with atoms that are not also gases. C. The noble gases are already stable, with full octets of valence electrons. D. The noble gases form bonds only with themselves because they have octets. A balloon that had a volume of 3.50 L at 25.0∘C is placed in a hot room at 40.0∘C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room? Use T 1V 1=T 2V 2. A. 2.19 L B. 3.33 L C. 3.68 L D. 5.60 L Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
13802	https://www.mashupmath.com/blog/translating-words-into-algebraic-expressions	Translating Words into Algebraic Expressions (Step-by-Step) — Mashup Math 0 items $0 Translating Words into Algebraic Expressions Welcome to this complete guide to translating words into algebraic expressions (also known as algebraic translation), where you will learn how to identify and apply key information, in the form of words and phrases, to accurately translate a given set of words into an algebraic expression involving both numbers and variables. Why is learning how to translate words into algebraic expressions a crucial skill that every math student must learn? Because it is often the case that math problems are expressed completely in words without any explicit use of numbers, expressions, or equations. In order to solve these types of math word problems, students have to be able to translate words into expressions or equations so they may model and solve such scenarios. Are you ready to learn everything there is to know about algebraic translation? The following free Translating Words into Algebraic Expressions lesson guide is a step-by-step tutorial that will teach you how to easily and accurately translate any given word phrase into a mat equation. How can you translate written expressions into numerical form? The key skill associated with algebraic translation is being able to rewrite mathematical situations expressed in words as a mathematical expression involving numbers, operations, and variables. Before we get to actually translating words into algebraic expressions, let's lay some important groundwork! Tip #1: Expressing Variables For example, what if we wanted to translate the phrase the sum of seven and five into an expression. It would be pretty easy to translate this phrase into 7 + 5 and your job would be done. But what if we changed the phrase to the sum of a number and five? How would our numerical expression change? Since a number could represent any value, we have to use a variable (since a variable can represent any value). In this case, you could translate the sum of a number and five into x + 5 where x represents a number. When using letters as variables in a math expression or equation, x is the most commonly chosen letter, but you can actually choose any letter to represent an unknown value. Example A: Translate the phrase ten plus a number into an algebraic expression. To complete this translation, we can break the given phrase down into three parts: I: ten ➔ 10 II: plus ➔ + III: a number ➔ x Now, you can translate ten into 10, plus into an addition sign, and a number into a variable leaving you with: ten plus a number ➔ 10 + x Tip #2: More Than/Less Than Now, let’s slightly change the words given in Example A as follows: Example B: Translate the phrase ten more than a number into an algebraic expression. You probably already know that more than is associated with addition so the sign is not going to change. But what about the order of the terms? Think about it this way: we have a number (some unknown value) and this phrase represents ten more than whatever that value is. So, in this case, you will start with the variable first and then add ten to it as follows: ten more than a number ➔ x+ 10 You would be correct to wonder whether or not the order of the terms matters in this example. Technically, it does not because addition is commutative. But what about subtraction, which is not commutative? See Also: The Commutative Property: Everything You Need to Know Example C: Translate the phrase six less than a number into an algebraic expression. Notice again that we are seeing the word than. You probably already know that less than is associated with subtraction so you already know what sign you will be using. This phrase represents a value that is six units smaller than whatever our unknown value is. So, to find that number, we would have to take our variable and subtract six from it as follows: six less than a number ➔ n-6 In cases like Example B and Example C, the second term comes first and the first term comes second (you have to switch the order). So, we can conclude that than is a switch word, which means that the operator in the middle stays the same, but the first term and the last term are switched. Look out for this relationship when you see the phrase more than or less than in words. Tip #3: Groupings and Parenthesis Let’s move onto another example… Example D: Translate the phrase the difference of three and a number into an algebraic expression. Translating this phrase into an expression should be pretty straightforward. Since difference means subtraction, we can easily perform the following algebraic translation: the difference of three and a number ➔ 3 - p Now, what if we changed this expression to the difference of three and twice a number plus one Example D: Translate the phrase the difference of three and twice a number plus one into an algebraic expression. So, now instead of 3 - p, we have to write the expression as 3 - the entire expression twice a number plus one, which we can call 2p + 1. Note that you will have to use parenthesis to enclose the entire expression twice a number plus one as follows: the difference of three and twice a number plus one ➔ 3 - (2p+1) So, whenever you are performing algebraic translations, you can use parenthesis to separate independent groupings. Translate Algebraic Expressions Practice Now that you understand some key elements of translating words into algebraic expressions, you are ready to practice on your own. Go ahead and translate the following words into algebraic expressions on your own and then check the answer key at the end of this post to see how you did! Practice Problems: Translate each phrase into an algebraic expression. 1.) nine times a number 2.) the sum of a number and twelve 3.) twice a number decreased by eleven 4.) twenty less than a number 5.) half a number plus two 6.) the quotient of five and a number 7.) five times the difference of a number and one 8.) the sum of sixteen and three times a number minus four Wait! Don’t scroll further until you’re ready to see the answer key. Answer Key: 1.) nine times a number ➔ 9x 2.) the sum of a number and twelve ➔ n + 12 3.) twice a number decreased by eleven ➔ 2y - 11 4.) twenty less than a number ➔ m - 20 5.) half a number plus two ➔ (x/2) + 2 6.) the quotient of five and a number ➔ 5 ÷ p 7.) five times the difference of a number and one ➔ 5(x-1) 8.) the sum of sixteen and three times a number minus four ➔ 16 + (3y - 4) How to Translate Algebraic Expressions Video Are you looking for more help with translating algebraic expressions? Check out our free step-by-step video lesson below: Keep Learning: Featured Reflection Over X Axis and Y Axis—Step-by-Step Guide Log Rules Explained! (Free Chart) Exponent Rules Explained! (Free Chart) How to Factor a Trinomial in 3 Easy Steps Calculating Percent Change in 3 Easy Steps What is Point-Slope Form in Math? 5 Point-Slope Form Examples with Simple Explanations Calculating Percent Decrease in 3 Easy Steps Where is the hundredths place value in math? How to Solve Compound Inequalities (in 3 Easy Steps) 3 Comments 3 Likes Share Comments (3) Most Liked Preview Post Comment… Royal-light Pending Awaiting Moderation · 0 Likes Can't I put my question, and AI will solve it for me Preview Post Reply Peta Pending Awaiting Moderation · 0 Likes Which of the following algebraic expressions will translate “the sum of eight and twice w”? Preview Post Reply Austin Charlie Pending Awaiting Moderation · 0 Likes times five less than twice X? Is it 2x-5? Preview Post Reply
13803	https://www.weizmann.ac.il/complex/falkovich/sites/complex.falkovich/files/uploads/statphys15short.pdf	Statistical Physics G. Falkovich More is diﬀerent (Anderson) Contents 1 Thermodynamics (brief reminder) 2 1.1 Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Legendre transform . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3 Stability of thermodynamic systems . . . . . . . . . . . . . . . 13 2 Basic statistical physics (brief reminder) 16 2.1 Microcanonical distribution . . . . . . . . . . . . . . . . . . . 16 2.2 Canonical distribution . . . . . . . . . . . . . . . . . . . . . . 19 2.3 Two-level system . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.4 Distribution in the phase space . . . . . . . . . . . . . . . . . 25 3 Entropy and information 26 3.1 Lyapunov exponent . . . . . . . . . . . . . . . . . . . . . . . . 27 3.2 Information theory approach . . . . . . . . . . . . . . . . . . . 33 1 This is the ﬁrst part of a graduate one-semester course. It brieﬂy remind what is supposed to be known from the undergraduate courses about thermo-dynamics and elementary statistical physics, using a bit more sophisticated language. Small-print parts can be skipped at ﬁrst reading. 1 Thermodynamics (brief reminder) Physics is an experimental science, and laws appear usually by induction: from particular cases to a general law and from processes to state functions. The latter step requires integration (to pass, for instance, from Newton equa-tion of mechanics to Hamiltonian or from thermodynamic equations of state to thermodynamic potentials). Generally, it is much easier to diﬀerentiate then to integrate and so deduction (or postulation approach) is usually much more simple and elegant. It also provides a good vantage point for further applications and generalizations. In such an approach, one starts from pos-tulating some function of the state of the system and deducing from it the laws that govern changes when one passes from state to state. Here such a deduction is presented for thermodynamics following the book H. B. Callen, Thermodynamics (John Wiley & Sons, NYC 1965). 1.1 Basic notions We use macroscopic description so that some degrees of freedom remain hidden. In mechanics, electricity and magnetism we dealt with the explic-itly known macroscopic degrees of freedom but in thermodynamics we deal with macroscopic manifestations of the hidden degrees of freedom. When detailed knowledge is unavailable, physicists use symmetries or conserva-tion laws. Thermodynamics studies restrictions on the possible properties of macroscopic matter that follow from the symmetries of the fundamental laws. Therefore, thermodynamics does not predict numerical values but rather sets inequalities and establishes relations among diﬀerent properties. The basic symmetry is invariance with respect to time shifts which gives energy conservation1. That allows one to introduce the internal energy E. 1Be careful trying to build thermodynamic description for biological or social-economic systems, since generally they are not time-invariant. For instance, living beings age and the amount of money is not always conserved. 2 Energy change generally consists of two parts: the energy change of macro-scopic degrees of freedom (which we shall call work) and the energy change of hidden degrees of freedom (which we shall call heat). To be able to measure energy changes in principle, we need adiabatic processes where there is no heat exchange. We wish to establish the energy of a given system in states independent of the way they are prepared. We call such states equilibrium, they are those that can be completely characterized by the static values of extensive parameters like energy E, volume V and mole number N (number of particles divided by the Avogadro number 6.02 × 1023). Other extensive quantities may include numbers of diﬀerent sorts of particles, electric and magnetic moments etc i.e. everything which value for a composite system is a direct sum of the values for the components. For a given system, any two equilibrium states A and B can be related by an adiabatic process either A →B or B →A, which allows to measure the diﬀerence in the internal energy by the work W done by the system. Now, if we encounter a process where the energy change is not equal to minus the work done by the system, we call the diﬀerence the heat ﬂux into the system: dE = δQ −δW . (1) This statement is known as the ﬁrst law of thermodynamics. The energy is a function of state so we use diﬀerential, but we use δ for heat and work, which aren’t diﬀerentials of any function as they refer to particular forms of energy transfer (not energy content). The basic problem of thermodynamics is the determination of the equilib-rium state that eventually results after all internal constraints are removed in a closed composite system. The problem is solved with the help of ex-tremum principle: there exists an extensive quantity S called entropy which is a function of the extensive parameters of any composite system. The values assumed by the extensive parameters in the absence of an internal constraint maximize the entropy over the manifold of constrained equilib-rium states. Since the entropy is extensive it is a homogeneous ﬁrst-order function of the extensive parameters: S(λE, λV, . . .) = λS(E, V, . . .). The entropy is a continuous diﬀerentiable function of its variables. This function (called also fundamental relation) is everything one needs to know to solve the basic problem (and other problems in thermodynamics as well). Since the entropy is generally a monotonic function of energy2 then S = 2This is not always so, we shall see in the second Chapter that the two-level system 3 S(E, V, . . .) can be solved uniquely for E(S, V, . . .) which is an equivalent fundamental relation. Indeed, assume (∂E/∂S)X > 0 and consider S(E, X) and E(S, X). Then3 ( ∂S ∂X ) E = 0 ⇒ ( ∂E ∂X ) S = −∂(ES) ∂(XS) ∂(EX) ∂(EX) = − ( ∂S ∂X ) E (∂E ∂S ) X = 0 . Diﬀerentiating the last relation one more time we get (∂2E/∂X2)S = −(∂2S/∂X2)E(∂E/∂S)X , since the derivative of the second factor is zero as it is at constant X. We thus see that the equilibrium is deﬁned by the energy minimum instead of the entropy maximum (very much like circle can be deﬁned as the ﬁgure of either maximal area for a given perimeter or of minimal perimeter for a given area). On the ﬁgure, unconstrained equilibrium states lie on the curve while all other states lie below. One can reach the state A either maximizing entropy at a given energy or minimizing energy at a given entropy: A S E One can work either in energy or entropy representation but ought to be careful not to mix the two. Experimentally, one usually measures changes thus ﬁnding derivatives (called equations of state). The partial derivatives of an extensive variable with respect to its arguments (also extensive parameters) are intensive pa-rameters4. For example, for the energy one writes ∂E ∂S ≡T(S, V, N) , ∂E ∂V ≡−P(S, V, N) ∂E ∂N ≡µ(S, V, N) , . . . (2) gives a counter-example as well as other systems with a ﬁnite phase space. 3An eﬃcient way to treat partial derivatives is to use jacobians ∂(u, v)/∂(x, y) = (∂u/∂x)(∂v/∂y) −(∂v/∂x)(∂u/∂y) and the identity (∂u/∂x)y = ∂(u, y)/∂(x, y). 4In thermodynamics we have only extensive and intensive variables (and not, say, surface-dependent terms ∝N 2/3) because we take thermodynamic limit N →∞, V →∞ keeping N/V ﬁnite. 4 These relations are called the equations of state and they serve as deﬁnitions for temperature T, pressure P and chemical potential µ while the respective extensive variables are S, V, N. From (2) we write dE = δQ −δW = TdS −PdV + µdN . (3) Entropy is thus responsible for hidden degrees of freedom (i.e. heat) while other extensive parameters describe macroscopic degrees of freedom. The derivatives (2) are deﬁned only in equilibrium. Therefore, δQ = TdS and δW = PdV −µdN for quasi-static processes i.e such that the system is close to equilibrium at every point of the process. A process can be considered quasi-static if its typical time of change is larger than the relaxation times (which for pressure can be estimates as L/c, for temperature as L2/κ, where L is a system size, c - sound velocity and κ thermal conductivity). Finite deviations from equilibrium make dS > δQ/T because entropy can increase without heat transfer. Let us give an example how the entropy maximum principle solves the basic problem. Consider two simple systems separated by a rigid wall which is impermeable for anything but heat. The whole composite system is closed that is E1 + E2 =const. The entropy change under the energy exchange, dS = ∂S1 ∂E1 dE1 + ∂S2 ∂E2 dE2 = dE1 T1 + dE2 T2 = ( 1 T1 −1 T2 ) dE1 , must be positive which means that energy ﬂows from the hot subsystem to the cold one (T1 > T2 ⇒∆E1 < 0). We see that our deﬁnition (2) is in agreement with our intuitive notion of temperature. When equilibrium is reached, dS = 0 which requires T1 = T2. If fundamental relation is known, then so is the function T(E, V ). Two equations, T(E1, V1) = T(E2, V2) and E1 + E2 =const completely determine E1 and E2. In the same way one can consider movable wall and get P1 = P2 in equilibrium. If the wall allows for particle penetration we get µ1 = µ2 in equilibrium. Both energy and entropy are homogeneous ﬁrst-order functions of its vari-ables: S(λE, λV, λN) = λS(E, V, N) and E(λS, λV, λN) = λE(S, V, N) (here V and N stand for the whole set of extensive macroscopic parame-ters). Diﬀerentiating the second identity with respect to λ and taking it at λ = 1 one gets the Euler equation E = TS −PV + µN . (4) 5 Let us show that there are only two independent parameters for a simple one-component system, so that chemical potential µ, for instance, can be found as a function of T and P. Indeed, diﬀerentiating (4) and comparing with (3) one gets the so-called Gibbs-Duhem relation (in the energy representation) Ndµ = −SdT + V dP or for quantities per mole, s = S/N and v = V/N: dµ = −sdT + vdP. In other words, one can choose λ = 1/N and use ﬁrst-order homogeneity to get rid of N variable, for instance, E(S, V, N) = NE(s, v, 1) = Ne(s, v). In the entropy representation, S = E 1 T + V P T −N µ T , the Gibbs-Duhem relation is again states that because dS = (dE + PdV − µdN)/T then the sum of products of the extensive parameters and the dif-ferentials of the corresponding intensive parameters vanish: Ed(1/T) + V d(P/T) −Nd(µ/T) = 0 . (5) One uses µ(P, T), for instance, when considering systems in the external ﬁeld. One then adds the potential energy (per particle) u(r) to the chemical potential so that the equilibrium condition is µ(P, T) + u(r) =const. Par-ticularly, in the gravity ﬁeld u(r) = mgz and diﬀerentiating µ(P, T) under T = const one gets vdP = −mgdz. Introducing density ρ = m/v one gets the well-known hydrostatic formula P = P0 −ρgz. For composite systems, the number of independent intensive parameters (thermodynamic degrees of freedom) is the number of components plus one. For example, for a mixture of gases, we need to specify the concentration of every gas plus temperature, which is common for all. Processes. While thermodynamics is fundamentally about states it is also used for describing processes that connect states. Particularly important questions concern performance of engines and heaters/coolers. Heat engine works by delivering heat from a reservoir with some higher T1 via some system to another reservoir with T2 doing some work in the process5. If the entropy of the hot reservoir decreases by some ∆S1 then the entropy of the cold one must increase by some ∆S2 ≥∆S1. The work ∆W is the diﬀerence between the heat given by the hot reservoir ∆Q1 = T1∆S1 and the heat absorbed by 5Think about how any real internal combustion engine works to appreciate the level of idealization achieved in distillation of that deﬁnition 6 the cold one ∆Q2 = T2∆S2 (assuming both processes quasi-static). Engine eﬃciency is the fraction of heat used for work that is ∆W ∆Q1 = ∆Q1 −∆Q2 ∆Q1 = 1 −T2∆S2 T1∆S1 ≤1 −T2 T1 . It is clear that maximal work is achieved for minimal entropy change ∆S2 = ∆S1, which happens for reversible (quasi-static) processes — if, for instance, the system is a gas which works by moving a piston then the pressure of the gas and the work are less for a fast-moving piston than in equilibrium. Similarly, refrigerator/heater is something that does work to transfer heat from cold to hot systems. The performance is characterized by the ra-tio of transferred heat to the work done. For the cooler, the eﬃciency is ∆Q2/∆W ≤T2/(T1 −T2), for the heater it is ∆Q1/∆W ≤T1/(T1 −T2). When the temperatures are close, the eﬃciency is large, as it requires almost no work to transfer heat. A speciﬁc procedure to accomplish reversible heat and work transfer is to use an auxiliary system which undergoes so-called Carnot cycle, where heat exchanges take place only at two temperatures. Engine goes through: 1) isothermal expansion at T1, 2) adiabatic expansion until temperature falls to T2, 3) isothermal compression until the entropy returns to its initial value, 4) adiabatic compression until the temperature reaches T1. The auxiliary system is connected to the reservoirs during isothermal stages: to the ﬁrst reservoir during 1 and to the second reservoir during 3. During all the time it is connected to our system on which it does work during 1 and 2, increasing the energy of our system, which then decreases its energy by working on the auxiliary system during 3 and 4. The total work is the area of the rectangle between the lines 1,3, the heat ∆Q1 is the area below the line 1. For heat transfer, one reverses the direction. T S T T2 P 1 1 2 3 4 4 1 3 2 Carnot cycle in T-S and P-V variables V Carnot cycle provides one with an operational method to measure the 7 ratio of two temperatures by measuring the engine eﬃciency6. Summary of formal structure. The fundamental relation (in energy rep-resentation) E = E(S, V, N) is equivalent to the three equations of state (2). If only two equations of state are given then Gibbs-Duhem relation may be integrated to obtain the third up to an integration constant; alternatively one may integrate molar relation de = Tds −Pdv to get e(s, v) again with an undetermined constant of integration. Example: consider an ideal monatomic gas characterized by two equations of state (found, say, experimentally with R ≃8.3 J/mole K ≃2 cal/mole K ): PV = NRT , E = 3NRT/2 . (6) The extensive parameters here are E, V, N so we want to ﬁnd the fundamental equation in the entropy representation, S(E, V, N). We write (4) in the form S = E 1 T + V P T −N µ T . (7) Here we need to express intensive variables 1/T, P/T, µ/T via extensive vari-ables. The equations of state (6) give us two of them: P T = NR V = R v , 1 T = 3NR 2E = 3R e . (8) Now we need to ﬁnd µ/T as a function of e, v using Gibbs-Duhem relation in the entropy representation (5). Using the expression of intensive via extensive variables in the equations of state (8), we compute d(1/T) = −3Rde/2e2 and d(P/T) = −Rdv/v2, and substitute into (5): d ( µ T ) = −3 2 R e de −R v dv , µ T = C −3R 2 ln e −R ln v , s = 1 T e + P T v −µ T = s0 + 3R 2 ln e e0 + R ln v v0 . (9) Here e0, v0 are parameters of the state of zero internal energy used to deter-mine the temperature units, and s0 is the constant of integration. 6Practical needs to estimate the engine eﬃciency during the industrial revolution led to the development of such abstract concepts as entropy 8 1.2 Legendre transform Let us emphasize that the fundamental relation always relates extensive quantities. Therefore, even though it is always possible to eliminate, say, S from E = E(S, V, N) and T = T(S, V, N) getting E = E(T, V, N), this is not a fundamental relation and it does not contain all the information. Indeed, E = E(T, V, N) is actually a partial diﬀerential equation (because T = ∂E/∂S) and even if it can be integrated the result would contain un-determined function. Still, it is easier to measure, say, temperature than entropy so it is convenient to have a complete formalism with intensive pa-rameters as operationally independent variables and extensive parameters as derived quantities. This is achieved by the Legendre transform: To pass from the relation Y = Y (X) to that in terms of P = ∂Y/∂X it is not enough to eliminate X and consider the function Y = Y (P), which determines the curve Y = Y (X) only up to a shift along X: X Y Y X For example, the same Y = P 2/4 correspond to the family of func-tions Y = (X + C)2 for arbitrary C. To ﬁx the shift one may consider the curve as the envelope of the family of the tangent lines characterized by the slope P and the position ψ of intercept of the Y -axis. The func-tion ψ(P) = Y [X(P)] −PX(P) completely deﬁnes the curve; here one sub-stitutes X(P) found from P = ∂Y (X)/∂X (which is possible only when ∂P/∂X = ∂2Y/∂X2 ̸= 0). The function ψ(P) is referred to as a Legen-dre transform of Y (X). From dψ = −PdX −XdP + dY = −XdP one gets −X = ∂ψ/∂P i.e. the inverse transform is the same up to a sign: Y = ψ + XP. In mechanics, we use the Legendre transform to pass from Lagrangian to Hamiltonian description. 9 Y XP X ψ P Y = Ψ + Diﬀerent thermodynamics potentials suitable for diﬀerent physical situations are obtained replacing diﬀerent extensive parameters by the re-spective intensive parameters. Free energy F = E −TS (also called Helmholtz potential) is that partial Legendre transform of E which replaces the entropy by the temperature as an independent variable: dF(T, V, N, . . .) = −SdT −PdV + µdN + . . .. It is particularly convenient for the description of a system in a thermal contact with a heat reservoir because then the temperature is ﬁxed and we have one variable less to care about. The maximal work that can be done under a constant temperature (equal to that of the reservoir) is minus the diﬀerential of the free energy. Indeed, this is the work done by the system and the thermal reservoir. That work is equal to the change of the total energy d(E + Er) = dE + TrdSr = dE −TrdS = d(E −TrS) = d(E −TS) = dF . In other words, the free energy F = E −TS is that part of the internal energy which is free to turn into work, the rest of the energy TS we must keep to sustain a constant temperature. The equilibrium state minimizes F, not absolutely, but over the manifold of states with the temperature equal to that of the reservoir. Indeed, consider F(T, X) = E[S(T, X), X]−TS(T, X), then (∂E/∂X)S = (∂F/∂X)T that is they turn into zero simultaneously. Also, in the point of extremum, one gets (∂2E/∂X2)S = (∂2F/∂X2)T i.e. both E and F are minimal in equilibrium. Monatomic gas at ﬁxed T, N has F(V ) = E −TS(V ) = −NRT ln V +const. If a piston separates equal amounts then the work done in changing the volume of a subsystem from V1 to V2 is ∆F = NRT ln[V2(V −V2)/V1(V −V1)]. Enthalpy H = E + PV is that partial Legendre transform of E which re-places the volume by the pressure dH(S, P, N, . . .) = TdS+V dP +µdN +. . .. It is particularly convenient for situation in which the pressure is maintained 10 constant by a pressure reservoir (say, when the vessel is open into atmo-sphere). Just as the energy acts as a potential at constant entropy and the free energy as potential at constant temperature, so the enthalpy is a poten-tial for the work done by the system and the pressure reservoir at constant pressure. Indeed, now the reservoir delivers pressure which can change the volume so that the diﬀerential of the total energy is d(E + Er) = dE −PrdVr = dE + PrdV = d(E + PrV ) = d(E + PV ) = dH . Equilibrium minimizes H under the constant pressure. On the other hand, the heat received by the system at constant pressure (and N) is the enthalpy change: δQ = dQ = TdS = dH. Compare it with the fact that the heat received by the system at constant volume (and N) is the energy change since the work is zero. One can replace both entropy and volume obtaining (Gibbs) thermody-namics potential G = E −TS + PV which has dG(T, P, N, . . .) = −SdT + V dP + µdN + . . . and is minimal in equilibrium at constant temperature and pressure. From (4) we get (remember, they all are functions of diﬀerent variables): F = −P(T, V )V + µ(T, V )N , H = TS + µN , G = µ(T, P)N . (10) When there is a possibility of change in the number of particles (because our system is in contact with some particle source having a ﬁxed chem-ical potential) then it is convenient to use the grand canonical potential Ω(T, V, µ) = E −TS −µN which has dΩ= −SdT −PdV −Ndµ. The grand canonical potential reaches its minimum under the constant temperature and chemical potential. Since the Legendre transform is invertible, all potentials are equivalent and contain the same information. The choice of the potential for a given physical situation is that of convenience: we usually take what is ﬁxed as a variable to diminish the number of eﬀective variables. Maxwell relations. Changing order of taking mixed second derivatives of a potential creates a class of identities known as Maxwell relations. For exam-ple, ∂2E/∂S∂V = ∂2E/∂V ∂S gives (∂P/∂S)V = −(∂T/∂V )S. That can be done for all three combinations (SV, SN, V N) possible for a simple single-component system and also for every other potential (F, H, G). Maxwell relations for constant N can be remembered with the help of the mnemonic 11 diagram with the sides labelled by the four common potentials ﬂanked by their respective natural independent variables. In the diﬀerential expression for each potential in terms of the natural variables arrow pointing away from the variable implies a positive sign while pointing towards the variable implies negative sign like in dE = TdS −PdV : V F T E S H P V S P T P S = G Maxwell relations are given by the corners of the diagram, for example, (∂V/∂S)P = (∂T/∂P)S etc. If we consider constant N then any fundamental relation of a single-component system is a function of only two variables and therefore have only three independent second derivatives. Traditionally, all derivatives are expressed via the three basic ones (those of Gibbs potential), the speciﬁc heat and the coeﬃcient of thermal expansion, both at a constant pressure, and isothermal compressibility: cP = T (∂S ∂T ) P = −T (∂2G ∂T 2 ) P , α = 1 V (∂V ∂T ) P , κT = −1 V (∂V ∂P ) T . In particular, the speciﬁc heat at constant volume is as follows: cV = T (∂S ∂T ) V = cP −TV α2 NκT . (11) That and similar formulas form a technical core of thermodynamics and the art of deriving them ought to be mastered. It involves few simple rules in treating partial derivatives: (∂X ∂Y ) Z= ( ∂Y ∂X )−1 Z , (∂X ∂Y ) Z ( ∂Y ∂W ) Z= ( ∂X ∂W ) Z, (∂X ∂Y ) Z (∂Y ∂Z ) X ( ∂Z ∂X ) Y=−1. An alternative (and more general) way to manipulate thermodynamic derivatives is to use jacobians and identity ∂(T, S)/∂(P, V ) = 1. Taking, say, S, V as independent variables, ∂(T, S) ∂(P, V ) = ∂(T, S) ∂(S, V ) ∂(S, V ) ∂(P, V ) = −(∂T/∂V )S (∂P/∂S)V = ESV EV S = 1 . 12 1.3 Stability of thermodynamic systems Consider entropy representation. Stationarity of equilibrium requires dS = 0 while stability requires d2S < 0. In particular, that means concavity of S(E, X). Indeed, for all ∆E one must have S(E+∆E, X)+S(E−∆E, X) ≤ 2S(E, X) otherwise our system can break into two halves with the energies E ± ∆E thus increasing total entropy. For ∆E →0 the stability require-ment means (∂2S/∂E2)X ≤0 ⇒(∂T/∂E)X ≥0 — increase of the energy must increase temperature. For the case X = V this can be also recast into (∂T/∂E)V = [∂(TV )/∂(EV )][∂(SV )/∂(SV )] = T −1(∂T/∂S)V = 1/cv ≥0 (adding heat to a stable system increases temperature). The same concav-ity requirement is true with respect to changes in other parameters X, in particular, (∂2S/∂V 2)E ≤0 ⇒(∂P/∂V )T ≤0 that is isothermal expansion must reduce pressure for the stable system. Considering both changes to-gether we must require SEE(∆E)2 + 2SEV ∆E∆V + SV V (∆V )2 ≤0. This quadratic form, SEE(∆E)2 + 2SEV ∆E∆V + SV V (∆V )2 = S−1 EE(SEE∆E + SEV ∆V )2 + (SV V −S2 EV S−1 EE)(∆V )2, has a deﬁnite sign if the determinant is positive: SEESV V −S2 EV ≥0. Manipulating derivatives one can show that this is equivalent to (∂P/∂V )S ≤0. Alternatively, one may consider the energy representation, here stability requires the energy minimum which gives ESS = T/cv ≥0, EV V = −(∂P/∂V )S ≥0. Considering both variations one can diagonalize d2E = ESS(dS)2 + EV V (dV )2 + 2ESV dSdV by intro-ducing the temperature diﬀerential dT = ESSdS + ESV dV so that 2d2E = E−1 SS(dT)2 + (EV V −E2 SV E−1 SS)(dV )2. It is thus clear that EV V −E2 SV E−1 SS = (∂2E/∂V 2)T = −(∂P/∂V )T and we recover all the same inequalities. Note that the pressure must decrease under both isothermal and adiabatic expan-sion. E E ∆ ∆ V Lines of constant entropy in unstable and stable cases ∆V ∆ The physical content of those stability criteria is known as Le Ch¨ atelier’s principle: if some perturbation deviates the system from a stable equilibrium 13 that induces spontaneous processes that reduce the perturbation. The third law of thermodynamics (Nernst theorem). It claims that S →0 as T →0. A standard argument is that since stability requires the positivity of the speciﬁc heat cv then the energy must monotonously increase with the temperature and zero temperature corresponds to the ground state. If the ground state is non-degenerate (unique) then S = 0. The ground can be degenerate yet generally that degeneracy grows slower than exponentially with N, then the entropy per particle is zero in the thermodynamic limit. While this argument is correct it is relevant only for temperatures less than the energy diﬀerence between the ﬁrst excited state and the ground state. As such, it has nothing to do with the third law established generally for much higher temperatures and related to the density of states as function of energy. Entropy goes to zero as T, T 3/2, T 3 for fermions, massive and massless bosons respectively, see Huang (Section 9.4) and L&L (Section 45) for more details. Phase transitions happen when some stability condition is not satisﬁed like in the region with (∂P/∂V )T > 0 as at the lowest isotherm in the below ﬁgure. When the pressure corresponds to the level NLC, it is clear that L is an unstable point and cannot be realized. But which stable point is realized, N or C? To get the answer, one must minimize the Gibbs potential since we have T and P ﬁxed. For one mole, it is the chemical potential which can be found integrating the Gibbs-Duhem relation, dµ(T, P) = −sdT + vdP, under the constant temperature: G = µ = ∫v(P)dP. The chemical potential increases up to the point (after E) with inﬁnite dV/dP. After that we move along the isotherm back having dP < 0 so that the integral decreases and then passes through another point of inﬁnite derivative and start to increase again. In other words, the third graph below represents three branches of the function µ(P) that has its derivative the function v(P) shown in the second graph. It is clear that to the intersection point D correspond to equal areas below and above the horizontal line on the ﬁrst graph.The pressure that corresponds to this point separates the absolute minimum at the left branch marked Q (solid-like) from that on the right one marked C (liquid-like). The dependence of volume on pressure is discontinuous along the isotherm. 14 P V V P P Q N C D E J L N Q C E D J L C D E J L N Q µ 15 2 Basic statistical physics (brief reminder) Here we describe two principal ways (microcanonical and canonical) to de-rive thermodynamics from statistical mechanics and introduce microscopic statistical description in the phase space. 2.1 Microcanonical distribution Consider a closed system with the energy E0. Boltzmann assumed that all microstates with the same energy have equal probability (ergodic hypothesis) which gives the microcanonical distribution: ρ(p, q) = Aδ[E(p, q) −E0] . (12) Usually one considers the energy ﬁxed with the accuracy ∆so that the mi-crocanonical distribution is ρ = { 1/Γ for E ∈(E0, E0 + ∆) 0 for E ̸∈(E0, E0 + ∆) , (13) where Γ is the volume of the phase space occupied by the system Γ(E, V, N, ∆) = ∫ E<H Nϵ/2 then the population of the higher level is larger than of the lower one (inverse population as in a laser) and the temperature is negative. Negative temperature may happen only in systems with the upper limit of energy levels and simply means that by adding energy beyond some level we actually decrease the entropy i.e. the number of accessible states. That example with negative temperature is to help you to disengage from the everyday notion of temperature and to get used to the physicist idea of temperature as the derivative of energy with respect to entropy. Available (non-equilibrium) states lie below the S(E) plot, notice that the entropy maximum corresponds to the energy minimum for positive tem-peratures and to the energy maximum for the negative temperatures part. A glance on the ﬁgure also shows that when the system with a negative tem-perature is brought into contact with the thermostat (having positive tem-perature) then our system gives away energy (a laser generates and emits light) decreasing the temperature further until it passes through inﬁnity to positive values and eventually reaches the temperature of the thermostat. That is negative temperatures are actually ”hotter” than positive. By itself though the system is stable since ∂2S/∂E2 = −N/L(N −L)ϵ2 < 0. Let us stress that there is no volume in S(E, N) that is we consider only subsystem or only part of the degrees of freedom. Indeed, real particles have kinetic energy unbounded from above and can correspond only to positive temperatures [negative temperature and inﬁnite energy give inﬁnite Gibbs factor exp(−E/T)]. Apart from laser, an example of a two-level system is spin 1/2 in the mag-netic ﬁeld H. Because the interaction between the spins and atom motions (spin-lattice relaxation) is weak then the spin system for a long time (tens of minutes) keeps its separate temperature and can be considered separately. External ﬁelds are parameters (like volume and chemical potential) that 22 determine the energy levels of the system. They are sometimes called gen-eralized thermodynamic coordinates, and the derivatives of the energy with respect to them are called respective forces. Let us derive the generalized force M that corresponds to the magnetic ﬁeld and determines the work done under the change of magnetic ﬁeld: dE(S, H) = TdS −MdH. Since the projection of every magnetic moment on the direction of the ﬁeld can take two values ±µ then the magnetic energy of the particle is ∓µH and E = −µ(N+ −N−)H. The force (the partial derivative of the energy with respect to the ﬁeld at a ﬁxed entropy) is called magnetization or magnetic moment of the system: M = − ( ∂E ∂H ) S = µ(N+ −N−) = Nµexp(µH/T) −exp(−µH/T) exp(µH/T) + exp(−µH/T) . (28) The derivative was taken at constant entropy that is at constant popula-tions N+ and N−. Note that negative temperature for the spin system corresponds to the magnetic moment opposite in the direction to the ap-plied magnetic ﬁeld. Such states are experimentally prepared by a fast re-versal of the magnetic ﬁeld. We can also deﬁne magnetic susceptibility: χ(T) = (∂M/∂H)H=0 = Nµ2/T, yet another second derivative that deter-mines the response and ﬂuctuations and will feature prominently in what follows. At weak ﬁelds and positive temperature, µH ≪T, (28) gives the formula for the so-called Pauli paramagnetism M Nµ = µH T . (29) Para means that the majority of moments point in the direction of the exter-nal ﬁeld. This formula shows in particular a remarkable property of the spin system: adiabatic change of magnetic ﬁeld (which keeps constant N+, N−and thus M) is equivalent to the change of temperature even though spins do not exchange energy. One can say that under the change of the value of the ho-mogeneous magnetic ﬁeld the relaxation is instantaneous in the spin system. This property is used in cooling the substances that contain paramagnetic impurities. For the entropy of the spin system to be preserved, one needs to change the ﬁeld slowly comparatively to the spin-spin relaxation and fast comparatively to the spin-lattice relaxation. The ﬁrst condition means that one cannot reach negative temperatures by adiabatically reversing magnetic 23 ﬁeld since the relaxation times of spins grow when ﬁeld decreases; indeed, negative temperatures must be reached through T →∞, not zero. In prac-tice, negative temperatures were reached (by Purcell, Pound and Ramsey in 1951) by fast reversal of the magnetic ﬁeld. To conclude let us treat the two-level system by the canonical approach where we calculate the partition function and the free energy: Z(T, N) = N ∑ L=0 CL N exp[−Lϵ/T] = [1 + exp(−ϵ/T)]N , (30) F(T, N) = −T ln Z = −NT ln[1 + exp(−ϵ/T)] . (31) We can now re-derive the entropy as S = −∂F/∂T and derive the (mean) energy and speciﬁc heat: ¯ E = Z−1 ∑ a Ea exp(−βEa) = −∂ln Z ∂β = T 2∂ln Z ∂T (32) = Nϵ 1 + exp(ϵ/T) , (33) C = dE dT = N exp(ϵ/T) [1 + exp(ϵ/T)]2 ϵ2 T 2 . (34) Here (32) is a general formula which we shall use in the future. Remark that even though canonical approach corresponds to a system in a thermostat, which necessary has positive temperature, all the formulas make sense at negative T too. Speciﬁc heat is one of the second derivatives of the thermodynamic po-tentials; such quantities characterize the response of the system to change of parameters and will feature prominently in the course. Speciﬁc heat tells us how much one raises the energy of the system when increasing the tem-perature by one degree (or, alternatively, how much energy one needs to increase the temperature by one degree). Speciﬁc heat of a two-level system turns into zero both at low temperatures (too small portions of energy are ”in circulation”) and at high temperatures (occupation numbers of two levels already close to equal so changing temperature does not change energy). C/N T/ε 1/2 2 24 A speciﬁc heat of this form characterized by a peak is observed in all systems with an excitation gap. More details can be found in Kittel, Section 24 and Pathria, Section 3.9. 2.4 Distribution in the phase space Here we introduce microscopic statistical description in the phase space and re-derive the canonical distribution. The phase space (p, q) has 6N dimen-sions. Instead of counting states we consider now evolution and deﬁne prob-ability for a subsystem to be in some ∆p∆q region of the phase space as the fraction of time it spends there: w = limT→∞∆t/T. Assuming that the probability to ﬁnd the subsystem within the volume dpdq is proportional to this volume, we introduce the statistical distribution in the phase space as density: dw = ρ(p, q)dpdq. By deﬁnition, the average with the statistical distribution is equivalent to the time average: ¯ f = ∫ f(p, q)ρ(p, q)dpdq = lim T→∞ 1 T ∫T 0 f(t)dt . (35) The main idea is that ρ(p, q) for a subsystem does not depend on the initial states of this and other subsystems so it can be found without actually solving equations of motion. We deﬁne statistical equilibrium as a state where macro-scopic quantities equal to the mean values. Assuming short-range forces we conclude that diﬀerent macroscopic subsystems interact weakly and are sta-tistically independent so that the distribution for a composite system ρ12 is factorized: ρ12 = ρ1ρ2. Now, we take the ensemble of identical systems starting from diﬀerent points in phase space. As coordinates and momenta change with time, each point moves in the phase space. Motion of all points of the ensem-ble constitutes a ﬂow with the velocity v = ( ˙ p, ˙ q), where the density at every point of the phase space changes according to the continuity equation: ∂ρ/∂t + div (ρv) = 0. However, more interesting is the density change along the trajectories, i.e. how the probability distribution changes with time for every given system. Such evolution is determined by the time derivative along the ﬂow: dρ(t, p, q)/dt = ∂ρ/∂t+∑ i( ˙ qi∂ρ/∂qi + ˙ pi∂ρ/∂pi = ∂ρ/∂t+(v·∇)ρ. According to the continuity equation, it is dρ(t, p, q)/dt = −ρdiv v. If the motion is considered for not very large time it is conservative and can be described by the Hamiltonian dynamics: ˙ qi = ∂H/∂pi and ˙ pi = −∂H/∂qi. Here the Hamiltonian generally depends on the momenta and coordinates 25 of the given subsystem and its neighbors. Hamiltonian ﬂow in the phase space is incompressible, it conserves area in each plane pi, qi and the total volume: div v = ∂˙ qi/∂qi + ∂˙ pi/∂pi = 0. That gives the Liouville theorem: dρ/dt = ∂ρ/∂t + (v · ∇)ρ = −ρdiv v = 0. The statistical distribution is thus conserved along the phase trajectories of any subsystem. As a result, equilibrium distribution ρ is an integral of motion. Consequently, it must be expressed solely via the integrals of motion. Since ln ρ is an additive quantity then it must be expressed linearly via the additive integrals of motions which for a general mechanical system are energy E(p, q), momentum P(p, q) and the momentum of momentum M(p, q): ln ρa = αa + βEa(p, q) + c · Pa(p, q) + d · M(p, q) . (36) Here αa is the normalization constant for a given subsystem while the con-stants β, c, d are the same for all subsystems to ensure additivity. Those seven constants (in 3 dimensions) are determined by the values of the seven integrals of motion for the whole system. We thus conclude that the additive integrals of motion is all we need to get the statistical distribution of a closed system (and any subsystem), those integrals replace all the enormous micro-scopic information. Considering system which neither moves nor rotates we are down to the single integral, energy. For any subsystem (or any system in the contact with thermostat) we get Gibbs’ canonical distribution ρ(p, q) = A exp[−βE(p, q)] . (37) See Landau & Lifshitz, Sects 1-4. 3 Entropy and information By deﬁnition, entropy of a closed system determines the number of available states (or, classically, phase volume). Assuming that system spends compa-rable time in diﬀerent available states we conclude that since the equilibrium must be the most probable state it corresponds to the entropy maximum. If the system happens to be not in equilibrium at a given moment of time [say, the energy distribution between the subsystems is diﬀerent from the most probable Gibbs distribution] then it is more probable to go towards equilibrium that is increasing entropy. This is a microscopic (probabilistic) interpretation of the second law of thermodynamics formulated by Clausius 26 in 1865. The probability maximum is very sharp in the thermodynamic limit since exp(S) grows exponentially with the system size. That means that for macroscopic systems the probability to pass into the states with lower entropy is so vanishingly small that such events are never observed. What often causes confusion here is that the dynamics (classical and quantum) of any given system is time reversible. The Hamiltonian evolution described in Sect 2.4 preserves the density in the phase space ρ(p, q), so how the entropy S = − ∫dpdqρ ln ρ can grow? To avoid the confusion, one must remember that we study the situations with incomplete knowledge of the system. That means that we know coordinates and momenta within some intervals, i.e. characterize the system not by a point in phase space but by a ﬁnite region there. Entropy growth is then related not to the trajectory of a single point in phase space but to the behavior of ﬁnite regions (i.e. sets of such points) or ensembles of systems. The entropy of a single point is zero. The necessity to consider ﬁnite regions follows from the insuﬃciency of information about the true state of the system. Consideration of ﬁnite regions is called coarse graining and it is the main feature of stat-physical approach responsible for the irreversibility of statistical laws. In this section we shall see how it works. 3.1 Lyapunov exponent The dynamical mechanism of the entropy growth is the separation of trajec-tories in phase space so that trajectories started from a small ﬁnite region are found in larger and larger regions of phase space as time proceeds. The relative motion is determined by the velocity diﬀerence between neighbor-ing points in the phase space: δvi = rj∂vi/∂xj = rjσij. Here x = (p, q) is the 6N-dimensional vector of the position and v = ( ˙ p, ˙ q) is the velocity in the phase space. The trace of the tensor σij is the rate of the volume change which must be zero according to the Liouville theorem (that is a Hamiltonian dynamics imposes an incompressible ﬂow in the phase space). We can de-compose the tensor of velocity derivatives into an antisymmetric part (which describes rotation) and a symmetric part (which describes deformation). We are interested here in deformation because it is the mechanism of the entropy growth. The symmetric tensor, Sij = (∂vi/∂xj + ∂vj/∂xi)/2, can be always transformed into a diagonal form by an orthogonal transformation (i.e. by the rotation of the axes), so that Sij = Siδij. Recall that for Hamiltonian motion, ∑ i Si = div v = 0, so that some components are positive, some are 27 negative. Positive diagonal components are the rates of stretching and nega-tive components are the rates of contraction in respective directions. Indeed, the equation for the distance between two points along a principal direction has a form: ˙ ri = δvi = riSi . The solution is as follows: ri(t) = ri(0) exp [∫t 0 Si(t′) dt′ ] . (38) For a time-independent strain, the growth/decay is exponential in time. One recognizes that a purely straining motion converts a spherical element into an ellipsoid with the principal diameters that grow (or decay) in time. Indeed, consider a two-dimensional projection of the initial spherical element i.e. a circle of the radius R at t = 0. The point that starts at x0, y0 = √ R2 −x2 0 goes into x(t) = eS11tx0 , y(t) = eS22ty0 = eS22t √ R2 −x2 0 = eS22t√ R2 −x2(t)e−2S11t , x2(t)e−2S11t + y2(t)e−2S22t = R2 . (39) The equation (39) describes how the initial circle turns into the ellipse whose eccentricity increases exponentially with the rate \|S11 −S22\|. In a multi-dimensional space, any sphere of initial conditions turns into the ellipsoid deﬁned by ∑6N i=1 x2 i (t)e−2Sit =const. t exp(S t) exp(S t) xx yy Figure 1: Deformation of a phase-space element by a permanent strain. Of course, as the system moves in the phase space, both the strain values and the orientation of the principal directions change, so that expanding direction may turn into a contracting one and vice versa. Since we do not want to go into details of how the system interacts with the environment, then we consider such evolution as a kind of random process. The question is whether averaging over all values and orientations gives a zero net result. It may seem counter-intuitive at ﬁrst, but in a general case an exponential stretching must persist on average and the majority of trajectories separate. 28 Physicists think in two ways: one in space and another in time (unless they are relativistic and live in a space-time). Let us ﬁrst look at separation of trajectories from a temporal perspective: even when the average rate of separation along a given direction Λi(t) = ∫t 0 Si(t′)dt′/t is zero, the average exponent of it is larger than unity (and generally growing with time): 1 T ∫T 0 dt exp [∫t 0 Si(t′)dt′ ] ≥1 . This is because the intervals of time with positive Λ(t) give more contribution into the exponent than the intervals with negative Λ(t). That follows from the concavity of the exponential function. In the simplest case, when −a < Λ < a, the average Λ is zero, while the average exponent is (1/2a) ∫−a a eΛdΛ = (ea −e−a)/2a > 1. Looking from a spatial perspective, consider the simplest example of a two-dimensional pure strain, which corresponds to an incompressible saddle-point ﬂow: vx = λx, vy = −λy. Here we have one expanding direction direction and one contracting direction, their rates being equal. The vector r = (x, y) which characterizes the distance between two close trajectories can look initially at any direction. The evolution of the vector components satisﬁes the equations ˙ x = vx and ˙ y = vy. Whether the vector is stretched or contracted after some time T depends on its orientation and on T. Since x(t) = x0 exp(λt) and y(t) = y0 exp(−λt) = x0y0/x(t) then every trajectory is a hyperbole. A unit vector initially forming an angle φ with the x axis will have its length [cos2 φ exp(2λT) + sin2 φ exp(−2λT)]1/2 after time T. The vector will be stretched if cos φ ≥[1 + exp(2λT)]−1/2 < 1/ √ 2, i.e. the fraction of stretched directions is larger than half. When along the motion all orientations are equally probable, the net eﬀect is stretching, proportional to the persistence time T. The net stretching and separation of trajectories is formally proved in mathe-matics by considering random strain matrix ˆ σ(t) and the transfer matrix ˆ W deﬁned by r(t) = ˆ W(t, t1)r(t1). It satisﬁes the equation d ˆ W/dt = ˆ σ ˆ W. The Liouville theo-rem tr ˆ σ = 0 means that det ˆ W = 1. The modulus r(t) of the separation vector may be expressed via the positive symmetric matrix ˆ W T ˆ W. The main result (Fursten-berg and Kesten 1960; Oseledec, 1968) states that in almost every realization ˆ σ(t), the matrix 1 t ln ˆ W T (t, 0) ˆ W(t, 0) tends to a ﬁnite limit as t →∞. In particular, its eigenvectors tend to d ﬁxed orthonormal eigenvectors fi. Geometrically, that 29 x x(T) x(0) y y(0) y(T) ϕ 0 Figure 2: The distance of the point from the origin increases if the angle is less than φ0 = arccos[1 + exp(2λT)]−1/2 > π/4. Note that for φ = φ0 the initial and ﬁnal points are symmetric relative to the diagonal: x(0) = y(T) and y(0) = x(T). precisely means than an initial sphere evolves into an elongated ellipsoid at later times. The limiting eigenvalues λi = lim t→∞t−1 ln \| ˆ Wfi\| (40) deﬁne the so-called Lyapunov exponents. The sum of the exponents is zero due to the Liouville theorem so there exists at least one positive exponent which gives stretching. Therefore, as time increases, the ellipsoid is more and more elongated and it is less and less likely that the hierarchy of the ellipsoid axes will change. Mathematical lesson to learn is that multiplying N random matrices with unit determinant (recall that determinant is the product of eigenvalues), one generally gets some eigenvalues growing (and some decreasing) exponentially with N. It is also worth remembering that in a random ﬂow there is always a probability for two trajectories to come closer. That probability decreases with time but it is ﬁnite for any ﬁnite time. In other words, majority of trajectories separate but some approach. The separating ones provide for the exponential growth of positive moments of the distance: E(a) = limt→∞t−1⟨ra(t)/ra(0)⟩> 0 for a > 0. However, approaching trajectories have r(t) decreasing, which guarantees that the moments with suﬃciently negative a also grow. Mention without proof that E(a) is a concave function, which evidently passes through zero, E(0) = 0. It must then have another zero which for isotropic random ﬂow in d-dimensional space can be shown to be a = −d, see home exercise. The probability to ﬁnd a ball turning into an exponentially stretching ellipse thus goes to unity as time increases. The physical reason for it is that 30 substantial deformation appears sooner or later. To reverse it, one needs to contract the long axis of the ellipse, that is the direction of contraction must be inside the narrow angle deﬁned by the ellipse eccentricity, which is less likely than being outside the angle. Randomly oriented deformations on average continue to increase the eccentricity. Armed with the understanding of the exponential stretching, we now re-turn to the dynamical foundation of the second law of thermodynamics. We assume that our ﬁnite resolution does not allow us to distinguish between the states within some square in the phase space. That square is our ”grain” in coarse-graining. In the ﬁgure below, one can see how such black square of initial conditions (at the central box) is stretched in one (unstable) direction and contracted in another (stable) direction so that it turns into a long nar-row strip (left and right boxes). Later in time, our resolution is still restricted - rectangles in the right box show ﬁnite resolution (this is coarse-graining). Viewed with such resolution, our set of points occupies larger phase volume (i.e. corresponds to larger entropy) at t = ±T than at t = 0. Time re-versibility of any particular trajectory in the phase space does not contradict the time-irreversible ﬁlling of the phase space by the set of trajectories con-sidered with a ﬁnite resolution. By reversing time we exchange stable and unstable directions (i.e. those of contraction and expansion), but the fact of space ﬁlling persists. We see from the ﬁgure that the volume and entropy increase both forward and backward in time. To avoid misunderstanding, note that usual arguments that entropy growth provides for time arrow are such: if we already observed an evolution that produces a narrow strip then its time reversal is contraction into a ball; but if we consider a narrow strip as an initial condition, it is unlikely to observe a contraction because of the narrow angle mentioned above. Therefore, being shown two movies, one with stretching, another with contraction we conclude that with probability close (but not exactly equal!) to unity the ﬁrst movie shows the true sequence of events, from the past to the future. t=T q p p q p q t=-T t=0 31 After the strip length reaches the scale of the velocity change (when one already cannot approximate the phase-space ﬂow by a linear proﬁle ˆ σr), strip starts to fold because rotation (which we can neglect for a ball but not for a long strip) is diﬀerent at diﬀerent parts of the strip. Still, however long, the strip continues locally the exponential stretching. Eventually, one can ﬁnd the points from the initial ball everywhere which means that the ﬂow is mixing, also called ergodic. Formal deﬁnition is that the ﬂow is called ergodic in the domain if the trajectory of almost every point (except possibly a set of zero volume) passes arbitrarily close to every other point. An equivalent deﬁnition is that there are no ﬁnite-volume subsets of the domain invariant with respect to the ﬂow except the domain itself. Ergodic ﬂow on an energy surface in the phase space provides for a micro-canonical distribution (i.e. constant), since time averages are equivalent to the average over the surface. While we can prove ergodicity only for relatively simple systems, like the gas of hard spheres, we believe that it holds for most systems of suﬃciently general nature (that vague notion can be make more precise by saying that the qualitative systems behavior is insensitive to small variations of its microscopic parameters). When the density spreads, entropy grows (as the logarithm of the volume occupied). If initially our system was within the phase-space volume ϵ6N, then its density was ρ0 = ϵ−6N inside and zero outside. After stretching to some larger volume Aϵ6N the entropy S = − ∫ρ ln ρdx has increased by ln A. The Lyapunov exponent determines the rate of the entropy growth. Two concluding remarks are in order. First, the notion of an exponential separation of trajectories put an end to the old dream of Laplace to be able to predict the future if only all coordinates and momenta are given. Even if we were able to measure all relevant phase-space initial data, we can do it only with a ﬁnite precision ϵ. However small is the indeterminacy in the data, it is ampliﬁed exponentially with time so that eventually ϵ exp(λT) is large and we cannot predict the outcome. Mathematically speaking, limits ϵ →0 and T →∞do not commute. Second, the above arguments did not use the usual mantra of thermodynamic limit, which means that even the systems with a small number of degrees of freedom need statistics for their description at long times if their dynamics has a positive Lyapunov exponent (which is generic) - this is sometimes called dynamical chaos. 32 3.2 Information theory approach Information is physical (Landauer) Here I brieﬂy re-tell the story of statistical physics using a diﬀerent lan-guage. It will let us see entropy in a new light. An advantage of using diﬀerent formulations is that it helps to understand things better and trig-gers diﬀerent intuition in diﬀerent people. Consider ﬁrst a simple problem in which we are faced with a choice among n equal possibilities (say, in which of n boxes a candy is hidden). How much we need to know to get the candy? Let us denote the missing information by I(n). Clearly, the information is an increasing function of n and I(1) = 0. If we have several independent problems then information must be additive. For example, consider each box to have m compartments: I(nm) = I(n)+ I(m). Now, we can write (Shannon, 1948) I(n) = I(e) ln n = k ln n (41) That it must be a logarithm is clear also from obtaining the missing informa-tion by asking the sequence of questions in which half we ﬁnd the box with the candy, one then needs log2 n of such questions and respective one-bit answers. We can easily generalize the deﬁnition (41) for non-integer rational numbers by I(n/l) = I(n) −I(l) and for all positive real numbers by con-sidering limits of the series and using monotonicity. So the single number of the lucky box with the candy brings the information k ln n. We used to think of information received through words and symbols. If we have an alphabet with n symbols then every symbol we receive is a choice out of n and brings the information k ln n. If symbols come independently then the message of the length N can potentially be one of nN possibilities so that it brings the information kN ln n. If all the 25 letters of the English alphabet were used with the same frequency then the word ”love” would bring the information equal to 4k ln 25 or 4 log2 25 bits. Here and below we assume that the receiver has no other prior knowledge on subjects like correlations between letters (for instance, everyone who knows English, can infer that there is only one four-letter word which starts with “lov...” so the last letter brings zero information for such people). 33 A A A A B E B B B Z Z Z Z L O V ... ... ... ... ... ... ... ... ... n N In reality though every letter brings even less information than k ln 25 since we know that letters are used with diﬀerent frequencies. Indeed, consider the situation when there is a probability wi assigned to each letter (or box) i = 1, . . . , n. It is then clear that diﬀerent letters bring diﬀerent information. When there is randomness, we evaluate the average information per symbol by repeating our choice, say, N times. As N →∞we know that candy in the i-th box in Nwi cases but we do not know the order in which diﬀerent possi-bilities appear. Total number of orders is N!/ Πi(Nwi)! and the information that we obtained from N symbols is IN = k ln ( N!/ Πi(Nwi)! ) ≈−Nk ∑ i wi ln wi + O(lnN) . (42) The missing information per symbol in the language coincides with the en-tropy (26): I(w1 . . . wn) = lim N→∞IN/N = −k n ∑ i=1 wi ln wi . (43) Incidentally for English language the information per symbol is − z ∑ i=a wi log2 wi ≈4.11 bits . The information (43) is zero for delta-distribution wi = δij; it is generally less than the information (41) and coincides with it only for equal probabil-ities, wi = 1/n, when the entropy is maximum. Indeed, equal probabilities we ascribe when there is no extra information, i.e. in a state of maximum ignorance. In this state, we get maximum information per symbol; any prior knowledge can reduce the information. Mathematically, the property I(1/n, . . . , 1/n) ≥I(w1 . . . wn) (44) is called convexity. It follows from the fact that the function of a single variable s(w) = −w ln w is strictly downward convex (concave) since its 34 second derivative, −1/w, is everywhere negative for positive w. For any concave function, the average over the set of points wi is less or equal to the function at the average value (so-called Jensen inequality): 1 n n ∑ i=1 s (wi) ≤s (1 n n ∑ i=1 wi ) . (45) −Wln W A (A+B)/2 B S[(A+B)/2]>[S(A)+S(B)]/2 W From here one gets the entropy inequality: I(w1 . . . wn) = n ∑ i=1 s (wi) ≤ns (1 n n ∑ i=1 wi ) = ns (1 n ) = I (1 n, . . . , 1 n ) . (46) The relation (45) can be proven for any concave function. Indeed, the concav-ity condition states that the linear interpolation between two points a, b lies everywhere below the function graph: s(λa+b−λb) ≥λs(a)+(1−λ)s(b) for any λ ∈[0, 1], see the Figure. For λ = 1/2 it corresponds to (45) for n = 2. To get from n = 2 to arbitrary n we use induction. For that end, we choose λ = (n −1)/n, a = (n −1)−1 ∑n−1 i=1 wi and b = wn to see that s (1 n n ∑ i=1 wi ) = s (n −1 n (n −1)−1 n−1 ∑ i=1 wi + wn n ) ≥n −1 n s ( (n −1)−1 n−1 ∑ i=1 wi ) + 1 ns (wn) ≥1 n n−1 ∑ i=1 s (wi) + 1 ns (wn) = 1 n n ∑ i=1 s (wi) . (47) In the last line we used the truth of (45) for n −1 to prove it for n. Note that when n →∞then (41) diverges while (43) may well be ﬁnite. We can generalize (43) for a continuous distribution by dividing into cells (that is considering a limit of discrete points). Here, diﬀerent choices of 35 variables to deﬁne equal cells give diﬀerent deﬁnitions of information. It is in such a choice that physics (or other speciﬁc knowledge) enters. Physics (quantum mechanics) requires that for Hamiltonian system the equal volumes in phase space contain equal number of states, so the measure is uniform in canonical coordinates; we then write the missing information in terms of the phase space density, which may also depend on time: I(t) = − ∫ ρ(p, q, t) ln[ρ(p, q, t)] dpdq . (48) It is maximal for the uniform distribution ρ = 1/Γ, I = ln Γ. If the density of the discrete points in the continuous limit is inhomogeneous, say m(x), then the proper generalization is I(t) = − ∫ ρ(x) ln[ρ(x)/m(x)] dx . It is invariant with respect to an arbitrary change of variables x →y(x) since ρ(y)dy = ρ(x)dx and m(y)dy = m(x)dx while (48) was invariant only with respect to canonical transformations (including a time evolution according to a Hamiltonian dynamics) that conserve the element of the phase-space volume. If we introduce the normalized distribution of points ρ′(x) = m(x)/Γ, then I(t) = ln V − ∫ ρ(x) ln[ρ(x)/ρ′(x)] dx . (49) The last term in (49) turns into zero when ρ and ρ′ coincide and thus presents some measure of the diﬀerence between the distributions. Mention brieﬂy the application of entropy in communication theory. In-equality (44) means that a communication channel transmitting bits (ones and zeros) on average can transmit no more than one unit of the information (43) per symbol. In other words, ∑z i=a wi log2 wi gives the minimum number of bits per symbol needed to transmit the ensemble of messages. We can say that the information content of a symbol number i is log2(1/wi), while the entropy is the mean information content per symbol. Note that less proba-ble symbols have larger information content, but they happen more rarely. The mean information content for a given letter, −w ln w, is maximal for w = 1/e. Diﬀerent probability of letters suggests a way of signal compres-sion by coding common letters by short sequences and infrequent letters by more lengthy combinations - lossless compressions like zip, gz and gif work this way (you may ﬁnd it interesting to know that jpeg, mpeg, mp3 and 36 telephone use lossy compression which removes information presumed to be unimportant for humans). Apart from restrictions imposed by the statistics of symbols to be trans-ferred, one also wish to characterize the quality of the channel. Note that in this context one can view measurements as messages about the value of the quantity we measure. Here, the message (measurement) A we receive gives the information about the event (quantity) B. That information must be related to the fact that having observed A increases the probability to have B comparing to the unconditional probability to observe B: I(A, B) = ln[P(B\|A)/P(B)] . Here P(B\|A) is the so-called conditional probability (of B in the presence of A). The conditional probability is related to the joint probability P(A, B) by the evident formula P(A, B) = P(B\|A)P(A), which allows one to write the information in a symmetric form I(A, B) = ln [ [P(B, A) P(A)P(B) ] . (50) When A and B are independent then the conditional probability is indepen-dent of A and information is zero. When they are dependent, P(B, A) ≥ P(A)P(B), so that that the information is always positive. If one is just interested in the channel as speciﬁed by P(B\|A) then one maxi-mizes I(A, B) over all choices of the source statistics P(B) and call it the channel capacity. Alternatively, one may wish to know how much information about B one obtains on average by measuring A. Summing over all possible B1, . . . , Bn and A1, . . . , Am we obtain Shannon’s “mutual information” used to evaluate the quality of communication systems (or measurements) I(A, B) = m ∑ i=1 n ∑ j=1 P(Ai, Bj) ln[P(Bj\|Ai)/P(Bj)] →I(Z, Y )= ∫ dzdyp(z, y) ln [p(z\|y) p(y) ] = ∫ dzdy p(z, y) ln [ p(z, y) p(z)p(y) ] . (51) Here we used p(z, y) = p(z\|y)p(y) - the probability to get y, z is the probability to get y times the probability to get z for this y. Note that (51) is the particular case of multidimensional (49), where one takes x = (y, z), ρ′ = p(z)p(y), that is mutual information measures the diﬀerence between the true joint distribution 37 and the distribution taken as if the quantities were statistically independent. It is straightforward to generalize it from the pair to many quantities. You probably noticed that (41,50) corresponds to the microcanonical de-scription (16) giving information/entropy as a logarithm of the number of states, while (43,49,51) corresponds to the canonical description (26) giving it as an average. An advantage of Shannon entropy (43,49,51) is that it is deﬁned for arbitrary distribution, not necessarily equilibrium. One can go further and deﬁne a free energy for any system in a contact with a thermostat having temperature T as F(ρ) = E(ρ) −TS(ρ), even when the distribution of the system itself is not equilibrium. Mutual information also allows to understand the second law of ther-modynamics from a diﬀerent perspective. Boltzmann considered the ther-modynamic entropy of the gas as a the sum of entropies of diﬀerent par-ticles ∑S(pi, qi), neglecting their correlations, i.e. the mutual information ∑ i S(pi, qi) −S(p1 . . . pn, q1, . . . qn) = I(p1 . . . pn, q1, . . . qn). That allowed him to establish H-theorem, that is the growth of the thermodynamic (uncorre-lated) entropy. Since the Liouville theorem guarantees that the true entropy S(p1 . . . pn, q1, . . . qn) does not change upon evolution, then the increase of the uncorrelated part must be compensated by the increase of the mutual information. In other words, one can replace the usual second law of thermo-dynamics by the law of conservation of the total entropy (or information), where the increase in the thermodynamic (uncorrelated) entropy is exactly compensated by the increase in correlations between particles expressed by the mutual information. The usual second law then results simply from our renunciation of all correlation knowledge, and not from any intrinsic behavior of dynamical systems. So far, we deﬁned information via the distribution. Now, we want to use the idea of information to get the distribution. Statistical mechanics is a systematic way of guessing, making use of incomplete information. The main problem is how to get the best guess for the probability distribution ρ(p, q, t) based on any given information presented as ⟨Rj(p, q, t)⟩= rj, i.e. as the expectation (mean) values of some dynamical quantities. Our distribution must contain the whole truth (i.e. all the given information) and nothing but the truth that is it must maximize the missing information I. This is to provide for the widest set of possibilities for future use, compatible with the 38 existing information. Looking for the maximum of I − ∑ j λj⟨Rj(p, q, t)⟩= ∫ ρ(p, q, t){ln[ρ(p, q, t)] − ∑ j λjRj(p, q, t)} dpdq , we obtain the distribution ρ(p, q, t) = Z−1 exp [ − ∑ j λjRj(p, q, t) ] , (52) where the normalization factor Z(λi) = ∫ exp [ − ∑ j λjRj(p, q, t) ] dpdq , can be expressed via the measured quantities by using ∂ln Z ∂λi = −ri . (53) For example, consider our initial ”candy-in-the-box” problem (think of an impurity atom in a lattice if you prefer physics to candies). Let us denote the number of the box with the candy j. Diﬀerent attempts give diﬀerent j (for impurity, think of X-ray with wavenumber k scattering on the lattice) but on average after many attempts we ﬁnd, say, ⟨cos(kj)⟩= 0.3. Then ρ(j) = Z−1(λ) exp[−λ cos(kj)] Z(λ) = n ∑ j=1 exp[λ cos(kj)] , ⟨cos(kj)⟩= d log Z/dλ = 0.3 . We can explicitly solve this for k ≪1 ≪kn when one can approximate the sum by the integral so that Z(λ) ≈nI0(λ) where I0 is the modiﬁed Bessel function. Equation I′ 0(λ) = 0.3I0(λ) has an approximate solution λ ≈0.63. Note in passing that the set of equations (53) may be self-contradictory or insuﬃcient so that the data do not allow to deﬁne the distribution or allow it non-uniquely. If, however, the solution exists then (48,52) deﬁne the missing information I{ri} which is analogous to thermodynamic entropy as a function of (measurable) macroscopic parameters. It is clear that I have a tendency to increase whenever a constraint is removed (when we measure less quantities Ri). 39 If we know the given information at some time t1 and want to make guesses about some other time t2 then our information generally gets less relevant as the distance \|t1 −t2\| increases. In the particular case of guessing the distribution in the phase space, the mechanism of loosing information is due to separation of trajectories described in Sect. 3. Indeed, if we know that at t1 the system was in some region of the phase space, the set of trajectories started at t1 from this region generally ﬁlls larger and larger regions as \|t1 −t2\| increases. Therefore, missing information (i.e. entropy) increases with \|t1 −t2\|. Note that it works both into the future and into the past. Information approach allows one to see clearly that there is really no contradiction between the reversibility of equations of motion and the growth of entropy. Yet there is one class of quantities where information does not age. They are integrals of motion. A situation in which only integrals of motion are known is called equilibrium. The distribution (52) takes the canonical form (36,37) in equilibrium. On the other hand, taking micro-canonical as constant over the constant-energy surface corresponds to the same approach of not adding any additional information to what is known (energy). From the information point of view, the statement that systems approach equilibrium is equivalent to saying that all information is forgotten except the integrals of motion. If, however, we possess the information about averages of quantities that are not integrals of motion and those averages do not coincide with their equilibrium values then the distribution (52) deviates from equilibrium. Examples are currents, velocity or temperature gradients like considered in kinetics. The concept of entropy as missing information8 allows one to understand that entropy does not really decrease in the system with Maxwell demon or any other information-processing device (indeed, if at the beginning one has an information on position or velocity of any molecule, then the entropy was less by this amount from the start; after using and processing the information the entropy can only increase). Consider, for instance, a particle in the box. If we know that it is in one half then entropy (the logarithm of available states) is ln(V/2). That also teaches us that information has thermodynamic (energetic) value: by placing a piston at the half of the box and allowing particle to hit and move it we can get the work T∆S = T ln 2 done. On the other hand, the law of energy conservation tells that to get such an 8that entropy is not a property of the system but of our knowledge about the system 40 information one must make a measurement whose minimum energetic cost is T∆S = T ln 2 (that was considered by Szilard in 1929 who also introduced ”bit” as a unit of information). Making measurement R one changes the distribution from ρ(x) to (generally non-equilibrium) ρ(x\|R), which decreases the entropy of the system by the mutual information (51): S(x\|R) −S(x) = − ∫ ρ(x\|R) ln ρ(x\|R) dxdR + ∫ ρ(x) ln ρ(x) dx = ∫ ρ(x, R) ln[ρ(x, R)/ρ(x)ρ(R)] dxdR . If such measurement does not change energy (like the knowledge in which half of the box the particles is), the entropy decrease increases the (non-equilibrium) free energy, so that the minimal work to perform such a mea-surement is F(ρ(x\|R)) −F(ρ(x)). More details can be found in Katz, Sects. 2-5, Sethna Sect. 5.3 and Kardar I, Problem 2.6. 41
13804	https://www.scribd.com/document/827439590/%E4%B8%AD%E8%80%83%E5%9F%B9%E4%BC%98%E4%B8%93%E9%A2%98%E7%BB%8F%E5%85%B8%E8%AE%B2%E4%B9%89-%E7%AC%AC2%E8%AE%B2-%E5%9E%82%E7%9B%B4%E5%B9%B3%E5%88%86%E7%BA%BF	中考培优专题经典讲义第2讲垂直平分线 \| PDF Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 452 views 12 pages 中考培优专题经典讲义第2讲垂直平分线本讲主要介绍了垂直平分线的性质及其应用，包括线段垂直平分线上的点到端点的距离相等的定理和逆定理。通过例题解析，展示了如何利用垂直平分线解决三角形相关问题。最后提供了巩固练习以加深理解。 Full description Uploaded by bingofalling AI-enhanced description Go to previous items Go to next items Download Save Save 中考培优专题经典讲义第2讲垂直平分线 For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save 中考培优专题经典讲义第2讲垂直平分线 For Later You are on page 1/ 12 Search Fullscreen 第 2 第第第第第第 1.垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂. PD 垂垂垂 AB 垂垂垂垂垂垂垂垂垂垂垂垂垂 PA 垂 PB 垂垂垂垂垂垂△ PAB 垂垂垂垂垂.垂垂垂垂垂 PA = PB 垂垂垂 P 垂 AB 垂垂垂垂垂垂垂. 第第第第第第第第 1 垂垂垂垂垂△ ABC 垂垂垂 D 垂 E 垂 F 垂垂垂 BC 垂 AB 垂 AC 垂. BD = CF 垂 BE = CD 垂 DG ⊥ EF 垂垂 G 垂垂 EG = FG .垂垂垂 AB = AC .垂垂垂垂垂垂 GD 垂 EF 垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂 DE 垂 DF 垂垂垂垂垂垂△ BDE 垂△ CFD 垂垂. 第第 2 垂垂垂垂垂 Rt △ ABC 垂垂∠ C =90°垂垂 D 垂 BC 垂垂垂 E 垂 AB 垂垂垂 DE ∥ AC 垂 AE =5垂 DE =2垂 DC =3垂垂垂 P 垂垂 A 垂垂垂垂垂 AC 垂垂垂 2 垂垂垂垂垂垂垂垂垂垂 C 垂垂垂垂垂垂垂垂垂 t 垂垂垂1垂垂垂 AC 垂垂= 垂垂2垂垂垂垂 EA 垂垂垂垂 Q 垂垂垂 ED = EQ 垂垂垂 DQ 垂 PE 垂垂 PE ⊥ DQ 垂垂垂垂 t 垂垂. adDownload to read ad-free 垂垂垂垂垂1垂 AC =6垂垂2垂垂 PE ⊥ DQ 垂垂垂垂 ED = EQ 垂垂垂 PE 垂垂垂垂 DQ 垂垂垂垂垂 PD 垂 PQ 垂垂垂垂 PD = PQ 垂垂垂垂 AP =2 t 垂垂垂 PC =6-2 t 垂 CD =3垂 EQ =2垂垂垂 AQ =3垂垂垂垂垂垂垂 Rt △ PCD 垂垂 PD 2 =3 2 +垂6-2 t 垂 2 垂垂 Rt △ PQF 垂垂 PQ 2 =垂垂 3 2 +垂6-2 t 垂 2 = 垂垂垂 .垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂 PE ⊥ DQ 垂垂垂垂 PE 垂垂∠ DEA 垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂+垂垂垂垂=垂垂垂垂垂垂垂垂△ AEP 垂垂垂垂垂垂垂垂垂 AP = AE =5垂垂 2 t =5垂 t = . 第第第第第第 1垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂 A .垂垂 B .垂垂 C .垂垂 D .垂垂 2垂垂△ AOB 垂垂垂垂垂垂 P 垂垂 P 垂 P 1 垂垂 OA 垂垂垂垂 P 垂 P 2 垂垂 BO 垂垂垂①垂△ OP 1 P 2 垂垂垂 A .垂垂垂垂垂 B .垂垂垂垂垂 C .垂垂垂垂垂 D .垂垂垂垂垂 ② 垂∠ AOB 垂垂垂垂垂垂垂垂△ OP 1 P 2 垂垂垂垂垂垂垂 3垂垂垂垂△ ABC 垂垂 AB 垂 AC 垂垂垂垂垂垂垂 BC 垂 D 垂 E 垂垂1垂垂∠ BAC =100°垂垂∠ DAE = 垂垂2垂垂∠ BAC =80°垂垂∠ DAE = 垂垂3垂垂∠ DAE =10°垂垂∠ BAC = 垂垂4垂垂△ ABC 垂垂垂垂 20垂△ ADE 垂垂垂垂 12垂垂 AB + AC = 垂垂5垂垂 AB = AC 垂垂∠ BAC =120°垂垂△ ADE 垂垂垂垂垂垂垂垂垂 adDownload to read ad-free 4垂垂垂垂垂垂△ ABC 垂垂垂垂 3垂 BO 垂 CO 垂垂垂∠ ABC 垂∠ ACB 垂垂垂垂垂垂 BO 垂 CO 垂垂垂垂垂垂垂 BC 垂 E 垂 F 垂垂 EF 垂垂垂.5垂垂垂垂垂垂垂垂△ ABC 垂 AB = BC =5垂 AC = 垂垂 BC 垂垂垂垂垂垂 P 垂垂垂垂垂垂 AB 垂垂垂垂垂垂垂垂垂 BP 垂垂垂 .6垂垂垂垂垂垂垂垂 AD 垂△ ABC 垂垂垂垂垂垂 DE ⊥ AB 垂 DF ⊥ AC 垂垂垂垂垂垂 E 垂 F .垂垂垂 AD 垂垂垂垂 EF .7垂△ ABC 垂垂 D 垂 BC 垂垂垂 DE ⊥ BC 垂垂∠ BAC 垂垂垂垂垂垂 E 垂 EF ⊥ AB 垂 F 垂 EG ⊥ AC 垂 G .垂垂垂 BF = CG .8垂垂垂垂△ ABC 垂垂垂 D 垂 BC 垂垂垂 AD 垂垂垂垂垂垂 EF 垂 BC 垂垂垂垂垂 F 垂垂∠ FAC =∠ B 垂垂垂垂 AD 垂垂∠ BAC .9垂垂垂垂垂△ ABC 垂垂 AB = AC 垂 D 垂垂垂垂垂垂垂垂垂△ DBC 垂垂垂垂垂垂.垂1垂垂垂垂垂垂 AD 垂垂垂垂 BC 垂垂2垂垂 AB 垂垂垂垂垂 AB 垂垂垂垂垂垂△ ABE 垂垂垂 DE 垂垂垂垂垂 DA 垂 DB 垂 DE 垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂. adDownload to read ad-free 10垂垂垂垂垂垂垂 y = ax 2 +2 ax + c 垂垂垂垂 x 垂垂垂垂垂 A 垂 B 垂垂垂垂 A 垂垂 B 垂垂垂垂垂垂 y 垂垂垂垂 C 垂垂垂垂 P 垂垂 C 垂0垂2垂垂 BC 垂垂垂垂垂垂垂垂 A 垂垂垂垂垂垂垂垂垂垂垂垂.11垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂垂 y = x +4 垂 x 垂垂垂垂 A 垂垂 y 垂垂垂垂 B 垂垂 P 垂垂 O 垂垂垂 OA 垂垂垂 1 垂垂垂垂垂垂垂垂垂 A 垂垂垂垂垂垂垂垂 A 垂垂垂垂垂垂垂垂垂垂 AO 垂垂垂垂 Q 垂 A 垂垂垂 AB 垂垂垂 1 垂垂垂垂垂垂垂垂垂 B 垂垂垂垂垂垂垂 P 垂 Q 垂垂垂垂 DE 垂垂垂垂垂垂 PQ 垂垂垂 PQ 垂垂 D 垂垂垂垂 QB - BO - OP 垂垂 E .垂 P 垂 Q 垂垂垂垂垂垂垂 Q 垂垂垂 B 垂垂垂垂垂垂垂 P 垂垂垂垂垂垂垂垂 P 垂 Q 垂垂垂垂垂垂 t 垂垂 t 0垂.垂1垂垂 Q 垂垂垂垂垂垂垂垂垂垂 t 垂垂垂垂垂垂垂垂垂2垂垂 t 垂垂垂垂垂垂垂 DE 垂垂垂 O .12垂垂垂 1垂垂垂垂 ABCD 垂垂 AB =4垂 BC =3垂垂 E 垂垂垂 CD 垂垂垂垂垂垂垂垂△ BCE 垂 BE 垂垂垂垂 C 垂垂垂垂垂 F .垂1垂垂垂 F 垂垂垂垂垂垂 AD 垂垂垂垂垂垂垂垂垂垂垂垂 CE 垂垂垂垂2垂垂垂 F 垂垂垂垂垂垂 AB 垂垂垂垂垂垂垂垂垂垂垂垂 CE 垂垂垂垂3垂垂垂垂 AF 垂垂垂 CD 垂垂 G 垂垂垂垂垂垂垂 CG 垂垂垂垂 .13垂垂垂垂垂垂垂垂垂垂垂垂 x 垂垂垂垂垂 A 垂-3垂0垂垂 B 垂-1垂0垂垂垂 y 垂垂垂垂垂 C 垂0垂3垂垂垂 P 垂垂垂垂垂垂垂垂垂垂 Q 垂垂垂垂垂4垂0垂.垂1垂垂垂垂垂垂垂垂垂垂垂垂垂2垂垂 OP // CQ 垂垂垂垂 P 垂垂垂垂垂3垂垂 M 垂 N 垂垂垂垂垂 AQ 垂 CQ 垂垂垂 M 垂垂垂 3 垂垂垂垂垂垂垂垂垂垂 A 垂垂 Q 垂垂垂垂垂垂垂 N 垂垂垂 1 垂垂垂垂垂垂垂垂垂垂 C 垂垂 Q 垂垂垂垂垂 M 垂 N 垂垂垂垂垂垂 Q 垂垂垂垂垂垂垂垂垂垂垂.垂垂垂垂垂垂 t 垂垂垂垂垂 PQ 垂垂垂垂垂垂 MN 垂垂垂垂垂垂垂 t 垂垂垂垂 P 垂垂垂. adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like 中考培优专题经典讲义第22讲构造圆问题 100% (1) 中考培优专题经典讲义第22讲构造圆问题 32 pages 圆的综合压轴题（学生版） 100% (1) 圆的综合压轴题（学生版） 18 pages 题型突破 (07) 二次函数与几何综合类问题 100% (1) 题型突破 (07) 二次函数与几何综合类问题 135 pages 专题5.17 作轴对称图形-将军饮马问题（基础篇）-（北师大版） No ratings yet 专题5.17 作轴对称图形-将军饮马问题（基础篇）-（北师大版） 32 pages GB 50017-2017 钢结构设计标准 No ratings yet GB 50017-2017 钢结构设计标准 534 pages 改斜归正秒杀直角类压轴题（全国通用）（学生版） No ratings yet 改斜归正秒杀直角类压轴题（全国通用）（学生版） 8 pages 上海初二第二学期课内压轴题（17） No ratings yet 上海初二第二学期课内压轴题（17） 9 pages 高思維數學訓練 (升級版) 解難策略工作紙答案 6下 No ratings yet 高思維數學訓練 (升級版) 解難策略工作紙答案 6下 2 pages 福建省厦门第一中学2023 2024学年高二上学期十二月月考数学试卷 No ratings yet 福建省厦门第一中学2023 2024学年高二上学期十二月月考数学试卷 12 pages 专题2二次函数与直角三角形问题（教师版含解析）挑战2023年中考数学压轴题之学霸秘笈大揭秘（全国通用） 100% (1) 专题2二次函数与直角三角形问题（教师版含解析）挑战2023年中考数学压轴题之学霸秘笈大揭秘（全国通用） 79 pages 2、初三数学抛物线与三角形、四边形专题学生版 100% (1) 2、初三数学抛物线与三角形、四边形专题学生版 16 pages 中考数学中的10大类辅助线问题全梳理 100% (1) 中考数学中的10大类辅助线问题全梳理 257 pages 专题3二次函数与等腰直角三角形问题（学生版）挑战2023年中考数学压轴题之学霸秘笈大揭秘（全国通用） 100% (1) 专题3二次函数与等腰直角三角形问题（学生版）挑战2023年中考数学压轴题之学霸秘笈大揭秘（全国通用） 12 pages 九上专题05 相似三角形中的动点问题（教师版） 100% (1) 九上专题05 相似三角形中的动点问题（教师版） 40 pages 中考培优专题经典讲义第1讲角平分线 100% (1) 中考培优专题经典讲义第1讲角平分线 12 pages 九年级数学中考复习第4章第1节相交线与平行线 (共45张ppt) 2 No ratings yet 九年级数学中考复习第4章第1节相交线与平行线 (共45张ppt) 2 45 pages 2020 2021学年北京各区初二下学期期末数学压轴题汇编 100% (1) 2020 2021学年北京各区初二下学期期末数学压轴题汇编 7 pages 李永乐高中数学 6解析几何 - 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13805	https://math.stackexchange.com/questions/1567346/how-to-take-an-integral-using-half-angle-trigonometric-substitution	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How to take an integral using half angle trigonometric substitution. Ask Question Asked Modified 9 years, 9 months ago Viewed 3k times 2 $\begingroup$ So i have this question which is asking to take the integral using a predefined trigonometric substitution which is $$u=\tan\frac{x}{2}$$ and the integral equation is $$\int\frac{\sin x\ dx}{(6\cos x-2)(3-2\sin x)}$$ How would i go on about this problem? Because to begin with i do not know how i would even use the given substitution method. Any help is appreciated thank you. calculus integration trigonometry Share edited Dec 9, 2015 at 11:32 Isura Manchanayake 64744 silver badges1515 bronze badges asked Dec 9, 2015 at 10:37 Carlos VCarlos V 46811 gold badge55 silver badges1313 bronze badges $\endgroup$ Add a comment \| 4 Answers 4 Reset to default 1 $\begingroup$ This substitution is used for integrals involving only trigonometric expressions. This method is very useful as it transforms the trigonometric integral into just rational integral. You should know how to write $\sin x, \cos x, \tan x$ in terms of $\tan \frac{x}{2}$ (try proving) $\sin x=\dfrac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$ $\cos x=\dfrac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$ $\tan x=\dfrac{2\tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$ To find $dx$ in terms of $du$, differentiate both sides. $u = \tan \dfrac{x}{2}$ $du = \sec^2 \dfrac{x}{2} \dfrac{1}{2} dx = (1+\tan^2 \dfrac{x}{2})\dfrac{1}{2} dx=(1+u^2)\dfrac{1}{2}dx \Rightarrow dx=\dfrac{2}{1+u^2}du$ For this example, The integral is converted to, $$\int\dfrac{\sin x}{(6cosx-2)(3-2sinx)}dx=\int\dfrac{\dfrac{2u}{1+u^2}}{(6\dfrac{1-u^2}{1+u^2}-2 )(3-2\dfrac{2u}{1+u^2})}\dfrac{2}{1+u^2}du$$ $$=\int\dfrac{4u}{(1+u^2)(6-6u^2-2-2u^2)(3+3u^2-4u)}du$$ $$=\int\dfrac{-4u}{(1+u^2)(8u^2-4)(3u^2-4u+3)}du$$ $$=\int\dfrac{3u^2-4u+3-(3u^2+3)}{(1+u^2)(8u^2-4)(3u^2-4u+3)}du$$ $$=\int\dfrac{du}{(1+u^2)(8u^2-4)}-\int\dfrac{3\ du}{(8u^2-4)(3u^2-4u+3)}$$ and integrate using partial fractions and substitute again $u=\tan \dfrac{x}{2}$. Look how nicely the trigonometric integral(which made no sense) is transformed to a rational integral(which can be solved at least by brute force). Try more examples. ( You will get to realize that some problems looks like can be solved using half angle substitutions but really can be solved without substitution. Example $$\int\dfrac{\sin x}{(\cos x -2)(2\cos x +3)}dx$$ ) Share edited Dec 9, 2015 at 11:28 answered Dec 9, 2015 at 11:20 Isura ManchanayakeIsura Manchanayake 64744 silver badges1515 bronze badges $\endgroup$ Add a comment \| 1 $\begingroup$ Consider a right triangle that has a side opposite angle $\frac{x}{2}$ of length $u=\tan\frac{x}{2}$ and an adjacent side of length 1. Obviously, $\tan \frac{x}{2} =\frac{opposite}{adjacent}= \frac{\tan\frac{x}{2}}{1}=\tan\frac{x}{2}=u$. What would the Hypotenuse be? Having done that, its a simple matter to find expressions for $\cos\frac{x}{2}$ and $\sin\frac{x}{2}$ (using the same triangle). Noting that $$\cos 2t=\cos^2 t - \sin^2t$$ and that $$\sin2t = 2\sin t\cos t$$ we can make the substitution $t=\frac{x}{2}$ to find expressions (in terms of $u$) for $\cos x$ and $\sin x$. Lastly we note that $$\frac{du}{dx}=\frac{d}{dx}\tan\frac{x}{2} = \frac{\sec^2\frac{x}{2}}{2}=\frac{\tan^2\frac{x}{2}+1}{2}=\frac{u^2+1}{2}$$ Putting all this information together yourself (as opposed to just parroting formulas) will give you a superior understanding of this substitution method. Share edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Dec 9, 2015 at 15:29 John JoyJohn Joy 8,07811 gold badge2525 silver badges3131 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ We have $$x=2arctan(u)$$, that gives $$dx=\frac{2}{u^2+1}du$$ Furthermore, we have $$cos(x)=\frac{1-u^2}{1+u^2}$$ and $$sin(x)=\frac{2u}{u^2+1}$$ Inserting those terms, you get a rational function of $u$. Try it from here. Share answered Dec 9, 2015 at 10:54 PeterPeter 86.8k1717 gold badges8787 silver badges238238 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ By letting $u=\tan\frac{x}2$, we get $$\int\frac{\sin x}{(6\cos x-2)(3-2\sin x)}d x=\int\frac{u d u}{(1-2u^2)(3u^2-4u+3)}\ =\frac{8}{49}\int\frac{(9u+4)du}{1-2u^2}+\frac{12}{49}\int\frac{(9u-8)du}{3u^2-4u+3},$$ the later two are both integration of rational functions. $$\int\frac{(9u+4)du}{1-2u^2}=9\int\frac{udu}{1-2u^2}+4\int\frac{du}{1-2u^2}\ =-\frac94\int\frac{d(1-2u^2)}{1-2u^2}+\sqrt2\int\left(\frac{1}{1-\sqrt2u}+\frac{1}{1-\sqrt2u}\right)du\ =-\frac94\ln\|1-2u^2\|+\sqrt2\ln\frac{1+\sqrt2u}{1-\sqrt2u}+C_1$$ $$\int\frac{(9u-8)du}{3u^2-4u+3}=\frac13\int\frac{9\left(u-\frac23\right)-2} {\left(u-\frac23\right)^2+\frac59}du=3\int\frac{\left(u-\frac23\right)} {\left(u-\frac23\right)^2+\frac59}du+\frac23\int\frac{du} {\left(u-\frac23\right)^2+\frac59}\ =\frac32\int\frac{d\left(u-\frac23\right)^2}{\left(u-\frac23\right)^2+\frac59} +\frac2{\sqrt5}\int\frac{d\frac3{\sqrt5}\left(u-\frac23\right)}{\left[\frac3{\sqrt5}\left(u-\frac23\right)\right]^2+1}\ =\frac32\ln\left(u^2-\frac43u+1\right)-\frac2{\sqrt5}\arctan\left[\frac3{\sqrt5}\left(u-\frac23\right)\right]+C_2$$ Share edited Dec 9, 2015 at 14:58 answered Dec 9, 2015 at 11:38 ybtang21cybtang21c 47122 silver badges99 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus integration trigonometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related Trigonometric substitution for $\int\frac{1}{x^2\sqrt{4-x^2}}dx$ 0 Integral of trigonometric function using substitution Integral of trig fraction using substitution? 1 Integral using trigonometric substitution 0 About Weierstrass / Tangent half-angle substitution Substitution for Trig Integral - GRE Math Subject Test 3 trigonometric integral with tangent 4 Evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution 3 Introducing a sine half-angle substitution 3 How to evaluate this indefinite integral by only using trigonometric substitution? 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Tools & Reference>Clinical Procedures Cricothyroidotomy (Cricothyrotomy) Updated: Dec 13, 2023 Author: Joshua E Markowitz, MD, FACEP, RDMS, SAIUM; Chief Editor: Zab Mosenifar, MD, FACP, FCCP more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Cricothyroidotomy (Cricothyrotomy) Sections Cricothyroidotomy (Cricothyrotomy) Overview Background Indications Contraindications Outcomes Show All Periprocedural Care Equipment Patient Preparation Show All Technique Cricothyroidotomy (Cricothyrotomy) Complications Show All Media Gallery;) References;) Overview Background An emergency surgical airway can be obtained by using one of several different methods, including the following: Open cricothyroidotomy Needle cricothyroidotomy with jet oxygenation Percutaneous cricothyroidotomy using the Seldinger technique A 2002 study of the National Emergency Airway Registry database found that only 0.56% (43 of 7712) of intubations required cricothyroidotomy (cricothyrotomy). [2, 3, 4] This percentage may be driven even lower by increasing adoption of rapid sequence intubation techniques, increased use of video-assisted intubation and other "difficult airway" devices, and increased prevalence of residency-trained emergency practitioners. [5, 6] However, some patients still require a surgical airway. [7, 8, 9, 10, 11] Some evidence has supported a trend toward open cricothyroidotomy and away from needle-based approaches, at least in settings where cricothyroidotomy is not frequently performed. [12, 13] Next: Indications Adults A cricothyroidotomy is indicated when a patients airway cannot be secured by using nonsurgical methods or when other devices or rescue techniques (ie, intubating laryngeal mask airway, fiberoptic scope, or lighted stylet) have failed or are unavailable. It is also indicated when an airway is required immediately in a patient who is not a candidate for orotracheal or nasotracheal intubation (see the video below)for instance, in the case of severe facial trauma. Fiberoptic-assisted tracheal intubation. Video courtesy of Therese Canares, MD, and Jonathan Valente, MD, Rhode Island Hospital, Brown University. View Media Gallery) In emergency settings, cricothyroidomy appears to give rise to fewer late complications than tracheostomy; however, current evidence is insufficient to support the use of emergency cricothyroidotomies as long-term airways, and thus, it is advisable to convert these cricothyroidotomies to tracheostomies in a timely manner. Children For children younger than 12 years, needle cricothyroidotomy with percutaneous transtracheal (jet) ventilation is the surgical airway of choice. Because a childs larynx and cricoid cartilage are very soft, mobile, and pliable, a surgical cricothyroidotomy is difficult in this setting. Previous Next: Contraindications An absolute contraindication for the creation of an emergency surgical airway is as follows: Patient younger than 12 years, unless the child is of teenage or adult size Relative contraindications for the creation of an emergency surgical airway include the following: Airway obstruction distal enough to the cricoid membrane that a cricothyroidotomy would not provide a secure airway with which to ventilate the patient Presence of a SHORT neck (ie, Surgery [history of prior neck surgery], Hematoma, Obesity, Radiation [evidence of radiation therapy], or Trauma/burns), making it difficult to locate the patients anatomic landmarks or causing an increased risk of further complications Tumor, infection, or abscess at the incision site Lack of operator expertise Previous Next: Outcomes Very little literature is available comparing surgical cricothyroidotomy with needle cricothyroidotomy. Some comparisons have been made between the use of a cricothyrotome kit and the rapid four-step surgical cricothyroidotomy, but they have demonstrated no significant difference in outcomes or complications. Kwon et al assessed the incidence and outcomes of cricothyroidotomy in a "cannot intubate, cannot oxygenate" (CICO) situation. A total of 10,187 tracheal intubations were attempted, and 23 patients received cricothyroidotomy (22 in the emergency department [ED] and one in the endoscopy room). The survival rate at hospital discharge was 47.8% (11/23). Aside from cases of cardiac arrest at admission, the survival rate was 62.5% (10/16). Cricothyroidotomy was successful in 17 patients (73.9%), nine (52.9%) of whom survived; it failed in six (26.1%), two (33.3%) of whom survived. After failed cricothyroidotomy, airways were secured via tracheal intubation, nasotracheal intubation, or tracheostomy. A systematic review and meta-analysis (32 studies) by Duan et al compared scalpel cricothyroidotomy (SCT) with puncture cricothyroidotomy (PCT) in the context of CICO. Overall success rate, first-time success rate after training, and time taken to perform the procedure were the primary outcomes, with complications a secondary outcome. SCT and PCT were not found to differ significantly with respect to overall success rate, first-time success rate, and complications; however, SCT had a significant advantage in terms of required procedural time. Previous Periprocedure References McKenna P, Desai NM, Tariq A, Morley EJ. Cricothyrotomy. Treasure Island, FL: StatPearls; 2023. [Full Text]. Bair AE, Filbin MR, Kulkarni RG, Walls RM. The failed intubation attempt in the emergency department: analysis of prevalence, rescue techniques, and personnel. J Emerg Med. 2002 Aug. 23 (2):131-40. [QxMD MEDLINE Link]. Siddiqui N, Arzola C, Friedman Z, Guerina L, You-Ten KE. Ultrasound Improves Cricothyrotomy Success in Cadavers with Poorly Defined Neck Anatomy: A Randomized Control Trial. Anesthesiology. 2015 Nov. 123 (5):1033-41. [QxMD MEDLINE Link]. Akulian JA, Yarmus L, Feller-Kopman D. The role of cricothyrotomy, tracheostomy, and percutaneous tracheostomy in airway management. Anesthesiol Clin. 2015 Jun. 33 (2):357-67. [QxMD MEDLINE Link]. Friedman Z, You-Ten KE, Bould MD, Naik V. Teaching lifesaving procedures: the impact of model fidelity on acquisition and transfer of cricothyrotomy skills to performance on cadavers. Anesth Analg. 2008 Nov. 107 (5):1663-9. [QxMD MEDLINE Link]. Turner JS, Stewart LK, Hybarger AC, Ellender TJ, Stepsis TM, Bartkus EA, et al. An investigation into emergency medicine resident cricothyrotomy competency: Is three the magic number?. AEM Educ Train. 2023 Dec. 7 (6):e10917. [QxMD MEDLINE Link]. [Guideline] Apfelbaum JL, Hagberg CA, Connis RT, Abdelmalak BB, Agarkar M, Dutton RP, et al. 2022 American Society of Anesthesiologists Practice Guidelines for Management of the Difficult Airway. Anesthesiology. 2022 Jan 1. 136 (1):31-81. [QxMD MEDLINE Link]. [Full Text]. [Guideline] DAS guidelines for management of unanticipated difficult intubation in adults 2015. Difficult Airway Society. Available at 2015; Accessed: November 29, 2023. McIntosh SE, Swanson ER, Barton ED. Cricothyrotomy in air medical transport. J Trauma. 2008 Jun. 64 (6):1543-7. [QxMD MEDLINE Link]. Liess BD, Scheidt TD, Templer JW. The difficult airway. Otolaryngol Clin North Am. 2008 Jun. 41 (3):567-80, ix. [QxMD MEDLINE Link]. Zhang X, Wang X, Gu Q, Zhang J, Wu R, Zhuang Q, et al. Evaluation of alternative airway management strategies in surgical repair of severe cranio-maxillofacial trauma. Minerva Chir. 2016 Dec. 71 (6):372-376. [QxMD MEDLINE Link]. Heymans F, Feigl G, Graber S, Courvoisier DS, Weber KM, Dulguerov P. Emergency Cricothyrotomy Performed by Surgical Airway-naive Medical Personnel: A Randomized Crossover Study in Cadavers Comparing Three Commonly Used Techniques. Anesthesiology. 2016 Aug. 125 (2):295-303. [QxMD MEDLINE Link]. [Full Text]. Mallows JL, Tyler PA. Randomized controlled trial comparing an open surgical technique and a Seldinger technique for cricothyrotomy performed on a simulated airway. AEM Educ Train. 2021 Aug. 5 (4):e10699. [QxMD MEDLINE Link]. Verschueren DS, Bell RB, Bagheri SC, Dierks EJ, Potter BE. Management of laryngo-tracheal injuries associated with craniomaxillofacial trauma. J Oral Maxillofac Surg. 2006 Feb. 64 (2):203-14. [QxMD MEDLINE Link]. Zasso FB, You-Ten KE, Ryu M, Losyeva K, Tanwani J, Siddiqui N. Complications of cricothyroidotomy versus tracheostomy in emergency surgical airway management: a systematic review. BMC Anesthesiol. 2020 Aug 27. 20 (1):216. [QxMD MEDLINE Link]. [Full Text]. Kwon YS, Lee CA, Park S, Ha SO, Sim YS, Baek MS. Incidence and outcomes of cricothyrotomy in the "cannot intubate, cannot oxygenate" situation. Medicine (Baltimore). 2019 Oct. 98 (42):e17713. [QxMD MEDLINE Link]. [Full Text]. Duan Q, Yang D, Gao H, Liu Q, Zhi J, Xu J, et al. Scalpel cricothyrotomy versus punctured cricothyrotomy in the context of the CICO crisis. A systematic review and Meta-analysis. Anaesth Crit Care Pain Med. 2023 Aug. 42 (4):101211. [QxMD MEDLINE Link]. [Full Text]. Lin J, Bellinger R, Shedd A, Wolfshohl J, Walker J, Healy J, et al. Point-of-Care Ultrasound in Airway Evaluation and Management: A Comprehensive Review. Diagnostics (Basel). 2023 Apr 25. 13 (9):[QxMD MEDLINE Link]. [Full Text]. Barkhuysen R, Merkx MA, van Damme PA, Buyne OR, van den Hoogen FJ. Acute upper airway failure and mediastinal emphysema following a wire-guided percutaneous cricothyrotomy in a patient with severe maxillofacial trauma. Oral Maxillofac Surg. 2008 May. 12 (1):35-8. [QxMD MEDLINE Link]. DeVore EK, Redmann A, Howell R, Khosla S. Best practices for emergency surgical airway: A systematic review. Laryngoscope Investig Otolaryngol. 2019 Dec. 4 (6):602-608. [QxMD MEDLINE Link]. [Full Text]. Moroco AE, Armen SB, Goldenberg D. Emergency Cricothyrotomy: A 10-Year Single Institution Experience. Am Surg. 2023 Apr. 89 (4):1243-1246. [QxMD MEDLINE Link]. Asai T. Emergency Cricothyrotomy: Toward a Safer and More Reliable Rescue Method in "Cannot Intubate, Cannot Oxygenate" Situation. Anesthesiology. 2015 Nov. 123 (5):995-6. [QxMD MEDLINE Link]. [Full Text]. Media Gallery Fiberoptic-assisted tracheal intubation. Video courtesy of Therese Canares, MD, and Jonathan Valente, MD, Rhode Island Hospital, Brown University. Surgical tracheostomy procedure. Video courtesy of Gauri Mankekar, MBBS, MS, PhD. Cricothyroidotomy (Seldinger technique). Video courtesy of Therese Canares, MD, and Jonathan Valente, MD, Rhode Island Hospital, Brown University. of 3 Tables Back to List Contributor Information and Disclosures Author Joshua E Markowitz, MD, FACEP, RDMS, SAIUM Senior Attending Physician, Department of Emergency Medicine, Emergency Medical Services Liaison, Assistant Lead for Disaster Preparedness, The Permanente Medical Group, Santa Clara Medical Group; Clinical Assistant Professor (Affiliate), Department of Emergency Medicine, Stanford University School of Medicine Joshua E Markowitz, MD, FACEP, RDMS, SAIUM is a member of the following medical societies: American College of Emergency Physicians, American Institute of Ultrasound in MedicineDisclosure: Nothing to disclose. Specialty Editor Board Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Nothing to disclose. Chief Editor Zab Mosenifar, MD, FACP, FCCP Geri and Richard Brawerman Chair in Pulmonary and Critical Care Medicine, Professor and Executive Vice Chairman, Department of Medicine, Medical Director, Women's Guild Lung Institute, Cedars Sinai Medical Center, University of California, Los Angeles, David Geffen School of MedicineZab Mosenifar, MD, FACP, FCCP is a member of the following medical societies: American College of Chest Physicians, American College of Physicians, American Federation for Medical Research, American Thoracic SocietyDisclosure: Nothing to disclose. Additional Contributors Michael R Filbin, MD, FACEP Clinical Instructor, Department of Emergency Medicine, Massachusetts General HospitalMichael R Filbin, MD, FACEP is a member of the following medical societies: American College of Emergency Physicians, Massachusetts Medical Society, Society for Academic Emergency MedicineDisclosure: Nothing to disclose. Acknowledgements The authors and editors of Medscape Drugs & Diseases gratefully acknowledge the assistance of Lars Grimm with the literature review and referencing for this article. Medscape Drugs & Diseases thanks Gauri Mankekar, MBBS, MS, PhD, Consulting Surgeon, Department of Otolaryngology, PD Hinduja National Hospital, India, for the video contribution to this article. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections Cricothyroidotomy (Cricothyrotomy) Overview Background Indications Contraindications Outcomes Show All Periprocedural Care Equipment Patient Preparation Show All Technique Cricothyroidotomy (Cricothyrotomy) Complications Show All Media Gallery;) References;) encoded search term (Cricothyroidotomy (Cricothyrotomy)) and Cricothyroidotomy (Cricothyrotomy) What to Read Next on Medscape Related Conditions and Diseases Cricothyroidotomy (Cricothyrotomy) Cricothyroidotomy Percutaneous Transtracheal Jet Ventilation Tracheostomy Epiglottitis Hanging Injuries and Strangulation DELETE - Facial Fracture Management in the Emergency Department News & Perspective Complications of Cricothyroidotomy Versus Tracheostomy in Emergency Surgical Airway Management Assessing Anesthesiology Residents' Out-of-the-Operating-Room (OOOR) Emergent Airway Management Drug Interaction Checker Pill Identifier Calculators Formulary 2002 80241-overview Procedures Procedures Cricothyroidotomy (Cricothyrotomy) 2002 1830008-overview Procedures Procedures Cricothyroidotomy 2002 1413327-overview Procedures Procedures Percutaneous Transtracheal Jet Ventilation
13807	https://mathematica.stackexchange.com/questions/105352/imaginary-part-of-complex-number-nothing-works	simplifying expressions - Imaginary part of complex number, nothing works? - Mathematica Stack Exchange Join Mathematica By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Imaginary part of complex number, nothing works? Ask Question Asked 9 years, 8 months ago Modified9 years, 8 months ago Viewed 855 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I want to find the imaginary part of the complex number (a+ib). say I have a function in the following form: mathematica ((-2 + 4 I) Cos[0.0628319 z] + (2 - I) Sin[0.0628319 z])/ ((1 + 2 I) Cos[0.0628319 z] + (4 + 2 I) Sin[0.0628319 z]) where z z is a real number variable. I want to reform the function into (a+ib) form. attempt: if I do Im, then it wil give me the following output: mathematica Im[((-2 + 4 I) Cos[0.0628319 z] + (2 - I) Sin[ 0.0628319 z])/((1 + 2 I) Cos[0.0628319 z] + (4 + 2 I) Sin[ 0.0628319 z])] but this is not what I want. simplifying-expressions complex Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Jan 31, 2016 at 22:01 MarcoB 68k 19 19 gold badges 98 98 silver badges 205 205 bronze badges asked Jan 31, 2016 at 21:49 randyrandy 121 2 2 bronze badges 2 1 Consider tugging // ComplexExpand // Simplify on the end.kirma –kirma 2016-01-31 21:53:12 +00:00 Commented Jan 31, 2016 at 21:53 4 If you know that z is a real number, then do ComplexExpand[Im[ ... ]].march –march 2016-01-31 21:53:22 +00:00 Commented Jan 31, 2016 at 21:53 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. You're not tryng to get the imaginary part of a number. You are trying to get the imaginary part of an expression. Let's say we define your expression to be equivalent to eq: mathematica eq==(2 + 4 I) Cos[0.0628319 z] + (2 - I) Sin[0.0628319 z])/ ((1 + 2 I) Cos[0.0628319 z] + (4 + 2 I) Sin[0.0628319 z]) Solving the above for z gives 4 solutions. Lets say we take one of them and try a full simplify on it. We would get a fraction whose denominator has a real and an imaginary part. (I'm working via the cloud so I'll paste the result as InputForm: mathematica ArcCos[((5.322525763080274 - 9.01827480722634I) - (20.816270607258474 - 2.3058885477431663I)eq)/ Sqrt[(159. + 288.I) + eq((-392. + 16.I) + 481.eq)]] This expression cannot be split between the real and imaginary parts because they are intertwined. Mathematica would give you a nice z=a+bI but it can't. Therefore it uses Im[...] because it cannot decompose it any further. Try assigning value ranges for eq and finding the limits on z for the different values. Do bear in mind that eq is by definition a complex number too. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Feb 1, 2016 at 6:33 Mr.Wizard 275k 34 34 gold badges 604 604 silver badges 1.5k 1.5k bronze badges answered Feb 1, 2016 at 6:15 ManuelManuel 91 3 3 bronze badges 1 Please don't forget to format your code blocks in markdown; see editing help for details.Mr.Wizard –Mr.Wizard 2016-02-01 06:33:51 +00:00 Commented Feb 1, 2016 at 6:33 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. ```mathematica func[z_] := ((-2 + 4 I) Cos[0.0628319 z] + (2 - I) Sin[ 0.0628319 z])/((1 + 2 I) Cos[0.0628319 z] + (4 + 2 I) Sin[ 0.0628319 z]) ComplexExpand@func Show[ Plot[ Im@func[z], {z, 0, 100}, PlotStyle -> Blue ], Plot[ Re@func[z], {z, 0, 100}, PlotStyle -> Red ], PlotRange -> All ] (1.2 + 1.6 I) ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Feb 1, 2016 at 7:30 answered Feb 1, 2016 at 6:38 Baran CimenBaran Cimen 1,194 1 1 gold badge 8 8 silver badges 19 19 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. 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13808	https://www.open.edu/openlearn/science-maths-technology/what-chemical-compounds-might-be-present-drinking-water/content-section-3.2.2	Skip to main content The Open University Accessibility hub Guest user / Sign out Study with The Open University Accessibility hub Close OpenLearn will be unavailable from 8am to 10.30am on Wednesday 3 September due to scheduled maintenance. Close OpenLearn will be unavailable from 8am to 10.30am on Wednesday 3 September due to scheduled maintenance. Create account / Sign in The Open University Accessibility hub Guest user / Sign out Study with The Open University Accessibility hub Close OpenLearn will be unavailable from 8am to 10.30am on Wednesday 3 September due to scheduled maintenance. Menu My OpenLearn Profile Personalise your OpenLearn profile, save your favourite content and get recognition for your learning Create account / Sign in About this free course Become an OU student Download this course Share this free course Course content Course content What chemical compounds might be present in drinking water? Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available. Create account / Sign inMore free courses 3.2.2 Nomenclature of oxoacids Traditionally oxoacid formulas are written with hydrogen first, which conceals the fact that hydrogen is often bonded to oxygen. The rarely used formula (OH)3PO is a more meaningful formulation of phosphoric acid than H3PO4. The nomenclature is further complicated by the fact that the inorganic acids and their organic derivatives also have different common names; for instance, phosphorous acid in inorganic chemistry becomes phosphonic acid for its organic derivatives. To help you through this topic the prefixes and suffixes are in bold italics and the most common name is always used. Historically, where the central element forms oxoacids in two oxidation numbers: the higher state is indicated by the suffix -ic the lower state is indicated by the suffix -ous. For example phosphoric acid, H3PO4, and phosphorous acid, H3PO3, have phosphorus oxidation numbers +5 and +3 respectively. If more than two oxidation numbers are involved, the prefixes per- and hypo- are used as well: per- denotes the highest oxidation number hypo- denotes the lowest oxidation number. For example the oxoacids of chlorine are shown in Table 4. Table 4 The oxoacids of chlorine. \| Formula \| Oxidation number \| Name \| \| HClO4 \| +7 \| perchloric \| \| HClO3 \| +5 \| chloric \| \| HClO2 \| +3 \| chlorous \| \| HClO \| +1 \| hypochlorous \| Oxoanions derived from -ic acids are given the ending -ate and from -ous acids are given the ending -ite. To practise, name the following anions OCl−, ClO4− and NO2−. The ions are hypochlorite, perchlorate and nitrite. Condensed forms of oxoacids are also distinguished by prefixes: ortho- refers to the 'monomeric', or most highly hydroxylated, form meta- refers to the 'polymeric', or least highly hydroxylated, form di- and tri- in this context logically refer to 'dimers' or 'trimers'. The oxoanions follow this labelling convention, for example orthophosphates, PO43−, diphosphates, P2O74−, and triphosphates, P3O105−. Additionally, the prefix cyclo- or catena- distinguish cyclic from linear condensed anions, respectively. Previous 3.2.1 Polyacids Next 3.2.3 Prediction of formulas Share on Facebook Share on Twitter Share on LinkedIn Share via Email
13809	https://www.scirp.org/journal/paperinformation?paperid=134184	Published Time: 2024-06-28T00:00:00+00:00 Newton’s Law of Universal Gravitation Explained by the Theory of Informatons Login Login切换导航 Home Articles Journals Books News About Services Submit Home Journals Article Journal of High Energy Physics, Gravitation and Cosmology>Vol.10 No.3, July 2024 Newton’s Law of Universal Gravitation Explained by the Theory of Informatons () Antoine Acke Retired Professor Kaho Sint-Lieven, Now Faculty of Engineering Technology, KU Leuven, Ghent Campus, Gent, Belgium. DOI:10.4236/jhepgc.2024.103056PDFHTMLXML 206 Downloads 1,239 ViewsCitations Abstract In the context of classical physics, Newton’s law of universal gravitation describes the attraction between two mass particles separated in space. In the same context a vector field E g E g , that is not associated with anything substantial, has been introduced as the entity that mediates in the gravitational interactions. In this article, we will show that E g E g is the mathematical quantity that—at the macroscopic level—fully characterizes the medium that makes the interaction between particles at rest possible. We identify that medium as “the gravitational field”. To define the nature of the gravitational field, we will start from the hypothesis that a material object manifests itself in space by the emission—at a rate proportional to its rest mass—of mass and energy less granular entities that—relative to an inertial reference frame—are rushing away with the speed of light and that are carriers of information referring to the position of their emitter (“g-information”). Because they transport nothing else than information, we call these entities “informatons”. We will show that the expanding cloud of g-information created by the continuous emission of informatons by a mass particle at rest can be fully characterized by the vector field E g E g , which implies that that cloud can be identified as the gravitational field of the particle. We will also show that the gravitational interaction between mass particles can be explained as the response of a particle to the disturbance of the symmetry of its “proper” gravitational field by the field that, in its direct vicinity, is created and maintained by other mass particles. Keywords Gravity, Gravitational Field, Gravitational Interaction, Informatons Share and Cite: FacebookTwitterLinkedInSina WeiboShare Acke, A. (2024) Newton’s Law of Universal Gravitation Explained by the Theory of Informatons. Journal of High Energy Physics, Gravitation and Cosmology, 10, 918-929. doi: 10.4236/jhepgc.2024.103056. 1. Introduction Daily contact with the things on hand confronts us with their substantiality. An object is not just form, it is also matter. It takes space, it eliminates emptiness. The amount of matter within the contours of a physical body is called its mass. The mass of an object manifests itself when it interacts with other objects. A fundamental form of interaction between objects is “gravitation”. Material objects (“masses”) attract each other and, if they are free, they move to each other. In the framework of the classical theory of fields(“Newtonian gravity”), the gravitational interactions are described by introducing the field concept. Each material object manifests its substantiality by creating and maintaining a vector field, characterized by the vectoral quantity E g E g that has a value at every point of space and time and is thus—relative to an inertial reference frame (IRF) O—regarded as a function of space and time coordinates. And each object in that field experiences a tendency to accelerate. The field theory considers E g E g as the mathematical entity that mediates in the gravitational interactions. Although the classical theory of fields describes the gravitational phenomena in a correct and coherent manner, it doesn’t create clarity about the physical nature of gravity: the gravitational field is considered as a purely mathematical construction. In what follows we develop the idea that, if masses can influence each other “at a distance”, they must in one way or another exchange data. We assume that each mass emits information regarding its magnitude and its position, and that it is able to “interpret” the information emitted by its neighbors. In this way we propose a physical foundation of Newtonian gravity by introducing information as the substance of a gravitational field. We start from the idea that a material object at rest relative to an IRF O manifests itself in space by the emission—at a rate proportional to its rest mass—of mass and energy less granular entities that are rushing away with the speed of light and are carrying information regarding the position (“g-information”) of their emitter. Because they transport nothing else than information, we call these entities “informatons”. In that context, the gravitational field of a material object will be understood as an expanding cloud of informatons, that forms an indivisible whole with that object. In the postulate of the emission of informatons, we define an informaton by its attributes and determine the rules that govern the emission by a point mass that is anchored in an IRF O. A direct consequence of that postulate is that a mass particle at rest in O, and by extension any material object at rest, is the source of an expanding cloud of informatons, that—at an arbitrary point P—is characterised by the density of the flow of g-information at that point. That vectoral quantity can be identified with E g E g , the vector field that—according to classical physics—mediates in the gravitational interactions, which implies that we can refer to the cloud of informatons as the gravitational field in O. Finally we explain the gravitational interaction between mass particles at rest relative to an IRF O(Newton’s law of universal gravitation) as the response of an object to the disturbance of the symmetry of its “proper” gravitational field by the field that, in its direct vicinity, is created and maintained by other masses. 2. Preliminary Definitions A material body occupies space, its surface encloses matter. The amount of matter within its contours is called its mass. According to the field theory, any material body is the source of a gravitational field that at a sufficiently large distance is independent of the form of the body. This “far field” can be calculated by reducing the body to a mathematical point in which all the mass is accumulated. Such a point is called a “mass particle” or a “point mass” and it will be graphically represented by a little sphere. If we can calculate the gravitational field generated by a point mass, integral calculus delivers the methods to calculate the gravitational field generated by any material body. This justifies the fact that we in the first instance focus on the emission of informatons by a mass particle. The phenomena that are the subject of this article are situated in spacetime: they are located in “space” and dated in “time”. 1) In the context of the theory of informatons space is conceived as a three-dimensional, homogeneous, isotropic, unlimited and empty continuum. This continuum is called the “Euclidean space” because what geometrically is possible in that space, is determined by the Euclidean geometry. By anchoring a standardized Cartesian coordinate system to a reference body, an observer can—relative to that reference body—localize each point by three coordinates x, y, z. 2) In the same context we define time as the monotonically increasing real quantity t that is generated by a standard clock 1. In a Cartesian coordinate system a standard clock links to each event a “moment”—this is a specific value of t — and to each duration a “period” or “time interval”—this is a specific increase of t. The introduction of time makes it possible for the observer to express, in an objective manner, the chronological order of events in a Cartesian coordinate system. A Cartesian coordinate system together with a standard clock is called a “re f erence frame”. We represent a reference frame as OXYZ(T), shortly as O. A reference frame is called an “inertial reference frame” if light propagates rectilinear (in the sense of the Euclidean geometry) with constant speed everywhere in the empty space linked to that frame. This definition implies that the space linked to an inertial reference frame is a homogeneous, isotropic, unlimited and empty continuum in which the Euclidean geometry is valid. 3. The Concept of Gravitational Information Newton’s law of universal gravitation may be expressed as follows: The force between any two particles having masses m 1 and m 2 separated by a distance r is an attraction working along the line joining the particles and has a magnitude F=G⋅m 1⋅m 2 r 2 F=G⋅m 1⋅m 2 r 2 where G is a universal constant having the same value for all pairs of particles. This law expresses the basic fact of gravitation, namely that two masses are interacting “at-a-distance”: they exert forces on one another even though they are not in contact. According to Newton’s law F B F B , the force exerted by a particle A — with mass m 1—on a particle B—with mass m—is pointing to the position of A and has a magnitude: F B=(G⋅m 1 r 2)⋅m F B=(G⋅m 1 r 2)⋅m The orientation of this force and the fact that it is directly proportional to the mass of A and inversely proportional to the square of the distance from A to B,implies that particle B must receive information about the presence in space of particle A: particle A must send information to B about its position and about its mass. This conclusion is independent of the position and the mass of B; so we can generalize it and posit that A particle manifests itself in space by emitting information about its mass and about its position. We consider that type of information as a substantial element of nature and call it “gravitational information” or “g-information”. We assume that g-information is transported by mass and energy less granular entities that rush through space with the speed of light (c). These grains of g-information are called informatons. 4. The Postulate of the Emission of Informatons A material object manifests its presence in space by continuously emitting informatons. The emission of informatons by a material object anchored in an IRF O, is governed by the “postulate of the emission of informatons”. A. The emission of informatons by a particle at rest is governed by the following rules: 1) The emission is uniform in all directions of space, and the informatons d i verge with the speed of light (c = 3 × 10 8 m/s ) along radial trajectories relative to the position of the emitter. 2)N˙=d N d t N˙=d N d t,the rate at which a particle emits informat ons 2 , is ti me indepe n dent and proportional to the rest mass m 0 of the emitter. So there is a constant K so that: N˙=K⋅m 0 N˙=K⋅m 0 3) The constant K is equal to the ratio of the square of the speed of light (c)to the Planck constant (h): K=c 2 h=1.36×10 50 kg−1⋅s−1 K=c 2 h=1.36×10 50 kg−1⋅s−1 B. We call the essential attribute of an informaton its g-index. The g-index of an informaton refers to information about the position of its emitter and equals the elementary quantity of g-information. It is represented by a vectoral quantity s g s g : 1) s g s g points to the position of the emitter. 2) The elementary quantity of g-information is: s g=1 K⋅η 0=h η 0⋅c 2=6.18×10−60 m 3⋅s−1 s g=1 K⋅η 0=h η 0⋅c 2=6.18×10−60 m 3⋅s−1 where η 0=1 4⋅π⋅G=1.19×10 9 kg⋅s 2⋅m−3 η 0=1 4⋅π⋅G=1.19×10 9 kg⋅s 2⋅m−3 , G being the gravitational constant. Rule A.1 is the expression of the hypothesis that the space is a homogenous and isotropic continuum in which the gravitational phenomena are travelling with the speed of light. Rule A.2 posits that the rate at which a particle emits informatons is a measure for its rest mass and rule A.3 implies the fact that, when a particle absorbs (emits) a photon h⋅ν h⋅ν , its rest mass is increasing (decreasing) with an amount h⋅ν c 2 h⋅ν c 2 while its emission rate is increasing (decreasing) with an amount ν ν . Rule B.1 and rule B.2 respectively express the facts that the gravitational field of a particle always points to the position of the source of that field and that the gravitational force between any two particles depends on a universal constant G. To summarize, each material object manifests itself in space by the emission of informatons, it is a source of informatons. Informatons are grains of g-information and, as such, the constituent elements of gravitational fields. In the context of the postulate informatons are completely defined by their g-index s g s g . We will represent an informaton as a quasi-infinitely small sphere, moving with velocity c c and carrying a vector s g s g . 5. The Gravitational Field of a Particle at Rest Figure 1.The emission of an informaton by a particle. In Figure 1 we consider a mass particle with rest mass m 0 that is anchored at the origin of an inertial reference frame O. According to the postulate it continuously emits informatons in all directions of space. The informatons that with velocity c=c⋅r r=c⋅e r c=c⋅r r=c⋅e r pass near a fixed point P, defined by the position vector r r , are characterised by their g-index s g s g : s g=−1 K⋅η 0⋅r r=−1 K⋅η 0⋅e r s g=−1 K⋅η 0⋅r r=−1 K⋅η 0⋅e r The rate at which the point mass emits g-information is the product of the rate at which it emits informatons with the elementary g-information quantity: N˙⋅s g=m 0 η 0 N˙⋅s g=m 0 η 0 Of course, this is also the rate at which it sends g-information through any closed surface that surrounds m 0: it is the intensity of the g-information-flow through any closed surface that encloses m 0. The emission of informatons fills the space around m 0 with an expanding cloud of g-information. This cloud has the shape of a sphere whose surface moves away from the center O — the position of the point mass—with the speed of light . 1)Within that cloud there is a stationary state. Because for each spatial region the inflow of g-information equals the outflow, each spatial region contains an unchanging number of informatons and thus a constant quantity of g-information. Moreover, the orientation of the g-indices of the informatons passing near a fixed point is always the same. 2)That cloud can be identified with a continuum. Each spatial region contains a very large number of informatons: the g-information is like continuously spread over the volume of the region. The cloud of g-information surrounding O can be identified as the gravit a tional field or the g-field of the point mass m 0. Without interruption “countless” informatons are rushing through any—even a very small—surface in the gravitational field: we can describe the motion of g-information through a surface as a continuous flow of g-information. We know already that the intensity of the flow of g-information through a closed surface that surrounds O is expressed as: N˙⋅s g=m 0 η 0 N˙⋅s g=m 0 η 0 If the closed surface is a sphere with radius r, the intensity of the flow per unit area is given by: m 0 4⋅π⋅r 2⋅η 0 m 0 4⋅π⋅r 2⋅η 0 This is the density of the flow of g-information at any point P at a distance r from m 0(Figure 1). This quantity is, together with the orientation of the g-indices of the informatons that are passing near P, characteristic for the gravitational field at that point. Thus, at a point P, the gravitational field of the point mass m 0 is unambiguously defined by the vectoral quantity E g E g : E g=N˙4⋅π⋅r 2⋅s g=−m 0 4⋅π⋅η 0⋅r 2⋅e r=−m 0 4⋅π⋅η 0⋅r 3⋅r E g=N˙4⋅π⋅r 2⋅s g=−m 0 4⋅π⋅η 0⋅r 2⋅e r=−m 0 4⋅π⋅η 0⋅r 3⋅r This quantity is the gravitational field strength or the g-field strength or the g-field (m∙s−2). At any point of the gravitational field of the point mass m 0, the orientation of E g E g corresponds to the orientation of the g-indices of the informatons which are passing near that point. And the magnitude of E g E g is the density of the g-information flow at that point (the rate per unit area at which g-information at P flows through an elementary surface perpendicular to the direction of E g E g ). Let us note that, in the case under consideration, E g E g is opposite to the direction of movement of the informatons. Finally, let us consider a surface-element d S at P (Figure2(a)). Its orientation and magnitude are completely determined by the surface-vector d S d S (Figure 2(b)). By −d Φ G−d Φ G , we represent the rate at which g-information flows through d S in the sense of the positive normal e n e n and we call the scalar quantity d Φ G d Φ G defined as d Φ G=E g⋅d S=E g⋅d S⋅cos α d Φ G=E g⋅d S=E g⋅d S⋅cos α the elementary g-flux through d S.(m 3∙s−2). Figure2.The elementary g-flux through a surface-element. For an arbitrary closed surface S that surrounds m 0, the outward flux (which we obtain by integrating the elementary contributions d Φ g d Φ g over S) must be equal to the rate at which the mass emits g-information. Thus: Φ G=∯E g⋅d S=−m 0 η 0 Φ G=∯E g⋅d S=−m 0 η 0 This is Gauss’s law in the case of a mass particle at rest. Gausse’s law is the expression of the conservation of g-information. 6. The Gravitational Field of a Set of Particles at Rest We consider a set of particles with rest masses m 1,⋯,m i,⋯,m n m 1,⋯,m i,⋯,m n that are anchored in an inertial reference frame O. At an arbitrary point P, the flows of g-information who are emitted by the distinct masses are defined by the gravitational fields E g 1,⋯,E g i,⋯,E g n E g 1,⋯,E g i,⋯,E g n . −d Φ g−d Φ g , the rate at which g-information flows through a surface-element d S at P in the sense of the positive normal, is the sum of the contributions of the distinct masses: −d Φ G=∑i=1 n−(E g i⋅d S)=−(∑i=1 n E g i)⋅d S=−E g⋅d S−d Φ G=∑i=1 n−(E g i⋅d S)=−(∑i=1 n E g i)⋅d S=−E g⋅d S So, the effective density of the flow of g-information at P(the effective g-field)is completely defined by: E g=∑i=1 n E g i E g=∑i=1 n E g i We conclude: At a point in space, the g-field of a set of point masses at rest is completely d e fined by the vectoral sum of the g-fields caused by the distinct masses. Let us remark that the orientation of the effective g-field has no longer a relation with the direction in which the passing informatons are moving. One easily shows that the outward g-flux through a closed surface in the g-field of a set of anchored point masses only depends on the surrounded masses m in: −∯E g⋅d S=m i n η 0−∯E g⋅d S=m i n η 0 This is Gauss’s law in the case of a set of mass particles at rest. It is the expression of the conservation of g-information. 7. The Gravitational Field of a Mass Continuum at Rest We call an object in which the matter in a time independent manner is spread over the occupied volume, a mass continuum. At each point Q in such a continuum, the accumulation of mass is defined by the (mass) density ρ G ρ G . To define this scalar quantity one considers the mass d m of a volume element d V that contains Q. The accumulation of mass in the vicinity of Q is defined by: ρ G=d m d V ρ G=d m d V A mass continuum—anchored in an inertial reference frame—is equivalent to a set of infinitely many infinitesimal small mass elements d m. The contribution of each of them to the field strength at an arbitrary point P is d E g d E g . E g E g , the effective g-field at P, is the result of the integration over the volume of the continuum of all these contributions. It is evident that the outward g-flux through a closed surface S only depends on the mass enclosed by that surface (the enclosed volume is V): −∯S E g⋅d S=1 η 0⋅∭V ρ G⋅d V−∯S E g⋅d S=1 η 0⋅∭V ρ G⋅d V This is Gauss’s law in the case of a mass continuum. It is the expression of the conservation of g-information. That relation is equivalent with (theorem of Ostrogradsky ): d i v E g=−ρ G η 0 d i v E g=−ρ G η 0 Furthermore, one can show that in any matter free point r o t E g=0 r o t E g=0 , what implies the existence of a gravitational potential function V g for which: E g=−g r a d V g E g=−g r a d V g 8. The Gravitational Field of Objects at Rest A mass particle at rest, a set of mass particles at rest and a mass continuum at rest are the sources of gravitational fields that are completely characterized by the time independent vectoral quantity E g E g . The magnitude of this quantity is the rate per unit area at which g-information at an arbitrary point flows through an elementary surface perpendicular to the direction of E g E g . The constituent element of these fields is the “informaton” and their substance is “g-information”. This implies that gravitational fields are granular, that they continuously regenerate, that they show fluctuations, that they expand with the speed of light, that gravitational phenomena propagate with that speed and that there is conservation of g-information. 9. The Gravitational Interaction between Mass Particles at Rest We consider a set of mass particles anchored in an IRF O. They create and maintain a gravitational field that at each point of the space linked to O is completely determined by the vector E g E g . Each particle is “immersed” in a cloud of g-information. At every point, except at its own position, each particle contributes to the construction of that cloud. Let us consider the particle with rest mass m 0 anchored at P. If the other particles were not there, then m 0 would be at the center of a perfectly spherical cloud of g-information. In reality this is not the case: the emission of g-information by the other particles is responsible for the disturbance of that “characteristic symmetry” of the proper g-field of m 0. Because E g E g at P represents the intensity of the flow of g-information send to P by the other particles, the extent of disturbance of the characteristic symmetry in the immediate vicinity of m 0 is determined by E g E g at P. If it was free to move, particle m 0 could restore the characteristic symmetry of the g-information cloud in its immediate vicinity by accelerating with an amount a=E g a=E g . Indeed, accelerating this way has the effect that the extern field disappears in the origin of the reference frame anchored to m 0. If it accelerates with an amount a=E g a=E g , m 0 would become “blind” for the g-information send to its immediate vicinity by the other particles, it would “see” only its proper spherical g-information cloud. So, from the point of view of a particle at rest at a point P in a gravitational field E g E g , the characteristic symmetry of the g-information cloud in its immediate vicinity is conserved if it accelerates with an amount a=E g a=E g . A particle that is anchored in a gravitational field cannot accelerate. In that case it tends to move. This insight is expressed in the following postulate: A particle anchored at a point in a gravitational field is subjected to a tende n cy to move in the direction defined by E g E g , the g-field at that point. Once the anchorage is broken, the mass acquires a vectoral acceleration a→a→that equals E g E g . 10. The Gravitational Force—The Force Concept A particle m 0,anchored at a point P in a gravitational field, experiences an action because of that field, an action that is compensated by the anchorage. 1) That action is proportional to the extent to which the characteristic symmetry of the proper gravitational field of m 0 in the immediate vicinity of P is disturbed by the extern g-field, thus to the value of E g E g at P. 2) It depends also on the magnitude of m 0. Indeed, the g-information cloud created and maintained by m 0 is more compact as m 0 is greater. That implies that the disturbing effect on the spherical symmetry around m 0 by the extern g-field E g E g is smaller when m 0 is greater. Thus, to impose the acceleration a=E g a=E g , the action of the gravitational field on m 0 must be greater as m 0 is greater. We can conclude that the action that tends to accelerate a particle in a gravitational field must be proportional to E g E g —the g-field to which the particle is exposed—and to m 0—the rest mass of the particle. We represent that action by F G F G and we call this vectoral quantity “the force developed by the g-field on the particle” or the gravitational force on m 0. We define it by the relation: F G=m 0⋅E g F G=m 0⋅E g A particle anchored at a point P cannot accelerate, which implies that the effect of the anchorage must compensate the gravitational force. It cannot be otherwise than that the anchorage exerts an action on that particle that is exactly equal and opposite to the gravitational force. That action is called a reaction force. Between the gravitational force on a particle with rest mass m 0 and the local field strength exists the following relationship: E g=F G m 0 E g=F G m 0 So, the acceleration imposed to the mass by the gravitational force is: a=F G m 0 a=F G m 0 Considering that the gravitational force is nothing but a special force, we can conclude that this relation can be generalized. The relation between a force F F and the acceleration a a that it imposes to a free particle with rest mass m 0 is: F=m 0⋅a F=m 0⋅a 11. Newton’s Law of Universal Gravitation Figure 3.The gravitational interaction between two particles at rest. In Figure 3 we consider two particles with (rest) masses m 1 and m 2 anchored at the points P 1 and P 2 in an inertial reference frame. 1) m 1 creates and maintains a gravitational field that at P 2 is defined by the g-field: E g 2=−m 1 4⋅π⋅η 0⋅R 2⋅e 12 E g 2=−m 1 4⋅π⋅η 0⋅R 2⋅e 12 If m 2 was free, according to the postulate of the gravitational interaction it would accelerate with an amount a a : a=E g 2 a=E g 2 So the gravitational field of m 1 exerts a “gravitational force” on m 2: F 12=m 2⋅a=m 2⋅E g 2=−m 1⋅m 2 4⋅π⋅η 0⋅R 2⋅e 12 F 12=m 2⋅a=m 2⋅E g 2=−m 1⋅m 2 4⋅π⋅η 0⋅R 2⋅e 12 In a similar manner we find F 21 F 21 : F 21=−m 1⋅m 2 4⋅π⋅η 0⋅R 2⋅e 21=−F 12 F 21=−m 1⋅m 2 4⋅π⋅η 0⋅R 2⋅e 21=−F 12 This is the mathematical expression of “Newton’s law of universal gravitation” : The force between any two particles having masses m 1 and m 2 separated by a distance R is an attraction acting along the line joining the particles and has the magnitude F=G⋅m 1⋅m 2 R 2=1 4 π η 0⋅m 1⋅m 2 R 2 F=G⋅m 1⋅m 2 R 2=1 4 π η 0⋅m 1⋅m 2 R 2 G=1 4 π η 0 G=1 4 π η 0 is a universal constant having the same value for all pairs of particles. 12. Conclusions The phenomenon described by Newton’s law of universal gravitation can perfectly be explained by the hypothesis that g-information, i.e. information carried by informatons,is the substance of the medium that the interaction between mass particles separated in space makes possible. On the macroscopic level, that medium, the “gravitational field”, manifests itself as the vector field E g E g that—according to classical physics—mediates the gravitational interactions. Each mass particle is the source of a gravitational field: it creates and maintains a cloud of g-information that, when the particle is at rest, at an arbitrary point P is completely defined by the vector field E g E g . E g, the magnitude at P, is the density of the flow of g-information at that point (the rate per unit area at which g-information at P flows through an elementary surface perpendicular to the direction of E g E g .) A mass particle with rest mass m 0 in a gravitational field E g E g generated by other particles is subjected to a tendency to accelerate with an amount a=E g a=E g . The gravitational field exerts a force F F on it: F=m 0⋅a F=m 0⋅a . 13. Epilogue What precedes this can be expanded to the interaction between moving particles . In the follow-up article “The gravitational interaction between moving mass particles explained by the theory of informatons” we deduce from the postulate of the emission of informatons that the gravitational field of a moving mass particle is a dual entity always having a field- and an induction-component simultaneously created by their common sources: time-variable masses and mass flows and that the gravitational interaction is the effect of the fact that an object in a gravitational field tends to become “blind” for that field by accelerating according to a Lorentz-like law. NOTES 1 The operation of a standard clock is based on the counting of the successive cycles of a periodic process that is generated by a device inside the clock. 2 We neglect the possible stochastic nature of the emission, that is responsible for noise on the quantities that characterize the gravitational field. So, N˙N˙ is the average emission rate. Conflicts of Interest The author declares no conflicts of interest regarding the publication of this paper. References Resnick, D. and Halliday, R. (1970) Fundamentals of Physics. John Wiley & Sons. Ohanian, H.C. (1985) Physics. W. W. Norton & Company, Inc. Angot, A. (1957) Compléments de Mathematiques. Editions de la Revue d’Optique. Acke, A. (1990) Beginselen van de informatonentheorie. Uitgeverij Nevelland. Acke, A. (2008) Gravitatie en elektromagnetisme. Uitgeverij Nevelland. 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13810	https://www.nagwa.com/en/videos/817109238048/	Question Video: Finding the Number of Negative Terms in a Given Arithmetic Sequence Mathematics • Second Year of Secondary School Find the number of negative terms in the arithmetic sequence (−40, −33, −26, ...). Video Transcript Find the number of negative terms in the arithmetic sequence negative 40, negative 33, negative 26, and so on. An arithmetic sequence has first term 𝑎 and common difference 𝑑. The first term in our sequence is negative 40. Therefore, 𝑎 equals negative 40. To get from negative 40 to negative 33, we add seven. The same is true to get from negative 33 to negative 26. We can therefore say that 𝑑 is equal to seven or positive seven. The general term of any arithmetic sequence is given by 𝑎 plus 𝑛 minus one multiplied by 𝑑. In this question, we are looking for the number of negative terms. These are the terms that will be less than zero. Our general term equation must be less than zero. Substituting in our values of 𝑎 and 𝑑 gives us negative 40 plus 𝑛 minus one multiplied by seven is less than zero. We can distribute the parentheses or expand the brackets by multiplying seven by 𝑛 and seven by negative one. This gives us negative 40 plus seven 𝑛 minus seven is less than zero. Adding 40 and seven to both sides of the inequality gives us seven 𝑛 is less than 47. Dividing both sides by seven gives us 𝑛 is less than 47 over seven. As 47 divided by seven is six remainder five, this can be written as a mixed number as six and five-sevenths. 𝑛 must be a whole number or integer value. Therefore, there are six terms that are negative. We could’ve found these by continuing to add seven to each of the terms in the sequence. Negative 26 add seven is negative 19. Repeating this twice more gives us negative 12 and negative five. The next number in the sequence would be two, which is positive. This once again proves that we have six negative terms in the sequence negative 40, negative 33, negative 26, and so on. Lesson Menu Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company Content Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
