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1,312	Here are some ways to do just that: Here are some ways you can teach your children to appreciate differences in a diverse world. According to koodakpress، Kids aren’t born to hate. Intolerance is learned. And in the same way, kids can be taught to be sensitive, understanding, empathetic and tolerant. Although it’s certainly never too late to begin, the sooner we start, the better chance we have of preventing intolerant attitudes from taking hold of our children’s minds and hearts. After all, if our kids are to have any chance of living harmoniously in our multiethnic, global 21st century, it is critical that we raise them to be tolerant, kind and inclusive. Here are some ways to do just that: Confront your prejudices. The first step to nurturing tolerance is to examine your own biases. Chances are that you are communicating those attitudes and might not even know you are tainting your kids’ views. Reflect on your upbringing and your own parents’ prejudices. Might any biases remain with you today that you are projecting to your child? Make a conscious attempt to temper those biases so that they don’t prejudice your child. Model tolerance. Traits such as tolerance and compassion are caught as well as taught, and that’s exactly why we need to practice what we preach. Convey to your child how strongly you believe that all people – regardless of race, gender, religion, age, ability, sexual orientation, economic background, appearance or culture – should be treated with respect and dignity. Broaden horizons. Encourage your child to have contact with individuals and make friends of different races, cultures, ages, genders, abilities and beliefs. We’re more likely to empathize with those who are like us: our gender, age, income or race, for example. But stepping outside our comfort zones and mingling with and befriending people who are different from us builds tolerance and conveys to our kids that they should expand their own social horizons. Look for commonalities, too. We are more likely to empathize with those who are “like us.” This doesn’t mean overlooking differences. Rather, when your child points out how another child is different from them, it’s about also showing them things they have in common. Like if your child notices that another kid looks different or has beliefs that your child doesn’t share, take the time to discuss similarities as well, such as that they like the same subjects in school. Talk about race and exclusion. Kids are naturally curious, so you should expect questions about differences, but how you respond can create or prevent stereotypes from forming. Answer questions simply and honestly though some may seem embarrassing or even taboo. Suppose your child says, “Sally is a girl and shouldn’t play football!” You might say, “Girls can play the same sports boys do. Sally likes to play football, so she should play it.” Or perhaps your child asks why the color of other kids’ skin is different from hers. You could explain that “skin comes in lots of different colors just like eyes come in different colors.” Expose your child to diverse experiences and ideas. Inexperience, especially if combined with incomplete information, can lead children to have fears or insecurities about others and develop stereotypes. To help your child respect different perspectives, expose her to toys, dolls, food, music, customs, videos and games from an early age that represent a wide range of multicultural groups. Encourage your child to participate in social and community activities or to visit museums which promote cross-cultural programs, diversity, resistance to hate groups and other actions that nurture tolerance. Studies show that only a small percent of children’s books feature people of color. So expose your kids to literature that features positive images of various cultures and help them recognize that even though people look different on the outside, we’re all the same on the inside. For younger kids, check out “Same, Same But Different” by Jenny Sue Kostecki-Shaw (India); “How My Family Came to Be: Daddy, Papa and Me” by Andrew Aldrich, about the author’s experience as an African-American child adopted by white, gay couple; “The Girl Who Loved Wild Horses,” by Paul Goble (Native American). Older kids, check out, “Children of the River,” by Linda Crew (Cambodia); “Dear Mrs. Parks,” by Rosa Parks; and “The Invisible Thread” by Yoshiko Uchida (Japan). Refuse to allow discriminatory comments. When you hear prejudicial comments, verbalize your displeasure. Such stereotypes can address gender, race, age, lifestyles, beliefs, appearance, abilities, religion and culture and are always sweeping generalities that can become entrenched beliefs. Your child needs to hear your discomfort so that she knows your values: “That’s disrespectful and I won’t listen to disrespect” or “That’s a biased comment, and I don’t want to hear it.” It also models a response kids can imitate if prejudicial comments are made in their presence. Teach kids how to be inclusive. If we want our kids to include others, we need teach them how. Show your child how to meet new people. Model how to introduce yourself and invite an acquaintance to sit with you. Let your child see how you start conversations and encourage others. Then help your child practice those skills at home so he can use them in the real world. Stress empathy and talk about exclusion. Talk to your child about the effects of deliberately excluding someone because they are different. “How would you feel? How do you think she feels? What could you do to make her feel better?” Role play with your child what to do or say if your child sees it happen. “Suppose a new student comes to school who is of a different ethnic group than your friends. You want to get to know him better, but your friends don’t want to include him because he looks different. What will you do? What should you do?”<\|endoftext\|>	3.984375
378	How do you factor c^3 +f^3? $\left(c + f\right) \left({c}^{2} - c f + {f}^{2}\right)$ Explanation: the formula for factoring sum of two cubes is $\left({a}^{3} + {b}^{3}\right) = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$ therefore we let a=c and b=f and we have ${c}^{3} + {f}^{3} = \left(c + f\right) \left({c}^{2} - c f + {f}^{2}\right)$ God bless....I hope the explanation is useful... Feb 26, 2016 ${c}^{3} + {f}^{3} = \left(c + f\right) \left({c}^{2} - c f + f 2\right)$ Explanation: ${c}^{3} + {f}^{3}$ represents a sum of cubes, ${a}^{3} + {b}^{3}$, where $a = c$ and $b = f$. The formula for the factorization of a sum of cubes is ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$. Substitute $c \mathmr{and} f$ into the formula. ${c}^{3} + {f}^{3} = \left(c + f\right) \left({c}^{2} - c f + {f}^{2}\right)$<\|endoftext\|>	4.53125
2,574	Courses # Ratio And Proportion - MCQ 3 ## 20 Questions MCQ Test Quantitative Techniques for CLAT \| Ratio And Proportion - MCQ 3 Description This mock test of Ratio And Proportion - MCQ 3 for Quant helps you for every Quant entrance exam. This contains 20 Multiple Choice Questions for Quant Ratio And Proportion - MCQ 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Ratio And Proportion - MCQ 3 quiz give you a good mix of easy questions and tough questions. Quant students definitely take this Ratio And Proportion - MCQ 3 exercise for a better result in the exam. You can find other Ratio And Proportion - MCQ 3 extra questions, long questions & short questions for Quant on EduRev as well by searching above. QUESTION: 1 ### Divide Rs.2340 into three parts, such that first part be double that of second part and second part be 1/3 of the third part.Find the Third part amount? Solution: Answer – B.Rs.1170 Explanation : First: Second: Third = 2:1:3 Third part = 32340/6 = 1170 QUESTION: 2 ### The ratio of income of A and B is 2:3. The sum of their expenditure is Rs.8000 and the amount of savings of A is equal to the amount of expenditure of B.What is the their ratio of sum of income to their sum of savings? Solution: Answer -A.5:3 Explanation : 2I-E + E = 8000 I = 4000 Sum of their Income = 5I = 54000 = 20,000 Sum of their Savings = 20000-8000 = 12,000 20000:12000 = 5:3 QUESTION: 3 ### There are 2 containers of equal capacity. The ratio of milk to water in the first container is 4:5 and in the second container is 3:7.If they are mixed up then the ratio of milk to water in the mixture will be Solution: Answer – D.67:113 Explanation : 4+5 = 9=> 40:50 3+7 = 10=> 27:63 40+27 : 50:63 = 67:113 QUESTION: 4 There are two numbers. When 25% of the first number is added to the second number, the resultant number is 1.5times th first number.What is the ratio of 1st number to the 2nd number ? Solution: Answer – C.4:5 Explanation : A+25/100 + B = 1.5A A/4 + B = 15A/10 10A+40B/40 =60A/40 10A+40B = 60A 50A = 40B A/B = 4/5 QUESTION: 5 A bag contains 10p,25p and Rs50p coins in the ratio of 5:2:1 respectively. If the total money in the bag is Rs.120.Find the number of 25p coins in that bag? Solution: Answer – A.160 Explanation : 105 : 252 : 501 = 50:50:50 = 1:1:1 120/3 = Rs.40 Rs. 1 = 4 Rs.40 = 440 = 160 coins QUESTION: 6 The ratio of Ganesh’s age and his mother’s age is 5:12.The difference of their ages is 21.The ratio of their ages after 4 years will be Solution: Answer – D.19:40 Explanation : 12x – 5x = 21 7x = 21 X = 3 5:12 = 15:36 After 4 years = 19:40 QUESTION: 7 The ratio of students of three classes is 2:3:4. If 15 students are increased in each classes then their ratio turns into 13:18:23. What was the total number of students in all the three classes originally ? Solution: Answer – C.225 Explanation : 50:75:100 15 students increased 65:90:115 => 13:18 :23 Total no of students = 50+75+100 = 225 QUESTION: 8 Ravi and Govind have money in the ratio 5 : 12 and Govind and Kiran also have money in the same ratio 5 : 12. If Ravi has Rs. 500, Kiran has Solution: Answer – B.Rs.2880 Explanation : Ravi : Kiran = 5/12 5/12 = 25/144 Kiran = 144500/25 = 2880 QUESTION: 9 A town with a population of 1000 has provision for 30days, after 10 days 600 more men added, how long will the food last at the same rate ? Solution: Answer – C.12 ½ days Explanation : 100020/1600 = 12 1/2 days QUESTION: 10 A man spends Rs.2480 to buy lunch box Rs.120 each and bottles at Rs.80 each,What will be the ratio of maximum number of bottles to lunch box are bought ? Solution: Answer – A.13:12 Explanation : Check the ans using option 1380+ 12120 = 1040+1440 = 2480 QUESTION: 11 Three cars travel same distance with speeds in the ratio 2 : 4 : 7. What is the ratio of the times taken by them to cover the distance? Solution: B) 14 : 7 : 4 Explanation: s = d/t Since distance is same, so ratio of times: 1/2 : 1/4 : 1/7 = 14 : 7 : 4 QUESTION: 12 Section A and section B of 7th class in a school contains total 285 students.Which of the following can be a ratio of the ratio of the number of boys and number of girls in the class? Solution: B) 10 : 9 Explanation: The number of boys and girls cannot be in decimal values, so the denominator should completely divide number of students (285). Check each option: 6+5 = 11, and 11 does not divide 285 completely. 10+9 = 19, and only 19 divides 285 completely among all. QUESTION: 13 180 sweets are divided among friends A, B, C and D in which B and C are brothers also such that sweets divided between A and B are in the ratio 2 : 3, between B and C in the ratio 2 : 5 and between C and D in ratio 3 : 4. What is the number of sweets received by the brothers together? Solution: B) 84 Explanation: A/B = N1/D1 B/C = N2/D2 C/D = N3/D3 A : B : C : D = N1N2N3 : D1N2N3 : D1D2N3 : D1D2D3 A/B = 2/3 B/C = 2/5 C/D = 3/4 A : B : C : D 223 : 323 : 353 : 354 4 : 6 : 15 : 20 B and C together = [(6+15)/(4+6+15+20)] * 180 QUESTION: 14 Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys? Solution: B) 54.8% Explanation: Total students in both = 6x+11x = 17x Boys in class 4 = (60/100)6x = 360x/100 Boys in class 5 = (52/100)11x = 572x/100 So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x % of boys = [9.32x/17x] * 100 QUESTION: 15 Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture? Solution: E) 5 : 7 Explanation: Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg Total brass = 30+40 = 70 kg So copper in mixture is (50+70) – 70 = 50 kg So copper to brass = 50 : 70 QUESTION: 16 Ratio of A and B is in the ratio 5 : 8. After 6 years, the ratio of ages of A and B will be in the ratio 17 : 26. Find the present age of B. Solution: A) 72 Explanation: A/B = 5/8 , A+6/B+6 = 17/26 Solve both, B = 72 QUESTION: 17 A bag contains 25p, 50p and 1Re coins in the ratio of 2 : 4 : 5 respectively. If the total money in the bag is Rs 75, find the number of 50p coins in the bag. Solution: D) 40 Explanation: 2x, 4x, 5x (25/100)2x + (50/100)4x + 15x = 75 x = 10, so 50 p coins = 4x = 40 QUESTION: 18 A is directly proportional to B and also directly proportional to C. When B = 6 and C = 2, A = 24. Find the value of A when B = 8 and C = 3. Solution: D) 48 Explanation: A directly proportional B, A directly proportional to C: A = kB, A = kC Or A = kBC When B = 6 and C = 2, A = 24: 24 = k62 k = 2 Now when B = 8 and C = 3: A = 283 QUESTION: 19 A is directly proportional to B and also inversely proportional to the square of C.When B = 16 and C = 2, A = 36. Find the value of A when B = 32 and C = 4. Solution: C) 18 Explanation: A = kB, A = k/C2 Or A = kB/ C2 When B = 16 and C = 2, A = 36: 36 = k16/ 22 k = 9 Now when B = 32 and C = 4: A = 932/ 42 QUESTION: 20 A is directly proportional to the inverse of B and also inversely proportional to C. When B = 36 and C = 9, A = 42. Find the value of A when B = 64 and C = 21. Solution: A) 24 Explanation: A = k√B, A = k/C Or A = k√B/C When B = 36 and C = 9, A = 42: 42 = k√36/9 k = 63 Now when B = 64 and C = 21: A = 63√64/21<\|endoftext\|>	4.5625
1,567	RhizangiidaeStephen D. Cairns The rhizangiids are known from the Lower Cretaceous (about 135 million years ago) to the Recent, and are widespread in the tropical and temperate regions of all oceans. All species live on hard substrates on the continental shelf (0-200 m), but at least one genus (Culicia) is known from as deep as 238 m. All species are azooxanthellate; however, one (Astrangia poculata) contains populations that are also zooxanthellate (Peters et al., 1988). Among the azooxanthellate Scleractinia, the rhizangiids have most successfully occupied the neritic, hard bottom, temperate niche. All Recent rhizangiids are colonial and firmly attached, but because they rarely form large colonies, they do not contribute to shallow or deeper-water reef structure. A total of 33 Recent species is known in four genera (Cairns et al., 1999). This family is sometimes incorrectly referred to as the Astrangiidae Verrill, 1869. Primarily colonial, azooxanthellate coralla formed by extratentacular budding, resulting in plocoid or reptoid colonies. Corallites usually small and low, united basally by stolons or a thin basal coenosteum. Septa composed of one fan system of simple or compound trabeculae, resulting in finely dentate septal margins. Endotheca present. Left: Culicia hoffmeisteri : Calice of an individual corallite from a colonial skeleton collected from Victoria, Australia (depth 7 m). Diameter of corallite 5.5 mm. Photograph from Cairns and Parker, 1992. Copyright © 1995 South Australian Museum. Right: Oulangia bradleyi: Calice of an individual corallite from a colonial skeleton collected at the Galapagos (depth 14 m). Diameter of corallite 12.2 mm (from Cairns, 1991). No phylogenetic analysis has been performed on the genera within this family. The monophyletic nature of the family is doubtful. For instance, Chevalier (1987) included Polycyathus in this family, and Verheij and Best (1987) included both Polycyathus and Paracyathus as rhizangiids, whereas Cairns and Zibrowius (1997) considered both of these genera to be caryophylliids, as well as transferring Colangia and Phyllangia from the Rhizangiidae to the Caryophylliidae. Recent molecular analyses (Romano and Cairns, 2000) place Polycyathus and Paracyathus with the Oculinidae, but as yet no nucleic acids from rhizangiids have been sequenced. Cairns, S. D. 1991. A revision of the ahermatypic Scleractinia of the Galápagos and Cocos Islands. Smithsonian Contributions to Zoology, 504: 32 pp. Cairns, S. D. 1994. Scleractinia of the temperate North Pacific. Smithsonian Contributions to Zoology, 557: 150 pp. Cairns, S. D. 1995. The marine fauna of New Zealand: Scleractinia (Cnidaria: Anthozoa). New Zealand Oceanographic Institute Memoirs, 103, 210 pp. Cairns, S. D., B. W. Hoeksema, and J. van der Land. 1999. Appendix: List of Extant Stony Corals. Attol Research Bulletin, 459:13-46. Cairns, S. D. and S. A. Parker. 1992. Review of the Recent Scleractinia (Stony Corals) of South Australia, Victoria and Tasmania. Records of the South Australian Museum, Monograph Series, 3: 82 pp. Cairns, S. D. and H. Zibrowius. 1997. Cnidaria, Anthozoa: Azooxanthellate Scleractinia from the Philippine and Indonesian regions. Mémoires du Muséum National d'Histoire Naturelle, 172: 27-243. Chevalier, J.-P. 1987. Ordre des Scléractiniares: Systematique. Pp. 679-753 In: Grassé (editor) Traité de Zoologie, 3(3) Masson, Paris. Peters, E. C. et al. 1988. Nomenclature and biology of Astrangia poculata (Cnidaria: Anthozoa). Proceedings of the Biological Society of Washington, 101(2): 234-250. Romano, S. L. and S. D. Cairns. 2000. Molecular phylogenetic hypotheses for the evolution of scleractinian corals. Bull. Mar. Sci., 67(3): 1043-1068. Verheij, E. and M. B. Best. 1987. Notes on the genus Polycyathus Duncan, 1876 and a description of three new scleractinian corals from the Indo-Pacific. Zoologische Mededelingen, 61(12): 147-154. Wells, J. W. 1956. Scleractinia. Pp. F328-F444 In: Moore, R. C (editor) Treatise on Invertebrate Paleontology, Part F: Coelenterata. University of Kansas Press, Lawrence. Creation of this page was supported by US National Science Foundation grants DEB95-21819 and DEB 99-78106 (in the program PEET - Partnerships to Enhance Expertise in Taxonomy) to Daphne G. Fautin, grant DEB99-78086 (in the program PEET) to Stephen D. Cairns, and grant OCE 00-03970 (in NOPP, the National Oceanographic Partnership Program) to D.G.F. and Robert W. Buddemeier. Technical assistance was rendered by Adorian Ardelean. The author welcomes the opportunity to identify specimens from this family, and offers to incorporate them into the collections of the National Museum of Natural History, Smithsonian, unless their return is requested. Correspondence regarding this page should be directed to Stephen D. Cairns at Page copyright © 2002 Page: Tree of Life Rhizangiidae Authored by . Stephen D. Cairns. The TEXT of this page is licensed under the Creative Commons Attribution-NonCommercial License - Version 3.0. Note that images and other media featured on this page are each governed by their own license, and they may or may not be available for reuse. Click on an image or a media link to access the media data window, which provides the relevant licensing information. For the general terms and conditions of ToL material reuse and redistribution, please see the Tree of Life Copyright Policies. - First online 28 October 2002 Citing this page: Cairns, Stephen D. 2002. Rhizangiidae http://tolweb.org/Rhizangiidae/19083/2002.10.28 in The Tree of Life Web Project, http://tolweb.org/. Version 28 October 2002.<\|endoftext\|>	3.71875
1,117	How is Bernoulli’s equation used in differential equations? When n = 0 the equation can be solved as a First Order Linear Differential Equation. When n = 1 the equation can be solved using Separation of Variables. and turning it into a linear differential equation (and then solve that). What is Bernoulli’s rule? In fluid dynamics, Bernoulli’s principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid’s potential energy. This requires that the sum of kinetic energy, potential energy and internal energy remains constant. What is the differential of an equation? In mathematics, a differential equation is an equation that relates one or more functions and their derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. How do you solve first order differential equations? Here is a step-by-step method for solving them:Substitute y = uv, and. Factor the parts involving v.Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)Solve using separation of variables to find u.Substitute u back into the equation we got at step 2. What is linear equation in differential equation? In mathematics, a linear differential equation is a differential equation that is defined by a linear polynomial in the unknown function and its derivatives, that is an equation of the form. where , , and are arbitrary differentiable functions that do not need to be linear, and. How do you solve a Riccati differential equation? If the coefficients in the Riccati equation are constants, this equation can be reduced to a separable differential equation. The solution is described by the integral of a rational function with a quadratic function in the denominator: y′=ay+by2+c,⇒dydx=ay+by2+c,⇒∫dyay+by2+c=∫dx. How do you solve a second order differential equation? Second Order Differential EquationsHere we learn how to solve equations of this type: d2ydx2 + pdydx + qy = 0.Example: d3ydx3 + xdydx + y = ex We can solve a second order differential equation of the type: d2ydx2 + P(x)dydx + Q(x)y = f(x) Example 1: Solve. d2ydx2 + dydx − 6y = 0. Example 2: Solve. Example 3: Solve. Example 4: Solve. Example 5: Solve. You might be interested: Schrodinger equation derivation How do you do separable differential equations? The method for solving separable equations can therefore be summarized as follows: Separate the variables and integrate.Example 1: Solve the equation 2 y dy = ( x 2 + 1) dx.Example 2: Solve the equation.Example 3: Solve the IVP.Example 4: Find all solutions of the differential equation ( x 2 – 1) y 3 dx + x 2 dy = 0. What is Bernoulli’s Theorem and its application? Bernoulli’s theorem is the principle of energy conservation for ideal fluids in steady, or streamline, flow and is the basis for many engineering applications. Where is Bernoulli’s principle used? Bernoulli’s principle can be applied to many everyday situations. For example, this principle explains why airplane wings are curved along the top and why ships have to steer away from each other as they pass. The pressure above the wing is lower than below it, providing lift from underneath the wing. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one point along its flow. Because the Bernoulli equation is equal to a constant at all points along a streamline, we can equate two points on a streamline. How difficult is differential equations? Don’t be surprised to know that Differential Equations is really not too difficult as feared, or widely imagined. All you need, for 98% of the entirety of ODE (Ordinary Differential Equations), is how to integrate. How do you solve differential equations examples? Example 5y’ = 5. as a differential equation:dy = 5 dx. Integrating both sides gives:y = 5x + K. Applying the boundary conditions: x = 0, y = 2, we have K = 2 so:y = 5x + 2. Releated Convert to an exponential equation How do you convert a logarithmic equation to exponential form? How To: Given an equation in logarithmic form logb(x)=y l o g b ( x ) = y , convert it to exponential form. Examine the equation y=logbx y = l o g b x and identify b, y, and x. Rewrite logbx=y l o […] H2o2 decomposition equation What does h2o2 decompose into? Hydrogen peroxide can easily break down, or decompose, into water and oxygen by breaking up into two very reactive parts – either 2OHs or an H and HO2: If there are no other molecules to react with, the parts will form water and oxygen gas as these are more stable […]<\|endoftext\|>	4.4375
695	# Which polygon has no of sides diagonal? ## Which polygon has no of sides diagonal? Pentagon is a diagonal having the number of diagonals equal to the number of sides. ## What is the rule of diagonals in polygons? The number of diagonals in a polygon that can be drawn from any vertex in a polygon is three less than the number of sides. To find the total number of diagonals in a polygon, multiply the number of diagonals per vertex (n – 3) by the number of vertices, n, and divide by 2 (otherwise each diagonal is counted twice). Do irregular polygons have diagonals? Since the irregular polygons have the same number of vertices, and the number of diagonals depends only on the number of vertices, not the lengths of the sides, the number of diagonals would be exactly the same for irregular polygons as for regular polygons with the same number of sides. Are diagonals equal in regular polygon? Since all regular polygons have all angles of equal measure, to obtain to the measure of each angle in a polygon with vertices we can simply divide the sum of the measures of all interior angles by . As we already noticed, diagonals in a regular polygon do not intersect at one point. ### What polygon is not regular? An irregular polygon is any polygon that is not a regular polygon. It can have sides of any length and each interior angle can be any measure. They can be convex or concave, but all concave polygons are irregular since the interior angles cannot all be the same. ### Do polygons have to have equal sides? A polygon can have a certain number of sides, but the sides do not necessarily have to be the same length. Both of the polygons below are pentagons because they both have five angles and sides, but look at the differences. What is not a type of polygon? To be a polygon, a flat, closed shape must use only line segments to create its sides. So a circle or any shape that has a curve is not a polygon. What are the maximum number of diagonals in a polygon? 2 diagonals. An octagon has. 20 diagonals. A polygon ‘s diagonals are line segments from one corner to another (but not the edges). The number of diagonals of an n-sided polygon is: n (n − 3) / 2. #### How many connected sides does a polygon have? Thus, a figure having many sides is called a polygon. Its accepted mathematical definition is any figure that has more than two sides and which forms a closed figure is called polygon. Hence, polygon can have three and many more but finite sides. A three-sided polygon is called a triangle. The four-sided quadrilateral is a polygon. #### How many sides and angles does every polygon have? The simplest polygon, the triangle, has three interior angles (and three exterior angles!) so it also has three sides . You can have polygons with so many sides, they appear to be circles even when they are not. How do you find diagonals in a polygon? The number of diagonals in a polygon that can be drawn from any vertex in a polygon is three less than the number of sides. To find the total number of diagonals in a polygon, multiply the number of diagonals per vertex (n – 3) by the number of vertices, n, and divide by 2 (otherwise each diagonal is counted twice).<\|endoftext\|>	4.59375
384	Building knowledge of Native American history No, they're not lifesize homes - they're models of homes that represent those built by Native Americans. It's part of a fifth grade project in which students are learning about this culture that's been an important part of American history. The students are learning about Native Americans in their classes, and they're taking this knowledge home with them as they do research on a tribe of their choosing and create a home model that represents the type of home built by the tribe. The students also have to determine what environment the tribe lived in, such as woodlands, and create a setting to place the home in. The students created their projects at home with parental assistance. They worked on their projects for about three weeks. The students are encouraged to use building materials found in nature or around the house, rather than using prefabricated materials. Students drummed up their imagination and used such materials as cotton balls, twigs, tree bark, stones, sugar cubes, paper bags, popsicle sticks, marshmallows, and other items. Along with constructing the model homes, students complete written assignments based on their research. Teachers say the students enjoy creating the homes and showing them to classmates. They like to see how their fellow classmates constructed their homes. Younger students look forward to being in fifth grade so they can have a chance to participate. Students from the elementary school come by each year to take a gander at the creations. While it provides lots of fun, the project also provides lessons on what life was like for Native Americans hundreds of years ago. They learn about the resources and tools the Native Americans used to build the homes, which are quite different from what's used today. The students are finding that life for the tribes certainly wasn't the same as it is today. "There were no Walmarts," said Andrew Batzer.<\|endoftext\|>	3.71875
141	This model provides a bridge between concrete manipulatives and abstract numeric symbols. If you laminated the number paths, the students could mark out problems on their number path and then erase it. For comparison problems it may be useful to have two number paths so students can easily see how the two quantities compare. Use this model to lay the foundation for the use of the bar model and the number line to solve problems with a larger range of numbers in the later grades. The link above takes you directly to the information on number paths, but the Mathematically Minded website has a number of printable resources that are worth checking out. http://www.mathematicallyminded.com/book1.html<\|endoftext\|>	3.953125
2,892	# Rectangle Rectangle Rectangle Edges and vertices 4 Schläfli symbol { } × { } Coxeter diagram Symmetry group Dihedral (D2), [2], (*22), order 4 Dual polygon rhombus Properties convex, isogonal, cyclic Opposite angles and sides are congruent In Euclidean plane geometry, a rectangle is a quadrilateral with four right angles. It can also be defined as an equiangular quadrilateral, since equiangular means that all of its angles are equal (360°/4 = 90°). It can also be defined as a parallelogram containing a right angle. A rectangle with four sides of equal length is a square. The term oblong is occasionally used to refer to a non-square rectangle.[1][2][3] A rectangle with vertices ABCD would be denoted as ABCD. The word rectangle comes from the Latin rectangulus, which is a combination of rectus (as an adjective, right, proper) and angulus (angle). A crossed rectangle is a crossed (self-intersecting) quadrilateral which consists of two opposite sides of a rectangle along with the two diagonals.[4] It is a special case of an antiparallelogram, and its angles are not right angles. Other geometries, such as spherical, elliptic, and hyperbolic, have so-called rectangles with opposite sides equal in length and equal angles that are not right angles. Rectangles are involved in many tiling problems, such as tiling the plane by rectangles or tiling a rectangle by polygons. ## Characterizations A convex quadrilateral is a rectangle if and only if it is any one of the following:[5][6] • a parallelogram with at least one right angle • a parallelogram with diagonals of equal length • a parallelogram ABCD where triangles ABD and DCA are congruent • a quadrilateral with four right angles • a convex quadrilateral with successive sides a, b, c, d whose area is .[7]:fn.1 • a convex quadrilateral with successive sides a, b, c, d whose area is [7] ## Classification A rectangle is a special case of both parallelogram and trapezoid. A square is a special case of a rectangle. A rectangle is a special case of a parallelogram in which each pair of adjacent sides is perpendicular. A parallelogram is a special case of a trapezium (known as a trapezoid in North America) in which both pairs of opposite sides are parallel and equal in length. A trapezium is a convex quadrilateral which has at least one pair of parallel opposite sides. • Simple: The boundary does not cross itself. • Star-shaped: The whole interior is visible from a single point, without crossing any edge. ### Alternative hierarchy De Villiers defines a rectangle more generally as any quadrilateral with axes of symmetry through each pair of opposite sides.[8] This definition includes both right-angled rectangles and crossed rectangles. Each has an axis of symmetry parallel to and equidistant from a pair of opposite sides, and another which is the perpendicular bisector of those sides, but, in the case of the crossed rectangle, the first axis is not an axis of symmetry for either side that it bisects. Quadrilaterals with two axes of symmetry, each through a pair of opposite sides, belong to the larger class of quadrilaterals with at least one axis of symmetry through a pair of opposite sides. These quadrilaterals comprise isosceles trapezia and crossed isosceles trapezia (crossed quadrilaterals with the same vertex arrangement as isosceles trapezia). ## Properties ### Symmetry A rectangle is cyclic: all corners lie on a single circle. It is equiangular: all its corner angles are equal (each of 90 degrees). It is isogonal or vertex-transitive: all corners lie within the same symmetry orbit. It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). ### Rectangle-rhombus duality The dual polygon of a rectangle is a rhombus, as shown in the table below.[9] RectangleRhombus All angles are equal. All sides are equal. Alternate sides are equal. Alternate angles are equal. Its centre is equidistant from its vertices, hence it has a circumcircle. Its centre is equidistant from its sides, hence it has an incircle. Two axes of symmetry bisect opposite sides. Two axes of symmetry bisect opposite angles. Diagonals are equal in length. Diagonals intersect at equal angles. • The figure formed by joining, in order, the midpoints of the sides of a rectangle is a rhombus and vice versa. ### Miscellaneous The two diagonals are equal in length and bisect each other. Every quadrilateral with both these properties is a rectangle. A rectangle is rectilinear: its sides meet at right angles. A rectangle in the plane can be defined by five independent degrees of freedom consisting, for example, of three for position (comprising two of translation and one of rotation), one for shape (aspect ratio), and one for overall size (area). Two rectangles, neither of which will fit inside the other, are said to be incomparable. ## Formulae The formula for the perimeter of a rectangle The area of a rectangle is the product of the length and width. If a rectangle has length and width • it has area , • it has perimeter , • each diagonal has length , • and when , the rectangle is a square. ## Theorems The isoperimetric theorem for rectangles states that among all rectangles of a given perimeter, the square has the largest area. The midpoints of the sides of any quadrilateral with perpendicular diagonals form a rectangle. A parallelogram with equal diagonals is a rectangle. The Japanese theorem for cyclic quadrilaterals[10] states that the incentres of the four triangles determined by the vertices of a cyclic quadrilateral taken three at a time form a rectangle. The British flag theorem states that with vertices denoted A, B, C, and D, for any point P on the same plane of a rectangle:[11] For every convex body C in the plane, we can inscribe a rectangle r in C such that a homothetic copy R of r is circumscribed about C and the positive homothety ratio is at most 2 and .[12] ## Crossed rectangles A crossed (self-intersecting) quadrilateral consists of two opposite sides of a non-self-intersecting quadrilateral along with the two diagonals. Similarly, a crossed rectangle is a crossed quadrilateral which consists of two opposite sides of a rectangle along with the two diagonals. It has the same vertex arrangement as the rectangle. It appears as two identical triangles with a common vertex, but the geometric intersection is not considered a vertex. A crossed quadrilateral is sometimes likened to a bow tie or butterfly. A three-dimensional rectangular wire frame that is twisted can take the shape of a bow tie. A crossed rectangle is sometimes called an "angular eight". The interior of a crossed rectangle can have a polygon density of ±1 in each triangle, dependent upon the winding orientation as clockwise or counterclockwise. A crossed rectangle is not equiangular. The sum of its interior angles (two acute and two reflex), as with any crossed quadrilateral, is 720°.[13] A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: • Opposite sides are equal in length. • The two diagonals are equal in length. • It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). ## Other rectangles A saddle rectangle has 4 nonplanar vertices, alternated from vertices of a cuboid, with a unique minimal surface interior defined as a linear combination of the four vertices, creating a saddle surface. This example shows 4 blue edges of the rectangle, and two green diagonals, all being diagonal of the cuboid rectangular faces. In spherical geometry, a spherical rectangle is a figure whose four edges are great circle arcs which meet at equal angles greater than 90°. Opposite arcs are equal in length. The surface of a sphere in Euclidean solid geometry is a non-Euclidean surface in the sense of elliptic geometry. Spherical geometry is the simplest form of elliptic geometry. In elliptic geometry, an elliptic rectangle is a figure in the elliptic plane whose four edges are elliptic arcs which meet at equal angles greater than 90°. Opposite arcs are equal in length. In hyperbolic geometry, a hyperbolic rectangle is a figure in the hyperbolic plane whose four edges are hyperbolic arcs which meet at equal angles less than 90°. Opposite arcs are equal in length. ## Tessellations The rectangle is used in many periodic tessellation patterns, in brickwork, for example, these tilings: ## Squared, perfect, and other tiled rectangles A rectangle tiled by squares, rectangles, or triangles is said to be a "squared", "rectangled", or "triangulated" (or "triangled") rectangle respectively. The tiled rectangle is perfect[14][15] if the tiles are similar and finite in number and no two tiles are the same size. If two such tiles are the same size, the tiling is imperfect. In a perfect (or imperfect) triangled rectangle the triangles must be right triangles. A rectangle has commensurable sides if and only if it is tileable by a finite number of unequal squares.[14][16] The same is true if the tiles are unequal isosceles right triangles. The tilings of rectangles by other tiles which have attracted the most attention are those by congruent non-rectangular polyominoes, allowing all rotations and reflections. There are also tilings by congruent polyaboloes. ## References 1. "Archived copy" (PDF). Archived from the original (PDF) on 2014-05-14. Retrieved 2013-06-20. 2. Definition of Oblong. Mathsisfun.com. Retrieved 2011-11-13. 3. Oblong – Geometry – Math Dictionary. Icoachmath.com. Retrieved 2011-11-13. 4. Coxeter, Harold Scott MacDonald; Longuet-Higgins, M.S.; Miller, J.C.P. (1954). "Uniform polyhedra". Philosophical Transactions of the Royal Society of London. Series A. Mathematical and Physical Sciences. The Royal Society. 246 (916): 401–450. doi:10.1098/rsta.1954.0003. ISSN 0080-4614. JSTOR 91532. MR 0062446. 5. Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. A Study of Definition", Information Age Publishing, 2008, pp. 34–36 ISBN 1-59311-695-0. 6. Owen Byer; Felix Lazebnik; Deirdre L. Smeltzer (19 August 2010). Methods for Euclidean Geometry. MAA. pp. 53–. ISBN 978-0-88385-763-2. Retrieved 2011-11-13. 7. Josefsson Martin (2013). "Five Proofs of an Area Characterization of Rectangles" (PDF). Forum Geometricorum. 13: 17–21. 8. An Extended Classification of Quadrilaterals (An excerpt from De Villiers, M. 1996. Some Adventures in Euclidean Geometry. University of Durban-Westville.) 9. de Villiers, Michael, "Generalizing Van Aubel Using Duality", Mathematics Magazine 73 (4), Oct. 2000, pp. 303-307. 10. Cyclic Quadrilateral Incentre-Rectangle with interactive animation illustrating a rectangle that becomes a 'crossed rectangle', making a good case for regarding a 'crossed rectangle' as a type of rectangle. 11. Hall, Leon M. & Robert P. Roe (1998). "An Unexpected Maximum in a Family of Rectangles" (PDF). Mathematics Magazine. 71 (4): 285–291. JSTOR 2690700. 12. Lassak, M. (1993). "Approximation of convex bodies by rectangles". Geometriae Dedicata. 47: 111. doi:10.1007/BF01263495. 13. Stars: A Second Look. (PDF). Retrieved 2011-11-13. 14. R.L. Brooks, C.A.B. Smith, A.H. Stone and W.T. Tutte (1940). "The dissection of rectangles into squares". Duke Math. J. 7 (1): 312–340. doi:10.1215/S0012-7094-40-00718-9. 15. J.D. Skinner II, C.A.B. Smith and W.T. Tutte (November 2000). "On the Dissection of Rectangles into Right-Angled Isosceles Triangles". Journal of Combinatorial Theory, Series B. 80 (2): 277–319. doi:10.1006/jctb.2000.1987. 16. R. Sprague (1940). "Ũber die Zerlegung von Rechtecken in lauter verschiedene Quadrate". Journal für die reine und angewandte Mathematik. 182: 60–64. • Weisstein, Eric W. "Rectangle". MathWorld. • Definition and properties of a rectangle with interactive animation. • Area of a rectangle with interactive animation.<\|endoftext\|>	4.4375
1,838	# Probability Distributions ### Variables A variable whose value depends upon a chance experiment is called a random variable. Suppose that a person is asked who that person is closest to: their mother or their father. The random variable of this experiment is the boolean variable whose possibilities are {Mother, Father}. A continuous random variable is a variable whose possible outcomes are part of a continuous data set. The random variable that represents the height of the next person who walks in the room is a continuous random variable while the random variable that represents the number rolled on a six sided die is not a continuous random variable. A random variable that is not continuous is called a discreet random variable. ### Probability Distributions Example $$\PageIndex{1}$$ Suppose we toss two dice. We will make a table of the probabilities for the sum of the dice. The possibilities are : 2,3,4,5,6,7,8,9,10,11,12. $$x$$ $$P(x)$$ 2 3 4 5 6 7 8 9 10 11 12 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 Exercise $$\PageIndex{1}$$ Suppose that you buy a raffle ticket for $5. If 1,000 tickets are sold and there are 10 third place winners of$25, three second place winners of $100 and 1 grand prize winner of$2,000, construct a probability distribution table. Do not forget that if you have the $25 ticket, you will have won$20. ### Value (Mean) Example $$\PageIndex{2}$$: Insurance We when we buy insurance in black jack we lose the insurance bet if the dealer does not have black jack and win twice the bet if the dealer does have black jack. Suppose you have $20 wagered and that you have a king and a 9 and the dealer has an ace. Should you buy insurance for$10? Solution We construct a probability distribution table $$x$$ $$P(x)$$ -10 34/49 20 15/49 (There are 49 cards that haven't been seen and 15 are 10JKQ (jacks, kings and queens) and the other 34 are non tens.) We define the expected value = $$S \times P(x)$$ We calculate: $-10(34/49) + 20(15/49) = -40/49$ Hence the expected value is negative so that we should not buy insurance. What if I am playing with my wife. My cards are 2 and a 6 and my wife's are 7 and 4. Should I buy insurance? We have: $$x$$ $$P(x)\ -10 31/47 20 16/47 We calculate: $-10(31/47) + 20(16/47) = 10/47 = 0.21$ Hence my expected value is positive so that I should buy insurance. ### Standard Deviation We compute the standard deviation for a probability distribution function the same way that we compute the standard deviation for a sample, except that after squaring $x - m$$, we multiply by $$P(x)$$. Also we do not need to divide by $$n - 1$$. Consider the second insurance example: $$x$$ $$P(x)$$ $$x - \overline{x}$$ $$(x - \overline{x}$$^2$ -10 31/47 -10.21 104 20 16/47 19.79 392 Hence the variance is $104(31/47) + 392(16/47) = 202$ and the standard deviation is the square root of the variance, which is14.2. ### Combining Distributions If we have two distributions with independent random variables $$x$$ and $$y$$ and if $$a$$ and $$b$$ are constants then if $$L = a + bx$$ and $$W = ax + by$$ then 1. $$m_L = a + bm$$ 2. $$\sigma_L^2 = b^2s^2$$ 3. $$\sigma_L = \|b\|\, s$$ 4. $$m_W = a\,m_x + b\,m_y$$ 5. $$\sigma_W^2 = a^2s_1^2 + b^2s_2^2$$ 6. $$\sigma_W = \sqrt{a^2\, \sigma_x^2 + b^2\, \sigma_y^2}$$ Example $$\PageIndex{3}$$ Gamblers who played both black jack and craps were studied and it was found that the average amount of black playing per weekend was 7 hours with a standard deviation of 3 hours. The average amount of craps play was 4 hour with a standard deviation of 2 hours. What is the mean and standard deviation for the total amount of gaming? Solution Here $$a$$ and $$b$$ are 1 and 1. The mean is just $7 + 4 = 11$ and the standard deviation is just $\sqrt{3^2 + 2^2} = \sqrt{13}$ Example $$\PageIndex{4}$$ If each player spends about $100 per hour on black jack and$200 per hour on craps, what will be the mean and standard deviation for the amount of money that the casino wins per person? Solution Here a and b are 100 and 200. The mean is $100(7) + 200(4) = 1,500$ and the standard deviation is $\sqrt{(100^2)(3^2)+(200^2)(2^2)}=100\sqrt{17}$ Example $$\PageIndex{5}$$ If the players spend \$150 on the hotel, find the mean and standard deviation of the total amount of money that the players spend. Here $L = 150 + x$ where $$x$$ is the result from part B. Hence the mean is $150 + 1500 = 1,650$ and the standard deviation is the same as part B since the coefficient is 1. ### The Binomial Distribution There is a type of distribution that occurs so frequently that it has a special name. We call a distribution a binomial distribution if all of the following are true 1. There are a fixed number of trials, $$n$$, which are all independent. 2. The outcomes are Boolean, such as True or False, yes or no, success or failure. 3. The probability of success is the same for each trial. For a binomial distribution with $$n$$ trials with the probability of success $$p$$ and failure $$q$$, we have $P(r \text { successes}) = C_{n,r}\, p^r \,q^{n-r}$ Example $$\PageIndex{6}$$ Suppose that each time you take a free throw shot, you have a 25% chance of making it. If you take 15 shots, what is the probability of making exactly 5 of them. Solution We have $$n = 15$$, $$r = 5$$, $$p = 0.25$$, and $$q = 0.75$$ Compute $C_{15,5}\, 0.25^5 \,0.75^{10} = 0.165$ There is a 16.5 % chance of making exactly 5 shots. Example $$\PageIndex{7}$$ What is the probability of making fewer than 3 shots? Solution The possible outcomes that will make this happen are 2 shots, 1 shot, and 0 shots. Since these are mutually exclusive, we can add these probabilities. $C_{15,2} \, 0.25^2\, 0.75^{13} + C_{15,1}\, 0.25^1\, 0.75^{14} + C_{15,0}\, 0.25^0 \,0.75^{15}$ $= 0.156 + 0.067 + 0.013 = 0.236$ There is a 24 % chance of sinking fewer than 3 shots.<\|endoftext\|>	4.46875