13811	https://www.quora.com/How-do-I-find-the-length-of-each-side-of-a-triangle-on-a-graph	How to find the length of each side of a triangle on a graph - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In mathematics triangle sides distance formula coordinate plane graphing (functions) perimeter of triangles coordinate systems triangles properties of triangles 5 How do I find the length of each side of a triangle on a graph? All related (32) Sort Recommended Rebecca Ball Former Retired Solicitor · Author has 5.9K answers and 3.1M answer views ·Updated 1y By using Pythagoras’ theorem. each side is the hypoteneuse of a right triangle whose other two sides have lengths equal to the difference between the x co-ordinates of their terminal vertices and the difference between the y co-ordinates of their terminal vertices. Hence, if triangle ABC has the following vertices: A = (1,2) B = (3,4) C = (7, 1) Then AB^2 is 2 squared + 2 squared, so AB is root 8 BC^2 is 4 squared + 3 squared so BC is root 25 or 5 CA^2 is 6 squared + 1 squared so CA is root 37 Upvote · 9 1 9 1 Promoted by SavingsPro.org Mark Bradley Economist ·9mo What are the dumbest financial mistakes most Americans make? Where do I start? I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits. Here are the biggest mistakes people are making and how to fix them: Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up. Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one Continue Reading Where do I start? I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits. Here are the biggest mistakes people are making and how to fix them: Not having a separate high interest savings account: Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it. Plus with rates above 5.00%, the interest you can earn compared to most banks really adds up. Here is a list of the top savings accounts available today. Deposit $5 before moving on because this is one of the biggest mistakes and easiest ones to fix. Overpaying on car insurance: You’ve heard it a million times before, but the average American family still overspends by $417/year on car insurance. If you’ve been with the same insurer for years, chances are you are one of them. Pull up Coverage.com, a free site that will compare prices for you, answer the questions on the page, and it will show you how much you could be saving. That’s it. You’ll likely be saving a bunch of money. Here’s a link to give it a try. Not using a debt relief program: People with $10K in credit card debt can get significant reductions by using a debt relief program. Don’t suffer alone, here’s a quick 2 minute quiz that will tell you if you qualify. Not earning free money while investing: Most people think investing is complicated or for the rich. Not anymore. There are platforms that literally give you free money just to get started. Deposit as little as $25, and you could get up to $1000 in bonus funds. This is the siteI recommend to my friends. How to get started Hope this helps! Here are the links to get started: Have a separate savings account Stop overpaying for car insurance Finally get out of debt Start investing with a free bonus Upvote · 4.7K 4.7K 1K 1K 999 163 Hilmar Zonneveld Translator (1985–present) · Upvoted by Klaus Ole Kristiansen , M.Sc. Mathematics, University of Copenhagen (1992) · Author has 58.6K answers and 18.8M answer views ·4y Related How do I find the side lengths of a triangle? Either you measure them, Or you calculate them, based on some other information available. Here are some examples of the second option. You know the lengths of two sides, and you know that the triangle has a right angle. You use the Pythagorean Theorem to calculate the third side. Similarly, if you know two sides and one angle, or one side and two angles, you can calculate the remaining sides and angles - though in some cases, there is more than one solution. If you know the triangle’s three angles, you don’t have enough information to calculate the length of the sides. Thus, the answer really depen Continue Reading Either you measure them, Or you calculate them, based on some other information available. Here are some examples of the second option. You know the lengths of two sides, and you know that the triangle has a right angle. You use the Pythagorean Theorem to calculate the third side. Similarly, if you know two sides and one angle, or one side and two angles, you can calculate the remaining sides and angles - though in some cases, there is more than one solution. If you know the triangle’s three angles, you don’t have enough information to calculate the length of the sides. Thus, the answer really depends on what information you are given. Upvote · 9 3 Assistant Bot · 10mo To find the length of each side of a triangle plotted on a graph, you can use the distance formula. The distance between two points (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) in a Cartesian coordinate system is given by: d=√(x 2−x 1)2+(y 2−y 1)2 d=(x 2−x 1)2+(y 2−y 1)2 Here’s how you can apply this to find the lengths of the sides of a triangle: Identify the vertices of the triangle: Let’s say the vertices of the triangle are A(x 1,y 1)A(x 1,y 1), B(x 2,y 2)B(x 2,y 2), and C(x 3,y 3)C(x 3,y 3). Calculate the lengths of the sides: Length of side AB: A B=√(x 2−x 1)2+(y 2−y 1)2 A B=(x 2−x 1)2+(y 2−y 1)2 Length of side BC: BC=\sqrt{(x_3-x_ BC=\sqrt{(x_3-x_ Continue Reading To find the length of each side of a triangle plotted on a graph, you can use the distance formula. The distance between two points (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) in a Cartesian coordinate system is given by: d=√(x 2−x 1)2+(y 2−y 1)2 d=(x 2−x 1)2+(y 2−y 1)2 Here’s how you can apply this to find the lengths of the sides of a triangle: Identify the vertices of the triangle: Let’s say the vertices of the triangle are A(x 1,y 1)A(x 1,y 1), B(x 2,y 2)B(x 2,y 2), and C(x 3,y 3)C(x 3,y 3). Calculate the lengths of the sides: Length of side AB: A B=√(x 2−x 1)2+(y 2−y 1)2 A B=(x 2−x 1)2+(y 2−y 1)2 Length of side BC: B C=√(x 3−x 2)2+(y 3−y 2)2 B C=(x 3−x 2)2+(y 3−y 2)2 Length of side CA: C A=√(x 1−x 3)2+(y 1−y 3)2 C A=(x 1−x 3)2+(y 1−y 3)2 3. Substitute the coordinates: Plug in the coordinates of the vertices into the formulas to compute the lengths of each side. Example Suppose the vertices of the triangle are: A(1,2)A(1,2) B(4,6)B(4,6) C(7,2)C(7,2) Calculating the lengths: Length of A B A B: A B=√(4−1)2+(6−2)2=√3 2+4 2=√9+16=√25=5 A B=(4−1)2+(6−2)2=3 2+4 2=9+16=25=5 Length of B C B C: B C=√(7−4)2+(2−6)2=√3 2+(−4)2=√9+16=√25=5 B C=(7−4)2+(2−6)2=3 2+(−4)2=9+16=25=5 Length of C A C A: C A=√(1−7)2+(2−2)2=√(−6)2+0 2=√36=6 C A=(1−7)2+(2−2)2=(−6)2+0 2=36=6 Summary The lengths of the sides of the triangle are: A B=5 A B=5 B C=5 B C=5 C A=6 C A=6 You can use this method for any triangle on a graph by identifying the coordinates of its vertices and applying the distance formula. Upvote · Related questions More answers below How do I find the length of a triangle? How can we find the length of one side using two sides in a triangle? How do you find the side length of a triangle? What is the shortest side length of a triangle? How do you find the length of the third side of a triangle? Roy Narten M.Sc (Mechanical Engineering), former Engineering Instructor · Author has 2.2K answers and 4.8M answer views ·4y Related How do I solve to get the side lengths of this triangle? How do I solve to get the side lengths of this triangle? For any triangle the Law of Sines (or the Sine Law) is: a s i n A=b s i n B=c s i n C a s i n A=b s i n B=c s i n C For your question: 2 x+5 s i n 60=x+6 s i n 45=3 y−12 s i n 75 2 x+5 s i n 60=x+6 s i n 45=3 y−12 s i n 75 Solve this first to get x : 2 x+5 s i n 60=x+6 s i n 45 2 x+5 s i n 60=x+6 s i n 45 or (2 x+5)(s i n 45)=(x+6)s i n 60(2 x+5)(s i n 45)=(x+6)s i n 60 x[2 s i n 45−s i n 60]=6 s i n 60−5 s i n 45 x[2 s i n 45−s i n 60]=6 s i n 60−5 s i n 45 x=6 s i n 60−5 s i n 45 2 s i n 45−s i n 60=3.029 x=6 s i n 60−5 s i n 45 2 s i n 45−s i n 60=3.029 Then solve the latter part of the sine law to get y: x+6 s i n 45=3 y−12 s i n 75 x+6 s i n 45=3 y−12 s i n 75 Continue Reading How do I solve to get the side lengths of this triangle? For any triangle the Law of Sines (or the Sine Law) is: a s i n A=b s i n B=c s i n C a s i n A=b s i n B=c s i n C For your question: 2 x+5 s i n 60=x+6 s i n 45=3 y−12 s i n 75 2 x+5 s i n 60=x+6 s i n 45=3 y−12 s i n 75 Solve this first to get x : 2 x+5 s i n 60=x+6 s i n 45 2 x+5 s i n 60=x+6 s i n 45 or (2 x+5)(s i n 45)=(x+6)s i n 60(2 x+5)(s i n 45)=(x+6)s i n 60 x[2 s i n 45−s i n 60]=6 s i n 60−5 s i n 45 x[2 s i n 45−s i n 60]=6 s i n 60−5 s i n 45 x=6 s i n 60−5 s i n 45 2 s i n 45−s i n 60=3.029 x=6 s i n 60−5 s i n 45 2 s i n 45−s i n 60=3.029 Then solve the latter part of the sine law to get y: x+6 s i n 45=3 y−12 s i n 75 x+6 s i n 45=3 y−12 s i n 75 Upvote · 9 3 9 1 Ahmad Rezaei Math teacher from 2000 to 2016 (2000–present) · Author has 2.1K answers and 1.2M answer views ·3y Related How do I calculate the length of the triangle? mBC=(yB - yC)/(xB -xC) mBC=(7+4)/(3 - 6)= - 11/3 Now we are going to write the equation of the side BC : A line that passes thfough the poing B by the slope m = - 11/3 y - yB=m(x - xB) y - 7= - 11/3(x -3) Mutiply both sides by 3 : 11x+3y -54=0 The height of AH is : h=\|ax1+by1+c\| / [(a^2+b^2)^1/2 h=\|( -2)(11)+(3)(5) - 54\| /(121+9)^1/2 h=61 / (130)^1/2 h=5/35 units Upvote · 9 5 9 2 Promoted by Spokeo Spokeo - People Search \| Dating Safety Tool Dating Safety and Cheater Buster Tool ·Apr 16 Is there a way to check if someone has a dating profile? Originally Answered: Is there a way to check if someone has a dating profile? Please be reliable and detailed. · Yes, there is a way. If you're wondering whether someone has a dating profile, it's actually pretty easy to find out. Just type in their name and click here 👉 UNCOVER DATING PROFILE. This tool checks a bunch of dating apps and websites to see if that person has a profile—either now or in the past. You don’t need to be tech-savvy or know anything complicated. It works with just a name, and you can also try using their email or phone number if you have it. It’s private, fast, and really helpful if you’re trying to get some peace of mind or just want to know what’s out there. 🔍 HERE IS HOW IT WORK Continue Reading Yes, there is a way. If you're wondering whether someone has a dating profile, it's actually pretty easy to find out. Just type in their name and click here 👉 UNCOVER DATING PROFILE. 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ALSO HERE ARE OTHER HELPFUL TOOLS: Dating Research Tool – Search a large database to learn more about who you’re dating. Who’s Texting Your Partner – Discover who your partner is texting or calling, including their name, age, location, and social profiles. Verify People Tool– Confirm if someone is really who they say they are. Find Social Profiles – Locate someone's social media and dating profiles. People Search Directory– Look up someone's phone number and contact details. Dating Safety Check – Review your date’s background to help keep you safe. Upvote · 2.7K 2.7K 99 46 Vaman Kulkarni Ph.D Space and Plasma in Physics, Gujarat University (Graduated 1975) · Author has 4.8K answers and 1.8M answer views ·4y Related How do I find the side lengths of a triangle? I do not know that this is the correct answer. Triangle has three sides and three angles. Therefore you need minimum three information. Two sides and the angle between them, one side the angle at its two ends. Without these it is difficult answer your question. Upvote · Related questions More answers below What happens to the length of a triangle's sides when all three sides change? How do you calculate the average length of a triangle's sides when given only their lengths? What is the length of each side of ∆RAN? How do you find the length of a triangle given two sides? If you know the two side lengths of a right triangle, how can you find the third side length? Barry Gehm Former Asst Prof. Of Chemistry/Biochemistry at Lyon College (2003–2024) · Author has 13.4K answers and 16.3M answer views ·May 6 Related How can I calculate the length of one side of a triangle knowing the length of the other sides? Thanks. You can’t, without more information, such as the angle between the two sides whose length you know. Consider an analog clock. The lengths of the hour and minute hands are fixed, but the distance between their tips is constantly changing. Upvote · Sponsored by Intuit Accuracy and scale for every return, all year Streamline complex returns and reduce errors. ProConnect supports high-volume teams beyond April. Learn More 999 138 Haresh Sagar Studied Science&Mathematics (Graduated 1988) · Author has 6.1K answers and 6.8M answer views ·3y Related How do I calculate the length of the triangle? A=(−2,5)A=(−2,5), B=(3,7)B=(3,7), C=(6,−4)C=(6,−4) To find height of triangle from vertex A A to base B C B C, let's first derive equation of B C B C. (7−−4)x+(6−3)y+(3)(−4)−(7)(6)=0(7−−4)x+(6−3)y+(3)(−4)−(7)(6)=0 11 x+3 y−54=0 11 x+3 y−54=0 Shortest distance from B C B C to vertex A A, A D=11(−2)+3(5)−54√11 2+3 2 A D=11(−2)+3(5)−54 11 2+3 2 =−61√130≈−5.35=−61 130≈−5.35 Dimensions are positive, so height A D A D is approximately 5.35 5.35 units. Continue Reading A=(−2,5)A=(−2,5), B=(3,7)B=(3,7), C=(6,−4)C=(6,−4) To find height of triangle from vertex A A to base B C B C, let's first derive equation of B C B C. (7−−4)x+(6−3)y+(3)(−4)−(7)(6)=0(7−−4)x+(6−3)y+(3)(−4)−(7)(6)=0 11 x+3 y−54=0 11 x+3 y−54=0 Shortest distance from B C B C to vertex A A, A D=11(−2)+3(5)−54√11 2+3 2 A D=11(−2)+3(5)−54 11 2+3 2 =−61√130≈−5.35=−61 130≈−5.35 Dimensions are positive, so height A D A D is approximately 5.35 5.35 units. Upvote · 9 5 Atilla Garay Studied at Bachelor of Science in Mechanical Engineering · Author has 84 answers and 91.6K answer views ·1y Related How do you use angles in a triangle to find the lengths of the sides? First off you’ll need to know at least one side length. If you just have 3 angles then you’ll have an infinite number of similar triangles. Then you can use the sine identity. Thus A/sine a = B/sine b = C/sine c where all upper case side lengths are opposite the lower case angles. Upvote · Promoted by Grammarly Grammarly Great Writing, Simplified ·Updated 2y How can I effectively edit my own writing? So, you think you’ve drafted a tweet, an email, a short story, or even a novel. These are different forms of communication, but the process of bringing them to fruition has a necessary, sometimes overlooked step: editing! Unless you’re a professional writer, it’s unlikely that you have an editor who can review your writing regularly. Here are some tips to help you review your own work. Give your writing some space. Have you ever felt a mix of pure relief and joy when you’ve finished a draft of something? Don’t downplay that feeling and the ability to walk away from your work before you start ed Continue Reading So, you think you’ve drafted a tweet, an email, a short story, or even a novel. These are different forms of communication, but the process of bringing them to fruition has a necessary, sometimes overlooked step: editing! Unless you’re a professional writer, it’s unlikely that you have an editor who can review your writing regularly. Here are some tips to help you review your own work. Give your writing some space. Have you ever felt a mix of pure relief and joy when you’ve finished a draft of something? Don’t downplay that feeling and the ability to walk away from your work before you start editing it. You may need minutes, hours, or days, but once you sit back down with what you originally had on the page, you’ll have the thrill of looking at it with fresh eyes. You’ll notice errors you may not have seen the first time. You’ll come to new realizations about its overall tone and structure. If it’s a text or email, maybe you only need a few minutes away from it. If it’s a story or essay, perhaps you’ll need longer. Regardless of what type of work it is, it will help your writing tremendously. Don’t use overachieving synonyms. Looking at your work for the second, third, or fourth time around may inspire you to spice up your language with longer, more uncommon words. There’s nothing wrong with having a thesaurus nearby, but try to limit the repetition of long, pretentious-feeling words so your work flows well and doesn’t feel too bogged down. At the end of the day, you want it to feel true to you and the message you’re conveying. Remember who the reader is. Don’t forget your own voice as the writer—but don’t forget who your reader is. Many writers get too close to their work; editing is a chance to try to get out of your own head. Who is your ideal reader? What do you want them to take away from the writing? It’s a unique time to step in their shoes, to make sure your communication is as effective as you’d like it to be. Kill your darlings. Don’t be scared to remove chunks of your work, even if it feels precious to you. If it’s a passage that’s really tough to part with, try saving it somewhere else, so you can return to it later in your piece or for another work. Use Grammarly. Last but not least, Grammarly has countless resources for editing your work. Our writing assistant helps you find areas of your writing that are unclear or too wordy, as well as help you find mistakes you might not have caught. Editing may feel tedious, but it’s just as important as writing itself. For an extra pair of editing eyes on everything you write, download the free Grammarly for Windows and Mac today. Upvote · 1.1K 1.1K 999 227 99 32 George Ivey Former Math Professor at Gallaudet University · Author has 22.6K answers and 2.3M answer views ·May 6 Related How can I calculate the length of one side of a triangle knowing the length of the other sides? Thanks. You CAN’T. It also depends on the angle between the two known sides. The “cosine law”: if a and b are the length of two sides of a triangle and the angle between them is θ θ, then the length of the third side, c, is given by c 2=a 2+b 2−2 a b c o s(θ)c 2=a 2+b 2−2 a b c o s(θ). Upvote · Gary Ward MaEd in Education&Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.8K answers and 7.4M answer views ·3y Related How do you calculate the distances of the orthocentre of the triangle from the vertices? How do you calculate the distances of the orthocenter of the triangle from the vertices? First you must find where the orthocenter is located and it is easier to do If you place the longest side (AC in this example) on the x-axis with the left end at the origin. Bs coordinates can be found in different ways depending on the information given. If the length of AB is known and angle BAC then B(AB · cos <BAC, AB · sin <BAC. x = B’s x-coordinate is one altitude right away and also the x-coordinate of the orthocenter. Find the slope of another side, AB in this case. (Simply By / Bx) Note that you can g Continue Reading How do you calculate the distances of the orthocenter of the triangle from the vertices? First you must find where the orthocenter is located and it is easier to do If you place the longest side (AC in this example) on the x-axis with the left end at the origin. Bs coordinates can be found in different ways depending on the information given. If the length of AB is known and angle BAC then B(AB · cos <BAC, AB · sin <BAC. x = B’s x-coordinate is one altitude right away and also the x-coordinate of the orthocenter. Find the slope of another side, AB in this case. (Simply By / Bx) Note that you can go directly to step 4 and bypass this. The line with the negative reciprocal of this slope(-Bx /By) passing through point C (10, 0) will contain the altitude and intersect the other altitude at the orthocenter. Use the point-slope equation (y - 0 = -(4/4)(x - 10) and plug in the x-value of the orthocenter to find the y-value of the orthocenter. (y = -(4) + 10 = 6; so O(4,6) The third line of altitude is unnecessary. Once you have the coordinates of the orthocenter use the distance equation three times. This is one way to do it and rather easy if the above steps are followed. In this case: AO = √[(4–0)² + (6–0)²] = √(52) = 2√(13) BO = √[(4–4)² + (4–6)²] = √4 = 2 CO = √[(10–4)√ + (0–6)²] = √(72) = 6√2 AO = 2√(13); BO = 2; CO = 6√2 are the distances. Upvote · 9 7 9 1 9 1 Peter Hoffman Author has 6.3K answers and 570.2K answer views ·May 6 Related How can I calculate the length of one side of a triangle knowing the length of the other sides? Thanks. It is totally obvious, to any half-conscious 12 year old, that you cannot without additional information. Join two straight sticks with a hinge and rotate back-and-forth to get every length from 0 to the total of the sticks. Try to think before making a fool of yourself here. Upvote · Mariano Llancaman Studied Mathematics at University of Chile · Author has 95 answers and 225.9K answer views ·5y Related How do you find the length of the third side of a triangle? It depends on what information you have. The way the question is presented, I assume you already know the lenghts of 2 sides of the triangle. However this information is simply not enough to find out the length of the last side: In this diagram¯¯¯¯¯¯¯¯¯¯¯¯¯A 1 B 1=¯¯¯¯¯¯¯¯¯¯¯¯¯A 2 B 2 A 1 B 1¯=A 2 B 2¯ and ¯¯¯¯¯¯¯¯¯¯¯¯A 1 C 1=¯¯¯¯¯¯¯¯¯¯¯¯A 2 C 2 A 1 C 1¯=A 2 C 2¯, but ¯¯¯¯¯¯¯¯¯¯¯¯B 1 C 1≠¯¯¯¯¯¯¯¯¯¯¯¯B 2 C 2 B 1 C 1¯≠B 2 C 2¯. If the length of the two known sides are a a and b b, then the length of the last side could be anything from 0 0 to a+b a+b. If aside from the lengths of 2 sides (a a and b b), you also have the angle these sides form when they intersect (θ θ), t Continue Reading It depends on what information you have. The way the question is presented, I assume you already know the lenghts of 2 sides of the triangle. However this information is simply not enough to find out the length of the last side: In this diagram¯¯¯¯¯¯¯¯¯¯¯¯¯A 1 B 1=¯¯¯¯¯¯¯¯¯¯¯¯¯A 2 B 2 A 1 B 1¯=A 2 B 2¯ and ¯¯¯¯¯¯¯¯¯¯¯¯A 1 C 1=¯¯¯¯¯¯¯¯¯¯¯¯A 2 C 2 A 1 C 1¯=A 2 C 2¯, but ¯¯¯¯¯¯¯¯¯¯¯¯B 1 C 1≠¯¯¯¯¯¯¯¯¯¯¯¯B 2 C 2 B 1 C 1¯≠B 2 C 2¯. If the length of the two known sides are a a and b b, then the length of the last side could be anything from 0 0 to a+b a+b. If aside from the lengths of 2 sides (a a and b b), you also have the angle these sides form when they intersect (θ θ), then you can figure out the length of the last side (c c) using the law of cosines: c 2=a 2+b 2−2 a b c o s(θ)c 2=a 2+b 2−2 a b c o s(θ) Upvote · 9 2 9 1 Related questions How do I find the length of a triangle? How can we find the length of one side using two sides in a triangle? How do you find the side length of a triangle? What is the shortest side length of a triangle? How do you find the length of the third side of a triangle? What happens to the length of a triangle's sides when all three sides change? How do you calculate the average length of a triangle's sides when given only their lengths? What is the length of each side of ∆RAN? How do you find the length of a triangle given two sides? If you know the two side lengths of a right triangle, how can you find the third side length? How can you find the area of a triangle using only its three vertices (without any side lengths)? How do you find the missing side length for each interior angle in a triangle? How can we find the vertices of a triangle if the mid points of their sides are given? How can you find an unknown side length of a triangle if given two known lengths and their area? How do I find the length of one side of this triangle? Related questions How do I find the length of a triangle? How can we find the length of one side using two sides in a triangle? How do you find the side length of a triangle? What is the shortest side length of a triangle? How do you find the length of the third side of a triangle? What happens to the length of a triangle's sides when all three sides change? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13812	https://www.youtube.com/watch?v=ShAtryxVNg0	Common Core Geometry.Unit #7.Lesson #4.Similarity eMATHinstruction 54700 subscribers 117 likes Description 23458 views Posted: 2 Aug 2017 In this lesson, we look at the definition of similarity through similarity transformations, i.e. those that involve a dilation and possible rigid motions. This definition is then used to solve for missing sides in similar triangles Transcript: [Music] hello and welcome to another Common Core geometry lesson by emath instruction my name is Kirk Wier and today we'll be doing unit 7 lesson number four on similarity now I've been kind of hinting at similarity all through unit number seven if you've been joining us um and today we really formally introduce the idea and this is something you definitely saw in e8th grade mathematics the idea of similarity is is very simple it's the idea that you've got two geometric objects that have the same shape but possibly and almost always different sizes and today what we'll do is we'll be getting into the technical definition of what makes figures similar how to justify that they're similar and things like that so let's jump right into the lesson all right similar two fig similar figures two figures are similar and we're going to use that symbol if all pairs of corresponding sides are proportional we'll talk about what that means and the corresponding angles are of equal measure okay so let's just take a look at a couple triangles right so here we've got a triangle that has sides of 3 4 and 5 and another triangle with sides of 9 12 and 15 so first let's talk about what it means for the corresponding sides to be proportional and what that means is if I take the 12 which is the side that corresponds to the four right and I divide them I get three if I take the nine which corresponds to the three and I divide them I get three if I take the 15 and the five and I divide them I get three so in other words 12 / 4 is equal to the 9 / 3 that almost looks like a 9 which is equal to the 15 / five this is what it means for the sides to be proportional to one another and that probably looks similar to that whole scale factor issue that we've been talking about and then the other piece is that the angles have to be equal all right 37° 37° 53° 53° 90° 90° all right so two figures will be similar if all the angles inside the figure are the same as the angles inside of the other figure and the side lengths are proportional to one another which means when you divide two side lengths that are in the same place on the figure corresponding parts okay now aren't congruent but are proportional right so 3 9 / 3 = 15 / 5 = 12 / 4 all right that's just the definition of similar figures okay so let's keep going exercise number one in the diagram below we know that triangle ABC is similar to Triangle DF use tracing paper to verify all corresponding angle pairs pairs corresponding angle pairs between the two triangles are the same measure all right so what I'd like you to do is kind of what I've done take your tracing paper out and Trace over triangle ABC ABC take a moment to do that all right this is similar to something that we did in another lesson and now what we're going to do is we're going to verify all the angles of this triangle are equal to all the angles of this triangle and we'll do that very simply by just kind of doing this oh yeah look at that lies on top there right same deal that angle is the same and let's see if I can do this without turning my head at all and that angle is the same right so it's easy to now verify that all three angles of the smaller triangle are congruent to all three angles of the larger triangle now letter B says solve for the missing side lengths of both triangles show your work or explain your reasoning okay this gets into the idea that the side lengths of similar triangles or similar figures in general are proportional to one another okay so this is sort of how we would do it we find two side lengths that are in the same spot on a given triangle so the 14 and the Seven are corresponding sides of these two similar triangles and what we can now say is if I take that 14 and I divide it by seven that's going to be be the same as if I take this side EF or Fe and I divide it by this side that corresponds to it which is8 and because Fe is the only thing I don't know here I can easily solve for now you probably already realize something very simple which is that right 14 / 7 is 2 so what this means is given that this is two I'll just kind of swing it around there all right I can just multiply both sides by 8 and Fe must be 16 all right now likewise right if I want to know the length of ab what I can do is I can do something like I could stick with the 14 and the 7 I could say that 14 / 7 is equal to 20 / AB right and then I could solve this particular proportion again simplifying that 14 / 7 into 2 or I could just do cross multiplication as well right in fact if I didn't have such a nice simplification and I just wanted to cross multiply right I could say 14 a is equal to 7 20 which is 140 then I could divide both sides by 14 and a b was equal to 10 all right now if you know anything about similarity right it's all about all the sides on this triangle being some multiple of the sides of this triangle so right away you may have been able to say well that one's seven and that one's 14 therefore all the side length on this triangle will be twice the length of all of them on this triangle and then it's very easy then we can just say well if that's eight then that has to be 16 and if that's 20 that has to be 10 most of the time the dilation constant okay cuz that's what it really is the dilation constant won't be this nice number like two or three or something obvious it might be more challenging in which case the kind of cross multiplying is a no fail way to figure it out all right but again it's all about being able to set up these proportions cross multiplying and solving for a single missing side speaking about dilations letter C says give a dilation of ABC that would guarantee that its image a prime B Prime C Prime is congruent to DF how do you know it's congruent all right so let's talk about dilating ABC okay well let's dilate triangle ABC by a factor of k = 2 and really any Center any Center really doesn't matter all I need is the image of this triangle to be congruent to this one I don't need it to lie on top of this one I just need it to be congruent to it well once we do that right what will happen is that we will then have a Prime C Prime B Prime right we would have um this to be 14 this would be 16 right and what we would know you know even after given all the other work that we have what we would now know is we would know that all the angles of these two triangles were the same right and all the side lengths would also be the same right they would all be the same length in fact we would know since we knew back in the first part of the problem that this is 10 that this would now be 20 right we also learned in the last problem that this was 16 right yeah and so we could justify it in many different ways we could now say that by the side side side theorem these two triangles would now be congruent we could say by the side angle side theorem by the angle side angle theorem we could do a lot of things and in fact if we didn't even have let's say these two other things listed if we just had these two it would be enough by the angle side angle theorem it would be enough by the angle side angle theorem to know that the two triangles were congruent because remember dilations don't change angle measurements so this angle would be the same as this angle this angle would be the same as this angle right and there therefore we could say ah I got an angle equal to an angle a side equal to a side and an angle equal to an angle so the two triangles would have to be congruent now this is important because it's going to lead us to the technical definition of similarity and let's take a look at that all right so what's called or it involves what are known as similarity Transformations and let me read through this and let's talk about it two geometric figures are similar if there exists a similarity transformation that will map one figure onto the other that should sound familiar one figure onto the other this transformation is either a dilation alone or a dilation in combination with one or more rigid motions I want to just kind of Backtrack on this a little bit this is very very similar to the definition of congruence all right back from unit 2 we saw that two figures were congruent if there was a sequence of one or more rigid motions that would map one figure on top of the other then we kind of got out of that when we looked at you know side side side side angle side angle side angle things like that but ultimately speaking from a transformation perspective if we could use either a translation a rotation a reflection or a combination of those three to map one figure on top of the other then they were congruent now what we've got is if a combination of rigid motions with a dilation Maps one figure on top of the other then they're similar all right so similarity is all about dilations it's all about dilations okay um and again it's kind of a neat idea these two triangles are similar all right so if somehow I was able to take this triangle and let's say translate it so that a prime you know came over here and landed on D and then I was able to stretch the triangle out now one thing that's a little bit unfortunate about this program is that the dilation Center is always at the corner of this rectangular box so the center of dilation is always up here which is a little bit tricky but notice with that translation a dilation and then I guess another translation back right I was able to map this particular triangle on top of this one but make no doubt about it when I have this geometric figure and I'm doing this kind of thing with it what I'm doing is I'm dilating it with this as its Center whoops I guess I didn't really want to do that all right and therefore it allows me to kind of see that these two things are similar figures and that's what we're going to be working with actually for the rest of this lesson is just how can we identify and describe these similarity Transformations that will map one figure onto another one and therefore prove that the two figures are similar so let's um let's take a look at that now these can be fairly challenging and we're going to start off with one that's kind of tough let's take a look at this exercise two given that triangle ABC and DF below are similar with congruent angles marked give a similarity transformation that would map triangle ABC onto triangle DF using tracing paper to help be as specific as possible okay be as specific as possible all right and again this can be rather challenging especially if you don't have tracing paper to work with so pause for a moment take out some tracing paper Trace out triangle ABC like I've done I would even suggest ma like writing on point a point B Point C and then we'll take it from there all right let's take a look and let's let's talk about how we're going to do this all right so step one all right I'm going to translate one of I'm going to translate the whole triangle but I'm going to translate it in a way that's going to take one of the three vertices and map it so it lies on top of the corresponding one now for me the one that's most obvious is to transform it so that or translate it so that b Prime lands on C so translate triangle AB C so that b Maps 2 e now technically that means that we've got B Prime sitting here all right so I've taken this whole thing and I've kind of moved it over there now what I'd like to do notice how A and D are the same right with that one kind of angle Arc Mark there what I'd like to do is I'd like to rotate this thing up okay so I'm going to rotate and at this point it's really rotate triangle a prime B Prime C Prime about e so that and I know it didn't rotate it about e but so that it would now be a prime B Prime lies on Ed all right now now notice how similar this figure is now to those figures that we've been looking at all along with dilations right and we're now at the point where we've got a dilation right or where we should use a dilation now what I'd like is to stretch this red triangle out right so now I'm going to dilate I'm going to dilate triangle a prime B Prime C Prime with a center at e right I don't want that to move and now I have to give a dilation Factor now remember what I really want is I I sort of want this length to stretch up this length to stretch up so that it's that length right so what is that a little bit with a center at E and A dilation constant of k equals remember new divided by old and that would be the new length would be e divided by the old length which is actually ab and you could say able Prime d double Prime that would be okay as well everything up to now has been a rigid motion so it hasn't changed lengths now you might say well why did I why did I take length a d and divide it by length AB or sorry Ed and divide it by length AB as opposed to something else because I could have also gone with let's say EF / BC and the answer is it could be any of them all right as long as you take the newer length and you divide it by an older length that would be okay so Ed divided by AB would be fine let's see what else would be okay what else would be okay would be EF / BC or you could even say DF / AC okay any of those right will be the same constant because that's the idea of similarity all right it would stretch out and it would stretch out right eventually what would happen um again this is going to change my center of dilation but is that those two would sit right on top of each other okay so a series of rigid motions okay up with by a dilation and the idea is always when you've got that series of rigid motions is you eventually want a picture that kind of looks like this right where essentially the two triangles the two similar triangles are oriented in the same way and then you use that vertex that's been matched up as the center of your dilation then take as your dilation constant a new length that you want divided by an old length that you have they've got to be the corresponding side lengths and you have that similarity transformation it's going to be kind of tricky all right there's no doubt about it a little bit harder than those um rigid motion transformations to prove congruence let me step out of the way go ahead and write down anything you need to and then we'll move on and we'll do another one of these all right let's do it okay another one right away exercise number three in the following diagram it's known that bcn and ACM and ABC is similar to MNC give a similarity transformation that would map ABC onto MCN use tracing paper to help visualize the rigid motions okay same idea I want to somehow map ABC onto MNC using a similarity transformation now this is kind of critical right bcn and ACM these seem like they're they're kind of like tossin all that's really doing is telling you that these are collinear and these three points are collinear all right so let's try to get these things to match up now before we did it we began with a translation here I can go a little bit easier than that actually I can simply rotate triangle ABC about C by 180° I can just do this all right and if I had everything perfect on this picture it would just kind of line up like that and notice we already are kind of in that situation that took us two steps in the last problem to get to this one's actually a little bit easier right I'm already at a point where I'm kind of ready to dilate this thing I'm ready to shrink it down so it lies right on top of that so the second piece is the dilation now I'm going to dilate triangle a prime B Prime C Prime okay but what I'd like you to do now is tell me or try to figure out what is the center of dilation and what should I use as the scaling Factor all right let's talk about it well the center is no question with a center at C and A scaling factor of and here's where you have lots of choices again all right lots and lots and lots of choices we just have to go old length divided by new length so for instance I could take the length of BC which is really this length right here the length of BC all right um that's remember we always want to do new divided by old maybe even write that down new divided by old and the new length that I'd want BC to be would be NC or CN and then we would divide it by the old length BC all right so if I took the the new length that I want it to be NC or CN and I divide it by the old length BC or CB right that will then compress it down so that this length here becomes this length here and all the other ones will kind of follow along because all these corresponding sides are proportional to one another all right but those aren't our only choices okay another thing that you could do is you could say well I could take uh length MN all right that's my new length and I could divide it by my old length length AB now you only need one of these I'm just showing all the different choices and finally we could also do the length of CM or MC and we could divide it by the length of AC that scaling factor which is very important you'll certainly be tested on it repeatedly that scaling Factor will always be the newer length the triangle you're trying to transform into divided by an older length the triangle that you started with since we're starting with this larger triang triangle and going to this smaller one we always want to take a length off of this smaller one and divide it by a length on that larger one when we do that eventually what would happen is we would get two triangles maybe get two triangles that are congruent something like that all right similarity Transformations rigid body motions combined with dilations now a side note you could do the dilation first all right I always tend to do the dilation last I make sure I get it so that the two triangles are oriented in the same way and they share a Vertex they share one of the three points and then I use that point that shared Point as the center of dilation and then stretch it out using a constant that's a new length divided by an old length all right let's keep moving along ah the coordinate plane love being back into the coordinate plane there we can count we can measure we can see all sorts of stuff so let's take a look at this problem given triangle ABC is similar to Triangle DF on the grid shown what is the scaling Factor needed to dilate ABC so that it is congruent to DF show how you arrived at your answer so in this one we can actually come up with a hard number it's going to be okay my scaling factor is three or it's 1/3 or something like that see if you can figure out what the scaling factor is okay remember to dilate ABC so this is the old to get DF and that's the new go ahead think about that for a minute all right let's go through it well all you really have to do is come up with two sides that correspond to one another and divide their lengths taking the new length and dividing it by the older length two of the three lengths are very very convenient one of them is not and that's that hypotenuse for me I would simply take the length of De which looks to be 6 units long and divide it by the length of ab which is 4 units long and when I do that I would get a dilation constant of 1.5 or three Hales you could leave it as in fraction form you would get exactly the same thing if you took the length of EF which is 12 and you divided it by the length of BC which is 8 and that is also 3es or 1.5 again new divided by old if we were going the other way around by the way if they had said ah well what dilation constant would you have to use to do to make DF congruent to ABC then both of these fractions would be reversed and we'd get a dilation factor of 2/3 or 666 repeated 67 if you rounded it to a reasonable amount remember that direction matters because are we taking a smaller object and making it larger or vice versa let's take a look at letter B specify a similarity transformation that would map ABC onto DF all right well let me get rid of all this mess so that we can kind of look at this a little bit easier so now I want that step one step two step three thing why don't you go ahead and take a shot at this okay there's many different correct answers we'll take a look at it at one of them all right let's go through it um I'm going to even try to trace out ABC to see if I have got a working um model that I can trace with hopefully that'll work a lovely all right so let's do it um I want to do exactly what I did before I want to get two vertices matching up that I know have the same angle and probably the easiest thing to do is get B to match up with e so first thing I'm going to do is I'm going to translate I'm going to translate triangle ABC so that b maps to e ah so this thing is going to come up kind of like that right all right that's not bad now what I'd like to do though is I'd like to have it in the same orientation so I want to reflect that so the second thing I'm going to do is I'm going to reflect triangle a prime B Prime C Prime across what well you could describe it in many different ways I want to I want to reflect it across this vertical line so you could give that equation you could say I I'm going to cross the line xal 3 that's that vertical line or you could say or you could just say across EF right you don't even have to use the equation even though we're in the coordinate grid you could just say I'm going to reflect it across EF let's let's do that hopefully it will allow me to um let's flip this thing left right all right and that would end up whoops making it look kind of like that I won't waste your time by trying to get it exactly but now again look we've got the two triangles in exactly the same orientation so now all we need is that final dilation piece let me bring it up just a little bit and now let's dilate triangle a prime B Prime C Prime with a center at e and we've already figured out the constant and a scale factor huh there's my magic red scale factor of k equal 1.5 all right I like the coordinate grid because often times we can give that scale factor in a precise numerical form we don't just have to say oh it's de divided by BC or something like that we can literally say okay I did the division and I got 1.5 let me step out of the way for a second so you can take a look at this and then we'll we'll move on and to do more all right let's take a look at the last problem here we go exercise number five in the diagram below ab and C are collinear as well as a d and e it's also known that BD is parallel to c e all right I'm going to mark that on my diagram right away BD is parallel to C give them a dilation that would map ABD on to AC all right this is actually pretty simple why don't you go ahead and do it all right the reason that this is fairly simple is that we don't have to give any rigid body motions to get the triangles in the same orientation they're already sitting there and hopefully to get ABD the smaller triangle to map on to AC the larger triangle we would just simply do this use a center at a and a scale factor factor and this is where you got some Choice a scale factor of k equals new divided by old so I could say AC / AB or a e ided a d or C whoops divided by BD okay any of those ratios where I have taken the newer length in this case the larger length of the larger triangle and divided it by a length of the smaller triangle we'll get the job done we'll never know exactly what that is numerically because we don't have any dimensions on this picture now the last part of this triangle is pretty the last part of this triangle the last part of this problem is pretty challenging let's take a look at it it says Circle the proportion that is incorrect based on the fact that triangle ABD is similar to Triangle AC e all right now one thing I'm going to do is I'm going to draw these two triangles separately not a bad idea ever and then we'll kind of move the thing up a bit so smaller triangle is ABD all but none the larger triangle is AC and I can't even make a silly pun out of that one certainly not a ace okay so let's take a look right which of these proportions is incorrect well let's take a look at it it says AC divided by AB so what would that be let's I could kind of think of that as maybe move this down here and my letters as well okay so AC / AB um that's large / small equals c / BD that's large / small that's fine okay that's you know this larger side divided by the corresponding smaller side this larger side divided by this corresponding smaller side so that that one's fine now the angles the measure of angle a this thing divided by the measure of angle ABD so this one divided by this one or better yet this one they're the same angle is equal to the measure of angle a divided by the me measure of angle AC e well keep in mind that all the angles in the two triangles are the same all the corresponding angle pairs so a / AB D is equal to a / a c e e a C is the same because it's sort of single Arc ID double Arc equals single Arc ID double Arc right here it's not that the angles are proportional to another the angles are literally equal so you can really track that with the arcs if I've kind of got this measured out nicely single Arc / double Arc is equal to single Arc / double Arc so that works fine too let's take a look at number three number three says a d / AE right A D / AE that's going to be small ID AE large is equal to AC uhoh large / AB small uh no no the orders got to stay the same so if I'm taking a length from the small triangle dividing it by a corresponding length of the large triangle then I'd have to have a length from the small triangle divided by a length from the large triangle so this is actually the correct choice because it's an incorrect ratio we could look back at this as well right if we looked at this one a e c that's sort of a a double arc angle divided by a c e whoops wait wait a e c no that's triple uh triple Arc divided by AC c e double A DB triple divided by a BD double all right so that one's fine as well we don't really need to do that once we knew this one was wrong all right and again that can be a little bit tricky because we don't have numbers we don't have ratios some people may still be a little bit uncomfortable with ratios and Division and things like that but remember these proportions always have to stay the same small divid large has to be small ID large right because that's that dilation constant not with the angles but with the side lengths and that dilation constant multiplies all the sides and changes them all right so let's wrap it up similarity the idea that two figures have the same shape but different sizes what does it mean to have the same shape it means that all the angles are the same and all corresp oning side lengths are in proportion to one another so if I've got two triangles all the angles are the same but if one side in the larger triangle is twice the length of another side in the smaller triangle then all the sides have to be twice the length Okay so we then introduce the idea of a similarity transformation which is very similar to a congruence in other words I've got to use transformations to map one of these figures typically a triangle on top of the other the difference now is not only do we have the three classic rigid motions translations rotations and Reflections but we then throw in a dilation in order to stretch or compress one of these figures so that it lies directly on top of the other all right we'll work a little bit more with this in the next lesson for now I'd like to thank you for joining me for another Common Core geometry lesson by emath instruction my name is Kirk Wier and until next time keep thinking and keep solving problems [Music]
13813	https://news.ycombinator.com/item?id=26047603	Righty tighty, lefty loosey, right? This phrase always irked me, because when ... \| Hacker News Hacker Newsnew \| past \| comments \| ask \| show \| jobs \| submitlogin jstanleyon Feb 6, 2021 \| parent \| context \| favorite \| on: I told a cat food company how much they’d save by ... > Righty tighty, lefty loosey, right? This phrase always irked me, because when you're rotating something, it's not moving either left or right. The top of it moves right, the bottom moves left. Is it tightening or loosening? Further, the only time a phrase might be needed to work out which way to tighten something is when it's unusually-positioned, for example upside down or facing away from you, and in those cases "righty tighty, lefty loosey" gives exactly the wrong answer (to the extent that it gives an answer at all, since "clockwise" and "right" are not the same thing). I prefer the phrase "clockwise moves the thing you're turning away from you, anti-clockwise moves it towards you", which I'm hoping will catch on. tzson Feb 6, 2021 \| next[–] I'd guess it is related to the conventional way steering wheels work--turn clockwise to steer the vehicle right, counterclockwise to steer it left. Hence, clockwise gets associated with right and counterclockwise with left. I have no idea why steering wheels work that way. If I had to guess it would because most steering wheels are below eye level so it makes sense that you move the top of the wheel in the direction you want to go. Note that if you are holding the top half of a steering wheel that follows this convention, then when you turn the vehicle the centriwhatever force on you will tend to oppose the force you are using to turn the wheel. If the steering while used the opposite convention, that force would reinforce your force. That positive feedback would make it a lot easier to lose control during the turn. LiberatedLlamaon Feb 6, 2021 \| parent \| next[–] Without power steering, holding the steering wheel at the top gives you more leverage than holding it at the bottom. When holding the wheel at the top, turning right is a matter of moving your hands to the right. throwaway316943on Feb 13, 2021 \| parent \| prev \| next[–] ITT people who have never used a wrench. lostloginon Feb 6, 2021 \| prev \| next[–] > Righty tighty, lefty loosey I remember an old mechanic sarcastically saying this when I was a kid and as as he worked on an old car that had wheel nuts that did up the other way on one side of the car. It really pissed him off. A quick search seems to show that this was a thing. kelnoson Feb 6, 2021 \| parent \| next[–] I read the first answer there, but it didn't answer the related question: why don't all cars have that weird setup now? Is there a modern design feature that we have nowadays that allows for normal threading on the left-side wheels that keeps them from coming loose? jstanleyon Feb 7, 2021 \| root \| parent \| next[–] I think it's: > Improvements in metallurgy, changes in brake design, first to finned drums and then to disks, and changes in wheel nut architecture, from flat washer to conical design have effectively reduced the problem. leetcrewon Feb 6, 2021 \| prev \| next[–] I feel like this is the sort of thing that actually makes less sense when you develop explicit reasoning skills. as a child, "righty tighty, lefty loosey" made perfect sense. it never occurred to me that right/left could be relative to the bottom (from my perspective) of a jar lid. only when I started learning about math and coordinate systems did I think "right/left relative to what?". wlon Feb 6, 2021 \| prev \| next[–] I've adopted the right-hand rule: The screw goes the direction my thumb points when my right hand closes. And if it's something left-handed like a bike pedal, it's easy to just use the other hand. glipticon Feb 6, 2021 \| prev \| next[–] I prefer "clockwise lockwise". tshaddoxon Feb 6, 2021 \| prev \| next[–] The phrase is talking about the top surface of the circle as the circle is facing you, since that’s normally where your vantage point is when you’re turning something with your hand. You have to think things through more carefully if the circle is oriented differently (imagine reaching through a broken window to turn a doorknob). Dylan16807on Feb 6, 2021 \| prev \| next[–] > This phrase always irked me, because when you're rotating something, it's not moving either left or right. The top of it moves right, the bottom moves left. Is it tightening or loosening? And the answer is "You look at the top, of course, why would it mean the bottom?" davchanaon Feb 6, 2021 \| prev[–] Think of the vantage point of observer standing (& his land under his feet also rotating with circle causing him to always rotate along the rotation.) at the center of that circle. ACW will always be right moving to him, CW will be left. The other is, follow the fingers not thumb. Guidelines \| FAQ \| Lists \| API \| Security \| Legal \| Apply to YC \| Contact Search:
13814	https://www.quora.com/Man-has-to-do-a-piece-of-work-in-three-days-he-completes-1-5-of-it-on-yhe-first-day-and-5-8-of-the-remainder-on-the-second-day-what-fraction-of-the-piece-of-work-is-done-on-the-third-day	Man has to do a piece of work in three days.he completes 1/5 of it on yhe first day and 5/8 of the remainder on the second day.what fraction of the piece of work is done on the third day? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Labor and Employment Problem Solving Skills Fractions and Decimals Time and Work Arithmetical Calculations Arithmetic Fractions (mathematics) Logical Problem Solving 5 Man has to do a piece of work in three days.he completes 1/5 of it on yhe first day and 5/8 of the remainder on the second day.what fraction of the piece of work is done on the third day? All related (34) Sort Recommended Assistant Bot · Sep 6 Total work = 1. Work done on day 1 = 1/5. Remainder after day 1 = 1 − 1/5 = 4/5. Work done on day 2 = 5/8 of the remainder = (5/8) × (4/5) = 4/8 = 1/2. Remainder after day 2 (work done on day 3) = 1 − (day1 + day2) = 1 − (1/5 + 1/2). Compute 1/5 + 1/2 = (2/10 + 5/10) = 7/10, so day 3 = 1 − 7/10 = 3/10. Answer: 3/10. Upvote · Related questions More answers below If 3/5 of work is done on the first day, 2/3 of the remainder is completed on the second day, and 7/8 of what remained is done, what fraction of work still remains to be done? On the first day, a man did 3/35 of a job. On the next day, he completed another 2/5 of the job. What fraction of the work was left undone? Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? 6 workmen are employed to finish a certain work in 48 days but it is found that in 24 days, only 1/3 of the work was done. How many more men must be taken in order to do the piece of work in time? 20 workers can finish a work in 30 days. After how many days should 5 workers leave the job so the work is completed in 35 days? Judy Owen Former Retired secondary Maths Teacher at Telford School ·3y He completes !/5 +5/84/5 =8/40 +20/40 = 28/40 completed in 2 days. This leaves 12/40 to complete on the third day. Simplify this fraction to 3/10. QED Upvote · 9 1 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Catalino Lansangan Former Civil and Structural Inspector at Stantec Consulting (2010–2016) · Author has 206 answers and 347.5K answer views ·3y Work done: 1st day=1/5, 2nd day=5/8(4/5)=1/2, Total work done for 2 days=1/5+1/2=(2+5)/10=7/10 Remaining balance of work obliged to be completed on 3rd day: 10/10–7/10=3/10 (answer for the fraction of work done on the third day to complete because he has to complete a piece of work in 3 days. Upvote · Derek McNeil Studied at University of Waterloo · Author has 3.5K answers and 8.9M answer views ·2y Related Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? If 3/5 of the work is done on the first day, then 2/5 of the work is remaining. Of that remaining work, 3/4 is done the second day, leaving 1/4 remaining. Of that remaining work, 7/8 of the work is done on the third day, leaving 1/8 remaining. So, at the end of the first day, 2/5 of the total work is remaining. At the end of the second day, 1/4 of that 2/5 is remaining. At the end of the third day, 1/8 of that 1/4 of the original 2/5 is remaining. When dealing with fractions in math, “of” means to multiply. For example, 1/2 of 1/2 means 1/2 x 1/2. So, “1/8 of 1/4 of 2/5” means “1/8 x 1/4 x 2/5”. To mu Continue Reading If 3/5 of the work is done on the first day, then 2/5 of the work is remaining. Of that remaining work, 3/4 is done the second day, leaving 1/4 remaining. Of that remaining work, 7/8 of the work is done on the third day, leaving 1/8 remaining. So, at the end of the first day, 2/5 of the total work is remaining. At the end of the second day, 1/4 of that 2/5 is remaining. At the end of the third day, 1/8 of that 1/4 of the original 2/5 is remaining. When dealing with fractions in math, “of” means to multiply. For example, 1/2 of 1/2 means 1/2 x 1/2. So, “1/8 of 1/4 of 2/5” means “1/8 x 1/4 x 2/5”. To multiply fractions, you multiply the numerators together to provide the numerator of the fraction that is the product of the fractions. So, 1 x 1 x 2, which equals 2. Then you multiply the denominators to provide the denominator of the product fraction. So, 8 x 4 x 5, which equals 160. So, the product of the three fractions is 2/160. This can be simplified by dividing both the numerator and denominator of the fraction by 2, to give us the fraction 1/80. Therefore, 1/80 of the total work is left to be done at the end of the third day. Upvote · 9 2 Related questions More answers below Asha can complete 1/10 of a work in 3 days. How long will she take to do 2/5 of the work? 10 men, 6 women and 8 children can complete a piece of work in 24 days .in how many days can 45 women alone complete the same piece of work if 45 men alone complete it in 45 days? 7 men and 16 women can do a piece of work in 15 days. 5 men and 8 women will complete the work in how many days? A can do a work in 15 days and B in 20 days. If they work on it together for 5 days, then what is the fraction of the work that is left? 4 men can complete a piece of work in 18 days. 4 women can complete the same piece of work in 28 days. In how many days will 9 men and 12 women together complete the same piece of work? Samson Golohor Lived in Warri (2021–0) · Author has 358 answers and 256.2K answer views ·Feb 19 Related If 3/5 of work is done on the first day, 2/3 of the remainder is completed on the second day, and 7/8 of what remained is done, what fraction of work still remains to be done? To determine the fraction of work remaining after three days, First Day On the first day, 3/5 of the work is completed. The remaining work after the first day is: 1 - 3/5 = 2/5 Work Done on the Second Day On the second day, 2/3 of the remaining work 2/5 is completed. The work done on the second day 2/3 x 2/5 = 4/15 The remaining work after the second day is 2/5 - 4/15 = 6/15 - 4/15}= 2/15 Work Done on the Third Day On the third day, 7/8 of the remaining work 2/15 is completed. The work done on the third day is: 7/8 x 2/15} = 14/120} = 7/60 The remaining work after the third day is: 2/15} - 7/60 = 8/60 - 7 Continue Reading To determine the fraction of work remaining after three days, First Day On the first day, 3/5 of the work is completed. The remaining work after the first day is: 1 - 3/5 = 2/5 Work Done on the Second Day On the second day, 2/3 of the remaining work 2/5 is completed. The work done on the second day 2/3 x 2/5 = 4/15 The remaining work after the second day is 2/5 - 4/15 = 6/15 - 4/15}= 2/15 Work Done on the Third Day On the third day, 7/8 of the remaining work 2/15 is completed. The work done on the third day is: 7/8 x 2/15} = 14/120} = 7/60 The remaining work after the third day is: 2/15} - 7/60 = 8/60 - 7/60} = 1/60 The fraction of work that still remains to be done is: 1/60 Upvote · Pardha Saradhi Mandadi Arbitrator and Mediator at Self Employeed Professional (2014–present) · Author has 12.6K answers and 4.7M answer views ·3y Fraction of work done on first day=1/5 remainig work=1–1/5=(5–1)/5=4/5 Fraction of work done on first day=5/84/5=4/8=1/2 Fraction of work done on first 2 days=1/5+1/2=(2+5)/10=7/10 Fraction of work done on third day=1–7/10=(10–7)/10=3/10 Upvote · Sponsored by JetBrains Write better C++ code with less effort. Boost your efficiency with refactorings, code analysis, unit test support, and an integrated debugger. Download 999 897 Jayanta Mukherjee B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990) · Author has 43.6K answers and 11.1M answer views ·5y Related If 3/5 of work is done on the first day, 2/3 of the remainder is completed on the second day, and 7/8 of what remained is done, what fraction of work still remains to be done? On first day, (3 / 5) job was done. On second day [{1 - (3 / 5)} (2 / 3)] = (4 / 15) job was done. So, after second day {1 - (3 / 5) - (4 / 15)} = (2 / 15) job is left. On third day (2 / 15) (7 / 8) = (7/ 60) job is done. So, after third day {(2 / 15) - (7 / 60)} = (1 / 60) of total work remains to be done. Upvote · 9 4 Laura Kay Posey Education, K-6th from SWT (Graduated 1966) · Author has 6.5K answers and 2M answer views ·2y Related Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? ANSWER: 1 25/100 = 1.25 = 1 1/4 or 5/4 3/5, 1ST DAY, 3/4 remains on 2ND DAY, 7/8 of any/all 3rd DAY, what fraction left 5/5 = 100% A. 1st Day 3/5 of 100 = 3/5 × 100/1 = 300/5 = 60 B. 100 --60 = 40 [ 40/100= 4 / 10= 2 / 5 ] C. 2 nd Day 3/4 of 40 = 3/4 × 40/ 1 = 120 / 4 = 30 D. 40 -- 30 = 10 ( 10 /100 = 1/10] E. 3rd Day 7 / 8 of 10 = 7/8 × 10/1 = 70/8 70 / 8 = 8 6/8 = 8 3/4 = 8.75 F. 10 — 8.75 = 1.25 10.00 —8.75 = 1.25 = 1.25 = 1 25/100 = 1 1/4 ( 3 /5 of 100 = 100 - 60 = 40, 1st Day. ) ( 3 / 4 of 40 = 40 — 30 = 10, 2 nd Day. ) ( 7/8 of 10 = 10 — 8 3/4 = 3 rd Day ) FRACTION WORK Continue Reading ANSWER: 1 25/100 = 1.25 = 1 1/4 or 5/4 3/5, 1ST DAY, 3/4 remains on 2ND DAY, 7/8 of any/all 3rd DAY, what fraction left 5/5 = 100% A. 1st Day 3/5 of 100 = 3/5 × 100/1 = 300/5 = 60 B. 100 --60 = 40 [ 40/100= 4 / 10= 2 / 5 ] C. 2 nd Day 3/4 of 40 = 3/4 × 40/ 1 = 120 / 4 = 30 D. 40 -- 30 = 10 ( 10 /100 = 1/10] E. 3rd Day 7 / 8 of 10 = 7/8 × 10/1 = 70/8 70 / 8 = 8 6/8 = 8 3/4 = 8.75 F. 10 — 8.75 = 1.25 10.00 —8.75 = 1.25 = 1.25 = 1 25/100 = 1 1/4 ( 3 /5 of 100 = 100 - 60 = 40, 1st Day. ) ( 3 / 4 of 40 = 40 — 30 = 10, 2 nd Day. ) ( 7/8 of 10 = 10 — 8 3/4 = 3 rd Day ) FRACTION WORK REMAINING 10 — 8 3 / 4, for work on another day 10 = 9 4 / 4 —— 8 3 / 4 = === 1 1/4 Upvote · Sponsored by MailerLite Your free AI assistant completes time-consuming tasks in seconds. Once you connect your AI tool to MailerLite, you can start asking questions and giving it tasks. Learn More 9 1 Janet Heberling Lives in San Francisco, CA (2022–present) · Author has 21.5K answers and 9.4M answer views ·2y Related Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? (1−3 5)(1−3 4)(1−7 8)=2 5⋅1 4⋅1 8=1 80(1−3 5)(1−3 4)(1−7 8)=2 5⋅1 4⋅1 8=1 80 Upvote · 9 1 Chris Crocker BS in Chemisrty&Biology, University of Providence (Graduated 2004) ·2y Related Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? I look at this in decimals first. Lets start with 1.0000. 3/5 is the same as 6/10 or 0.6000, so subtract 0.6000 from 1.0000 and the balance of work to complete after day one is 0.4000 (40% remain). On day two, 3/4 of the work is completed. So, this is 0.3000 subtracted from 0.4000 leaving 0.1000 at the end of day two. On day three, 7/8 of the remaining is completed, which is 0.1000 - 0.0875. This leaves 0.0125 or 1.25% of the work remaining after day three. Since 0.0125 is equal to 125/10000, or 25/2000, or 5/400, or any multiple of these you could imagine such as (125/10000) x 7/7= 875/70000, Continue Reading I look at this in decimals first. Lets start with 1.0000. 3/5 is the same as 6/10 or 0.6000, so subtract 0.6000 from 1.0000 and the balance of work to complete after day one is 0.4000 (40% remain). On day two, 3/4 of the work is completed. So, this is 0.3000 subtracted from 0.4000 leaving 0.1000 at the end of day two. On day three, 7/8 of the remaining is completed, which is 0.1000 - 0.0875. This leaves 0.0125 or 1.25% of the work remaining after day three. Since 0.0125 is equal to 125/10000, or 25/2000, or 5/400, or any multiple of these you could imagine such as (125/10000) x 7/7= 875/70000, would answer the question. However the smallest reduced fraction is 1/80. Upvote · Sponsored by State Bank of India Stay Informed!! Stay Protected!! Our Contact Centre calls only from numbers beginning with 1600 or 140 series. Learn More 99 46 Naga Prabha Former Income Tax Officer at Income Tax Department, Government of India (1982–2020) · Author has 4.8K answers and 481.3K answer views ·3y Related A person completed 4/9 part of work on 1st day and 3/9 part of work on 2nd day. How much of part of the work does that person still have to complete on third day? 1st day work =4/9 2nd day work=3/9 Remaining work =1-(4/9+3/9)=1-7/9=2/9 3rd day work to be completed =2/9 Upvote · Richard Chinn MS from University of South Florida (Graduated 1989) · Author has 57 answers and 518.8K answer views ·2y Related Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? At the end of the first day, 3/5 is done and 2/5 remains. At the end of the second day, 3/4 x 2/5 is done and 1/4 x 2/5 = 2/20 = 1/10 remains At the end of the third day, 1/10 x 7/8 is done and 1/10 x 1/8 remains = 1/80 Upvote · Louis M. Rappeport B.S. from University of California, Berkeley · Author has 7.7K answers and 6.7M answer views ·1y Related Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? Given: 3/5ths of the job is done on the first day, leaving 2/5ths behind. 3/4ths of the remaining work is then done, leaving 1/4 x 2/5=1/10 of the original job undone The third day, 7/8ths of the remaining work is done, leaving 1/8 x 1/10, or 1/80th of the job unfinished. Upvote · Related questions If 3/5 of work is done on the first day, 2/3 of the remainder is completed on the second day, and 7/8 of what remained is done, what fraction of work still remains to be done? On the first day, a man did 3/35 of a job. On the next day, he completed another 2/5 of the job. What fraction of the work was left undone? Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? 6 workmen are employed to finish a certain work in 48 days but it is found that in 24 days, only 1/3 of the work was done. How many more men must be taken in order to do the piece of work in time? 20 workers can finish a work in 30 days. After how many days should 5 workers leave the job so the work is completed in 35 days? Asha can complete 1/10 of a work in 3 days. How long will she take to do 2/5 of the work? 10 men, 6 women and 8 children can complete a piece of work in 24 days .in how many days can 45 women alone complete the same piece of work if 45 men alone complete it in 45 days? 7 men and 16 women can do a piece of work in 15 days. 5 men and 8 women will complete the work in how many days? A can do a work in 15 days and B in 20 days. If they work on it together for 5 days, then what is the fraction of the work that is left? 4 men can complete a piece of work in 18 days. 4 women can complete the same piece of work in 28 days. In how many days will 9 men and 12 women together complete the same piece of work? A person can do 3/4 of a work in 12 days. In how many days can he finish 1/8 of the work? A can do 3/4 of the work in 12 days. In how many days can he complete 1/8 of the work? 4 men can complete a piece of work in 2 days. 4 women can complete the same piece of work in 4 days. Whereas 5 children can complete a piece of work in 4 days. If 2 men, 4 women, and 10 children work together, how many days can the work be completed? If a can do a piece of work in 12 days, how many days does he take to do 1/3 of work? 8 men can complete a piece of work in 20 days. 8 women can complete the same work in 32 days. In how many days will 5 men and 8 women together complete the same work? Related questions If 3/5 of work is done on the first day, 2/3 of the remainder is completed on the second day, and 7/8 of what remained is done, what fraction of work still remains to be done? On the first day, a man did 3/35 of a job. On the next day, he completed another 2/5 of the job. What fraction of the work was left undone? Three-fifth of work is done on the first day. On the second day, 3/4 of the remainder is completed. If the third day 7/8 of what remained is done, what fraction of work remains to be done? 6 workmen are employed to finish a certain work in 48 days but it is found that in 24 days, only 1/3 of the work was done. How many more men must be taken in order to do the piece of work in time? 20 workers can finish a work in 30 days. After how many days should 5 workers leave the job so the work is completed in 35 days? Asha can complete 1/10 of a work in 3 days. How long will she take to do 2/5 of the work? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
13815	https://math.stackexchange.com/questions/191711/how-many-rows-and-columns-are-in-an-m-x-n-matrix	matrices - How many rows and columns are in an m x n matrix? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How many rows and columns are in an m x n matrix? Ask Question Asked 13 years ago Modified1 year, 8 months ago Viewed 186k times This question shows research effort; it is useful and clear 53 Save this question. Show activity on this post. A simple question: By definition, does an m x n matrix have m rows and n columns, or is it vice versa? matrices Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Sep 6, 2012 at 2:15 Anderson GreenAnderson Green 1,053 8 8 gold badges 17 17 silver badges 26 26 bronze badges 3 Yes it's always "{number of rows} by {number of columns}"Colonel Panic –Colonel Panic 2015-02-18 16:15:56 +00:00 Commented Feb 18, 2015 at 16:15 2 You can name the variables how you like though. Curiously "m by n matrix" is about twice as common as "n by m matrix" in Google search results.Colonel Panic –Colonel Panic 2015-02-18 16:19:07 +00:00 Commented Feb 18, 2015 at 16:19 2 @ColonelPanic, that's probably because for a matrix A A operating on an n n dimensional vector x x (i.e. A x=y) y is m dimensional. In other words, it puts the input dimension before the output dimension alphabetically.Shep –Shep 2015-04-03 01:42:26 +00:00 Commented Apr 3, 2015 at 1:42 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 49 Save this answer. Show activity on this post. An m×n matrix has m rows and n columns. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 13, 2015 at 21:52 Zhanxiong 15.2k 2 2 gold badges 29 29 silver badges 61 61 bronze badges answered Sep 6, 2012 at 2:16 JamesJames 1,320 5 5 gold badges 21 21 silver badges 22 22 bronze badges 9 can you provide a reference/citation for this?Anderson Green –Anderson Green 2012-09-06 02:16:47 +00:00 Commented Sep 6, 2012 at 2:16 3 All the textbooks i have read (both cs and math) have used this notation. For example, Strang's Introduction to Linear Algebra 4th.James –James 2012-09-06 02:17:54 +00:00 Commented Sep 6, 2012 at 2:17 You said "almost all". Were there any exceptions?Anderson Green –Anderson Green 2012-09-06 02:19:24 +00:00 Commented Sep 6, 2012 at 2:19 1 Sorry, I meant all.James –James 2012-09-06 02:20:12 +00:00 Commented Sep 6, 2012 at 2:20 1 @IvanBalashov In Numpy the first dimension is the row, not the column.bfontaine –bfontaine 2016-10-22 07:36:26 +00:00 Commented Oct 22, 2016 at 7:36 \|Show 4 more comments This answer is useful 1 Save this answer. Show activity on this post. I suggest you always to check the notation on the book which you are using. I found sometimes this notation with different meaning. In advanced books, for example. Even the notation for linear maps as matrices. Sometimes they write x T. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Sep 6, 2012 at 2:28 SigurSigur 6,680 3 3 gold badges 27 27 silver badges 47 47 bronze badges 2 What does xT refer to in this case?Anderson Green –Anderson Green 2012-09-06 02:36:02 +00:00 Commented Sep 6, 2012 at 2:36 It is the notation for the image of x by the linear map T. Usually we write T(x) or T x.Sigur –Sigur 2012-09-06 02:38:50 +00:00 Commented Sep 6, 2012 at 2:38 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Always check and make sure you have the right convention for the occasion. Usually m x n is rows x columns. I like to remember this as being in REVERSE alphabetical order - Rows by Columns, or R first then C. However, in Boyce & DiPrima's book "Elementary Differential Equations and Boundary Value Problems" an m x n matrix has m vertical columns and n horizontal rows. However, when addressing elements within a matrix, it's the opposite. The element "a sub i,j" references the element in the ith row and jth column. Lesson? Always check to make sure you have the correct convention! Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jun 4, 2015 at 21:56 SeanSean 21 1 1 bronze badge 2 1 so much for the "universal language of mathematics" :(Robert Lugg –Robert Lugg 2019-01-21 17:32:48 +00:00 Commented Jan 21, 2019 at 17:32 @RobertLugg exactly. x columns by y rows. i columns by j rows. Then algebra with the 'hold my beer' goes with m rows by n columns. I almost have feelings about this...AndrewBenjamin –AndrewBenjamin 2023-04-19 13:15:34 +00:00 Commented Apr 19, 2023 at 13:15 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Yes... It's m-rows and n-Columns. Here is an example, how you can generate and read a matrix in JavaScript :) ``` let createMatrix = (m, n) => { let [row, column] = , rowColumn = m n for (let i = 1; i <= rowColumn; i++) { column.push(i) if (i % n === 0) { row.push(column) column = [] } } return row } let setColorForEachElement = (matrix, colors) => { let row = matrix.map(row => { let column = row.map((column, key) => { return { number: column, color: colors[key] } }) return column }) return row } const colors = ['red', 'green', 'blue', 'purple', 'brown', 'yellow', 'orange', 'grey'] const matrix = createMatrix(6, 8) const colorApi = setColorForEachElement(matrix, colors) let table ='' colorApi.forEach(row => { table+='' row.forEach(column => table +=<td style='background: ${column.color};'>${column.number}<td> ) table+='' }) document.write(table); ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 13, 2019 at 23:21 answered Mar 13, 2019 at 23:08 Driton HaxhiuDriton Haxhiu 11 3 3 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Sometimes one can forget which number maps to row or column numbers. However, we don't need remember it. Just remember the Matrix multiplication: For Matrix A, B, and R = A × B, we have A (m × p) × B (p × n) = R (m × n) Now from the definition of multiplication, you will easily find that a (m × n) matrix has m rows and n columns. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 11, 2024 at 4:08 Robin HsuRobin Hsu 101 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions matrices See similar questions with these tags. 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13816	https://pubchem.ncbi.nlm.nih.gov/compound/Ammonium-bisulfide	Ammonium bisulfide \| NH4SH \| CID 25515 - PubChem An official website of the United States government Here is how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. NIH National Library of Medicine NCBI PubChem About Docs Submit Contact Search PubChem compound Summary Ammonium bisulfide PubChem CID 25515 Structure Primary Hazards Laboratory Chemical Safety Summary (LCSS) Datasheet Molecular Formula H 4 N.HS NH 4 SH H 5 NS (NH 4)HS Synonyms Ammonium bisulfide Ammonium hydrosulfide Ammonium sulfhydrate Ammonium hydrogen sulfide 12124-99-1 View More... Molecular Weight 51.11 g/mol Computed by PubChem 2.2 (PubChem release 2025.04.14) Component Compounds CID 402 (Hydrogen Sulfide) CID 222 (Ammonia) Dates Create: 2005-08-08 Modify: 2025-09-13 Description Ammonium hydrosulfide, solution is a clear, yellowish liquid. Kept basic with NaOH, as acid will release hydrogen sulfide gas. Technical grade is 40-44%. Used in photography, textiles, synthetic flavors, coloring brasses, bronzes, and iron control. CAMEO Chemicals 1 Structures 1.1 2D Structure Structure Search Get Image Download Coordinates Chemical Structure Depiction Full screen Zoom in Zoom out PubChem 1.2 3D Status Conformer generation is disallowed since mixture or salt PubChem 1.3 Crystal Structures COD Number 1010249 Associated Article West, C D. The Crystal Structures of Some Alkali Hydrosulfides and Monosulfides.. Zeitschrift fuer Kristallographie, Kristallgeometrie, Kristallphysik, Kristallchemie (-144,1977) 1934;88:97-115. Crystal Structure Depiction Hermann-Mauguin space group symbol P 4/n m m :1 Hall space group symbol P 4ab 2ab -1ab Space group number 129 a 6.011 Å b 6.011 Å c 4.009 Å α 90 ° β 90 ° γ 90 ° Z 2 Z' 0.125 Component 1 x [NH4+] (Ammonium) Component 1 x [SH-] (Hydrosulfide) Crystallography Open Database (COD) 2 Names and Identifiers 2.1 Computed Descriptors 2.1.1 IUPAC Name azanium;sulfanide Computed by Lexichem TK 2.7.0 (PubChem release 2025.04.14) PubChem 2.1.2 InChI InChI=1S/H3N.H2S/h1H3;1H2 Computed by InChI 1.07.2 (PubChem release 2025.04.14) PubChem 2.1.3 InChIKey HIVLDXAAFGCOFU-UHFFFAOYSA-N Computed by InChI 1.07.2 (PubChem release 2025.04.14) PubChem 2.1.4 SMILES [NH4+].[SH-] Computed by OEChem 2.3.0 (PubChem release 2025.04.14) PubChem 2.2 Molecular Formula H 4 N.HS Australian Industrial Chemicals Introduction Scheme (AICIS) NH 4 SH CAMEO Chemicals H 5 NS (NH 4)HS ILO-WHO International Chemical Safety Cards (ICSCs) H 5 NS Computed by PubChem 2.2 (PubChem release 2025.04.14) PubChem 2.3 Other Identifiers 2.3.1 CAS 12124-99-1 Australian Industrial Chemicals Introduction Scheme (AICIS); CAMEO Chemicals; CAS Common Chemistry; ChemIDplus; EPA Chemicals under the TSCA; EPA DSSTox; European Chemicals Agency (ECHA); FDA Global Substance Registration System (GSRS); ILO-WHO International Chemical Safety Cards (ICSCs); New Zealand Environmental Protection Authority (EPA) 2.3.2 Deprecated CAS 581801-02-7, 7725-91-9 ChemIDplus; EPA Chemicals under the TSCA 2.3.3 European Community (EC) Number 235-184-3 European Chemicals Agency (ECHA) 2.3.4 UNII A824D6LXMB FDA Global Substance Registration System (GSRS) 2.3.5 UN Number 2683 (AMMONIUM HYDROSULFIDE, SOLUTION) CAMEO Chemicals 2.3.6 DSSTox Substance ID DTXSID20894064 EPA DSSTox 2.3.7 ICSC Number 1035 ILO-WHO International Chemical Safety Cards (ICSCs) 2.3.8 Wikidata Q197523 Wikidata 2.3.9 Wikipedia Ammonium hydrosulfide Wikipedia 2.4 Synonyms 2.4.1 Depositor-Supplied Synonyms Ammonium bisulfide Ammonium hydrosulfide Ammonium sulfhydrate Ammonium hydrogen sulfide 12124-99-1 Ammonium mercaptan Ammonium hydrogensulphide Monoammonium sulfide Ammonium sulfide ((NH4)(SH)) Sirnik amonny Sirnik amonny [Czech] EINECS 235-184-3 A824D6LXMB AMMONIUM BISULFIDE [MI] DTXSID20894064 RefChem:557670 DTXCID301324107 235-184-3 azanium;sulfanide Ammonium sulfide (NH4SH) UNII-A824D6LXMB Schwefelammonium Ammonium hydrogensulfide Q197523 PubChem 3 Chemical and Physical Properties 3.1 Computed Properties Property Name Property Value Reference Property Name Molecular Weight Property Value 51.11 g/mol Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Hydrogen Bond Donor Count Property Value 2 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Hydrogen Bond Acceptor Count Property Value 1 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Rotatable Bond Count Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Exact Mass Property Value 51.01427034 Da Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Monoisotopic Mass Property Value 51.01427034 Da Reference Computed by PubChem 2.2 (PubChem release 2025.04.14) Property Name Topological Polar Surface Area Property Value 2 Å² Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Heavy Atom Count Property Value 2 Reference Computed by PubChem Property Name Formal Charge Property Value 0 Reference Computed by PubChem Property Name Complexity Property Value 0 Reference Computed by Cactvs 3.4.8.18 (PubChem release 2025.04.14) Property Name Isotope Atom Count Property Value 0 Reference Computed by PubChem Property Name Defined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Atom Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Defined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Undefined Bond Stereocenter Count Property Value 0 Reference Computed by PubChem Property Name Covalently-Bonded Unit Count Property Value 2 Reference Computed by PubChem Property Name Compound Is Canonicalized Property Value Yes Reference Computed by PubChem (release 2025.04.14) PubChem 3.2 Experimental Properties 3.2.1 Physical Description Ammonium hydrosulfide, solution is a clear, yellowish liquid. Kept basic with NaOH, as acid will release hydrogen sulfide gas. Technical grade is 40-44%. Used in photography, textiles, synthetic flavors, coloring brasses, bronzes, and iron control. CAMEO Chemicals White to yellow hygroscopic solid; [ICSC] Decomposes when moisture or >0 deg C; [Hawley] Technical grade is 40-44% and kept basic with sodium hydroxide; [CAMEO] Solution has strong odor of rotten eggs and ammonia; [CHRIS] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases HYGROSCOPIC WHITE-TO-YELLOW CRYSTALS WITH CHARACTERISTIC ODOUR. ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.2 Solubility Solubility in water, g/100ml: 128.1 ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.3 Vapor Pressure 390.0 [mmHg] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Vapor pressure, kPa at 22Â °C: 52 ILO-WHO International Chemical Safety Cards (ICSCs) 3.2.4 Decomposition room temperature. This produces toxic and corrosive gases including ammonia and hydrogen sulfide. Reacts with acids. This produces hydrogen sulfide and sulfur oxides. Reacts violently with oxidants. This generates fire or explosion hazard. ILO-WHO International Chemical Safety Cards (ICSCs) 3.3 Chemical Classes Other Classes ->Sulfur Compounds Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 4 Related Records 4.1 Related Compounds with Annotation Follow these links to do a live 2D search or do a live 3D search for this compound, sorted by annotation score. This section is deprecated (see the neighbor discontinuation help page for details), but these live search links provide equivalent functionality to the table that was previously shown here. PubChem 4.2 Component Compounds CID 402 (Hydrogen Sulfide) CID 222 (Ammonia) PubChem 4.3 Related Compounds Mixtures, Components, and Neutralized Forms Count 2 Similar Compounds (2D) View in PubChem Search Similar Conformers (3D) View in PubChem Search PubChem 4.4 Substances 4.4.1 PubChem Reference Collection SID 505553399 PubChem 4.4.2 Related Substances Same Count 31 PubChem 4.4.3 Substances by Category PubChem 5 Chemical Vendors PubChem 6 Use and Manufacturing 6.1 Uses EPA CPDat Chemical and Product Categories The Chemical and Products Database, a resource for exposure-relevant data on chemicals in consumer products, Scientific Data, volume 5, Article number: 180125 (2018), DOI:10.1038/sdata.2018.125 EPA Chemical and Products Database (CPDat) Sources/Uses Used in lubricants; [Merck Index] Used in photography, textiles, synthetic flavors, for coloring brasses and bronzes, and for iron control (soda ash production); [CAMEO] Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Industrial Processes with risk of exposure Photographic Processing [Category: Other] Textiles (Printing, Dyeing, or Finishing) [Category: Industry] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases Activities with risk of exposure Applying metallic patinas [Category: Hobbies] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 6.2 General Manufacturing Information EPA TSCA Commercial Activity Status Ammonium sulfide ((NH4)(SH)): ACTIVE EPA Chemicals under the TSCA 7 Safety and Hazards 7.1 Hazards Identification 7.1.1 GHS Classification 1 of 2 items View All Pictogram(s) Signal Danger GHS Hazard Statements H312 (31.3%): Harmful in contact with skin [Warning Acute toxicity, dermal] H314 (100%): Causes severe skin burns and eye damage [Danger Skin corrosion/irritation] H400 (100%): Very toxic to aquatic life [Warning Hazardous to the aquatic environment, acute hazard] Precautionary Statement Codes P260, P264, P273, P280, P301+P330+P331, P302+P352, P302+P361+P354, P304+P340, P305+P354+P338, P316, P317, P321, P362+P364, P363, P391, P405, and P501 ECHA C&L Notifications Summary Aggregated GHS information provided per 115 reports by companies from 3 notifications to the ECHA C&L Inventory. Each notification may be associated with multiple companies. Information may vary between notifications depending on impurities, additives, and other factors. The percentage value in parenthesis indicates the notified classification ratio from companies that provide hazard codes. Only hazard codes with percentage values above 10% are shown. For more detailed information, please visit ECHA C&L website. European Chemicals Agency (ECHA) 7.1.2 Hazard Classes and Categories Acute Tox. 4 (31.3%) Skin Corr. 1B (100%) Aquatic Acute 1 (100%) European Chemicals Agency (ECHA) View More... 7.1.3 Health Hazards Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: May cause toxic effects if inhaled or ingested. Contact with substance may cause severe burns to skin and eyes. Fire will produce irritating, corrosive and/or toxic gases. Vapors may cause dizziness or asphyxiation, especially when in closed or confined areas. Runoff from fire control or dilution water may cause environmental contamination. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals 7.1.4 Fire Hazards Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: Flammable/combustible material. May be ignited by heat, sparks or flames. Vapors may form explosive mixtures with air. Vapors may travel to source of ignition and flash back. Most vapors are heavier than air. They will spread along the ground and collect in low or confined areas (sewers, basements, tanks, etc.). Vapor explosion hazard indoors, outdoors or in sewers. Those substances designated with a (P) may polymerize explosively when heated or involved in a fire. Runoff to sewer may create fire or explosion hazard. Containers may explode when heated. Many liquids will float on water. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals Combustible. Many reactions may cause fire or explosion. Gives off irritating or toxic fumes (or gases) in a fire. ILO-WHO International Chemical Safety Cards (ICSCs) 7.1.5 Hazards Summary Crystals decompose at room temperature forming hydrogen sulfide (may be fatal) and ammonia; A severe skin irritant; [Merck Index] A severe skin, eye, and respiratory tract irritant; Can be absorbed through skin; [ICSC] Produces hydrogen sulfide on mixing with water; A reducing agent; [CAMEO] Aqueous solution: Causes severe eye burns and severe skin irritation; Inhalation of 500 ppm for 30 minutes causes headache, dizziness, and bronchial pneumonia; 600 ppm for 30 minutes can be fatal; Can be absorbed through skin resulting in hydrogen sulfide poisoning; [CHRIS] Causes mydriasis, dyspnea, and respiratory stimulation in lethal-dose animal studies (dermal, intraperitoneal, intravenous, and oral routes); [RTECS] See Hydrogen sulfide. Merck Index - O'Neil MJ, Heckelman PE, Dobbelaar PH, Roman KJ (eds). The Merck Index, An Encyclopedia of Chemicals, Drugs, and Biologicals, 15th Ed. Cambridge, UK: The Royal Society of Chemistry, 2013. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 7.2 First Aid Measures Inhalation First Aid Fresh air, rest. Refer for medical attention. ILO-WHO International Chemical Safety Cards (ICSCs) Skin First Aid Remove contaminated clothes. Rinse skin with plenty of water or shower. Refer for medical attention . ILO-WHO International Chemical Safety Cards (ICSCs) Eye First Aid First rinse with plenty of water for several minutes (remove contact lenses if easily possible), then refer for medical attention. ILO-WHO International Chemical Safety Cards (ICSCs) Ingestion First Aid Rinse mouth. Refer for medical attention . ILO-WHO International Chemical Safety Cards (ICSCs) 7.2.1 First Aid Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: Refer to the "General First Aid" section. Specific First Aid: For corrosives, in case of contact, immediately flush skin or eyes with running water for at least 30 minutes. Additional flushing may be required. In case of burns, immediately cool affected skin for as long as possible with cold water. Do not remove clothing if adhering to skin. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals 7.3 Fire Fighting Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: Some of these materials may react violently with water. SMALL FIRE: Dry chemical, CO2, water spray or alcohol-resistant foam. LARGE FIRE: Water spray, fog or alcohol-resistant foam. If it can be done safely, move undamaged containers away from the area around the fire. Dike runoff from fire control for later disposal. Do not get water inside containers. FIRE INVOLVING TANKS, RAIL TANK CARS OR HIGHWAY TANKS: Fight fire from maximum distance or use unmanned master stream devices or monitor nozzles. Cool containers with flooding quantities of water until well after fire is out. Withdraw immediately in case of rising sound from venting safety devices or discoloration of tank. ALWAYS stay away from tanks in direct contact with flames. For massive fire, use unmanned master stream devices or monitor nozzles; if this is impossible, withdraw from area and let fire burn. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals Use foam, powder, carbon dioxide. In case of fire: keep drums, etc., cool by spraying with water. ILO-WHO International Chemical Safety Cards (ICSCs) 7.4 Accidental Release Measures 7.4.1 Isolation and Evacuation Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: IMMEDIATE PRECAUTIONARY MEASURE: Isolate spill or leak area for at least 50 meters (150 feet) in all directions. SPILL: Increase the immediate precautionary measure distance, in the downwind direction, as necessary. FIRE: If tank, rail tank car or highway tank is involved in a fire, ISOLATE for 800 meters (1/2 mile) in all directions; also, consider initial evacuation for 800 meters (1/2 mile) in all directions. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals 7.4.2 Spillage Disposal Evacuate danger area! Consult an expert! Personal protection: chemical protection suit including self-contained breathing apparatus. Ventilation. Remove all ignition sources. Sweep spilled substance into covered air-tight containers. ILO-WHO International Chemical Safety Cards (ICSCs) 7.5 Handling and Storage 7.5.1 Nonfire Spill Response Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: ELIMINATE all ignition sources (no smoking, flares, sparks or flames) from immediate area. All equipment used when handling the product must be grounded. Do not touch or walk through spilled material. Stop leak if you can do it without risk. Prevent entry into waterways, sewers, basements or confined areas. A vapor-suppressing foam may be used to reduce vapors. Absorb with earth, sand or other non-combustible material. For hydrazine, absorb with DRY sand or inert absorbent (vermiculite or absorbent pads). Use clean, non-sparking tools to collect absorbed material. LARGE SPILL: Dike far ahead of liquid spill for later disposal. Water spray may reduce vapor, but may not prevent ignition in closed spaces. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals 7.5.2 Safe Storage Separated from acids, oxidants and food and feedstuffs. Cool. Dry. Well closed. Keep in a well-ventilated room. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6 Exposure Control and Personal Protection 7.6.1 Inhalation Risk A harmful contamination of the air can be reached very quickly on evaporation of this substance at 20Â °C. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.2 Effects of Short Term Exposure The substance is severely irritating to the eyes, skin and respiratory tract. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.3 Personal Protective Equipment (PPE) Excerpt from ERG Guide 132 [Flammable Liquids - Corrosive]: Wear positive pressure self-contained breathing apparatus (SCBA). Wear chemical protective clothing that is specifically recommended by the manufacturer when there is NO RISK OF FIRE. Structural firefighters' protective clothing provides thermal protection but only limited chemical protection. (ERG, 2024) 2024 Emergency Response Guidebook, CAMEO Chemicals 7.6.4 Fire Prevention NO open flames. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.5 Exposure Prevention STRICT HYGIENE! IN ALL CASES CONSULT A DOCTOR! ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.6 Inhalation Prevention Use local exhaust or breathing protection. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.7 Skin Prevention Protective gloves. Protective clothing. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.8 Eye Prevention Wear face shield or eye protection in combination with breathing protection. ILO-WHO International Chemical Safety Cards (ICSCs) 7.6.9 Ingestion Prevention Do not eat, drink, or smoke during work. ILO-WHO International Chemical Safety Cards (ICSCs) 7.7 Stability and Reactivity 7.7.1 Air and Water Reactions Water soluble, with release of hydrogen sulfide, especially when acidified. CAMEO Chemicals 7.7.2 Reactive Group Bases, Strong Sulfides, Inorganic Bases, Weak Water and Aqueous Solutions CAMEO Chemicals 7.7.3 Reactivity Profile AMMONIUM HYDROSULFIDE, SOLUTION is a reducing agent. Reacts with oxidizing agents, including inorganic oxoacids, organic peroxides and epoxides. Reacts vigorously with acids to release hydrogen sulfide gas. Evolves poisonous ammonia on contact with strong bases. CAMEO Chemicals 7.8 Transport Information 7.8.1 DOT Label Corrosive Poison Flammable Liquid CAMEO Chemicals 7.8.2 Packaging and Labelling Airtight. Do not transport with food and feedstuffs. ILO-WHO International Chemical Safety Cards (ICSCs) 7.9 Regulatory Information The Australian Inventory of Industrial Chemicals Chemical: Ammonium sulfide, (NH4)(SH) Australian Industrial Chemicals Introduction Scheme (AICIS) REACH Registered Substance Status: Cease Manufacture Update: 27-05-2018 European Chemicals Agency (ECHA) REACH Restricted Substance Restricted substance: Ammonium hydrogen sulphide EC: 235-184-3 Restriction condition document: PDF link European Chemicals Agency (ECHA) New Zealand EPA Inventory of Chemical Status Ammonium hydrosulfide: Does not have an individual approval but may be used under an appropriate group standard New Zealand Environmental Protection Authority (EPA) 7.10 Other Safety Information Chemical Assessment IMAP assessments - Soluble Sulfide Salts: Human health tier II assessment Australian Industrial Chemicals Introduction Scheme (AICIS) 8 Toxicity 8.1 Toxicological Information 8.1.1 Exposure Routes The substance can be absorbed into the body through the skin, by inhalation and by ingestion. ILO-WHO International Chemical Safety Cards (ICSCs) 8.1.2 Signs and Symptoms Inhalation Exposure Cough. Laboured breathing. Shortness of breath. Sore throat. ILO-WHO International Chemical Safety Cards (ICSCs) Skin Exposure Redness. Pain. ILO-WHO International Chemical Safety Cards (ICSCs) Ingestion Exposure Abdominal cramps. Abdominal pain. Burning sensation in the mouth. Diarrhoea. Nausea. Vomiting. ILO-WHO International Chemical Safety Cards (ICSCs) 8.1.3 Adverse Effects Other Poison - Chemical Asphyxiant Dermatotoxin - Skin burns. Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 8.1.4 Acute Effects ChemIDplus 8.2 Ecological Information 8.2.1 ICSC Environmental Data This substance may be hazardous to the environment. Special attention should be given to aquatic organisms. ILO-WHO International Chemical Safety Cards (ICSCs) 9 Associated Disorders and Diseases Associated Occupational Diseases with Exposure to the Compound Asphyxiation, chemical [Category: Acute Poisoning] Haz-Map, Information on Hazardous Chemicals and Occupational Diseases 10 Literature 10.1 Consolidated References PubChem 10.2 Springer Nature References Springer Nature 10.3 Thieme References Thieme Chemistry 10.4 Chemical Co-Occurrences in Literature PubChem 10.5 Chemical-Disease Co-Occurrences in Literature PubChem 10.6 Chemical-Organism Co-Occurrences in Literature PubChem 11 Patents 11.1 Depositor-Supplied Patent Identifiers PubChem Link to all deposited patent identifiers PubChem 11.2 WIPO PATENTSCOPE Patents are available for this chemical structure: PATENTSCOPE (WIPO) 11.3 Chemical Co-Occurrences in Patents PubChem 11.4 Chemical-Disease Co-Occurrences in Patents PubChem 11.5 Chemical-Gene Co-Occurrences in Patents PubChem 11.6 Chemical-Organism Co-Occurrences in Patents PubChem 12 Classification 12.1 ChemIDplus ChemIDplus 12.2 CAMEO Chemicals CAMEO Chemicals 12.3 UN GHS Classification GHS Classification (UNECE) 12.4 EPA CPDat Classification EPA Chemical and Products Database (CPDat) 12.5 EPA DSSTox Classification EPA DSSTox 12.6 EPA TSCA and CDR Classification EPA Chemicals under the TSCA 12.7 EPA Substance Registry Services Tree EPA Substance Registry Services 12.8 Chemicals in PubChem from Regulatory Sources PubChem 13 Information Sources Filter by Source Australian Industrial Chemicals Introduction Scheme (AICIS)LICENSE Ammonium sulfide, (NH4)(SH) Ammonium sulfide, (NH4)(SH) CAMEO ChemicalsLICENSE CAMEO Chemicals and all other CAMEO products are available at no charge to those organizations and individuals (recipients) responsible for the safe handling of chemicals. However, some of the chemical data itself is subject to the copyright restrictions of the companies or organizations that provided the data. AMMONIUM HYDROSULFIDE, SOLUTION CAMEO Chemical Reactivity Classification CAS Common ChemistryLICENSE The data from CAS Common Chemistry is provided under a CC-BY-NC 4.0 license, unless otherwise stated. Ammonium hydrosulfide ChemIDplusLICENSE Ammonium bisulfide ChemIDplus Chemical Information Classification EPA Chemicals under the TSCALICENSE Ammonium sulfide ((NH4)(SH)) EPA TSCA Classification EPA DSSToxLICENSE Ammonium bisulfide CompTox Chemicals Dashboard Chemical Lists European Chemicals Agency (ECHA)LICENSE Use of the information, documents and data from the ECHA website is subject to the terms and conditions of this Legal Notice, and subject to other binding limitations provided for under applicable law, the information, documents and data made available on the ECHA website may be reproduced, distributed and/or used, totally or in part, for non-commercial purposes provided that ECHA is acknowledged as the source: "Source: European Chemicals Agency, Such acknowledgement must be included in each copy of the material. ECHA permits and encourages organisations and individuals to create links to the ECHA website under the following cumulative conditions: Links can only be made to webpages that provide a link to the Legal Notice page. Ammonium hydrogensulphide Ammonium hydrogensulphide (EC: 235-184-3) FDA Global Substance Registration System (GSRS)LICENSE Unless otherwise noted, the contents of the FDA website (www.fda.gov), both text and graphics, are not copyrighted. They are in the public domain and may be republished, reprinted and otherwise used freely by anyone without the need to obtain permission from FDA. Credit to the U.S. Food and Drug Administration as the source is appreciated but not required. AMMONIUM BISULFIDE ILO-WHO International Chemical Safety Cards (ICSCs)LICENSE Creative Commons CC BY 4.0 AMMONIUM BISULFIDE New Zealand Environmental Protection Authority (EPA)LICENSE This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International licence. 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EPA Chemical and Products Database (CPDat)LICENSE Ammonium bisulfide EPA CPDat Classification Hazardous Chemical Information System (HCIS), Safe Work AustraliaAmmonium sulfide, (NH4)(SH) Springer Nature Thieme ChemistryLICENSE The Thieme Chemistry contribution within PubChem is provided under a CC-BY-NC-ND 4.0 license, unless otherwise stated. WikidataLICENSE CCZero ammonium bisulfide Wikipediaammonium hydrosulfide PubChem Chemicals in PubChem from Regulatory Sources GHS Classification (UNECE)GHS Classification EPA Substance Registry ServicesLICENSE EPA SRS List Classification PATENTSCOPE (WIPO)SID 403402404 Cite Download CONTENTS Title and Summary 1 Structures Expand this menu 2 Names and Identifiers Expand this menu 3 Chemical and Physical Properties Expand this menu 4 Related Records Expand this menu 5 Chemical Vendors 6 Use and Manufacturing Expand this menu 7 Safety and Hazards Expand this menu 8 Toxicity Expand this menu 9 Associated Disorders and Diseases 10 Literature Expand this menu 11 Patents Expand this menu 12 Classification Expand this menu 13 Information Sources Connect with NLM Twitter Facebook YouTube National Library of Medicine 8600 Rockville Pike, Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov
13817	https://www.cs.cmu.edu/~fp/courses/15317-f00/handouts/prop.pdf	Chapter 2 Propositional Logic The goal of this chapter is to develop the two principal notions of logic, namely propositions and proofs. There is no universal agreement about the proper foundations for these notions. One approach, which has been particularly suc-cessful for applications in computer science, is to understand the meaning of a proposition by understanding its proofs. In the words of Martin-L¨ of [ML96, Page 27]: The meaning of a proposition is determined by [...] what counts as a veriﬁcation of it. In this chapter we apply Martin-L¨ of’s approach, which follows a rich philo-sophical tradition, to explain the basic propositional connectives. We will see later that universal and existential quantiﬁers and types such as natural num-bers, lists, or trees naturally ﬁt into the same framework. 2.1 Judgments and Propositions The cornerstone of Martin-L¨ of’s foundation of logic is a clear separation of the notions of judgment and proposition. A judgment is something we may know, that is, an object of knowledge. A judgment is evident if we in fact know it. We make a judgment such as “it is raining”, because we have evidence for it. In everyday life, such evidence is often immediate: we may look out the window and see that it is raining. In logic, we are concerned with situation where the evidence is indirect: we deduce the judgment by making correct inferences from other evident judgments. In other words: a judgment is evident if we have a proof for it. The most important judgment form in logic is “A is true”, where A is a proposition. In order to reason correctly, we therefore need a second judgment form “A is a proposition”. But there are many others that have been studied extensively. For example, “A is false”, “A is true at time t” (from temporal Draft of August 31, 2000 6 Propositional Logic logic), “A is necessarily true” (from modal logic), “program M has type τ” (from programming languages), etc. Returning to the ﬁrst two judgments, let us try to explain the meaning of conjunction. We write A prop for the judgment “A is a proposition” and A true for the judgment “A is true” (presupposing that A prop). Given propositions A and B, we want to form the compound proposition “A and B”, written more formally as A ∧B. We express this in the following inference rule: A prop B prop ∧F A ∧B prop This rule allows us to conclude that A ∧B prop if we already know that A prop and B prop. In this inference rule, A and B are schematic variables, and ∧F is the name of the rule (which is short for “conjunction formation”). The general form of an inference rule is J1 . . . Jn name J where the judgments J1, . . ., Jn are called the premises, the judgment J is called the conclusion. In general, we will use letters J to stand for judgments, while A, B, and C are reserved for propositions. Once the rule of conjunction formation (∧F) has been speciﬁed, we know that A ∧B is a proposition, if A and B are. But we have not yet speciﬁed what it means, that is, what counts as a veriﬁcation of A ∧B. This is accomplished by the following inference rule: A true B true ∧I A ∧B true Here the name ∧I stands for “conjunction introduction”, since the conjunction is introduced in the conclusion. We take this as specifying the meaning of A∧B completely. So what can be deduce if we know that A∧B is true? By the above rule, to have a veriﬁcation for A ∧B means to have veriﬁcations for A and B. Hence the following two rules are justiﬁed: A ∧B true ∧EL A true A ∧B true ∧ER B true The name ∧EL stands for “left conjunction elimination”, since the conjunction in the premise has been eliminated in the conclusion. Similarly ∧ER stands for “right conjunction elimination”. We will later see what precisely is required in order to guarantee that the formation, introduction, and elimination rules for a connective ﬁt together cor-rectly. For now, we will informally argue the correctness of the elimination rules. Draft of August 31, 2000 2.2 Hypothetical Judgments 7 As a second example we consider the proposition “truth” written as ⊤. ⊤F ⊤prop Truth should always be true, which means its introduction rule has no premises. ⊤I ⊤true Consequently, we have no information if we know ⊤true, so there is no elimi-nation rule. A conjunction of two propositions is characterized by one introduction rule with two premises, and two corresponding elimination rules. We may think of truth as a conjunction of zero propositions. By analogy it should then have one introduction rule with zero premises, and zero corresponding elimination rules. This is precisely what we wrote out above. 2.2 Hypothetical Judgments Consider the following derivation, for some arbitrary propositions A, B, and C: A ∧(B ∧C) true ∧ER B ∧C true ∧EL B true Have we actually proved anything here? At ﬁrst glance it seems that cannot be the case: B is an arbitrary proposition; clearly we should not be able to prove that it is true. Upon closer inspection we see that all inferences are correct, but the ﬁrst judgment A ∧(B ∧C) has not been justiﬁed. We can extract the following knowledge: From the assumption that A ∧(B ∧C) is true, we deduce that B must be true. This is an example of a hypothetical judgment, and the ﬁgure above is an hypothetical derivation. In general, we may have more than one assumption, so a hypothetical derivation has the form J1 · · · Jn . . . J where the judgments J1, . . ., Jn are unproven assumptions, and the judgment J is the conclusion. Note that we can always substitute a proof for any hypoth-esis Ji to eliminate the assumption. We call this the substitution principle for hypotheses. Draft of August 31, 2000 8 Propositional Logic Many mistakes in reasoning arise because dependencies on some hidden as-sumptions are ignored. When we need to be explicit, we write J1, . . ., Jn ⊢J for the hypothetical judgment which is established by the hypothetical derivation above. We may refer to J1, . . ., Jn as the antecedents and J as the succedent of the hypothetical judgment. One has to keep in mind that hypotheses may be used more than once, or not at all. For example, for arbitrary propositions A and B, A ∧B true ∧ER B true A ∧B true ∧EL A true ∧I B ∧A true can be seen a hypothetical derivation of A ∧B true ⊢B ∧A true. With hypothetical judgments, we can now explain the meaning of implication “A implies B” or “if A then B” (more formally: A ⊃B). First the formation rule: A prop B prop ⊃F A ⊃B prop Next, the introduction rule: A ⊃B is true, if B is true under the assumption that A is true. u A true . . . B true ⊃Iu A ⊃B true The tricky part of this rule is the label u. If we omit this annotation, the rule would read A true . . . B true ⊃I A ⊃B true which would be incorrect: it looks like a derivation of A ⊃B true from the hypothesis A true. But the assumption A true is introduced in the process of proving A ⊃B true; the conclusion should not depend on it! Therefore we label uses of the assumption with a new name u, and the corresponding inference which introduced this assumption into the derivation with the same label u. As a concrete example, consider the following proof of A ⊃(B ⊃(A ∧B)). u A true w B true ∧I A ∧B true ⊃Iw B ⊃(A ∧B) true ⊃Iu A ⊃(B ⊃(A ∧B)) true Draft of August 31, 2000 2.2 Hypothetical Judgments 9 Note that this derivation is not hypothetical (it does not depend on any assump-tions). The assumption A true labeled u is discharged in the last inference, and the assumption B true labeled w is discharged in the second-to-last inference. It is critical that a discharged hypothesis is no longer available for reasoning, and that all labels introduced in a derivation are distinct. Finally, we consider what the elimination rule for implication should say. By the only introduction rule, having a proof of A ⊃B true means that we have a hypothetical proof of B true from A true. By the substitution principle, if we also have a proof of A true then we get a proof of B true. A ⊃B true A true ⊃E B true This completes the rule concerning implication. With the rules so far, we can write out proofs of simple properties con-cerning conjunction and implication. The ﬁrst expresses that conjunction is commutative—intuitively, an obvious property. u A ∧B true ∧ER B true u A ∧B true ∧EL A true ∧Iu B ∧A true ⊃I (A ∧B) ⊃(B ∧A) true When we construct such a derivation, we generally proceed by a combination of bottom-up and top-down reasoning. The next example is a distributivity law, allowing us to move implications over conjunctions. This time, we show the partial proofs in each step. Of course, other sequences of steps in proof constructions are also possible. . . . (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true First, we use the implication introduction rule bottom-up. u A ⊃(B ∧C) true . . . (A ⊃B) ∧(A ⊃C) true ⊃Iu (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true Draft of August 31, 2000 10 Propositional Logic Next, we use the conjunction introduction rule bottom-up. u A ⊃(B ∧C) true . . . A ⊃B true u A ⊃(B ∧C) true . . . A ⊃C true ∧I (A ⊃B) ∧(A ⊃C) true ⊃Iu (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true We now pursue the left branch, again using implication introduction bottom-up. u A ⊃(B ∧C) true w A true . . . B true ⊃Iw A ⊃B true u A ⊃(B ∧C) true . . . A ⊃C true ∧I (A ⊃B) ∧(A ⊃C) true ⊃Iu (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true Note that the hypothesis A true is available only in the left branch, but not in the right one: it is discharged at the inference ⊃Iw. We now switch to top-down reasoning, taking advantage of implication elimination. u A ⊃(B ∧C) true w A true ⊃E B ∧C true . . . B true ⊃Iw A ⊃B true u A ⊃(B ∧C) true . . . A ⊃C true ∧I (A ⊃B) ∧(A ⊃C) true ⊃Iu (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true Now we can close the gap in the left-hand side by conjunction elimination. Draft of August 31, 2000 2.3 Disjunction and Falsehood 11 u A ⊃(B ∧C) true w A true ⊃E B ∧C true ∧EL B true ⊃Iw A ⊃B true u A ⊃(B ∧C) true . . . A ⊃C true ∧I (A ⊃B) ∧(A ⊃C) true ⊃Iu (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true The right premise of the conjunction introduction can be ﬁlled in analo-gously. We skip the intermediate steps and only show the ﬁnal derivation. u A ⊃(B ∧C) true w A true ⊃E B ∧C true ∧EL B true ⊃Iw A ⊃B true u A ⊃(B ∧C) true v A true ⊃E B ∧C true ∧ER C true ⊃Iv A ⊃C true ∧I (A ⊃B) ∧(A ⊃C) true ⊃Iu (A ⊃(B ∧C)) ⊃((A ⊃B) ∧(A ⊃C)) true 2.3 Disjunction and Falsehood So far we have explained the meaning of conjunction, truth, and implication. The disjunction “A or B” (written as A ∨B) is more diﬃcult, but does not require any new judgment forms. A prop B prop ∨F A ∨B prop Disjunction is characterized by two introduction rules: A∨B is true, if either A or B is true. A true ∨IL A ∨B true B true ∨IR A ∨B true Now it would be incorrect to have an elimination rule such as A ∨B true ∨EL? A true because even if we know that A∨B is true, we do not know whether the disjunct A or the disjunct B is true. Concretely, with such a rule we could derive the Draft of August 31, 2000 12 Propositional Logic truth of every proposition A as follows: u B true ⊃Iu B ⊃B true w B ⊃B true ∨IR A ∨(B ⊃B) true ∨EL? A true ⊃Iw (B ⊃B) ⊃A true ⊃E A true Thus we take a diﬀerent approach. If we know that A ∨B is true, we must consider two cases: A true and B true. If we can prove a conclusion C true in both cases, then C must be true! Written as an inference rule: A ∨B true u A true . . . C true w B true . . . C true ∨Eu,w C true Note that we use once again the mechanism of hypothetical judgments. In the proof of the second premise we may use the assumption A true labeled u, in the proof of the third premise we may use the assumption B true labeled w. Both are discharged at the disjunction elimination rule. Let us justify the conclusion of this rule more explicitly. By the ﬁrst premise we know A ∨B true. The premises of the two possible introduction rules are A true and B true. In case A true we conclude C true by the substitution principle and the second premise: we substitute the proof of A true for any use of the assumption labeled u in the hypothetical derivation. The case for B true is symmetric, using the hypothetical derivation in the third premise. Because of the complex nature of the elimination rule, reasoning with dis-junction is more diﬃcult than with implication and conjunction. As a simple example, we prove the commutativity of disjunction. . . . (A ∨B) ⊃(B ∨A) true We begin with an implication introduction. u A ∨B true . . . B ∨A true ⊃Iu (A ∨B) ⊃(B ∨A) true Draft of August 31, 2000 2.3 Disjunction and Falsehood 13 At this point we cannot use either of the two disjunction introduction rules. The problem is that neither B nor A follow from our assumption A∨B! So ﬁrst we need to distinguish the two cases via the rule of disjunction elimination. u A ∨B true v A true . . . B ∨A true w B true . . . B ∨A true ∨Ev,w B ∨A true ⊃Iu (A ∨B) ⊃(B ∨A) true The assumption labeled u is still available for each of the two proof obligations, but we have omitted it, since it is no longer needed. Now each gap can be ﬁlled in directly by the two disjunction introduction rules. u A ∨B true v A true ∨IR B ∨A true w B true ∨IL B ∨A true ∨Ev,w B ∨A true ⊃Iu (A ∨B) ⊃(B ∨A) true This concludes the discussion of disjunction. Falsehood (written as ⊥, some-times called absurdity) is a proposition that should have no proof! Therefore there are no introduction rules, although we of course have the standard forma-tion rule. ⊥F ⊥prop Since there cannot be a proof of ⊥true, it is sound to conclude the truth of any arbitrary proposition if we know ⊥true. This justiﬁes the elimination rule ⊥true ⊥E C true We can also think of falsehood as a disjunction between zero alternatives. By analogy with the binary disjunction, we therefore have zero introduction rules, and an elimination rule in which we have to consider zero cases. This is precisely the ⊥E rule above. From this is might seem that falsehood it useless: we can never prove it. This is correct, except that we might reason from contradictory hypotheses! We will see some examples when we discuss negation, since we may think of the proposition “not A” (written ¬A) as A ⊃⊥. In other words, ¬A is true precisely if the assumption A true is contradictory because we could derive ⊥true. Draft of August 31, 2000 14 Propositional Logic 2.4 Summary Judgments. A prop A is a proposition A true Proposition A is true Propositional Constants and Connectives. The following table summa-rizes the introduction and elimination rules for the propositional constants (⊤, ⊥) and connectives (∧, ⊃, ∨). We omit the straightforward formation rules. Introduction Rules Elimination Rules A true B true ∧I A ∧B true A ∧B true ∧EL A true A ∧B true ∧ER B true ⊤I ⊤true no ⊤E rule u A true . . . B true ⊃Iu A ⊃B true A ⊃B true A true ⊃E B true A true ∨IL A ∨B true B true ∨IR A ∨B true A ∨B true u A true . . . C true w B true . . . C true ∨Eu,w C true no ⊥I rule ⊥true ⊥E C true Draft of August 31, 2000 Bibliography [ML96] Per Martin-L¨ of. On the meanings of the logical constants and the jus-tiﬁcations of the logical laws. Nordic Journal of Philosophical Logic, 1(1):11–60, 1996. Draft of August 31, 2000
13818	http://wkretype.bdimg.com/view/9d73bdbcc77da26925c5b0bd	一元二次方程应用题分类 - 百度文库新建文档新建表格新建上传至我的资料库转为在线文档上传至创作中心在文库发布个人作品上传首页助手最近知识库我的收藏我的下载我的上传我的购买查看更多文件资料库文档瘦身格式转换 PDF合并 PDF拆分 PDF加水印图片转文字工具辅助模式已关闭查看更多前往帮助中心 > 联系我们免费下载PC客户端免广告格式转换离线浏览文档管理下载客户端扫码使用文库APP 新建文档新建表格新建上传至我的资料库转为在线文档上传至创作中心在文库发布个人作品上传最近编辑我的收藏我的下载大家都在搜开通会员登录后查看会员权益开通会员每日抽奖专属权益精选内容前往会员中心抽奖登录相关文档集共15篇 1 一元二次方程应用题分类 3.8 1,659下载 1.12W阅读 2 一元二次方程应用题分类 3.8 31下载 1,041阅读 3 一元二次方程应用题分类 3.6 83下载 479阅读 4 一元二次方程应用题分类 2.1 39下载 226阅读 5 一元二次方程应用题分类 2.1 292下载 1,188阅读 6 一元二次方程应用题分类 4.1 5,456下载 1.66W阅读 7 一元二次方程应用题——分类 3.5 42下载 151阅读 8 一元二次方程应用题分类 3.4 1下载 14阅读 9 一元二次方程应用题分类 3.0 4下载 32阅读 10 一元二次方程应用题分类 3.7 4下载 227阅读 11 一元二次方程应用题分类 2.1 29下载 239阅读 12 一元二次方程应用题分类 3.9 22下载 192阅读 13 一元二次方程应用题分类 4.0 79下载 271阅读 14 一元二次方程应用题分类 4.9 401下载 1,457阅读 15 一元二次方程应用题分类 4.1 55下载 209阅读编辑文档您的皮肤设置未保存，开通VIP保存设置查看更多15项会员特权不保存开通文库VIP 一元二次方程应用题分类一元二次方程应用题分类 2011-10-20 dlzylh AI编辑插入正文思源黑体样式更多选择插入表格的行列表格插入分栏分栏图片柱状图折线图饼状图图表公式链接引用分割线颜色分割线高亮块代码块 Markdown导入复制剪切删除行间距取消确定查找替换上一个下一个取消确定一元二次方程应用题分类数字问题 1 、有一个两位数，它的个位上的数字与十位上的数字之和是 6 ，如果把它的个位数字与十位数字调换位置，所得的两位数乘以原来的两位数所得的积等于 1008 ，求调换位置后得到的两位数。面积问题 2 、用一块长 80cm ，宽 60cm 的薄钢片，在四个角上截去四个相同的边长为 Xcm 的小正方形，然后做成底面积为 1500cm 2 的无盖的长方形盒子，求 X 的值。 3 、如图，在长为 32m ，宽为 20m 的矩形耕地上，修筑同样宽的三条道路，把耕地分成大小不等的六块作实验田，要使试验田面积为 570m 2 ，道路的宽应为多少？增长率问题：变化前数量 × （ 1  x ） n ＝变化后数量 4 、某校 2003 年捐款 1 万元给希望工程，以后每年都捐款，计划到 2005 年共捐款 4.75 万元，问该校捐款的平均年增长率是多少？ 5 、某新华书店计划第一季度共发行图书 122 万册，其中一月份发行图书 32 万册，二、三月份平均每月增长率相同，求二、三月份各应发行图书多少万册？销售问题：售价—进价 = 利润，一件商品的利润×销售量 = 总利润，单价×销售量 = 销售额 6 、某商场销售一批名牌衬衫，平均每天可售出 20 件，每件盈利 40 元，为了扩大销售量增加盈利，尽快减少库存，商场决定采取适当的降价措施，经调查发现，如果每件衬衫每降价 1 元，商场平均每天可多售 2 件，如果商场平均每天要盈利 1200 元，每件衬衫应降价多少元？ 7 、某商店如果将进货价格为 8 元的商品按每件 10 元售出，每天可销售 200 件，现采取提高售价，减少进货量的方法，增加利润，已知这种商品每涨价 0.5 元，其销售量就减少 10 件，问应将售价定为多少元时可赚利润 720 元？ 8. 某西瓜经营户以 2 元 / 千克的价格购进一批小型西瓜，以 3 元 / 千克的价格出售，每天可售出 200 千克。为了促销，该经营户决定降价销售。经调查发现，这种小西瓜每降价 0.1 元 / 千克，每天可多售出 40 千克。另外，每天的房租等固定成本共 24 元。该经营户要想每天盈利 200 元，则应将每千克的小型西瓜的售价降低多少元？相互问题 9 ．（ 1 ）参加一次聚会的每两人都握了一次手 , 所有人共握手 66 次 , 有多少人参加聚会 ? (2) 要组织一场篮球联赛 , 赛制为单循环形式 , 即每两队之间都赛一场 , 计划安排 28 场比赛 , 应邀请多少个球队参加比赛 ? (3) 初三毕业晚会时每人互相送照片一张 , 一共要 90 张照片 , 有多少人 ? （ 4 ）有一人患了流感，经过两轮传染后共有 121 人患了流感，每轮传染中平均一个人传染了几个人？利息、利率问题 10. 某人将 2000 元人民币按一年定期存入银行，到期后支取 1000 元用于购物，剩下的 1000 元及应得利息又全部按一年定期存入银行，若存款的利率不变，到期后本金和利息共 1320 元，求这种存款方式的年利率。（利息税为 20% ）移动到查看全部包含“”的文档搜索复制发送到手机翻译复制搜索发送到手机翻译复制搜索翻译 <> 等0人已复制此文本文字已复制您已超出最大复制字数限制，请减少字数重新尝试已为您复制内容开通VIP，立即解锁文本内容开通VIP VIP会员尊享权益超出发送上限，加入VIP即可继续发送未开通VIP 选择文档内容扫一扫，立即发送到手机开通VIP，享无限制发送与复制特权享VIP专享文档下载特权赠共享文档下载特权 100W篇文档免费专享超过1万字无法发送，您可减少字数多次发送选择内容扫一扫立即发送到手机 VIP享无限制发送特权以下结果由提供：× 百度翻译移动到开通VIP 享海量文档+AI助手权益最低仅需¥2.00开通VIP 下一篇：一元二次方程应用题分类 3.8 1041阅读版权说明：本文档由用户提供并上传，收益归属内容提供方，若内容存在侵权，请进行举报或认领页数说明：当前展示页数非原始文档页数，原始文档共2页全屏阅读举报认领此文档是否涉及以下问题？色情、淫秽、低俗信息涉嫌违法犯罪时政信息不实广告或垃圾信息此文档是否涉及侵权？盗版或侵权未获得权利人的合法授权、侵犯个人隐私或泄露单位商业机密处理此类举报需要您提供更多信息，请按引导上传相关材料，这样会有助您成功举报。取消提交若您为此篇文档的原创作者，提交权利人真实信息及权属证明材料。我们承诺：材料信息经核实后，将于工作日24小时之内将此文档以VIP专享或免费文档的形式转至您名下，后续被认领成功的文档所产生的收益均归您所有。认领文档一元二次方程应用题分类实名信息当前账号暂未入驻，个人作者入驻机构入驻请入驻文库创作者平台后刷新当前页面认领理由权属材料上传权属证明材料可上传多个附件，支持png、jpg、pdf、word等格式，附件内容需要包含： ①毕业证/工作证/曾经的发表记录/版权证明/版权声明书等原文档版权证明 ②身份证或营业执照等立即认领联系电话 400-921-7005 如您为此篇文档作者，可提供证明材料申请认领，获取版权文档收益并下架侵权文档：发送申请邮件审核（1-2个工作日）认领结果答复申请邮箱： wenku-renling@baidu.com 复制证明材料： 1、侵权文档文库链接地址 2、您的版权文档文库链接地址：如尚无，请先上传 3、身份证明材料：身份证正面照 / 护照身份页照片 4、版权证明材料：作品认证证书 / 作品登记证书 / 已署名的出版物专属客服： QQ 800049878 电话 400-921-7005 您的举报提交成功！您的举报已收到，我们会尽快进行核实处理。关闭添加文库智能助手到电脑桌面安装后可在桌面快捷访问文库，轻松搜索海量文档和使用AI创作能力立即添加分享批量下载(15篇) 全部文档请确认浏览器允许下载，下载时自动消耗特权该文档需单独付费，不支持批量下载 1 一元二次方程应用题分类已编辑 2 一元二次方程应用题分类已编辑 3 一元二次方程应用题分类已编辑 4 一元二次方程应用题分类已编辑 5 一元二次方程应用题分类已编辑 6 一元二次方程应用题分类已编辑 7 一元二次方程应用题——分类已编辑 8 一元二次方程应用题分类已编辑 9 一元二次方程应用题分类已编辑 10 一元二次方程应用题分类已编辑 11 一元二次方程应用题分类已编辑 12 一元二次方程应用题分类已编辑 13 一元二次方程应用题分类已编辑 14 一元二次方程应用题分类已编辑 15 一元二次方程应用题分类已编辑单篇下载 1000+下载下载百度文库客户端畅享免费海量文档｜一键制作汇报总结下载客户端下载客户端转PDF 更多操作引用智能助手全屏探索新建指南文档难找？试试AI为您生成的内容吧：高校，即高等院校，是指提供高智能PPT 智能写作智能图表 DeepSeek-R1 联网搜索畅享AI生成内容购买会员畅享智能创作 AI生成PPT PPT高级模板 AI帮你写 AI扩写 AI润色 AI续写 AI修订无限复制无限导出 6亿VIP文档查看更多权益换个话题重新开始吧，新建对话 AI辅助生成PPT 支持多种方式一键智能生成专业级PPT 去试试拖拽文件开始上传文档大小：不超过200MB 文档类型：doc、docx、ppt、pptx、pdf、xls、xlsx 限时抢购: 01 时 00 分 00 秒公式常用符号常用公式常用符号希腊字母分数微分根式角标极限对数三角函数积分运算大型运算括号取整数组矩阵清除输出区域取消确定
13819	https://help.synthesisplatform.net/weibull_alta9/life_data_analysis_plots.htm	Life Data Analysis Plots Show Weibull++ Standard Folio> Life Data Analysis Plots Related Topics: Weibull++ Standard Folio Life Data Analysis Plots Weibull++ includes multiple plot types you can use to visualize the results of your life data analysis. You can create plots by choosing Life Data > Analysis > Plot or by clicking the icon on the Main page of the control panel. The scaling, setup, exporting and confidence bounds settings are similar to the options available for all other Weibull++ plot sheets. To learn how to customize a plot, see Plot Setup. Tip: Weibull++ includes two additional plots you could use across all types of data folios: the overlay plot, which allows you to compare different data sets or distributions; and the side-by-side plot, which allows you to display different plots of a single data set all in a single window for easy comparison. The following is a description of the different types of plots that can be created in a Weibull++ standard folio: The Probability plot shows the trend in the probability of failure over time. The plotting positions of the data points are determined by the failure/suspension times in the data set (x-axis) and their corresponding unreliability estimates (y-axis). Note: Unlike the probability plots for other distributions, the y-axis in an exponential probability plot always indicates the reliability instead of the unreliability. This tradition arose from the time when probability plotting was performed "by hand." The exponential reliability model starts with R = 1 at T = 0 (or gamma). Thus, if the unreliability were plotted, the axis would start at Q = 1 - R = 0, which is not possible, given that the y-axis scale is logarithmic. When the parameters have been calculated using rank regression analysis, the regression line is fitted to the data points on the plot in order to obtain the distribution parameters. Therefore, the plot can also be used to compare how different distributions fit a particular data set. The closer the regression line tracks the points on the plots, the better the fit. In contrast, the maximum likelihood estimation (MLE) method obtains the solution of the line from the likelihood function, not by the plotting positions of the data points. Therefore, the line is not expected to track the points on the plot, and the plot should not be used to evaluate the fit of a distribution when using MLE. The next figures show the rank regression analysis of single data set using a Weibull distribution and a lognormal distribution. As you can see, the probability plot shows that the Weibull distribution presents the better fit to this particular data set. The Reliability vs. Time plot shows the reliability values over time, capturing trends in the product’s failure behavior. The Unreliability vs. Time plot shows the probability of failure of the product over time. The pdf plot shows the probability density function of the data over time, allowing you to visualize the distribution of the data set. The Failure Rate vs. Time plot shows the failure rate function of the product over time. The Contour Plot shows a 3D surface on a 2D plot. The plot represents the 3D surface by plotting constant z slices, called contours, on a 2D format. The shape of a contour is determined by plotting the parameters of the data set (x- and y-axis) at the position where a given confidence level occurs (z-axis). You can use the overlay plot feature to superimpose two contour plots from two different data sets at the same confidence level in order to determine the level at which the two data sets are statistically different. If the contour lines overlap, then the data sets do not show a statistically significant difference at that confidence level. (See Contour Plots.) The Failures/Suspensions Histogram is a bar chart that shows the number of failures/suspensions that fall into different time intervals. This allows you to evaluate the frequency distribution of the failures and suspensions in the data set. The chart has two display options: Probability Density Values displays a chart where the height of each bar is proportional to the frequency of occurrence of failures in each interval. Select the Superimpose pdf check box to display the pdf line on the same chart. Failures displays a chart where the height of each bar is proportional to the number of failures in each interval. Select the Show suspensions check box to display a bar chart of the number of suspensions in each interval. You have the option to adjust the time intervals to get a good indication of how the data points are distributed. Note that if the histogram has intervals that are, in theory, infinitesimally small, the chart will result in a plot of the probability density function (pdf). The Failures/Suspensions Pie chart displays the ratio and proportion of failures and suspensions to the whole data set. The Failures/Suspensions Timeline plots are similar to horizontal bar charts. The values of the failures and suspensions are plotted on the y-axis and the time periods are on the x-axis. All the lines are anchored to the y-axis, giving a common point of measurement. © 1992-2015. ReliaSoft Corporation. ALL RIGHTS RESERVED.