18,765	## Big Ideas Math Book 3rd Grade Answer Key Chapter 1 Understand Multiplication and Division Big Ideas Math Book 3rd Grade Answer Key Chapter 1 Understand Multiplication and Division answer key is accessible here. This answer key is useful for students who are preparing for their examinations and can download this pdf for free of cost. In this chapter each and every step was explained in detail which helps students to understand easily. Big Ideas Math Book 3rd Grade Answer Key Chapter 1 explains different types of questions on multiplications, divisions. ## Big Ideas Math Book 3rd Grade Answer Key Chapter 1 Understand Multiplication and Division Multiplication and Division chapter comprises of the themes of equal groups to multiply, number lines to multiply, array to multiply, multiply in any order, and many more other topics. Those topics were being set up by the numerical specialists as indicated by the most recent release. Look down this page to get the answers for all the inquiries. Tap the connection to look at the subjects shrouded in this chapter Understand Multiplication and Division. Lesson 1 Use Equal Groups to Multiply Lesson 2 Use Number Lines to Multiply Lesson 3 Use Arrays to Multiply Lesson 4 Multiply in Any Order Lesson 5 Divide: Size of Equal Groups Lesson 6 Divide: Number of Equal Groups Lesson 7 Use Number Lines to Divide ### Lesson 1.1 Use Equal Groups to Multiply Explore and Grow Question 1. Put 24 counters into equal groups. Draw to show your groups. 24 counters in four equal groups Explanation: To keep 24 counters into an equal group, we will find the factors of 24 and then we can put 24 counters into equal groups. The factors of 24 are 2,3,4,6,8,12. Now we can pick any of the numbers from the factors of 24 and then put these 24 counters in equal groups. Let’s take four equal groups and arrange 24 counters in four equal groups as shown in the below image. Question 2. Put 24 counters into a different number of equal groups. Draw to show your groups. 7 counters in the first group, 8 counters in the second group, 6 counters in the third group, and 4 counters in the fourth group. Explanation: In this, we will take some number of groups and in that, we will place some different number of counters in that group. In the given image, we will take four groups, and we will place 7 counters in the first group, 8 counters in the second group, 6 counters in the third group, and 4 counters in the fourth group. Question 3. Structure Compare your models. How are the models the same? How are they different? In the first model, we can see that there is an equal number of counters in each group, and in the second model, we can see the different number of counters in each group. They are the same with an equal number of groups. Show and Grow Use the model to complete the statements. Question 1. ____ groups of ____ ____ + ____ +____ + ____ = ____ ____ × ____ = ____ 4 groups of 5, 5+5+5+5= 20, 4×5= 20. Explanation: In the above image, we can see 4 groups and in each group, 5 counters are there, and the product is 4×5= 20. And we can also write by adding 5+5+5+5= 20. There are 4 groups of 5. Question 2. ____ groups of ____ ____ + ____ = ____ ____ × ____ = ____ 2 groups of 6, 6+6= 12, 2×6= 12. Explanation: As we can see there are 2 groups of 6 which are two groups with the number 6 in the box. So the product of the two groups is 2×6= 12 and the sum of the two groups is 6+6= 12. Apply and Grow: Practice Question 3. Use the model to complete the statements. ____ groups of ____ ____ + ____ = ____ ____ × ____ = ____ 2 groups of 9 9+9= 18, 2×9= 18. Explanation: In the above image, we can see 2 groups with 9 counters in each group, and the product is 2×9= 18. By adding these two groups we will get 18, which is 9+9= 18. Draw equal groups. Then complete the equations. Question 4. 4 groups of 2 ____ + ____ + ____ + ____ =____ ____ × ____ = ____ 4 groups of 2 2+2+2+2= 8, 4×2= 8. Explanation: In the above image, we can see 4 groups with 2 counters in each group, and the product is 4×2= 8. By adding these four groups we will get 8, which is 2+2+2+2= 18. Question 5. 3 groups of 5 ____ + ____ + ____ = ____ ____ × ____ = ____ 3 groups of 5, 5+5+5= 15, 3×5= 15. Explanation: In the above image, we can see 3 groups with 5 counters in each group, and the product is 3×5= 15. By adding these three groups we will get 15, which is 5+5+5= 15. Write the addition equation as a multiplication equation. Question 6. 8 + 8 + 8 = 24 3×8= 24. Explanation: To write the addition equation as a multiplication equation in the above we can see 3 times 8 which can be written as 3×8= 24. Question 7. 7 + 7 + 7 + 7 + 7 = 35 5×7= 35. Explanation: To write the addition equation as a multiplication equation in the above we can see 5 times 7 which can be written as 5×7= 35. Question 8. You Be The Teacher Newton says he circled 3 groups of 4 counters. Is he correct? Explain. No, Newton is not correct. Explanation: In the above image, we can see 4 groups of 3 counters. So Newton is not correct. Question 9. DIG DEEPER! You wash 5 cars. How many times do you wash? Explanation: As the car has 4 tires and there are 5 cars to wash, so there are 5 groups of 4. The number of tires to wash is 5×4= 20. So there are 20 tires to wash. Think and Grow: Modeling Real Life Question 1. Complete the statement: ______ groups of ________ Multiplication equation: You buy _____ cards in all. 6 groups of 10 6×10= 60. Explanation: As there are 6 packs of 10 trading cards, which means 6 groups of 10. So 6×10= 60 cards can we buy in all. Show and Grow Question 10. You buy 8 packs of 4 highlighters. How many highlighters do you buy in all? 8 groups of 4 8×4= 32 Explanation: As there 8 packs of 4 highlighters, which means 8 groups of 4. The number of highlighters bought is 8×4= 32. Question 11. DIG DEEPER! Explanation: In this query, we can see 5 bracelets have 3 beads which are 5×3= 15, and a friend makes 6 bracelets that have 2 beads which are 6×2= 12. So the total number of beads used in all is 15+12= 27 beads. ### Use Equal Groups to Multiply Homework & Practice 1.1 Question 1. Use the model to complete the statements. ___ groups of ____ ___ + ___ + ___ = ___ ___ × ___ = ___ 3 groups of 4, 4+4+4= 12, 4×3= 12. Explanation: We can see in the above image there are 3 groups with 4 counter in each group, which means 3 groups of 4. The sum is 4+4+4= 12 and the product is 4×3= 12. Draw equal groups. Then complete the equations. Question 2. 2 groups of 8 ___ + ___ = ___ ___ × ___ = ___ 8+8= 16 2×8= 16. Explanation: 2 groups of 8 means, there are 2 groups with 8 counters in each group. So the sum is 8+8= 16 and the product is 2×8= 16. Question 3. 5 groups of 3 ___ + ___ + ___ + ___ + ___ = ___ ___ × ___ = ___ 3+3+3+3+3= 15, 5×3= 15. Explanation: 5 groups of 3 means, there are 5 groups with 3 counters in each group. So the sum is 3+3+3+3+3= 15, and the product is 5×3= 15. Write the addition equation as a multiplication equation. Question 4. 3 + 3 + 3 + 3 + 3 + 3 = 18 6×3= 18. Explanation: To write the addition equation as a multiplication equation in the above we can see 6 times 3, which can be written as 6×3= 18. Question 5. 2 + 2 + 2 + 2 + 2 + 2 + 2 = 14 7×2= 14. Explanation: To write the addition equation as a multiplication equation in the above we can see 7 times 2, which can be written as 7×2= 14. Question 6. DIG DEEPER! You have 16 action figures. Can you put an equal number of figures on 3 shelves? Explain No, we cannot put an equal number of figures on 3 shelves. Explanation: As given that we have 16 action figures and asked to determine whether it is possible to an equal number of figures on 3 shelves. As factors of 16 are 1,2,4,8,16 as 16 is not divisible by 3, so we cannot put an equal number of figures on 3 shelves. Question 7. Which One Doesn’tBelong? Which one does not belong with the other two? 2 groups of 3 don’t belong with the other two. Explanation: As 2 groups of 3 means in 2 groups, there will be 3 counters, which does not belong to the other two. Because in the above image we can see 3 groups of 2, which means 3 groups with 2 counters in each group which is equal to 2+2+2= 6. Question 8. Modeling Real Life You make 7 gift bags for your friends. Each gift bag has 3 pom-pom pets. How many pom-pom pets are there in all? 21 pom-pom pets. Explanation: As there are 7 gift bags and each gift bag has 3 pom-pom pets, so the number of pom-pom pets are 7×3= 21. Question 9. DIG DEEPER Newton has 2 stacks of 5 books. Descartes has 3 stacks of 4 books. How many books do they have in all? 22 books. Explanation: As Newton has 2 stacks of 5 books, which is 2×5= 10 books and Descartes has 3 stacks of 4 books which is 3×4= 12 books. So the number of books is 10+12= 22books. Review & Refresh Question 10. 50 + 30 = ____ 80 Explanation: By adding 50+30 we will get 80. Question 11. 27 + 40 = ______ 67 Explanation: By adding 27+40 we will get 67. Question 12. 19 + 20 = _____ 39 Explanation: By adding 19+20 we will get 39. ### Lesson 1.2 Use Number Lines to Multiply Explore and Grow Question 1. Find the sums. Use each sum as the missing addend in the next equation. Model the problems on the number line. 0+3= 3, 3+3= 6, 6+3= 9, 9+3= 12. Explanation: First, we must begin at 0 and then skip the count by 3’s then make jumps until you get 12. So 3 x 4 = 12. Reasoning How can you use a number line to help you find 4 × 3? 4×3= 12. Explanation: To find 4×3 using the number line, we will make a jump from 0 to 4 for 3 times. Think and Grow: Multiplication and Number Lines Example Find 3 × 4 3 × 4 means 3 groups of 4. Number of jumps: _____ Size of each jump: ______ Start at 0. Skip count by 4s three times. 3 × 4 ____ 3 × 4 = 12 Number of jumps: 3 Size of each jump: 4. Explanation: To represent the given value in the number line, we will start from 0 and skip the count by 4 three times. So the number of jumps is 3 and the size of each jump is 4. Show and Grow Question 1. Find 2 × 4 Number of jumps: _____ Size of each jump: ______ 2 × 4 = _____ 2 × 4 = 8, Number of jumps: 2 Size of each jump: 4 Explanation: To find the product of 2×4 using the number line, we will start from zero and then skip the count by 4s two times. So the number of jumps is 2 and the size of each jump is 4. Question 2. Fing 6 × 3 Number of jumps: _____ Size of each jump: ______ 6 × 3 = _____ 6 × 3= 18 Number of jumps: 6 Size of each jump: 3 Explanation: To find the product of 6 × 3 using the number line, we will start from zero and then skip the count by 3s six times. So the number of jumps is 6 and the size of each jump is 3. Apply and Grow: Practice Question 3. Find 3 × 5 Number of jumps: _____ Size of each jump: ______ 3 × 5 = _____ 3 × 5= 15 Number of jumps: 3 Size of each jump: 5 Explanation: To find the product of 3 × 5 using the number line, we will start from zero and then skip the count by 5s three times. So the number of jumps is 3 and the size of each jump is 5. Question 4. Fing 5 × 4 Number of jumps: _____ Size of each jump: ______ 5 × 4 = _____ 5 × 4= 20 Number of jumps: 5 Size of each jump: 4 Explanation: To find the product of 5 × 4 using the number line, we will start from zero and then skip the count by 4s five times. So the number of jumps is 5 and the size of each jump is 4. Question 5. Find 3 × 8 Number of jumps: _____ Size of each jump: ______ 3 × 8= 24 Number of jumps: 8 Size of each jump: 3 Explanation: To find the product of 3 × 8 using the number line, we will start from zero and then skip the count by 3s eight times. So the number of jumps is 8 and the size of each jump is 3. Question 6. Structure Draw jumps to show 4 groups of 6 and 6 groups of 4. Think: How are they the same? How are they different? 4 groups of 6 4×6= 24 Number of jumps: 6 Size of each jump: 4 6 groups of 4 6×4= 24 Number of jumps: 4 Size of each jump: 6 The product of 4 groups of 6 and 6 groups of 4 is 24 this is the same in 4 groups of 6 and 6 groups of 4. The number of jumps and the size of the jumps is different. Explanation: To find the product of 4 × 6 using the number line, we will start from zero and then skip the count by 4s six times. So the number of jumps is 6 and the size of each jump is 4. To find the product of 6 × 4 using the number line, we will start from zero and then skip the count by 6s four times. So the number of jumps is 4 and the size of each jump is 6. Think and Grow: Modeling Real Life A group of lions is called a pride. There are 2 prides in a savanna. Each pride has 9 lions. How many lions are there in all? Model: There are ____ lions in all. 2×9= 18 lions. Explanation: As there are 2 prides in the savanna and each pride has 9 lions, the total number of lions is 2×9= 18 lions. To represent this in the number line, we will start from 0 then skip the count by 9 twice. So the number of jumps is 2 and the size of the jump is 9. Show and Grow Question 7. There are 3 bike racks at a park. Each bike rack has 4 bikes. How many bikes are there in all? 3×4= 12 bikes. Explanation: As there are 3 bike racks and each rack has 4 bikes, so the number of bikes are there is 3×4= 12. To represent this in the number line, we will start from 0 then skip the count by 4 three times. So the number of jumps is 3 and the size of the jump is 4. Question 8. DIG DEEPER! You dig 8 holes. You plant 2 flower bulbs in each hole. You have5 bulbs left. How many flower bulbs did you have to start? 3×5= 15 flower bulbs. Explanation: The number of holes dug is 8 and in each hole, 2 flower bulbs are planted, so the number of flower bulbs planted is 2×5= 10 and still, there are 5 bulbs left. So the total number of flower bulbs is 10+5= 15. To represent this in the number line, as there are 5 bulbs added then it will be 3 groups of 5 which is 3×5= 15. So we will start from 0 then skip the count by 5 three times and the number of jumps is 3 and the size of the jump is 5. ### Use Number Lines to Multiply Homework & Practice 1.2 Question 1. Find 3 × 6 Number of jumps: _____ Size of each jump: ______ 3 × 6 = _____ 3 × 6= 18 Number of jumps: 3 Size of each jump: 6. Explanation: To represent this in the number line, we will start from 0 then skip the count by 6 three times. So the number of jumps is 3 and the size of the jump is 6. Question 2. Find 4 × 5 Number of jumps: _____ Size of each jump: ______ 4 × 5 = _____ 4 × 5= 20 Number of jumps: 4 Size of each jump: 5. Explanation: To represent this in the number line, we will start from 0 then skip the count by 5 four times. So the number of jumps is 4 and the size of the jump is 5. Question 3. Structure Complete the multiplication equations in two different ways. Model each equation on the number line. ____ × ____ = 12 2×6= 12 Explanation: To represent this in the number line, we will start from 0 then skip the count by 6 two times. So the number of jumps is 2 and the size of the jump is 6. ____ × ____ = 12 4×3= 12 Explanation: To represent this in the number line, we will start from 0 then skip the count by 3 four times. So the number of jumps is 4 and the size of the jump is 3. Question 4. Writing Explain how you can use a number line to ﬁnd 5 × 3. 5×3= 15. Explanation: To represent this in the number line, we will start from 0 then skip the count by 3 five times. So the number of jumps is 5 and the size of the jump is 3. Question 5. Modeling Real Life You have6 boxes of blueberry muffins. Each box has 4 muffins. How many muffins do you have in all? 6×4= 24. Explanation: As we have 6 boxes of blueberry muffins with 4 muffins in each box, which means 6×4 = 24 muffins do we have it all. Question 6. DIG DEEPER! You ﬁll 8 pages of a photo album. Each page has 3 photos. You have one photo left. How many photos did you have to start? 25 photos. Explanation: The total number of pages in a photo album is 8 and each page contains 3 photos, which means 8×3= 24, so there will be a total of 24 photos. Now 1 photo left, which means 24+1= 25. So there will be a total of 25 photos. Review & Refresh Question 7. 9 + 8 + 2 = _____ 19. Explanation: On adding 9+8+2 we will get the sum as 19. Question 8. 6 + 5 + 3 = _____ 14. Explanation: On adding 6+5+3 we will get the sum as 14. Question 9. 7 + 4 + 7 = ______ ### Lesson 1.3 Use Arrays to Multiply Explore and Grow Question 1. Put 24 counters into equal rows. Draw your model. 6×4= 24. rows and 4 columns. Explanation: To keep 24 counters into equal rows, we will find the factors of 24 and then we can put 24 counters into equal rows. The factors of 24 are 2,3,4,6,8,12. Now we can pick any of the numbers from the factors of 24 and then put these 24 counters in equal groups. Let’s take six equal rows and arrange 24 counters in four equal rows as shown in the below image. Question 2. Put 24 counters into a different number of equal rows. Draw your model. Question 3. Structure Compare your models. How are the models the same? How are they different? Think and Grow: Multiplication and Arrays An array is a group of objects organized into rows and columns. Each row has the same number of objects. Example How many counters are there in all? Show and Grow Question 1. How many counters are there in all? 5 rows and 4 columns, 5+5+5+5= 20, 5×4= 20. The total number of counters is 20. Explanation: From the above image, we can see there are 5 rows and 4 columns. The product is 5×4= 20 and the sum is 5+5+5+5= 20. Question 2. Draw an array to multiply 6 × 3. 6 × 3 = ______ 6 × 3 = 18 Explanation: To draw an array of 6 × 3, we will place 6 rows and 3 columns. By that, we can form an array of 6×3 which is 18. Apply and Grow: Practice Draw an array to multiply. Question 3. 4 × 8 = _____ 4 × 8 = 32 Explanation: To draw an array of 4 × 8, we will place 4 rows and 8 columns. By that, we can form an array of 4×8 which is 32. Question 4. 3 × 9 = ____ 3×9= 27 Explanation: To draw an array of 3×9, we will place 3 rows and 9 columns. By that, we can form an array of 3×9 which is 27. Question 5. 7 × 3 = _____ 7 × 3 = 21 Explanation: To draw an array of 7×3, we will place 7 rows and 3 columns. By that, we can form an array of 7×3 which is 21. Question 6. 6 × 5 = _____ 6×5= 30 Explanation: To draw an array of 6×5, we will place 6 rows and 5 columns. By that, we can form an array of 6×5 which is 30. Question 7. Number Sense Newton has a 2 × 10 array of baseballs. He adds another row. How many baseballs does he add? Write a multiplication equation for his new array. ____ × _____ = _____ 3×10= 30. Explanation: As Newton has an array of 2 × 10 of baseballs which are 20 baseballs as he adds another row, which means 3×10 of baseballs which are 30 baseballs. So he adds 30 and 20 which is 10 baseballs Newton was added and the multiplication equation for his new array is 3×10 which is 30 baseballs. Question 8. DIG DEEPER! Use 6 counters to make as many different arrays as possible using all of the counters. Draw the arrays. Then write a multiplication equation for each array. 2×3= 6, 3×2= 6. Explanation: The different possible arrays for 6 counters are 2×3= 6 which has 2 rows and 3 columns and 3×2= 6 which has 3 rows and 2 columns. This array consists of 2 rows and 3 columns This array consists of 3 rows and 2 columns. Think and Grow: Modeling Real Life A phone has 6 rows of apps with 4 apps in each row. How many apps are on the phone? Draw: Equation: There are _____ apps on the phone. 6×4= 24 Explanation: As a phone has 6 rows of apps with 4 apps in each row, The number of apps on the phone are 6×4= 24. Show and Grow Question 9. Your classroom has 3 rows of desks with 10 desks in each row. How many desks are in your classroom? 3×10= 30 desks. Explanation: The number of rows in the classroom is 3 rows with 10 desks in each row, which means 3×10= 30 desks in the classroom. Question 10. DIG DEEPER! A square array has an equal number of rows and columns. A farmer has 9 corn seeds to plant in a square array. Draw the square array the farmer can use to plant all of the seeds. How many rows and columns are there? 3×3= 9. Explanation: As the farmer has 9 corn seed to plant in a square array, so the possible square array is with 3 rows and 3 columns, which is 3×3= 9. ### Use Arrays to Multiply Homework & Practice 1.3 Question 1. How many counters are there in all? 4×9= 36 counters. 4 rows, 9 columns. 9+9+9+9= 36. Explanation: In the above image, we can see 4 rows and 9 columns which makes 36 counters. The product is 4×9= 36 and the sum is 9+9+9+9= 36. Draw an array to multiply. Question 2. 9 × 2 = _____ 9×2= 18. Explanation: This array contains 9 rows and 2 columns which makes 18 counters. The product is 9×2= 18 and the sum is 9+9= 18. Question 3. 4 × 5 = _____ 4×5= 20. Explanation: This array contains 4 rows and 5 columns which makes 20 counters. The product is 4×5= 20 and the sum is 5+5+5+5= 20. Question 4. YOU BE THE TEACHER Descartes has 24 counters. He says he can use all the counters to make an array with 3 rows. Is he correct? Explain. With 3 rows and 8 columns, 24 counters can be made. Explanation: Yes, he is correct. With 3 rows and 8 columns, he can make 24 counters. The product of the array is 3×8= 24 and the sum of the array is 8+8+8= 24. Question 5. Number Sense Newton has a 4 × 8 array of dominoes. He adds 2 more rows. How many dominoes does he add? Write a multiplication equation for his new array. _____ × ____ = _____ 6×8= 48. Explanation: As Newton has a 4×8 array of dominoes which is 32 dominoes and he adds 2 more rows, which is 4+2= 6. Then the dominoes will be a 6×8 array. And the multiplication equation for Newton’s new array is 6×8= 48. So the number of dominoes added is 48-32 which is 16 dominoes he was added. Question 6. Modeling Real Life An art teacher hangs 2 rows of paintings with 10 paintings in each row. How many paintings does she hang? 20 paintings. Explanation: As an art teacher hangs 2 rows of paintings in each row, so the number of paintings is 2×10= 20 paintings. Question 7. DIG DEEPER! A museum has 16 shark teeth to display in a square array. Draw the square array the museum can use to display all of the teeth. How many rows and columns are there? There are 4 rows and 4 columns. Explanation: As the museum has 16 shark teeth to display in a square array, so the array can be written as 4×4= 16 which has 4 rows and 4 columns. Review & Refresh Complete the equation Question 8. 3 + 8 = 8 + ____ 3+8= 8+3 Explanation: By the commutative property of addition, we can change the order of the addends which does not change the sum. So 3+8= 8+3. Question 9. 10 + 0 = ____ + 10 10+0= 0+10. Explanation: By the commutative property of addition, we can change the order of the addends which does not change the sum. So 10+0= 0+10. Question 10. 6 + ____ = 7 + 6 6+7= 7+6. Explanation: By the commutative property of addition, we can change the order of the addends which does not change the sum. So 6+7= 7+6. Question 11. ____ + 8 = 8 + 9 9+8= 8+9. Explanation: By the commutative property of addition, we can change the order of the addends which does not change the sum. So 9+8= 8+9. ### Lesson 1.4 Multiply in Any Order Explore and Grow Question 1. Write the multiplication equation for the array. Turn your paper and write the equation for the array. 5×3= 15. Explanation: In the above image, we can see 5 rows and 3 columns. So the product of the array is 5×3= 15. Structure Compare the equations. How are they the same? How are they different? Think and Grow: Commutative Property of Multiplication Ina multiplication equation, the numbers that are multiplied are called factors. The answer is called the product. Commutative Property of Multiplication: Changing the order of factors does not change the product. Example Complete the statements Show and Grow Question 1. Complete the statements 5×2= 2×5. Explanation: In the first image, we can see 5 rows and 2 columns, this means 5 rows of 2. And in the second image, we can 2 rows and 5 columns, this means 2 rows of 5. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 5×2= 2×5. Question 2. Draw an array to show the Commutative Property of Multiplication. Complete the statements. ____ × ____ = _____ _____ × ____ = _____ So, ____ × _____ = ____ × ____ 2×3= 6, 3×2= 6. So, 2×3= 3×2. Explanation: In the above image, we can see 2 rows and 3 columns, this means 2 rows of 3. And we can also write as 3 rows and 2 columns as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 2×3= 3×2. Apply and Grow: Practice Draw an array to show the Commutative Property of Multiplication. Complete the statements. Question 3. 1×4= 4, 4×1= 4. So, 1×4= 4×1. Explanation: In the above image, we can see 1 row and 4 columns. Which makes 1 row of 4. And we can also write as 4 rows and 1 column as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 1×4= 4×1. Question 4. 2×6= 12, 6×2= 12, 2×6= 6×2. Explanation: In the above image, we can see 2 rows and 6 columns. Which makes 2 rows of 6. And we can also write as 6 rows and 2 columns as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 2×6= 6×2. Question 5. 4×5= 20, 5×4= 20. So, 4×5= 5×4. Explanation: In the above image, we can see 4 rows and 5 columns. Which makes 4 rows of 5. And we can also write as 5 rows and 4 columns as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 4×5= 5×4. Complete the equation. Question 6. 8 × 3 = 3 × ____ 8×3= 24, 3×8= 24. So, 8×3= 3×8. Explanation: By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can also write as 3 rows and 8 columns as we can see in the given below image which is 8×3= 3×8. Question 7. 10 × 2 = ____ × 10 10×2 = 20, 2×10= 20. So, 10×2= 2×10. Explanation: By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can also write as 2 rows and 10 columns as we can see in the given below image which is 10×2= 2×10. Question 8. 1 × ____ = 9 × 1 1×9= 9, 9×1= 9. So, 1×9= 9×1. Explanation: By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can also write as 9 rows and 1 column which is 1×9= 9×1. Question 9. Structure Which shape completes the equation? Moon shape. Explanation: By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, the missing shape is the moon. Think and Grow: Modeling Real Life Your friend makes 7 rows of 6 stickers. You want to put the same number of stickers into 6 rows. How many stickers do you put in each row? Explain. You put ____ stickers in each row. Explain. 7×6= 42, 6×7= 42. So, 7×6= 6×7. You can put 7 stickers in each row. Explanation: The number of stickers made is 7 rows of 6 stickers and want to put the same number of stickers into 6 rows. So, by the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can make 6 rows of 7 stickers, which has 6 rows and 7 columns. Show and Grow Question 10. Your friend makes 9 rows of 4 award ribbons. You want to put the same number of award ribbons into 4 rows. How many award ribbons do you put in each row? Explain 9×4= 36, 4×9= 36. So, 9×4= 4×9. So 9 award ribbons can be put in each row. Explanation: Given, 9 rows of 4 award ribbons and we need to put the same number of award ribbons into 4 rows. So by the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can make 4 rows of 9 award ribbons, which has 4 rows and 9 columns and 9 award ribbons can be put in each row. Question 11. DIG DEEPER! You have 2 rows of 8 toy cars. Your friend has 5 rows of 2 toy cars. How can you use the Commutative Property of Multiplication to find how many rows your friend needs to add so that you both have the same number of toy cars? Explanation: ### Multiply in Any Order Homework & Practice 1.4 Question 1. Complete the statements. 3 rows of 6, 3×6= 18. 6 rows of 3 6×3= 18, So, 3×6= 6×3. Explanation: In the first image, we can see 3 rows and 6 columns, this means 3 rows of 6. And in the second image, we can 6 rows and 3 columns, this means 6 rows of 3. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 3×6= 6×3. Question 2. Draw an array to show the Commutative Property of Multiplication. Complete the statements. 7×4= 28, 4×7= 28. So, 7×4= 4×7 Explanation: In the image, we can see 7 rows and 4 columns, this means 7 rows of 4. And by the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. To draw an array by the Commutative Property of Multiplication, we will take 4 rows of 7, which means 4 rows and 7 columns as in the below image. So, 7×4= 4×7. Which two arrays can you use to show the Commutative Property of Multiplication? The first image and third image shows the commutative property of multiplication. Explanation: From the above image, the first and third arrays show the Commutative property of multiplication. Which states that the order in which we multiply the numbers does not change the product. As in the first image, we can see 4 rows of 3 which have 4 rows and 3 columns and in the third image, we can see 3 rows of 4 which has 3 rows and 4 columns. Question 4. Precision Write two equations that show the Commutative Property of Multiplication. ____ × ____ = ____ × _____ ____ × _____ = ____ × ____ 5×6= 6×5, 2×3= 3×2. Explanation: The Commutative property of multiplication states that the order in which we multiply the numbers does not change the product. For example, if we take 5×6= 30 then the commutative property is 6×5= 30. So 5×6= 6×5. Question 5. Modeling Real Life A computer lab has 6 rows of 5 computers. A technology teacher wants to rearrange the computers into5 rows. How many computers does the teacher put in each row? Explain. 6 computers do the teacher put in each row. Explanation: As a computer lab has 6 rows of 5 computers, and the technology teacher rearranged the computers into 5 rows, so the teacher will put 6 computers by commutative property as it states that the order in which we multiply the numbers does not change the product. So 6×5= 5×6. Question 6. DIG DEEPER! You have 6 rows of 4 pennies. Your friend has 2 rows of 6 pennies. How many rows does your friend need to add so that you both have the same number of pennies? So, a friend needs 2 rows to add then both will have the same number of pennies. Explanation: There are 6 rows of 4 pennies, which is 6 rows and 4 columns and the friend has 2 rows of 6 pennies which is 2 rows and 6 columns. So my friend needs to add 2 rows and then it will be 2 rows+2 rows= 4 rows and there will be 6 columns. So, by the commutative property, as it states that the order in which we multiply the numbers does not change the product. So, 6×4= 4×6. Review & Refresh Question 7. Newton hits a ball 5 fewer times than Descartes does. Newton hits the ball 9 times. How many times does Descartes hit the ball? Explanation: ### Lesson 1.5 Divide: Size of Equal Groups Explore and Grow Question 1. Put 18 counters in 6 equal groups. Draw to show your groups Number of counters in each group: _____ The number of counters in each group is 3 counters. Explanation: To put 18 counters in 6 equal groups, we will divide 18 by 6 then the result is 3. So, we can put counters in 3 counters in 6 equal groups as shown in the given below image. Question 2. Put 18 counters in 3 equal groups. Draw to show your groups Number of counters in each group: _____ The number of counters in each group is 6. Explanation: To put 18 counters in 3 equal groups, we will divide 18 by 3 then the result is 6. So, we can put counters in 6 counters in 3 equal groups as shown in the given below image. Structure How does changing the number of equal groups change the number of counters in each group? Explanation: Think and Grow: Using Equal Groups to Divide Division is an operation that gives the size of equal groups or the number of equal groups. When you know the total number of objects and the number of equal groups, you can divide to find the size of each group.3 equal groups Example Divide 12 counters into. How many counters are in each group? Show and Grow Question 1. Divide 15 counters into5 equal groups. How many counters are in each group? 15 ÷ 5 = ____ The number of counters placed in 5 equal groups is 3 counters. Explanation: There are 15 counters and we need to place those 15 counters in 5 equal groups. So we will divide 15 by 5, 15÷5= 3. So we will pace 3 counters in 5 equal groups. Use the tape diagram to model the equation. The number of counters placed in 5 equal groups is 3 counters. Explanation: There are 15 counters and we need to place those 15 counters in 5 equal groups. So we will divide 15 by 5, 15÷5= 3. So we will pace 3 counters in 5 equal groups. Apply and Grow: Practice Question 2. Divide 30 counters into 6 equal groups. How many counters are in each group? 30 ÷ 6 = ____ The number of counters placed in 6 equal groups is 5 counters. AS 30÷6= 5. Explanation: There are 30 counters and we need to place those 30 counters in 6 equal groups. So we will divide 30 by 6, 30÷6= 5. So we will pace 5 counters in 6 equal groups. Use the tape diagram to model the equation. The number of counters placed in 6 equal groups is 5 counters. AS 30÷6= 5. Explanation: There are 30 counters and we need to place those 30 counters in 6 equal groups. So we will divide 30 by 6, 30÷6= 5. So we will pace 5 counters in 6 equal groups. Question 3. Divide 16 counters into 2 equal groups. How many counters are in each group? 16 ÷ 2 = ____ 16÷2= 8. Explanation: There are 16 counters and we need to place those 16 counters in 2 equal groups. So we will divide 16 by 2, 16÷2= 8. So we will pace 8 counters in 2 equal groups. Question 4. Divide 9 counters into 3 equal groups. How many counters are in each group? 9 ÷ 3 = ____ 9 ÷ 3= 3. Explanation: There are 9 counters and we need to place those 9 counters in 3 equal groups. So we will divide 9 by 3, 9÷3= 3. So we will pace 3 counters in 3 equal groups. Question 5. Structure Write the division equation that matches the tape diagram. ____ ÷ ____ = ____ 18÷2= 9. Explanation: There are 18 counters and we need to place those 18 counters in 2 equal groups. So we will divide 18 by 2, 18÷2= 9. So we will pace 9 counters in 2 equal groups. Question 6. DIG DEEPER! Newton has a tennis ball collection. He can divide the balls into3 equal groups with none left over. He can also divide the balls into4 equal groups with none left over. How many tennis balls does he have? Think and Grow: Modeling Real Life You have 30 seashells. You put an equal number of seashells in 5 bags. How many seashells are in each bag? Model: Division equation: There are _____ seashells in each bag. Division equation: 30÷5= 6. There are 6 seashells in each bag. Explanation: As there are 30 seashells and we need to put an equal number of seashells in 5 bags, we will divide 30÷5= 6. So we will 6 seashells in each bag. Show and Grow Question 7. You have 28 rocks. You put an equal number of rocks in 4 piles. How many rocks are in each pile? 28÷4= 7. Explanation: There are 28 rocks and we should put an equal number of rocks in 4 piles, we will divide 28 by 4. So 28÷4= 7. So there will be 7 rocks in each pile. Question 8. DIG DEEPER! Newton and Descartes each have40 quarters. Newton puts his quarters into5 equal groups. Descartes puts his quarters into4 equal groups. Who has more quarters in each group? Descartes has more quarters than Newton. Explanation: As both Newton and Descartes have 40 quarters and Newton puts his quarters into 5 equal groups, which means 40÷5= 8. Newton put his 8 quarters into 5 equal groups and Descartes puts his quarters into 4 equal groups, which means 40÷4= 10. Descartes put his 10 quarters into 4 equal groups. So, Descartes has more quarters than Newton. ### Divide: Size of Equal Groups Homework & Practice 1.5 Question 1. Divide 16 counters into 4 equal groups. How many counters are in each group? 16 ÷ 4 = ____ 16÷4= 4. Explanation: There are 16 counters and we need to place those 16 counters in 4 equal groups. So we will divide 16 by 4, 16÷4= 4. So we will pace 4 counters in 4 equal groups. Use the tape diagram to model the equation. 16÷4= 4. Explanation: There are 16 counters and we need to place those 16 counters in 4 equal groups. So we will divide 16 by 4, 16÷4= 4. So we will pace 4 counters in 4 equal groups. Question 2. Divide 28 counters into 7 equal groups. How many counters are in each group? 28 ÷ 7 = ____ 28 ÷ 7= 4. Explanation: There are 28 counters and we need to place those 28 counters in 7 equal groups. So we will divide 28 by 7, 28 ÷ 7= 4. So we will pace 4 counters in 7 equal groups. Question 3. Divide 27 counters into 3 equal groups. How many counters are in each group? 27 ÷ 3 = ____ 27 ÷ 3 = 9. Explanation: There are 27 counters and we need to place those 27 counters in 3 equal groups. So we will divide 27 by 3, 27 ÷ 3= 9. So we will pace 9 counters in 3 equal groups. Question 4. YOU BE THE TEACHER Newton says you divide 18 counters into 3 equal groups. Is he correct? Yes, Newton is correct. Explanation: Yes, Newton is correct. As we can see in the above image he divided 18 counters into 3 equal groups with different numbers of counters in each group. Question 5. Precision A class has 14 boys and 18 girls. Can the teacher divide the class equally into 4 groups with no students remaining? Explain. The teacher places 8 students in 4 equal groups with no students remaining. Explanation: As a class has 14 boys and 18 girls, so the total number of students is 14+18= 32 students. As the teacher divided the class equally into 4 groups, 32÷4= 8. So the teacher places 8 students in each group with no students remaining. Question 6. Modeling Real Life You have 14 erasers. You and your friend share them equally. How many erasers do you and your friend each get? 7 erasers will get each. Explanation: There are 14 erasers and those erasers are shared equally among two of them, so we will divide 14 by 2 which is 14÷2= 7. So 7 erasers will get each. Question 7. DIG DEEPER! Newton and Descartes each have 42 glow-in-the-dark stickers. Newton divides into 6 equal groups. Descartes divides his into7 equal groups. Who has more stickers in each group? Newton has more stickers than Descartes in each group. Explanation: As Newton and Descartes, each have 42 glow-in-the-dark stickers and Newton divides them into 6 equal groups, which is 42÷6= 7. So Newton has 7 glow-in-the-dark stickers in 6 equal groups. And Descartes divides his into7 equal groups which is 42÷7= 6. So Descartes has 6 glow-in-the-dark stickers in 7 equal groups. Newton has more stickers than Descartes in each group. Review & Refresh Question 8. 41 Explanation: On adding 26+15 we will get 41. Question 9. 87. Explanation: On adding 32+55 we will get 87. Question 10. 61. Explanation: On adding 49+12 we will get 61. Question 11. 92. Explanation: On adding 24+68 we will get 92. ### Lesson 1.6 Divide: Number of Equal Groups Explore and Grow Question 1. Put 24 counters in equal groups of 4. Draw to show your groups. Number of groups: ______ The number of groups is 6. Explanation: To put 24 counters in equal groups of 4, we will divide 24 by 4, 24÷4= 6. So we will put 4 counters in 6 equal number of group. Question 2. Put 24 counters in equal groups of 6. Draw to show your groups. Number of groups: ______ The number of groups is 4. Explanation: To put 24 counters in equal groups of 6, we will divide 24 by 6, 24÷6= 4. So we will put 6 counters in 4 equal number of group. Structure How does changing the size of the groups change the number of equal groups? Think and Grow: Using Equal Groups to Divide When you know the total number of objects and the size of each group, you can divide to find the number of equal groups. Example Divide 12 counters into. How many groups are there? Show and Grow Question 1. Divide 10 counters into groups of 2. How many groups are there? 10 ÷ 2 = ____ 10 ÷ 2 = 5. Explanation: To put 10 counters in equal groups of 2, we will divide 10 by 2, 10÷2= 5. So we will put 2 counters in 5 equal number of group. Use the tape diagram to model the equation. 10÷2= 5. Explanation: To put 10 counters in equal groups of 2, we will divide 10 by 2, 10÷2= 5. So we will put 2 counters in 5 equal number of group. Question 2. Divide 24 counters into groups of 4. How many groups are there? 24 ÷ 4 = ____ 24÷4= 6. Explanation: To put 24 counters in equal groups of 4, we will divide 24 by 4, 24÷4= 6. So we will put 4 counters in 6 equal number of group. Use the tape diagram to model the equation. 24÷4= 6. Explanation: To put 24 counters in equal groups of 4, we will divide 24 by 4, 24÷4= 6. So we will put 4 counters in 6 equal number of group. Apply and Grow: Practice Question 3. Divide 30 counters into groups of 6. How many groups are there? 30÷6= Use the tape diagram to model the equation. 30÷6= 5. Explanation: To put 30 counters in equal groups of 6, we will divide 30 by 6, 30÷6= 5. So we will put 5 counters in 6 equal number of group. Question 4. Divide 15 counters into groups of 5. How many groups are there? 15 ÷ 5 = _____ 15÷5= 3. Explanation: To put 15 counters in equal groups of 5, we will divide 15 by 5, 15÷5= 3. So we will put 3 counters in 5 equal number of group. Question 5. Divide 16 counters into groups of 4. How many groups are there? 16 ÷ 4 = _____ 16÷4= 4. Explanation: To put 16 counters in equal groups of 4, we will divide 16 by 4, 16÷4= 4. So we will put 4 counters in 4 equal number of group. Question 6. Structure You want to bake as many loaves of banana bread as possible with 12 eggs. Each loaf of bread requires 2 eggs. Which models can you use to ﬁnd how many loaves of bread you can make? Explanation: Think and Grow: Modeling Real life A florist uses 35 roses to make bouquets. Each bouquet has 7 roses. How many bouquets does the florist make? Equation: Model: The florist makes ______ bouquets Equation: 35÷7= 5 Explanation: As the florist has 35 roses to make bouquets and each bouquet has35 7 roses, so the florist can make 35÷7= 5 bouquets. Show and Grow Question 7. A farmer puts 48 eggs into cartons. He puts 6 eggs in each carton. How many cartons does he use? 8 cartons. Explanation: As the farmer puts 48 eggs into cartons and in each carton, there are 6 eggs. So the farmer uses 48÷6= 8 cartons. Question 8. DIG DEEPER! Newton uses his subway pass 3 times each day. Descartes uses his pass 2 times each day. Who will use all of his rides first? Explain. Would your answer change if Newton and Descartes both use their passes 3 times each day? Explain. Newton will finish his rides first then Descartes. Descartes will use all of his rides first if Descartes use passes 3 times each day. Explanation: As Newton has 24 subway rides left and he uses subway pass 3 times each day, so the number of rides left are 24-3= 21. Descartes has 18 subway rides and he uses his pass 2 times each day, so the number of rides left are 18-2= 16. By using repeated subtraction which starts from 24 as given and then we will subtract 3 until we get 0. 24-3= 21 21-3= 18 18-3= 15 15-3= 12 12-3= 9 9-3= 6 6-3= 3 3-3=0. So, Newton will complete the subway pass by 8 times. 18-2= 16 16-2= 14 14-2= 12 12-2= 10 10-2= 8 8-2= 6 6-2= 4 4-2= 2 2-2= 0. And Descartes will complete the subway pass by 9 times. So Newton will use all his rides first. And if Descartes uses 3 times of his pass then 18-3= 15 15-3= 12 12-3= 9 9-3= 6 6-3= 3 3-3= 0. Descartes will complete the subway pass by 6 times, So Descartes will complete his pass first. ### Divide: Number of Equal Groups Homework & Practice 1.6 Question 1. Divide 28 counters into groups of 4. How many groups are there? 28 ÷ 4 = ____ 28 ÷ 4 =7 Explanation: To put 28 counters in equal groups of 4, we will divide 28 by 4, 28÷4= 7. So we will put 4 counters in 7 equal number of group. Use the tape diagram to model the equation. 28÷4= 7. Explanation: To put 28 counters in equal groups of 4, we will divide 28 by 4, 28÷4= 7. So we will put 4 counters in 7 equal number of group. Question 2. Divide 25 counters into groups of 5. How many groups are there? 25 ÷ 5 = _____ 25 ÷ 5 = 5. Explanation: To put 25 counters in equal groups of 5, we will divide 25 by 5, 25÷5= 5. So we will put 5 counters in 5 equal number of group. Question 3. Divide 12 counters into groups of 6. How many groups are there? 12 ÷ 6 = _____ 12 ÷ 6 = 2. Explanation: To put 12 counters in equal groups of 6, we will divide 12 by 6, 12÷6= 2. So we will put 2 counters in 6 equal number of group. Question 4. Writing Write and solve a problem in which you need to find the number of equal groups. Divide 22 counters into groups of 11. How many groups are there? Explanation: To put 22 counters in equal groups of 11, we will divide 22 by 11, 22÷11= 2. So we will put 2 counters in 11 equal number of group. Question 5. Reasoning Your classroom has 30 chairs that need to be stacked with 5 chairs in each stack. Your teacher already made 2 stacks. How many stacks of chairs still need to be made? The number of stacks of chairs that still need to be made is 4. Explanation: As a classroom has 30 chairs and that needed to be stacked with 5 chairs in each stack, so 30÷5= 6 stacks. And the teacher already made 2 stacks, which means 6-2= 4. So the number of stacks of chairs that still need to be made are 4 stacks. Question 6. DIG DEEPER! You have more than 30 and fewer than 40 piñata toys. You divide them into groups with 8 in each group. How many groups do you make? There will be 4 groups with 8 pinata toys in each group. Explanation: As there are more than 30 and fewer than 40 pinata toys, and we need to divide them into groups with 8 in each group. So we need to choose the number between 30-40. As we can see fewer than 40 and more than 30, so we will choose 32 as it will divide them into groups with 8 in each group, 32÷8= 4. So there will be 4 groups with 8 pinata toys in each group. Question 7. A street vendor puts 42 apples into baskets. She puts 6 apples in each basket. How many baskets does she use? Explanation: As a street vendor puts 42 apples into baskets and she puts 6 apples in each basket, so she needs 42÷6= 7 baskets. Question 8. DIG DEEPER! Newton and Descartes are at an amusement park. Newton uses 2 tickets to ride each roller coaster. Descartes uses 3 tickets to ride at the front of each roller coaster. Who runs out of tickets first? Explain. Newton Descartes, They run out of tickets at the same time Newton will run out of tickets first, as he has fewer tickets. Explanation: As Newton has 14 coaster tickets and uses 2 tickets to ride each roller coaster, so there will 14-2= 12 tickets remaining. And Descartes has 21 coaster tickets and uses 3 tickets to ride, so the remaining tickets are 21-3= 18 tickets. Newton will run out of tickets first, as he has fewer tickets. Review & Refresh Question 9. 96-58= 38, 38+58= 96. Explanation: On subtracting 96-58 we will get 38 and by adding 58 and 38 we will get 96. So by adding we can check your answer. Question 10. 48-32= 16, 16+32= 48. Explanation: On subtracting 48-32 we will get 16 and by adding 32 and 16 we will get 48. So by adding we can check your answer. ### Lesson 1.7 Use Number Lines to Divide Explore and Grow Question 1. Find the difference. Use each difference as the starting number in the next equation. Model the problems on the number line. 12-3= 9, 9-3= 6, 6-3= 3, 3-3= 0. Explanation: To represent a number line, we will use repeated subtraction which starts from 12 as given and then we will subtract 3 until we get 0. So the number of jumps is 4 and the size of the jump is 3. Question 2. Structure How can you use a number line to help you find 12 ÷ 3? 12÷3= 4. Explanation: To use the number line for the given value 12÷3 which is 4, we will start from 0 then skip the count by 4 three times. So the number of jumps is 4 and the size of the jump is 3. And we can also use the subtraction method, which starts with 12 and subtract 3 until we get 0. Think and Grow: Number Lines and Repeated Subtraction Example Find 20 ÷ 4. Another Way: Use repeated subtraction. Start with 20. Subtract 4 until you reach 0. Show and Grow Complete the equations. Question 1. 24 ÷ 6 = ______ 24 ÷ 6 = 4. Explanation: By using the number line, we will countback by 6s from 24 until we reach 0, 24÷6= 4, so there are 4 groups of 6. Question 2. 16 ÷ 8 = _____ 16 – 8 = _____ ____ – 8 = 0 16 ÷ 8 = 3, 16-8= 8, 8-8= 0. Explanation: We will use repeated subtraction which starts from 16 as given and then we will subtract 8 until we get 0. So the number of jumps is 2 and the size of the jump is 8. 16 ÷ 8 = 3, 16-8= 8, 8-8= 0. Question 3. 27 ÷ 9 = _____ 27 – 9 = _____ ____ – 9 = ____ _____ – 9 = 0 27 ÷ 9 = 3, 27-9= 18, 18-9= 9, 9-9= 0. Explanation: We will use repeated subtraction which starts from 27 as given and then we will subtract 9 until we get 0. So the number of jumps is 3 and the size of the jump is 9. 27 ÷ 9 = 3, 27-9= 18, 18-9= 9, 9-9= 0. Apply and Grow: Practice Complete the equations. Question 4. 25 ÷ 5 = _____ 25 ÷ 5 = 5. Explanation: By using the number line, we will countback by 5s from 25 until we reach 0, 25÷5= 5, so there are 5 groups of 5. Question 5. 21 – 7 = _____ ____ – 7 = _____ _____ – 7 = 0 ____ ÷ _____ = ____ 21-7= 14, 14-7= 7, 7-7= 0, 21÷7= 3. Explanation: we will use repeated subtraction which starts from 21 as given and then we will subtract 7 until we get 0. 21-7= 14, 14-7= 7, 7-7= 0, 21÷7= 3. Question 6. 36 – 9 = _____ ____ – 9 = ____ ____ – 9 = ____ _____ – 9 = 0 ____ ÷ ____ = ____ 36-9= 27, 27-9= 18, 18-9= 9, 9-9= 0. 36÷9= 4. Explanation: we will use repeated subtraction which starts from 36 as given and then we will subtract 9 until we get 0. 36-9= 27, 27-9= 18, 18-9= 9, 9-9= 0. 36÷9= 4. Question 7. YOU BE THE TEACHER Descartes uses a number line to find 18 ÷ 2. Is he correct? Explain. Explanation: No, Descartes is not correct. As 18÷2= 9 and here the given result 8. So Descartes is incorrect. Think and Grow: Modeling Real life Each age group is divided into teams with 6 players on each team. Each team receives a trophy at the end of the season. How many trophies are needed? Age Group Number of Players 6 – 7 Years old 18 8 – 9 Years old 24 Division equations: _____ trophies are needed 7 trophies are needed. Explanation: As each age group is divided into teams with 6 players on each team and each team receives a trophy at the end of the season, so the number of trophies needed for age group 6-7 years old is 18÷6= 3 trophies. And for the age group 8-9 years old is 24÷6= 4 trophies. Show and Grow Question 8. Each age group is put into cabins with 8 campers in each cabin. How many cabins are needed? Age Group Number of Campers 5 – 7 Years old 32 8 – 10 Years old 24 The two age groups are combined into one group. Does the total number of cabins that are needed change? Explain The number of cabins needed for the age group of 5-7 years is 4 cabins and the number of cabins needed for the age group of 8-10 years old is 3 cabins. There will be no change in the total number of cabins. Explanation: For the age group of 5-7 years old number of cabins needed are 32÷8= 4 cabins. And for the age group 8- 10 years old number of cabins needed are 24÷8= 3 cabins. If two age groups are combined into one group then the total number of campers are 24+32= 56. So the total number of cabins needed are 56÷8= 7 cabins. So there will be no change. ### Use Number Lines to Divide Homework & Practice 1.7 Complete the equations Question 1. 9 ÷ 3 = _____ 9÷3= 3. Explanation: By using the number line, we will countback by 3s from 9 until we reach 0, 9÷3= 3, so there are 3 groups of 3. Question 2. 20 – 5 = ____ ____ – 5 = ____ ____ – 5 = ____ ____ – 5 = 0 ____ ÷ ____ = ____ 20-5= 15, 15-5= 10, 10-5= 5, 5-5= 0, 20÷5= 4. Explanation: we will use repeated subtraction which starts from 20 as given and then we will subtract 5 until we get 0. 20-5= 15, 15-5= 10, 10-5= 5, 5-5= 0, 20÷5= 4. Question 3. 10 – 2 = ____ ____ – 2 = ____ ____ – 2 = ____ ____ – 2 = ____ ____ – 2 = 0 ____ ÷ ____ = ____ 10-2 = 8, 8-2= 6, 6-2= 4, 4-2= 2, 2-2= 0, 10÷5= 5. Explanation: we will use repeated subtraction which starts from 10 as given and then we will subtract 2 until we get 0. 10-2 = 8, 8-2= 6, 6-2= 4, 4-2= 2, 2-2= 0, 10÷5= 5. Question 4. YOU BE THE TEACHER Descartes uses repeated subtraction to ﬁnd 15 ÷ 5. Is he correct? Explain. Explanation: As Descartes uses repeated subtraction which starts from 15 as given and then we will subtract 5 until we get 0. So Descartes is correct. Question 5. DIG DEEPER! Find Newton’s missing number. Explain how you solved. 18-6= 12, 12-6= 6, 6-6 = 0. Explanation: Here, given subtract 6 from a number 3 times, and reach 0, so we will multiply 6 by 3 and we will get a number, which is 6×3= 18. So the number is 18 if we subtract 6, 3 times we will reach 0. 18-6= 12, 12-6= 6, 6-6 = 0. Question 6. Modeling Real Life Each age group is divided into groups of 7 swimmers. How many groups are there in each age group? Age Group Number of Swimmers 6 – 8 Years old 28 9 – 11 Years old 14 The two age groups are combined into one group. Does the total number of groups change? Explain. By combining two age groups into one group the total number of groups does not change, as the total number of groups and the combined two age group is equal. Explanation: As each age was divided into groups of 7 swimmers, so the number of groups in the 6-8 years old age group is 28÷7= 4 groups. And the number of groups in the 9-11 years old age group is 14÷7= 2 groups. And the total number of groups is 4+2= 6. As the two age groups are combined into one group, so 28+14= 42. So the total number of groups will be 42÷7= 6 groups. Review & Refresh Question 7. Circle the values of the underlined digit. 8 tens. Explanation: The value of 8 in the digit 581 is 8 tens. ### Understand Multiplication and Division Performance Task Question 1. a. Your science teacher gives you 71 picture cards to sort into categories, living and nonliving. You sort 11 cards into the nonliving category. How many cards do you sort into the living category? b. You sort the living cards into 2 categories, plants, and animals. The numbers of cards in each category are equal. How many cards are in the animal category? c. You divide the animal cards into 6 equal groups. How many animal cards are in each group? d. The animals in 5 of the groups have backbones, and the animals in the other group do not have backbones. How many more cards have animals with backbones? Explain. a) 60. b) 30 cards in the animal category. c) 5 animal cards. d) Explanation: a) As there are 71 picture cards and 11 of them are sorted into the non-living category, so the remaining cards in the living category are 71-11= 60. b) As the living cards are 60 and those are divided into two equal categories 60÷2= 30, which is 30 plants and 30 animal cards. So there are 30 cards in the animal category. c) To divide the animal cards into 6 equal groups, we will divide 30 by 6 and there will be 30÷6= 5 animal cards are in each group. d) ### Understand Multiplication and Division Activity Hooray Array! Getting Started: Fill in your board with each number from the Number List. You may write each number in any square. Each square can only have one number. Directions: 1. Choose a player to be the caller. The caller selects a Hooray Array Equation Card and reads the equation. 2. All players solve the equation and place a counter on the answer. Cover only 1 number per turn. 3. Repeat the process with players taking turns as the caller. 4. The winner is the ﬁrst player who creates a 3 × 3 array on the board and yells, “HOORAY ARRAY!” ### Understand Multiplication and Division Chapter practice 1.1 Use Equal Groups to Multiply Question 1. Use the model to complete the statements. 2 groups of 3, 3+3= 6, 2×3= 6. Explanation: Here, we can see 2 groups with 3 counters in each group. So, 2 groups of 3 which is 6, which means 2×3= 6. And if we add 2 groups of the counter, we will get the same result, which is 3+3= 6. Draw equal groups. Then complete the equations. Question 2. 3 groups of 6 3 groups of 6, 2+2+2= 6, 3×6= 18. Explanation: Here, we can see 3 groups with 6 counters in each group. So, 3 groups of 6 which are 18, which means 3×6= 18. And if we add 3 groups of the counter, we will get the same result, which is 2+2+2= 6. Question 3. 4 groups of 5 4 groups of 5, 5+5+5+5= 20, 4×5= 20. Explanation: Here, we can see 4 groups with 5 counters in each group. So, 4 groups of 5 which is 20, which means 4×5= 20. And if we add 4 groups of the counter, we will get the same result, which is 5+5+5+5= 20. 1.2 Use Number Lines to Multiply Question 4. Find 8 × 3 Number of jumps: ___ Size of each jump: ___ 8 × 3 = ____ 8×3= 24. Explanation: Here, we will start at 0 and then we will skip count by 3s eight times. So, the number of jumps is 8, and the size of each jump is 3. which is 8×3= 24. 1.3 Use Arrays to Multiply Draw an array to multiply. Question 5. 2 × 8 = ___ 2 × 8 = 16. Explanation: To draw an array of 2×8, we will take 2 rows and 8 columns to build an array. Question 6. 7 × 4 = ___ 7 × 4 = 28 Explanation: To draw an array of 7×4, we will take 7 rows and 4 columns to build an array. Question 7. YOU BE THE TEACHER Newton has 32 counters. He says that he can use all the counters to make an array with 6 rows. Is he correct? Explain. Answer: No, Newton is not correct. Explanation: No, as Newton has 32 counters which will not make an array with 6 rows. As 32 is not divisible 6, so he is not correct. 1.4 Multiply in Any Order Question 8. Draw an array to show the Commutative Property of Multiplication. Complete the statements. ____ × ____ = ____ ____ × ____ = ____ So, ___ × ___ = ____ × ___ 9×5 = 45, 5×9 = 45, So, 9×5 = 5×9. Explanation: In the above image, we can see 9 rows of 5, which means 9×5 = 45, so by the commutative property of multiplication, 9×5 = 5×9, which is 45. Complete the equation. Question 9. 4 × 10 = 10 × ___ Explanation: By the commutative property of multiplication, which means changing the order of factors which does not change the product. So, 4 × 10 = 10 × 4. Question 10. 3 × 9 = ____ × 3 Explanation: By the commutative property of multiplication, which means changing the order of factors which does not change the product. So, 3 × 9 = 9 × 3. Question 11. 8 × ___ = 4 × 8. Explanation: By the commutative property of multiplication, which means changing the order of factors which does not change the product. So, 8 × 4= 4 × 8. 1.5 Divide: Size of Equal Groups Question 12. Divide 27 counters into 3 equal groups. How many counters are in each group? Use the tape diagram to model the equation. 27 ÷ 3 = ___ Answer: 9 counters in each group. Explanation: Given Counters is 27, now, we will divide 27 counters into groups of 3, which means 27÷3= 9. On dividing 27 counters by 3, we will get the result as 9 groups. Question 13 Divide 32 counters into4 equal groups. How many counters are in each group? 32 ÷ 4 = ___ Answer: 8 counters in each group. Explanation: Given Counters is 32, now, we will divide 32 counters into groups of 4, which means 32÷4= 8. On dividing 32 counters by 4, we will get the result as 8 counters in each group. Question 14. Divide 12 counters into4 equal groups. How many counters are in each group? 12 ÷ 4 = ___ Answer: 3 counters in each group. Explanation: Given Counters is 12, now, we will divide 12 counters into groups of 4, which means 12÷4= 3. On dividing 12 counters by 4, we will get the result as 3 counters in each group. 1.6 Divide: Number of Equal Groups Question 15. Divide 15 counters into groups of 3. How many groups are there? 15 ÷ 3 = ___ Explanation: Given Counters is 15, now, we will divide 15 counters into groups of 3 which means 15÷3= 5. On dividing 15 counters by 3, we will get the result in 5 groups. Question 16. Divide 20 counters into groups of 2. How many groups are there? 20 ÷ 2 = __ Explanation: Given Counters is 20, now, we will divide 20 counters into groups of 2, which means 20÷2= 10. On dividing 20 counters by 2, we will get the result as 10 groups. Question 17. Modeling Real Life Newton and Descartes are trying a new music app. Newton uses 4 credits a day to hear songs without commercials. Descartes uses 2 credits a day to hear songs with commercials. Who runs out of credits ﬁrst? Explain. Newton will run out of credits first than Descartes. No, they will not run out of credits at the same time. Explanation: The total credits do Newton has in a new music app are 24 and he uses 4 credits a day to hear songs without commercials, so the remaining credits are 24-4= 20. And the total credits Descartes had are 14 and he uses 2 credits a day to hear songs with commercials, which is 14-2= 12. we will use repeated subtraction which starts from 24 as given and then we will subtract 4 until we get 0 which is 24-4= 20 20-4= 16 16-4= 12 12-4= 8 8-4= 4 4-4= 0 so, Newton will run out of credits by his 6 uses. And we will use repeated subtraction for Descartes’s credits, which is 14-2= 12 12-2= 10 10-2= 8 8-2= 6 6-2= 4 4-2= 2 2-2= 0 so, Descartes will run out of credits by his 7 uses. So Newton will run out of credits earlier than Descartes. 1.7 Use Number Lines to Divide Complete the equations. Question 18. 28 ÷ 7 = ___ 28 ÷ 7 = 4. Explanation: By using the number line, we will countback by 7s from 28 until we reach 0, 28÷7= 4, so there are 4 groups of 7. Question 19. 18 – 9 = ___ ___ – 9 = 0 ___ ÷ ___ = ___ 18-9= 0, 9-9= 0 18÷9= 2 Explanation: We will use repeated subtraction which starts with 18 as given and then we will subtract 9 until we get 0. So the number of jumps is 2 and the size of the jump is 9. Question 20. 12 – 4 = ___ ___ – 4 = ___ ___ – 4 = 0 ___ ÷ ___ = ___<\|endoftext\|>	4.90625
501	USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # Is 8 Divisible By 2? • Yes, 8 is divisible by 2. It will leave no comma spot. • Divisibilty rule for 2 is: Units are divisible by two if the last digit is even. Even numbers for 2 are (0,2,4,6,8). ## What Is 8 Divided By 2 • Eight divided by two is 4. Math: 8÷2=4 ## Rules Of Divisibility For 2 • First, take any number (for this example it will be 376) and note the last digit in the number, discarding the other digits. Then take that digit (6) while ignoring the rest of the number and determine if it is divisible by 2. If it is divisible by 2, then the original number is divisible by 2. • Example: 376 (The original number). 37 6 (Take the last digit). 6÷2 = 3 (Check to see if the last digit is divisible by 2) 376÷2 = 188 (If the last digit is divisible by 2, then the whole number is divisible by 2). ## Mathematical Information About Numbers 8 • About Number 8. The octahedron is one of the five platonic bodies. A polygon with eight sides is an octagon. In computer technology we use a number system on the basis of eight, the octal system. Eight is the first real cubic number, if one disregards 1 cube. It is also the smallest composed of three prime number. Every odd number greater than one, raised to the square, resulting in a multiple of eight with a remainder of one. The Eight is the smallest Leyland number. ## What are divisibility rules? A divisibility rule is a shorthand way of determining whether a given number is divisible by a fixed divisor without performing the division, usually by examining its digits. Although there are divisibility tests for numbers in any radix, and they are all different, this article presents rules and examples only for decimal numbers. For divisors with multiple rules, the rules are generally ordered first for those appropriate for numbers with many digits, then those useful for numbers with fewer digits. © Mathspage.com \| Privacy \| Contact \| info [at] Mathspage [dot] com<\|endoftext\|>	4.4375
910	## AREA OF SQUARE A square is a four-sided closed figure where the lengths of all the four sides will be equal and each vertex angle will be right angle or 90o as shown below. ## Formula for Area of a Square Let s be the length of each side of a square. Then, the formula for area of a square : Area = s2 Let d be the length of each diagonal of a square. Then, the formula for area of a square : Area = 1/2 ⋅ d2 Example 1 : Find the area of the square having side length 24 cm. Solution : When the length of a side is given, formula for area of a square : = s2 Substitute 24 for s. = 242 = 576 So, area of the square is 576 square cm. Example 2 : If the area of a square is 64 square inches, then find the length of each side. Solution : Area of the square = 64 in2 s2 = 64 Find positive square root on both sides. √s2 = √64 s = 8 So, the length of each side of the square is 8 inches. Example 3 : The square has side length 250 cm. Find its area in square meter. Solution : When the length of a side is given, formula for area of a square : = s2 Substitute 250 for s. = 2502 = 62500 cm2 ----(1) We know 100 cm = 1 m Square both sides. (100 cm)2 = (1 m)2 1002 cm2 = 12 m2 10000 cm2 = 1 m2 Therefore, to convert centimeter square into meter square, we have to divide by 10000. (1)----> Area of the square = 62500 cm2 Divide the right side by 10000 to convert cm2 into m2. Area of the square = (62500/10000) m2 6.25 m2 So, the area of the square is 6.25 square meter. Example 4 : If the length of each diagonal is 2√2 cm, then find its area. Solution : When the length of a diagonal is given, formula for area of a square : = 1/2 ⋅ d2 Substitute 2√2 for d. = 1/2 ⋅ (2√2)2 Simplify. = 1/2 ⋅ (4 ⋅ 2) = 1/2 ⋅ (8) = 4 So, the area of the square is 4 square cm. Example 5 : If the lengths of the diagonals of two squares are in the ratio 2 : 5. then find the ratio of their areas. Solution : From the ratio 2 : 5, let the diagonals of two squares be 2x and 5x respectively. When the length of a diagonal is given, formula for area of a square : = 1/2 ⋅ d2 Area of 1st square= 1/2 ⋅ (2x)2= 1/2 ⋅ (4x2)= 4x2 / 2 Area of 2nd square= 1/2 ⋅ (5x)2= 1/2 ⋅ (25x2)= 25x2 / 2 Ratio of the areas : = (4x2 / 2) : (25x2 / 2) Multiply each term of the ratio by 2. 4x2 : 25x2 Divide each term by x2. 4 : 25 So, the ratio of the areas of two squares is 4 : 25. Kindly mail your feedback to [email protected] ## Recent Articles 1. ### Order of Operations Worksheet Jul 24, 24 03:02 AM Order of Operations Worksheet 2. ### Order of Operations Jul 24, 24 03:01 AM Order of Operations - Concept - Solved Examples<\|endoftext\|>	4.75
992	Stained Glass Alignments to Content Standards: 7.G.B.4 7.EE.B.3 The students in Mr. Rivera's art class are designing a stained-glass window to hang in the school entryway. The window will be 2 feet tall and 5 feet wide. They have drawn the design below: They have raised \$100 for the materials for the project. The colored glass costs \$5 per square foot and the clear glass costs \$3 per square foot. The materials they need to join the pieces of glass together costs 10 cents per foot and the frame costs \$4 per foot. Do they have enough money to cover the costs of the materials they will need to make the window? IM Commentary The purpose of this task is for students to find the area and perimeter of geometric figures whose boundaries are segments and fractions of circles and to combine that information to calculate the cost of a project. The shape of the regions in the stained glass window are left purposefully unspecified, as one component skill of modeling with mathematics (MP4) is for students to make simplifying assumptions themselves. Given the precision needed for these estimates, assuming the curves in the design are arcs of a circle is not only reasonable, it is the most expedient assumption to make as well. What is important is that students recognize they are making this assumption and are explicit about it. The question of whether the students have to pay for the scraps of glass left over from cutting out the shapes can be dealt with in different ways. In reality, if they had to buy the glass at a store, the glass would likely come in square or rectangular sheets and they would need to buy more than they were going to use. But exactly how much extra material they would have to buy depends on how the raw materials are sold, so without additional information, it would be hard to determine that without doing some research into how stained glass is sold. Alternatively, the art teacher might already have the materials and just wants his students to stay within a certain budget for the materials they use, knowing that the scraps can be used for future student projects. In any case, this task can provide the springboard for a good classroom discussion around issues that students need to think about when modeling with mathematics. Solution There are many ways to do this. Here is one: Assume that the students only have to pay for the glass they use and not the scraps that they would cut away. That means we need to figure out the area of the colored glass and the area of the clear glass as well as the total length of the seams between the panels of glass. First, we need to find the area of the clear glass and the area of the colored glass. The entire rectangle is 2 feet by 5 feet. Assuming that the curves are all parts of a circle with a 1 foot diameter, there are five 1 by 2 foot rectangles with either 4 half-circles or 2 half circles and 4 quarter circles of clear glass. That means there are 2 full circles of clear glass in each 1 by 2 foot rectangle. Thus, there are 10 complete circles of clear glass, each with a 1 foot diameter (or a $\frac12$ foot radius). Then area of the entire window is 10 square feet, and the area of the clear glass is $$10\times \pi (\frac12)^2= \frac52\pi$$ or approximately 7.9 square feet. That means the area of colored glass is approximately 10 - 7.9 = 2.1 square feet. Now we need to find the total length of the "seams" between the pieces of glass. Again, there are 10 circles. Their total circumference is $$10\times \pi\times1$$ which is about 31.4 feet. There are also four 2-foot straight "seams." So all together there are about 39.4 feet of "seams." The frame is 2+2+5+5 = 14 feet. The cost for the clear glass is $7.9 \times 3 = 23.70$ dollars. The cost for the colored glass is $2.1 \times 5 = 10.50$ dollars. The cost for the materials for the seams is $39.4 \times 0.10 = 3.94$ dollars. The cost of the frame is $14\times4=56$ dollars. The total cost of the materials is $23.70+10.50+3.94+56\approx 94$ dollars. So if these assumptions are accurate, they have just enough money to buy the materials. If they need to pay for the scraps or if they break pieces as they go, they don't have much wiggle room.<\|endoftext\|>	4.65625
1,327	RD Sharma Solutions -Ex-5.1, (Part - 2) Factorization Of Algebraic Expressions, Class 9, Maths # RD Sharma Solutions -Ex-5.1, (Part - 2) Factorization Of Algebraic Expressions, Class 9, Maths - Notes \| Study RD Sharma Solutions for Class 9 Mathematics - Class 9 1 Crore+ students have signed up on EduRev. Have you? Q20 . x2−y2−4xz+4z2 SOLUTION : On rearranging the terms = x2−4xz+4z2−y2 = (x)2−2×x×2z+(2z)2−y2 Using the identity x2−2xy+y2=(x−y)2 = (x−2z)2−y2 Using the identity p2−q2= ( p + q )( p – q ) = (x−2z+y)(x−2z−y) ∴ x2−y2−4xz+4z= (x−2z+y)(x−2z−y) Q21 . SOLUTION : Splitting the middle term , Q22 . SOLUTION : Splitting the middle term, = Q23 . SOLUTION : Splitting the middle term, Q24 . SOLUTION : Splitting the middle term, Q25 . SOLUTION : Splitting the middle term, Q26 . SOLUTION : Splitting the middle term, Q27 . SOLUTION : Splitting the middle term, Q28 . SOLUTION : Using the identity (x−y)2=x2+y2−2xy Q29 . SOLUTION : Splitting the middle term, Q30 . SOLUTION : Splitting the middle term, Q31 . 9(2a−b)2−4(2a−b)−13 SOLUTION : Let 2a – b = x = 9x2−4x−13 Splitting the middle term, = 9x2−13x+9x−13 = x(9x−13)+1(9x−13) = (9x−13)(x+1) Substituting x = 2a – b = [9(2a−b)−13](2a−b+1) = (18a−9b−13)(2a−b+1) ∴ 9(2a−b)2−4(2a−b)−13 = (18a−9b−13)(2a−b+1) Q 32 . 7(x−2y)2−25(x−2y)+12 SOLUTION : Let x-2y = P = 7P2−25P+12 Splitting the middle term, = 7P2−21P−4P+12 = 7P(P−3)−4(P−3) = (P−3)(7P−4) Substituting P = x – 2y = (x−2y−3)(7(x−2y)−4) = (x−2y−3)(7x−14y−4) ∴ 7(x−2y)2−25(x−2y)+12 = (x−2y−3)(7x−14y−4) Q33 . 2(x+y)2−9(x+y)−5 SOLUTION : Let x+y = z = 2z2−9z−5 Splitting the middle term, = 2z2−10z+z−5 = 2z(z−5)+1(z−5) = (z−5)(2z+1) Substituting z = x + y = (x+y−5)(2(x+y)+1) = (x+y−5)(2x+2y+1) ∴ 2(x+y)2−9(x+y)−5 = (x+y−5)(2x+2y+1) Q34 . Give the possible expression for the length & breadth of the rectangle having 35y2−13y−12 as its area. SOLUTION : Area is given as 35y2−13y−12 Splitting the middle term, Area = 35y2+218y−15y−12 = 7y(5y+4)−3(5y+4) = (5y+4)(7y−3) We also know that area of rectangle = length ×breadth ∴ Possible length = (5y+4) and breadth=(7y−3) Or possible length = (7y−3) and breadth= (5y+4) Q35 . What are the possible expression for the cuboid having volume 3x2−12x. SOLUTION : Volume = 3x2−12x = 3x(x−4) = 3×x(x−4) ∴ Possible expression for dimensions of cuboid are = 3 , x , (x−4) The document RD Sharma Solutions -Ex-5.1, (Part - 2) Factorization Of Algebraic Expressions, Class 9, Maths - Notes \| Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics. All you need of Class 9 at this link: Class 9 ## RD Sharma Solutions for Class 9 Mathematics 91 docs Use Code STAYHOME200 and get INR 200 additional OFF ## RD Sharma Solutions for Class 9 Mathematics 91 docs ### Top Courses for Class 9 Track your progress, build streaks, highlight & save important lessons and more! , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ;<\|endoftext\|>	4.53125
835	# Find expressions for the quadratic functions whose graphs are shown. • Sep 7th 2009, 05:40 PM zweevu Find expressions for the quadratic functions whose graphs are shown. Find expressions for the quadratic functions whose graphs are shown. http://www.webassign.net/scalcet/1-2-8.gif I know how to do f(x) but I'm stuck on g(x). • Sep 7th 2009, 05:46 PM Prove It Quote: Originally Posted by zweevu Find expressions for the quadratic functions whose graphs are shown. http://www.webassign.net/scalcet/1-2-8.gif I know how to do f(x) but I'm stuck on g(x). First, note that the y-intercept is 1. So $g(x) = ax^2 + bx + 1$. Using the other two points, we will have two equations in two unknowns to solve simultanously. Equation 1: $2 = a(-2)^2 + b(-2) + 1$ $4a - 2b + 1 = 2$ $4a - 2b = 1$. Equation 2: $-2.5 = a(1)^2 + b(1) + 1$ $a + b + 1 = -2.5$ $a + b = -1.5$. So your system of equations is... $4a - 2b = 1$ $a + b = -1.5$. Solve for a and b. • Sep 7th 2009, 05:51 PM pickslides The second function has the form $y = ax^2+bx+c$ The y-intercept c is 1 as shown in the picture so $y = ax^2+bx+1$ Now use points (-2,2) $y = ax^2+bx+1$ $2 = a(-2)^2+b(-2)+1$ ...(1) and (1,-2.5) $y = ax^2+bx+1$ $-2.5 = a(1)^2+b(1)+1$ ...(2) and solve the sytem • Sep 7th 2009, 06:15 PM zweevu The question is looking g(x) = ... what would I type in for the answer. I got +/- 1/6 for a and b though. Edit: NVM You plug them into the original equation. Thanks! • Sep 9th 2009, 03:23 PM pickslides You have $g(x) = y = ax^2+bx+c$ With $c = 1 , b = \frac{-7}{6}$ and $a = \frac{-1}{3}$ can you find $g(x)$ ? • Sep 10th 2009, 06:43 AM OliverQ We assume that: $g(x) = ax^2 + bx + c$ and the graph of the function got through 3 points, they are: $(-2, 2), (0, 1), (1, -2.5)$ $g(-2) = 4a - 2b + c = 2;$ $g(0) = c = 1$ $g(1) = a + b +c = -2.5$ So: $a = -1; b = -2.5; c = 1;$ $g(x) = -x^2 - 2.5x + 1$ • Sep 10th 2009, 03:52 PM kyleu03 can anyone figure out f(x) ??? • Sep 10th 2009, 04:07 PM pickslides $f(x) = 2(x-3)^2$<\|endoftext\|>	4.46875
982	Quick summary: In this lesson, students will explore issues associated with the disposal of mobile phones and other digital devices. They will then consider how the practices of a circular economy could be applied as a way to resolve or reduce the impact of these issues. The lesson concludes with students creating a poster to encourage others to dispose of their phones in ways that are beneficial to people and the environment. This lesson has been developed as part of Planet Ark’s National Recycling Week, which aims to raise awareness around the benefits of using recycled products. Take part in the Schools Recycling Right Challenge for Planet Ark’s National Recycling Week. Register your lesson or other activities today! - Students understand what a circular economy is. - Students understand negative consequences of our disposal of digital devices. - Students are able to present and share information for the purpose of informing and persuading others. 21st century skills: Australian Curriculum Mapping Year 7 English: - Plan, draft and publish imaginative, informative and persuasive texts, selecting aspects of subject matter and particular language, visual, and audio features to convey information and ideas (ACELY1725) - Use a range of software, including word processing programs, to confidently create, edit and publish written and multimodal texts (ACELY1728) - Identify and explore ideas and viewpoints about events, issues and characters represented in texts drawn from different historical, social and cultural contexts (ACELT1619). Year 8 English: - Create imaginative, informative and persuasive texts that raise issues, report events and advance opinions, using deliberate language and textual choices, and including digital elements as appropriate (ACELY1736) - Use a range of software, including word processing programs, to create, edit and publish texts imaginatively (ACELY1738). Year 7 & 8 Digital Technology: - Define and decompose real-world problems taking into account functional requirements and economic, environmental, social, technical and usability constraints (ACTDIP027) - Evaluate how student solutions and existing information systems meet needs, are innovative, and take account of future risks and sustainability (ACTDIP031). Relevant parts of Year 7 English achievement standards: Students understand how the selection of a variety of language features can influence an audience. They understand how to draw on personal knowledge, textual analysis and other sources to express or challenge a point of view. They create texts showing how language features and images from other texts can be combined for effect. Students create structured and coherent texts for a range of purposes and audiences. They make presentations and contribute actively to class and group discussions, using language features to engage the audience. When creating and editing texts they demonstrate understanding of grammar, use a variety of more specialised vocabulary and accurate spelling and punctuation. Relevant parts of Year 8 English achievement standards: Students understand how the selection of language features can be used for particular purposes and effects. They explain the effectiveness of language choices they make to influence the audience. Through combining ideas, images and language features from other texts, students show how ideas can be expressed in new ways. Students create texts for different purposes, selecting language to influence audience response. They make presentations and contribute actively to class and group discussions, using language patterns for effect. When creating and editing texts to create specific effects, they take into account intended purposes and the needs and interests of audiences. They demonstrate understanding of grammar, select vocabulary for effect and use accurate spelling and punctuation. Relevant parts of Year 7 & 8 Digital Technologies achievement standards: By the end of Year 8, students explain how social, ethical, technical and sustainability considerations influence the design of innovative and enterprising solutions to meet a range of present and future needs. They explain how the features of technologies influence design and production decisions. Topic: National Recycling Week, Sustainability, Waste, Consumption Unit of work: National Recycling Week – Year 7 & 8 Time required: 90 mins. Level of teacher scaffolding: Medium – support students to understand linear and circular economies through discussion. Resources required: Student Worksheets – one copy per student. News article – War on waste: why we should recycle our old mobile phones (one per pair). Device capable of presenting a video to the class. Device capable of accessing the internet to create a digital poster (one each, or one per pair). Headphones (one set per student). Promoting a Circular Economy Task Sheet. Keywords: economy, circular economy, linear economy, reuse, recycle, sustainable, ethical, welfare, mobile phones, waste, National Recycling Week. Cool Australia’s curriculum team continually reviews and refines our resources to be in line with changes to the Australian Curriculum.<\|endoftext\|>	4.125
1,743	In this enquiry we will be finding out all about how to sort animals out using keys, the difference between vertebrates and invertebrates, what micro organisms are how plants and animals are different. Use the links, gallery, video and information below to do some research for yourself. Use the game below by clicking on the picture to look at how to sort animals into different types. What are vertebrates? Vertebrates are animals that have a backbone or spinal column, also called vertebrae. These animals include fish, birds, mammals, amphibians, and reptiles. There are currently around 65,000 known species of vertebrate animals. This sounds like a lot, but vertebrates are only around 3% of all the animals on Earth. Most of the animal species are invertebrates. - Fish – Fish are animals that live in the water. They have gills that allow them to breathe under water. Different species of fish may live in fresh water or salt water. Some examples of fish include the brook trout, the great white shark, lionfish, and the swordfish. - Birds – Birds are animals that have feathers, wings, and lay eggs. Many, but not all, birds can fly. Some examples of bird species include the bald eagle, the cardinal, the flamingo, ostriches, and the red-tailed hawk. - Mammals – Mammals are warm-blooded animals that nurse their young with milk and have fur or hair. Some examples of mammals include humans, dolphins, giraffes, horses, and spotted hyenas. - Amphibians – Amphibians are cold-blooded animals. They start out their lives living in the water with gills just like fish. Later they develop lungs and can move to dry land. Amphibians include frogs, toads, newts, and salamanders. - Reptiles – Reptiles are cold-blooded animals which lay eggs. Their skin is covered with hard and dry scales. Reptile species include alligators, crocodiles, snakes, lizards, and turtles. Cold-blooded and Warm-blooded Vertebrate animals can be either warm-blooded or cold-blooded. A cold-blooded animal cannot maintain a constant body temperature. The temperature of their body is determined by the outside surroundings. Cold-blooded animals will move around during the day between the shade and the sun to warm up or cool down. Cold-blooded animals are ectothermic, which means outside heat. Reptiles, amphibians, and fish are all cold-blooded. Warm-blooded animals are able to regulate their internal temperature. They can sweat or pant to cool off and have fur and feathers to help keep them warm. Warm-blooded animals are called endothermic, meaning “heat inside”. Only birds and mammals are warm-blooded. Did you know? The smallest vertebrate is thought to be a tiny frog called the Paedophryne amauensis. It only grows to about 0.3 inches long. The largest is the blue whale, which can grow to over 100 feet long and 400,000 pounds. Fun Facts about Vertebrates - The only mammals that lay eggs are monotremes such as the platypus and spiny anteater. - There are reptiles that live on every continent except Antarctica. - Most fish have skeletons make of bone, they are called bony fish. Other fish have skeletons made of cartilage. These include sharks and rays. - Frogs can breathe through their skin. - The shortest childhood of any mammal is the hooded seal. They are considered adults when they are just four days old. - Vertebrates tend to be much more intelligent than invertebrates. (The above is from http://www.ducksters.com/animals/vertebrates.php) What are invertebrates? Invertebrates are animals that do not have backbones, also called vertebrae or spinal bones. Invertebrates as a group do not have a specific classification. Mammals, amphibians, reptiles, fish, and birds all have vertebrae. This might seem like a lot of the animals you know, but all these animals make up less than 4% of the total animals species. This means that over 96% of all the animal species on Earth are invertebrates. What are some invertebrate animals? - Marine Invertebrates – There are a wide variety of interesting ocean animals that are invertebrates. These include sponges, corals, jellyfish, anemones, and starfish. - Mollusks – Mollusks have a soft body that is covered by an outer layer called a mantle. Many mollusks live inside a shell, but not all of them. Some examples of mollusks include squid, snails, slugs, octopuses, and oysters. - Crustaceans -Crustaceans are a type of arthropod, meaning that they have jointed legs. They also have an exoskeleton (their bones are on the outside like a shell). Some examples of crustaceans are crabs, lobster, shrimp, and barnacles. - Worms – The term “worm” is not a scientific word, but is often used to refer to invertebrate animals that don’t have legs. Worms may live in the soil, in the water, or even inside other animals as parasites. Some examples include the tapeworm, the leech, and the earthworm. - Insects – Insects are part of the Earth’s largest animal phylum, the arthropods. There are over 1 million species of insects including such animals as the grasshopper, dragonfly, yellow jacket, butterfly, and praying mantis. - Spiders, Centipedes and Scorpions – These animals are all part of the arthropod phylum. Spiders and scorpions are arachnids because they have eight legs. Centipedes and millipedes are myriapods and have lots of legs. Some myriapods have as many as 750 legs. Some example species include the tarantula and black widow, which are both spiders. Did you know? The largest of the invertebrates is the colossal squid. It can grow to over 40 feet long and weigh over 1,000 pounds. The longest invertebrate is the ribbon worm which can grow to 180 feet long. The smallest invertebrate is the rotifer, or wheel animal, which can be as small as 50um. Way too small to see with just your eyes. Fun Facts about Invertebrates - Around 23% of all marine organisms are mollusks. - The only hard body part of an octopus is a hooked beak at the end of its tentacles. - Some invertebrates, such as echinoderms, do not have heads. - There are likely millions of invertebrates living in your house right now. They are called dust mites and you can’t see them. - When a crustacean outgrows its shell, it sheds the shell and grows a new one. - Lobsters, crabs, and shrimp all have 10 legs. The front two legs have pincers they can use to catch food and fight off predators. - Some scorpion mothers protect their young by carrying them on their backs. - Centipedes are carnivores which eat insects and worms. They have a poisonous bite to help them kill their prey. Millipedes are herbivores who eat plants and rotting material. Can you spot which animals are Vertebrates and which are invertebrates in the gallery below? What have you learnt? [Spider_Single_Video track=”9″ theme_id=”3″ priority=”0″] What are microorganisms? Microorganisms are very tiny living things. They are so small that you need a microscope to see them. Microorganisms are all around us, in the air, in our bodies and in water. Some microorganisms are harmful to us, but others are helpful to us. Click here or here to find out more! There are three types of microorganism: Play some games below by clicking on the pictures<\|endoftext\|>	3.796875
1,459	Hong Kong Stage 1 - Stage 3 # Solve problems involving 4 operations Lesson Let's write down some of the things we have discovered about addition and multiplying on the number line. ## Absolute value The absolute value of a number is a measure of its size or magnitude. On the number line the absolute value of a number is its distance from the the number $0$0. Distances have no sign attached to them, and neither do absolute values. We use small vertical lines to surround a number to indicate its absolute value, so that the absolute value of minus three is written as $\|-3\|$\|3\|. Its value is simply $3$3. The value of $\|3\|$\|3\| is also $3$3, because $3$3 and $-3$3 are an equal distance away from $0$0 on the number line. The meaning of the two absolute values $\|-3\|$\|3\| and $\|2\|$\|2\| are illustrated here on this number line: ## Operating with negative numbers We will use the numbers $6$6, $-6$6, $2$2, and $-3$3 to illustrate what happens when we add, subtract or multiply using signed numbers in mathematical sentences. The same rules apply for any number on the number line. The sentence $3+(-3)$3+(3) changes to $3-3$33 and so the answer is $0$0 The sentence $-3+(-3)$3+(3) changes to $-3-3$33 and so the answer is $-6$6 The sentence $6-(-6)$6(6) changes to $6+6$6+6 and so the answer is $12$12 The sentence $-6-(-6)$6(6) changes to $-6+6$6+6 and so the answer is $0$0 #### Multiplication The sentence $6\times(-3)$6×(3) becomes $-18$18, because the product of a $+$+ and a $-$ is always a $-$. The sentence $-6\times(-3)$6×(3) becomes $18$18, because the product of a $-$ and a $-$ is always a $+$+ The sentence $-6\times3$6×3 becomes $-18$18, because the product of a $+$+ and a $-$ is always a $-$. The sentence $6\times3$6×3 becomes $18$18, because, well, thats always been the answer! #### Division Division works in the same way as multiplication. The division of a positive number by a negative number, or a negative number by a positive number is always negative. so $6\div(-2)$6÷(2) is $-3$3 and $-6\div2$6÷2 is $-3$3 Division of any negative number by a negative number is always positive, so $-6\div(-2)$6÷(2) becomes $3$3. And, as usual, division of a positive number by a positive number is positive! Of course, dividing two integers can lead to numbers that are not integers. So $-3\div-6$3÷6 becomes $\frac{1}{2}$12 ## The order of operations Mathematicians use an important convention when simplifying number sentences. A convention is an agreed way of interpreting things so that we don't confuse meaning. For example, we might think that a sentence like $2+(-3)\times6-(-2)$2+(3)×6(2) is worked out from left to right so that we might simplify it as: $2+(-3)\times6-(-2)$2+(−3)×6−(−2) $=$= $-1\times6-(-2)$−1×6−(−2) $=$= $-6-(-2)$−6−(−2) $=$= $-6+2$−6+2 $=$= $-4$−4 We tend to think this is correct, because we are used to reading words in a sentence from left to right. But in mathematics, ... "convention not is this the". In fact we have a strict and agreed order to reading certain sentences. This includes working within brackets first, then simplifying divisions and multiplications from left to write, and then finally doing the additions and subtractions, also from left to right. The sentence above, $2+(-3)\times6-(-2)$2+(3)×6(2) is correctly worked out as: $2+(-3)\times6-(-2)$2+(−3)×6−(−2) $=$= $2+(-18)-(-2)$2+(−18)−(−2) $=$= $2-18+2$2−18+2 $=$= $-16+2$−16+2 $=$= $-14$−14 Here is another interesting example involving a fraction where the line in the fraction (called a vinculum) acts as a bracket, so that the numbers in the numerator are simplified first: $\frac{6-(-4)}{-5}$6−(−4)−5 $=$= $\frac{6+4}{-5}$6+4−5 $=$= $\frac{10}{-5}$10−5 $=$= $-2$−2 #### Worked Examples ##### Question 1 For the following expressions, $q$q is a negative number, $r$r is a negative number, and $t$t is a positive number. We want to decide whether each expression simplifies to a positive or negative number, or if it is not possible to determine. 1. The sign of $\frac{q}{r-t}$qrt is: Not possible to determine A Positive B Negative C 2. The sign of $q+t$q+t is: Not possible to determine A Positive B Negative C 3. The sign of $r\left(q-t\right)$r(qt) is: Negative A Positive B Not possible to determine C ##### Question 2 An investment loses $\$2220022200. If this loss is shared equally among six, use a negative integer to describe the loss for each person. ##### Question 3 Consider the following phrase: The quotient of $-3$3 and the sum of $7$7 and $6$6 . 1. Without simplifying the result, translate this sentence into a mathematical expression. 2. Evaluate the expression.<\|endoftext\|>	4.875