13820	https://byjus.com/physics/bulk-modulus-of-elasticity-definition-formula-questions/	What is Bulk Modulus of Elasticity? Bulk modulus of elasticity is defined as the proportion of volumetric stress linked to the volumetric strain of definite material, although the deformation of the material is within the elastic limit. Bulk modulus is basically a numerical constant that measures and defines the elastic properties of a solid or fluid when pressure is applied on all the surfaces. Bulk Modulus Formulas Mathematically we can express bulk modulus of elasticity as- Bulk modulus of elasticity (K) =P/Volumetric strain = K = (P/ (- ΔV/V)) Where, P = pressure applied to the body, ΔV represents the change in the volume of the body due to the applied stress and V is the original volume of the body. In the mathematical expression, a negative sign indicates that due to increase in the value of the pressure applied, the volume decreases. This is the same as, the change in pressure divided by the change in volume divided by initial volume: Bulk Modulus (K) = (p1 – p0) / [(V1 – V0) / V0] Here, initial pressure and volume will be p0 and V0 and p1 and V1 are the pressure and volume measured upon compression. Hence, in terms of pressure and density bulk modulus elasticity may also be expressed as K = (p1 – p0) / [(ρ1 – ρ0) / ρ0] Here, ρ0 and ρ1 are the initial and final density values. SI Unit of Bulk Modulus of Elasticity According to the above formula, the bulk modulus of elasticity units will be the same as the unit of pressure because the volumetric strain is unitless, (N/m2) is the SI units of pressure. Hence, Pascal (Pa) or Newton per square metre (N/m2) will be the SI units of bulk modulus of elasticity. Megapascal (MPa), Kilopascal (Kpa) are some other units of bulk modulus of elasticity too. Dimensions of Bulk Modulus of Elasticity Since, volumetric strain is dimensionless and the dimension of pressure is the same as that of force per unit area which is [ML−1T−2]. Therefore, the dimension of the bulk modulus of elasticity will be equal to the dimension of the pressure that is [ML−1T−2]. Important Questions on Bulk Modulus of Elasticity 1) Bulk Modulus of elasticity is the ratio of? a) Compressive stress to linear strain b) Compressive stress to volumetric strain c) Tensile stress to volumetric strain d) Tensile stress to linear strain Correct Option: (b) Explanation: Bulk Modulus k is associated with the decrease in volume per unit volume and the compression of a liquid. Therefore, it is the ratio of compressive stress to the volumetric strain. 2) The exact relationship between compressibility (β) and Bulk Modulus (k) is which one among the following? a) β = 1/k b) β = k c) β = 3k d) β = k/4 Correct Option: (a) Explanation: β = 1/k is the correct relation because the compressibility β of a liquid is defined as the ratio of volumetric strain to the compressive stress while Bulk Modulus is the ratio of compressive stress to volumetric strain. 3) For an incompressible fluid, what will be the value of Bulk Modulus of elasticity? a) Infinity b) Unity c) Zero d) very low Correct Option (a) Explanation: k = 1/β, where k is Bulk Modulus of elasticity and β is compressibility. Therefore, β=0, for an incompressible fluid, hence the value of k will be infinity. 4) Which among the following is the dimension of compressibility? a) [M1L1T-1]. b) [M1L1T-2]. c) [M-1L1T2]. d) [M-1L1T-2]. Correct Option: (c) Explanation: Since, k = 1/β, where β= compressibility and k = Bulk Modulus of elasticity and the dimensions of k is [ML-1T-2] 5) Which among the following is the unit of compressibility? a) m2/N b) m/N c) m3/N d) Compressibility is unitless Correct Option: (a) Explanation: Since, k (Bulk Modulus of elasticity) = Compressive Stress/Volumetric Stress, therefore, the unit of bulk modulus is N/m2 and hence, the unit of compressibility will be m2/N. 6) Which one of these won’t have the same unit as the others? a) Bulk Modulus b) Force c) Pressure d) Stress Correct Option: (b) Explanation: N/m2 is the SI unit of Bulk Modulus, Stress and Pressure but the unit of Force is N. 7) The Bulk Moduli of 3 fluids, that is, a, b and c are p1, p2 and p3, respectively. If p1 > p2 > p3, which liquid will have the maximum compressibility? a) liquid a b) liquid b c) liquid c d) they’ll have equal compressibility Correct Option: (c) Explanation: p = 1=β, where p= Bulk Modulus of elasticity and β= compressibility. If p1 > p2 > p3, then β1 < β2 < β3. Thus, liquid c will have the highest compressibility. 8) Which one of the following is the correct relationship between young’s modulus (E), bulk modulus (K) and Poisson’s ratio (µ)? a) E=3K(1-2µ) b) E=2K(1-2µ) c) E=2K(1-2µ) d) E=2K(1-3µ) Correct Option: (a) Explanation: relationship between young’s modulus (E), bulk modulus (K) and Poisson’s ratio (µ) will be given asE=3K(1-2µ) 9) Factor of safety is the ratio of? a) Ultimate stress / Permissible stress b) Compressive stress / Ultimate stress c) Tensile stress / Permissible stress d) Ultimate stress / Shear stress Correct Option: (a) Explanation: Factor of safety is defined as the ratio of ultimate stress to the permissible stress, therefore the correct answer will be option a. 10) The bulk modulus of elasticity a) Increases with pressure b) Decreases with pressure c) Is independent of temperature d) Is independent of pressure and temperature Correct Option: (a) Explanation: Since, the bulk modulus of elasticity is directly proportional to the change of pressure, therefore as the pressure increases the bulk modulus of elasticity increases. Practice Questions Define bulk modulus of elasticity. Differentiate between bulk modulus and density of a liquid. Calculate the bulk modulus of a body that experiences a change in pressure of 6104N/m2 and its volume goes from 5 cm3 to 4.9 cm3. Define Poisson’s ratio. Define young’s modulus of elasticity. \| Related Articles Relation between Young’s Modulus and Bulk Modulus Shearing Stress Shear Modulus Of Rigidity \| Suggested Videos Stay tuned to BYJU’S and Fall in Love with Learning! Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs Register with BYJU'S & Watch Live Videos
13821	https://www.doubtnut.com/qna/642564605	In the binomial expansion of (1+x)n , coefficients of the fifth, sixth and seventh terms are in A.P. find all the values of n for which this can happen. More from this Exercise To solve the problem, we need to find the values of n for which the coefficients of the fifth, sixth, and seventh terms in the binomial expansion of (1+x)n are in Arithmetic Progression (A.P.). 1. Identify the Terms: The r-th term in the binomial expansion of (1+x)n is given by: Tr=(nr−1)xr−1 Therefore, the coefficients of the fifth, sixth, and seventh terms are: - Fifth term: T5=(n4) - Sixth term: T6=(n5) - Seventh term: T7=(n6) 2. Set Up the A.P. Condition: For the coefficients to be in A.P., the condition is: 2⋅(n5)=(n4)+(n6) 3. Use the Binomial Coefficient Formula: We can express the binomial coefficients using the formula: (nk)=n!k!(n−k)! Thus, we have: (n4)=n!4!(n−4)!,(n5)=n!5!(n−5)!,(n6)=n!6!(n−6)! 4. Substituting the Coefficients: Substitute the coefficients into the A.P. condition: 2⋅n!5!(n−5)!=n!4!(n−4)!+n!6!(n−6)! Dividing through by n! (assuming n≥6): 2⋅15!(n−5)!=14!(n−4)!+16!(n−6)! 5. Simplifying the Equation: Multiply through by 5!(n−5)! to eliminate the denominators: 2=5!4!⋅1(n−4)!+5!6!⋅1(n−6)! This simplifies to: 2=5⋅1(n−4)!+16(n−6)! 6. Finding a Common Denominator: The common denominator is 6(n−4)!(n−6)!: 2⋅6(n−4)!(n−6)!=30(n−6)!+(n−4)! This leads to: 12(n−4)(n−6)=30+(n−4)(n−5) 7. Rearranging the Equation: Rearranging gives: 12n−48=30+n2−9n+20 Simplifying leads to: n2−21n+18=0 8. Factoring the Quadratic: Factoring gives: (n−7)(n−14)=0 Thus, the solutions are: n=7orn=14 Final Answer: The values of n for which the coefficients of the fifth, sixth, and seventh terms are in A.P. are n=7 and n=14. To solve the problem, we need to find the values of n for which the coefficients of the fifth, sixth, and seventh terms in the binomial expansion of (1+x)n are in Arithmetic Progression (A.P.). Identify the Terms: The r-th term in the binomial expansion of (1+x)n is given by: Tr=(nr−1)xr−1 Therefore, the coefficients of the fifth, sixth, and seventh terms are: Fifth term: T5=(n4) Sixth term: T6=(n5) Seventh term: T7=(n6) Set Up the A.P. Condition: For the coefficients to be in A.P., the condition is: 2⋅(n5)=(n4)+(n6) Use the Binomial Coefficient Formula: We can express the binomial coefficients using the formula: (nk)=n!k!(n−k)! Thus, we have: (n4)=n!4!(n−4)!,(n5)=n!5!(n−5)!,(n6)=n!6!(n−6)! Substituting the Coefficients: Substitute the coefficients into the A.P. condition: 2⋅n!5!(n−5)!=n!4!(n−4)!+n!6!(n−6)! Dividing through by n! (assuming n≥6): 2⋅15!(n−5)!=14!(n−4)!+16!(n−6)! Simplifying the Equation: Multiply through by 5!(n−5)! to eliminate the denominators: 2=5!4!⋅1(n−4)!+5!6!⋅1(n−6)! This simplifies to: 2=5⋅1(n−4)!+16(n−6)! Finding a Common Denominator: The common denominator is 6(n−4)!(n−6)!: 2⋅6(n−4)!(n−6)!=30(n−6)!+(n−4)! This leads to: 12(n−4)(n−6)=30+(n−4)(n−5) Rearranging the Equation: Rearranging gives: 12n−48=30+n2−9n+20 Simplifying leads to: n2−21n+18=0 Factoring the Quadratic: Factoring gives: (n−7)(n−14)=0 Thus, the solutions are: n=7orn=14 Final Answer: The values of n for which the coefficients of the fifth, sixth, and seventh terms are in A.P. are n=7 and n=14. Topper's Solved these Questions Explore 248 Videos Explore 77 Videos Similar Questions In the binomial expansion of (a+b)n , coefficients of the fourth and thirteenth terms are equal to each other. Find n . In the binomial expansion of (1+x)m+n, prove that the coefficients of xmandxn are equal. If in the expansion of (1+x)n the coefficients of 14th, 15th and 16th terms are in A.P. then n= (A) 12 (B) 23 (C) 27 (D) 34 In the expansion of (1+x)n the binomial coefficients of three consecutive terms are respectively 220, 495 and 792 find the value of n. If the coefficients of 2nd, 3rd and 4th terms in the expansion of(1+x)n are in A.P., then find the value of n. If the constant term in the binomial expansion of (x2−1x)n,n∈N is 15, then find the value of n. If the constant term in the binomial expansion of (x2−1x)n,n∈N is 15, then find the value of n. If the coefficients of 2nd, 3rd and 4th terms in the expansion of (1+x)2n are in A.P. Then find the value of n. If in the expansion of (1+x)n the coefficient of three consecutive terms are 56,70 and 56, then find n and the position of the terms of these coefficients. Write down the fourth term in the binomial expansion of (px+1x)n. If this term is independent of x, find the value of n. 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13822	https://www.dummies.com/article/academics-the-arts/math/calculus/how-to-solve-limits-by-conjugate-multiplication-192309/	Meet Earth’s mightiest heroes! Avengers For Dummies is here! Order your copy today. Book & Article Categories Collections Custom Solutions Dummies AI Main Menu Book & Article Categories Technology Academics & The Arts Home, Auto, & Hobbies Body, Mind, & Spirit Business, Careers, & Money Dummies AI Main Menu Book & Article Categories Technology Academics & The Arts Home, Auto, & Hobbies Body, Mind, & Spirit Business, Careers, & Money Dummies AI Main Menu Collections Explore all collections BYOB (Be Your Own Boss) Be a Rad Dad Career Shifting Contemplating the Cosmos For Those Seeking Peace of Mind For the Aspiring Aficionado For the Budding Cannabis Enthusiast For the College Bound For the Exam-Season Crammer For the Game Day Prepper Dummies AI Home Academics & The Arts Articles Math Articles Calculus Articles How to Solve Limits by Conjugate Multiplication By No items found. Updated 2016-03-26 21:20:35 From the book No items found. Share Download E-Book Personal Finance For Dummies Explore Book Calculus II Workbook For Dummies Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego Download E-Book Personal Finance For Dummies Explore Book Calculus II Workbook For Dummies Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego To solve certain limit problems, you’ll need the conjugate multiplication technique. When substitution doesn’t work in the original function — usually because of a hole in the function — you can use conjugate multiplication to manipulate the function until substitution does work (it works because your manipulation plugs up the hole). Try this method for fraction functions that contain square roots. Conjugate multiplication rationalizes the numerator or denominator of a fraction, which means getting rid of square roots. Try substitution. Multiply the numerator and denominator by the conjugate of the expression containing the square root. The conjugate of a two-term expression is just the same expression with subtraction switched to addition or vice versa. The product of conjugates is always the square of the first thing minus the square of the second thing. 3. Cancel the (x – 4) from the numerator and denominator. 4. Now substitution works. This rationalizing process plugged the hole in the original function. And you see that the answer to the limit problem is the height of the hole. About This Article This article is from the book: No items found. About the book author: No items found. This article can be found in the category: Calculus No items found. Get a Subscription
13823	https://steeltubeinstitute.org/resources/hss-base-plate-design-for-axial-compression-and-bending-moment/	Skip to content HSS Base Plate Design for Axial Compression and Bending Moment By Mike Manor, PE, MLSE Technical Consultant, Steel Tube Institute September, 2024 Download PDF of Article All steel columns need a base plate to transfer the axial and/or flexural forces from the building above into the supporting foundation. Designing base plates for HSS columns is very similar to base plate design for wide flange columns with a few key differences in the specific details. AISC Design Guide 1 Third Edition (AISC DG1) explains the design process for various base plate loadings extremely well for wide flanges, but it only provides a brief paragraph discussion of the design process modifications for HSS. The AISC 16th Edition Manual Companion includes Example K.9 for an axially loaded square HSS column, but current standards do not provide examples for round HSS columns or any HSS columns subject to moment loading. Thus, design examples for HSS base plates are difficult to find. STI provides a great introductory overview of HSS baseplates, “Axially Loaded HSS Column to Base Plate Connections”, which includes a discussion on axial compression loading, axial tensile loading, fabrication, and welding design. To continue the discussion of HSS baseplate design, this article will explore the design of HSS column base plates under axial compression and bending moment. Geometry Design of HSS base plates begins with consideration of the geometry and available space for the base plate itself, the supporting foundations, and the anchor rod layout. Anchor rods provide a positive connection to the foundation in cases of uplift, and they must be positioned to allow both adequate space for installation and sufficient concrete cover for the loads present. See Figure 2 for several typical anchor rod layout options specific to HSS columns. For axial compression alone, the location of the anchor rods is not critical in base plate design. However, when uplift or overturning moment exists, the anchor location will play a role in determining the base plate thickness. Compression Loading Turning the focus to compression loading of HSS base plates, the main assumption for design is to evenly distribute the compression load under the base plate on the supporting concrete. AISC 360-16 equations J8-1 and J8-2 provide the capacity of the concrete with the second equation accounting for the effects of concrete confinement. The design of the steel base plate is based on the pressure from the total concrete capacity pushing upward on the plate with an even distribution per the AISC 16th Edition Steel Manual (AISC 16th Edition) and AISC DG1. The thickness is then designed for the flexure developed through cantilever action of the plate portion beyond the footprint of the HSS column. The maximum cantilever distance, m or n (See Figure 3 and AISC 16th Edition Equations 14-2 and 14-3), is the distance from end of plate to the critical flexural bending section of the plate and is used to determine the required base plate thickness. Note that the closed shape of HSS columns precludes the need to consider the λn’ distance that is necessary for the open shape of wide flanges where yield lines can form in the area between the flanges and web. Rectangular HSS columns have two advantages over wide flange columns. First, HSS columns are more efficient for axial loading, resulting in less column steel weight when comparing the capacity at tall unbraced lengths. See the STI HSS and WF Column Selection Guide for more information. Secondly, the coefficient for the critical and cantilever distance equations is 0.95 at rectangular HSS for both orthogonal directions, unlike WF sections which use a 0.8 coefficient for dimension n parallel to the flanges. When comparing WF and HSS with similar nominal dimensions, the smaller cantilever distance means a reduction in the bending moment and thus thinner base plates in many cases for less total steel weight. This can result in a significant tonnage difference in a large building with many columns. Round HSS is similar to rectangular HSS in that the cantilever coefficients are the same in the two orthogonal directions; however, the coefficient is 0.8 rather than 0.95. The use of these coefficients for HSS is intended to use the same values as wide flange for simplicity while providing a conservative approximation for the distance to the critical bending section. Uplift Loading Steel columns can also experience load cases of net uplift over either the entire base plate (axial tension) or on one side only for base plates under large moment (tension component at anchor rods). The tension means that the base plate must be designed for downward flexure which is in the opposite direction than assumed for axial compression as discussed above. The base plate in this case is no longer experiencing an evenly distributed pressure as the load is instead concentrated at the anchor rods. The base plate flexure occurs along either a straight or curved path for rectangular or round HSS respectively. There are two methods to design the base plate thickness for flexure due to uplift, from which the worst case will control. Check flexure due to the cantilever of the entire base plate beyond the column critical bending section using the entire base plate width with a moment arm equal to the distance 𝑥 in Figure 4a or 4c. Check flexure due to a critical bending section located 45° from the column corner. Refer to Figure 4a and 4b. This method is used when the anchor rods are located near the corner of the base plate, which is common for HSS columns. However, the difficult part of this method is determining the location of the flexural failure plane and the associated effective width of the plate. The assumption is that the base plate failure plane occurs at the corner of the HSS. First, start with a line from the anchor rod hole to the corner of the column as the flexural moment arm. Second, the effective width is determined by drawing two more lines from the anchor rod hole 45° each way from the first line until they intersect the assumed failure plane. See the “Designing Column Base Plates for Uplift” article (pp 27-28) in Structure Magazine for an example of this procedure. Alternatively, a viable option for base plates under tension is the method employed for HSS end plate tension splices. STI has many resources available for this condition of tension loading and similar conditions under flexural loading as well. See the following list of references for connection information and design examples for these alternate methods: Round HSS Bolted End-Plate Connections Under Axial Tension Loading Round HSS Bolted End-Plate Connections Under Bending Moment HSS Design Manual Volume 4 (Bolted Truss End Plate Splice Section) AISC Design Guide 24: Hollow Structural Section Connections Second Edition (Ch. 7) HSS Poles HSS Column Base Plate Design Examples Following are two design examples that demonstrate how to apply the base plate design methodology to HSS columns as presented in AISC Design Guide 1 Third Edition and Part 14 of the AISC 16th Edition Manual. The first example is a round HSS column with only an axial compression load. The second example is a rectangular HSS column with a bending moment combined with an axial compression load to demonstrate the effects of the interaction on the base plate design. Example #1: Round HSS Column Given: HSS Material: ASTM A500 Gr. C (Fy = 50 ksi) HSS 12×0.375 Column Base Plate Material: ASTM A572 Gr. 50 (Fy = 50 ksi) Factored Axial Load Pu = 500 k f‘c = 3 ksi Conservatively assume no increase from concrete confinement, thus A1 = A2 HSS Geometry D = 12 in tnominal = 0.375 in tdesign = 0.349 in Determine the required base plate area and optimize the base plate dimensions: For determining the base plate cantilever distance of round HSS, use 0.8D. Round up N = B = 18 in Per assumption given A1 = A2 = NB = (18 in)(18 in) = 324 in2 l = max⁡(m,n) = 4.2 in (Note: λn‘ is not used for HSS per AISC DG1) Use 18in x 18in x 1 1/4 in A572 Gr. 50 base plate. Example #2: Rectangular HSS Column With Bending Moment Given: HSS Material: ASTM A500 Gr. C (Fy = 50 ksi) HSS 16x12x1/2 Column Base Plate Material: ASTM A572 Gr. 50 (Fy = 50 ksi) Factored Axial Load Pu = 376 k Factored Moment Load Mu = 300 k ∙ ft = 3600 in ∙ k f‘c = 4 ksi Conservatively assume no increase from concrete confinement, thus A1 = A2 Eight anchor rods (4 each side) HSS Geometry HHSS = 16 in BHSS = 12 in tnominal = 0.5 in tdesign = 0.465 in Choose a trial base plate size based on locating anchor rods with dedge = 2″ steel edge distance and 2” clearance from the face of the column for installation, thus 4” beyond the HSS column in each direction. N = HHSS + (2)(4in) = 24 in B = BHSS + (2)(4in) = 20 in Determine the magnitude of the moment relative to the axial load: Since e > ecrit’ the loading is considered a large moment, and the assumed baseplate geometry needs to be verified that a solution is possible. Verify if the following inequality is true 484 ≥ 333 Inequality is true Determine the length of bearing along the base plate parallel to the moment by solving the quadratic equation: Since 34.29 in > N and N > 9.71 in > 0, use Y = 9.71 in The total tension at anchor rods is: Tu = qmaxY – Pu = (44.2 k/in)(9.71 in) – 376 k = 53.3 k [AISC DG1 Equation 4-55] Assuming four anchor rods on the tension side, the anchor rods need to be designed for: Determine the thickness required for the base plate. For cantilever distance of rectangular HSS columns, use a multiplier of 0.95 for both m and n. max(m, n) = 4.4 in (Note: If n was larger than m, n would be substituted in the following) Y = 9.71 in > m = 4.4 in (Compression zone extends beyond the base plate cantilever under the column) Base Plate Yielding Limit at Bearing Interface: Base Plate Yielding Limit at Tension Interface: tp(req) = max(1.38 in, 0.73 in) = 1.38 in (Bearing interface controls) Use 24 in x 20 in x 1 1/2 in A572 Gr. 50 base plate. Finite Element Analysis The previous portions of this article provide base plate design information and design examples based on hand calculation methods. However, software packages for steel connections now exist that include finite element analysis (FEA) specifically developed for design by practicing engineers. One option is Idea StatiCa that provides FEA for steel connections in accordance with the AISC Specification. Following is a brief demonstration of the FEA output from Idea StatiCa for comparison with the two examples above (see Figure 5). For the calculations above in Example 1, the assumption for a base plate under compression loading with a round HSS column per DG1 guidance is to assume a straight line critical flexural bending section (Figure 3c) based on the diameter of the column. This is a conservative assumption for adapting the rectangular column procedure to the round column shape for ease of hand calculation. However, engineering judgement intuitively would expect more of a circular flexural bending section where the highest base plate stresses would occur. From the FEA, Figure 6a is a map of the base plate stresses showing that indeed the highest stresses develop in a circular pattern. The color-coding scale in Figure 6a was developed by analyzing a 1-1/4” thick 50 ksi base plate as determined by Example 1 and shows a maximum plate stress in the range of 29 ksi. The maximum upper end of the scale comes from applying the AISC 360-16 Section F1 LRFD factor φ = 0.90 to the base plate yield stress: φFy = (0.90)(50 ksi) = 45 ksi. For comparison, the example has a demand capacity ratio (DCR) = t2min/t2 = 1.102/1.252 = 0.77 while the FEA has a DCR = 29 ksi/45 ksi = 0.64 showing the hand calculation to be conservative as expected. When using FEA for steel connection design, strain is also a calculated measure of connection acceptability. A common general assumption is to allow a maximum 5% plastic strain limit. Figure 6b shows the plastic strain for the round HSS column and base plate. In this example, the maximum strain is only 0.1% < 5%, but it is all located within the column rather than the base plate. For this round HSS example, the FEA demonstrates through both stress and strain that the 1-1/4” base plate is conservative. By maximizing the FEA, Idea StatiCa shows that a 1” A572 Gr. 50 ksi plate is adequate for the loading in Example 1 thus providing a savings for the base plate steel weight. Turning to Example 2, similar comparisons can be made between the hand calculations and the FEA for the rectangular HSS with large moment as shown in Figure 7. Figure 7a indicates that the entire compression flange of the HSS column plus a portion of each sidewall has reached the maximum yield stress and thus some plastic strain. However, looking at Figure 7b, the base plate has a much lower maximum flexural stress in the range of 34 ksi where the corners of the rectangular HSS provide the highest plate compression. Note that the stress in the plate is less than the maximum 45 ksi allowed and thus there is no plastic strain in the base plate. The hand calculated DCR is = t2min/t2 = 1.382/1.52 = 0.85 and the FEA DCR = 34 ksi/45 ksi = 0.76. By again maximizing the FEA, Idea StatiCa determines that a 1-1/4” A572 Gr. 50 ksi base plate is adequate for the loading rather than the 1-1/2” thickness from the hand calculations again providing a steel savings. Conclusion While base plate design for HSS members shares similarities with that of wide flange columns, there are key distinctions that engineers must account for, particularly in calculating the cantilever distances for flexural bending. The closed shape of HSS columns, especially rectangular ones, generally results in shorter cantilever distances, which can translate to more economical base plates, particularly in structures with numerous columns. Furthermore, finite element analysis (FEA) software provides valuable tools for optimizing base plate designs, offering a more precise understanding of stress and strain distribution. By leveraging these design resources, engineers can not only ensure structural integrity but also achieve potential savings in steel usage. This efficiency, combined with HSS columns’ performance under axial loads, especially at large unbraced lengths, extends the benefits of HSS members from the column itself to the supporting base plate. References STI Article: “Axially Loaded HSS Column to Base Plate Connections” by Kim Olson AISC Publication: “Design Guide 1 Third Edition: Base Connection Design for Steel Structures” AISC Publication: “360-22: Specification for Structural Steel Buildings” AISC Publication: “Steel Construction Manual 16th Edition” AISC/STI/AISI Publication: “Hollow Structural Sections Connection Manual” Structure Magazine Article (pp 27-28): “Designing Column Base Plates for Uplift” CISC Publication: “Hollow Structural Section Connections and Trusses – A Design Guide”
13824	https://www.doubtnut.com/qna/209193891	Find the domain of defination the following functions. f(x)=arccos(2x1+x) More from this Exercise The correct Answer is:−1/3≤x≤1 To find the domain of the function f(x)=arccos(2x1+x), we need to determine the values of x for which the function is defined. The function arccos(y) is defined for y in the interval (−1,1). Therefore, we need to ensure that: −1≤2x1+x≤1 We will break this down into two inequalities and solve them step by step. Step 1: Solve the first inequality 2x1+x≥−1 To solve this inequality, we first rearrange it: 2x1+x+1≥0 This simplifies to: 2x+1+x1+x≥0 which simplifies to: 3x+11+x≥0 Next, we find the critical points by setting the numerator and denominator to zero: 1. 3x+1=0 gives x=−13 2. 1+x=0 gives x=−1 Now, we will test the intervals determined by these critical points: (−∞,−1), (−1,−13), and (−13,∞). - For x<−1 (e.g., x=−2): 3(−2)+11+(−2)=−6+1−1=−5−1=5(positive) - For −1<x<−13 (e.g., x=−0.5): 3(−0.5)+11+(−0.5)=−1.5+10.5=−0.50.5=−1(negative) - For x>−13 (e.g., x=0): 3(0)+11+0=11=1(positive) Thus, the solution for the first inequality is: x∈(−∞,−1)∪(−13,∞) Step 2: Solve the second inequality 2x1+x≤1 Rearranging this inequality gives: 2x1+x−1≤0 This simplifies to: 2x−(1+x)1+x≤0 which simplifies to: x−11+x≤0 Again, we find the critical points: 1. x−1=0 gives x=1 2. 1+x=0 gives x=−1 Now, we will test the intervals determined by these critical points: (−∞,−1), (−1,1), and (1,∞). - For x<−1 (e.g., x=−2): −2−1−2+1=−3−1=3(positive) - For −1<x<1 (e.g., x=0): 0−10+1=−11=−1(negative) - For x>1 (e.g., x=2): 2−12+1=13(positive) Thus, the solution for the second inequality is: x∈(−1,1) Step 3: Find the intersection of the two solutions Now we need to find the intersection of the two sets: 1. From the first inequality: (−∞,−1)∪(−13,∞) 2. From the second inequality: (−1,1) The intersection is: (−13,1) Conclusion Thus, the domain of the function f(x)=arccos(2x1+x) is: (−13,1) To find the domain of the function f(x)=arccos(2x1+x), we need to determine the values of x for which the function is defined. The function arccos(y) is defined for y in the interval (−1,1). Therefore, we need to ensure that: −1≤2x1+x≤1 We will break this down into two inequalities and solve them step by step. Step 1: Solve the first inequality 2x1+x≥−1 To solve this inequality, we first rearrange it: 2x1+x+1≥0 This simplifies to: 2x+1+x1+x≥0 which simplifies to: 3x+11+x≥0 Next, we find the critical points by setting the numerator and denominator to zero: 3x+1=0 gives x=−13 1+x=0 gives x=−1 Now, we will test the intervals determined by these critical points: (−∞,−1), (−1,−13), and (−13,∞). For x<−1 (e.g., x=−2): 3(−2)+11+(−2)=−6+1−1=−5−1=5(positive) For −1<x<−13 (e.g., x=−0.5): 3(−0.5)+11+(−0.5)=−1.5+10.5=−0.50.5=−1(negative) For x>−13 (e.g., x=0): 3(0)+11+0=11=1(positive) Thus, the solution for the first inequality is: x∈(−∞,−1)∪(−13,∞) Step 2: Solve the second inequality 2x1+x≤1 Rearranging this inequality gives: 2x1+x−1≤0 This simplifies to: 2x−(1+x)1+x≤0 which simplifies to: x−11+x≤0 Again, we find the critical points: x−1=0 gives x=1 1+x=0 gives x=−1 Now, we will test the intervals determined by these critical points: (−∞,−1), (−1,1), and (1,∞). For x<−1 (e.g., x=−2): −2−1−2+1=−3−1=3(positive) For −1<x<1 (e.g., x=0): 0−10+1=−11=−1(negative) For x>1 (e.g., x=2): 2−12+1=13(positive) Thus, the solution for the second inequality is: x∈(−1,1) Step 3: Find the intersection of the two solutions Now we need to find the intersection of the two sets: From the first inequality: (−∞,−1)∪(−13,∞) From the second inequality: (−1,1) The intersection is: (−13,1) Conclusion Thus, the domain of the function f(x)=arccos(2x1+x) is: (−13,1) Topper's Solved these Questions Explore 4 Videos Explore 7 Videos Explore 12 Videos Explore 6 Videos Explore 17 Videos Similar Questions Find the domain of defination the following functions. f(x)=√3−x+cos−1(3−2x5)+ln(2\|x\|−3) Find the domain of defination the following functions. f(x)=√1−sinxlog5(1−4x2)+cos−1(1−{x}). where {x} is the fractional part of x. 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13825	https://en.wikipedia.org/wiki/Binomial_test	Jump to content Search Contents (Top) 1 Usage 2 Common use 3 Large samples 4 Example 5 In statistical software packages 6 See also 7 References 8 Further reading 9 External links Binomial test Deutsch Español Euskara Italiano Nederlands 日本語 Polski Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Test of statistical significance \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Binomial test" – news · newspapers · books · scholar · JSTOR (November 2016) (Learn how and when to remove this message) \| Binomial test is an exact test of the statistical significance of deviations from a theoretically expected distribution of observations into two categories using sample data. Usage [edit] A binomial test is a statistical hypothesis test used to determine whether the proportion of successes in a sample differs from an expected proportion in a binomial distribution. It is useful for situations when there are two possible outcomes (e.g., success/failure, yes/no, heads/tails), i.e., where repeated experiments produce binary data. If one assumes an underlying probability between 0 and 1, the null hypothesis is For a sample of size , we would expect successes. The formula of the binomial distribution gives the probability of those samples instead producing successes: Suppose that we want to test the alternative hypothesis i.e., we suspect that the actual probability of success is lower than . Then the -value of our experiment would be computed using a one-tailed test; specifically, we compute the probability of seeing an outcome as extreme as, or more extreme (i.e., less likely), than (where is defined as the number of successes in the trials of our experiment): An analogous computation can be done if we're testing if using the summation of the range from to instead. Calculating a -value for a two-tailed test is slightly more complicated, since a binomial distribution isn't symmetric if . This means that we can't just double the -value from the one-tailed test. Recall that we want to consider events that are as extreme, or more extreme, than the one we've seen, so we should consider the probability that we would see an event that is as, or less, likely than . Let denote all such events. Then the two-tailed -value is calculated as, Common use [edit] One common use of the binomial test is the case where the null hypothesizes that two categories occur with equal frequency (), such as a coin toss. Tables are widely available to give the significance observed numbers of observations in the categories for this case. However, as the example below shows, the binomial test is not restricted to this case. When there are more than two categories, and an exact test is required, the multinomial test, based on the multinomial distribution, must be used instead of the binomial test. Most common measures of effect size for Binomial tests are Cohen's h or Cohen's g. Large samples [edit] For large samples such as the example below, the binomial distribution is well approximated by convenient continuous distributions, and these are used as the basis for alternative tests that are much quicker to compute, such as Pearson's chi-squared test and the G-test. However, for small samples these approximations break down, and there is no alternative to the binomial test. The most usual (and easiest) approximation is through the standard normal distribution, in which a z-test is performed of the test statistic , given by where is the number of successes observed in a sample of size and is the probability of success according to the null hypothesis. An improvement on this approximation is possible by introducing a continuity correction: For very large , this continuity correction will be unimportant, but for intermediate values, where the exact binomial test doesn't work, it will yield a substantially more accurate result. In notation in terms of a measured sample proportion , null hypothesis for the proportion , and sample size , where and , one may rearrange and write the z-test above as by dividing by in both numerator and denominator, which is a form that may be more familiar to some readers. Example [edit] Suppose we have a board game that depends on the roll of one die and attaches special importance to rolling a 6. In a particular game, the die is rolled 235 times, and 6 comes up 51 times. If the die is fair, we would expect 6 to come up times. We have now observed that the number of 6s is higher than what we would expect on average by pure chance had the die been a fair one. But, is the number significantly high enough for us to conclude anything about the fairness of the die? This question can be answered by the binomial test. Our null hypothesis would be that the die is fair (probability of each number coming up on the die is 1/6). To find an answer to this question using the binomial test, we use the binomial distribution : with pmf . As we have observed a value greater than the expected value, we could consider the probability of observing 51 6s or higher under the null, which would constitute a one-tailed test (here we are basically testing whether this die is biased towards generating more 6s than expected). In order to calculate the probability of 51 or more 6s in a sample of 235 under the null hypothesis we add up the probabilities of getting exactly 51 6s, exactly 52 6s, and so on up to probability of getting exactly 235 6s: If we have a significance level of 5%, then this result (0.02654 < 5%) indicates that we have evidence that is significant enough to reject the null hypothesis that the die is fair. Normally, when we are testing for fairness of a die, we are also interested if the die is biased towards generating fewer 6s than expected, and not only more 6s as we considered in the one-tailed test above. In order to consider both the biases, we use a two-tailed test. Note that to do this we cannot simply double the one-tailed p-value unless the probability of the event is 1/2. This is because the binomial distribution becomes asymmetric as that probability deviates from 1/2. There are two methods to define the two-tailed p-value. One method is to sum the probability that the total deviation in numbers of events in either direction from the expected value is either more than or less than the expected value. The probability of that occurring in our example is 0.0437. The second method involves computing the probability that the deviation from the expected value is as unlikely or more unlikely than the observed value, i.e. from a comparison of the probability density functions. This can create a subtle difference, but in this example yields the same probability of 0.0437. In both cases, the two-tailed test reveals significance at the 5% level, indicating that the number of 6s observed was significantly different for this die than the expected number at the 5% level. In statistical software packages [edit] Binomial tests are available in most software used for statistical purposes. E.g. In R the above example could be calculated with the following code: binom.test(51, 235, 1/ 6, alternative = "less") (one-tailed test) binom.test(51, 235, 1/ 6, alternative = "greater") (one-tailed test) binom.test(51, 235, 1/ 6, alternative = "two.sided") (two-tailed test) In Java using the Apache Commons library: new BinomialTest(). binomialTest(235, 51, 1.0 / 6, AlternativeHypothesis. LESS_THAN) (one-tailed test) new BinomialTest(). binomialTest(235, 51, 1.0 / 6, AlternativeHypothesis. GREATER_THAN) (one-tailed test) new BinomialTest(). binomialTest(235, 51, 1.0 / 6, AlternativeHypothesis. TWO_SIDED) (two-tailed test) In SAS the test is available in the Frequency procedure PROC FREQ 0.1666670.05;RUN; In SPSS the test can be utilized through the menu Analyze > Nonparametric test > Binomial npar tests /binomial (.5) = node1 node2. In Python, use SciPy's binomtest: + scipy. stats. binomtest(51, 235,1.0/ 6, alternative = "greater") (one-tailed test) + scipy. stats. binomtest(51, 235,1.0/ 6, alternative ="two-sided") (two-tailed test) In MATLAB, use myBinomTest, which is available via Mathworks' community File Exchange website. myBinomTest will directly calculate the p-value for the observations given the hypothesized probability of a success. [pout]= myBinomTest(51, 235, 1/ 6) (generally two-tailed, but can optionally perform a one-tailed test). In Stata, use bitest. In Microsoft Excel, use Binom.Dist. The function takes parameters (Number of successes, Trials, Probability of Success, Cumulative). The "Cumulative" parameter takes a boolean True or False, with True giving the Cumulative probability of finding this many successes (a left-tailed test), and False the exact probability of finding this many successes. See also [edit] Wikiversity has learning resources about Binomial test p-value Cohen's g Cohen's h Lady tasting tea experiment References [edit] ^ Howell, David C. (2007). Statistical methods for psychology (6. ed.). Belmont, Calif.: Thomson. ISBN 978-0495012870. Further reading [edit] Dougherty, Edward R. (1990). "Testing a Proportion". Probability and Statistics for the Engineering, Computing, and Physical Sciences. Englewood Cliffs: Prentice Hall. pp. 417–423. ISBN 0-13-711995-X. External links [edit] Binomial Probability Calculator "The binomial test". www.graphpad.com. 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13826	https://stackoverflow.com/questions/16025620/finding-the-line-along-the-intersection-of-two-planes	Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Collectives¢ on Stack Overflow Find centralized, trusted content and collaborate around the technologies you use most. Learn more about Collectives Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Finding the line along the intersection of two planes Ask Question Asked Modified 7 years, 5 months ago Viewed 8k times 10 I am trying to draw the line formed by the intersections of two planes in 3D, but I am having trouble understanding the math, which has been explained here and here. I tried to figure it out myself, but the closest that I got to a solution was a vector pointing along the same direction as the intersection line, by using the cross product of the normals of the planes. I have no idea how to find a point on the intersection line, any point would do. I think that this method is a dead end. Here is a screenshot of this attempt: I tried to use the solution mentioned in this question, but it has a dead link to the original explanation, and the equation didn't work for me (it has unbalanced parentheses, which I tried to correct below). var planeA = new THREE.Plane((new THREE.Vector3(0, 0, 1)).normalize(), 100); var planeB = new THREE.Plane((new THREE.Vector3(1, 1, 1)).normalize(), -100); var x1 = planeA.normal.x, y1 = planeA.normal.y, z1 = planeA.normal.z, d1 = planeA.constant; var x2 = planeB.normal.x, y2 = planeB.normal.y, z2 = planeB.normal.z, d2 = planeB.constant; var point1 = new THREE.Vector3(); point1.x = 0; point1.z = (y2 / y1) (d1 - d2) / (z2 - z1 y2 / y1); point1.y = (-z1 point1.z - d1) / y1; var point2 = new THREE.Vector3(); point2.x = 1; point2.z = (y2 / y1) (x1 point2.x + d1) - (x2 point2.x - d2) / (z2 - z1 y2 / y1); point2.y = (-z1 point2.z - x1 point2.x - d1) / y1; console.log(point1, point2); output: THREE.Vector3 {x: -1, y: NaN, z: NaN, ¦} THREE.Vector3 {x: 1, y: Infinity, z: -Infinity, ¦} expected output: A point along the intersection where x = 0, and Another point on the same line where x = 1 If someone could point me to a good explanation of how this is supposed to work, or an example of a plane-plane intersection algorithm, I would be grateful. algorithm math 3d three.js linear-algebra Share Improve this question edited Apr 11, 2018 at 2:04 CÅur 39k2525 gold badges206206 silver badges281281 bronze badges asked Apr 15, 2013 at 22:17 Dan RossDan Ross 3,70944 gold badges3535 silver badges6161 bronze badges 1 If you just let x = 0 in both plane equations you can find y and z corresponding to that x. Repeat this for x = 1. user1329187 – user1329187 2013-04-17 15:11:00 +00:00 Commented Apr 17, 2013 at 15:11 Add a comment \| 4 Answers 4 Reset to default 7 Here is an implementation of a solution for plane-plane intersections described at . Essentially, you first use the cross product of the normals of the planes to find the direction of a line in both planes. Secondly, you use some algebra on the implicit equation of the planes (P . n + d = 0 where P is some point on the plane, n is the normal and d is the plane constant) to solve for a point which is on the intersection of the planes and also on one of the x=0, y=0 or z=0 planes. The solution is then the line described by a point and a vector. I was using three.js version 79 / Algorithm taken from See the section 'Intersection of 2 Planes' and specifically the subsection (A) Direct Linear Equation / function intersectPlanes(p1, p2) { // the cross product gives us the direction of the line at the intersection // of the two planes, and gives us an easy way to check if the two planes // are parallel - the cross product will have zero magnitude var direction = new THREE.Vector3().crossVectors(p1.normal, p2.normal) var magnitude = direction.distanceTo(new THREE.Vector3(0, 0, 0)) if (magnitude === 0) { return null } // now find a point on the intersection. We use the 'Direct Linear Equation' // method described in the linked page, and we choose which coordinate // to set as zero by seeing which has the largest absolute value in the // directional vector var X = Math.abs(direction.x) var Y = Math.abs(direction.y) var Z = Math.abs(direction.z) var point if (Z >= X && Z >= Y) { point = solveIntersectingPoint('z', 'x', 'y', p1, p2) } else if (Y >= Z && Y >= X){ point = solveIntersectingPoint('y', 'z', 'x', p1, p2) } else { point = solveIntersectingPoint('x', 'y', 'z', p1, p2) } return [point, direction] } / This method helps finding a point on the intersection between two planes. Depending on the orientation of the planes, the problem could solve for the zero point on either the x, y or z axis / function solveIntersectingPoint(zeroCoord, A, B, p1, p2){ var a1 = p1.normal[A] var b1 = p1.normal[B] var d1 = p1.constant var a2 = p2.normal[A] var b2 = p2.normal[B] var d2 = p2.constant var A0 = ((b2 d1) - (b1 d2)) / ((a1 b2 - a2 b1)) var B0 = ((a1 d2) - (a2 d1)) / ((a1 b2 - a2 b1)) var point = new THREE.Vector3() point[zeroCoord] = 0 point[A] = A0 point[B] = B0 return point } var planeA = new THREE.Plane((new THREE.Vector3(0, 0, 1)).normalize(), 100) var planeB = new THREE.Plane((new THREE.Vector3(1, 1, 1)).normalize(), -100) var [point, direction] = intersectPlanes(planeA, planeB) Share Improve this answer answered Jul 18, 2016 at 13:20 user2013483user2013483 7122 silver badges22 bronze badges 1 Comment user585776 user585776 This code is not working! i posted it in a jsfiddle and dra the planes and a vector along the returned intersect, its completly somewhere else: jsfiddle.net/ksfoLp1d/6 How did this get 7 upvotes? 5 When I have problems like this, I usually let a symbolic algebra package (Mathematica in this case) deal with it. After typing In:= n1={x1,y1,z1};n2={x2,y2,z2};p={x,y,z}; In:= Solve[n1.p==d1&&n2.p==d2,p] and simplifying and substituting x=0 and x=1, I get d2 z1 - d1 z2 d2 y1 - d1 y2 Out= {{{y -> -------------, z -> ----------------}}, y2 z1 - y1 z2 -(y2 z1) + y1 z2 d2 z1 - x2 z1 - d1 z2 + x1 z2 > {{y -> -----------------------------, y2 z1 - y1 z2 d2 y1 - x2 y1 + (-d1 + x1) y2 > z -> -----------------------------}}} -(y2 z1) + y1 z2 Share Improve this answer answered Apr 15, 2013 at 23:15 David EisenstatDavid Eisenstat 65.7k77 gold badges6666 silver badges126126 bronze badges 5 Comments Dan Ross Dan Ross That sounds like a great tool for problems like this, but I am unfamiliar with symbolic algebra packages. To tell you the truth, I am even having a hard time understanding the output when you did it for me :p I checked out Octave, and decided that it would take about as much time to learn the tool as it would to research a solution to my problem the old fashioned way. I would love to explore Octave more in the future though, maybe while taking taking an online course in linear algebra. I hope Khan has lessons! Dan Ross Dan Ross Thank you for showing me a great tool for future geometry problem solving. I am new to 3D programming, and I can see that there will be a lot more problems like this in my future. David Eisenstat David Eisenstat @DanRoss Unless Octave has features I don't know about, you want something more like Sage. David Eisenstat David Eisenstat Also, you might want to learn about projective geometry to reduce the number of degenerate cases you have to deal with (e.g., line is parallel to the x-axis). Dan Ross Dan Ross Yes, cases. An article recommended choosing the axis with the largest absolute value in the cross product of the plane normals. I chose the x axis arbitrarily. Projective geometry you call it, I will have to google that. 4 It is easy to let three.js solve this for you. If you were to express your problem in matrix notation m x = v Then the solution for x is x = inverse( m ) v We'll use a 4x4 matrix for m, because three.js has an inverse() method for the Matrix4 class. var x1 = 0, y1 = 0, z1 = 1, d1 = 100; var x2 = 1, y2 = 1, z2 = 1, d2 = -100; var c = 0; // the desired value for the x-coordinate var v = new THREE.Vector4( d1, d2, c, 1 ); var m = new THREE.Matrix4( x1, y1, z1, 0, x2, y2, z2, 0, 1, 0, 0, 0, 0, 0, 0, 1 ); var minv = new THREE.Matrix4().getInverse( m ); v.applyMatrix4( minv ); console.log( v ); The x-component of v will be equal to c, as desired, and the y- and z-components will contain the values you are looking for. The w-component is irrelevalent. Now, repeat for the next value of c, c = 1. three.js r.58 Share Improve this answer answered Apr 17, 2013 at 21:57 WestLangleyWestLangley 105k1111 gold badges288288 silver badges285285 bronze badges 1 Comment user585776 user585776 executing this code returns { x: 100, y: -100, z: 0, w: 1 }, so x is not equal to c. And when i play around this always seems to return in x and y what is in d1 and d2 and in z whats in c and never changes no matter how i change the vectors of the plane creation. How is this the accepted answer? Here is a jsfiddle where i was trying this out: jsfiddle.net/0vzbjeco 3 Prerequisites Recall that to represent a line we need a vector describing its direction and a point through which this line goes. This is called parameterized form: line_point(t) = t (point_2 - point_1) + point_1 where point_1 and point_2 are arbitrary points through which the line goes, and t is a scalar which parameterizes our line. Now we can find any point line_point(t) on the line if we put arbitrary t into the equation above. NOTE: The term (point_2 - point_1) is nothing, but a vector describing the direction of our line, and the term point_1 is nothing, but a point through which our line goes (of course point_2) would also be fine to use too. The Algorithm Find the direction direction of the intersection line by taking cross product of plane normals, i.e. direction = cross(normal_1, normal_2). Take any plane, for example the first one, and find any 2 distinct points on this plane: point_1 and point_2. If we assume that the plane equation is of the form a1 x + b1 y + c1 z + d1 = 0, then to find 2 distinct points we could do the following: y1 = 1 z1 = 0 x1 = -(b1 + d1) / a1 y2 = 0 z2 = 1 x2 = -(c1 + d1) / a1 where point_1 = (x1, y1, z1) and point_2 = (x2, y2, z2). 3. Now that we have 2 points, we can construct the parameterized representation of the line lying on this first plane: line_point(t) = t (point_2 - point_1) + point_1, where line_point(t) describes any point on this line, and t is just an input scalar (frequently called parameter). 4. Find the intersection point intersection_point of the line line_point(t) and the second plane a2 x + b2 y + c2 z + d2 = 0 by using the standard line-plane intersection algorithm (pay attention to the Algebraic form section as this is all you need to implement line-plane intersection, if you haven't done so already). 5. Our intersection line is now found and can be constructed in parameterized form as usual: intersection_line_point(s) = s direction + intersection_point, where intersection_line_point(s) describes any point on this intersection line, and s is parameter. NOTE: I didn't read this algorithm anywhere, I've just devised it from the top of my head based on my knowledge of linear algebra. That doesn't mean that it doesn't work, but it might be possible that this algorithm can be optimized further. Conditioning When 2 normal vectors normal_1 and normal_2 are almost collinear this problem gets extremely ill-conditioned. Geometrically it means that the 2 planes are almost parallel to each other and determining the intersection line with acceptable precision becomes impossible in finite-precision arithmetic which is floating-point arithmetic in this case. Share Improve this answer edited Apr 16, 2013 at 0:48 answered Apr 15, 2013 at 23:51 Alexander ShukaevAlexander Shukaev 17k88 gold badges7272 silver badges8989 bronze badges 2 Comments Dan Ross Dan Ross Well, I got to Step 2, and found a problem: point_1 (red ball) and point_2 (green ball) were not on the first plane, they were on either side of it pic. It is quite possible that I did not properly understand your explanation, 3D math is a new subject for me, and I didn't do much linear algebra in school. I did finally find a good article about intersection algorithms, and it seems that I have made a working script from it's "Direct Linear Equation" solution. Alexander Shukaev Alexander Shukaev You're obviously doing it wrong. Once again, on the step 2 we want to choose any plain of these 2 (because it simply does not matter which) and find any line that lies on that plane. To determine a line all we have to do is simple take 2 arbitrary points and that plane (again doesn't matter which points exactly, it only matters that they are on that plane). So we pick a simple choice to do little computations, and when these 2 points are computed - our line (lying on the chosen plane) can be fully determined by the line_point(t) equation from the step 3. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm math 3d three.js linear-algebra See similar questions with these tags. 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13827	https://www.iso.org/standard/10706.html	ISO 4724:1984 - Oil of cedarwood, Virginia (Juniperus virginiana Linnaeus) Skip to main content Applications OBP English español français русский Menu Standards SectorsHealth IT & related technologies Management & services Security, safety & risk Transport Energy Diversity & inclusion Environmental sustainability Food & agriculture Materials Building & construction Engineering About ISO Insights & newsInsightsAll insights Healthcare Artificial intelligence Climate change Transport Cybersecurity Quality management Renewable energy Occupational health and safety NewsExpert talk Standards world Media kit Taking part Store Search Cart Reference number ISO 4724:1984 © ISO 2025 ISO 4724:1984 Oil of cedarwood, Virginia (Juniperus virginiana Linnaeus) Edition 1 1984-10 Withdrawn ISO 4724:1984 10706 ~~ISO 4724:1984~~Oil of cedarwood, Virginia (Juniperus virginiana Linnaeus) Withdrawn (Edition 1, 1984) New version available: ISO 4724:2004 Abstract The Standard specifies the following characteristics of the oil. Appearance: Fairly viscous liquid, sometimes containing crystals. Colour: Almost colourless to pale yellow. Odour: Characteristic, woody, pleasantly sweet and balsamic, recalling the wood of the tree. Relative density at 20/20 C: 0,941 - 0,970. Refractive index at 20 °C: 1,5040 - 1,5080. Optical rotation at 20 °C: Range from - 38 to - 14 . Miscibility with 95 % (V/V) ethanol at 20 °C: 1 : 5. Cedrol content by chromatography: Maximum 14 % (m/m). General information Status :Withdrawn Publication date :1984-10 Stage : Withdrawal of International Standard [95.99] Edition :1 Number of pages :2 Technical Committee: ISO/TC 54 ICS: 71.100.60 RSSupdates Life cycle Now Withdrawn ISO 4724:1984 Stage: 95.99 00 Preliminary 10 Proposal 20 Preparatory 30 Committee 30.00 1975-09-01 Committee draft (CD) registered 30.99 1981-02-01 CD approved for registration as DIS 40 Enquiry) 40.00 1983-01-01 DIS registered 40.20 1983-04-01 DIS ballot initiated: 12 weeks 40.60 1983-11-01 Close of voting 50 Approval) 50.00 1984-06-01 Final text received or FDIS registered for formal approval 50.20 1984-10-01 Proof sent to secretariat or FDIS ballot initiated: 8 weeks 50.60 1984-10-01 Close of voting. Proof returned by secretariat 60 Publication) 60.60 1984-10-01 International Standard published 90 Review) 90.92 1993-11-30 International Standard to be revised 95 Withdrawal) 95.99 2004-06-15 Withdrawal of International Standard Revised by Published ISO 4724:2004 Got a question? Check out our Help and Support Store Store ICS 71 71.100 71.100.60 ISO 4724:1984 Sitemap Standards Benefits Popular standards Conformity assessment SDGs Sectors Health IT & related technologies Management & services Security, safety & risk Transport Energy Environmental sustainability Materials About ISO What we do Structure Members Events Strategy Insights & news Insights All insights Healthcare Artificial intelligence Climate change Transport News Expert talk Standards world Media kit Taking part Who develops standards Deliverables Get involved Collaborating to accelerate effective climate action Resources Drafting standards Store Store Publications and products ISO name and logo Privacy Notice Copyright Cookie policy Media kit Jobs Help and support Making liveseasier,saferandbetter. Sign up for email updates © All Rights Reserved All ISO publications and materials are protected by copyright and are subject to the user’s acceptance of ISO’s conditions of copyright. Any use, including reproduction requires our written permission. All copyright requests should be addressed to copyright@iso.org. With the exception of the content available through the ISO Open datapage and subject to the terms contained therein, no ISO content may be used for any machine learning and/or artificial intelligence and/or similar technologies, including but not limited to accessing or using it to (i) train data for large language or similar models, or (ii) prompt or otherwise enable artificial intelligence or similar tools to generate responses. We are committed to ensuring that our website is accessible to everyone. If you have any questions or suggestions regarding the accessibility of this site, please contact us. Add to cart
13828	https://chemistry.stackexchange.com/questions/59050/significant-figures-in-conversion-factors	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Significant figures in conversion factors? Ask Question Asked Modified 7 years, 7 months ago Viewed 25k times 2 $\begingroup$ Example: I have $\pu{64 cm^{3}}$ of milk. How much is that in gallons? When I use the conversion factors $\frac{\pu{1 mL}}{\pu{1 cm^3}}$ $\frac{\pu{1L}}{\pu{1000 mL}}$ $\frac{\pu{1.0567 qt}}{\pu{1 L}}$ $\frac{\pu{1 gal}}{\pu{4 qt}}$. I know the final answer should only have two significant figures, but how many significant figures should the intermediate conversion factors have? I have heard that: conversion factors should have one more sig fig than the least precise measurement. (So three in this case) But I also heard that we shouldn't round anything until the end. (So use all 5, and round after conversions are complete) The reason why I got confused is because of my textbook: significant-figures Share edited Feb 19, 2018 at 2:22 pentavalentcarbon 7,55066 gold badges5151 silver badges7979 bronze badges asked Sep 14, 2016 at 20:07 eromoderomod 30333 gold badges55 silver badges1414 bronze badges $\endgroup$ 9 1 $\begingroup$ check your conversion factor. $1 \text{cm}^3$ is a milliliter which is much much less than a quart. $\endgroup$ MaxW – MaxW 2016-09-14 20:14:28 +00:00 Commented Sep 14, 2016 at 20:14 1 $\begingroup$ With a modern calculator, I wouldn't round anything until the end. When doing calculations with a slide rule or log tables, then things were a bit different. $\endgroup$ MaxW – MaxW 2016-09-14 20:15:51 +00:00 Commented Sep 14, 2016 at 20:15 $\begingroup$ This is case dependent. You need to realize what kind of mathematical object are you deal with. $\endgroup$ user1420303 – user1420303 2016-09-14 20:52:47 +00:00 Commented Sep 14, 2016 at 20:52 $\begingroup$ @MaxW, so is my textbook wrong in rounding in this example? $\endgroup$ eromod – eromod 2016-09-15 23:28:25 +00:00 Commented Sep 15, 2016 at 23:28 1 $\begingroup$ As I said, if I had a calculator, I wouldn't round anything until the end. If I had to do all the calculations by hand, then I'd be inclined to round intermediate values. I'm not going to multiple two 8-digit numbers by hand and then round to 2 digits. $\endgroup$ MaxW – MaxW 2016-09-16 03:36:56 +00:00 Commented Sep 16, 2016 at 3:36 \| Show 4 more comments 2 Answers 2 Reset to default 1 $\begingroup$ Use significant figures as much as you can in intermediate conversion factors,and then round off the final answer to two significant figures,using more significant figures in intermediate conversion factors will lead to a accurate answer. I let you conclude. Share answered Sep 15, 2016 at 13:25 Vidyanshu MishraVidyanshu Mishra 1,41722 gold badges1818 silver badges3535 bronze badges $\endgroup$ 2 $\begingroup$ that was my initial impression too, but have a look at my textbook example I attached. $\endgroup$ eromod – eromod 2016-09-15 23:26:57 +00:00 Commented Sep 15, 2016 at 23:26 $\begingroup$ you should use as much as sig.fig.,i have seen many examples and solved many exersises regarding this topic,and have never seen rounding off before the final answer unless significant figures are too much and cannot be taken under arithmatic operations. $\endgroup$ Vidyanshu Mishra – Vidyanshu Mishra 2016-09-16 05:12:24 +00:00 Commented Sep 16, 2016 at 5:12 Add a comment \| 1 $\begingroup$ I'm not sure where your conversion factor came from as it is incorrect but regarding the correct factor: Because the inch is defined using the metre, and the gallon defined using the inch, there is an exact conversion between them1 US gallon is exactly 3.785411784 L. Thus, like any exact quantity or conversion factor, we treat the gallon to litre conversion factor as having infinite significant figures. Because we have defined it exactly, there is no implicit uncertainty in the number. There are other conversion factors that are not exact and do actually have uncertainties, but for most practical purposes, these have far more significant figures than a normal measured quantity. See this question for more details. Share edited Apr 13, 2017 at 12:57 CommunityBot 1 answered Sep 14, 2016 at 20:57 Michael DM DrydenMichael DM Dryden 13.2k4343 silver badges5858 bronze badges $\endgroup$ 5 $\begingroup$ conversion factors were wrong, I updated them. But what is the conclusion though? How many sig figs should I use? That link says I should look it up in the CRC Handbook. That what you recommend too? $\endgroup$ eromod – eromod 2016-09-14 21:50:46 +00:00 Commented Sep 14, 2016 at 21:50 $\begingroup$ As my answer says, the conversion factor is exact. $\endgroup$ Michael DM Dryden – Michael DM Dryden 2016-09-15 03:16:07 +00:00 Commented Sep 15, 2016 at 3:16 $\begingroup$ So if my measured quantity is $\ 64 cm^{3}$ , how many significant figures should I use in the qt/L conversion? All of them because its an exact conversion? Or just three sig figs (one more that the two I have to work with) $\endgroup$ eromod – eromod 2016-09-15 03:27:54 +00:00 Commented Sep 15, 2016 at 3:27 $\begingroup$ @eromod The conversion factor is only exact if you use that exact conversion factor. For example, an inch is defined as 2.54 cm. If you use that number, it has infinite precision. However, if you say "ahh, 2.5 is close enough" then the conversion factor only has 2 significant digits. Likewise here. If you use the exact conversion factor, there's infinite precision. If you round it, it has the rounded amount of precision. $\endgroup$ R.M. – R.M. 2016-09-15 22:19:43 +00:00 Commented Sep 15, 2016 at 22:19 $\begingroup$ @ R.M. But in my textbook, they rounded the conversion factor...I added a screenshot $\endgroup$ eromod – eromod 2016-09-15 23:24:58 +00:00 Commented Sep 15, 2016 at 23:24 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions significant-figures See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked Should durations of time be considered to determine final significant figures of answer? 4 Do all conversion factors have an infinite number of significant figures? 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13829	https://artofproblemsolving.com/downloads/printable_post_collections/3003271.pdf?srsltid=AfmBOoparTk9Hr28JMst-BX3SrFNU4nUDMhff58nXjC5_YiG_4Dcs8v2	AoPS Community 2012 EMCC Exeter Math Club Competition 2012 www.artofproblemsolving.com/community/c3003271 by parmenides51 – Team Round – p1. The longest diagonal of a regular hexagon is 12 inches long. What is the area of the hexagon, in square inches? p2. When Al and Bob play a game, either Al wins, Bob wins, or they tie. The probability that Al does not win is 23 , and the probability that Bob does not win is 34 . What is the probability that they tie? p3. Find the sum of the a + b values over all pairs of integers (a, b ) such that 1 ≤ a < b ≤ 10 .That is, compute the sum (1 + 2) + (1 + 3) + (1 + 4) + ... + (2 + 3) + (2 + 4) + ... + (9 + 10) . p4. A 3 × 11 cm rectangular box has one vertex at the origin, and the other vertices are above the x-axis. Its sides lie on the lines y = x and y = −x. What is the y-coordinate of the highest point on the box, in centimeters? p5. Six blocks are stacked on top of each other to create a pyramid, as shown below. Carl re-moves blocks one at a time from the pyramid, until all the blocks have been removed. He never removes a block until all the blocks that rest on top of it have been removed. In how many dif-ferent orders can Carl remove the blocks? png p6. Suppose that a right triangle has sides of lengths pa + b√3,p3 + 2 √3, and p4 + 5 √3,where a, b are positive integers. Find all possible ordered pairs (a, b ). p7. Farmer Chong Gu glues together 4 equilateral triangles of side length 1 such that their edges coincide. He then drives in a stake at each vertex of the original triangles and puts a rubber band around all the stakes. Find the minimum possible length of the rubber band. p8. Compute the number of ordered pairs (a, b ) of positive integers less than or equal to 100 ,such that a b − 1 is a multiple of 4. ©2023 AoPS Incorporated 1AoPS Community 2012 EMCC p9. In triangle ABC , ∠C = 90 o. Point P lies on segment BC and is not B or C. Point I lies on segment AP . If ∠BIP = ∠P BI = ∠CAB = mo for some positive integer m, find the sum of all possible values of m. p10. Bob has 2 identical red coins and 2 identical blue coins, as well as 4 distinguishable buck-ets. He places some, but not necessarily all, of the coins into the buckets such that no bucket contains two coins of the same color, and at least one bucket is not empty. In how many ways can he do this? p11. Albert takes a 4 × 4 checkerboard and paints all the squares white. Afterward, he wants to paint some of the square black, such that each square shares an edge with an odd number of black squares. Help him out by drawing one possible configuration in which this holds. (Note: the answer sheet contains a 4 × 4 grid.) p12. Let S be the set of points (x, y ) with 0 ≤ x ≤ 5, 0 ≤ y ≤ 5 where x and y are integers. Let T be the set of all points in the plane that are the midpoints of two distinct points in S. Let U be the set of all points in the plane that are the midpoints of two distinct points in T . How many distinct points are in U ? (Note: The points in T and U do not necessarily have integer coordinates.) p13. In how many ways can one express 6036 as the sum of at least two (not necessarily posi-tive) consecutive integers? p14. Let a, b, c, d, e, f be integers (not necessarily distinct) between −100 and 100 , inclusive, such that a + b + c + d + e + f = 100 . Let M and m be the maximum and minimum possible values, respectively, of abc + bcd + cde + def + ef a + f ab + ace + bdf. Find Mm . p15. In quadrilateral ABCD , diagonal AC bisects diagonal BD . Given that AB = 20 , BC = 15 , CD = 13 , AC = 25 , find DA .PS. You had better use hide for answers. Collected here ( community/c5h2760506p24143309 ). – Guts Round – Round 1 ©2023 AoPS Incorporated 2AoPS Community 2012 EMCC p1. Ravi has a bag with 100 slips of paper in it. Each slip has one of the numbers 3, 5, or 7 written on it. Given that half of the slips have the number 3 written on them, and the average of the values on all the slips is 4.4, how many slips have 7 written on them? p2. In triangle ABC , point D lies on side AB such that AB ⊥ CD . It is given that CD BD = 12 , AC = 29 , and AD = 20 . Find the area of triangle BCD . p3. Compute (123 + 4)(123 + 5) − 123 · 132 .Round 2 p4. David is evaluating the terms in the sequence an = ( n + 1) 3 − n3 for n = 1 , 2, 3, .... (that is, a1 = 2 3 − 13 , a2 = 3 3 − 23, a3 = 4 3 − 33, and so on). Find the first composite number in the sequence. (An positive integer is composite if it has a divisor other than 1 and itself.) p5. Find the sum of all positive integers strictly less than 100 that are not divisible by 3. p6. In how many ways can Alex draw the diagram below without lifting his pencil or retracing a line? (Two drawings are different if the order in which he draws the edges is different, or the direction in which he draws an edge is different). png Round 3 p7. Fresh Mann is a 9th grader at Euclid High School. Fresh Mann thinks that the word vertices is the plural of the word vertice. Indeed, vertices is the plural of the word vertex. Using all the letters in the word vertice, he can make m 7-letter sequences. Using all the letters in the word vertex, he can make n 6-letter sequences. Find m − n. p8. Fresh Mann is given the following expression in his Algebra 1 class: 101 − 102 = 1 . Fresh Mann is allowed to move some of the digits in this (incorrect) equation to make it into a correct equation. What is the minimal number of digits Fresh Mann needs to move? p9. Fresh Mann said, ”The function f (x) = ax 2 + bx + c passes through 6 points. Their x-coordinates are consecutive positive integers, and their y-coordinates are 34 , 55 , 84 , 119 , 160 , ©2023 AoPS Incorporated 3AoPS Community 2012 EMCC and 207 , respectively.” Sophy Moore replied, ”You’ve made an error in your list,” and replaced one of Fresh Mann’s numbers with the correct y-coordinate. Find the corrected value. Round 4 p10. An assassin is trying to find his target’s hotel room number, which is a three-digit positive integer. He knows the following clues about the number: (a) The sum of any two digits of the number is divisible by the remaining digit. (b) The number is divisible by 3, but if the first digit is removed, the remaining two-digit number is not. (c) The middle digit is the only digit that is a perfect square. Given these clues, what is a possible value for the room number? p11. Find a positive real number r that satisfies 4 + r3 9 + r6 = 15 − r3 − 19 + r6 . p12. Find the largest integer n such that there exist integers x and y between 1 and 20 inclusive with 21 19 − xy < 1 n . PS. You had better use hide for answers. Last rounds have been posted here ( com/community/c4h2784267p24464980 ). Collected here ( community/c5h2760506p24143309 ). – Round 5 p13. A unit square is rotated 30 o counterclockwise about one of its vertices. Determine the area of the intersection of the original square with the rotated one. p14. Suppose points A and B lie on a circle of radius 4 with center O, such that ∠AOB = 90 o.The perpendicular bisectors of segments OA and OB divide the interior of the circle into four regions. Find the area of the smallest region. p15. Let ABCD be a quadrilateral such that AB = 4 , BC = 6 , CD = 5 , DA = 3 , and ∠DAB =90 o. There is a point I inside the quadrilateral that is equidistant from all the sides. Find AI .Round 6 ©2023 AoPS Incorporated 4AoPS Community 2012 EMCC The answer to each of the three questions in this round depends on the answer to one of the other questions. There is only one set of correct answers to these problems; however, each question will be scored independently, regardless of whether the answers to the other questions are correct. p16. Let C be the answer to problem 18 . Compute 1 − 122 1 − 132 ... 1 − 1 C2 . p17. Let A be the answer to problem 16 . Let P QRS be a square, and let point M lie on segment P Q such that M Q = 7 P M and point N lie on segment P S such that N S = 7 P N . Segments M S and N Q meet at point X. Given that the area of quadrilateral P M XN is A − 12 , find the side length of the square. p18. Let B be the answer to problem 17 and let N = 6 B. Find the number of ordered triples (a, b, c ) of integers between 0 and N − 1, inclusive, such that a + b + c is divisible by N .Round 7 p19. Let k be the units digit of 7777777 \| {z } Seven 7s . What is the largest prime factor of the number consisting of k 7’s written in a row? p20. Suppose that E = 7 7 , M = 7 , and C = 777 . The characters E, M, C, C are arranged randomly in the following blanks. ... × ... × ... × ... Then one of the multiplication signs is chosen at random and changed to an equals sign. What is the probability that the resulting equation is true? p21 . During a recent math contest, Sophy Moore made the mistake of thinking that 133 is a prime number. Fresh Mann replied, ”To test whether a number is divisible by 3, we just need to check whether the sum of the digits is divisible by 3. By the same reasoning, to test whether a number is divisible by 7, we just need to check that the sum of the digits is a multiple of 7, so 133 is clearly divisible by 7.” Although his general principle is false, 133 is indeed divisible by 7.How many three-digit numbers are divisible by 7 and have the sum of their digits divisible by 7?Round 8 ©2023 AoPS Incorporated 5AoPS Community 2012 EMCC p22. A look-and-say sequence is defined as follows: starting from an initial term a1, each subse-quent term ak is found by reading the digits of ak−1 from left to right and specifying the number of times each digit appears consecutively. For example, 4 would be succeeded by 14 (”One four.”), and 31337 would be followed by 13112317 (”One three, one one, two three, one seven.”) If a1 is a random two-digit positive integer, find the probability that a4 is at least six digits long. p23. In triangle ABC , ∠C = 90 o. Point P lies on segment BC and is not B or C. Point I lies on segment AP , and ∠BIP = ∠P BI = ∠CAB . If AP BC = k, express IP CP in terms of k. p24. A subset of {1, 2, 3, ..., 30 } is called delicious if it does not contain an element that is 3 times another element. A subset is called super delicious if it is delicious and no delicious set has more elements than it has. Determine the number of super delicious subsets. PS. You sholud use hide for answers. First rounds have been posted here ( com/community/c4h2784267p24464980 ). Collected here ( community/c5h2760506p24143309 ). – Individual Accuracy – p1. An 18 oz glass of apple juice is 6% sugar and a 6oz glass of orange juice is 12% sugar. The two glasses are poured together to create a cocktail. What percent of the cocktail is sugar? p2. Find the number of positive numbers that can be expressed as the difference of two integers between −2 and 2012 inclusive. p3. An annulus is defined as the region between two concentric circles. Suppose that the inner circle of an annulus has radius 2 and the outer circle has radius 5. Find the probability that a randomly chosen point in the annulus is at most 3 units from the center. p4. Ben and Jerry are walking together inside a train tunnel when they hear a train approaching. They decide to run in opposite directions, with Ben heading towards the train and Jerry heading away from the train. As soon as Ben finishes his 1200 meter dash to the outside, the front of the train enters the tunnel. Coincidentally, Jerry also barely survives, with the front of the train exiting the tunnel as soon as he does. Given that Ben and Jerry both run at 1/9 of the train’s speed, how long is the tunnel in meters? p5. Let ABC be an isosceles triangle with AB = AC = 9 and ∠B = ∠C = 75 o. Let DEF be another triangle congruent to ABC . The two triangles are placed together (without overlapping) ©2023 AoPS Incorporated 6AoPS Community 2012 EMCC to form a quadrilateral, which is cut along one of its diagonals into two triangles. Given that the two resulting triangles are incongruent, find the area of the larger one. p6. There is an infinitely long row of boxes, with a Ditto in one of them. Every minute, each existing Ditto clones itself, and the clone moves to the box to the right of the original box, while the original Ditto does not move. Eventually, one of the boxes contains over 100 Dittos. How many Dittos are in that box when this first happens? p7. Evaluate 26 + 36 + 998 + 26 · 36 + 26 · 998 + 36 · 998 + 26 · 36 · 998 . p8. There are 15 students in a school. Every two students are either friends or not friends. Among every group of three students, either all three are friends with each other, or exactly one pair of them are friends. Determine the minimum possible number of friendships at the school. p9. Let f (x) = p 2x + 1 + 2 √x2 + x. Determine the value of 1 f (1) + 1 f (1) + 1 f (3) + ... + 1 f (24) . p10. In square ABCD , points E and F lie on segments AD and CD , respectively. Given that ∠EBF = 45 o, DE = 12 , and DF = 35 , compute AB .PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ). – Individual Speed – 20 problems for 20 minutes. p1. Evaluate = 12·3·4 + 13·4·5 . p2. A regular hexagon and a regular n-sided polygon have the same perimeter. If the ratio of the side length of the hexagon to the side length of the n-sided polygon is 2 : 1 , what is n? p3. How many nonzero digits are there in the decimal representation of 2 · 10 · 500 · 2500 ? p4. When the numerator of a certain fraction is increased by 2012 , the value of the fraction increases by 2. What is the denominator of the fraction? ©2023 AoPS Incorporated 7AoPS Community 2012 EMCC p5. Sam did the computation 1 − 10 · a + 22 , where a is some real number, except he messed up his order of operations and computed the multiplication last; that is, he found the value of (1 − 10) · (a + 22) instead. Luckily, he still ended up with the right answer. What is a? p6. Let n! = n · (n − 1) · ·· 2 · 1. For how many integers n between 1 and 100 inclusive is n! divisible by 36 ? p7. Simplify the expression q 3·27 3 27 ·33 p8. Four points A, B, C, D lie on a line in that order such that AB CB = AD CD . Let M be the midpoint of segment AC . If AB = 6 , BC = 2 , compute M B · M D . p9. Allan has a deck with 8 cards, numbered 1, 1, 2, 2, 3, 3, 4, 4. He pulls out cards without re-placement, until he pulls out an even numbered card, and then he stops. What is the probability that he pulls out exactly 2 cards? p10. Starting from the sequence (3 , 4, 5, 6, 7, 8, ... ), one applies the following operation repeat-edly. In each operation, we change the sequence (a1, a 2, a 3, ..., a a1−1, a a1 , a a1+1 , ... ) to the sequence (a2, a 3, ..., a a1 , a 1, a a1+1 , ... ). (In other words, for a sequence starting with x, we shift each of the next x − 1 term to the left by one, and put x immediately to the right of these numbers, and keep the rest of the terms unchanged. For example, after one operation, the sequence is (4 , 5, 3, 6, 7, 8, ... ), and after two operations, the sequence becomes (5 , 3, 6, 4, 7, 8, ... ). How many operations will it take to obtain a sequence of the form (7 , ... ) (that is, a sequence starting with 7)? p11. How many ways are there to place 4 balls into a 4 × 6 grid such that no column or row has more than one ball in it? (Rotations and reflections are considered distinct.) p12. Point P lies inside triangle ABC such that ∠P BC = 30 o and ∠P AC = 20 o. If ∠AP B is a right angle, find the measure of ∠BCA in degrees. p13. What is the largest prime factor of 93 − 43? ©2023 AoPS Incorporated 8AoPS Community 2012 EMCC p14. Joey writes down the numbers 1 through 10 and crosses one number out. He then adds the remaining numbers. What is the probability that the sum is less than or equal to 47 ? p15. In the coordinate plane, a lattice point is a point whose coordinates are integers. There is a pile of grass at every lattice point in the coordinate plane. A certain cow can only eat piles of grass that are at most 3 units away from the origin. How many piles of grass can she eat? p16. A book has 1000 pages numbered 1, 2, ... , 1000 . The pages are numbered so that pages 1 and 2 are back to back on a single sheet, pages 3 and 4 are back to back on the next sheet, and so on, with pages 999 and 1000 being back to back on the last sheet. How many pairs of pages that are back to back (on a single sheet) share no digits in the same position? (For example, pages 9 and 10 , and pages 89 and 90 .) p17. Find a pair of integers (a, b ) for which 10 a a! = 10 b b! and a < b . p18. Find all ordered pairs (x, y ) of real numbers satisfying ( −x2 + 3 y2 − 5x + 7 y + 4 = 0 2x2 − 2y2 − x + y + 21 = 0 p19. There are six blank fish drawn in a line on a piece of paper. Lucy wants to color them either red or blue, but will not color two adjacent fish red. In how many ways can Lucy color the fish? p20. There are sixteen 100 -gram balls and sixteen 99 -gram balls on a table (the balls are visibly indistinguishable). You are given a balance scale with two sides that reports which side is heav-ier or that the two sides have equal weights. A weighing is defined as reading the result of the balance scale: For example, if you place three balls on each side, look at the result, then add two more balls to each side, and look at the result again, then two weighings have been performed. You wish to pick out two different sets of balls (from the 32 balls) with equal numbers of balls in them but different total weights. What is the minimal number of weighings needed to ensure this? PS. You should use hide for answers. Collected here ( community/c5h2760506p24143309 ). ©2023 AoPS Incorporated 9
13830	https://www.cuemath.com/algebra/singleton-set/	LearnPracticeDownload Singleton Set Singleton set is a set containing only one element. The singleton set is of the form A = {a}, and it is also called a unit set. The singleton set has two subsets, which is the null set, and the set itself. Let us learn more about the properties of singleton set, with examples, FAQs. \| \| \| --- \| \| 1. \| What Is A Singleton Set? \| \| 2. \| Properties Of Singleton Set \| \| 3. \| Examples On Singleton Set \| \| 4. \| Practice Questions \| \| 5. \| FAQs On Singleton Set \| What Is A Singleton Set? A singleton set is a set containing only one element. The singleton set is of the form A = {a}, Where A represents the set, and the small alphabet 'a' represents the element of the singleton set. Since a singleton set has only one element in it, it is also called a unit set. The cardinal number of a singleton set is one. The number of subsets of a singleton set is two, which is the empty set and the set itself with the single element. The number of singleton sets that are subsets of a given set is equal to the number of elements in the given set. The set A = {a, e, i , o, u}, has 5 elements. Hence the set has five singleton sets, {a}, {e}, {i}, {o}, {u}, which are the subsets of the given set. Properties Of Singleton Set The following are some of the important properties of a singleton set. The singleton set has only one element in it. The cardinality of a singleton set is one. The singleton set has two subsets. Null set is a subset of every singleton set. The two subsets of a singleton set are the null set, and the singleton set itself. The powerset of a singleton set has a cardinal number of 2 Related Topics The following topics help in a better understanding of singleton set. Disjoint Sets Union of Sets Empty Set Superset Complement of a Set Examples on Singleton Set Example 1: Find the subsets of the set A = {1, 3, 5, 7, 11} which are singleton sets. Solution: The given set is A = {1, 3, 5, 7, 11}. The given set has 5 elements and it has 5 subsets which can have only one element and are singleton sets. Therefore the five singleton sets which are subsets of the given set A is {1}, {3}, {5}, {7}, {11}. 2. Example 2: Find the powerset of the singleton set {5}. Solution: The given singleton set is A = {5}. The two possible subsets of this singleton set are { }, {5}. The power set can be formed by taking these subsets as it elements. Powerset of A = {{ }, {5}}. Therefore the powerset of the singleton set A is {{ }, {5}}. View Answer > go to slidego to slide Breakdown tough concepts through simple visuals. Math will no longer be a tough subject, especially when you understand the concepts through visualizations. Book a Free Trial Class Practice Questions on Singleton Set Check Answer > go to slidego to slide FAQs on Singleton Set What Is Singleton Set? A set containing only one element is called a singleton set. The singleton set is of the form A = {a}. Since a singleton set has only one element in it, it is also called a unit set. The number of subsets of a singleton set is two, which is the empty set and the set itself with the single element. What Is The Cardinal Number Of A Singleton Set? The cardinal number of a singleton set is 1. The singleton set has only one element in it. What Is The Other Name Of Singleton Set? The singleton set has only one element, and hence a singleton set is also called a unit set. What Are The Subsets Of Singleton Set? The singleton set has two sets, which is the null set and the set itself. For a set A = {a}, the two subsets are { }, and {a}. 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13831	https://www.symbolab.com/solver/step-by-step/%5Csin(2%CE%B8)=2%5Csin(%CE%B8)%5Ccos(%CE%B8)	Symbolab for Chrome Snip & solve on any website Practice More \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| x^2 \| x^{\msquare} \| \log_{\msquare} \| \sqrt{\square} \| \nthroot[\msquare]{\square} \| \le \| \ge \| \frac{\msquare}{\msquare} \| \cdot \| \div \| x^{\circ} \| \pi \| \| \left(\square\right)^{'} \| \frac{d}{dx} \| \frac{\partial}{\partial x} \| \int \| \int_{\msquare}^{\msquare} \| \lim \| \sum \| \infty \| \theta \| (f\:\circ\:g) \| f(x) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| x^2 \| x^{\msquare} \| \log_{\msquare} \| \sqrt{\square} \| \nthroot[\msquare]{\square} \| \le \| \ge \| \frac{\msquare}{\msquare} \| \cdot \| \div \| x^{\circ} \| \pi \| \| \left(\square\right)^{'} \| \frac{d}{dx} \| \frac{\partial}{\partial x} \| \int \| \int_{\msquare}^{\msquare} \| \lim \| \sum \| \infty \| \theta \| (f\:\circ\:g) \| f(x) \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| ▭\:\longdivision{▭} \| \times \twostack{▭}{▭} \| + \twostack{▭}{▭} \| - \twostack{▭}{▭} \| \left( \| \right) \| \times \| \square\frac{\square}{\square} \| Special Offer Double the Tools. One Smart Bundle. qb-banner-title sin(2θ)=2sin(θ)cos(θ) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| x^2 \| x^{\msquare} \| \log_{\msquare} \| \sqrt{\square} \| \nthroot[\msquare]{\square} \| \le \| \ge \| \frac{\msquare}{\msquare} \| \cdot \| \div \| x^{\circ} \| \pi \| \| \left(\square\right)^{'} \| \frac{d}{dx} \| \frac{\partial}{\partial x} \| \int \| \int_{\msquare}^{\msquare} \| \lim \| \sum \| \infty \| \theta \| (f\:\circ\:g) \| f(x) \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- \| \| x^2 \| x^{\msquare} \| \log_{\msquare} \| \sqrt{\square} \| \nthroot[\msquare]{\square} \| \le \| \ge \| \frac{\msquare}{\msquare} \| \cdot \| \div \| x^{\circ} \| \pi \| \| \left(\square\right)^{'} \| \frac{d}{dx} \| \frac{\partial}{\partial x} \| \int \| \int_{\msquare}^{\msquare} \| \lim \| \sum \| \infty \| \theta \| (f\:\circ\:g) \| f(x) \| \| \| \| \| \| \| \| \| \| --- --- --- \| - \twostack{▭}{▭} \| \lt \| 7 \| 8 \| 9 \| \div \| AC \| \| + \twostack{▭}{▭} \| \gt \| 4 \| 5 \| 6 \| \times \| \square\frac{\square}{\square} \| \| \times \twostack{▭}{▭} \| \left( \| 1 \| 2 \| 3 x \| \| ▭\:\longdivision{▭} \| \right) \| . \| 0 \| = \| + \| y \| Number Line step-by-step \sin(2θ)=2\sin(θ)\cos(θ) en Please add a message. Message received. Thanks for the feedback.
13832	https://mytextbookswheniwantem.files.wordpress.com/2016/03/physics-chapter-22-heat-engines-entropy-and-the-second-law-of-thermodynamics1.pdf	625 chapter 22 Heat Engines, Entropy, and the Second Law of Thermodynamics 22.1 Heat Engines and the Second Law of Thermodynamics 22.2 Heat Pumps and Refrigerators 22.3 Reversible and Irreversible Processes 22.4 The Carnot Engine 22.5 Gasoline and Diesel Engines 22.6 Entropy 22.7 Entropy and the Second Law 22.8 Entropy on a Microscopic Scale The first law of thermodynamics, which we studied in Chapter 20, is a statement of conservation of energy and is a special-case reduction of Equation 8.2. This law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both. Although the first law of thermodynamics is very important, it makes no distinction between processes that occur spontaneously and those that do not. Only certain types of energy conversion and energy transfer processes actually take place in nature, however. The second law of thermodynamics, the major topic in this chapter, establishes which processes do and do not occur. The following are examples of A Stirling engine from the early nineteenth century. Air is heated in the lower cylinder using an external source. As this happens, the air expands and pushes against a piston, causing it to move. The air is then cooled, allowing the cycle to begin again. This is one example of a heat engine, which we study in this chapter. (© SSPL/The Image Works) 626 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics processes that do not violate the first law of thermodynamics if they proceed in either direction, but are observed in reality to proceed in only one direction: • When two objects at different temperatures are placed in thermal contact with each other, the net transfer of energy by heat is always from the warmer object to the cooler object, never from the cooler to the warmer. • A rubber ball dropped to the ground bounces several times and eventually comes to rest, but a ball lying on the ground never gathers internal energy from the ground and begins bouncing on its own. • An oscillating pendulum eventually comes to rest because of collisions with air mol-ecules and friction at the point of suspension. The mechanical energy of the system is converted to internal energy in the air, the pendulum, and the suspension; the reverse conversion of energy never occurs. All these processes are irreversible; that is, they are processes that occur naturally in one direction only. No irreversible process has ever been observed to run backward. If it were to do so, it would violate the second law of thermodynamics.1 22.1 Heat Engines and the Second Law of Thermodynamics A heat engine is a device that takes in energy by heat2 and, operating in a cyclic process, expels a fraction of that energy by means of work. For instance, in a typical process by which a power plant produces electricity, a fuel such as coal is burned and the high-temperature gases produced are used to convert liquid water to steam. This steam is directed at the blades of a turbine, setting it into rotation. The mechanical energy associated with this rotation is used to drive an electric genera-tor. Another device that can be modeled as a heat engine is the internal combustion engine in an automobile. This device uses energy from a burning fuel to perform work on pistons that results in the motion of the automobile. A heat engine carries some working substance through a cyclic process during which (1) the working substance absorbs energy by heat from a high-temperature energy reservoir, (2) work is done by the engine, and (3) energy is expelled by heat to a lower-temperature reservoir. As an example, consider the operation of a steam engine (Fig. 22.1), which uses water as the working substance. The water in a boiler absorbs energy from burning fuel and evaporates to steam, which then does work by expanding against a piston. After the steam cools and condenses, the liquid water produced returns to the boiler and the cycle repeats. It is useful to represent a heat engine schematically as in Active Figure 22.2. The engine absorbs a quantity of energy \|Q h\| from the hot reservoir. For the mathemati-cal discussion of heat engines, we use absolute values to make all energy transfers Lord Kelvin British physicist and mathematician (1824–1907) Born William Thomson in Belfast, Kelvin was the first to propose the use of an absolute scale of temperature. The Kelvin temperature scale is named in his honor. Kelvin’s work in thermodynamics led to the idea that energy cannot pass spontaneously from a colder object to a hotter object. J-L Charmet/Science Photo Library/Photo Researchers, Inc. 1Although a process occurring in the time-reversed sense has never been observed, it is possible for it to occur. As we shall see later in this chapter, however, the probability of such a process occurring is infinitesimally small. From this viewpoint, processes occur with a vastly greater probability in one direction than in the opposite direction. 2We use heat as our model for energy transfer into a heat engine. Other methods of energy transfer are possible in the model of a heat engine, however. For example, the Earth’s atmosphere can be modeled as a heat engine in which the input energy transfer is by means of electromagnetic radiation from the Sun. The output of the atmospheric heat engine causes the wind structure in the atmosphere. Figure 22.1 A steam-driven loco-motive obtains its energy by burning wood or coal. The generated energy vaporizes water into steam, which powers the locomotive. Modern locomotives use diesel fuel instead of wood or coal. Whether old-fashioned or modern, such locomotives can be modeled as heat engines, which extract energy from a burning fuel and convert a fraction of it to mechanical energy. © Andy Moore/Photolibrary/Jupiterimages 22.1 \| Heat Engines and the Second Law of Thermodynamics 627 by heat positive, and the direction of transfer is indicated with an explicit positive or negative sign. The engine does work Weng (so that negative work W 5 2Weng is done on the engine) and then gives up a quantity of energy \|Q c\| to the cold reservoir. Because the working substance goes through a cycle, its initial and final internal energies are equal: DEint 5 0. Hence, from the first law of thermodynamics, DEint 5 Q 1 W 5 Q 2 Weng 5 0, and the net work Weng done by a heat engine is equal to the net energy Q net transferred to it. As you can see from Active Figure 22.2, Q net 5 \|Q h\| 2 \|Q c\|; therefore, Weng 5 \|Q h\| 2 \|Q c\| (22.1) The thermal efficiency e of a heat engine is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature during the cycle: e ; Weng 0 Q h 0 5 0 Q h 0 2 0 Q c 0 0 Q h 0 5 1 2 0 Q c 0 0 Q h 0 (22.2) You can think of the efficiency as the ratio of what you gain (work) to what you give (energy transfer at the higher temperature). In practice, all heat engines expel only a fraction of the input energy Q h by mechanical work; consequently, their efficiency is always less than 100%. For example, a good automobile engine has an efficiency of about 20%, and diesel engines have efficiencies ranging from 35% to 40%. Equation 22.2 shows that a heat engine has 100% efficiency (e 5 1) only if \|Q c\| 5 0, that is, if no energy is expelled to the cold reservoir. In other words, a heat engine with perfect efficiency would have to expel all the input energy by work. Because efficiencies of real engines are well below 100%, the Kelvin–Planck form of the second law of thermodynamics states the following: It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the per-formance of an equal amount of work. This statement of the second law means that during the operation of a heat engine, Weng can never be equal to \|Q h\| or, alternatively, that some energy \|Q c\| must be rejected to the environment. Every heat engine must have some energy exhaust. Figure 22.3 is a schematic diagram of the impossible “perfect” heat engine. Quick Quiz 22.1 The energy input to an engine is 3.00 times greater than the work it performs. (i) What is its thermal efficiency? (a) 3.00 (b) 1.00 (c) 0.333 (d) impossible to determine (ii) What fraction of the energy input is expelled to the cold reservoir? (a) 0.333 (b) 0.667 (c) 1.00 (d) impossible to determine Thermal efficiency of a heat W engine Pitfall Prevention 22.1 The First and Second Laws Notice the distinction between the first and second laws of thermody-namics. If a gas undergoes a one-time isothermal process, then DEint 5 Q 1 W 5 0 and W 5 2Q. Therefore, the first law allows all energy input by heat to be expelled by work. In a heat engine, however, in which a substance undergoes a cyclic process, only a portion of the energy input by heat can be expelled by work accord-ing to the second law. Qh Q c Hot reservoir at Th Cold reservoir at Tc Heat engine t Weng Energy Qh enters the engine. Energy Q c leaves the engine. The engine does work Weng. Schematic representation of a heat engine. ACTIVE FIGURE 22.2 Q h Hot reservoir at Th Cold reservoir at Tc Heat engine t Weng An impossible heat engine Figure 22.3 Schematic diagram of a heat engine that takes in energy from a hot reservoir and does an equivalent amount of work. It is impossible to construct such a per-fect engine. 628 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics Example 22.1 The Efficiency of an Engine An engine transfers 2.00 3 103 J of energy from a hot reservoir during a cycle and transfers 1.50 3 103 J as exhaust to a cold reservoir. (A) Find the efficiency of the engine. SOLUTION Conceptualize Review Active Figure 22.2; think about energy going into the engine from the hot reservoir and splitting, with part coming out by work and part by heat into the cold reservoir. Categorize This example involves evaluation of quantities from the equations introduced in this section, so we catego-rize it as a substitution problem. Find the efficiency of the engine from Equation 22.2: e 5 1 2 0 Q c 0 0 Q h 0 5 1 2 1.50 3 103 J 2.00 3 103 J 5 0.250, or 25.0% Find the work done by the engine by taking the differ-ence between the input and output energies: Weng 5 \|Q h\| 2 \|Q c\| 5 2.00 3 103 J 2 1.50 3 103 J 5 5.0 3 102 J (B) How much work does this engine do in one cycle? SOLUTION WHAT IF? Suppose you were asked for the power output of this engine. Do you have sufficient information to answer this question? Answer No, you do not have enough information. The power of an engine is the rate at which work is done by the engine. You know how much work is done per cycle, but you have no information about the time interval associated with one cycle. If you were told that the engine operates at 2 000 rpm (revolutions per minute), however, you could relate this rate to the period of rotation T of the mechanism of the engine. Assuming there is one thermodynamic cycle per revolution, the power is P 5 Weng T 5 5.0 3 102 J 1 1 2 000 min2 a1 min 60 s b 5 1.7 3 104 W 22.2 Heat Pumps and Refrigerators In a heat engine, the direction of energy transfer is from the hot reservoir to the cold reservoir, which is the natural direction. The role of the heat engine is to pro-cess the energy from the hot reservoir so as to do useful work. What if we wanted to transfer energy from the cold reservoir to the hot reservoir? Because that is not the natural direction of energy transfer, we must put some energy into a device to be successful. Devices that perform this task are called heat pumps and refrigerators. For example, homes in summer are cooled using heat pumps called air conditioners. The air conditioner transfers energy from the cool room in the home to the warm air outside. In a refrigerator or a heat pump, the engine takes in energy \|Q c\| from a cold reservoir and expels energy \|Q h\| to a hot reservoir (Active Fig. 22.4), which can be accomplished only if work is done on the engine. From the first law, we know that the energy given up to the hot reservoir must equal the sum of the work done and the energy taken in from the cold reservoir. Therefore, the refrigerator or heat pump transfers energy from a colder body (for example, the contents of a kitchen refrigerator or the winter air outside a building) to a hotter body (the air in the kitchen or a room in the building). In practice, it is desirable to carry out this pro-cess with a minimum of work. If the process could be accomplished without doing any work, the refrigerator or heat pump would be “perfect” (Fig. 22.5). Again, the Q h Q c Hot reservoir at Th Cold reservoir at Tc Heat pump W Energy Q h is expelled to the hot reservoir. Energy Q c is drawn from the cold reservoir. Work W is done on the heat pump. Schematic representation of a heat pump. ACTIVE FIGURE 22.4 22.2 \| Heat Pumps and Refrigerators 629 existence of such a device would be in violation of the second law of thermodynam-ics, which in the form of the Clausius statement3 states: It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work. In simpler terms, energy does not transfer spontaneously by heat from a cold object to a hot object. Work input is required to run a refrigerator. The Clausius and Kelvin–Planck statements of the second law of thermodynam-ics appear at first sight to be unrelated, but in fact they are equivalent in all respects. Although we do not prove so here, if either statement is false, so is the other.4 In practice, a heat pump includes a circulating fluid that passes through two sets of metal coils that can exchange energy with the surroundings. The fluid is cold and at low pressure when it is in the coils located in a cool environment, where it absorbs energy by heat. The resulting warm fluid is then compressed and enters the other coils as a hot, high-pressure fluid. There it releases its stored energy to the warm surroundings. In an air conditioner, energy is absorbed into the fluid in coils located in a building’s interior; after the fluid is compressed, energy leaves the fluid through coils located outdoors. In a refrigerator, the external coils are behind or underneath the unit (Fig. 22.6). The internal coils are in the walls of the refrig-erator and absorb energy from the food. The effectiveness of a heat pump is described in terms of a number called the coefficient of performance (COP). The COP is similar to the thermal efficiency for a heat engine in that it is a ratio of what you gain (energy transferred to or from a reservoir) to what you give (work input). For a heat pump operating in the cool-ing mode, “what you gain” is energy removed from the cold reservoir. The most effective refrigerator or air conditioner is one that removes the greatest amount of energy from the cold reservoir in exchange for the least amount of work. There-fore, for these devices operating in the cooling mode, we define the COP in terms of \|Q c\|: COP 1cooling mode2 5 energy transferred at low temperature work done on heat pump 5 0 Q c 0 W (22.3) A good refrigerator should have a high COP, typically 5 or 6. In addition to cooling applications, heat pumps are becoming increasingly pop-ular for heating purposes. The energy-absorbing coils for a heat pump are located outside a building, in contact with the air or buried in the ground. The other set of coils are in the building’s interior. The circulating fluid flowing through the coils absorbs energy from the outside and releases it to the interior of the building from the interior coils. In the heating mode, the COP of a heat pump is defined as the ratio of the energy transferred to the hot reservoir to the work required to transfer that energy: COP 1heating mode2 5 energy transferred at high temperature work done on heat pump 5 0 Qh 0 W (22.4) If the outside temperature is 25°F (24°C) or higher, a typical value of the COP for a heat pump is about 4. That is, the amount of energy transferred to the building is about four times greater than the work done by the motor in the heat pump. As the outside temperature decreases, however, it becomes more difficult for the heat pump to extract sufficient energy from the air and so the COP decreases. There-fore, the use of heat pumps that extract energy from the air, although satisfactory in 3First expressed by Rudolf Clausius (1822–1888). 4See an advanced textbook on thermodynamics for this proof. Q h Q c Q c Hot reservoir at Th Cold reservoir at Tc Heat pump An impossible heat pump Figure 22.5 Schematic diagram of an impossible heat pump or refriger-ator, that is, one that takes in energy from a cold reservoir and expels an equivalent amount of energy to a hot reservoir without the input of energy by work. The coils on the back of a refrigerator transfer energy by heat to the air. Figure 22.6 The back of a house-hold refrigerator. The air surround-ing the coils is the hot reservoir. © Cengage Learning/Charles D. Winters 630 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics moderate climates, is not appropriate in areas where winter temperatures are very low. It is possible to use heat pumps in colder areas by burying the external coils deep in the ground. In that case, the energy is extracted from the ground, which tends to be warmer than the air in the winter. Quick Quiz 22.2 The energy entering an electric heater by electrical trans-mission can be converted to internal energy with an efficiency of 100%. By what factor does the cost of heating your home change when you replace your electric heating system with an electric heat pump that has a COP of 4.00? Assume the motor running the heat pump is 100% efficient. (a) 4.00 (b) 2.00 (c) 0.500 (d) 0.250 Example 22.2 Freezing Water A certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500 W. A sample of water of mass 500 g and temperature 20.0°C is placed in the freezer compartment. How long does it take to freeze the water to ice at 0°C? Assume all other parts of the refrigerator stay at the same temperature and there is no leakage of energy from the exterior, so the operation of the refrigerator results only in energy being extracted from the water. SOLUTION Conceptualize Energy leaves the water, reducing its temperature and then freezing it into ice. The time interval required for this entire process is related to the rate at which energy is withdrawn from the water, which, in turn, is related to the power input of the refrigerator. Categorize We categorize this example as one that combines our understanding of temperature changes and phase changes from Chapter 20 and our understanding of heat pumps from this chapter. Analyze Use the power rating of the refrigerator to find the time interval Dt required for the freez-ing process to occur: P 5 W Dt S Dt 5 W P Use Equation 22.3 to relate the work W done on the heat pump to the energy \|Q c\| extracted from the water: Dt 5 \|Q c \| P 1COP2 Use Equations 20.4 and 20.7 to substitute the amount of energy \|Q c\| that must be extracted from the water of mass m: Dt 5 0 mc DT 1 L f Dm 0 P 1COP2 Recognize that the amount of water that freezes is Dm 5 2m because all the water freezes: Dt 5 0 m1c DT 2 L f2 0 P 1COP2 Subsitute numerical values: Dt 5 0 10.500 kg2 3 14 186 J/kg ? °C2 1220.0°C2 2 3.33 3 105 J/kg4 0 1500 W2 15.002 5 83.3 s Finalize In reality, the time interval for the water to freeze in a refrigerator is much longer than 83.3 s, which suggests that the assumptions of our model are not valid. Only a small part of the energy extracted from the refrigerator interior in a given time interval comes from the water. Energy must also be extracted from the container in which the water is placed, and energy that continuously leaks into the interior from the exterior must be extracted. 22.3 \| Reversible and Irreversible Processes 631 22.3 Reversible and Irreversible Processes In the next section, we will discuss a theoretical heat engine that is the most effi-cient possible. To understand its nature, we must first examine the meaning of reversible and irreversible processes. In a reversible process, the system undergo-ing the process can be returned to its initial conditions along the same path on a PV diagram, and every point along this path is an equilibrium state. A process that does not satisfy these requirements is irreversible. All natural processes are known to be irreversible. Let’s examine the adiabatic free expansion of a gas, which was already discussed in Section 20.6, and show that it cannot be reversible. Consider a gas in a thermally insulated container as shown in Figure 22.7. A membrane separates the gas from a vacuum. When the membrane is punctured, the gas expands freely into the vacuum. As a result of the puncture, the system has changed because it occupies a greater volume after the expansion. Because the gas does not exert a force through a displacement, it does no work on the surroundings as it expands. In addition, no energy is transferred to or from the gas by heat because the container is insulated from its surroundings. Therefore, in this adiabatic process, the system has changed but the surroundings have not. For this process to be reversible, we must return the gas to its original volume and temperature without changing the surroundings. Imagine trying to reverse the process by compressing the gas to its original volume. To do so, we fit the container with a piston and use an engine to force the piston inward. During this process, the surroundings change because work is being done by an outside agent on the system. In addition, the system changes because the compression increases the tempera-ture of the gas. The temperature of the gas can be lowered by allowing it to come into contact with an external energy reservoir. Although this step returns the gas to its original conditions, the surroundings are again affected because energy is being added to the surroundings from the gas. If this energy could be used to drive the engine that compressed the gas, the net energy transfer to the surroundings would be zero. In this way, the system and its surroundings could be returned to their ini-tial conditions and we could identify the process as reversible. The Kelvin–Planck statement of the second law, however, specifies that the energy removed from the gas to return the temperature to its original value cannot be completely converted to mechanical energy in the form of the work done by the engine in compressing the gas. Therefore, we must conclude that the process is irreversible. We could also argue that the adiabatic free expansion is irreversible by relying on the portion of the definition of a reversible process that refers to equilibrium states. For example, during the sudden expansion, significant variations in pres-sure occur throughout the gas. Therefore, there is no well-defined value of the pressure for the entire system at any time between the initial and final states. In fact, the process cannot even be represented as a path on a PV diagram. The PV diagram for an adiabatic free expansion would show the initial and final conditions as points, but these points would not be connected by a path. Therefore, because the intermediate conditions between the initial and final states are not equilibrium states, the process is irreversible. Although all real processes are irreversible, some are almost reversible. If a real process occurs very slowly such that the system is always very nearly in an equilib-rium state, the process can be approximated as being reversible. Suppose a gas is compressed isothermally in a piston–cylinder arrangement in which the gas is in thermal contact with an energy reservoir and we continuously transfer just enough energy from the gas to the reservoir to keep the temperature constant. For exam-ple, imagine that the gas is compressed very slowly by dropping grains of sand onto a frictionless piston as shown in Figure 22.8. As each grain lands on the piston and compresses the gas a small amount, the system deviates from an equilibrium state, but it is so close to one that it achieves a new equilibrium state in a relatively short time interval. Each grain added represents a change to a new equilibrium state, but Pitfall Prevention 22.2 All Real Processes Are Irreversible The reversible process is an idealiza-tion; all real processes on the Earth are irreversible. Vacuum Gas at Ti Insulating wall Membrane Figure 22.7 Adiabatic free expan-sion of a gas. Energy reservoir The gas is compressed slowly as individual grains of sand drop onto the piston. Figure 22.8 A method for com-pressing a gas in a reversible isother-mal process. 632 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics the differences between states are so small that the entire process can be approx-imated as occurring through continuous equilibrium states. The process can be reversed by slowly removing grains from the piston. A general characteristic of a reversible process is that no dissipative effects (such as turbulence or friction) that convert mechanical energy to internal energy can be present. Such effects can be impossible to eliminate completely. Hence, it is not surprising that real processes in nature are irreversible. 22.4 The Carnot Engine In 1824, a French engineer named Sadi Carnot described a theoretical engine, now called a Carnot engine, that is of great importance from both practical and theo-retical viewpoints. He showed that a heat engine operating in an ideal, reversible cycle—called a Carnot cycle—between two energy reservoirs is the most efficient engine possible. Such an ideal engine establishes an upper limit on the efficien-cies of all other engines. That is, the net work done by a working substance taken through the Carnot cycle is the greatest amount of work possible for a given amount of energy supplied to the substance at the higher temperature. Carnot’s theorem can be stated as follows: No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. To prove the validity of this theorem, imagine two heat engines operating between the same energy reservoirs. One is a Carnot engine with efficiency eC, and the other is an engine with efficiency e, where we assume e . eC. Because the cycle in the Carnot engine is reversible, the engine can operate in reverse as a refrigera-tor. The more efficient engine is used to drive the Carnot engine as a Carnot refrig-erator. The output by work of the more efficient engine is matched to the input by work of the Carnot refrigerator. For the combination of the engine and refrigerator, no exchange by work with the surroundings occurs. Because we have assumed the engine is more efficient than the refrigerator, the net result of the combination is a transfer of energy from the cold to the hot reservoir without work being done on the combination. According to the Clausius statement of the second law, this pro-cess is impossible. Hence, the assumption that e . eC must be false. All real engines are less efficient than the Carnot engine because they do not operate through a reversible cycle. The efficiency of a real engine is further reduced by such practical difficulties as friction and energy losses by conduction. To describe the Carnot cycle taking place between temperatures Tc and Th, let’s assume the working substance is an ideal gas contained in a cylinder fitted with a movable piston at one end. The cylinder’s walls and the piston are thermally non-conducting. Four stages of the Carnot cycle are shown in Active Figure 22.9, and the PV diagram for the cycle is shown in Active Figure 22.10. The Carnot cycle con-sists of two adiabatic processes and two isothermal processes, all reversible: 1. Process A S B (Active Fig. 22.9a) is an isothermal expansion at temperature Th. The gas is placed in thermal contact with an energy reservoir at tem-perature Th. During the expansion, the gas absorbs energy \|Q h\| from the reservoir through the base of the cylinder and does work WAB in raising the piston. 2. In process B S C (Active Fig. 22.9b), the base of the cylinder is replaced by a thermally nonconducting wall and the gas expands adiabatically; that is, no energy enters or leaves the system by heat. During the expansion, the temperature of the gas decreases from Th to Tc and the gas does work WBC in raising the piston. Sadi Carnot French engineer (1796–1832) Carnot was the first to show the quantitative relationship between work and heat. In 1824, he published his only work, Reflections on the Motive Power of Heat, which reviewed the industrial, political, and economic importance of the steam engine. In it, he defined work as “weight lifted through a height.” J.-L. Charmet/Science Photo Library/ Photo Researchers, Inc. Pitfall Prevention 22.3 Don’t Shop for a Carnot Engine The Carnot engine is an idealiza-tion; do not expect a Carnot engine to be developed for commercial use. We explore the Carnot engine only for theoretical considerations. 22.4 \| The Carnot Engine 633 3. In process C S D (Active Fig. 22.9c), the gas is placed in thermal contact with an energy reservoir at temperature Tc and is compressed isothermally at temperature Tc. During this time, the gas expels energy \|Q c\| to the reser-voir and the work done by the piston on the gas is WCD. 4. In the final process D S A (Active Fig. 22.9d), the base of the cylinder is replaced by a nonconducting wall and the gas is compressed adiabatically. The temperature of the gas increases to Th, and the work done by the pis-ton on the gas is WDA. The thermal efficiency of the engine is given by Equation 22.2: e 5 1 2 0 Q c 0 0 Q h 0 In Example 22.3, we show that for a Carnot cycle, 0 Q c 0 0 Q h 0 5 Tc Th (22.5) Hence, the thermal efficiency of a Carnot engine is e C 5 1 2 Tc Th (22.6) Efficiency of a Carnot engine W a c b d Cycle Q 0 Q 0 Energy reservoir at Th Q h Energy reservoir at Tc Q c A S B The gas undergoes an isothermal expansion. C S D The gas undergoes an isothermal compression. B S C The gas undergoes an adiabatic expansion. D S A The gas undergoes an adiabatic compression. Thermal insulation Thermal insulation The Carnot cycle. The letters A, B, C, and D refer to the states of the gas shown in Active Figure 22.10. The arrows on the piston indicate the direction of its motion during each process. ACTIVE FIGURE 22.9 PV diagram for the Carnot cycle. The net work done Weng equals the net energy transferred into the Car-not engine in one cycle, \|Q h\| 2 \|Q c\|. ACTIVE FIGURE 22.10 V P Weng D B Qh Th Tc Q c C A The work done during the cycle equals the area enclosed by the path on the PV diagram. 634 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics This result indicates that all Carnot engines operating between the same two tem-peratures have the same efficiency.5 Equation 22.6 can be applied to any working substance operating in a Carnot cycle between two energy reservoirs. According to this equation, the efficiency is zero if Tc 5 Th, as one would expect. The efficiency increases as Tc is lowered and Th is raised. The efficiency can be unity (100%), however, only if Tc 5 0 K. Such reservoirs are not available; therefore, the maximum efficiency is always less than 100%. In most practical cases, Tc is near room temperature, which is about 300 K. Therefore, one usually strives to increase the efficiency by raising Th. Theoretically, a Carnot-cycle heat engine run in reverse constitutes the most effective heat pump possible, and it determines the maximum COP for a given com-bination of hot and cold reservoir temperatures. Using Equations 22.1 and 22.4, we see that the maximum COP for a heat pump in its heating mode is COPC 1heating mode2 5 0 Q h 0 W 5 0 Q h 0 0 Q h 0 2 0 Q c 0 5 1 1 2 0 Q c 0 0 Q h 0 5 1 1 2 Tc Th 5 Th Th 2 Tc The Carnot COP for a heat pump in the cooling mode is COPC 1cooling mode2 5 Tc Th 2 Tc As the difference between the temperatures of the two reservoirs approaches zero in this expression, the theoretical COP approaches infinity. In practice, the low temperature of the cooling coils and the high temperature at the compressor limit the COP to values below 10. Quick Quiz 22.3 Three engines operate between reservoirs separated in temperature by 300 K. The reservoir temperatures are as follows: Engine A: Th 5 1 000 K, Tc 5 700 K; Engine B: Th 5 800 K, Tc 5 500 K; Engine C: Th 5 600 K, Tc 5 300 K. Rank the engines in order of theoretically possible effi-ciency from highest to lowest. 5For the processes in the Carnot cycle to be reversible, they must be carried out infinitesimally slowly. Therefore, although the Carnot engine is the most efficient engine possible, it has zero power output because it takes an infinite time interval to complete one cycle! For a real engine, the short time interval for each cycle results in the working substance reaching a high temperature lower than that of the hot reservoir and a low temperature higher than that of the cold reservoir. An engine undergoing a Carnot cycle between this narrower temperature range was analyzed by F. L. Curzon and B. Ahlborn (“Efficiency of a Carnot engine at maximum power output,” Am. J. Phys. 43(1), 22, 1975), who found that the efficiency at maximum power output depends only on the reservoir temperatures Tc and Th and is given by eC-A 5 1 2 (Tc/Th)1/2. The Curzon–Ahlborn efficiency eC-A provides a closer approximation to the efficiencies of real engines than does the Carnot efficiency. Example 22.3 Efficiency of the Carnot Engine Show that the ratio of energy transfers by heat in a Carnot engine is equal to the ratio of reservoir temperatures, as given by Equation 22.5. SOLUTION Conceptualize Make use of Active Figures 22.9 and 22.10 to help you visualize the processes in the Carnot cycle. Categorize Because of our understanding of the Carnot cycle, we can categorize the processes in the cycle as isothermal and adiabatic. 22.3 cont. 22.4 \| The Carnot Engine 635 Finalize This last equation is Equation 22.5, the one we set out to prove. Analyze For the isothermal expansion (process A S B in Active Fig. 22.9), find the energy transfer by heat from the hot reservoir using Equation 20.14 and the first law of thermodynamics: 0 Q h 0 5 0 DEint 2 WAB 0 5 0 0 2 WAB 0 5 nRTh ln VB VA In a similar manner, find the energy transfer to the cold reservoir during the isothermal compression C S D: 0 Q c 0 5 0 DEint 2 WCD 0 5 0 0 2 WCD 0 5 nRTc ln VC VD Divide the second expression by the first: (1) 0 Q c 0 0 Q h 0 5 Tc Th ln 1VC/VD2 ln 1VB/VA2 Apply Equation 21.20 to the adiabatic processes B S C and D S A: ThVB g21 5 TcVC g21 ThVA g21 5 TcVD g21 Divide the first equation by the second: aVB VA b g21 5 aVC VD b g21 (2) VB VA 5 VC VD Substitute Equation (2) into Equation (1): 0 Q c 0 0 Q h 0 5 Tc Th ln 1VC/VD2 ln 1VB/VA2 5 Tc Th ln 1VC/VD2 ln 1VC/VD2 5 Tc Th Example 22.4 The Steam Engine A steam engine has a boiler that operates at 500 K. The energy from the burning fuel changes water to steam, and this steam then drives a piston. The cold reservoir’s temperature is that of the outside air, approximately 300 K. What is the maximum thermal efficiency of this steam engine? SOLUTION Conceptualize In a steam engine, the gas pushing on the piston in Active Figure 22.9 is steam. A real steam engine does not operate in a Carnot cycle, but, to find the maximum possible efficiency, imagine a Carnot steam engine. Categorize We calculate an efficiency using Equation 22.6, so we categorize this example as a substitution problem. Substitute the reservoir temperatures into Equation 22.6: eC 5 1 2 Tc Th 5 1 2 300 K 500 K 5 0.400 or 40.0% continued This result is the highest theoretical efficiency of the engine. In practice, the efficiency is considerably lower. WHAT IF? Suppose we wished to increase the theoretical efficiency of this engine. This increase can be achieved by raising Th by DT or by decreasing Tc by the same DT. Which would be more effective? Answer A given DT would have a larger fractional effect on a smaller temperature, so you would expect a larger change in efficiency if you alter Tc by DT. Let’s test that numerically. Raising Th by 50 K, corresponding to Th 5 550 K, would give a maximum efficiency of e C 5 1 2 Tc Th 5 1 2 300 K 550 K 5 0.455 636 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics 22.4 cont. Decreasing Tc by 50 K, corresponding to Tc 5 250 K, would give a maximum efficiency of e C 5 1 2 Tc Th 5 1 2 250 K 500 K 5 0.500 Although changing Tc is mathematically more effective, often changing Th is practically more feasible. 22.5 Gasoline and Diesel Engines In a gasoline engine, six processes occur in each cycle; they are illustrated in Active Figure 22.11. In this discussion, let’s consider the interior of the cylinder above the piston to be the system that is taken through repeated cycles in the engine’s opera-tion. For a given cycle, the piston moves up and down twice, which represents a four-stroke cycle consisting of two upstrokes and two downstrokes. The processes in the cycle can be approximated by the Otto cycle shown in the PV diagram in Active Figure 22.12. In the following discussion, refer to Active Figure 22.11 for the pictorial representation of the strokes and Active Figure 22.12 for the significance on the PV diagram of the letter designations below: 1. During the intake stroke (Active Fig. 22.11a and O S A in Active Figure 22.12), the piston moves downward and a gaseous mixture of air and fuel is drawn into the cylinder at atmospheric pressure. That is the energy input part of the cycle: energy enters the system (the interior of the cylinder) by matter transfer as potential energy stored in the fuel. In this process, the volume increases from V2 to V1. This apparent backward numbering is The intake valve opens, and the air– fuel mixture enters as the piston moves down. The piston moves up and compresses the mixture. The piston moves up and pushes the remaining gas out. The spark plug fires and ignites the mixture. The hot gas pushes the piston downward. Intake Compression Spark Power Release Exhaust Air and fuel Exhaust Spark plug Piston a b c d e f The exhaust valve opens, and the residual gas escapes. The four-stroke cycle of a conventional gasoline engine. The arrows on the piston indicate the direc-tion of its motion during each process. ACTIVE FIGURE 22.11 22.5 \| Gasoline and Diesel Engines 637 based on the compression stroke (process 2 below), in which the air–fuel mixture is compressed from V1 to V2. 2. During the compression stroke (Active Fig. 22.11b and A S B in Active Fig. 22.12), the piston moves upward, the air–fuel mixture is compressed adiabatically from volume V1 to volume V2, and the temperature increases from TA to TB. The work done on the gas is positive, and its value is equal to the negative of the area under the curve AB in Active Figure 22.12. 3. Combustion occurs when the spark plug fires (Active Fig. 22.11c and B S C in Active Fig. 22.12). That is not one of the strokes of the cycle because it occurs in a very short time interval while the piston is at its highest position. The combustion represents a rapid energy transformation from potential energy stored in chemical bonds in the fuel to internal energy associated with molecular motion, which is related to temperature. During this time interval, the mixture’s pressure and temperature increase rapidly, with the temperature rising from TB to TC. The volume, however, remains approxi-mately constant because of the short time interval. As a result, approxi-mately no work is done on or by the gas. We can model this process in the PV diagram (Active Fig. 22.12) as that process in which the energy \|Q h\| enters the system. (In reality, however, this process is a conversion of energy already in the cylinder from process O S A.) 4. In the power stroke (Active Fig. 22.11d and C S D in Active Fig. 22.12), the gas expands adiabatically from V2 to V1. This expansion causes the temperature to drop from TC to TD. Work is done by the gas in pushing the piston down-ward, and the value of this work is equal to the area under the curve CD. 5. Release of the residual gases occurs when an exhaust valve is opened (Active Fig. 22.11e and D S A in Active Fig. 22.12). The pressure suddenly drops for a short time interval. During this time interval, the piston is almost stationary and the volume is approximately constant. Energy is expelled from the interior of the cylinder and continues to be expelled dur-ing the next process. 6. In the final process, the exhaust stroke (Active Fig. 22.11e and A S O in Active Fig. 22.12), the piston moves upward while the exhaust valve remains open. Residual gases are exhausted at atmospheric pressure, and the vol-ume decreases from V1 to V2. The cycle then repeats. If the air–fuel mixture is assumed to be an ideal gas, the efficiency of the Otto cycle is e 5 1 2 1 1V1/V22 g21 1Otto cycle2 (22.7) where V1/V2 is the compression ratio and g is the ratio of the molar specific heats CP/CV for the air–fuel mixture. Equation 22.7, which is derived in Example 22.5, shows that the efficiency increases as the compression ratio increases. For a typi-cal compression ratio of 8 and with g 5 1.4, Equation 22.7 predicts a theoretical efficiency of 56% for an engine operating in the idealized Otto cycle. This value is much greater than that achieved in real engines (15% to 20%) because of such effects as friction, energy transfer by conduction through the cylinder walls, and incomplete combustion of the air–fuel mixture. Diesel engines operate on a cycle similar to the Otto cycle, but they do not employ a spark plug. The compression ratio for a diesel engine is much greater than that for a gasoline engine. Air in the cylinder is compressed to a very small volume, and, as a consequence, the cylinder temperature at the end of the compression stroke is very high. At this point, fuel is injected into the cylinder. The temperature is high enough for the air–fuel mixture to ignite without the assistance of a spark plug. Diesel engines are more efficient than gasoline engines because of their greater compression ratios and resulting higher combustion temperatures. PV diagram for the Otto cycle, which approximately represents the pro-cesses occurring in an internal com-bustion engine. ACTIVE FIGURE 22.12 V P C Qh B D TC Qc Adiabatic processes V2 V1 O A TA 638 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics Example 22.5 Efficiency of the Otto Cycle Show that the thermal efficiency of an engine operating in an idealized Otto cycle (see Active Figs. 22.11 and 22.12) is given by Equation 22.7. Treat the working substance as an ideal gas. SOLUTION Conceptualize Study Active Figures 22.11 and 22.12 to make sure you understand the working of the Otto cycle. Categorize As seen in Active Figure 22.12, we categorize the processes in the Otto cycle as isovolumetric and adiabatic. Analyze Model the energy input and output as occur-ring by heat in processes B S C and D S A. (In reality, most of the energy enters and leaves by matter transfer as the air–fuel mixture enters and leaves the cylinder.) Use Equation 21.8 to find the energy transfers by heat for these processes, which take place at constant volume: B S C \|Q h\| 5 nCV (TC 2 TB) D S A \|Q c\| 5 nCV (TD 2 TA) Substitute these expressions into Equation 22.2: (1) e 5 1 2 0 Q c 0 0 Q h 0 5 1 2 TD 2 TA TC 2 TB Apply Equation 21.20 to the adiabatic processes A S B and C S D: A S B TAVA g21 5 TBVB g21 C S D TCVC g21 5 TDVD g21 Solve these equations for the temperatures TA and TD, noting that VA 5 VD 5 V1 and VB 5 VC 5 V2: (2) TA 5 TB aVB VA b g21 5 TB a V2 V1 b g21 (3) TD 5 TC aVC VD b g21 5 TC a V2 V1 b g21 Subtract Equation (2) from Equation (3) and rearrange: (4) TD 2 TA TC 2 TB 5 a V2 V1 b g21 Substitute Equation (4) into Equation (1): e 5 1 2 1 1V1/V22 g21 Finalize This final expression is Equation 22.7. 22.6 Entropy The zeroth law of thermodynamics involves the concept of temperature, and the first law involves the concept of internal energy. Temperature and internal energy are both state variables; that is, the value of each depends only on the thermody-namic state of a system, not on the process that brought it to that state. Another state variable—this one related to the second law of thermodynamics—is entropy S. In this section, we define entropy on a macroscopic scale as it was first expressed by Clausius in 1865. Entropy was originally formulated as a useful concept in thermodynamics. Its importance grew, however, as the field of statistical mechanics developed because the analytical techniques of statistical mechanics provide an alternative means of interpreting entropy and a more global significance to the concept. In statisti-cal mechanics, the behavior of a substance is described in terms of the statistical behavior of its atoms and molecules. An important finding in these studies is that Pitfall Prevention 22.4 Entropy Is Abstract Entropy is one of the most abstract notions in physics, so follow the discussion in this and the subse-quent sections very carefully. Do not confuse energy with entropy. Even though the names sound similar, they are very different concepts. 22.6 \| Entropy 639 isolated systems tend toward disorder, and entropy is a measure of this disorder. For example, consider the molecules of a gas in the air in your room. If half the gas molecules had velocity vectors of equal magnitude directed toward the left and the other half had velocity vectors of the same magnitude directed toward the right, the situation would be very ordered. Such a situation is extremely unlikely, however. If you could view the molecules, you would see that they move haphazardly in all directions, bumping into one another, changing speed upon collision, some going fast and others going slowly. This situation is highly disordered. The cause of the tendency of an isolated system toward disorder is easily explained. To do so, let’s distinguish between microstates and macrostates of a system. A microstate is a particular configuration of the individual constituents of the sys-tem. For example, the description of the ordered velocity vectors of the air mole-cules in your room refers to a particular microstate, and the more likely haphazard motion is another microstate. A macrostate is a description of the system’s condi-tions from a macroscopic point of view. For a thermodynamic system, macrostates are described by macroscopic variables such as pressure, density, and temperature. For any given macrostate of the system, a number of microstates are possible. Let’s first consider some nonthermodynamic systems for simplicity. For example, the macrostate of a 4 on a pair of dice can be formed from the possible microstates 1–3, 2–2, and 3–1. The macrostate of 2 has only one microstate, 1–1. It is assumed all microstates are equally probable. When all possible macrostates are examined, however, it is found that macrostates associated with disorder have far more possi-ble microstates than those associated with order. Therefore, 4 is a more disordered macrostate for two dice than 2 because there are three microstates for a 4 and only one for a 2. There is only one microstate associated with the macrostate of a royal flush in a poker hand of five spades, laid out in order from ten to ace (Fig. 22.13a). Figure 22.13b shows another poker hand. The macrostate here is “worthless hand.” The particular hand (the microstate) in Figure 22.13b is as equally probable as the hand in Figure 22.13a. There are, however, many other hands similar to that in Figure 22.13b; that is, there are many microstates that also qualify as worthless hands. The more microstates that belong to a particular macrostate, the higher the probability that macrostate will occur. The macrostate of a royal flush in spades is ordered, of low probability, and of high value in poker. The macrostate of a worthless hand is disordered, of high probability, and of low poker value. Quick Quiz 22.4 (a) Suppose you select four cards at random from a standard deck of playing cards and end up with a macrostate of four deuces. How many microstates are associated with this macrostate? (b) Suppose you pick up two cards and end up with a macrostate of two aces. How many micro-states are associated with this macrostate? We can also imagine ordered macrostates and disordered macrostates in physi-cal processes, not just in games of dice and poker. The result of a dice throw or a poker hand stays fixed once the dice are thrown or the cards are dealt. Physical systems, on the other hand, are in a constant state of flux, changing from moment to moment from one microstate to another. Based on the relationship between the probability of a macrostate and the number of associated microstates, we therefore see that the probability of a system moving in time from an ordered macrostate to a disordered macrostate is far greater than the probability of the reverse because there are more microstates in a disordered macrostate. The original formulation of entropy in thermodynamics involves the transfer of energy by heat during a reversible process. Consider any infinitesimal process in which a system changes from one equilibrium state to another. If dQ r is the amount of energy transferred by heat when the system follows a reversible path between the Figure 22.13 (a) A royal flush has low probability of occurring. (b) A worthless poker hand, one of many. a © Cengage Learning/George Semple b © Cengage Learning/George Semple 640 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics states, the change in entropy dS is equal to this amount of energy for the reversible process divided by the absolute temperature of the system: dS 5 dQ r T (22.8) We have assumed the temperature is constant because the process is infinitesimal. Because entropy is a state variable, the change in entropy during a process depends only on the endpoints and therefore is independent of the actual path followed. Consequently, the entropy change for an irreversible process can be determined by calculating the entropy change for a reversible process that connects the same initial and final states. The subscript r on the quantity dQ r is a reminder that the transferred energy is to be measured along a reversible path even though the system may actually have followed some irreversible path. When energy is absorbed by the system, dQ r is posi-tive and the entropy of the system increases. When energy is expelled by the system, dQ r is negative and the entropy of the system decreases. Notice that Equation 22.8 does not define entropy but rather the change in entropy. Hence, the meaningful quantity in describing a process is the change in entropy. To calculate the change in entropy for a finite process, first recognize that T is generally not constant during the process. Therefore, we must integrate Equa-tion 22.8: DS 5 3 f i dS 5 3 f i dQ r T (22.9) As with an infinitesimal process, the change in entropy DS of a system going from one state to another has the same value for all paths connecting the two states. That is, the finite change in entropy DS of a system depends only on the properties of the initial and final equilibrium states. Therefore, we are free to choose a par-ticular reversible path over which to evaluate the entropy in place of the actual path as long as the initial and final states are the same for both paths. This point is explored further in Section 22.7. Quick Quiz 22.5 An ideal gas is taken from an initial temperature Ti to a higher final temperature Tf along two different reversible paths. Path A is at constant pressure, and path B is at constant volume. What is the rela-tion between the entropy changes of the gas for these paths? (a) DSA . DSB (b) DSA 5 DSB (c) DSA , DSB Change in entropy for an X infinitesimal process Change in entropy X for a finite process Example 22.6 Change in Entropy: Melting A solid that has a latent heat of fusion Lf melts at a temperature Tm. Calculate the change in entropy of this substance when a mass m of the substance melts. SOLUTION Conceptualize Imagine placing the substance in a warm environment so that energy enters the substance by heat. The process can be reversed by placing the substance in a cool environment so that energy leaves the substance by heat. The mass m of the substance that melts is equal to Dm, the change in mass of the higher-phase (liquid) substance. Categorize Because the melting takes place at a fixed temperature, we categorize the process as isothermal. Analyze Use Equation 20.7 in Equation 22.9, noting that the temperature remains fixed: DS 5 3 dQ r T 5 1 Tm 3 dQ r 5 Q r Tm 5 L f Dm Tm 5 L fm Tm Finalize Notice that Dm is positive so that DS is positive, representing that energy is added to the ice cube. 22.7 \| Entropy and the Second Law 641 22.6 cont. WHAT IF? Suppose you did not have Equation 22.9 available to calculate an entropy change. How could you argue from the statistical description of entropy that the changes in entropy should be positive? Answer When a solid melts, its entropy increases because the molecules are much more disordered in the liquid state than they are in the solid state. The positive value for DS also means that the substance in its liquid state does not spon-taneously transfer energy from itself to the warm surroundings and freeze because to do so would involve a spontaneous increase in order and a decrease in entropy. Let’s consider the changes in entropy that occur in a Carnot heat engine that operates between the temperatures Tc and Th. In one cycle, the engine takes in energy \|Q h\| from the hot reservoir and expels energy \|Q c\| to the cold reservoir. These energy transfers occur only during the isothermal portions of the Carnot cycle; therefore, the constant temperature can be brought out in front of the integral sign in Equation 22.9. The integral then simply has the value of the total amount of energy transferred by heat. Therefore, the total change in entropy for one cycle is DS 5 0 Q h 0 Th 2 0 Q c 0 Tc where the minus sign represents that energy is leaving the engine. In Example 22.3, we showed that for a Carnot engine, 0 Q c 0 0 Q h 0 5 Tc Th Using this result in the previous expression for DS, we find that the total change in entropy for a Carnot engine operating in a cycle is zero: DS 5 0 Now consider a system taken through an arbitrary (non-Carnot) reversible cycle. Because entropy is a state variable—and hence depends only on the properties of a given equilibrium state—we conclude that DS 5 0 for any reversible cycle. In gen-eral, we can write this condition as C dQ r T 5 0 1reversible cycle2 (22.10) where the symbol r indicates that the integration is over a closed path. 22.7 Entropy and the Second Law By definition, a calculation of the change in entropy for a system requires infor-mation about a reversible path connecting the initial and final equilibrium states. To calculate changes in entropy for real (irreversible) processes, remember that entropy (like internal energy) depends only on the state of the system. That is, entropy is a state variable, and the change in entropy depends only on the initial and final states. You can calculate the entropy change in some irreversible process between two equilibrium states by devising a reversible process (or series of reversible processes) between the same two states and computing DS 5 edQ r/T for the reversible pro-cess. In irreversible processes, it is important to distinguish between Q, the actual energy transfer in the process, and Q r, the energy that would have been transferred by heat along a reversible path. Only Q r is the correct value to be used in calculat-ing the entropy change. 642 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics If we consider a system and its surroundings to include the entire Universe, the Universe is always moving toward a higher-probability macrostate, corresponding to greater disorder. Because entropy is a measure of disorder, an alternative way of stating this behavior is as follows: The entropy of the Universe increases in all real processes. This statement is yet another wording of the second law of thermodynamics that can be shown to be equivalent to the Kelvin-Planck and Clausius statements. When dealing with a system that is not isolated from its surroundings, remember that the increase in entropy described in the second law is that of the system and its surroundings. When a system and its surroundings interact in an irreversible process, the increase in entropy of one is greater than the decrease in entropy of the other. Hence, the change in entropy of the Universe must be greater than zero for an irreversible process and equal to zero for a reversible process. Ultimately, because real processes are irreversible, the entropy of the Universe should increase steadily and eventually reach a maximum value. At this value, the Universe will be in a state of uniform temperature and density. All physical, chemical, and biologi-cal processes will have ceased at this time because a state of perfect disorder implies that no energy is available for doing work. This gloomy state of affairs is sometimes referred to as the heat death of the Universe. Quick Quiz 22.6 True or False: The entropy change in an adiabatic process must be zero because Q 5 0. Entropy Change in Thermal Conduction Let’s now consider a system consisting of a hot reservoir and a cold reservoir that are in thermal contact with each other and isolated from the rest of the Universe. A process occurs during which energy Q is transferred by heat from the hot reservoir at temperature Th to the cold reservoir at temperature Tc. The process as described is irreversible (energy would not spontaneously flow from cold to hot), so we must find an equivalent reversible process. Because the temperature of a reservoir does not change during the process, we can replace the real process for each reservoir with a reversible, isothermal process in which the same amount of energy is trans-ferred by heat. Consequently, for a reservoir, the entropy change does not depend on whether the process is reversible or irreversible. Because the cold reservoir absorbs energy Q, its entropy increases by Q/Tc. At the same time, the hot reservoir loses energy Q, so its entropy change is 2Q/Th. Because Th . Tc, the increase in entropy of the cold reservoir is greater than the decrease in entropy of the hot reservoir. Therefore, the change in entropy of the system (and of the Universe) is greater than zero: DSU 5 Q Tc 1 2Q Th . 0 Suppose energy were to transfer spontaneously from a cold object to a hot object, in violation of the second law. This impossible energy transfer can be described in terms of disorder. Before the transfer, a certain degree of order is associated with the different temperatures of the objects. The hot object’s molecules have a higher average energy than the cold object’s molecules. If energy spontaneously transfers from the cold object to the hot object, the cold object becomes colder over a time interval and the hot object becomes hotter. The difference in average molecular energy becomes even greater, which would represent an increase in order for the system and a violation of the second law. In comparison, the process that does occur naturally is the transfer of energy from the hot object to the cold object. In this process, the difference in average Entropy statement X of the second law of thermodynamics 22.8 \| Entropy on a Microscopic Scale 643 molecular energy decreases, which represents a more random distribution of energy and an increase in disorder. Entropy Change in a Free Expansion Let’s again consider the adiabatic free expansion of a gas occupying an initial vol-ume Vi (Fig. 22.14). In this situation, a membrane separating the gas from an evacu-ated region is broken and the gas expands to a volume Vf. This process is irrevers-ible; the gas would not spontaneously crowd into half the volume after filling the entire volume. What are the changes in entropy of the gas and of the Universe during this process? The process is neither reversible nor quasi-static. As shown in Section 20.6, the initial and final temperatures of the gas are the same. To apply Equation 22.9, we cannot take Q 5 0, the value for the irreversible process, but must instead find Q r; that is, we must find an equivalent reversible path that shares the same initial and final states. A simple choice is an isothermal, reversible expansion in which the gas pushes slowly against a piston while energy enters the gas by heat from a reservoir to hold the temperature constant. Because T is constant in this process, Equation 22.9 gives DS 5 3 f i dQ r T 5 1 T 3 f i dQ r For an isothermal process, the first law of thermodynamics specifies that e f i dQ r is equal to the negative of the work done on the gas during the expansion from Vi to Vf, which is given by Equation 20.14. Using this result, we find that the entropy change for the gas is DS 5 nR ln a Vf Vi b (22.11) Because Vf . Vi, we conclude that DS is positive. This positive result indicates that both the entropy and the disorder of the gas increase as a result of the irreversible, adiabatic expansion. It is easy to see that the gas is more disordered after the expansion. Instead of being concentrated in a relatively small space, the molecules are scattered over a larger region. Because the free expansion takes place in an insulated container, no energy is transferred by heat from the surroundings. (Remember that the isothermal, revers-ible expansion is only a replacement process used to calculate the entropy change for the gas; it is not the actual process.) Therefore, the free expansion has no effect on the surroundings, and the entropy change of the surroundings is zero. 22.8 Entropy on a Microscopic Scale As we have seen, entropy can be approached by relying on macroscopic concepts. Entropy can also be treated from a microscopic viewpoint through statistical analy-sis of molecular motions. Let’s use a microscopic model to investigate once again the free expansion of an ideal gas, which was discussed from a macroscopic point of view in Section 22.7. In the kinetic theory of gases, gas molecules are represented as particles moving randomly. Suppose the gas is initially confined to the volume Vi shown in Figure 22.14. When the membrane is removed, the molecules eventually are distributed throughout the greater volume Vf of the entire container. For a given uniform dis-tribution of gas in the volume, there are a large number of equivalent microstates, and the entropy of the gas can be related to the number of microstates correspond-ing to a given macrostate. Let’s count the number of microstates by considering the variety of molecular locations available to the molecules. Let’s assume each molecule occupies some Vacuum Gas at Ti in volume Vi Insulating wall Membrane When the membrane is ruptured, the gas will expand freely and irreversibly into the full volume. Figure 22.14 Adiabatic free expan-sion of a gas. The container is ther-mally insulated from its surround-ings; therefore, Q 5 0. 644 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics microscopic volume Vm. The total number of possible locations of a single molecule in a macroscopic initial volume Vi is the ratio wi 5 Vi/Vm, which is a huge number. We use wi here to represent either the number of ways the molecule can be placed in the initial volume or the number of microstates, which is equivalent to the num-ber of available locations. We assume the probabilities of a molecule occupying any of these locations are equal. As more molecules are added to the system, the number of possible ways the molecules can be positioned in the volume multiplies. For example, if you consider two molecules, for every possible placement of the first, all possible placements of the second are available. Therefore, there are wi ways of locating the first molecule, and for each way, there are wi ways of locating the second molecule. The total num-ber of ways of locating the two molecules is wiwi 5 wi 2. Neglecting the very small probability of having two molecules occupy the same location, each molecule may go into any of the Vi/Vm locations, and so the number of ways of locating N molecules in the volume becomes Wi 5 wi N 5 1Vi /Vm2 N. (Wi is not to be confused with work.) Similarly, when the volume is increased to Vf, the number of ways of locating N molecules increases to Wf 5 wf N 5 1Vf /Vm2 N. The ratio of the number of ways of placing the molecules in the volume for the initial and final configurations is Wf Wi 5 1Vf /Vm2 N 1Vi/Vm2 N 5 a Vf Vi b N Taking the natural logarithm of this equation and multiplying by Boltzmann’s con-stant gives k B ln a Wf Wi b 5 k B ln a Vf Vi b N 5 nNA k B ln a Vf Vi b where we have used the equality N 5 nNA. We know from Equation 19.11 that NAkB is the universal gas constant R; therefore, we can write this equation as k B lnWf 2 k B lnWi 5 nR ln a Vf Vi b (22.12) From Equation 22.11, we know that when a gas undergoes a free expansion from Vi to Vf, the change in entropy is Sf 2 Si 5 nR ln a Vf Vi b (22.13) Notice that the right sides of Equations 22.12 and 22.13 are identical. Therefore, from the left sides, we make the following important connection between entropy and the number of microstates for a given macrostate: S ; k B ln W (22.14) The more microstates there are that correspond to a given macrostate, the greater the entropy of that macrostate. As discussed previously, there are many more micro-states associated with disordered macrostates than with ordered macrostates. There-fore, Equation 22.14 indicates mathematically our earlier statement that entropy is a measure of disorder. Although our discussion used the specific example of the free expansion of an ideal gas, a more rigorous development of the statistical inter-pretation of entropy would lead us to the same conclusion. We have stated that individual microstates are equally probable. Because there are far more microstates associated with a disordered macrostate than with an ordered macrostate, however, a disordered macrostate is much more probable than an ordered one. Entropy (microscopic X definition) 22.8 \| Entropy on a Microscopic Scale 645 Conceptual Example 22.7 Let’s Play Marbles! Suppose you have a bag of 100 marbles of which 50 are red and 50 are green. You are allowed to draw four marbles from the bag according to the following rules. Draw one marble, record its color, and return it to the bag. Shake the bag and then draw another marble. Continue this process until you have drawn and returned four marbles. What are the possible macrostates for this set of events? What is the most likely macrostate? What is the least likely macrostate? SOLUTION Because each marble is returned to the bag before the next one is drawn and the bag is then shaken, the prob-ability of drawing a red marble is always the same as the probability of drawing a green one. All the possible microstates and macrostates are shown in Table 22.1. As this table indicates, there is only one way to draw a macrostate of four red marbles, so there is only one microstate for that macrostate. There are, however, four possible microstates that correspond to the macrostate of one green marble and three red marbles, six micro-states that correspond to two green marbles and two red marbles, four microstates that correspond to three green marbles and one red marble, and one microstate that corresponds to four green marbles. The most likely, and most dis-ordered, macrostate—two red marbles and two green marbles—corresponds to the largest number of microstates. The least likely, most ordered macrostates—four red marbles or four green marbles—correspond to the smallest number of microstates. Let’s explore this concept by considering 100 molecules in a container. At any given moment, the probability of one molecule being in the left part of the con-tainer shown in Active Figure 22.15a as a result of random motion is 1 2. If there are two molecules as shown in Active Figure 22.15b, the probability of both being in the left part is 1 1 22 2, or 1 in 4. If there are three molecules (Active Fig. 22.15c), the probability of them all being in the left portion at the same moment is 1 1 22 3, or 1 in 8. For 100 independently moving molecules, the probability that the 50 fastest ones will be found in the left part at any moment is 1 1 22 50. Likewise, the probability that the remaining 50 slower molecules will be found in the right part at any moment is 1 1 22 50. Therefore, the probability of finding this fast–slow separation as a result of random motion is the product 1 1 22 501 1 22 50 5 1 1 22 100, which corresponds to about 1 in 1030. When this calculation is extrapolated from 100 molecules to the number in 1 mol of gas (6.02 3 1023), the ordered arrangement is found to be extremely improbable! b c a (a) One molecule in a container has a 1-in-2 chance of being on the left side. (b) Two molecules have a 1-in-4 chance of being on the left side at the same time. (c) Three molecules have a 1-in-8 chance of being on the left side at the same time. ACTIVE FIGURE 22.15 Possible Results of Drawing Four Marbles from a Bag Total Number of Macrostate Possible Microstates Microstates All R RRRR 1 1G, 3R RRRG, RRGR, RGRR, GRRR 4 2G, 2R RRGG, RGRG, GRRG, 6 RGGR, GRGR, GGRR 3G, 1R GGGR, GGRG, GRGG, RGGG 4 All G GGGG 1 TABLE 22.1 646 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics Example 22.8 Adiabatic Free Expansion: One Last Time Let’s verify that the macroscopic and microscopic approaches to the calculation of entropy lead to the same conclusion for the adiabatic free expansion of an ideal gas. Suppose an ideal gas expands to four times its initial volume. As we have seen for this process, the initial and final temperatures are the same. (A) Using a macroscopic approach, calculate the entropy change for the gas. SOLUTION Conceptualize Look back at Figure 22.14, which is a diagram of the system before the adiabatic free expansion. Imagine breaking the membrane so that the gas moves into the evacuated area. The expansion is irreversible. Categorize We can replace the irreversible process with a reversible isothermal process between the same initial and final states. This approach is macroscopic, so we use a thermodynamic variable, in particular, the volume V. Analyze Use Equation 22.11 to evaluate the entropy change: DS 5 nR ln a Vf Vi b 5 nR ln a4Vi Vi b 5 nR ln 4 Analyze The number of microstates available to a single molecule in the initial volume Vi is wi 5 Vi/Vm. Use this number to find the number of available microstates for N molecules: Wi 5 wi N 5 a Vi Vm b N Find the number of available microstates for N mol-ecules in the final volume Vf 5 4Vi: Wf 5 a Vf Vm b N 5 a4Vi Vm b N Use Equation 22.14 to find the entropy change: DS 5 k B ln Wf 2 k B ln Wi 5 k B ln a Wf Wi b 5 k B ln a4Vi Vi b N 5 k B ln 14N 2 5 Nk B ln 4 5 nR ln 4 (B) Using statistical considerations, calculate the change in entropy for the gas and show that it agrees with the answer you obtained in part (A). SOLUTION Categorize This approach is microscopic, so we use variables related to the individual molecules. Finalize The answer is the same as that for part (A), which dealt with macroscopic parameters. WHAT IF? In part (A), we used Equation 22.11, which was based on a reversible isothermal process connecting the ini-tial and final states. Would you arrive at the same result if you chose a different reversible process? Answer You must arrive at the same result because entropy is a state variable. For example, consider the two-step pro-cess in Figure 22.16: a reversible adiabatic expansion from Vi to 4Vi (A S B) during which the temperature drops from T1 to T2 and a reversible isovolumetric process (B S C) that takes the gas back to the initial temperature T1. During the reversible adiabatic process, DS 5 0 because Q r 5 0. V P Vi 4Vi B C A T1 T2 Figure 22.16 (Example 22.8) A gas expands to four times its initial volume and back to the initial tempera-ture by means of a two-step process. \| Summary 647 22.8 cont. For the reversible isovolumetric process (B S C), use Equation 22.9: DS 5 3 f i dQ r T 5 3 T1 T2 nCVdT T 5 nCV ln aT1 T2 b Find the ratio of temperature T1 to T2 from Equation 21.20 for the adiabatic process: T1 T2 5 a4Vi Vi b g21 5 142 g21 Substitute to find DS: DS 5 nCV ln 142 g21 5 nCV 1g 2 12 ln 4 5 nCV aCP CV 2 1b ln 4 5 n1CP 2 CV2 ln 4 5 nR ln 4 and you do indeed obtain the exact same result for the entropy change. Summary Definitions The thermal efficiency e of a heat engine is e ; Weng 0 Q h 0 5 0 Q h 0 2 0 Q c 0 0 Q h 0 5 1 2 0 Q c 0 0 Q h 0 (22.2) From a microscopic viewpoint, the entropy of a given macrostate is defined as S ; kB ln W (22.14) where kB is Boltzmann’s constant and W is the number of micro-states of the system corresponding to the macrostate. In a reversible process, the system can be returned to its initial conditions along the same path on a PV diagram, and every point along this path is an equilibrium state. A process that does not satisfy these requirements is irreversible. Concepts and Principles A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work. The net work done by a heat engine in carrying a working substance through a cyclic process (DEint 5 0) is Weng 5 \|Q h\| 2 \|Q c\| (22.1) where \|Q h\| is the energy taken in from a hot res-ervoir and \|Q c\| is the energy expelled to a cold reservoir. Two ways the second law of thermodynamics can be stated are as follows: • It is impossible to construct a heat engine that, operat-ing in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work (the Kelvin–Planck statement). • It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work (the Clausius statement). continued 648 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics 5. Consider cyclic processes completely characterized by each of the following net energy inputs and outputs. In each case, the energy transfers listed are the only ones occur-ring. Classify each process as (a) possible, (b) impossible according to the first law of thermodynamics, (c) impos-sible according to the second law of thermodynamics, or (d) impossible according to both the first and second laws. (i) Input is 5 J of work, and output is 4 J of work. (ii) Input is 5 J of work, and output is 5 J of energy transferred by heat. (iii) Input is 5 J of energy transferred by electrical trans-mission, and output is 6 J of work. (iv) Input is 5 J of energy transferred by heat, and output is 5 J of energy transferred by heat. (v) Input is 5 J of energy transferred by heat, and output is 5 J of work. (vi) Input is 5 J of energy transferred by heat, and output is 3 J of work plus 2 J of energy trans-ferred by heat. 6. A compact air-conditioning unit is placed on a table inside a well-insulated apartment and is plugged in and turned on. What happens to the average temperature of the apart-ment? (a) It increases. (b) It decreases. (c) It remains con-stant. (d) It increases until the unit warms up and then decreases. (e) The answer depends on the initial tempera-ture of the apartment. 1. A steam turbine operates at a boiler temperature of 450 K and an exhaust temperature of 300 K. What is the maximum theoretical efficiency of this system? (a) 0.240 (b) 0.500 (c) 0.333 (d) 0.667 (e) 0.150 2. An engine does 15.0 kJ of work while exhausting 37.0 kJ to a cold reservoir. What is the efficiency of the engine? (a) 0.150 (b) 0.288 (c) 0.333 (d) 0.450 (e) 1.20 3. A refrigerator has 18.0 kJ of work done on it while 115 kJ of energy is transferred from inside its interior. What is its coefficient of performance? (a) 3.40 (b) 2.80 (c) 8.90 (d) 6.40 (e) 5.20 4. Of the following, which is not a statement of the second law of thermodynamics? (a) No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely to do work. (b) No real engine operating between two energy reservoirs can be more efficient than a Car-not engine operating between the same two reservoirs. (c) When a system undergoes a change in state, the change in the internal energy of the system is the sum of the energy transferred to the system by heat and the work done on the system. (d) The entropy of the Universe increases in all natural processes. (e) Energy will not spontaneously transfer by heat from a cold object to a hot object. Carnot’s theorem states that no real heat engine operating (irreversibly) between the temperatures Tc and Th can be more efficient than an engine operat-ing reversibly in a Carnot cycle between the same two temperatures. The thermal efficiency of a heat engine operating in the Carnot cycle is eC 5 1 2 Tc Th (22.6) The change in entropy dS of a system during a process between two infinitesimally separated equilibrium states is dS 5 dQ r T (22.8) where dQ r is the energy transfer by heat for the system for a reversible process that connects the initial and final states. The change in entropy of a system during an arbitrary process between an initial state and a final state is DS 5 3 f i dQ r T (22.9) The value of DS for the system is the same for all paths connecting the initial and final states. The change in entropy for a system undergoing any reversible, cyclic process is zero, and when such a process occurs, the entropy of the Universe remains constant. The second law of thermodynamics states that when real (irreversible) processes occur, the degree of disorder in the system plus the surroundings increases. When a process occurs in an isolated system, the state of the system becomes more disordered. The measure of disorder in a system is called entropy S. Therefore, yet another way the second law can be stated is as follows: • The entropy of the Universe increases in all real processes. Objective Questions denotes answer available in Student Solutions Manual/Study Guide \| Conceptual Questions 649 except D, the volume changes by a factor of 2. All five pro-cesses are reversible. Rank the processes according to the change in entropy of the gas from the largest positive value to the largest-magnitude negative value. In your rankings, display any cases of equality. 10. Assume a sample of an ideal gas is at room temperature. What action will necessarily make the entropy of the sample increase? (a) Transfer energy into it by heat. (b) Trans-fer energy into it irreversibly by heat. (c) Do work on it. (d) Increase either its temperature or its volume, with-out letting the other variable decrease. (e) None of those choices is correct. 11. The arrow OA in the PV diagram shown in Figure OQ22.11 represents a reversible adiabatic expansion of an ideal gas. The same sample of gas, starting from the same state O, now undergoes an adiabatic free expansion to the same final volume. What point on the diagram could represent the final state of the gas? (a) the same point A as for the reversible expansion (b) point B (c) point C (d) any of those choices (e) none of those choices 7. The second law of thermodynamics implies that the coef-ficient of performance of a refrigerator must be what? (a) less than 1 (b) less than or equal to 1 (c) greater than or equal to 1 (d) finite (e) greater than 0 8. A thermodynamic process occurs in which the entropy of a system changes by 28 J/K. According to the second law of thermodynamics, what can you conclude about the entropy change of the environment? (a) It must be 18 J/K or less. (b) It must be between 18 J/K and 0. (c) It must be equal to 18 J/K. (d) It must be 18 J/K or more. (e) It must be zero. 9. A sample of a monatomic ideal gas is contained in a cylin-der with a piston. Its state is represented by the dot in the PV diagram shown in Figure OQ22.9. Arrows A through E represent isobaric, isothermal, adiabatic, and isovolumet-ric processes that the sample can undergo. In each process P C D B A E V Figure OQ22.9 P V O B A C Figure OQ22.11 Conceptual Questions denotes answer available in Student Solutions Manual/Study Guide 1. What are some factors that affect the efficiency of automo-bile engines? 