655	Osaka - Solar cells will play a key role in shifting to a renewable economy. Organic photovoltaics (OPVs) are a promising class of solar cells, based on a light-absorbing organic molecule combined with a semiconducting polymer. OPVs are made from inexpensive, lightweight materials, and benefit from good safety as well as easy production. However, their power conversion efficiencies (PCEs) - the ability to convert light into electricity - are still too low for full-scale commercialization. The PCE depends on both the organic and the polymer layer. Traditionally, chemists have experimented with different combinations of these by trial-and-error, leading to a lot of wasted time and effort. Now, a team of Osaka University researchers has used computer power to automate the search for well-matched solar materials. In the future, this could lead to vastly more efficient devices. The study was reported in The Journal of Physical Chemistry Letters. "The choice of polymer affects several properties, like short-circuit current, that directly determine the PCE," study first author Shinji Nagasawa explains. "However, there's no easy way to design polymers with improved properties. Traditional chemical knowledge isn't enough. Instead, we used artificial intelligence to guide the design process." Informatics can make sense of large, complex datasets by detecting statistical trends that elude human experts. The team gathered data on 1,200 OPVs from around 500 studies. Using Random Forest machine learning, they built a model combining the band gap, molecular weight, and chemical structure of these previous OPVs, together with their PCE, to predict the efficiency of potential new devices. Random Forest uncovered an improved correlation between the properties of the materials and their actual performance in OPVs. To exploit this, the model was used to automatically "screen" prospective polymers for their theoretical PCE. The list of top candidates was then whittled down based on chemical intuition about what can be synthesized in practice. This strategy led the team to create a new, previously untested polymer. In the event, a practical OPV based on this first try proved less efficient than expected. However, the model provided useful insights into the structure-property relationship. Its predictions could be improved by including more data, such as the polymers' solubility in water, or the regularity of their backbone. "Machine learning could hugely accelerate solar cell development, since it instantaneously predicts results that would take months in the lab," co-author Akinori Saeki says. "It's not a straightforward replacement for the human factor - but it could provide crucial support when molecular designers have to choose which pathways to explore." Osaka University was founded in 1931 as one of the seven imperial universities of Japan and now has expanded to one of Japan's leading comprehensive universities. The University has now embarked on open research revolution from a position as Japan's most innovative university and among the most innovative institutions in the world according to Reuters 2015 Top 100 Innovative Universities and the Nature Index Innovation 2017. The university's ability to innovate from the stage of fundamental research through the creation of useful technology with economic impact stems from its broad disciplinary spectrum.<\|endoftext\|>	3.671875
634	# intersections • Jan 8th 2008, 07:38 PM chibiusagi intersections Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point • Jan 8th 2008, 07:46 PM Jhevon Quote: Originally Posted by chibiusagi Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point we can find how many zeros the function has by simply solving for them. set $(x - 2) \left( x^2 + 2x + 6 \right) = 0$ $\Rightarrow x - 2 = 0$ or $x^2 + 2x + 6 = 0$ $\Rightarrow x = 2$ or $x = \frac {-2 \pm \sqrt{-20}}2$ clearly the quadratic has no real roots, thus the only root is $x = 2$ to find the tangent line, use the point-slope form. $y - y_1 = m(x - x_1)$ here, $m = f'(2)$ and $(x_1,y_1) = (2,0)$ • Jan 8th 2008, 07:50 PM topsquark Quote: Originally Posted by chibiusagi Prove that the curve y=(x-2)(x^2+2x+6) crosses the x axis at on point only and find the equation of the tengent at that point $y = (x - 2)(x^2 + 2x + 6)$ When this crosses the x-axis we have y = 0. So to find these x values: $(x - 2)(x^2 + 2x + 6) = 0$ So either $x - 2 = 0 \implies x = 2$ or $x^2 + 2x + 6 = 0$ Note, however, that the quadratic factor here has no real zeros. So no real zeros are obtained by solving this equation. Thus the curve only crosses the x-axis once. Now to find the equation of the tangent. $y^{\prime}(x) = 3x^2 + 2$ So at x = 2 the slope of the tangent to the curve is $y^{\prime}(2) = 14$. So we need the equation of a line with a slope of 14 that passes through the point (2, 0). $y = 14x + b$ $0 = 14 \cdot 2 + b \implies b = -28$ So the tangent line at (2, 0) is $y = 14x - 28$. -Dan<\|endoftext\|>	4.40625
732	Canadian and World Studies In Canadian and world studies, students develop skills, knowledge and understanding, and attitudes that will serve them both inside and outside the classroom, including in the world of work and as responsible citizens in the various communities to which they belong. Read the complete curriculum document Geography – Grade 9 The Grade 9 geography courses provide students with opportunities to explore these three aspects of geography as they investigate geographic issues in Canada. In these courses, students will examine issues relating to interactions between physical processes and people living in Canada; changing populations in this country; economic and environmental sustainability; and interconnections between Canada and the global community. History – Grade 10 History involves the study of diverse individuals, groups, and institutions as well as significant events, developments, and issues in the past. The Grade 10 history courses provide students with an overview of Canadian history from the eve of World War I to the present. Civics – Grade 10 Politics involves the study of how societies are governed, how policy is developed, how power is distributed, and how citizens take public action. The Grade 10 course Civics and Citizenship focuses on civics, a branch of politics that explores the rights and responsibilities of citizens, the processes of public decision making, and ways in which citizens can act for the common good within communities at the local, national, and/or global level. Grades 11 & 12 Read the complete curriculum document Economics is about making choices, as individuals and as a society, about how best to use limited resources. An understanding of fundamental concepts, models, and methods of inquiry associated with economics can help us make informed decisions about how to allocate resources to address people’s unlimited needs and wants. The Grade 11 and 12 economics courses provide students with opportunities to develop an understanding of different economic systems and institutions and to assess the ability of those systems and institutions to satisfy people’s needs and wants. In the Grade 11 and 12 geography courses, students will develop their ability to apply both the geographic inquiry process and the concepts of geographic thinking. They apply this process and these concepts as they investigate a wide range of geographic issues and deepen their awareness of interconnections between Canadian and global issues. History involves the study of diverse individuals, groups, and institutions as well as significant events, developments, and issues in the past. The Grade 11 and 12 history courses provide students with opportunities to study many aspects of Canadian and world history, from early societies to the present. These courses convey a sense of the dynamic nature of Canadian and world history. The law is about society’s efforts to promote fairness and justice. It involves formal rules that are enforced and adjudicated by institutions. The law shapes politics, the economy, and society in many ways as it attempts to mediate relationships between people. The Grade 11 and 12 law courses provide students with opportunities to develop an understanding of the historical and philosophical foundations of our legal system. Canadian and World Studies Students develop an understanding of the relevance of law in everyday life and of their rights and responsibilities within the Canadian legal system. What they learn about the law will also help students understand why laws change over time. The Grade 11 and 12 politics courses provide opportunities for students to investigate a range of issues of political importance at the local, national, and global level. Students will explore issues related to policy making, political engagement, the distribution of power, human rights, and international relations. Students are encouraged to explore political thought, to clarify their own values and positions relating to political issues, and to explore ways in which they can contribute to political change.<\|endoftext\|>	3.71875
1,108	# 7/14/2015 12:41 AM6.4 - Dividing Polynomials (Long Division)1 Polynomial Division SECTION 6.4 LONG DIVISION and Synthetic Division. ## Presentation on theme: "7/14/2015 12:41 AM6.4 - Dividing Polynomials (Long Division)1 Polynomial Division SECTION 6.4 LONG DIVISION and Synthetic Division."— Presentation transcript: 7/14/2015 12:41 AM6.4 - Dividing Polynomials (Long Division)1 Polynomial Division SECTION 6.4 LONG DIVISION and Synthetic Division 7/14/2015 12:41 AM 6.4 - Dividing Polynomials (Long Division)2 The methods In this section, we will look at two methods to divide polynomials:In this section, we will look at two methods to divide polynomials: –long division (similar to arithmetic long division) –synthetic division (a quicker, short-hand method) Steps for Long DivisionSteps for Long Division –Multiply the answer by the divisor and then subtract Subtracting involves multiplying by -1Subtracting involves multiplying by -1 –Repeat process until it can not be done –Leftover is remainder (Just like division) 7/14/2015 12:41 AM 6.4 - Dividing Polynomials (Long Division)3 Example 1 Divide (x 2 – 5x + 4) ÷ (x – 1)Divide (x 2 – 5x + 4) ÷ (x – 1) Rewrite in long division form... 7/14/2015 12:41 AM 6.4 - Dividing Polynomials (Long Division)4 Example 2 Divide (x 3 – 28x – 48) ÷ (x + 4)Divide (x 3 – 28x – 48) ÷ (x + 4) 7/14/2015 12:41 AM 6.4 - Dividing Polynomials (Long Division)5 Example 3 Divide (2x 2 + 3x – 4) ÷ (x – 2)Divide (2x 2 + 3x – 4) ÷ (x – 2) Rewrite in long division form... 7/14/2015 12:41 AM 6.4 - Dividing Polynomials (Long Division)6 Example 4 Divide (x 3 – 6) ÷ (x – 1)Divide (x 3 – 6) ÷ (x – 1) 7/14/2015 12:41 AM 6.4 - Dividing Polynomials (Long Division)7 Example 5 Divide (12x 4 - 5x 2 – 3) ÷ (3x 2 + 1)Divide (12x 4 - 5x 2 – 3) ÷ (3x 2 + 1) Rewrite in long division form... 7/14/2015 12:41 AM 6.3 Polynomials and Polynomial Functions H 8 Synthetic Division Synthetic Division allows omitting all variables and exponents 7/14/2015 12:41 AM9.4 Polynomials and Polynomial Functions H 9 Steps Steps for Synthetic Division: 1.Identify the divisor and reverse the sign of the constant term. Write the coefficients of the polynomial in standard form. 2.Bring down the first coefficient 3.Multiply the first coefficient by the new divisor, identify the result under the next coefficient and add. 4.Repeat the steps of multiplying and adding until the remainder is found 5.Go backwards from the remainder and assign variables 7/14/2015 12:41 AM 6.3 Polynomials and Polynomial Functions H 10 Example 1 Divide (x 3 – 13x + 12) ÷ (x + 4) by synthetic div. 7/14/2015 12:41 AM 6.3 Polynomials and Polynomial Functions H 11 Example 1 Divide (x 3 – 13x + 12) ÷ (x + 4) –4 · 1 = –4 –4 · –4 = 16 –4 · 3 = –12 7/14/2015 12:41 AM 6.3 Polynomials and Polynomial Functions H 12 Example 1 Divide (x 3 – 13x + 12) ÷ (x + 4) 7/14/2015 12:41 AM9.4 Polynomials and Polynomial Functions H 13 Example 2 Divide (x 3 – 2x 2 – 5x + 6) ÷ (x + 2) using synthetic division 7/14/2015 12:41 AM9.4 Polynomials and Polynomial Functions H 14 Example 3 Divide (x 3 – 3x 2 – 5x – 25) ÷ (x – 5) using synthetic division Download ppt "7/14/2015 12:41 AM6.4 - Dividing Polynomials (Long Division)1 Polynomial Division SECTION 6.4 LONG DIVISION and Synthetic Division." Similar presentations<\|endoftext\|>	4.59375
275	Tin is an essential metal in the creation of tin bronzes, and its acquisition was an important part of ancient cultures from the Bronze Age onward. Its use began in the Middle East and the Balkans around 3000 BC. Tin is a relatively rare element in the Earth's crust, with about 2 parts per million (ppm), compared to iron with 50,000 ppm, copper with 70 ppm, lead with 16 ppm, arsenic with 5 ppm, silver with 0.1 ppm, and gold with 0.005 ppm. Ancient sources of tin were therefore rare, and the metal usually had to be traded over very long distances to meet demand in areas which lacked tin deposits. The importance of tin to the success of Bronze Age cultures and the scarcity of the resource offers a glimpse into that time period's trade and cultural interactions, and has therefore been the focus of intense archaeological studies. However, a number of problems have plagued the study of ancient tin such as the limited archaeological remains of placer mining, the destruction of ancient mines by modern mining operations, and the poor preservation of pure tin objects due to tin disease or tin pest. These problems are compounded by the difficulty in provenancing tin objects and ores to their geological deposits using isotopic or trace element analyses. Current archaeological debate is concerned with the origins of tin in the earliest Bronze Age cultures of the Near East.<\|endoftext\|>	3.96875
3,839	A linear gradient is a gradient that fades from one color to another over a line (as opposed to a radial gradient that has a circular or elliptical shape). Concepts and Terminology A gradient is drawn into a box with the dimensions of the concrete object size, referred to as the gradient box. However, the gradient itself has no intrinsic dimensions. A linear gradient is defined by an axis called the gradient line. The colors making up the gradient are then placed along that line. The line can be horizontal, vertical, or diagonal, taking any angle value. The direction of the line and the colors specified are all defined in the function, but we’ll get to these after covering the concept of the gradient line. The gradient image is constructed by creating an “infinite canvas” and painting it with lines perpendicular to the gradient line, with the color of each line being the color of the gradient line where the two intersect. This produces a smooth fade from each color to the next, progressing in the specified direction. To better understand this, please have a look at the following illustration: The gradient line is the line along which the colors are going to be painted. The colors you specify will start at the starting point and end at the ending point. The first color your specify will be the color of the starting point, and the final color you specify will be the color of the ending point. Between the starting point and the ending point, the points will be painted by the intermediate colors between the first and last color. If you specify multiple colors like three, four, or more, those colors will also be spread along the gradient line, starting with the first, going through the intermediate ones, and ending with the last color in the list. Notice how, though the starting point and ending point are outside of the element, they’re positioned precisely right so that the gradient starts with the pure first color exactly at the starting corner, and is the pure ending color exactly at the opposite corner. That’s intentional, and will always be true for linear gradients. Notice also how the two perpendicular lines—the starting line and ending line—are positioned. Every point on the starting line is going to be of the starting color. Every point on the ending line is the ending color. In between these two perpendicular lines, imagine a set of similar perpendicular lines, each line colored the same color as its point of intersection with the gradient line. Creating Linear Gradients: Specifying a Gradient Line When you want to create your own linear gradient, you need to start by specifying the gradient line. You can do this by specifying an angle value which specifies the angle of the line inside the element. You can also use a keyword value to specify the direction of the gradient line. Possible keywords are to top (which means that the gradient line will be vertical, and will start from the bottom up), to bottom (the opposite of to right (which means that the line will be horizontal and will start from the left and progress to the right), and to left (which is the opposite of The gradient line will determine how the gradient is laid out inside the element. The following is an interactive demo forked from this demo by Ana Tudor. In this demo, you can click on any of the points on the circle. Each point specifies an angle value, and when you click on it, the gradient line (the blue line) and its corresponding starting and ending lines (pink and black lines, respectively) will be repositioned so you can see how they react to the angle value you choose. The gradient image inside the element will also be updated for every angle value. The angles start at zero degrees at the y-axis, and move clockwise. You can also see the syntax of the linear-gradient() function get updated with the value of the angle as you click them. The first value inside the linear-gradient() function specifies the angle or direction of the gradient line. Creating Linear Gradients: Specifying Color Stops A gradient can be made up of as many colors as you want. In addition to an angle/direction, the linear-gradient() function accepts a list of comma-separated color stops. A color stop is a color value followed by an optional stop position. linear-gradient(angle/direction, color stop, color stop, ...); A very simple linear gradient can be created by specifying only two color stops: linear-gradient(to right, yellow, purple); The gradient will start from the left and proceed to the right (notice the to right keyword). It starts with the yellow color and fades smoothly until it becomes purple at the end point of the gradient line. Using a different angle value, you can change the direction of the gradient. However, in this section we want to focus on the color stops, so we’ll keep using the to right direction. Now, suppose you want the gradient to contain three or four colors (or even more!). It can look like so: linear-gradient(to right, yellow, #009966, purple); When three colors are specified, they are distributed along the gradient line such that the starting point at 0% is pure yellow (in our example), the 50% point along the gradient line is pure #009966 (which is a green color), and the 100% point (ending point) is pure purple. The points lying between 0% and 50% are the intermediate colors you pass through when you’re moving from yellow to #009966. Similarly, the points between the 50% color stop and the 100% position, are the intermediate colors between #009966 and purple. If you specify four colors, then the 0% color stop will be the first color, the ~33% color stop will be the second color, the ~66% stop will be the third color, and then the 100% will be the fourth color. In addition to specifying the colors you want to have in your gradient, you can control where each color starts instead of leaving it to the browser to distribute the colors evenly. If we were to go back to the above three-color example, and change the start position of the green color to be 20% instead of the default 50%: linear-gradient(to right, yellow, #009966 20%, purple); we would get the following result: As you can see in the above image, the distance between pure yellow (0%) and pure #009966 decreases, because the line is pure #009966 at 20% instead of 50%. Controlling where a color starts is useful, but what about creating “sharp” color transitions? For example, what if I want the line to be pure green for more than just one point on the line? The key to getting complete sections of the same color is to specify the start position of a color at one point where we want that section to start, and then specify another color stop for the same color at the point where we want the section to end. For example, suppose we want the green color #009966 to start at 20%, and we want the gradient line to remain green until 80%, instead of starting to fade into purple. The gradient function would look like the following: linear-gradient(to right, yellow, #009966 20%, #009966 80%, purple); The result of the above function looks like this: The area between 0% and 20% fades from pure yellow to #009966 as it would by default. Then, the section between 20% and 80% maintains a #009966 color. After that, the color starts fading again to purple. Using this concept, we can create “sharp” color stops. Suppose we have the following gradient function: linear-gradient(to right, yellow, yellow 20%, #009966 20%, #009966 80%, purple 80%, purple); The gradient will look like the following: So what exactly happened there? Well, we’ve specified six color stops: 0%, 20%, 20% again, then 80%, 80% again, and then the final 100%. For each of these stops, we’ve specified a color. Then, the area between these points will be made up of the intermediate colors. So, if we were to take the first section, for example, between 0% and 20%: we have yellow at 0%, and then yellow again at 20%; this means that the color will fade from yellow to yellow which, in other words, means that the entire section will be yellow. Then, at 20% again we start with #009966 (this could have been specified as 21%, too). The browser will fade the color between #009966 at 20%, and then the next color stop at 80%, which happens to be also #009966. Therefore, the entire section between 20% and 80% will be #009966 green. The same concept applies to the last purple section. Many examples can be created using different colors and a different number of color stops. For example, the following is a simple rainbow created using the above concept of “sharp” color stops. Note that the color stop position can be set in percentages or absolute length values (like pixels, for example, or em), and that the color stop position must come after the color value. So, the following syntax is invalid: linear-gradient(to right, red, red 14.3%, 14.3% orange, orange 28.6%, 28.6% yellow, yellow 42.9%, 42.9% green, green 57.2%, 57.2% blue, blue 71.5%, 71.5% indigo, indigo 85.8%, 85.8% violet, violet); linear-gradient() function accepts any valid <color> value. This means that you can set the colors using any of the possible color functions like hsl(), among others. See the <color> entry for more possible values. <color> value also means that you can use a transparent color as well. For example, the following creates a linear gradient that starts with a color, is transparent in the middle, and the ends with another color: linear-gradient(to right, yellow, transparent, purple); Notes about Color Stops Color-stops are points placed along the line defined by the gradient line at the beginning of the rule. Color-stops must be specified in order. Percentages refer to the length of the gradient line, with 0% being at the starting point and 100% being at the ending point. Lengths are measured from the starting point in the direction of the ending point. Color-stops are usually placed between the starting point and ending point, but that’s not required; the gradient line extends infinitely in both directions, and a color-stop can be placed at any position on the line. At each color-stop, the line is the color of the color-stop. Between two color-stops, the line’s color is linearly interpolated between the colors of the two color-stops, with the interpolation taking place in premultiplied RGBA space. Before the first color-stop, the line is the color of the first color-stop. After the last color-stop, the line is the color of the last color-stop. The following steps must be applied by user agents in order to process the list of color-stops. They offer some great insight into how gradients work. After applying these rules, all color-stops will have a definite position and they will be in ascending order: If the first color-stop does not have a position, set its position to 0%. If the last color-stop does not have a position, set its position to 100%. For example, this: linear-gradient(red, white 20%, blue); is equivalent to: linear-gradient(red 0%, white 20%, blue 100%); If a color-stop has a position that is less than the specified position of any color-stop before it in the list, set its position to be equal to the largest specified position of any color-stop before it. For example, this: linear-gradient(red 80px, white 0px, black, blue 100px); linear-gradient(red 80px, white 80px, black 90px, blue 100px); If any color-stop still does not have a position, then, for each run of adjacent color-stops without positions, set their positions so that they are evenly spaced between the preceding and following color-stops with positions. For example, this: linear-gradient(red -50%, white, blue); is equivalent to: linear-gradient(red -50%, white 25%, blue 100%); Trivia & Notes Note that, even though a gradient is made up of colors, it is not a <color> value, and therefore can’t be used as one. Repeating Gradients (Tiling them) A linear gradient created using linear-gradient() cannot be tiled by itself. It’s usually painted across the entire area of the element it is applied to, unless you specify otherwise using a property like background-size for example. To create repeating linear gradients, you can use the repeating-linear-gradient() function. Meanwhile, you can also fake a repeating gradient using the background-repeat properties to create patterns. Creating Patterns Using Linear Gradients Gradients are images. They can be used as background images. And just like background images, they can be sized, positioned, and tiled as a background image. This allows us to create some really nice patterns. We can also use multiple linear gradients as multiple backgrounds on an element. Because creating the actual patterns is outside the scope of this entry, we suggest you have a look at the following two gradient patterns galleries by Lea Verou and Bennett Feely, respectively: - CSS3 Patterns Gallery - New CSS3 Gradients Possibilities (makes use of the CSS Both of these galleries allow you to view the code for each pattern. You can play with the code to see how the patterns are created and customize them to your will. Animating Linear Gradients Linear gradients are not animatable except in IE10+. All other browsers don’t allow you to animate a gradient. Usually, an illusion of gradient color transitions is created by transitioning the position of the gradient image on an element. You can read more about this in the article Animating CSS3 Gradients by Louis Lazaris. <linear-gradient> = linear-gradient( [ [ <angle> \| to <side-or-corner> ] ,]? <color-stop>[, <color-stop>]+ ) /* where.. / <side-or-corner> = [left \| right] \|\| [top \| bottom] / and / <color-stop> = <color> [ <percentage> \| <length> ]? <angle>value that specifies the direction of the gradient. 0degpoints upward, and positive angles represent clockwise rotation, so 90degpoints toward the right. This argument can specify either a side or a corner that are used to specify the angle of the gradient. Possible side values are: left. If a side is specified, it is preceded by the keyword to. For example, to top, which resolves to an angle of to rightwhich is equivalent to to bottomwhich resolves to a value of to leftwhich is equal to an angle value of Possible corner values are: bottom right, and bottom left. Each of these keywords is also preceded by the tokeyword. Resulting corner values are to top left, to top right, to bottom right, and to bottom left. When a corner value is used, it is translated into an angle value that allows the ending point to be in the same quadrant as the specified corner, and so that the line defined by the ending-point and the corner is perpendicular to the gradient line. (See interactive demo in the description section at the beginning of this entry.) This means that the ending line will be positioned exactly so that the last color of the gradient will be applied to the specified corner of the element. The starting line will be positioned so that it’s symmetric to the ending line with respect to the center of the element. <color>value followed by an optional color stop position. The color stop position can be set in percentage values or fixed lengthvalues such as em, among others. Color stops are positioned along the gradient line, and are used to specify the points along that line that colors start at. The following are all simple examples of linear gradients created using The following creates a striped background image using a linear gradient made up of two colors: black and white. The effect is achieved by using the background-size property and specifying the width of the gradient to any width you want, depending on how thin or thick you want the stripes to be. The gradient will then be repeated because the default value of the background-size property is repeat. Play with the values of the background-size property and try using background-repeat: no-repeat to see how the gradient is sized. Play with the values of the angle/direction and color stops of the following demo to see how the gradient changes.View this demo on the Codrops Playground Method of defining a linear or radial color gradient as a CSS image. W3C Candidate Recommendation Supported from the following versions: Mobile / Tablet denotes prefix required. Stats from caniuse.com Notes: Old & New Syntax and Prefixes linear-gradient() syntax described in this entry is the latest and newest one, and the stable one. Before this syntax was stabilized, there were other syntaxes used among browsers, in addition to different prefixes. For a recap of the history of the syntax and how it has evolved to the current one, you can check the MDN entry. The support table also shows in which browser versions prefixes are needed.<\|endoftext\|>	3.9375
905	Jülich, 8 December 2016. An international team of scientists has succeeded in making further improvements to the lifetime of superconducting quantum circuits. An important prerequisite for the realization of high-performance quantum computers is that the stored data should remain intact for as long as possible. The researchers, including Jülich physicist Dr. Gianluigi Catelani, have developed and tested a technique that removes unpaired electrons from the circuits. These are known to shorten the qubit lifetime (to be published online by the journal Science today, DOI: 10.1126/science.aah5844). Quantum computers could one day achieve significantly higher computing speeds than conventional digital computers in performing certain types of tasks. Superconducting circuits belong to the most promising candidates for implementing quantum bits, known as qubits, with which quantum computers can store and process information. The high error rates associated with previously available qubits have up to now limited the size and efficiency of quantum computers. Dr. Gianluigi Catelani of the Peter Grünberg Institute (PGI-2) in Jülich, together with his colleagues has now found a way to prolong the time in which the superconducting circuits are able to store a "0" or a "1" without errors. Beside Catelani, the team comprises researchers working in the USA (Massachusetts Institute of Technology, Lincoln Laboratory, and the University of California, Berkeley), Japan (RIKEN), and Sweden (Chalmers University of Technology). When superconducting materials are cooled below a material-specific critical temperature, electrons come together to form pairs; then current can flow without resistance. However, so far it has not been possible to build superconducting circuits in which all electrons bundle together. Single electrons remain unpaired and are unable to flow without resistance. Due to these so-called quasiparticles, energy is lost and this limits the length of time that the circuits can store data. Researchers have now developed and tested a technique that can temporarily remove unpaired electrons away from the circuit; with the help of microwave pulses, they are in effect "pumped out". This results in a three-fold improvement in the lifespan of the qubits. "The technique can in principle be put to immediate use for all superconducting qubits", explained Catelani, who, as a theoretical physicist has contributed to the analysis and interpretation of the experimental data. However, he emphasised that the lifespan of qubits is only one of many hurdles in the development of complex quantum computers. Moreover, the new technique means that the quasiparticles are not permanently removed, but flow back again and again. The scientists have another solution ready to solve this problem: the pumping technique can be combined with another method that permanently traps the quasiparticles. Catelani, together with his colleagues from Jülich and Yale, has already analysed and tested such a quasiparticle "trap". Their results were published in September in the journal Physical Review B (DOI: 10.1103/PhysRevB.94.104516). Suppressing relaxation in superconducting qubits by quasiparticle pumping; Simon Gustavsson et al.; Science (to be published online on 8. Dec. 2016), DOI: 10.1126/science.aah5844 (available from 8:00 pm CET). Illustration of the filtering of unwanted quasiparticles (red spheres) from a stream of superconducting electron pairs (blue spheres) using a microwave-driven pump. Copyright: Philip Krantz, Krantz NanoArt Forschungszentrum Jülich: http://www. Research at the Peter Grünberg Institute, Theoretical Nanoelectronics (PGI-2): http://www. Press release from 17.4.2014 "Quantum computing: 50-year-old prediction confirmed": http://www. Dr. Gianluigi Catelani, Peter Grünberg Institute, Theoretical Nanoelectronics (PGI-2), Forschungszentrum Jülich, Germany Phone: +49 2461 61-9360 Angela Wenzik, Science Journalist, Forschungszentrum Jülich, Germany Phone: +49 2461 61-6048,<\|endoftext\|>	3.75
457	White out the speech bubbles on several comic strips and photocopy them for the class to use. Have groups of students decide what might be going on in each frame of the comic strip. After they finish, encourage them to share their ideas with the class, as well as why they made those inferences from the pictures. Inferences vs. Facts It is important for students to differentiate between facts and inferences, and this inference activity will help them to do just that. After reading a story, make a two-column chart on the board with the headings “Fact" and “Inference." Then write various facts or inferences on sentence strips and have students put each sentence into the appropriate column. Talk to them about the difference between the facts and the inferences. Make sure that they understand that you can point to a fact in the text, whereas with an inference, you can point to something in the text that seems to hint at the inference. Make sure that students are able to point at the sentence in the text that helps them make that inference. Guess the Definition Show students how to use inferences to understand unfamiliar vocabulary words. For example, write several sentences on the board such as “I didn’t want to abseculate again this winter. Last time I did it I broke my arm going down a steep hill." Make a list of facts that students know about the nonsense word “abseculate" from reading the sentences, such as the facts that it can be done in the winter and it involves hills. Then have students come up with inferences that they can make about the word abseculate, such as the ideas that it probably requires snow and involves going very quickly. Help students to extend these inference activities to real-life applications by choosing a real sentence in a text that contains a difficult word that can be understood from context. Have students use the same process to try to infer what the word might mean. This post is part of the series: Reading Strategy Lesson Plans: Making Inferences and Drawing Conclusions - Inference Games and Activities - Lesson Plans on Reading Strategies: Drawing Conclusions - Activities for Drawing Conclusions - Teaching Students to Make Inferences<\|endoftext\|>	4.46875
505	# Thread: Angles of a triangle 1. ## Angles of a triangle Prove that if $A,B,C$ are angles of a triangle , then : $ \begin{array}{l} 1)\cos A + cosB + \cos C \le \frac{3}{2} \\ 2)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8} \\ \end{array} $ 2. 1) $2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+\cos C\leq\frac{3}{2}$ $2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\leq\frac{3}{2}$ $2\sin^2\frac{C}{2}-2\cos\frac{A-B}{2}\sin\frac{C}{2}+\frac{1}{2}\geq 0$ Let $f(x)=2x^2-2\cos\frac{A-B}{2}x+\frac{1}{2}$ The discriminant is $\Delta=4\left(\cos^2\frac{A-B}{2}-1\right)\leq 0\Rightarrow f(x)\geq 0, \ \forall x\in\mathbb{R}$ 2) Is the same inequality because $\cos A+\cos B+\cos C=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1$ 3. A shorter solution for 1. It's enough to show that it's true for $A,B,C\in[0,\frac{\pi}{2}]$. Now to the proof for this case. Notice that $\cos x$ is a concave on $[0,\frac{\pi}{2}]$ so by Jensens inequality: $\frac{\cos A + \cos B + \cos C}{3} \leq \cos \left (\frac{A+B+C}{3} \right )=\cos \frac{\pi}{3} = \frac{1}{2}$ with equality iff $A=B=C=\frac{\pi}{3}$.<\|endoftext\|>	4.375
511	# Cluster estimation Cluster estimation can be used to estimate sums and products when the numbers you are adding or multiplying cluster near or is close in value to a single number Example # 1: Estimate 699 + 710 + 695 + 705 + 694 + 715 Carefully examine all the numbers above. You should notice that they all cluster around 700 Therefore, 700 + 700 + 700 + 700 + 700 + 700 will give us a good estimate for the answer. 700 × 6 = 4,200 is a good estimate of the sum In fact, the real answer 699 + 710 + 695 + 705 + 694 + 715 = 4,218. Thus 4,200 is indeed close to the real answer Example # 2: Estimate 257 + 247 + 255 + 245 Carefully examine all the numbers above. You should notice that they all cluster around 250 Therefore, 250 + 250 + 250 + 250 will give us a good estimate for the answer. 250 × 4 = 1,000 is a good estimate of the sum In fact, 257 + 247 + 255 + 245 = 1004. Thus 1000 is indeed close to the real answer Example # 3: Estimate 23 × 18 × 22 × 17 This time, you are estimating a multiplication problem. However,you will still use cluster estimation to estimate the product. Just notice that all numbers above cluster around 20 Therefore, 20 × 20 × 20 × 20 will give us a good estimate for the answer. Multiply the 2s to get 16. Then, just put four zeros after 16 to get 160,000 The real answer is 23 × 18 × 22 × 17 = 154,836. It is not a very good estimate, but we can live with that In general, addition will give better estimates Example # 4: Estimate 8 × 11 × 12 Just notice that all numbers above cluster around 10 Therefore, 10 × 10 × 10 will give us a good estimate for the answer. 10 × 10 × 10 = 1000 The real answer is 8 × 11 × 12 = 1056. Fun math game: Destroy numbered balls by adding to 10<\|endoftext\|>	4.53125
2,101	# How to choose a statistical test?8 min read 0 Shares If you are from a non-statistical background it is essentially the most complicated aspect of statistics, are always the basic statistical tests, and how to choose statistical test? In this article I have tried to mark out the distinction between the common statistical tests, the use of null value hypothesis in these tests by outlining the circumstances below which a particular test needs to be used. ## Null Hypothesis and Testing Before going to the difference between different test and choose a statistical test you must understand what a null hypothesis is. A null hypothesis tells that no significant difference exists in a set of given observations. Null: Given two sample means are equal. Alternate: Given two sample means usually are not equal For rejecting a null hypothesis, a test statistic is calculated. This test-statistic is then compared with a critical value and if it is greater than the critical value then the hypothesis is rejected. “In the theoretical aspect, hypothesis tests are based on the notion of critical regions: the null hypothesis is rejected if the test statistic falls in the critical region. The critical values are the boundaries of the critical region. If the test is one-sided then there will be just one critical value, but in other cases (like a two-sided t-test) there will be two” ## Critical Value A critical value is a point on the scale of the test statistic past which we reject the null hypothesis, and, is derived from the level of significance $alpha$ of the test. Critical value tells us, what’s the likelihood of two sample means belonging to the same distribution. Higher, the critical value means lower the likelihood of two samples belonging to the same distribution. The common critical value for a two-tailed test is 1.96, which relies on the fact that 95% of the area of a normal distribution is within 1.96 standard deviations of the mean. Critical values can be used to do hypothesis testing in the following ways: 1. Calculate test statistic 2. Calculate critical values based mostly on significance level alpha 3. Compare test statistic with critical values. If the test statistic is lower than the critical value, accept the hypothesis or else reject the hypothesis. Before I move ahead with different statistical tests it’s crucial to understand the difference between a sample and a population. In statistics “population” refers to the overall set of observations that may be made. For eg, if I would like to calculate the average height of people currently on the earth, “population” would be the “total number of people actually present on the earth”. A sample, alternatively, is a set of data collected from a pre-defined process. As in the example above, it will likely be a small group of individuals chosen randomly from some parts of the earth. To draw inferences from a sample by validating a hypothesis it’s crucial that the sample is random. For instance, if we choose individuals randomly from all areas(Asia, America, Europe, Africa and so forth.)on earth, our estimate can be close to the actual estimate and will be assumed as a sample mean, whereas if we make selection let’s say solely from the United States, then our average height estimate is not going to be correct however would solely represent the data of a selected area (United States). Such a sample is then known as a biased sample and isn’t a representative of “population”.Another important aspect to understand in statistics is “distribution”. When “population” is infinitely massive it’s inconceivable to validate any hypothesis by calculating the mean value or test parameters on your entire population. In such instances, a population is assumed to be of some kind of distribution. The commonest types of distributions are Binomial, Poisson and Discrete. However, there are numerous different types which are given in detail here. The determination of distribution type is important to decide the critical value and test to be chosen to validate any hypothesis. Now, once you are clear on the concept of population, sample, and distribution you will be able to understand different kinds of test and the distribution types for which they’re used. ## Relationship between p-value, critical value and test statistic The critical value is some extent beyond which we reject the null hypothesis. Alternatively, P-value is defined as the probability to the right of respective statistics (Z, T or chi). The advantage of using p-value is that it calculates a probability estimate, that you can test at any desired level of significance by comparing this probability instantly with the significance level. For e.g., assume Z-value for a selected experiment comes out to be 1.67 which is greater than the critical value at 5% which is 1.64. Now to check for a different significance level of 1% a new critical value is to be calculated. However, if you calculate the p-value for 1.67 it comes to be 0.047. You can use this p-value to reject the hypothesis at 5% significance level since 0.047 < 0.05. But with an extra stringent significance level of 1%, the hypothesis can be accepted since 0.047 > 0.01. The important point to observe here is that there isn’t any double calculation required. ## Z-test In a z-test, the sample is assumed to be normally distributed. A z-score is calculated with population parameters resembling “population mean” and “population standard deviation” and it is used to validate the hypothesis that the sample drawn belongs to the same population. Null: Given two sample means are equal Alternate: Given two sample means usually are not equal The statistics used for this hypothesis testing known as z-statistic, the score for which is calculated as: $z=\frac{(x-\mu)}{sigma/sqrt{n}}$ where, x= sample mean μ = population mean σ / √n = population standard deviation If the test statistic is lower than the critical value, accept the hypothesis or else reject the hypothesis. ## T-test A t-test is used to compare the mean of two given samples. Like a z-test, a t-test also assumes the sample is normally distributed. A t-test is used when the population parameters (mean and standard deviation) are usually not known.There are three variations of t-test: 1. Independent samples t-test which compares mean for two groups 2. Paired sample t-test which compares means from the identical group at different times 3. One sample t-test which tests the mean of a single group against a known mean. The statistic for this hypothesis testing is known as t-statistic, the score for which is calculated as: $t=\frac{x1-x2}{\sigma/sqrt{n1}+ \sigma/sqrt{n2}}$ where x1 = mean of sample 1 x2 = mean of sample 2 n1 = size of sample 1 n2 = size of sample 2 There are a number of variations of t-test which you can find here. ## ANOVA ANOVA also referred to as analysis of variance, is used to compare multiple (three or more) samples with a single test. There are 2 main flavours of ANOVA. • One-way ANOVA: It is used to compare the difference between the three or more samples/groups of a single independent variable. • MANOVA: MANOVA allows us to test the effect of one or more independent variable on two or more dependent variables. In addition, MANOVA also can detect the difference in co-relation between dependent variables given the groups of independent variables. The hypothesis being tested in ANOVA is Null: All pairs of samples are identical i.e. all sample means are equal Alternate: At least one pair of samples is considerably different The statistics used to measure the significance, in this case, is known as F-statistics. The F value is calculated using the below formula: $f=\frac{{((SSE1 - SSE2)/m)}}{{SSE2/n-k}}$ where, SSE = residual sum of squares m = number of restriction k = number of independent variables There are a number of tools out there such as SPSS, R packages, Excel and so forth. to perform ANOVA on a given sample. ## Chi-Square Test Chi-square test is used to compare categorical variables. There are two kinds of chi-square test. 1. The goodness of fit test, which determines if a sample matches the population. 2. A chi-square fit test for 2 independent variables and is used to compare two variables in a contingency table to test if the data fits. • Small chi-square value implies that data fits • A high chi-square value implies that data doesn’t match. Null: Variable A and Variable B are independent. Alternate: Variable A and Variable B aren’t independent. The statistic used to measure significance, in this case, is known as the chi-square statistic. The formula used for calculating the statistic is: $x^2 = \sum_{[Or,c - Er,c]}2 / Er,c$ where Or,c = observed frequency count at level r of Variable A and level c of Variable B Er,c = expected frequency count at level r of Variable A and level c of Variable B Key Takeaway As one can see from the above examples, in all of the tests a statistic is being compared with an important value to accept or reject a hypothesis. However, the statistic and approach to calculating it differ depending on the type of variable, the number of samples being analyzed and if the population parameters are known. #### Thus relying upon such factors and appropriate test and the null hypothesis is chosen. If you liked this article, you might also want to read Confidence Interval for Population Mean and Central Limit Theorem as well. Do you any tips to add Let us know in the comments. Please subscribe to our mailing list for weekly updates. You can also find us on Instagram and Facebook. 0 Shares ##### bySubhro Kar Been in the realm with the professionals of the IT industry. I am passionate about Coding, Blogging, Web Designing and deliver creative and useful content for a wide array of audience. This site uses Akismet to reduce spam. Learn how your comment data is processed. Share via<\|endoftext\|>	4.625