2. A steam-driven turbine is one major component of an elec-tric power plant. Why is it advantageous to have the tem-perature of the steam as high as possible? 3. Does the second law of thermodynamics contradict or cor-rect the first law? Argue for your answer. 4. “The first law of thermodynamics says you can’t really win, and the second law says you can’t even break even.” Explain how this statement applies to a particular device or pro-cess; alternatively, argue against the statement. 5. Is it possible to construct a heat engine that creates no thermal pollution? Explain. 6. (a) Give an example of an irreversible process that occurs in nature. (b) Give an example of a process in nature that is nearly reversible. 7. The device shown in Figure CQ22.7, called a thermoelec-tric converter, uses a series of semiconductor cells to trans-form internal energy to electric potential energy, which we will study in Chapter 25. In the photograph on the left, both legs of the device are at the same temperature and no electric potential energy is produced. When one leg is at a higher temperature than the other as shown in the pho-tograph on the right, however, electric potential energy is produced as the device extracts energy from the hot reser-voir and drives a small electric motor. (a) Why is the differ-ence in temperature necessary to produce electric poten-tial energy in this demonstration? (b) In what sense does this intriguing experiment demonstrate the second law of thermodynamics? Figure CQ22.7 Courtesy of PASCO Scientific Company 650 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics 12. (a) If you shake a jar full of jelly beans of different sizes, the larger beans tend to appear near the top and the smaller ones tend to fall to the bottom. Why? (b) Does this process violate the second law of thermodynamics? 13. The energy exhaust from a certain coal-fired electric generating station is carried by “cooling water” into Lake Ontario. The water is warm from the viewpoint of living things in the lake. Some of them congregate around the outlet port and can impede the water flow. (a) Use the the-ory of heat engines to explain why this action can reduce the electric power output of the station. (b) An engineer says that the electric output is reduced because of “higher back pressure on the turbine blades.” Comment on the accuracy of this statement. 8. Discuss three different common examples of natural processes that involve an increase in entropy. Be sure to account for all parts of each system under consideration. 9. Discuss the change in entropy of a gas that expands (a) at constant temperature and (b) adiabatically. 10. Suppose your roommate cleans and tidies up your messy room after a big party. Because she is creating more order, does this process represent a violation of the second law of thermodynamics? 11. “Energy is the mistress of the Universe, and entropy is her shadow.” Writing for an audience of general readers, argue for this statement with at least two examples. Alternatively, argue for the view that entropy is like an executive who instantly determines what will happen, whereas energy is like a bookkeeper telling us how little we can afford. (Arnold Sommerfeld suggested the idea for this question.) Problems Section 22.1 Heat Engines and the Second Law of Thermodynamics 1. An engine absorbs 1.70 kJ from a hot reservoir at 277°C and expels 1.20 kJ to a cold reservoir at 27°C in each cycle. (a) What is the engine’s efficiency? (b) How much work is done by the engine in each cycle? (c) What is the power output of the engine if each cycle lasts 0.300 s? 2. The work done by an engine equals one-fourth the energy it absorbs from a reservoir. (a) What is its thermal efficiency? (b) What fraction of the energy absorbed is expelled to the cold reservoir? 3. A heat engine takes in 360 J of energy from a hot reservoir and performs 25.0 J of work in each cycle. Find (a) the effi-ciency of the engine and (b) the energy expelled to the cold reservoir in each cycle. 4. A gun is a heat engine. In particular, it is an internal com-bustion piston engine that does not operate in a cycle, but comes apart during its adiabatic expansion process. A cer-tain gun consists of 1.80 kg of iron. It fires one 2.40-g bul-let at 320 m/s with an energy efficiency of 1.10%. Assume the body of the gun absorbs all the energy exhaust—the other 98.9%—and increases uniformly in temperature for a short time interval before it loses any energy by heat into the environment. Find its temperature increase. 5. A particular heat engine has a mechanical power output of 5.00 kW and an efficiency of 25.0%. The engine expels 8.00 3 103 J of exhaust energy in each cycle. Find (a) the energy taken in during each cycle and (b) the time interval for each cycle. 6. A multicylinder gasoline engine in an airplane, operat-ing at 2.50 3 103 rev/min, takes in energy 7.89 3 103 J and exhausts 4.58 3 103 J for each revolution of the crankshaft. (a) How many liters of fuel does it consume in 1.00 h of operation if the heat of combustion of the fuel is equal to 4.03 3 107 J/L? (b) What is the mechanical power output of the engine? Ignore friction and express the answer in horsepower. (c) What is the torque exerted by the crank-shaft on the load? (d) What power must the exhaust and cooling system transfer out of the engine? 7. Suppose a heat engine is connected to two energy reser-voirs, one a pool of molten aluminum (660°C) and the other a block of solid mercury (238.9°C). The engine runs denotes asking for quantitative and conceptual reasoning denotes symbolic reasoning problem denotes Master It tutorial available in Enhanced WebAssign denotes guided problem denotes “paired problems” that develop reasoning with symbols and numerical values The problems found in this chapter may be assigned online in Enhanced WebAssign 1. denotes straightforward problem; 2. denotes intermediate problem; 3. denotes challenging problem 1. full solution available in the Student Solutions Manual/Study Guide 1. denotes problems most often assigned in Enhanced WebAssign; these provide students with targeted feedback and either a Master It tutorial or a Watch It solution video. shaded \| Problems 651 17. What is the coefficient of performance of a refrigerator that operates with Carnot efficiency between temperatures 23.00°C and 127.0°C? 18. Why is the following situation impossible? An inventor comes to a patent office with the claim that her heat engine, which employs water as a working substance, has a thermody-namic efficiency of 0.110. Although this efficiency is low compared with typical automobile engines, she explains that her engine operates between an energy reservoir at room temperature and a water–ice mixture at atmospheric pressure and therefore requires no fuel other than that to make the ice. The patent is approved, and working proto-types of the engine prove the inventor’s efficiency claim. 19. A heat engine is being designed to have a Carnot efficiency of 65.0% when operating between two energy reservoirs. (a) If the temperature of the cold reservoir is 20.0°C, what must be the temperature of the hot reservoir? (b) Can the actual efficiency of the engine be equal to 65.0%? Explain. 20. An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy \|Q c\| is taken in from a cold reservoir and energy \|Q h\| is rejected to a hot reservoir. (a) Show that the work that must be sup-plied to run the refrigerator or heat pump is W 5 Th 2 Tc Tc 0 Q c 0 (b) Show that the coefficient of performance (COP) of the ideal refrigerator is COP 5 Tc Th 2 Tc 21. What is the maximum possible coefficient of perfor-mance of a heat pump that brings energy from outdoors at 23.00°C into a 22.0°C house? Note: The work done to run the heat pump is also available to warm the house. 22. How much work does an ideal Carnot refrigerator require to remove 1.00 J of energy from liquid helium at 4.00 K and expel this energy to a room-temperature (293-K) environment? 23. If a 35.0%-efficient Carnot heat engine (Active Fig. 22.2) is run in reverse so as to form a refrigerator (Active Fig. 22.4), what would be this refrigerator’s coefficient of performance? 24. A Carnot heat engine operates between tempera-tures Th and Tc. (a) If Th 5 500 K and Tc 5 350 K, what is the efficiency of the engine? (b) What is the change in its efficiency for each degree of increase in Th above 500 K? (c) What is the change in its efficiency for each degree of change in Tc? (d) Does the answer to part (c) depend on Tc? Explain. 25. An ideal gas is taken through a Carnot cycle. The isother-mal expansion occurs at 250°C, and the isothermal com-pression takes place at 50.0°C. The gas takes in 1.20 3 103 J of energy from the hot reservoir during the isothermal expansion. Find (a) the energy expelled to the cold reser-voir in each cycle and (b) the net work done by the gas in each cycle. by freezing 1.00 g of aluminum and melting 15.0 g of mer-cury during each cycle. The heat of fusion of aluminum is 3.97 3 105 J/kg; the heat of fusion of mercury is 1.18 3 104 J/kg. What is the efficiency of this engine? Section 22.2 Heat Pumps and Refrigerators 8. A refrigerator has a coefficient of performance equal to 5.00. The refrigerator takes in 120 J of energy from a cold reservoir in each cycle. Find (a) the work required in each cycle and (b) the energy expelled to the hot reservoir. 9. During each cycle, a refrigerator ejects 625 kJ of energy to a high-temperature reservoir and takes in 550 kJ of energy from a low-temperature reservoir. Determine (a) the work done on the refrigerant in each cycle and (b) the coeffi-cient of performance of the refrigerator. 10. A heat pump has a coefficient of performance of 3.80 and operates with a power consumption of 7.03 3 103 W. (a) How much energy does it deliver into a home during 8.00 h of continuous operation? (b) How much energy does it extract from the outside air? 11. A freezer has a coefficient of performance of 6.30. It is advertised as using electricity at a rate of 457 kWh/yr. (a) On average, how much energy does it use in a single day? (b) On average, how much energy does it remove from the refrigerator in a single day? (c) What maximum mass of water at 20.0°C could the freezer freeze in a single day? Note: One kilowatt-hour (kWh) is an amount of energy equal to running a 1-kW appliance for one hour. 12. A heat pump has a coefficient of performance equal to 4.20 and requires a power of 1.75 kW to operate. (a) How much energy does the heat pump add to a home in one hour? (b) If the heat pump is reversed so that it acts as an air conditioner in the summer, what would be its coefficient of performance? Section 22.3 Reversible and Irreversible Processes Section 22.4 The Carnot Engine 13. One of the most efficient heat engines ever built is a coal-fired steam turbine in the Ohio River valley, operating between 1 870°C and 430°C. (a) What is its maximum theo-retical efficiency? (b) The actual efficiency of the engine is 42.0%. How much mechanical power does the engine deliver if it absorbs 1.40 3 105 J of energy each second from its hot reservoir? 14. A heat engine operates between a reservoir at 25.0°C and one at 375°C. What is the maximum efficiency possible for this engine? 15. A Carnot engine has a power output of 150 kW. The engine operates between two reservoirs at 20.0°C and 500°C. (a) How much energy enters the engine by heat per hour? (b) How much energy is exhausted by heat per hour? 16. A Carnot engine has a power output P. The engine operates between two reservoirs at temperature Tc and Th. (a) How much energy enters the engine by heat in a time interval Dt? (b) How much energy is exhausted by heat in the time interval Dt? 652 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics pressures, volumes, and temperatures as you fill in the fol-lowing table: P V T A 1 400 kPa 10.0 L 720 K B C 24.0 L D 15.0 L (b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A S B, B S C, C S D, and D S A. (c) Calcu-late the efficiency Wnet/\|Q h\|. (d) Show that the efficiency is equal to 1 2 TC/TA, the Carnot efficiency. 31. A heat pump used for heating shown in Figure P22.31 is essentially an air conditioner installed backward. It extracts energy from colder air outside and deposits it in a warmer room. Suppose the ratio of the actual energy entering the room to the work done by the device’s motor is 10.0% of the theoretical maximum ratio. Determine the energy entering the room per joule of work done by the motor given that the inside temperature is 20.0°C and the outside temperature is 25.00°C. 26. An electric power plant that would make use of the temperature gradient in the ocean has been proposed. The system is to operate between 20.0°C (surface-water temperature) and 5.00°C (water temperature at a depth of about 1 km). (a) What is the maximum efficiency of such a system? (b) If the electric power output of the plant is 75.0 MW, how much energy is taken in from the warm reservoir per hour? (c) In view of your answer to part (a), explain whether you think such a system is worthwhile. Note that the “fuel” is free. 27. Argon enters a turbine at a rate of 80.0 kg/min, a tem-perature of 800°C, and a pressure of 1.50 MPa. It expands adiabatically as it pushes on the turbine blades and exits at pressure 300 kPa. (a) Calculate its temperature at exit. (b) Calculate the (maximum) power output of the turn-ing turbine. (c) The turbine is one component of a model closed-cycle gas turbine engine. Calculate the maximum efficiency of the engine. 28. Suppose you build a two-engine device with the exhaust energy output from one heat engine supplying the input energy for a second heat engine. We say that the two engines are running in series. Let e1 and e2 represent the efficiencies of the two engines. (a) The overall efficiency of the two-engine device is defined as the total work out-put divided by the energy put into the first engine by heat. Show that the overall efficiency e is given by e 5 e1 1 e2 2 e1e2 What If? For parts (b) through (e) that follow, assume the two engines are Carnot engines. Engine 1 operates between temperatures Th and Ti. The gas in engine 2 varies in temperature between Ti and Tc. In terms of the tempera-tures, (b) what is the efficiency of the combination engine? (c) Does an improvement in net efficiency result from the use of two engines instead of one? (d) What value of the intermediate temperature Ti results in equal work being done by each of the two engines in series? (e) What value of Ti results in each of the two engines in series having the same efficiency? 29. An electric generating station is designed to have an electric output power of 1.40 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 110°C. (a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature Th. (b) If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change? (c) Find the exhaust power for Th 5 800°C. (d) Find the value of Th for which the exhaust power would be only half as large as in part (c). (e) Find the value of Th for which the exhaust power would be one-fourth as large as in part (c). 30. At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume of 10.0 L, and a temperature of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown Outside Tc Q c Q h Inside Th Heat pump Figure P22.31 32. An ideal (Carnot) freezer in a kitchen has a constant tem-perature of 260 K, whereas the air in the kitchen has a con-stant temperature of 300 K. Suppose the insulation for the freezer is not perfect but rather conducts energy into the freezer at a rate of 0.150 W. Determine the average power required for the freezer’s motor to maintain the constant temperature in the freezer. Section 22.5 Gasoline and Diesel Engines Note: For problems in this section, assume the gas in the engine is diatomic with g 5 1.40. 33. In a cylinder of an automobile engine, immedi-ately after combustion the gas is confined to a volume of 50.0 cm3 and has an initial pressure of 3.00 3 106 Pa. The piston moves outward to a final volume of 300 cm3, and the gas expands without energy transfer by heat. (a) What is the final pressure of the gas? (b) How much work is done by the gas in expanding? 34. A gasoline engine has a compression ratio of 6.00. (a) What is the efficiency of the engine if it operates in an idealized Otto cycle? (b) What If? If the actual efficiency is 15.0%, what fraction of the fuel is wasted as a result of friction and energy transfers by heat that could be avoided in a revers- \| Problems 653 41. A 2.00-L container has a center partition that divides it into two equal parts as shown in Figure P22.41. The left side contains 0.044 0 mol of H2 gas, and the right side contains 0.044 0 mol of O2 gas. Both gases are at room temperature and at atmospheric pressure. The partition is removed, and the gases are allowed to mix. What is the entropy increase of the system? ible engine? Assume complete combustion of the air–fuel mixture. 35. An idealized diesel engine operates in a cycle known as the air-standard diesel cycle shown in Figure P22.35. Fuel is sprayed into the cylinder at the point of maximum com-pression, B. Combustion occurs during the expansion B S C, which is modeled as an isobaric process. Show that the efficiency of an engine operating in this idealized die-sel cycle is e 5 1 2 1 g a TD 2 TA TC 2 TB b Adiabatic processes A B C D P V Qh Qc V2 VB V1 VA VC Q Figure P22.35 Section 22.6 Entropy Section 22.7 Entropy and the Second Law 36. An ice tray contains 500 g of liquid water at 0°C. Calculate the change in entropy of the water as it freezes slowly and completely at 0°C. 37. A Styrofoam cup holding 125 g of hot water at 100°C cools to room temperature, 20.0°C. What is the change in entropy of the room? Neglect the specific heat of the cup and any change in temperature of the room. 38. Two 2.00 3 103-kg cars both traveling at 20.0 m/s undergo a head-on collision and stick together. Find the change in entropy of the surrounding air resulting from the collision if the air temperature is 23.0°C. Ignore the energy carried away from the collision by sound. 39. A 70.0-kg log falls from a height of 25.0 m into a lake. If the log, the lake, and the air are all at 300 K, find the change in entropy of the air during this process. 40. A 1.00-mol sample of H2 gas is contained in the left side of the container shown in Figure P22.40, which has equal volumes on the left and right. The right side is evacuated. When the valve is opened, the gas streams into the right side. (a) What is the entropy change of the gas? (b) Does the temperature of the gas change? Assume the container is so large that the hydrogen behaves as an ideal gas. Valve Vacuum H2 Figure P22.40 0.044 0 mol O2 0.044 0 mol H2 Figure P22.41 42. How fast are you personally making the entropy of the Uni-verse increase right now? Compute an order-of-magnitude estimate, stating what quantities you take as data and the values you measure or estimate for them. 43. When an aluminum bar is connected between a hot reser-voir at 725 K and a cold reservoir at 310 K, 2.50 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir. In this irreversible process, calculate the change in entropy of (a) the hot reservoir, (b) the cold reservoir, and (c) the Universe, neglecting any change in entropy of the aluminum rod. 44. When a metal bar is connected between a hot reservoir at Th and a cold reservoir at Tc, the energy transferred by heat from the hot reservoir to the cold reservoir is Q. In this irreversible process, find expressions for the change in entropy of (a) the hot reservoir, (b) the cold reservoir, and (c) the Universe, neglecting any change in entropy of the metal rod. 45. The temperature at the surface of the Sun is approximately 5 800 K, and the temperature at the surface of the Earth is approximately 290 K. What entropy change of the Uni-verse occurs when 1.00 3 103 J of energy is transferred by radiation from the Sun to the Earth? Section 22.8 Entropy on a Microscopic Scale 46. If you roll two dice, what is the total number of ways in which you can obtain (a) a 12 and (b) a 7? 47. Prepare a table like Table 22.1 by using the same proce-dure (a) for the case in which you draw three marbles from your bag rather than four and (b) for the case in which you draw five marbles rather than four. 48. (a) Prepare a table like Table 22.1 for the following occur-rence. You toss four coins into the air simultaneously and then record the results of your tosses in terms of the num-bers of heads (H) and tails (T) that result. For example, HHTH and HTHH are two possible ways in which three heads and one tail can be achieved. (b) On the basis of your table, what is the most probable result recorded for a toss? In terms of entropy, (c) what is the most ordered mac-rostate, and (d) what is the most disordered? Additional Problems 49. The energy absorbed by an engine is three times greater than the work it performs. (a) What is its thermal efficiency? 654 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics ety of applications ever since, including the solar power application discussed on the cover of this textbook. Fuel is burned externally to warm one of the engine’s two cylin-ders. A fixed quantity of inert gas moves cyclically between the cylinders, expanding in the hot one and contracting in the cold one. Figure P22.57 represents a model for its thermodynamic cycle. Consider n moles of an ideal mon-atomic gas being taken once through the cycle, consisting of two isothermal processes at temperatures 3Ti and Ti and two constant- volume processes. Let us find the efficiency of this engine. (a) Find the energy transferred by heat into the gas during the isovolumetric process AB. (b) Find the energy transferred by heat into the gas during the isother-mal process BC. (c) Find the energy transferred by heat into the gas during the isovolumetric process CD. (d) Find the energy transferred by heat into the gas during the iso-thermal process DA. (e) Identify which of the results from parts (a) through (d) are positive and evaluate the energy input to the engine by heat. (f) From the first law of ther-modynamics, find the work done by the engine. (g) From the results of parts (e) and (f), evaluate the efficiency of the engine. A Stirling engine is easier to manufacture than an internal combustion engine or a turbine. It can run on burning garbage. It can run on the energy transferred by sunlight and produce no material exhaust. Stirling engines are not currently used in automobiles due to long startup times and poor acceleration response. (b) What fraction of the energy absorbed is expelled to the cold reservoir? 50. A steam engine is operated in a cold climate where the exhaust temperature is 0°C. (a) Calculate the theoretical maximum efficiency of the engine using an intake steam temperature of 100°C. (b) If, instead, superheated steam at 200°C is used, find the maximum possible efficiency. 51. Find the maximum (Carnot) efficiency of an engine that absorbs energy from a hot reservoir at 545°C and exhausts energy to a cold reservoir at 185°C. 52. Every second at Niagara Falls, some 5.00 3 103 m3 of water falls a distance of 50.0 m. What is the increase in entropy of the Universe per second due to the falling water? Assume the mass of the surroundings is so great that its tempera-ture and that of the water stay nearly constant at 20.0°C. Also assume a negligible amount of water evaporates. 53. Energy transfers by heat through the exterior walls and roof of a house at a rate of 5.00 3 103 J/s 5 5.00 kW when the interior temperature is 22.0°C and the outside temperature is 25.00°C. (a) Calculate the electric power required to maintain the interior temperature at 22.0°C if the power is used in electric resistance heaters that convert all the energy transferred in by electrical transmission into internal energy. (b) What If? Calculate the electric power required to maintain the interior temperature at 22.0°C if the power is used to drive an electric motor that operates the compressor of a heat pump that has a coefficient of performance equal to 60.0% of the Carnot-cycle value. 54. In 1993, the U.S. government instituted a require-ment that all room air conditioners sold in the United States must have an energy efficiency ratio (EER) of 10 or higher. The EER is defined as the ratio of the cooling capacity of the air conditioner, measured in British thermal units per hour, or Btu/h, to its electrical power requirement in watts. (a) Convert the EER of 10.0 to dimensionless form, using the conversion 1 Btu 5 1 055 J. (b) What is the appropri-ate name for this dimensionless quantity? (c) In the 1970s, it was common to find room air conditioners with EERs of 5 or lower. State how the operating costs compare for 10 000-Btu/h air conditioners with EERs of 5.00 and 10.0. Assume each air conditioner operates for 1 500 h during the summer in a city where electricity costs 17.0¢ per kWh. 55. An airtight freezer holds n moles of air at 25.0°C and 1.00 atm. The air is then cooled to 218.0°C. (a) What is the change in entropy of the air if the volume is held constant? (b) What would the entropy change be if the pressure were maintained at 1.00 atm during the cooling? 56. Suppose an ideal (Carnot) heat pump could be con-structed for use as an air conditioner. (a) Obtain an expres-sion for the coefficient of performance (COP) for such an air conditioner in terms of Th and Tc. (b) Would such an air conditioner operate on a smaller energy input if the difference in the operating temperatures were greater or smaller? (c) Compute the COP for such an air conditioner if the indoor temperature is 20.0°C and the outdoor tem-perature is 40.0°C. 57. In 1816, Robert Stirling, a Scottish clergyman, patented the Stirling engine, which has found a wide vari-Isothermal processes P V Vi 2Vi Ti 3Ti A B C D Figure P22.57 58. A firebox is at 750 K, and the ambient temperature is 300 K. The efficiency of a Carnot engine doing 150 J of work as it transports energy between these constant- temperature baths is 60.0%. The Carnot engine must take in energy 150 J/0.600 5 250 J from the hot reservoir and must put out 100 J of energy by heat into the environment. To fol-low Carnot’s reasoning, suppose some other heat engine S could have an efficiency of 70.0%. (a) Find the energy input and exhaust energy output of engine S as it does 150 J of work. (b) Let engine S operate as in part (a) and run the Carnot engine in reverse between the same reservoirs. The output work of engine S is the input work for the Carnot refrigerator. Find the total energy transferred to or from the firebox and the total energy transferred to or from the environment as both engines operate together. (c) Explain how the results of parts (a) and (b) show that the Clausius statement of the second law of thermodynamics is violated. (d) Find the energy input and work output of engine S as it puts out exhaust energy of 100 J. Let engine S operate as in part (c) and contribute 150 J of its work output to running \| Problems 655 63. A power plant, having a Carnot efficiency, produces 1.00 GW of electrical power from turbines that take in steam at 500 K and reject water at 300 K into a flowing river. The water downstream is 6.00 K warmer due to the output of the power plant. Determine the flow rate of the river. 64. A power plant, having a Carnot efficiency, produces electric power P from turbines that take in energy from steam at temperature Th and discharge energy at tempera-ture Tc through a heat exchanger into a flowing river. The water downstream is warmer by DT due to the output of the power plant. Determine the flow rate of the river. 65. A sample consisting of n moles of an ideal gas under-goes a reversible isobaric expansion from volume Vi to vol-ume 3Vi. Find the change in entropy of the gas by calculat-ing e f i dQ/T, where dQ 5 nCP dT. 66. An athlete whose mass is 70.0 kg drinks 16.0 ounces (454 g) of refrigerated water. The water is at a tempera-ture of 35.0°F. (a) Ignoring the temperature change of the body that results from the water intake (so that the body is regarded as a reservoir always at 98.6°F), find the entropy increase of the entire system. (b) What If? Assume the entire body is cooled by the drink and the average specific heat of a person is equal to the specific heat of liquid water. Ignoring any other energy transfers by heat and any meta-bolic energy release, find the athlete’s temperature after she drinks the cold water given an initial body temperature of 98.6°F. (c) Under these assumptions, what is the entropy increase of the entire system? (d) State how this result com-pares with the one you obtained in part (a). 67. A 1.00-mol sample of an ideal monatomic gas is taken through the cycle shown in Figure P22.67. The process A S B is a reversible isothermal expansion. Calculate (a) the net work done by the gas, (b) the energy added to the Carnot engine in reverse. Find (e) the total energy the firebox puts out as both engines operate together, (f) the total work output, and (g) the total energy transferred to the environment. (h) Explain how the results show that the Kelvin–Planck statement of the second law is violated. Therefore, our assumption about the efficiency of engine S must be false. (i) Let the engines operate together through one cycle as in part (d). Find the change in entropy of the Universe. (j) Explain how the result of part (i) shows that the entropy statement of the second law is violated. 59. Review. This problem complements Problem 84 in Chapter 10. In the operation of a single-cylinder internal combus-tion piston engine, one charge of fuel explodes to drive the piston outward in the power stroke. Part of its energy output is stored in a turning flywheel. This energy is then used to push the piston inward to compress the next charge of fuel and air. In this compression process, assume an original volume of 0.120 L of a diatomic ideal gas at atmo-spheric pressure is compressed adiabatically to one-eighth of its original volume. (a) Find the work input required to compress the gas. (b) Assume the flywheel is a solid disk of mass 5.10 kg and radius 8.50 cm, turning freely with-out friction between the power stroke and the compression stroke. How fast must the flywheel turn immediately after the power stroke? This situation represents the minimum angular speed at which the engine can operate without stalling. (c) When the engine’s operation is well above the point of stalling, assume the flywheel puts 5.00% of its maximum energy into compressing the next charge of fuel and air. Find its maximum angular speed in this case. 60. A biology laboratory is maintained at a constant temper-ature of 7.00°C by an air conditioner, which is vented to the air outside. On a typical hot summer day, the outside temperature is 27.0°C and the air-conditioning unit emits energy to the outside at a rate of 10.0 kW. Model the unit as having a coefficient of performance (COP) equal to 40.0% of the COP of an ideal Carnot device. (a) At what rate does the air conditioner remove energy from the labora-tory? (b) Calculate the power required for the work input. (c) Find the change in entropy of the Universe produced by the air conditioner in 1.00 h. (d) What If? The outside tem-perature increases to 32.0°C. Find the fractional change in the COP of the air conditioner. 61. A heat engine operates between two reservoirs at T2 5 600 K and T1 5 350 K. It takes in 1.00 3 103 J of energy from the higher-temperature reservoir and performs 250 J of work. Find (a) the entropy change of the Universe DSU for this process and (b) the work W that could have been done by an ideal Carnot engine operating between these two reservoirs. (c) Show that the difference between the amounts of work done in parts (a) and (b) is T1 DSU. 62. A 1.00-mol sample of a monatomic ideal gas is taken through the cycle shown in Figure P22.62. At point A, the pressure, volume, and temperature are Pi, Vi, and Ti, respectively. In terms of R and Ti, find (a) the total energy entering the system by heat per cycle, (b) the total energy leaving the system by heat per cycle, and (c) the efficiency of an engine operating in this cycle. (d) Explain how the efficiency compares with that of an engine operating in a Carnot cycle between the same temperature extremes. D A B C P Pi 3Pi Vi 2Vi V Q2 Q4 Q3 Q1 2Pi Q Figure P22.62 5 Isothermal process 1 10 50 V (liters) B C A P (atm) Figure P22.67 656 CHAPTER 22 \| Heat Engines, Entropy, and the Second Law of Thermodynamics Challenge Problems 71. A 1.00-mol sample of an ideal gas (g 5 1.40) is carried through the Carnot cycle described in Active Figure 22.10. At point A, the pressure is 25.0 atm and the temperature is 600 K. At point C, the pressure is 1.00 atm and the temper-ature is 400 K. (a) Determine the pressures and volumes at points A, B, C, and D. (b) Calculate the net work done per cycle. 72. The compression ratio of an Otto cycle as shown in Active Figure 22.12 is VA/VB 5 8.00. At the beginning A of the compression process, 500 cm3 of gas is at 100 kPa and 20.0°C. At the beginning of the adiabatic expansion, the temperature is TC 5 750°C. Model the working fluid as an ideal gas with g 5 1.40. (a) Fill in this table to follow the states of the gas: T (K) P (kPa) V (cm3) A 293 100 500 B C 1 023 D (b) Fill in this table to follow the processes: Q W DEint A S B B S C C S D D S A ABCDA (c) Identify the energy input \|Q h\|, (d) the energy exhaust \|Q c\|, and (e) the net output work Weng. (f) Calculate the thermal efficiency. (g) Find the number of crankshaft rev-olutions per minute required for a one-cylinder engine to have an output power of 1.00 kW 5 1.34 hp. Note: The ther-modynamic cycle involves four piston strokes. the gas by heat, (c) the energy exhausted from the gas by heat, and (d) the efficiency of the cycle. (e) Explain how the efficiency compares with that of a Carnot engine oper-ating between the same temperature extremes. 68. A system consisting of n moles of an ideal gas with molar specific heat at constant pressure CP undergoes two reversible processes. It starts with pressure Pi and vol-ume Vi, expands isothermally, and then contracts adiabati-cally to reach a final state with pressure Pi and volume 3Vi. (a) Find its change in entropy in the isothermal process. (The entropy does not change in the adiabatic process.) (b) What If? Explain why the answer to part (a) must be the same as the answer to Problem 65. (You do not need to solve Problem 65 to answer this question.) 69. A sample of an ideal gas expands isothermally, doubling in volume. (a) Show that the work done on the gas in expanding is W 5 2nRT ln 2. (b) Because the inter-nal energy Eint of an ideal gas depends solely on its tem-perature, the change in internal energy is zero during the expansion. It follows from the first law that the energy input to the gas by heat during the expansion is equal to the energy output by work. Does this process have 100% efficiency in converting energy input by heat into work output? (c) Does this conversion violate the second law? Explain. 70. Why is the following situation impossible? Two samples of water are mixed at constant pressure inside an insulated container: 1.00 kg of water at 10.0°C and 1.00 kg of water at 30.0°C. Because the container is insulated, there is no exchange of energy by heat between the water and the envi-ronment. Furthermore, the amount of energy that leaves the warm water by heat is equal to the amount that enters the cool water by heat. Therefore, the entropy change of the Universe is zero for this process.
13833	https://www.thoughtco.com/ounces-to-grams-conversion-example-609317	Converting Ounces to Grams Skip to content Menu Home Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Search Close Search the site GO Science, Tech, Math Science Math Social Sciences Computer Science Animals & Nature Humanities History & Culture Visual Arts Literature English Geography Philosophy Issues Languages English as a Second Language Spanish French German Italian Japanese Mandarin Russian Resources For Students & Parents For Educators For Adult Learners About Us Contact Us Editorial Guidelines Privacy Policy Science, Tech, Math› Science› Chemistry› Converting Ounces to Grams Print Chocolate.Dave King Dorling Kindersley, Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. Learn about ourEditorial Process Updated on May 28, 2019 Close This worked example problem demonstrates how to convert ounces to grams. This is a common type of mass unit conversion problem. One of the most common practical reasons to know how to do this conversion is for recipes, so let's start with a food example: Ounces To Grams Problem A chocolate bar weighs 12 ounces. What is its weight in grams? Solution One of the easiest ways to solve this problem is to use the pound to kilogram conversion. If you like in a country where both units are used, this is a useful conversion to know. Start by converting ounces into pounds. Then convert the pounds into kilograms. All that remains is to move the decimal point three places to the right to convert kilograms into grams. Here are the conversions you need to know: 16 oz = 1 lb 1 kg = 2.2 lbs 1000 g = 1 kg You are solving for "x" numbers of grams. First, convert ounces into pounds. The next part of the solution converts pounds to kilograms, while the final section converts kilograms to grams. Note how units cancel each other out, so all you are left with is grams. x g = 12 oz x g = 12 oz x (1 lb/16 oz) x (1 kg/2.2 lb) x (1000 g/1 kg) x g = 340.1 g Answer The 12 oz chocolate bar weighs 340.1 g. Cite this Article Format mlaapachicago Your Citation Helmenstine, Anne Marie, Ph.D. "Converting Ounces to Grams." ThoughtCo, Jun. 25, 2024, thoughtco.com/ounces-to-grams-conversion-example-609317.Helmenstine, Anne Marie, Ph.D. (2024, June 25). Converting Ounces to Grams. Retrieved from Helmenstine, Anne Marie, Ph.D. "Converting Ounces to Grams." ThoughtCo. (accessed September 29, 2025). copy citation Sponsored Stories Do More with Your Browser - Download Chrome for Windows - The Official Browser of Google google.com Jeff Bezos Says the 1-Hour Rule Makes Him Smarter. 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13834	https://math.stackexchange.com/questions/83023/reverse-a-mathematics-calculation	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Reverse a mathematics calculation Ask Question Asked Modified 13 years, 10 months ago Viewed 3k times 0 $\begingroup$ I have a formula to do the calculation from 7 set of numbers to 4 digits, for example: 04, 05, 19, 21, 22, 31, 13 ====> 4382 These 7 sets of numbers are ranging from 01 to 45, each set will only exists once. Calculation are as follow: Sum up 1st to 6th number, and times with 2 (4 + 5 + 19 + 21 + 22 + 31) 2 = 204 2. Take the result of step 1, minus 1st number and 6th number 204 - 4 - 31 = 169 3. Take the result of step 2, plus with the 7th number 169 + 13 = 182 From here, take the last 2 digits of the result from step 3 > 82 Sum up the 4th and 5th number 21 + 22 = 43 From here, take the last digit of the result from step 4 > 3 Sum up the 2nd and 3rd number 5 + 19 =24 From here, take the last digit of the result from step 5 > 4 There you get the final result of 4382 If any result from any step above contains 0 (zero), take that as a value in one of the digit into the final result. So anyone has any idea if I have "4382" with me, can you get back the 7 sets of numbers? Not necessary must be exactly the same as I've written on top, as long as the 7 sets of numbers can produce the final result into "4382". Really need a master to gimme some clues. Thanks. 1st to 6th number must be in a sequence from smallest to largest number in the range of 01 to 45. It's a game that comes out with this calculation, and I need to find out the way to reverse it. elementary-number-theory Share edited Nov 17, 2011 at 11:59 J. M. ain't a mathematician 76.7k88 gold badges222222 silver badges347347 bronze badges asked Nov 17, 2011 at 4:38 Renise SachikoRenise Sachiko $\endgroup$ 1 2 $\begingroup$ Do you mind sharing the origin of this problem? It's very ill-posed, so I'm curious as to how it arose. $\endgroup$ PengOne – PengOne 2011-11-17 05:05:11 +00:00 Commented Nov 17, 2011 at 5:05 Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ Your problem sets out to solve the following system of modular equations: x1 + 2x2 + 2x3 + 2x4 + 2x5 + x6 + x7 = A (mod 100) x4 + x5 = B (mod 10) x2 + x3 = C (mod 10) The first equivalence comes from steps 1-3 above, the second from step 4 and the third from step 5. The basic math that you need to solve a system of equivalences is the Chinese Remainder Theorem. This system will not have a unique solution. To see how redundant it is, count the possibilities for the sequence (x1,x2,x3,x4,x5,x6,x7) and (C,B,A) separately. For the former, x1,x6,x7 are interchangeable as are x4,x5 and x2,x3, so we count them separately. There are (45+3-1 choose 3) ways to choose x1,x6,x7 each between 01 and 45 (remember to choose with repetition), and (45+2-1 choose 2) ways to choose each of the other pairs. This gives a total of (47 choose 3)(46 choose 2)^2 = 16 782 525 ways of choosing the xs. For the 4-digit sequence, there are 10 000 possibilities. The number above is far larger, so certainly there is a lot of repetition. If you want to pin down just one x sequence that yields a given CBA sequence, then you can simplify the problem by choosing arbitrary values for x6,x7,x5,x3 and then using the CRT to solve for x1,x4,x2 (which may or may not exist). For example, suppose you take x3 = 19, x5 = 22, x6 = 31, x7 = 13 Then you solve for the remaining xs by solving this system x4 + 22 = 3 (mod 10) x2 + 19 = 4 (mod 10) x1 + 2x2 + 38 + 2x4 + 44 + 31 + 13 = 82 (mod 100) The first two equations are easy, though the solutions are not unique x4 = 1 (mod 10) => x4 = 1, 11, 21, 31, or 41 x2 = 5 (mod 10) => x2 = 5, 15, 25, 35, or 45 Adding in the final equation to solve for x1 and pin down x2,x4 gives x1 + 2(x2 + x4) = 56 (mod 100) which has many solutions, among them x1 = 4, x2 = 5, x4 = 21, but also x1 = 44, x2 = 5, x4 = 1 as well as many others. Share answered Nov 17, 2011 at 4:51 PengOnePengOne 18911 silver badge22 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ The easiest way to find a solution to this would be to reverse your operations. However, this is actually not possible with the given algorithm. Take, for example your last step, " take the last digit of the result of the sum of the 2nd and 3rd number". since there are multiple possible combinations of numbers that will give you this result, the step is irreversible. If your last digit is 4 as in your example, the 2nd and 3rd number could be any of the following: 21 + 23 = 44, 23 + 21 = 24, 32 + 02 = 34, 02 + 32 = 34, etc. What do you need this for? Hopefully not some sort of home-baked encryption/security mechanism :) Share answered Nov 17, 2011 at 4:53 tjsciencetjscience $\endgroup$ 1 $\begingroup$ @PengOne Not sure how this has anything to do with the Chinese Remainder Theorem, but even it is does, the question was how to get back to the 7 number sequence from the 4 digit number, which your answer does not mention. $\endgroup$ tjscience – tjscience 2011-11-17 05:00:40 +00:00 Commented Nov 17, 2011 at 5:00 Add a comment \| You must log in to answer this question. 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13835	https://www.hopkinsmedicine.org/health/conditions-and-diseases/infectious-esophagitis	Skip to Main Content Health Infectious Esophagitis Esophagitis is swelling and irritation of your esophagus. The esophagus is the tube you use to swallow. It connects the back of your throat to your stomach. The most common cause of swelling and irritation of the esophagus is stomach acid that flows back into your esophagus. But infections can also cause this swelling and irritation. Fungi, yeast, viruses, and bacteria can all set off the condition, called infectious esophagitis. Anyone can get it, but you are more likely to develop it if your immune system is weakened. Causes Infectious esophagitis can be caused by fungi, yeast, viruses, and bacteria. Symptoms These are symptoms of infectious esophagitis: Pain when swallowing Difficulty swallowing Mouth pain Chest pain Nausea or vomiting Chills or fever Risk factors People with a normal immune system are unlikely to get infectious esophagitis. If you have a medical condition or are undergoing treatment that weakens your immune system, you could be at risk. These conditions put you at risk: HIV/AIDS Cancer treatments, including chemotherapy and radiation treatments Diabetes Bone marrow or stem-cell transplant treatment Medicines that depress the immune system, such as steroids or medicines taken after an organ transplant Long-term antibiotic use Medicines that limit how much stomach acid you produce Alcohol abuse Advanced age can also make you more likely to get it. Diagnosis Your healthcare provider may suspect infectious esophagitis if you have symptoms of esophagitis along with a condition that weakens the immune system. To make a diagnosis, your healthcare provider may order certain tests: Endoscopy. During this outpatient procedure, the gastroenterologist passes a thin, flexible scope through your mouth to examine your esophagus. He or she might take swabs and scrapings to find the cause of an infection if he or she sees signs, such as white patches, fluid-filled blisters, or sores in your esophagus. Blood work. Your healthcare provider may test your blood for viruses that can cause infectious esophagitis, such as herpes simplex virus. Treatment If you have a healthy immune system, your infection may clear on its own without treatment. How infectious esophagitis is treated often depends on the cause: Esophagitis caused by a fungus called Candida. This fungus may be treated with an antifungal medicine called fluconazole or other similar medicines. Viral esophagitis may be treated with antiviral medicines, such as acyclovir. Bacterial esophagitis may be treated with broad-spectrum antibiotics. These are medicines that work against many types of bacteria. Sometimes acid blockers are used along with other treatments. Complications Complications are unusual unless you have a condition or disease that weakens your immune system. Complications may include: Infection that spreads to other parts of your body Scar tissue that forms in the esophagus and causes a narrowing Ulcers in the esophagus that lead to bleeding A hole in the esophagus called a perforation or fistula When to call the healthcare provider Call your healthcare provider if you have any questions about your medicines or any other aspect of your treatment. Let your healthcare provider know right away if you have: Increased difficulty swallowing Pain with swallowing Symptoms of infection, such as chills or fever Chest pain or difficulty breathing Living with infectious esophagitis While you are recovering from infectious esophagitis, work closely with your healthcare provider and keep all your follow-up appointments. If you have ongoing symptoms of painful or difficult swallowing, your healthcare provider may suggest that you take these steps: Stop smoking. Avoid alcohol and caffeine. Avoid over-the-counter medicines that may irritate your esophagus, such as aspirin or ibuprofen. Avoid foods or beverages that give you heartburn. Lose weight if you are overweight. Eat smaller meals more often. Avoid eating for 3 hours before you go to bed. Avoid sleeping in a flat position. Elevate the head of your bed several inches. Specializing In: Indigestion Gastroenterology Gastroenterology and Hepatology Pediatric Gastroenterology, Nutrition & Hepatology (Johns Hopkins All Children's Hospital) Find Additional Treatment Centers at: Howard County Medical Center Sibley Memorial Hospital Suburban Hospital Related Gas in the Digestive Tract Understanding Viral Gastroenteritis Understanding Fecal Incontinence Lactose Intolerance Request an Appointment Find a Doctor Find a Doctor Related Gas in the Digestive Tract Understanding Viral Gastroenteritis Understanding Fecal Incontinence Related Topics Stomach and Gut
13836	https://www.aafp.org/pubs/afp/issues/2024/0600/salivary-gland-disorders.pdf	Scheduled maintenance is planned for September 26–29. You may experience brief interruptions during this time. MICHAEL J. KIM, MD, ANNA MILLIREN, DO, AND DENNIS J. GEROLD, JR., MD Am Fam Physician. 2024;109(6):550-559 Author disclosure: No relevant financial relationships. The major salivary glands are the paired parotid, submandibular, and sublingual glands. Salivary gland disorders can affect the glandular tissue or its excretory system. The parotid glands are the largest and produce aqueous serous secretions that are less immunogenic. They are more susceptible to infections and neoplasms. The submandibular glands produce mucinous secretions that are high in calcium and phosphate salts through a long submandibular duct that flows against gravity. The submandibular glands are responsible for more than 80% of salivary stones. Sialadenitis can be acute or chronic and caused by bacterial, viral, and obstructive etiologies; the most common bacteria is Staphylococcus aureus. The most common viral etiologies in children are mumps (globally) and juvenile recurrent parotitis (in vaccinated populations). Sialadenosis is a chronic asymptomatic enlargement of the salivary glands due to systemic disease. Sialolithiasis causes up to 50% of salivary gland disorders. It is associated with salivary stasis and inflammation caused by dehydration, malnutrition, medications, or chronic illness. Obstruction is also caused by trauma, stenosis, and mucoceles. Neoplasms are rare and typically benign, but they warrant referral and imaging with ultrasonography, computed tomography, or magnetic resonance sialography. Most disorders are managed with conservative measures by treating the underlying etiology, optimizing predisposing factors, controlling pain, and increasing salivary flow with sialagogues, hydration, massage, warm compresses, oral hygiene, and medication adjustment. Sialendoscopy is a gland-sparing technique that can treat obstructive and nonobstructive disorders. (Am Fam Physician. 2024;109(6):550-559. Copyright © 2024 American Academy of Family Physicians.) Subscribe Issue Access Article Only Mehanna H, McQueen A, Robinson M, et al. Salivary gland swellings. Clin Otolaryngol. 2013;38(1):58-65. Kraaij S, Karagozoglu KH, Forouzanfar T, et al. Salivary stones: symptoms, aetiology, biochemical composition and treatment. Br Dent J. 2014;217(11):E23-E23. Hammett JT, Walker C. Sialolithiasis. StatPearls. Updated September 26, 2022. Accessed August 13, 2023. Fattahi TT, Lyu PE, Van Sickels JE. Management of acute suppurative parotitis. J Oral Maxillofac Surg. 2002;60(4):446-448. Ugga L, Ravanelli M, Pallottino AA, et al. Diagnostic work-up in obstructive and inflammatory salivary gland disorders. Acta Otorhinolaryngol Ital. 2017;37(2):83-93. Wood J, Toll EC, Hall F, et al. Juvenile recurrent parotitis: review and proposed management algorithm. Int J Pediatr Otorhinolaryngol. 2021;142:110617. Mandel L. Salivary gland disorders. Med Clin North Am. 2014;98(6):1407-1449. Geiger JL, Ismaila N, Beadle B, et al. Management of salivary gland malignancy: ASCO guideline. J Clin Oncol. 2021;39(17):1909-1941. Peraza A, Gómez R, Beltran J, et al. Mucoepidermoid carcinoma. An update and review of the literature. J Stomatol Oral Maxillofac Surg. 2020;121(6):713-720. Alves LDB, de Melo AC, Farinha TA, et al. A systematic review of secretory carcinoma of the salivary gland: where are we?. Oral Surg Oral Med Oral Pathol Oral Radiol. 2021;132(4):e143-e152. Ogle OE. Salivary gland diseases. Dent Clin North Am. 2020;64(1):87-104. Armstrong MA, Turturro MA. Salivary gland emergencies. Emerg Med Clin North Am. 2013;31(2):481-499. Wilson KF, Meier JD, Ward PD. Salivary gland disorders. Am Fam Physician. 2014;89(11):882-888. Lau RK, Turner MD. Viral mumps: increasing occurrences in the vaccinated population. Oral Surg Oral Med Oral Pathol Oral Radiol. 2019;128(4):386-392. Riley P, Glenny AM, Hua F, et al. Pharmacological interventions for preventing dry mouth and salivary gland dysfunction following radiotherapy. Cochrane Database Syst Rev. 2017(7):CD012744. Sánchez Barrueco A, González Galán F, Alcalá Rueda I, et al. Incidence and risk factors for radioactive iodine-induced sialadenitis. Acta Otolaryngol. 2020;140(11):959-962. Negrini S, Emmi G, Greco M, et al. Sjögren's syndrome: a systemic auto-immune disease. Clin Exp Med. 2022;22(1):9-25. Puxeddu I, Capecchi R, Carta F, et al. Salivary gland pathology in IgG4-related disease: a comprehensive review. J Immunol Res. 2018;2018:6936727. Bowers EMR, Schaitkin B. Management of mucoceles, sialoceles, and ranulas. Otolaryngol Clin North Am. 2021;54(3):543-551. Key S, Chia C, Hasan Z, et al. Systematic review of prognostic factors in carcinoma ex pleomorphic adenoma. Oral Oncol. 2022;133:106052. Abdel Razek AAK, Mukherji S. Imaging of sialadenitis. Neuroradiol J. 2017;30(3):205-215. Skálová A, Hyrcza MD, Leivo I. Update from the 5th edition of the World Health Organization classification of head and neck tumors: salivary glands. Head Neck Pathol. 2022;16(1):40-53. Wolff A, Joshi RK, Ekström J, et al. A guide to medications inducing salivary gland dysfunction, xerostomia, and subjective sialorrhea: a systematic review sponsored by the World Workshop on Oral Medicine VI. Drugs R D. 2017;17(1):1-28. Friedman E, Cai Y, Chen B. Imaging of major salivary gland lesions and disease. Oral Maxillofac Surg Clin North Am. 2023;35(3):435-449. Haran S, Kazi S, Caldera S. Presentation is key to diagnosing salivary gland disorders. J Fam Pract. 2019;68(8):E1-E7. Fisher J, Monette DL, Patel KR, et al. COVID-19 associated parotitis. Am J Emerg Med. 2021;39:254.e1-254.e3. Stafford JA, Moore CA, Mark JR. Acute sialadenitis associated with 2017–2018 influenza A infection: a case series. Laryngoscope. 2018;128(11):2500-2502. Lim ZY, Ang AXY, Cross GB. COVID-19 associated parotitis. IDCases. 2021;24:e01122. Nkuna T, Maharaj S, Hari K. Benign lymphoepithelial cyst of parotid glands in HIV infected patients on anti-retroviral therapy: a narrative review. Indian J Otolaryngol Head Neck Surg. 2023;75(2):547-556. Lee E, Badger C, Thakkar PG. Otorhinolaryngology manifestations of systemic illness. Med Clin North Am. 2021;105(5):871-883. Davis AB, Hoffman HT. Management options for sialadenosis. Otolaryngol Clin North Am. 2021;54(3):605-611. Troeltzsch M, Pache C, Probst FA, et al. Antibiotic concentrations in saliva: a systematic review of the literature, with clinical implications for the treatment of sialadenitis. J Oral Maxillofac Surg. 2014;72(1):67-75. Erkul E, Gillespie MB. Sialendoscopy for non-stone disorders: the current evidence. Laryngoscope Investig Otolaryngol. 2016;1(5):140-145. Prasad RS. Parotid gland imaging. Otolaryngol Clin North Am. 2016;49(2):285-312. Diebold S, Overbeck M. Soft tissue disorders of the mouth. Emerg Med Clin North Am. 2019;37(1):55-68. Thomas WW, Douglas JE, Rassekh CH. Accuracy of ultrasonography and computed tomography in the evaluation of patients undergoing sialendoscopy for sialolithiasis. Otolaryngol Head Neck Surg. 2017;156(5):834-839. Goncalves M, Schapher M, Iro H, et al. Value of sonography in the diagnosis of sialolithiasis: comparison with the reference standard of direct stone identification. J Ultrasound Med. 2017;36(11):2227-2235. Kim DH, Kang JM, Kim SW, et al. Utility of ultrasonography for diagnosis of salivary gland sialolithiasis: a meta-analysis. Laryngoscope. 2022;132(9):1785-1791. Purcell YM, Kavanagh RG, Cahalane AM, et al. The diagnostic accuracy of contrast-enhanced CT of the neck for the investigation of sialolithiasis. AJNR Am J Neuroradiol. 2017;38(11):2161-2166. Bachesk AB, Bin LR, Iwaki IV, et al. Ranula in children: retrospective study of 25 years and literature review of the plunging variable. Int J Pediatr Otorhinolaryngol. 2021;148:110810. Jain P, Jain R, Morton RP, et al. Plunging ranulas: high-resolution ultrasound for diagnosis and surgical management. Eur Radiol. 2010;20(6):1442-1449. Isa AY, Hilmi OJ. An evidence based approach to the management of salivary masses. Clin Otolaryngol. 2009;34(5):470-473. Strychowsky JE, Sommer DD, Gupta MK, et al. Sialendoscopy for the management of obstructive salivary gland disease: a systematic review and meta-analysis. Arch Otolaryngol Head Neck Surg. 2012;138(6):541-547. Gillespie MB, O'Connell BP, Rawl JW, et al. Clinical and quality-of-life outcomes following gland-preserving surgery for chronic sialadenitis. Laryngoscope. 2015;125(6):1340-1344. Fabie JE, Kompelli AR, Naylor TM, et al. Gland-preserving surgery for salivary stones and the utility of sialendoscopes. Head Neck. 2019;41(5):1320-1327. Ramsha A, Keskool P, Ongard S, et al. Outcome of the management of salivary gland diseases by sialendoscopy: a university hospital's experience. J Oral Maxillofac Surg. 2023;81(3):344-349. Agarwal AK, Kanekar SG. Imaging of submandibular and sublingual salivary glands. Neuroimaging Clin N Am. 2018;28(2):227-243. Afzelius P, Nielsen MY, Ewertsen C, et al. Imaging of the major salivary glands. Clin Physiol Funct Imaging. 2016;36(1):1-10. Schmidt RL, Hall BJ, Wilson AR, et al. A systematic review and meta-analysis of the diagnostic accuracy of fine-needle aspiration cytology for parotid gland lesions. Am J Clin Pathol. 2011;136(1):45-59. McQuone SJ. Acute viral and bacterial infections of the salivary glands. Otolaryngol Clin North Am. 1999;32(5):793-811. Salum FG, Medella-Junior FAC, Figueiredo MAZ, et al. Salivary hypofunction: an update on therapeutic strategies. Gerodontology. 2018;35(4):305-316. Brook I. Acute bacterial suppurative parotitis: microbiology and management. J Craniofac Surg. 2003;14(1):37-40. Di Pietrantonj C, Rivetti A, Marchione P, et al. Vaccines for measles, mumps, rubella, and varicella in children. Cochrane Database Syst Rev. 2021(11):CD004407. Marlow MA, Marin M, Moore K, et al. CDC guidance for use of a third dose of MMR vaccine during mumps outbreaks. J Public Health Manag Pract. 2020;26(2):109-115. Su SB, Chang HL, Chen KT. Current status of mumps virus infection: epidemiology, pathogenesis, and vaccine. Int J Environ Res Public Health. 2020;17(5):1686. Nouraei SAR, Ismail Y, Ferguson MS, et al. Analysis of complications following surgical treatment of benign parotid disease. ANZ J Surg. 2008;78(3):134-138. Benaim E, Fan T, Dash A, et al. Common characteristics and clinical management recommendations for juvenile recurrent parotitis: a 10-year tertiary center experience. OTO Open. 2022;6(1) ): 2473974X221077874. Garcia Garcia B, Dean Ferrer A, Diaz Jimenez N, et al. Bilateral parotid sialadenosis associated with long-standing bulimia: a case report and literature review. J Maxillofac Oral Surg. 2018;17(2):117-121. Moorthy A, Bachalli PS, Krishna S, et al. Sialendoscopic management of obstructive salivary gland pathology: a retrospective analysis of 236 cases. J Oral Maxillofac Surg. 2021;79(7):1474-1481. Van Cleemput T, Vanpoecke J, Coropciuc R, et al. Sialendoscopy: a four-year single center experience. J Oral Maxillofac Surg. 2021;79(11):2285-2291. Badash I, Raskin J, Pei M, et al. Contemporary review of submandibular gland sialolithiasis and surgical management options. Cureus. 2022;14(8):e28147. Kolomvos N, Kalfarentzos E, Papadogeorgakis N. Surgical treatment of plunging ranula: report of three cases and review of literature. Oral Maxillofac Surg Cases. 2019;5(1):100098. Continue Reading More in AFP More in PubMed Copyright © 2024 by the American Academy of Family Physicians. This content is owned by the AAFP. A person viewing it online may make one printout of the material and may use that printout only for his or her personal, non-commercial reference. This material may not otherwise be downloaded, copied, printed, stored, transmitted or reproduced in any medium, whether now known or later invented, except as authorized in writing by the AAFP. See permissions for copyright questions and/or permission requests. Copyright © 2025 American Academy of Family Physicians. All Rights Reserved.
13837	https://www.scientificlib.com/en/Mathematics/Geometry/SimsonLine.html	Simson line . Simson line In geometry, given a triangle and a point on its circumcircle, the intersections formed when lines are constructed from the point perpendicular to each of the triangle's sides are collinear. The line through these points is the Simson line, named for Robert Simson. The converse is also true; if the feet of the perpendiculars dropped from a point to the sides of the triangle are collinear, then the point is on the circumcircle. The Simson line of a point is just the pedal triangle of it; the case when that pedal triangle degenerates to a line. The Simson line LNM of the triangle ABC. Properties The Simson line of a vertex of the triangle is the altitude of the triangle dropped from that vertex, and the Simson line of the point diametrically opposite to the vertex is the side of the triangle opposite to that vertex. If P and P' are points on the circumcircle, then the angle between the Simson lines of P and P' is half the angle of the arc PP'. In particular, if the points are diametrically opposite, their Simson lines are perpendicular and in this case the intersection of the lines is on the nine-point circle. Let H denote the orthocenter of the triangle ABC, then the Simson line of P bisects the segment PH in a point that lies on the nine-point circle. Given two triangles with the same circumcircle, the angle between the Simson lines of a point P on the circumcircle for both triangles doesn't depend of P. Proof of existence The method of proof is to show that . PBCA is a cyclic quadrilateral, so . PNMA is a cyclic quadrilateral (Thales' theorem), so . Hence . Now PLBN is cyclic, so . Therefore . See also Pedal triangle Simson, Robert Links Simson Line Simson Line at Wolfram Mathworld Geometry Undergraduate Texts in Mathematics Graduate Texts in Mathematics Graduate Studies in Mathematics Mathematics Encyclopedia Retrieved from " All text is available under the terms of the GNU Free Documentation License Hellenica World - Scientific Library × Custom Search
13838	http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html	\| \| \| \| --- \| Early Photoelectric Effect Data Electrons ejected from a sodium metal surface were measured as an electric current. Finding the opposing voltage it took to stop all the electrons gave a measure of the maximum kinetic energy of the electrons in electron volts. The minimum energy required to eject an electron from the surface is called the photoelectric work function. The threshold for this element corresponds to a wavelength of 683 nm. Using this wavelength in the Planck relationship gives a photon energy of 1.82 eV. Further analysis \| \| \| Table of photoelectric effect work functions \| \| Index Millikan reference Photoelectric effect \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| --- \| \| Early Photoelectric Effect Data Planck hypothesis \| Index Millikan reference Photoelectric effect \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| \| \| \| \| --- --- --- \| The Planck HypothesisIn order to explain the frequency distribution of radiation from a hot cavity (blackbody radiation) Planck proposed the ad hoc assumption that the radiant energy could exist only in discrete quanta which were proportional to the frequency. This would imply that higher modes would be less populated and avoid the ultraviolet catastrophe of the Rayleigh-Jeans Law. The quantum idea was soon seized to explain the photoelectric effect, became part of the Bohr theory of discrete atomic spectra, and quickly became part of the foundation of modern quantum theory. \| \| \| \| \| --- --- \| \| Calculation \| Interaction of radiation with matter \| Electromagnetic spectrum \| Basic quantum processes \| \| \| \| Are there limits on the frequency of a photon? \| \| Index Photoelectric effect \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| \| \| \| --- --- --- \| \| Photon Energies for EM Spectrum \| \| \| --- \| \| Planck hypothesis \| Interaction of radiation with matter \| \| \| \| --- \| \| Energies in electron volts \| Thermal energy for comparison \| \| Index \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| \| \| \| --- --- --- \| \| Photons: The Quanta of Light According to the Planck hypothesis, all electromagnetic radiation is quantized and occurs in finite "bundles" of energy which we call photons. The quantum of energy for a photon is not Planck's constant h itself, but the product of h and the frequency. The quantization implies that a photon of blue light of given frequency or wavelength will always have the same size quantum of energy. For example, a photon of blue light of wavelength 450 nm will always have 2.76 eV of energy. It occurs in quantized chunks of 2.76 eV, and you can't have half a photon of blue light - it always occurs in precisely the same sized energy chunks. But the frequency available is continuous and has no upper or lower bound, so there is no finite lower limit or upper limit on the possible energy of a photon. On the upper side, there are practical limits because you have limited mechanisms for creating really high energy photons. Low energy photons abound, but when you get below radio frequencies, the photon energies are so tiny compared to room temperature thermal energy that you really never see them as distinct quantized entities - they are swamped in the background. Another way to say it is that in the low frequency limits, things just blend in with the classical treatment of things and a quantum treatment is not necessary. \| \| \| \| \| --- --- \| \| Calculation \| Interaction of radiation with matter \| Electromagnetic spectrum \| Basic quantum processes \| \| Index Photoelectric effect \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \|
13839	https://mathoverflow.net/questions/51725/a-kind-of-character-sum-concerning-legendre-symbols	Skip to main content A kind of character sum concerning Legendre symbols Ask Question Asked Modified 14 years, 7 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. It is known that ∑p≤x(qp)=o(π(x)) for any q which is not a square. Is there some references on such a character sum (summation over the moduli)? Of course, by quadratic reciprocity law, it can be transformed to consider the following sum ∑p≤x(pq). By Perron's formula and some results of Dirichlet L-functions, we can of course obtain an upper bound. I want to know whether there is certain elementary proof. nt.number-theory Share CC BY-SA 2.5 Improve this question Follow this question to receive notifications asked Jan 11, 2011 at 1:51 arithboyarithboy 23011 silver badge88 bronze badges 6 Also true when q=1 ? – Luis H Gallardo Commented Jan 11, 2011 at 8:31 Sorry q=1 is a square ! – Luis H Gallardo Commented Jan 11, 2011 at 8:33 What are your objections to using Perron's formula? – Micah Milinovich Commented Jan 11, 2011 at 16:04 For example, to estimate the related sum ∑n≤xΛ(n)χ(n), Perron's formula allows us to calculate certain mean value of L-functions to obtain the upper bound of the character sum – arithboy Commented Jan 12, 2011 at 2:10 I found we can apply Dirichlet's PNT in arithmetic progressions to get an easier proof. However, this is not elementary and direct enough. – arithboy Commented Jan 12, 2011 at 2:13 \| Show 1 more comment 2 Answers 2 Reset to default This answer is useful 2 Save this answer. Show activity on this post. In fact, ∑p≤x(pq)=∑amodq(aq)π(x;q,a). For sufficiently large x, we have π(x;q,a)=1φ(q)(1+o(1))π(x), where the o constant depends on q, thus ∑p≤x(pq)=1φ(q)∑amodq(aq)(1+o(1))π(x)=o(π(x)). Of course, I expect a more elementary proof, which doesn't rely on the PNT in arithmetic progressions. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Jan 13, 2011 at 7:08 arithboyarithboy 23011 silver badge88 bronze badges 2 But what makes you think such a thing is possible? The question is so close to PNT-for-AP, especially in the case q=−1 in the answer I posted, that it's hard to imagine an elementary solution that doesn't amount to an elementary proof of PNT-for-AP. And if what you're really after is an elementary proof of PNT-for-AP, well, then that's what you should really be asking for. – Gerry Myerson Commented Jan 13, 2011 at 11:51 Dear Myerson, I quite agree with what you have said. This problem must be related to the distribution of prime numbers, in AP for instance. However, I am looking for an alternative proof for the estimate, not relying on PNT in AP, maybe this requires certain subtle facts on primes. BTW, for the elementary proof for PNT in AP, I know there is one due to Selberg, are there any others? – arithboy Commented Jan 13, 2011 at 13:50 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Let's take the case where q=−1. Then your sum is the difference between the number of primes up to x that are 4n+1 and the number that are 4n−1. I suspect that information on that difference, of the strength you require, is available only via the Prime Number Theorem for arithmetic progressions. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Jan 13, 2011 at 6:11 Gerry MyersonGerry Myerson 40.7k1010 gold badges191191 silver badges256256 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions nt.number-theory See similar questions with these tags. Related 4 Estimate on an exponential sum 7 short character sums averaged on the character 5 Is there a Poisson Summation formula for imprimitive Dirichlet characters? 20 Sum of the reciprocals of radicals 6 Sum over characters 2 Seeking a sharp bound for a Dirichlet character sum Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
13840	https://www.grc.nasa.gov/www/k-12/airplane/pranmyer.html	Prandtl-Meyer Angle + Text Only Site + Non-Flash Version + Contact Glenn As an object moves through a gas, the gas molecules are deflected around the object. If the speed of the object is much less than the speed of sound of the gas, the density of the gas remains constant and the flow of gas can be described by conserving momentum and energy. As the speed of the objects increases towards the speed of sound, we must consider compressibility effects on the gas. The density of the gas varies locally as the gas is compressed by the object. When an object moves faster than the speed of sound, and there is an abrupt decrease in the flow area, shock waves are generated. If the flow area increases, however, a different flow phenomenon is observed. If the increase is abrupt, we encounter a centered expansion fan. There are some marked differences between shock waves and expansion fans. Across a shock wave, the Mach number decreases, the static pressure increases, and there is a loss of total pressure because the process is irreversible. Through an expansion fan, the Mach number increases, the static pressure decreases and the total pressure remains constant. Expansion fans are isentropic. The calculation of the expansion fan involves the use of the Prandtl-Meyer function. This function is derived from conservation of mass, momentum, and energy for very small (differential) deflections. The Prandtl-Meyer function is denoted by the Greek letter nu on the slide and is a function of the Mach numberM and the ratio of specific heatsgam of the gas. nu = {sqrt[(gam+1)/(gam-1)]} atan{sqrt[(gam-1)(M^2 - 1)/(gam+1)]} - atan{sqrt[M^2 -1]} where atan is the trigonometric inverse tangent function. It is also written as shown on the slide tan^-1. The meaning of atan can be explained by these two equations: atan(a) = b tan(b) = a As mentioned above, the Mach number of a supersonic flow increases through an expansion fan. The amount of the increase depends on the incoming Mach number and the angle of the expansion. The physical interpretation of the Prandtl-Meyer function is that it is the angle through which you must expand a sonic (M=1) flow to obtain a given Mach number. The value of the Prandtl-Meyer function is therefore called the Prandtl-Meyer angle. Here's a JavaScript program which solves for the Prandtl-Meyer angle . Isentropic Flow Calculator ========================== Gamma Compute Input Output Mach Mach Angle P-M Ang p/pt T/Tt rho/rhot q/p A/A Wcor/A You select an input variable by using the choice button labeled Input Variable. Next to the selection, you then type in the value of the selected variable. When you hit the red COMPUTE button, the output values change. The default input variable is the Mach number, and by varying Mach number you can see the effect on Prandtl-Meyer angle. You can also select Prandtl-Meyer angle as an input, and see its effect on the flow variables downstream of the turning. If you are an experienced user of this calculator, you can use a sleek version of the program which loads faster on your computer and does not include these instructions. You can also download your own copy of the program to run off-line by clicking on this button: Activities: Guided Tours Compressible Aerodynamics: Isentropic Flow Calculator: Multiple Shock Wave Simulator: Navigation .. Beginner's Guide Home Page + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility CertificationEditor: Nancy Hall NASA Official: Nancy Hall Last Updated: May 13 2021 + Contact Glenn
13841	https://www.studocu.com/en-us/document/stanford-university/integral-calculus/roots-of-unity-filter-1-saa/113835243	Roots of Unity Filter - MATH 101 Lecture Notes - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Roots of Unity Filter - MATH 101 Lecture Notes sa,a Original title: Roots of Unity Filter 1 - sa,a Course Integral Calculus (MATH 20) 317 documents University Stanford University Academic year:2023/2024 Uploaded by: AM Aniket Mangalampalli Stanford University 0 followers 1 Uploads1 upvotes Follow Recommended for you 27 Integral Calculus - Lecture notes - 1 - 11 Integral Calculus Lecture notes 95% (73) 2 Lecture notes, lecture 1 Integral Calculus Lecture notes 100% (6) Comments Please sign in or register to post comments. Report Document Students also viewed The Mathematical Association of Victoria 2018 Df530x maintenance manual 2017 en Gene Linkage & Chromosome Mapping Study Guide 7 Math 10550: Supplemental Problems for Exam 3 (Nov 15, 2011) Math 10550 - Practice Final Exam with Questions and Answers Honors Algebra 2 Final Exam KEY: Direct & Inverse Variation Practice Related documents MATH 486478072 - Quiz on Arithmetic Sequences and Patterns 256622620 Grade 5 Math Worksheets: Fractions MCQ & SAQ Mathematics Intervention Program (469898145): Strategies & Evaluation MATH5133 - Summary Notes by Emma (Submission on Nov 11, 2020) Math Concepts Review & Practice - Intermediate Stats & Algebra Asimptote - Teorii și Exerciții (Bac2010) Preview text Roots of Unity Filter Jessica Yan June 15, 2024 1 Introduction Roots of unity filter is a method used to find the sum of every nth coefficient of a polynomial. This can be helpful in generating functions or counting problems such as finding the number of ways to choose a multiple of five number of balls from a basket. Generating Functions Probability can be represented by a polynomial called a generating function, where each term represents one event. f (x) = p 0 xq 0 + p 1 xq 1 + ⋯ pi is the probability the event will occur and qi is the event itself. The nth roots of unity are the solutions to xn = 1. nth roots of unity x = e 2 πkin , k = 0 , 1 , 2 ⋯n − 1 Geometrically, they are equally spaced out on the unit circle and form a regular polygon when connected. Roots of Unity Filter Given P (x) = a 0 + a 1 x + a 2 x 2 + ⋯, ∑ k= 0 akn = P ( 1 ) + P (w) + P (ω 2 ) + ⋯ + P (ωn− 1 ) n , where ω is an nth root of unity, e 2 nπi . We will consider a term and whether its degree is a multiple of n or not. When a term’s exponent is a multiple of n, the summation for that term is akn( 1 kn + ωkn + ω 2 kn + ⋯ + ω(n− 1 )kn), by definition, ω is an nth root of unity, so ωn = 1 , and the expression simplifies to akn( 1 + 1 + ⋯ + 1 ·„„„„„„„„„„„„„„„„„„„„„„„„„„„„„‚„„„„„„„„„„„„„„„„„„„„„„„„„„„„„¶ n ) = nakn. Adding these for all our possible values of k, we get n ∑k= 0 akn. When the power is not a multiple of n, the summation would be am( 1 m + ωm + ω 2 m + ⋯ + ω(n− 1 )m), Contrary to the first case, we can use the sum of geometric series formula, since the denominator would be nonzero by definition. We would get am(ωmn − 1 ) ωm − 1 = 0. Therefore, the coefficients we don’t want are multiplied by 0 and eliminated, and we are left with n times of the LHS of the formula, finishing the proof. 2 Examples Here are a couple of examples where we can apply roots of unity filter! Example 1. Find the sum of coefficients of a real polynomial P (x) given that P (− 1 ) = 2 , P (i) = 3 , P (−i) = 5 , and the sum of coefficients of terms with degree multiple of 4 is 9. The sum of coefficients of a polynomial is also P ( 1 ). We can now use roots of unity filter because 1 , − 1 , i, and −i are the fourth roots of unity. P ( 1 ) + P (− 1 ) + P (i) + P (−i) 4 = 9 Ô⇒ P ( 1 ) + 2 + 3 + 5 4 = 9 Ô⇒ P ( 1 ) = 26. Example 1. Compute ( 2024 1 ) + ( 2024 4 ) + ⋯ + ( 2024 2023 ) We recognize these terms in the binomial expansion P (x) = (x + 1 ) 2024 = ( 2024 0 )x 0 + ( 2024 1 )x 1 + ⋯ + ( 2024 2023 )x 2023 + ( 2024 2024 )x 2024. The sum we want isn’t the sum of every multiple of three, but we can shift the polynomial to our liking. Q(x) = x 2 (x + 1 ) 2024 = 0 x 0 + 0 x 1 + ( 2024 0 )x 2 + ( 2024 1 )x 3 + ⋯ + ( 2024 2023 )x 2025 + ( 2024 2024 )x 2026. Now, we can apply roots of unity filter to sum the coefficients of terms with degrees divisible by 3. Roots of Unity Filter - MATH 101 Lecture Notes Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 5 0 Save Roots of Unity Filter - MATH 101 Lecture Notes Course: Integral Calculus (MATH 20) 317 documents University: Stanford University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 5 0 Save Ro ots of Unit y Filter Jessica Y an June 15,2024 1 In tro duction Ro ots of unity filter is a method used to find the su m of ev ery n th c o efficien t of a p oly nomial.This can b e helpful in generating functions or counting problems suc h as fin ding the n umber of w ays to c ho ose a multiple of fiv e num b er of balls from a basket. Generating F unctions Probability can be represented b y a p olynomial called a generating func tion,where each term represents one ev ent. f(x)=p 0 x q 0+p 1 x q 1+⋯ p i is the probability the ev ent will occur and q i is the even t itself. The n th ro ots of unity are the solutions to x n=1. n th ro ots of un it y x=e 2 πk i n,k=0,1,2⋯n−1 Geometrically,they are equally spaced out on the unit circle and form a re gular p olygon when connected. Ro ots of Unity Filter Giv en P(x)=a 0+a 1 x+a 2 x 2+⋯, ∑ k=0 a kn=P(1)+P(w)+P(ω 2)+⋯+P(ω n−1) n, where ω is an n th ro ot of unity,e 2 πi n. W e will consider a term and whether its degree is a multiple of n or not. When a term’s exp onent is a m ultiple of n,the summation for that term is a kn(1 k n+ω kn+ω 2 k n+⋯+ω(n−1)kn), b y definition,ω is an n th ro ot of u nit y,so ω n=1,and the expression simplifies to a kn(1+1+⋯+1 ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ n )=na kn. Adding these for all our p ossible v alues of k,we get n∑k=0 a kn. When the p ow er is not a multiple of n,the summation w ould b e a m(1 m+ω m+ω 2 m+⋯+ω(n−1)m), Con trary t o the first case,we can use the sum of geometric series form ula,since the denominator w ould b e nonzero by definition.W e w ould get a m(ω mn−1) ω m−1=0. Therefore,the co efficients w e don’t wan t are m ultiplied by 0 and eliminated,and we are left with n times of the LHS of the formula,finishing the proof. 2 Examples Here are a couple of examples where we can apply roots of unity filter! Example 1.1 Find the sum of co efficients of a real polynomial P(x)given that P(−1)=2,P(i)=3, P(−i)=5,and the sum of co efficients of terms with degree m ultiple of 4 is 9. The sum of co efficients of a polynomial is also P(1).W e can now use ro ots of unity filter because 1,−1,i,and−i are the fourth ro ots of unity. P(1)+P(−1)+P(i)+P(−i) 4=9 Ô⇒P(1)+2+3+5 4=9 Ô⇒P(1)=26. Example 1.2 Compute (2024 1)+(2024 4)+⋯+(2024 2023) W e recognize these terms in th e binomial expansion P(x)=(x+1)2024=(2024 0)x 0+(2024 1)x 1+⋯+(2024 2023)x 2023+(2024 2024)x 2024. The sum we w an t isn’t the sum of every m ultipl e of three,but we can shift the polynomial to our liking. Q(x)=x 2(x+1)2024=0 x 0+0 x 1+(2024 0)x 2+(2024 1)x 3+⋯+(2024 2023)x 2025+(2024 2024)x 2026. No w,w e can apply ro ot s of unit y filt er to sum the co efficients of terms with degrees divisible b y 3. Q(1)+Q(ω)+Q(ω 2) 3 =1(1+1)2024+ω 2(1+ω)2024+ω 4(1+ω 2)2024 3 =2 2024+ω 2(−ω 2)2024+ω 4(−ω)2024 3 =2 2024+ω 4050+ω 2028 3 =2 2024+2 3. W e can actually generalize these kinds of sums. Sum of Com binations F or integer c≤b≤a, ∑ i≥1(a bi+c)=1 b∑ ω b=1(1+ω)a ω a−c 3 More Problems Problem 2.1(2021 AMC 12A#15):A choir director m ust select a group of singers from among his 6 tenors and 8 basses.The only requirements are that the difference betw een the num b er of tenors and basses must be a multiple of 4,and the group m ust hav e at least one singer.Let N b e the num b er of groups that can b e selected.What is the remainder when N is divided by 100? Problem 2.2(2018 AIME I#12)F or every subset T of U={1,2,3,...,18},let s(T)b e the sum of the elements of T,with s(∅)defined to be 0.If T is chosen at random among all subsets of U,the probability that s(T)is divisible b y 3 is m n,where m and n are relatively prime positive in tegers.Find m Problem 2.3(2020 CMIMC Combinatorics#5):Sev en cards num b ered 1 through 7 lay stac ked in a pile in ascending order from top to b ottom(1 on top,7 on bottom).A sh uffle inv olves pic king a random card of the six not curr ently on top and putting it on top.The relative order of all the other cards remains unchanged.Find the probability that,after 10 sh uffles,6 is higher in the pile than 3. 1 out of 3 Share Download Download More from:Integral Calculus(MATH 20) More from: Integral CalculusMATH 20Stanford University 317 documents Go to course 27 Integral Calculus - Lecture notes - 1 - 11 Integral Calculus Lecture notes 95% (73) 2 Lecture notes, lecture 1 Integral Calculus Lecture notes 100% (6) 21 Quaternion Rotations of 3D Vectors for Com S 477/577 Final Exam Notes Integral Calculus Lecture notes 100% (4) 4 Syllabus for Math 104: Intro to Analysis - Spring 2024 Integral Calculus Lecture notes 100% (3) More from: Integral CalculusMATH 20Stanford University317 documents Go to course 27 Integral Calculus - Lecture notes - 1 - 11 Integral Calculus 95% (73) 2 Lecture notes, lecture 1 Integral Calculus 100% (6) 21 Quaternion Rotations of 3D Vectors for Com S 477/577 Final Exam Notes Integral Calculus 100% (4) 4 Syllabus for Math 104: Intro to Analysis - Spring 2024 Integral Calculus 100% (3) 29 Unit 6 Homework Keys: Integration & Accumulation of Change Integral Calculus 100% (2) 9 3D geometry & triangles solutions Integral Calculus 100% (2) Recommended for you 27 Integral Calculus - Lecture notes - 1 - 11 Integral Calculus Lecture notes 95% (73) 2 Lecture notes, lecture 1 Integral Calculus Lecture notes 100% (6) 27 Integral Calculus - Lecture notes - 1 - 11 Integral Calculus 95% (73) 2 Lecture notes, lecture 1 Integral Calculus 100% (6) Students also viewed The Mathematical Association of Victoria 2018 Df530x maintenance manual 2017 en Gene Linkage & Chromosome Mapping Study Guide 7 Math 10550: Supplemental Problems for Exam 3 (Nov 15, 2011) Math 10550 - Practice Final Exam with Questions and Answers Honors Algebra 2 Final Exam KEY: Direct & Inverse Variation Practice Related documents MATH 486478072 - Quiz on Arithmetic Sequences and Patterns 256622620 Grade 5 Math Worksheets: Fractions MCQ & SAQ Mathematics Intervention Program (469898145): Strategies & Evaluation MATH5133 - Summary Notes by Emma (Submission on Nov 11, 2020) Math Concepts Review & Practice - Intermediate Stats & Algebra Asimptote - Teorii și Exerciții (Bac2010) Get homework AI help with the Studocu App Open the App English United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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13842	https://mathoverflow.net/questions/185407/clustering-of-periodic-points-for-a-polynomial-iteration-of-mathbbc	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Clustering of periodic points for a polynomial iteration of $\mathbb{C}$ Ask Question Asked Modified 10 years, 11 months ago Viewed 374 times 6 $\begingroup$ Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial map of degree $q > 1$. Consider $E_n \subset \mathbb{C}$ the set of periodic points with period (dividing) $n$; generally, $\|E_n\| = q^n$. Since all the peridic points are limited to a bounded subset of $\mathbb{C}$, we know that there are always two among them within a distance of $O(q^{-n/2})$. Question. (i) How close can two distinct elements of $E_n$ get? In the asymptotic where $n \to \infty$ along a sequence, can it happen that $E_n$ contains a pair of points within a distance of $o(1/q^n)$? (ii) Given any point $z_0 \in \mathbb{C}$ and $C$, does the disk $\|z-z_0\| < C/q^n$ contain only $O_{C,z_0}(1)$ points from $E_n$? complex-dynamics Share Improve this question asked Oct 26, 2014 at 2:58 Vesselin DimitrovVesselin Dimitrov 13.9k33 gold badges5757 silver badges9797 bronze badges $\endgroup$ Add a comment \| 3 Answers 3 Reset to default 9 $\begingroup$ Consider $f(x) = x^2 - 2$. This has the property that $f(2 \cos (x)) = 2 \cos(2x)$, so that $f_n(2 \cos (x)) = 2 \cos(2^n x)$ (where $f_n$ is $f$ iterated $n$ times). Thus $E_n = {2 \cos (x): \cos(x) = \cos(2^n x)}$, and the two greatest members of $E_n$ are both greater than $2 \cos(\pi 2^{1-n})\approx 2 - \pi^2 2^{2-2n}$ Share Improve this answer answered Oct 26, 2014 at 4:16 Robert IsraelRobert Israel 54.7k11 gold badge7979 silver badges156156 bronze badges $\endgroup$ 1 1 $\begingroup$ Thank you! This example clarifies a lot. I think it would be interesting to know also if $O(q^{-2n})$ is as small as the distance can get. $\endgroup$ Vesselin Dimitrov – Vesselin Dimitrov 2014-10-26 04:41:56 +00:00 Commented Oct 26, 2014 at 4:41 Add a comment \| 6 $\begingroup$ The answer to both of your questions is negative, even if you replace $q^{-n}$ by any other sequence $(a_n)$ of natural numbers. The reason is, broadly speaking, that you may have of course have maps having multiple periodic points, and under small perturbations these will yield maps having a cycle of potentially large period nearby. A little bit more precisely, let us consider the family of quadratic polynomials, $f_c(z) = z^2+c$. The following is well-known: Proposition. Suppose that $f_c$ has a periodic point $z_0$ of period $n$, with corresponding multiplier $\mu = (f_c^n)'(z_0) = e^{2\pi i p/q}$, for some $p,q\in\mathbb{N}$. Then, for any neighbourhood $U$ of $z_0$, there is a parameter $c'$ (arbitrarily close to $c$) such that $f_{c'}$ has a periodic point $z'$ of period $nq$ such that $f_{c'}^{kn}(z')\in U$ for $k\geq 0$, and such that $z'$ itself has multiplier $\mu' = (f_{c'}^{nq})'(z') = e^{2\pi i /q'}$ for some (large) integer $q'$. (In addition, the map $f_{c'}$ also has a (repelling) periodic point of period $n$ in $U$.) (What happens is that the parameter $c$ is on the boundary of two components of the interior of the Mandelbrot set, one consisting of points having an attracting orbit of period $n$ and one having an attracting orbit of period $nq$. The perturbation is obtained by passing a little bit along the boundary of the latter component.) Now start e.g. with the map $f_{1/4}$ and $z=1/2$, and apply this proposition inductively. Take a limit of the resulting parameters (taking care to make each perturbation small enough as to not destroy the features already constructed). Clearly in this way we can find a quadratic polynomial having periodic points $(z_j)_{j\geq 0}$ of periods $(n_j)$, where $$ n_j = \prod_{k=1}^j q_j$$. (Here $q_j$ is some rapidly increasing sequence of positive integers obtained in the construction.) Given any sequence $(\varepsilon_n)$, we can inductively carry out the construction so that $q_j$ points of the orbit of $z_j$ are within distance $\varepsilon_{n_j}$ of each other. Letting $\varepsilon_n$ tend to zero faster than $a_n$, the claim follows. Share Improve this answer answered Oct 26, 2014 at 21:13 Lasse RempeLasse Rempe 6,62811 gold badge2929 silver badges4848 bronze badges $\endgroup$ 4 $\begingroup$ Thank you for explaining this! On the other hand, if $f$ is defined over $\bar{\mathbb{Q}}$, we have a diophantine bound and we may not take $a_n$ to decay too fast. Thus the limit value of the parameter $c$ must be transcendental. $\endgroup$ Vesselin Dimitrov – Vesselin Dimitrov 2014-10-26 22:00:15 +00:00 Commented Oct 26, 2014 at 22:00 1 $\begingroup$ @VesselinDimitrov that's an interesting point. The construction is a generalization of the Feigenbaum point (where we make a sequence of period 2 bifurcations, and take the limit). In this case, I believe that it is not known whether the limit parameter is transcendental. (More generally, one could ask whether every infinitely renormalizable parameter is transcendental.) $\endgroup$ Lasse Rempe – Lasse Rempe 2014-10-26 22:38:35 +00:00 Commented Oct 26, 2014 at 22:38 $\begingroup$ Take, say, $a_n := \exp(-5^{n})$. If $c \in \bar{\mathbb{Q}}$, then the points of period $n$ are algebraic numbers of degree $O(2^n)$ and bounded (absolute, multiplicative Weil) height. Then their pairwise differences are algebraic numbers of degrees $O(4^n)$ and still of bounded height. But a non-zero algebraic number of degree $d$ and absolute multiplicative height $H$ is at least $H^{-d}$ in absolute value: this is the Liouville diophantine bound. $\endgroup$ Vesselin Dimitrov – Vesselin Dimitrov 2014-10-26 23:04:09 +00:00 Commented Oct 26, 2014 at 23:04 1 $\begingroup$ @VesselinDimitrov Yes, that's what I took your first comment to mean. Interestingly, using recent work on 'near-parabolic renormalization', it could be possible to quantify the above construction, and hence - combined with your observation - give estimates on the combinatorics (i.e., the numbers $q_j$) that ensure that the resulting parameter is transcendental. That could be a publishable result, although depending on the technical work needed it may or may not be worthwhile to work it out. Davoud Cheraghi (Imperial College) is a world expert on this type of argument. $\endgroup$ Lasse Rempe – Lasse Rempe 2014-10-27 09:22:38 +00:00 Commented Oct 27, 2014 at 9:22 Add a comment \| 5 $\begingroup$ While Robert Israel's answer is correct, $x^2-2$ is a very exceptional case. In a way, how close do they get is not the interesting question: rather, what gets strange is how far apart do the closest ones remain? We know the periodic points are dense in the Julia set, but in the case of weird ones (like the ones with Cremer points, or even some with Siegel disks where the disk itself is very 'deep' within the Julia set, as measured by the external rays), the periodic points tend to avoid certain parts of the Julia set as long as possible. This is what causes the 'inverse method' of rendering images of Julia sets to be so bad for those cases. Share Improve this answer answered Oct 26, 2014 at 14:52 Jacques CaretteJacques Carette 12k44 gold badges4646 silver badges8484 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions complex-dynamics See similar questions with these tags. 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13843	https://pixel-druid.com/articles/median-minimizes-l1-norm	A Universe of Sorts § Median minimizes L1 norm Consider the meadian of x s[1..N]xs[1..N]x s[1..N]. We want to show that the median minimizes the L1 norm L 1(y)=∑i∣x s[i]−y∣L_1(y) = \sum_i \|xs[i] - y\|L 1(y)=∑i∣x s[i]−y∣. If we differentiate L 1(y)L_1(y)L 1(y) with respect to y y y, we get: d L 1(y)/y=∑i−sign(x s[i]−y) d L_1(y)/y = \sum_i - \texttt{sign}(xs[i] - y) d L 1(y)/y=i∑−sign(x s[i]−y) Recall that d(∣x∣)/d x=sign(x)d(\|x\|)/dx = \texttt{sign}(x)d(∣x∣)/d x=sign(x) Hence, the best y y y to minimize the L 1 L_1 L 1 norm is the value that makes the sum of the signs ∑i sign(x s[i]−y)\sum_i \texttt{sign}(xs[i] - y)∑isign(x s[i]−y) minimal. The median is perfect for this optimization problem. When the list has an odd number of elements, say, 2 k+1 2k + 1 2 k+1, k k k elements will have sign −1-1−1, the middle element will have sign 0 0 0, and the k k k elements after will have sign +1+1+1. The sum will be 0 0 0 since half of the −1-1−1 and the +1+1+1 cancel each other out. Similar things happen for even, except that we can get a best total sign distance of +1+1+1 using either of the middle elements. § Proof 2: Math.se has a nice picture proof abot walking from left to right. § Proof 3: Consider the case where x s xs x s has only two elements, with x s<x sxs < xsx s<x s. Then the objective function to minimize the L1 norm, ie, to minimize ∣x s−y∣+∣x s−y∣\|xs - y\| + \|xs - y\|∣x s−y∣+∣x s−y∣. This is satisfied by any point in between x sxsx s and x sxsx s. In the general case, assume that x s<x s⋯<x s[N]xs < xs \dots < xs[N]x s<x s⋯<x s[N]. Pick the smallest number x sxsx s and the largest number x s[N]xs[N]x s[N]. We have that any y y y between x sxsx sand x s[N]xs[N]x s[N] satisfies the condition. Now, drop off x sxsx s and x s[N]xs[N]x s[N], knowing that we must have y∈[x s,x s[N]]y \in [xs, xs[N]]y∈[x s,x s[N]]. Recurse. At the end, we maybe left with a single element x s[k]xs[k]x s[k]. In such a case, we need to minimize ∣x s[k]−y∣\|xs[k] - y\|∣x s[k]−y∣. That is, we set x s[k]=y xs[k] = y x s[k]=y. On the other hand, we maybe left with two elements. In this case, any point between the two elements is a legal element. We may think of this process as gradually "trapping" the median between the extremes, using the fact that that any point y∈[l,r]y \in [l, r]y∈[l,r] minimizes ∣y−l∣+∣y−r∣\|y - l\| + \|y - r\|∣y−l∣+∣y−r∣! Taken from math.se Newer ৪ Blog ৪ Older
13844	https://ocw.mit.edu/courses/18-02sc-multivariable-calculus-fall-2010/42a8b344ffc58e1df8b56144d245fdf1_MIT18_02SC_we_7_comb.pdf	Equation of a Plane 1. Later we will return to the topic of planes in more detail. Here we will content ourself with one example. Find the equation of the plane containing the three points P1 = (1, 3, 1), P2 = (1, 2, 2), P3 = (2, 3, 3). Answer: The − − − → − −→ vectors P1P2 and P1 − P3 are in the plane, so i j k − − − → N = P1P2 P − − 1 − P → × 3 = 0 −1 1 = i(−2) − j(−1) + k(1) = (−2, 1, 1). 1 0 2 is orthogonal to the plane. Now for any point P = (x, y, z) in the plane, the vector P − − 1 → P is also in the plane and is therefore orthogonal to N. Expressing this with the dot product we get − − → N · P1P = 0 ⇔ (−2, 1, 1) · (x − 1, y − 3, z − 1) = 0 ⇔ −2(x − 1) + (y − 3) + (z − 1) = 0 ⇔ −2x + y + z = 2. P2 P = (x, y, z) N P3 P1 The equation of the plane is −2x + y + z = 2. You should check that the three points P1, P2, P3 do, in fact, satisfy this equation. The standard terminology for the vector N is to call it a normal to the plane. MIT OpenCourseWare 18.02SC Multivariable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit:
13845	https://artofproblemsolving.com/wiki/index.php/1985_IMO_Problems/Problem_5?srsltid=AfmBOooNi-4CmuLbB1wxcCiO_4jMl92c54C9g1YNkTjk7IbDR5NM0vPJ	Art of Problem Solving 1985 IMO Problems/Problem 5 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1985 IMO Problems/Problem 5 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1985 IMO Problems/Problem 5 Contents 1 Problem 2 Solution 3 Solution 2 4 Solution 3 (No Miquel's point) 5 Solution 4 (Sparrow solution) 6 See Also Problem A circle with center passes through the vertices and of the triangle and intersects the segments and again at distinct points and respectively. Let be the point of intersection of the circumcircles of triangles and (apart from ). Prove that . Solution is the Miquel Point of quadrilateral , so there is a spiral similarity centered at that takes to . Let be the midpoint of and be the midpoint of . Thus the spiral similarity must also send to and so is cyclic. is also cyclic with diameter and thus must lie on the same circumcircle as , , and so . Solution 2 Let and be the circumcircles and circumcenters of respectively. Let is cyclic The radius of is Let and be midpoints of and respectively. is the Miquel Point of quadrilateral so is cyclic. is trapezium as desired. vladimir.shelomovskii@gmail.com, vvsss Solution 3 (No Miquel's point) Consider and , they are similar because = , and also . Now draw , and intersecting at ; , at . Naturally bisects , and bisects . We claim , because Thus , this implies . Obviously BMPQ is cyclic, and so is BPOQ. Finally, we have . (by gougutheorem) Solution 4 (Sparrow solution) Let and be the circumcircle of circumcenter and radius of Let and be the circumcircle of circumcenter and radius of is cyclic, so is antiparallel We use Sparrow’s Lemma 3A for circle and get that point lies on altitude of Let be the point on opposite is isogonal to lies on bisector is parallelogram is parallelogram. Let vladimir.shelomovskii@gmail.com, vvsss See Also Miquel's point 1985 IMO (Problems) • Resources Preceded by Problem 41•2•3•4•5•6Followed by Problem 6 All IMO Problems and Solutions Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
13846	https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion?srsltid=AfmBOordr98Rimi-HgzEdizwe5y2BxsKQ3zaN7QZtKxnRx1i_O1QqxPE	Art of Problem Solving Principle of Inclusion-Exclusion - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Principle of Inclusion-Exclusion Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Principle of Inclusion-Exclusion The Principle of Inclusion-Exclusion (abbreviated PIE) provides an organized method/formula to find the number of elements in the union of a given group of sets, the size of each set, and the size of all possible intersections among the sets. Contents [hide] 1 Important Note(!) 2 Application 2.1 Two Set Example 2.2 Three Set Example 2.3 Four Set Example 2.3.1 Problem 2.3.2 Solution 2.4 Five Set Example 2.4.1 Problem 2.4.2 Solution 3 Statement 4 Proof 5 Remarks 6 Examples 7 See also Important Note(!) When using PIE, one should understand how to strategically overcount and undercount, in the end making sure every element is counted once and only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. Application Here, we will illustrate how PIE is applied with various numbers of sets. Two Set Example Assume we are given the sizes of two sets, and , and the size of their intersection, . We wish to find the size of their union, . To find the union, we can add and . In doing so, we know we have counted everything in at least once. However, some things were counted twice. The elements that were counted twice are precisely those in . Thus, we have that: . Three Set Example Assume we are given the sizes of three sets, and , the size of their pairwise intersections, , and , and the size their overall intersection, . We wish to find the size of their union, . Just like in the Two Set Example, we start with the sum of the sizes of the individual sets . We have counted the elements which are in exactly one of the original three sets once, but we've obviously counted other things twice, and even other things thrice! To account for the elements that are in two of the three sets, we first subtract out . Now we have correctly accounted for them since we counted them twice originally, and just subtracted them out once. However, the elements that are in all three sets were originally counted three times and then subtracted out three times. We have to add back in . Putting this all together gives: . Four Set Example Problem Six people of different heights are getting in line to buy donuts. Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to back. (2015 ARML I10) Solution Let be the event that the first, second, and third people are in ordered height, be the event that the second, third, and fourth people are in ordered height, be the event that the third, fourth, and fifth people are in ordered height, and be the event that the fourth, fifth and sixth people are in ordered height. By a combination of complementary counting and PIE, we have that our answer will be . Now for the daunting task of evaluating all of this. For , we just choose people and there is only one way to put them in order, then ways to order the other three guys for . Same goes for , , and . Now, for , that's just putting four guys in order. By the same logic as above, this is . Again, would be putting five guys in order, so . is just choosing guys out of , then guys out of for . Now, is just the same as , so , is so , and is so . Moving on to the next set: is the same as which is , is ordering everybody so , is again ordering everybody which is , and is the same as so . Finally, is ordering everybody so . Now, lets substitute everything back in. We get a massive expression of . Five Set Example Problem There are five courses at my school. Students take the classes as follows: 243 take algebra. 323 take language arts. 143 take social studies. 241 take biology. 300 take history. 213 take algebra and language arts. 264 take algebra and social studies. 144 take algebra and biology. 121 take algebra and history. 111 take language arts and social studies. 90 take language arts and biology. 80 take language arts and history. 60 take social studies and biology. 70 take social studies and history. 60 take biology and history. 50 take algebra, language arts, and social studies. 50 take algebra, language arts, and biology. 50 take algebra, language arts, and history. 50 take algebra, social studies, and biology. 50 take algebra, social studies, and history. 50 take algebra, biology, and history. 50 take language arts, social studies, and biology. 50 take language arts, social studies, and history. 50 take language arts, biology, and history. 50 take social studies, biology, and history. 20 take algebra, language arts, social studies, and biology. 15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history. 10 take algebra, social studies, biology, and history. 10 take language arts, social studies, biology, and history. 5 take all five. None take none. How many people are in my school? Solution Let A be the subset of students who take Algebra, L-languages, S-Social Studies, B-biology, H-history, M-the set of all students. We have: Thus, there are people in my school. Statement If are finite sets, then: . Proof We prove that each element is counted once. Say that some element is in sets. Without loss of generality, these sets are We proceed by induction. This is obvious for If this is true for we prove this is true for For every set of sets not containing with size there is a set of sets containing with size In PIE, the sum of how many times these sets are counted is There is also one additional set of sets so is counted exactly once. Remarks Sometimes it is also useful to know that, if you take into account only the first sums on the right, then you will get an overestimate if is odd and an underestimate if is even. So, , , , and so on. Examples 2011 AMC 8 Problems/Problem 6 2017 AMC 10B Problems/Problem 13 2005 AMC 12A Problems/Problem 18 2001 AIME II Problems/Problem 9 2002 AIME I Problems/Problem 1 2020 AIME II Problems/Problem 9 2001 AIME II Problems/Problem 2 2017 AIME II Problems/Problem 1 See also Combinatorics Overcounting Retrieved from " Category: Combinatorics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
13847	https://www.nsta.org/science-and-children/science-and-children-february-2020/q-whats-difference-between-evaporation-and?srsltid=AfmBOoodNikpBJAd_AQ70ndjJBHOEyvPRAMy05z45gEiyRmDkGcGRdjO	Science 101 Q: What’s the difference between evaporation and boiling? Breadcrumb Science and Children—February 2020 (Volume 57, Issue 6) By Matt Bobrowsky Share Start a Discussion A: To complement last month’s article about condensation, we will discuss evaporation (the opposite of condensation) and the closely related phenomenon of boiling. Both evaporation and boiling involve changing a liquid to a gas, but there are a number of differences between them. EVAPORATION AND BOILING DIFFERENCES Speed: Evaporation is a slower process and boiling is faster. Try this: In the weeks leading up to a lesson on the water cycle (including evaporation and condensation), set a tall glass of water on a shelf, and have students measure the height of the water every few days and keep a table showing the changing level (Figure 1). If students don’t already know what’s going to happen, don’t tell them; let them discover it. They can also graph the height of the water over time. Then when you’re ready to talk about evaporation, ask, “What happened to the water that was in the glass? Where did it go?” (This investigation could also be done with students reading the amount of water remaining in a graduated cylinder, but you might prefer to have students practice making measurements with the ruler.) In contrast, you can boil water on a hot plate and watch the water level lower in a matter of minutes, rather than the days and weeks it takes for water to evaporate. Temperature required: A liquid will evaporate at any temperature above freezing. In the previous example, the water was at room temperature, and it slowly evaporated. If the water was warmer, it would have evaporated faster. In contrast, boiling occurs only when the liquid reaches a certain temperature, which we call the boiling point. The boiling point of water at sea level is 100°C (212°F). Bubbles: Evaporation does not involve the formation of bubbles. When liquid evaporates, individual molecules leave the liquid and become part of the air. In Figure 2, the dots represent molecules. The ones that leave the liquid and become part of the gas (air) are evaporating. The ones that go from gas to liquid are condensing. When the water in your glass was evaporating, molecules must have been moving from the water to the air more often than molecules were moving from the air to the water. So there was a net transfer of water molecules from the liquid in the glass to the air. As you can see from Figure 2, this process does not involve any bubbles. In contrast, when you boil water, the liquid is changing to a gas so rapidly that bubbles of water vapor form. Here’s a think-pair-share question for students: When you boil water and see bubbles, is there anything in the bubbles? If so, what are the bubbles made of? Some students might think those are air bubbles. But now you understand that they’re not air bubbles; they’re bubbles of water vapor—water that has changed from a liquid to a gas. Is there an easy way to demonstrate that the bubbles contain water molecules? Yes! Bring a pot of water to a rolling boil. Put a lid on it for about 20 seconds, and then remove the lid and look at the inside of it. It’s covered with drops of water from those bubbles that condensed on the cooler lid (Figure 3). Location of the change to a gas: You’ve probably noticed that when you heat water to bring it to boiling, bubbles first form on the bottom of the pot. (Students wearing safety goggles can observe this close up.) That’s because, at first, the bottom of the pot is the only place where it’s hot enough to heat the water to the boiling point and change it to a gas. But once you have a rolling boil, the entire pot of water is at the boiling point (100°C), and bubbles form throughout the bulk of the water. Evaporation, on the other hand, occurs only at the surface of the water. (Recall Figure 2, showing the evaporating molecules leaving the surface of the water.) Source of energy: Boiling usually requires an external source of energy, such as the burner under the pot of water in which you’re boiling your eggs. Evaporation, however, uses the energy already in the liquid. If you have a puddle of water, it has some heat energy, which usually just came from the environment. The heat in that water results in some molecules moving fast enough to escape into the air, that is, evaporate. No additional source of energy is required for evaporation, and the water does not need to reach the boiling point to evaporate. As we’ve seen, water will evaporate at room temperature. What you just read implies that evaporation, but not boiling, is a natural process. Your puddle of water or the water on your hair that you just washed will evaporate without you doing anything special. Just wait, and it dries. But boiling does not usually happen naturally. We have to deliberately heat the liquid to get it to boil. Temperature change of the liquid: While water is boiling, its temperature remains constant at 100°C. A rolling boil doesn’t make the water any hotter than a medium boil. Your eggs will cook just as fast either way. On the other hand, evaporation of water will cool the water—and any surface that the water is evaporating off of. That’s why if you wait to dry off after you step out of the shower, you feel cold. The evaporating water molecules carry away heat from your skin. This is also why you perspire on a hot, summer day. The additional moisture on your skin results in more evaporation, which cools your skin. So don’t wipe that sweat off your brow; let it evaporate, and you’ll feel cooler! Why did I say, “at sea level”? Because at high altitudes, the air pressure is lower. With lower air pressure, there’s less of a “lid” on the water, and it can change to a gas at a lower temperature. Any cooks out there? I just picked up my old copy of Joy of Cooking, in which the authors include how cooking instructions must be modifi ed at high altitudes. For example, in the soup section, they say, “Above 2,500 feet, soups need longer cooking periods than called for in the regular recipes, as the liquids boil at a lower temperature.” There’s so much cool science in cooking! Remind me to do an article on that. FACTORS AFFECTING RATE OF EVAPORATION Ask students, “What can we do to make water evaporate faster?” Here are some things that affect how fast a liquid will evaporate: Heat. Clothes drying on a clothesline will dry faster on a summer day than in winter. Wind. If you step out of an outdoor swimming pool when the wind is blowing, you feel colder because the wind causes the water to evaporate faster from your skin, carrying away heat energy from your skin faster, leaving your skin colder (Figure 4). This effect is also the cause of the “wind chill factor” that we hear about in weather reports on cold, winter days. You weren’t in a swimming pool, but your skin always has some moisture on it. The wind causes that moisture to evaporate faster, carrying away more heat from your skin. That’s why it’s important to stay covered up on cold, windy days. Humidity. Water evaporates faster when the air is dry. When the air is dry, there are fewer water molecules in the air returning to the liquid, so the water evaporates faster. When the air has many water molecules in it (i.e., it is humid), like on a rainy day, evaporation is slower because more water molecules return to the liquid. Try this investigation: Put an equal amount of water in three identical jars. Leave one uncovered, cover another one with aluminum foil, and cover the last jar with a tightly fitting lid (Figure 5). Based on what we’ve learned, what would you expect to see after a few days to a week? The tighter the lid, the less evaporation. But why, exactly? Students might suggest that, with the lid on a jar, the water molecules have no place to go. You can respond, “Yes, but can’t the water molecules still go into the air above the water in the jar?” They can, but that will result in a lot more water molecules in the air in the jar (higher humidity) and a lot more water molecules returning to the liquid (as in Figure 2). It quickly gets to the point where the number of water molecules returning to the liquid is equal to the number leaving the liquid, so there’s no net change in the amount of liquid water (i.e., no evaporation). That’s a much more sophisticated explanation than the lid simply holding in the water. You probably won’t need to go into that much detail with your students, but you’ll have a lot more confidence in your content knowledge if you understand the process in-depth. Surface area. Increasing surface area will provide more surface from which water can evaporate. So a wet towel will dry faster if it’s spread out rather than left folded or bunched up. To summarize, evaporation is slower, occurs only from the surface of the liquid, does not produce bubbles, and leads to cooling. Boiling is faster, can occur throughout the liquid, produces lots of bubbles, and does not result in cooling. Between last month’s article and this one, you’re ready to discuss the water cycle. In its simplest form, the water cycle is how water changes from being water vapor (a gas) to liquid water (condensation) and then back to a gas (evaporation). Holy cow, it just hit me how many real-life applications we found for this one simple topic. There’s so much to discover in our day-to-day activities. That’s why I always say, “Never stop learning!” Matt Bobrowsky is the lead author of the NSTA Press book series, Phenomenon-Based Learning: Using Physical Science Gadgets & Gizmos. You can let him know if there’s a science concept that you would like to hear more about. Contact him at: DrMatt@msb-science.com Labs Physical Science Elementary You may also like NSTA Kids Click here to view video of Dear Ms. Guadalupe: Letters to My Librarian... NSTA Press Book NOW AVAILABLE FOR PRESALE (TO SHIP EARLY OCTOBER)! Two young explorers embark on an exciting journey in search of the perfect planet. As they travel through outer space, they discover and visit incredible exoplanets—from scorching gas giants to ... NSTA Kids NOW AVAILABLE FOR PRESALE (TO SHIP EARLY OCTOBER)! Two young explorers embark on an exciting journey in search of the perfect planet. As they travel through outer space, they discover and visit incredible exoplanets—from scorching gas giants to ... NSTA Kids NOW AVAILABLE FOR PRESALE (TO SHIP MID OCTOBER)! Two young explorers embark on an exciting journey in search of the perfect planet. As they travel through outer space, they discover and visit incredible exoplanets—from scorching gas giants to mi... National Science Teaching Association 405 E Laburnum Avenue Ste 3 Richmond, VA 23222 (T) 703.524.3646 (F) 703.243.7177
13848	https://community.deeplearning.ai/t/minimizing-variance/863152	Minimizing variance Why minimizing the variance of f(w) is the same as maximizing f(w)? image1124×874 79 KB hi @flyunicorn the screenshot you shared nowhere mention minimize variance of f(w) is same as maximizing f(w)? can you point out which sentence in the screenshot mentions.your aforementioned statement. Regards DP It is here. Because the question is about maximizing but then in the solution, the aim becomes to minimizing. I think there is some link missing in my understanding. image1120×864 79 KB There are two different quantities discussed there. f(w) and L(w) are different functions. To maximize the profit, you minimize the cost, right? Although what they are doing is a bit more subtle than that. The loss function L(w) is the variance of f(w) values. The part that they just say we have to take for granted and they don’t explain or prove is that minimizing that definition of L(w) will maximize the profit. Or at least will give you the best chance of maximizing the profit, given that we can’t exactly predict the future price moves. Related topics \| Topic \| \| Replies \| Views \| Activity \| --- --- \| Programming Assignment - Choice of Loss Function Calculus for Machine Learning and Data Science week-module-1 \| 4 \| 440 \| June 14, 2023 \| \| C2_W1_Assignment why not minimise total cost instead of loss function Calculus for Machine Learning and Data Science week-module-1 \| 3 \| 283 \| December 11, 2023 \| \| Question about computing variance for normalization Improving Deep Neural Networks: Hyperparameter tun coursera-platform \| 4 \| 564 \| June 24, 2021 \| \| Optimizing Functions of One Variable: Cost Minimization Calculus for Machine Learning and Data Science \| 3 \| 127 \| February 13, 2023 \| \| Week 2 loss function Neural Networks and Deep Learning coursera-platform \| 1 \| 579 \| May 23, 2021 \| Powered by Discourse, best viewed with JavaScript enabled
13849	https://www.pinterest.com/pin/days-and-months--420312577727620609/	Skip to content Search for esl vocabulary vocabulary games for kids printable crossword puzzles test for kids unscramble words esl vocabulary When autocomplete results are available use up and down arrows to review and enter to select. Touch device users, explore by touch or with swipe gestures. Log in Sign up Explore More about this Pin Board containing this Pin Days and months 2.8k Pins 3d Related interests Esl Vocabulary Vocabulary Games For Kids Printable Crossword Puzzles Test For Kids Unscramble Words Word Search Puzzles Printables Kids Worksheets Printables Vocabulary Games English Activities Read it Save englishwsheets.com Months ESL Vocabulary Worksheets ESL Printable Months Vocabulary Worksheets, Picture Dictionaries, Matching Exercises, Word Search and Crossword Puzzles, Missing Letters in Words and Unscramble the Words Exercises, Multiple Choice Tests, Flashcards, Vocabulary Learning Cards, ESL Fidget Spinner and Dominoes Games ...more Englishwsheets.com Comments Add a comment More to explore More to explore More about this Pin 1 Like Board containing this Pin ingles 185 Pins 4w Related interests Months Of The Year Word Search Printable Months Of The Year Word Search Months Word Search For Kids Months Word Search Printable Months Word Search Pdf New Year Activities Printable Months Of The Year Activity Wordsearch For Kids WORD SEARCH - months More about this Pin 8 Likes 24 Shares Board containing this Pin English Act 71 Pins 3y Related interests Classroom Month Learning Activity Educational Months Activity Sheet Months Of The Year Worksheet Months Activities Worksheets Months Activities Months Of The Year Activities Printable Months Of The Year Teaching Months To Kids Year 1 English ESL Months of the Year Vocabulary Cards and Worksheet Activities More about this Pin 133 Likes 298 Shares Board containing this Pin Classroom ideas 281 Pins 1d Related interests Color By Grammar Colouring Worksheet English Time English Activities For Kids Understanding Emotions English Worksheets For Kids Kids English English Classroom English Verbs MANDALA ENGLISH VOCABULARY GREAT! COLOUR THE MANDALA ACCORDING TO INSTRUCTIONS. You are signed out Sign in to get the best experience Continue with email By continuing, you agree to Pinterest's Terms of Service and acknowledge you've read our Privacy Policy. Notice at collection. Log in to see more Email Email Password Use 8 or more letters, numbers and symbols Forgot your password? Log in OR Use QR code Facebook login is no longer available Update login method Not on Pinterest yet? Sign up Are you a business? Get started here! By continuing, you agree to Pinterest's Terms of Service and acknowledge you've read our Privacy Policy. Notice at collection. Months ESL Vocabulary Worksheets ===============
13850	https://www.geeksforgeeks.org/java/abstraction-in-java-2/	Abstraction in Java Last Updated : 28 Jul, 2025 Suggest changes 339 Likes Abstraction in Java is the process of hiding internal implementation details and showing only essential functionality to the user. It focuses on what an object does rather than how it does it. Key features of abstraction Abstraction hides the complex details and shows only essential features. Abstract classes may have methods without implementation and must be implemented by subclasses. By abstracting functionality, changes in the implementation do not affect the code that depends on the abstraction. How to Achieve Abstraction in Java? Java provides two ways to implement abstraction, which are listed below: Abstract Classes (Partial Abstraction) Interface (100% Abstraction) Real-Life Example of Abstraction The television remote control is the best example of abstraction. It simplifies the interaction with a TV by hiding all the complex technology. We don't need to understand how the tv internally works, we just need to press the button to change the channel or adjust the volume. Java ```` // Working of Abstraction in Java abstract class Geeks { abstract void turnOn(); abstract void turnOff(); } // Concrete class implementing the abstract methods class TVRemote extends Geeks { @Override void turnOn() { System.out.println("TV is turned ON."); } @Override void turnOff() { System.out.println("TV is turned OFF."); } } // Main class to demonstrate abstraction public class Main { public static void main(String[] args) { Geeks remote = new TVRemote(); remote.turnOn(); remote.turnOff(); } } ```` // Working of Abstraction in Java // Working of Abstraction in Java abstract class Geeks {abstract class Geeks abstract void turnOn(); abstract void turnOn abstract void turnOff(); abstract void turnOff } // Concrete class implementing the abstract methods // Concrete class implementing the abstract methods class TVRemote extends Geeks {class TVRemote extends Geeks @Override @Override void turnOn() {void turnOn System.out.println("TV is turned ON."); System out println"TV is turned ON." } @Override @Override void turnOff() {void turnOff System.out.println("TV is turned OFF."); System out println"TV is turned OFF." } } // Main class to demonstrate abstraction // Main class to demonstrate abstraction public class Main {public class Main public static void main(String[] args) {public static void main String args Geeks remote = new TVRemote(); Geeks remote = new TVRemote remote.turnOn(); remote turnOn remote.turnOff(); remote turnOff } } Output ``` TV is turned ON. TV is turned OFF. ``` Explanation: Geeks is abstract class defining turnOn() and turnOff() methods. TVRemote class implements the abstract methods with specific logic. Main class uses Geeks remote = new TVRemote(); to interact without knowing the internal implementation. Abstract class An abstract class is a way to achieve abstraction in Java. It is declared using the abstract keyword and can contain both abstract methods non abstract methods. Abstract classes cannot be instantiated directly and are meant to be extended by subclasses. Besides abstraction, abstract classes also allow code reusability through shared behavior and state. Consider a classic “shape” example, perhaps used in a computer-aided design system or game simulation. The base type is “shape” and each shape has a color, size, and so on. From this, specific types of shapes are derived(inherited)-circle, square, triangle, and so on — each of which may have additional characteristics and behaviors. For example, certain shapes can be flipped. Some behaviors may be different, such as when you want to calculate the area of a shape. The shape hierarchy shows both the similarities that all shapes share and the differences that makes each one unique. Example: This program defines an abstract class Shape with an abstract method area() and a concrete method getColor(), demonstrating partial abstraction. It shows how an abstract class can have constructors and both implemented and unimplemented methods. Java ```` abstract class Shape { String color; // these are abstract methods abstract double area(); public abstract String toString(); // abstract class can have the constructor public Shape(String color) { System.out.println("Shape constructor called"); this.color = color; } // this is a concrete method public String getColor() { return color; } } class Circle extends Shape { double radius; public Circle(String color, double radius) { // calling Shape constructor super(color); System.out.println("Circle constructor called"); this.radius = radius; } @Override double area() { return Math.PI Math.pow(radius, 2); } @Override public String toString() { return "Circle color is " + super.getColor() + "and area is : " + area(); } } class Rectangle extends Shape { double length; double width; public Rectangle(String color, double length, double width) { // calling Shape constructor super(color); System.out.println("Rectangle constructor called"); this.length = length; this.width = width; } @Override double area() { return length width; } @Override public String toString() { return "Rectangle color is " + super.getColor() + "and area is : " + area(); } } public class Test { public static void main(String[] args) { Shape s1 = new Circle("Red", 2.2); Shape s2 = new Rectangle("Yellow", 2, 4); System.out.println(s1.toString()); System.out.println(s2.toString()); } } ```` abstract class Shape {abstract class Shape String color; String color // these are abstract methods // these are abstract methods abstract double area(); abstract double area public abstract String toString(); public abstract String toString // abstract class can have the constructor // abstract class can have the constructor public Shape(String color) public Shape String color { System.out.println("Shape constructor called"); System out println "Shape constructor called" this.color = color; this color = color } // this is a concrete method // this is a concrete method public String getColor() { return color; } public String getColor return color } class Circle extends Shape {class Circle extends Shape double radius; double radius public Circle(String color, double radius) public Circle String color double radius { // calling Shape constructor // calling Shape constructor super(color); super color System.out.println("Circle constructor called"); System out println "Circle constructor called" this.radius = radius; this radius = radius } @Override double area() @Override double area { return Math.PI Math.pow(radius, 2); return Math PI Math pow radius 2 } @Override public String toString() @Override public String toString { return "Circle color is " + super.getColor() return "Circle color is " + super getColor + "and area is : " + area(); +"and area is : " + area } } class Rectangle extends Shape {class Rectangle extends Shape double length; double length double width; double width public Rectangle(String color, double length, public Rectangle String color double length double width) double width { // calling Shape constructor // calling Shape constructor super(color); super color System.out.println("Rectangle constructor called"); System out println "Rectangle constructor called" this.length = length; this length = length this.width = width; this width = width } @Override double area() { return length width; } @Override double area return length width @Override public String toString() @Override public String toString { return "Rectangle color is " + super.getColor() return "Rectangle color is " + super getColor + "and area is : " + area(); +"and area is : " + area } } public class Test {public class Test public static void main(String[] args) public static void main String args { Shape s1 = new Circle("Red", 2.2); Shape s1 = new Circle "Red"2.2 Shape s2 = new Rectangle("Yellow", 2, 4); Shape s2 = new Rectangle "Yellow" 2 4 System.out.println(s1.toString()); System out println s1 toString System.out.println(s2.toString()); System out println s2 toString } } Example: This program demonstrates abstraction using an abstract class Animal with an abstract method makeSound(). Subclasses Dog and Cat provide specific implementations, and objects are accessed via the abstract reference type. Java ```` // Abstract Class declared abstract class Animal { private String name; public Animal(String name) { this.name = name; } public abstract void makeSound(); public String getName() { return name; } } // Abstracted class class Dog extends Animal { public Dog(String name) { super(name); } public void makeSound() { System.out.println(getName() + " barks"); } } // Abstracted class class Cat extends Animal { public Cat(String name) { super(name); } public void makeSound() { System.out.println(getName() + " meows"); } } // Driver Class public class Geeks { // Main Function public static void main(String[] args) { Animal myDog = new Dog("ABC"); Animal myCat = new Cat("XYZ"); myDog.makeSound(); myCat.makeSound(); } } ```` // Abstract Class declared // Abstract Class declared abstract class Animal {abstract class Animal private String name; private String name public Animal(String name) { public Animal String name this.name = name; this name = name } public abstract void makeSound(); public abstract void makeSound public String getName() { public String getName return name; return name } } // Abstracted class // Abstracted class class Dog extends Animal {class Dog extends Animal public Dog(String name) { public Dog String name super(name); super name } public void makeSound() public void makeSound { System.out.println(getName() + " barks"); System out println getName + " barks" } } // Abstracted class // Abstracted class class Cat extends Animal {class Cat extends Animal public Cat(String name) { public Cat String name super(name); super name } public void makeSound() public void makeSound { System.out.println(getName() + " meows"); System out println getName + " meows" } } // Driver Class // Driver Class public class Geeks {public class Geeks ``` ``` // Main Function // Main Function public static void main(String[] args) public static void main String args { Animal myDog = new Dog("ABC"); Animal myDog = new Dog "ABC" Animal myCat = new Cat("XYZ"); Animal myCat = new Cat "XYZ" myDog.makeSound(); myDog makeSound myCat.makeSound(); myCat makeSound } } Output ``` ABC barks XYZ meows ``` Interface Interfaces is a blueprint of a class used to achieve 100% abstraction in Java. It can contain abstract methods and constants but no method bodies (except default and static methods from Java 8 onward). Implementation: To implement an interface we use the keyword “implements” with class. Example: Below is the Implementation of Abstraction using Interface. Java ```` // Define an interface named Shape interface Shape { double calculateArea(); // Abstract method for // calculating the area } // Implement the interface // in a class named Circle class Circle implements Shape { private double r; // radius // Constructor for Circle public Circle(double r) { this.r = r; } // Implementing the abstract method // from the Shape interface public double calculateArea() { return Math.PI r r; } } // Implement the interface in a // class named Rectangle class Rectangle implements Shape { private double length; private double width; // Constructor for Rectangle public Rectangle(double length, double width) { this.length = length; this.width = width; } // Implementing the abstract // method from the Shape interface public double calculateArea() { return length width; } } // Main class to test the program public class Main { public static void main(String[] args) { // Creating instances of Circle and Rectangle Circle c = new Circle(5.0); Rectangle rect = new Rectangle(4.0, 6.0); System.out.println("Area of Circle: " + c.calculateArea()); System.out.println("Area of Rectangle: " + rect.calculateArea()); } } ```` // Define an interface named Shape // Define an interface named Shape interface Shape {interface Shape double calculateArea(); // Abstract method for double calculateArea // Abstract method for // calculating the area // calculating the area } // Implement the interface // Implement the interface // in a class named Circle // in a class named Circle class Circle implements Shape {class Circle implements Shape private double r; // radius private double r // radius // Constructor for Circle // Constructor for Circle public Circle(double r) { public Circle double r this.r = r; this r = r } // Implementing the abstract method // Implementing the abstract method // from the Shape interface // from the Shape interface public double calculateArea() public double calculateArea { return Math.PI r r; return Math PI r r } } // Implement the interface in a // Implement the interface in a // class named Rectangle // class named Rectangle class Rectangle implements Shape {class Rectangle implements Shape private double length; private double length private double width; private double width // Constructor for Rectangle // Constructor for Rectangle public Rectangle(double length, double width) public Rectangle double length double width { this.length = length; this length = length this.width = width; this width = width } // Implementing the abstract // Implementing the abstract // method from the Shape interface // method from the Shape interface public double calculateArea() { public double calculateArea return length width; return length width } } // Main class to test the program // Main class to test the program public class Main {public class Main public static void main(String[] args) public static void main String args { ``` ``` // Creating instances of Circle and Rectangle // Creating instances of Circle and Rectangle Circle c = new Circle(5.0); Circle c = new Circle5.0 Rectangle rect = new Rectangle(4.0, 6.0); Rectangle rect = new Rectangle4.06.0 System.out.println("Area of Circle: " System out println"Area of Circle: " + c.calculateArea()); + c calculateArea System.out.println("Area of Rectangle: " System out println"Area of Rectangle: " + rect.calculateArea()); + rect calculateArea } } Output ``` Area of Circle: 78.53981633974483 Area of Rectangle: 24.0 ``` Advantages of Abstraction Abstraction makes complex systems easier to understand by hiding the implementation details. Abstraction keeps different part of the system separated. Abstraction maintains code more efficiently. Abstraction increases the security by only showing the necessary details to the user. Disadvantages of Abstraction It can add unnecessary complexity if overused. May reduce flexibility in implementation. Makes debugging and understanding the system harder for unfamiliar users. Overhead from abstraction layers can affect performance. Abstract Classes and Abstract Methods \| Abstract Classes \| Abstract Methods \| --- \| \| An abstract class is a class that is declared with an abstract keyword. \| An abstract method is a method that is declared without implementation. \| \| An abstract class may have both abstract methods (methods without implementation) and concrete methods (methods with implementation) \| An abstract method must always be redefined in the subclass, thus making overriding compulsory or making the subclass itself abstract. \| \| Any class that contains one or more abstract methods must also be declared with an abstract keyword. \| An abstract method is a method that is declared without an implementation (i.e., without a body) and is meant to be overridden in a subclass. \| Common Mistakes to Avoid The common mistakes that can occur and we should avoid when working with Abstraction in Java are listed below: Not Implementing Abstract Methods: Always make sure that the abstract methods are implemented in the concrete subclass. Overusing Abstraction: Avoid making everything abstract when it’s not required. Use abstraction only when it enhances the design. Inconsistent Method Signatures in Subclasses: When you override abstract methods, please make sure the method signature matches exactly, any mistake can cause errors. 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13851	https://my.clevelandclinic.org/health/treatments/21533-skin-care-basics-and-tips	Abu Dhabi\|Canada\|Florida\|London\|Nevada\|Ohio\| Home/ Health Library/ Medical Treatments/ Skin Care Basics and Tips AdvertisementAdvertisement Skin Care Basics and Tips Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services. Policy Care at Cleveland Clinic Get Dermatology Care Make an Appointment Advertisement Advertisement Advertisement Advertisement ContentsTreatment DetailsAdditional Details Treatment Details What should I know about caring for dry skin? Dry skin is defined as flaking or scaling of the skin when there is no evidence of dermatitis (inflammation). It appears most often on the shins, hands, and sides of the abdomen, and can be associated with itching. Dry skin is more common during the winter months, when humidity is low, and improves in the summer time. Some people also have a genetic (inherited) tendency to develop dry skin. In addition, elderly people tend to have more trouble with dry skin because of the natural changes in skin that occur as we get older. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services. Policy Treatment is important because excessively dry skin can lead to dermatitis or eczema. Dry skin may be prevented or treated by: Taking lukewarm baths or showers as opposed to using hot water. Limiting baths/showers to 5 to 10 minutes. Applying a moisturizer right after drying off from a shower or washing your hands. Using a moisturizing body soap and hand soap. Using heavier creams or ointments during the winter months and lighter lotions in the summer. If the above treatments do not improve the condition of the dry skin, it is possible that the flaking is a sign of underlying dermatitis (which is also called eczema). There are different types of dermatitis that may cause dry, itchy, flaking skin. They include: Seborrheic dermatitis: A red, scaly, mildly itchy rash on the scalp, eyebrows, and sides of the nose in areas that contain many oil glands. Allergic contact dermatitis: A rash that results when the skin comes in contact with a substance that causes an allergic reaction, such as poison ivy. Allergic contact dermatitis of the hands often causes scaling, redness, and vesicles (blisters filled with fluid) on the palms or fingers. Atopic dermatitis:Long-lasting type of dermatitis usually starting in childhood that tends to run in families. It also may cause excessively dry, itchy skin on the face and body. Athlete's foot: Dry flaking skin on the soles and/or sides of the feet and between toes, caused by a fungus. Advertisement What should I know about facial skin care for mature skin? Photoaging refers to the damage that is done to the skin from prolonged exposure to UV radiation over a person's lifetime. Examples of skin changes associated with photoaging include: Roughness. Wrinkling. Irregular pigmentation (coloration). Inelasticity. Enlarged sebaceous (oil) glands). Precancerous and cancerous lesions. Sunscreens and sun protection are important to prevent further progression of photoaging. Smoking has also been shown to speed up aging of skin, so smoking cessation is important for good skin health. In addition, a well-balanced diet allows the skin to get the nutrition it needs to help repair ongoing damage from the sun and other environmental elements. Many topical (applied to the skin) non-prescription and prescription products are currently available for anti-aging purposes, including: Retinoids: Retinoids are prescription (tretinoin, tazarotene) and nonprescription (adapalene) medications first developed to treat acne. They improve skin texture and color when used over an extended period of time. They exfoliate the skin (removes a dead layer of skin cells), help even pigmentation, and minimize fine lines. Many people can benefit from using retinoids at bedtime. Side effects of retinoids include dryness, skin peeling, and redness. You may be able to minimize these side effects by using the cream every other night and then gradually increasing the frequency to every night, as tolerated. Retinoids can also makes the skin more sensitive to ultraviolet rays from the sun; therefore, the use of a broad-spectrum sunscreen each morning is recommended. Alpha hydroxy acids (AHAs): Alpha hydroxy acids (glycolic, lactic, tartaric, and citric acids) are found as ingredients of numerous skin products. In the United States alone, there are approximately 185 manufacturers of products containing AHAs. Creams and lotions with AHA may help with fine lines, irregular pigmentation, age spots, and may help decrease enlarged pores. Side effects of AHAs include mild irritation and sun sensitivity. For that reason, sunscreen should also be used every morning. Beta hydroxy acid (salicylic acid): Salicylic acid has been studied for its effect on photoaged skin. It exfoliates skin, and can improve skin texture and color. It penetrates oil-laden hair follicle openings and, as a result, also helps improve acne. There are many over-the-counter products available that contain salicylic acid. What should I know about facial skin care for acne-prone skin? If you are prone to acne, choose a cleanser specially formulated for acne. These products often contain salicylic acid or benzoyl peroxide, which help to clear acne. Clean your face gently, as trauma to the acne breakouts may worsen acne or cause scarring. Avoid harsh mechanical scrubbing of skin and picking lesions. If you need to use a moisturizer, use light, non-comedogenic moisturizers, which do not cause acne. Women should use an oil-free non-comedogenic foundation, as heavy makeup or other cosmetic products may block pores and worsen acne. Care at Cleveland Clinic Get Dermatology Care Make an Appointment Additional Details What should I know about protecting skin from the sun? Protecting your skin from the sun is important because the sun emits ultraviolet radiation (UVR). Over time, UVR exposure causes many changes in the skin, including wrinkles, discoloration, age spots, benign (non-cancerous) growths, and pre-cancerous and cancerous growths. In fact, most skin cancers are related to sun exposure.UVR consists of two main subtypes: UVB and UVA. UVB rays are responsible for sunburns and tanning. UVA rays are believed to be responsible for photoaging (the damage that occurs to the skin from many years of exposure to the sun). Both types have been linked to cancer.Most sunscreen products prevent sunburns by blocking UVB rays. Newer sunscreen products are also successful in blocking UVA rays. For that reason, sun protection recommendations emphasize certain behaviors, as well as the use of sunscreens. The recommendations include: Advertisement Avoiding midday sun (between 10 a.m. and 3 p.m.). Wearing wide-brimmed hats, long-sleeved shirts, and pants. Using a generous amount of sunscreen and reapplying it frequently (every 2 to 3 hours). Using sunscreens that have a sun protection factor (SPF) greater than 30, and have UVA and UVB coverage. Avoiding tanning beds (tanning beds are now considered a carcinogen, capable of causing cancer). Advertisement Care at Cleveland Clinic Every day, people see your skin, hair and nails. At Cleveland Clinic, our expert and caring dermatology team will make sure they’re healthy and strong. Get Dermatology Care Make an Appointment Medically Reviewed Last reviewed on 05/04/2021. Learn more about the Health Library and our editorial process. AdvertisementAdvertisement Ad Appointments 216.444.5725 Appointments & Locations Request an Appointment Rendered: Sun Sep 14 2025 01:06:37 GMT+0000 (Coordinated Universal Time)
13852	https://ufhealth.org/conditions-and-treatments/familial-lipoprotein-lipase-deficiency	Familial Lipoprotein Lipase Deficiency - UF Health Skip to main content Update Location Close My location Use your current location or add an address to show providers, locations, and services closest to you. Use My Location Address, City, or ZIP code (352) 733-0111My UFHealth UF Health Home Search Search Close Menu Find a Doctor Conditions & Treatments Locations Patients & Visitors Clinical Trials Community Billing Insurance Events Make a gift Careers Volunteer Education Research About (352) 733-0111My UFHealth Close Conditions and Treatments Familial Lipoprotein Lipase Deficiency Familial Lipoprotein Lipase Deficiency Menu Open Menu Familial Lipoprotein Lipase Deficiency Menu Close On this page Collapse navigation Definition Alternative Names Causes Symptoms Exams and Tests Treatment Support Groups Outlook (Prognosis) Possible Complications When to Contact a Medical Professional Prevention Gallery References Related specialties Definition Familial lipoprotein lipase deficiency is a group of rare genetic disorders in which a person lacks a protein needed to break down fat molecules. The disorder causes a large amount of fat to build up in the blood. Alternative Names Type I hyperlipoproteinemia; Familial chylomicronemia; Familial LPL deficiency Causes Familial lipoprotein lipase deficiency is caused by a defective gene that is passed down through families. People with this condition lack an enzyme called lipoprotein lipase. Without this enzyme, the body cannot break down fat from digested food. Fat particles called chylomicrons build up in the blood. Risk factors include a family history of lipoprotein lipase deficiency. The condition usually first develops during infancy or childhood. Symptoms Symptoms may include any of the following: Abdominal pain (may appear as colic in infants) Loss of appetite Nausea, vomiting Pain in the muscles and bones Enlarged liver and spleen Failure to thrive in infants Fatty deposits in the skin (xanthomas) High triglyceride levels in the blood Pale retinas and white-colored blood vessels in the retinas Chronic inflammation of the pancreas Yellowing of the eyes and skin (jaundice) Exams and Tests Your health care provider will perform a physical examination and ask about your symptoms. Blood tests will be done to check cholesterol and triglyceride levels. Sometimes, a special blood test is done after you are given blood thinners through a vein. This test looks for lipoprotein lipase activity in your blood. Genetic tests may be done. Treatment Treatment aims to control the symptoms and blood triglyceride levels with a very low-fat diet. Your provider will likely recommend that you eat no more than 20 grams of fat per day to prevent the symptoms from coming back. Twenty grams of fat is equal to one of the following: Two 8-ounce (240 milliliters) glasses of whole milk 4 teaspoons (9.5 grams) of margarine 4 ounces (113 grams) serving of meat The average American diet has a fat content of up to 45% of total calories. Fat-soluble vitamins A, D, E, and K and mineral supplements are recommended for people who eat a very low-fat diet. You may want to discuss your diet needs with your provider and a registered dietitian. Pancreatitis that is related to lipoprotein lipase deficiency responds to treatments for that disorder. Support Groups These resources can provide more information on familial lipoprotein lipase deficiency: National Organization for Rare Disorders -- rarediseases.org/rare-diseases/familial-lipoprotein-lipase-deficiency NIH Genetics Home Reference -- ghr.nlm.nih.gov/condition/familial-lipoprotein-lipase-deficiency Outlook (Prognosis) People with this condition who follow a very low-fat diet can live into adulthood. Possible Complications Pancreatitis and recurrent episodes of abdominal pain may develop. Xanthomas are not usually painful unless they are rubbed a lot. When to Contact a Medical Professional Contact your provider for screening if someone in your family has lipoprotein lipase deficiency. Genetic counseling is recommended for anyone with a family history of this disease. Prevention There is no known prevention for this rare, inherited disorder. Awareness of risks may allow early detection. Following a very low-fat diet can improve the symptoms of this disease. Gallery The coronary arteries supply blood to the heart muscle itself. Blood supply through these arteries is critical for the heart. Coronary artery disease usually results from the build-up of fatty material and plaque, a condition called atherosclerosis. As the coronary arteries narrow, the flow of blood to the heart can slow or stop, causing chest pain (stable angina), shortness of breath, heart attack, or other symptoms. References Genest J, Mora S, Libby P. Lipoprotein disorders and cardiovascular disease. In: Libby P, Bonow RO, Mann DL, Tomaselli GF, Bhatt DL, Solomon SD, eds. Braunwald's Heart Disease: A Textbook of Cardiovascular Medicine. 12th ed. Philadelphia, PA: Elsevier; 2022:chap 27. Semenkovich CF, Goldberg IJ. Disorders of lipid metabolism. In: Melmed S, Auchus RJ, Goldfine AB, Koenig RJ, Rosen CJ, eds. Williams Textbook of Endocrinology. 14th ed. Philadelphia, PA: Elsevier; 2020:chap 41. Last reviewed May 12, 2023 by Sandeep K. Dhaliwal, MD, board-certified in Diabetes, Endocrinology, and Metabolism, Springfield, VA. Also reviewed by David C. Dugdale, MD, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team.. Related specialties Family Medicine Hematology University of Florida HealthUniversity of Florida (352) 733-0111My UFHealthMake a Gift Patients & Visitors About UF Health Locations Medical Records Billing, Costs & Insurances Visiting UF Health Health Professionals Referrals, Consults & Transfers Continuing Education Credentialing Labs Research & Education Colleges Research Centers & Institutes Libraries Research Support Get Involved Careers Volunteer Employees UF Health Bridge St Johns Intranet VPN News & Events UF Health StoriesNewsroomEvents Connect with UF Health FacebookTwitterInstagramYouTubeLinkedIn Report a site issueDisclaimer & Permitted UseNondiscrimination and AccessibilitySecurity and PrivacyPrice TransparencyThis site uses Google Analytics Updated: May 27, 2025 © 2025 UF Health
13853	https://mathworld.wolfram.com/Point-SlopeForm.html	Point-Slope Form -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Geometry Line Geometry Lines Point-Slope Form The point-slope form of a line through the point with slope in the Cartesian plane is given by See also Intercept Form, Line, Slope-Intercept Form, Two-Point Form Explore with Wolfram\|Alpha More things to try: lines Cayley lines point slope form of 7x-3y=50 Cite this as: Weisstein, Eric W. "Point-Slope Form." From MathWorld--A Wolfram Resource. Subject classifications Geometry Line Geometry Lines About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
13854	https://edia.app/worksheets/precalculus/arithmetic_and_geometric_sequences/geometric_series	Precalculus > Arithmetic & geometric sequences Sum of geometric series This Precalculus arithmetic & geometric sequences worksheet generates free practice problems on sum of geometric series. Sum of geometric series Worksheet Name: Date:Period: 1. Find the sum of the first 6 terms of the geometric sequence. 3 , − 6 , 12 , − 24 , 48 ,... 2. Find the sum. k 0 ∑ 7 − 3 ( − 2 1 ) k 3. State whether the infinite series converges or diverges. 8 − 2 + 2 1 − 8 1 + 32 1 − 128 1 + ... Converges Diverges 4. Find the sum of the infinite series. 4 − 2 3 + 16 9 − 128 27 + 1024 81 − ... 5. State whether the infinite series converges or diverges. n 1 ∑ ∞ 64 ( 4 5 ) n Converges Diverges 6. Choose all of the following series that diverge. 3 − 2 + 3 4 − 9 8 + 27 16 − 81 32 + ... − 6 − 1 + 4 + 9 + 14 + 19 + ... 54 + 36 + 24 + 16 + 3 32 + 9 64 + ... − 2 + 4 + 10 + 16 + 22 + 28 + ... 7. Choose all of the following series that converge. n 1 ∑ ∞ − 5 n + 10 n 1 ∑ ∞ 6 ( − 3 4 ) n n 1 ∑ ∞ 5 n − 8 n 1 ∑ ∞ 4 n − 4 8. Write the repeating decimal as a fraction. 0. 17 Free resources at edia.app © 2025 Edia Learning, Inc. All Rights Reserved. 9. Find the sum of the first 6 terms of the geometric sequence. a n = − 5 ( 3 1 ) n − 1 10. A company is offering a new subscription service. For promotional purposes, the company decides to offer the service at a discounted rate for the first year. The cost of the subscription starts at $115.00 in the 1 st month, $86.25 in the 2 nd month, $64.69 in the 3 rd month, and so on. Each month the cost is 75% of the previous month's cost. The promotion lasts 11 months. How much will a customer pay in total? The total cost can be found using the formula for the sum of a finite geometric series: S n = 1 − r a 1 ( 1 − r n ) Round your answer to the nearest whole number. Free resources at edia.app © 2025 Edia Learning, Inc. All Rights Reserved. Sum of geometric series Answer key 1. Find the sum of the first 6 terms of the geometric sequence. 3 , − 6 , 12 , − 24 , 48 ,... S 6 = − 63 2. Find the sum. k 0 ∑ 7 − 3 ( − 2 1 ) k S − 128 255 3. State whether the infinite series converges or diverges. 8 − 2 + 2 1 − 8 1 + 32 1 − 128 1 + ... Converges Diverges 4. Find the sum of the infinite series. 4 − 2 3 + 16 9 − 128 27 + 1024 81 − ... S 11 32 5. State whether the infinite series converges or diverges. n 1 ∑ ∞ 64 ( 4 5 ) n Converges Diverges 6. Choose all of the following series that diverge. 3 − 2 + 3 4 − 9 8 + 27 16 − 81 32 + ... − 6 − 1 + 4 + 9 + 14 + 19 + ... 54 + 36 + 24 + 16 + 3 32 + 9 64 + ... − 2 + 4 + 10 + 16 + 22 + 28 + ... 7. Choose all of the following series that converge. n 1 ∑ ∞ − 5 n + 10 n 1 ∑ ∞ 6 ( − 3 4 ) n n 1 ∑ ∞ 5 n − 8 n 1 ∑ ∞ 4 n − 4 8. Write the repeating decimal as a fraction. 0. 17 99 17 Free resources at edia.app © 2025 Edia Learning, Inc. All Rights Reserved. 9. Find the sum of the first 6 terms of the geometric sequence. a n = − 5 ( 3 1 ) n − 1 S 6 = − 243 1820 10. A company is offering a new subscription service. For promotional purposes, the company decides to offer the service at a discounted rate for the first year. The cost of the subscription starts at $115.00 in the 1 st month, $86.25 in the 2 nd month, $64.69 in the 3 rd month, and so on. Each month the cost is 75% of the previous month's cost. The promotion lasts 11 months. How much will a customer pay in total? The total cost can be found using the formula for the sum of a finite geometric series: S n = 1 − r a 1 ( 1 − r n ) Round your answer to the nearest whole number. 441 Free resources at edia.app © 2025 Edia Learning, Inc. All Rights Reserved. 1. Find the sum of the first 6 terms of the geometric sequence. 3 , − 6 , 12 , − 24 , 48 ,... 2. Find the sum. k 0 ∑ 7 − 3 ( − 2 1 ) k 3. State whether the infinite series converges or diverges. 8 − 2 + 2 1 − 8 1 + 32 1 − 128 1 + ... Converges Diverges 4. Find the sum of the infinite series. 4 − 2 3 + 16 9 − 128 27 + 1024 81 − ... 5. State whether the infinite series converges or diverges. n 1 ∑ ∞ 64 ( 4 5 ) n Converges Diverges 6. Choose all of the following series that diverge. 3 − 2 + 3 4 − 9 8 + 27 16 − 81 32 + ... − 6 − 1 + 4 + 9 + 14 + 19 + ... 54 + 36 + 24 + 16 + 3 32 + 9 64 + ... − 2 + 4 + 10 + 16 + 22 + 28 + ... 7. Choose all of the following series that converge. n 1 ∑ ∞ − 5 n + 10 n 1 ∑ ∞ 6 ( − 3 4 ) n n 1 ∑ ∞ 5 n − 8 n 1 ∑ ∞ 4 n − 4 8. Write the repeating decimal as a fraction. 0. 17 9. Find the sum of the first 6 terms of the geometric sequence. a n = − 5 ( 3 1 ) n − 1 10. A company is offering a new subscription service. For promotional purposes, the company decides to offer the service at a discounted rate for the first year. The cost of the subscription starts at $115.00 in the 1 st month, $86.25 in the 2 nd month, $64.69 in the 3 rd month, and so on. Each month the cost is 75% of the previous month's cost. The promotion lasts 11 months. How much will a customer pay in total? The total cost can be found using the formula for the sum of a finite geometric series: S n = 1 − r a 1 ( 1 − r n ) Round your answer to the nearest whole number. 1. Find the sum of the first 6 terms of the geometric sequence. 3 , − 6 , 12 , − 24 , 48 ,... S 6 = − 63 2. Find the sum. k 0 ∑ 7 − 3 ( − 2 1 ) k S − 128 255 3. State whether the infinite series converges or diverges. 8 − 2 + 2 1 − 8 1 + 32 1 − 128 1 + ... Converges Diverges 4. Find the sum of the infinite series. 4 − 2 3 + 16 9 − 128 27 + 1024 81 − ... S 11 32 5. State whether the infinite series converges or diverges. n 1 ∑ ∞ 64 ( 4 5 ) n Converges Diverges 6. Choose all of the following series that diverge. 3 − 2 + 3 4 − 9 8 + 27 16 − 81 32 + ... − 6 − 1 + 4 + 9 + 14 + 19 + ... 54 + 36 + 24 + 16 + 3 32 + 9 64 + ... − 2 + 4 + 10 + 16 + 22 + 28 + ... 7. Choose all of the following series that converge. n 1 ∑ ∞ − 5 n + 10 n 1 ∑ ∞ 6 ( − 3 4 ) n n 1 ∑ ∞ 5 n − 8 n 1 ∑ ∞ 4 n − 4 8. Write the repeating decimal as a fraction. 0. 17 99 17 9. Find the sum of the first 6 terms of the geometric sequence. a n = − 5 ( 3 1 ) n − 1 S 6 = − 243 1820 10. A company is offering a new subscription service. For promotional purposes, the company decides to offer the service at a discounted rate for the first year. The cost of the subscription starts at $115.00 in the 1 st month, $86.25 in the 2 nd month, $64.69 in the 3 rd month, and so on. Each month the cost is 75% of the previous month's cost. The promotion lasts 11 months. How much will a customer pay in total? The total cost can be found using the formula for the sum of a finite geometric series: S n = 1 − r a 1 ( 1 − r n ) Round your answer to the nearest whole number. 441 Worksheets related to Sum of geometric series View all Determine if sequences are arithmetic or geometric View worksheet → Arithmetic sequences (recursive) View worksheet → Arithmetic sequences (explicit) View worksheet → Sum of arithmetic sequence View worksheet → Geometric sequences (recursive) View worksheet → Geometric sequences (explicit) View worksheet → Sum of geometric series View worksheet → Constructing functions from points View worksheet → Constructing functions from word problems View worksheet → Recursive formulas for arithmetic sequences View worksheet → Recursive formulas for geometric sequences View worksheet → Converting between recursive and explicit formulas for arithmetic sequence View worksheet → Converting between recursive and explicit formulas for geometric sequence View worksheet → Building arithmetic sequences from a context View worksheet → Building geometric sequences from a context View worksheet →
13855	https://brainly.com/question/8891449	[FREE] A kangaroo wants to arrange the twelve numbers from 1 to 12 in a circle such that any neighboring numbers - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +55,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +31,8k Ace exams faster, with practice that adapts to you Practice Worksheets +8,4k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified A kangaroo wants to arrange the twelve numbers from 1 to 12 in a circle such that any neighboring numbers always differ by either 1 or 2. Which of the following pairs of numbers have to be neighbors? 1 See answer Explain with Learning Companion NEW Asked by SupahPiggeh • 02/24/2018 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 56284 people 56K 5.0 3 Upload your school material for a more relevant answer Start by laying out the 12. The neighbors of 12 must be 11 and 10. The neighbor of 11 must be 9, and the neighbor of 10 must be 8, and so on and so forth, until you get a unique arrangement. Answered by jijishero •61 answers•56.3K people helped Thanks 3 5.0 (2 votes) 1 Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 56284 people 56K 5.0 2 Inorganic Coordination Chemistry - Kai Landskron The Basics of General, Organic, and Biological Chemistry - David W Ball A Practicum in Behavioral Economics - Arthur J Caplan Upload your school material for a more relevant answer The pairs of numbers that must be neighbors when arranged in a circle such that neighboring numbers differ by 1 or 2 include (11, 12) and (10, 11). This analysis follows the relationship rules of numbers, considering their proximity in circular arrangement and the conditions given. Explanation To solve the problem of arranging the numbers from 1 to 12 in a circle with the condition that neighboring numbers differ by only 1 or 2, we can start by analyzing the circle's structure and the properties of the numbers themselves. Understanding the Differences: The neighboring numbers must differ by 1 or 2. This means that if a number is n, its potential neighbors can be: n - 2 n - 1 n + 1 n + 2 Identifying Neighbors for Each Number: We can set up a relationship between the numbers: 1 can only be neighbors with 2 and 3. 2 can be with 1, 3, and 4. 3 with 1, 2, 4, and 5. 4 with 2, 3, 5, and 6. … and so on until 12. Analyzing Boundary Numbers: Looking at the boundary numbers: Number 12 can only be neighbors with 11 and 10. Number 11 can be with 10, 12, and 9, meaning 11 is directly connected with 10 and 12 as stated. Conclusion: From this pattern, we can see that pairs must be neighbors to satisfy the circle's conditions, which are based on the wrap-around nature of the circle. This leads to the following necessary pairs that must be adjacent: (1, 2) (2, 3) (11, 12) (10, 11) These established relationships enable relations that proceed through the entire sequence from 1 to 12, ensuring compliance with the given constraints. Examples & Evidence For example, if we take the numbers 10, 11, and 12, we find that number 12 neighbors only 11 and 10, complying with their difference conditions. Similarly, with the arrangement involving number 1, number 1 only neighbors 2 and 3. This illustrates how numbers must connect based on their difference constraints and ensures a valid circular arrangement. This conclusion can be verified through logical deduction of each number's potential neighbors and checking all arrangements to note the consistent neighbor pairs throughout. Each number's possible neighbors will always hold true, reflecting the ability to arrange the numbers in compliance with circle conditions. Thanks 2 5.0 (1 vote) Advertisement SupahPiggeh has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 3 The average value of two positive numbers is 30% less than one of the two numbers. By which percentage is average value bigger than the other number? Community Answer 4.5 20 Ben frank is an electrician who installs lighting fixtures. He charges (in dollars) his clients 15h + 40f where h is the number of hours worked and f is the number of fixtures installed. How much does Ben Frank charge when works 7 hours andbinstalls 9 fixtures? Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? New questions in Mathematics 24 b sec 5×4 Choose an equation that would be used to solve 0=−x 2+10 x−8. Solve the equation to find where the trestle meets ground level. Enter your answers from the nearest tenth. The approximate antler length L (in inches) of a deer buck can be modeled by L=9 3 t+15 where t is the age in years of the buck. If a buck has an antler length of 36 inches, what is its age? Cholesterol levels (in m g/d L) were collected from a random sample of 24 patients two days after they had a heart attack. \| Cholesterol Levels (in mg/dL) \| \| 186 \| \| 224 \| \| 280 \| \| 236 \| \| 142 \| \| 226 \| \| 244 \| \| 282 \| \| 236 \| \| 220 \| \| 278 \| \| 272 \| \| 234 \| \| 276 \| \| 310 \| \| 288 \| \| 242 \| \| 280 \| \| 294 \| \| 282 \| \| 160 \| \| 288 \| \| 206 \| \| 266 \| For the data shown above, find the following. Do not round any of your answers. a) Find the 5-number summary: Evaluate 1 F 4+9 C 2 in base Sixteen. 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13856	http://ceemrr.com/Geometry1/Diagonals/Diagonals_print.html	Diagonals in Quadrilaterals Main Theorems The diagonals of a quadrilateral can determine whether it is a parallelogram, a rectangle, a rhombus, etc.. We will list and prove the main theorems here. Theorem 1: If the diagonals of a quadrilateral bisect each other then the quadrilateral is a parallelogram. Proof: Let the quadrilateral be ABCD with diagonals AC and BD intersecting at P: Then AP = PC and DP = PB since the diagonals bisect each other. Now consider triangles APD and CPB. The vertical angles APD and CPB are equal, and we have pairs of sides that are equal (AP = PC and DP = PB), so these triangles are congruent by SAS. As a consequence, the corresponding angles, DAP and BCP are equal. But these are alternate interior angles for lines AD and BC with transversal AC. So this proves side AD is parallel to side BC. Similarly, if we consider triangles APB and CPD, they are congruent by the same reasoning, so corresponding angles BAP and DCP are equal. These are alternate interior angles for lines AB and DC, so those lines are also parallel. Therefore ABCD is a parallelogram, by definition. Theorem 2: If the diagonals of a quadrilateral bisect each other and have the same length, then the quadrilateral is a rectangle. Proof: Since the diagonals bisect each other, we already know (from Theorem 1) that it is a parallelogram, so all we need to prove is that the angles at the vertices are right angles. Again let the quadrilateral be ABCD with diagonals AC and BD intersecting at P. Since the diagonals bisect each other, P is the midpoint of both diagonals. That is, AP = PB and CP = PD. But the diagonals are also of the same length, so AP + PB = CP + PD, and by substitution this gives us AP + AP = CP + CP, or 2AP= 2CP. That is, AP = CP. Likewise, PB = PD. Consequently, all 4 triangles APD, APB, CPD, and BPC are isosceles. So angles PAD and PDA are congruent, angles PBC and PCB are congruent, angles PAB and PBA are congruent, and angles PDC and PCD are congruent. We already saw in the proof of Theorem 1 that triangles APD and CPB are congruent, as are triangles APB and CPD. By angle addition, it follows that the 4 angles of the quadrilateral (angles ABC, BCD, CDA, and DAB) are all equal. But the angles of a quadrilateral add to 360o, and therefore each of these 4 angles must be 90o. Theorem 3: If the diagonals of a quadrilateral bisect each other and are perpendicular then the quadrilateral is a rhombus. Proof: Again let the quadrilateral be ABCD with diagonals AC and BD intersecting at P: Since they bisect each other and are perpendicular, triangles APB, BPC, CPD, and DPA are right triangles. They are all congruent by SAS. For example, triangle APB is congruent to triangle CPB because they share a common side BD, sides AP and CP are congruent (since P is the midpoint of AC), and the included angles are both right angles. Since they are all congruent, their third sides (the hypotenuse of each) are congruent (CPCTC). A rhombus is a quadrilateral in which all four sides are congruent, so ABCD is a rhombus. Theorem 4: If the diagonals of a quadrilateral are perpendicular and one bisects the other, then the quadrilateral is a kite. That is, if the quadrilateral is ABCD with diagonals AC and BC intersecting at P, and if AP = CP, then AB = BC and AD = BD: Proof: In this case, triangles APB and CPB are congruent (by SAS), and triangles APD and CPD are congruent. Therefore AB = BC and AD = CD. Narrowing the Type When a quadrilateral is known to be of certain special type, then additional properties of the diagonals can narrow the type. The following theorems demonstrate this: Theorem 5: If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Proof: Let the parallelogram be ABCD with congruent diagonals AC and BD. Consider the overlapping triangles ADC and BCD. Since opposite sides of a parallelogram are congruent, AD = BC. Since the diagonals of the parallelogram are congruent, AC = BD, and the overlapping triangles have a common side, DC. Therefore they are congruent by SSS. So angle ADC and BCD are congruent. But these are same-side interior angles for parallel lines AD and BC with transversal DC. Since same-side interior angles add to 180o, each must be 90o, so the parallelogram is a rectangle. This theorem is often used by carpenters to check a door or window to see if it is really rectangular. First the carpenter measures the opposite sides. If they are the same length, then he measures the diagonals. If they too are the same length, then he knows the angles are right angles. Theorem 6: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. Proof: Let the parallelogram be ABCD with perpendicular diagonals AC and BD intersecting at P: The diagonals of a parallelogram bisect each other, so triangles APB and CPB are congruent by SAS. Therefore the corresponding parts, sides AB and CB are congruent. Likewise, triangles BPC and DPC are congruent, so sides BC and DC are congruent, and similarly sides AD and CD are congruent. So all 4 sides are congruent, which makes the parallelogram a rhombus. There are other theorems which could be stated, but the main ideas revolve around congruent triangles formed by the diagonals and sides of the quadrilateral.
13857	https://thecontentauthority.com/blog/tenuity-vs-tenuousness	Tenuity vs Tenuousness: Deciding Between Similar Terms Skip to Content Home Grammar Capitalization Definitions Idioms Parts of Speech Word Lists Word Usage Blog About Write For Us Login Tenuity vs Tenuousness: Deciding Between Similar Terms Home » Grammar » Word Usage Are you confused about the difference between tenuity and tenuousness? You’re not alone. These two words are often used interchangeably, but they actually have distinct meanings. In this article, we’ll explore the definitions of tenuity and tenuousness, and clarify which of the two is the correct word to use in different contexts. Let’s define the terms. Tenuity refers to the quality of being thin, fine, or delicate. It can also refer to the rarity or scarcity of something. Tenuousness, on the other hand, refers to the quality of being weak, flimsy, or insubstantial. It can also refer to the lack of clarity or certainty in a situation. next stay CC Settings Off Arabic Chinese English French German Hindi Portuguese Spanish Font Color white Font Opacity 100%Font Size 100%Font Family Arial Text Shadow none Background Color black Background Opacity 50%Window Color black Window Opacity 0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25% 200%175%150%125%100%75%50% Arial Georgia Garamond Courier New Tahoma Times New Roman Trebuchet MS Verdana None Raised Depressed Uniform Drop Shadow White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% White Black Red Green Blue Yellow Magenta Cyan 100%75%50%25%0% So, which of these words is the proper one to use? It depends on the context. If you’re describing something that is physically thin or delicate, tenuity is the correct word to use. For example, you might say that a spider’s web has a remarkable tenuity. If you’re describing something that is weak or insubstantial, tenuousness is the correct word. For example, you might say that a politician’s argument has a tenuous connection to the truth. Now that we’ve clarified the difference between tenuity and tenuousness, let’s explore some common situations where people often confuse the two words. Define Tenuity Tenuity is a term that refers to the state or quality of being thin, weak, or insubstantial. It is often used to describe something that lacks density or substance, such as a flimsy piece of paper or a wispy cloud in the sky. Tenuity can also be used to describe a lack of strength or durability, such as a fragile piece of glass that is easily broken. Define Tenuousness Tenuousness, on the other hand, refers to the state or quality of being thin, fine, or delicate. It is similar to tenuity in that it can describe something that lacks substance or strength, but it often carries a connotation of fragility or vulnerability. For example, a tenuous relationship might be one that is easily broken or strained, while a tenuous argument might be one that is based on flimsy or weak evidence. Overall, tenuousness is a term that is often used to describe something that is delicate or precarious, while tenuity is more often used to describe something that is simply thin or insubstantial. How To Properly Use The Words In A Sentence Choosing the right word can make all the difference in the clarity and impact of your writing. When it comes to tenuity and tenuousness, it’s important to understand the nuances of each word to use them correctly in context. How To Use Tenuity In A Sentence Tenuity refers to the quality of being thin, rare, or delicate. It can also refer to the lack of substance or strength. Here are some examples of how to use tenuity in a sentence: The tenuity of the spider’s web made it easy to tear. His argument lacked tenuity and was easily refuted. The tenuity of the atmosphere at high altitudes can cause breathing difficulties. How To Use Tenuousness In A Sentence Tenuousness, on the other hand, refers to the quality of being flimsy, weak, or uncertain. It can also refer to a lack of clarity or definition. Here are some examples of how to use tenuousness in a sentence: The tenuousness of their relationship was evident in their lack of communication. There was a tenuousness to her argument that left many questions unanswered. The tenuousness of the bridge made me nervous as I crossed it. By understanding the subtle differences between tenuity and tenuousness, you can choose the right word to convey your meaning with precision and impact. More Examples Of Tenuity & Tenuousness Used In Sentences As we delve deeper into the differences between tenuity and tenuousness, it’s important to examine how they are used in sentences. Here are some examples: Examples Of Using Tenuity In A Sentence The tenuity of the spider’s web was impressive. His argument lacked tenuity and was easily refuted. The tenuity of the ice made it unsafe to walk on. The tenuity of the thread made it difficult to sew the button back on. She was impressed by the tenuity of the artist’s brushstrokes. The tenuity of the wire made it easy to bend into shape. The tenuity of the material made it ideal for use in the aerospace industry. The tenuity of the plot made the movie difficult to follow. The tenuity of the excuse was obvious to everyone in the room. The tenuity of the bridge made it unsafe for heavy traffic. Examples Of Using Tenuousness In A Sentence The tenuousness of the relationship was evident from the lack of communication. The tenuousness of the peace agreement was a cause for concern. The tenuousness of her grasp on reality was becoming increasingly apparent. The tenuousness of the evidence led to the case being dismissed. The tenuousness of the connection between the two events was tenuous at best. The tenuousness of the situation made it difficult to make a decision. The tenuousness of the thread made it easy to break. The tenuousness of the bridge made it unsafe for heavy traffic. The tenuousness of the argument was exposed under cross-examination. The tenuousness of the theory was revealed by further research. Common Mistakes To Avoid When it comes to using tenuity and tenuousness, people often make the mistake of using these two words interchangeably. However, there is a distinct difference between the two words that should not be overlooked. Here are some common mistakes to avoid: Using Tenuousness When You Mean Tenuity One of the most common mistakes people make is using tenuousness when they actually mean tenuity. Tenuousness refers to something that is weak or flimsy, while tenuity refers to something that is thin or fine. For example, saying “the tenuousness of the argument” would be incorrect if you meant to convey that the argument was thin or fine. Instead, you should use tenuity and say “the tenuity of the argument.” Using Tenuity When You Mean Tenuousness Conversely, some people use tenuity when they actually mean tenuousness. Tenuity refers to something that is thin or fine, while tenuousness refers to something that is weak or flimsy. For example, saying “the tenuity of the bridge” would be incorrect if you meant to convey that the bridge was weak or flimsy. Instead, you should use tenuousness and say “the tenuousness of the bridge.” Using Both Words Interchangeably Another common mistake is using both words interchangeably, without regard for their distinct meanings. This can lead to confusion and miscommunication. To avoid this mistake, it’s important to understand the difference between tenuousness and tenuity, and to use the appropriate word depending on the context. Tips For Avoiding These Mistakes To avoid these common mistakes, here are some tips: Take the time to understand the distinct meanings of tenuousness and tenuity Think carefully about the context in which you are using these words Consider using a thesaurus to find alternative words that may be more appropriate Context Matters When it comes to choosing between the words tenuity and tenuousness, context is key. Both words refer to the quality of being thin, weak, or insubstantial, but they are not always interchangeable. The choice between these two words depends on the context in which they are used. Examples Of Different Contexts Let’s take a look at some examples of different contexts and how the choice between tenuity and tenuousness might change: \| Context \| Example Sentence \| Preferred Word \| --- \| Physics \| The tenuity of the atmosphere decreases with altitude. \| Tenuity \| \| Art \| The tenuousness of the brushstrokes gives the painting a dreamlike quality. \| Tenuousness \| \| Business \| The company’s financial situation is tenuous at best. \| Tenuousness \| \| Language \| The tenuousness of the argument was exposed by the opposing viewpoint. \| Tenuousness \| In physics, tenuity is often used to describe the thinness or rarity of a gas or atmosphere. In this context, tenuousness would not be appropriate. On the other hand, in art, tenuousness is often used to describe something that is delicate or ethereal. In this context, tenuity would not be appropriate. In business, tenuousness is often used to describe a situation that is uncertain or unstable. In this context, tenuity would not be appropriate. Similarly, in language, tenuousness is often used to describe an argument or idea that is weak or unconvincing. In this context, tenuity would not be appropriate. Overall, the choice between tenuity and tenuousness depends on the context in which they are used. It is important to consider the nuances of each word and choose the one that best fits the intended meaning. Exceptions To The Rules While the rules for using tenuity and tenuousness are generally straightforward, there are a few exceptions to keep in mind. 1. Contextual Usage Depending on the context, one word may be more appropriate than the other. For example, tenuousness may be more suitable when discussing abstract concepts or intangible ideas, while tenuity may be a better fit for physical objects or materials. Example 1: The tenuousness of the political situation made it difficult to predict the outcome of the election. Example 2: The tenuity of the thread made it prone to snapping under pressure. 2. Regional Variations There may be regional variations in the usage of tenuity and tenuousness. For instance, in British English, tenuousness is more commonly used than tenuity. Example: The tenuousness of the argument was apparent to everyone in the room. 3. Technical Jargon In technical jargon, tenuity may refer to the ratio of the density of a substance to that of a reference substance, while tenuousness may refer to the thinness or scarcity of a gas or fluid. \| Tenuity \| Tenuousness \| --- \| \| The tenuity of the gas mixture was calculated using a specialized tool. \| The tenuousness of the atmosphere at high altitudes can cause breathing difficulties. \| By keeping these exceptions in mind, you can use tenuity and tenuousness more effectively and accurately in your writing. Practice Exercises One of the best ways to improve your understanding and use of tenuity and tenuousness is to practice using them in sentences. Here are some exercises to help you do just that: Exercise 1: Choose the correct word (tenuity or tenuousness) to complete each sentence: The ____ of the rope made it difficult to climb. The ____ of his argument was impressive. The ____ of the spider’s web was amazing. The ____ of the bridge was questionable. The ____ of the book’s plot was hard to follow. Answer Key: tenuousness tenuity tenuity tenuousness tenuousness Exercise 2: Write a sentence using each of the following words: tenuity tenuousness Answer Key: The tenuity of the ice made it dangerous to walk on. The tenuousness of their relationship was evident in their lack of communication. By practicing with these exercises, you can improve your understanding and use of tenuity and tenuousness in your writing and communication. Conclusion In conclusion, understanding the difference between tenuity and tenuousness is crucial for effective communication. While both words pertain to thinness or lack of substance, tenuity specifically refers to the quality of being thin or rarefied, whereas tenuousness refers to the lack of strength or stability. It is important to use these words correctly in order to avoid confusion and miscommunication. Additionally, this article highlights the importance of paying attention to word choice and grammar in order to effectively convey meaning. Key Takeaways: Tenuity refers to the quality of being thin or rarefied. Tenuousness refers to the lack of strength or stability. Using these words correctly is important for effective communication. Paying attention to word choice and grammar is crucial for conveying meaning. As language is constantly evolving, it is important to continue learning and expanding our knowledge of grammar and language use. By doing so, we can improve our communication skills and avoid misunderstandings. Shawn Manaher Shawn Manaher is the founder and CEO of The Content Authority. He’s one part content manager, one part writing ninja organizer, and two parts leader of top content creators. You don’t even want to know what he calls pancakes. 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13858	https://www.youtube.com/watch?v=4vsngbLrm6Y	Quadratic Inequality in One Variable with Case Analysis Todd Leinweber 288 subscribers 51 likes Description 17135 views Posted: 13 Apr 2013 9 comments Transcript: okay so this is an example of a quadratic inequality in one variable that I'm going to solve um through the use of a case analysis so before I can do that I'm going to have to factor it so what I'm going to do is just going to in this equation I just don't like the way that it's uh set up here I don't really like negative uh coefficients on my x s I just I'm just not used to factoring those so what I'm going to do is put everything put everything to the other side so that my coefficient on x^2 is positive um and then I can go ahead and Factor this so if I decompose that middle term into those two numbers 15 and minus one um I'll be able to factor it so I can see here that uh a 3X -1 and an x + 5 are my two factors so the X intercepts for this quadratic equation are therefore X = to A3 and X = -5 so if I want to do a case analysis I'm going to have to draw a number line so essentially what I have here is two factors 3x - 1 x + 5 so two factors so if I knew their uh if I knew the sign of each ter term I could decide whether the Y values um after you input the the X I can decide whether the yv values are positive or negative and this is what I'm looking for in my uh inequality so for example if I take a a test point in this in this range like a minus 6 for example and I go ahead and plug uh plug that into plug that into this term minus 6 and see what happens and I plug that into- 6 into that term and see what happens what I'm going to get is uh two negative values multiplied by each other and what's going to happen is that's going to turn out positive so I know that the Y values in this in this data range okay the Y values are going to be positive testing for uh in between minus5 and 1/3 I'll just choose zero uh that's easiest and then you can see that I'll have a minus uh from this term if I plug in zero and I'll have a positive from this term if I plug in zero so negative time positive and that's going to come out negative so the Y values are going to be uh negative in this in this data range here finally for the last uh section here uh X greater than a 3 I'll just pick a value of two for x or one for that matter one will be fine so the first term if I plug in a one that'll be positive and if I plug in and a one here that will also be positive so here I have positive times positive and that of course will come out positive so here I have a I have proved um on this number line these these range of X values and uh they're corresponding outputs in y positive for X less than -5 negative for X between -5 and 1/3 and positive for X greater than 1/3 got
13859	http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/jackie2.html	4 times as many or 4 times more? - Math Central ← BACKPRINT+ TEXT SIZE –SEARCHHOME Math Central Quandaries & Queries Question from Jackie: Dear Sir , Gratefully I would like to extend my big thanks to you for spending time attending to my query.It is indeed very helpful! I have a problem with the semantics and syntax of phrasing the conclusion can be made for the below statement Given : Here are 3 squares and 4 sets of 3 circles. I wonder it is right to write in the below manner to represent the following Conclusion that can be made from the above given information: 1. There are 4 times as many circles as there are squares, 2. There are 4 times fewer square than circles; 3. There are 4 times more circles than squares. 4. there are 1/4 as many squares as circles 5. I wonder whether it make sense to say ; there are 1/4 fewer squares than circles or say there are 1/4 times fewer squares than circles? ( do they belong to the additive or multiplicative situation) 6. there are 9 more circles than squares 7. there are 9 fewer squares than circles I would like to thank you in anticipation. Jackie Jackie, [Three squares, twelve circles.] There are 4 times as many circles as there are squares, This is definitely right. There are 4 times fewer square than circles; There are 4 times more circles than squares. and 3. are commonly used but confuse the difference with the ratio. I would try to avoid them except perhaps in very informal speech. there are 1/4 as many squares as circles Definitely correct. I wonder whether it make sense to say ; there are 1/4 fewer squares than circles or say there are 1/4 times fewer squares than circles? ( do they belong to the additive or multiplicative situation) Both of these are bad, and if they mean anything mean that the number of squares was 3/4 of the number of circles (1/4 fewer). AVOID this one at all costs. there are 9 more circles than squares there are 9 fewer squares than circles and 7. are both correct. Note also that percentages are used slightly differently. Their customary use is to describe changes or subsets. We can say "5% of the men are sailors" just as we would say "1/20 of the men are sailors" but using them to describe disjoint groups is harder: "The number of men is 80% of the number of women." The normal way to use percentages here is to describe differences, specifically describing the size of one group in terms of the difference as a percentage of the other group. However, as a result of this, percentages should be avoided in circumstances where this is not a natural way to look at things. [8 boys,10 girls] "There are 25% more girls/ There are 20% fewer boys" will usually be correctly understood [5 boys, 20 girls] "There are 300% more girls" is likely to be misunderstood as "There are three times as many girls" [5 boys, 15 girls] "There are 75% fewer boys" is likely to be misunderstood as "The number of boys is 75% that of girls" [15 boys, 20 girls] Percentages are best kept for small proportions and changes. -RD Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
13860	https://zhuanlan.zhihu.com/p/602127327	直答首发于高中数学笔记切换模式高中数学-常见求和公式咖啡和茶健康快乐最重要！别总执着得不到。收录于 · 高中数学笔记 11 人赞同了该文章目录收起前言 1.常见求和公式及证明 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 结语前言我们都知道，等差数列和等比数列是有固定公式的，求和非常便捷，但是常见的一些内容中有些可以直接利用，有些却不行，本次为大家列举一些常见的求和公式。 1.常见求和公式及证明 1.1. \sum\limits_{k=1}^n k = \frac{n(n+1)}{2} \sum\limits_{k=1}^n k = 1+2+3+...+n = \frac{n(n+1)}{2} 1.2. \sum\limits_{k=1}^n (2k-1) = n^2 这个本质上还是等差数列，就可以用等差数列求和公式： \frac{n(a_1 + a_n)}{2} ，当我们知道第一项和最后一项和项的个数时，我们优先使用这个公式。 \sum\limits_{k=1}^n (2k-1) = 1 + 3 +5 +...+ (2n-1) \ = \frac{n (1+2n-1)}{2} \ = n^2 1.3. \sum\limits_{k=1}^n (k^2) = \frac{n(n+1)(2n+1)}{6} 这个公式比较难证明，因为需要你熟悉立方相关的式子。本式需要构拟出各个平方的数相加，故我们创造这样的数列： (n+1)^3 - n^3 = 3n^2 +3n + 1 n^3 - (n-1)^3 = 3(n-1)^2 +3(n-1) + 1 (n-1)^3 - (n-2)^3 = 3(n-2)^2 +3(n-2) + 1 ... 我们把左边的式子和右边的式子相加，就可以发现我们要求的公式了： (n+1)^3 - 1 = 3 [n^2 + (n-1)^2 + ... + 1] + 3 [n + (n-1) + ... + 1] + n 这里，我们为了简便，我们把要求的 [n^2 + (n-1)^2 + ... + 1] 简写为 S，那么就可以继续推导： (n+1)^3 - 1 = 3S + 3 [n + (n-1) + ... + 1] + n \ 3S = (n+1)^3 - 1 - 3 [n + (n-1) + ... + 1] -n \ 3S = n^3 + 3n^2 + 3n + 1 -1 - 3 [\frac{n(n+1)}{2}] - n \ 3S = n^3 + 3n^2 + 3n -\frac{3}{2} n^2 - \frac{3}{2}n - n \ 3S = n^3 + \frac{3}{2} n^2 + \frac{1}{2}n \ 3S = \frac{2n^3 + 3n^2 + n}{2} \ S = \frac{2n^3 + 3n^2 + n}{6} 由于这个公式比较难背，而且代入变量n的时候，万一是个复杂的式子，求立方毕竟不方便，我们通常进行因式分解： S = \frac{2n^3 + 3n^2 + n}{6} \ S = \frac{ n (2n^2 + 3n + 1)}{6} \ S = \frac{ n (n+1) (2n+1)}{6} \ 最后的这个公式如何记忆呢？因为推导过公式，要先记住分母肯定是个6，因为左边3S右边分母是2，最后是个6。分子呢？上边是个常见的求和公式里的 n(n+1) ，最后一个要乘以的分子的因式恰好是 n+n+1 = 2n+1 。大家可以通过这两个线索去记忆。 1.4. \sum\limits_{k=1} ^n k^3 = [ \frac{n(n+1)}{2} ]^2 本公式属于记忆好记忆，但是推导困难。本式其实就是先求和再平方，也就是： \sum\limits_{k=1} ^n k^3 = ( \sum\limits_{k=1}^n k)^2 。这样做的好处是，代数部分（也就是变量部分），很早就推导完了，后面就是把数字进行算术计算。推导过程和上面的公式类似，了解即可。我们还是先强行构造一个数列： (n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1 \ n^4 - (n-1)^4 = 4(n-1)^3 + 6(n-1)^2 + 4(n-1) + 1 \ ... 等式两边相加得： (n+1)^4 - 1 = 4 [n^3 + (n-1)^3 + ... +1 ] + 6 ( 1^2 + 2^2 + ... + n^2) + 4 (1+2+...+n) + n 我们令 S = n^3 + (n-1)^3 + ... +1 ，得： (n+1)^4 - 1 = 4 S + 6 ( 1^2 + 2^2 + ... + n^2) + 4 (1+2+...+n) + n 4 S = (n+1)^4 - 1 - 6 （\frac{n(n+1)(2n+1)}{6}）- 4[\frac{n(n+1)}{2}] -n S = [ \frac{n(n+1)}{2} ]^2 1.5. \sum \limits_{k=1}^n k(k+1) = \frac{n(n+1)(n+2)}{3} 本式直接求和即可： a_n = n(n+1) = n^2 +n a_1 = 1^2 + 1 \ a_2 = 2^2 + 2 \ ... \ a_n = n^2 +n 两边相加，易得： S_n = (1^2 + 2^2 + ... + n^2) + (1+2+...+ n) \ S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \ S_n = \frac{n(n+1)(2n+1)+3n(n+1)}{6} \ S_n = \frac{(n+1)(2n^2 +n +3n)}{6} \ S_n = \frac{(n+1)(2n^2 + 4n)}{6} \ S_n = \frac{(n+1)(n^2+2n)}{3} \ S_n = \frac{n(n+1)(n+2)}{3} 如何记忆呢？你就记住这个求和的时候，会比 \sum\limits_{k=1}^n (k^2) = \frac{n(n+1)(2n+1)}{6} 多一个 \frac{n(n+1)}{2} ，所以分母就变成3了。 1.6. \sum\limits_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1} \sum\limits_{k=1}^n \frac{1}{k(k+1)} \ = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + ... + \frac{1}{n \times (n+1)} \ = 1-\frac{1}{2} + \frac{1}{2} -\frac{1}{3} + ... + \frac{1}{n} - \frac{1}{n+1} \ = \frac{n}{n+1} 结语综上，我们发现公式1.3在求其他公式时频繁出现，故公式1.3的记忆尤为重要。大家可以在评论区相互答疑。编辑于 2023-01-31 22:04・北京高中高中数学写下你的评论... 还没有评论，发表第一个评论吧关于作者咖啡和茶健康快乐最重要！别总执着得不到。回答 67 文章 68 关注者 3,146 推荐阅读 # 从数列求和谈数学基本思维能力(1) 许兴华数学 # 呆哥数学数列合集——数列通项公式的6种求法类型4和5【9】高中数学李老师 # 高中数学系列之数列(二) Zephy... 发表于摸鱼与划水... # 高中数学，等差、等比数列及其应用——257页高考复习专题数列是高中数学的重要内容，近三年考纲要求，等差、等比数列都是C级要求，考试题多以中等及以上难度为主。填空题常常考查等差、等比数列的通项公式、前n项和公式及等差、等比数列的性质，… 阿七酱想来知乎工作？请发送邮件到 jobs@zhihu.com 未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App
13861	https://www.youtube.com/watch?v=cE22xKvi2fI	Solving Word Problems Involving Inequalities - Example 1 Patrick J 1400000 subscribers 524 likes Description 131054 views Posted: 26 Jun 2010 Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) !! Solving Word Problems Involving Inequalities - Example 1 49 comments Transcript: all right in this video I'm going to do an example of a word problem involving some inequalities so suppose we've got a a widget Factory um so whatever the heck a widget is but suppose a factory is making something and they've got a fixed operating cost of $3600 a day uh plus costs it costs $140 for them to make a widget suppose they can sell these things for $4.20 each and we want to know what's the least number of these uh these things these widgets M must they sell a day in order to make a profit okay so a couple different ways you know maybe you could go about approaching this but I think the basic idea really to get started here you know in order to make a profit to me it says your revenue or your profit um Revenue uh that to me must be greater than your um costs if if you're going to make a profit um if it was equal to then you're exactly breaking even so I'm going to say it's got to be strictly greater um well let's think about the revenue you know so so if you're selling these objects suppose um we say uh we'll let just to Define our variables let's let X be the uh number of widgets sold okay so since we're selling them for $4 20 each your profit would be 420 uh time x you just multiply the the price you earn right we get 420 for each one we're going to multiply that by the number sold and that would give us uh our our Revenue well our costs we know that we've got to pay $3,600 a day maybe they're renting a big Factory um you know electricity Etc but then they also have costs I mean if you're making you know if you're making an object there's certainly uh costs involved in making those and they cost $140 per each widget produced so now the idea is i' I've got a nice little inequality setup that I can hopefully solve without too much trouble so it says we have 4.20 X and that's going to be greater than we want that to be greater than 3600 + 1.4x so I'm going to subtract off uh the 1 4X from both sides so uh $4.20 if we take away $2 uh4 uh I think I said $4.20 take away $140 uh that's going to be 280 and again that needs to be uh now bigger than 3600 and now I'm just going to divide both sides by 2.80 2.80 this to me um now pretty much gets us there so I'm going to use a calculator here uh 36 600 / 2.80 I'm getting 128.57 okay so now this is obviously where we have to interpret things a little bit you know assumingly you know you can only make a whole number of objects so it it basically says you have to make 1,285 71 or more so to me that would tell me that we would have to really make you know 1,28 86 um or more widgets uh to make a profit
13862	https://fiveable.me/computational-algebraic-geometry/unit-1/algebraic-sets-geometric-interpretation/study-guide/cg2QOyjkD8cEfmzZ	printables 🌿Computational Algebraic Geometry Unit 1 Review 1.2 Algebraic sets and geometric interpretation 🌿Computational Algebraic Geometry Unit 1 Review 1.2 Algebraic sets and geometric interpretation Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 APA 🌿Computational Algebraic Geometry Unit & Topic Study Guides 1.1 Historical overview and fundamental concepts 1.2 Algebraic sets and geometric interpretation 1.3 Connections between algebra and geometry Algebraic sets are the building blocks of algebraic geometry, representing solutions to polynomial equations. They bridge algebra and geometry, allowing us to study geometric objects through their defining equations and vice versa. In this part of the chapter, we'll explore how algebraic sets are defined, their properties, and their geometric interpretations. We'll see how these concepts lay the foundation for understanding more complex ideas in algebraic geometry. Algebraic sets and their properties Definition and basic properties An algebraic set is the solution set of a system of polynomial equations over a field It is a subset of affine or projective space defined by the vanishing of a collection of polynomials The Zariski topology on affine or projective space is defined by taking algebraic sets as the closed sets This topology is coarser than the Euclidean topology Irreducibility and dimension An algebraic set is irreducible if it cannot be written as the union of two proper algebraic subsets Irreducible algebraic sets are the building blocks of algebraic geometry The dimension of an algebraic set is the maximum length of a chain of irreducible subsets Length is defined as the number of strict inclusions The dimension of an irreducible algebraic set equals the Krull dimension of its coordinate ring The singular locus of an algebraic set is the subset of points where the Jacobian matrix of the defining equations does not have full rank The complement of the singular locus is called the smooth locus Algebraic sets can be classified and studied based on properties such as reducibility, singularity, and dimension Geometric interpretation of algebraic sets Visualization in various dimensions Algebraic sets in affine or projective space can be visualized as geometric objects The geometry of an algebraic set reflects the properties of its defining equations In the affine plane (dimension 2), algebraic sets are curves defined by polynomial equations in two variables Examples include lines, parabolas, ellipses, hyperbolas, and more complicated curves In affine 3-space, algebraic sets are surfaces defined by polynomial equations in three variables Examples include planes, spheres, cylinders, and other surfaces that can be described algebraically In higher dimensions, algebraic sets can represent hypersurfaces, curves, or other geometric objects Visualization becomes more challenging, but techniques from algebraic geometry still apply Singularities and smooth points Singular points of an algebraic set correspond to geometric singularities Examples include cusps, nodes, or self-intersections The smooth locus represents the non-singular part of the geometric object Techniques such as projections, intersections, and parametrizations can be used to study the geometric properties of algebraic sets These techniques relate the algebraic descriptions to their geometric counterparts Algebraic sets vs defining equations Ideals and coordinate rings An algebraic set is defined by a collection of polynomial equations The structure of these equations determines the geometric properties of the algebraic set The ideal of an algebraic set is the set of all polynomials that vanish on the set This ideal captures the algebraic relations satisfied by the points of the set The coordinate ring of an algebraic set is the quotient of the polynomial ring by the ideal of the set It encodes the algebraic functions on the set and reflects its geometric properties Nullstellensatz and operations on algebraic sets The Nullstellensatz establishes a correspondence between radical ideals and algebraic sets Every radical ideal is the ideal of some algebraic set, and conversely, the ideal of an algebraic set is always radical Operations on algebraic sets correspond to operations on their defining ideals Union corresponds to sum, intersection to intersection, and complement to quotient The prime decomposition of an ideal corresponds to the irreducible decomposition of the corresponding algebraic set This allows studying reducible algebraic sets in terms of their irreducible components Construction and manipulation of algebraic sets Constructing algebraic sets Constructing algebraic sets involves finding polynomial equations that define the desired geometric object This can be done by using algebraic techniques or by geometric reasoning Parametrization is a technique for describing algebraic sets using rational functions It allows representing the points of an algebraic set in terms of a smaller number of parameters Computational tools and techniques Elimination theory provides methods for eliminating variables from a system of polynomial equations This can be used to project an algebraic set onto a lower-dimensional space or to compute the intersection of algebraic sets The resultant and discriminant are algebraic tools for studying the common solutions of polynomial equations They can be used to determine the existence and multiplicity of intersection points Gröbner bases are a powerful computational tool for solving systems of polynomial equations They provide a systematic way to manipulate and simplify the defining equations of an algebraic set Transformations and real algebraic sets Algebraic sets can be transformed using maps between affine or projective spaces These maps can be defined by polynomials and can be used to study the relationships between different algebraic sets Real algebraic sets are algebraic sets defined over the real numbers They have additional geometric properties and can be studied using techniques from real algebraic geometry Examples include the real Nullstellensatz and semi-algebraic sets
13863	https://www.studypug.com/calculus-help/u-substitution	Master U-Substitution: Simplify Complex Integrals \| StudyPug Courses Sign InTry Free Home Calculus Integration Techniques U-Substitution U-Substitution: The Key to Solving Complex Integrals Unlock the power of U-Substitution in calculus. Learn to simplify and solve challenging integrals with our comprehensive guide, practice problems, and real-world applications. Get the most by viewing this topic in your current grade. Pick your course now. Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 0:00 Loaded: 0% Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Picture-in-Picture Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Now Playing:U substitution– Example 0 Intros Introduction to u-Substitution ⋅\cdot⋅ What is u u u-Substitution? ⋅\cdot⋅Exercise: Find ∫(5 x 4−6 x)cos⁡(x 5−3 x 2)d x\int (5x^4-6x) \cos (x^5-3x^2)dx∫(5 x 4−6 x)cos(x 5−3 x 2)d x. How to pick "u u u"? How to verify the final answer? Examples Integrate: Polynomial Functions ∫−7 x(6 x 2+1)10 d x \int-7x(6x^2+1)^{10}dx ∫−7 x(6 x 2+1)10 d x Integrate: Radical Functions ∫x 6 3(8−5 x 7)2 d x \int\frac{x^6}{{^3}\sqrt{(8-5x^7)^2}}dx ∫3(8−5 x 7)2x 6d x ∫6−3 x \int\sqrt{6-3x} ∫6−3 xd x dx d x Integrate: Exponential Functions ∫e 2 x d x \int e^{2x}dx ∫e 2 x d x Integrate: Logarithmic Functions ∫(ln⁡x)3 x d x \int \frac{(\ln x)^3}{x}dx ∫x(l n x)3d x ∫d x x ln⁡x \int \frac{dx}{x \ln x} ∫x l n x d x Integrate: Trigonometric Functions ∫sin⁡3 x cos⁡x d x\int \sin ^3 x \cos x\; dx ∫sin 3 x cos x d x ∫sec⁡2 x(tan⁡x−1)100 d x\int \sec ^2 x(\tan x-1)^{100}\;dx ∫sec 2 x(tan x−1)100 d x Not-So-Obvious U-Substitution ∫x 3−8 x 5 d x \int \sqrt{x^3-8}x^5dx ∫x 3−8x 5 d x ∫3 1+x 2 x 5 d x \int {^3}\sqrt{1+x^2}x^5dx ∫3 1+x 2x 5 d x ∫1+x 1+x 2 d x \int \frac{1+x}{1+x^2}dx ∫1+x 2 1+xd x ∫cot⁡x\int \cot x ∫cot x d x dx d x Evaluate Definite Integrals in Two Methods Evaluate: ∫−1 2 6−3 x d x\int_{-1}^{2} \sqrt{6-3x} dx ∫−1 26−3 xd x 1. Introduction to definite integrals. 2. Method 1: evaluate the definite integral in terms of "x x x". 3. Method 2: evaluate the definite integral in terms of "u u u". 4. Method 1 VS. Method 2. Evaluate Definite Integrals Evaluate: ∫0 π 3 sin⁡θ cos⁡2 θ d θ\int_{0}^{\frac{\pi}{3}} \frac{\sin \theta}{\cos ^2 \theta}d \theta∫0 3 πc o s 2 θ s i n θd θ Definite Integral: Does Not Exist (DNE) Evaluate: ∫1 5 d x(x−3)2\int_{1}^{5} \frac{dx}{(x-3)^2}∫1 5(x−3)2 d x View All Practice Now Practicing:U Substitution 1 Free to Join! StudyPug is a learning help platform covering math and science from grade 4 all the way to second year university. Our video tutorials, unlimited practice problems, and step-by-step explanations provide you or your child with all the help you need to master concepts. On top of that, it's fun — with achievements, customizable avatars, and awards to keep you motivated. 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Unlock more options the more you use StudyPug. Try Free U-Substitution Jump to:NotesConceptFAQsPrerequisites Notes You may start to notice that some integrals cannot be integrated by normal means. Therefore, we introduce a method called U-Substitution. This method involves substituting ugly functions as the letter "u", and therefore making our integrands easier to integrate. We will use this technique to integrate many different functions such as polynomial functions, irrational functions, trigonometric functions, exponential functions and logarithmic functions. We will also integrate functions with a combination of different types of functions. Pre-requisite: Differential Calculus –"Chain Rule" Integral Calculus –"Antiderivatives" Note: The main challenge in using the u−S u b s t i t u t i o n u-Substitution u−S u b s t i t u t i o n is to think of an appropriate substitution. Question: how to choose u u u? Answer: choose u u u to be some function in the integrand whose differential also occurs! hint: u u u is usually the inside of a function, for example: the inside a power function: (u)10( u )^{10}(u)10 the inside a radical function: u\sqrt{u}u the inside of an exponential function: e u e^u e u the inside of a logarithmic function: ln⁡\ln ln? (u)(u) (u) the inside of a trigonometric function: sin⁡\sin sin(u)(u)(u) Concept Introduction to U-Substitution U-Substitution is a powerful technique in calculus used for integrating complex functions. The introduction video provides a crucial foundation for understanding this concept, making it an essential starting point for students. U-Substitution, also known as substitution integration, involves replacing parts of the integrand with a new variable 'u' to simplify the integration process. This method is particularly useful when dealing with composite functions or expressions that involve products of functions and their derivatives. By strategically choosing a substitution, complex integrals can often be transformed into more manageable forms. U-Substitution is a fundamental tool in the integration toolkit, allowing mathematicians and students to solve a wide range of problems that would otherwise be challenging or impossible using basic integration techniques. Mastering U-Substitution opens doors to solving more advanced integration problems and is a crucial skill for anyone studying calculus or higher mathematics. FAQs 1. What is U-Substitution and why is it important in calculus? U-Substitution is an integration technique used to simplify complex integrals by substituting a part of the integrand with a new variable, typically 'u'. It's important because it allows us to solve integrals that would be difficult or impossible using basic integration methods. This technique is particularly useful for integrals involving composite functions or products of functions and their derivatives. 2. How do I know when to use U-Substitution? U-Substitution is typically appropriate when you notice a composite function in the integrand and its derivative (or part of it) appears as a factor. For example, in e^(2x) 2dx, e^(2x) is a composite function, and its derivative (2e^(2x)) partially appears as the factor 2. This makes it a good candidate for U-Substitution. 3. What are common mistakes to avoid when using U-Substitution? Common mistakes include choosing an inappropriate substitution, forgetting to change the limits of integration for definite integrals, omitting the du/dx term when substituting, and not including the constant of integration for indefinite integrals. Always double-check your work and verify your solution by differentiating the result. 4. Can U-Substitution be used for all types of integrals? While U-Substitution is versatile, it's not suitable for all integrals. It works well for many polynomial, trigonometric, exponential, and logarithmic functions, especially when they appear as composite functions. However, some integrals may require other techniques like integration by parts or trigonometric substitution. 5. How does U-Substitution relate to real-world applications? U-Substitution has practical applications in various fields. In physics, it's used to calculate work done by variable forces. Engineers use it in fluid dynamics problems. Economists apply it to analyze consumer and producer surplus. Biologists use it in population growth models. These applications demonstrate how U-Substitution helps solve complex real-world problems across different disciplines. Prerequisites Understanding U-Substitution in calculus requires a solid foundation in several key mathematical concepts. One of the most fundamental prerequisites is composite functions. These functions are essential because U-Substitution often involves recognizing and manipulating composite functions to simplify complex integrals. Another crucial skill is determining trigonometric functions given their graphs. This ability helps in identifying suitable substitutions, especially when dealing with trigonometric integrals. Similarly, familiarity with the derivative of inverse trigonometric functions is vital, as these often appear in U-Substitution problems. The derivative of logarithmic functions and understanding the natural log (ln) are also important. These concepts frequently arise in U-Substitution problems, particularly when dealing with exponential or logarithmic expressions. Likewise, knowledge of the derivative of exponential functions is essential for tackling a wide range of U-Substitution scenarios. While it may seem unrelated at first, experience with solving 3 variable systems of equations by substitution can enhance your problem-solving skills and substitution techniques, which are directly applicable to U-Substitution in calculus. Understanding the applications of polynomial functions provides context for many U-Substitution problems, as these often involve polynomial expressions. Additionally, the ability to determine the equation of a polynomial function can be helpful in recognizing suitable substitutions and in verifying your results. Mastering these prerequisite topics will significantly enhance your ability to apply U-Substitution effectively. Each concept builds upon the others, creating a robust foundation for tackling complex integration problems. By understanding how these topics interconnect and relate to U-Substitution, you'll be better equipped to recognize patterns, choose appropriate substitutions, and solve a wide variety of calculus problems with confidence. 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13864	https://zhuanlan.zhihu.com/p/639986160	【鸢尾花书系列】矩阵力量-Chapter2向量运算（二） - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册【鸢尾花书系列】矩阵力量-Chapter2向量运算（二）切换模式【鸢尾花书系列】矩阵力量-Chapter2向量运算（二）小狗贝贝Baby 学习编程的可爱小狗~从零开始学习知识系列~ 9 人赞同了该文章目录 2.6 向量内积：结果为标量向量内积 (inner product)，又叫标量积 (scalar product)、点积 (dot product)、点乘注意向量内积的运算结果为标量，而非向量给定如下 a 和 b 两个等行数列向量： 33 列向量 a 和 b 的内积定义如下： 34 运算法则就是对应项相乘同样地，适用于两个等列数的行向量计算内积向量内积也是一种“向量 → 标量”的运算规则例子图 16所示，两个列向量 a 和 b 的内积为： \boldsymbol{a} \cdot \boldsymbol{b}=\left[\begin{array}{l} 4 \ 3 \end{array}\right] \cdot\left[\begin{array}{c} 5 \ -2 \end{array}\right]=4 \times 5+3 \times(-2)=14 图 16. a 和 b 两个平面向量常用的向量内积性质： 36 注意以下几个向量内积运算和 Σ 求和运算的关系： 37 其中， 38 几何视角如图 16所示，向量内积相当于两个向量的模 (L 2范数) 与它们之间夹角余弦值三者之积： 39 注意，上式中 θ 代表向量 a 和 b 的“相对夹角” a 的 L 2范数也可以通过向量内积求得： 40 柯西-施瓦茨不等式从向量内积式子推导，cos θ 的取值范围为 [−1, 1] a 和 b 内积取值范围如下： 42 可视化：图 17所示为 7 个不同向量夹角状态 \| 内容 \| 结论 \| --- \| \| θ = 0°时，cosθ = 1 \| a 和 b 同向，此时向量内积最大 \| \| θ = 180°时，cosθ = −1 \| a 和 b 反向，向量内积最小 \| \| θ = 90°时，cosθ = 0 \| 两者向量内积为 0 \| 多维向量a 和 b 向量内积为 0，我们称 a 和 b 正交 (orthogonal)。本书上一章提到，正交是线性代数的概念，是垂直的推广。有了以上分析，我们就可以引入一个重要的不等式——柯西-施瓦茨不等式 (Cauchy–Schwarz inequality)： 44 \left \| a \cdot{b} \right \|\le \left \\| a \right \\| \left \\| b \right \\| 用尖括号来表达向量内积: \left \langle a,b \right \rangle^2\le\left \langle a,a \right \rangle \left \langle b,b \right \rangle 在 $R^n$ 空间中，上述不等式等价于 48 余弦定理高考必考内容！当然我们是偏向可视化理解余弦定理 (law of cosines)： 49 其中，a、b 和 c 为三角形的三边的边长图 18. 余弦定理推导下面，我们来用余弦定理推导： 1.将三角形三个边视作向量，将三个向量长度代入 (49)，可以得到： 50 2.向量 a 和 b 之差为向量 c： c=a-b 3.等式左右分别和自身计算向量内积，得到如下等式： c\cdot{c}=\left ( a-b \right )\cdot \left ( a-b \right ) 4.将上式展开，代入得到： 55 2.7 向量夹角：反余弦通过向量内积推导，可以得到非零向量 a 和 b 夹角余弦值： 64 通过反余弦，可以得到向量 a 和 b 夹角： 65 上式代表向量 a 和 b 之间的“相对角度” 极坐标将向量放在极坐标中解释向量夹角余弦值。给定向量 a 和 b 坐标如下： 66 向量 a 和 b 在极坐标中各自的角度为 θ_a 和 θ_b 角度 θ_a 和 θ_b 的正弦和余弦可以通过下式计算得到： 67 注意上式中，向量a,b与单位向量内积 θ_a 和 θ_b 相当于绝对角度图 19. 极坐标中解释向量夹角根据角的余弦和差恒等式，cos(θ) 可以由 θ_a 和 θ_b 正、余弦构造： 68 将 (67) 代入 (68) 得到： 69 单位向量给定两个非 0 向量 a 和 b，首先计算它们各自方向上的单位向量： 70 两个单位向量的内积就是夹角的余弦值 71 正交单位向量平面直角坐标系中 e1和 e2分别代表为沿着横轴、纵轴的单位向量。它们相互正交，也就是向量内积为 0： 72 在一个平面上，单位向量 e1、e2相互垂直，它俩“张起”的方方正正的网格，就是标准直角坐标系，具体如图 20 (a) 所示。图 20. 向量 a 在三个不同的正交直角坐标系中位置 v1、v2构造如图 20 (b) 所示直角坐标系。 w1、w2也可以构造如图 20 (c) 所示直角坐标系平面上可以存在无数个直角坐标系平面上，成对正交单位向量有无数组，比如图 21所示平面两组正交单位向量：图 21. 两组正交单位向量 [e1, e2]、[v1, v2]、[w1, w2] 都叫做的规范正交基 (orthonormal basis) [e1, e2] ——标准正交基 (standard basis) 而且大家很快就会发现 [e1, e2] 旋转一定角度可以得到 [v1, v2]、[w1, w2] 2.8 余弦相似度和余弦距离余弦相似度机器学习中有一个重要的概念，叫做余弦相似度 (cosine similarity)。余弦相似度用向量夹角的余弦值度量样本数据的相似性。定义用 k(x, q) 来表达 x 和 q 两个列向量的余弦相似度，定义如下： 74 这不是一模一样吗？那么同理，余弦相似度取值范围在 [−1, +1] 之间余弦距离定义余弦距离定义基于余弦相似度，用 d(x, q) 来表达 x 和 q 两个列向量的余弦距离，具体定义如下： 75 \| 对比 \| \| \| --- \| 欧几里得距离 \| L 2范数，是一种最常见的距离度量 \| L 2范数的取值范围为 [0, +∞) \| \| 余弦距离 \| 一种常见的距离度量 \| 余弦距离的取值范围为 [0, 2] \| 鸢尾花例子我们选取其中4组数组图 22. 鸢尾花的四个样本数据给出鸢尾花四个样本数据 x (1) 和 x (2) 两个样本对应的鸢尾花都是 setosa 这一亚属 x (51)样本对应的鸢尾花为 versicolor 这一亚属； x (101) 样本对应的鸢尾花为 virginica 这一亚属。计算向量余弦距离： x (1) 与 x (2) 76 同理，可以计算得到 x (1) 和 x (51) ，x (1) 和 x (101) 两个余弦距离： 77 总结 \| 主体 \| 表现 \| 相似程度 \| --- \| x (1) 和 x (2) 【setosa 亚属】 \| 余弦距离较近 \| 较为相似 \| \| x (1) 和 x (101) 【setota ， virginica 亚属】 \| 余弦距离较远 \| 不相似 \| 思考鸢尾花数据有 150 个数据点，任意两个数据点可以计算得到一个余弦相似度。因此成对余弦相似度有 11175 个，大家想想该怎么便捷计算、存储这些数据呢？我们从图像入手，联系之前学过的去均值，将向量起点移动到原点，然后再计算余弦距离，并比较结果差异图 23. 向量起点移到鸢尾花数据质心 2.9 向量积：结果为向量向量积 (vector product) 也叫叉乘 (cross product)，向量积结果为向量。也就是说，向量积一种“向量 → 向量”的运算规则。 a 和 b 向量积，记做 a × b。a × b 作为一个向量，我们需要了解它的方向和大小两个成分方向 a × b 方向分别垂直于向量 a 和 b，即 a × b 垂直于向量 a 和 b 构成平面图 24. a × b 垂直于向量 a 和 b 构成平面向量 a 和 b 以及 a × b 三者关系可以用右手法则判断，如图 25所示。图 25. 向量叉乘右手定则图 25这幅图中，我们可以看到 a × b 和 b × a 方向相反。大小 a × b 模，也就是 a × b 向量积大小，通过下式获得： 78 其中 θ 为向量 a 和 b 夹角几何角度从几何角度，向量积的模 \|\|a × b\|\| 相当于图中平行四边形的面积。图 26. a × b 向量积的几何含义正交向量之间的叉乘如图所示图 27. 三维空间正交单位向量基底之间关系图(a)中空间直角坐标系中三个正交向量 e1 (i) (横轴正方向)、e2 (j) (纵轴正方向) 和e3 (k) (竖轴正方向) 向量叉乘关系存在如下关系： 79 图(b) 展示以上三个等式中 i、j 和 k 前后顺序关系。若调换上式叉乘元素顺序，结果反向，对应以下三个运算式： 80 特别的，向量与自身叉乘等于 0 向量，比如： 81 叉乘运算常见性质： 82 任意两个向量的叉乘在三维直角坐标系中，用 i、j 和 k 表达向量 a 和 b： 83 整理向量 a 和 b 叉乘，如下： 84 2.10 逐项积：对应元素分别相乘元素乘积 (element-wise multiplication)，也称为阿达玛乘积 (Hadamard product) 或逐项积(piecewise product) 逐项积指的是两个形状相同的矩阵，对应元素相乘得到同样形状的矩阵向量是一种特殊矩阵，阿达玛乘积也适用于向量。图 28. 向量逐项积运算图 28给出的是从数据角度看向量逐项积运算给定如下 a 和 b 两个等行数列向量： 87 列向量 a 和 b 的逐项积定义如下： 88 逐项积是一种“向量 → 向量”的运算规则。 2.11向量张量积：张起网格面张量积 (tensor product) 又叫克罗内克积 (Kronecker product)，两个列向量 a 和 b 张量积 a ⊗ b定义如下： 89 向量张量积是一种“向量 → 矩阵”的运算规则向量 a 和其自身张量积 a ⊗ a 结果为方阵： 90 张量积常见性质: 91 几何视角从几何图像角度解释向量的张量积，图 29所示为：图 29. 从几何角度解释向量张量积向量 a 和 b 相当于两个维度上的支撑框架，两者的张量积则“张起”一个网格面 a ⊗ b 当我们关注 b 方向时，网格面沿同一方向的每一条曲线都类似 b，唯一的差别是高度上存在一定比例的缩放，这个缩放比例就是 ai。ai 是向量 a 中的某一个元素。举个例子观察 (89)，利用矩阵乘法展开，发现 a ⊗ b 可以写成两种形式： 93 上式中，第一种形式相当于，a 先按不同比例 (bj) 缩放得到 bja，再左右排列。第二种形式相当于，bT先按不同比例 (ai) 缩放得到 ai b T，再上下叠加。张量积更加底层的认知还没有，只能先记住定义和他的作用啦。张量入门每个分量只有一个下标，因为每个分量只由一个基向量构成（one basis vector per component），所以向量也称为[1阶张量]（Tensors of rank 1）。相应的，标量（scalar）也称为[0阶张量]（Tensors of rank 0），因为标量没有方向，因此也就不存在基向量，可以说标量的每个分量是由0个基向量构成的。更高阶的张量想象在该物体里有一个平面，这个平面的朝向需要用一个向量来表示，为了表示该向量需要引入1组（3个）基向量；在每个平面上有一个力，这个力则需要用第二个向量来表示，这样对于第一组中每个基向量又引入了第2组（3个）基向量与之组合。如果想要表示所有的平面与平面上的力的组合，需要9个分量，每个分量有2个下标（index）来表示该分量由哪两个基向量组合构成。 Axx表示在法线为x方向的平面上的方向为x方向的力。这9个分量与9个基向量共同组成了[2阶张量] 升维继续提高维度，现在每个分量有3个下标，所有的下标组合共有3 3 3=27个，故共有27组基向量，不同基向量对应一个分量总结：什么是张量以及为什么张量这么有用呢？张量是一种表示物理量的方式，这个方式就是用基向量与分量组合表示物理量（Combination of basis vector and component）。由于基向量可以有丰富的组合，张量可以表示非常丰富的物理量。此外，张量所描述的物理量是不随观察者或者说参考系而变化的，当参考系变化时（其实就是基向量变化），其分量也会相应变化，最后结果就是基向量与分量的组合（也就是张量）保持不变。考虑到张量有如此强大的表示能力，又不随观察者不同而变化，能够有效的表示宇宙间的万物，Lillian R. Lieber给了张量一个形象的称呼the fact of the universe. 内容来自：怎么通俗地理解张量？发布于 2023-06-27 12:32・河南矩阵向量书籍阅读赞同 92 条评论分享喜欢收藏申请转载写下你的评论... 2 条评论默认最新 merry 写的好棒！！！给我帮助非常大，谢谢作者啦~~~ 2023-10-23 · 广东回复1 小狗贝贝Baby 作者 2023-10-24 · 河南回复喜欢关于作者小狗贝贝Baby 学习编程的可爱小狗~从零开始学习知识系列~ 回答 27文章 141关注者 529 关注他发私信推荐阅读从向量空间的角度来理解方程组有无解的问题 ==================== 在开始之前，我们需要明确方程组可以转化成一组列向量的线性组合。什么意思呢？我们以下面一个例子进行介绍： x_1+2x_2+x_3 = 1 \ 2x_1+3x_2+3x_3 = 3 \ x_1+3x_2+x_3=3 可转化成如下形… marsg...发表于AutoM... 线性代数-1.向量 ========= 钱辰发表于Re:从零...向量与向量组的线性相关性线性代数 ================= 我们向量要研究什么？实际上我们在现实中是研究向量与向量之间的关系，而在关系中最重要的是向量与向量之间能否线性表示。我们在数学中首次理解的向量一般都是二维向量吧，一般是一个直角… 盛寒发表于线性代数线性代数的本质(1)--向量是什么 ================= 初瞳想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
13865	https://www.vocabulary.com/lists/52238	Censure words - Vocabulary List \| Vocabulary.com Compete in the Vocabulary Bowl: 192 countries, 50 states, countless prizes. Register now!X SKIP TO CONTENT Log inSign up Dictionary Vocabulary Lists VocabTrainer™ Lists by Grade Literature Non-Fiction Textbooks & Curricula Test Prep Current Events Roots & Affixes Just For Fun Speeches Historical Documents OthersNew list Speeches Historical Documents Censure words Symorne P. (Virgin Islands (British)) Share Copy link 3 words 0 learners Learn words with Flashcards and other activities Definition first Print Flashcards censure /ˈsɛntʃər/ /ˈsɛnʃə/ n.harsh criticism or disapproval excommunication /ˌɛkskəmˌjunəˈkeɪʃən/ n.cutting a person off from a religious society reprimand /ˌrɛprəˈmænd/ /ˈrɛprɪmænd/ n.an act or expression of criticism and censure Continue learning with flashcards! Log inSign up Shuffle Previous 1/3 Next View fullscreen Other learning activities Practice Answer a few questions about each word. Use this to prep for your next quiz!Vocabulary Jam Compete with other teams in real time to see who answers the most questions correctly!Spelling Bee Test your spelling acumen. Read the definition, listen to the word and try spelling it! Teaching tools Quiz Create and assign quizzes to your students to test their vocabulary.Assign activities Assign learning activities including Practice, Vocabulary Jams and Spelling Bees to your students, and monitor their progress in real-time. Full list of words from this list: words onlydefinitions & notes censureharsh criticism or disapproval excommunicationcutting a person off from a religious society reprimandan act or expression of criticism and censure Created on May 18, 2011 Sign up now (it’s free!) Whether you’re a teacher or a learner, Vocabulary.com can put you or your class on the path to systematic vocabulary improvement. Get started Learn with us Learner subscriptions Vocabulary lists Dictionary Test Prep Join a Vocabulary Jam Commonly confused words Word of the day Teach with us For educators For schools and districts How it works Success stories Research Professional development Contact sales Resources Help articles/FAQ Teaching resources Learner resources ESL/ELL resources Grade level resources IPA Pronunciation Contact support Leaderboards Vocabulary Bowl Today's leaders Weekly leaders Monthly leaders About Our Mission Blog Tell us what you think Privacy Policy Terms of Use Comprehensive K-12 personalized learning Immersive learning for 25 languages Trusted tutors ready to help in 300+ subjects 35,000 worksheets, games, and lesson plans Marketplace for millions of educator-created resources Fun educational games for kids Spanish-English dictionary, translator, and learning Diccionario inglés-español, traductor y sitio de aprendizaje Fast and accurate language certification French-English dictionary, translator, and learning Copyright © 2025 Vocabulary.com, Inc. A division of IXL Learning • All Rights Reserved. Learn with us Learner subscriptions Vocabulary lists Dictionary Test Prep Join a Vocabulary Jam Commonly confused words Word of the day Teach with us For educators For schools and districts How it works Success stories Research Professional development Contact sales Resources Help articles/FAQ Teaching resources Learner resources ESL/ELL resources Grade level resources IPA Pronunciation Contact support Leaderboards Vocabulary Bowl Today's leaders Weekly leaders Monthly leaders About Our Mission Blog Tell us what you think Privacy Policy Terms of Use My Account Log inSign up My Account Log Out My Learning My Proficiency Report My Profile Schools & Teachers My Classes Assignments & Activities My Lists Find a List to Learn... Create a New List... My Progress Words I'm Learning My Trouble Words Words I've Mastered My Achievements User Administration User Authentication My Account Comprehensive K-12 personalized learning Immersive learning for 25 languages Trusted tutors ready to help in 300+ subjects 35,000 worksheets, games, and lesson plans Marketplace for millions of educator-created resources Fun educational games for kids Spanish-English dictionary, translator, and learning Diccionario inglés-español, traductor y sitio de aprendizaje Fast and accurate language certification French-English dictionary, translator, and learning Copyright © 2025 Vocabulary.com, Inc. A division of IXL Learning • All Rights Reserved.
13866	https://www.quora.com/How-much-work-is-done-against-gravity-to-lift-a-10-kg-object-2-0-m-off-the-ground-Show-your-work-and-the-formula-you-are-using	Something went wrong. Wait a moment and try again. Displacement of Object Scientific Problems Potential Energy Gravity (physics) Work (physics) Physics Problem Solving Force (physics) 5 How much work is done against gravity to lift a 10 kg object 2.0 m off the ground? Show your work and the formula you are using. · To calculate the work done against gravity when lifting an object, you can use the formula: W=F×d where: - W is the work done (in joules), - F is the force applied (in newtons), and - d is the distance moved in the direction of the force (in meters). When lifting an object against gravity, the force F is equal to the weight of the object, which can be calculated using the formula: F=m×g where: - m is the mass of the object (in kilograms), - g is the acceleration due to gravity (approximately 9.81m/s2 on the surface of the Earth). Step 1: Calculate the force Given: - Mass To calculate the work done against gravity when lifting an object, you can use the formula: W=F×d where: - W is the work done (in joules), - F is the force applied (in newtons), and - d is the distance moved in the direction of the force (in meters). When lifting an object against gravity, the force F is equal to the weight of the object, which can be calculated using the formula: F=m×g where: - m is the mass of the object (in kilograms), - g is the acceleration due to gravity (approximately 9.81m/s2 on the surface of the Earth). Step 1: Calculate the force Given: - Mass m=10kg - Acceleration due to gravity g=9.81m/s2 F=m×g=10kg×9.81m/s2=98.1N Step 2: Calculate the work done Given: - Distance d=2.0m W=F×d=98.1N×2.0m=196.2J The work done against gravity to lift a 10 kg object 2.0 m off the ground is 196.2 joules. Was this worth your time? This helps us sort answers on the page. Absolutely not Related questions How is work done by gravity on an incline? What is the formula? Why is work done by gravity when a body is lifted up by a height h = -mgh? A person lifts a 4.3 kg centimeter block a vertical distance of 1.10 m and then carries the block horizontally a distance of 10.0 m. How much work is done by the person? How much work is done by gravity? How much work is done against gravity on a 40 kg object to a height of 0.6 meters? How much work is done by you when you lift a 9.0 kg object 0.50 meters off of the ground? Geoff Mangum Self-study 1973-2019. · Author has 3.8K answers and 3.1M answer views · Updated 3y Work is defined as force moving mass a distance when the line of force is the same as the line of motion. The formula is W = F•distance (often “s”, but when speaking about vertical Forces and motions, distance is “h” or height). So W = F(gravity) • h The unit of Work is Force Newtons times distance in meters, so N-m. 1 N moves 1 kg of mass 1 meter along the same line as the force. 1 N-m is also call Work is defined as force moving mass a distance when the line of force is the same as the line of motion. The formula is W = F•distance (often “s”, but when speaking about vertical Forces and motions, distance is “h” or height). So W = F(gravity) • h The unit of Work is Force Newtons times distance in meters, so N-m. 1 N moves 1 kg of mass 1 meter along the same line as the force. 1 N-m is also called 1 Joule. The “force of gravity” F(g) is from F(g) = m•g, where g is the acceleration of gravity on Earth’s surface (9.8 m/s^2). Combining, W = m•g•h The problem specifies moving 10 kg 2 meters up. Gravity makes the weight F(g) = 10 kg • 9.8 m/s^2, so the weight is 98 Newtons. The force that lifts this mass is the weight and the work is weight times distance. W = 98 N • 2 m = 196 N-m or Joules. The other way to do this is to define Work as vertical change of gravitational energy by the lifting. Like stretching a spring by lifting against the attraction of gravity, raising a mass adds to its “Potential Energy” (PE). PE = gravity force • height above the surface (since g, the acceleration of Earth’s gravity, is the rate that applies at the surface). If a mass on the surface of Earth has zero height and zero PE above the surface, and the mass is lifted 2 meters, the PE... 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I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Igor PhD in Mechanical Engineering, Technion - Israel Institute of Technology (Graduated 1997) · Author has 3.1K answers and 1.3M answer views · Updated 3y W = F s = m g s = 10 9.8 2 = 196 Joules Here W is the work, F is the gravity force, m = 10 kg is the mass, g = 9.8 m/s^2 is the gravity acceleration, and s = 2 m is the displacement Gary Sales Works at University of Massachusetts, Lowell · Author has 461 answers and 730.5K answer views · 6y Related How is work done by gravity on an incline? What is the formula? Related questions How much work is done against gravity in lifting a 6.5-kg weight through a distance of 21 cm? A boy of mass 50kg runs up a step of height 3.0m. What is the work done against gravity? Is work done under gravity negative? A container with a mass of 10 kg is lifted to a height of 8m. How much work is done against gravity? An object weighing 20 N moves horizontally toward the right at a distance of 5.0 m. What is the work done on the object by the force of gravity? Inesiko Ambrose Lives in Kampala, Uganda · 3y Work done = force acting on 10kg object distance travelled. Force = mass acceleration due to gravity Force = 109.81 Force = 98.1N So work done = 98.12 Work done = 196.2Nm Robert Nadaskay HS Mathematics/Physics Teacher/PE Teacher · 5y Related A man lifts 10 boxes each of mass 12 kg to a height of 3 above the ground. What is the work done by the man against gravity? Work is defined the product of a force exerted upon an object and the distance the object moves. Typically, scientists measure force in Newtons (mass x acceleration) and distance in meters. Your height does not have a unit, but I’ll assume you mean 3 meters? So to determine force exerted, multiply the mass of the box (12 kg) by acceleration due to gravity (-9.8 m/s^2) to get -117.6 Newtons exerted. The multiply force x distance (-117.6 N x 3 m) to get 352.8 Joules (N x m) of work for each box. (Work is a scalar quantity, so the negative indicating work against gravity is unnecessary.) Since ther Work is defined the product of a force exerted upon an object and the distance the object moves. Typically, scientists measure force in Newtons (mass x acceleration) and distance in meters. Your height does not have a unit, but I’ll assume you mean 3 meters? So to determine force exerted, multiply the mass of the box (12 kg) by acceleration due to gravity (-9.8 m/s^2) to get -117.6 Newtons exerted. The multiply force x distance (-117.6 N x 3 m) to get 352.8 Joules (N x m) of work for each box. (Work is a scalar quantity, so the negative indicating work against gravity is unnecessary.) Since there are 10 boxes, he must do 3,528 J of work. Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. 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Units are left as an exercise for the student. Eduardo Dequilla B.S. Physics from Ateneo de Davao University (Graduated 2005) · Author has 8.2K answers and 12.4M answer views · 5y The work done in lifting the 10 kg object to a height of 2.0 m is computed by multiplying the mass of 10 kg by its height of 2.0 m by the acceleration due to gravity of 9.8 m/s^2. In symbols. the formula is Work = m g h. Your answer will be in joules or Newton meters. I do believe you can do the computations yourself. Promoted by Grammarly Grammarly Great Writing, Simplified · Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. 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The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Steve Johnson BS in Physics, North Dakota State University (Graduated 1966) · Author has 5.7K answers and 2.3M answer views · 5y There are 2 approaches, in this case it is the same calculations, but for different reasons. Work = forcedistance. The force will have to be its weight in our gravity field or mg. So, Work = mgdistance Considering conservation of energy, work equals the change in gravitational potential energy. And GPE = mgh. Franklin Veaux Lives in Portland, OR · Author has 62K answers and 1,296.3M answer views · 2y Related How much work is done when a 30 kg child is lifted off the floor at 2 m? This is a simple calculation. First, calculate the amount of work in joules it takes to lift one kilogram one meter. Next, multiply that by 2, since you’re lifting this baby 2 meters, which is high enough for it to die if you drop it, you monster. Then multiply that by 30, since we’re talking about a 30-kilogram baby. Then, add in the amount of work it takes to grab the baby away from its mother, since there’s no way any mother will let some stranger just take a baby to do this, you monster. (In order to do this calculation, we will need to know more information, like the mass of the mother, the h This is a simple calculation. First, calculate the amount of work in joules it takes to lift one kilogram one meter. Next, multiply that by 2, since you’re lifting this baby 2 meters, which is high enough for it to die if you drop it, you monster. Then multiply that by 30, since we’re talking about a 30-kilogram baby. Then, add in the amount of work it takes to grab the baby away from its mother, since there’s no way any mother will let some stranger just take a baby to do this, you monster. (In order to do this calculation, we will need to know more information, like the mass of the mother, the height of the mother, what kind of shoes the mother is wearing, whether or not the mother knows aikido, and whether or not the mother is armed.) Next, calculate how many meters away you’ll need to run to escape the mother’s rage, and multiply that by the energy cost of crossing that distance. Remember to add the baby’s mass to your own, since you’ll be carrying it! Finally, when you’ve done that, calculate how much energy you’ll expend dialing your lawyer on your smartphone. (For the sake of this assignment, assume you have your lawyer on speed dial, you monster, and don’t consider the energy cost of the phone call itself, since that’s handled by the phone’s battery.) The numerical answer is trivial and will be left as an exercise to the reader. John Halloran Studied Physics & Mathematics · Updated 6y Related How does gravity work? Gravity is caused by the curving of space-time, as predicted in Einstein's theory of general relativity. Everything has a gravitational pull, but it is more noticeable for large masses, since it curves space-time more. Use this as a thought experiment. You have a trampoline and a bowling ball, with the trampoline being space-time and the bowling ball being a large mass such as earth. You put the the bowling ball on the trampoline and the area in which the ball is placed sinks downwards, creating sort of a funnel towards it. If you then put a ping pong ball on the trampoline near the ball, the Gravity is caused by the curving of space-time, as predicted in Einstein's theory of general relativity. Everything has a gravitational pull, but it is more noticeable for large masses, since it curves space-time more. Use this as a thought experiment. You have a trampoline and a bowling ball, with the trampoline being space-time and the bowling ball being a large mass such as earth. You put the the bowling ball on the trampoline and the area in which the ball is placed sinks downwards, creating sort of a funnel towards it. If you then put a ping pong ball on the trampoline near the ball, the ping pong ball will fall towards it. That is how the effect of gravity works. This also demonstrates how smaller gravitational pulls work. The ping pong ball makes the trampoline sink, but only a little. Same thing for larger masses. The images above are both pictures of the bowling ball experiment that I described. The image right here is an actual representation of general relativity. John Barrow Former Taught Physics at UK Secondary School (1982–2010) · Author has 647 answers and 409.4K answer views · Apr 8 Related What is the formula for calculating the work done by gravity when lifting an object against its weight in physics? The formula the OP asks about is … work done = mass gravity height OR work done = m g h. m is the mass of the object being lifted. g is the acceleration due to gravity h is the vertical distance (height) through which the object is moved. This equation is derived from the definition of work done = force distance. When moving upwards against the pull of gravity the force here is the force needed to lift the weight of the object, m g. The distance is the height, h. Ergo, work done = (mg) h = m g h Rob Williams PhD in Physics, Keele University (Graduated 1990) · Author has 247 answers and 858.7K answer views · 3y Related Why is work done when lifting an object = weight times height (gravity, work, physics)? This is an interesting question with a subtle answer. We're taught that Work is defined as force times distance. When dropping an object it is easy to get the work done because we get the force using Newton's second law, F=ma, and substitute in the acceleration due to gravity at the Earth's surface, g, to get F=mg. Distance is just the height through which the object falls, h. So we get work done is mgh. As mg is called "weight" we get the work done is weight time height. Lifting an object is arguably more interesting. Work is actually net force times distance. The net force is your upwards forc This is an interesting question with a subtle answer. We're taught that Work is defined as force times distance. When dropping an object it is easy to get the work done because we get the force using Newton's second law, F=ma, and substitute in the acceleration due to gravity at the Earth's surface, g, to get F=mg. Distance is just the height through which the object falls, h. So we get work done is mgh. As mg is called "weight" we get the work done is weight time height. Lifting an object is arguably more interesting. Work is actually net force times distance. The net force is your upwards force minus gravity. We could make this arbitrarily small. We could lift the object really slowly with the upward force being only slightly bigger than gravity. If the net force is close to zero then the work would be zero regardless of how high we lift it. We could lift an object slowly using little work and then drop it to get more work back. Free energy! Clearly not right. We have to think a little deeper about what work is. It is trying to tell us about changes in kinetic energy under the action of a force. Work is that energy change. In a potential field (like gravity) if we don't apply an external force, the field itself applies a force equal to the gradient of the field. Gravity supplies the force and accelerates objects downwards. But what if you just moved an object in a potential field without changing its motion? No change in kinetic energy. This is the case here when lifting an object very slowly. Work can be force time distance for kinetic energy changes but it can also just be the change in potential energy if the object changes location and not its motion. We expect this to be the same work as dropping an object but can we formally confirm this? We start with the difference in potential energy at the surface and at a height h above the surface. R is the radius of the Earth, M is the mass of the Earth, and G is Newton’s gravitational constant. E=GMmR−GMmR+h To get the 2 terms with the same denominator we multiply the first by (R+h)/(R+h) and the second by R/R (both equal to 1 of course): E=GMm(R+h)R(R+h)−GMmRR(R+h) We can now combine the terms, expand the numerator, and simplify: E=GMm(R+h)−GMmRR(R+h) E=GMmR+GMmh−GMmRR(R+h) E=GMmhR(R+h) One more trick by multiplying by R/R again: E=GMmhRR2(R+h) We recognize GMR2 as g. So we can write it as: E=mghRR+h E is mgh, weight, but with the factor R/(R+h). If h is much smaller than R then this can be considered 1 and so we get E=mgh, as desired. If you feel uncomfortable about that then just remember we made the same approximation when thinking about work done by gravitational force by assuming g is a constant for all small h. In reality g, and therefore force, decreases with height a tiny amount. We often forget we are using that assumption. So, as expected, the work lifting an object is the same as dropping it under free-fall, with value mgh, weight times height. As long as height is small. Related questions How is work done by gravity on an incline? What is the formula? Why is work done by gravity when a body is lifted up by a height h = -mgh? A person lifts a 4.3 kg centimeter block a vertical distance of 1.10 m and then carries the block horizontally a distance of 10.0 m. How much work is done by the person? How much work is done by gravity? How much work is done against gravity on a 40 kg object to a height of 0.6 meters? How much work is done by you when you lift a 9.0 kg object 0.50 meters off of the ground? How much work is done against gravity in lifting a 6.5-kg weight through a distance of 21 cm? A boy of mass 50kg runs up a step of height 3.0m. What is the work done against gravity? Is work done under gravity negative? A container with a mass of 10 kg is lifted to a height of 8m. How much work is done against gravity? An object weighing 20 N moves horizontally toward the right at a distance of 5.0 m. What is the work done on the object by the force of gravity? Why is work done when lifting an object = weight times height (gravity, work, physics)? Why is work done against the force of gravity negative? How do you calculate work done by gravity when an object falls a distance "h"? While lifting a ball to some height h, then work done by us = mgh and work of gravity = m-gh hence total work = 0 but this is contradictory HELP ? How much work is done at 10 kg mass? 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13867	https://www.ncbi.nlm.nih.gov/books/NBK470553/	An official website of the United States government The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Gram-Positive Bacteria Omeed Sizar; Stephen W. Leslie; Chandrashekhar G. Unakal. Author Information and Affiliations Authors Omeed Sizar1; Stephen W. Leslie2; Chandrashekhar G. Unakal3. Affiliations 1 St. Lucie Medical Center 2 Creighton University School of Medicine 3 The University of the West Indies Last Update: May 30, 2023. Continuing Education Activity Gram-positive organisms have highly variable growth and resistance patterns. The SCOPE project (Surveillance and Control of Pathogens of Epidemiologic Importance) found that in those with an underlying malignancy, gram-positive organisms accounted for 62 percent of all bloodstream infections in 1995 and 76 percent in 2000 while gram-negative organisms accounted for 22 percent in 1995 and 14 percent in 2000. This activity reviews the evaluation and management of gram-positive bacterial infections and explains the role of the interprofessional team in improving care for affected patients. Objectives: Explain how to evaluate for a gram-positive bacterial infection. Identify common infections caused by gram-positive bacteria. Describe treatment strategies for gram-positive bacterial infections. Outline interprofessional team strategies to improve care coordination and communication to provide quality care to patients with gram-positive bacterial infections. Access free multiple choice questions on this topic. Introduction Health professionals need to understand the important difference between gram-positive and gram-negative bacteria. Gram-positive bacteria are bacteria classified by the color they turn in the staining method. Hans Christian Gram developed the staining method in 1884. The staining method uses crystal violet dye, which is retained by the thick peptidoglycan cell wall found in gram-positive organisms. This reaction gives gram-positive organisms a blue color when viewed under a microscope. Although gram-negative organisms classically have an outer membrane, they have a thinner peptidoglycan layer, which does not hold the blue dye used in the initial dying process. Other information used to differentiate bacteria is the shape. Gram-positive bacteria comprise cocci, bacilli, or branching filaments. Etiology Gram-positive cocci include Staphylococcus (catalase-positive), which grows clusters, and Streptococcus (catalase-negative), which grows in chains. The staphylococci further subdivide into coagulase-positive (S. aureus) and coagulase-negative (S. epidermidis and S. saprophyticus) species. Streptococcus bacteria subdivide into Strep. pyogenes (Group A), Strep. agalactiae (Group B), enterococci (Group D), Strep viridans, and Strep pneumonia. Gram-positive bacilli (rods) subdivide according to their ability to produce spores. Bacillus and Clostridia are spore-forming rods while Listeria and Corynebacterium are not. Spore-forming rods that produce spores can survive in environments for many years. Also, the branching filament rods encompass Nocardia and actinomyces. Gram-positive organisms have a thicker peptidoglycan cell wall compared with gram-negative bacteria. It is a 20 to 80 nm thick polymer while the peptidoglycan layer of the gram-negative cell wall is 2 to 3 nm thick and covered with an outer lipid bilayer membrane. Epidemiology Bloodstream infection mortality rates have increased by 78% in just two decades. Gram-positive organisms have highly variable growth and resistance patterns. The SCOPE project (Surveillance and Control of Pathogens of Epidemiologic Importance) found that gram-positive organisms in those with an underlying malignancy accounted for 62% of all bloodstream infections in 1995 and 76% in 2000 while gram-negative organisms accounted for 22% and 14% of infections for these years. Pathophysiology Gram-positive cocci: Staphylococcus aureus is a gram-positive, catalase-positive, coagulase-positive cocci in clusters. S. aureus can cause inflammatory diseases, including skin infections, pneumonia, endocarditis, septic arthritis, osteomyelitis, and abscesses. S. aureus can also cause toxic shock syndrome (TSST-1), scalded skin syndrome (exfoliative toxin, and food poisoning (enterotoxin). Staphylococcus epidermidis is a gram-positive, catalase-positive, coagulase-negative cocci in clusters and is novobiocin sensitive. S. epidermidis commonly infects prosthetic devices and IV catheters producing biofilms. Staphylococcus saprophyticus is novobiocin resistant and is a normal flora of the genital tract and perineum. S. saprophyticus accounts for the second most common cause of uncomplicated urinary tract infection (UTI). Streptococcus pneumoniae is a gram-positive, encapsulated, lancet-shaped diplococci, most commonly causing otitis media, pneumonia, sinusitis, and meningitis. Streptococcus viridans consist of Strep. mutans and Strep mitis found in the normal flora of the oropharynx commonly cause dental carries and subacute bacterial endocarditis (Strep. sanguinis). Streptococcus pyogenes is a gram-positive group A cocci that can cause pyogenic infections (pharyngitis, cellulitis, impetigo, erysipelas), toxigenic infections (scarlet fever, necrotizing fasciitis), and immunologic infections (glomerulonephritis and rheumatic fever). ASO titer detects S. pyogenes infections. Streptococcus agalactiae is a gram-positive group B cocci that colonize the vagina and is found mainly in babies. Pregnant women need screening for Group-B Strep (GBS) at 35 to 37 weeks of gestation. Enterococci is a gram-positive group D cocci found mainly in the colonic flora and can cause biliary tract infections and UTIs. Vancomycin-resistant enterococci (VRE) are an important cause of nosocomial infections. Gram-positive rods: Clostridia is a gram-positive spore-forming rod consisting of C. tetani, C. botulinum, C. perfringens, and C. difficile. C. difficile is often secondary to antibiotic use (clindamycin/ampicillin), PPI use, and recent hospitalization. Treatment involves primarily with oral vancomycin. Bacillus anthracis is a gram-positive spore-forming rod that produces anthrax toxin resulting in an ulcer with a black eschar. Bacillus cereus is a gram-positive rod that can be acquired from spores surviving under-cooked or reheated rice. Symptoms include nausea, vomiting, and watery non-bloody diarrhea. Corynebacterium diphtheria is a gram-positive club-shaped rod that can cause pseudomembranous pharyngitis, myocarditis, and arrhythmias. Toxoid vaccines prevent diphtheria. Listeria monocytogenes is a gram-positive rod acquired by the ingestion of cold deli meats and unpasteurized dairy products or by vaginal transmission during birth. Listeria can cause neonatal meningitis, meningitis in immunocompromised patients, gastroenteritis, and septicemia. Treatment includes ampicillin. History and Physical It is important to identify patients with sepsis and order necessary blood cultures and labs. Physical Bullous impetigo Draining sinus tracts Erythema Fever Murmur if endocarditis is present Petechiae if toxic shock syndrome is present Superficial abscesses Warmth Evaluation Once a gram-positive organism infection is suspected, these laboratory studies are useful: CBC Electrolytes Blood cultures Pro-calcitonin level Echocardiogram if endocarditis is suspected Joint aspiration if a septic joint is suspected Treatment / Management Penicillin was the first antibiotic ever introduced during World War II by Alexander Fleming in 1928. Penicillin does not cover Staph or Enterococcus but used mainly streptococcal infections. The penicillinase-resistant organisms (nafcillin, oxacillin, cloxacillin, dicloxacillin) cover Staph (MSSA) and Strep. Anti-pseudomonal penicillins include piperacillin and ticarcillin effective against gram-positive, gram-negative, pseudomonas, and anaerobes. Carbapenems cover gram-positives, gram negatives, and anaerobes. Trimethoprim/sulfamethoxazole, clindamycin, and doxycycline are oral antibiotics used for mild to moderate MRSA infections. It is important to note that trimethoprim/sulfamethoxazole increases warfarin levels leading to increased INR. Vancomycin, linezolid, daptomycin, and tigecycline cover moderate to severe community and hospital-acquired MRSA. Vancomycin requires renal dosing with trough levels between 15 to 20. Linezolid is an option if a patient is allergic to vancomycin. CBC needs to be checked weekly to avoid bone marrow suppression, neutropenia, thrombocytopenia, and anemia. Linezolid, daptomycin, and tigecycline are options to treat vancomycin-resistant enterococci. Differential Diagnosis Bacteremia Bronchiectasis imaging Chemical burns Diarrhea Electrical injuries in emergency Medicine Emergent management of acute otitis Emergent management of thermal burns Empyema imaging Fever in the infant and toddler Fever without a focus Henoch-schonlein purpura Hospital-acquired infections Ingrown nails Necrotizing enterocolitis imaging Prognosis The prognosis following infection with gram-positive organisms is variable. The highest mortality rates are in elderly persons who tend to have suppressed immune systems and less physiologic reserve. Enhancing Healthcare Team Outcomes Health professionals, including doctors, nurses, and pharmacists, need to be aware of risk factors to treat patients with selected antibiotics properly. Pharmacists need to accurately monitor vancomycin trough levels to avoid mortality in patients with Staph aureus. They also need to review medication for dose and interactions and counsel patients about finishing all prescribed antibiotics. Infection control nurses evaluate nosocomial infections and implement appropriate policies. An interprofessional approach will produce the best outcomes. [Level 5] Outcomes: Screening for methicillin-resistant Staphylococcus aureus (MRSA) risk factors enhance infection control. MRSA risk factors include patients who are above age 65, urinary catheter, previous antibiotic treatment past three months, trauma, and those admitted from a long-term facility. [Level 5] Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Figure Gram Stain of Staphylococcus aureus Contributed by Scott Jones, MD References 1. : National Nosocomial Infections Surveillance (NNIS) system report, data summary from January 1992-April 2000, issued June 2000. Am J Infect Control. 2000 Dec;28(6):429-48. [PubMed: 11114613] 2. : Wisplinghoff H, Seifert H, Wenzel RP, Edmond MB. Current trends in the epidemiology of nosocomial bloodstream infections in patients with hematological malignancies and solid neoplasms in hospitals in the United States. Clin Infect Dis. 2003 May 01;36(9):1103-10. [PubMed: 12715303] 3. : Righi E, Carnelutti A, Bassetti M. Current role of oxazolidinones and lipoglycopeptides in skin and soft tissue infections. Curr Opin Infect Dis. 2019 Apr;32(2):123-129. [PubMed: 30664028] 4. : Zamoner W, Prado IRS, Balbi AL, Ponce D. Vancomycin dosing, monitoring and toxicity: Critical review of the clinical practice. Clin Exp Pharmacol Physiol. 2019 Apr;46(4):292-301. [PubMed: 30623980] 5. : Rogalla D, Bomar PA. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Jul 4, 2023. Listeria Monocytogenes. [PubMed: 30521259] 6. : Rostkowska KA, Szymanek-Pasternak A, Simon KA. Spontaneous bacterial peritonitis - therapeutic challenges in the era of increasing drug resistance of bacteria. Clin Exp Hepatol. 2018 Dec;4(4):224-231. [PMC free article: PMC6311748] [PubMed: 30603669] 7. : Gashaw M, Berhane M, Bekele S, Kibru G, Teshager L, Yilma Y, Ahmed Y, Fentahun N, Assefa H, Wieser A, Gudina EK, Ali S. Emergence of high drug resistant bacterial isolates from patients with health care associated infections at Jimma University medical center: a cross sectional study. Antimicrob Resist Infect Control. 2018;7:138. [PMC free article: PMC6245755] [PubMed: 30479751] 8. : Bolia IK, Tsiodras S, Chloros GD, Kaspiris A, Sarlikiotis T, Savvidou OD, Papagelopoulos PJ. A Review of Novel Antibiotic Regimens for the Treatment of Orthopedic Infections. Orthopedics. 2018 Nov 01;41(6):323-328. [PubMed: 30452066] 9. : Callejo-Torre F, Eiros Bouza JM, Olaechea Astigarraga P, Coma Del Corral MJ, Palomar Martínez M, Alvarez-Lerma F, López-Pueyo MJ. Risk factors for methicillin-resistant Staphylococcus aureus colonisation or infection in intensive care units and their reliability for predicting MRSA on ICU admission. Infez Med. 2016 Sep 01;24(3):201-9. [PubMed: 27668900] : Disclosure: Omeed Sizar declares no relevant financial relationships with ineligible companies. : Disclosure: Stephen Leslie declares no relevant financial relationships with ineligible companies. : Disclosure: Chandrashekhar Unakal declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK470553PMID: 29261915 Share Views PubReader Print View Cite this Page Sizar O, Leslie SW, Unakal CG. Gram-Positive Bacteria. [Updated 2023 May 30]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. In this Page Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Enhancing Healthcare Team Outcomes Review Questions References Related information PMC PubMed Central citations PubMed Links to PubMed Similar articles in PubMed Gram Staining.[StatPearls. 2025] Gram Staining. Tripathi N, Zubair M, Sapra A. StatPearls. 2025 Jan Deeply branching Bacillota species exhibit atypical Gram-negative staining.[Microbiol Spectr. 2024] Deeply branching Bacillota species exhibit atypical Gram-negative staining. Choi JK, Poudel S, Yee N, Goff JL. Microbiol Spectr. 2024 Oct 3; 12(10):e0073224. Epub 2024 Aug 20. Review Use of the gram stain in microbiology.[Biotech Histochem. 2001] Review Use of the gram stain in microbiology. Beveridge TJ. Biotech Histochem. 2001 May; 76(3):111-8. Studies in gram staining.[Stain Technol. 1975] Studies in gram staining. Adams E. Stain Technol. 1975 Jul; 50(4):227-31. Review Microwave-accelerated cytochemical stains for the image analysis and the electron microscopic examination of light microscopy diagnostic slides.[Scanning. 1993] Review Microwave-accelerated cytochemical stains for the image analysis and the electron microscopic examination of light microscopy diagnostic slides. Hanker J, Giammara B. Scanning. 1993 Mar-Apr; 15(2):67-80. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Gram-Positive Bacteria - StatPearls Gram-Positive Bacteria - StatPearls Your browsing activity is empty. Activity recording is turned off. 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13868	https://www.circuitspecialists.com/blog/quick-guide-on-how-to-taking-measurements-with-an-oscilloscope/?srsltid=AfmBOoqmsomcDlF4dHHLeHESQpBR3RFUJMpRCNqnBuYNqyJbbFuwmdsW	Quick Guide on How to Taking Measurements with an Oscilloscope Skip to content Simply Smarter Circuitry Blog How To's, Reviews, and Product News from Circuit Specialists HQ. Home Announcements DIY 3D Printing Reviews About Us Shop Online All PostsAnnouncementsDIYElectronics Industry Quick Guide on How to Taking Measurements with an Oscilloscope February 14, 2022Khang A technician needs to take measurements with an Oscilloscope on a daily so here is the Quick Guide on How to Taking Measurements with an Oscilloscope. Furthermore, an automotive engineer uses an oscilloscope to correlate analog data from sensors with serial data from the engine control unit. In addition, computer engineers use oscilloscopes to measure the consumption of microchips. Table of Content What is an oscilloscope? Basic operation. AC Amplitude. AC Frequency. DC Voltage Signal. Conclusion. What is an oscilloscope? For instance, a digital storage oscilloscope is an electronic device used to view electrical signals. Secondly, it consists of a display screen, inputs, and several controls, mainly used for taking measurements. Hantek DSO5202P 200MHz 2 Channel Basic operation For taking measurements with an oscilloscope, Firstly, you plug the electrical signal you’d like to view into one of the oscilloscope’s inputs of which there are typically two, labeled A and B. Noted: when you first switch on the oscilloscope, the signal won’t be visible until you adjust two settings: volts/division and time/division (or timebase). Secondly, when measuring the vertical scale, the volts/division determines the number of volts for each vertical division. The time/division controls the horizontal scale. The amount of time each horizontal division shows is commensurately changed when you adjust the time/division. Adjust these two settings until the signal is clearly displayed on the oscilloscope’s screen. For more detail on how to operate the oscilloscope, you can read thisarticleby theUniversity of Nebraska. Oscilloscope Basic Understanding AC Amplitude For taking measurements with an Oscilloscope for the alternating current (AC) amplitude you start by plugging the AC signal into one of the oscilloscope’s inputs before optimizing the signal. The AC signal will oscillate and resemble a sine wave. You’ll measure the signal’s amplitude by counting the number of vertical divisions between the signal’s highest and lowest points (i.e. its peak and trough). Finally, you can get the amplitude in volts by multiplying the number of vertical divisions by your volts/division setting. Hantek2D72 70 MHz Oscilloscope, Waveform Generator & Digital Multimeter AC Frequency For instance, if you’d like to measure the alternating current frequency you should plug the AC signal into one of the inputs on yourdigital oscilloscopeand optimize the signal. Count the number of horizontal divisions from one high point to the next (i.e. peak to peak) of your oscillating signal. Next, you’ll multiply the number of horizontal divisions by the time/division to find the signal’s period. You can calculate the signal’s frequency with this equation: frequency=1/period. Ground Loop DC Voltage Signal In order to take measurements with an Oscilloscope for a direct current (DC) signal’s voltage, you first turn on your oscilloscope without connecting the input signal. (Note that a DC signal will be flat on your oscilloscope’s display.) Place the oscilloscope line over the zero volt level with the vertical position setting. Then plug the DC signal path into one of your oscilloscope’s inputs. After plugging in the signal you will notice the oscilloscope line shift on the vertical axis. You’ll count the number of vertical divisions that your oscilloscope line shifts and multiply the vertical divisions by the volts/division to find the DC signal voltage. DC Voltage Signal Conclusion A technician needs to take measurements with an Oscilloscope on a daily so here is the Quick Guide on How to Taking Measurements with an Oscilloscope. If your oscilloscope is battery powered or your device under test is isolated from Earth ground (for example, it is also battery powered or you are using a two-pronged wall adapter), then you do not have to worry about shorting your power supply to the ground. SDS1204X-E 200MHz 4 Channel Digital Super Phosphor Oscilloscope If you are interested in getting an oscilloscope and don’t know where to start. You can read this Quick Guide on How to buy an Oscilloscope 2022. Through-hole assembly? Am I too old? 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13869	https://math.stackexchange.com/questions/187255/determining-the-value-of-h-that-makes-a-linear-system-consistent	Determining the value of $h$ that makes a linear system consistent. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Determining the value of h h that makes a linear system consistent. Ask Question Asked 13 years, 1 month ago Modified9 years ago Viewed 36k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I'm just beginning linear algebra at university and have a teacher who moves very fast and has pre-done slides so i can't actually see the problem worked out, he just talks it out. On top of this, he's also from China and heavily accented, making him hard to understand. Anyway, i have an augmented matrix, and i want the values of h h that make it consistent: [1 2 h−8−5 6][1 h−5 2−8 6] and quite frankly, i'm not sure just how to start. I tried eliminating the 1 in the second row, but that made the second line [0 h+4−8][0 h+4−8] and i'm not even sure if that's the right direction or even allowed. Thanks in advance. linear-algebra Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 26, 2012 at 23:38 Américo Tavares 39.2k 14 14 gold badges 110 110 silver badges 252 252 bronze badges asked Aug 26, 2012 at 22:01 BMEdwards37BMEdwards37 105 1 1 gold badge 4 4 silver badges 13 13 bronze badges 4 1 Is it clear to you what "consistent" means, or even what we mean by an "augmented matrix"?akkkk –akkkk 2012-08-26 22:13:04 +00:00 Commented Aug 26, 2012 at 22:13 6 I don't see the relevance of the ethnicity of your professor. I would wager he spends time to create those slides so that you read them.James S. Cook –James S. Cook 2012-08-26 23:02:58 +00:00 Commented Aug 26, 2012 at 23:02 I don't see the relevance of looking so much into it, I was just stating that I felt behind because he was from another country and heavily accented and I was having a hard time keeping up. The only difference was that I off-handedly mentioned where he was from.BMEdwards37 –BMEdwards37 2012-09-11 16:17:51 +00:00 Commented Sep 11, 2012 at 16:17 Get used to moving fast. It's just the way things are. At the level of mathematics that you are in, they expect you to keep up or get out. I've had many instructors with foreign accents. The trick is to study in such a way that you already know what they are talking about. It makes a huge difference. Since you don't know what 'consistent' means, my guess is you don't know your way around a text book. Your instructor isn't your problem. You might also want to ask your instructor if you can get copies of his notes after the lecture.Steven Alexis Gregory –Steven Alexis Gregory 2016-09-21 07:29:31 +00:00 Commented Sep 21, 2016 at 7:29 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. A linear system is inconsistent is if it represents a contradiction, for instance the system [0 3 0−2−10 1][0 0−10 3−2 1] is inconsistent because the first line represents a linear equation 0 x+0 y=−10 0 x+0 y=−10, i.e. 0=−10 0=−10, which is a contradiction. Geometrically, when you solve a 2x2 linear system, you are finding the intersection between a pair of lines. If you reach a contradiction, like the system above, then your lines do not intersect, i.e. they must be parallel. If you are being asked this question, you have probably already covered Gauss-Jordan ellimination. Inconsistencies in linear systems can be readily identified if the system is brought to reduced row echelon form (can you see why?), so I would start with that. The steps are simple: [1 2 h−8−5 6][1 h−5 2−8 6] Multiply the second row by 1/2 1/2: [1 1 h−4−5 3][1 h−5 1−4 3] Subtract the second row from the first: [0 1 h+4−4−8 3][0 h+4−8 1−4 3] Without even proceeding further, it is obvious that one way for the system to be inconsistent is if the first line is 0 0\|−8 0 0\|−8, since this would be equivalent to saying 0 x+0 y=−8 0 x+0 y=−8, that is 0=−8 0=−8, a contradiction. The first row would have this form only if h=−4 h=−4, so h=−4 h=−4 makes the system inconsistent. Now it is pretty clear at this point that no other value of h h would make the system inconsistent, and after you are comfortable with Gauss-Jordan elimination this fact would be apparent to you as well, though you should really try to understand why first. So let's say h≠−4 h≠−4. Then we can multiply the first row by 1 h+4 1 h+4: [0 1 1−4−8 h+4 3][0 1−8 h+4 1−4 3] And now add 4 times the first row to the second: ⎡⎣⎢0 1 1 0−8 h+4 3−32 h+4⎤⎦⎥[0 1−8 h+4 1 0 3−32 h+4] To really be precise, you can swap the two rows: ⎡⎣⎢1 0 0 1 3−32 h+4−8 h+4⎤⎦⎥[1 0 3−32 h+4 0 1−8 h+4] Thus for any value of h h other than −4−4, we can solve the system - there is no way to make the system displayed above have a row which looks like 0 0\|c 0 0\|c, for any non-zero number c c. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Sep 21, 2016 at 7:13 user99914 answered Aug 27, 2012 at 0:28 mboratkomboratko 4,603 28 28 silver badges 36 36 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. A simpler solution is based on a theorem that any system A x=b A x=b is consistent iff rank of [A∣b][A∣b] equal to rank of A A. To compute rank of A A perform elimination on A A to get: (1 0 0−8−2 h)(1 0 0−8−2 h) Hence rank(A)=1 rank(A)=1 if h=−4 h=−4 and rank(A)=2 rank(A)=2 otherwise. To compute rank of [A∣b][A∣b] perform elimination on [A∣b][A∣b] to get: (1 0 0 1 3 h−20 h+4−8 h+4)(1 0 3 h−20 h+4 0 1−8 h+4) So for values other that h=−4 h=−4 we have rank([A∣b])=2 rank([A∣b])=2. Comparing two ranks, we have a consistent system other than h=−4 h=−4. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Aug 27, 2012 at 0:56 user2468 user2468 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. So it is consistent whenever there is at least one solution. That means that the two lines you have cannot be parallel to each other. Multiply the first row by 2 2, and you get [2,2 h,−10][2,2 h,−10]. The lines will be parallel for the equations m 1 x+n 1 y=a m 1 x+n 1 y=a and m 2 x+n 2 y=b m 2 x+n 2 y=b if m 1=m 2 m 1=m 2 and n 1=n 2 n 1=n 2. In this case, m 1=2,m 2=2,n 1=2 h,n 2=−8 m 1=2,m 2=2,n 1=2 h,n 2=−8. Since m 1=m 2,n 1≠n 2 m 1=m 2,n 1≠n 2, so 2 h≠−8,h≠−4 2 h≠−8,h≠−4. Note that there are an infinite number of solutions (aka consistent) if m 1=m 2,n 1=n 2,m 1=m 2,n 1=n 2, and a=b a=b. Otherwise, you do not have to worry about the a a and b b values. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 26, 2012 at 22:28 answered Aug 26, 2012 at 22:16 Sidd SingalSidd Singal 3,502 3 3 gold badges 25 25 silver badges 33 33 bronze badges 5 Wow, I was going in the complete wrong direction. I'm probably going to have another question here soon, but let me look at it first now after seeing this and i'll see if it helps, these variables are really throwing me off. Thank you though, and keep an eye out for another question here in about 20 minutes :p BMEdwards37 –BMEdwards37 2012-08-26 22:28:01 +00:00 Commented Aug 26, 2012 at 22:28 @BMEdwards37 Sorry I messed up with the negatives, but i fixed my answer. You can think of an augmented matrix as a system of equations, where the each column represents a different dimension, and the last column represents a constant. I was wrong in using "x" and "y" as dimensions. Usually, people would use X 1,X 2,X 3...X n X 1,X 2,X 3...X n Sidd Singal –Sidd Singal 2012-08-26 22:30:43 +00:00 Commented Aug 26, 2012 at 22:30 @BMEdwards37 Wait, I want to let you know that you are not going in the wrong direction. You ended up with [0 h+4 -8], which is good! The only way to make sure a system is consistent is that ALL of the values cannot equal 0 (nevermind the last column). In your example, since the first column is 0, then the second column CANNOT be 0, so h+4≠0 h+4≠0. If all of the columns except the last equal 0, then there are no solutions. If all columns including the last equal 0, then there are infinite solutions =)Sidd Singal –Sidd Singal 2012-08-26 22:35:57 +00:00 Commented Aug 26, 2012 at 22:35 I understood your meaning, didn't catch the negatives either. I'm going to post another question because i'm not even sure what it's asking, i appreciate the help.BMEdwards37 –BMEdwards37 2012-08-26 22:37:27 +00:00 Commented Aug 26, 2012 at 22:37 Ahh i see, i wasn't so far off. Thanks.BMEdwards37 –BMEdwards37 2012-08-26 22:38:36 +00:00 Commented Aug 26, 2012 at 22:38 Add a comment\| You must log in to answer this question. Protected question. To answer this question, you need to have at least 10 reputation on this site (not counting the association bonus). The reputation requirement helps protect this question from spam and non-answer activity. Start asking to get answers Find the answer to your question by asking. 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13870	https://calcresource.com/solid-poly-prism.html	Properties of prism with regular polygonal base - calculator \| calcresource CALC RESOURCE Calculation Tools & Engineering Resources × Custom Search Sort by: Relevance Relevance Date CalculatorsResourcesFeedbackTerms of use Cancel Print Jump to -Calculator -Definitions Table of Contents -Calculator -Definitions -Geometry Share this See also Cylinder properties Cone properties Cube properties Pyramid properties Pyramid with polygonal base properties Hex-prism properties All solids All Geometric Shapes x Calculation options: Number of digits:6 10 x notation for big numbers:- [x] Properties of Prism (with polygonal base) By Dr. Minas E. Lemonis, PhD - Updated: January 16, 2021 Home>Solids>Polygonal prism This tool calculates the basic geometric properties of a right prism, with a regular polygonal base. Enter the number of base edges 'N' and the shape dimensions 'a' and 'h' below. The calculated results will have the same units as your input. Please use consistent units for any input. N = a = h = Geometric properties: Volume = Surface area = Base area = Lateral surface area = Number of faces = Number of edges = Number of vertices = ADVERTISEMENT Table of Contents -Calculator -Definitions -Geometry Share this Definitions Geometry The prism is a solid object enclosed by two parallel planar polygonal bases and a lateral prismatic surface. Cube is a prism while cylinder can be considered a prism with infinite number of lateral faces. With a polygonal base, having n sides/vertices, the number of faces, edges and vertices (NF, NE, NV respectively) is given by the formulas: The volume of a prism is given by the formula: where the surface area of the base and h the height of the prism. For a regular polygon with n edges, the base area is given by: where the length of an edge of the base regular polygon. The surface area of one lateral face of the prism, is equal to: Since there are n lateral faces and two bases, the total surface area of the n base prism is: See also Cylinder properties Cone properties Cube properties Pyramid properties Pyramid with polygonal base properties Hex-prism properties All solids All Geometric Shapes Connect with us: About Website calcresource offers online calculation tools and resources for engineering, math and science. Read more about us here. Short disclaimer Although the material presented in this site has been thoroughly tested, it is not warranted to be free of errors or up-to-date. The author or anyone else related with this site will not be liable for any loss or damage of any nature. For the detailed terms of use click here. Help us Send your feedbackAdd to Favorites/BookmarkLink to this siteLink to this pageBack to top Copyright © 2015-2024, calcresource. All rights reserved.