363	Both plants and animals exchange carbon with their environment until they die.Afterward, the amount of the radioactive isotope carbon-14 in their remains decreases. Researchers can measure the amount of these trapped electrons to establish an age. But to use any trapped charge method, experts first need to calculate the rate at which the electrons were trapped. Sometimes only one method is possible, reducing the confidence researchers have in the results. “They’re based on ‘it’s that old because I say so,’ a popular approach by some of my older colleagues,” says Shea, laughing, “though I find I like it myself as I get more gray hair.” Kidding aside, dating a find is crucial for understanding its significance and relation to other fossils or artifacts. Methods fall into one of two categories: relative or absolute. Before more precise absolute dating tools were possible, researchers used a variety of comparative approaches called relative dating. These methods — some of which are still used today — provide only an approximate spot within a previously established sequence: Think of it as ordering rather than dating.This includes factoring in many variables, such as the amount of radiation the object was exposed to each year.These techniques are accurate only for material ranging from a few thousand to 500,000 years old — some researchers argue the accuracy diminishes significantly after 100,000 years.Researchers can first apply an absolute dating method to the layer.They then use that absolute date to establish a relative age for fossils and artifacts in relation to that layer. Anything below the Taupo tephra is earlier than 232; anything above it is later.Biostratigraphy: One of the first and most basic scientific dating methods is also one of the easiest to understand.<\|endoftext\|>	4.15625
303	SPARK COIL, OR RUHMKORFF COIL The coil is an induction coil, which is a form of transformer designed to produce high voltage sparks across the spark gap above the coil. This principle is used in cars to produce the sparks in the spark plugs. Ruhmkorff coils have been made that can produce sparks up to 1.2 metres long. Heinrich Daniel Ruhmkorff developed the coil in the 1850s from designs of some earlier experimenters. It is another device operating on the principle of electro-magnetic induction. Two separate coils are wound on to a magnetic iron core. One is a low voltage coil, the other a high voltage coil. The low voltage coil has relatively few turns. It is connected to a DC source, typically a battery, by the two terminals at the left hand end of the machine. The high voltage coil has very many turns. It is connected to the spark gap through the two terminals on the top of the machine As the electro-magnetic induction effect of a transformer requires a change of current for the effect, the current to the low voltage coil is rapidly switched on and off by an interrupting device in the battery circuit. A capacitor, usually fitted in the base of the machine, increases the effectiveness of the machine. The induction coil has been used for a number of purposes. Early radio transmitters, called spark transmitters, used this device to produce the broadcast signal.<\|endoftext\|>	3.765625
3,034	# Recursion with Fibonacci numbers Fibonacci numbers are definitely special and not only in mathematics but in other areas as well. They are sometimes considered having a mysterious effect on life. One can decide if they believe it's true or not. We will create Fibonacci numbers with the concept of recursion in this article. ## The problem I got this puzzle from the great math website projecteuler.net: Find the sum of the even Fibonacci numbers that are less than 4000000. Fear not if you don’t know anything of Fibonacci or his numbers, the next section will explain what you need to know to succeed in this challenge. ## Background Fibonacci was an Italian mathematician, who lived in the 13th century. He was very talented and created a famous number sequence we will be working with below. Fibonacci numbers make up a sequence. A sequence is nothing else but a series of numbers coming after each other by some rule. The members of the sequence are called terms. The rule for the Fibonacci sequence is that each term equals the sum of the previous two terms. Obviously, the sequence needs to start somewhere; otherwise we wouldn’t have anything to add. It’s agreed that the first two terms are equal to 1. So the third member of the sequence is going to be 1 + 1 = 2, the 4th term is 1 (2nd term) + 2 (3rd term) = 3, the 5th term equals 2 (3rd term) + 3 (4th term) = 5 and so on. Let’s list the first few terms then: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … ## Breaking down the problem One way of solving the problem is to get a pen and a paper, write down the terms of the sequence until 4000000, and then add them together. You don’t want to do that, if so, good luck with it. Instead, let’s create an algorithm for the problem. As always, it’s a good idea to sit back and think about the problem first, then break it down into smaller, more manageable parts. ### Repeating the same process First, notice that we always create the next term by adding the last two terms. It’s a repeating process. So if we had something (a function) that calculates the sum of two numbers and then we could somehow keep repeating and applying it on the next two numbers, we would get the terms of the sequence one by one. The process of repeatedly invoking a function is called recursion. ### Upper limit We also need a threshold, in our case it’s 4000000. At this stage it could be an argument to our function to make it configurable and reusable. ### Find even members Once we have found the Fibonacci numbers (or even during the process?), we need to find the evens, so some kind of filtering will be necessary. ### Add them up This is probably the easiest part. The even numbers should be marked and be added up. ### Edge cases It’s always a good idea to think about the extremes. Edge cases can often act as source of bugs in applications. Still, it’s not always easy to find all possibilities for an error, but one should strive to cover as many as possible. Life (and users particularly) are very creative and bugs will occur however good you do (goodbye, perfectionism!). What can be some edge cases for this problem? The terms of the sequence must be all positive integers. Or the user enters a string instead of a number. We need to handle these when we write the code. ## Assigning each part to JS Creating Fibonacci sequence is a separate task, sort of a component, so I would create a function just for this part. Because the given definition of the Fibonacci sequence is recursive, I will invoke the function in itself. I will also handle some of the edge cases here. The upper limit will be the argument of the function, so it’s going to be reusable. If I change my mind and want to solve the same problem for Fibonacci terms less than 1000 instead of 4000000, I can easily do so. I also need to know when I reach the upper limit; iteration over positive integers will be of use here. Finally, I need to find the even Fibonacci numbers, using the remainder (%) operator and have to add them together. ## Solving the problem Let’s do it! ### Creating the Fibonacci function First, we create the function that generates the Fibonacci numbers and surprisingly call it `fibonacci`: ``````const fibonacci = (n) => { if (n === 1) return 1; if (n === 2) return 2; return fibonacci(n - 1) + fibonacci(n - 2); }; `````` This function will return the nth Fibonacci number. Let’s have a closer look at it! Because the definition of the sequence is to add the previous two terms to get the next one, we need to define the first two terms. They must be hardcoded so that we can calculate the Fibonacci numbers from the third term onwards. (Note: The original problem defines 2 as the second term, so I’ll stick with this. Many resources define both the first and the second term as being equal to 1. This small difference won’t influence either the end result or the way the terms are calculated.) The `return` statement is the recursion part. We add the last and the second last terms here. If you log `fibonacci(5)`, you’ll get 8, which is the 5th Fibonacci term. Indeed: 1, 2, 3, 5 and 8. The function doesn’t seem to do any magic, so how come that it still works? When the function invoked with the parameter of 5, the two conditions will not trigger any actions. So we’ll get `fibonacci(4) + fibonacci(3)` invoked, since this is what we return when we call the function. When `fibonacci(3)` is invoked, again, `fibonacci(2) + fibonacci(1)` gets returned and invoked, and the conditions set in the `if` statements will trigger the actions here. `fibonacci(2)` equals 2 and `fibonacci(1)` equals 1, so add them up to get 3. The `fiboancci(4)` part of the addition returns `fibonacci(3) + fibonacci(2)`, and we have seen this before: `fibonacci(3)` is equal to 3, `fibonacci(2)` evaluates to 2. If we add them up, we’ll get 3 + 3 + 2 = 8. So the magic is that when we call a function recursively, it will “breed” more of the same functions, they get invoked again and they will “breed” even more functions and so on. Eventually these functions will get down to the default 1 and 2 values. If we add these many 1s and 2s, we’ll get our Fibonacci term. This wasn’t easy! ### Adding up the Fibonacci numbers We create a function called `addEvenFibNumbers` for this part. The function will accept one argument, and this will be the threshold (the number the user enters, which is 4000000 in the original example.) This is a good place to handle some edge cases that can lead to errors: ``````const addEvenFibNumbers = (threshold) => { const limit = Number(threshold); if (!limit \|\| !Number.isInteger(limit) \|\| limit < 0) { return 'Please enter a positive integer!'; } }; `````` First, the parameter is converted to a number, using the `Number` object and is stored in a variable called `limit`. We then check for the edge cases. If the entered value (e.g. the string “hello”) cannot be converted to a number, we’ll get `Nan`, which is a falsy value, so the `!limit` condition will be true, and our gentle error message is returned. If it can be converted to a number but the number is not an integer, the message will appear again. Similarly, we don’t want to work with negative numbers (we can’t in this context anyway), so let the user know what they should do next time. Now that we only have positive integers, we can start iterating over the numbers from 1 to the `limit` and check for even Fibonacci numbers. This is the function itself: ``````const addEvenFibNumbers = (threshold) => { const limit = Number(threshold); if (!limit \|\| !Number.isInteger(limit) \|\| limit < 0) { return 'Please enter a positive integer!'; } let sum = 0; let n = 1; while (fibonacci(n) < limit) { if (fibonacci(n) % 2 === 0) sum += fibonacci(n); n++; } return sum; }; `````` We use a `while` loop here. `while` loops need to be used with care, as it’s easy to run into infinite loops. Luckily, we have an upper limit (4000000 in this example), so the loop will finish once `fibonacci(n)` reaches this limit (i.e. when the actual Fibonacci term exceeds 4000000). We need to filter out the even Fibonacci numbers using the `%` remainder operator, and when we find one, we simply add that number to the sum, which initially equals 0. We return the `sum` in the end. Now, if we log `addEvenFibNumbers(4e6)` (or `addEvenFibNumbers(4000000)`), the result is displayed (`4613732`). ## Alternative way of defining Fibonacci function As you might have noticed, it took some time for your computer to calculate the final answer. It might have even been a noticeable delay between pressing Enter and getting the result. (If you haven’t noticed any delay, try entering a bigger number in `addEvenFibNumbers`). The reason for this is your computer needs to take a lot of operations to compute the result. It means that we are probably better off finding another way of calculating Fibonacci terms, a way that requires less resource and effort from the computer. Let’s think about it for a second. When we created `fibonacci`, we “translated” the definition of the sequence into JavaScript. What if we tried to translate the behavior of the terms? What do I mean by behavior? Have a look at the first few terms: 1, 2, 3, 5, 8, 13, 21, … If we take for example the 5th term as current term (let’s call this step 1), which is, the last term is 5 and the second last term. When we move on to the 6th term (13) as current (step 2), the last term becomes 8 (the current of step 1) and the second last term becomes 5 (the last term of step 1). So we always “slide” these three elements (second last, last and current) by one to the right as we calculate the terms of the sequence. Can we use this principle to determine Fibonacci numbers from the point of the current term? Absolutely! Here’s how: ``````const fibonacciAlternative = (n) => { let secondLastTerm = 1; let lastTerm = 2; if (n === 1) return secondLastTerm; if (n === 2) return lastTerm; for (let i = 3; i <= n; i++) { // the current Fib number becomes the last term as the iteration moves on lastTerm = lastTerm + secondLastTerm; // line 1 // and the second last term will become the PREVIOUS last term secondLastTerm = lastTerm - secondLastTerm; // line 2 } return lastTerm; }; `````` The first term we need to calculate is the 3rd, so we need to define the first two here as well. The first term is 1; this will be the `secondLastTerm` from the point of the 3rd term. Similarly, if we take the 3rd term as current, the `lastTerm` will be the 2nd. We return them if the parameter is 1 or 2, no change in this compared to the previous code. The magic starts with the `for` loop. The iteration starts at 3(remember, this is the first term to be calculated) and goes until and including `n`. Here comes the hard part. In terms of the current Fibonacci number (which we want to calculate), we need to add the last and second last terms, this is straightforward, it comes from the definition. But at the same time, we prepare `lastTerm` for the computation of the next Fibonacci number, and make it equal to the current term (line 1), which we return when the iteration finishes. (At this point there will be no need to calculate the next Fibonacci number, so the value of `lastTerm` will be the actual Fibonacci number.) Similarly, we need to slide `secondLastTerm` as well, and give it the value of the previous `lastTerm` (line 2). The value of `lastTerm` will be the new value calculated in line 1, so to get the previous value of `lastTerm`, we simply subtract what we added to it in line 1. All we have to do now is to replace `fibonacci` with `fibonacciAlternative` in `addEvenFibNumbers`, and compare the computation time. You will see the difference. Well, that wasn’t easy. Reread this section if it wasn’t clear. And again. This is what I do when I come across something new and difficult. It will click. ## Conclusion This is a nice little problem to solve. Recursion is much worse in terms of performance than simple `for` loops. The greater the number we work with is, the more visible the difference is between each method. It would be a good challenge to write the `addEvenFibNumbers` function in a more functional way using `filter` (instead of the `if` statement) and `reduce` (instead of the continuous addition). If you have time, feel free to work on it. Enjoy coding!<\|endoftext\|>	4.75
2,932	# Construction Related to Segment and Triangles Geometry is a skill of the eyes, hands and that of the mind. Any construction in geometry involves drawing a diagram based on provided data. In this module, we will learn a few skills regarding construction related to the line segment and triangles. We will also revise a few properties of geometry while doing construction. ## Construction Related to Line Segments You must have seen many times that a point divides a line segment in a particular ratio. In this module, we will learn how to divide a given line segment in the desired ratio x:y without using a ruler. ### Dividing a given line segment in a particular ratio Let’s say we have been given a line segment AB which we want to divide it by a ratio of 3:2. Here, x = 3 and y = 2. To construct such a line segment, follow the steps as given below: ### Steps of Construction 1. Draw any ray AZ making an acute angle with AB. 2. Locate 5 points (x + y = 3 + 2 = 5), say, A1, A2, A3, A4, and A5 using a compass on AZ such that these points are equally distanced from each other. 1. Open the compass of a specific width and make an arc on the ray AZ keeping the needle on the point A. The point where the arc intersects with AZ is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the steps 3 more times making AA1 = A1A2 = A2A3 = A3A4 = A4A5. 3. Join A5 with B making a line segment A5B. (See Figure 1) 4. Since we need to divide the line segment in 3:2, draw a line through the point A3 which is parallel to A5B by making ∠AA3C = ∠AA5B such that A3C intersects AB at the point C. 1. To construct a parallel line using Angle Copy Method, adjust the width of the compass to roughly half of the length of A4A5. Keeping the needle at point A5, make an arc cutting the ray AZ and the line segment A5B. This is arc 1. 2. Make a similar arc from point A3 using the same compass width. This is arc 2. 3. Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AZ, the pencil should be on the point where arc 1 meets the line segment A5B. 4. Keeping the same width, cut the arc 2 making an intersecting point G. 5. Make a line segment from A3 through G meeting the line segment AB at C. The line segment A3C \|\| A5B. 5. Therefore, AC : CB = 3 : 2. (See Figure 1) Figure 1: AC : CB = 3 : 2 ### Rationale and Proof Since A3C \|\| A5B ∴ $$\frac{AA_3}{A_3A_5}$$ = $$\frac{AC}{CB}$$ [Triangle Proportionality TheoremIf a line parallel to one side of a triangle intersects the other two sides, it divides those sides proportionally] $$\frac{AA_3}{A_3A_5}$$ = $$\frac{3}{2}$$ [By construction] ∴ $$\frac{AC}{CB}$$ = $$\frac{3}{2}$$ ## Alternate Method Following is the alternative method to divide a given line segment AB into a said ratio (same as above, that is, 3:2). ### Steps of Construction 1. Draw any ray AG making an acute angle with AB. 2. Draw a ray BH parallel to AG by making ∠ABH equal to ∠BAG using the same method as described above in steps from 4(a) to 4(e). 3. Locate the pointsA1, A2, and A3 (x = 3) on AG and B1, B2 (y = 2) on BH such that AA1 = A1A2 = A2A3 = BB1 = B1B2 using the same step as described in 2(a) above. 4. Join A3B2. Let it intersect AB at a point C (See Figure 2). 5. Then AC : CB = 3 : 2 Figure 2: AC : CB = 3 : 2 (Alternate method) ### Rationale and Proof Here, ∠CAA3 = ∠CBB2 [By construction] ∠ACA3 = ∠BCB2 [Vertically opposite angles are always equal] ∠AA3C = ∠BB2C [Alternate interior angles within two parallel lines are always equal] ∴ ∆ACA3 ~ ∆BCB2 [By AAA criteria of triangle similarity] ⇒ $$\frac{AC}{BC}$$ = $$\frac{CA_3}{CB_2}$$ = $$\frac{AA_3}{BB_2}$$ Now, using construction $$\frac{AA_3}{BB_2}$$ = $$\frac{3}{2}$$ ∴ $$\frac{AC}{BC}$$ = $$\frac{3}{2}$$ ## Construction Related to Triangles We have studied about similar triangles and their properties in previous classes. Have you ever wondered how these similar triangles are designed? Let’s take a look how. ### Constructing a triangle similar to a given triangle There are a few ways by which we can draw a triangle similar to the given one. Here, we shall study how to construct the same using the Scale Factor method without the help of a ruler. Scale Factor is the ratio of two corresponding sides of the similar triangles. For example, let’s say that ∆ABC is similar to ∆DEF (∆ABC ~ ∆DEF) wherein AB = 3, BC = 4, and AC = 5; whereas DE = 6, EF = 8 and DF = 10. Here, the Scale Factor of two corresponding sides is $$\frac{AB}{DE}$$ = $$\frac{BC}{EF}$$ = $$\frac{AC}{DF}$$ = $$\frac{1}{2}$$ So, let’s see how we can construct a triangle similar to the given triangle (∆PQR) with a scale factor of $$\frac{3}{4}$$, that is, to construct a smaller triangle (∆P′QR′) compared to ∆PQR wherein the length of the sides of ∆P′QR′ will be 3/4th to that of ∆PQR. ### Steps of Construction using Scale Factor 1. Draw any ray QX making an acute angle with QR on the side opposite to the vertex P. 2. Locate 4 points Q1, Q2, Q3, and Q4 on QX so that QQ1 = Q1Q2= Q2Q3 = Q3Q4 using the same method as described above in step 2(a) (Dividing a line segment in a given ratio). Note: We located 4 points because 4 is the greater value in the Scale Factor ratio of $$\frac{3}{4}$$ 3. Considering the given triangle, ∆PQR, join a line Q4R. 4. Draw a line through Q3 parallel to Q4R using the same method as described above in steps from 4(a) to 4(e) (Dividing a line segment in a given ratio) intersecting the line QR at R′. This gives us Q3R′ \|\| Q4R and ∠QQ3R′ = ∠QQ4R (See Figure 3). 5. Similarly, draw a line through R′ parallel to the line PR intersecting the line PQ at a point, P′. This gives us P′R′ \|\| PR and ∠P′R′Q = ∠PRQ (See Figure 3). 6. Subsequently, ∆P′QR′ is the required triangle similar to ∆PQR with Scale Factor of $$\frac{3}{4}$$. Figure 3: ∆ P′QR′ ~ ∆PQR ### Rationale and Proof Since Q3R′ \|\| Q4R and QQ3 is proportionate to QQ4 in the ratio of 3:4 (by construction) ∴ $$\frac{QQ_3}{QQ_4}$$ = $$\frac{QR’}{QR}$$ = $$\frac{3}{4}$$ [By Triangle Proportionality Theorem] Here, ∠P′QR′ = ∠PQR [Common angle] ∠P′R′Q = ∠PRQ [By Construction] ∠R′P′Q = ∠RPQ [If two corresponding angles of two triangles are equal, the third angle will always be equal] ∴ ∆P′QR′ ~ ∆PQR [By AAA criteria of triangle similarity] Since ∆P′QR′ ~ ∆PQR and $$\frac{QR’}{QR}$$ = $$\frac{3}{4}$$ ∴ ∆P′QR′ ~ ∆PQR with Scale Factor of $$\frac{3}{4}$$. ## Solved Examples for You Question 1: Given a line segment AB. Find a point, describing the steps, which divides the line segment in 2:1 without using a ruler. 1. Draw any ray AX, making an acute angle with AB. 2. Locate 3 points (x + y = 2 + 1 = 3), say, A1, A2, and Ausing a compass on AX such that these points are equally distanced from each other. 1. Open the compass of a specific width and make an arc on the ray AX keeping the needle on the point A. The point where the arc intersects with AX is now A1. Repeat the step keeping the same width from A1 making a point A2. Repeat the step one more times making AA1 = A1A2 = A2A3. 3. Join A3 with B making a line segment A3B. (See Figure A) 4. Since we need to divide the line segment in 2:1, draw a line through the point A2 which is parallel to A3 by making ∠AA2B = ∠AA3B intersecting AB at the point C. 1. To construct a parallel line, adjust the width of the compass to roughly half of the length of A2A3. Keeping the needle at point A3, make an arc cutting the ray AX and the line segment A3B. This is arc 1. 2. Make a similar arc from point A2 using the same compass width. This is arc 2. 3. Now, adjust the compass width to the arc 1 such that keeping the needle on the point where arc 1 intersects AX, the pencil should be on the point where arc 1 meets the line segment A3B. 4. Keeping the same width, cut the arc 2 making an intersecting point O. 5. Make a line segment from A2 through O meeting the line segment AB at C. The line segment A2C \|\| A3B. 5. This point C divides AB as AC : CB = 2 : 1. (See Figure A) Figure A. AC : CB = 2 : 1 Question 2: What do you mean by segment in maths? Answer: Segment refers to the part of a line that connects two points. It happens to the shortest distance that exists between the two points. Furthermore, segment has a length. The word “segment” is important because a line can go on continuously without end. In contrast, a line segment has definite endpoints. Question 3: Explain what is meant by segments of circle? Answer: A chord of a circle divides a circle into two parts. These parts are known as the segments of circle. Question 4: How can one measure a line segment? Answer: A line segment is nothing more than a part of a line. In order to measure the length of a line segment, put the segment’s endpoint on the ruler’s zero mark. They must see where it ends. In inches, one must round to the nearest ¼, while in centimeters; one must measure to the nearest centimeter. Question 5: How can one find out the midpoint of a segment? Answer: One can find out the midpoint of a segment by dividing the segment’s length by 2 and counting that value from any endpoint. Share with friends ## Customize your course in 30 seconds ##### Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started<\|endoftext\|>	4.875
924	Courses Courses for Kids Free study material Offline Centres More Store # A particle moves in the x-y plane according to rule $x = a\cos \omega t$ and $y = a\sin \omega t$. The particle follows(A) An elliptical path (B) A circular path (C) A parabolic path (D) A straight line path inclined equally to x and y axes. Last updated date: 17th Jun 2024 Total views: 52.8k Views today: 0.52k Verified 52.8k+ views Hint: The equations given are parametric equations of a particular shape. We need to compare the equation with the parametric equations of the given options. Formula used: In this solution we will be using the following formulae; $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ Cartesian equation of an ellipse, with semi major axis $a$ and semi minor axis $b$. $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$ Cartesian equation of a circle, with radius $a$. $x = a\cos \omega t$ and $y = b\sin \theta$ this is the parametric equation of an ellipse. $x = a\cos \theta$ and $y = a\sin \theta$ this is the parametric equation of a circle Complete Step-by-Step Solution: A particle is said to move in the x y plane according to the rule $x = a\cos \omega t$ and also $y = a\sin \omega t$ This is a parametric equation. We shall compare it to the parametric equation of each of the options. Option A is given an ellipse. The parametric equation of an ellipse centred at the origin is given as $x = a\cos \omega t$ and $y = b\sin \theta$ where $a$ is the semi major axis and $b$ is the semi minor axis. (this can be interchanged, depending on the orientation of the ellipse) Hence, by observation, we see that this is not the equation is not that of an ellipse. Option B is given as a circle. The parametric equation of a circle is given as $x = a\cos \theta$ and $y = a\sin \theta$ where $a$ is the radius. Hence by observation, we can see that the equation above is exactly like the equation of a circle where $\theta = \omega t$ Hence, the particle traces out a circular path Thus, the correct option is B. Note: Alternatively, the equation can be converted to Cartesian form or standard form. This can be done as follows, Squaring both x and y equation, we have ${x^2} = {\left( {a\cos \omega t} \right)^2} = {a^2}{\cos ^2}\omega t$, and for y coordinate, ${y^2} = {\left( {a\sin \omega t} \right)^2} = {a^2}{\sin ^2}\omega t$ Now, we add the two equations together, such that we get ${x^2} + {y^2} = {a^2}{\cos ^2}\omega t + {a^2}{\sin ^2}\omega t$ By factorising out ${a^2}$, we get ${x^2} + {y^2} = {a^2}\left( {{{\cos }^2}\omega t + {{\sin }^2}\omega t} \right)$ Then, since${\cos ^2}\omega t + {\sin ^2}\omega t = 1$, we have ${x^2} + {y^2} = {a^2}$ or $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}}} = 1$ And this is precisely the Cartesian equation of a circle. An ellipse would have been, $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$<\|endoftext\|>	4.46875
12,121	# 3 Random Variables Let $$S$$ be the sample space of an experiment. A random variable is a function from $$S$$ to the real line, which is typically denoted by a capital letter. Suppose $$X$$ is a random variable. The expression $P(X \in (a,b))$ denotes the probability that the random variable takes values in the open interval $$(a,b)$$. This can be done by computing $$P(\{s\in S: a < X(s) < b \})$$. (Remember, $$X:S\to {\Bbb R}$$ is a function.) Example Suppose that three coins are tossed. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\},$ and all eight outcomes are equally likely, each occurring with probability 1/8. Now, suppose that the number of heads is observed. That corresponds to the random variable $$X$$ which is given by: $X(HHH) = 3 \\ X(HHT) = X(HTH) = X(THH) = 2 \\ X(TTH) = X(THT) = X(HTT) = 1\\ X(TTT) = 0$ In order to compute probabilities, we could use $P(X = 2) = P(\{HHT, HTH, THH\}) = \frac{3}{8}.$ We will only work explicitly with the sample spaces this one time in this textbook, and we will not always define the sample space when we are defining random variables. It is easier, more intuitive and (for the purposes of this book) equivalent to just understand $$P(a < X < b)$$ for all choices of $$a < b$$. In order to understand such probabilities, we will split into two cases. ## 3.1 Discrete Random Variables A discrete random variable is a random variable that can only take on values that are integers, or more generally, any discrete subset of $${\Bbb R}$$. Discrete random variables are characterized by their probability mass function (pmf) $$p$$. The pmf of a random variable $$X$$ is given by $$p(x) = P(X = x)$$. This is often given either in table form, or as an equation. Example Let $$X$$ denote the number of Heads observed when a coin is tossed three times. $$X$$ has the following pmf: $\begin{array}{l\|cccc} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 0.125 & 0.375 & 0.375 & 0.125 \end{array}$ An alternative description of $$p$$ would be $p(x) = {3 \choose x} (.5)^3 \qquad x = 0,\ldots,3$ In this context we assume that $$p$$ is zero for all values not mentioned; both in the table version and in the formula version. Probability mass functions satisfy the following properties: Theorem 3.1 Let $$p$$ be the probability mass function of $$X$$. 1. $$p(n) \ge 0$$ for all $$n$$. 2. $$\sum_n p(n) = 1$$. ### 3.1.1 Expected Values of Discrete Random Variables The expected value of a random variable is, intuitively, the average value that you would expect to get if you observed the random variable more and more times. For example, if you roll a single six-sided die, you would the average to be exactly half-way in between 1 and 6; that is, 3.5. The definition is $E[X] = \sum_x x p(x)$ where the sum is taken over all possible values of the random variable $$X$$. Example Let $$X$$ denote the number of heads observed when three coins are tossed. The pmf of $$X$$ is given by $$p(x) = {3 \choose x} (0.5)^x$$, where $$x = 0,\ldots,3$$. The expected value of $$X$$ is $0 \times .125 + 1 \times .375 + 2 * .375 + 3 * .125 = 1.5.$ ## 3.2 Continuous Random Variables A continuous random variable $$X$$ is a random variable for which there exists a function $$f$$ such that whenever $$a \le b$$ (including $$a = -\infty$$ or $$b = \infty$$) $P(a \le X \le b) = \int_a^b f(x)\, dx$ The function $$f$$ in the definition of a continuous random variable is called the probability density function (pdf) of $$X$$. The cumulative distribution function (cdf) associated with $$X$$ is the function $F(x) = P(X \le x) = \int_{-\infty}^x f(x)\, dx$ By the fundamental theorem of calculus, $$F$$ is a continuous function, hence the name continuous rv. The function $$F$$ is sometimes referred to as the distribution function of $$X$$. One major difference between discrete rvs and continuous rvs is that discrete rv’s can take on only countably many different values, while continuous rvs typically take on values in an interval such as $$[0,1]$$ or $$(-\infty, \infty)$$. Another major difference is that for continuous random variables, $$P(X = a) = 0$$ for all real numbers $$a$$. Theorem 3.2 Let $$X$$ be a random variable with pdf $$f$$ and cdf $$F$$. 1. $$f(x) \ge 0$$ for all $$x$$. 2. $$\int f(x)\, dx = 1$$. 3. $$\frac d{dx} F = f$$. Example Suppose that $$X$$ has pdf $$f(x) = e^{-x}$$ for $$x > 0$$. Find $$P(1 \le X \le 2)$$. By definition, $P(1\le X \le 2) = \int_1^2 e^{-x}\, dx = -e^{-x}\Bigl\|_1^2 = e^{-1} - e^{-2} = .233$ ### 3.2.1 Expected value of a continuous random variable The expected value of $$X$$ is $E[X] = \int x f(x)\, dx$ Example Find the expected value of $$X$$ when its pdf is given by $$f(x) = e^{-x}$$ for $$x > 0$$. We compute $E[X] = \int_0^\infty x e^{-x} \, dx = \left(-xe^{-x} - e^{-x}\right)\Bigr\|_0^\infty = 1$ (Recall: to integrate $$xe^{-x}$$ you use integration by parts.) ## 3.3 Expected value of a function of an rv: Variance More generally, one can compute the expected value of a function of a random variable. The formula is $E\left[g(X)\right] = \begin{cases} \sum g(x) p(x) & X {\rm \ \ discrete} \\ \int g(x) f(x)\, dx & X {\rm \ \ continuous}\end{cases}$ Example Compute $$E[X^2]$$ for $$X$$ that has pmf $$p(x) = {3 \choose x} (.5)^3$$, $$x = 0,\ldots,3$$. We compute $E[X^2] = 0^2\times .125 + 1^2 \times .375 + 2^2 \times .375 + 3^2 \times .125 = 3$ Example Compute $$E[X^2]$$ for $$X$$ that has pdf $$f(x) = e^{-x}$$, $$x > 0$$. We compute $E[X^2] = \int_0^\infty x^2 e^{-x}\, dx = \left(-x^2 e^{-x} - 2x e^{-x} - 2 e^{-x}\right)\Bigl\|_0^\infty = 2$ (Here, again, we use integration by parts.) One of the most important examples of expected values of functions of a random variable is the variance of $$X$$, which is defined by $$E[(X - \mu)^2]$$, where $$\mu = E[X]$$. ### 3.3.1 Linearity of expected values We have the following: $E[aX + bY] = aE[X] + bE[Y]$ The reason this is true is because of linearity of integration and summation. We will not provide details. We also have $E[c] = c$ This is because $$E[c] = \int c f(x)\, dx = c\int f(x)\, dx = c$$ (a similar computation holds for discrete random variables). Applying linearity of expected values to the definition of variance yields: $E[(X - \mu)^2] = E[X^2 - 2\mu X + \mu^2] = E[X^2] - 2\mu E[X] + \mu^2 = E[X^2] - 2\mu^2 + \mu^2 = E[X^2] - \mu^2.$ This formula for the variance of an rv is often easier to compute than from the definition. Example Compute the variance of $$X$$ if the pdf of $$X$$ is given by $$f(x) = e^{-x}$$, $$x > 0$$. We have already seen that $$E[X] = 1$$ and $$E[X^2] = 2$$. Therefore, the variance of $$X$$ is $$E[X^2] - E[X]^2 = 2 - 1 = 1$$. Note that the variance of an rv is always positive (in the French sense1), as it is the integral (or sum) of a positive function. Finally, the standard deviation of an rv $$X$$ is the square root of the variance of $$X$$. Definition 3.1 The variance of a random variable $$X$$ is ${\rm Var}(X) = = E[(X - \mu)^2] = E[X^2] - E[X]^2$ The standard deviation of $$X$$ is the square root of the variance. The standard deviation is easier to interpret in many cases than the variance. For many distributions, about 95% of the values will lie within 2 standard deviations of the mean. (What do we mean by “about”? Well, 85% would be about 95%. 15% would not be about 95%. It is a very vague rule of thumb. If you want something more precise, see Chebychev’s Theorem, which says in particular that the probability of being more than 2 standard deviations away from the mean is at most 25%.) Sometimes, you know that the data you collect will likely fall in a certain range of values. For example, if you are measuring the height in inches of 100 randomly selected adult males, you would be able to guess that your data will very likely lie in the interval 60-84. You can get a rough estimate of the standard deviation by taking the expected range of values and dividing by 6. (Here, we are using the heuristic that “nearly all” data will fall within three standard deviations of the mean.) This can be useful as a quick check on your computations. Theorem 3.3 Let $$X$$ and $$Y$$ be independent random variables, and let $$a$$ and $$b$$ be constants. Then, ${\rm Var}(aX + bY) = a^2{\rm Var}(X) + b^2{\rm Var}(Y)$ We will not prove this theorem, but let’s see why the easier statement $${\rm Var}(aX) = a^2{\rm Var}(X)$$ is true: \begin{align} {\rm Var}(aX) =& E[(aX)^2] - E[aX]^2 = a^2E[X^2] = (aE[X])^2\\ =&a^2\bigl(E[X^2] - E[X]^2) = a^2{\rm Var}(X) \end{align} ## 3.4 Special Discrete Random Variables In this section, we discuss the binomial, geometric and poisson random variables, and their implementation in R. In order to understand the binomial and geometric rv’s, we will consider the notion of Bernoulli trials. A Bernoulli trial is an experiment that can result in two outcomes, which we will denote as “Success” and “Failure”. The probability of a success will be denoted $$p$$. A typical example would be tossing a coin and considering heads to be a success, where $$p = .5$$. ### 3.4.1 Binomial Random Variable Several random variables consist of repeated independent Bernoulli trials, with common probability of success $$p$$. Definition 3.2 The binomial random variable counts the number of successes observed when $$n$$ independent Bernoulli trials are formed, each of which has probability of success $$p$$. If $$X$$ is binomial with number of trials $$n$$ and probability of success $$p$$, then $P(X = x) = {n \choose x} p^x(1 - p)^{n-x}$ Here, $$P(X = x)$$ is the probability that we observe $$x$$ successes. Note that $\sum_x p(x) = \sum_x P(X = x) = \sum_x {n \choose x} p^x(1 - p)^{n-x} = \bigl(p + (1 - p)\bigr)^n = 1$ by the binomial expansion theorem. Example Suppose 100 dice are thrown. What is the probability of observing 10 or fewer sixes? We assume that the results of the dice are independent and that the probability of rolling a six is $$p = 1/6.$$ Then, the probability of observing 10 or fewer sixes is $P(X \le 10) = \sum_{j=0}^{10} P(X = j)$ where $$X$$ is binomial with parameters $$n = 100$$ and $$p = 1/6$$. The R command for computing this is sum(dbinom(0:10, 100, 1/6)) or pbinom(10,100,1/6). We will discuss in detail the various binom commands in Section 3.7. Theorem 3.4 Let $$X$$ be a binomial random variable with $$n$$ trials and probability of success $$p$$. Then • The expected value of $$X$$ is $$np$$. • The variance of $$X$$ is $$np(1-p)$$ #### 3.4.1.1 Proof (optional) Let $$X$$ be a binomial random variable with $$n$$ trials and probability of success $$p$$. We show $E[X] = \sum_x x p(x) = \sum_{x=0}^n x {n \choose x} p^x (1 - p)^{n - x} = np$ We first note that that $$x = 0$$ term doesn’t contribute to the sum, so $E[X] = \sum_{x = 1}^n x {n \choose x} p^x (1 - p)^{n - x}$ Now, since $$x {n \choose x} = n {{n-1} \choose {x-1}}$$, we get $\sum_{x = 1}^n x {n \choose x} p^x (1 - p)^{n - x} = \sum_{x = 1}^n n {{n-1} \choose {x-1}} p^x (1 - p)^{n - x}.$ We factor out an $$n$$ from the second sum, and re-index to get $E[X] = n \sum_{x = 0}^{n-1} {{n-1} \choose {x}} p^{x+1} (1 - p)^{n - 1 - x} = np \sum_{x = 0}^{n-1} {{n-1} \choose {x}} p^{x} (1 - p)^{n - 1 - x}$ Finally, we note that ${{n-1} \choose {x}} p^{x} (1 - p)^{n - 1 - x}$ is the pmf of a binomial rv with parameters $$n-1$$ and $$p$$, and so $$\sum_{x = 0}^{n-1} {{n-1} \choose {x}} p^{x} (1 - p)^{n - 1 - x} = 1$$, and $E[X] = np.$ To compute the variance, use $${\rm Var}(X) = E[X^2] - E[X]^2 = E[X^2] - n^2p^2$$. The interested student can find $$E[X^2]$$ by using the identity $$E[X^2] = E[X(X - 1)] + E[X]$$ and computing $$E[X(X - 1)]$$ using a method similar to that which was used to compute $$E[X]$$ above. Here are some sample plots of the pmf of a binomial rv for various values of $$n$$ and $$p= .5$$2. Here are some with $$n = 100$$ and various $$p$$. ### 3.4.2 Geometric Random Variable Definition 3.3 A geometric random variable counts the number of failures in Bernoulli trials until the first success occurs. A geometric random variable can take on values $$0,1,2,\dotsc$$. The pmf of a geometric rv is given by $p(x) = p(1-p)^{x} \qquad x = 0,\ldots$ Note that $$\sum_x p(x) = 1$$ by the sum of a geometric series formula. Theorem 3.5 Let $$X$$ be a geometric random variable with probability of success $$p$$. Then • The expected value of $$X$$ is $$\frac{(1-p)}{p}$$. • The variance of $$X$$ is $$\frac{(1-p)}{p^2}$$ Example A die is tossed until the first 6 occurs. What is the probability that it takes 4 or more tosses? We let $$X$$ be a geometric random variable with probability of success $$1/6.$$ Since $$X$$ counts the number of failures before the first success, the problem is asking us to compute $P(X \ge 3) = \sum_{j=3}^\infty P(X = j)$ One approach is to approximate the infinite sum of the pdf dgeom: sum(dgeom(3:1000, 1/6)) ## [1] 0.5787037 Better is to use the built-in cdf function pgeom: pgeom(2, 1/6, lower.tail = FALSE) ## [1] 0.5787037 Setting lower.tail = FALSE forces R to compute $$P(X > 2) = P(X \ge 3)$$ since $$X$$ is discrete. Another approach is to apply rules of probability: $$P(X \ge 3) = 1 - P(X < 3) = 1 - P(X \le 2)$$. Then with R: 1 - pgeom(2,1/6) ## [1] 0.5787037 or 1 - sum(dgeom(0:2, 1/6)) ## [1] 0.5787037 All of these show there is about a 0.58 probability that it will take four or more tosses to roll a six. Does this sound reasonable? The expected number of tosses (failures) before one obtains a 6 is $$\frac{1-p}{p} = \frac{5/6}{1/6} = 5$$, so $$P(X \ge 3) \approx 0.58$$ is not surprising at all. The standard deviation of the number of tosses is $$\sqrt{\frac{5/6}{1/6^2}} = \sqrt{30} \approx 5.5$$. Using the rule of thumb that “most” data lies within two standard deviations of the mean, we would be surprised (but perhaps not shocked) if it took more than $$5 + 2\sqrt(30) \approx 16$$ rolls before we finally rolled a 6. The probability of it taking more than 16 rolls is $$P(X \ge 16) \approx .054$$. #### 3.4.2.1 Proof that the mean of a geometric rv is $$\frac{1-p}{p}$$ (Optional) Let $$X$$ be a geometric rv with probability of success $$p$$. We show that $$E[X] = \frac{1-p}{p}$$. We must compute $E[X] = \sum_{x = 0}^\infty x p(1 - p)^x = (1 - p)p \sum_{x=1}^\infty x(1-p)^{x-1}$ Let $$q = 1- p$$ and see \begin{align} E[X] &= q( 1-q) \sum_{x=1}^\infty xq^{x-1} = q(1 - q) \sum_{x=1}^\infty \frac {d}{dq} q^x \\ &= q(1 - q) \frac {d}{dq} \sum_{x=1}^\infty q^x = q(1- q) \frac{d}{dq} (1 - q)^{-1} = \frac{q(1-q)}{(1- q)^2}\\ &=\frac{q}{1-q} = \frac{1-p}{p} \end{align*} For the variance, the interested reader can compute $$E[X^2]$$ using $$E[X^2] = E[X(X - 1)] + E[X]$$ and a similar method as that which was used in the above computation. Here are some plots of pmfs with various $$p$$. ### 3.4.3 Poisson Random Variables This section discusses the Poisson random variable, which comes from a Poisson process. Suppose events occur spread over time span $$[0,T]$$. If the occurrences of the event satisfy the following properties, then the events form a Poisson process: 1. The probability of an event occurring in a time interval $$[a,b]$$ depends only on the length of the interval $$[a,b]$$. 2. If $$[a,b]$$ and $$[c,d]$$ are disjoint time intervals, then the probability that an event occurs in $$[a,b]$$ is independent of whether an event occurs in $$[c,d]$$. (That is, knowing that an event occurred in $$[a,b]$$ does not change the probability that an event occurs in $$[c,d]$$.) 3. Two events cannot happen at the same time. (Formally, we need to say something about the probability that two or more events happens in the interval $$[a, a + h]$$ as $$h\to 0$$.) 4. The probability of an event occuring in a time interval $$[a,a + h]$$ is roughly $$\lambda h$$, for some constant $$\lambda$$. Property (4) says that events occur at a certain rate, which is denoted by $$\lambda$$. Definition 3.4 A Poisson random variable with rate $$\lambda$$ counts the number of events that occur in the Poisson process. The pmf of $$X$$ when $$X$$ is Poisson with rate $$\lambda$$ is $p(x) = \frac 1{x!} \lambda^x e^{-\lambda}, \qquad x = 0,1,\ldots$ Note that the Taylor series for $$e^t$$ is $$\sum_{n=0}^\infty \frac 1{n!} t^n$$, which can be used to show that $$\sum_x p(x) = 1$$. Theorem 3.6 The mean and variance of a Poisson random variable are both $$\lambda$$. Example Suppose a typist makes typos at a rate of 3 typos per page. What is the probability that they will make exactly 5 typos on a two page sample? Here, we assume that typos follow the properties of a Poisson rv. It is not clear that they follow it exactly. For example, if the typist has just made a mistake, it is possible that their fingers are no longer on home position, which means another mistake is likely, violating the independence rule (2). Nevertheless, assuming the typos are Poisson is a reasonable approximation. The rate at which typos occur per two pages is 6, so we use $$\lambda = 6$$ in the formula for Poisson. We get $P(X = 5) = \frac{6^5}{5!} e^{-6} \approx 0.16.