13871	https://www.ncbi.nlm.nih.gov/books/NBK567777/	An official website of the United States government The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Endometriosis Eleni S. Tsamantioti; Heba Mahdy. Author Information and Affiliations Authors Eleni S. Tsamantioti1; Heba Mahdy. Affiliations 1 Karolinska Institute Last Update: January 23, 2023. Continuing Education Activity Endometriosis is a chronic estrogen-dependent chronic condition characterized by the ectopic implantation of functional tissue lining the uterus (endometrial glands and stroma) outside of the uterine cavity. Endometriosis, a word derived from the Greek endo ‘'inside'', metra ''uterus'' and osis ‘' disease,'' remains to some extent vague, with the most common clinical symptoms being pelvic pain and infertility. Most frequently, endometrial tissue is found in ovaries, resulting in the formation of chocolate cysts. Still, it can also be found in the Fallopian tubes, uterosacral ligaments, the gastrointestinal tract, and less often in the pleura, pericardium, or the central nervous system. This activity illustrates the evaluation and treatment of endometriosis and reviews the role of the healthcare team in managing patients afflicted with endometriosis. Objectives: Identify the risk factors for developing endometriosis and summarize the epidemiology of the disease. Evaluate the history and physical exam findings associated with endometriosis and describe the available ways to evaluate the disease. Differentiate the treatment and management options available for endometriosis. Communicate the importance of interprofessional team collaboration to improve the outcomes of endometriosis. Access free multiple choice questions on this topic. Introduction Endometriosis is a chronic gynecologic disease characterized by the development and presence of histological elements like endometrial glands and stroma in anatomical positions and organs outside the uterine cavity. The main clinical manifestations of the disease are chronic pelvic pain and impaired fertility. The localization of endometriosis lesions can vary, with the most commonly involved focus of the disease being the ovaries, followed by the posterior broad ligament, the anterior cul-de-sac, the posterior cul-de-sac, and the uterosacral ligament. Endometriotic nodules also affect the intestinal tract and the urinary system, like the ureter, the bladder, and the urethra. Nevertheless, endometriosis is not limited to the pelvis but can damage extra pelvic structures like the pleura, the pericardium, or the central nervous system. The main theories utilized to explain the pathogenesis of endometriosis are Sampson’s theory, the coelomic metaplastic theory, the stem cell theory, the Müllerian remnant theory, and the vascular and lymphatic metastasis theory. Etiology There are several developed theories about the etiology of endometriosis based on the logical sequelae relating the severity of symptoms to the stage of the disease. However, none of these proposed models can fully explain the range of clinical manifestations of the disease. The most widely plausible theory is Sampson's assumption, stating that the viable cells in the peritoneal fluid with retrograde menstruation can implant, grow, and infiltrate the peritoneal cavity. Retrograde menstruation is a term used to describe the reflux of the blood backward from the Fallopian tubes to the peritoneum during menstruation. It is a phenomenon quite common in a large proportion of women of reproductive age. An alternative to Sampson's theory is the coelomic metaplastic theory suggested by Meyer. This theory relies on the ability of parietal peritoneum epithelium to differentiate into endometrial tissue, probably under the stimuli of cytokines and growth factors of the endometrial stroma. The hypothesis of metaplasia could justify the occurrence of endometriosis in women without a uterus or a lack of endometria, such as female patients with Mayer-Rokitansky-Küster-Hauser syndrome or the rare cases of males who have endometriosis. To explain the occurrence of endometriosis in the cul-de-sac and uterosacral ligaments, the main used theory is the Müllerian remnant theory, suggesting that atypical migration or differentiation of these remnants could imitate endometriotic tissue in the poster pelvic floor. Another theory developed to explain endometriosis is the lymphatic and vascular metastasis theory. This theory proposes that endometrial tissue can infiltrate the lymphatics and vasculature and, through them, get transferred to remote foci like the brain or the pleura or retroperitoneal locations. The stem cell theory could also partially explain the pathogenesis and establishment of endometriosis. Growing evidence has supported that the endometrial stem cells might be responsible for the development and progression of endometriosis. Endometrial progenitor cells have not only been recognized in menstrual blood but have also been identified as clonogenic cells in the endometrial lesions. The mechanism of generation and establishing endometriosis might be by the retrograde discarding of the endometrial stem cells in the pelvic cavity either with neonatal uterine bleeding or with menstruation after menarche. Moreover, the ambiguous pathogenesis of endometriosis led the investigators to research the role of oxidative stress, inflammatory elements, and reactive oxygen species as long as genetic and epigenetic factors. Oxidative stress is caused by the imbalance between the reactive oxygen species and the body's antioxidant ability. Reactive oxygen species can damage several components of the cell's nucleic acids and proteins. If the enzymatic and non-enzymatic cellular antioxidant capacity is decreased, reactive oxygen species are not eliminated from the cells and can be the leading cause of endometriosis. Besides, the inability of just 1 theory to explain the pathogenesis of endometriosis could be ascribed to the complex interaction between the expression of genes involved in endometriosis, inflammatory reactions, and the disrupted hormone response to these stimuli. Endometriotic lesions present as a chronic local inflammatory disease and involve alterations in cellular immunity and the expression of inflammatory cytokines. According to previous research, patients with endometriosis have elevated serum levels of many pro-inflammatory cytokines like IL-1, IL-6, and IL-8, resulting in chemotaxis, recruitment, and activation of peritoneal macrophages and proliferation of monocytes. The surgical excision of the endometriotic lesions leads to a decrease in the serum levels of the interleukins, providing evidence that their local production is the reason for the systematic inflammatory reaction. Tumor necrosis factor-alpha (TNF-a) in the peritoneal fluid, produced by the peritoneal macrophages, amplifies the inflammatory response. The role of steroid hormones in the pathogenesis of endometriosis is also indisputable. Estrogen is the main hormone responsible for the propagation and expansion of the endometrium. The increased action of aromatase, mainly in the deep infiltrating endometriosis, leads to locally increased estrogen activity. On the other hand, the inability of progesterone to have an antagonistic action with estrogen in the endometrial tissue is a determining factor for establishing endometriosis. The epigenetic changes are also an undeniable factor in the pathogenesis of endometriosis, and this is demonstrated by the fact that not only women of reproductive age suffer from the disease but also adolescents and younger women with a family history. Specific genetic loci have also been identified for the initiation of the disease, making some women more prone to the disease than others. However, the genetic profile is not yet fully understood. Epidemiology The determination of the epidemiological measurement features sometimes becomes very difficult since a lot of women often remain asymptomatic and go undiagnosed. The exact prevalence cannot be easily defined since the definitive diagnosis of the disease requires a laparoscopic examination. Endometriosis is estimated to affect approximately 10% to 15% of women of reproductive age, whereas this prevalence increases by up to 70% in women with chronic pelvic pain. In the U.S., according to a more recent survey of the National Hospital Discharge Survey, 11.2% of all women between 18 and 45 years old hospitalized for genitourinary causes were diagnosed with endometriosis, and approximately 10.3% of the women who have undergone gynecologic surgeries have endometriosis. Endometriosis is a disease with a high burden in Europe, and it has been shown that in Europe, the average annual cost for every patient, including health care and loss of productivity, was approximately €10,000, whereas, in the U.S., the health care cost is 63% higher than the average woman. The diagnosis of endometriosis in the majority of women is often delayed. Thus, women unavoidably suffer from the pain and the long-term effects of this debilitating disease, including infertility. In women with infertility, endometriosis is up to 50%, whereas, in adolescents, the incidence of endometriosis is reported to be 47% of those who have experienced laparoscopy for pelvic pain. Several predisposing factors have been linked with the risk of developing endometriosis. Early age at menarche (age below 11 years old), shorter duration of menstrual periods(less than 27 days), heavy menstrual bleeding (menorrhagia), and nulliparity increase the risk for endometriosis, indicating and ascertaining the hypothesis that endometriosis is closely linked with the hormonal status of a woman(high estrogen levels and low progesterone). On the contrary, there are protective factors against endometriosis, mainly by lowering the inflammatory process or decreasing estrogen levels in the body. Parity, protracted breastfeeding, current oral contraceptive use, tubal ligation, and smoking are related to a decreased risk for endometriosis. Tubal ligation has been assumed to decrease the risk of endometriosis by hampering the menstrual reflux to reach the pelvic cavity. Prolonged breastfeeding seems to suppress the development of endometriosis through postpartum amenorrhea but also with other mechanisms. Contraceptive use and parity hamper the occurrence of ovulation. Whenever ovulation occurs, it is followed by inflammation and release of pro-inflammatory cytokines, including TNF-a, interleukin 1 (IL-1), and interleukin 6 (IL-6). These cytokines enhance cell proliferation and oxidative stress and lead to increased production of leukotrienes and prostaglandins. Additionally, primary amenorrhea, physical activity, and other dietary and lifestyle habits such as increased intake of omega-3 fatty acids and smoking are associated with a decreased risk for endometriosis. The fact that smoking has an inverse association with the risk of endometriosis attracts a lot of attention. Smoking has a catastrophic effect on almost every aspect of health, but this is not the case with endometriosis. Pathophysiology The understanding of the pathophysiology of endometriosis remains incomplete in many aspects, and there is not a coherent suggested theory to explain all different types of endometriosis, integrating the epigenetic, genetic, immunological, and environmental data. Of the proposed pathogenic theories, the most plausible is Sampson’s theory, suggesting that with retrograde menstruation, viable cells, and menstrual fragments can migrate through the Fallopian tubes, infiltrate into the peritoneal cavity, and then proliferate and cause chronic inflammation. The fact that retrograde menstruation is a phenomenon caused in a large proportion of women of reproductive age, but not all of them suffer from endometriosis, indicates that retrograde menstruation is not enough by itself to cause endometriosis, indicates that there are also other factors contributing to the generation of disease. Other theories, such as the coelomic metaplastic theory and the vascular and lymphatic metastatic theory, are necessary to explain some forms of endometriosis. The role of oxidative stress and ROS, together with genetic, epigenetic, and environmental factors, must be integrated for a more complete picture of the pathogenesis of endometriosis. History and Physical The peak of the disease incidence is greater in women between 25 and 29 years old and lowest in women over 44 years old, whereas the prevalence in whites is higher than in African-Americans. The clinical presentation of the disease differs in women and may be unexpected not only in the presentation but also in the duration. Clinicians usually suspect and are more likely to diagnose the disease in females presenting with the typical symptomatology such as dyspareunia, namely painful sexual intercourse, pelvic pain during menstruation (dysmenorrhea), pain in the urination (dysuria), defecation (dyschezia), or infertility. The pain is usually characterized as chronic, cyclic, and progressive (exacerbating over time). Furthermore, some women suffering from endometriosis experience allodynia, a phenomenon when, even with the application of a nonpainful stimulus, an intolerable painful reaction is released. This condition indicates neuropathic pain. In the case of deep infiltrating endometriosis, the neural damage is caused by the invasion of endometrial stromal cells and mediators such as serotonin, histamine, prostaglandins, and nerve growth factor. These agents are released from mast cells, activated macrophages, and leukocytes, directly damaging the sensory nerve fibers and circulating in the peritoneal fluid. Three subtypes of endometriosis often overlap with each other. These are the superficial peritoneal lesions, the ovarian endometrioma, and the deep infiltrating endometriosis. Notably, the degree of clinical manifestations of the patient is not directly associated with the extent of the disease or the size of endometriosis lesions. Patients with bowel endometriosis often present with a wide range of gastrointestinal symptoms like diarrhea, constipation, abdominal pain, or bloating and could mimic other clinical conditions like inflammatory bowel disease or irritable bowel syndrome. Rectovaginal endometriosis is a very severe deep infiltrating form of the disease involving the vagina, the rectum, and the rectovaginal septum. Most often, it can be presented with bowel irritation, dyspareunia, dysmenorrhea, dyschezia, and rectal bleeding coinciding with menstrual bleeding. Finally, a symptom with great concern is infertility, which often leads clinicians to suspect endometriosis, even in asymptomatic patients. The endometriotic lesions in ovaries are described as ovarian endometriomas or as pseudocysts. The frequency of ovarian endometriomas accounts for approximately 17% to 44% of women with endometriosis, are bilateral in 50% of the cases, and are twice more frequent in the left than in the right ovary due to anatomic variables. They differentiate from the common ovarian cysts by the fact that the endometriotic tissue bleeds inside of the endometriotic cyst, is a well-defined extra ovarian hematoma, and is surrounded by a duplicated ovarian parenchyma, lined by endometrial stroma, epithelium, and glands. Chocolate fluid accumulates, covering the wall and filling in the cyst, containing old degenerated blood products, like hemosiderin-filled macrophages and pigmented histiocytes, resulting in a chocolate-like appearance in laparoscopic imaging. Deep infiltrating endometriosis refers to endometriotic lesions that infiltrate the peritoneum more than 5 mm and cause severe symptomatology. Deep endometriosis can involve additionally the bladder, the ureter, and the bowel, more frequently the rectovaginal septum, and less commonly the sigmoid. Superficial peritoneal lesions are, in most cases, an accidental finding. Still, they sporadically can cause a thickening or hemorrhage of the mucosa or even be the reason for a pathological cervicovaginal smear. The histologic diagnosis of endometriosis could be hindered or obscured by a non-typical appearance of the endometrial glands or even the stroma caused by inflammation, edema, or hemorrhage. Evaluation The disease diagnosis is usually delayed an average of 4 to 11 years from the onset of symptoms. This phenomenon occurs in low- and middle-income countries and wealthy societies with universal healthcare access. This delay is attributed to the non-existence of a pathognomonic test or biomarker to detect the disease but also to the diversity of symptoms that could be considered physiologic responses during menstruation (like pain and discomfort), and the wide range of reported symptoms overlap with other gastrointestinal or gynecological causes. To diagnose endometriosis properly, the physician should start by taking a detailed history and performing a gynecological physical examination. Positive family history, pelvic pain, benign ovarian cysts, pelvic surgeries, and infertility issues enhance and alarm physicians to diagnose endometriosis. Physical examination reveals variable findings, depending on the location and size of the endometriotic lesion. Tenderness on vaginal examination, palpable nodules in the posterior fornix, adnexal masses, and immobility of the uterus are diagnostically indicating findings of endometriosis. Nevertheless, the absence of physical findings cannot exclude the diagnosis of endometriosis. The gold standard diagnostic tool remains laparoscopy, combined with an abdominal cavity exploration and a histological biopsy. The lesions can differ in size and color. They can appear as red, white, or clear vesicular. Black "powder burns" or "gunshot" lesions are brown or black endometriotic tissue. The role of histological confirmation is a bit controversial since macroscopically detected endometriotic lesions cannot always be verified histologically, and vice versa, namely in a macroscopically normal peritoneum, can be detected endometriosis tissue. Direct visualization of the endometriotic lesion alone without a histological confirmation lacks enough diagnostic value since, to a great extent, it relies on the surgical skills and capability of the physician. Because diagnostic laparoscopy is an invasive procedure, it entails risks associated with any invasive process and could lead, albeit rare, to severe complications. Hence, there is a need for a diagnostic transition from surgical to non-surgical options, which could also contribute to a decrease in the time between the onset of symptoms and the diagnosis. Several other low-invasive diagnostic methods have been evaluated regarding their diagnostic value, including magnetic resonance imaging and transvaginal ultrasound. Transvaginal ultrasound is a diagnostic tool with high sensitivity and specificity for ovarian endometriomas and permits a wide exploration of the pelvic cavity. Endometriomas appear as homogenous formations with a classic ground-glass appearance and low-level internal echoes. Transvaginal ultrasound could moreover facilitate the diagnosis of deep infiltrating endometriosis located in the rectovaginal septum, uterosacral ligaments, pouch of Douglas, and vaginal wall. To assess whether specific serum, tissue, and urine biomarkers could assist diagnostically for endometriosis, CA125 is elevated in patients with endometriosis. Still, this test could not stand as a single diagnostic test. This is because CA125 can be increased in several pathological conditions except for endometriosis and cannot define the location of endometriotic lesions. Various miRNAs have been found to upregulate or downregulate specific genes and play a significant role in the pathogenesis of infertility and endometriosis. Still, since many results are contradictory, more research is necessary to elucidate these agents' role further. Treatment / Management The treatment of endometriosis is broadly categorized into 2 main categories: pharmacological and surgical. Currently, there is no specific drug that could inhibit the progress of the disease other than hormonal and non-hormonal agents used to alleviate the symptoms and increase fertility rates. An empirical medical therapy can be instituted for women with endometriosis symptoms (ie, pelvic pain) even without histological confirmation of the disease. It should be highlighted that since endometriosis is a chronic disease, the treatment is mainly used to alleviate symptoms and not ultimately for the cure. Providers should always remember that response to the therapy does not verify the diagnosis of endometriosis. Many physicians treat women who suffer from persistent pain empirically after taking a detailed history, conducting a thorough physical examination, and excluding other pathologies, even though medical treatment does not improve fertility. The first-line pharmacological therapy proposed for the management of endometriosis consists of non-steroidal anti-inflammatory drugs, progestins, or combined hormonal contraceptives. Combined hormonal contraceptives can be administered either cyclically or continuously, and they exert their effect by inhibiting follicular development, lowering the levels of LH and FSH, and leading to decidualization and atrophy of the human endometrium. Oral contraceptives are generally well-tolerated, affordable drugs, but in case of discontinuance, they have a high possibility of endometriosis-related pain recurrence. Progestins are another option for the medical treatment of endometriosis. Progestins demonstrate their therapeutic action by inhibiting ovulation and creating a hypoestrogenic milieu. By binding directly to the progesterone receptors in the endometrium, they cause decidualization and atrophy of endometriotic implants. They can also relieve symptoms by decreasing peritoneal inflammation. Weight gain and acne are potential side effects associated with the administration of progestins. Lastly, non-steroidal anti-inflammatory drugs have been evaluated as very effective against endometriosis-related pain. They are supported as first-line medical therapy due to their easy accessibility as over-the-counter drugs and the low adverse-effect profile. A second-line medical empiric treatment is a 3-month trial of gonadotropin-releasing hormone analogs for the suppression of endometriosis-related symptoms. The continuous administration of the gonadotropin-releasing hormone acts by binding to pituitary receptors and downregulating the pituitary-ovarian axis. This results in pituitary desensitization, a fall in the levels of LH and FSH, anovulation, hypoestrogenism, and endometrial atrophy. The only concerns regarding this therapy are related to the side effects of hypogonadism, including bone loss, hot flashes, vaginal dryness, and headache. Danazol, an androgen used to alleviate endometriosis-related symptoms, leads to atrophy of endometriotic implants by hindering LH surge and lowering estrogen levels. On the other hand, testosterone levels increase, and side effects like hirsutism, irreversible deepening of the voice, or acne can appear. Surgical treatment could also be adopted as a potential management option, even though it entails several risks related to the complications. The major advantage of surgical treatment compared to pharmaceutical options is the ability to enhance fertility capability, but it can also provide pain relief simultaneously . Surgery should be considered in cases of superficial endometriosis unresponsive or with contradictions to medical treatment. When laparoscopic surgery is performed, the recommendation is to excise all the endometriotic lesions and adhesions. By the ablation of the endometriotic tissue, the local inflammatory milieu decreases in the pelvic cavity, thus increasing the chances of conception. On the other hand, the management of ovarian endometriomas is still challenging and poses a dilemma to clinicians. The reason behind this is the cumulative research evidence that shows that the removal of the capsule of the cyst can result in a decrease in ovarian reserve and follicular loss. However, cystectomy is preferable to cyst drainage or ablation due to its greater effectiveness in pain relief and the lower recurrence rate. The final decision about which treatment approach should be followed should be taken collaboratively with the patient after a detailed explanation of all the possible risks and benefits related to each treatment option. Differential Diagnosis The most common symptoms of endometriosis are infertility and chronic cyclic pelvic pain; therefore, other conditions should be ruled out from the differential diagnosis. Chronic pelvic pain is a usual symptom derived from pathologic conditions of the urologic, reproductive, and gastrointestinal systems. Pelvic inflammatory disease, adhesions, endometritis, primary dysmenorrhea, and second dysmenorrhea due to adenomyosis, myoma, and cervical stenosis should be excluded from the pathologies of the genital system. Other diseases that should be excluded come from the gastrointestinal system, like irritable bowel syndrome, constipation, inflammatory bowel disease, and the urinary system, such as interstitial cystitis or chronic urinary inflammation. Finally, the physician should also rule out neurologic and psychosomatic causes leading to chronic pelvic pain. Prognosis Patients with endometriosis have fewer chances for childbearing and a higher risk for miscarriages and ectopic pregnancies than those without the disease. Additionally, endometriotic lesions can spontaneously regress in approximately one-third of affected women who are not receiving any treatment. According to different estimates, endometriosis recurrence rates after surgery vary between 6 and 67%. Potential risk factors predicting a possible recurrence have not been completely clarified, but the recurrent endometriotic lesions may arise from de novo cells or residual endometriotic tissue. Medical treatment can be proven effective, but in 5% to 59% of patients, the pain continues to exist at the end of the therapy. Even in cases of treatment cessation, pain recurrence has been reported in a percentage of 17% to 34%. Complications The main complications of endometriosis include infertility or subfertility, chronic pain, and other debilitating persistent symptoms. These are wide in range, including dysmenorrhea, dyspareunia, and dyschezia. Still, endometriosis can also cause a decrease in the quality of life of patients, complications of surgical procedures, anatomical abnormalities due to possible adhesions, bowel or/and bladder dysfunction, but in the case of ovarian endometriomas could even lead to the development of cancer. The role of endometriosis in infertile women has been debated for a long time, with literature research showing that infertile women are 6 up to 8-fold more likely to suffer from endometriosis than fertile females. Even though several mechanisms have been proposed to explain this phenomenon, no consensus has been reached in the scientific community about the exact mechanism behind infertility in endometriosis. Chronic pelvic pain is closely associated with endometriosis since endometriosis has been the cause in 71% to 87% of women suffering from chronic pelvic pain. This disease can negatively affect the health-related quality of life and crucially hinder social, emotional, and sexual well-being as well as other domains like daily routines, family planning, efficacy, or productivity of patients in the working environment. Additionally, patients with endometriosis have higher stress levels, report worse quality of sleep, and lower physical activity compared to healthy women. Bowel dysfunction like constipation or other digestive problems can appear in females who have endometriosis as a result of the inflammatory process of irritation of the gastrointestinal system rather than due to the involvement of endometriotic nodules affecting the rectum since patients undergoing surgical treatment for rectal endometriosis might continue experiencing those displeasing symptoms. Although primarily endometriosis was considered a benign condition, later research findings showed a higher risk for ovarian cancer in patients with endometriosis. Deterrence and Patient Education Recurrence is common in patients who discontinue therapy or even after the surgery. Patients should be encouraged to continue with medical therapy for several months, even though they might present some adverse effects. Females who undergo surgical treatment should adhere to the doctor’s post-surgical instructions to avoid and minimize postoperative complications such as infections. Patients should be encouraged to visit the health care specialist for further evaluation in case of a recurrence. Nevertheless, women with a diagnosis of the disease should be aware of possible complications of the disease, predominantly the risk for infertility and chronic pelvic pain. Enhancing Healthcare Team Outcomes Endometriosis remains a disease with high comorbidity, even though extensive research has been conducted and new medical treatments have been developed. Healthcare professionals need to join hands and work collectively when they need to work with patients with suspected cases of endometriosis. It is not random that the mean delay of a definitive diagnosis of endometriosis is approximately 10 years. Public health interventions should be introduced and promote awareness among females of childbearing age for signs and symptoms of endometriosis; thus, they could seek medical care in an earlier stage of the disease. Clinicians should work as a team regarding the management of endometriosis to achieve the most effective treatment and avoid surgical complications. Additionally, healthcare professionals engaged with the management of endometriosis should be informed about new drugs, their indications or adverse effects to implement a successful treatment, and their attention should be drawn to the fact that the symptomatology of the disease has no direct relationship with the extent of the disease. The treatment plan should be personalized and patient-centered, chosen basically by the individual characteristics of each patient. The symptoms, age, and childbearing desire are essential factors that should be communicated with the patient and within the health care team. Endometriosis is a chronic condition. Therefore, the main goals of the healthcare team should be the alleviation of symptoms, the minimization of pain recurrence, and the safety of the patient. Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. References 1. : Jenkins S, Olive DL, Haney AF. Endometriosis: pathogenetic implications of the anatomic distribution. Obstet Gynecol. 1986 Mar;67(3):335-8. [PubMed: 3945444] 2. : Macer ML, Taylor HS. Endometriosis and infertility: a review of the pathogenesis and treatment of endometriosis-associated infertility. Obstet Gynecol Clin North Am. 2012 Dec;39(4):535-49. [PMC free article: PMC3538128] [PubMed: 23182559] 3. : Koninckx PR, Barlow D, Kennedy S. Implantation versus infiltration: the Sampson versus the endometriotic disease theory. Gynecol Obstet Invest. 1999;47 Suppl 1:3-9; discussion 9-10. [PubMed: 10087422] 4. : Matsuura K, Ohtake H, Katabuchi H, Okamura H. Coelomic metaplasia theory of endometriosis: evidence from in vivo studies and an in vitro experimental model. Gynecol Obstet Invest. 1999;47 Suppl 1:18-20; discussion 20-2. [PubMed: 10087424] 5. : Konrad L, Dietze R, Kudipudi PK, Horné F, Meinhold-Heerlein I. Endometriosis in MRKH cases as a proof for the coelomic metaplasia hypothesis? Reproduction. 2019 Aug;158(2):R41-R47. [PubMed: 30978694] 6. : Jabr FI, Mani V. An unusual cause of abdominal pain in a male patient: Endometriosis. Avicenna J Med. 2014 Oct;4(4):99-101. [PMC free article: PMC4183904] [PubMed: 25298953] 7. : Vercellini P, Viganò P, Somigliana E, Fedele L. Endometriosis: pathogenesis and treatment. 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[PubMed: 12861175] 18. : Missmer SA, Hankinson SE, Spiegelman D, Barbieri RL, Marshall LM, Hunter DJ. Incidence of laparoscopically confirmed endometriosis by demographic, anthropometric, and lifestyle factors. Am J Epidemiol. 2004 Oct 15;160(8):784-96. [PubMed: 15466501] 19. : Agarwal SK, Chapron C, Giudice LC, Laufer MR, Leyland N, Missmer SA, Singh SS, Taylor HS. Clinical diagnosis of endometriosis: a call to action. Am J Obstet Gynecol. 2019 Apr;220(4):354.e1-354.e12. [PubMed: 30625295] 20. : Borghese B, Santulli P, Marcellin L, Chapron C. [Definition, description, clinicopathological features, pathogenesis and natural history of endometriosis: CNGOF-HAS Endometriosis Guidelines]. Gynecol Obstet Fertil Senol. 2018 Mar;46(3):156-167. [PubMed: 29540335] 21. : Chapron C, Bourret A, Chopin N, Dousset B, Leconte M, Amsellem-Ouazana D, de Ziegler D, Borghese B. Surgery for bladder endometriosis: long-term results and concomitant management of associated posterior deep lesions. Hum Reprod. 2010 Apr;25(4):884-9. [PubMed: 20129993] 22. : Remorgida V, Ferrero S, Fulcheri E, Ragni N, Martin DC. Bowel endometriosis: presentation, diagnosis, and treatment. Obstet Gynecol Surv. 2007 Jul;62(7):461-70. [PubMed: 17572918] 23. : Moawad NS, Caplin A. Diagnosis, management, and long-term outcomes of rectovaginal endometriosis. Int J Womens Health. 2013 Nov 08;5:753-63. [PMC free article: PMC3825702] [PubMed: 24232977] 24. : Bastos BB, Fonseca EKUN, Yamauchi FI, Baroni RH. Chocolate cyst with ground glass appearance in endometriomas. Abdom Radiol (NY). 2017 Dec;42(12):2957-2958. [PubMed: 28676998] 25. : Gałczyński K, Jóźwik M, Lewkowicz D, Semczuk-Sikora A, Semczuk A. Ovarian endometrioma - a possible finding in adolescent girls and young women: a mini-review. J Ovarian Res. 2019 Nov 07;12(1):104. [PMC free article: PMC6839067] [PubMed: 31699129] 26. : De Cicco C, Corona R, Schonman R, Mailova K, Ussia A, Koninckx P. Bowel resection for deep endometriosis: a systematic review. BJOG. 2011 Feb;118(3):285-91. [PubMed: 21040395] 27. : Clement PB. The pathology of endometriosis: a survey of the many faces of a common disease emphasizing diagnostic pitfalls and unusual and newly appreciated aspects. Adv Anat Pathol. 2007 Jul;14(4):241-60. [PubMed: 17592255] 28. : Vercellini P, Trespidi L, De Giorgi O, Cortesi I, Parazzini F, Crosignani PG. Endometriosis and pelvic pain: relation to disease stage and localization. Fertil Steril. 1996 Feb;65(2):299-304. [PubMed: 8566252] 29. : Hickey M, Ballard K, Farquhar C. Endometriosis. BMJ. 2014 Mar 19;348:g1752. [PubMed: 24647161] 30. : Kiesel L, Sourouni M. Diagnosis of endometriosis in the 21st century. Climacteric. 2019 Jun;22(3):296-302. [PubMed: 30905186] 31. : Kim JH, Han E. Endometriosis and Female Pelvic Pain. Semin Reprod Med. 2018 Mar;36(2):143-151. [PubMed: 30566980] 32. : Platteeuw L, D'Hooghe T. Novel agents for the medical treatment of endometriosis. Curr Opin Obstet Gynecol. 2014 Aug;26(4):243-52. [PubMed: 24978852] 33. : Wellbery C. Diagnosis and treatment of endometriosis. Am Fam Physician. 1999 Oct 15;60(6):1753-62, 1767-8. [PubMed: 10537390] 34. : Jeng CJ, Chuang L, Shen J. A comparison of progestogens or oral contraceptives and gonadotropin-releasing hormone agonists for the treatment of endometriosis: a systematic review. Expert Opin Pharmacother. 2014 Apr;15(6):767-73. [PubMed: 24588662] 35. : Magon N. Gonadotropin releasing hormone agonists: Expanding vistas. Indian J Endocrinol Metab. 2011 Oct;15(4):261-7. [PMC free article: PMC3193774] [PubMed: 22028996] 36. : Kho RM, Andres MP, Borrelli GM, Neto JS, Zanluchi A, Abrão MS. Surgical treatment of different types of endometriosis: Comparison of major society guidelines and preferred clinical algorithms. Best Pract Res Clin Obstet Gynaecol. 2018 Aug;51:102-110. [PubMed: 29545114] 37. : Wozniak S. Chronic pelvic pain. Ann Agric Environ Med. 2016 Jun 02;23(2):223-6. [PubMed: 27294622] 38. : Hjordt Hansen MV, Dalsgaard T, Hartwell D, Skovlund CW, Lidegaard O. Reproductive prognosis in endometriosis. A national cohort study. Acta Obstet Gynecol Scand. 2014 May;93(5):483-9. [PubMed: 24617701] 39. : Harrison RF, Barry-Kinsella C. Efficacy of medroxyprogesterone treatment in infertile women with endometriosis: a prospective, randomized, placebo-controlled study. Fertil Steril. 2000 Jul;74(1):24-30. [PubMed: 10899492] 40. : Selçuk I, Bozdağ G. Recurrence of endometriosis; risk factors, mechanisms and biomarkers; review of the literature. J Turk Ger Gynecol Assoc. 2013;14(2):98-103. [PMC free article: PMC3881735] [PubMed: 24592083] 41. : Becker CM, Gattrell WT, Gude K, Singh SS. Reevaluating response and failure of medical treatment of endometriosis: a systematic review. Fertil Steril. 2017 Jul;108(1):125-136. [PMC free article: PMC5494290] [PubMed: 28668150] 42. : Bulletti C, Coccia ME, Battistoni S, Borini A. Endometriosis and infertility. J Assist Reprod Genet. 2010 Aug;27(8):441-7. [PMC free article: PMC2941592] [PubMed: 20574791] 43. : Bloski T, Pierson R. Endometriosis and Chronic Pelvic Pain: Unraveling the Mystery Behind this Complex Condition. Nurs Womens Health. 2008 Oct;12(5):382-95. [PMC free article: PMC3096669] [PubMed: 18837717] 44. : Marinho MCP, Magalhaes TF, Fernandes LFC, Augusto KL, Brilhante AVM, Bezerra LRPS. Quality of Life in Women with Endometriosis: An Integrative Review. J Womens Health (Larchmt). 2018 Mar;27(3):399-408. [PubMed: 29064316] 45. : Roman H, Bridoux V, Tuech JJ, Marpeau L, da Costa C, Savoye G, Puscasiu L. Bowel dysfunction before and after surgery for endometriosis. Am J Obstet Gynecol. 2013 Dec;209(6):524-30. [PubMed: 23583209] 46. : Králíčková M, Laganà AS, Ghezzi F, Vetvicka V. Endometriosis and risk of ovarian cancer: what do we know? Arch Gynecol Obstet. 2020 Jan;301(1):1-10. [PubMed: 31745637] 47. : Mehedintu C, Plotogea MN, Ionescu S, Antonovici M. Endometriosis still a challenge. J Med Life. 2014 Sep 15;7(3):349-57. [PMC free article: PMC4233437] [PubMed: 25408753] : Disclosure: Eleni Tsamantioti declares no relevant financial relationships with ineligible companies. : Disclosure: Heba Mahdy declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK567777PMID: 33620854 Share Views PubReader Print View Cite this Page Tsamantioti ES, Mahdy H. Endometriosis. [Updated 2023 Jan 23]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. In this Page Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Related information PMC PubMed Central citations PubMed Links to PubMed Similar articles in PubMed Indocyanine Green-Assisted Retrograde Ureterolysis in Robotic Transvaginal NOTES for the Management of Stage IV Endometriosis with Obliterated Cul-de-sac.[J Minim Invasive Gynecol. 2023] Indocyanine Green-Assisted Retrograde Ureterolysis in Robotic Transvaginal NOTES for the Management of Stage IV Endometriosis with Obliterated Cul-de-sac. Guan X, Guan Z, Sunkara S, Thigpen B. J Minim Invasive Gynecol. 2023 Apr; 30(4):266-267. Epub 2023 Feb 9. Lower Genitourinary Trauma.[StatPearls. 2025] Lower Genitourinary Trauma. Tullington JE, Blecker N. StatPearls. 2025 Jan Advanced laparoscopic surgery for the removal of rectovaginal septum endometriotic or adenomyotic nodules.[Baillieres Clin Obstet Gynaeco...] Advanced laparoscopic surgery for the removal of rectovaginal septum endometriotic or adenomyotic nodules. Donnez J, Nisolle M. Baillieres Clin Obstet Gynaecol. 1995 Dec; 9(4):769-74. Review The Antigen-Processing Pathway via Major Histocompatibility Complex I as a New Perspective in the Diagnosis and Treatment of Endometriosis.[Arch Immunol Ther Exp (Warsz)....] Review The Antigen-Processing Pathway via Major Histocompatibility Complex I as a New Perspective in the Diagnosis and Treatment of Endometriosis. Nowak I, Bochen P. Arch Immunol Ther Exp (Warsz). 2024 Jan 1; 72(1). Epub 2024 Mar 13. Review Endometriosis: interaction of immune and endocrine systems.[Semin Reprod Med. 2003] Review Endometriosis: interaction of immune and endocrine systems. Seli E, Arici A. Semin Reprod Med. 2003 May; 21(2):135-44. 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13872	https://www.osti.gov/servlets/purl/2491452	This manuscript has been authored by UT -Battelle LLC under contract DE -AC05 -00OR22725 with the US Department of Energy (DOE). The US government retains and the publisher, by accepting the article for publication, acknowledges that the US government retains a nonexclusive, paid -up, irrevocable, worldwide license to publish or reproduce the published form of this manuscript, or allow others to do so, for U S government purposes. DOE will provide public access to these results of federally sponsored research in accordance with the DOE Public Access Plan ( - public -access -plan ). Nuclide Importance to Decay Heat in Advanced Reactors ’ Spent Fuel Transportation and Storage Germina Procop 1, Yves Robert 1 1 Oak Ridge National Laboratory , Oak Ridge , TN ilasg@ornl.gov [Placeholder for Digital Object Identifier (DOI) to be added by ANS] INTRODUCTION The energy released by the decay of radionuclides in irradiated fuel after its discharge from a nuclear reactor is an important parameter for design, safety, and licensing analyses of spent nuclear fuel storage, transportation, and repository systems. For light -water reactor (LWR) spent fuel, well -validated computational methods and tools are available to provide reliable e stima tes of this energy release, which is generally known as decay heat . Validation of the computational tools and associated nuclear data has benefitted from a wealth of knowledge acquired from decades of LWR operational experience and large sets of experimental data that serve as a validation ba sis. However , for advanced reactors, there is a significant gap in experience and available validation basis. Although the foundational physics remains the same, advanced reactor technologies are significantly different from LWRs with respect to materials, design s, and physics behavior. The se technologies are based on different reactor types with either a thermal or a fast neutron spectrum, diverse fuel forms, various coolant materials, fuel s with higher initial enrichments and higher target discharge burnups. Consequently, the computational tools needed to accurately predict the nuclear transmutation and decay during reactor operation and post -discharge i n advanced reactor fuels , as well as the characteristics of the fuel discharged from these reactors , have different requirements compared to those for LWRs . To improve the knowledge of fuel characteristics for fuel irradiated in advanced reactors and their impact on the back end of the fuel cycle applications , a concerted effort was initiated earlier in 2024 at Oak Ridge National Laboratory (ORNL). The primary goal of this effort is to investigate, for irradiated fuel s representative of several advanced reactor types, the nuclide importance to criticality safety, thermal analysis , and shielding related to transportation and interim storage. A team of several analysts with expertise in neutron transmutation and decay physics, criticality safety, and radiation transport and shielding is involved in this activity. Another motivation for this effort is to augment and provide a parallel to a previous study on nuclide importance to LWR spent fuel transportation and interim storage. This earlier study, performed by ORNL and funded by the Nuclear Regulatory Commission (NRC), aimed to improve the technical basis for identify ing those spent fuel characteristics that may adversely affect the accuracy of the computational tools and data, and for support ing the assessment of the potential impact on existing NRC guidance. The r esults of this study were documented in the technical report NUREG/CR -6700 published in 2001 . Th is report continues to be used as a reference by analysts who investiga te potential design s for advanced reactor spent fuel transportation, storage, and shielding, in lieu of an appropriate reference that is not yet available. The basis and the approach used to determine the nuclide inventory as a function of cooling time post -irradiation for fuels representative of several advanced reactor configurations are summarized herein, along with preliminary results for nuclide importance to decay heat for a specific fuel type . FRAMEWORK Th e current study relies on and benefits from the strong foundation resulting from a collaborative project initiated by the NRC in 2019 , between Sandia National Laboratory (Sandia ) and ORNL . Through this collaboration, the two national laboratories have demonstrated the capabilities of the MELCOR and SCALE code systems to support thermal hydraulics, accident progression, source terms, and consequence analyses for several non -LWR reference plants. The considered non -LWR system s, which are represe ntative of several advanced reactor technologies being actively pursued for development and deployment by the US industry and with government support, includ e the following : pebble - bed high -temperature gas -cooled reactors (HTGRs), pebble - bed fluoride salt –cooled high -temperatures reactors (FHRs), molten salt –fueled reactors (MSRs), heat pipe reactors (HPRs), and sodium -cooled fast reactors (SFRs). The role of SCALE in the Sandia –ORNL collaboration was to provide MELCOR with the initial and boundary conditions needed to simulate accidents, including nuclide inventories, decay heat, power profiles, and reactivity coefficients . All the input and output files generated with SCALE, along with the list of corresponding references — technical reports, journal or conference papers —were shared with the nuclear science and engineering community via a public repository , which is available at -lwr -models -vol3 .The nuclide inventories at discharge for the considered non -LWR reactors and fuels are available as binary files in the afore mentioned repository, along with information in sufficient detail for an independent user to understand the meaning of and apply the provided data to a different application of interest. These nuclide inventories at discharge serve as input to the current study on nuclide importance to advanced reactors ’ spent fuel transportation and storage , with all the reactor types —HTGR, FHR, HPR , MSR, and SFR — that were the focus o f the previous non -LWR project being represented in the current study. APPROACH Decay heat is driven by the nuclide inventory in the fuel at the end of irradiation. If the inventory is available at the end of an irradiation time , then determining decay heat as a function of cooling time is straightforward. The evolution of the nuclide inventory and the corresponding decay heat as a function of cooling is determined in this study via ORIGEN decay simulations. The ORIGEN nuclear transmutation and decay code is the key component of all depletion capabilities in the SCALE code system: it s erves as a depletion solver in SCALE ’s lattice physics codes (i.e., TRITON, Polaris), is the engine of the ORIGAMI tool for rapid fuel depletion using pre -generated libraries , and it can also be used as a standalone depletion/decay code. To better facilitate the generation of nuclide inventories as a function of cooling time , for the different types of advanced reactor fuel s considered herein and in the appropriate format for subsequent use in criticality safety and shielding models, Python scrip ts are being developed to automate the process. This automation , in progress at the time of this writing, includes (a) interrogation of inventory files that are available in binary format to extract the inventory at a desired burnup for use as input in an ORIGEN decay simulation; (b) creation and running of ORIGEN decay inputs for diffe rent fuel burnup s and cooling times; and (c) post -processing of the resulting files. Inventory data as a function of cooling time are sufficient as fuel material input for the nuclide importance to decay heat and criticality safety analyses planned for this study. For the planned shielding analyses, detailed neutron and gamma source and spectral information is also necessary as input. The se shielding analyses will be based on an on -the -fly shielding analysis method, which has been demonstrated to perform adequately for LWR spent fuel shielding applications [ 6]. The automation script s under development address these additional input needs. The FHR spent fuel case served as a test bed for testing and optimizing the workflow of the automation. The irradiated fuel inventory data available for this reactor type includes the average inventory for a fuel pebble as a function of its multiple passes through the core, along with the inventory corresponding to an equilibrium core. During reactor operation at equilibrium , the core consists of a mixture of fuel pebbles with different irradiation histories and at different burnups. A f uel pebble travels on average eight times through the core before it has reached the target discharge burnup of 180 GWd/MTU and is replaced by a fresh fuel pebble . The generation of fuel inventory for this equilibrium core was based on the SCALE Leap -In method for Cores at Equilibrium (SLICE), with details available elsewhere [5, 7]. PRELIMINARY RESULTS Nuclide inventories were determined for the FHR irradiated fuel as a function of cooling time between 0.5 and 100 years and for different burnup values, corresponding to a different number of a pebble passes through the core . The corresponding decay heat curves are illustrated in Fig. 1. As expected, the v ariation of the decay heat for the FHR spent fuel is similar to tha t for an LWR; it decreases exponentially with increasing cooling time for a given burnup and increases with increasing burnup for a given cooling time. Fig. 1. Decay heat vs. cooling time for FHR spent fuel . Although t he decay heat absolute values for the FHR and LWR fuels are not too different for a same burnup and cooling time , the specific decay heat of FHR fuel at discharge is expected to be much higher than that of LWR fuel , because the discharge burnup for FHR fuel is significantly higher than that of typical LWR fuel . In th e FHR fuel case, the decay heat of fuel at discharge burnup is on the order of ~10 7 and decreases to ~10 3 W/MTU at 100 yr cooling. In contrast , LWR fuel typically varies between 10 4 and 10 6 at discharge time to a few hundred W/MTU after 100 yr of cooling. The top ten nuclides ranked by their relative contribution to the total decay heat for FHR spent fuel at 180 GW d/MT U burnup and for cooling times of 1, 10, and 100 yr are listed in Table I. As is well known for LWRs, and seen here for FHR spent fuel, fission products are the top contributors to decay heat at shorter cooling times; at long cooling times, while the fission products decay out, the actinides become top contributors. For the specific nuclides shown in Table I, the aggregate contribution to the total decay heat of the fission products is 91% at 1 yr and decreases to 76% at 10 yr and 62% at 100 yr. Meanwhile, the relative contribution to decay heat for the actinides increases from 4% at 1 yr to 37% at 100 yr cooling. Generally, most of these nuclides overlap with those in the top ten contributors to decay heat for LWR spent fuel for a given cooling time in the 1 to 100 yr range, though their order in the list of top ten may slightly differ. Table I. Nuclide contribution to decay heat for FHR fuel t=1yr 6.5 ×10 4W/MTU t=10yr 6.9 ×10 3W/MTU t=100yr 1.5 ×10 3W/MTU nuclide % nuclide % nuclide % Pr -144 39.9 Y-90 28.5 Am -241 33.9 Rh -106 22.2 Ba -137m 25.1 Pu -238 23.2 Cs -134 13.5 Pu -238 10.1 Y-90 15.2 Y-90 3.7 Cm -244 8.8 Ba -137m 14.7 Ce -144 3.6 Cs -137 7.2 Cs -137 4.2 Ba -137m 3.3 Cs -134 6.3 Sr -90 3.2 Nb -95 3.2 Sr -90 6.0 Pu -240 2.4 Cm -242 2.5 Am -241 3.2 Pu -239 1.4 Zr -95 1.5 Eu -154 1.8 Cm -244 1.3 Cm -244 1.3 Kr -85 0.7 Am -243 0.3 remainder 5.4 remainder 2.2 remainder 0.2 For transportation and storage considerations, in addition to the decay heat per MTU, approaching this from the perspective of the corresponding volume would help optimize the envisioned transportation and storage systems. For FHR fuel, with each pebble containing ~1.5g U initial, the decay heat for a pebble with 180 GWd/MTU burnup would be ~20 W at discharge and decrease to 0.2 W at 0.5 yr, 0.01 W at 10 yr, and 0.002 W at 100 yr of cooling. The number of pebbles corresponding to 1MTU is 6.7 ×10 5, and the associated volume is 9.4 m3; considering a sphere packing fraction of 60%, the volume occupied by the pebbles is ~15.7m 3. In comparison, for PWR 17 ×17 fuel, for which the initial U load in a fuel assembly is ~0.5 MTU, the volume corresponding to 1 MTU is ~0. 4 m3. For the same amount of MTU, the fuel pebbles occupy ~ 40 times the volume of corresponding PWR spent fuel . FURTHER STEPS At this early stage , in an overall effort to investigate the nuclide importance of criticality safety, decay heat, and shielding analyses for advanced reactors ’ spent fuel transportation and storage , the focus is on establishing a consistent input basis for all the analysis models involved. It includes streamlining the generation of the required nuclide inventory and radiation sources input data as a function of burnup and cooling time . Five types of irradiated fuels are considered, which are representative of HTGR, FHR, MSR, HPR, and SFR configurations, and for which nuclide inventories at discharge are available. The overall effort is planned to be finalized in 2025 and result in the publication of a NUREG/CR report and other conference and journal publications. ACKNOWLEDGMENTS This work was funded by the U.S. Department of Energy /Nuclear Regulatory Commission, Criticality Safety for HALEU (DNCSH) program. REFERENCES I. C. GAULD, J. C. RYMAN , “ Nuclide Importance to Criticality Safety, Decay Heating, and Source Terms Related to Transport and Interim Storage of High -Burnup Fuel ,” NUREG/CR -6700, U.S. Nuclear Regulatory Commission (200 1). “NRC Non -Light Water Reactor (Non -LWR) Vision and Strategy, Volume 3: Computer Code Development Plans for Severe Accident Progression, Source Term, and Consequence Analysis,” Technical report ML20030A178, Rev.1, U.S. Nuclear Regulatory Commission (2020). L. L. HUMPHRIES, B. A. BEENY, F. GELBARD, D. L. LOUIE, J. PHILLIPS, R. C. SCHMIDT, N. E. BIXLER, “MELCOR Computer Code Manuals Vol. 1: Primer and Users’ Guide Version 2.2.18019,” SAND2021 -0726, Sandia National La boratories (2021). W. A. WIESELQUIST, R. A. LEFEBVRE, eds. , “ SCALE 6.3.1 User Manual ,” ORNL/TM -SCALE -6.3.1 , Oak Ridge National Laboratory (2023). . R. BOSTELMANN, S. SKUTNIK, A. SHAW, D. HARTANTO, W. WIESELQUIST, “SCALE 6.3 Modeling Strategies for Reactivity, Nuclide Inventory, and Decay Heat of Non -LWRs”, ORNL/TM -2024/3213, Oak Ridge National Laboratory (2024). G. RADULESCU, K. BANERJEE, L. P. MILLER, “Demonstration of the On -the -Fly Shielding Analysis Method”, ORNL/ SPR -2021/ 1913 , Oak Ridge National Laboratory (202 1). . F. BOSTELMANN, C. CELIK, R. F. KILE, W. A. WIESELQUIST, “SCALE Analysis of a Fluoride Salt - Cooled High -Temperature Reactor in Support of Severe Accident Analysis”, ORNL/TM -2021/2273, Oak Ridge National Laboratory (2022). 75.
13873	https://www.youtube.com/watch?v=4HymwFByREg	Atwood Machine - Acceleration, Net Force, Time to Reach Ground Greg Clements 5290 subscribers Description 13 views Posted: 23 Jul 2025 This video briefly explains the concept of Atwood's machine. A string passes over a pulley. Unequal masses are attached to the ends of the string. The acceleration of the system, tension in the string, and time for the larger mass to hit the floor are calculated with step-by-step solutions. The result is evaluated to see if it is reasonable. Introductory Physics Prof. Greg Clements Transcript: This video solves an Atwoods machine problem. This Atwoods machine. The pulley at the top has a string around it and the pulley is massless. The pulley has no friction. So, it's not going to enter into our calculations. This particular problem, there's a 510 g mass on the right side at the end of the string and 220 g on the left side at the end of the string. Uh the distance above the floor is labeled for both of those objects. Um and we're going to do some calculations here on the dynamics and motion of u this system. So let's get started with that. We want to calculate four things. Want to calculate the acceleration of the 510 g mass. There is an imbalance in the masses. So the system will be accelerating the acceleration of the 220 g mass. Want to know the tension in the string and I want to know the time for the 510 g mass to hit the floor. U the 510 g its weight is going to dominate the uh forces in the sit in this uh calculation. So the 510 grams is going to move towards the floor. Well, the acceleration of the 510 g mass, we're going to consider the whole system here. Uh, so both the 510 g and the 220 g object. They're going to be in our system and we're going to use net force equals mass time acceleration. How do we get some forces involved here? Well, the 510 g object has a weight to it. Mg so minus 5.03 newtons. I'm using minus because the force is downward. And in this problem, I'm electing to use up as the positive direction. The 220 g object has a weight of minus2.158 newtons. The total mass of the system is if we add these two grams together we get 73 kilg doing a conversion and the net force u I hope you would agree that the weight of the 510 g object and the weight of the 220 g object these weights these forces do not u enhance the acceleration they don't work together they're opposing each other Now in my drawing they're in the same direction but the pulley the pulley changes the direction of force. So these two weights end up opposing each other and to get the net force we must subtract them and we get a net force of minus2.845 newtons. Um calculating the acceleration then net force divided by the total mass of the system. It is important use the total mass of the system not just the 510 gram mass and we come up with minus 3.897 m/s squared. Rounding this to my typical three significant figures, we get minus 3.9 m/s squared. That's the acceleration of the 510 g object towards the floor. So it's a negative number. What about the acceleration of the 200 g object? Well, just think about it. You don't have to do a calculation. It's tied with a string to the 510 g object. This string does not stretch. That's typical physics string. And thus, it shares the motion. It shares the acceleration number with a 520 20 g object. The only difference is the 220 g object is going to be accelerating upward. uh the 510 g object on the right goes down. The string pulls the 220 g object up. So plus 3.9 m/s squared. What about the tension in the string? This tension in the string, this is an internal force in the problem. So I did not have to use it in the net force calculation. Net force only deals with external forces. But now we're going to take one object at a time. And uh I'm going to first consider the 510 g object. And our system is just going to be this object. We're going to ignore the 220 g object for a minute. The system is only the 510 g object and the forces that apply to it attached to it. So tension that's upward in my drawing. So that's a positive. And then the weight is a negative number minus 5.0. 003 newtons. That's the net force acting on the 510 g object. Its mass in kilogram. So use standard units is 0.51. Change the 510 g to 0.51 kg. And previously we calculated its acceleration minus 3.9. Um and you should pause the video now use your calculator and uh verify uh the numbers here. I'm adding 5.00. 03 newtons to both sides, multiplying 0.51 -3.9 and ending up with a tension of 3.01 Newtons. 3.01 Newtons. Um, is this reasonable? Well, I'd say it is reasonable because the tension upward of three roughly 3 newtons is smaller than the weight downward of 5 Newtons. This object will go down. The 510 g object will move downward. Uh the tension is not enough to keep it stable or move it upward. What about the 220 g object? Now let's just focus in on 220 g. The tension in the string acting upward on it here. Its weight acting downward. So that gives us our net force calculation. And then ma and again I'm focusing on just on one object. So I don't use the total system mass. I just use the 0 22 kilograms and its acceleration is a positive. It's accelerating upward. Again, pause your uh video and pull out your calculator. And I get 3.02 Newtons. I'm going to claim uh it's it is the same as 3.01 Newtons. The only reason the number appears a little different is I've rounded here instead of keeping more digits in the calculation. So, this is reasonable. We're getting the same uh value for the tension on both sides. There's only one tension number in this string. So, that's reasonable to get the same number. What about the time for the 510 g object to move down to the floor? Well, it's accelerating. So, we're going to use a kinematic equation here. Let's write down what we know. We know our starting position is 1.3 m up above the floor. We're going to end this motion when we hit the floor. So y will be 0 m. We're going to start from rest. That was given in initially in the problem. So the initial y velocity is zero. We do not know the final velocity. Um not going to bother to calculate it either. But the final velocity when the object just touches the floor. That number is unknown. We know the acceleration. We do not know the time. That's what we're trying to calculate. So if you think about the four kinematic equations, the third kinematic equation, at least what I call the third one, is what we need. It does not have the final velocity symbol in it, and it has the t number that we want to calculate. So uh y - y, the final y is 0. The initial y 1.3 m. The initial velocity in the y direction is zero. We're starting from rest. and then 0.5 the acceleration and t ^2. Again, pause this, pull out your calculator, and I come up with t ^2 of 6666. A lot of sixes on my display. And taking a square root of both sides, I find that the time for the 510 g object to move from rest down to the floor is 0816 seconds. Is this reasonable? Well, suppose we cut the string and just let the 510 g object do a freef fall from 1.3 m above the floor. Uh, when it's in freef fall, its acceleration towards the floor is 9.81 m. And I'm not bothering at the minus sign on both sides here. But, uh, 1.3 m, the object moves down is 0.5 9.81 and t ^2. Again, it's this third kinematic equation. and I come up with 0.515 seconds, a smaller time. And this is reasonable when we're in the Atwoods machine configuration and there's a string attached to the 510 g object and over on the other side to 220 g. Uh the 510 g is not in freefall. It is uh pulling this 220 g object upward. Consequently, it takes more time to get down to the ground. So, reasonable result. So, there's that was machine problem. Net force is important. It is crucial that you decide what objects are in your system and write down your net force for your system. Don't use any internal forces. Uh as we started off here, the tension of the string was an internal force. For more examples of physics uh problems and some short physics lectures, I have an annotated list that includes astronomy videos as well. gplemens.com. Everything's free here. You're not allowed to resell it, but uh you can use it all for free. You don't register. You don't give me your name. You don't give me your email address, uh etc. And my website does not have commands in it that would put cookies on your computer. So don't worry about that. Keep working physics problems and ask your instructor if you have questions.
13874	https://www.thesaurus.com/browse/complies	Advertisement Skip to verb (1) Advertisement complies verb as in abide by, follow agreement or instructions Strongest matches Strong matches Weak matches Advertisement From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. Advertisement Advertisement Advertisement Browse Follow us Get the Word of the Day every day! By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies.
13875	https://programs.iowadnr.gov/bionet/Fish/Species/72	River & Stream Biological Monitoring Fish and Benthic Macroinvertebrate Surveys Physical Habitat Assessments Rock Bass Ambloplites rupestris Family Centrarchidae (Sunfishes) Tolerance Sensitive Trophic Class Top Carnivore Is Exotic to Iowa? False Is Lithophilous Spawner? False Is Hybrid? False State Listing Status Not Listed External Databases Iowa DNR Fisheries - Fish Species Description Integrated Taxonomic Information System Assessment Program Statistics This species was found at 99 bioassessment sites, 3 rapid fish bioassessment sites, 70 fisheries assessment sites, and 0 fisheries presence-only assessment sites. In total, it was collected at 166 distinct sites, or 10.7% of the 1554 total sites monitored by the bioassessment program. It is the 41st most commonly collected species. The Rock Bass was collected in 233 bioassessment sampling sessions and 331 fisheries assessment sessions. It was present in 3 rapid bioassessment sessions and 0 presence-only sessions. The biological assessment program has collected a total of 8,131 individual Rock Bass specimens, ranking it the #34 most collected fish. Species Characteristics A stout, heavy-bodied sunfish with a large mouth, which extends beyond mid-eye when the mouth is closed. The spiny dorsal fin and soft dorsal fin are broadly connected, but without a notch. The dorsal fin is much longer and more noticeable than the anal fin. Six anal fin spines and 12 dorsal fin spines distinguish this fish from all other sunfish. The body is olive with brassy reflections and dark mottlings along the sides. The breast and belly are whitish, and the lower side has spots that form horizontal lines. There is brown mottling and faint banding on the anal, dorsal and tail fins. The pectoral fins are rounded, set low and are amber-colored. Unable to retrieve photo ribbon at this time Species Distribution Maps Site and Time HUC12 Watershed HUC12 watersheds where this species has been found
13876	https://www.khanacademy.org/science/health-and-medicine/advanced-hematologic-system/bleeding-and-impaired-hemostasis/v/coagulation-cascade	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
13877	https://jnnp.bmj.com/content/96/7/690	Changes in the prognosis of CADASIL over time: a 23-year study in 555 individuals \| Journal of Neurology, Neurosurgery & Psychiatry Skip to main content Intended for healthcare professionals Subscribe Log InMoreLog in via Institution Log in via OpenAthens ### Log in using your username and password For personal accounts OR managers of institutional accounts Username Password Forgot your log in details?Register a new account? Forgot your user name or password? 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BMJ Journals You are here Home Archive Volume 96,Issue 7 Changes in the prognosis of CADASIL over time: a 23-year study in 555 individuals Email alerts Article Text Article menu Article Text Article info Citation Tools Share Rapid Responses Article metrics Alerts Article Text Article info Citation Tools Share Rapid Responses Article metrics Alerts PDF PDF + Supplementary Material Cerebrovascular disease Original research Changes in the prognosis of CADASIL over time: a 23-year study in 555 individuals Nontapat Sukhonpanich1,2, Fatemeh Koohi1, Amy A Jolly1, S Markus1 1 Stroke Research Group, Department of Clinical Neurosciences, University of Cambridge, Cambridge, Cambridgeshire, UK 2 Department of Medicine, Faculty of Medicine Siriraj Hospital, Mahidol University, Bangkok, Thailand Correspondence to Professor Hugh S Markus; hsm32@medschl.cam.ac.uk Abstract Background Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) is the most common monogenic form of stroke and is associated with early-onset stroke and dementia. Whether its clinical phenotype is becoming milder with better risk factor treatments and other care improvements is unknown. In a large longitudinal CADASIL cohort, we determined whether the prognosis has changed over 23 years. Methods Patients were identified from the Cambridge CADASIL register and the UK Familial stroke study. Change in age at stroke over the time of recruitment was determined using linear mixed-effects model, and the impact of genetic and vascular risk factors on stroke and dementia risk was further evaluated using Cox proportional hazard regression. Results A total of 555 patients with CADASIL were recruited between 2001 and 2023. The age of stroke onset significantly increased over time (p<0.001), with the mean age of stroke onset for patients recruited before 2016 (n=265) at 46.7±9.2 years and 51.6±9.5 years for those recruited since 2016 (n=290). Patients recruited since 2016 had lower risks of both stroke (HR 0.36, 95% CI 0.26 to 0.50, p<0.001) and dementia (HR 0.43, 95% CI 0.19 to 0.99, p=0.046) after adjusting for sex, hypertension history, smoking status, epidermal growth factor-like repeat position and calendar effect. Conclusions The clinical phenotype of CADASIL is improving. While this may be partly explained by reduced vascular risk factors such as smoking and the identification of milder cases, differences persisted after controlling for risk factors and mutation sites. These updated risk estimates should be used when counselling patients with CADASIL on prognosis. CADASIL STROKE CEREBROVASCULAR DISEASE GENETICS Data availability statement Data are available upon reasonable request. This is an open access article distributed in accordance with the Creative Commons Attribution 4.0 Unported (CC BY 4.0) license, which permits others to copy, redistribute, remix, transform and build upon this work for any purpose, provided the original work is properly cited, a link to the licence is given, and indication of whether changes were made. See: Statistics from Altmetric.com See more details Posted by 9 X users On 1 Facebook pages 6 readers on Mendeley Supplementary materials Request Permissions If you wish to reuse any or all of this article please use the link below which will take you to the Copyright Clearance Center’s RightsLink service. You will be able to get a quick price and instant permission to reuse the content in many different ways. Request permissions CADASIL STROKE CEREBROVASCULAR DISEASE GENETICS WHAT IS ALREADY KNOWN ON THIS TOPIC Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) is the most common monogenic form of stroke. Initial reports suggested stroke occurred in the 40s but the perception of many clinicians is that the disease may be becoming milder, with increasing numbers of patients remaining stroke and disability-free at later ages. WHAT THIS STUDY ADDS In this large, longitudinal prospectively recruited CADASIL cohort, the age onset of stroke in patients with CADASIL increased significantly over the time of recruitment. Patients recruited since 2016 had lower risk of both stroke and dementia. Additionally, survival analysis showed that patients who stopped smoking at any time had a significantly higher probability of remaining stroke and dementia-free. HOW THIS STUDY MIGHT AFFECT RESEARCH, PRACTICE OR POLICY This study provides updated risk estimates which can be used when counselling patients with CADASIL on prognosis. It emphasises the importance of treating cardiovascular risk factors in CADASIL. Introduction Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) is the most common monogenic form of stroke, causing early-onset lacunar stroke and dementia. Early reports described a mean age of onset of stroke in the 40s and that the disease was highly penetrant, with most patients with CADASIL suffering from disability and dementia by late middle age.1–5 However, recently, the perception of many clinicians is that the disease may be becoming milder, with increasing numbers of patients remaining stroke and disability-free at later ages. Over the last decade, in developed countries, there has been a marked decline in stroke incidence, perhaps reflecting better treatment of cardiovascular risk factors.6 7 For example, a study based in England noted that between 2000–2003 and 2012–2015, the age-adjusted incidence of ischaemic stroke decreased by 43%.8 Although CADASIL is a monogenic disorder, conventional cardiovascular risk factors alter its severity. Both smoking and hypertension have been shown to have a marked effect on the CADASIL phenotype, with stroke occurring approximately 10 years earlier in active smokers.9 10 It is possible that an improvement in the CADASIL phenotype may have occurred, paralleling the reduced incidence seen in sporadic stroke. It is also possible that as the disease has become increasingly recognised, milder and less penetrant cases have been diagnosed, which would result in an improvement in the CADASIL phenotype. To determine whether there have indeed been changes in the severity of the CADASIL phenotype over time, we studied a prospectively recruited cohort of patients with CADASIL recruited over 23 years to determine whether the age of onset of first stroke and dementia had changed. We then determined whether any change was explained by alterations in risk factor profiles or other factors, including mutation location. Methods Participants and recruitment Patients were studied from two prospective UK CADASIL disease registers: the UK CADASIL National Referral Service at Cambridge (2001–2023) and the UK Familial Small Vessel Disease (SVD) study, which prospectively recruited patients from six neurological centres in England between 2016 and 2023. A total of 555 patients with CADASIL were included in this study; data on 200 have already been reported in a previous study.9 The inclusion criterion was a diagnosis of CADASIL on genetic testing, with a typical cysteine-changing mutation. All patients gave written informed consent for their data to be used for research. Clinical parameters Clinical details were prospectively collected using a standardised proforma. Clinical details recorded included the presence and age at onset of stroke, migraine, psychiatric symptoms, encephalopathic episodes, seizures and dementia. Stroke was defined as a focal neurological deficit lasting more than 24 hours, accompanied by neuroradiological evidence of an ischaemic or haemorrhagic lesion. Migraine with or without aura was defined by the third edition of the International Classification of Headache Disorders (ICHD-3).11 History of any psychiatric symptoms was recorded. Depression was defined by a history of low mood requiring medical treatment or psychotherapy. CADASIL encephalopathy was described as an acute reversible encephalopathic episode without any other organic causes, lasting more than 24 hours and warranting a hospital admission.12 Epileptic seizures were only recorded if they were not related to an encephalopathic episode. Dementia was defined when it was previously diagnosed by a neurologist or psychiatrist or following the Diagnostic and Statistical Manual of Mental Disorders Text Revision, Fourth Edition (DSM-IV-TR) criteria.13 The onset of symptoms was defined as the age when the first event of each symptom occurred. Cardiovascular risk factors were recorded. Hypertension was defined as an elevated systolic and/or diastolic blood pressure ≥140 and 90 mm Hg or the use of antihypertensive treatment. Hypercholesterolaemia was defined as an elevated total cholesterol level ≥5.2 mmol/L or on a lipid-lowering agent. Diabetes mellitus was defined as having previously diagnosed with diabetes, the use of an anti-diabetic drug or having a fasting plasma glucose ≥7.0 mmol/L. Smoking was categorised as current, ex and never smoking. The age at onset of risk factor was defined as the age at which the disease was diagnosed or treatment was started. Statistical analysis Demographic and clinical characteristics of patients were presented as mean, SD, median and IQR for continuous variables and as frequency (%) for categorical variables. As it has been previously demonstrated that patients with mutations in epidermal growth factor-like repeat (EGFr) domain 1–6 exhibit more severe disease,14 mutation sites were categorised into two groups: mutations affecting EGFr domain 1–6 and those affecting 7–34. To explore changes in age at first stroke over time, we applied a linear mixed-effects model, using age at first stroke as the dependent variable and the year of recruitment as the independent variable. The mixed-effects model specification was employed to account for likely correlations among data from multiple members of a single family and calendar time. In the analysis, adjustments were made for sex, hypertension, smoking status and EGFr domain by treating them as fixed effects. Additionally, control for the variation arising from calendar time was implemented by considering the birth cohort (10-year intervals) as a random effect. The ‘lmer’ function of the ‘lme4’ R package was used to fit the model. This analysis was not performed for age at dementia due to the small number of patients with dementia in the cohort. We additionally divided patients into two groups of approximately equal size based on recruitment year: those recruited before 2016 and from 2016 to 2023. This cut-off was decided prior to data analysis. Differences between groups were analysed using χ 2 test and Student’s t-test, depending on variable types. We compared age at onset of first stroke and dementia using time-to-event analysis. This allowed the inclusion of data from individuals who had not yet experienced an event (stroke or dementia) at the time of the last follow-up. Analysis was performed using Kaplan-Meier survival analysis, and survival time comparisons were made using the log-rank test. Cox proportional hazard regression was used to evaluate the impact of other independent variables on the risk of stroke and dementia. Because age from birth served as the timescale, we conducted a Cox model stratified by birth cohort (10-year intervals) to account for calendar effects. Recruitment period, sex, hypertension, smoking status and EGFr domain were considered in the model. To assess whether the changes in age at stroke and dementia were accounted by which variables, we created four different Cox models controlling for different sets of variables, including only sex (model 1), sex and mutation site (model 2), sex and vascular risk factors (model 3), and sex, mutation site and vascular risk factors (model 4). Nagelkerke’s pseudo-R 2 was calculated to compare the variance explained by each model. A two-sided p value <0.05 was considered statistically significant, with no adjustment for multiple comparisons. Kaplan-Meier curves were created using ‘survminer’ package. All statistical analyses were carried out using the R software V.4.3.2. Results Clinical spectrum of 555 patients with CADASIL and their risk profile Five hundred fifty-eight patients with CADASIL were recruited. All were heterozygotes except for one whose mutation was compound heterozygous and one was homozygous. These two cases and one with incomplete data were omitted, leaving 555 in the analysis. Of these, 526 (94.8%) were symptomatic, and 29 (5.2%) were asymptomatic diagnosed on predictive testing. The mean age at recruitment was 49.2±11.9 years, and the mean age at first presentation in those with symptoms was 31.7±14.4 years. All patients had cysteine-altering NOTCH3 mutations; 433 (78.0%) were in EGFr domain 1–6. Clinical features and prevalence of cardiovascular risk factors are summarised in table 1, and risk factor profiles for each clinical feature are provided in online supplemental table 1. The distribution of age at the onset of migraine, stroke and dementia is shown in online supplemental figure 1. Supplemental material [jnnp-2024-334823supp001.pdf] View this table: View inline View popup Table 1 Demographic and clinical characteristics of patients by recruitment year Differences in CADASIL characteristics in patients recruited before and since 2016 There were 265 and 290 patients with CADASIL recruited before 2016 and since 2016, respectively. Patients recruited since 2016 were more likely to be recruited at an older age (50.5±12.0 vs 47.8±11.7, p=0.007), less likely to have a history of stroke (40.0% vs 50.9%, p=0.012), more likely to have a history of psychiatric symptoms (49.6% vs 38.1%, p=0.045), less likely to smoke at the time of stroke (14.1% vs 30.2%, p<0.001) and less likely to have a mutation in EGFr 1–6 (71.4% vs 85.3%, p<0.001). The mean age at onset of stroke was 46.7±9.2 and 51.6±9.5 years, respectively, in patients recruited before and since 2016 (p<0.001), and 57.3±7.9 and 60.2±8.7 years, respectively, for dementia (p=0.203) (figure 1). Differences in baseline characteristics, clinical features and other risk factor profiles at stroke are summarised in table 1. Download figure Open in new tab Download powerpoint Figure 1 Comparison of the distribution of age at onset of stroke and dementia stratified by recruitment year before and from 2016. (A) The mean age at onset stroke was 46.7±9.2 and 51.6±9.5 years in patients recruited before and from 2016 (p<0.001), and (B) 57.3±7.9 and 60.2±8.7 years for dementia (p=0.203). Stroke Two hundred fifty-one (45.2%) individuals had a history of stroke, of whom 246 (98.0%) experienced a lacunar ischaemic stroke and 10 (4.0%) experienced intracerebral haemorrhage; 4 (1.6%) experienced both stroke subtypes. The mean age at stroke onset was 49.0±9.6 years, and stroke was the first presenting feature in 75 (14.3%) patients. In those with stroke, there was a gradual increase in age at first stroke between 2001 and 2023 (figure 2), which was significant after controlling for birth cohort effect, sex and EGFr position (p<0.001) and remained significant after additionally controlling for hypertension and smoking status (p<0.001). The only risk factor associated with age at stroke onset was current smoking (p=0.009), which decreased age at stroke onset (online supplemental table 2 and figure 2,3). Download figure Open in new tab Download powerpoint Figure 2 Predicted means of age at stroke onset over years of recruitment to the clinic. The points show the age at stroke for each patient. Figure 3A shows the stroke-free survival probability of patients recruited since 2016 compared with those recruited before 2016. Kaplan-Meier analysis indicated that age of onset of first stroke was significantly higher in those recruited since 2016 (p <0.001). The median stroke-free survival time was 60.0 years (95% CI 58.0 to 64.0) in those recruited since 2016 and 53.0 years (95% CI 51.0 to 55.0) in those recruited before 2016. When further stratifying with smoking status, the difference in survival probability still remained significant in all groups (p<0.001). The median survival time was 55.0 (95% CI 53.0 to 61.0) and 61.0 (95% CI 59.0 to 65.0) years in never or former smokers recruited before and after 2016 and was 49.0 (95% CI 47.0 to 51.0) and 51.0 (95% CI 47.0 to 65.0) in current smokers recruited before and after 2016, respectively (online supplemental figure 4). Download figure Open in new tab Download powerpoint Figure 3 Kaplan-Meier survival estimates. Comparison of (A) stroke-free and (B) dementia-free survival in patients recruited since 2016 and those recruited before 2016. The median stroke-free survival time was 60.0 years (95% CI 58.0 to 64.0) among those recruited since 2016 and 53.0 years (95% CI 51.0 to 55.0) among those recruited before 2016. Median dementia-free survival time was 77.0 years (95% CI 74.0 to 77.0) in those recruited since 2016 and 68.0 years (95% CI 65.0 to 72.0). The results of the Cox proportional hazard model, stratifying by birth cohort (10-year intervals), are presented in table 2. Patients recruited after 2016 had a lower stroke risk when controlling for sex (model 1: HR 0.28, 95% CI 0.20 to 0.38, p<0.001); after further adjusting for mutation site (model 2: HR 0.30; 95% CI 0.22 to 0.42; p<0.001); vascular risk factors (model 3: HR 0.33; 95% CI 0.24 to 0.45; p<0.001), and mutation site and vascular risk factors (model 4: HR 0.36; 95% CI 0.26 to 0.50; p<0.001). In the fully adjusted model 4, current smoking (HR 2.09; 95% CI 1.56 to 2.82; p<0.001), male sex (HR 1.61; 95% CI 1.25 to 2.06; p<0.001) and EGFr domain 1–6 (HR 1.64; 95% CI 1.18 to 1.30; p=0.004) were predictors of younger age at first stroke. Regarding this result, we further stratified patients with the EGFr group in the Kaplan-Meier analysis, and the difference in survival probability was still significantly different in all groups (p<0.001) (online supplemental figure 5). View this table: View inline View popup Table 2 Multivariable Cox proportional hazard model stratifying by birth cohort for stroke Dementia Fifty-four (9.7%) individuals had a diagnosis of dementia with a mean age at onset of 58.6±8.3 years. Of these, 35 (64.8%) had a history of stroke prior to dementia diagnosis. Figure 3B shows the dementia-free survival probability of patients recruited before and since 2016. On Kaplan-Meier estimates, those recruited since 2016 had a significantly older age at dementia onset (p=0.004). Median dementia-free survival time was 77.0 years (95% CI 74.0 to 77.0) in those recruited since 2016 and 68.0 years (95% CI 65.0 to 72.0) in the other group. When further stratifying with smoking status, the difference in survival probability still remained significant in all groups (p<0.001). The median survival time was 71.0 (95% CI 67.0 to 72.0) and 77 (95% CI 74.0 to 77.0) years in never or former smokers recruited before and after 2016, and was 64.0 (95% CI 62.0 to 64.0) and 65.0 (95% CI 60.0 to 68.0) in current smokers recruited before and after 2016, respectively (online supplemental figure 6). The results of the Cox proportional hazard model, stratifying by birth cohort (10-year intervals), are presented in table 3. The risk of dementia was lower in those recruited since 2016 after controlling for sex (model 1: HR 0.38, 95% CI 0.17 to 0.85, p=0.019); after further adjusting for EGFr site (model 2: HR 0.41; 95% CI 0.18 to 0.93; p=0.032); vascular risk factors (model 3: HR 0.39; 95% CI 0.17 to 0.87; p=0.021) and mutation site and vascular risk factors (model 4: HR 0.43; 95% CI 0.19 to 0.99; p=0.046). In the fully adjusted model 4, predictors of dementia were male sex (HR 2.65; 95% CI 1.51 to 4.63; p<0.001) and current smoking (HR 3.67; 95% CI 1.82 to 7.48; p<0.001). View this table: View inline View popup Table 3 Multivariable Cox proportional hazard model stratifying by birth cohort for dementia Discussion Our results, from a cohort of CADASIL subjects prospectively recruited over a 23-year period, show a marked improvement in the clinical phenotype with increasing age at the onset of the first stroke and dementia. We have demonstrated that a number of factors, including smoking and the location of the mutation, affect the disease severity, but neither fully explained the reduction in disease severity observed in our cohort. In longitudinal time trend analysis, we demonstrated that the age onset of first stroke in subjects suffering from stroke had increased progressively and significantly. Including data from the whole cohort using a time-to-event analysis, we showed that subjects recruited since 2016 had a median stroke-free survival time 7 years later than those recruited before 2016 (60.0 years vs 53.0 years). Early studies reported the mean age of stroke in CADASIL at the 40s (41.2–46.1 years).1 3 4 More recent reports suggest that it may have increased to the late 40s or early 50s (49.3–52.0 years).15–18 Our data suggest that the observed age of onset is not considerably later (49.0 years) and provides useful up-to-date information for counselling on stroke risk to CADASIL mutation carriers. We found a similar reduction in disease severity when dementia was considered, with time-to-event analysis showing the median dementia-free survival time increased to 77 years in those recruited from 2016, compared with 68 years in those recruited before 2016. There are several explanations for this improvement in prognosis, including reduced cardiovascular risk factor exposure and diagnosis of milder cases as awareness of the disease increases. Cardiovascular risk factors have been shown to aggravate the disease phenotype. For example, smoking, hypertension and diabetes have been associated with increasing stroke risk in CADASIL,9 19 20 while blood pressure, even across the normal range, has been associated with increased brain atrophy.21 It is possible that a reduction in risk factor exposure and better treatment of risk factors could explain part of the improvement in disease severity. Studies in the UK have showed that control of some cardiovascular risk factors has improved over this period. The proportion of adults with untreated hypertension (systolic blood pressure ≥140 mm Hg and/or diastolic ≥90 mm Hg, not currently taking medication for blood pressure) decreased from 2003 to 2019 for both men (20% to 14%) and women (16% to 11%).22 However, the only risk factor we found to have a significant reduction during the duration of this study was smoking. Although there was an association of smoking with age of onset of both stroke and dementia, this explained little of the change in age of onset of both stroke and dementia over time. We approximated the variance explained by smoking to be 3–4% (0.034 for stroke and 0.038 for dementia) by comparing the difference in pseudo-R 2 values between models with and without smoking. Interestingly, we found that despite smoking in the past, patients recruited at any time who stopped smoking experienced stroke and dementia at a significantly older age, emphasising the importance of smoking cessation at any time. An alternative explanation is that milder cases have been diagnosed as disease awareness has increased. Early genetic testing for CADASIL tended to screen only those exons most commonly affected and exclude more distal exons.23 More recent data have shown that mutations in the proximal EGFr 1–6 are associated with an earlier age of onset of stroke and more severe disease.14 24 Furthermore, analyses of sequencing databases have shown that the frequency of typical CADASIL-type mutations is much more common in the general population than expected from the epidemiological estimates of the frequency of CADASIL.10 25 26 Therefore, it is possible that increasing numbers of milder cases, perhaps with more distal mutations, have been diagnosed, which have accounted for the milder disease severity currently observed. Supporting this hypothesis, an increasing number of distal (EGFr 7 and beyond) mutations were recruited over time (before 2016, 14.7%; after 2016, 28.6%). However, while EGFr 1–6 was a risk factor for earlier onset of stroke, entering it in the multivariable analysis had only little effect on the strength of association between time of recruitment and age of stroke onset. Therefore, the reduction in disease severity is not fully explained by changes in the number of subjects with distal EGFr. However, we cannot exclude the diagnosis of milder cases resulting in the improvement in prognosis we have seen over time. Hence, neither cardiovascular risk factors nor mutation sites can explain most of the change in disease severity over time, and its cause remains uncertain. Lifestyle interventions such as exercise, alcohol consumption and the Mediterranean diet, which are currently more widely known, could also play a role in the recent improvements seen in CADASIL phenotypes. Despite this, our study has an important implication in clinical practice. It provides up-to-date information with which to counsel patients with CADASIL and mutation carriers on likely prognosis and suggests this is more favourable than previously believed, especially in those who carry mutations affected more distal EGFr and those who do not smoke or have quit smoking. Most importantly, it offers strong evidence supporting lifestyle modifications, such as smoking cessation or control of vascular risk factors, which might improve the disease prognosis. Our study has several strengths. By studying a large cohort of subjects recruited over a long time period, we were able to have the power to detect time trends. All data were collected prospectively using the same definition for disease and risk factors throughout. Using linear mixed-effects and stratified Cox regression models strengthened the validity of the findings by accounting for potential correlations and variations arising from the same families and birth cohort. However, it also has limitations. Being an observational study in nature, it is difficult to demonstrate a direct causal relationship between variables and the delay in stroke onset. We used a diagnosis of definite dementia diagnosed by a dementia expert, but not all patients have such an assessment so the number of cases may be underestimated and the date of dementia onset can be difficult to accurately define. In addition, the power of the analysis in patients with dementia was limited due to the low prevalence of dementia in the cohort. Some other factors that might affect the risk of stroke and dementia, such as socioeconomic status,27 were not included in the analysis. In addition, due to the low prevalence of some vascular risk factors, such as diabetes, there was not enough power to investigate its effect on age at stroke onset or stroke and dementia risk, therefore they were omitted. We did not have data on treatment of risk factors during follow-up, and an intervention was not incorporated into the analytical models, making it impossible to determine the effect of treatment on the change in CADASIL prognosis. Further prospective studies with documentation of treatment changes are needed. Lastly, these results are from the UK, and need replicating in other populations. Data from the other continents do suggest that risk factors are associated with a more severe phenotype in other populations, for example, including Taiwan.20 However, recent data have also reported that stroke incidence is rising in many low- and middle-income countries (LMICs), in contrast to the fall seen in many higher income countries such as the UK. This could make trends in the severity of CADASIL phenotype differ across countries, and particularly between the UK and LMICs.7 Conclusion In summary, our study has shown that the prognosis of CADASIL is improving, as evidenced by the age at first stroke and dementia is increasing over time. While this might partly be explained by the reduction of smoking in later recruited patients and the inclusion of milder cases, the association remained significant after controlling for mutation position and cardiovascular risk factors. Our study provides definitive data which can be used in counselling patients on CADASIL prognosis, although it needs replicating in other populations. Data availability statement Data are available upon reasonable request. Ethics statements Patient consent for publication Not applicable. Ethics approval The study was approved by the East of England Cambridge Central Research Ethics Committee (16/EE/0118). Participants gave informed consent to participate in the study before taking part. Acknowledgments The authors would like to thank all CADASIL patients who participated in the UK Familial SVD study and the recruiting centres: Cambridge University Hospital NHS Foundation Trust: Markus H. S., Edwards H., Jolly A. and Henthorn M.; Leeds Teaching Hospital NHS Trust: Hassan A. and Waugh D.; Royal Hallamshire Hospital, Sheffield: Harkness K., Howe J., Edwards M. and Richard E.; St. George’s Healthcare NHS Trust, London: Khan U., Ghalata R., Stratton S. and Williams R.; University College London Hospitals NHS Foundation Trust: Werring D., Banara A. and Scheherazade F. References ↵ Tournier-Lasserve E, Joutel A, Melki J, et al . Cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy maps to chromosome 19q12. Nat Genet 1993;3:256–9. doi:10.1038/ng0393-256 OpenUrlCrossRefPubMedWeb of ScienceGoogle Scholar ↵ Chabriat H, Vahedi K, Iba-Zizen MT, et al . 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Statistical analysis: FK, N. Writing first draft: NS, HSM. Revising paper: All authors. Guarantor: HSM. Funding The UK Familial SVD study is funded by a British Heart Foundation Program grant (RG/F/22/110052). Infrastructural support was provided by the Cambridge British Heart Foundation Centre of Research Excellence (RE/18/1/34212) and the Cambridge University Hospitals National Institute for Health and Care Research (NIHR) Biomedical Research Centre (NIHR203312). NS was supported by the Prince Mahidol Award Youth Programme scholarship from the Prince Mahidol Foundation, Thailand. Competing interests None declared. Provenance and peer review Not commissioned; externally peer reviewed. Supplemental material This content has been supplied by the author(s). It has not been vetted by BMJ Publishing Group Limited (BMJ) and may not have been peer-reviewed. Any opinions or recommendations discussed are solely those of the author(s) and are not endorsed by BMJ. BMJ disclaims all liability and responsibility arising from any reliance placed on the content. Where the content includes any translated material, BMJ does not warrant the accuracy and reliability of the translations (including but not limited to local regulations, clinical guidelines, terminology, drug names and drug dosages), and is not responsible for any error and/or omissions arising from translation and adaptation or otherwise. Read the full text or download the PDF: Subscribe Log in Log in via Institution Log in via OpenAthens Log in using your username and password For personal accounts OR managers of institutional accounts Username Password Forgot your log in details?Register a new account? Forgot your user name or password? 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13878	https://www.youtube.com/watch?v=89EmbkLUE1Q	Exterior Angle Theorem Lesson for High School Geometry Lindsay Bowden - Secondary Math Videos 3090 subscribers 4 likes Description 821 views Posted: 5 Dec 2024 Welcome to the Exterior Angle Theorem for High School Geometry! In this lesson, you’ll learn about the Exterior Angle Theorem, which states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. 🔑 What you'll learn: ✔ Definition of Exterior Angles in Triangles ✔ The Exterior Angle Theorem and Its Proof ✔ Using the Theorem to Solve for Missing Angle Measures ✔ Application Problems Involving Exterior Angles ➡️ Find this lesson here: 💡 For additional resources, check out my educational store: 🎥 Other helpful video links: ✔ Slope of Parallel and Perpendicular Lines: ✔ Congruent Triangles: ✔ Special Angle Pairs Proofs: 🎓 About Lindsay Bowden: With a Bachelors of Science in Education, M.Ed in Curriculum and Instruction, and Ed.S in Curriculum and Instruction, I specialize in creating engaging secondary math resources tailored for on-level and special education classrooms. Check out my full curriculum bundles and additional lessons that will make your math class a breeze! 🌐 Connect with me: ● Website: ● Facebook: ● Instagram: Transcript: welcome to this lesson on the exterior angle theorem the exterior angle theorem says in a triangle an exterior angle is equal to the sum so that means to add of the two non-adjacent that means not next to interior angles so basically if i want to find the measure of this exterior angle exterior means it's outside the triangle that would be this the measure b c d then i can add the two interior angles but not this one because that's adjacent we want the ones that are not adjacent so i can add this one a and this one b all right so let's try some examples so number one we have an a missing exterior angle here so that is equal to the sum of these two interior angles that are not touching they're not adjacent to so 59 plus 75 which would be 134 degrees so that means this angle measure here is 134 degrees all right number two we're given this exterior angle 139 but we're missing one of the interior angles so i know that i can add these up but i'm solving backwards here so i need to subtract 74 from both sides that would give me 65 degrees so this angle here is 65 degrees all right number three we have some an algebraic expression here so i always start with the exterior angle always write that first 142 is equal to and then add up find the sum of the two interior angles you don't have to write the parentheses or the degree measure all right so combine like terms 78 plus 2x i'm going to subtract 78 on both sides then divide by 2 so that would be 32 and then i'm just going to substitute 32 in for this angle here 2 times 32 plus 1 that would be 65 degrees all right you can also find the other interior angle of course just subtract from 180 so let's go back to number two and do that so 180 minus 74 minus 65 is 41. or even faster these are supplementary because they form a line so you could just take 180 minus 139. same thing here 180 minus 142 38 easy all right number four exterior angle equals the sum of the two interior angles remember that's 90 degrees so minus 90 minus 90. so that would be 40 degrees so this is 40 which would make this 50 of course all right and then number five we have a isosceles triangle here which means these base angles are going to be the same so 4x equals 55 plus 55 add those together and then divide by four all right and then four times 27.5 that would be 110. and this would be 70. again i got this angle by just taking 180 minus 110 because these are supplementary they form a line they're a linear pair all right and then the last one triangle abc has the following angle measures i'm going to draw this just so we make sure we get the right exterior angle so i'm just going to make this 20 so let's do abc bca is 64. i didn't really draw that to scale it's okay a segment extends from triangle abc creating bcd so let me move this c find the measure of acd so we need to find this so basically this is going to be equal to the two non-adjacent angles so we're not going to use 64 here we can and i'll show you that at the end but let's use actually the exterior angle theorem that we've been learning so that would be 20 plus then we need to find the measure of angle a so i'm just going to take 180 minus 64 minus 20 to find that third angle which is 96. all right so that exterior angle equals 96 plus 20 which is 116. now you may be thinking um wait there's a faster way and there is you can just take this linear pair here so this is 64 then it just has to be 180 minus 116. you can also do that okay you can pause the video now and complete your practice and then come back when you are done and we will go over the proof all right the exterior angle proof so let's prove why this works so we have a triangle here with angles one two and three and then an exterior angle of 4 so that will be given number 2 says the measure of angle 1 plus 2 plus 3 equals 180 and that is the triangle sum theorem or you can just put the three angles and the three interior angles and triangle equal 180 and then 3 plus 4 equals 180 that's a linear pair here they form a line number four we have this really long statement but this section came from here this section came from here so basically we're saying these two expressions these two statements are equal and the reason we can say that is because they're both equal to 180. so if two things are equal to the same thing we can say they're equal to each other that's called the transitive property all right and then from number four to number five it looks very similar but if you look for just a second there's just something missing and what's missing is the measure of angle three the reason we can take that away is because we have it on both sides so we can subtract it from both sides without changing the value of the equation that's called the subtraction property of equality because we subtracted it from both sides all right you can go ahead and stop the video and complete your practice
13879	https://theengineeringmindset.com/density-of-metals/	Published Time: 2015-05-18T21:28:55+00:00 Density of Metals - The Engineering Mindset Home Electrical Controls HVACR Mechanical Energy Merch Shop Search The Engineering Mindset Home Electrical Controls HVACR Mechanical Energy Merch Shop HomeMechanicalMaterial PropertiesDensity of Metals Mechanical Material Properties Density of Metals Density of common metals By Paul Evans May 18, 2015 8 FacebookTwitterPinterestWhatsApp The typical density of common metals can be found listed in the table below. To see how density of materials is calculated, click here. If you find this article useful then please like, share and comment. \| Material \| Density (ρ) kg/m 3 \| --- \| \| Aluminum \| 2,705 \| \| Brass \| 8,587 \| \| Cast Iron \| 7300 \| \| Copper \| 8,944 \| \| Gold \| 19,320 \| \| Iron \| 7,860 \| \| Lead \| 11,343 \| \| Mercury \| 13,570 \| \| Mild Steel \| 7,850 \| \| Platinum \| 21,425 \| \| Silver \| 10,497 \| \| Stainless Steel \| 7,982 \| \| Tin \| 7,260 \| \| Titanium \| 4,520 \| \| Tungsten \| 19,450 \| \| Zinc \| 7,068 \| TAGS Aluminum Brass Cast Iron Copper densities density density of engineering Gold Iron kg/m3 Lead Mercury Mild Steel Platinum Silver Stainless Steel Tin Titanium Tungsten Zinc FacebookTwitterPinterestWhatsApp Previous articleArc length, how to calculate Next articleDensity explained! Paul Evans RELATED ARTICLESMORE FROM AUTHOR Pump Head Pressure Basics How Multispeed pumps work Resistance Temperature Detector Basics 8 COMMENTS Density explained! \| The Engineering MindsetMay 18, 2015 At 10:01 pm[…] news! we have already listed the typical density of common metals, just click here to view the table. (opens in a new […] Reply Uddin NegamOct 2, 2018 At 7:12 amIt is really useful for me to recover of memorize. It will helpful for engineering works. Reply jamilaNov 5, 2020 At 1:54 pmdear all, are you sure that the units of the pressure mentioned here are kg/m3. I read in other websites that its is g/cm3. so there should be a factor 1000 added here? thanks j. Reply 4. Bob SmithJul 21, 2021 At 1:22 pmThe table has an error; gold and tungsten are essentially the same density. They are so similar that a gold-plated tungsten bar would be indistinguishable from a solid gold bar, unless someone were to drill into it, where the hardness of tungsten would immediately indicate it is a fake gold bar. Reply 5. TinaNov 7, 2021 At 11:27 amThanks ppl, that’s really useful. Tina Reply 6. MuhammadDec 10, 2021 At 11:01 amYes, Jamila it is measured in g/cm³. The were been converted to kilo meaning it was multipled by 10³. For you to convert it to cm just divided the value by 10². Example Iron it was written as 7860kg/m³, simply divide 7860 by 10² =78.6g/cm³ Reply 7. YusufMar 5, 2022 At 10:33 amI want the density of Aluzinc Reply 8. Thomas M LittlejohnAug 29, 2022 At 4:57 pmCan the density of a pure metal such as .9999 silver be changed by forging or casting? Reply LEAVE A REPLY Cancel reply Please enter your comment! Please enter your name here You have entered an incorrect email address! Please enter your email address here [x] Save my name, email, and website in this browser for the next time I comment. Δ Support more content Found the tutorials super useful? Support our efforts to make even more engineering content You'll like these too! Refrigeration Design Software Paul Evans-Jul 1, 20192 Chiller Flow Rate Measurement Paul Evans-Jun 5, 20194 AC and DC Electricity Paul Evans-Dec 15, 20221 DC Motor Explained Paul Evans-Apr 21, 20204 Circuit Breakers (MCB’s) Paul Evans-Nov 27, 20240 R718 (Water) Pressure Enthalpy Chart Paul Evans-Apr 14, 20190 Recent Topics How to ACTUALLY Use an Oscilloscope (Beginner-Friendly Guide!) How Electricity Works- For Visual Learners Power Factor Explained – Your Electricity Bill Money Drain (Reactive Power) Clamp Meter Ground Faults Latest Content How to ACTUALLY Use an Oscilloscope (Beginner-Friendly Guide!) Paul Evans-Apr 28, 20250 Learn how to use an oscilloscope like a pro! In this ultimate beginner’s guide, we’ll break down what an oscilloscope is, how it works,... How Electricity Works- For Visual Learners Feb 26, 2025 Power Factor Explained – Your Electricity Bill Money Drain (Reactive Power) Feb 17, 2025 Clamp Meter Dec 11, 2024 Ground Faults Nov 27, 2024 Scroll Compressors Nov 27, 2024 How MOSFET Works- Ultimate guide, understand like a PRO Nov 27, 2024 Circuit Breakers (MCB’s) Nov 27, 2024 Simplify. Teach. Inspire. About Contact Us Disclaimer Contact The Engineering Mindset Team Cookies Policy ©
13880	https://www.cut-the-knot.org/Curriculum/Calculus/DistanceToLine.shtml	Site What's new Content page Front page Index page About Privacy policy Help with math Subjects Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry Distance From a Point to a Straight Line On this page we'll derive an engaging formula for the distance from a point to a straight line. If a line L is given by its general equation Ax + By + C = 0 and a point P = (u, v) is given in the plane, then the distance dist(P, L) from the point to the line is determined by (2) dist(P, L) = \|Au + Bv + C\| / \|\|n\|\|, where n = (A, B) is the vector of the coefficients, normal to the line. \|\|n\|\| is the norm of the vector. If the equation is normalized, i.e., if A2 + B2 = 1, (2) is simplified: (2') dist(P, L) = \|Au + Bv + C\|. To see why this is so, observe that (1) tells us that a straight line is a level curve f(x, y) = 0 of the linear function (3) f(x, y) = Ax + By + C. (For any function f of any number of arguments, a level surface is the locus of points where f takes on the same value f = c, say. For functions of two arguments, the locii are naturally called level curves.) For the function (3), the level curves Ax + By + C = c are all straight lines parallel to the one given by (1) because all of them are perpendicular to the same normal vector n = (A, B). (In the applet below, a straight line is defined by two points, each of which can be dragged independently causing a rotation around the other point. The line can also be dragged parallel to itself at any other point. There is an additional point with the distance to the line indicated in blue. The point can also be dragged. Note: the applet gives you a precision of 0 to 6 digits. The equalities therefore are not quite exact.) \| \| \| What if applet does not run? \| The distance from a point to a line (or to any set, for that matter) is defined as the minimum distance between the given point and the points on the line: dist(P, L) = min(dist(P, r)), over all r = (x, y) belong to line L. The minimum always exists for a straight line, but may not exist for other sets. For a straight line, the minimum is achieved at the foot of the perpendicular from the given point to the line. It is noteworthy that all points on one of the level lines have the same distance to any other level line, because the lines are parallel and the distances are measured via the perpendiculars. This is how the normal to the line comes into the play. The function does not change its value along the level lines. All the change comes along the direction of the normal n = (A, B)! Assume m = n/\|\|n\|\| is the unit vector normal to the level lines of the given function: m = (A/\|\|n\|\|, B/\|\|n\|\|). Pick a point (x0, y0) and define (x1, y1) = (x0, y0) + m. What is the difference between f(x1, y1) and f(x0, y0)? Since the function f is linear we have \| \| \| --- \| \| f(x1, y1) - f(x0, y0) \| = (Ax1 + By1 + C) - (Ax0 + By0 + C) \| \| \| = A·A/\|\|n\|\| + B·B/\|\|n\|\| \| \| \| = A2/\|\|n\|\| + B2/\|\|n\|\| \| \| \| = (A2 + B2)/\|\|n\|\| \| \| \| = \|\|n\|\|2/ \|\|n\|\| \| \| \| = \|\|n\|\|. \| That is, every step by one unit in the direction normal to the line change the value of the function by the norm of its normal vector? This is true for any linear function and for any form of its equation. In particular, if the equation is normalized to start with, a unit step in the direction normal to the line changes its value by exactly 1!. With every unit step, the values on the level lines grow by 1 in the direction of the normal, but they decrease by 1 in the opposite direction. The line (1) divides the plane into two halves. In one, which is pointed to by the normal n = (A, B), f takes on positive values; in the other f is negative throughout. Once this simple behavior has been observed, let d = dist(P, L). Define now (4) (x, y) = (u, v) - d(±m). where we choose the sign "+" if P lies in the "positive half-plane". We choose "-" otherwise. Since d is the distance from P = (u, v) to the line, and the distance from a point to the line is measured along the normal, i.e. m, the point (x, y) in (4) lies on the line: f(x, y) = 0. This entails a sequence of steps: \| \| \| --- \| \| 0 \| = f(x, y) \| \| \| = A(u - d·(±A/\|\|n\|\|)) + B(v - d·(±B/\|\|n\|\|)) + C \| \| \| = (Au + Bv + C) - d·[±(A2/\|\|n\|\| + B2/\|\|n\|\|)] \| \| \| = f(u, v) - [±d\|\|n\|\|]. \| In other words, (5) d = ± f(u, v) / \|\|n\|\|, where the sign is chosen so as to make ± f(u, v) positive. If d is considered the signed distance, i.e., if d comes with a sign minus if f(u, v) is negative, the formula is simplified to (5') d = f(u, v) / \|\|n\|\|. In any event, we can conclude that (2) is indeed true. If a straight line is given by an implicit equation n·(r - r0) = 0, we can similarly set f(r) = n·(r - r0) for r not necessarily on the line. Then the distance d from a point P to the line can be found from d = \| n·(P - r0)\| / \|\|n\|\| = \| f(P)/\|\|n\|\| \|. \| \| \| Related materialRead more... \| \| \| \| - Cavalieri's Principle \| \| - Derivative of Sine and Cosine \| \| - Estimating Circumference of a Circle \| \| - Maximum Volume of a Cut Off Box \| \| - Mistrust Intuition of the Infinite \| \| - Naturally Discontinuous Functions \| \| - Rolle's and The Mean Value Theorems \| \| - Function, Derivative and Integral \| \| - Area of a Circle by Rabbi Abraham bar Hiyya Hanasi \| \| - Schwarz Lantern \| \| - Two Circles and a Limit \| \| - Deceptive Appearances \| \| - Problem 4010 from Crux Mathematicorum \| \| \| \|Contact\| \|Front page\| \|Contents\| \|Algebra\| Copyright © 1996-2018 Alexander Bogomolny
13881	https://www.cuemath.com/numbers/gcf-of-36-and-99/	GCF of 36 and 99 GCF of 36 and 99 is the largest possible number that divides 36 and 99 exactly without any remainder. The factors of 36 and 99 are 1, 2, 3, 4, 6, 9, 12, 18, 36 and 1, 3, 9, 11, 33, 99 respectively. There are 3 commonly used methods to find the GCF of 36 and 99 - prime factorization, long division, and Euclidean algorithm. \| \| \| --- \| \| 1. \| GCF of 36 and 99 \| \| 2. \| List of Methods \| \| 3. \| Solved Examples \| \| 4. \| FAQs \| What is GCF of 36 and 99? Answer: GCF of 36 and 99 is 9. Explanation: The GCF of two non-zero integers, x(36) and y(99), is the greatest positive integer m(9) that divides both x(36) and y(99) without any remainder. Methods to Find GCF of 36 and 99 Let's look at the different methods for finding the GCF of 36 and 99. GCF of 36 and 99 by Listing Common Factors There are 3 common factors of 36 and 99, that are 1, 3, and 9. Therefore, the greatest common factor of 36 and 99 is 9. GCF of 36 and 99 by Prime Factorization Prime factorization of 36 and 99 is (2 × 2 × 3 × 3) and (3 × 3 × 11) respectively. As visible, 36 and 99 have common prime factors. Hence, the GCF of 36 and 99 is 3 × 3 = 9. GCF of 36 and 99 by Euclidean Algorithm As per the Euclidean Algorithm, GCF(X, Y) = GCF(Y, X mod Y) where X > Y and mod is the modulo operator. Here X = 99 and Y = 36 Therefore, the value of GCF of 36 and 99 is 9. ☛ Also Check: GCF of 36 and 99 Examples Example 1: The product of two numbers is 3564. If their GCF is 9, what is their LCM? Solution: Given: GCF = 9 and product of numbers = 3564 ∵ LCM × GCF = product of numbers ⇒ LCM = Product/GCF = 3564/9 Therefore, the LCM is 396. Example 2: Find the GCF of 36 and 99, if their LCM is 396. Solution: ∵ LCM × GCF = 36 × 99 ⇒ GCF(36, 99) = (36 × 99)/396 = 9 Therefore, the greatest common factor of 36 and 99 is 9. Example 3: For two numbers, GCF = 9 and LCM = 396. If one number is 99, find the other number. Solution: Given: GCF (y, 99) = 9 and LCM (y, 99) = 396 ∵ GCF × LCM = 99 × (y) ⇒ y = (GCF × LCM)/99 ⇒ y = (9 × 396)/99 ⇒ y = 36 Therefore, the other number is 36. go to slidego to slidego to slide Book a Free Trial Class FAQs on GCF of 36 and 99 What is the GCF of 36 and 99? The GCF of 36 and 99 is 9. To calculate the GCF (Greatest Common Factor) of 36 and 99, we need to factor each number (factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36; factors of 99 = 1, 3, 9, 11, 33, 99) and choose the greatest factor that exactly divides both 36 and 99, i.e., 9. What are the Methods to Find GCF of 36 and 99? There are three commonly used methods to find the GCF of 36 and 99. If the GCF of 99 and 36 is 9, Find its LCM. GCF(99, 36) × LCM(99, 36) = 99 × 36 Since the GCF of 99 and 36 = 9 ⇒ 9 × LCM(99, 36) = 3564 Therefore, LCM = 396 ☛ GCF Calculator What is the Relation Between LCM and GCF of 36, 99? The following equation can be used to express the relation between LCM and GCF of 36 and 99, i.e. GCF × LCM = 36 × 99. How to Find the GCF of 36 and 99 by Long Division Method? To find the GCF of 36, 99 using long division method, 99 is divided by 36. The corresponding divisor (9) when remainder equals 0 is taken as GCF. How to Find the GCF of 36 and 99 by Prime Factorization? To find the GCF of 36 and 99, we will find the prime factorization of the given numbers, i.e. 36 = 2 × 2 × 3 × 3; 99 = 3 × 3 × 11. ⇒ Since 3, 3 are common terms in the prime factorization of 36 and 99. Hence, GCF(36, 99) = 3 × 3 = 9 ☛ What are Prime Numbers?
13882	https://www.adhischools.com/blog/guide-dual-agency-real-estate	Understanding Dual Agency: A Guide for New Real Estate Agents Press Option+1 for screen-reader mode, Option+0 to cancelAccessibility Screen-Reader Guide, Feedback, and Issue Reporting \| New window DRE Statutory Sponsor ID #S0348 \| CE #6404 Join Login Get My License Renew My License Crash Course Find a Session Resources Blogs Success Stories Real Estate Practice Questions How to get a Real Estate License How to Become a Broker FAQ Become a Real Estate Agent Blogs Real Estate Practice Questions How to get a Real Estate License How to Become a Broker FAQ Should I Become a Real Estate Agent Join Login AdhiSchools Blog Understanding Dual Agency: A Guide for New Real Estate Agents ByKartik Subramaniam June 03, 2024 Reading Time : 3 minutes Navigating the Complexities of Dual Agency: A Comprehensive Guide for New Real Estate Agents Dual agency, is a scenario where a single real estate agent or brokerage represents both the buyer and seller in a real estate transaction, is a multifaceted practice fraught with ethical and legal complexities. While it offers a streamlined approach, it demands careful navigation due to potential conflicts of interest. A thorough understanding of its nuances is essential for new real estate agents venturing into this arena. Unraveling the Dual Agency Landscape Dual agency occurs when one agent or multiple agents from the same brokerage represent both the buyer and the seller in a real estate deal. This situation requires the agent(s) to carefully balance both parties' interests fairly and without favoring one side over the other. For instance, consider a situation where an agent from XYZ Realty is representing both John, the seller, and Jane, the buyer, in the sale of a home. The agent must navigate this dual agency with extreme care. They need to provide impartial advice to both parties, ensuring that both John and Jane feel their interests are being respected and protected. For example, when discussing the price, the agent must not disclose how low John would go or how high Jane would go, but rather help them reach a fair agreement based on market data and individual circumstances. Legal Status: A Patchwork of Regulations The legal status of dual agency varies significantly from state to state, creating a patchwork of regulations across the United States. In several states, dual agency is expressly prohibited due to the perceived conflicts of interest it entails, such as in Alaska, Colorado, Florida, Kansas, Maryland, Texas, Vermont, and Wyoming. In these jurisdictions, real estate agents are forbidden from representing both the buyer and seller in the same transaction, as this is seen as a potential threat to the fairness and integrity of the process. Ethical Considerations: A Moral Compass for Agents Understanding the ethical dimensions of dual agency is crucial for new real estate agents. Agents must prioritize full disclosure, openly communicating their role and the nature of dual agency to both parties. This includes discussing potential conflicts of interest and ensuring that both parties fully comprehend the implications of the arrangement. Maintaining Confidentiality Keeping confidentiality is crucial. Agents must protect private information from both the buyer and the seller, making sure that no sensitive details are shared improperly between the two. This requires careful communication and record-keeping, as well as an understanding of the legal and ethical rules about private information. Example: Imagine an agent representing both a buyer who wants to pay as little as possible and a seller who hints at accepting a lower price due to an urgent need to sell. The agent must not reveal the seller’s urgency to the buyer to secure a lower offer, as doing so would breach confidentiality and favor the buyer’s interests over the seller’s. Navigating the Dual Agency Landscape: Practical Strategies For agents who find themselves in a dual-agency situation, several strategies can help navigate this complex terrain. Transparency, obtaining written consent, and maintaining open communication channels are paramount. Regular communication with both parties can help identify and address potential issues early on, minimizing the risk of conflicts or misunderstandings. Maintaining fairness and impartiality is a continuous effort in dual agency. This may involve seeking external advice or mediation in situations where conflicts arise. Ultimately, dual agency in real estate is a nuanced and complex practice that demands careful consideration. New real estate agents must be aware of the legal and ethical dimensions involved and the potential challenges and risks. By prioritizing transparency, fairness, and open communication, agents can successfully navigate dual agency while upholding their fiduciary duties to the buyer and seller. Love, Kartik Join Our Newsletter Subscribed Please wait Live Stream Courses NOW AVAILABLE Are you ready to get your real estate license? I'm ready Recent Posts The Hidden Risk in Choosing a Real Estate School What’s the Perfect Age to Get Your Real Estate License? (It Might Surprise You) Should You Get a Real Estate License? 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13883	https://ocw.mit.edu/courses/6-041-probabilistic-systems-analysis-and-applied-probability-fall-2010/resources/lecture-20-the-central-limit-theorem/	Lecture 20: Central Limit Theorem \| Probabilistic Systems Analysis and Applied Probability \| Electrical Engineering and Computer Science \| MIT OpenCourseWare Browse Course Material Syllabus Calendar Readings Lecture Notes Video Lectures Recitations Tutorials Assignments Exams Course Info Instructor Prof. John Tsitsiklis Departments Electrical Engineering and Computer Science As Taught In Fall 2010 Level Undergraduate Topics Engineering Systems Engineering Mathematics Discrete Mathematics Probability and Statistics Learning Resource Types assignment_turned_in Problem Sets with Solutions grading Exams with Solutions theaters Lecture Videos Download Course menu search Give Now About OCW Help & Faqs Contact Us searchGIVE NOWabout ocwhelp & faqscontact us 6.041 \| Fall 2010 \| Undergraduate Probabilistic Systems Analysis and Applied Probability Menu More Info Syllabus Calendar Readings Lecture Notes Video Lectures Recitations Tutorials Assignments Exams Video Lectures Lecture 20: Central Limit Theorem Description: In this lecture, the professor discussed central limit theorem, Normal approximation, 1/2 correction for binomial approximation, and De Moivre–Laplace central limit theorem. Instructor: John Tsitsiklis Transcript Download video Download transcript Lecture Slides Central Limit Theorem (PDF) Course Info Instructor Prof. John Tsitsiklis Departments Electrical Engineering and Computer Science As Taught In Fall 2010 Level Undergraduate Topics Engineering Systems Engineering Mathematics Discrete Mathematics Probability and Statistics Learning Resource Types assignment_turned_in Problem Sets with Solutions grading Exams with Solutions theaters Lecture Videos Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Accessibility Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare close Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Stay Here Continue
13884	http://www.tihu.com.cn/which-ux-research-methods.html	如何正确的选择用户体验研究方法？ - 客户体验管理 - 鹈鹕顾问首页我们的服务我们的观点我们的客户联系我们如何正确的选择用户体验研究方法？ 2018-12-25 鹈鹕全面客户体验管理 \| 译者：马振江摘要：如今的用户体验研究方法能够解决非常广泛的问题。我们将 20 种研究方法呈现在一个典型产品开发过程的三维图表中，方便大家更好的了解何时使用哪种用户研究方法。用户体验领域有很多可用的研究方法，从以往经过验证的实验室可用性研究，到最近开发的一些方法，如在线用户体验评估，等等。虽然在一个特定的项目中使用全部的方法是不现实的，但是几乎所有的项目都会用到多种研究方法，综合形成见解和结论。不幸的是，许多设计团队只使用他们熟悉的一两种方法。问题的关键是，这些研究方法的选择和使用时机（ what to do when）。为了更好地理解什么时候使用哪种方法，可以沿着一个具有以下坐标轴的三维图来查看它们：态度 Vs. 行为定性 Vs. 定量使用场景态度（Attitudinal）Vs. 行为（Behavioral）维度这种区别可以通过对比「人们说了什么」和「人们做了什么」来概括（通常两者非常不同）。态度研究的目的一般是了解或测量人们的意见，因此被大量用在市场部门，比如了解消费者对品牌或某种产品/概念的看法。虽然大多数可用性研究应该更多依赖于行为，但是让被研究者自我陈述（self-reported information，即主动表达自己的态度）的方法仍然对设计人员非常有用。例如，卡片分类法（Card sorting）可以洞察用户对信息空间的心理模型，并帮助确定产品、应用程序或网站的最佳信息架构。使用多种调查方法来收集自我陈述的数据，对态度进行测量和分类，有助于跟踪或发现需要解决的重要问题。由于各种原因，焦点小组（Focus groups ）在可用性方面往往用处不大，但它能直观的反映人们在群体环境中对品牌或产品概念的看法。维度的另一端，主要关注行为的方法聚焦于理解人们对产品或服务「做了什么」。例如，A/B 测试（ A/B testing）将站点设计的变化展示给随机访问者，但保持其他所有因素不变，以便查看不同设计方案对用户行为的影响，而眼动测试（Eyetracking）则试图了解用户在视觉上如何与界面设计交互。在这两个维度之间是我们最常用的两种方法：可用性研究（Usability studies）和实地研究（Field studies）。它们会同时收集态度陈述和行为数据，可以朝两个维度的任何一端移动，尽管通常建议偏向行为方面。定性（Qualitative） Vs. 定量（Quantitative）维度狭义上的定性，被认为是“开放的”，有点像调研问卷中的开放式问题那样。本文讨论的定性与这种狭义理解有很大不同。本质上定性的研究是在直接观察行为或态度的基础上，生成相关的数据。而在定量研究中，有关行为或态度的数据是通过测量或工具（如调查/分析工具）间接收集的。例如，在实地研究和可用性研究中，研究人员直接观察人们如何使用技术（或不使用技术）来满足他们的需求。这使他们能够随时提出问题，探究行为，甚至可能调整研究方案，以更好实现调研目标。定性方法对数据的分析通常不是数学上的。相比之下，定量方法的洞察力通常来自数学分析，因为数据收集工具（例如，调查工具或 web 服务器日志）捕获了大量的数据，很容易进行数字化处理。由于这种差异，定性方法更适合回答关于「为什么」或「如何解决」之类的问题，而定量方法适合解决「有多少」之类的数量或程度问题。拥有这样的数字有助于区分资源的优先级，使决策者关注影响最大的问题。产品的使用场景第三个维度与参与者如何使用调研的产品或服务有关。可以描述为：产品在正常场景下使用产品的脚本化/流程化使用在研究期间不使用产品以上的混合场景让参与者在自然场景下使用产品，目标是减少来自研究的干扰，以便尽可能接近现实地理解用户行为或态度。这种方式更有效，但缺乏对研究主题的控制。许多实地调研都试图做到这一点，尽管总是存在一些观察偏差。拦截调查（Intercept surveys）、数据挖掘（Data mining）或其他分析技术就是这方面的定量例子。对产品使用进行脚本化处理（或指定产品的使用顺序），可以将重心聚焦在一些明确的使用方向上。比如，脚本化测试可以把使用重点放在新设计的产品流程方面，而正常场景下的使用则很难做到主题引导。根据研究目标的不同，脚本编写的程度可能会有很大的差异。例如，基准测试研究（Benchmarking study）通常是非常严格脚本化的，并且在本质上更加定量，因此它可以产生可靠的可用性度量数据。不使用产品的研究是为了了解比可用性更广泛的问题，例如对品牌或文化行为的研究。混合场景，是为了实现调研目标，创造一种新的产品形态。例如，作为产品体验的一部分，参与式设计方法（Participatory-design methods）允许用户与设计元素进行交互，甚至重新排列设计元素，观察用户使用行为，以便了解当前的解决方案怎样能更好满足用户需求，以及用户为什么做出某些选择。概念测试方法（Concept-testing methods）会采用一个低保真的产品或服务原型，主要目的是了解用户是否想要或需要该产品或服务（而不是体验产品的细节）。图表中的大多数方法可以沿着一个或多个维度移动，有些方法甚至在同一个研究中也是如此，通常是为了满足多个目标。例如，实地研究可以关注人们说什么（现场访谈）或他们做什么（实地观察）；需求度研究（Desirability studies）和卡片分类有定性和定量两种版本；眼动测试可以是脚本化的，也可以是非脚本化的。产品开发的阶段（时间维度）在选择研究方法时需要考虑的另一个重要因素是产品开发阶段及其相关目标。策略阶段：在产品开发的初始阶段，通常会考虑未来的新想法和机会。这一阶段的用户研究方法可以多种多样。执行阶段：确定设计方向之后，在持续优化过程中，最终会到了需要做出决定的时刻：『做还是不做』。这个阶段主要是定型性研究，有助于发现设计方案存在的问题，降低执行决策风险。评估阶段：当产品或服务被足够多的用户使用，这时你就可以开始衡量效果了。这是典型的总结性研究，可能是针对产品本身的历史数据或竞争产品的对比。下表总结了这些目标，并列出了与这些目标相关的典型研究方法：艺术还是科学？虽然许多用户体验研究方法都有其科学实践的根源，但是它们的目的并不是纯科学的，仍然需要调整以满足调查者的需求。这就是为什么这里的方法的描述是一般的指导方针，而不是死板的分类。最后，你的工作是否成功，将取决于它对改善网站或产品的用户体验有多大的影响。这些分类是为了帮助你在正确的时间做出最好的选择。 20 种用户研究方法简介实验室可用性测试（Usability-Lab Studies）：参与者被带入实验室，与研究员一对一，按照给定的场景，完成测试产品或服务相关的任务和使用细节。实地调查（Ethnographic Field Studies）：研究员在实验室之外的自然环境中会见参与者，观察参与者在正常场景下使用产品或服务。参与式设计（Participatory Design）：为参与者提供设计元素或创意材料，用具体的方式构建他们的理想体验，从而了解何种设计方式对他们最重要。焦点小组（Focus Groups）：3-12 名参与者组成一个小组，引导他们按照设定的主题讨论，并通过讨论和答题来获得口头和书面的反馈。访谈法（Interviews）：针对研究主题，研究员和参与者一对一的深度讨论。眼动测试（Eyetracking）：在参与者执行任务或与网站、APP、物理产品或环境交互时，通过一台安装好的眼动测试仪精确跟踪他们都看了哪些地方。可用性基准测试（Usability Benchmarking）：使用精确和预定的性能度量，对几个参与者进行严格脚本化的可用性研究。有主持的远程可用性研究（Moderated Remote Usability Studies）：使用屏幕共享软件和远程控制能力的工具，远程进行的可用性研究。无主持的远程跟踪研究（Unmoderated Remote Panel Studies）：由受过训练的参与者组成的小组，参与者在自己的个人设备上安装了视频记录和数据收集软件，他们在使用网站或产品时，说出自己的体验并记录，以便研究人员或公司立即回放和分析。概念测试（Concept Testing）：向参与者展示一个具有新概念或新产品核心本质（价值主张）的粗略原型，以确定这个产品或服务是否满足目标受众的需要；它可以是一对一的，也可以是一对多，可以是面对面的，也可以在线完成。日记/录像研究（Diary/Camera Studies）：给参与者一种方式（日记或照相机）来记录和描述他们生活中与产品或服务相关的方面，或者仅仅是对目标受众最重要的方面；日记研究通常是纵向的，只能针对参与者容易记录的数据进行。顾客反馈（Customer Feedback）：通常通过一个反馈链接、按钮、表单或邮件，以开放或封闭的问题形式提供给用户。需求度研究（Desirability Studies）：向参与者提供不同的视觉设计备选方案，让他们将每一个备选方案与一组属性相关联，而这些属性可从一个封闭的清单中选择。这个研究可以是定性的，也可以是定量的。卡片分类法（Card Sorting）：一个定性或定量方法，要求用户对类目进行编组，并对每组进行分类。这个方法可以了解用户的心理模型，帮助网站进行信息架构的构建和优化。点击流分析（Clickstream Analysis）：分析用户在使用网站或软件产品时点击和查看页面的行为记录。这要求网站具备动作记录能力，或在应用程序上收集被授权的数据。 A/B 测试（也叫做多变量测试、实时测试、小范围测试）：随机分配一组用户与不同的设计方案交互，测量这些方案对用户行为的影响，科学地测试网站或程序上不同设计的效果。无主持的用户体验研究（Unmoderated UX Studies）：一种定量或定性的自动化方法，它使用专门的研究工具（通过安装在参与者计算机/浏览器上的软件）捕捉参与者的行为和态度（通过嵌入的调查问题）。通常会给参与者一个网站或原型，要求参与者完成某种目标或场景。真实意图研究（True-Intent Studies）：随机询问网站访问者在进入网站时的目标或意图，测量他们随后的行为，并询问他们在离开网站时是否成功实现了目标。拦截调查（Intercept Surveys）：用户在使用网站或应用程序时触发的调查。邮件调查（Email Surveys）：通过发送电子邮件的方式招募参与者，或邀请参与填写调查问卷。原文：作者：Christian Rohrer 本文为作者原创翻译，欢迎转发分享。转载时需在文章开头注明作者和“来源：鹈鹕全面客户体验管理（微信号：CEM-tihu）”，文字颜色为黑色，且不得修改原文内容。欢迎小伙伴投稿合作，具体请联系：易女士 Yiml@tihu.com.cn 鹈鹕顾问 © 2024 Pelican Consulting Co.,Ltd. © 版权所有北京鹈鹕管理咨询有限公司 ICP证：京ICP备05044559号-1
13885	https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity?srsltid=AfmBOooOYz0rsHX9NL7ZAX8S8vg93Cn78CndrIsfNY_SV2FB14Qrjh4F	Art of Problem Solving Combinatorial identity - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Combinatorial identity Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Combinatorial identity Contents [hide] 1 Pascal's Identity 1.1 Proof 1.2 Alternate Proofs 2 Vandermonde's Identity 2.1 Video Proof 2.2 Combinatorial Proof 2.3 Algebraic proof 3 Hockey-Stick Identity 3.1 Proof 4 Another Identity 4.1 Hat Proof 4.2 Proof 2 5 Even Odd Identity 6 Examples 7 See also Pascal's Identity Pascal's Identity is a very important formula for olympiad math and it states that for any positive integers and . Here, is the binomial coefficient . This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things. Proof If then and so the result is trivial. So assume . Then Alternate Proofs Here, we prove this using committee forming. Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways. Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways. But we already know they can be picked in ways, so Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, . Vandermonde's Identity Vandermonde's Identity states that , which can be proven combinatorially by noting that any combination of objects from a group of objects must have some objects from group and the remaining from group . Video Proof Combinatorial Proof Think of the right hand side as picking people from men and women. Think of the left hand side as picking men from the total men and picking women from the total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true. ~avn Algebraic proof For all The coefficients of on both sides must be the same, so using the Binomial Theorem, Hockey-Stick Identity For . This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical. Proof Inductive Proof This identity can be proven by induction on . Base Case Let . . Inductive Step Suppose, for some , . Then . Algebraic Proof It can also be proven algebraically with Pascal's Identity, . Note that , which is equivalent to the desired result. Combinatorial Proof 1 Imagine that we are distributing indistinguishable candies to distinguishable children. By a direct application of Balls and Urns, there are ways to do this. Alternatively, we can first give candies to the oldest child so that we are essentially giving candies to kids and again, with Balls and Urns, , which simplifies to the desired result. Combinatorial Proof 2 We can form a committee of size from a group of people in ways. Now we hand out the numbers to of the people. We can divide this into disjoint cases. In general, in case , , person is on the committee and persons are not on the committee. This can be done in ways. Now we can sum the values of these disjoint cases, getting Algebraic Proof 2 Apply the finite geometric series formula to : Then expand with the Binomial Theorem and simplify: Finally, equate coefficients of on both sides: Since for , , this simplifies to the hockey stick identity. Algebraic Proof 3 Consider the number of solutions to the equation ++++++.......+ ≤ N where each is a non-negative integer for 1≤i≤r. METHOD 1 We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer. Therfore, ++++++.......+ = 0,1,2......N. Consider each case seperately. ++++++.......+ =0 by Stars-and-bars the equation has solutions. ++++++.......+ =1 by Stars-and-bars the equation has solutions. ++++++.......+ =2 by Stars-and-bars the equation has solutions. ........... ++++++.......+ =N by Stars-and-bars the equation has solutions. Hence, the equation has + + +.... = (where k=N+r-1) SOLUTIONS. METHOD 2 Since, ++++++.......+ ≤ N Therefore we may say ++++++.......+ = N -m where m is another non-negative integer. 0 ≤++++++.......+ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction. Now, ++++++.......+ +m = N. Again by Stars-and-bars this has solutions. Therefore, the equation has = solutions(As N+r-1 =k). Since, both methods yeild the same answer ⇒ = . Taking r-1= p redirects to the honeystick identity. ~SANSGANKRSNGUPTA Another Identity Hat Proof We have different hats. We split them into two groups, each with k hats: then we choose hats from the first group and hats from the second group. This may be done in ways. Evidently, to generate all possible choices of hats from the hats, we must choose hats from the first and the remaining hats from the second ; the sum over all such is the number of ways of choosing hats from . Therefore , as desired. Proof 2 This is a special case of Vandermonde's identity, in which we set . Even Odd Identity Examples 1986 AIME Problem 11 2000 AIME II Problem 7 2013 AIME I Problem 6 (Solution 2) 2015 AIME I Problem 12 2018 AIME I Problem 10 2020 AIME I Problem 7 2016 AMC 10A Problem 20 2021 AMC 12A Problem 15 1981 IMO Problem 2 2022 AIME I Problem 12 See also Pascal's Triangle Combinatorics Pascal's Identity Retrieved from " Categories: Theorems Combinatorics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
13886	https://www.youtube.com/watch?v=AiN8TLIkQwU	direct proof 2: Prove that the product of perfect squares is a perfect square logic maths gotserved 61300 subscribers 202 likes Description 17408 views Posted: 17 Jul 2014 Math Tutorials Links Website www.mathgotserved.com Algebra Foundations Converting /Translating Verbal to Expressions Equations Order of Operations: Adding and Subtracting Integers Rationals: Multiplying and Dividing Rationals Integers The Distributive Property how to Equations Solve One Step Equations Solve Multi step equations solving how to Absolute Value Equations how to solve Mixture Problem how to solve Combined Work Word Problems Literal Equations How to Solve formulas Inequalities Graphing single variable inequalities Solve One Step Inequalities how to Solve two step inequalities Solving Multi Step Inequalities Solve Proportions solving Graphing how to Graph Linear Functions How to Graph Linear Inequalities Graph Absolute Value Equations Solve Systems of Equations by Graphing Systems of Equations by elimination Systems of Equations by substitution Systems of Linear Inequatlities The quotient property of exponents Domain and Range Logarithms and Exponents: 1) Graphing Exponential Functions: 2) Graphing Logarithmic Functions: 3) Laws of Exponents: 4) Condensing Logarithms: 5) Change of base Precalculus Entire Collection Vectors Induction Discrete Math Proofs Matrices Unit Circle Identities Conic Sections Laws of Sines Cosines Ap Calculus Collection Ap calculus Limits Ap Calculus Derivatives Ap calculus Implicit Differentiation Optimization Extrema Maximum Minimum Solids of Revolution Definite Integral Integration Integration by Parts U-Substitution Integration Technique Related Rates Tabular Integration TESTS Accuplacer ACT CBEST NY Regents Praxis Ap Calculus MC Ap Calculus FRQ Thea TI 83 89 Texas INstruments Calculator tutorials Formula Sheet Collection Subscribe Here For more cool math videos visit our site at or To expand logarithms, you can use the following rules: Product rule: The product rule of logarithms states that log base b of (x y) is equal to the sum of log base b of x and log base b of y. Therefore, to expand a logarithm of a product, you can split it into the sum of logarithms of each factor. For example: log base 2 of (x y) = log base 2 of x + log base 2 of y Quotient rule: The quotient rule of logarithms states that log base b of (x / y) is equal to the difference of log base b of x and log base b of y. Therefore, to expand a logarithm of a quotient, you can split it into the difference of logarithms of the numerator and denominator. For example: log base 2 of (x / y) = log base 2 of x - log base 2 of y Power rule: The power rule of logarithms states that log base b of (x^y) is equal to y times log base b of x. Therefore, to expand a logarithm of a power, you can multiply the exponent by the logarithm of the base. For example: log base 2 of (x^3) = 3 log base 2 of x Change of base formula: The change of base formula allows you to change the base of a logarithm. It states that log base b of x can be expressed as log base a of x divided by log base a of b. Therefore, to expand a logarithm with a different base, you can use the change of base formula to express it in terms of a logarithm with a known base. For example: log base 2 of x = log base 10 of x / log base 10 of 2 It's important to remember that when expanding logarithms, you may need to simplify or combine terms afterwards to get the expression into a more useful or simplified form. 6 comments Transcript: good day students uh welcome to direct proof number two um do not forget to visit Matt good serve.com for access to other tutorials such as this ranging from algebra all the way to calculus so let's go ahead and write down the um proof we about to carry out for our number two direct proof so for number two um we asked to prove the following proof prove that um if n and M are perfect squares they perfect squares then um their product then their product NM is also a perfect square so this is what we're going to prove prove today okay so um first thing we're going to do let's take a look at NM we're told that they're perfect squares now how can we capture that statement in a mathematical form can we write an equation using this fact and the answer is yes all right so um since so this is our proof right here want to let me delate it so you know where I'm starting from prove since n and M are perfect squares perfect squares U what does that mean since n and M are perfect squares uh then um then n is equal to a s and m is equal to b² for some integers integers A and B okay so any number that's a square can be expressed as an integer Square for example if you have 25 25 is a perfect square because it can be expressed as 5 squar and you notice that five is an integer so this is um uh what it is to be a square by definition okay so let's go ahead and write that down since um since n and M are perfect squares then n is equal a square and M is equal B squ for some integers um AB um by definition definition of a square all right so by definition we can U make these two statements now we have an equation that we can use to represent to find what the value of NM is is all right so n m is simply going to be a s b s okay and then we can write this as a a B B using the properties of exponents now what can we do we know that multiplication commute so we can write this as a Time B um time a B commutative property commutative property right there of multiplication and then um we can then associate them in this order AB time AB okay using the associative property of in of of um multiplication if you associate the multiplicands um you do not change the value of the product it doesn't matter how you you associate them you still get the same answer so we can associate them in this fashion Now using the properties of exponents we multiply to identical bases what do you do you add the B the um Powers right so this is the same thing as AB squ so NM is equal to AB squ now what do we know about AB that's the next question since A and B are integers then what do we know about the product the product is going to be an integer also by closure right so since A and B e are integers integers then AB um is also an integer because um integers are closed on the multiplication where you multiply integers you get an integer okay a is also an integer so let's make a substitution here first of all I like to declare the variable I'll use in making my substitution let s equals AB all right for uh some integer s for some integer s so if I make that substitution of the value of ab which is s into this equation and M here I will have n m is equal to instead of ab is equal to s² so if n m is equal to s² for some integer s what does this uh what is the this is a definition of what this is the definition of a perfect square okay so um n m is a perfect square by definition okay by definition we can conclude that NM is in fact a perfect square so that concludes our proof thanks so much for taking the time to watch this presentation we really appreciate it feel free to subscribe to our channel for updates to other um math tutorial such as this and do post a comment to let us know what you think about this tutorial more Clips can be found on M serve.com thanks again for watching and have a wonderful day
13887	https://brighterly.com/math/hexagon/	Math tutors / Knowledge Base / Hexagon – Definition, Types, Properties, Examples, FAQs Hexagon – Definition, Types, Properties, Examples, FAQs Jo-ann Caballes Updated on January 13, 2024 Table of Contents Geometry is an essential part of basic math that every kid will encounter from grade to grade. At first, kids get acquainted with simple basic shapes like circles, triangles, rectangles, squares, and the like. However, as time passes, they encounter more complicated shapes like hexagons. If you teach kids about the hexagon, you must learn a few things about it. This article contains all the information you need to know about the hexagon. What is a Hexagon? A hexagon is a 6 sided polygon which can also be described as a closed 6 sided shape where all the sides are straight and equal, as well as the angles. While you may not see hexagons every day and everywhere, they exist in some real-life objects’ forms which you may come across occasionally. You can find the hexagon shape in honeycombs and stop signs. In some cases, the hexagons come in the middle of other things, and you may have to look twice to see them; for example, soccer balls, floor tiles, and nuts and bolts. If your kid asks, “What does a hexagon look like?”, you can search for the items attached to the hexagon definition to get a physical representation of the shape. Types of hexagon The types of hexagon are differentiated according to their sides and angles. There are two significant hexagons, regular and irregular, and others. Here are some of the hexagons in the world of geometry: Regular hexagon A regular hexagon is a hexagon where all 6 sides are of the same length and all angles are equal to 120 degrees, with the sum of all interior angles being 720 degrees. The sum of the exterior angles for any polygon is 360 degrees, so each exterior angle of the regular hexagon is 60 degrees. You would find such hexagons everywhere, in the stop sign, soccer balls, and honeycombs. Irregular hexagon An irregular hexagon is the one where all the sides and angles are unequal. Unlike a regular hexagon where all the sides and angles are equal, an irregular hexagon can have angles of different sizes and sides of different lengths. You can find irregular hexagons in objects like rocks or buildings with designs that you will describe as asymmetrical. However, irregular hexagons are not seen in many everyday objects, so if you need to show a visual representation of an irregular hexagon to a child, you might need to search harder. Some types of hexagons are not usually discussed in early mathematics because kids do not have to use them until they advance in their grades. These types of hexagons are as follows: Concave hexagon Convex hexagon Symmetrical hexagon Asymmetrical hexagon You do not need to worry about these hexagons for now as kids will not be learning them. Classification of hexagons based on their angles You can classify hexagons according to the size of their interior angles, and here are these categories: Acute hexagon An acute hexagon is the one in which all 6 internal angles are less than 90 degrees. You can find acute hexagon angles in snowflakes and architectural designs of some buildings. Obtuse hexagons In an obtuse hexagon, all 6 internal angles are greater than 90 degrees but lesser than 180 degrees. Because of their irregular shapes, you will not find obtuse hexagons in many places. But some people use them as patterns in artworks. Right hexagons In a right hexagon, all of the six angles are just equal to 90 degrees. So, another name for right-angle hexagons is rectangular or 90-degree hexagons. You can find the right hexagons in tile designs, artwork, and architecture. Properties of hexagons Before you can solve a problem involving a hexagon, you must first know the properties of a hexagon, and here are some of them: A hexagon has 6 sides and 6 angles. A hexagon has 6 vertices where all the sides meet. The sum of the interior of a hexagon is always equal to 720 degrees. In a regular hexagon, the 6 interior angles measure and are equal to 120 degrees. This sum can change with irregular angles. The size of the exterior angles is always 60 degrees. A regular hexagon has six lines of symmetry which you can use to divide the hexagon into 6 congruent parts. The hexagon’s perimeter is the sum of the length of the six sides. The area of a hexagon is 3√3s2/2, where s is the length of one side. A hexagon has 9 diagonals, defined as line segments connecting nonadjacent vertices. Hexagon sides The hexagon sides are 6 in number. In a regular hexagon, all the sides are of equal length, while in an irregular hexagon, some of the sides, at least two, will be of different lengths. The sum of all of the hexagon’s sides is the hexagon’s perimeter. If you know the length of one of the sides and the perimeter of a regular hexagon, you can get the length of the rest by dividing the perimeter by 6. Note that this formula above may not work if you are dealing with irregular hexagons. Hexagon angles Hexagon angles are 6 in number. In a regular hexagon, all 6 angles are equal to 120. You can calculate each angle by dividing the total sum of all the angles by six (720÷6). The process is a little different if you are dealing with an irregular hexagon. You have to divide the hexagon into triangles and then calculate the angle of each triangle, which naturally should be 180 degrees. Alternatively, you can use (n-2)x180 degrees, with n representing the number of sides; this is the sum of interior angles for any polygon. At the end of the calculation, regardless of the angles’ measures, the sum must be 720 degrees. Perimeter of a hexagon The perimeter of a hexagon is the total length of all 6 sides of the hexagon. If you know the length of one side of a regular hexagon, you can calculate the perimeter using p=6s, with s being one side of the hexagon and p being the perimeter. In an irregular hexagon, you will need to add up each side to get the sum which would now be the perimeter. Area of a hexagon The formula to calculate the area of a regular hexagon is A = (3√3 / 2) × s². A is the area of a hexagon, and s is the length of the hexagon’s sides. Alternatively, you can use the formula A = (3√3 / 2) × a² if you have something called the apothem. The apothem is the distance from the center of the hexagon to the midpoint of one of its sides. In this formula, A stands for the apothem of length. You can only use these two formulas if you are working with a regular hexagon; if you are calculating the area for an irregular hexagon, you must break the hexagon into smaller shapes and calculate their areas individually. Solved examples on hexagon Here are some solved examples on a hexagon that would help you understand better how the formulas work: Example 1: Find the perimeter of a regular hexagon with a side length of 6 cm. Solution: A regular hexagon has six equal sides, so you get the perimeter of the hexagon by multiplying 6 by the length of one of its sides. Perimeter of hexagon = 6 x side length = 6 x 6 cm = 36 cm Therefore, the perimeter of the regular hexagon is 36 cm. Example 2: Find the area of a regular hexagon with an apothem length of 8 cm. Solution: The formula for finding the area of a regular hexagon is: Area of hexagon = 3 x √3 x (apothem length)² ÷ 2 Given, apothem length = 8 cm Area of hexagon = 3 x √3 x (8 cm)² ÷ 2 = 3 x √3 x 64 cm² ÷ 2 = 96√3 cm² Therefore, the area of the regular hexagon is 96√3 cm². Example 3:The distance between two opposite vertices of a regular hexagon is 12 cm. Find its side length. Solution: S is the side length of the regular hexagon. A regular hexagon’s distance between two opposite vertices equals twice the apothem length. Therefore, 2a = 12 cm, where a is the apothem length. Dividing both sides by 2, we get: a = 6 cm The apothem length can also be calculated using the formula a = s√3/2, where s is the side length. Substituting a = 6 cm and solving for s, we get: 6 = s√3/2 s = 12/√3 cm Therefore, the side length of the regular hexagon is 12/√3 cm (or approximately 6.93 cm). FAQ What are the angles of a hexagon? The angles of a hexagon are 6 angles, 120 degrees each for regular hexagons and different measures for irregular angles, but all must get a sum of 720 degrees. How many sides does a hexagon have? A hexagon has 6 sides. How many types of hexagons are there? There are two significant types of hexagons, regular and irregular hexagons. What is the sum of all interior angles of a hexagon? The sum of all interior angles of a hexagon is 720 degrees. How many diagonals does a hexagon have? A hexagon has 9 diagonals. Does a hexagon always have equal sides? Regular hexagons have equal sides; irregular hexagons may have different sides’ lengths. Jo-ann Caballes 13 articles As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable. Table of Contents What is a Hexagon? Types of hexagon Regular hexagon Irregular hexagon Classification of hexagons based on their angles Acute hexagon Obtuse hexagons Right hexagons Properties of hexagons Hexagon sides Hexagon angles Perimeter of a hexagon Area of a hexagon Solved examples on hexagon FAQ What are the angles of a hexagon? How many sides does a hexagon have? How many types of hexagons are there? What is the sum of all interior angles of a hexagon? How many diagonals does a hexagon have? Does a hexagon always have equal sides? Problems with Geometry? Does your child struggle with understanding geometry basics? Try learning with an online tutor. Choose kid's grade Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Is your child having difficulties with mastering geometry fundamentals? An online tutor could provide the necessary help. Book free lesson Related math Parallelogram – Definition with Examples Brighterly is here to make math concepts easy for kids to understand and apply. As part of our knowledge base series on shapes, we’re exploring parallelograms. There are unique properties of a parallelogram that are exciting to explore – and we’ll do just that here. In this article, we share the parallelogram geometry definition with […] Read more 56000 in Words We write the number 56000 in words as “fifty-six thousand”. It’s fifty-six sets of one thousand each. If you have fifty-six thousand crayons, it means you have fifty-six thousand crayons in total. Thousands Hundreds Tens Ones 56 0 0 0 How to Write 56000 in Words? To write 56000 in words, we see that it […] Read more 29500 in Words The number 29500 is written as “twenty-nine thousand five hundred” in words. It is five hundred more than twenty-nine thousand. Imagine you have twenty-nine thousand five hundred balloons; that means you have twenty-nine thousand balloons, plus five hundred more. Thousands Hundreds Tens Ones 29 5 0 0 How to Write 29500 in Words? To write […] Read more Close a child’s math gaps with a tutor! Book a free demo lesson with our math tutor and see your kid fill math gaps with interactive lessons Book demo lesson Get full test results See Your Child’s Test Results Enter your name and email to view your child’s test results now! Parent’s name Child’s grade Choose kid's grade Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Parent’s email Submit & View results
13888	https://ods.od.nih.gov/factsheets/VitaminD-HealthProfessional/	Vitamin D - Health Professional Fact Sheet Skip navigation links U.S. Department of Health & Human Services HHS National Institutes of Health NIH Division of Program Coordination, Planning, and Strategic Initiatives DPCPSI Información en español español Strengthening Knowledge and Understanding of Dietary Supplements Health InformationHealth Information Health Information Supplement Fact Sheets Supplement FAQ What You Need To Know About Supplements Videos Información en español Dictionary of Terms Nutrient Recommendations For Health Professionals Consumer Awareness & Protection My Dietary Supplement and Medicine Record News & EventsNews & Events News and Events Headlines Media Contacts Newsletters Announcements and News Releases Seminars, Conferences & Workshops Programs & ActivitiesPrograms & Activities Programs & Activities Analytical Methods and Reference Materials (AMRM) Botanical Research Centers (CARBON) Population Studies Resilience & Health Studies Dietary Supplements Research Database (CARDS) Dietary Supplement Label Database (DSLD) Other Databases Dietary Supplement Research Practicum Scholars Program Evidence-based Reviews Grants & FundingGrants & Funding Grants & Funding Co-funding Program Co-funded Research Portfolio Product Integrity Information Grants & Co-funding FAQ About ODSAbout ODS About ODS Mission, Origin, and Mandate Strategic Plan 2025-2029 ODS Leadership NIH Dietary Supplement Research Coordinating Committee Staff Publications Staff Presentations Budget Timeline Contact Us Search the ODS website Submit Search Search the ODS website Submit Search Share ); "Share this page on X") ); "Share this page on Facebook") ); "Share this page on Pinterest") Home>Health Information>Dietary Supplement Fact Sheets>Vitamin D>Vitamin D - Health Professional Vitamin D Fact Sheet for Health Professionals Expand All Have a question? Ask ODS Join the ODS Email List Consumer Datos en español Health Professional Other Resources Table of Contents Introduction Recommended Intakes Sources of Vitamin D Vitamin D Intakes and Status Vitamin D Deficiency Groups at Risk of Vitamin D Inadequacy Vitamin D and Health Health Risks from Excessive Vitamin D Interactions with Medications Vitamin D and Healthful Diets References Disclaimer This is a fact sheet intended for health professionals. For a general overview, see ourconsumer fact sheet. For information on vitamin D and COVID-19, see Dietary Supplements in the Time of COVID-19. Introduction Vitamin D (also referred to as calciferol) is a fat-soluble vitamin that is naturally present in a few foods, added to others, and available as a dietary supplement. It is also produced endogenously when ultraviolet (UV) rays from sunlight strike the skin and trigger vitamin D synthesis. Vitamin D obtained from sun exposure, foods, and supplements is biologically inert and must undergo two hydroxylations in the body for activation. The first hydroxylation, which occurs in the liver, converts vitamin D to 25-hydroxyvitamin D [25(OH)D], also known as calcidiol. The second hydroxylation occurs primarily in the kidney and forms the physiologically active 1,25-dihydroxyvitamin D [1,25(OH)2D], also known as calcitriol [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Vitamin D promotes calcium absorption in the gut and maintains adequate serum calcium and phosphate concentrations to enable normal bone mineralization and to prevent hypocalcemic tetany (involuntary contraction of muscles, leading to cramps and spasms). It is also needed for bone growth and bone remodeling by osteoblasts and osteoclasts [1-31. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010. Norman AW, Henry HH. Vitamin D. In: Erdman JW, Macdonald IA, Zeisel SH, eds. Present Knowledge in Nutrition, 10th ed. Washington DC: Wiley-Blackwell, 2012. Jones G. Vitamin D. In: Ross AC, Caballero B, Cousins RJ, Tucker KL, Ziegler TR, eds. Modern Nutrition in Health and Disease, 11th ed. Philadelphia: Lippincott Williams & Wilkins, 2014.]. Without sufficient vitamin D, bones can become thin, brittle, or misshapen. Vitamin D sufficiency prevents rickets in children and osteomalacia in adults. Together with calcium, vitamin D also helps protect older adults from osteoporosis. Vitamin D has other roles in the body, including reduction of inflammation as well as modulation of such processes as cell growth, neuromuscular and immune function, and glucose metabolism [1-31. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010. Norman AW, Henry HH. Vitamin D. In: Erdman JW, Macdonald IA, Zeisel SH, eds. Present Knowledge in Nutrition, 10th ed. Washington DC: Wiley-Blackwell, 2012. Jones G. Vitamin D. In: Ross AC, Caballero B, Cousins RJ, Tucker KL, Ziegler TR, eds. Modern Nutrition in Health and Disease, 11th ed. Philadelphia: Lippincott Williams & Wilkins, 2014.]. Many genes encoding proteins that regulate cell proliferation, differentiation, and apoptosis are modulated in part by vitamin D. Many tissues have vitamin D receptors, and some convert 25(OH)D to 1,25(OH)2D. In foods and dietary supplements, vitamin D has two main forms, D 2 (ergocalciferol) and D 3 (cholecalciferol), that differ chemically only in their side-chain structures. Both forms are well absorbed in the small intestine. Absorption occurs by simple passive diffusion and by a mechanism that involves intestinal membrane carrier proteins [44. Silva MC, Furlanetto TW. Intestinal absorption of vitamin D: A systematic review. Nutr Rev 2018;76:60-76. [PubMed abstract]]. The concurrent presence of fat in the gut enhances vitamin D absorption, but some vitamin D is absorbed even without dietary fat. Neither aging nor obesity alters vitamin D absorption from the gut [44. Silva MC, Furlanetto TW. Intestinal absorption of vitamin D: A systematic review. Nutr Rev 2018;76:60-76. [PubMed abstract]]. Assessing vitamin D status Serum concentration of 25(OH)D is the main indicator of vitamin D status. However, the serum concentrations of 25(OH)D that are associated with vitamin D deficiency have not been definitively identified. The Food and Nutrition Board at the National Academies of Sciences, Engineering, and Medicine states that levels of 50 nmol/L (20 ng/mL) or more are sufficient for most people, and that the risk of deficiency increases at serum concentrations of less than 30 nmol/L (12 ng/mL). Serum concentration of 25(OH)D is currently the main indicator of vitamin D status. It reflects vitamin D produced endogenously and that obtained from foods and supplements [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. In serum, 25(OH)D has a fairly long circulating half-life of 15 days [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Serum concentrations of 25(OH)D are reported in both nanomoles per liter (nmol/L) and nanograms per milliliter (ng/mL). One nmol/L is equal to 0.4 ng/mL, and 1 ng/mL is equal to 2.5 nmol/L. Assessing vitamin D status by measuring serum 25(OH)D concentrations is complicated by the considerable variability of the available assays (the two most common ones involve antibodies or chromatography) used by laboratories that conduct the analyses [55. Sempos CT, Heijboer AC, Bikle DD, Bollerslev J, Bouillon R, Brannon PM, et al. Vitamin D assays and the definition of hypovitaminosis D. Results from the First International Conference on Controversies in Vitamin D. Br J Clin Pharmacol 2018;84:2194-207. [PubMed abstract],66. LeFevre ML. Screening for vitamin deficiency in adults: U.S. Preventive Services Task Force recommendation statement. Ann Intern Med 2015;162:133-40. [PubMed abstract]]. As a result, a finding can be falsely low or falsely high, depending on the assay used and the laboratory. The international Vitamin D Standardization Program has developed procedures for standardizing the laboratory measurement of 25(OH)D to improve clinical and public health practice [55. Sempos CT, Heijboer AC, Bikle DD, Bollerslev J, Bouillon R, Brannon PM, et al. Vitamin D assays and the definition of hypovitaminosis D. Results from the First International Conference on Controversies in Vitamin D. Br J Clin Pharmacol 2018;84:2194-207. [PubMed abstract],7-107. Brooks SPJ, Sempos CT. The importance of 25-hydroxyvitamin D assay standardization and the Vitamin D Standardization Program. Journal of AOAC International 2017;100:1223-4. Taylor CL, Sempos CT, Davis CD, Brannon PM. Vitamin D: moving forward to address emerging science. Nutrients 2017, 9, 1308; doi:10.3390/mu9121308. [PubMed abstract] Sempos CT, Binkley N. 25-hydroxyvitamin D assay standardisation and vitamin D guidelines paralysis. Public Health Nutrition 2020;23:1153-64. [PubMed abstract] Office of Dietary Supplements, National Institutes of Health. Vitamin D Standardization Program (VDSP).]. In contrast to 25(OH)D, circulating 1,25(OH)2D is generally not a good indicator of vitamin D status because it has a short half-life measured in hours, and serum levels are tightly regulated by parathyroid hormone, calcium, and phosphate [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Levels of 1,25(OH)2D do not typically decrease until vitamin D deficiency is severe [22. Norman AW, Henry HH. Vitamin D. In: Erdman JW, Macdonald IA, Zeisel SH, eds. Present Knowledge in Nutrition, 10th ed. Washington DC: Wiley-Blackwell, 2012.]. Although 25(OH)D functions as a biomarker of exposure, the extent to which 25(OH)D levels also serve as a biomarker of effect on the body (i.e., relating to health status or outcomes) is not clear [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,33. Jones G. Vitamin D. In: Ross AC, Caballero B, Cousins RJ, Tucker KL, Ziegler TR, eds. Modern Nutrition in Health and Disease, 11th ed. Philadelphia: Lippincott Williams & Wilkins, 2014.]. Researchers have not definitively identified serum concentrations of 25(OH)D associated with deficiency (e.g., rickets), adequacy for bone health, and overall health. After reviewing data on vitamin D needs, an expert committee of the Food and Nutrition Board (FNB) at the National Academies of Sciences, Engineering, and Medicine (NASEM) concluded that people are at risk of vitamin D deficiency at serum 25(OH)D concentrations less than 30 nmol/L (12 ng/mL; see Table 1 for definitions of deficiency and inadequacy) [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Some people are potentially at risk of inadequacy at 30 to 50 nmol/L (12–20 ng/mL). Levels of 50 nmol/L (20 ng/mL) or more are sufficient for most people. The FNB also noted that serum concentrations greater than 125 nmol/L (50 ng/mL) can be associated with adverse effects [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.] (Table 1). The Endocrine Society has not identified 25(OH)D concentrations associated with vitamin D sufficiency, insufficiency, and deficiency and does not recommend routine testing of 25(OH)D concentrations in healthy individuals [1111. Demay MB, Pittas AG, Bikle DD, Diab DL, Kiely ME, et al. Vitamin D for the Prevention of Disease: An Endocrine Society Clinical Practice Guideline. J Clin Endocrinol Metab. 2024 Jul 12;109(8):1907-1947. [PubMed abstract],1212. Shah VP, Nayfeh T, Alsawaf Y, Saadi S, Farah M, et al. A Systematic Review Supporting the Endocrine Society Clinical Practice Guidelines on Vitamin D. J Clin Endocrinol Metab. 2024 Jul 12;109(8):1961-1974. [PubMed abstract]]. Table 1: Serum 25-Hydroxyvitamin D [25(OH)D] Concentrations and Health [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]\| nmol/L \| ng/mL \| Health status \| --- \| <30 \| <12 \| Associated with vitamin D deficiency, which can lead to rickets in infants and children and osteomalacia in adults \| \| 30 to <50 \| 12 to <20 \| Generally considered inadequate for bone and overall health in healthy individuals \| \| ≥50 \| ≥20 \| Generally considered adequate for bone and overall health in healthy individuals \| \| >125 \| >50 \| Linked to potential adverse effects, particularly at >150 nmol/L (>60 ng/mL) \| \| Serum concentrations of 25(OH)D are reported in both nanomoles per liter (nmol/L) and nanograms per milliliter (ng/mL). One nmol/L = 0.4 ng/mL, and 1 ng/mL = 2.5 nmol/L. \| Download this table as a CSV file Optimal serum concentrations of 25(OH)D for bone and general health have not been established because they are likely to vary by stage of life, by race and ethnicity, and with each physiological measure used [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,1313. Holick MF. Vitamin D deficiency. N Engl J Med 2007;357:266-81. [PubMed abstract],1414. Brown LL, Cohen B, Tabor D, Zappala G, Maruvada P, Coates PM. The vitamin D paradox in Black Americans: A systems-based approach to investigating clinical practice, research, and public health—expert panel meeting report. BMC Proceedings, 2018;12(Suppl 6):6. [PubMed abstract]]. In addition, although 25(OH)D levels rise in response to increased vitamin D intake, the relationship is nonlinear [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. The amount of increase varies, for example, by baseline serum levels and duration of supplementation. Recommended Intakes The Food and Nutrition Board at the National Academies of Sciences, Engineering, and Medicine has established Recommended Dietary Allowances and Adequate Intakes for vitamin D. These values range from 15 to 20 mcg (600–800 IU) for adults and from 10 to 15 mcg (400–600 IU) for infants, children, and adolescents, depending on age. Intake recommendations for vitamin D and other nutrients are provided in the Dietary Reference Intakes (DRIs) developed by expert committees of NASEM [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. DRI is the general term for a set of reference values used for planning and assessing nutrient intakes of healthy people. These values, which vary by age and sex, include the following: Recommended Dietary Allowance (RDA): Average daily level of intake sufficient to meet the nutrient requirements of nearly all (97%–98%) healthy individuals; often used to plan nutritionally adequate diets for individuals Adequate Intake (AI): Intake at this level is assumed to ensure nutritional adequacy; established when evidence is insufficient to develop an RDA Estimated Average Requirement (EAR): Average daily level of intake estimated to meet the requirements of 50% of healthy individuals; usually used to assess the nutrient intakes of groups of people and to plan nutritionally adequate diets for them; can also be used to assess the nutrient intakes of individuals Tolerable Upper Intake Level (UL): Maximum daily intake unlikely to cause adverse health effects The FNB established RDAs for vitamin D to indicate daily intakes sufficient to maintain bone health and normal calcium metabolism in healthy people. RDAs for vitamin D are listed in both micrograms (mcg) and International Units (IU); 1 mcg vitamin D is equal to 40 IU (Table 2). Even though sunlight is a major source of vitamin D for some people, the FNB based the vitamin D RDAs on the assumption that people receive minimal sun exposure [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. For infants, the FNB developed AIs based on the amount of vitamin D that maintains serum 25(OH)D levels above 20 ng/mL (50 nmol/L) and supports bone development. Table 2: Recommended Dietary Allowances (RDAs) for Vitamin D in Micrograms (mcg) and International Units (IU) [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]\| Age \| Male \| Female \| Pregnancy \| Lactation \| --- --- \| 0–6 months \| 10 mcg (400 IU) \| 10 mcg (400 IU) \| \| \| \| 7–12 months \| 10 mcg (400 IU) \| 10 mcg (400 IU) \| \| \| \| 1–3 years \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| \| \| \| 4–8 years \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| \| \| \| 9–13 years \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| \| \| \| 14–18 years \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| \| 19–50 years \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| \| 51–70 years \| 15 mcg (600 IU) \| 15 mcg (600 IU) \| \| \| \| >70 years \| 20 mcg (800 IU) \| 20 mcg (800 IU) \| \| \| \| Adequate Intake (AI) \| Download this table as a CSV file Many other countries around the world and some professional societies have somewhat different guidelines for vitamin D intakes [1515. Bouillon R. Comparative analysis of nutritional guidelines for vitamin D. Nat Rev Endocrinol 2017;13:466-79. [PubMed abstract]]. These differences are a result of an incomplete understanding of the biology and clinical implications of vitamin D, different purposes for the guidelines (e.g., for public health in a healthy population or for clinical practice), and/or the use in some guidelines of observational studies in addition to randomized clinical trials to establish recommendations [99. Sempos CT, Binkley N. 25-hydroxyvitamin D assay standardisation and vitamin D guidelines paralysis. Public Health Nutrition 2020;23:1153-64. [PubMed abstract],1515. Bouillon R. Comparative analysis of nutritional guidelines for vitamin D. Nat Rev Endocrinol 2017;13:466-79. [PubMed abstract]]. For example, the United Kingdom Scientific Advisory Committee on Nutrition recommends intakes of 10 mcg (400 IU)/day for individuals age 4 years and older [1616. Scientific Advisory Committee on Nutrition. Vitamin D and Health. 2016.]. The Endocrine Society recommends routine vitamin D supplementation for children and teens age 1 to 18 years, people who are pregnant, adults with pre-diabetes, and adults age 75 years and older, but not for healthy adults age 19 to 74 [1111. Demay MB, Pittas AG, Bikle DD, Diab DL, Kiely ME, et al. Vitamin D for the Prevention of Disease: An Endocrine Society Clinical Practice Guideline. J Clin Endocrinol Metab. 2024 Jul 12;109(8):1907-1947. [PubMed abstract],1212. Shah VP, Nayfeh T, Alsawaf Y, Saadi S, Farah M, et al. A Systematic Review Supporting the Endocrine Society Clinical Practice Guidelines on Vitamin D. J Clin Endocrinol Metab. 2024 Jul 12;109(8):1961-1974. [PubMed abstract]]. The Endocrine Society does not recommend specific doses but notes that all individuals should adhere to the RDA. Sources of Vitamin D Food Vitamin D is found naturally in a few foods, such as the flesh of fatty fish, fish liver oils, beef liver, egg yolks, and cheese. In American diets, fortified foods (e.g., milk, breakfast cereals) provide most of the vitamin D. Few foods naturally contain vitamin D. The flesh of fatty fish (such as trout, salmon, tuna, and mackerel) and fish liver oils are among the best sources [1717. Roseland JM, Phillips KM, Patterson KY, Pehrsson PR, Taylor CL. Vitamin D in foods: An evolution of knowledge. Pages 41-78 in Feldman D, Pike JW, Bouillon R, Giovannucci E, Goltzman D, Hewison M, eds. Vitamin D, Volume 2: Health, Disease and Therapeutics, Fourth Edition. Elsevier, 2018.,11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. An animal’s diet affects the amount of vitamin D in its tissues. Beef liver, egg yolks, and cheese have small amounts of vitamin D, primarily in the form of vitamin D 3 and its metabolite 25(OH)D 3. Mushrooms provide variable amounts of vitamin D 2 [1717. Roseland JM, Phillips KM, Patterson KY, Pehrsson PR, Taylor CL. Vitamin D in foods: An evolution of knowledge. Pages 41-78 in Feldman D, Pike JW, Bouillon R, Giovannucci E, Goltzman D, Hewison M, eds. Vitamin D, Volume 2: Health, Disease and Therapeutics, Fourth Edition. Elsevier, 2018.]. Some mushrooms available on the market have been treated with UV light to increase their levels of vitamin D 2. In addition, the Food and Drug Administration (FDA) has approved UV-treated mushroom powder as a food additive for use as a source of vitamin D 2 in food products [1818. U.S. Food and Drug Administration. Food additives permitted for direct addition to food for human consumption; vitamin D2 mushroom powder. Federal Register 2020;85:41916-20.]. Very limited evidence suggests no substantial differences in the bioavailability of vitamin D from various foods [1919. Borel P, Caillaud D, Cano NJ. Vitamin D bioavailability: State of the art. Crit Rev Food Sci Nutr 2015;55:1193-205. [PubMed abstract]]. Animal-based foods typically provide some vitamin D in the form of 25(OH)D in addition to vitamin D 3. The impact of this form on vitamin D status is an emerging area of research. Studies show that 25(OH)D appears to be approximately five times more potent than the parent vitamin for raising serum 25(OH)D concentrations [1717. Roseland JM, Phillips KM, Patterson KY, Pehrsson PR, Taylor CL. Vitamin D in foods: An evolution of knowledge. Pages 41-78 in Feldman D, Pike JW, Bouillon R, Giovannucci E, Goltzman D, Hewison M, eds. Vitamin D, Volume 2: Health, Disease and Therapeutics, Fourth Edition. Elsevier, 2018.,2020. Taylor CL, Patterson KY, Roseland JM, Wise SA, Merkel JM, Pehrsson PR, Yetley EA. Including food 25-hydroxyvitamin D in intake estimates may reduce the discrepancy between dietary and serum measures of vitamin D status. J Nutr 2014;144:654-9. [PubMed abstract],2121. Cashman KD, Seamans KM, Lucey AJ, Stocklin E, Weber P, Kiely M, Hill TR. Relative effectiveness of oral 25-hydroxyvitamin D3 and vitamin D3 in raising wintertime serum 25-hydroxyvitamin D in older adults. Am J Clin Nutr 2012;95:1350-6. [PubMed abstract]]. One study found that when the 25(OH)D content of beef, pork, chicken, turkey, and eggs is taken into account, the total amount of vitamin D in the food is 2 to 18 times higher than the amount in the parent vitamin alone, depending on the food [2020. Taylor CL, Patterson KY, Roseland JM, Wise SA, Merkel JM, Pehrsson PR, Yetley EA. Including food 25-hydroxyvitamin D in intake estimates may reduce the discrepancy between dietary and serum measures of vitamin D status. J Nutr 2014;144:654-9. [PubMed abstract]]. Fortified foods provide most of the vitamin D in American diets [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,2222. Calvo MS, Whiting SJ, Barton CN. Vitamin D fortification in the United States and Canada: Current status and data needs. Am J Clin Nutr 2004;80:1710S-6S. [PubMed abstract]]. For example, almost all of the U.S. milk supply is voluntarily fortified with about 3 mcg/cup (120 IU), usually in the form of vitamin D 3 [2323. Yetley EA. Assessing the vitamin D status of the US population. Am J Clin Nutr 2008;88:558S-64S. [PubMed abstract]]. In Canada, milk must be fortified with 0.88 to 1.0 mcg/100 mL (35–40 IU), and the required amount for margarine is at least 13.25 mcg/100 g (530 IU). Other dairy products made from milk, such as cheese and ice cream, are not usually fortified in the United States or Canada. Plant milk alternatives (such as beverages made from soy, almond, or oats) are often fortified with similar amounts of vitamin D to those in fortified cow's milk (about 3 mcg [120 IU]/cup); the Nutrition Facts label lists the actual amount [2424. U.S. Food and Drug Administration. Vitamin D for milk and milk alternatives. January 4, 2018.]. Ready-to-eat breakfast cereals often contain added vitamin D, as do some brands of orange juice, yogurt, margarine, and other food products. The United States mandates the fortification of infant formula with 1 to 2.5 mcg/100 kcal (40–100 IU) vitamin D; 1 to 2 mcg/100 kcal (40–80 IU) is the required amount in Canada [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. A variety of foods and their vitamin D levels per serving are listed in Table 3. Table 3: Vitamin D Content of Selected Foods [2525. U.S. Department of Agriculture, Agricultural Research Service. FoodData Central.]\| Food \| Micrograms (mcg) per serving \| International Units (IU) per serving \| Percent DV \| --- --- \| \| Cod liver oil, 1 tablespoon \| 34.0 \| 1,360 \| 170 \| \| Trout (rainbow), farmed, cooked, 3 ounces \| 16.2 \| 645 \| 81 \| \| Salmon (sockeye), cooked, 3 ounces \| 14.2 \| 570 \| 71 \| \| Mushrooms, white, raw, sliced, exposed to UV light, ½ cup \| 9.2 \| 366 \| 46 \| \| Milk, 2% milkfat, vitamin D fortified, 1 cup \| 2.9 \| 120 \| 15 \| \| Soy, almond, and oat milks, vitamin D fortified, various brands, 1 cup \| 2.5–3.6 \| 100–144 \| 13–18 \| \| Ready-to-eat cereal, fortified with 10% of the DV for vitamin D, 1 serving \| 2.0 \| 80 \| 10 \| \| Sardines (Atlantic), canned in oil, drained, 2 sardines \| 1.2 \| 46 \| 6 \| \| Egg, 1 large, scrambled \| 1.1 \| 44 \| 6 \| \| Liver, beef, braised, 3 ounces \| 1.0 \| 42 \| 5 \| \| Tuna fish (light), canned in water, drained, 3 ounces \| 1.0 \| 40 \| 5 \| \| Cheese, cheddar, 1.5 ounce \| 0.4 \| 17 \| 2 \| \| Mushrooms, portabella, raw, diced, ½ cup \| 0.1 \| 4 \| 1 \| \| Chicken breast, roasted, 3 ounces \| 0.1 \| 4 \| 1 \| \| Beef, ground, 90% lean, broiled, 3 ounces \| 0 \| 1.7 \| 0 \| \| Broccoli, raw, chopped, ½ cup \| 0 \| 0 \| 0 \| \| Carrots, raw, chopped, ½ cup \| 0 \| 0 \| 0 \| \| Almonds, dry roasted, 1 ounce \| 0 \| 0 \| 0 \| \| Apple, large \| 0 \| 0 \| 0 \| \| Banana, large \| 0 \| 0 \| 0 \| \| Rice, brown, long-grain, cooked, 1 cup \| 0 \| 0 \| 0 \| \| Whole wheat bread, 1 slice \| 0 \| 0 \| 0 \| \| Lentils, boiled, ½ cup \| 0 \| 0 \| 0 \| \| Sunflower seeds, roasted, ½ cup \| 0 \| 0 \| 0 \| \| Edamame, shelled, cooked, ½ cup \| 0 \| 0 \| 0 \| \| DV = Daily Value. The FDA developed DVs to help consumers compare the nutrient contents of foods and dietary supplements within the context of a total diet. The DV for vitamin D is 20 mcg (800 IU) for adults and children age 4 years and older [2626. U.S. Food and Drug Administration. Food labeling: Revision of the Nutrition and Supplement Facts labels. Federal Register 81(103):33742-33999. 2016.]. The labels must list vitamin D content in mcg per serving and have the option of also listing the amount in IUs in parentheses. Foods providing 20% or more of the DV are considered to be high sources of a nutrient, but foods providing lower percentages of the DV also contribute to a healthful diet. \| \| Vitamin D is in the yolk. \| Download this table as a CSV file The U.S. Department of Agriculture’s (USDA’s) FoodData Central lists the nutrient content of many foods and provides a comprehensive list of foods containing vitamin D arranged by nutrient content and by food name. However, FoodData Central does not include the amounts of 25(OH)D in foods. Sun exposure People can get some of their daily vitamin D through exposure to sunlight, although the season, time of day, length of day, cloud cover, melanin content of the skin, and other factors can affect ultraviolet radiation exposure and vitamin D synthesis. Most people in the world meet at least some of their vitamin D needs through exposure to sunlight [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Type B UV (UVB) radiation with a wavelength of approximately 290 to 320 nanometers penetrates uncovered skin and converts cutaneous 7-dehydrocholesterol to previtamin D 3, which in turn becomes vitamin D 3. Season, time of day, length of day, cloud cover, smog, skin melanin content, and sunscreen are among the factors that affect UV radiation exposure and vitamin D synthesis. Older people and people with dark skin are less able to produce vitamin D from sunlight [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. UVB radiation does not penetrate glass, so exposure to sunshine indoors through a window does not produce vitamin D [2727. Hossein-nezhad A, Holick MF. Vitamin D for health: A global perspective. Mayo Clin Proc 2013;88:720-55. [PubMed abstract]]. The factors that affect UV radiation exposure, individual responsiveness, and uncertainties about the amount of sun exposure needed to maintain adequate vitamin D levels make it difficult to provide guidelines on how much sun exposure is required for sufficient vitamin D synthesis [1515. Bouillon R. Comparative analysis of nutritional guidelines for vitamin D. Nat Rev Endocrinol 2017;13:466-79. [PubMed abstract],2828. U.S. Department of Health and Human Services. The Surgeon General's Call to Action to Prevent Skin Cancer. Washington, DC: U.S. Dept of Health and Human Services, Office of the Surgeon General; 2014.]. Some expert bodies and vitamin D researchers suggest, for example, that approximately 5 to 30 minutes of sun exposure, particularly between 10 a.m. and 4 p.m., either daily or at least twice a week to the face, arms, hands, and legs without sunscreen usually leads to sufficient vitamin D synthesis [1313. Holick MF. Vitamin D deficiency. N Engl J Med 2007;357:266-81. [PubMed abstract],1515. Bouillon R. Comparative analysis of nutritional guidelines for vitamin D. Nat Rev Endocrinol 2017;13:466-79. [PubMed abstract],2828. U.S. Department of Health and Human Services. The Surgeon General's Call to Action to Prevent Skin Cancer. Washington, DC: U.S. Dept of Health and Human Services, Office of the Surgeon General; 2014.]. Moderate use of commercial tanning beds that emit 2% to 6% UVB radiation is also effective [1313. Holick MF. Vitamin D deficiency. N Engl J Med 2007;357:266-81. [PubMed abstract],2929. Holick MF. Vitamin D: the underappreciated D-lightful hormone that is important for skeletal and cellular health. Curr Opin Endocrinol Diabetes 2002;9:87-98.]. However, despite the importance of the sun for vitamin D synthesis, limiting skin exposure to sunlight and UV radiation from tanning beds is prudent [2828. U.S. Department of Health and Human Services. The Surgeon General's Call to Action to Prevent Skin Cancer. Washington, DC: U.S. Dept of Health and Human Services, Office of the Surgeon General; 2014.]. UV radiation is a carcinogen, and UV exposure is the most preventable cause of skin cancer. Federal agencies and national organizations advise taking photoprotective measures to reduce the risk of skin cancer, including using sunscreen with a sun protection factor (SPF) of 15 or higher, whenever people are exposed to the sun [2828. U.S. Department of Health and Human Services. The Surgeon General's Call to Action to Prevent Skin Cancer. Washington, DC: U.S. Dept of Health and Human Services, Office of the Surgeon General; 2014.,3030. Weisberg P, Scanlon KS, Li R, Cogswell ME. Nutritional rickets among children in the United States: review of cases reported between 1986 and 2003. Am J Clin Nutr 2004;80:1697S-705S. [PubMed abstract]]. Sunscreens with an SPF of 8 or more appear to block vitamin D-producing UV rays. In practice, however, people usually do not apply sufficient amounts of sunscreen, cover all sun-exposed skin, or reapply sunscreen regularly. Their skin probably synthesizes some vitamin D, even with typically applied sunscreen amounts [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,2828. U.S. Department of Health and Human Services. The Surgeon General's Call to Action to Prevent Skin Cancer. Washington, DC: U.S. Dept of Health and Human Services, Office of the Surgeon General; 2014.]. Dietary supplements Vitamin D is present in dietary supplements as either vitamin D 2 or vitamin D 3. Both can raise the serum level of 25(OH)D. However, research shows that vitamin D 3 increases serum 25(OH)D levels to a greater extent than vitamin D 2 and can maintain those higher levels for longer periods of time. Dietary supplements can contain vitamins D 2 or D 3. Vitamin D 2 is manufactured using UV irradiation of ergosterol in yeast, and vitamin D 3 is typically produced with irradiation of 7-dehydrocholesterol from lanolin obtained from the wool of sheep [1313. Holick MF. Vitamin D deficiency. N Engl J Med 2007;357:266-81. [PubMed abstract],3131. Hirsch AL. Industrial Aspects of Vitamin D. In: Feldman D, Pike JW, Adams JS, eds. Vitamin D. 3rd ed. Academic Press; 2011:73-93.]. An animal-free version of vitamin D 3 sourced from lichen is also available [3232. National Institutes of Health. Dietary Supplement Label Database. 2020.]. People who avoid all animal-sourced products can contact dietary supplement manufacturers to ask about their sourcing and processing techniques. Both vitamins D 2 and D 3 raise serum 25(OH)D levels, and they seem to have equivalent ability to cure rickets [44. Silva MC, Furlanetto TW. Intestinal absorption of vitamin D: A systematic review. Nutr Rev 2018;76:60-76. [PubMed abstract]]. In addition, most steps in the metabolism and actions of vitamins D 2 and D 3 are identical. However, most evidence indicates that vitamin D 3 increases serum 25(OH)D levels to a greater extent and maintains these higher levels longer than vitamin D 2, even though both forms are well absorbed in the gut [33-3633. Tripkovic L, Lambert H, Hart K, Smith CP, Bucca G, Penson S, et al. Comparison of vitamin D2 and vitamin D3 supplementation in raising serum 25-hydroxyvitamin D status: A systematic review and meta-analysis. Am J Clin Nutr 2012;95:1357-64. [PubMed abstract] Lehmann U, Hirche F, Stangl GI, Hinz K, Westphal S, Dierkes J. Bioavailability of vitamin D2 and D3 in healthy volunteers, a randomised placebo-controlled trial. J Clin Endocrin Metab 2013;98:4339-45. [PubMed abstract] Logan VF, Gray AR, Peddie MC, Harper MJ, Houghton LA. Long-term vitamin D3 supplementation is more effective than vitamin D2 in maintaining serum 25-hydroxyvitamin D status over the winter months. Br J Nutr 2013;109:1082-8. [PubMed abstract] Tripkovic L, Wilson LR, Hart K, Johnsen S, de Lusignan S, Smith CP, et al. Daily supplementation with 15 µg vitamin D2 compared with vitamin D3 to increase wintertime 25-hydroxyvitamin D status in healthy South Asian and white European women: A 12-wk randomized, placebo-controlled food-fortification trial. Am J Clin Nutr 2017;106:481-90. [PubMed abstract]]. Some studies have used dietary supplements containing the 25(OH)D 3 form of vitamin D. Per equivalent microgram dose, 25(OH)D 3 is three to five times as potent as vitamin D 3 [3737. Graeff-Armas LA, Bendik I, Kunz I, Schoop R, Hull S, Beck M. Supplemental 25-hydroxycholecalciferol is more effective than cholecalciferol in raising serum 25-hydroxyvitamin D concentrations in older adults. J Nutr 2020;150:73-81. [PubMed abstract],3838. Quesada-Gomez JM, Bouillon R. Is calcifediol better than cholecalciferol for vitamin D supplementation? Osteoporos Int 2018;29:1697-1711. [PubMed abstract]]. However, no 25(OH)D 3 dietary supplements appear to be available to consumers on the U.S. market at this time [3232. National Institutes of Health. Dietary Supplement Label Database. 2020.]. Vitamin D Intakes and Status According to data from the National Health and Nutrition Examination Survey, most people in the United States consume less than the recommended amounts of vitamin D. However, evidence suggests that the majority of people have sufficient serum concentrations of vitamin D based on the thresholds set by the Food and Nutrition Board. Sun exposure is one of the reasons why serum levels of 25(OH)D are generally higher than would be predicted on the basis of vitamin D dietary intakes alone. Most people in the United States consume less than recommended amounts of vitamin D. An analysis of data from the 2015–2016 National Health and Nutrition Examination Survey (NHANES) found that average daily vitamin D intakes from foods and beverages were 5.1 mcg (204 IU) in men, 4.2 mcg (168 IU) in women, and 4.9 mcg (196 IU) in children age 2 to 19 years [3939. Percent reporting and mean amounts of selected vitamins and minerals food and beverages and dietary supplements by gender and age, in the United States, 2015-2016. What We Eat in America, NHANES 2015-2016. 2019.]. In fact, 2013–2016 NHANES data showed that 92% of men, more than 97% of women, and 94% of people age 1 year and older ingested less than the EAR of 10 mcg (400 IU) of vitamin D from food and beverages [4040. Usual nutrient intake from foods and beverages, by gender and age. What We Eat in America, NHANES 2013-2016. 2019.]. The analysis of 2015–2016 data also showed that 28% of all individuals age 2 years and older in the United States took a dietary supplement containing vitamin D [3939. Percent reporting and mean amounts of selected vitamins and minerals food and beverages and dietary supplements by gender and age, in the United States, 2015-2016. What We Eat in America, NHANES 2015-2016. 2019.]. In addition, 26% of participants age 2 to 5 years and 14% of those age 6 to 11 years took supplements; rates increased with age from 10% of those age 12 to 19 years to 49% of men and 59% of women age 60 and older. Total vitamin D intakes were three times higher with supplement use than with diet alone; the mean intake from foods and beverages alone for individuals age 2 and older was 4.8 mcg (192 IU) but increased to 19.9 mcg (796 IU) when dietary supplements were included. Some people take very high doses of vitamin D supplements. In 2013–2014, an estimated 3.2% of the U.S. adult population took supplements containing 100 mcg (4,000 IU) or more vitamin D [4141. Rooney MR, Harnack L, Michos ED, Ogilvie RP, Sempos CT, Lutsey PL. Trends in use of high-dose vitamin D supplements exceeding 1000 or 4000 International Units daily, 1999-2014. JAMA 2017;317:2448-50. [PubMed abstract]]. One might expect a large proportion of the U.S. population to have vitamin D inadequacy on the basis of vitamin D intakes from foods, beverages, and even dietary supplements. However, comparing vitamin D intakes to serum 25(OH)D levels is problematic. One reason is that sun exposure affects vitamin D status, so serum 25(OH)D levels are usually higher than would be predicted on the basis of vitamin D dietary intakes alone [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Another reason is that animal foods contain some 25(OH)D. This form of vitamin D is not included in intake surveys and is considerably more potent than vitamins D 2 or D 3 at raising serum 25(OH)D levels [4242. Taylor CL, Roseland JM, Coates PM, Pehrsson PR. The emerging issue of 25-hydroxyvitamin D in foods. J Nutr 2016;146:855-6. [PubMed abstract]]. An analysis of NHANES 2011–2014 data on serum 25(OH)D levels found that most people in the United States age 1 year and older had sufficient vitamin D intakes according to the FNB thresholds [4343. Herrick KA, Storandt RJ, Afful J, Pfeiffer CM, Schleicher RL, Gahche JJ, Potischman N. Vitamin D status in the United States, 2011-2014. Am J Clin Nutr 2019;110:150-7. [PubMed abstract]]. However, 18% were at risk of inadequacy (levels of 30–49 nmol/L [12–19.6 ng/mL]), and 5% were at risk of deficiency (levels below 30 nmol/L [12 ng/mL]). Four percent had levels higher than 125 nmol/L (50 ng/mL). Proportions at risk of deficiency were lowest among children age 1 to 5 years (0.5%), peaked at 7.6% in adults age 20 to 39 years, and fell to 2.9% among adults age 60 years and older; patterns were similar for risks of inadequacy. Rates of deficiency varied by race and ethnicity: 17.5% of non-Hispanic Blacks were at risk of vitamin D deficiency, as were 7.6% of non-Hispanic Asians, 5.9% of Hispanics, and 2.1% of non-Hispanic White people. Again, the pattern was similar for the risk of inadequacy. Vitamin D status in the United States remained stable in the decade between 2003–2004 and 2013–2014. Vitamin D Deficiency Vitamin D deficiency is more common among people who are lactose intolerant, have a milk allergy, or follow an ovo-vegetarian or vegan diet. Deficiency may also occur in people who have limited exposure to sunlight, those whose kidneys cannot convert 25(OH)D to its active form, or those who cannot absorb vitamin D efficiently in the digestive tract. Vitamin D deficiency can manifest as rickets in children and as osteomalacia in adolescents and adults. People can develop vitamin D deficiency when usual intakes are lower over time than recommended levels, exposure to sunlight is limited, the kidneys cannot convert 25(OH)D to its active form, or absorption of vitamin D from the digestive tract is inadequate. Diets low in vitamin D are more common in people who have milk allergy or lactose intolerance and those who consume an ovo-vegetarian or vegan diet [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. In children, vitamin D deficiency is manifested as rickets, a disease characterized by a failure of bone tissue to become properly mineralized, resulting in soft bones and skeletal deformities [4444. Elder CJ, Bishop NJ. Rickets. Lancet 2014;383:1665-76. [PubMed abstract]]. In addition to bone deformities and pain, severe rickets can cause failure to thrive, developmental delay, hypocalcemic seizures, tetanic spasms, cardiomyopathy, and dental abnormalities [4545. Munns CF, Shaw N, Kiely M, Specker BL, Thacher TD, Ozono K, et al. Global consensus recommendations on prevention and management of nutritional rickets. J Clin Endocrinol Metab 2016;101:394-415. [PubMed abstract],4646. Uday S, Hogler W. Nutritional rickets and osteomalacia in the twenty-first century: Revised concepts, public health, and prevention strategies. Curr Osteoporos Rep 2017;15:293-302. [PubMed abstract]]. Prolonged exclusive breastfeeding without vitamin D supplementation can cause rickets in infants, and, in the United States, rickets is most common among breastfed Black infants and children [4747. Weisberg P, Scanlon KS, Li R, Cogswell ME. Nutritional rickets among children in the United States: Review of cases reported between 1986 and 2003. Am J Clin Nutr 2004;80:1697S-705S. [PubMed abstract]]. In one Minnesota county, the incidence rate of rickets in children younger than 3 years in the decade beginning in 2000 was 24.1 per 100,000 [4848. Thacher TM, Fischer PR, Tebben PJ, Singh RJ, Cha SS, Maxson JA, Yawn BP. Increasing incidence of nutritional rickets: A population-based study in Olmsted County, Minnesota. Mayo Clin Proc 2013;88:176-83. [PubMed abstract]]. Rickets occurred mainly in Black children who were breastfed longer, were born with low birthweight, weighed less, and were shorter than other children. The incidence rate of rickets in the infants and children (younger than 7) seen by 2,325 pediatricians throughout Canada was 2.9 per 100,000 in 2002–2004, and almost all patients with rickets had been breastfed [4949. Ward LM, Gaboury I, Ladhani M, Zlotkin S. Vitamin D-deficiency rickets among children in Canada. CMAJ 2007;177:161-6. [PubMed abstract]]. The fortification of milk (a good source of calcium) and other staples, such as breakfast cereals and margarine, with vitamin D beginning in the 1930s along with the use of cod liver oil made rickets rare in the United States [2828. U.S. Department of Health and Human Services. The Surgeon General's Call to Action to Prevent Skin Cancer. Washington, DC: U.S. Dept of Health and Human Services, Office of the Surgeon General; 2014.,5050. Rajakumar K. Vitamin D, cod-liver oil, sunlight, and rickets: A historical perspective. Pediatrics 2003;112:e132-5. [PubMed abstract]]. However, the incidence of rickets is increasing globally, even in the United States and Europe, especially among immigrants from African, Middle-Eastern, and Asian countries [5151. Creo AL, Thacher TD, Pettifor JM, Strand MA, Ficsher PR. Nutritional rickets around the world: An update. Paediatr Int Child Health 2017;37:84-98. [PubMed abstract]]. Possible explanations for this increase include genetic differences in vitamin D metabolism, dietary preferences, and behaviors that lead to less sun exposure [4545. Munns CF, Shaw N, Kiely M, Specker BL, Thacher TD, Ozono K, et al. Global consensus recommendations on prevention and management of nutritional rickets. J Clin Endocrinol Metab 2016;101:394-415. [PubMed abstract],4646. Uday S, Hogler W. Nutritional rickets and osteomalacia in the twenty-first century: Revised concepts, public health, and prevention strategies. Curr Osteoporos Rep 2017;15:293-302. [PubMed abstract]]. In adults and adolescents, vitamin D deficiency can lead to osteomalacia, in which existing bone is incompletely or defectively mineralized during the remodeling process, resulting in weak bones [4646. Uday S, Hogler W. Nutritional rickets and osteomalacia in the twenty-first century: Revised concepts, public health, and prevention strategies. Curr Osteoporos Rep 2017;15:293-302. [PubMed abstract]]. Signs and symptoms of osteomalacia are similar to those of rickets and include bone deformities and pain, hypocalcemic seizures, tetanic spasms, and dental abnormalities [4545. Munns CF, Shaw N, Kiely M, Specker BL, Thacher TD, Ozono K, et al. Global consensus recommendations on prevention and management of nutritional rickets. J Clin Endocrinol Metab 2016;101:394-415. [PubMed abstract]]. Screening for vitamin D status is becoming a more common part of the routine laboratory bloodwork ordered by primary-care physicians, irrespective of any indications for this practice [66. LeFevre ML. Screening for vitamin deficiency in adults: U.S. Preventive Services Task Force recommendation statement. Ann Intern Med 2015;162:133-40. [PubMed abstract],52-5452. Rockwell M, Kraak V, Hulver M, Epling J. Clinical management of low vitamin D: A scoping review of physicians' practices. Nutrients 2018 Apr 16;10(4). pii: E493. doi: 10.3390/nu10040493. [PubMed abstract] Taylor CL, Thomas PR, Aloia JF, Millard PS. Questions about vitamin D for primary care practice: Input from an NIH conference. Am J Med 2015;128:1167-70. [PubMed abstract] Taylor CL, Rosen CJ, Dwyer JT. Considerations in dietetic counseling for vitamin D. J Acad Nutr Diet 2019;119:901-9. [PubMed abstract]]. No studies have examined whether such screening for vitamin D deficiency results in improved health outcomes [5555. Agency for Healthcare Research and Quality. Screening for vitamin D deficiency: Systematic review for the U.S. Preventive Services Task Force recommendation. Evidence Synthesis Number 118. AHRQ-Pub No. 13-05183-EF-1. June 2014.]. The U.S. Preventive Services Task Force (USPSTF) found insufficient evidence to assess the benefits and harms of screening for vitamin D deficiency in asymptomatic adults [66. LeFevre ML. Screening for vitamin deficiency in adults: U.S. Preventive Services Task Force recommendation statement. Ann Intern Med 2015;162:133-40. [PubMed abstract]]. It added that no national professional organization recommends population screening for vitamin D deficiency. Groups at Risk of Vitamin D Inadequacy Certain groups of people are more likely than others to have inadequate vitamin D status. These include breastfed infants, older adults, people with limited sun exposure, people with dark skin, people with conditions that limit fat absorption, and people with obesity or those who have undergone gastric bypass surgery. Obtaining sufficient vitamin D from natural (nonfortified) food sources alone is difficult. For many people, consuming vitamin D-fortified foods and exposing themselves to some sunlight are essential for maintaining a healthy vitamin D status. However, some groups might need dietary supplements to meet their vitamin D requirements. The following groups are among those most likely to have inadequate vitamin D status. Breastfed infants Consumption of human milk alone does not ordinarily enable infants to meet vitamin D requirements, because it provides less than 0.6 to 2.0 mcg/L (25 to 78 IU/L) [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,5656. Picciano MF. Nutrient composition of human milk. Pediatr Clin North Am 2001;48:53-67. [PubMed abstract],5757. Wagner CL, Greer FR, American Academy of Pediatrics Section on Breastfeeding, American Academy of Pediatrics Committee on Nutrition. Prevention of rickets and vitamin D deficiency in infants, children, and adolescents. Pediatrics 2008;122:1142-52. [PubMed abstract]]. The vitamin D content of human milk is related to the mother’s vitamin D status; studies suggest that the breastmilk of mothers who take daily supplements containing at least 50 mcg (2,000 IU) vitamin D 3 have higher levels of the nutrient [5757. Wagner CL, Greer FR, American Academy of Pediatrics Section on Breastfeeding, American Academy of Pediatrics Committee on Nutrition. Prevention of rickets and vitamin D deficiency in infants, children, and adolescents. Pediatrics 2008;122:1142-52. [PubMed abstract],5858. Dawodu A, Tsang RC. Maternal vitamin D status: Effect on milk vitamin D content and vitamin D status of breastfeeding infants. Adv Nutr 2012;3:353-61. [PubMed abstract]]. Although UVB exposure can produce vitamin D in infants, the American Academy of Pediatrics (AAP) advises parents to keep infants younger than 6 months out of direct sunlight, dress them in protective clothing and hats, and apply sunscreen on small areas of exposed skin when sun exposure is unavoidable [5959. Davis CD, Dwyer JT. The 'sunshine vitamin': benefits beyond bone? J Natl Cancer Inst 2007;99:1563-5. [PubMed abstract]]. The AAP recommends 10 mcg (400 IU)/day vitamin D supplements for exclusively and partially breastfed infants starting shortly after birth and lasting until they are weaned and consume at least 1,000 mL/day vitamin D-fortified formula or whole milk [5757. Wagner CL, Greer FR, American Academy of Pediatrics Section on Breastfeeding, American Academy of Pediatrics Committee on Nutrition. Prevention of rickets and vitamin D deficiency in infants, children, and adolescents. Pediatrics 2008;122:1142-52. [PubMed abstract]]. The AAP also recommends 10 mcg (400 IU)/day supplemental vitamin D for all infants who are not breastfed and ingest less than 1,000 mL/day vitamin D-fortified formula or milk. An analysis of NHANES 2009–2016 data found that only 20.5% of breastfed infants and 31.1% of infants who were not breastfed ingested these recommended amounts of supplements [6060. Simon AE, Ahrens KA. Adherence to vitamin D intake guidelines in the United States. Pediatrics 2020;145:e20193574. [PubMed abstract]]. Older adults Older adults are at increased risk of developing vitamin D insufficiency, partly because the skin's ability to synthesize vitamin D declines with age [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,6161. Chalcraft JR, Cardinal LM, Wechsler PJ, Hollis BW, Gerow KG, Alexander BM, et al. Vitamin D synthesis following a single bout of sun exposure in older and younger men and women. Nutrients 2020; 12, 2237; doi:10.3390/nu12082237. [PubMed abstract]]. In addition, older adults are likely to spend more time than younger people indoors, and they might have inadequate dietary intakes of the vitamin [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. People with limited sun exposure Homebound individuals; people who wear long robes, dresses, or head coverings for religious reasons; and people with occupations that limit sun exposure are among the groups that are unlikely to obtain adequate amounts of vitamin D from sunlight [6262. Sowah D, Fan X, Dennett L, Hagtvedt R, Straube S. Vitamin D levels and deficiency with different occupations: A systematic review. BMC Public Health 2017;17:519. [PubMed abstract]]. The use of sunscreen also limits vitamin D synthesis from sunlight. However, because the extent and frequency of sunscreen use are unknown, the role that sunscreen may play in reducing vitamin D synthesis is unclear [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. People with dark skin Greater amounts of the pigment melanin in the epidermal layer of the skin result in darker skin and reduce the skin’s ability to produce vitamin D from sunlight [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Black Americans, for example, typically have lower serum 25(OH)D levels than White Americans. However, whether these lower levels in persons with dark skin have significant health consequences is not clear [1414. Brown LL, Cohen B, Tabor D, Zappala G, Maruvada P, Coates PM. The vitamin D paradox in Black Americans: A systems-based approach to investigating clinical practice, research, and public health—expert panel meeting report. BMC Proceedings, 2018;12(Suppl 6):6. [PubMed abstract]]. Those of African American ancestry, for example, have lower rates of bone fracture and osteoporosis than do Whites (see the section below on bone health and osteoporosis). People with conditions that limit fat absorption Because vitamin D is fat soluble, its absorption depends on the gut’s ability to absorb dietary fat [44. Silva MC, Furlanetto TW. Intestinal absorption of vitamin D: A systematic review. Nutr Rev 2018;76:60-76. [PubMed abstract]]. Fat malabsorption is associated with medical conditions that include some forms of liver disease, cystic fibrosis, celiac disease, Crohn’s disease, and ulcerative colitis [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,6363. Pappa HM, Bern E, Kamin D, Grand RJ. Vitamin D status in gastrointestinal and liver disease. Curr Opin Gastroenterol 2008;24:176-83. [PubMed abstract]]. In addition to having an increased risk of vitamin D deficiency, people with these conditions might not eat certain foods, such as dairy products (many of which are fortified with vitamin D), or eat only small amounts of these foods. Individuals who have difficulty absorbing dietary fat might therefore require vitamin D supplementation [6363. Pappa HM, Bern E, Kamin D, Grand RJ. Vitamin D status in gastrointestinal and liver disease. Curr Opin Gastroenterol 2008;24:176-83. [PubMed abstract]]. People with obesity or who have undergone gastric bypass surgery Individuals with a body mass index (BMI) of 30 or more have lower serum 25(OH)D levels than individuals without obesity. Obesity does not affect the skin’s capacity to synthesize vitamin D. However, greater amounts of subcutaneous fat sequester more of the vitamin [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. People with obesity might need greater intakes of vitamin D to achieve 25(OH)D levels similar to those of people with normal weight [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,6464. Drincic A, Fuller E, Heaney RP, Armas LAG. 25-hydroxyvitamin D response to graded vitamin D3 supplementation among obese adults. J Clin Endocrinol Metab 2013;98:4845-51. [PubMed abstract],6565. Ekwaru JP, Zwicker JD, Holick MF, Giovannucci E, Veugelers PJ. The importance of body weight for the dose response relationship of oral vitamin D supplementation and serum 25-hydroxyvitamin D in healthy volunteers. PLOS ONE 2014;9:e111265. [PubMed abstract]]. Individuals with obesity who have undergone gastric bypass surgery can also become vitamin D deficient. In this procedure, part of the upper small intestine, where vitamin D is absorbed, is bypassed, and vitamin D that is mobilized into the bloodstream from fat stores might not raise 25(OH)D to adequate levels over time [6666. Chakhtoura M, Rahme M, Fuleihan E-H. Vitamin D metabolism in bariatric surgery. Endocrinol Metab Clin North Am 2017;46:947-82. [PubMed abstract],6767. Peterson L, Zeng X, Caufield-Noll CP, Schweitzer MA, Magnuson TH, Steele KE. Vitamin D status and supplementation before and after bariatric surgery: A comprehensive literature review. Surg Obes Relat Dis 2016;12:693-702. [PubMed abstract]]. Various expert groups—including the American Association of Metabolic and Bariatric Surgery, The Obesity Society, and the British Obesity and Metabolic Surgery Society—have developed guidelines on vitamin D screening, monitoring, and replacement before and after bariatric surgery [6666. Chakhtoura M, Rahme M, Fuleihan E-H. Vitamin D metabolism in bariatric surgery. Endocrinol Metab Clin North Am 2017;46:947-82. [PubMed abstract],6868. Chakhtoura MT, Nakhoul N, Akl EA, Mantzoros CS, El Hajj Guleihan GA. Guidelines on vitamin D replacement in bariatric surgery? Identification and systematic appraisal. Metabolism 2016;65:586-97. [PubMed abstract]] Vitamin D and Health The FNB committee that established DRIs for vitamin D found that the evidence was inadequate or too contradictory to conclude that the vitamin had any effect on a long list of potential health outcomes (e.g., on resistance to chronic diseases or functional measures), except for measures related to bone health. Similarly, in a review of data from nearly 250 studies published between 2009 and 2013, the Agency for Healthcare Research and Quality concluded that no relationship could be firmly established between vitamin D and health outcomes other than bone health [6969. Newberry SJ, Chung M, Shekelle PG, Booth MS, Liu JL, Maher AR, et al. Vitamin D and calcium: A systematic review of health outcomes (update). Evidence Report/Technology Assessment No. 217. (Prepared by the Southern California Evidence-based Practice Center under Contract No. 290- 2012-00006-I.) AHRQ Publication No. 14-E004-EF. Rockville, MD: Agency for Healthcare Research and Quality. September 2014.]. However, because research has been conducted on vitamin D and numerous health outcomes, this section focuses on seven diseases, conditions, and interventions in which vitamin D might be involved: bone health and osteoporosis, cancer, cardiovascular disease (CVD), depression, multiple sclerosis (MS), type 2 diabetes, and weight loss. Most of the studies described in this section measured serum 25(OH)D levels using various methods that were not standardized by comparing them to the best methods. Use of unstandardized 25(OH)D measures can raise questions about the accuracy of the results and about the validity of conclusions drawn from studies that use such measures and, especially, from meta-analyses that pool data from many studies that use different unstandardized measures [55. Sempos CT, Heijboer AC, Bikle DD, Bollerslev J, Bouillon R, Brannon PM, et al. Vitamin D assays and the definition of hypovitaminosis D. Results from the First International Conference on Controversies in Vitamin D. Br J Clin Pharmacol 2018;84:2194-207. [PubMed abstract],99. Sempos CT, Binkley N. 25-hydroxyvitamin D assay standardisation and vitamin D guidelines paralysis. Public Health Nutrition 2020;23:1153-64. [PubMed abstract],7070. Sempos CT, Carter GD, Binkley NC. 25-hydroxyvitamin D assays: Standardization, guidelines, problems, and interpretation. Pages 939-57 in Feldman D, Pike JW, Bouillon R, Giovannucci E, Goltzman D, Hewison M, eds. Vitamin D, Volume 1: Biochemistry, Physiology and Diagnostics, Fourth Edition. Elsevier, 2018.]. More information about assay standardization is available from the Vitamin D Standardization Program webpage. Bone health and osteoporosis Osteoporosis is characterized by low bone mass and the deterioration of bone tissue, which increases bone fragility and the risk of fractures. Clinical trials have shown that vitamin D and calcium supplements may increase bone mineral density in some postmenopausal women and older men, but it is not clear whether they reduce falls and fracture rates. In addition, the results of studies that have evaluated the effects of supplemental vitamin D on muscle strength and the rate of decline in muscle function have been inconsistent. Bone is constantly being remodeled. However, as people age—and particularly in women during menopause—bone breakdown rates overtake rates of bone building. Over time, bone density can decline, and osteoporosis can eventually develop [7171. Jin, J. Vitamin D and calcium supplements for preventing fractures. JAMA 2018;319:1630. [PubMed abstract]]. More than 53 million adults in the United States have or are at risk of developing osteoporosis, which is characterized by low bone mass and structural deterioration of bone tissue that increases bone fragility and the risk of bone fractures [7272. National Institutes of Health Osteoporosis and Related Bone Diseases National Resource Center.Osteoporosis Overview.]. About 2.3 million osteoporotic fractures occurred in the United States in 2015 [7373. Hansen D, Bazell C, Pelizzari P, Pyenson B. Medicare cost of osteoporotic fractures. Milliman research report, August 2019.]. Osteoporosis is, in part, a long-term effect of calcium and/or vitamin D insufficiency, in contrast to rickets and osteomalacia, which result from vitamin D deficiency. Osteoporosis is most often associated with inadequate calcium intakes, but insufficient vitamin D intakes contribute to osteoporosis by reducing calcium absorption [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Bone health also depends on support from the surrounding muscles to assist with balance and postural sway and thereby reduce the risk of falling. Vitamin D is also needed for the normal development and growth of muscle fibers. In addition, inadequate vitamin D levels can adversely affect muscle strength and lead to muscle weakness and pain (myopathy) [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. Most trials of the effects of vitamin D supplements on bone health also included calcium supplements, so isolating the effects of each nutrient is difficult. In addition, studies provided different amounts of nutrients and used different dosing schedules. Clinical trial evidence on older adults Among postmenopausal women and older men, many clinical trials have shown that supplements of both vitamin D and calcium result in small increases in bone mineral density throughout the skeleton [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,7474. Chung M, Balk EM, Brendel M, Ip S, Lau J, Lee J, et al. Vitamin D and calcium: A systematic review of health outcomes. Evidence Report/Technology Assessment No. 183 prepared by the Tufts Evidence-based Practice Center under Contract No. 290-2007-10055-I. AHRQ Publication No. 09-E015. Rockville, MD: Agency for Healthcare Research and Quality, 2009.]. They also help reduce fracture rates in institutionalized older people. However, the evidence on the impact of vitamin D and calcium supplements on fractures in community-dwelling individuals is inconsistent. The USPSTF evaluated 11 randomized clinical trials of vitamin D and/or calcium supplementation in a total of 51,419 healthy, community-dwelling adults age 50 years and older who did not have osteoporosis, vitamin D deficiency, or prior fractures [7575. U.S. Preventive Services Task Force. Vitamin D, calcium, or combined supplementation for the primary prevention of fractures in community-dwelling adults. US Preventive Services Task Force recommendation statement. JAMA 2018;319:1592-9. [PubMed abstract],7676. Kahwati LC, Weber RP, Pan H, Gourlay M, LeBlanc E, Coker-Schwimmer M, Viswanathan M. Vitamin D, calcium, or combined supplementation for the primary prevention of fractures in community-dwelling adults: Evidence report and systematic review for the US Preventive Services Task Force. JAMA 2018;319:1600-12. [PubMed abstract]]. It concluded that the current evidence was insufficient to evaluate the benefits and harms of supplementation to prevent fractures. In addition, the USPSTF recommended against supplementation with 10 mcg (400 IU) or less of vitamin D and 1,000 mg or less of calcium to prevent fractures in this population, but it could not determine the balance of benefits and harms from higher doses. The USPSTF also reviewed the seven published studies on the effects of vitamin D supplementation (two of them also included calcium supplementation) on the risk of falls in community-dwelling adults age 65 years or older who did not have osteoporosis or vitamin D deficiency. It concluded with moderate certainty that vitamin D supplementation does not reduce the numbers of falls or injuries, such as fractures, resulting from falls [7777. Guirguis-Blake JM, Michael YL, Perdue LA, Coppola EL, Beil TL. Interventions to prevent falls in older adults: Updated evidence report and systematic review for the US Preventive Services Task Force. JAMA 2018;319:1705-16. [PubMed abstract],7878. U.S. Preventive Services Task Force. Interventions to prevent falls in community-dwelling older adults. US Preventive Services Task Force recommendation statement. JAMA 2018;319:1696-1704. [PubMed abstract]]. Another recent systematic review also found that vitamin D and calcium supplements had no beneficial effects on fractures, falls, or bone mineral density [7979. Bolland MJ, Grey A, Avenell A. Effects of vitamin D supplementation on musculoskeletal health: A systematic review, meta-analysis, and trial sequential analysis. Lancet Diabetes Endocrinol 2018;6:847-58. [PubMed abstract],8080. Gallagher JC. Vitamin D and bone density, fractures, and falls: The end of the story? Lancet Diabetes Endocrinol 2018;6:834-5. [PubMed abstract]]. In contrast, a meta-analysis of six trials in 49,282 older adults found that daily vitamin D (10 or 20 mcg [400 IU or 800 IU]/day) and calcium (800 or 1,200 mg/day) supplementation for a mean of 5.9 years reduced the risk of any fracture by 6% and of hip fracture by 16% [8181. Yao P, Bennett D, Mafham M, Lin X, Chen Z, Armitage J, Clarke R. Vitamin D and calcium for the prevention of fracture: A systematic review and meta-analysis. JAMA Network Open 2019;2(12):e1917789. doi: 10.1001/jamanetworkopen.2019.17789.]. One systematic review and meta-analysis of 11 randomized, controlled trials published through 2018 of vitamin D supplementation alone (10–20 mcg [400–800 IU]/day or more at least every week or as rarely as once a year) for 9 months to 5 years found that the supplements provided no protection from fractures in 34,243 older adults [8181. Yao P, Bennett D, Mafham M, Lin X, Chen Z, Armitage J, Clarke R. Vitamin D and calcium for the prevention of fracture: A systematic review and meta-analysis. JAMA Network Open 2019;2(12):e1917789. doi: 10.1001/jamanetworkopen.2019.17789.]. More recently, a 2022 ancillary study of the Vitamin D and Omega-3 Trial (VITAL; described in the Cancer section below) investigated whether supplemental vitamin D 3 (50 mcg [2,000 IU]/day) would lower the risk of fractures in 25,871 generally healthy men age 50 years and older and women age 55 years and older over a median follow-up of 5.3 years [8282. LeBoff MS, Chou SH, Ratliff KA, Cook NR, Khurana B, Kim E, et al. Supplemental vitamin D and incident fractures in midlife and older adults. N Eng J Med 2022;387:299-309. [PubMed abstract]]. The mean age of all participants was 67.1 years; 50.6% were women and 20.2% were Black. Most participants were vitamin D sufficient; at baseline, only 2.4% had serum 25(OH)D levels less than 30 nmol/L (12 ng/mL), and 12.9% less than 50 nmol/L (20 ng/mL). Vitamin D supplementation did not lower the risk of total fractures, hip fractures, or nonvertebral fractures as compared with placebo. No substantial between-group differences in fracture incidence were found by race, ethnic group, BMI, age, baseline 25(OH)D levels, or whether participants took supplemental calcium, were at high fracture risk, or had a history of fragility fractures. Vitamin D supplements for bone health in minority populations Bone mineral density, bone mass, and fracture risk are correlated with serum 25(OH)D levels in White Americans and Mexican Americans, but not in Black Americans [1414. Brown LL, Cohen B, Tabor D, Zappala G, Maruvada P, Coates PM. The vitamin D paradox in Black Americans: A systems-based approach to investigating clinical practice, research, and public health—expert panel meeting report. BMC Proceedings, 2018;12(Suppl 6):6. [PubMed abstract],8383. Aloia JF, Talwar SA, Pollack S, Yeh J. A randomized controlled trial of vitamin D3 supplementation in African American women. Arch Intern Med 2005;165:1618-23. [PubMed abstract]]. Factors such as adiposity, skin pigmentation, vitamin D binding protein polymorphisms, and genetics contribute to differences in 25(OH)D levels between Black and White Americans. One clinical trial randomized 260 Black women age 60 years and older (mean age 68.2 years) to receive 60 to 120 mcg (2,400 to 4,800 IU) per day vitamin D 3 supplementation to maintain serum 25(OH)D levels above 75 nmol/L (30 ng/mL) for 3 years [8484. Aloia JF, Rubinova R, Fazzari M, Islam S, Mikhail M, Ragolia L. Vitamin D and falls in older African American women: The PODA randomized clinical trial. J Am Geriatr Soc 2019;67:1043-49. [PubMed abstract]]. The results showed no association between 25(OH)D levels or vitamin D dose and the risk of falling in the 184 participants who completed the study. In fact, Black Americans might have a greater risk than White Americans of falls and fractures with daily vitamin D intakes of 50 mcg (2,000 IU) or more [1414. Brown LL, Cohen B, Tabor D, Zappala G, Maruvada P, Coates PM. The vitamin D paradox in Black Americans: A systems-based approach to investigating clinical practice, research, and public health—expert panel meeting report. BMC Proceedings, 2018;12(Suppl 6):6. [PubMed abstract]]. Furthermore, the bone health of older Black American women does not appear to benefit from raising serum 25(OH)D levels beyond 50 nmol/L (20 ng/mL) [8484. Aloia JF, Rubinova R, Fazzari M, Islam S, Mikhail M, Ragolia L. Vitamin D and falls in older African American women: The PODA randomized clinical trial. J Am Geriatr Soc 2019;67:1043-49. [PubMed abstract]]. Vitamin D supplements and muscle function Studies examining the effects of supplemental vitamin D on muscle strength and on rate of decline in muscle function have had inconsistent results [5555. Agency for Healthcare Research and Quality. Screening for vitamin D deficiency: Systematic review for the U.S. Preventive Services Task Force recommendation. Evidence Synthesis Number 118. AHRQ-Pub No. 13-05183-EF-1. June 2014.]. One recent clinical trial, for example, randomized 78 frail and near-frail adults age 65 years and older to receive 20 mcg (800 IU) vitamin D 3, 10 mcg 25(OH)D, or placebo daily for 6 months. The groups showed no significant differences in measures of muscle strength or performance [8585. Vaes AMM, Tieland M, Toussaint N, Nilwik R, Verdijk LB, van Loon LJC, de Groot CPGM. Cholecalciferol or 25-hydroxycholecalciferol supplementation does not affect muscle strength and physical performance in prefrail and frail older adults. J Nutr 2018;148:712-20. [PubMed abstract]]. Another study randomized 100 community-dwelling men and women age 60 years and older (most were White) with serum 25(OH)D levels of 50 nmol/L (20 ng/ml) or less to 800 IU vitamin D 3 or placebo for 1 year [8686. Shea MK, Fielding RA, Dawson-Hughes B. The effect of vitamin D supplementation on lower-extremity power and function in older adults: a randomized controlled trial. Am J Clin Nutr 2019;109:369-79. [PubMed abstract]]. Participants in the treatment group whose serum 25(OH)D level was less than 70 nmol/L (28 ng/ml) after 4 months received an additional 800 IU/day vitamin D 3. Despite increasing serum 25(OH)D levels to an average of more than 80 nmol/L (32 ng/ml), vitamin D supplementation did not affect lower-extremity power, strength, or lean mass. Conclusions about vitamin D supplements and bone health All adults should consume recommended amounts of vitamin D and calcium from foods and supplements if needed. Older women and men should consult their health care providers about their needs for both nutrients as part of an overall plan to maintain bone health and to prevent or treat osteoporosis. Cancer Some evidence suggests that vitamin D might inhibit carcinogenesis and slow tumor progression. However, most research has found no relationship between serum 25(OH)D levels and risk of cancer. The results of clinical trials have generally failed to show that vitamin D supplementation with or without calcium supplementation reduces the incidence of cancer. Adequate or higher 25(OH)D levels might reduce cancer mortality rates, but more research is needed to determine the effects of vitamin D supplementation in people with cancer. Laboratory and animal studies suggest that vitamin D might inhibit carcinogenesis and slow tumor progression by, for example, promoting cell differentiation and inhibiting metastasis. Vitamin D might also have anti-inflammatory, immunomodulatory, proapoptotic, and antiangiogenic effects [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.,8787. Manson JE, Bassuk SS, Buring JE. Vitamin D, calcium, and cancer: Approaching daylight? JAMA 2017;317:1217-8. [PubMed abstract]]. Observational studies and clinical trials provide mixed evidence on whether vitamin D intakes or serum levels affect cancer incidence, progression, or mortality risk. Total cancer incidence and mortality Some observational studies show associations between low serum levels of 25(OH)D and increased risks of cancer incidence and death. In a meta-analysis of 16 prospective cohort studies in a total of 137,567 participants who had 8,345 diagnoses of cancer, 5,755 participants died from cancer [8888. Yin L, Ordonez-Mena JM, Chen T, Schottker B, Arndt V, Brenner H. Circulating 25-hydroxyvitamin D serum concentration and total cancer incidence and mortality: A systematic review and meta-analysis. Preventive Medicine 2013;57:753-64. [PubMed abstract]]. A 50 nmol/L (20 ng/mL) increase in 25(OH)D levels was associated with an 11% reduction in total cancer incidence rates and, in women but not men, a 24% reduction in cancer mortality rates. A meta-analysis of prospective studies that evaluated the association between serum 25(OH)D levels and cancer incidence (8 studies) or cancer mortality (16 studies) found that cancer risk decreased by 7% and cancer mortality rates decreased by 2% with each 20 nmol/L (8 ng/mL) increase in serum 25(OH)D levels [8989. Han J, Guo X, Yu X, Liu S, Cui X, Zhang B, Liang H. 25-hydroxyvitamin D and total cancer incidence and mortality: A meta-analysis of prospective cohort studies. Nutrients 2019;11,2295; doi:10.3390/nu11102295. [PubMed abstract]]. Importantly, not all observational studies found higher vitamin D status to be beneficial, and the studies varied considerably in study populations, baseline comorbidities, and measurement of vitamin D levels. Clinical trial evidence provides some support for the observational findings. For example, three meta-analyses of clinical trial evidence found that vitamin D supplementation does not affect cancer incidence but does significantly reduce total cancer mortality rates by 12% to 13% [90-9290. Keum N, Giovannucci E. Vitamin D supplements and cancer incidence and mortality: A meta-analysis. British Journal of Cancer 2014;111:976-80. [PubMed abstract] Keum N, Lee DH, Greenwood DC, Manson JE, Giovannucci E. Vitamin D supplementation and total cancer incidence and mortality: A meta-analysis of randomized controlled trials. Ann Oncol 2019;30:733-43. [PubMed abstract] Bjelakovic G, Gluud LL, Nikolova D, Whitfield K, Krstic G, Wetterslev J, Gluud C. Vitamin D supplementation for prevention of cancer in adults. Cochrane Database Syst Rev 2014; 23(6):CD007469. doi: 10.1002/14651858.CD007469.pub2. [PubMed abstract]]. In the most recent meta-analysis, 10 randomized clinical trials (including the VITAL trial described below) that included 6,537 cancer cases provided 10 to 50 mcg (400–2,000 IU) vitamin D 3 daily (six trials) or 500 mcg (20,000 IU)/week to 12,500 mcg (500,000 IU)/year boluses of vitamin D 3 (four trials) [9191. Keum N, Lee DH, Greenwood DC, Manson JE, Giovannucci E. Vitamin D supplementation and total cancer incidence and mortality: A meta-analysis of randomized controlled trials. Ann Oncol 2019;30:733-43. [PubMed abstract]]. The study reports included 3 to 10 years of follow-up data. The vitamin D supplements were associated with serum 25(OH)D levels of 54 to 135 nmol/L (21.6–54 ng/mL). Vitamin D supplementation reduced cancer mortality rates by 13%, and most of the benefit occurred with daily supplementation. The VITAL clinical trial that investigated the effects of vitamin D supplementation on the primary prevention of cancer in the general population gave 50 mcg (2,000 IU)/day vitamin D 3 supplements with or without 1,000 mg/day marine omega-3 fatty acids or a placebo for a median of 5.3 years [9393. Manson JE, Cook NR, Lee I-M, Christen W, Bassuk S, Mora S, et al. Vitamin D supplements and prevention of cancer and cardiovascular disease. N Engl J Med 2019:380:33-44. [PubMed abstract]]. The study included 25,871 men age 50 years and older and women age 55 years and older who had no history of cancer, and most had adequate serum 25(OH)D levels at baseline. Rates of breast, prostate, and colorectal cancer did not differ significantly between the vitamin D and placebo groups. However, normal-weight participants had greater reductions in cancer incidence and mortality rates than those with overweight or obesity. The Women's Health Initiative (WHI) clinical trial randomized 36,282 postmenopausal women to receive 10 mcg (400 IU) vitamin D 3 plus 1,000 mg calcium daily or a placebo for a mean of 7 years [9494. Brunner RL, Wactawski-Wende J, Caan BJ, Cochrane BB, Chlebowski RT, et al. The effect of calcium plus vitamin D on risk for invasive cancer: results of the Women's Health Initiative (WHI) calcium plus vitamin D randomized clinical trial. Nutr Cancer. 2011;63(6):827-41. [PubMed abstract]]. Results showed no effect of supplemental vitamin D and calcium on cancer incidence or mortality during the 7-year trial. Similarly, the supplements did not affect cancer incidence over a long-term median follow-up of 22.3 years, but they did reduce cancer mortality by 7% over this period of time [9595. Thomson CA, Aragaki AK, Prentice RL, Stefanick ML, Manson JE, et al. Long-Term Effect of Randomization to Calcium and Vitamin D Supplementation on Health in Older Women : Postintervention Follow-up of a Randomized Clinical Trial. Ann Intern Med. 2024 Apr;177(4):428-438. [PubMed abstract]]. A few studies have examined the effect of vitamin D supplementation on specific cancers. Below are brief descriptions of studies of vitamin D and its association with, or effect on, breast, colorectal, lung, pancreatic, and prostate cancers. Breast cancer Some observational studies support an inverse association between 25(OH)D levels and breast cancer risk and mortality, but others do not [96-9996. McNamara M, Rosenberger KD. The significance of vitamin D status in breast cancer: A state of the science review. J Midwifery Womens Health 2019;64:276-88. [PubMed abstract] O'Brien KM, Sandler DP, Taylor JA, Weinberg CR. Serum vitamin D and risk of breast cancer within five years. Environ Health Perspect 2017;125(7):077004. [PubMed abstract] Skaaby T, Husemoen LLN, Thuesen BH, Pisinger C, Jorgensen T, Roswall N, et al. Prospective population-based study of the association between serum 25-hydroxyvitamin-D levels and the incidence of specific types of cancer. Cancer Epidemiol Biomarkers Prev 2014;23:1220-9. [PubMed abstract] Yao S, Kwan ML, Ergas IJ, Roh JM, Cheng T-YD, Hong C-C, et al. Association of serum level of vitamin D at diagnosis with breast cancer survival: A case-cohort analysis in the Pathways Study. JAMA Oncol 2017;3:351-7. [PubMed abstract]]. In the WHI clinical trial described above, vitamin D 3 and calcium supplements did not reduce breast cancer incidence, and 25(OH)D levels at the start of the study were not associated with breast cancer risk [100100. Chlebowski RT, Johnson KC, Kooperberg C, Pettinger M, Wactawski-Wende J, Rohan T, et al. Calcium plus vitamin D supplementation and the risk of breast cancer. J Natl Cancer Inst 2007;100:1581-91. [PubMed abstract]]. In a subsequent investigation for 4.9 years after the study's end, women who had taken the vitamin D and calcium supplements (many of whom continued to take them) had an 18% lower risk of in situ (noninvasive) breast cancer [101101. Cauley JA, Chlebowski RT, Wactawski-Wende J, Robbins JA, Rodabough RJ, Chen Z, et al. Calcium plus vitamin D supplementation and health outcomes five years after active intervention ended: The Women's Health Initiative. J Womens Health 2013:22,915-29. [PubMed abstract]]. However, women with vitamin D intakes higher than 15 mcg (600 IU)/day at the start of the trial and who received the supplements experienced a 28% increased risk of invasive (but not in situ) breast cancer. Colorectal cancer A large case-control study included 5,706 individuals who developed colorectal cancer and whose 25(OH)D levels were assessed a median of 5.5 years from blood draw to cancer diagnosis and 7,105 matched controls [102102. McCullough ML, Zoltick ES, Weinstein SJ, Fedirko V, Wang M, Cook NR, et al. Circulating vitamin D and colorectal cancer risk: An international pooling project of 17 cohorts. J Natl Cancer Inst 2019;111:158-69. [PubMed abstract]]. The results showed an association between 25(OH)D levels lower than 30 nmol/L (12 ng/mL) and a 31% higher colorectal cancer risk. Levels of 75 to less than 87.5 nmol/L (30 to less than 35 ng/mL) and 87.5 to less than 100 nmol/L (35 to less than 40 ng/mL) were associated with a 19% and 27% lower risk, respectively. The association was substantially stronger in women. In the WHI clinical trial described above, vitamin D 3 and calcium supplements had no effect on rates of colorectal cancer [103103. Wactawski-Wende J, Kotchen JM, Anderson GL, Assaf AR, Brunner RL, O'Sullivan MJ, et al. Calcium plus vitamin D supplementation and the risk of colorectal cancer. N Engl J Med 2006;354:684-96. [PubMed abstract]]. In a subsequent investigation for 4.9 years after the study's end, women who had taken the vitamin D and calcium supplements (many of whom continued to take them) still had the same colorectal cancer risk as those who received placebo [101101. Cauley JA, Chlebowski RT, Wactawski-Wende J, Robbins JA, Rodabough RJ, Chen Z, et al. Calcium plus vitamin D supplementation and health outcomes five years after active intervention ended: The Women's Health Initiative. J Womens Health 2013:22,915-29. [PubMed abstract]]. Another study included 2,259 healthy individuals age 45 to 75 years who had had one or more serrated polyps (precursor lesions to colorectal cancer) that had been removed [104104. Crockett SD, Barry EL, Mott LA, Ahnen DJ, Robertson DJ, Anderson JC, et al. Calcium and vitamin D supplementation and increased risk of serrated polyps: Results from a randomised clinical trial. Gut. 2019 Mar;68(3):475-486. [PubMed abstract]]. These participants were randomized to take 25 mcg (1,000 IU) vitamin D 3, 1,200 mg calcium, both supplements, or a placebo daily for 3 to 5 years, followed by an additional 3 to 5 years of observation after participants stopped the treatment. Vitamin D alone did not significantly affect the development of new serrated polyps, but the combination of vitamin D with calcium increased the risk almost fourfold. The VITAL trial found no association between vitamin D supplementation and the risk of colorectal adenomas or serrated polyps [105105. Song M, Lee IM, Manson JE, Buring JE, Dushkes R, Gordon D, et al. No association between vitamin D supplementation and risk of colorectal adenomas or serrated polyps in a randomized trial. Clin Gastroeterol Hepatol 2020. [PubMed abstract]]. Lung cancer A study of cohorts that included 5,313 participants who developed lung cancer and 5,313 matched controls found no association between serum 25(OH)D levels and risk of subsequent lung cancer, even when the investigators analyzed the data by sex, age, race and ethnicity, and smoking status [106106. Muller DC, Hodge AM, Fanidi A, Albanes D, Mai XM, Shu XO, et al. No association between circulating concentrations of vitamin D and risk of lung cancer: An analysis in 20 prospective studies in the Lung Cancer Cohort Consortium (LC3). Ann Oncol 2018;29:1468-75. [PubMed abstract]]. Pancreatic cancer One study comparing 738 men who developed pancreatic cancer to 738 matched controls found no relationship between serum 25(OH)D levels and risk of pancreatic cancer [107107. van Duijnhoven FJB, Jenab M, Hveem K, Siersema PD, Fedirko V, Duell EJ, et al. Circulating concentrations of vitamin D in relation to pancreatic cancer risk in European populations. Int J Cancer 2018;142:1189-201. [PubMed abstract]]. Another study that compared 200 male smokers in Finland with pancreatic cancer to 400 matched controls found that participants in the highest quintile of 25(OH)D levels (more than 65.5 nmol/L [26.2 ng/mL]) had a threefold greater risk of developing pancreatic cancer over 16.7 years than those in the lowest quintile (less than 32 nmol/L [12.8 ng/mL]) [108108. Stolzenberg-Solomon RZ, Vieth R, Azad A, Pietinen P, Taylor PR, Virtamo J, et al. A prospective nested case-control study of vitamin D status and pancreatic cancer risk in male smokers. Cancer Res 2006;66:10213-9. [PubMed abstract]]. An investigation that pooled data from 10 studies of cancer in 12,205 men and women found that concentrations of 25(OH)D greater than 75 nmol/L (30 ng/mL) but less than 100 nmol/L (40 ng/mL) did not reduce the risk of pancreatic cancer. However, the results did show an increased risk of pancreatic cancer with 25(OH)D levels of 100 nmol/L (40 ng/mL) or above [109109. Helzlsouer KJ for the VDPP Steering Committee. Overview of the Cohort Consortium Vitamin D Pooling Project of Rarer Cancers. Am J Epidemiol 2010;172:4-9. [PubMed abstract]]. Prostate cancer Research to date provides mixed evidence on whether levels of 25(OH)D are associated with the development of prostate cancer. Several studies published in 2014 suggested that high levels of 25(OH)D might increase the risk of prostate cancer. For example, a meta-analysis of 21 studies that included 11,941 men with prostate cancer and 13,870 controls found a 17% higher risk of prostate cancer for participants with higher levels of 25(OH)D [110110. Xu Y, Shao X, Yao Y, Xu L, Chang L, Jiang Z, Lin Z. Positive association between circulating 25-hydroxyvitamin D levels and prostate cancer risk: New findings from an updated meta-analysis. J Cancer Res Clin Oncol 2014;140:1465-77. [PubMed abstract]]. What constituted a higher level varied by study but was typically at least 75 nmol/L (30 ng/mL). In a cohort of 4,733 men, of which 1,731 had prostate cancer, those with 25(OH)D levels of 45 to 70 nmol/L (18–28 ng/mL) had a significantly lower risk of the disease than men with either lower or higher values [111111. Kristal AR, Till C, Song X, Tangen CM, Goodman PJ, Neuhauser ML, et al. Plasma vitamin D and prostate cancer risk: Results from the Selenium and Vitamin E Cancer Prevention Trial. Cancer Epidemiol Biomarkers Prev 2014;23:1494-504. [PubMed abstract]]. This U-shaped association was most pronounced for men with the most aggressive forms of prostate cancer. A case-control analysis of 1,695 cases of prostate cancer and 1,682 controls found no associations between 25(OH)D levels and prostate cancer risk [112112. Schenk JM, Till CA, Tangen CM, Goodman PJ, Song X, Torkko KC, et al. Serum 25-hydroxyvitamin D concentrations and risk of prostate cancer: Results from the Prostate Cancer Prevention Trial. Cancer Epidemiol Biomarkers Prev 2014;23:1484-93. [PubMed abstract]]. However, higher serum 25(OH)D levels (at a cut point of 75 nmol/L [30 ng/mL]) were linked to a modestly higher risk of slow-growth prostate cancer and a more substantial lower risk of aggressive disease. Since 2014, however, several published studies and meta-analyses have found no relationship between 25(OH)D levels and prostate cancer risk [113113. Heath AK, Hodge AM, Ebeling PR, Eyles DW, Kvaskoff D, Buchanan DD, et al. Circulating 25-hydroxyvitamin D concentration and risk of breast, prostate, and colorectal cancers: The Melbourne Collaborative Cohort Study. Cancer Epidemiol Biomarkers Prev 2019;28:900-8. [PubMed abstract],114114. Jiang X, Dimou NL, Al-Dabhani K, Lewis SJ, Martin RM, Haycock PC, et al. Circulating vitamin D concentrations and risk of breast and prostate cancer: A Mendelian randomization study. International Journal of Epidemiology 2019;48:1416-24. [PubMed abstract]]. For example, an analysis was conducted of 19 prospective studies that provided data on prediagnostic levels of 25(OH)D for 13,462 men who developed prostate cancer and 20,261 control participants [115115. Travis RC, Perez-Cornago A, Appleby PN, Albanes D, Joshu CE, Lutsey PL, et al. A collaborative analysis of individual participant data from 19 prospective studies assesses circulating vitamin D and prostate cancer risk. Cancer Res 2019;79:274-85. [PubMed abstract]]. Vitamin D deficiency or insufficiency did not increase the risk of prostate cancer, and higher 25(OH)D concentrations were not associated with a lower risk. Several studies have examined whether levels of 25(OH)D in men with prostate cancer are associated with a lower risk of death from the disease or from any cause. One study included 1,119 men treated for prostate cancer whose plasma 25(OH)D levels were measured 4.9 to 8.6 years after their diagnosis. Among the 198 participants who died (41 deaths were due to prostate cancer), 25(OH)D levels were not associated with risk of death from prostate cancer or any cause [116116. Nair-Shalliker V, Bang A, Egger S, Clements M, Gardiner RA, Kricker A, et al. Post-treatment levels of plasma 25- and 1,25-dihydroxy vitamin D and mortality in men with aggressive prostate cancer. Scientific Reports 2020;10:7736. [PubMed abstract]]. However, a meta-analysis of seven cohort studies that included 7,808 men with prostate cancer found higher 25(OH)D levels to be significantly associated with lower mortality rates from prostate cancer or any other cause [117117. Song Z-y, Yao Q, Zhuo Z, Ma Z, Chen G. Circulating vitamin D level and mortality in prostate cancer patients: A dose-response meta-analysis. Endocrine Connections 2018;7:R294-303. [PubMed abstract]]. A dose-response analysis found that each 20 nmol/L [8 ng/mL] increase in 25(OH)D was associated with a 9% lower risk of both all-cause and prostate cancer-specific mortality. For men with prostate cancer, whether vitamin D supplementation lengthens cancer-related survival is not clear. A meta-analysis of three randomized controlled trials in 1,273 men with prostate cancer found no significant differences in total mortality rates between those receiving vitamin D supplementation (from 10 mcg [400 IU]/day for 28 days to 45 mcg [1,800 IU] given in three doses total at 2-week intervals) and those receiving a placebo [118118. Shahvazi S, Soltani S, Ahmadi SM, de Souza RJ, Salehi-Abargouei A. The effect of vitamin D supplementation on prostate cancer: A systematic review and meta-analysis of clinical trials. Horm Metab Res 2019;51:11-21. [PubMed abstract]]. Conclusions about vitamin D and cancer The USPSTF stated that, due to insufficient evidence, it was unable to assess the balance of benefits and harms of supplemental vitamin D to prevent cancer [119119. US Preventive Services Task Force; Mangione CM, Barry MJ, Nicholson WK, Cabana M, et al. Vitamin, Mineral, and Multivitamin Supplementation to Prevent Cardiovascular Disease and Cancer: US Preventive Services Task Force Recommendation Statement. JAMA. 2022 Jun 21;327(23):2326-2333. [PubMed abstract]]. Taken together, studies to date do not indicate that vitamin D with or without calcium supplementation reduces the incidence of cancer, but adequate or higher 25(OH)D levels might reduce cancer mortality rates. Further research is needed to determine whether vitamin D inadequacy increases cancer risk, whether greater exposure to the nutrient can prevent cancer, and whether some individuals could have an increased risk of cancer because of their vitamin D status over time. Cardiovascular disease Vitamin D has been linked to heart health and the risk of cardiovascular disease and vitamin D deficiency is associated with vascular dysfunction, arterial stiffening, left ventricular hypertrophy, and hyperlipidemia. While data from observational studies support an association between higher serum levels of 25(OH)D and a lower risk of cardiovascular disease incidence and mortality, clinical trials have not confirmed these results; several trials have reported that vitamin D supplementation did not reduce the risk of cardiovascular disease, even among people with low 25(OH)D status. Vitamin D helps regulate the renin-angiotensin-aldosterone system (and thereby blood pressure), vascular cell growth, and inflammatory and fibrotic pathways [120120. Kassi E, Adamopoulos C, Basdra EK, Papavassiliou AG. Role of vitamin D in atherosclerosis. Circulation 2013;128:2517-31. [PubMed abstract]]. Vitamin D deficiency is associated with vascular dysfunction, arterial stiffening, left ventricular hypertrophy, and hyperlipidemia [121121. Mheid IA, Quyyumi AA. Vitamin D and cardiovascular disease: Controversy unresolved. J Am Coll Cardiol 2017;70:89-100. [PubMed abstract]]. For these reasons, vitamin D has been linked to heart health and risk of CVD. Observational studies support an association between higher serum 25(OH)D levels and a lower risk of CVD incidence and mortality. For example, a meta-analysis included 34 observational studies that followed 180,667 participants (mean age greater than 50 years) for 1.3 to more than 32 years. The results showed that baseline serum 25(OH)D levels were inversely associated with total number of CVD events (including myocardial infarction, ischemic heart disease, heart failure, and stroke) and mortality risk [122122. Zhang R, Li B, Gao X, Tian R, Pan Y, Jiang Y, et al. Serum 25-hydroxyvitamin D and the risk of cardiovascular disease: Dose-response meta-analysis of prospective studies. Am J Clin Nutr 2017;105:810-9. [PubMed abstract]]. Overall, the risk of CVD events was 10% lower for each 25 nmol/L (10 ng/mL) increase in serum 25(OH)D. Another large observational study that followed 247,574 adults from Denmark for 0 to 7 years found that levels of 25(OH)D that were low (about 12.5 nmol/L [5 ng/mL]) and high (about 125 nmol/L [50 ng/mL]) were associated with a greater risk of mortality from CVD, stroke, and acute myocardial infarction [123123. Durup D, Jorgensen HL, Christensen J, Tjonnland A, Olsen A, Halkjaer J, et al. A reverse J-shaped association between serum 25-hydroxyvitamin D and cardiovascular disease mortality: The CopD study. J Clin Endorcinol Metab 2015;100:2339-46. [PubMed abstract]]. Other meta-analyses of prospective studies have found associations between lower vitamin D status measured by serum 25(OH)D levels or vitamin D intakes and an increased risk of ischemic stroke, ischemic heart disease, myocardial infarction, and early death [124124. Brondum-Jacobsen P, Benn M, Jensen GB, Nordestgaard BG. 25-hydroxyvitamin D levels and risk of ischemic heart disease, myocardial infarction, and early death: Population-based study and meta-analyses of 18 and 17 studies. Arterioscler Thromb Vasc Biol 2012;32:2794-802. [PubMed abstract],125125. Zhou R, Wang M, Huang H, Li W, Hu Y, Wu T. Lower vitamin D status is associated with an increased risk of ischemic stroke: A systematic review and meta-analysis. Nutrients 2018; 10, 277;doi:10.3390/nu10030277. [PubMed abstract]]. In contrast to the observational studies, clinical trials have provided little support for the hypothesis that supplemental vitamin D reduces the risk of CVD or CVD mortality. For example, a 3-year trial in New Zealand randomized 5,110 adults (mean age 65.9 years) to a single dose of 5,000 mcg (200,000 IU) vitamin D 3 followed by 2,500 mcg (100,000 IU) each month or a placebo for a median of 3.3 years [126126. Scragg R, Stewart AW, Waayer D, Lawes CMM, Toop L, Sluyter J, et al. Effect of monthly high-dose vitamin D supplementation on cardiovascular disease in the Vitamin D Assessment Study: A randomized clinical trial. JAMA Cardiol 2017;2:608-16. [PubMed abstract]]. Vitamin D supplementation had no effect on the incidence rate of myocardial infarction, angina, heart failure, arrhythmia, arteriosclerosis, stroke, venous thrombosis, or death from CVD. Similarly, the VITAL clinical trial described above found that vitamin D supplements did not significantly decrease rates of heart attacks, strokes, coronary revascularization, or deaths from cardiovascular causes [9393. Manson JE, Cook NR, Lee I-M, Christen W, Bassuk S, Mora S, et al. Vitamin D supplements and prevention of cancer and cardiovascular disease. N Engl J Med 2019:380:33-44. [PubMed abstract]]. Moreover, the effects did not vary by baseline serum 25(OH)D levels or whether participants took the trial’s omega-3 supplement in addition to vitamin D. However, another clinical trial designed to investigate bone fracture risk found that 20 mcg (800 IU)/day vitamin D 3 (with or without calcium) or a placebo in 5,292 adults age 70 years and older for a median of 6.2 years offered protection from cardiac failure, but not myocardial infarction or stroke [127127. Ford JA, MacLennan GS, Avenell A, Bolland M, Grey A, Witham M. Cardiovascular disease and vitamin D supplementation: Trial analysis, systematic review, and meta-analysis. Am J Clin Nutr 2014;100:746-55. [PubMed abstract]]. In the WHI clinical trial described above, daily supplementation with 10 mcg (400 IU) vitamin D 3 and 1,000 mg calcium did not affect cardiovascular disease risk during the 7-year trial [128128. Hsia J, Heiss G, Ren H, Allison M, Dolan NC, et al. Calcium/vitamin D supplementation and cardiovascular events. Circulation. 2007 Feb 20;115(7):846-54. [PubMed abstract]] or during a long-term median follow-up of 22.3 years [9595. Thomson CA, Aragaki AK, Prentice RL, Stefanick ML, Manson JE, et al. Long-Term Effect of Randomization to Calcium and Vitamin D Supplementation on Health in Older Women : Postintervention Follow-up of a Randomized Clinical Trial. Ann Intern Med. 2024 Apr;177(4):428-438. [PubMed abstract]]. However, during this long-term follow-up period, women who received the supplements had a 6% higher risk of death from cardiovascular disease than those who received placebo [9595. Thomson CA, Aragaki AK, Prentice RL, Stefanick ML, Manson JE, et al. Long-Term Effect of Randomization to Calcium and Vitamin D Supplementation on Health in Older Women : Postintervention Follow-up of a Randomized Clinical Trial. Ann Intern Med. 2024 Apr;177(4):428-438. [PubMed abstract]]. High serum cholesterol levels and hypertension are two of the main risk factors for CVD. The data on supplemental vitamin D and cholesterol levels are mixed, as shown in one meta-analysis of 41 clinical trials in a total of 3,434 participants (mean age 55 years). The results of this analysis showed that 0.5 mcg (20 IU) to 214 mcg (8,570 IU)/day vitamin D supplementation (mean of 2,795 IU) for 6 weeks to 3 years reduced serum total cholesterol, low-density lipoprotein cholesterol, and triglyceride levels, but not high-density lipoprotein cholesterol levels [129129. Dibaba DT. Effect of vitamin D supplementation on serum lipid profiles: A systematic review and meta-analysis. Nutr Rev 2019;77:890-902. [PubMed abstract]]. Studies of the effects of vitamin D supplements on hypertension have also had mixed findings. In one meta-analysis of 46 clinical trials that included 4,541 participants, vitamin D supplements (typically 40 mcg [1,600 IU]/day or less) for a minimum of 4 weeks had no significant effects on systolic or diastolic blood pressure [130130. Beveridge LA, Struthers AD, Khan F, Jorde R, Scragg R, Macdonald HM, et al. Effect of vitamin D supplementation on blood pressure: A systematic review and meta-analysis incorporating individual patient data. JAMA Intern Med 2015;175:745-54. [PubMed abstract]]. In contrast, another meta-analysis of 30 clinical trials in 4,744 participants (mean age 54.5 years) that administered 5 mcg (200 IU) to 300 mcg (12,000 IU)/day vitamin D 3 for a mean of 5.6 months showed that more than 20 mcg (800 IU)/day significantly reduced systolic and diastolic blood pressure in normal-weight participants who had hypertension [131131. Golzarand M, Shab-Bidar S, Koochakpoor G, Speakman JR, Djafarian K. Effect of vitamin D3 supplementation on blood pressure in adults: An updated meta-analysis. Nutr Metab Cardiovasc Dis 2016;26:663-73. [PubMed abstract]]. However, more than 20 mcg (800 IU)/day vitamin D 3, when taken with calcium supplements, significantly increased blood pressure in participants with overweight and obesity. Another meta-analysis of genetic studies in 146,581 participants (primarily adults) found that a low vitamin D status increased blood pressure and hypertension risk in people with genetic variants associated with low endogenous production of 25(OH)D [132132. Vimaleswaran KS, Cavadino A, Berry DJ, Jorde R, Dieffenbach AK, Lu C, et al. Association of vitamin D status with arterial blood pressure and hypertension risk: A mendelian randomisation study. Lancet Diabetes-Endocrinol 2014;2:719-29. [PubMed abstract]]. Overall, clinical trials show that vitamin D supplementation does not reduce CVD risk, even for people with low 25(OH)D status (below 20 nmol/L [12 ng/mL]) at baseline [9393. Manson JE, Cook NR, Lee I-M, Christen W, Bassuk S, Mora S, et al. Vitamin D supplements and prevention of cancer and cardiovascular disease. N Engl J Med 2019:380:33-44. [PubMed abstract],126126. Scragg R, Stewart AW, Waayer D, Lawes CMM, Toop L, Sluyter J, et al. Effect of monthly high-dose vitamin D supplementation on cardiovascular disease in the Vitamin D Assessment Study: A randomized clinical trial. JAMA Cardiol 2017;2:608-16. [PubMed abstract]]. Depression Vitamin D may be involved in the pathophysiology of depression, and low levels of 25(OH)D have been associated with a higher risk of depression. However, vitamin D supplementation has not been shown to prevent depression or treat depressive symptoms in clinical trials. Vitamin D receptors are present on neurons and glia in areas of the brain that are thought to be involved in the pathophysiology of depression [133133. Anglin RES, Samaan Z, Walter SD, McDonald SD. Vitamin D deficiency and depression in adults: Systematic review and meta-analysis. The British Journal of Psychiatry 2013;202:100-7. [PubMed abstract]]. A systematic review and meta-analysis of 14 observational studies that included a total of 31,424 adults (mean age ranging from 27.5 to 77 years) found an association between deficient or low levels of 25(OH)D and depression [133133. Anglin RES, Samaan Z, Walter SD, McDonald SD. Vitamin D deficiency and depression in adults: Systematic review and meta-analysis. The British Journal of Psychiatry 2013;202:100-7. [PubMed abstract]]. Clinical trials, however, do not support these findings. For example, a meta-analysis of nine trials with a total of 4,923 adult participants diagnosed with depression or depressive symptoms found no significant reduction in symptoms after supplementation with vitamin D [134134. Gowda U, Mutowo MP, Smith BJ, Wluka AE, Renzaho AMN. Vitamin D supplementation to reduce depression in adults: Meta-analysis of randomized controlled trials. Nutrition 2015;31:421-9. [PubMed abstract]]. The trials administered different amounts of vitamin D (ranging from 10 mcg [400 IU]/day to 1,000 mcg [40,000 IU]/week). They also had different study durations (5 days to 5 years), mean participant ages (range, 22 years to 75 years), and baseline 25(OH)D levels; furthermore, some but not all studies administered concurrent antidepressant medications. Three trials conducted since that meta-analysis also found no effect of vitamin D supplementation on depressive symptoms. One trial included 206 adults (mean age 52 years) who were randomized to take a bolus dose of 2,500 mcg (100,000 IU) vitamin D 3 followed by 500 mcg (20,000 IU)/week or a placebo for 4 months [135135. Jorde R, Kubiak J. No improvement in depressive symptoms by vitamin D supplementation: Results from a randomised controlled trial. Journal of Nutrition Science 2018;7:1-7. [PubMed abstract]]. Most participants had minimal or mild depression, had a low mean baseline 25(OH) level of 33.8 nmol/L (13.5 ng/mL), and were not taking antidepressants. The second trial included 155 adults age 60 to 80 years who had clinically relevant depressive symptoms, no major depressive disorder, and serum 25(OH)D levels less than 50 to 70 nmol/L (20 to 28 ng/mL) depending on the season; in addition, they were not taking antidepressants [136136. de Koning EJ, Lips P, Penninx BWJH, Elders PJM, Heijboer AC, den Heijer M, et al. Vitamin D supplementation for the prevention of depression and poor physical function in older persons: The D-Vitaal study, a randomized clinical trial. Am J Clin Nutr 2019;110:1119-30. [PubMed abstract],137137. Jorde R, Grimnes G. Vitamin D: No cure for depression. Am J Clin Nutr 2019;110:1043-4. PMID: 31504098 [PubMed abstract]]. Participants were randomized to take either 30 mcg (1,200 IU)/day vitamin D 3 or a placebo for 1 year. In the VITAL trial described above, 16,657 men and women 50 years of age and older with no history of depression and 1,696 with an increased risk of recurrent depression (that had not been medically treated for the past 2 years) were randomized to take 50 mcg (2,000 IU)/day vitamin D 3 (with or without fish oil) or a placebo for a median of 5.3 years [138138. Okereke OI, Reynolds III CF, Mischoulon D, Chang G, Vyas CM, Cook NR, et al. Effect of long-term vitamin D3 supplementation vs placebo on risk of depression or clinically relevant depressive symptoms and on change in mood scores: A randomized clinical trial. JAMA 2020;324:471-80. [PubMed abstract]]. The groups showed no significant differences in the incidence and recurrent rates of depression, clinically relevant depressive symptoms, or changes in mood scores. Overall, clinical trials did not find that vitamin D supplements helped prevent or treat depressive symptoms or mild depression, especially in middle-age to older adults who were not taking prescription antidepressants. No studies have evaluated whether vitamin D supplements may benefit individuals under medical care for clinical depression who have low or deficient 25(OH)D levels and are taking antidepressant medication. Multiple sclerosis Epidemiological and genetic studies have found an association between multiple sclerosis and low 25(OH)D levels. Observational studies suggest that adequate vitamin D concentrations might reduce the risk of contracting multiple sclerosis, slow the disease’s progression, and decrease the risk of relapse. Clinical trial evidence is limited, so experts cannot draw conclusions about whether vitamin D can help prevent multiple sclerosis, but vitamin D supplementation does not appear to help mitigate the signs and symptoms of the disease or reduce relapse rates. MS is an autoimmune disease of the central nervous system that damages the myelin sheath surrounding and protecting nerve cells in the brain and spinal cord. This damage hinders or blocks messages between the brain and body, leading to clinical features, such as vision loss, motor weakness, spasticity, ataxia, tremor, sensory loss, and cognitive impairment [139139. MedLinePlus. Multiple sclerosis. 2020.,140140. Jagannath VA, Filippini G, Di Pietrantonj C, Asokan GV, Robak EW, Whamond L, Robinson SA. Vitamin D for the management of multiple sclerosis (review). Cochrane Database of Systematic Reviews 2018, issue 9, Art. No.: CD008422. DOI: 10.1002/14651858.CD008422.pub3. [PubMed abstract]]. Some people with MS eventually lose the ability to write, speak, or walk. The geographical distribution of MS around the world is unequal. Few people near the equator develop the disease, whereas the prevalence is higher further north and south. This uneven distribution has led to speculation that lower vitamin D levels in people who have less sunlight exposure might predispose them to the disease [140140. Jagannath VA, Filippini G, Di Pietrantonj C, Asokan GV, Robak EW, Whamond L, Robinson SA. Vitamin D for the management of multiple sclerosis (review). Cochrane Database of Systematic Reviews 2018, issue 9, Art. No.: CD008422. DOI: 10.1002/14651858.CD008422.pub3. [PubMed abstract]]. Many epidemiological and genetic studies have shown an association between MS and low 25(OH)D levels before and after the disease begins [140140. Jagannath VA, Filippini G, Di Pietrantonj C, Asokan GV, Robak EW, Whamond L, Robinson SA. Vitamin D for the management of multiple sclerosis (review). Cochrane Database of Systematic Reviews 2018, issue 9, Art. No.: CD008422. DOI: 10.1002/14651858.CD008422.pub3. [PubMed abstract]]. Observational studies suggest that adequate vitamin D levels might reduce the risk of contracting MS and, once MS is present, decrease the risk of relapse and slow the disease's progression [141141. Sintzel MB, Rametta M, Reder AT. Vitamin D and multiple sclerosis: A comprehensive review. Neurol Ther 2018;7:59-85. [PubMed abstract]]. One study, for example, tested 25(OH)D levels in 1,092 women in Finland an average of 9 years before their MS diagnosis and compared their outcomes with those of 2,123 similar women who did not develop MS [142142. Munger K, Hongell K, Aivo J, Soilu-Hanninen M, Surcel H-M, Ascherio A. 25-hydroxyvitamin D deficiency and risk of MS among women in the Finnish Maternity Cohort. Neurology 2017;89: 1578-83. [PubMed abstract]]. More than half the women who developed MS had deficient or insufficient vitamin D levels. Women with 25(OH)D levels of less than 30 nmol/L (12 ng/mL) had a 43% higher MS risk than women with levels of 50 nmol/L (20 ng/mL) or higher. Among the women with two or more serum 25(OH)D samples taken before diagnosis (which reduced random measurement variation), a 50 nmol/L increase in 25(OH)D was associated with a 41% reduced risk of MS, and 25(OH)D levels less than 30 nmol/L were associated with an MS risk that was twice as high as levels of 50 nmol/L or higher. Two earlier prospective studies of similar design—one in the United States with 444 non-Hispanic White individuals [143143. Munger KL, Levin LI, Hollis BW, Howard NS, Ascherio A. Serum 25-hydroxyvitamin D levels and risk of multiple sclerosis. JAMA 2006;296:2832-8. [PubMed abstract]] and the other with 576 individuals in northern Sweden [144144. Salzer J, Hallmans G, Nystrom M, Stenlund H. Wadell G, Sundstrom P. Vitamin D as a protective factor in multiple sclerosis. Neurology 2012;79:2140-5. [PubMed abstract]]—found that levels of 25(OH)D greater than 99.1 nmol/L (39.6 ng/mL) and at least 75 nmol/L (30 ng/mL), respectively, were associated with a 61% to 62% lower risk of MS. No clinical trials have examined whether vitamin D supplementation can prevent the onset of MS, but several have investigated whether supplemental vitamin D can help manage the disease. A 2018 Cochrane Review analyzed 12 such trials that had a total of 933 participants with MS; the reviewers judged all of these trials to be of low quality [140140. Jagannath VA, Filippini G, Di Pietrantonj C, Asokan GV, Robak EW, Whamond L, Robinson SA. Vitamin D for the management of multiple sclerosis (review). Cochrane Database of Systematic Reviews 2018, issue 9, Art. No.: CD008422. DOI: 10.1002/14651858.CD008422.pub3. [PubMed abstract]]. Overall, vitamin D supplementation, when compared with placebo administration, had no effect on relevant clinical outcomes, such as recurrent relapse or worsened disability. Experts have reached no firm consensus on whether vitamin D can help prevent MS given the lack of clinical trial evidence [145145. Marrie RA, Beck CA. Preventing multiple sclerosis: To (take) vitamin D or not to (take) vitamin D? Neurology 2017;89:1538-9. [PubMed abstract]]. In addition, studies have not consistently shown that vitamin D supplementation tempers the signs and symptoms of active MS or reduces rates of relapse. Type 2 diabetes Observational studies have found an inverse association between vitamin D status and blood glucose levels. However, vitamin D supplementation does not appear to help maintain glucose homeostasis, reduce the risk of progression from prediabetes to type 2 diabetes, or help manage the disease, particularly in vitamin D-replete individuals. Vitamin D plays a role in glucose metabolism. It stimulates insulin secretion via the vitamin D receptor on pancreatic beta cells and reduces peripheral insulin resistance through vitamin D receptors in the muscles and liver [146146. Li X, Liu Y, Zheng Y, Wang P, Zhang Y. The effect of vitamin D supplementation on glycemic control in type 2 diabetes patients: A systematic review and meta-analysis. Nutrients 2018; 10, 375; doi:10.3390/nu10030375 [PubMed abstract]]. Vitamin D might be involved in the pathophysiology of type 2 diabetes through its effects on glucose metabolism and insulin signaling as well as its ability to reduce inflammation and improve pancreatic beta-cell function [147147. Mousa A, Naderpoor N, Teede H, Scragg R, de Courten, B. Vitamin D supplementation for improvement of chronic low-grade inflammation in patients with type 2 diabetes: A systematic review and meta-analysis of randomized controlled trials. Nutr Rev 2018;76:380-94. [PubMed abstract],148148. Pittas A, Dawson-Hughes B, Sheehan P, Ware JH, Knowler WC, Aroda VR, et al. Vitamin D supplementation and prevention of type 2 diabetes. N Engl J Med 2019;381:520-30. [PubMed abstract]]. Observational studies have linked lower serum 25(OH)D levels to an increased risk of diabetes, but their results might have been confounded by the fact that many participants were overweight or had obesity and were therefore more predisposed to developing diabetes and having lower 25(OH)D levels [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. A review of 71 observational studies in adults with and without type 2 diabetes from 16 countries found a significant inverse relationship between vitamin D status and blood sugar levels in participants who did and did not have diabetes [149149. Rafiq S, Jeppesen PB. Is hypovitaminosis D related to incidence of type 2 diabetes and high fasting glucose level in healthy subjects: A systematic review and meta-analysis of observational studies. Nutrients 2018, 10, 59; doi:10.3390/nu10010059. [PubMed abstract]]. In contrast to observational studies, clinical trials provide little support for the benefits of vitamin D supplementation for glucose homeostasis. One trial included 65 adult men and women (mean age 32 years) with overweight or obesity who were otherwise healthy, did not have diabetes, and had low serum vitamin D levels (at or below 50 nmol/L [20 ng/mL]) [150150. Mousa A, Naderpoor N, de Courten MPJ, Teede H, Kellow N, Walker K, et al. Vitamin D supplementation has no effect on insulin sensitivity or secretion in vitamin D-deficient, overweight or obese adults: A randomized placebo-controlled trial. Am J Clin Nutr 2017;105:1372-81. [PubMed abstract]]. The investigators randomly assigned participants to receive either a bolus oral dose of 2,500 mcg (100,000 IU) vitamin D 3 followed by 100 mcg (4,000 IU)/day or a placebo for 16 weeks. In the 54 participants who completed the study, vitamin D supplementation did not improve insulin sensitivity or insulin secretion in comparison with placebo. One systematic review and meta-analysis evaluated 35 clinical trials that included 43,407 adults with normal glucose tolerance, prediabetes, or type 2 diabetes who received a median of 83 mcg (3,332 IU)/day vitamin D supplements or placebo for a median of 16 weeks [151151. Seida JC, Mitri J, Colmers IN, Majumdar SR, Davidson MB, Edwards AL, et al. Effect of vitamin D3 supplementation on improving glucose homeostasis and preventing diabetes: A systematic review and meta-analysis. J Clin Endocrinol Metab 2014;99:3551-60. [PubMed abstract]]. Vitamin D had no significant effects on glucose homeostasis, insulin secretion or resistance, or hemoglobin A1c levels (a measure of average blood sugar levels over the previous 2–3 months), irrespective of the study population, vitamin D dose, or trial quality. Several trials have investigated whether vitamin D supplementation can prevent the transition from prediabetes to diabetes in patients with adequate 25(OH)D levels, and all have had negative results. In a trial in Norway, 511 men and women age 25 to 80 years (mean age 62 years) with prediabetes received 500 mcg (20,000 IU) vitamin D 3 or a placebo each week for 5 years [152152. Jorde R, Sollid ST, Svartberg J, Schirmer H, Joakimsen RM, Njolstad I, et al. Vitamin D 20 000 IU per week for five years does not prevent progression from prediabetes to diabetes. J Clin Endocrinol Metab 2016;101:1647-55. [PubMed abstract]]. The results showed no significant differences in rates of progression to type 2 diabetes; in serum glucose, insulin, or hemoglobin A1c levels; or in measures of insulin resistance. At baseline, participants had an adequate mean serum 25(OH)D level of 60 nmol/L (24 ng/mL). The largest trial to date of vitamin D supplements for diabetes prevention randomized 2,423 men and women age 25 years and older (mean age 60 years) with prediabetes and overweight or obesity (mean BMI of 32.1) to 100 mcg (4,000 IU)/day vitamin D 3 or placebo for a median of 2.5 years [148148. Pittas A, Dawson-Hughes B, Sheehan P, Ware JH, Knowler WC, Aroda VR, et al. Vitamin D supplementation and prevention of type 2 diabetes. N Engl J Med 2019;381:520-30. [PubMed abstract]]. Most participants (78%) had adequate serum levels of vitamin D at baseline (at least 50 nmol/L [20 ng/mL]). Vitamin D did not significantly prevent the development of diabetes in comparison with placebo. However, a post hoc analysis showed a 62% lower incidence of diabetes among participants with low baseline serum 25(OH)D levels (less than 30 nmol/L [12 ng/mL]) who took the vitamin D supplement than among those who took the placebo [148148. Pittas A, Dawson-Hughes B, Sheehan P, Ware JH, Knowler WC, Aroda VR, et al. Vitamin D supplementation and prevention of type 2 diabetes. N Engl J Med 2019;381:520-30. [PubMed abstract],153153. Pittas A, Dawson-Hughes B, Staten M. The authors reply: Vitamin D supplementation and prevention of type 2 diabetes. N Engl J Med 2019;381:1785-6. [PubMed abstract]]. Studies have also assessed the value of vitamin D supplementation for managing diabetes, and they have found that the vitamin offers limited benefits. One meta-analysis of 20 clinical trials compared the effects of 0.5 mcg (20 IU)/day to 1,250 mcg (50,000 IU)/week vitamin D supplementation for 2 to 6 months with those of placebo on glycemic control in 2,703 adults from around the world who had diabetes [146146. Li X, Liu Y, Zheng Y, Wang P, Zhang Y. The effect of vitamin D supplementation on glycemic control in type 2 diabetes patients: A systematic review and meta-analysis. Nutrients 2018; 10, 375; doi:10.3390/nu10030375 [PubMed abstract]]. The vitamin D reduced insulin resistance to a small but significant degree, especially in people taking more than 50 mcg (2,000 IU)/day who were vitamin D deficient at baseline, had good glycemic control, did not have obesity, and were of Middle Eastern ethnicity. However, the supplementation had no significant effects on fasting blood glucose, hemoglobin A1c, or fasting insulin levels. Clinical trials to date provide little evidence that vitamin D supplementation helps maintain glucose homeostasis, reduces the risk of progression from prediabetes to type 2 diabetes, or helps manage the disease, particularly in vitamin D-replete individuals. Weight loss Although there is some evidence that higher body weights are associated with lower vitamin D status, the available research does not support the use of vitamin D supplements to promote weight loss. Observational studies indicate that higher body weights are associated with lower vitamin D status, and individuals with obesity frequently have marginal or deficient circulating 25(OH)D levels [154154. Earthman CP, Beckman LM, Masodkar K, Sibley SD. The link between obesity and low circulating 25-hydroxyvitamin D concentrations: considerations and implications. Int J Obes (Lond) 2012;36:387-96. [PubMed abstract]]. However, clinical trials do not support a cause-and-effect relationship between vitamin D and weight loss. A systematic review and meta-analysis of 15 weight-loss intervention studies that used caloric restriction, exercise, or both, but not necessarily vitamin D supplementation or other treatments, found that people who lost weight had significantly greater increases in serum 25(OH)D levels than those who maintained their weight [155155. Mallard SR, Howe AS, Houghton LA. Vitamin D status and weight loss: A systematic review and meta-analysis of randomized and nonrandomized controlled weight-loss trials. Am J Clin Nutr 2016;104:1151-9. [PubMed abstract]]. In another study, 10 mcg (400 IU)/day vitamin D and 1,000 mg/day calcium supplementation slightly, but significantly, reduced weight gain amounts in comparison with placebo in postmenopausal women, especially those with a baseline total calcium intake of less than 1,200 mg/day [156156. Caan B, Neuhouser M, Aragaki A, Lewis CB, Jackson R, LeBoff MS, et al. Calcium plus vitamin D supplementation and the risk of postmenopausal weight gain. Arch Intern Med 2007;167:893-902. [PubMed abstract]]. However, a meta-analysis of 12 vitamin D supplementation trials (including 5 in which body composition measurements were primary outcomes) found that vitamin D supplements without calorie restriction did not affect body weight or fat mass when the results were compared with those of placebo [157157. Pathak K, Soares MJ, Calton EK, Zhao Y, Hallett J. Vitamin D supplementation and body weight status: A systematic review and meta-analysis of randomized controlled trials. Obes Rev 2014;15:528-37. [PubMed abstract]]. Overall, the available research suggests that consuming higher amounts of vitamin D or taking vitamin D supplements does not promote weight loss. Health Risks from Excessive Vitamin D Vitamin D toxicity can cause hypercalcemia, hypercalciuria, and high serum 25(OH)D concentrations; in extreme cases, it may lead to renal failure, calcification of soft tissues, cardiac arrhythmias, and death. Vitamin D toxicity is almost always a result of excessive intakes of vitamin D through supplements. Taking calcium supplements in combination with vitamin D supplements may increase the risk of certain adverse effects. The tolerable upper intake level for vitamin D ranges from 25 to 100 mcg (1,000–4,000 IU), depending on age. Excess amounts of vitamin D are toxic. Because vitamin D increases calcium absorption in the gastrointestinal tract, vitamin D toxicity results in marked hypercalcemia (total calcium greater than 11.1 mg/dL, beyond the normal range of 8.4–10.2 mg/dL), hypercalciuria, and high serum 25(OH)D levels (typically >375 nmol/l [150 ng/mL]) [158158. Galior K, Grebe S, Singh R. Development of vitamin D toxicity from overcorrection of vitamin D deficiency: A review of case reports. Nutrients 2018, 10, 953. [PubMed abstract]]. Hypercalcemia, in turn, can lead to nausea, vomiting, muscle weakness, neuropsychiatric disturbances, pain, loss of appetite, dehydration, polyuria, excessive thirst, and kidney stones. In extreme cases, vitamin D toxicity causes renal failure, calcification of soft tissues throughout the body (including in coronary vessels and heart valves), cardiac arrhythmias, and even death. Vitamin D toxicity has been caused by consumption of dietary supplements that contained excessive vitamin D amounts because of manufacturing errors, that were taken inappropriately or in excessive amounts, or that were incorrectly prescribed by physicians, [158-160158. Galior K, Grebe S, Singh R. Development of vitamin D toxicity from overcorrection of vitamin D deficiency: A review of case reports. Nutrients 2018, 10, 953. [PubMed abstract] Auguste BL, Avila-Casado C, Bargman JM. Use of vitamin D drops leading to kidney failure in a 54-year-old man. CMAJ 2019;191:E390-4. [PubMed abstract] Vogiatzi MG, Jacobson-Dickman E, DeBoer MD. Vitamin D supplementation and risk of toxicity in pediatrics: A review of current literature. J Clin Endocrinol Metab 2014;99:1132-41. [PubMed abstract]]. Experts do not believe that excessive sun exposure results in vitamin D toxicity because thermal activation of previtamin D 3 in the skin gives rise to various non-vitamin D forms that limit formation of vitamin D 3. Some vitamin D 3 is also converted to nonactive forms [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. However, frequent use of tanning beds, which provide artificial UV radiation, can lead to 25(OH)D levels well above 375 to 500 nmol/L (150–200 ng/mL) [161-163161. Singh P, Trivedi N. Tanning beds and hypervitaminosis D: A case report. Ann Intern Med 2014;160:810-1. [PubMed abstract] Laurent MR, Gielen E, Pauwels S, Vanderschueren D, Bouillon R. Hypervitaminosis D associated with tanning bed use: A case report. Ann Intern Med 2017;166:155-6. [PubMed abstract] Perez-Castrillon JL, Vega G, Abad L, Sanz A, Chaves J, Hernandez G, Duenas A. Effects of atorvastatin on vitamin D levels in patients with acute ischemic heart disease. Am J Cardiol 2007;99:903-5. [PubMed abstract]]. The combination of high intakes of calcium (about 2,100 mg/day from food and supplements) with moderate amounts of vitamin D (about 19 mcg [765 IU]/day from food and supplements) increased the risk of kidney stones by 17% over 7 years among 36,282 postmenopausal women who were randomly assigned to take 1,000 mg/day calcium and 10 mcg (400 IU)/day vitamin D or a placebo [164164. Jackson RD, LaCroix AZ, Gass M, Wallace RB, Robbins J, Lewis CE, et al. Calcium plus vitamin D supplementation and the risk of fractures. N Engl J Med 2006;354:669-82. [PubMed abstract]]. However, other, shorter (from 24 weeks to 5 years) clinical trials of vitamin D supplementation alone or with calcium in adults found greater risks of hypercalcemia and hypercalciuria, but not of kidney stones [165165. Malihi Z, Lawes CMM, Wu Z, Huang Y, Waayer D, Toop L, et al. Monthly high-dose vitamin D supplementation does not increase kidney stone risk or serum calcium: Results from a randomized controlled trial. Am J Clin Nutr 2019;109:1578-87. [PubMed abstract],166166. Malihi Z, Wu Z, Stewart AW, Lawes CMM, Scragg R. Hypercalcemia, hypercalciuria, and kidney stones in long-term studies of vitamin D supplementation: A systematic review and meta-analysis. Am J Clin Nutr 2016;104:1039-51. [PubMed abstract]]. The FNB established ULs for vitamin D in 2010 (Table 4) [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]. While acknowledging that signs and symptoms of toxicity are unlikely at daily intakes below 250 mcg (10,000 IU), the FNB noted that even vitamin D intakes lower than the ULs might have adverse health effects over time. The FNB recommended avoiding serum 25(OH)D levels above approximately 125 to 150 nmol/L (50–60 ng/mL), and it found that even lower serum levels (approximately 75–120 nmol/L [30–48 ng/mL]) are associated with increases in rates of all-cause mortality, risk of cancer at some sites (e.g., pancreas), risk of cardiovascular events, and number of falls and fractures among older adults. Table 4: Tolerable Upper Intake Levels (ULs) for Vitamin D in Micrograms (mcg) and International Units (IU) [11. Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010.]\| Age \| Male \| Female \| Pregnancy \| Lactation \| --- --- \| 0–6 months \| 25 mcg (1,000 IU) \| 25 mcg (1,000 IU) \| \| \| \| 7–12 months \| 38 mcg (1,500 IU) \| 38 mcg (1,500 IU) \| \| \| \| 1–3 years \| 63 mcg (2,500 IU) \| 63 mcg (2,500 IU) \| \| \| \| 4–8 years \| 75 mcg (3,000 IU) \| 75 mcg (3,000 IU) \| \| \| \| 9–13 years \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| \| \| \| 14–18 years \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| \| 19–50 years \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| \| 51–70 years \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| \| \| \| >70 years \| 100 mcg (4,000 IU) \| 100 mcg (4,000 IU) \| \| \| Download this table as a CSV file Interactions with Medications Vitamin D supplements may interact with medications, and some medications may affect vitamin D levels. These medications include orlistat, statins, steroids, and thiazide diuretics. Vitamin D supplements may interact with several types of medications. A few examples are provided below. Individuals taking these and other medications on a regular basis should discuss their vitamin D intakes and status with their health care providers. Orlistat The weight-loss drug orlistat (Xenical and alli), together with a reduced-fat diet, can reduce the absorption of vitamin D from food and supplements, leading to lower 25(OH)D levels [167-170167. Gotfredsen A, Westergren Hendel H, Andersen T. Influence of orlistat on bone turnover and body composition. Int J Obes Relat Metab Disord 2001;25:1154-60. [PubMed abstract] James WP, Avenell A, Broom J, Whitehead J. A one-year trial to assess the value of orlistat in the management of obesity. Int J Obes Relat Metab Disord 1997;21:S24-30. [PubMed abstract] McDuffie JR, Calis KA, Booth SL, Uwaifo GI, Yanovski JA. Effects of orlistat on fat-soluble vitamins in obese adolescents. Pharmacotherapy 2002;22:814-22. [PubMed abstract] Robien K, Oppeneer SJ, Kelly JA, Hamilton-Reeves JM. Drug-vitamin D interactions: A systematic review of the literature. Nutr Clin Pract 2013;28:194-208. [PubMed abstract]]. Statins Statin medications reduce cholesterol synthesis. Because endogenous vitamin D is derived from cholesterol, statins may also reduce vitamin D synthesis [170170. Robien K, Oppeneer SJ, Kelly JA, Hamilton-Reeves JM. Drug-vitamin D interactions: A systematic review of the literature. Nutr Clin Pract 2013;28:194-208. [PubMed abstract]]. In addition, high intakes of vitamin D, especially from supplements, might reduce the potency of atorvastatin (Lipitor), lovastatin (Altoprev and Mevacor), and simvastatin (FloLipid and Zocor), because these statins and vitamin D appear to compete for the same metabolizing enzyme [170-173170. Robien K, Oppeneer SJ, Kelly JA, Hamilton-Reeves JM. Drug-vitamin D interactions: A systematic review of the literature. Nutr Clin Pract 2013;28:194-208. [PubMed abstract] Schwartz JB. Effects of vitamin D supplementation in atorvastatin-treated patients: A new drug interaction with an unexpected consequence. Clin Pharmacol Ther 2009;85:198-203. [PubMed abstract] Perez-Castrillon JL, Vega G, Abad L, Sanz A, Chaves J, Hernandez G, Duenas A. Effects of atorvastatin on vitamin D levels in patients with acute ischemic heart disease. Am J Cardiol 2007;99:903-5. [PubMed abstract] Aloia JF, Li-Ng M, Pollack S. Statins and vitamin D. Am J Cardiol 2007;100:1329. [PubMed abstract]]. Steroids Corticosteroid medications, such as prednisone (Deltasone, Rayos, and Sterapred), are often prescribed to reduce inflammation. These medications can reduce calcium absorption and impair vitamin D metabolism [174-176174. Buckley LM, Leib ES, Cartularo KS, Vacek PM, Cooper SM. Calcium and vitamin D3 supplementation prevents bone loss in the spine secondary to low-dose corticosteroids in patients with rheumatoid arthritis. A randomized, double-blind, placebo-controlled trial. Ann Intern Med 1996;125:961-8. [PubMed abstract] de Sevaux RGL, Hoitsma AJ, Corstens FHM, Wetzels JFM. Treatment with vitamin D and calcium reduces bone loss after renal transplantation: a randomized study. J Am Soc Nephrol 2002;13:1608-14. [PubMed abstract] Lukert BP, Raisz LG. Glucocorticoid-induced osteoporosis: pathogenesis and management. Ann Intern Med 1990;112:352-64. [PubMed abstract]]. In the NHANES 2001–2006 survey, 25(OH)D deficiency (less than 25 nmol/L [10 ng/mL]) was more than twice as common among children and adults who reported oral steroid use (11%) than in nonusers (5%) [177177. Skversky AL, Kumar J, Abramowitz MK, Kaskel FJ, Melamed ML. Association of glucocorticoid use and low 25-hydroxyvitamin D levels: Results from the National Health and Nutrition Examination Survey (NHANES): 2001-2006. J Clin Endocrinol Metab 2011;96:3838-45. [PubMed abstract]]. Thiazide diuretics Thiazide diuretics (e.g., Hygroton, Lozol, and Microzide) decrease urinary calcium excretion. The combination of these diuretics with vitamin D supplements (which increase intestinal calcium absorption) might lead to hypercalcemia, especially among older adults and individuals with compromised renal function or hyperparathyroidism [170170. Robien K, Oppeneer SJ, Kelly JA, Hamilton-Reeves JM. Drug-vitamin D interactions: A systematic review of the literature. Nutr Clin Pract 2013;28:194-208. [PubMed abstract],178178. Drinka PJ, Nolten WE. Hazards of treating osteoporosis and hypertension concurrently with calcium, vitamin D, and distal diuretics. J Am Geriatr Soc 1984;32:405-7. [PubMed abstract],179179. Crowe M, Wollner L, Griffiths RA. Hypercalcaemia following vitamin D and thiazide therapy in the elderly. Practitioner 1984;228:312-3. [PubMed abstract]]. Vitamin D and Healthful Diets In general, a person’s nutritional needs should be met primarily through foods. Fortified foods and dietary supplements may be useful in cases where it is not possible to meet the needs for specific nutrients through food alone, especially during certain life stages. The Dietary Guidelines for Americans offers a general description of healthy dietary patterns. The federal government's 2020–2025 Dietary Guidelines for Americans notes that "Because foods provide an array of nutrients and other components that have benefits for health, nutritional needs should be met primarily through foods....some cases, fortified foods and dietary supplements are useful when it is not possible otherwise to meet needs for one or more nutrients (e.g., during specific life stages such as pregnancy)." For more information about building a healthy dietary pattern, refer to the Dietary Guidelines for Americans and the USDA's MyPlate. The Dietary Guidelines for Americans describes a healthy dietary pattern as one that Includes a variety of vegetables; fruits; grains (at least half whole grains); fat-free and low-fat milk, yogurt, and cheese; and oils. Milk, many ready-to-eat cereals, and some brands of yogurt and orange juice are fortified with vitamin D. Cheese naturally contains small amounts of vitamin D. Vitamin D is added to some margarines. Includes a variety of protein foods such as lean meats; poultry; eggs; seafood; beans, peas, and lentils; nuts and seeds; and soy products. Fatty fish, such as salmon, tuna, and mackerel, are very good sources of vitamin D. Beef liver and egg yolks have small amounts of vitamin D. Limits foods and beverages higher in added sugars, saturated fat, and sodium. Limits alcoholic beverages. Stays within your daily calorie needs. References Institute of Medicine, Food and Nutrition Board. Dietary Reference Intakes for Calcium and Vitamin D. Washington, DC: National Academy Press, 2010. Norman AW, Henry HH. Vitamin D. In: Erdman JW, Macdonald IA, Zeisel SH, eds. Present Knowledge in Nutrition, 10th ed. Washington DC: Wiley-Blackwell, 2012. Jones G. Vitamin D. In: Ross AC, Caballero B, Cousins RJ, Tucker KL, Ziegler TR, eds. Modern Nutrition in Health and Disease, 11th ed. Philadelphia: Lippincott Williams & Wilkins, 2014. Silva MC, Furlanetto TW. Intestinal absorption of vitamin D: A systematic review. Nutr Rev 2018;76:60-76. [PubMed abstract] Sempos CT, Heijboer AC, Bikle DD, Bollerslev J, Bouillon R, Brannon PM, et al. Vitamin D assays and the definition of hypovitaminosis D. Results from the First International Conference on Controversies in Vitamin D. Br J Clin Pharmacol 2018;84:2194-207. [PubMed abstract] LeFevre ML. 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[PubMed abstract] Pittas A, Dawson-Hughes B, Sheehan P, Ware JH, Knowler WC, Aroda VR, et al. Vitamin D supplementation and prevention of type 2 diabetes. N Engl J Med 2019;381:520-30. [PubMed abstract] Rafiq S, Jeppesen PB. Is hypovitaminosis D related to incidence of type 2 diabetes and high fasting glucose level in healthy subjects: A systematic review and meta-analysis of observational studies. Nutrients 2018, 10, 59; doi:10.3390/nu10010059. [PubMed abstract] Mousa A, Naderpoor N, de Courten MPJ, Teede H, Kellow N, Walker K, et al. Vitamin D supplementation has no effect on insulin sensitivity or secretion in vitamin D-deficient, overweight or obese adults: A randomized placebo-controlled trial. Am J Clin Nutr 2017;105:1372-81. [PubMed abstract] Seida JC, Mitri J, Colmers IN, Majumdar SR, Davidson MB, Edwards AL, et al. Effect of vitamin D3 supplementation on improving glucose homeostasis and preventing diabetes: A systematic review and meta-analysis. J Clin Endocrinol Metab 2014;99:3551-60. [PubMed abstract] Jorde R, Sollid ST, Svartberg J, Schirmer H, Joakimsen RM, Njolstad I, et al. Vitamin D 20 000 IU per week for five years does not prevent progression from prediabetes to diabetes. J Clin Endocrinol Metab 2016;101:1647-55. [PubMed abstract] Pittas A, Dawson-Hughes B, Staten M. The authors reply: Vitamin D supplementation and prevention of type 2 diabetes. N Engl J Med 2019;381:1785-6. [PubMed abstract] Earthman CP, Beckman LM, Masodkar K, Sibley SD. The link between obesity and low circulating 25-hydroxyvitamin D concentrations: considerations and implications. Int J Obes (Lond) 2012;36:387-96. [PubMed abstract] Mallard SR, Howe AS, Houghton LA. Vitamin D status and weight loss: A systematic review and meta-analysis of randomized and nonrandomized controlled weight-loss trials. Am J Clin Nutr 2016;104:1151-9. [PubMed abstract] Caan B, Neuhouser M, Aragaki A, Lewis CB, Jackson R, LeBoff MS, et al. Calcium plus vitamin D supplementation and the risk of postmenopausal weight gain. Arch Intern Med 2007;167:893-902. [PubMed abstract] Pathak K, Soares MJ, Calton EK, Zhao Y, Hallett J. Vitamin D supplementation and body weight status: A systematic review and meta-analysis of randomized controlled trials. Obes Rev 2014;15:528-37. [PubMed abstract] Galior K, Grebe S, Singh R. Development of vitamin D toxicity from overcorrection of vitamin D deficiency: A review of case reports. Nutrients 2018, 10, 953. [PubMed abstract] Auguste BL, Avila-Casado C, Bargman JM. Use of vitamin D drops leading to kidney failure in a 54-year-old man. CMAJ 2019;191:E390-4. [PubMed abstract] Vogiatzi MG, Jacobson-Dickman E, DeBoer MD. Vitamin D supplementation and risk of toxicity in pediatrics: A review of current literature. J Clin Endocrinol Metab 2014;99:1132-41. [PubMed abstract] Singh P, Trivedi N. Tanning beds and hypervitaminosis D: A case report. Ann Intern Med 2014;160:810-1. [PubMed abstract] Laurent MR, Gielen E, Pauwels S, Vanderschueren D, Bouillon R. Hypervitaminosis D associated with tanning bed use: A case report. Ann Intern Med 2017;166:155-6. [PubMed abstract] Perez-Castrillon JL, Vega G, Abad L, Sanz A, Chaves J, Hernandez G, Duenas A. Effects of atorvastatin on vitamin D levels in patients with acute ischemic heart disease. Am J Cardiol 2007;99:903-5. [PubMed abstract] Jackson RD, LaCroix AZ, Gass M, Wallace RB, Robbins J, Lewis CE, et al. Calcium plus vitamin D supplementation and the risk of fractures. N Engl J Med 2006;354:669-82. [PubMed abstract] Malihi Z, Lawes CMM, Wu Z, Huang Y, Waayer D, Toop L, et al. Monthly high-dose vitamin D supplementation does not increase kidney stone risk or serum calcium: Results from a randomized controlled trial. Am J Clin Nutr 2019;109:1578-87. [PubMed abstract] Malihi Z, Wu Z, Stewart AW, Lawes CMM, Scragg R. Hypercalcemia, hypercalciuria, and kidney stones in long-term studies of vitamin D supplementation: A systematic review and meta-analysis. Am J Clin Nutr 2016;104:1039-51. [PubMed abstract] Gotfredsen A, Westergren Hendel H, Andersen T. Influence of orlistat on bone turnover and body composition. Int J Obes Relat Metab Disord 2001;25:1154-60. [PubMed abstract] James WP, Avenell A, Broom J, Whitehead J. A one-year trial to assess the value of orlistat in the management of obesity. Int J Obes Relat Metab Disord 1997;21:S24-30. [PubMed abstract] McDuffie JR, Calis KA, Booth SL, Uwaifo GI, Yanovski JA. Effects of orlistat on fat-soluble vitamins in obese adolescents. Pharmacotherapy 2002;22:814-22. [PubMed abstract] Robien K, Oppeneer SJ, Kelly JA, Hamilton-Reeves JM. Drug-vitamin D interactions: A systematic review of the literature. Nutr Clin Pract 2013;28:194-208. [PubMed abstract] Schwartz JB. Effects of vitamin D supplementation in atorvastatin-treated patients: A new drug interaction with an unexpected consequence. Clin Pharmacol Ther 2009;85:198-203. [PubMed abstract] Perez-Castrillon JL, Vega G, Abad L, Sanz A, Chaves J, Hernandez G, Duenas A. Effects of atorvastatin on vitamin D levels in patients with acute ischemic heart disease. Am J Cardiol 2007;99:903-5. [PubMed abstract] Aloia JF, Li-Ng M, Pollack S. Statins and vitamin D. Am J Cardiol 2007;100:1329. [PubMed abstract] Buckley LM, Leib ES, Cartularo KS, Vacek PM, Cooper SM. Calcium and vitamin D3 supplementation prevents bone loss in the spine secondary to low-dose corticosteroids in patients with rheumatoid arthritis. A randomized, double-blind, placebo-controlled trial. Ann Intern Med 1996;125:961-8. [PubMed abstract] de Sevaux RGL, Hoitsma AJ, Corstens FHM, Wetzels JFM. Treatment with vitamin D and calcium reduces bone loss after renal transplantation: a randomized study. J Am Soc Nephrol 2002;13:1608-14. [PubMed abstract] Lukert BP, Raisz LG. Glucocorticoid-induced osteoporosis: pathogenesis and management. Ann Intern Med 1990;112:352-64. [PubMed abstract] Skversky AL, Kumar J, Abramowitz MK, Kaskel FJ, Melamed ML. Association of glucocorticoid use and low 25-hydroxyvitamin D levels: Results from the National Health and Nutrition Examination Survey (NHANES): 2001-2006. J Clin Endocrinol Metab 2011;96:3838-45. [PubMed abstract] Drinka PJ, Nolten WE. Hazards of treating osteoporosis and hypertension concurrently with calcium, vitamin D, and distal diuretics. J Am Geriatr Soc 1984;32:405-7. [PubMed abstract] Crowe M, Wollner L, Griffiths RA. Hypercalcaemia following vitamin D and thiazide therapy in the elderly. Practitioner 1984;228:312-3. [PubMed abstract] Disclaimer This fact sheet by the National Institutes of Health (NIH) Office of Dietary Supplements (ODS) provides information that should not take the place of medical advice. 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13889	https://www.leonschools.net/cms/lib/FL01903265/Centricity/Domain/3141/ch.%205%20answers.pdf	Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 7 Name Class Date 5-1 Standardized Test Prep Midsegments of Triangles Gridded Response Solve each exercise and enter your answer on the grid provided. In kRST, U is the midpoint of RS, V is the midpoint of ST, and W is the midpoint of TR. 1. What is the length of RS? 2. What is the value of x? 3. What is the value of y? 4. What is the length of UW? 5. What is the length of UV ? S 2x V 11 T 3y W 15.9 R 12 U Answers 1. 2. 3. 4. 5. 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 a 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 a 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 a 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 a 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 a 2 4 5 . 5 5 . 3 1 1 1 5 9 . Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 17 Name Class Date 5-2 Standardized Test Prep Perpendicular and Angle Bisectors Multiple Choice For Exercises 1–6, choose the correct letter. Use the fi gure at the right. 1. Which ray is a bisector of /ABC? BC) BA) BD) BF) 2. What is GH? 5 10 15 25 3. What is the value of y? 2 4 16 20 4. What is m/DBE? 20 30 40 50 5. What is m/ABE? 20 30 40 60 6. If m/FBA 5 7x 1 6y, what is m/FBA? 40 44 47 60 Short Response 7. Construct the bisector of /ABC. D C E B F A I H G 3x 10y (8y 4) 5x 10 A C B B H H A D H Student construction is accurate and complete, including three arcs and a bisector labeled with two points. The construction is incorrect or there are key elements, such as one of the arcs, missing. There is no ray added to the diagram. Answers may vary. Sample: D Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 27 Name Class Date 5-3 Standardized Test Prep Bisectors in Triangles Multiple Choice For Exercises 1–5, choose the correct letter. Use the fi gure below. 1. Which point is the incenter of nABC? X T R Y 2. Which point is the circumcenter of nABC? X T R Y 3. Which segment is an angle bisector of nABC? BX SX AS RZ 4. Which segment is a perpendicular bisector of nABC? BW SB AS RZ 5. If RC 5 x 1 3 and RA 5 3x 2 3, what is the value of x? 3 6 7 9 Extended Response 6. Draw nABC. Construct three angle bisectors. Use the point of concurrency to construct the inscribed circle. B Z C R X W T Y S A B H A A Student draws kABC, constructs three angle bisectors, and constructs the inscribed circle. Student draws kABC, constructs two angle bisectors and the inscribed circle. Student submits an incomplete but accurate drawing. Student drawing is inaccurate. Student fails to submit a drawing or constructs no angle bisectors. I Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 37 Name Class Date 5-4 Standardized Test Prep Medians and Altitudes Multiple Choice For Exercises 1–5, choose the correct letter. 1. Z is the centroid of nABC. If AZ 5 12, what is ZY? 6 12 9 18 2. What is the best description of AB? altitude perpendicular bisector median angle bisector 3. What is the best description of P? incenter centroid circumcenter orthocenter Use kXYZ for Exercises 4 and 5. 4. Which is an altitude of nXYZ? AZ XB XY ZY 5. Which is a median of nXYZ? AZ XY BX YZ Short Response 6. M is the centroid of nQRS, and QM 5 22x 1 10y. What expressions can you write for MV and QV? A B C X Y Z Z X A B M N O P X Y A B D Z Y R Q S O M V A F D F B MV 5 11x 1 5y AND QV 5 33x 1 15y MV 5 11x 1 5y OR QV 5 33x 1 15y No correct response is given. Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 47 Name Class Date 5-5 Standardized Test Prep Indirect Proof Multiple Choice For Exercises 1–5, choose the correct letter. 1. Which two statements contradict each other? II. nABC is scalene. III. nABC is right. II. nABC is isosceles. IV. nABC is acute. I and II I and IV II and III II and IV 2. nMNO is equilateral. Which is a contradiction to this statement? nMON is equilateral. m/M 5 120 nMNO is acute. m/M 5 60 3. What is the fi rst step of the following indirect proof? Given: Th e side lengths of a triangle are 4, 4, and 6. Prove: Th e triangle is not a right triangle. Assume the triangle is a right triangle. Assume the triangle is obtuse. Assume the side lengths are not 4, 4, and 6. Assume the side lengths are 4, 5, and 6. 4. MN 5 PQ. Which is a contradiction to this statement? MN 6 PQ MN > PQ MN ' PQ MN is not congruent to PQ. 5. What is the fi rst step of an indirect proof of the statement: A number x is not divisible by 5? Assume x is not divisible by 5. Assume x is divisible by 2. Assume x is divisible by 5. Assume x is prime. Short Response 6. What is the fi rst step of an indirect proof of the following statement? Explain. If a number ends in 0, then it is divisible by 5. A A B H Assume a number that ends in 0 is not divisible by 5. This is the opposite of what I want to prove. Assume the opposite of what I want to prove OR Assume a number that ends in 0 is not divisible by 5. No assumption or explanation is given. I Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 57 Name Class Date 5-6 Standardized Test Prep Inequalities in One Triangle Multiple Choice For Exercises 1–6, choose the correct letter. 1. Which of the following could be lengths of sides of a triangle? 11, 15, 27 13, 14, 32 16, 19, 34 33, 22, 55 2. nABC has the following angle measures: m/A 5 120, m/B 5 40, and m/C 5 20. Which lists the sides in order from shortest to longest? CB, BA, AC AC, BA, CB BA, AC, CB CB, AC, BA 3. nRST has the following side lengths: RS 5 7, ST 5 13, and RT 5 19. Which lists the angles in order from smallest to largest? /R, /S, /T /S, /T, /R /T, /S, /R /T, /R, /S 4. A triangle has side lengths 21 and 17. Which is a possible length for the third side? 2 4 25 39 5. Look at nLMN. Which lists the angles in order from the smallest to the largest? /L, /M, /N /N, /M, /L /M, /N, /L /M, /L, /N 6. Algebra What are the possible lengths for x, the third side of a triangle, if two sides are 13 and 7? 6 , x , 20 7 , x , 13 6 # x # 20 7 # x # 13 Short Response 7. What is the relationship between a and y? Explain. N L M 13 7 7.1 x y a a S y; the measure of an exterior angle of a triangle is greater than the measure of a remote interior angle (Corollary to the Triangle Exterior Angle Theorem). a S y OR the measure of an exterior angle of a triangle is greater than the measure of a remote interior angle (Corollary to the Triangle Exterior Angle Theorem). No relationship is stated nor explanation given. C G D D F H Prentice Hall Geometry • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved. 67 Name Class Date 5-7 Standardized Test Prep Inequalities in Two Triangles Multiple Choice For Exercises 1–5, choose the correct letter. 1. At which time is the distance between the tip of the hour hand and the tip of the minute hand on a clock the greatest? 12:00 12:10 1:30 5:25 2. What is the range of possible values for x? 2 3 , x , 24 0 , x , 48 3 2 , x , 24 x . 24 3. Which inequality relates BC and XY? BC , XY BC 5 XY BC . XY BC $ XY 4. Four pairs of identical scissors lie on a table. Scissors 1 is opened 308, scissors 2 is opened 298, scissors 3 is opened 598, and scissors 4 is opened 748. In which pair of scissors is the distance between the tips of the scissor blades greatest? scissors 1 scissors 2 scissors 3 scissors 4 5. In nABC and nDEF, AB 5 DE, CA 5 FD, and BC , EF. Which of the following must be true? m/B , m/E m/C , m/F m/A , m/D m/B 5 m/E Short Response 6. What value must x be greater than, and what value must x be less than? B D C E 45 (2x 3) 16 15 B C A Y X Z 64 55 56 (4x 12) 17 18.5 18 17 R U S T C G A I B x must be greater than 3 AND x must be less than 17. x must be greater than 3 OR x must be less than 17. No correct value for x is given.