$ Using R, and the pdf dpois: dpois(5,6) ## [1] 0.1606231 There is a 0.16 probability of making exactly five typos on a two-page sample. Often, when modeling a count of something, you need to choose between binomial, geometric, and Poisson. The binomial and geometric random variables both come from Bernoulli trials, where there is a sequence of individual trials each resulting in success or failure. In the Poisson process, events are spread over a time interval, and appear at random. Here some plots of pmfs with various means. ### 3.4.4 Negative Binomial (Optional) Example Suppose you repeatedly roll a fair die. What is the probability of getting exactly 14 non-sixes before getting your second 6? As you can see, this is an example of repeated Bernoulli trials with $$p = 1/6$$, but it isn’t exactly geometric because we are waiting for the second success. This is an example of a negative binomial random variable. More generally, suppose that we observe a sequence of Bernoulli trials with probability of success prob. If $$X$$ denotes the number of failures x before the nth success, then $$X$$ is a negative binomial random variable with parameters n and p. The probability mass function of $$X$$ is given by $p(x) = {{x + n - 1}\choose{x}} p^n (1 - p)^x, \qquad x = 0,1,2\ldots$ The mean of a negative binomial is $$n p/(1 - p)$$, and the variance is $$n p /(1 - p)^2$$. The root R function to use with negative binomials is nbinom, so dnbinom is how we can compute values in R. The function dnbinom uses prob for $$p$$ and size for $$n$$ in our formula. So, to continue with the example above, the probability of obtaining exactly 14 non-sixes before obtaining 2 sixes is: dnbinom(x = 14, size = 2, prob = 1/6) ## [1] 0.03245274 Note that when size = 1, negative binomial is exactly a geometric random variable, e.g. dnbinom(x = 14, size = 1, prob = 1/6) ## [1] 0.01298109 dgeom(14, prob = 1/6) ## [1] 0.01298109 ### 3.4.5 Hypergeometric (Optional) Consider the experiment which consists of sampling without replacement from a population that is partitioned into two subgroups - one subgroup is labeled a “success” and one subgroup is labeled a “failure”. The random variable that counts the number of successes in the sample is an example of a hypergeometric random variable. To make things concrete, we suppose that we have $$m$$ successes and $$n$$ failures. We take a sample of size $$k$$ (without replacement) and we let $$X$$ denote the number of successes. Then $P(X = x) = \frac{{m \choose x} { n \choose {k - x}} }{{m + n} \choose k}$ We also have $E[X] = k (\frac{m}{m + n})$ which is easy to remember because $$k$$ is the number of samples taken, and the probability of a success on any one sample (with no knowledge of the other samples) is $$\frac {m}{m + n}$$. The variance is similar to the variance of a binomial as well, $V(X) = k \frac{m}{m+n} \frac {n}{m+n} \frac {m + n - k}{m + n - 1}$ but we have the “fudge factor” of , which means the variance of a hypergeometric is less than that of a binomial. In particular, when $$m + n = k$$, the variance of $$X$$ is 0. Why? When $$m + n$$ is much larger than $$k$$, we will approximate a hypergeometric random variable with a binomial random variable with parameters $$n = m + n$$ and $$p = \frac {m}{m + n}$$. Finally, the R root for hypergeometric computations is hyper. In particular, we have the following example: Example 15 US citizens and 20 non-US citizens pass through a security line at an airport. Ten are randomly selected for further screening. What is the probability that 2 or fewer of the selected passengers are US citizens? In this case, $$m = 15$$, $$n = 20$$, and $$k = 10$$. We are looking for $$P(X \le 2)$$, so sum(dhyper(x = 0:2, m = 15, n = 20, k = 10)) ## [1] 0.08677992 ## 3.5 Special Continuous Random Variables In this section, we discuss the uniform, exponential, and normal random variables. There are many others that are useful, and we will cover them as needed in the rest of the text. ### 3.5.1 Uniform Random Variable A uniform random $$X$$ over the interval $$(a,b)$$ satisfies the property that for any interval inside of $$(a,b)$$, the probability that $$X$$ is in that interval depends only on the length of the interval. The pdf is given by $f(x) = \begin{cases} \frac{1}{b -a} & a \le x \le b\\0&{\rm otherwise} \end{cases}$ The expected value of $$X$$ is $$\int_a^b \frac x{b - a} \, dx = \frac{b + a}2$$, and the variance is $$\sigma^2 = \frac{b^2 - a^2}{12}$$. Example A random number is chosen between 0 and 10. What is the probability that it is bigger than 7, given that it is bigger than 6? $P(X > 7\ \|\ X > 6) = \frac{P(X > 7 \cap X > 6)}{P(X > 6)} = \frac{P(X > 7)}{P(X > 6)} =\frac{3/10}{4/10} = \frac{3}{4}.$ ### 3.5.2 Exponential Random Variable An exponential random variable $$X$$ with rate $$\lambda$$ has pdf $f(x) = \lambda e^{-\lambda x}, \qquad x > 0$ Exponential rv’s are useful for, among other things, modeling the waiting time until the first occurrence in a Poisson process. So, the waiting time until an electronic component fails, or until a customer enters a store could be modeled by exponential random variables. The mean of an exponential random variable is $$\mu = 1/\lambda$$ and the variance is $$\sigma^2 = 1/\lambda^2$$. Here are some plots of pdf’s with various rates. ### 3.5.3 Normal Random Variable A normal random variable with mean $$\mu$$ and standard deviation $$\sigma$$ has pdf given by $f(x) = \frac 1{\sigma\sqrt{2\pi}} e^{-(x - \mu)^2/(2\sigma^2)}$ The mean of a normal rv is $$\mu$$ and the standard deviation is $$\sigma$$. We will see many uses for normal random variables throughout the book, but for now let’s just compute some probabilities when $$X \sim {\rm Normal}(\mu = 2, \sigma = 2)$$. 1. $$P(X \le 4)$$ = pnorm(4,mean = 2,sd = 4) 2. $$P(0 \le X \le 2)$$ = pnorm(2,2,4) - pnorm(0,2,4). 3. Find the value of $$q$$ such that $$P(X \le q) = .75$$. $$q$$ = qnorm(.75,2,4). We get about 4.7; we can check it with pnorm(4.7,2,4). Here are some plots with fixed mean 0 and various sd, followed by fixed sd and various means. ## 3.6 Independent Random Variables We say that two random variables, $$X$$ and $$Y$$, are independent if knowledge of the outcome of $$X$$ does not give probabilistic information about the outcome of $$Y$$ and vice versa. As an example, let $$X$$ be the amount of rain (in inches) recorded at Lambert Airport on a randomly selected day in 2017, and let $$Y$$ be the height of a randomly selected person in Botswana. It is difficult to imagine that knowing the value of one of these random variables could give information about the other one, and it is reasonable to assume that the rvs are independent. On the other hand, if $$X$$ and $$Y$$ are the height and weight of a randomly selected person in Botswana, then knowledge of one variable could well give probabilistic information about the other. For example, if you know a person is 70 inches tall, it is very unlikely that they weigh 12 pounds. We would like to formalize that notion by saying that whenever $$E, F$$ are subsets of $${\mathbb R}$$, $$P(X\in E\|Y\in F) = P(X \in E)$$. There are several issues with formalizing the notion of independence that way, so we give a definition that is somewhat further removed from the intuition. The random variables $$X$$ and $$Y$$ are independent if 1. For all $$x$$ and $$y$$, $$P(X = x, Y = y) = P(X = x) P(Y = y)$$ if $$X$$ and $$Y$$ are discrete. 2. For all $$x$$ and $$y$$, $$P(X \le x, Y \le Y) = P(X \le x) P(Y \le y)$$ if $$X$$ and $$Y$$ are continuous. For our purposes, we will often be assuming that random variables are independent. ## 3.7 Using R to compute probabilities For all of the random variables that we have mentioned so far (and many more!), R has built in capabilities of computing probabilities. The syntax is broken down into two pieces: the root and the prefix. The root determines which random variable that we are talking about, and here are the names of the ones that we have covered so far: 1. binom is binomial 2. geom is geometric 3. pois is Poisson 4. unif is uniform 5. exp is exponential 6. norm is normal The available prefixes are 1. p computes the cumulative distribution 2. d computes pdf or pmf 3. r samples from the rv 4. q quantile function For now, we will focus on the prefixes p and d. Example Let $$X$$ be binomial with $$n = 10$$ and $$p = .3$$. 1. Compute $$P(X = 5)$$. We are interested in the pmf of a binomial rv, so we will use the R command dbinom as follows: dbinom(5, 10, .3) ## [1] 0.1029193 and we see that answer is about 1/10. 1. Compute $$P(1 \le X \le 5)$$. We want to sum $$P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$$. We could do that the long way as follows: dbinom(5, 10, .3) + dbinom(4, 10, .3) + dbinom(3, 10, .3) + dbinom(2, 10, .3) + dbinom(1, 10, .3) ## [1] 0.9244035 But, that is going to get old very quickly. Note that dbinom also allows vectors as arguments, so we can compute all of the needed probabilities like this: dbinom(1:5, 10, .3) ## [1] 0.1210608 0.2334744 0.2668279 0.2001209 0.1029193 And, all we need to do is add those up: sum(dbinom(1:5, 10 , .3)) ## [1] 0.9244035 Check that we got the same answer. Yep. Example Let $$X$$ be a normal random variable with mean 1 and standard deviation 5. 1. Compute $$P(X \le 1)$$. It is not easy to use the same technique as before, because we would need to compute $\int_{-\infty}^1 {\text {dnorm(x,1,5))}} \, dx$ But, note that the probability we are trying to compute is exactly the cdf, so we want to use pnorm. pnorm(1,1,5) ## [1] 0.5 1. Compute $$P(1 \le X \le 5)$$. Here, we compute the probability that $$X$$ is less than or equal to five, and subtract the probability that $$X$$ is less than 1. Since $$X$$ is continuous, that is the same as computing $P(X \le 5) - P(X \le 1)$ pnorm(5,1,5) - pnorm(1,1,5) ## [1] 0.2881446 Note to use p + a discrete rv, you will need to make adjustments to the above technique, e.g. if $$X$$ is binomial(10, .3) and you want to compute $$P(1\le X \le 5)$$, you would use pbinom(5,10,.3) - pbinom(0,10,.3) ## [1] 0.9244035 ## 3.8 A ggplot interlude In this short section, we introduce the basics of plotting using ggplot. We will go into more details later, but for now, we want to know how to plot scatterplots and lines using ggplot. The package ggplot2 is not part of the base R package, and needs to be installed on your machine. If it has not already been installed, then you need to do so (you can use the Install button under the Packages tab in RStudio, or see this) and use the command library(ggplot2) to be able to use it. If you have never used ggplot before, it works differently than other graphing utilities that I have seen. The basic idea is that the call of ggplot() sets up the default data that will be plotted and the basic aesthetic mappings. For our purposes in this section, the first argument of ggplot will be a data frame that we wish to visualize, and the second argument will be aes(x = , y = ). You will need to tell ggplot which variable you want to be considering as the independent variable (x) and which is the dependent variable (y). For example, consider the mtcars data set. If we wish to plot mpg versus disp, we would start by typing ggplot(mtcars, aes(x = disp, y = mpg)). Run this now. As you can see, ggplot has set up the axes according to disp and mpg with reasonable values, but no data has been plotted. This is because we haven’t told ggplot what we want to do with the data. Do we want a histogram? A scatterplot? A heatmap? Well, let’s assume we want a scatterplot here. We need to add the scatterplot to the pre-existing plot. We do that as follows: library(ggplot2) #Only need to do once ggplot(mtcars, aes(x = disp, y = mpg)) + geom_point() Later, we will see that geom_point() has its own aesthetic mappings and options, and we can combine things in more complicated and beautiful ways. But for now, we just want the basic scatterplot. In order to graph a function, we would create a data frame with the $$x$$ values and $$y$$ values that we wish to plot, and use geom_line(), as follows. Example Plot the pdf of a normal rv with mean 1 and standard deviation 2. xvals <- seq(-4,6,.01) plotdata <- data.frame(x = xvals, y = dnorm(xvals, 1, 2)) ggplot(plotdata, aes(x = x, y = y)) + geom_line() Example Plot the pmf of a geometric random variable with probability of success $$1/3$$. xvals <- 0:20 plotdata <- data.frame(x = xvals, y = dgeom(xvals, 1/3)) ggplot(plotdata, aes(x, y)) + geom_bar(stat = "identity") The reason we have to add stat = "identity" to the above plot is that geom_bar() normally plots the number of observances of the x-values versus the x-values. In this case, however, we already have the y-values that we want plotted. We are using geom_bar() just to get the form of the graph to look like this, so we tell geom_bar() to use the y-values in the aesthetic mapping by including stat = "identity". ## 3.9 Summary Here is a list of the (non-optional) random variables that we introduced in this section, together with pmf/pdf, expected value, variance and root R function. RV PMF/PDF Range Mean Variance R Root Binomial $${{n}\choose {x}} p^x(1 - p)^{n - x}$$ $$0\le x \le n$$ $$np$$ $$np(1 - p)$$ binom Geometric $$p(1-p)^{x}$$ $$x \ge 0$$ $$\frac{1-p}{p}$$ $$\frac{1-p}{p^2}$$ geom Poisson $$\frac {1}{x!} \lambda^x e^{-\lambda}$$ $$x \ge 0$$ $$\lambda$$ $$\lambda$$ pois Uniform $$\frac{1}{b - a}$$ $$a \le x \le b$$ $$\frac{a + b}{2}$$ $$\frac{b^2 - a^2}{12}$$ unif Exponential $$\lambda e^{-\lambda x}$$ $$x \ge 0$$ $$1/\lambda$$ $$1/\lambda^2$$ exp Normal $$\frac 1{\sigma\sqrt{2\pi}} e^{(x - \mu)^2/(2\sigma^2)}$$ $$-\infty < x < \infty$$ $$\mu$$ $$\sigma^2$$ norm ## 3.10 Exercises 1. Let $$X$$ be a discrete random variable with probability mass function given by $p(x) = \begin{cases} 1/4 & x = 0 \\ 1/2 & x = 1\\ 1/8 & x = 2\\ 1/8 & x = 3 \end{cases}$ 1. Verify that $$p$$ is a valid probability mass function. 2. Find the mean and variance of $$X$$. 3. Find $$P(X \ge 2)$$. 4. Find $$P(X \ge 2\ \|\ X \ge 1)$$. 2. Plot the pdf and cdf of a uniform random variable on the interval $$[0,1]$$. 3. Compare the cdf and pdf of an exponential random variable with rate $$\lambda = 2$$ with the cdf of an exponential rv with rate 1/2. (If you wish to read ahead in the section on plotting, you can learn how to put both plots on the same axes, with different colors.) 4. Compare the pdfs of three normal random variables, one with mean 1 and standard deviation 1, one with mean 1 and standard deviation 10, and one with mean -4 and standard deviation 1. 5. Let $$X$$ be a normal rv with mean 1 and standard deviation 2. 1. Find $$P(a \le X \le a + 2)$$ when $$a = 3$$. 2. Sketch the graph of the pdf of $$X$$, and indicate the region that corresponds to your answer in the previous part. 3. Find the value of $$a$$ such that $$P(a \le X \le a + 2)$$ is the largest. 6. Let $$X$$ be an exponential rv with rate $$\lambda = 1/4$$. 1. What is the mean of $$X$$? 2. Find the value of $$a$$ such that $$P(a \le X \le a + 1)$$ is maximized. Is the mean contained in the interval $$[a, a+1]$$? 7. Let $$X$$ be a random variable with pdf $$f(x) = 3(1 - x)^2$$ when $$0\le x \le 1$$, and $$f(x) = 0$$ otherwise. 1. Verify that $$f$$ is a valid pdf. 2. Find the mean and variance of $$X$$. 3. Find $$P(X \ge 1/2)$$. 4. Find $$P(X \ge 1/2\ \|\ X \ge 1/4)$$. 8. Is there a function which is both a valid pdf and a valid cdf? If so, give an example. If not, explain why not. 9. (Memoryless Property) Let $$X$$ be an exponential random variable with rate $$\lambda$$. If $$a$$ and $$b$$ are positive numbers, then $P(X > a + b\ \|\ X > b) = P(X > a)$ 1. Explain why this is called the memoryless property. 2. Show that for an exponential rv $$X$$ with rate $$\lambda$$, $$P(X > a) = e^{-a\lambda}$$. 3. Use the result in (b) to prove the memoryless property for exponential random variables. 10. Let $$X$$ be a Poisson rv with mean 3.9. 1. Create a plot of the pmf of $$X$$. 2. What is the most likely outcome of $$X$$? 3. Find $$a$$ such that $$P(a \le X \le a + 1)$$ is maximized. 4. Find $$b$$ such that $$P(b \le X \le b + 2)$$ is maximized. 11. For each of the following descriptions of a random variable, indicate whether it can best be modeled by binomial, geometric, Poisson, uniform, exponential or normal. Answer the associated questions. Note that not all of the experiments yield rv’s that are exactly of the type listed above, but we are asking about reasonable modeling. 1. Let $$Y$$ be the random variable that counts the number of sixes which occur when a die is tossed 10 times. What type of random variable is $$Y$$? What is $$P(Y=3)$$? What is the expected number of sixes? What is $${\rm Var}(Y)$$? 2. Let $$U$$ be the random variable which counts the number of accidents which occur at an intersection in one week. What type of random variable is $$U$$? Suppose that, on average, 2 accidents occur per week. Find $$P(U=2)$$, $$E(U)$$ and $${\rm Var}(U)$$. 3. Suppose a stop light has a red light that lasts for 60 seconds, a green light that lasts for 30 seconds and a yellow light that lasts for 5 seconds. When you first observe the stop light, it is red. Let $$X$$ denote the time until the light turns green. What type of rv would be used to model $$X$$? What is its mean? 4. Customers arrive at a teller’s window at a uniform rate of 5 per hour. Let $$X$$ be the length in minutes of time that the teller has to wait until they see their first customer after starting their shift. What type of rv is $$X$$? What is its mean? Find the probability that the teller waits less than 10 minutes for their first customer. 5. A coin is tossed until a head is observed. Let $$X$$ denote the total number of tails observed during the experiment. What type of rv is $$X$$? What is its mean? Find $$P(X \le 3)$$. 6. Let $$X$$ be the recorded body temperature of a healthy adult in degrees Fahrenheit. What type of rv is $$X$$? Estimate its mean and standard deviation, based on your knowledge of body temperatures. 12. Suppose you turn on a soccer game and see that the score is 1-0 after 30 minutes of play. Let $$X$$ denote the time (in minutes from the start of the game) that the goal was scored. What type of rv is $$X$$? What is its mean? 13. (Requires optional sections) Prove that the mean of a Poisson rv with parameter $$\lambda$$ is $$\lambda$$. 14. Roll two ordinary dice and let $$X$$ be their sum. Draw the pmf and cmf for X. Compute the mean and standard deviation of $$X$$. 15. Suppose you roll two ordinary dice. Calculate the expected value of their product. 16. Go to the Missouri lottery Pick 3 web page http://www.molottery.com/pick3/pick3.jsp and compute the expected value of these bets: 1. $1 Front Pair 2.$1 Back Pair 3. $6 6-way combo 4.$3 3-way combo 5. $1 1-off 17. Suppose you take a 20 question multiple choice test, where each question has four choices. You guess randomly on each question. What is your expected score? What is the probability you get 10 or more questions correct? 18. Steph Curry is a 91% free throw shooter. Suppose he shoots 10 free throws in a game. What is his expected number of shots made? What is the probability that he makes at least 8 free throws? 19. Suppose that 55% of voters support Proposition A. 1. You poll 200 voters. What is the expected number that support the measure? 2. What is the margin of error for your poll (two standard deviations)? 3. What is the probability that your poll claims that Proposition A will fail? 4. How large a poll would you need to reduce your margin of error to 2%? 20. The charge $$e$$ on one electron is too small to measure. However, one can make measurements of the current $$I$$ passing through a detector. If $$N$$ is the number of electrons passing through the detector in one second, then $$I = eN$$. Assume $$N$$ is Poisson. Show that the charge on one electron is given by $$\frac{{\rm Var}(I)}{E(I)}$$. 21. Climbing rope will break if pulled hard enough. Experiments show that 10.5mm Dynamic nylon rope has a mean breaking point of 5036 lbs with a standard deviation of 122 lbs. Assume breaking points of rope are normally distributed. 1. Sketch the distribution of breaking points for this rope. 2. What proportion of ropes will break with 5000 lbs of load? 3. At what load will 95% of all ropes break? 22. There exist naturally occurring random variables that are neither discrete nor continuous. Suppose a group of people is waiting for one more person to arrive before starting a meeting. Suppose that the arrival time of the person is exponential with mean 4 minutes, and that the meeting will start either when the person arrives, or after 5 minutes, whichever comes first. Let $$X$$ denote the length of time the group waits before starting the meeting. 1. Find $$P(0 \le X \le 4)$$. 2. Find $$P(X = 5)$$. 23. A roulette wheel has 38 slots and a ball that rolls until it falls into one of the slots, all of which are equally likely. Eighteen slots are black numbers, eighteen are red numbers, and two are green zeros. If you bet on “red”, and the ball lands in a red slot, the casino pays you your bet, otherwise the casino wins your bet. 1. What is the expected value of a$1 bet on red? 2. Suppose you bet $1 on red, and if you win you “let it ride” and bet$2 on red. What is the expected value of this plan? 24. One (questionable) roulette strategy is called bet doubling. You bet $1 on red, and if you win, you pocket the$1. If you lose, you double your bet so you are now betting $2 on red, but have lost$1. If you win, you win $2 for a$1 profit, which you pocket. If you lose again, you double your bet to $4 (having already lost$3). Again, if you win, you have $1 profit, and if you lose, you double your bet again. This guarantees you will win$1, unless you run out of money to keep doubling your bet. 1. Say you start with a bankroll of $127. How many bets can you lose in a row without losing your bankroll? 2. If you have a$127 bankroll, what is the probability that bet doubling wins you $1? 3. What is the expected value of the bet doubling strategy with a$127 bankroll? 4. If you play the bet doubling strategy with a \$127 bankroll, how many times can you expect to play before you lose your bankroll? 1. That is, the variance is greater than or equal to zero. 2. You don’t need to know the R code to generate this yet, but here it is: library(dplyr) library(ggplot2) binomdata <- data.frame(x = rep(0:75, 4)) binomdata <- mutate(binomdata, n = c(rep(10, 76),rep(20,76),rep(50,76),rep(100,76)), pmf = dbinom(x,n,p=.5)) ggplot(binomdata, aes(x = x, y = pmf)) + geom_bar(stat = "identity") + facet_wrap(~n, labeller = "label_both")<\|endoftext\|>	4.5625
333	30 September, 05:47 # Kwamee created a coffee blend for his cafe by mixing Kona beans and Fuji beans. Kona beans cost \$11 per pound and Fuji beans cost \$7.50 per pound. He bought a total of 23 pounds of coffee beans and it cost \$197. Write a system of equations that can be used to determine how many pounds of Kona and Fuji beans Kwamee bought. +5 1. 30 September, 09:47 0 Let's use K for Kona and F for Fuji. The system of equations has to be a balanced system. For example, you can't mix the number of pounds of beans with the cost for each because pounds and dollars are different and you can only combine like terms ... pounds with pounds and dollars with dollars. So let's start with the number of pounds. Since we don't know how much of each he bought we have the 2 unknowns, F and K, but we DO know that he bought 23 pounds total. So the first equation is K + F = 23 Now let's see what we can do with the dollars. Again, we don't know how much he bought of each kind of coffee, but we do know that Kona beans cost \$11 per pound and that Fuji beans cost \$7.50 per pound, and we know that he spent a total of \$197. So let's set that up: 11K + 7.50F = 197 Those are your 2 equations. It doesn't say you need to solve them, so you're done.<\|endoftext\|>	4.40625
612	Maths Resources GCSE Worksheets Subtracting Fractions Worksheet • Section 1 of the subtracting fractions worksheet contains 20+ skills-based subtracting fractions questions, in 3 groups to support differentiation • Section 2 contains 3 applied subtracting fractions questions with a mix of word problems and deeper problem solving questions • Section 3 contains 3 foundation and higher level GCSE exam style subtracting fractions problems • Answers and a mark scheme for all subtracting fractions questions are provided • Questions follow variation theory with plenty of opportunities for students to work independently at their own level • All questions created by fully qualified expert secondary maths teachers • Suitable for GCSE maths revision for AQA, OCR and Edexcel exam boards • This field is for validation purposes and should be left unchanged. You can unsubscribe at any time (each email we send will contain an easy way to unsubscribe). To find out more about how we use your data, see our privacy policy. Subtracting fractions at a glance We can subtract fractions using different methods. In order to subtract simple fractions with the same denominator, we just subtract the numerators and leave the denominator the same. For example, frac{3}{5} - frac{1}{5} = frac{2}{5} . If two fractions have different denominators, we need to rewrite them as equivalent fractions with a common denominator before we can subtract them. To write fractions with a common denominator, the first step is to find a common multiple of the denominators. This could be the least common multiple or any other common multiple. We then need to find equivalent fractions to the originals with denominators equal to the common multiple. Once we have two fractions with like denominators, we can subtract the numerators, leaving the denominators the same. Mixed numbers can be subtracted in the same way as proper fractions. It is often easiest to write the mixed number as an improper fraction before following the same process as above. Once the fractions have been subtracted, any improper fraction can be written back as a whole number and a fraction. Looking forward, students can then progress to additional adding and subtracting fractions worksheets and other number worksheets, for example a multiplying and dividing decimals worksheet or an order of operations worksheet. For more teaching and learning support on Number our GCSE maths lessons provide step by step support for all GCSE maths concepts. Do you have GCSE students who need additional support? There will be students in your class who require individual attention to help them achieve their target GCSE maths grade. In a class of 30, it’s not always easy to provide. Help your students feel confident with exam-style questions and the strategies they’ll need to answer them correctly with personalised online one to one tutoring from Third Space Learning Lessons are selected to provide support where each student needs it most, and specially-trained GCSE maths tutors adapt the pitch and pace of each lesson. This ensures a personalised revision programme that raises grades and boosts confidence. GCSE Revision Programme<\|endoftext\|>	4.625
1,104	# Calculus Applets Prev Home Next ## Volumes of Revolution We have seen how to find the area between two curves by finding the formula for the area of a thin rectangular slice, then integrating this over the limits of integration. We can use the same strategy to find the volume that is swept out by an area between two curves when the area is revolved around an axis. Try the following: 1. The applet initially shows a yellow area that is bounded by the two curves f (x) = 2 and g(x) = 0, and extends from x = 0 to x = 2. If this yellow rectangle is revolved around the horizontal x axis, the result is a cylinder. A small rectangular slice of the yellow area is shown in dark gray. If this is revolved around the x axis, it sweeps out a disk, which is also shown in a lighter gray. The formula for the volume of a circular cylinder is V = π r² h. In this case, the height h is the thickness of the disc, which we will call dx. The radius is just the height of the yellow rectangle, which is a constant 2. So the volume of the gray disc slice is π 2²dx = 4πdx. So the total volume of the cylinder is , which we can verify directly from the formula for the volume of a cylinder. 2. Select the second example from the drop down menu. Here the only change is that f (x) = x. Move the x slider and notice that the radius of our slice now changes. In this case, the radius is just the value of f for a particular x, which is just x. So the integral becomes , which you can also verify using the geometry formula for the volume of a cone. 3. Select the third example from the drop down menu, showing f (x) = Move the x slider to see how the radius of the disc changes as x changes. Here, the radius of the disc is x², so the integral becomes . In summary, when the axis of revolution and the lower function are both the x axis, the integral is . 4. Select the fourth example, showing a cylinder with a hole. This is the first example again, but now g(x) = 1. The yellow rectangle doesn't extend all the way to the axis of revolution, so there is a hole in the cylinder. Our gray disc has become a washer. The easiest way to think about this is to find the area of the cylinder, then find the area of the hole, then subtract. We know the integral for the area of the outer cylinder (see example 1, above). The integral for the hole is similar, but with a smaller radius. Hence the problem becomes: , where R ("big R") is the outer radius and r ("little r") is the radius of the hole. For this problem, we get: . 5. Select the fifth example, showing a cylinder with a cone-shaped hole. Drag the x slider to see how the washer changes as x changes. Big R is just 2, as before, but now little r (the radius of the hole) changes and is g(x) = x. So the integrals are: . 6. Select the sixth example, showing an odd shape that is revolved around the x axis. Move the x slider to get a feel for the shape that is generated when the yellow area is revolved. Using our two integrals we get: . 7. Select the seventh example, showing a different odd shape. This is the same functions and axis as the previous example, but now the limits of integration have shifted. Note that this changes which function is big R and which is little r. If you just use the same formula as the previous problem with new limits, you will get a negative volume. To make the volume come out positive, we need to change big R to be the function that is furthest from the axis of revolution. Hence the integrals are: . 8. Select the eigth example, showing the original odd shape, but now the axis of revolution has shifter to y = 1. Move the x slider to get a feel for the shape of the volume. Big R in this case is the distance from the axis to g(x), so R = 1 -x². Similarly, little r (the radius of the hole) is the distance from the axis to f (x) so r = 1 - x. Hence the integrals become: . Notice the use of parentheses to make sure that R and r get squared properly. 9. Select the ninth example. This shows the original odd shape, but now it it revolved around the vertical axis. Note that the functions are now functions of y instead of x. Move the y slider to see how the washer changes. The overall shape is sort of a bowl. The integrals are: . We can revolve areas around vertical axis, just like around horizontal axis. Notice that the volume comes out different when compared to example 6 (the same shape revolved around the x axis). 10. Select the tenth example. You can enter your own functions, pick an axis, and change the limits. You can also click the inverse check box to make the axis vertical, but then you also have to rewrite your function definitions in terms of y instead of x.<\|endoftext\|>	4.53125
533	# How often does a power of 3 occur compared to power of 6 I've been trying to solve this many different ways but I am completely stumped. How does one go about this? What I'm trying to figure out is something like this... (n is a natural number) How many numbers are there smaller than the n-th power of 3 ($$3^n$$), that are powers of 6? ($$6^n$$) I've been trying to figure out an equation for this without success. Example: For the 2nd power of 3 ($$3^2$$), we have 2 occurences of a power of 6. (1 and 6). For the 4th power of 3 ($$3^4$$), we have 3 occurences of a power of 6. (1, 6 and 36) Is there a formula that can be derived to establish this relationship? • Hint: use logs. – lulu Apr 19, 2019 at 20:29 The number of integers of the form $$6^n$$ below m is equal to $$\lfloor \log_6{m} \rfloor + 1$$. So the number of powers of 6 below the nth power of 3 is equal to $$\lfloor \log_6{3^n}\rfloor + 1$$. You can interpret this intuitively as being the number of times we must multiply $$1$$ by $$3$$ before we reach $$m$$, or just overshoot it. If we overshoot it, then the floor function ensures that we only calculate the number of powers that actually lie below $$m$$. We add one to each of these expressions because we want to include $$1$$ as a power of $$6$$. This is based on your example where you include $$1$$ as a power, but it’s worth noting that this potentially contradicts your original description of $$6^n$$ ‘where $$n$$ is a natural number’ as the natural numbers are often, although not always, assumed to begin at $$1$$. Of course, you can add $$1$$ or not, depending on what you want. • It is pretty common to consider $0$ to be a natural number. Unfortunately it is also pretty common to consider $0$ not to be a natural number. In general just the appearance of the term "natural number" is not enough to tell you which meaning the writer had in mind. Apr 19, 2019 at 21:08<\|endoftext\|>	4.5625
910	In the mountainous cloud forests of Borneo, nothing goes to waste, and that includes waste itself. Here, the largest carnivorous plant on Earth has evolved to become a sort of customised outhouse for local tree shrews (Tupaia montana). With a toilet seat perfectly contoured to the mammal's backside, this huge, jug-shaped pitcher plant (Nepenthes rajah) survives not by swallowing insects, as do many of its close relatives, but by devouring nutrient-rich faeces. It's one of the most peculiar and puzzling plants on Earth, with a pitcher so large it can hold two litres of water and glands so busy they produce huge quantities of nectar. Yet it's only in the past decade, more than 150 years after the plant's discovery, that scientists have truly understood the purpose of its sweet juice. By using nectar to tempt the tree shrews, researchers say this pitcher plant is basically "a toilet with a feeding station". Video observations reveal tree shrews jumping onto the pitcher plants and licking nectar from their leafy lids, all the while marking their territory with little droppings. You can see this in action in the David Attenborough video below. These 3 pitcher plants grow where insects are scarce & so evolved 'toilets' for shrews, which excrete in the pitchers, fertilising the plant pic.twitter.com/EBvN9K7t8R— Chris Thorogood (@thorogoodchris1) January 8, 2017 In a region where mosquitoes are scarce, this carnivorous plant appears to have followed a unique evolutionary path. Unlike other pitcher plants, N. rajah is particularly sturdy, with a shrew-shaped orifice and a non-slippery rim, allowing the shrews to stick around and feed for longer. At the same time, the shape of this pitcher plant's funnel also ensures the shrew's poo ends up in the bowl, whether by gravity or rain runoff. As strange as it sounds, the plant's appetite is not an anomaly. Another pitcher plant has developed a similarly mutual relationship with bats. N. hemsleyana provides a shelter for the woolly bat (Kerivoula hardwicki) to roost in during the day, and in return feasts on the bat's droppings. Meanwhile, the same shrews have been found toileting in other carnivorous pitcher plants, including N. lowii and N. macrophylla. But while the ground-based N. lowii has effectively lost the capacity to trap animals, both the other species can still capture arthropod prey, as well as feed on shrew poo. "These 'tree shrew toilets' have concave, reflexed lids oriented to position the animal for optimal faeces capture," explains Chris Thorogood, an evolutionary geneticist and plant taxonomist at the University of Oxford, in a blog post. "Tree shrew toilets may have evolved from pitchers with broad pitcher mouths adapted for enhanced water or leaf litter capture." While shrew poo may not sound like the most appetising meal, it is rich with nutrients that pitcher plants need to survive, like nitrogen and phosphorous. In fact, recent research has found that shrew faeces account for "57–100 percent of foliar nitrogen" in the N. lowii species alone. It's a win-win situation for everyone involved, but there are still many details that need to be teased out. For instance, how exactly does the tree shrew know when the pitcher plant is ready to hold it? A newly-formed pitcher takes days to become sturdy enough for visitors, yet most in the wild show little signs of damage, indicating that the tree shrew is a patient feeder. "One possible explanation for this is that the plants signal the tree shrews to indicate whether or not individual pitchers are "open for business," one study hypothesises. For example, in N. lowii, the inner surface turns to a solid, dark purple when the plant matures, potentially letting tree shrews know that it is ready to be landed on. "The fact that the aerial pitchers are so tailored to the shape and activities of the tree shrew suggests this has been going on for a long time," ecologist Jonathan Moran of Royal Roads University in British Columbia told Live Science in 2009.<\|endoftext\|>	3.6875
569	Watching an octopus change colors is mesmerizing. Many of us have spent an entire dive hypnotized by the flowing colors of an octopus or one of its sister cephalopods. Octopuses (not octopi) are sometimes called the chameleons of the sea. But, as you’ll read below, a chameleon’s camouflage game is weak compared to that of the mighty octo. In this article, we’ll cover how an octopus changes color and why…but it may leave you with more questions than answers. Why Does an Octopus Change Color? Cephalopods (octopuses, squid and cuttlefish) change their appearance to attract a mate, hide, or send a warning message. The notorious blue-ringed octopus flashes iridescent blue brings to warn potential predators to stay away. It’s nature’s version of a “Don’t Touch” sign. How do they do it? Cephalopods have something called chromatophores under their skin. Using a combination of pigment, nerves and muscles, the animal manipulates their chromatophores to change their external appearance. According to Wikipedia, chromatophores change based on “neuronal activation.” Once the animal’s brain gives a signal, the color change literally ripples through the animal’s body. Chameleons, on the other hand, change color gradually. It can take up to several minutes for their transformation complete. Maybe we ought to call chameleons jungle octopuses, or dirt squids. They Can Do More Than Just Change Color Changing color is just one way an octopus can transform its appearance. These clever critters can also modify their skin texture to mimic rocks, sand, coral heads, or other landscape elements by altering the papillae on their skin. A mimic octopus can impersonate the shape of other sea creatures in addition to changing color. Watch the video below to see how the mimic octopus can look like a flat fish, lion fish and sea snake: Are Octopuses “Colorblind?” An octopus uses its eyes to determine what color and pattern to mimic, but they lack some of the receptors in their eyes that humans use to see color. Scientists aren’t sure why, but as you may remember from the Open Water course, an object that is red on the surface may appear brown or black at depth. A team of scientists at the University of California Berkeley believe octopuses and other cephalopods may use focus (rather than cones) to determine color rather than seeing the color as humans do.<\|endoftext\|>	3.859375
189	In each section of this book, a particular drumming style is described. Exercises are then presented to develop technique and fluency in the style being taught. Recorded examples of: a) the exercises, b) beats, and c) beats with rhythm section accompaniment are presented on the CD to further illucidate the style. A drum part (chart ) written in the specific style is presented at the end of each section. The examples on the CD are record a) with drums and a rhythm section and b) with rhythm section alone, so that students can add their own rhythms. This format gives the teacher a progressive set of elementary charts to use in class. The charts do not include difficult or tricky rhythmic figures. They are presented so that the student can study them and hear their interpretations on the CD. Students can then experiment with their own interpretations by playing along with the selections as recorded with or without drums.<\|endoftext\|>	3.890625
1,063	Astronomically Speaking #2, published in the York University Gazette, November 1999 Peer into the sky towards the constellation Cygnus, the Swan, near the horizon this month. One of the most interesting objects up there is one that, no matter how hard you try, you will never see. Up amongst the stars, hidden from us by its own gravity, is a black hole, and it's the closest one to the Earth at just one thousand light years away. Black holes are perhaps the strangest objects in the universe. To understand them, we should first talk a little about Einstein. Early this century, Albert Einstein developed his theory of gravity known as General Relativity. Before Einstein, space and time were always part of the background, the playing field on which the games of physics were held. Objects moved through space, changing their position against the regular heart-beats of time. Relativity says, however, that space and time, tÎogether known as space-time, aren’t just labels for where an object is found. Space-time itself is a real dynamic entity. In the vicinity of a massive object like a star, space-time is curved. Curved space-time isn't easy to imagine, even for physicists. Try this: imagine you've got a nice big, soft king-sized bed. If you roll a bowling ball across the bed, the ball sinks down a bit, deforming the bed's surface into a curved dip as it goes. Place a bowling ball in the middle of the bed, and roll a marble across the mattress — the marble will, of course, follow the dip of the mattress down towards the bowling ball, along a curved path. The mattress is a bit like space-time, the bowling ball like a massive object, such as a star. The mass of a star warps the space-time around it, and any object coming near the star will follow a curved path through the curved space-time. The Earth does this as it orbits the Sun; it traces out an ellipse in the space curved by the enormous mass of our nearest star. Einstein became a bit of a celebrity, however, for suggesting that even light itself gets bent as it passes a heavy object. Careful observations have verified this effect. They have shown that a distant star seems to shift position in the sky a little when the Sun passes in front of it - the light coming from the star is bent by the Sun's gravity, and the amount of the bending is exactly the amount predicted by Einstein’s General Relativity. Headlines shouted "Einstein Proven Correct!", and his face began to appear in newspapers and magazines, and on coffee mugs and t-shirts. What's all this to do with black holes? Well, what if you had an object so massive that passing rays of light would bend so much they would spiral down towards the object, never to escape? Any light trying to leave the surface of this object would be pulled back. And if light can't get away, then nothing else could either, since nothing in the universe can travel faster than light. Such things are perfectly acceptable results of General Relativity. Theorists dubbed them 'black holes' since light can't escape to show us what they look like. You really wouldn't want to stumble across one of these things. The incredible gravitational pull of the black hole wouldn’t just drag you towards it, it would do very odd things to your body in the process. For one thing, the gravitational pull on your feet would be stronger than on your head. This is true if you're standing on the surface of the Earth too, the pull towards the ground is ever so slightly greater at your feet than up at your head. We don't notice this because the difference is incredibly small. But near a black hole, your feet would drop away and your body would stretch out like so much spaghetti. So how do we know there really is a black hole in Cygnus? That's the thing about black holes: they're black, we can't see them directly. We can, however, tell they're around by looking for the effects of their gravity on nearby stars. Astronomers observed that a large star in the constellation Cygnus appeared to be orbiting around some other, very heavy object. When they looked to find the other object, though, there was nothing there. Just as odd, the star was emitting huge blasts of x-ray energy, as if it's surface was being ripped apart and dragged away. Astrophysicists calculated that these effects must be due to a black hole very near to that star, and dubbed the system "Cygnus X-1". The "X" is for the X-rays bursting from it, and the “1" means it was the first black hole they found. So as you gaze up into the sky this month, just think what a wonderful, poetic twist it is, that deep within the constellation of Cygnus, in the very heart of that graceful white swan, lies one of the darkest and strangest denizens of the sky.<\|endoftext\|>	3.75
292	# How Many Terms of the Geometric Progression 1 + 4 + 16 + 64 + …….. Must Be Added to Get Sum Equal to 5461? - Mathematics Sum How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461? #### Solution Given G.P. : 1 + 4 + 16 + 64 +........... Here, first term, a = 1 common ratio, r = 4/1=4 (r > 1) Let the number of terms to be added = n Then, Sn = 5461 =>(a(r^n-1))/(r-1)=5461 => (1(4^"th"-1))/(4-1)=5461 =>(4^"th"-1)/3=5461 ⇒ 4th - 1 = 16383 ⇒ 4th = 16384 ⇒ n = 7 Hence, required number of terms = 7 Concept: Simple Applications - Geometric Progression Is there an error in this question or solution? #### APPEARS IN Selina Concise Maths Class 10 ICSE Chapter 11 Geometric Progression Exercise 11 (D) \| Q 2 \| Page 161<\|endoftext\|>	4.4375