13890	https://math.stackexchange.com/questions/2766710/arithmetic-progression-of-primes-of-length-7	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Arithmetic Progression of Primes of length 7 Ask Question Asked Modified 2 years, 5 months ago Viewed 957 times 3 $\begingroup$ Determine the least possible value of the largest term in an arithmetic progression of seven distinct primes. I don't have a great understanding of how to start the problem. I have seen approaches of searching for sequences with a difference between terms of $(2\cdot3\cdot5\cdot7)$ and then seeing what the largest number is. How would you go about starting this question? Is it possible to bound the largest term? Edit - commentators mention that we can search for multiples that differ by 30 instead of 210. Why is this the case? algebra-precalculus elementary-number-theory prime-numbers contest-math Share edited Nov 21, 2018 at 12:45 Klangen 5,51955 gold badges3737 silver badges7676 bronze badges asked May 4, 2018 at 16:51 AbeAbe 1,04055 silver badges2020 bronze badges $\endgroup$ 8 6 $\begingroup$ Have you seen this already? $\endgroup$ B. Mehta – B. Mehta 2018-05-04 16:58:06 +00:00 Commented May 4, 2018 at 16:58 3 $\begingroup$ The primes in question would be a sequence like $p,p+k,p+2k,\dots,p+6k$. By the pigeonhole principle, one of these must be divisible by $7$, or else $k$ must be divisible by $7$ (same with $2,3,5$). If we start with $p=7,k=30$ we almost make it, except for $187=11\cdot 17$. This means that we have to use $k=210=2\cdot3\cdot5\cdot7$... $\endgroup$ abiessu – abiessu 2018-05-04 17:04:33 +00:00 Commented May 4, 2018 at 17:04 $\begingroup$ @abiessu, We can try multiples of $30$ below $210$ as well, and there is in fact a lower possible value than the one you suggest. $\endgroup$ B. Mehta – B. Mehta 2018-05-04 17:18:51 +00:00 Commented May 4, 2018 at 17:18 2 $\begingroup$ You could start with $-113$ or lower and go up in steps of $30$¦ somewhere you need a positivity criterion. $\endgroup$ Macavity – Macavity 2018-05-05 04:38:45 +00:00 Commented May 5, 2018 at 4:38 1 $\begingroup$ By the way, here is an example I found: $58n^2+1$. It is not a prime arithmetic progression, really, because it includes an exponent greater than $1$, but it does have length $7$ from $n=1$. How I found it, just trial and error. Another more complex one with length $7$ is $$811+10\sum_{k=0}^n(5k+1)$$ from $n=1$. Believe it or not, they are not too difficult to find. Go here to factorize integers and see whether or not they are prime $\longrightarrow$ alpertron.com.ar/ECM.HTM $\endgroup$ Mr Pie – Mr Pie 2018-05-11 02:08:23 +00:00 Commented May 11, 2018 at 2:08 \| Show 3 more comments 1 Answer 1 Reset to default 2 $\begingroup$ I had found an AP which has all prime numbers for now: 7, 157, 307, 457, 607, 757, 907. (This IS the actual answer too, which future me will say in a few lines.) The common difference is 150. So, the answer is lesser than or equal to 907. If you wanted a more procedural method, then maybe you could find things like: The AP cannot start with any number lesser than 7 (say k), since the (k+1)th term (k + kd) = k(d + 1) is definitely divisible by k. The AP has to start with a prime greater than or equal to 7 (an implication of 1.) The common difference cannot be odd (It should be a 2 multiple). Any prime (greater than or equal to 7) mod 3 gives 1 or 2. So, if the common difference is not a 3 multiple, then at least one term would be a 3 multiple. Example: If the starting prime is of the form 3k + 1, and the common difference is of the form 3m + 2, then the second term would be a 3 multiple. You can also generalize the fourth statement to other numbers. Edit: Wait, it IS 907! You can prove this easily. Remember how I said you can generalize the fourth statement? Well, you can do that with 5 too. (That is, the common difference is a 5 multiple.) So, the common difference HAS to be a 2 3 5 = 30 multiple. After trying with 7 from common difference 30 to 150, you can see that common difference 150 works (With final number as 907). Now, we have to try with other primes. But, when we try with 11, we always get a 7 multiple in between. Why is that? Well, 7 is of the form 7m, and 30 is of the form 7k + 2. So, only the 8th term (7m + (7k + 2) 7) can be a 7 multiple. But, 11 is of the form 7m + 4, and 30 is of the form 7k + 2. So, the 5th term would be a 7 multiple. (7m + 4 + 35k + 10 = 7m + 35k + 14). The same happens to all 30 multiples from 30 to 180. It doesn't have a 7 multiple only when the common difference is a 7 multiple (210 is the least value for common difference now). The least possible end term is 11 + 210 7 = 1481, way higher than 907. And, the following primes are not 7 multiples either, so they also can have minimum common difference as 210, which still has a higher end term. So, we can see that 907 is the answer! (Note: Please notify me if I am wrong, since I can at least see what mistake I did in my calculations.) Share edited Apr 11, 2023 at 12:25 answered Apr 11, 2023 at 11:43 Ajaiy PAjaiy P 2944 bronze badges $\endgroup$ 1 $\begingroup$ This looks right to me, if a bit rambly. Also most of the observations are also in the comments to the question, as is an OEIS link that confirms the answer $907$. $\endgroup$ ronno – ronno 2023-04-11 15:10:44 +00:00 Commented Apr 11, 2023 at 15:10 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus elementary-number-theory prime-numbers contest-math See similar questions with these tags. 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13891	https://cglab.ca/~discmath/graphs-introduction.html	Discrete Mathematics Study Center Discrete Mathematics Study Center HomeCourse NotesExercisesMock ExamAbout Graphs Graphs can be used to model problems from virtually any field. A graph is a pair (V,E)(V,E) where V V is a set called the vertex set and E E is a set called the edge set. Each edge in E E describes how vertices in V V are connected. Here is an example of a graph: This particular graph is a simple graph: no edge connects a vertex to itself only one edge between two vertices In this graph, edges don't have a direction. We say that the graph is undirected. Therefore, edges can be represented as sets consisting of two vertices. For the above graph, V={San Francisco,Los Angeles,Denver,Chicago,Detroit,New York,Washington}V={San Francisco,Los Angeles,Denver,Chicago,Detroit,New York,Washington} E={{San Francisco,Denver},{San Francisco,Los Angeles},{Los Angeles,Denver},{Denver,Chicago},{Chicago,Detroit},{Detroit,New York},{New York,Washington},{Washington,Chicago},{Chicago,New York}}E={{San Francisco,Denver},{San Francisco,Los Angeles},{Los Angeles,Denver},{Denver,Chicago},{Chicago,Detroit},{Detroit,New York},{New York,Washington},{Washington,Chicago},{Chicago,New York}} Sometimes, we might want to allow multiple edges between vertices. Such graphs are called multigraphs: Edges are still undirected; we can represent them as sets of two vertices. However, now these sets can appear more than once. We say that if there are m m distinct edges between u u and v v, the edge {u,v}{u,v} has multiplicitym m. Multigraphs are used, for example, to model redundant connections in a network. We might also want to relax the restriction that there are no edges between a vertex and itself. Such edges are called self-loops and a graph that contains self-loops is called a pseudograph. Self-loops model such things as loopbacks in networks. We can also have directed versions of these graphs, where edges only go in one direction. Here is an example of a directed multigraph: Directed edges can be represented as an ordered pair (u,v)(u,v) instead of a set {u,v}{u,v}. Directed edges model such things as single duplex lines or one-way streets in road networks. There are many uses of graphs: social networks (Facebook, etc.) Hollywood graph (Six Degrees of Kevin Bacon) Web graph Representing Graphs We need a way of representing graphs if we want to perform operations on them. We could simply list the vertices and edges, but that is a lot of work and it is hard to extract much information from that representation. Here are two alternatives: adjacency list: For each vertex, list the vertices that are connected to that vertex by an edge. Such vertices are said to be adjacent. (This works for both directed and undirected graphs, even if they contain loops.) adjacency matrix: one row and one column for each vertex v 1,v 2,…,v n v 1,v 2,…,v n, row i i column j j is 1 1 if the edge is in E E, 0 0 otherwise. (For this to work, you have to fix some ordering on the vertices.) For simple undirected graphs, the adjacency matrix is symmetric (a i j=a j i a i j=a j i and the main diagonal is all 0 0 s (a i i=0 a i i=0) since no self-loops are allowed. In general, we can allow for multigraphs by using the multiplicities as entries instead of just 0 0 or 1 1. Adjacency and Degree Let G=(V,E)G=(V,E) be a graph. two vertices are adjacent (are neighbours) in G G if u u and v v are endpoints of some edge in G G if edge e e connects u u and v v, we say e e is incident onu u (and v v) or incident withu u and v v the degree of a vertex in an undirected graph is the number of edges incident with it, denoted by deg v deg⁡v. If a self-loop is present, it is counted twice! For example: The Handshaking Theorem says that, for a graph G=(V,E)G=(V,E), we always have ∑v∈V deg(v)=2\|E\|∑v∈V deg⁡(v)=2\|E\| For a directed edge (u,v)(u,v) in a directed graph, u u is the initial vertex and v v is the terminal vertex or end vertex. the number of incoming edges to v v (that is, the number of edges with v v as terminal) is denoted deg−(v)deg−⁡(v) and called the in-degree of v v the number of outgoing edges from u u (that is, the number of edges with u u as initial) is denoted deg+(u deg+⁡(u and called the out-degree of u u Note that ∑v∈V deg−(v)=∑v∈V deg+(v)=\|E\|∑v∈V deg−⁡(v)=∑v∈V deg+⁡(v)=\|E\|. Some Special Graphs There are a few (types of) graphs with special names: the complete graph on n n vertices, K n K n, has n n vertices, each of which is connected to all other vertices. From left to right, we have K 1,K 2,K 3,K 4,K 5 K 1,K 2,K 3,K 4,K 5: the cycle on n≥3 n≥3 vertices, C n C n, has vertices v 1,v 2,…,v n v 1,v 2,…,v n and edges {v 1,v 2},{v 2,v 3},…,{v n−1,v n},{v n,v 1}{v 1,v 2},{v 2,v 3},…,{v n−1,v n},{v n,v 1}. From left to right, we have C 3,C 4,C 5 C 3,C 4,C 5: the wheel on n≥3 n≥3 vertices, W n W n, takes C n C n and adds one vertex connected to all the other vertices. From left to right, we have W 3,W 4,W 5 W 3,W 4,W 5: a bipartite graphG=(V,E)G=(V,E) partitions V V into V 1 V 1 and V 2 V 2 such that V 1∪V 2=V V 1∪V 2=V and V 1∩V 2=∅V 1∩V 2=∅, and every edge in E E has one endpoint in V 1 V 1 and one endpoint in V 2 V 2. This is the same as assigning one of two colours to every vertex such that no adjacent vertices have the same colour. The complete bipartite graphK m,n K m,n has partitions of size m m and n n and every element in one partition is connected to every element of the other partition. Here are K 2,3 K 2,3 and K 3,3 K 3,3: Subgraphs A subgraph of a graph G=(V,E)G=(V,E) is a graph H=(W,F)H=(W,F) such that W⊆V W⊆V and F⊆E F⊆E. A subgraph H H of G G is a proper subgraph of G G if H≠G H≠G. For example, on the left we have K 5 K 5 and on the right is a subgraph of K 5 K 5. A subgraph is spanning if it contains all vertices of the original graph. Connectivity Sometimes, we want to know if two vertices in a graph are connected by a sequence of edges that might visit other vertices on the way (for example, can two computers on a network communicate?) A path is a sequence of edges that begins at a vertex and travels from vertex to vertex along edges of the graph. More formally, a path of length n≥0 n≥0 from vertex u u to vertex v v in G G is a sequence of n n edges e 1,e 2,…,e n e 1,e 2,…,e n of G G such that e 1={x 0=u,x 1},e 2={x 1,x 2},…,e n={x n−1,x n}e 1={x 0=u,x 1},e 2={x 1,x 2},…,e n={x n−1,x n}. if the graph is simple, we can just use the vertex sequence to label the path the path is a circuit if u=v u=v the path passes through vertices x 1,x 2,…,x n−1 x 1,x 2,…,x n−1 and traverses edges e 1,e 2,…,e n e 1,e 2,…,e n a path is simple if it does not traverse an edge more than once For example: An undirected graph is connected if there is a path between every two distinct vertices in the graph. For example, the graph on the left is connected but the graph on the right is disconnected. The different parts that are maximally connected are called connected components. One special type of connected (sub)graph is a tree, which is a connected undirected graph with no simple circuits. For example, the graph on the left is a tree; the graph in the center is not a tree because it contains a circuit; and the graph on the right is not a tree because it is not connected. Because trees are "simple" in structure, we often want to find a spanning subgraph that is a tree: In the graph above, the red edges form a spanning subgraph because every vertex appears exactly once. It is a tree because it is connected and contains no circuits. Returning to subgraphs in general, we note that sometimes removing a single vertex or edge would cause a graph to become disconnected: a cut vertex is a vertex whose removal disconnects the remaining graph (note that any edges incident on the removed vertex are removed too) a cut edge is an edge whose removal disconnects the remaining graph For example, consider the following graph: The cut vertices are b,c,e b,c,e, and the cut edges are {a,b},{c,e}{a,b},{c,e}. We can also talk about connectivity in directed graphs: a directed graph is strongly connected if there is a path from a a to b b and from b b to a a for every pair of vertices a a and b b a directed graph is weakly connected if the graph is connected when you ignore the directions of the edges (that is, the "underlying undirected graph") For example, the graph on the left is both strongly and weakly connected, while the graph on the right is only weakly connected since there is no path from a a to b b.
13892	https://faculty-legacy.arch.tamu.edu/anichols/index_files/courses/arch614/arch614old/NS26-1footings.pdf	ARCH 614 Note Set 26.1 S2012abn 1 Foundation Design Notation: a = name for width dimension A = name for area b = width of retaining wall stem at base = width resisting shear stress bo = perimeter length for two-way shear in concrete footing design B = spread footing or retaining wall base dimension in concrete design cc = shorthand for clear cover d = effective depth from the top of a reinforced concrete member to the centroid of the tensile steel = name for diameter e = eccentric distance of application of a force (P) from the centroid of a cross section f = symbol for stress c f  = concrete design compressive stress Fhorizontal-resisting = total force resisting horizontal sliding Fsliding = total sliding force Fx = force in the x direction F.S. = shorthand for factor of safety hf = height of a concrete spread footing H = height of retaining wall HA = horizontal force due to active soil pressure ld = development length for reinforcing steel L = name for length or span length M = moment due to a force Mn = nominal flexure strength with the steel reinforcement at the yield stress and concrete at the concrete design strength for reinforced concrete beam design Moverturning = total overturning moment Mresisting = total moment resisting overturning about a point Mu = maximum moment from factored loads for LRFD beam design n = name for number N = name for normal force to a surface o = point of overturning of a retaining wall, commonly at the “toe” p = pressure pA = active soil pressure P = name for axial force vector = force due to a pressure PD = dead load axial force PL = live load axial force Pu = factored axial force q = soil bearing pressure qa = allowable soil bearing stress in allowable stress design, as is qallowable qg = gross soil bearing pressure qnet = net allowed soil bearing pressure, as is qn qu = ultimate soil bearing strength in allowable stress design = factored soil bearing capacity in concrete footing design from load factors, as is qnu R = name for reaction force vector SF = shorthand for factor of safety t = thickness of retaining wall stem at top T = name of a tension force V = name for volume Vc = shear force capacity in concrete Vu = factored shear for reinforced concrete design w = name for width wu = load per unit length on a beam from load factors W = name for force due to weight x = horizontal distance y = the distance in the y direction from a reference axis to the centroid of a shape  = resistance factor c  = density or unit weight of concrete s  = density or unit weight of soil  = pi (3.1415 radians or 180)  = reinforcement ratio in concrete beam design = As/bd  = coefficient of static friction ARCH 614 Note Set 26.1 S2012abn 2 Foundations A foundation is defined as the engineered interface between the earth and the structure it supports that transmits the loads to the soil or rock. The design differs from structural design in that the choices in material and framing system are not available, and quality of materials cannot be assured. Foundation design is dependent on geology and climate of the site. Soil Mechanics Soil is another building material and the properties, just like the ones necessary for steel and concrete and wood, must be known before designing. In addition, soil has other properties due to massing of the material, how soil particles pack or slide against each other, and how water affects the behavior. The important properties are  specific weight (density)  allowable soil pressure  factored net soil pressure – allowable soil pressure less surcharge with a factor of safety  shear resistance  backfill pressure  cohesion & friction of soil  effect of water  settlement  rock fracture behavior Structural Strength and Serviceability There are significant serviceability considerations with soil. Soils can settle considerably under foundation loads, which can lead to redistribution of moments in continuous slabs or beams, increases in stresses and cracking. Excessive loads can cause the soil to fail in bearing and in shear. The presence of water can cause soils to swell or shrink and freeze and thaw, which causes heaving. Fissures or fault lines can cause seismic instabilities. A geotechnical engineer or engineering service can use tests on soil bearings from the site to determine the ultimate bearing capacity, qu. Allowable stress design is utilized for soils because of the variability do determine the allowable bearing capacity, qa = qu/(safety factor). Values of qa range from 3000 – 4000 psi for most soils, while clay type soils have lower capacities and sandy soils to rock have much higher capacities. slip zone punched wedge ARCH 614 Note Set 26.1 S2012abn 3 active (trying to move wall) passive (resists movement) Soil acts somewhat like water, in that it exerts a lateral pressure because of the weight of the material above it, but the relationship is not linear. Soil can have an active pressure from soil behind a retaining wall and a passive pressure from soil in front of the footing. Active pressure is typically greater than passive pressure. Foundation Materials Typical foundation materials include:  plain concrete  reinforced concrete  steel  wood  composites, ie. steel tubing filled with concrete Foundation Design Generalized Design Steps Design of foundations with variable conditions and variable types of foundation structures will be different, but there are steps that are typical to every design, including: 1. Calculate loads from structure, surcharge, active & passive pressures, etc. 2. Characterize soil – hire a firm to conduct soil tests and produce a report that includes soil material properties 3. Determine footing location and depth – shallow footings are less expensive, but the variability of the soil from the geotechnical report will drive choices 4. Evaluate soil bearing capacity – the factor of safety is considered here 5. Determine footing size – these calculations are based on working loads and the allowable soil pressure 6. Calculate contact pressure and check stability 7. Estimate settlements 8. Design the footing structure – design for the material based on applicable structural design codes which may use allowable stress design, LRFD or limit state design (concrete). ARCH 614 Note Set 26.1 S2012abn 4 Shallow Foundation Types Considered simple and cost effective because little soil is removed or disturbed. Spread footing – A single column bears on a square or rectangular pad to distribute the load over a bigger area. Wall footing – A continuous wall bears on a wide pad to distribute the load. Eccentric footing – A spread or wall footing that also must resist a moment in addition to the axial column load. Combined footing – Multiple columns (typically two) bear on a rectangular or trapezoidal shaped footing. Unsymmetrical footing – A footing with a shape that does not evenly distribute bearing pressure from column loads and moments. It typically involves a hole or a non-rectangular shape influenced by a boundary or property line. Strap footing – A combined footing consisting of two spread footings with a beam or strap connecting the slabs. The purpose of this is to limit differential settlements. Mat foundation – A slab that supports multiple columns. The mat can be stiffened with a grid or grade beams. It is typically used when the soil capacity is very low. Deep Foundation Types Considerable material and excavation is required, increasing cost and effort. Retaining Walls – A wall that retains soil or other materials, and must resist sliding and overturning. Can have counterforts, buttresses or keys. Basement Walls – A wall that encloses a basement space, typically next to a floor slab, and that may be restrained at the top by a floor slab. Piles – Next choice when spread footings or mats won’t work, piles are used to distribute loads by end bearing to strong soil or friction to low strength soils. Can be used to resist uplift, a moment causing overturning, or to compact soils. Also useful when used in combination to control settlements of mats or slabs. Drilled Piers – Soil is removed to the shape of the pier and concrete is added. Caissons –Water and possibly wet soil is held back or excavated while the footing is constructed or dropped into place. Rp a p a f A P   end bearing Rs =ƒ(adhesion) P 0  P R friction P T N tapered friction P uplift/tension Pile Types ARCH 614 Note Set 26.1 S2012abn 5 p w RIGID footing on sand RIGID footing on clay IDEAL stress Loads and Stresses Bearing loads must be distributed to the soil materials, but because of their variability and the stiffness of the footing pad, the resulting stress, or soil pressure, is not necessarily uniform. But we assume it is for design because dealing with the complexity isn’t worth the time or effort. The increase in weight when replacing soil with concrete is called the overburden. Overburden may also be the result of adding additional soil to the top of the excavation for a retaining wall. It is extra uniformly distributed load that is considered by reducing the allowable soil pressure (instead of increasing the loads), resulting in a net allowable soil pressure, qnet: In order to design the footing size, the actual stress P/A must be less than or equal to the allowable pressure: Design Stresses The result of a uniform pressure on the underside of a footing is identical to a distributed load on a slab over a column when looked at upside down. The footing slab must resist bending, one-way shear and two-way shear (punching). Stresses with Eccentric Loading Combined axial and bending stresses increase the pressure on one edge or corner of a footing. We assume again a linear distribution based on a constant relationship to settling. If the pressure combination is in tension, this effectively means the contact is gone between soil and footing and the pressure is really zero. To avoid zero pressure, the eccentricity must stay within the kern. The maximum pressure must not exceed the net allowable soil pressure. If the contact is gone, the maximum pressure can be determined knowing that the volume of the pressure wedge has to equal the column load, and the centroid of the pressure wedge coincides with the effective eccentricity. Wedge volume is 2 wpx V  where w is the width, p is the soil pressure, and x is the wedge length (3a), so wx N or wx P p 2 2  (and N M or P M e  and a=½ width - e) ) ( h q q s c f allowable net     net q A P  one-way shear two-way shear M P ARCH 614 Note Set 26.1 S2012abn 6 Overturning is considered in design such that the resisting moment from the soil pressure (equivalent force at load centroid) is greater than the overturning moment, M, by a factor of safety of at least 1.5 where Mresist = average resultant soil pressure x width x location of load centroid with respect to column centroid Moverturning = P x e Combined Footings The design of combined footing requires that the centroid of the area be as close as possible to the resultant of the two column loads for uniform pressure and settling. Retaining Walls The design of retaining walls must consider overturning, settlement, sliding and bearing pressure. The water in the retained soil can significantly affect the loading and the active pressure of the soil. The lateral force acting at a height of H/3 is determined from the active pressure, pA, (in force/cubic area) as: Overturning is considered the same as for eccentric footings: where Mresist = summation of moments about “o” to resist rotation, typically including the moment due to the weight of the stem and base and the moment due to the passive pressure. Moverturning = moment due to the active pressure about “o”. Sliding must also be avoided: where: Fhorizontal-resist = summation of forces to resist sliding, typically including the force from the passive pressure and friction (F=N where  is a constant for the materials in contact and N is the normal force to the ground acting down and shown as R). Fsliding = sliding force as a result of active pressure. 5 1 g overturnin . M M SF resist   P1 P2 R = P1+P2 y o Fx R W Fresist 2 5 1 g overturnin    . M M SF resist 2 25 1     . F F SF sliding resist horizontal HA pA H/3 2 2 H p H A A  ARCH 614 Note Set 26.1 S2012abn 7 For sizing, some rules of thumbs are:  footing size, B  reinforced concrete, B  2/5 - 2/3 wall height (H)  footing thickness, hf  1/12 - 1/8 footing size (B)  base of stem, b  1/10 - 1/12 wall height (H+hf)  top of stem, t  12 inches Example 1 (page 533) H B hf t b w PD = 200k PL = 300k 15” square column hf hf Assume the soil has a density of 90 lb/ft3 ARCH 614 Note Set 26.1 S2012abn 8 ARCH 614 Note Set 26.1 S2012abn 9 Example 2 For the 16 in. thick 8.5 ft. square reinforced concrete footing carrying 150 kips dead load and 100 kips live load on a 24 in. square column, determine if the footing thickness is adequate for 4000 psi . A 3 in. cover is required with concrete in contact with soil. Also determine the moment for reinforced concrete design. SOLUTION: 1. Find design soil pressure: A P q u u  Pu = 1.2D + 1.6L = 1.2 (150 k) + 1.6 (100 k) = 340 k 2 ) 5 . 8 ( 340 ft k qu  = 4.71 k/ft2 2. Evaluate one-way shear at d away from column face (Is Vu < Vc?) d = hf – c.c. – distance to bar intersection presuming #8 bars: d = 16 in. – 3 in. (soil exposure) - 1 in. x (1 layer of #8’s) = 12 in. Vu = total shear = qu (edge area) Vu on a 1 ft strip = qu (edge distance) (1 ft) Vu = 4.71 k/ft2 [(8.5 ft – 2 ft)/2 – (12 in.)(1 ft/12 in.)] (1 ft) = 10.6 k Vc = one-way shear resistance = 2 c f bd for a one foot strip, b = 12 in. Vc = 0.75(2 4000 psi)(12 in.)(12 in.) = 13.7 k > 10.6 k OK 3. Evaluate two-way shear at d/2 away from column face (Is Vu < Vc?) bo = perimeter = 4 (24 in. + 12 in.) = 4 (36 in.) = 144 in Vu = total shear on area outside perimeter = Pu – qu (punch area) Vu = 340 k – (4.71 k/ft2)(36 in.)2(1 ft/12 in.)2 = 297.6 kips Vc = two-way shear resistance = 4 c f bod = 0.75(4 4000 psi)(144 in.)(12 in.) = 327.9 k > 297.6 k OK 4. Design for bending at column face Mu = wuL2/2 for a cantilever. L = (8.5 ft – 2 ft)/2 = 3.25 ft, and wu for a 1 ft strip = qu (1 ft) Mu = 4.71 k/ft2(1 ft)(3.25 ft)2/2 = 24.9 k-ft (per ft of width) To complete the reinforcement design, use b =12 in. and trial d = 12 in., choose , determine As, find if Mn > Mu..... ARCH 614 Note Set 26.1 S2012abn 10 Example 3 Determine the depth required for the group of 4 friction piles having 12” diameters if the column load is 100 kips and the frictional resistance is 400 lbs/ft2. SOLUTION: The downward load is resisted by a friction force. Friction is determined by multiplying the friction resistance (a stress) by the area: SKIN fA F  The area of n cylinders is: ) 2 2 ( L d n ASKIN   Our solution is to set P  F and solve for length: ) lb k ( ) in ft ( L ) in )( )( ( k piles ft lb 1000 1 12 1 2 12 2 4 400 100 2     pile ft L 9 . 19  Example 4 Determine the depth required for the friction & bearing pile having a 36” diameter if the column load is 300 kips, the frictional resistance is 600 lbs/ft2 and the end bearing pressure allowed is 8000 psf. SOLUTION: The downward load is resisted by a friction force and a bearing force, which can be determined from multiplying the bearing pressure by the area in contact: TIP SKIN qA fA F   The area of a circle is: 4 2 d ATIP   Our solution is to set P  F and solve for length: ) lb k ( ) in ft ( ) in ( ) lb k ( ) in ft ( L ) in )( ( k ft lb ft lb 1000 1 12 1 4 36 8000 1000 1 12 1 2 36 2 600 300 2 2 2 2         ft . L 1 43  Example 5 Determine the factor of safety for overturning and sliding on the 15’ retaining wall, 16” wide stem, 10’ base, 16” heigh base, when the equivalent fluid pressure is 30 pcf, the weight of the stem of the footing is 4 kips, the weight of the pad is 5 kips, the passive pressure is ignored for this design, and the friction coefficient for sliding is 0.58. The center of the stem is located 3’ from the toe. 15’ 10’ 16” 3’ 16” P = 100 kips P = 300 kips
13893	https://www.coursera.org/courses?query=epidemics	For Individuals For Businesses For Universities For Governments Skip to main content Browse Epidemics Epidemics Courses Online Learn about epidemics and their impact on public health. Understand the spread, control, and prevention of infectious diseases. Skip to search results Filter by Subject Language Required The language used throughout the course, in both instruction and assessments. Learning Product Required Learn from top instructors with graded assignments, videos, and discussion forums. Get in-depth knowledge of a subject by completing a series of courses and projects. Level Required Duration Required Skills Required Subtitles Required Educator Required Explore the Epidemics Course Catalog Status: Preview Preview T The University of Hong Kong ### Epidemics Skills you'll gain: Infectious Diseases, Epidemiology, Public Health and Disease Prevention, Public Health, Healthcare Ethics, Community Health, Emergency Response, Research, Immunology, Microbiology, Risk Analysis, Media and Communications 4.9 Rating, 4.9 out of 5 stars · 232 reviews Beginner · Course · 1 - 3 Months Status: Preview Preview T The Pennsylvania State University ### Epidemics - the Dynamics of Infectious Diseases Skills you'll gain: Infectious Diseases, Epidemiology, Public Health and Disease Prevention, Microbiology, Public Health, Pathology, Immunology, Preventative Care, Sanitation, Social Determinants Of Health, Social Sciences, Community Health, Molecular, Cellular, and Microbiology, Medical Science and Research, Biology, Drug Development, Social Network Analysis, Case Studies 4.8 Rating, 4.8 out of 5 stars · 1.7K reviews Mixed · Course · 1 - 3 Months Status: Free Trial Free Trial J Johns Hopkins University ### Outbreaks and Epidemics Skills you'll gain: Epidemiology, Infectious Diseases, Risk Analysis, Public Health, Investigation, Case Studies, Laboratory Testing, Statistical Analysis, Research Methodologies, Statistics, Report Writing 4.8 Rating, 4.8 out of 5 stars · 763 reviews Beginner · Course · 1 - 4 Weeks Status: Preview Preview P Politecnico di Milano ### One Health: Pandemic preparedness, prevention, and response Skills you'll gain: Public Health and Disease Prevention, Public Health, Environmental Science, Infectious Diseases, Preventative Care, Health Policy, Health Care Procedure and Regulation, Policty Analysis, Research, and Development, Climate Change Adaptation, Governance, Food Safety and Sanitation, Law, Regulation, and Compliance, Case Studies 4.6 Rating, 4.6 out of 5 stars · 44 reviews Intermediate · Course · 1 - 3 Months Status: Preview Preview D Dartmouth College ### Reflections from 40 Years Fighting International Epidemics Skills you'll gain: Infectious Diseases, Epidemiology, Infection Control, Public Health and Disease Prevention, Public Health, Healthcare Ethics, Health Care, Health Systems, Health Policy, Medical Science and Research, Liberal Arts, Cultural Sensitivity, Case Studies, Storytelling Beginner · Course · 1 - 3 Months Status: Preview Preview J Johns Hopkins University ### Investigating Epidemics like COVID-19: An Analyst's Guide Skills you'll gain: Epidemiology, Health Disparities, Spatial Data Analysis, Data Visualization, Public Health, Geographic Information Systems, Data Analysis, Statistical Analysis, Infectious Diseases, Risk Analysis, Microsoft Excel Intermediate · Course · 1 - 4 Weeks What brings you to Coursera today? Status: Free Trial Free Trial I Imperial College London ### Epidemiology for Public Health Skills you'll gain: Epidemiology, Diagnostic Tests, Research Design, Public Health and Disease Prevention, Biostatistics, Clinical Research, Public Health, Preventative Care, Data Collection, Research Methodologies, Program Evaluation, Risk Analysis, Quantitative Research, Health Policy, Correlation Analysis, Statistical Analysis, Statistical Methods, Research, Sample Size Determination, Data Analysis 4.8 Rating, 4.8 out of 5 stars · 1.2K reviews Beginner · Specialization · 1 - 3 Months Status: Free Trial Free Trial J Johns Hopkins University ### Epidemiology in Public Health Practice Skills you'll gain: Epidemiology, Public Health, Data Presentation, Public Health and Disease Prevention, Trend Analysis, Geographic Information Systems, Health Policy, Data Visualization Software, Chronic Diseases, GIS Software, Health Systems, Infectious Diseases, Biostatistics, Risk Analysis, Investigation, Health Care, Health Disparities, Community Health, Data Analysis, Statistics 4.7 Rating, 4.7 out of 5 stars · 4.5K reviews Beginner · Specialization · 3 - 6 Months Status: New New Status: Free Trial Free Trial J Johns Hopkins University ### Drug Development and Pharmacoepidemiology Skills you'll gain: Drug Development, Clinical Trials, Pharmaceuticals, Pharmacology, Pharmacotherapy, Clinical Research, Pharmacy, Patient Safety, Medical Prescription, Epidemiology, Research Design, Contraindication, Health Policy, Research Methodologies, Medical Science and Research, Drug Interaction, Science and Research, Program Evaluation, Public Health, Data Analysis 4.7 Rating, 4.7 out of 5 stars · 51 reviews Intermediate · Specialization · 3 - 6 Months Status: Free Trial Free Trial I Imperial College London ### Foundations in Virology and Vaccinology Skills you'll gain: Immunology, Infectious Diseases, Molecular Biology, Epidemiology, Microbiology, Public Health, Drug Development, Biotechnology, Medical Science and Research, Climate Change Programs, Cell Biology, Health Disparities, Biology, Clinical Trials, Laboratory Research, Pharmacology, Public Health and Disease Prevention, New Product Development, Emerging Technologies, Manufacturing Processes 4.9 Rating, 4.9 out of 5 stars · 69 reviews Intermediate · Specialization · 3 - 6 Months Status: Preview Preview S Stanford University ### Stanford Introduction to Food and Health Skills you'll gain: Cooking, Meal Planning And Preparation, Nutrition and Diet, Health Education, Food and Beverage, Behavioral Health, Chronic Diseases, Health Informatics 4.7 Rating, 4.7 out of 5 stars · 34K reviews Beginner · Course · 1 - 3 Months Status: Free Trial Free Trial I Imperial College London ### Global Disease Masterclass: Communicable Diseases Epidemiology, Intervention and Prevention Skills you'll gain: Infectious Diseases, Epidemiology, Public Health, Microbiology, Health Disparities, Community Health, Social Determinants Of Health, Preventative Care, Health Care, Health Policy, Public Policies, Trend Analysis, Immunology 4.7 Rating, 4.7 out of 5 stars · 183 reviews Beginner · Course · 1 - 4 Weeks Epidemics learners also search Health Public Health Epidemiology Advanced Health Health Policy Health Economics Global Health Health And Nutrition 1 2 3 4 … 9 In summary, here are 10 of our most popular epidemics courses Epidemics: The University of Hong Kong Epidemics - the Dynamics of Infectious Diseases: The Pennsylvania State University Outbreaks and Epidemics: Johns Hopkins University One Health: Pandemic preparedness, prevention, and response: Politecnico di Milano Reflections from 40 Years Fighting International Epidemics: Dartmouth College Investigating Epidemics like COVID-19: An Analyst's Guide: Johns Hopkins University Epidemiology for Public Health: Imperial College London Epidemiology in Public Health Practice: Johns Hopkins University Drug Development and Pharmacoepidemiology: Johns Hopkins University Foundations in Virology and Vaccinology: Imperial College London Frequently Asked Questions about Epidemics Epidemics refer to the occurrence of a higher-than-normal number of cases of a particular disease within a specific population or geographic area. It typically represents the spread of a disease or infection beyond the levels normally expected for that population. Epidemics can occur for various reasons, such as the introduction of a new pathogen, a decrease in immunity levels within the population, or changes in the behavior or environment of the affected individuals. It is crucial to understand and effectively respond to epidemics, as they can cause significant public health concerns and impact communities worldwide.‎ To study Epidemics, you would need to learn the following skills: Epidemiology: This is the study of the patterns, causes, and effects of diseases in populations. Understanding epidemiology is crucial for investigating epidemics, managing outbreaks, and implementing effective public health measures. Biostatistics: It is necessary to have a strong foundation in biostatistics to analyze and interpret epidemiological data accurately. This skill helps in designing studies, calculating disease prevalence, and determining the significance of findings. Public Health: Knowledge of public health principles and practices is essential when dealing with epidemics. This includes studying concepts of disease prevention, health promotion, outbreak management, and interventions for disease control. Infectious Diseases: Developing an understanding of various infectious diseases, including their transmission, symptoms, and prevention strategies, is crucial for comprehending epidemics. This knowledge aids in identifying risk factors, implementing preventive measures, and planning appropriate responses. Data Analysis: Proficiency in data analysis techniques and tools is important for effective management of epidemics. Skills such as data collection, cleaning, visualization, and interpretation are necessary to make informed decisions based on epidemiological findings. Research Methodology: Obtaining knowledge in research methodologies enables you to conduct studies related to epidemics. This includes designing research protocols, collecting data, analyzing findings, and drawing appropriate conclusions to contribute to the field of epidemiology. Communication: Effective communication skills are essential for sharing information about epidemics with various stakeholders, including health professionals, policymakers, and the general public. Clear and concise communication helps in disseminating important messages, raising awareness, and promoting disease prevention strategies. Remember that studying epidemics is a multidisciplinary field, and acquiring a combination of skills from different domains would help you gain a comprehensive understanding of the subject matter.‎ With Epidemics skills, you can pursue various career paths in the healthcare and public health sectors. Here are some job roles you can consider: Epidemiologist: Conduct research, analyze data, and study patterns of diseases to prevent and control their spread. Infectious Disease Specialist: Diagnose and treat infectious diseases, develop management plans, and provide preventive measures. Public Health Advisor: Advise on public health policies, develop and implement strategies to control epidemics, and educate communities on disease prevention. Healthcare Data Analyst: Analyze health data to identify patterns and trends related to epidemics, contributing to the development of effective prevention and control measures. Epidemiology Research Associate: Assist epidemiologists in designing and conducting research studies, collecting and analyzing data, and preparing reports. Outbreak Investigator: Investigate disease outbreaks, identify their sources and modes of transmission, and take appropriate action to minimize their impact. Health Educator: Develop and deliver educational programs on epidemics, promoting healthy behaviors and providing information on disease prevention and management. Biostatistician: Apply statistical methods to analyze and interpret health data related to epidemics, providing valuable insights for epidemiological research and policy development. Remember, these are just a few examples, and there are numerous opportunities in public health and related fields for individuals with epidemics skills.‎ People who are interested in public health, epidemiology, and disease prevention are best suited for studying epidemics. They should have a strong scientific background and an analytical mindset to understand the complexities of disease transmission, outbreak investigation, and control measures. Additionally, individuals who are compassionate, empathetic, and have a desire to make a positive impact on public health would find studying epidemics fulfilling.‎ Here are some topics that you can study related to Epidemics: Epidemiology: Learn about the study of how diseases spread within populations, including patterns, causes, and control measures. Public Health: Explore topics like disease prevention, health promotion, risk assessment, and community health interventions during epidemics. Virology: Gain knowledge about viruses, their characteristics, structure, replication, and how they contribute to the occurrence and spread of epidemics. Infectious Diseases: Understand different types of infectious diseases, their causes, symptoms, transmission modes, treatments, and how they can lead to epidemics. Outbreak Investigation: Learn about the process of investigating and managing outbreaks, including surveillance methods, data analysis, and response strategies. Pandemic Preparedness and Response: Study various aspects of preparing for and responding to pandemics, such as crisis communication, emergency planning, resource management, and healthcare system resilience. Global Health: Explore the impact of epidemics on a global scale, including the social, economic, and political factors influencing disease control and prevention efforts worldwide. Infection Control: Understand the principles and practices of preventing and controlling infections in healthcare settings, which is crucial during epidemics. Disease Modeling and Data Analysis: Learn about mathematical modeling techniques and data analysis methods used to estimate disease spread and predict epidemic outcomes. Healthcare Ethics: Examine the ethical considerations and challenges related to epidemics, such as allocation of scarce resources, triage protocols, and public health measures. Remember, these topics are just a starting point, and there are various levels of study available, ranging from introductory courses to advanced academic degrees.‎ Online Epidemics courses offer a convenient and flexible way to enhance your knowledge or learn new Epidemics refer to the occurrence of a higher-than-normal number of cases of a particular disease within a specific population or geographic area. It typically represents the spread of a disease or infection beyond the levels normally expected for that population. Epidemics can occur for various reasons, such as the introduction of a new pathogen, a decrease in immunity levels within the population, or changes in the behavior or environment of the affected individuals. It is crucial to understand and effectively respond to epidemics, as they can cause significant public health concerns and impact communities worldwide. skills. Choose from a wide range of Epidemics courses offered by top universities and industry leaders tailored to various skill levels.‎ When looking to enhance your workforce's skills in Epidemics, it's crucial to select a course that aligns with their current abilities and learning objectives. Our Skills Dashboard is an invaluable tool for identifying skill gaps and choosing the most appropriate course for effective upskilling. For a comprehensive understanding of how our courses can benefit your employees, explore the enterprise solutions we offer. Discover more about our tailored programs at Coursera for Business here.‎ Other topics to explore Arts and Humanities 338 courses 1095 courses Computer Science 668 courses Data Science 425 courses Information Technology 145 courses Health 471 courses Math and Logic 70 courses Personal Development 137 courses Physical Science and Engineering 413 courses Social Sciences 401 courses Language Learning 150 courses
13894	https://www.statology.org/confidence-intervals-ti-84-calculator/	How to Calculate Confidence Intervals on a TI-84 Calculator by Zach Bobbitt Posted on A confidence interval (C.I.)is a range of values that is likely to include a population parameter with a certain degree of confidence. This tutorial explains how to calculate the following confidence intervals on a TI-84 calculator: Confidence interval for a population mean; σ known Confidence interval for a population mean; σ unknown Confidence interval for a population proportion Example 1: C.I. for a population mean; σ known Find a 95% confidence interval for a population mean, given the following information: sample mean x = 14 sample size n = 35 population standard deviation = 4 Step 1: Choose Z Interval. Press Stat and then scroll over to TESTS. Highlight 7:ZIntervaland press Enter. Step 2: Fill in the necessary information. The calculator will ask for the following information: Inpt:Choose whether you are working with raw data (Data) or summary statistics (Stats). In this case, we will highlight Stats and press ENTER. σ: The population standard deviation. We will type 4 and press ENTER. x: The sample mean. We will type 14 and press ENTER. n: The sample size. We will type 35 and press ENTER. C-level:The confidence level We will type 0.95 and press ENTER. Lastly, highlight Calculate and press ENTER. Step 3: Interpret the results. Once you press ENTER, the 95% confidence interval for the population mean will be displayed: The 95% confidence interval for the population mean is (12.675, 15.325). Example 2: C.I. for a population mean; σ unknown Find a 95% confidence interval for a population mean, given the following information: sample mean x = 12 sample size n = 19 sample standard deviation = 6.3 Step 1: Choose T Interval. Press Stat and then scroll over to TESTS. Highlight 8:TIntervaland press Enter. Step 2: Fill in the necessary information. The calculator will ask for the following information: Inpt:Choose whether you are working with raw data (Data) or summary statistics (Stats). In this case, we will highlight Stats and press ENTER. x: The sample mean. We will type 12 and press ENTER. Sx: The sample standard deviation. We will type 6.3 and press ENTER. n: The sample size. We will type 19 and press ENTER. C-level:The confidence level We will type 0.95 and press ENTER. Lastly, highlight Calculate and press ENTER. Step 3: Interpret the results. Once you press ENTER, the 95% confidence interval for the population mean will be displayed: The 95% confidence interval for the population mean is (8.9635, 15.037). Example 3: C.I. for a population proportion Find a 95% confidence interval for a population proportion, given the following information: number of “successes” (x) = 12 number of trials (n) = 19 Step 1: Choose 1 Proportion Z Interval. Press Stat and then scroll over to TESTS. Highlight1-PropZIntand press Enter. Step 2: Fill in the necessary information. The calculator will ask for the following information: x:The number of successes. We will type 12 and press ENTER. n: The number of trials. We will type 19 and press ENTER. C-level:The confidence level We will type 0.95 and press ENTER. Lastly, highlight Calculate and press ENTER. Step 3: Interpret the results. Once you press ENTER, the 95% confidence interval for the population proportion will be displayed: The 95% confidence interval for the population proportion is (0.41468, 0.84848). Zach Bobbitt Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations. Leave a Reply Cancel reply
13895	https://www.intmath.com/exponents-radicals/2-fractional-exponent-laws.php	Interactive Mathematics Thank you for booking, we will follow up with available time slots and course plans. On this page Exponents and Radicals 1. Simplifying Expressions with Integral Exponents Fractional Exponents 3. Simplest Radical Form 4. Addition and Subtraction of Radicals 5. Multiplication and Division of Radicals (Rationalizing the Denominator) 6. Equations with Radicals Related Sections Math Tutoring Need help? Chat with a tutor anytime, 24/7. Chat now Online Algebra Solver Solve your algebra problem step by step! Online Algebra Solver IntMath Forum Get help with your math queries: See Forum 2. Fractional Exponents Fractional exponents can be used instead of using the radical sign (√). We use fractional exponents because often they are more convenient, and it can make algebraic operations easier to follow. Fractional Exponent Laws The n-th root of a number can be written using the power 1/n, as follows: a^(1/n)=root(n)a Meaning: The n-th root of a when multiplied by itself n times, gives us a. a1/n × a1/n × a1/n × ... × a1/n = a [Multiply n times] Example 1 The cube root of 8 is 2 (since 2^3=8). We can write the cube root of 8 as: 8^(1//3) or root(3)8 The following 3 numbers are equivalent: 8^{1/3}=root(3)8=2 Example 2 The square root of a number can be written using the radical sign (√) or with exponent 1/2. The following are equivalent: sqrt(100)=100^(1/2)=10 Example 3 The 4-th root of 625 can be written as either: 6251/4 or equivalently, as root(4)625 Its value is 5, since 5^4= 625. So we could write: 625^(1/4) = root(4)625 = 5 Definitions Radicand The number under the radical is called the radicand (in Example 3, the number 625 is the radicand). Order/Index of the radical The number indicating the root being taken is called the order (or index) of the radical (in Example 3, the order is 4). These definitions are here so you know what your textbook is talking about. Raising the n-th root to the Power m If we need to raise the n-th root of a number to the power m (say), we can write this as: a^(m/n)=(root(n)a)^m This experssion means we need to "take the n-th root of the number a, then raise the result to the power m". With fractional exponents, we would write this as: (a^(1//n))^m Actually, we get the same final answer if we do it in the other order, "raise a to the power m, then find the n-th root of the result". That is, (a^m)^(1//n) But the first one is usually easier to do becuase finding the n-th root first gives us a smaller number, which is then easy to raise to the power m. Example 4 Evaluate 8^(2/3) Answer 8^(2/3)=(root(3)8)^2=(2)^2=4 First, we found the cube root of 8 and the answer was 2. Then we raised this result to the power 2, giving 4. Example 5 Simplify (8a^2b^4)^(1/3) Answer (8a^2b^4)^(1/3)=(8)^(1/3)(a^2)^(1/3)(b^4)^(1/3) =2a^(2/3)b^(4/3) In the first line, we used this rule from the last section: (am)n = amn That is, we took each item inside the brackets and raised them to the power 1/3. We can do this because each term is multiplied inside the bracket (if they were added or subtracted, we could not do this). When we expand this out, the only thing we can do is to find the cube root of 8, which is 2, and then just write the a and b parts with fractional powers. Example 6 Simplify a^(3text(/)4)a^(4text(/)5) Answer a^(3/4)a^(4/5)=a^(3/4+4/5)=a^(31/20) Example 7 Simplify ((4^(-3/2)x^(2/3)y^(-7/4))/(2^(3/2)x^(-1/3)y^(3/4)))^(2/3) Answer ((4^(-3/2)x^(2/3)y^(-7/4))/(2^(3/2)x^(-1/3)y^(3/4)))^(2/3) Don't panic when you see this one! In the first step, we move the top expressions with negative exponents to the bottom, and the bottom ones with negative exponents to the top. =((x^(2/3)x^(1/3))/(2^(3/2)4^(3/2)y^(3/4)y^(7/4)))^(2/3) Then we multiply terms with the same base (the x and y terms), by adding their indices. We can collect the 2 and 4 because they both have power 3/2. =((x^(2/3+1/3))/((2xx4)^(3/2)y^(3/4+7/4)))^(2/3) Next, we raise everything to the power 2/3, since that was the power outside the bracket. =(x/(8^(3/2)y^(10/4)))^(2/3) The final step is to tidy up the expression. =x^(2/3)/(8^((3/2xx2/3))y^((10/4xx2/3))) =x^(2/3)/(8y^(5/3)) Whew! Exercises Question 1: Evaluate 5^(1//2)5^(3//2) Answer Each item in this question has the same base (5), so we need to add the indices, as follows: 5^(1text(/)2)5^(3text(/)2) = 5^(1text(/)2+3text(/)2) =5^(4text(/)2) =5^2 =25 Question 2: Evaluate (1000^(1text(/)3))/(400^(-1text(/)2)) Answer On the top, the cube root of 1000 is 10. On the bottom of the fraction, we have a negative index. We need to recall the following 2 properties of fractions and negative indices: 1/a=a^(-1) If the number on the bottom has a negative index, for example negative 1, then this is the result: So 400^(−1//2 on the bottom becomes 400^(1//2 on the top. And 400^(1//2) =sqrt(400)= 20. Here's the complete answer: 1000^(1text(/)3)/(400^(-1text(/)2)) =10 xx 400^(1 text(/)2) =10 xx 20 =200 1. Simplifying Expressions with Integral Exponents 3. Simplest Radical Form Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.
13896	https://emedicine.medscape.com/article/958739-clinical	close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Pediatrics: General Medicine Pediatric Splenomegaly Clinical Presentation Updated: Jun 04, 2024 Author: Trisha Simone Natanya Tavares, MD; Chief Editor: Vikramjit S Kanwar, MBBS, MBA, MRCP(UK), FRCPCH, FAAP more...;) 9 Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Pediatric Splenomegaly Sections Pediatric Splenomegaly Overview Practice Essentials Pathophysiology Epidemiology Prognosis Patient Education Show All Presentation History Physical Causes Show All DDx Workup Approach Considerations Laboratory Studies Imaging Studies Histologic Findings Show All Treatment Approach Considerations Medical Care Surgical Care Consultations Activity Diet Show All Medication Medication Summary Vaccines Antibiotics Show All Follow-up Further Outpatient Care Complications Show All References;) Presentation History Despite the extensive differential diagnosis of splenomegaly, careful history taking and physical examination, along with a CBC and manual differential, often help to narrow down the list of likely causes. The history should include attention to the following important areas : Duration of known enlargement of the spleen Exposure to hepatotoxic agents or microorganisms resulting in hepatitis or portal hypertension Abdominal trauma that may cause splenic hematoma Signs of infection or known infections such as hepatitis, mononucleosis, malaria, or salmonellosis Inflammatory bowel disease Bone pain, fever, malaise, lethargy, pallor, bruising, weight loss, night sweats, or other findings that may indicate malignancy Jaundice suggestive of hepatobiliary disease The patient should also be assessed for the following: Complicated neonatal course - Eg, sepsis, hypotension Umbilical catheter thrombosis Hyperbilirubinemia, anemia - Eg, due to hereditary hemolysis Heart disease - Eg, congestive heart failure Past surgeries - Eg, leading to infection, thrombosis, portal hypertension and cholecystectomy due to cholelithiasis Anemia or transfusions - Eg, due to hematologic abnormalities and/or resulting in hepatitis Abdominal trauma - Possibly resulting in splenic pseudocyst or hematoma Travel - Possible exposure to malaria, leishmaniasis, schistosomiasis, trypanosomiasis, or other microorganisms Sexual behavior - Possible presence of sexually transmitted infections, including human immunodeficiency virus (HIV), cytomegalovirus (CMV), and hepatitis With regard to family history, carefully document the presence or absence of the following: Anemia Cholecystectomy Splenectomy - Eg, due to hemolytic anemia The patient’s ethnicity should also be determined: Mediterranean ethnicity - Increased incidence of thalassemia and glucose-6-phosphate dehydrogenase (G6PD) deficiency African ethnicity - Increased incidence of sickle cell anemia, G6PD deficiency, and hereditary pyropoikilocytosis European or American-Amish ancestry - Associated with pyruvate kinase deficiency and hereditary spherocytosis Ashkenazi Jewish ethnicity - Increased incidence of Gaucher disease and Niemann-Pick disease Asian ethnicity - Increased incidence of portal hypertension secondary to noncirrhotic portal fibrosis and increased incidence of G6PD deficiency Next: Physical Physical The patient should be examined in the supine or right lateral decubitus position. Palpation should start at the pubis and move toward the left upper quadrant to identify the medial and inferior border of the spleen. If the enlarged tip of the spleen is below the examiner's hands, he or she may not detect it. Likewise, light pressure should be used because the spleen can easily be displaced without the clinician identifying the organ’s edge. At times, the superior medial edge of the spleen is more readily palpated than the inferior margin. The characteristic downward movement of the spleen with inspiration can help in differentiating the spleen from other masses of the left upper quadrant. Percussion over the left lateral areas of the lower ribs may reveal splenomegaly that is not evident upon palpation. Percussion is particularly helpful in obese or crying patients. [48, 49, 50] Document weight and height with percentiles for age to identify growth abnormalities, and obtain vital signs, including heart rate and blood pressure. Identify tachycardia, which may be seen in anemia, and fever, which may be present in inflammatory and infectious conditions. Examine the patient supine with the hips and knees flexed. Place a pillow underneath the neck. Ensure relaxation of the abdominal musculature. Begin palpation at the iliac bone and palpate both lower quadrants. Percuss the lowest intercostal space in the left anterior axillary line; evaluate the space that is bound superiorly by the sixth rib, laterally by the axillary line, and inferiorly by the costal margin. Dullness to percussion in this area may indicate splenomegaly. Assess the patient’s general appearance to document ill appearance. This will guide evaluation but is nonspecific and may be seen in various conditions, including malignancy, chronic hemolysis, chronic infection, metabolic disease, liver disease, and inflammatory disease. Dermal findings These include the following: Pallor - Eg, due to anemia, which may indicate hemolysis, bone marrow infiltration, or hypersplenism Petechiae, purpura - Eg, due to thrombocytopenia, which may indicate bone marrow failure, autoimmune disorder, or hypersplenism Jaundice - Eg, due to hemolytic anemia or liver disease Exanthems - Eg, due to acute and chronic infections, systemic lupus erythematosus, rheumatoid arthritis, infective endocarditis, hemangiomata, histiocytosis, or immunodeficiency Head, eye, ear, nose, and throat findings These include the following: Icterus - Eg, due to hemolytic anemia or liver dysfunction Cherry red retinal spots, cloudy corneas - For example, due to lipid-storage diseases Respiratory and cardiovascular findings These include the following: Dyspnea, fatigue - Eg, due to anemia or congestive heart failure Cardiac murmur - Eg, due to anemia or endocarditis Gastrointestinal findings These include the following: Abdominal tenderness - Eg, due to gallstones, hepatitis, trauma, or acute splenomegaly Distention, prominent abdominal veins, ascites - Eg, due to liver disease Abnormal size or texture of the liver Musculoskeletal findings These include the following: Joint abnormalities - Eg, due to systemic lupus erythematosus, rheumatoid arthritis, or autoimmune inflammatory disease Bone abnormalities - Eg, due to storage diseases or osteopetrosis Neurologic abnormalities These include the following: Poor vision - Eg, due to osteopetrosis Uveitis, iritis - Eg, due to sarcoidosis or rheumatoid arthritis Inappropriate developmental milestones - Eg, due to storage diseases or other chronic illness Previous Next: Physical Causes Despite the numerous causes of splenomegaly (see Differentials), the spleen is rarely the primary site of disease. Splenomegaly is often categorized into the following groupings: Sequestration of blood cells - Such as in hemolytic conditions Proliferation due to infection or inflammation Deposition - Such as in Niemann-Pick and Gaucher disease and in some infections Infiltration due to granulomatous, histiocytic, lymphoproliferative, or malignant conditions Endowment - As caused by space-occupying lesions Splenic masses Hematomas of the spleen may develop after trauma, including birth trauma, and may occur in accessory spleens, as well as in the main spleen. Some splenic hematomas arise as complications of medical procedures, and they may also appear spontaneously as part of disease processes. Splenic hematomas can be associated with symptomatic bleeding and other complications and require evaluation and monitoring. Some cases may be managed medically, but resection is frequently required for splenic hematomas. [51, 19, 52] Splenic cysts are rare and are often discovered incidentally. Some patients with splenic cysts may report gastrointestinal complaints such as pain, nausea, altered bowel habits, flatulence, fullness, or emesis. Primary splenic cysts are the most common type of splenic cyst in children. The lesions are broadly categorized based on whether they are parasitic or non-parasitic in origin and by histology and etiology. Size, symptoms, and cause will determine the nature of any required treatment. There is a risk that a splenic cyst may become complicated by thrombocytopenia, bleeding, infection, or rupture. Splenectomy or partial splenectomy is sometimes performed. Hyperplasia The most common mechanism of pathologic splenomegaly in children is hyperplasia of the MPS. This is due to a variety of conditions that result in excessive antigenic stimulation, including infection and immune dysfunction, as well as hemolysis. [27, 54] Excessive antigenic stimulation due to infection is the cause of most cases of splenomegaly in children. Viral infections are the most frequent culprits, and the associated splenomegaly is usually transient and only mild to moderate in severity. Although Epstein-Barr virus (EBV) and CMV are well-known causes of splenomegaly, other, more common viral illnesses of childhood are the most frequent causes of pediatric infectious splenomegaly. (In addition, pediatric hepatosplenomegaly has been reported in patients with complications of severe acute respiratory syndrome coronavirus-2 [SARS-CoV-2] infection. ) Other common infectious etiologies include bacterial, protozoal, and fungal infections. In endemic areas, malaria and schistosomiasis are frequent causes of splenomegaly. Concomitant, generalized lymphadenopathy is common in many of these infectious conditions. Inflammation and hyperplasia due to collagen vascular diseases such as juvenile rheumatoid arthritis are relatively uncommon, but clinically significant, causes of splenomegaly. Malignancy Splenomegaly can be a presenting sign of neoplasia, being a key such feature in leukemia and lymphoma. Histiocytic disorders may also present with infiltration of the spleen. Metastasis to the spleen, which is uncommon in children, has been reported in neuroblastoma. Portal venous system abnormalities Obstructed venous blood flow of intrahepatic or extrahepatic etiology can cause splenomegaly. The most common causes include portal vein thrombosis, hepatic cirrhosis, and congestive heart failure. Children with extrahepatic portal venous obstruction, such as cavernous transformation, often present with splenomegaly as the primary manifestation of their disease. In an Italian multicenter, national study by Di Giorgio et al, children with noncirrhotic portal vein thrombosis were evaluated, and the condition was noted to be diagnosed subsequent to the detection of splenomegaly in 40% of patients. Deposition causing splenomegaly Storage diseases may result in splenomegaly, and in Gaucher and Niemann-Pick disease, it is often the first clinical manifestation. [58, 59, 60] Splenomegaly arises from the accumulation of abnormal lipids in splenic macrophages. A study by McGovern et al of adults and children with Niemann-Pick disease indicated that over the report’s follow-up period, the risk of death for patients with a history of severe splenomegaly or prior splenectomy was 10-fold that of individuals with moderate splenomegaly or intact spleens. With regard to a history of severe splenomegaly alone, the odds ratio for mortality was 6.0. Masses After trauma, palpable subcapsular hematomas may form in the spleen, which may eventually develop into clinically palpable pseudocysts. Patients with congenital true splenic cysts usually present with asymptomatic splenomegaly. Extramedullary hematopoiesis Although normally found only during the first 6 months of life, extramedullary hematopoiesis may occur in diseases associated with intense demand on the bone marrow for cell production. Thalassemia major, osteopetrosis, and idiopathic myelofibrosis are examples of this rare cause of splenomegaly. Hypersplenism Hypersplenism is a clinical syndrome in which cytopenias result from excessive splenic function and splenic hypertrophy. The pathologic action of the spleen, that is, the reduction of circulating blood elements, has been attributed to four possible mechanisms: excessive splenic phagocytic activity, splenic antibody formation that causes hematopoietic cell destruction, overactivity of splenic function, and sequestration. In patients with cirrhosis, abnormalities of cytokine production may contribute to the cytopenias noted. As the spleen enlarges, it can sequester erythrocytes, leukocytes, and platelets, resulting in cytopenias. Severe reductions in cell counts are unusual and should prompt a search for alternative etiologies. Venous obstruction is the most common cause of hypersplenism. Any increase in portal pressure is reflected in the splenic venous sinuses. This impairs blood flow out of the cords and results in the sequestration of blood cells and hypersplenism. Hypersplenism in children is most frequently caused by portal hypertension. Extrahepatic venous obstruction from portal vein thrombosis is the most common cause of increased portal pressures. In extrahepatic venous obstruction, hepatic function is normal. Intrahepatic venous obstruction is usually due to cirrhosis. Portal hypertension usually increases flow through minor collateral vessels between the portal circulation and the systemic circulation. Portal hypertension can result in recognizable dilatation of the superficial abdominal veins and esophageal varices. Patients with these varices may present with sudden and catastrophic GI hemorrhage. Splenic sequestration Splenic sequestration crisis is a specific form of acute hypersplenism in young children with sickle cell anemia. Children less than 6 years old can develop rapid splenic sequestration and splenomegaly with the consumption of large volumes of erythrocytes. They present with sudden weakness, dyspnea, and left-sided abdominal pain in addition to splenomegaly. Splenic sequestration is an emergency. Rapid death from hypovolemic shock can result. Treatment consists of fluids and erythrocyte transfusions. To prevent recurrences, splenectomy may be indicated. In most patients with sickle cell disease, the spleen eventually involutes, making sequestration no longer possible. Previous Differential Diagnoses References Ezeofor SN, Obikili EN, Anyanwu GE, Onuh AC, Mgbor SO. Sonographic assessment of the normal limits of the spleen in healthy school children in South-East Nigeria. 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Utria AF, Goffredo P, Keck K, Shelton JS, Shilyansky J, Hassan I. Laparoscopic Splenectomy: Has It Become the Standard Surgical Approach in Pediatric Patients?. J Surg Res. 2019 Aug. 240:109-114. [QxMD MEDLINE Link]. Van Der Veken E, Laureys M, Rodesch G, Steyaert H. Perioperative spleen embolization as a useful tool in laparoscopic splenectomy for simple and massive splenomegaly in children: a prospective study. Surg Endosc. 2016 Nov. 30 (11):4962-7. [QxMD MEDLINE Link]. Media Gallery of 0 Tables Back to List Contributor Information and Disclosures Author Trisha Simone Natanya Tavares, MD Contract Clinical Hematologist and Oncologist, Department of Pediatric Hematology and Oncology, Presbyterian Healthcare; Volunteer Hematology/Oncology Physician, St Vincent de Paul Medical Center Disclosure: Nothing to disclose. Specialty Editor Board Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Nothing to disclose. James L Harper, MD Associate Professor, Department of Pediatrics, Division of Hematology/Oncology and Bone Marrow Transplantation, Associate Chairman for Education, Department of Pediatrics, University of Nebraska Medical Center; Associate Clinical Professor, Department of Pediatrics, Creighton University School of Medicine; Director, Continuing Medical Education, Children's Memorial Hospital; Pediatric Director, Nebraska Regional Hemophilia Treatment Center James L Harper, MD is a member of the following medical societies: American Society of Pediatric Hematology/Oncology, American Federation for Clinical Research, Council on Medical Student Education in Pediatrics, Hemophilia and Thrombosis Research Society, American Academy of Pediatrics, American Association for Cancer Research, American Society of Hematology Disclosure: Nothing to disclose. Chief Editor Vikramjit S Kanwar, MBBS, MBA, MRCP(UK), FRCPCH, FAAP Professor Emeritus of Pediatrics, Albany Medical College; Chief of Pediatric Oncology, Homi Bhabha Cancer Hospital, Varanasi, India Vikramjit S Kanwar, MBBS, MBA, MRCP(UK), FRCPCH, FAAP is a member of the following medical societies: Children's Oncology Group, Indian Academy of Pediatrics, International Society of Pediatric Oncology Disclosure: Nothing to disclose. Additional Contributors J Martin Johnston, MD Associate Professor of Pediatrics, Mercer University School of Medicine; Director of Hematology/Oncology, The Children's Hospital at Memorial University Medical Center; Consulting Oncologist/Hematologist, St Damien's Pediatric Hospital J Martin Johnston, MD is a member of the following medical societies: American Academy of Pediatrics, American Society of Pediatric Hematology/Oncology, International Society of Paediatric Oncology Disclosure: Nothing to disclose. Richard H Sills, MD Professor of Pediatrics, State University of New York Upstate Medical University Richard H Sills, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Pediatrics, American Society of Hematology, American Society of Pediatric Hematology/Oncology Disclosure: Nothing to disclose. Mundeep K Kainth, DO, MPH, FAAP Attending Physician in Pediatric Infectious Diseases, Co-Lead of Antimicrobial Stewardship Program, Cohen Children's Medical Center; Assistant Professor, Donald and Barbara Zucker School of Medicine at Hofstra/Northwell; Instructor, Feinstein Institute for Medical Research, Northwell Health Mundeep K Kainth, DO, MPH, FAAP is a member of the following medical societies: American Academy of Pediatrics, Infectious Diseases Society of America, Pediatric Infectious Diseases Society Disclosure: Travel reimbursement for: Gilead. Alexander Gozman, MD Assistant Professor, Department of Pediatrics, Division of Hematology/Oncology, Albany Medical Center Alexander Gozman, MD is a member of the following medical societies: American Medical Association, American Society of Clinical Oncology, American Society of Hematology, American Society of Pediatric Hematology/Oncology, Children's Oncology Group Disclosure: Nothing to disclose. The authors and editors of Medscape Reference gratefully acknowledge the contributions of previous authors Wayne Hioe, MD, and Mundeep K Kainth, DO, to the development and writing of this article. encoded search term (Pediatric Splenomegaly) and Pediatric Splenomegaly What to Read Next on Medscape Log in or register for free to unlock more Medscape content Unlimited access to our entire network of sites and services Log in or Register
13897	https://pubs.acs.org/doi/10.1021/jp710447j	Nanosize Effects on Hydrogen Storage in Palladium \| The Journal of Physical Chemistry C Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Manage Preferences Recently Viewedclose modal ACS ACS Publications C&EN CAS Access through institution Log In Nanosize Effects on Hydrogen Storage in Palladium Cite Citation Citation and abstract Citation and references More citation options Share Share on Facebook X Wechat LinkedIn Reddit Email Bluesky Jump to Abstract Cited By Expand Collapse Back to top Close quick search form clear search J. Phys. Chem. 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Phys. Chem. C 2008, 112, 9, 3294-3299 ADVERTISEMENT Info Metrics The Journal of Physical Chemistry C Vol 112/Issue 9 Article Get e-Alerts Cite Citation Citation and abstract Citation and references More citation options Share Share on Facebook X WeChat LinkedIn Reddit Email Bluesky Jump to Abstract Cited By Expand Collapse Article February 12, 2008 Nanosize Effects on Hydrogen Storage in Palladium Click to copy article link Article link copied! Miho Yamauchi Ryuichi Ikeda Hiroshi Kitagawa Masaki Takata View Author Information View Author Information Department of Chemistry, Faculty of Science, Kyushu University, Hakozaki 6-10-1, Higashi-Ku, Fukuoka 812-8581, Japan, JST, PRESTO, 4-1-8, Kawaguchi-shi, Saitama 332-0012, Japan, JST, CREST, 4-1-8, Kawaguchi-shi, Saitama 332-0012, Japan, and Structural Science Laboratory, RIKEN SPring-8 Center, Sayo-gun, Hyogo 679-5148, Japan Access Through Access is not provided via Institution Name Loading Institutional Login Options... Access Through Your Institution Add or Change Institution Explore subscriptions for institutions Other Access Options The Journal of Physical Chemistry C Cite this: J. Phys. Chem. C 2008, 112, 9, 3294–3299 Click to copy citation Citation copied! Published February 12, 2008 Publication History Received 29 October 2007 Revised 12 December 2007 Published online 12 February 2008 Published in issue 1 March 2008 research-article Copyright © 2008 American Chemical Society Request reuse permissions Article Views 6511 Altmetric - Citations 354 Learn about these metrics close Article Views are the COUNTER-compliant sum of full text article downloads since November 2008 (both PDF and HTML) across all institutions and individuals. These metrics are regularly updated to reflect usage leading up to the last few days. Citations are the number of other articles citing this article, calculated by Crossref and updated daily.Find more information about Crossref citation counts. The Altmetric Attention Score is a quantitative measure of the attention that a research article has received online. Clicking on the donut icon will load a page at altmetric.com with additional details about the score and the social media presence for the given article. Find more information onthe Altmetric Attention Score and how the score is calculated. Abstract Click to copy section link Section link copied! The size dependencies of the hydrogen-storage properties in polymer-coated Pd nanoparticles with diameters of 2.6 ± 0.4 and 7.0 ± 0.9 nm were investigated by a measurement of hydrogen pressure-composition isotherms. Their storage capacities per constituent Pd atom in the particles decreased with decreasing particle size, whereas the hydrogen concentrations in the two kinds of nanoparticles were almost the same and 1.2 times as much, respectively, as that in bulk palladium after counting zero hydrogen occupancy on the atoms in the first surface layer of the particles. Furthermore, apparent changes in hydrogen absorption behavior with decreasing particle size were observed, that is, a narrowing of the two-phase regions of solid-solution and hydride phases, the lowering of the equilibrium hydrogen pressure, and a decrease in the critical temperature of the two-phase state. By analyzing the isotherms, we quantitatively determined the heat of formation (Δ H α→β) and the entropy change (Δ S α→β) in the hydride formation of the nanoparticle. Δ H α→β and Δ S α→β for the 2.6 ± 0.4 nm diameter Pd nanoparticle were −34.6 ± 0.61 kJ(H 2 mol)-1 and −83.1 ± 1.8 J(H 2 mol)-1 K-1, whereas for the 7.0 ± 0.9 nm diameter Pd nanoparticles the values were −31.0 ± 1.8 kJ(H 2 mol)-1 and −67.3 ± 5.1 J(H 2 mol)-1 K-1, respectively. These quantities gave us a prospective picture of hydrogen absorption in Pd nanoparticles and the peculiarities in the formation of a single nanometer-sized hydride. ACS Publications Copyright © 2008 American Chemical Society Subjects what are subjects Article subjects are automatically applied from the ACS Subject Taxonomy and describe the scientific concepts and themes of the article. Hydrogen Metal nanoparticles Nanoparticles Palladium Phase transitions Read this article To access this article, please review the available access options below. Recommended Access through Your Institution You may have access to this article through your institution. Your institution does not have access to this content. Add or change your institution or let them know you’d like them to include access. Access Through Recommend Publication Institution Name Loading Institutional Login Options... Access Through Your Institution Add or Change Institution Explore subscriptions for institutions Get instant access Purchase Access Read this article for 48 hours. Check out below using your ACS ID or as a guest. Purchase AccessRestore my guest access Recommended Log in to Access You may have access to this article with your ACS ID if you have previously purchased it or have ACS member benefits. Log in below. Login with ACS ID Purchase access Purchase this article for 48 hours $48.00 Add to cart Purchase this article for 48 hours Checkout Corresponding author. Address: Department of Chemistry, Faculty of Science, Kyushu University, Hakozaki 6-10-1, Higashi-Ku, Fukuoka 812-8581, Japan. Tel: +81-92-642-2319. Fax: +81-92-642-2570. E-mail: mhyamascc@mbox.nc.kyushu-u.ac.jp. † Kyushu University. ‡ JST, PRESTO. § JST, CREST. ‖ RIKEN SPring-8 Center. Cited By Click to copy section link Section link copied! Citation Statements beta Smart citations byscite.aiinclude citation statements extracted from the full text of the citing article. 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Progress in Materials Science2023, 131, 101004. Chirasmita Bhattacharya, Kamla D. Netam, Balaji R. Jagirdar. Multi-functional palladium–ruthenium nanocomposites: an approach towards semi-hydrogenation catalysis and hydrogen sorption. Physical Chemistry Chemical Physics2022, 24 (47) , 29043-29056. Komal C. Shrivastava, Ashok K. Pandey, Seemita Banerjee, A.K. Debnath, Sher Singh Meena, Amit P. Srivastava, V. Sudarsan. Hydrogen adsorption and interactions in self-reducing shell hosted palladium nanoparticles on magnetite support. International Journal of Hydrogen Energy2022, 47 (80) , 34128-34138. Load more citations Get e-Alerts Get e-Alerts The Journal of Physical Chemistry C Cite this: J. Phys. Chem. C 2008, 112, 9, 3294–3299 Click to copy citation Citation copied! Published February 12, 2008 Publication History Received 29 October 2007 Revised 12 December 2007 Published online 12 February 2008 Published in issue 1 March 2008 Copyright © 2008 American Chemical Society Request reuse permissions Article Views 6511 Altmetric - Citations 354 Learn about these metrics close Article Views are the COUNTER-compliant sum of full text article downloads since November 2008 (both PDF and HTML) across all institutions and individuals. These metrics are regularly updated to reflect usage leading up to the last few days. Citations are the number of other articles citing this article, calculated by Crossref and updated daily.Find more information about Crossref citation counts. The Altmetric Attention Score is a quantitative measure of the attention that a research article has received online. Clicking on the donut icon will load a page at altmetric.com with additional details about the score and the social media presence for the given article. Find more information onthe Altmetric Attention Score and how the score is calculated. Recommended Articles ### High-Performance Nanostructured Palladium-Based Hydrogen Sensors—Current Limitations and Strategies for Their Mitigation November 12, 2020 ACS Sensors Iwan Darmadi , Ferry Anggoro Ardy Nugroho ,and Christoph Langhammer ### Adsorption and Absorption Energies of Hydrogen with Palladium August 19, 2022 The Journal of Physical Chemistry C Michael Schwarzer , Nils Hertl , Florian Nitz , Dmitriy Borodin , Jan Fingerhut , Theofanis N. Kitsopoulos ,and Alec M. Wodtke ### On the Nature of Strong Hydrogen Atom Trapping Inside Pd Nanoparticles January 18, 2008 Journal of the American Chemical Society Hirokazu Kobayashi , Miho Yamauchi , Hiroshi Kitagawa , Yoshiki Kubota , Kenichi Kato ,and Masaki Takata ### Nanostructured Metal Hydrides for Hydrogen Storage October 2, 2018 Chemical Reviews Andreas Schneemann , James L. White , ShinYoung Kang , Sohee Jeong , Liwen F. 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13898	https://hifla.org/research/center-for-acoustic-neuroma-research/	Center for Acoustic Neuroma Research Research – Research Overview – Our Team – Research Centers – Active Studies – Research Labs – Research Initiatives Education – Education Overview – Temporal Bone Dissection Course – Temporal Bone Dissection Manual – Visiting Physicians Program – Fellows – NF2 Symposium – Grand Rounds – Online Neurotology Education (ONE) Grand Rounds – RESONANCE: Sound Solutions Symposium Global Hearing Health – Global Hearing Health Overview – Global Hearing Health Coordination Center (GHHCC) – Increasing Access – Providing Training – Coordinating Humanitarian Efforts – Addressing Policy Issues ×  House Institute Foundation House Children’s Hearing Center Hearing Health Centers Ear Clinic Neurosurgery Clinic  House Institute Foundation House Children’s Hearing Center Hearing Health Centers Ear Clinic Neurosurgery Clinic Donate About About Our Foundation History Why Hearing Health Leadership Our Work Resources Temporal Bone Dissection Manual House Calls Magazine Blog Sudden Sensorineural Hearing Loss For Professionals General Information on Sudden Sensorineural Hearing Loss Online Neurotology Education (ONE) Grand Rounds Support Us Donate Planned Giving Ways To Support Us Corporate Sponsors Contact About About Our Foundation History Why Hearing Health Leadership Our Work Resources Temporal Bone Dissection Manual House Calls Magazine Blog Sudden Sensorineural Hearing Loss For Professionals General Information on Sudden Sensorineural Hearing Loss Online Neurotology Education (ONE) Grand Rounds Support Us Donate Planned Giving Ways To Support Us Corporate Sponsors Contact Center for Acoustic Neuroma Research Home Research Center for Acoustic Neuroma Research The House Institute Foundation’s Center for Acoustic Neuroma Research is actively engaged in translational research that includes laboratory-based analysis of tumor tissue and the evaluation of innovative surgical techniques. The House Clinic researchers primarily work with patients living with NF type 2 who struggle from acoustic neuroma tumors that impact balance, the ability to hear, swallow, and speak, as well as eye and facial movement. About Neurofibromatosis and Acoustic Neuromas Neurofibromatosis (NF) is a genetic disorder that causes tumors to form on nerve tissue. There are two types of neurofibromatosis. Patients with NF type 1 often develop tumors on or close to the skin around the spine or in the brain. The House Clinic researchers primarily work with patients living with NF type 2, which often involves tumors in both ears and multiple simultaneous brain and spinal cord tumors. The most common tumor associated with NF type 2 is the acoustic neuroma. Acoustic Neuromas, also known as vestibular schwannomas, are slow-growing tumors of the balance and hearing nerves. As they grow, they can compress nerves for facial movement, swallowing, speaking, eye movement, hearing, and balance. HIF faculty in the Center for Acoustic Neuroma Research are actively engaged in translational research that includes laboratory-based analysis of tumor tissue and the evaluation of innovative surgical techniques. The research is guided by HIF’s commitment to improve the quality of life for patients living with acoustic neuromas and neurofibromatosis. Areas of Research The House Institute Foundation’s (HIF) Center for Acoustic Neuroma Research includes many accomplishments and studies that demonstrate the Center’s commitment to improving care for patients with acoustic neuromas or NF. The Center’s ongoing Natural History Study of NF2, initially funded by the Department of Defense, investigates growth patterns of NF2 tumors to allow for a better understanding of tumor progression and treatment timeline. As a major contribution to NF research, HIF developed the NF in vitro cell lines that are now used around the world to test drug therapies for NF. Finally, current research focuses on surgical techniques for acoustic neuromas. These include the trans-labyrinthine approach and the middle fossa approach, which can be used as a hearing preservation strategy or for decompression. Natural History Study The aim of this ongoing, longitudinal study is to understand the natural history of tumors and symptoms in patients with NF2. Due to the rarity of the disease and diverse presentation of symptoms in patients, little is known about the progression of NF2, and there are limited treatments. The Center intends to use clinical data, obtained over the course of the study, to discover common patterns and develop personalized treatments for NF2. The Natural History study will be key in fully understanding the cause, progression, and treatment of NF2, as well as aid concurrent studies of NF2 sequencing, and revolutionize treatment for NF2. Clinical Tumor Bank The Center has established a tumor bank that holds various tumor and nervous system tissues from patients with and without NF2. The clinical tumor bank allows for the identification of targeted treatment options for different NF2 tumor profiles. At the time of surgery, tumor tissue, ear tissues, serum and plasma, and cerebrospinal fluid samples are collected for further exploration. The clinical tumor bank is used for exploration of chemotherapies. Collected tumor tissue will be used to test against specific chemotherapeutic agents to discover the most effective agent for tumor growth arrest and shrinkage. Proteomics and Genomics of NF2 The Center is conducting proteomic and genomic analysis of NF tumor samples. Sequencing tumors will help to identify drivers of phenotype for NF2, and why tumors may respond more to some treatments than to others. Sequencing may also illuminate genetic factors that contribute to those outcomes. The Center will use these analyses, with a patient-driven approach, to create genetic and protein profiles for individual NF2 patients. Tumor sequencing may reveal secondary pathways and mutations that will respond to specific agents and aid treatment outcomes, driving clinical and research practices. Research Goals An ongoing initiative in the Center is mapping patients’ histories with NF2 and identifying patterns, specific NF2 mutations and pathways, and unique profiles to develop more personalized medicine for children and adults with NF2. Ultimately, the goal is to use the current Natural History data alongside tissue bank data to personalize medicine for NF2. The Center’s genomics and proteomics findings will work in conjunction with Natural History variables to better understand NF2 mutations and pathways that may have implications for groundbreaking clinical care. Natural History Study of NF2, initially funded by Department of Defense This study established the growth patterns of vestibular schwannomas in patients with NF2. This allows us to better understand how and when these tumors progress and when treatment is indicated. Development of House Ear Institute NF Cell Lines These in vitro cell lines are used around the world to test drug therapies for potential efficacy in NF. Pioneered the Auditory Brainstem Implant (ABI) This is the first central neural prosthesis for hearing restoration. ABI technology is critical when the cochlear nerve is damaged by tumors in NF2. Surgical techniques for vestibular schwannomas (acoustic neuromas) Physicians at the House Institute have developed surgical techniques for treating vestibular schwannomas in NF2 that are used throughout the world. These include the trans-labyrinthine approach and the middle fossa approach, which can be used as a hearing preservation strategy or for decompression. Our Team William H. Slattery, III MD, Board Chair – Read More Derald E. Brackmann MD – Neurotologist – Read More John W. House MD – Neurotologist Read More William M. Luxford MD – Neurotologist – Read More Kevin A. Peng MD, Research Committee Chair Read More Shemms Najjar Hearing Science Accelerator Coordinator– Stay Updated On Our Research Efforts First Name Last Name Email Submit LOCATION 1127 Wilshire Blvd., Suite 1620 Los Angeles, CA 90017  (213) 770-2187  hello@hifla.org Follow Follow Follow Follow Follow CLINICAL DEPARTMENTS Ear Clinic Hearing Aid Centers Neurosurgery Clinic Children’s Hearing Center SUPPORT US Donate Now Make a Monthly Gift Corporate Sponsors Planned Giving OUR STORY About Our Foundation History Why Hearing Health Careers Copyright © 2021 – The House Institute. All Rights Reserved. HIPAA Statement \| Privacy Policy \| Terms & Conditions \| Sitemap
13899	https://www.aiche.org/sites/default/files/cep/20161238_r.pdf	38 www.aiche.org/cep December 2016 CEP Back to Basics W hen I left university, I found that I needed addi-tional information to turn my theoretical knowl-edge of fluid mechanics into the practical knowl-edge required to specify a pump. Judging by the questions I see asked nearly every week on LinkedIn and elsewhere, I believe this is a problem shared by many engineers early in their careers. This article gives practical insight on how to specify a pump. Pump types Pumps can be used to move fluids, which flow from regions of high pressure to regions of low pressure, by increasing the pressure of the fluid. Before you purchase a pump, you must specify the type of pump and make sure it is capable of delivering a given flowrate at a given pressure. There are two main pump types: rotodynamic and positive-displacement. In a rotodynamic pump, a rotating impeller imparts energy to the fluid. The most common type of rotodynamic pump is the centrifugal pump (Figure 1). The amount of liquid that passes through the pump is inversely proportional to the pressure at the pump outlet. In other words, the outlet flowrate of a rotodynamic pump varies nonlinearly with pressure. In a positive-displacement (PD) pump, a discrete amount of fluid is trapped, forced through the pump, and discharged. A gear pump is an example of a PD pump (Figure 2). This pumping principle produces a pulsating flow, rather than a smooth flow. Its output flow tends to vary little with respect to the pressure at the pump outlet, because the moving displacement mechanism pushes the slug of liquid out at a constant rate. Most process pumps are rotodynamic pumps, so you need to know the required outlet pressure to specify the pump that will provide the required flow. Alhough cer-tain system head parameters are calculated the same way whether the driving force for flow is a pump or gravity, this article mainly addresses sizing concerns for rotodynamic pumps. Pump sizing Pump sizing involves matching the flow and pressure rating of a pump with the flowrate and pressure required for the process. The mass flowrate of the system is established on the process flow diagram by the mass balance. Achiev-ing this mass flowrate requires a pump that can generate a pressure high enough to overcome the hydraulic resistance of the system of pipes, valves, and so on that the liquid must travel through. This hydraulic resistance is known as the system head. In other words, the system head is the amount of pressure required to achieve a given flowrate in the system downstream of the pump. The system head is not a fixed quantity — the faster the liquid flows, the higher the system head becomes (for reasons to be discussed later). However, a curve, known as the system curve, can be drawn to show This article explains some of the core concepts behind pump sizing. Seán Moran Expertise Ltd. Pump Sizing: Bridging the Gap Between Theory and Practice Copyright © 2016 American Institute of Chemical Engineers (AIChE) CEP December 2016 www.aiche.org/cep 39 the relationship between flow and hydraulic resistance for a given system. Pump sizing, then, is the specification of the required outlet pressure of a rotodynamic pump (whose output flow varies nonlinearly with pressure) with a given system head (which varies nonlinearly with flow). Understanding system head The system head depends on properties of the system the pump is connected to — these include the static head and the dynamic head of the system. The static head is created by any vertical columns of liquid attached to the pump and any pressurized systems attached to the pump outlet. The static head exists under static conditions, with the pump switched off, and does not change based on flow. The height of fluid above the pump’s centerline can be determined from the plant layout drawing. The dynamic head varies dynamically with flowrate (and also with the degree of opening of valves). The dynamic head represents the inefficiency of the system — losses of energy as a result of friction within pipes and fittings and changes of direction. This ineffiency increases with the square of the average velocity of the fluid. Dynamic head can be further split into two parts. The frictional loss as the liquid moves along lengths of straight pipe is called the straight-run headloss, and the loss as a result of fluid passing through pipe fittings such as bends, valves, and so on is called the fittings headloss. Fully characterizing a hydraulic system is incredibly complex. Remember that in order to specify a pump, you only need to characterize the system well enough to choose a pump that will perform the job in question. How exact you need to be depends on where in the design process you are. If you are at the conceptual stage, you may be able to avoid specifying the pump at all, but experience suggests that you should use rules of thumb to specify certain parameters (such as superficial velocity) to prevent difficulties later. I also recommend designing the process so that it does not have two-phase flow. Two-phase flow is difficult to predict, and should be avoided in your design if at all possible — head losses can be one thousand times those for single-phase flow. Installing knock-out drums in the system and arranging pipework so that gases are not entrained in liquids can help mitigate two-phase flow. Superficial velocity is the same as average velocity and is the volumetric flowrate (in m3/sec, for example) divided by the pipe’s internal cross-sectional area (e.g., in m2). A very quick way to start the hydraulic calculations is to use the following superficial velocities: • pumped water-like fluids: <1.5 m/sec • gravity-fed water-like fluids: <1 m/sec • water-like fluids with settleable solids: >1, <1.5 m/sec • air-like gases: 20 m/sec Keeping the system within these acceptable ranges of superficial velocities, and avoiding two-phase flow, will typically produce sensible headlosses for the pipe lengths usually found in process plants. Determining frictional losses through fittings Dynamic, or friction, head is equal to the sum of the straight-run headloss and the fittings headloss. The fittings headloss is calculated by what is known as the k-value method. Each type of valve, bend, and tee has Discharge Suction Impeller Eye Impeller p Figure 1. In a centrifugal pump, a rotating impeller imparts energy to the liquid moving through the pump. Outlet Inlet Low-Pressure Fluid High-Pressure Fluid Fluid Carried Between Teeth and Case p Figure 2. A gear pump is a type of positive-displacement pump in which a discrete volume of fluid is trapped and then discharged. Copyright © 2016 American Institute of Chemical Engineers (AIChE) 40 www.aiche.org/cep December 2016 CEP Back to Basics a characteristic resistance coefficient, or k value, which can be found in Perry’s Handbook (1) and other sources (Table 1) (2). To use this method, count the number of valves on the piping and instrumentation diagram (P&ID), and the fittings, bends, and tees on the plant layout drawing for the relevant suction or delivery line. Multiply the number of each type of fitting by the corresponding k value, and add the k values for the various types of fittings to get the total k value. Use the total k value to calculate the headloss due to fittings: where hf is the fittings headloss in meters water gauge (mwg), k is the total k value, v is the superficial velocity (m/sec), and g is the acceleration due to gravity (9.81 m/sec2). Calculating straight-run headloss At a more-advanced stage of design, you might want to know a pump’s physical size to try out on a plant layout drawing. An easy way to determine the straight-run head-loss — the most difficult part of a headloss calculation — is to use a nomogram such as Figure 3 or a table. Pipe manu-facturers (and others) produce tables and nomograms that can be used to quickly look up headloss due to friction for liquids. To use the nomogram, use a ruler to draw a straight line through any pair of known quantities to determine unknown quantities. For example, for a 25-mm nominal-bore pipe with a flow velocity of 1 m/sec, the straight-run headloss is about 6 m per 100 m of pipe. So the headloss through 10 m of this pipe is around 0.6 mwg. At an early design stage, you often need to calculate the straight-run headloss multiple times. Rather than referring to a table or nomogram numerous times, it can be quicker to set up an Excel spreadsheet and use a formula to calculate the Darcy friction factor and headloss. Chemical engineering students are usually taught to find the Darcy friction factor using a Moody diagram, which is a summary of a large number of empirical experiments. You can use curve- fitting equations and software such as Excel to approximate the Moody diagram’s output. Don’t confuse the Darcy friction factor with the Fanning friction factor — the Darcy friction factor is by definition four times the Fanning friction factor. If you do decide to use a Moody diagram to find the friction factor, be aware of which friction factor is on the y-axis. I prefer the Colebrook-White approximation to calculate the Darcy friction factor. Although it is an approximation, it Table 1. Each type of pipe fitting has a resistance coefficient, or k value, that can be used to calculate the fittings headloss for the pump system (2). Fitting Type k Value Short-radius bends, for every 22.5 deg. allow 0.2 Long-radius bends, for every 22.5 deg. allow 0.1 Open isolation valve 0.4 Open control valve 10.8 Tee (flow from side branch) 1.2 Tee (flow straight-through) 0.1 Swing check non-return valve 1 Sharp entry 0.5 Internal Diameter, mm Flowrate, L/sec L/min Flow Velocity, m/sec Pressure Drop, m/100m 15 20 25 30 35 40 50 60 70 80 90 100 150 200 250 300 350 400 500 0.01 0.02 0.05 0.1 0.2 0.3 0.5 1 2 3 4 5 20 30 40 50 100 200 300 400 500 1000 2000 3000 4000 5000 0.4 1 2 3 4 5 20 30 40 50 100 200 300 400 500 1000 2000 3000 4000 5000 10 10000 20000 30000 40000 50000 100 200 300 m3/min 0.05 0.1 0.15 0.2 0.3 0.5 0.4 1 2 3 4 5 1.5 10 15 20 20 10 2 3 4 5 1 0.2 0.3 0.5 0.4 0.1 0.05 0.04 0.03 0.02 0.01 Approximate values only Water at 10°C p Figure 3. A piping nomogram, available from pipe manufacturers, can be used to estimate the straight-run headloss for a pump system. In the example shown by the red line, a 25-mm pipe with a flow velocity of 1 m/sec has a straight-run headloss of about 6 m per 100 m of pipe. Copyright image reproduced courtesy of Durapipe SuperFLO ABS technical data. Copyright © 2016 American Institute of Chemical Engineers (AIChE) CEP December 2016 www.aiche.org/cep 41 might be closer to the true experimental value than what the average person can read from a Moody diagram. The Colebrook-White approximation can be used to esti-mate the Darcy friction factor (fD) from Reynolds numbers greater than 4,000: where Dh is the hydraulic diameter of the pipe, ε is the sur-face roughness of the pipe, and Re is the Reynolds number: where ρ is the density of the fluid, D is the pipe internal diameter, and μ is the fluid dynamic viscosity. The Colebrook-White approximation can be used itera-tively to solve for the Darcy friction factor. The Goal Seek function in Excel does this quickly and easily. The Darcy-Weisbach equation states that for a pipe of uniform diameter, the pressure loss due to viscous effects (Δp) is proportional to length (L) and can be charac-terized by: This iterative approach allows you to calculate straight-run headloss to the degree of accuracy required for virtually any practical application. I recently came across a paper (3) that suggested there are other equations that provide more accurate results through curve-fitting than the Colebrook-White approxi-mation. If you are producing your own spreadsheet for this purpose, I suggest you look into the Zigrang and Sylvester (4) or Haaland equations (5) (Table 2). These equations also apply for Reynolds numbers greater than 4,000. Adding together the static head, the fittings headloss, and the straight-run headloss will give you the total head the pump needs to generate to overcome resistance and deliver the specified flowrate to the system. Suction head and net positive suction head Even at an early stage, I also recommend determining the pump’s required net positive suction head and calculat-ing the net positive suction head (NPSH), as they can affect much more than pump specification. The pump’s required net positive suction head takes into consideration the liquid’s vapor pressure to avoid cavitation in the pump. I recommend creating an Excel spreadsheet that uses the Antoine equation to estimate the vapor pressure of the liquid at the pump inlet and then calculate the NPSH at that vapor pressure. The Antoine equation may be expressed as: where Pv is vapor pressure of the liquid at the pump inlet, T is temperature, and A, B, and C are coefficients that can be obtained from the NIST database ( among other places. Table 3 shows an example for water. The net positive suction head is: where Po is the absolute pressure at the suction reservoir, ho is the reservoir liquid level relative to the pump center-line, and hSf is the headloss due to friction on the suction side of the pump. Note that NPSH is calculated differently for centrifugal and positive-displacement pumps, and that it varies with pump speed for positive-displacement pumps rather than with pressure as for centrifugal pumps. Equation 6 should only be used with centrifugal pumps. Article continues on next page Table 3. Vapor pressure for water at 30°C, calculated using the Antoine equation. Material A B C T, °C T, K Pv, bar Pv, Pa Water 5.40221 1,838.675 –31.737 30 303.15 0.042438 4,243.81 Table 2. These alternative curve-fitting equations can be used in lieu of the Colebrook-White equation to determine the Darcy friction factor. Equation Range Source e = 0.00004–0.05 (4) e = 0.000001–0.05 (5) Copyright © 2016 American Institute of Chemical Engineers (AIChE) 42 www.aiche.org/cep December 2016 CEP Back to Basics Determining pump power After the system head has been calculated, it can be used to calculate an approximate pump power rating for a centrif-ugal pump: where P is the pump power (kW), Q is the flowrate (m3/hr), H is the total pump head (m of fluid), and η is the pump efficiency (if you do not know the efficiency, use η = 0.7). The pump manufacturer provides the precise power rat-ings and motor size for the pump, but the electrical engineers need an approximate value of this (and pump location) early in the design process to allow them to size the power cables. You should err on the side of caution in this rating calculation (the electrical engineers will be much happier if you come back later to ask for a lower power rating than a higher one). In certain stages of design development, the preliminary drawings are modified to match likely hydraulic conditions across the design envelope. This may require you to do many approximate hydraulic calculations before the design has settled into a plausible form. After you have performed the hydraulic calculations, the pump and possibly the pipe sizes might need to be changed, as might the minimum and maximum operating pressures at certain points in the system. As the system design becomes more refined, there might even be a requirement to change from one pump type to another. Hydraulic networks The previous sections describe how to calculate the headloss through a single line, but what about the common situation where the process has branched lines, manifolds, and so on? When each branch handles a flow proportional to its headloss, and its headloss is proportional to the flow passing through it, producing an accurate model can become complex very quickly. My approach to this is to first simplify and then improve the design as much as possible with a few rules of thumb: • Avoid manifold arrangements that provide a straight-through path from the feed line to a branch. Entry perpendic-ular to branch direction is preferred. • Size manifolds such that the superficial velocity never exceeds 1 m/sec at the highest anticipated flowrate. • Specify progressively smaller manifold diameters to accommodate lower flows to downstream branches. • Include a small hydraulic restriction in the branch so the branch headloss is 10–100 times the headloss across the manifold. • Design-in passive flow equalization throughout the pip-ing system wherever possible by making branches hydrauli-cally equivalent. Perform headloss calculations for each section of the simplified plant design at expected flows to find the flow path with the highest headloss. Use the highest-headloss path to determine the required pump duty — calculate the pump duty at both the average flow with working flow equalization, and at full flow through a single branch. Usu-ally these do not differ much, and the more rigorous answer lies between them. Only if the two results of this approach are very different will I do a more rigorous (and time- consuming) analysis. If such a rigorous analysis is needed, I create an Excel spreadsheet based on the Hardy Cross method — a method for determining the flow in a pipe network when the flows within the network are unknown but the inputs and outputs are known — and solve for individual pipe flows. Excel’s Solver function can be used to find the change in flow that gives zero loop headloss. In the unlikely event that you have to do this, an explanation of how to carry out the method can be found in Ref. 6. There are many computer programs available to do these calculations. Pump curves A pump curve is a plot of outlet pressure as a function of flow and is characteristic of a certain pump. The most frequent use of pump curves is in the selection of centrifugal pumps, as the flowrate of these pumps varies dramatically with system pressure. Pump curves are used far less fre-quently for positive-displacement pumps. A basic pump curve plots the relationship between head and flow for a pump (Figure 4). On a typical pump curve, flowrate (Q) is on the horizon-tal axis and head (H) is on the vertical axis. The pump curve shows the measured relationship between these variables, so Pump Curve System Curve Duty Point Head, H, m Pressure, psi Flowrate, Q, L/sec Flowrate, Q, m3/hr p Figure 4. A basic pump curve plots pressure (or head) as a function of flowrate. Copyright © 2016 American Institute of Chemical Engineers (AIChE) CEP December 2016 www.aiche.org/cep 43 it is sometimes called a Q/H curve. The intersection of this curve with the vertical axis corresponds to the closed valve head of the pump. These curves are generated by the pump manufacturer under shop test conditions and ideally repre-sent average values for a representative sample of pumps. A plot of the system head over a range of flowrates, from zero to some value above the maximum required flow, is called the system curve. To generate a system curve, com-plete the system head calculations for a range of expected process flowrates. System head can be plotted on the same axes as the pump curve. The point at which the system curve and the pump curve intersect is the operating point, or duty point, of the pump. Remember that a system curve applies to a range of flows at a given system configuration. Throttling a valve in the system will produce a different system curve. If flow through the system will be controlled by opening and closing valves, you need to generate a set of curves that represent expected operating conditions, with a corresponding set of duty points. It is common to have efficiency, power, and NPSH plotted on the same graph (Figure 5). Each of these variables requires its own vertical axis. To obtain the pump efficiency at the duty point, draw a line vertically from the duty point to the efficiency curve, and then draw a horizontal line from there to the vertical axis that corresponds to efficiency. Sim-ilarly, to obtain the motor power requirement, draw a line down from the duty point to the motor duty curve. More sophisticated curves may include nested curves representing the flow/head relationship at different supply frequencies (i.e., the AC electrical supply’s frequency in Hz) or rotational speeds, with different impellers, or for different fluid densities. Curves for larger impellers or faster rotation lie above curves for smaller impellers or slower rotation, and curves for lower-density fluids lie above curves for higher- density fluids. A more-advanced pump curve might also incorporate impeller diameters and NPSH. Figure 6 depicts pump curves for four different impellers, ranging from 222 mm to 260 mm. Corresponding power curves for each impeller are shown on the bottom of the figure. The dashed lines in Figure 6 are efficiency curves. These curves can start to look a bit confusing, but the important point to keep in mind is that, just as in the simpler examples, flowrate is always on a common horizontal axis, and the corresponding value on any curve is vertically above or below the duty point. These more-advanced curves usually incorporate effi-ciency curves, and these curves define a region of highest efficiency. At the center of this region is the best efficiency point (BEP). Choose a pump that has an acceptable efficiency across the range of expected operating conditions. Note that we are not necessarily concerned with the entire design envelope — Pump Curve Efficiency System Curve Power Consumption NPSH Head, m 0 10 20 30 40 50 60 Efficiency, % 0 10 20 30 40 50 60 70 80 Power, kW 0 2 4 6 8 10 12 NPSH, m 0 2 4 6 8 10 12 Flowrate, m3/hr 10 20 30 40 50 60 70 80 p Figure 5. Efficiency, power, and net positive suction head can also be plotted on a pump curve. Original image courtesy of Grundfos. Head, m 80 70 60 50 40 30 20 0 20 40 60 80 100 120 140 160 180 0 20 40 60 80 100 120 140 160 180 Flowrate, m3/hr 0 4 8 12 16 20 24 28 32 36 40 Power, kW 0 2 4 6 8 10 12 NPSH, m NPSH 222 mm 235 mm 247 mm 260 mm 222 mm 235 mm 247 mm 260 mm 67.5% 66.8% 70.2% 71.7% 64% 67% 70% 64% 67%70% p Figure 6. A complex pump curve integrates efficiency, NPSH, and impeller diameters on one diagram. Copyright image reproduced courtesy of Grundfos. Copyright © 2016 American Institute of Chemical Engineers (AIChE) 44 www.aiche.org/cep December 2016 CEP Back to Basics it is not crucial to have high efficiency across all conceivable conditions, just the normal operating range. The optimal pump for your application will have a BEP close to the duty point. If the duty point is far to the right of a pump curve, well away from the BEP, it is not the right pump for the job. Even with the most cooperative pump supplier, some-times the curves that you need to make a pump selection may not be available. This is commonly the case if you want to use an inverter to control pump output based on speed. However, you can often generate acceptable pump curves using the curves you have and the following approxi-mate pump affinity relationships: where the subscript 1 designates an initial condition on a known pump curve and subscript 2 is some new condition. The NPSH relationship in Eq. 11 is more of an approx-imation than the others. The value of x lies in the range of –2.5 to +1.5, and y in the range of +1.5 to +2.5. Closing thoughts These are the basics of pump selection. A final word of advice: If you don’t understand what is presented here, or need to know more, I suggest that you talk to a pump sup-plier in private. Think twice before you post on social media to ask for advice on the basics of pump selection — the advice you receive may not be correct, and your post may reflect badly on you and your employer. SEÁN MORAN has had 25 years of experience in process plant design, troubleshooting, and commissioning. He was an associate professor and Coordinator of Design Teaching at the Univ. of Nottingham for four years, and is presently a visiting professor at the Univ. of Chester. He has written three books on process plant design for the Institution of Chemical Engineers. His professional practice now centers on acting as an expert witness in commercial disputes regarding process plant design issues, although he still has cause to put on a hardhat from time to time. He holds a master’s degree in biochemical engineering from Univ. College London. Literature Cited 1. Perry, R. H., and Green, D. W., “Perry’s Chemical Engineers’ Handbook,” 8th Ed., McGraw-Hill, New York, NY, p. 6-18 (2007). 2. Moran, S., “An Applied Guide to Process and Plant Design,” Butterworth-Heinemann Oxford, U.K. (2015). 3. Genić, S., et al., “A Review of Explicit Approximations of Colebrook’s Equation,” FME Transactions, 39, pp. 67–71 (June 2011). 4. Zigrang, D. J., and N. D. Sylvester, “Explicit Approximations to the Solution of Colebrook’s Friction Factor Equation,” AIChE Journal, 28 (3), pp. 514–515 (May 1982). 5. Haaland, S. E., “Simple and Explicit Formulas for the Friction Factor in Turbulent Flow,” Journal of Fluids Engineering, 105 (1), pp. 89–90 (1983). 6. Huddleston, D., et al., “A Spreadsheet Replacement for Hardy- Cross Piping System Analysis in Undergraduate Hydraulics,” Critical Transitions in Water and Environmental Resources Management, pp. 1–8 (2004). Nomenclature A, B, C = Antoine coefficients D = pipe internal diameter Dh = hydraulic diameter of the pipe fD = Darcy friction factor g = acceleration due to gravity (9.81 m/sec2) H = total system head hf = headloss due to fittings in meters water gauge (mwg) ho = reservoir liquid level relative to the pump centerline hSf = headloss due to friction on the suction side of the pump k = resistance coefficient of valves, fittings, bends, tees, etc. L = length of pipe NPSH = net positive suction head P = power (kW) Pv = vapor pressure of the liquid at the pump inlet Po = absolute pressure at the suction reservoir Q = flowrate Re = Reynolds number T = temperature v = superficial velocity Greek letters Δp = pressure loss due to viscous effects ε = surface roughness of the pipe η = pump efficiency μ = fluid dynamic viscosity (kg/(m-sec)) ρ = density of fluid (kg/m3) CEP Copyright © 2016 American Institute of Chemical Engineers (AIChE)

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