747	# 8 Fun Facts about Decimals Decimals are a way of representing numbers. They have a whole number part and a fractional (or decimal) part, which are separated by a decimal point. So that’s the basics covered. But here are some lesser known, and in some cases strange, facts about decimals: ## 1. The ‘dec’ in decimal means ten, and refers to the fact that each position in a decimal number corresponds to ten times more than the next position along. For example, the number 325.31 means 3 hundreds, 2 tens, 5 ones, 3 tenths and 1 hundredth. Humans decided to group in tens because that’s how many fingers/thumbs we have. It made counting and arithmetic a whole lot easier. Three-fingered aliens might well group in threes! ## 2. Have you ever been told that ‘to multiply by ten, just add a zero at the end’? If so, you’ve been misled! Here’s an example: 2.3 x10 is not 2.30, it is 23.0. What’s actually happening when you multiply by ten is that every digit shifts over one place to the left because there are now ten times as many of each digits. In our example, there are now 2 tens rather than 2 ones. ## 3. Some decimal expansions go on forever: for example, 1/3 = 0.333… where the ‘…’ means that the 3s never end. ## 4. 0.999…. = 1 This can’t be true, surely? It is, and we’re going to prove it: Let x = 0.999… Then 10x = 9.999… So 9x = 10x – x = 9.999… – 0.999… = 9 Which means x=1. Here’s another way to convince yourself: try plotting the value 0.999… on the number line. It has to be greater than 0.9, but also greater than 0.99, and 0.999 and so on. You don’t have any space to squeeze it before the 1 which must mean it is actually equal to 1! ## 5. Every fraction is a decimal. A fraction is one whole number divided by another (but we can’t divide by zero). Every fraction, small or large, positive or negative, can be written as a decimal. For example, 1/2 = 0.5, 1/3 = 0.333… and 1/7 = 0.142857142857… – where the ‘142857’ repeats forever! ## 6. But not every decimal is a fraction. Some numbers are so strange that they can’t be expressed as a fraction. The number π, for example, which we use to calculate circle measurements, has a decimal expansion that goes on forever and never repeats (it starts 3.14159). It cannot be written as a fraction. ## 7. Different countries and languages use different notation for the decimal point. For example, in Taiwan and Singapore the point is placed is placed mid-line, so 23·89 rather than 23.89. In many European countries, a comma is used instead: 23,89. To each their own! ## 8. Decimals are used in a range of real-world situations to represent different amounts. Money, weight and height all use decimal points. So does the Richter scale, which measures the strength of an earthquake.<\|endoftext\|>	4.625
513	Proposition: 6.19: Ratio of Areas of Similar Triangles (Proposition 19 from Book 6 of Euclid's “Elements”) Similar triangles are to one another in the squared1 ratio of (their) corresponding sides. • Let $ABC$ and $DEF$ be similar triangles having the angle at $B$ equal to the (angle) at $E$, and $AB$ to $BC$, as $DE$ (is) to $EF$, such that $BC$ corresponds to $EF$. • I say that triangle $ABC$ has a squared ratio to triangle $DEF$ with respect to (that side) $BC$ (has) to $EF$. Modern Formulation With $a:=\|\overline{AB}\|,$ $b:=\|\overline{BC}\|,$ $\gamma=\angle{ABC},$ $a':=\|\overline{DE}\|,$ $b':=\|\overline{EF}\|,$ and $\gamma'=\angle{DEF},$ this proposition states that if in two triangles $\triangle{ABC}$ $\triangle{DEF},$ the angles are equal $\gamma=\gamma'$ and the corresponding sides being the legs of these angles are proportional $\frac ab=\frac {a'}{b'}$ then $$\frac{\operatorname{area}\triangle{ABC}}{\operatorname{area}\triangle{DEF}}=\frac {b^2}{b'^2}.$$ Modern Proof: $$\begin{array}{rcll} \frac {a}{b}&=&\frac {a'}{b'}&\|\cdot \frac {b^2}{a'b'}\\ \frac {ab}{a'b'}&=&\frac {b^2}{b'^2}\\ \frac {\frac 12ab\sin(\gamma)}{\frac 12a'b'\sin(\gamma')}&=&\frac {b^2}{b'^2}&(\text{since }\gamma=\gamma')\\ \frac{\operatorname{area}\triangle{ABC}}{\operatorname{area}\triangle{DEF}}&=&\frac {b^2}{b'^2} \end{array}$$ Proofs: 1 Corollaries: 1 Corollaries: 1 2 Proofs: 3 4 Thank you to the contributors under CC BY-SA 4.0! Github: Footnotes 1. Literally, "double" (translator's note).<\|endoftext\|>	4.59375
407	Researchers at Lund University in Sweden have taken an important step on the road to understanding the underlying mechanism of how and why animals can feel pain in connection with cold or heat. However, according to the study, temperature is just one triggering factor – horseradish, mustard, cinnamon and wasabi have a similar effect. A few years ago, the research group produced a human receptor protein and tested it in an artificial cell membrane. Similar studies have now been conducted on a receptor of the malaria mosquito, and the results are consistent. All the evidence suggest that the basic underlying mechanism of the temperature sensor function is the same in insects as in vertebrates. The new study shows that the inherent ion channel in the mosquito receptor is activated by heat. It also shows that the first part of the receptor can be removed without destroying ion channel or the ability to react to temperature. So far, potential applications are relatively far off into the future. However, the researchers do know the areas in which the findings may be of significance: “Different ways of preventing activation of the receptor protein may lead to new drugs and treatments for pain and itching. The substances that activate can instead be developed into effective treatments, designed to deter specific insects that carry various infectious agents”, says Urban Johanson, professor at the Department of Chemistry. In this new study, the researchers produced two versions of the pure protein: one complete protein and one without the first half. Subsequently, they inserted the protein into an artificial cell membrane and measured the flow through individual ion channels — both at different temperatures, and after adding substances which can be found in, for instance, wasabi and cinnamon. Using spectroscopy, they were able to monitor the structural changes in the ion channel. “There are similarities in the structural changes, regardless of whether the change is caused by heat or wasabi. The molecular mechanism of how this happens is not yet clear, so we will now proceed with a more detailed study”, says Urban Johanson.<\|endoftext\|>	3.796875
1,409	# Electric Current – definition, equation, types and direction Electricity is one of the most essential things in our daily life. To deal with an electrical circuit, we need to know the physical quantities associated with it. Current, Voltage, resistance, power, etc. are among those quantities. Many scientists have given various laws of electricity and explained how to determine such physical quantities in a given circuit. Ohm’s law gives a relation between voltage, current and resistance. Kirchhoff’s laws help us to find electric current and node voltage at a junction point of a circuit. In this article, we’re going to explain what is an electric current, its equation, types, conventional direction and other things related to it. 1. What is an electric current? 2. Equation of current 3. Units and dimension of current 4. What is the conventional direction of electric current? 5. How can we determine the direction of current flow in a circuit? 6. Types of electric current ## What is electric current? Subatomic particles like electrons, protons and ions have electric charges. When these particles move, we call it a current. In simple words, the current flow is the flow of charged particles. It is symbolized by the alphabet ‘I‘. The electric current is defined as the amount of electric charges flows per unit time. Do you know what causes the flow of electric charges? It’s the electric potential difference or the voltage of a battery. ## Properties of electric current • It’s a fundamental physical quantity in S.I. system. • Flow of charged particles is responsible for electric current. • Current is a flow of energy. It has waveforms. • The direction of current is opposite to the direction of electron flow. ## The equation for electric current flow If Q amount of charge flows through the cross-section of a conductor in a time t, then the formula of electric current in the conductor is $\small\color{Blue}I = \frac{Q}{t}$………….(1) The magnitude of the electric charge of an electron is e = 1.6 × 10-19 Coulombs. If N number of electrons flows for a duration t, then the amount of charge flow in that duration is Q = eN. Then the above formula can be written as $\small\color{Blue}I = \frac{eN}{t}$…………(2) There is another equation of electric current that relates it to voltage and resistance. Ohm’s law of electricity gives this equation. If a voltage V is applied across a conductor of resistance R, then the amount of current flow through the conductor will be $\small\color{Blue}I = \frac{V}{R}$…………(3) ## Unit and Dimension of Current Electric current is one of the fundamental physical quantities in SI system. It has independent units and dimensional formula. The SI unit is Ampere and CGS unit is esu per second. Its dimension is [I] or [M0L0T0I]. From equation-(1), one can say that Coulomb per second (C.s-1) is another unit of current. ## Definition of 1 Ampere current One can define an ampere in two ways by using equation-(1) and equation-(3). In equation-(3), if you put V = 1 volt and R = 1 ohm, then the amount of current will become I = 1 ampere. Therefore, The amount of current flow through a conductor of resistance 1 ohm when a voltage of 1 volt is applied across the conductor is said to be 1-ampere current. Again, In equation-(1), if you put Q = 1 coulomb and t = 1 second, then the amount of current will become I = 1 ampere. Then, If 1 coulomb of electric charge flows through the cross-section of a conductor in 1 second, then the amount of current through the conductor will be 1 ampere. ## What is the conventional direction of electric current? Current is not a vector quantity, it’s a scalar. Still, it has a direction. The conventional direction of current flow is along the direction of the motion of positive charges. In most cases, free electrons are responsible for current flow. Therefore scientists mentioned its direction by comparing the direction of electron flow. They said, Electric current flows in the opposite direction of the flow of free electrons. Sounds like we can determine its direction only if we know the direction of electrons. But it is impossible to identify the direction of electron flow because we cannot see electrons any way. Then what to do? Don’t worry, we have some other methods to find it. ## How to find electric current in a circuit? There are some devices that can measure the magnitude and direction of current in a circuit. Its magnitude can be calculated by using the appropriate equation. ### How to measure the amount of current flow in a circuit? • An Ammeter can be used to measure the amount of current through a resistance. • Multimeter is widely used to measure both AC and DC current. • If the values of Voltage and resistances are given then the current can be calculated by using V = IR formula. ### How to determine the direction of current flow in a circuit? • Electric current flows from the positive terminal of a cell or battery to its negative terminal. Hence observing the terminals of a battery one can identify the direction of current in a mesh. In above circuit, the positive terminal of the battery is on the left side and hence the current flows clockwise. But this method cannot be applied to a complicated circuit having more than two or three loops. • For a complicated circuit, ammeters or multimeters can be used to determine the direction of current flow in various branches of the circuit. In this case, you have to observe the deflection of the pointer of the device. ## What are the different types of electric currents? According to the waveforms, there are three types of electric currents – • Direct current (DC) • Alternating current (3) • Pulsating current Direct current does not alter its direction or polarity and provides a steady amount of current flow through a circuit if the voltage is not charged. Alternating current changes its polarity periodically and produces sinusoidal or rectangular waveforms. Pulsating current is neither a DC nor an AC, its magnitude changes like a pulse but its polarity remains the same. ## Effects of electric current Electric current has wide uses. We cannot spend a week without electricity. It is used as a source of light, to charge mobiles, laptops, electric cars, etc. But we need to be careful while dealing with electricity. It has a huge bad effect. The current is responsible for electric shock. A sufficient amount of current can kill someone. Again short circuit can burn a house, office and other properties.<\|endoftext\|>	4.40625
1,498	SEMATHS.ORG Updated: 11 August 2021 04:30:00 AM # What is equal to 1/3? Fractions equivalent to 1/3 are 2/6, 3/9, 4/12, 5/15, Fractions equivalent to 1/4 are 2/8, 3/12, 4/16, 5/20, Fractions equivalent to 1/5 are 2/10, 3/15, 4/20, 5/25, Fractions equivalent to 2/5 are 4/10, 6/15, 8/20, 10/25, ## In this regard, what is 1/3 in a number? 1/3 = 0.33333333 with 3 keep repeating. If you want to round it to the nearest whole number, it is 0. ## Consequently, what is the equivalent of one third? One third is equivalent to the fraction: 1/3. Therefore, it is a third of an amount. Thirds are calculated by dividing by 3. ## Subsequently, question is, which of the following fractions is not equivalent to 1/3 *? #### How do you express 3/5 as a percentage? Convert 3/5 to a percentage. Now we have the answer: 60 %. #### How do you find the equivalent fraction of 1 3? You can multiply the numerator and denominator of 13 by any number and the resulting fraction would be equal to 13 so long as you are multiplying both by the same number. and so on and so forth. 13=26=39=412=515=927=1442=1751=50150.. #### What number is 40% of 80? Percentage Calculator: What is 40 percent of 80? = 32. #### What is 2 over 3 as a decimal? How to Write 3/2 as a Decimal? FractionDecimalPercentage 3/21.5150% 3/31100% 3/40.7575% 3/50.660% #### What is 3/5 as a percentage? To convert 3/5 to fraction Therefore, the solution is 60%. #### What is 3 2 as a percentage? Convert fraction (ratio) 3 / 2 Answer: 150% #### What is 1 and one third as a decimal? 1/3 in decimal form is 0.3333 (repeating). #### What is the simplified form of 7 3? 73 is already in the simplest form. It can be written as 2.333333 in decimal form (rounded to 6 decimal places). #### What number is 80 percent of 65? Percentage Calculator: What is 80 percent of 65? = 52. #### What is 3 and 3/4 as a decimal? Fraction to decimal conversion table FractionDecimal 3/40.75 1/50.2 2/50.4 3/50.6 #### What is 1/3 as a fraction? Example Values PercentDecimalFraction 12½%0.1251/8 20%0.21/5 25%0.251/4 331/3%0.3331/3 #### What is 7 3 as a decimal? 2.3333 is a decimal and 233.33/100 or 233.33% is the percentage for 7/3. How to Write 7/3 as a Decimal? FractionDecimalPercentage 7/32.3333233.33% 6/32200% 7/23.5350% 7/41.75175% #### How do you change one-third into a decimal? One-third means ONE divided into THREE parts. So, to find the decimal equivalent, do the division. 1 divided by 3 = 0.33333333 etc. #### What number is 10% of 80? Percentage Calculator: What is 10 percent of 80? = 8. #### What decimal is the same as 3 4? Fraction to Decimal Conversion Tables fraction = decimal 1/2 = 0.5 1/3 = 0.32/3 = 0.6 1/4 = 0.253/4 = 0.75 1/5 = 0.22/5 = 0.44/5 = 0.8 #### What number is 20% of 80? Percentage Calculator: What is 20 percent of 80? = 16. #### What is 3/8 as a percentage? Common Fractions with Decimal and Percent Equivalents FractionDecimalPercent 1/80.12512.5% 3/80.37537.5% 5/80.62562.5% 7/80.87587.5% #### What is 1 and 3/4 as a decimal? Explanation: To turn this into a decimal, divide 3 by 4 . Since it is −134 , the decimal is −1.75 . #### What is meant by 3 4? Example: 3/4 means: We have 3 parts. Each part is a quarter (1/4) of a whole. #### What is the fraction 3/4 equivalent to? Decimal and Fraction Conversion FractionEquivalent Fractions 2/34/68/12 1/42/84/16 3/46/812/16 1/52/104/20 #### What is the 5th equivalent fraction to 1 3? For example, the equivalent fraction of 1/3 is 5/15, because if we simplify 5/15, then the resulted fraction is the same. #### What is the fraction 2/3 equivalent to? Equivalent fraction of a given fraction is got by multiplying or dividing its numerator and denominator by the same whole number. For example, if we multiply the numerator and denominator of 2/3 by 4 we get. 2/3 = 2×4 / 3×4 = 8/12 which is an equivalent fraction of 2/3. #### What is 7 and 3/4 as a decimal? Explanation: This is a mixed number. 734 is between 7 and 8. 34 as a decimal is 0.75, so we just add this onto 7 to get 7.75. #### What number is 25% of 80? Percentage Calculator: What is 25 percent of 80? = 20. #### What number is 80% of 60? Percentage Calculator: What is 80 percent of 60? = 48. #### Can 3 2 be simplified? As you can see the fraction 3/2 can be written as 1 ½. These numbers are both the same value, but sometimes the answer will be need to be written as a mixed number to be considered fully reduced or simplified. To convert an improper fraction to a mixed number, follow these steps: Divide the numerator by the denominator.<\|endoftext\|>	4.375
673	LCM of 8 and also 14 is the smallest number amongst all usual multiples that 8 and also 14. The first few multiples that 8 and also 14 room (8, 16, 24, 32, 40, 48, . . . ) and (14, 28, 42, 56, 70, . . . ) respectively. There space 3 generally used techniques to find LCM that 8 and also 14 - by department method, through listing multiples, and by prime factorization. You are watching: Lowest common multiple of 8 and 14 1 LCM that 8 and also 14 2 List of Methods 3 Solved Examples 4 FAQs Answer: LCM of 8 and also 14 is 56. Explanation: The LCM of two non-zero integers, x(8) and y(14), is the smallest confident integer m(56) the is divisible by both x(8) and also y(14) without any type of remainder. The approaches to discover the LCM that 8 and also 14 are explained below. By department MethodBy Listing MultiplesBy element Factorization Method ### LCM that 8 and 14 by division Method To calculation the LCM the 8 and 14 by the department method, we will certainly divide the numbers(8, 14) by their prime determinants (preferably common). The product of these divisors offers the LCM that 8 and 14. Step 3: proceed the measures until only 1s are left in the last row. The LCM that 8 and also 14 is the product of all prime numbers on the left, i.e. LCM(8, 14) by department method = 2 × 2 × 2 × 7 = 56. ### LCM the 8 and also 14 by Listing Multiples To calculation the LCM of 8 and 14 by listing the end the typical multiples, we deserve to follow the given listed below steps: Step 1: perform a few multiples the 8 (8, 16, 24, 32, 40, 48, . . . ) and 14 (14, 28, 42, 56, 70, . . . . )Step 2: The common multiples from the multiples that 8 and also 14 are 56, 112, . . .Step 3: The smallest typical multiple that 8 and also 14 is 56. ∴ The least common multiple of 8 and 14 = 56. See more: What Two Items Must Be Equal For A Nuclear Equation To Be Balanced ? ### LCM of 8 and 14 by element Factorization Prime factorization of 8 and 14 is (2 × 2 × 2) = 23 and also (2 × 7) = 21 × 71 respectively. LCM the 8 and 14 deserve to be acquired by multiplying prime components raised to your respective highest power, i.e. 23 × 71 = 56.Hence, the LCM of 8 and 14 by prime factorization is 56.<\|endoftext\|>	4.71875
594	The U.S is a diverse nation- it has been shaped by many different cultures, ethnicities and religions. - In what ways were pre-Columbian Native American cultures distinct and similar to European cultures in the use of land and resources? - What are some of the ways Native American myths differ from the realities of Native American life, both in the past and in the present? - How can a study of Native cultures inform our lives today? Native American Instruction: A copy of the listening guide (available in the workbook) and a copy of the instructional PowerPoint used to learn about Native Americans can be accessed below: Native American Origins Homework: Due Sept 21st (B Day) 22nd (A Day) can be accessed from the link below: Native American Stories (shared at the beginning of each class period) can be accessed below: Native American Culture & Music Native American WalkAway Review & Remediation Native American Stereotypes: Stereotypes vs. Realities Day 1 This in class lecture was completed 9/18. Students who missed class can access the audio lecture on the page linked below (they will need to scroll 1/2 way down the page to access the audio presentation. Students should use the information within to complete the pages indicated during the lecture within their workbook. One of the main influences on the framers of the Constitution was the unwritten democratic constitution under which the Iroquois Confederacy had operated since the 16th century, according to a group of American Indians and scholars... For more information read the article below..... Web Studies: Native Americans Use the websites below to study Native Cultures. When finished write a list of 10 things you learned about Native Culture from the websites. Native Americans Before Columbus Published on Apr 8, 2013 Upon the arrival of Columbus in 1492 in the Carabean Islands, unknown to Columbus (and majority of the Eastern Hemisphere), he landed on Islands located in the middle of two huge continents now known has North America and South America that was teaming with huge Civilizations (that rivaled any in the world at that time) and thousands of smaller Nations and Tribes. With recent estimations, the population may have been over 100 million people that spanned from Alaska and Green Land, all the to the tip of southern South America. Please note the YouTube video posted below will not show in the classroom. A class room copy follows the YouTube video. Native Americans Before European Colonization for viewing in the classroom. Eastern Woodland Native Cultures Plains & California Native Cultures Plains: How did the horse change the life of Plains Indians? Explain. California: What were two characteristics of life within the Chumash Nation? How did their culture change with the coming of “Missions”? Explain. Northwest Natives: Additional Notes on Native Culture<\|endoftext\|>	4.1875
8,020	An Instruction Set Architecture (ISA) is part of the abstract model of a computer. It defines how software controls the processor. The Arm ISA allows you to write software and firmware that conforms to the Arm specifications. This mean that, if your software or firmware conforms to the specifications, any Arm-based processor will execute it in the same way. This guide introduces the A64 instruction set, used in the 64-bit Armv8-A architecture, also known as AArch64. We will not cover every single instruction in this guide. All instructions are detailed in the Arm Architecture Reference Manual (Arm ARM). Instead, we will introduce the format of the instructions, the different types of instruction, and how code written in assembler can interact with compiler-generated code. At the end of this guide, you can check your knowledge. You will have learned about the main classes of instructions, the syntax of data-processing instructions, and how the use of X registers affects instructions. The key outcome that we hope you will learn from this guide is to be able to explain how generated assembler code maps to C statements, when given a C program and the compiler output for it. Finally, this guide will show you how to write a function in assembler that can be called from C. Why you should care about the ISA As developers, you may not need to write directly in assembler in our day-to-day role. However, assembler is still important in some areas, such as the first stage boot software or some low-level kernel activities. Even if you are not writing assembly code directly, understanding what the instruction set can do, and how the compiler makes use of those instructions, can help you to write more efficient code. It can also help you to understand the output of the compiler. This can be useful when debugging. Instruction sets in the Armv8-A Armv8-A supports three instruction sets: A32, T32 and A64. The A64 instruction set is used when executing in the AArch64 Execution state. It is a fixed-length 32-bit instruction set. The ‘64’ in the name refers to the use of this instruction by the AArch64 Execution state. It does not refer to the size of the instructions in memory. The A32 and T32 instruction sets are also referred to as ‘ARM’ and ‘Thumb', respectively. These instruction sets are used when executing in the AArch32 Execution state. In this guide, we do not cover the A32 and T32 instruction sets. To find out more about these instruction sets, see the Related Information section of this guide. Instruction set resources Each version of the Arm architecture has its own Arm Architecture Reference Manual (Arm ARM), which can be found on the Arm Developer website. Every Arm ARM provides a detailed description of each instruction, including: - Encoding - the representation of the instruction in memory. - Arguments - inputs to the instruction. - Pseudocode - what the instruction does, as expressed in Arm pseudocode language. - Restrictions - when the instruction cannot be used, or the exceptions it can trigger. The instruction descriptions for A64 are also available in XML and HTML. The XML and HTML formats are useful if you need to refer to the instructions often. The XML and HTML formats can be found on the Arm Developer website. You can find a link in the Related information section of this guide. The XML can be downloaded as a compressed archive and the HTML can be viewed and searched using a web browser. Note: The information in the XML/HTML and the Arm ARM are taken from the same source but may be formatted slightly differently. Simple sequential execution The Arm architecture describes instructions following a Simple Sequential Execution (SSE) model. This means that the processor behaves as if the processor fetched, decoded and executed one instruction at a time, and in the order in which the instructions appeared in memory. In practice, modern processors have pipelines that can execute multiple instructions at once, and may do so out of order. This diagram shows an example pipeline for an Arm Cortex processor: You will remember that the architecture is a functional description. This means that it does not specify how an individual processor works. Each processor must behave consistently with the simple sequential execution model, even if it is re-ordering instructions internally. Most A64 instructions operate on registers. The architecture provides 31 general purpose registers. Each register can be used as a 64-bit X register ( X0..X30), or as a 32-bit W register ( W0..W30). These are two separate ways of looking at the same register. For example, this register diagram shows that W0 is the bottom 32 bits of W1 is the bottom 32 bits of For data processing instructions, the choice of W determines the size of the operation. Using X registers will result in 64-bit calculations, and using W registers will result in 32-bit calculations. This example performs a 32-bit integer addition: ADD W0, W1, W2 This example performs a 64-bit integer addition: ADD X0, X1, X2 W register is written, as seen in the example above, the top 32 bits of the 64-bit register are zeroed. There is a separate set of 32 registers used for floating point and vector operations. These registers are 128-bit, but like the general-purpose registers, can be accessed in several ways. Bx is 8 bits, Hx is 16 bits and so on to Qx which is 128 bits. The name you use for the register determines the size of the calculation. This example performs a 32-bit floating point addition: FADD S0, S1, S2 This example performs a 64-bit floating point addition: FADD D0, D1, D2 These registers can also be referred to as V registers. When the V form is used, the register is treated as being a vector. This means that it is treated as though it contains multiple independent values, instead of a single value. This example performs vector floating point addition: FADD V0.2D, V1.2D, V2.2D This example performs vector integer addition: ADD V0.2D, V1.2D, V2.2D We will look at vector instructions in more detail later in this guide. Here are some other registers in the A64 that you should know about: The zero registers, WZR, always read as 0 and ignore writes. You can use the stack pointer ( SP) as the base address for loads and stores. You can also use the stack pointer with a limited set of data-processing instructions, but it is not a regular general purpose register. Armv8-A has multiple stack pointers, and each one is associated with a specific Exception level. When SP is used in an instruction, it means the current stack pointer. The guide to the exception model explains how the stack pointer is selected. X30 is used as the Link Register and can be referred to as Note: Separate registers, ELR_ELx, are used for returning from exceptions. This is discussed in more detail in the guide to the exception model. The Program Counter ( PC) is not a general-purpose register in A64, and it cannot be used with data processing instructions. The PC can be read using: ADR Xd, . ADR instruction returns the address of a label, calculated based on the current location. Dot (‘.’) means 'here', so the shown instruction is returning the address of itself. This is equivalent to reading the PC. Some branch instructions, and some load/store operations, implicitly use the value of the Note: In the A32 and T32 instruction sets, the SP are general purpose registers. This is not the case in A64 instruction set. As well as general purpose registers, the architecture defines system registers. System registers are used to configure the processor and to control systems such as the MMU and exception handling. System registers cannot be used directly by data processing or load/store instructions. Instead, the contents of a system register need to be read into an X register, operated on, and then written back to the system register. There are two specialist instructions for accessing system registers: MRS Xd, <system register> reads the system register into MSR <system register> Xn to the system register. System registers are specified by name, for example MRS X0, SCTLR_EL1 System register names end with _ELx specifies the minimum privilege necessary to access the register. For example: requires EL1 or higher privilege. requires EL2 or higher privilege. requires EL3 privilege Attempting to access the register with insufficient privilege results in an exception. Note: Sometimes you will see _EL01. These are used as part of virtualization. Refer to the guide on virtualization for more information. Arithmetic and logic operations The basic format of logical and integer arithmetic instructions is: The parts of the instruction are as follows: - Operation. This defines what the instruction does. For example, ADDdoes addition and ANDperforms a logical AND. Scan be added to the operation to set flags. For example, stells the processor to update the ALUflags based on the result of instruction. We discuss ALUflags in the section on generating condition code. - Destination: The destination of the instruction is always a register, and specifies where the result of the operation is placed. Most instructions have a single destination register. A few instructions have two destination registers. When the destination is a Wregister, the upper 32 bits of the corresponding Xregister are set to 0. - Operand 1: This will always be a register. This is the first input to the instruction. - Operand 2: This will be a register or a constant, and is the second input to the instruction. When operand 2 is a register, it may include an optional shift. When operand 2 is a constant, it is encoded within the instruction itself. This means that the range of constants available is limited. You should be aware of a couple of special cases, such as the MOV moves a constant, or the contents of another register, into the register specified as the destination. MVN only require a single input operand, which can be either a register or a constant, as shown here: MOV X0, #1 X0 = 1 MVN W0, W1 W0 = ~W1 Floating-point operations follow the same format as integer data-processing instructions and use floating-point registers. Like with the integer data-processing instructions, the size of the operation determines the size of the register that is used. The operation part of a floating-point instruction always starts with an F. For example, this instruction sets H0 = H1 / H2 with half precision: FDIV H0, H1, H2 This instruction sets S0 = S1 + S2 with single precision: FADD S0, S1, S2 This instruction sets D0 = D1 - D2 with double precision: FSUB D0, D1, D2 Support for half precision (16 bit) was added in Armv8.2-A and is optional. Support for half-precision is reported by ID_AA64PFR0_EL1. Access to floating point registers can be trapped. This means that any attempt to use floating point registers will generate an exception. Trapping is discussed in more detail the exception model guide. Is floating point support optional? No. Support for floating point is mandatory in Armv8-A. The architecture specifies that it is required whenever a rich operating system, such as Linux, is used. You are technically permitted to omit floating point support, if you are running an entirely proprietary software stack. Most toolchains, including GCC and Arm Compiler 6, will assume floating point support. There are a set of instructions for manipulating bits within a register. This figure shows some examples: BFI instruction inserts a bit field into a register. In the preceding figure, BFI is taking a 6-bit field from the source register ( W0) and inserting it at bit position 9 in the destination register. UBFX extracts a bit field. In the preceding figure, UBFX is taking a 7-bit field from bit position 18 in the source register, and placing it in the destination register. Other instructions can reverse byte or bit order, as you can see in this figure: RBIT are particularly useful when you are handling data that is in a different endianness. Extension and saturation Sometimes it is necessary to convert data from one size to another. The SXTx (sign extend) and UXTx (unsign extend) instructions are available for this conversion. In this conversion, the x determines the size of the data being extended, as shown in this figure: In the first instruction, B means byte. It takes the bottom byte of W0 and sign extends it to 32 bits. UXTH is an unsigned extension of a halfword ( H). It takes the bottom 16 bits of W1 and zero extends it to 32 bits. The first two examples have W registers as a destination, meaning the extension is to 32 bits. The third example has an X register, meaning the sign extension is to 64 bits. Sub-register-sized integer data processing Some instructions perform saturating arithmetic. This means that if the result is larger or smaller than the destination can hold, then the result is set to the largest or smallest value of the destination's integer range. The data-processing instructions can efficiently handle 32-bit data and 64-bit data. In practice, you often see saturation instructions when handling sub-register calculations. Sub-register calculations are calculations of 16 bits or 8 bits. This table shows some examples of sub-register calculations in C and the generated assembler code: In the first example in the table, the 32-bit addition maps onto W registers and therefore can be handled easily. For the 16-bit examples in the table, an extra instruction is necessary. The third example in the table takes the 16-bit inputs, extends them to 32 bits, and then performs the addition. The sequence converts the 16-bit input to 32 bits, using: Then, this instruction performs the addition and saturates the result to signed 16 bits: , SXTH to the end of the operand list of the ADD operation causes the result to use saturating arithmetic. Because the destination is a W register, the ADD will saturate to a 16-bit integer range. We have seen that the MVN instructions copy the value from one register to another. Similarly, FMOV can be used to copy between floating-point and general purpose registers. FMOV copies the literal bit pattern between the registers. There are also instructions that can convert to the closest representation, as this figure shows: In this example, imagine that X0 contains the value 2 (positive integer 2): X0 = 0x0000_0000_0000_0002 Then, the following sequence is executed: FMOV D0, X0 SCVTF D1, X0 Both instructions “copy” X0 into a D register. However, the results are quite different: D0 = 0x0000_0000_0000_0002 = 9.88131e-324 D1 = 0x4000_0000_0000_0002 = 2.0 FMOV copied the literal bit pattern, which is a very different value when interpreted as a floating-point value. The SCVTF converted the value in X0 to the closest equivalent in floating-point. FCTxx can be used to convert a floating-point value to its closest integer representation. In this instance, different values of ' xx' control the rounding mode used. The A64 architecture also provides support for vector data processing. The two types of vector processing available are: Advanced SIMD, which is also known as NEON. Scalable Vector Extension, which is abbreviated to SVE. We will cover both types of vector processing in a later guide on vector programming. Note: The name Advanced SIMD derives from the existence of SIMD instructions that operated on regular 32-bit general-purpose registers in Armv6. In Armv7, the term Advanced SIMD was used for instructions that could operate on 128-bit vectors. The Armv6 style instructions do not exist in A64, but the naming convention remains. Loads and stores The basic load and store operations are: LDR (load) and STR (store). These operations transfer a single value between memory and the general-purpose registers. The syntax for these instructions is: LDR<Sign><Size> <Destination>, [<address>] STR<Size> <Destination>, [<address>] The size of the load or store is determined by the register type W and the <Size> field. X is used for 32 bits and W is used for 64 bits. For example, this instruction loads 32 bits from <address> into LDR W0, [<address>] This instruction loads 64 bits from <address> into LDR X0, [<address>] The <Size> field allows you to load a sub-register sized quantity of data. For example, this instruction stores the bottom byte ( W0 to <address>: STRB W0, [<address>] This instruction stores the bottom halfword ( W0 to <address>: STRH W0, [<address>] Finally, this instruction stores the bottom word ( X0 to <address>: STRW X0, [<address>] Zero and sign extension By default, when a sub-register-sized quantity of data is loaded, the rest of the register is zeroed, as shown in this figure: Note: Remember that whenever a W register is written, the top half of the X register is zeroed. Adding an S to the operation causes the value to be sign extended instead. How far the size extension goes depends on whether the target is a X register, as shown in this figure: The addresses for load and store instructions appear within the square brackets, as shown in this example: LDR W0, [X1] There are several addressing modes that define how the address is formed. - Base register - The simplest form of addressing is a single register. Base register is an Xregister that contains the full, or absolute, virtual address of the data being accessed, as you can see in this figure: - Offset addressing modes - An offset can be applied optionally to the base address, as you can see in this figure: In the preceding figure, X1 contains the base address and #12 is a byte offset from that address. This means that the accessed address is X1+12. The offset can be either a constant or another register. This type of addressing might be used for structs, for example. The compiler maintains a pointer to the base of struct using the offset to select different members. - Pre-index addressing modes - In the instruction syntax, pre-indexing is shown by adding an exclamation mark ! after the square brackets,as this figure shows: Pre-indexed addressing is like offset addressing, except that the base pointer is updated as a result of the instruction. In the preceding figure, X1 would have the value X1+12 after the instruction has completed. - Post-index addressing modes - With post-index addressing, the value is loaded from the address in the base pointer, and then the pointer is updated, as this figure shows: Post-index addressing is useful for popping off the stack. The instruction loads the value from the location pointed at by the stack pointer, and then moves the stack pointer on to the next full location in the stack. Load pair and store pair So far, we have discussed the load and store of a single register. A64 also has load ( LDP) and store pair ( These pair instructions transfer two registers to and from memory. The first instruction loads W3, and loads [X0 + 4] into LDP W3, W7, [X0] This second instruction stores [X4] and stores [X4 + 8]: STP D0, D1, [X4] Load and store pair instructions are often used for pushing, and popping off the stack. This first instruction pushes X1 onto the stack: STP X0, X1, [SP, #-16]! This second instruction pops X1 from the stack: LDP X0, X1, [SP], #16 Remember that in AArch64 the stack-pointer must be 128-bit aligned. Using floating point registers Loads and stores can also be carried out using the floating-point registers, as we will see here. The first instruction loads 64-bits from LDR D1, [X0] This second instruction stores 128-bits from [X0 + X1]: STR Q0, [X0, X1] Finally, this instruction loads a pair of 128-bit values from X5, then increments X5 by 256: LDP Q1, Q3, [X5], #256 There are some restrictions: - The size is specified by the register type only. - There is no option to sign extend loads. - The address must still be an Load and stores using floating-point registers can be found in unexpected cases. It is common for memcpy() type routines to use them. This is because the wider register means that fewer iterations are needed. Just because your code does not use floating-point values, don't assume that you won't need to use the floating-point registers. Ordinarily, a processor executes instructions in program order. This means that a processor executes instructions in the same order that they are set in memory. One way to change this order is to use branch instructions. Branch instructions change the program flow and are used for loops, decisions and function calls. The A64 instruction set also has some conditional branch instructions. These are instructions that change the way they execute, based on the results of previous instructions. Note: Armv8.3-A and Armv8.5-A introduced instructions to protect against return-orientated programming and jump-oriented programming. Loops and decisions In this section, we will examine how loops and decisions let you change the flow of your program code using branch instructions. There are two types of branch instructions: unconditional and conditional. Unconditional branch instructions The unconditional branch instruction B <label> performs a direct, PC-relative, branch to <label>. The offset from the current PC to the destination is encoded within the instruction. The range is limited by the space available within the instruction to record the offset and is +/-128MB. When you use BR <Xn>, BR performs an indirect, or absolute, branch to the address specified in Conditional branch instructions The conditional branch instruction B.<cond> <label> is the conditional version of the B instruction. The branch is only taken if the condition <cond> is true. The range is limited to +/-1MB. The condition is tested against the ALU flags stored in PSTATE and needs to be generated by a previous instruction such as a compare ( CBZ <Xn> <label> and CBNZ <Xn> <label> This instruction branches to Xn contains 0 ( CBZ), and branches to label if Xn does not contain 0 ( TBZ <Xn>, #<imm>, <label> and TBNZ <Xn>, #<imm>, <label> TBX works in a similar way to CBNZ, but tests the bit specified by Note: The direct, or PC-relative, branches store the offset to the destination within the instruction. The conditional branches have a smaller range. This is because some bits are needed to store the condition itself, which leaves fewer bits for the offset. Mapping these on to what you might write in C, this table shows examples of using branches for loops and decisions: \|C\|\|Typical output from a compiler\| Note: The labels shown in the output would not be created by a compiler. They are included here to aid readability. Generating condition code In Program flow - loops and decisions, we learned that the <cond> is tested against the ALU flags stored in PSTATE. The ALU flags are set as a side effect of data-processing instructions. To recap, an S at the end operation causes the ALU flags to be updated. This is an example of an instruction in which the ALU flags are not updated: ADD X0, X1, X2 This is an example of an instruction in which the ALU flags are updated with the use of ADDS X0, X1, X2 This method allows software to control when the flags are updated or not updated. The flags can be used by subsequent condition instructions. Let's take the following code as an example: SUBS X0, X5, #1 AND X1, X7, X9 SUBS instruction performs a subtract and updates the ALU flags. Then the AND instruction performs an and operation, and does not update the ALU flags. Finally, the B.EQ instruction performs a conditional branch, using flags set as result of the subtract. The flags are: - N - Negative - C - Carry - V - Overflow - Z - Zero Let’s take the Z flag as an example. If the result of the operation was zero, then the Z flag is set to 1. For example, here the Z flag will be set if X5 is 1, otherwise it will be cleared: SUBS X0, X5, #1 The condition codes map on to these flags and come in pairs. Let's take EQ (equal) and NE (not equal) as an example, and see how they map to the Z flag: The EQ code checks for Z==1. The NE code checks for Taking the following code as an example: SUBS W0, W7, W9 W0 = W7 - W9, In the first line, we have a subtract operation. In the second line, the flag is set if the result is zero. In the final line, there is a branch to label if 7==w9, the result of the subtraction will be zero, and the Z flag would have been set. Therefore, the branch to label will be taken if w9 are equal. In addition to the regular data-processing instructions, other instructions are available that only update the ALU flags: CMP - Compare TST - Test These instructions are aliases of other instructions. For example: CMP X0, X7 //an alias of SUBS XZR, X0, X7 TST W5, #1 //an alias of ANDS WZR, W5, #1 By using the Zero register as a destination, we are discarding the result of the operation and only updating the ALU flags. Conditional select instructions So far, we have seen examples that use branches to handle decisions. The A64 instruction set also provides conditional select instructions. In many cases, these instructions can be used as an alternative to branches. There are many variants, but the basic form is: CSEL Xd, Xn, Xm, cond This means that: if cond then Xd = Xn Xd = Xm You can see an example in this code: CMP W1, #0 CSEL W5, W6, W7, EQ Which gives the same result as: if (W1==0) then W5 = W6 W5 = W7 There are variants that combine another operation with the conditional select. For example, CSINC performs a select and addition: CSINC Xd, Xn, Xm, cond This means that: if cond then Xd = Xn Xd = Xm + 1 To just conditionally increment, you could write: CSINC X0, X0, X0, cond Which equates to: if cond then X0 = X0 X0 = X0 + 1 However, the architecture provides an alias, CINC, which is the equivalent to this command. Compilers choose the most efficient method to implement the functionality in your program. Compilers will often use a conditional select for small if ... else statements performing simple operations, because conditional selects can be more efficient than branches. Here are some simple if ... else examples that compare implementations using branches to equivalent implementations using conditional select instructions: In these types of examples, conditional selects have some advantages. The sequences are shorter and take the same number of instructions, regardless of the outcome. Importantly, conditional selects also remove the need to branch. In modern processors, this kind of branch can be difficult for the branch prediction logic to predict correctly. A mispredicted branch can have a negative effect on performance, it is better that you remove branches where possible. When calling a function or sub-routine, we need a way to get back to the caller when finished. Adding an L to the BR instructions turns them into a branch with link. This means that a return address is written into X30) as part of the branch. Note: The names X30 are interchangeable. An assembler, such as GNU GAS or armclang, will accept both. There is a specialist function return instruction, RET. This performs an indirect branch to the address in the link register. Together, this means that we get: Note: The figure shows the function foo() written in GAS syntax assembler. The keyword .global exports the symbol and .type indicates that the exported symbol is a function. Why do we need a special function return instruction? Functionally, BR LR would do the same job as RET tells the processor that this is a function return. Most modern processors, and all Cortex-A processors, support branch prediction. Knowing that this is a function return allows processors to more accurately predict the branch. Branch predictors guess the direction the program flow will take across branches. The guess is used to decide what to load into a pipeline with instructions waiting to be processed. If the branch predictor guesses correctly, the pipeline has the correct instructions and the processor does not have to wait for instructions to be loaded from memory. Procedure Call Standard The Arm architecture places few restrictions on how general purpose registers are used. To recap, integer registers and floating-point registers are general purpose registers.However, if you want your code to interact with code that is written by someone else, or with code that is produced by a compiler, then you need to agree rules for register usage. For the Arm architecture, these rules are called the Procedure Call Standard, or PCS. The PCS specifies: - Which registers are used to pass arguments into the function. - Which registers are used to return a value to the function doing the calling, known as the caller. - Which registers the function being called, which is known as the callee, can corrupt. - Which registers the callee cannot corrupt. Consider a function foo(), being called from The PCS says that the first argument is passed in X0, the second argument in X1, and so on up to X7. Any further arguments are passed on the stack. Our function, foo(), takes two arguments: a will be in b will be in W and not X? Because the arguments are a 32-bit type, and therefore we only need a Note: In C++, X0 is used to pass the implicit this pointer that points to the called function. Next, the PCS defines which registers can be corrupted, and which registers cannot be corrupted. If a register can be corrupted, then the called function can overwrite without needing to restore, as this table of PCS register rules shows: Parameter and Result Registers For example, the function foo() can use registers X15 without needing to preserve their values. However, if foo() wants to use X28 it must save them to stack first, and then restore from the stack before returning. Some registers have special significance in the PCS: XR- This is an indirect result register. If foo() returned a struct, then the memory for struct would be allocated by the caller, main() in the earlier example. XRis a pointer to the memory allocated by the caller for returning the struct. IP0 and IP1- These registers are intra-procedure-call corruptible registers. These registers can be corrupted between the time that the function is called and the time that it arrives at the first instruction in the function. These registers are used by linkers to insert veneers between the caller and callee. Veneers are small pieces of code. The most common example is for branch range extension. The branch instruction in A64 has a limited range. If the target is beyond that range, then the linker needs to generate a veneer to extend the range of the branch. FP- Frame pointer. X30is the link register ( LR) for function calls. Note: We previously introduced the ALU flags, which are used for conditional branches and conditional selects. The PCS says that the ALU flags do not need to be preserved across a function call. There is a similar set of rules for the floating-point registers: Parameter and Result Registers Sometimes it is necessary for software to request a function from a more privileged entity. This might happen when, for example, an application requests that the OS opens a file. In A64, there are special instructions for making such system calls. These instructions cause an exception, which allows controlled entry into a more privileged Exception level. SVC- Supervisor call Causes an exception targeting EL1. Used by an application to call the OS. HVC- Hypervisor call Causes an exception targeting EL2. Used by an OS to call the hypervisor, not available at EL0. SMC- Secure monitor call Causes an exception targeting EL3. Used by an OS or hypervisor to call the EL3 firmware, not available at EL0. If an exception is executed from an Exception level higher than the target exception level, then the exception is taken to the current Exception level. This means that an SVC at EL2 would cause exception entry to EL2. Similarly, an HVC at EL3 causes exception entry to EL3. This is consistent with the rule that an exception can never cause the processor to lose privilege. Here are some resources related to material in this guide: - Vectors: Neon guide (coming soon). - Vectors: SVE guide (coming soon). - Building an embedded image guide (coming soon). - Arm architecture and reference manuals (for information on the extensions to Armv8.3-A and Armv8.5-A instruction sets, vector data-processing instructions, Advance SIMD and SVE) - Arm community (ask development questions, and find articles and blogs on specific topics from Arm experts) - ARM Assembly Language by William Hohl (ISBN: 978-1-4398-06104) - Building an ELF Image for an Armv8-A Fixed Virtual Platform (blog post) - Changing Exception Level and Security State with an Armv8-A Fixed Virtual Platform (blog post) - Cortex-A Programmer's Guide - Retargeting and Enabling Exceptions with an ELF Image (blog post) Here are some resources related to topics in this guide: Instruction set resources Procedure Call Standard Using the Arm Instruction Set Architecture (ISA), you can write software or firmware that any Arm-based processor will execute in the same way, if that software or firmware conforms to the Arm specifications. In this guide, we introduced the A64 instruction set, which is used in Armv8-A AArch64. We introduced the format of the instructions, the different types of instruction, and how code written in assembler can interact with compiler-generated code. We explained the main classes of instructions, the syntax of data-processing instructions, and how the use of X registers affects instructions. Based on the material learned in this guide, you can explain how generated assembler code maps to C statements when given a C program and compiler output, and how to write a function in assembler that can be called from C. You will also understand how to find detailed descriptions for each instruction on the Arm Developer website, and concepts such as registers, data processing, program flow, and loads and stores. To keep learning about the Armv8-A architecture, see more in our series of guides. To check your knowledge of A64 assembler, try the ISA lab exercises (coming soon). The lab exercises require the Arm DS-5, Ultimate Edition. A 30-day evaluation version is available and can be used to complete the exercises.<\|endoftext\|>	4.0625
869	European researchers are developing a miniaturized instrument that could precisely measure carbon dioxide coming from cities and power plants. If it works, the device could fly aboard a constellation of small satellites starting in the late 2020s, helping to track daily fluctuations in greenhouse-gas emissions. Developers with the three-year, €3-million (US$3.5-million) project envision it complementing more expansive efforts to monitor CO2 from space, such as a proposed set of new Sentinel Earth-observing satellitesfrom the European Space Agency. If approved, those might also come online in the late 2020s. Several satellites currently monitor CO2 emissions, including Japan’s GOSAT, the United States’ Orbiting Carbon Observatory-2 (OCO-2), and China’s TanSat. But none of them launched with the explicit goal of tracking compliance with global treaties. In 2015, before the signing of the Paris accord to limit greenhouse-gas emissions, the European Commission began exploring how it could develop satellites to assess whether nations are abiding by their climate pledges. The new small sensor could play a part in that. “We want to improve the accuracy of monitoring anthropogenic CO2 emissions,” says Laure Brooker Lizon-Tati, an engineer with Airbus Defence and Space in Toulouse, France. She coordinates the project, called the Space Carbon Observatory (SCARBO), which is being developed by a consortium of eight European companies and research institutions. Team scientists will describe first results at a space-optics conference in Chania, Greece, on 10 and 11 October. The proposed Sentinel satellites would precisely measure greenhouse gases around the world. But they would not be able to make daily measurements above places of interest, such as cities. “This is where a constellation of tiny SCARBO systems could come into the game,” says Heinrich Bovensmann, a remote-sensing researcher at the University of Bremen in Germany. SCARBO satellites would weigh just 50 kilograms each, roughly one-tenth the mass of OCO-2 or TanSat. An estimated two dozen working together would be able to cover the globe once a week, but could fly over particular areas of interest once a day. Together they could monitor frequent changes in carbon emissions, such as morning and afternoon surges from an industrial area. But first, SCARBO scientists have to show that their plan can work. At its heart is a miniaturized spectrometer — no longer than an outstretched hand — that would detect CO2 concentrations in the air below. Fitting a spectrometer onto a small satellite requires shrinking optics and developing new methods for analysing carbon dioxide concentrations. “It’s a real challenge,” says Bovensmann. The scientists’ goal is to measure CO2 concentrations to an accuracy of less than 1 part per million at a resolution of 2 kilometres — comparable to the data collected by larger satellites now in orbit. “We want to prove the technology can achieve these types of measurements,” says Etienne Le Coarer of the University Grenoble-Alpes in Grenoble, France, which is building the instrument along with the ONERA French aerospace laboratory in Palaiseau. NASA’s Jet Propulsion Laboratory in Pasadena, California, has worked on a similar concept for miniaturized sensors, but using a different type of spectrometer. SCARBO scientists plan to test their instrument aboard a research aeroplane in 2020. It will fly alongside a Dutch-built instrument to study atmospheric aerosols, which are a major source of error when trying to measure greenhouse gases. The test will be the first time that aerosols and carbon dioxide are measured simultaneously to improve the quality of data on greenhouse-gas emissions, says Lizon-Tati. SCARBO is focusing on CO2 monitoring, although it would also be useful for tracking methane emissions, says Le Coarer. Several private efforts to monitor methane emissions cheaply from space are already under way, including a Canadian microsatellite that has been flying since 2016 and a planned small satellite from the Environmental Defense Fund, an advocacy group in New York City.<\|endoftext\|>	4.0625
6,692	Number and arithmetic skills in children with Down syndrome It is clear that arithmetic and number skills are areas of particular difficulty for individuals with Down syndrome. Studies of arithmetic development in typically developing children suggest that a pre-verbal “number sense” system and counting skills provide two critical foundations for the development of arithmetic. Studies of children with Down syndrome suggest that the development of both these foundational skills present difficulties for them, though these conclusions are based on relatively small samples of children. It would seem that further studies of arithmetic and number skills in children with Down syndrome, involving larger samples of children and broader ranges of measures, are badly needed. Down Syndrome Research and Practice Basic number skills, such as knowing how to count and solve simple arithmetic problems, are essential for everyday independent living. It is clear that arithmetic and number skills are areas of particular difficulty for individuals with Down syndrome. So far, we have quite limited understanding of the cognitive bases of the problems with number skills seen in children with Down syndrome. However, major advances have been made in research on the development of number skills in typically developing children and it appears that these advances offer the prospect of better understanding the problems seen in children with Down syndrome. Levels of attainment in arithmetic in children with Down syndrome Studies of arithmetic attainment in individuals with Down syndrome consistently report very low levels of attainment. Carr reported that more than half of her sample of 41 individuals aged 21 years, could only recognise numbers and count on the Vernon's arithmetic-mathematics test. Buckley and Sacks surveyed the number skills of 90 individuals with Down syndrome aged between 11 and 17 years and found that only 18% of the sample could count beyond 20 and only around half of the sample could solve simple addition problems. This pattern contrasts with the increasingly positive achievement levels in reading skills that children with Down syndrome are attaining (e.g. ref 3). Indeed the most consistent finding in the literature is that reading accuracy is significantly higher than arithmetic attainment[1-6]. Age equivalents on standardised number tests are typically reported to lag age equivalent reading scores by around two years in children with Down syndrome (e.g. ref 1). Brigstocke et al. report measures of arithmetic ability from a group of 49 children with Down syndrome. The sample ranged in age from 5:06 to 16:02, and had an average British Picture Vocabulary Scale (BPVS) standard score of 60 (range 39-91). Of this sample 45 children had measurable single word reading skills on the British Ability Scales (BAS) reading test (average standard score 67, range 55-115). However, only 27 of the sample could score on the BAS basic number skills test, and for these children their scores were very low (average standard score 62; range 55-111). It is clear therefore that in this sample, like others studied, number skills are much weaker than reading skills. In typically developing children education and age are strong predictors of arithmetic performance. Language skills are generally predictive of variations in arithmetic ability, and in line with this, children with specific language impairment typically have low arithmetic achievement despite average IQ. Nonverbal ability is also associated with arithmetic achievement in typical and atypical development (e.g. ref 10). Working memory and knowledge of number facts are also important predictors of arithmetic performance[11,12]. Number facts are learned from associations between the problem and the retrieved answer, which is a function of experience, and working memory capacity. These findings are particularly relevant given the cognitive profile typically noted in Down syndrome in which language skills[13,14] and verbal short-term memory[15,16] are weak relative to nonverbal abilities. In common with typical development, number skills in children with Down syndrome appear to improve with age [17,18,19] but this is not always the case and wide individual differences are noted in all studies. The relationship between achievement levels and mental age in Down syndrome is not clear so far. Findings are inconsistent (e.g. refs 17,20,21) but this may reflect the variety and generality of the measures used to assess number skills. Floor effects on standardised IQ tests can also be a problem (e.g. ref 22). Moreover, general ability is a wide measure and the mechanisms that govern the relationship between IQ and mathematical achievement are not clear even in typical development, which makes interpretation difficult. Language skills are related to achievement in number skills in children with Down syndrome (e.g. refs 5,23,24). For example in recently collected data there is a strong correlation between verbal ability measured by the BPVS and both single word reading standard scores (r = .69) and BAS arithmetic standard scores (r = .63). Education clearly has a positive influence on achievement levels in arithmetic. This is indicated by the success of small intervention studies (e.g. refs 19,25,26). Children with Down syndrome in mainstream schools have better attainments in arithmetic than those in special schools; but this is likely to be confounded with selection biases[1,17,27,28]. Nye et al. note that individual differences in response to an intervention using Numicon were related to the quality and quantity of teaching. The Numicon approach to teaching numbers skills and arithmetic[29,30] is based on a system of structured visual representation first developed by Catherine Stern which makes clear the stable order of the number system, and how different numbers are related. One of the key features of the scheme is it provides children with representations of whole numbers which are used to develop mental imagery of numbers, and makes an explicit connection between the preverbal number system, counting and arithmetical operations. Therefore, the scheme uses the perception of whole numbers to support mental arithmetic, rather than using counting as the basis for arithmetic, as is often the case in UK numeracy teaching. Whilst the scheme has been developed for all children to use it was thought to be particularly appropriate for trialling with children with Down syndrome as it complements their particular cognitive profile (e.g. having strengths in visual processing), and targets many of the areas of numeracy that they have difficulty with. The recent Portsmouth project followed the development of number skills in 16 children with Down syndrome (aged between 5 and 14 years) over 2.5 years whilst they were taught using Numicon, and their performance was compared to archive data of children with Down syndrome who had not used Numicon. A small but non-significant gain was seen in numerical performance (as measured by the BAS Basic Number Skills sub-test) in the children who had used Numicon compared to those who had not. Qualitative analysis of the children's profiles, including data from observations of lessons and non-standardised detailed number assessments indicated that Numicon is of particular benefit to children for developing both early numerical concepts and those who are starting to work with arithmetical operations. Regular use of the materials and creative adaptation of the scheme to meet the needs of the individual children were both found to be critical in effective implementation of the scheme. Whilst the main finding from the standardised measure was non-significant statistically, it should be noted that the was a wide range of gains seen in the children, and the gains made may still make a considerable different to children's numbers skills and resulting quality of life. In summary, individuals with Down syndrome find number skills very difficult in comparison to their ability to learn to read but respond positively to tuition. However, while studies investigating the cognitive correlates of general mathematic tests are a useful starting point, they give little insight into the underlying processes involved. The cognitive bases of arithmetic in normal development and the origins of mathematical difficulties In the last 20 years or so there has been a good deal of research concerned with understanding the cognitive bases and development of human numerical abilities. It appears from studies of animals and pre-verbal human infants that some basic numerical skills exist in the absence of language. This pre-verbal numerical system is probably somewhat imprecise and can only deal with small numbers of objects. Nevertheless, it has been suggested that such a preverbal "number sense" may form a foundation for more complex verbally elaborated number skills in humans[34,35]. The possible role of such a putative nonverbal number sense system in mature numerical processing remains controversial, but in one view, this nonverbal system provides the semantic underpinnings for understanding number since numbers, fundamentally, signify magnitudes. One other important skill that also develops in the pre-school years is counting. By the time children go to school they are generally proficient at counting, at least for numbers up to ten, and these counting skills form a foundation for the development of arithmetic skills. Counting is fundamentally a form of measurement, and one that is more flexible and precise than the form revealed in studies of animals' and infants' preverbal numerical abilities. Learning to perform basic addition, which is the earliest arithmetical skill to be taught in school, can be seen as a natural extension of counting. At first, children use a simple 'count all' strategy to solve addition problems. By the age of 6, most are using a 'counting on' strategy in which they start with the smaller number and count on from this (the min strategy). Later, as they learn the number bonds, they can begin to retrieve these automatically. Development involves a change in the mix of strategies that are used. Importantly, the creation, in long-term memory, of an association between the problem integers (e.g. 3+4) and the answer that is generated (7) requires practice in the execution of basic computations. With each execution, the probability of direct retrieval of that number fact or bond increases. This direct retrieval strategy is rapid and highly efficient but only develops after the child has performed many less automatic computations of the relevant sums. The possible cognitive bases of difficulties with arithmetic in children with Down syndrome Such studies of arithmetic in typically developing children suggest it is important to understand the integrity (or otherwise) of the pre-verbal number system in children with Down syndrome, and the development of counting skills, as these two skills appear to provide two of the foundations for the development of arithmetic in typically developing children. Preverbal numerical systems in children with Down syndrome Magnitude comparison tasks have proved a very useful paradigm for investigating number skills and cardinal number understanding in typical development. Some authors interpret the ability to discriminate between magnitudes as a behavioural indicator of the operation of a basic "number sense" (e.g. ref 34) that underlies later number skills. It has been suggested that difficulty judging between magnitudes may underlie the difficulties that typical children with dyscalculia have with mathematics (e.g. ref 35). In a typical numerical judgement task participants are presented with two stimuli (either digits, or arrays of dots differing in numerosity or squares differing in size) simultaneously on a computer screen and asked to indicate which is larger as quickly as possible. The simplicity and non-verbal requirements of the task make it ideal for individuals with Down syndrome. Findings from studies on numerical magnitude comparisons in typical adults and children have proved remarkably replicable. As Moyer and Landauer observed in their seminal study, the time required to compare the numerical magnitude of pairs of digits decreases as the numerical distance between stimuli increases (1 vs. 9 is a much easier judgement than 1 vs. 2). This is referred to as the symbolic distance effect (SDE). When the distance is held constant, discrimination of numbers becomes more difficult as their magnitude increases. This is referred to as the magnitude effect. This pattern of results mirrors that observed in comparison of physical magnitudes such as length and is the opposite pattern to that predicted if counting strategies were used. Recent work suggests that speed in making magnitude comparisons predicts individual differences in addition ability in typically developing children. This symbolic distance effect (SDE) has been observed across all ages from 6 years upwards, supporting the idea that the effect is relatively independent of educational influences and cognitive ability[38,39,40]. The hypothesis that it is independent of language and does not rely on counting is supported by findings that children with specific language impairment who have significant difficulties with the verbal count sequence demonstrate typical performance in numerical comparison tasks. Since individuals with Down syndrome are observed to have relatively preserved visuo-spatial abilities, and numerosity judgments are independent of language and general cognitive ability, this suggests that they will demonstrate normal performance on numerosity comparison tasks, provided they are sufficiently familiar with the count sequence and digits. Paterson, Girelli, Butterworth and Karmiloff-Smith investigated the distance effect in infants and older individuals with Down syndrome and Williams syndrome. They also administered a battery of number tasks hypothesised to rely on verbal abilities to the older groups. Eleven infants with Williams syndrome and 18 infants with Down syndrome, matched on chronological and mental age plus 16 mental age and 14 chronological age typically developing controls were tested on a preferential looking paradigm. Infants were familiarised with arrays of 2 objects. In the test phase they were presented simultaneously with one card displaying new objects but the familiar numerosity and one card with three objects i.e. a new numerosity. Cumulative looking times were measured. It was found that there was a significant difference in mean looking time between the familiar and novel numerosity in all groups except the Down syndrome group. This suggests that the infants with Down syndrome were unable to distinguish between 2 and 3 items. However performance in a numerosity comparison task with older individuals showed the reverse pattern. Eight older children and adults with William's syndrome, 7 with Down syndrome, 8 typically developing controls matched for mental age using the British Abilities Scales, and 8 typically developing controls matched for chronological age to the clinical groups took part in the experiment. Participants were asked to indicate the larger of two dot arrays presented simultaneously on a computer screen. Reaction times and accuracy were measured. The numerosity of the arrays varied from 2 to 9 and the numerical distance between the arrays was classified as small (a difference of 1 to 3) or large (a difference of 5 to 7). Although reaction times were slow in the Down syndrome group, individuals responded more quickly and more accurately to arrays that had a large difference between them than those that had a small distance between them. A significant effect in the same direction was noted in the control groups but this distance effect was not observed in the William's syndrome group. Analysis of errors revealed that the Williams syndrome group was the least accurate of all the groups. The results of this study support the conclusion that language skills do not support performance in magnitude comparison tasks. Participants also took part in a detailed battery of number tasks that assessed rote counting, dot and numeral seriation, matching dots to numerals and reading numerals aloud as well as single digit addition, subtraction and multiplication. Performance in the clinical groups was below that of the control participants who performed near ceiling. The William's syndrome group displayed considerable difficulties when compared to the Down syndrome group on all the tasks except rote counting from 1 to 20 and reading single digits where performance was good in both groups. Both groups found matching numerosities to Arabic numerals difficult. No correlations were found between performance in the dot comparison task and the number battery task except in the performance of the Down syndrome group on the matching dots to Arabic numerals task. This could suggest a link between the ability to discriminate numerosities and the ability to associate Arabic digits with their underlying quantity representation . An unpublished study conducted at York investigated the pattern of reaction times obtained by 16 children with Down syndrome, with a mean age of 13; 2 years (SD 24.44 months) and a receptive vocabulary level of above 5 years, on three computerised comparison tasks and a timed pencil and paper single digit addition task. Each computerised task comprised 54 trials and required participants to identify the larger of two simultaneously presented stimuli. There were three sets of stimuli: dot arrays (matched for surface area), Arabic digits and horizontal lines. The order in which these stimulus types were presented was counterbalanced between participants. Performance of the individuals with Down syndrome was compared to that obtained by typically developing children in Year 1 and Reception classes matched for receptive vocabulary level. The children in Year 1 demonstrated typical distance and magnitude effects in all tasks. The speed with which they made magnitude comparisons using line and digit stimuli correlated with their performance in the addition tasks (r=. 49). This correlation between comparison speed for numeric and physical stimuli and addition skills suggests numeric representations in this group are underpinned by analogue magnitudes representations, which in turn support addition skills. The children with Down syndrome and children in Reception also demonstrated typical distance and magnitude effects in all tasks, although five children with Down syndrome had to be excluded from RT analysis of the numeric comparison task because of high error rates on the task. Although group sizes were too small to make firm interpretations of the pattern of correlations achieved in these groups, intriguingly, speed in making magnitude comparisons using dot arrays was the only correlate with addition performance (reception: r=. 72; Down syndrome: r=. 69). This pattern suggests that children with Down syndrome may have typical representations of numerosity but raises the possibility that the ability to link digit representations to magnitudes may be immature in children with Down syndrome (as in much younger typically developing reception year children). The development of counting skills in children with Down syndrome The development of counting has been examined in some detail in individuals with Down syndrome. Counting is an important skill that is often claimed to underpin a number of later mathematic skills (e.g. ref 42) such as children's early attempts at addition. Counting involves not only learning the number words, their sequence and how to tag number words to individual objects, but also requires understanding of the cardinality principle. This refers to the fact that the final count word refers to an exact quantity - the cardinal value or magnitude of the set. The cardinality principle means that the order in which items are tagged is irrelevant to the cardinal value of the set. Understanding of the order-irrelevance principle is used to assess whether children understand the purpose as well as the procedure of counting. Gelman and Cohen reported the first detailed study of count production and understanding in ten children with Down syndrome with a mean chronological age of 10:06 years and mental ages ranging from 3:06 to 6:08years compared with younger typically developing children broadly matched for social economic status. All of the children with Down syndrome attended special school. Children were assessed on rote counting and object counting knowledge as well as a task designed to test knowledge of the order-irrelevant principle. Children were presented with a line of objects and asked to count them in a non-linear order. For example, they might be asked to label the middle object, "the one". The children with Down syndrome performed better than controls on rote counting and object counting but worse on the order irrelevance counting task. On this basis, the authors concluded that the children with Down syndrome performed rote counting with no conceptual understanding of number. However, the instructions for the order-irrelevant counting task and the feedback involved very complex language. In contrast, Caycho, Gunn and Siegel found no difference between 15 children with Down syndrome (mean chronological age of 9:07 years) and 15 typically developing children (mean age of 4:06 years) matched for receptive vocabulary level, on a simplified version of the Gelman and Cohen task. In this task the children presented with a row of items and asked to count them in a non-linear fashion but their finger was guided to the start item and they were told it was "one". The language and feedback used in the task were simplified. Caycho et al. concluded that conceptual understanding of counting is related to receptive vocabulary levels. A longitudinal study by Nye investigated performance of a group of children with Down syndrome and typically developing children matched for non-verbal mental age on a variety of counting tasks. A striking similarity was found between the counting skills of the Down syndrome group and the typically developing group matched for non-verbal mental age, both in terms of object counting and understanding of cardinality. While counting skills have not been found to be a particular problem for children with Down syndrome in previous research, what was particularly surprising here was how these skills developed in line with non-verbal mental ability (see ref 23). Even more surprising was the lack of a difference between the Down syndrome and typically developing groups in terms of their cardinal understanding; this would not have been predicted from previous research[20,23]. The only difference between the two matched groups was in count word vocabulary and sequence production, which were both significantly greater in the typically developing group, though by no means lacking in the Down syndrome group. However, any limitations that the children with Down syndrome had in production of the count word sequence did not seem to impact on their ability to count or give sets of objects, as evidenced by the lack of a difference between them and the typically developing children on these tasks. These seemingly positive findings of fledgling number skills in young children contrast strongly with the poor levels of achievement reported in older children although success was limited in these studies to very small arrays of objects (up to 18 items) so these positive results only extend to very basic skills, typically achieved by pre-school children. In order to progress to any form of number skills without the use of concrete props, understanding of the relative value of number and conceptual understanding of the number system that goes beyond the perceptual characteristics of a given array of items is essential. Conservation tasks that manipulate the surface characteristics of an array but keep the underlying value the same are often used to test this in typical development. The only studies to use traditional conservation tasks in Down syndrome were conducted by Lister and Lee and Lister, Leach and Riley[44,45]. They studied number and length understanding in 48 individuals with Down syndrome between 5 and 26 years. The tasks involved the subject creating or agreeing the initial equality of two stimuli. One of the stimuli was then transformed and the individual asked to judge whether the remaining quantities were still equal. None of the participants succeeded on all the conservation of length tasks, although five succeeded on all of the number conservation tasks. This is the pattern of development observed in typically developing children. No information is provided on the counting ability of the participants. Given the wide age range of the sample it is likely that this is a significant factor in performance. Consequently, no clear conclusions can be drawn about the understanding of conservation in Down syndrome without further research. It is clear that children with Down syndrome show severe difficulties in mastering basic number skills as assessed by tasks that include size and numerosity judgements, counting and simple arithmetic. There are suggestions that a pre-verbal "number sense" system may show atypical development in Down syndrome, but so far the group sizes studied preclude strong conclusions. It is clear that learning to count is difficult for children with Down syndrome, though there is no evidence that the development of counting follows a qualitatively different path to that seen in younger typically developing children. It appears that problems in the sphere of arithmetic show strong correlations with language skills in the Down syndrome population though such correlations may in part reflect limitations in children's ability to understand the arithmetic tasks they are required to complete. It would seem that studies of larger samples of children with Down syndrome that assess their pre-verbal number sense skills as well as counting and basic addition skills are badly needed. - Carr J. Six Weeks to Twenty-One Years Old: A Longitudinal Study of Children with Down's Syndrome and Their Families. Journal of Child Psychology and Psychiatry. 1988;29(4):407-431. - Buckley S, Sacks B. The adolescent with Down's syndrome: Life for the teenager and for the family. Portsmouth: Portsmouth Polytechnic; 1987. - Hulme C, Goetz K, Snowling M, Brigstocke S. The development of reading and language skills in children with Down syndrome. (in preparation). - Dunsden MI, Carter CO, Huntley RMC. Upper end range of intelligence in mongolism. Lancet. 1960;i:565-568. - Irwin K. The school achievement of children with Down's syndrome. New Zealand Medical Journal. 1989;102:11-13. - Pototzky C, Grigg AE. A revision of the prognosis in Mongolism. American Journal of Orthopsychiatry. 1942;72(3):303-510. - Brigstocke S, Hulme C, Goetz K, Snowling M. Number Skills in Children with Down Syndrome. (in preparation). - Geary DC. Children's Mathematical Development: Research and Practical Applications. Washington, DC: American Psychological Association; 1994. - Donlan C, Bishop D, Hitch G. Magnitude comparisons by children with specific language impairments; evidence of unimpaired symbolic processing. International Journal of Language and Communication Disorders. 1998;33(2):149-160. - Gobel S, Walsh V, Rushworth M. rTMS disrupts the representation of small numbers in supramarginal gyrus. Neuroimage. 2001;13:S409. - Lemaire P, Abdi H, Fayol M. Working memory and cognitive arithmetic: Evidence from the disruption of the associative confusion effect. European Journal of Cognitive Psychology. 1996;8:73-103. - Bull R, Scerif, G. Executive Functioning as a Predictor of Children's Mathematics Ability: Inhibition, Switching, and Working Memory. Developmental Neuropsychology. 2001;19(3):273-293. - Chapman R. Language development in children and adolescents with Down syndrome. Mental Retardation and Developmental Disabilities Research Review. 1997;3(4):307-312. - Fowler A. Language abilities in children with Down syndrome: evidence for a specific delay. In: Cicchetti D, Beeghley M, editors. Children with Down syndrome: A developmental perspective. Cambridge: Cambridge University Press; 1990. - Brock J, Jarrold C. Language Influences on Verbal STM Performance in Down syndrome; item and order recognition. Paper presented at the meeting of the Down Syndrome Symposium, Portsmouth; 2003. - Wang P, Bellugi U. Evidence from two genetic syndromes for a dissociation between verbal and visual-spatial short-term memory. Journal of Clinical Experimental Neuropsychology. 1994;16(2):317-22. - Sloper P, Cunningham CC, Turner S, Knussen C. Factors related to the academic attainments of children with Down's syndrome. British Journal of Educational Psychology. 1990;60:284-298. - Turner S, Alborz A. Academic attainments of children with Down's syndrome: A longitudinal study. British Journal of Educational Psychology. 2003;73:563-583. - Thorley BJ, Woods VM. Early number experiences for preschool Down's syndrome children. Australian Journal of Early Childhealth. 1979; 4(1): 15-20. - Gelman R, Cohen M. Qualitative differences in the way Down syndrome and normal children solve a novel counting task. In Nadel L, volume editor. The Psychobiology of Down Syndrome. Cambridge, Massachusetts: MIT Press; 1988. p. 51-101. - Cornwell A. Development of language, abstraction and numerical concept formation in Down's syndrome children. American Journal of Mental Deficiency. 1974;79(2):179-190. - Nye J, Clibbens J, Bird G. Numerical ability, general ability and language in children with Down syndrome. Down Syndrome Research and Practice. 1995;3(3):92-102. [Read Online ] - Caycho L, Gunn P, Siegal M. Counting by Children with Down Syndrome. American Journal of Mental Retardation. 1991;95(5), 575-583. - Porter J. Learning to count; a difficult task. Down Syndrome Research and Practice. 1999;6(2):85-94. [Read Online ] - Irwin K. Teaching children with Down syndrome to add by counting-on. Education and Treatment of Children. 1991;14(2):128-141. - Nye J, Buckley S, Bird G. Evaluating the Numicon system as a tool for teaching number skills to children with Down syndrome. Down Syndrome News and Update. 2005;5(1):2-13. [Read Online ] - Shepperdson B. Attainments in reading and number of teenagers and young adults with Down's syndrome. Down's Syndrome: Research and Practice. 1994;2:97-101. [Read Online ] - Laws G, Byrne A, Buckley S. Language and memory development in children with Down syndrome at mainstream and special schools: A comparison. Educational Psychology. 2000;20(4):447-457. - Atkinson R, Tacon R, Wing T. Numicon: Foundation Book. Brighton, UK: Numicon Ltd; 1999. - Atkinson R, Tacon R, Wing T. Numicon: Year 1 Teachers Book. Brighton, UK: Numicon Ltd; 2000. - Stern C. Children discover arithmetic. New York: Harper and Row; 1949. - Nye J, Buckley S, Bird G. Evaluating the Numicon system as a tool for teaching number skills to children with Down syndrome. Down Syndrome Research and Practice; submitted. - Elliot CD, Smith P, McCulloch K. British Ability Scales II. Windsor: NFER Nelson; 1997. - Dehaene S. The Number Sense. Oxford: Oxford University Press;1997. - Butterworth, B. The mathematical brain. London: Macmillan; 1999. - Moyer R, Landauer T. Time required for judgements of numerical inequality. Nature. 1967;215:1519. - Durand M, Hulme C, Larkin R, Snowling M. The cognitive foundations of reading and arithmetic skills in 7 to 10 year olds. Journal of Experimental Psychology. 2005;91:113-136. - Duncan E, McFarland C. Isolating the effects of symbolic distance and semantic congruity in comparative judgments. Memory and Cognition. 1980;8:612-633. - Sekuler R, Mierkiewicz D. Children's judgement of numerical inequality. Child Development. 1977;48:630-633. - Siegler R, Robinson M. The development of number understanding. In Reese HW, Lipsitt LP, editors. Advances in child development and behaviour. Vol 16. New York; Academic Press; 1982. - Paterson S, Girelli L, Butterworth B, Karmiloff-Smith, A. Are numerical impairments syndrome specific? Evidence from Williams syndrome and Down's syndrome. Journal of Child Psychology and Psychiatry. 2006;47(2):190-204. - Ashcraft MH. Cognitive Arithmetic: A Review of Data and Theory. Cognition. 1992;44:75-106. - Nye J. Numerical development in Children with Down syndrome: The role of parent-child interaction. Unpublished PhD thesis. Portsmouth: University of Portsmouth; 2003. - Lister C, Lee S. Process of development in understanding of length in individuals with Down syndrome. Early Child Development and Care. 1992;81:1-13. - Lister C, Leach C, Riley E. The development of understanding of quantity in children with Down's syndrome. Early Child Development and Care. 1989;49:57-66. Sophie Brigstocke and Charles Hulme are at the Department of Psychology, University of York, UK Joanna Nye is at the Department of Experimental Psychology, University of Bristol, UK Correspondence to Sophie Brigstocke • e-mail: [email protected] Paper prepared from presentations and discussions at the Down Syndrome Research Directions Symposium 2007, Portsmouth, UK. The symposium was hosted by Down Syndrome Education International in association with the Anna and John J Sie Foundation, Denver. Major sponsors also included the Down Syndrome Foundation of Orange County, California and the National Down Syndrome Society of the USA. Information about the symposium can be found at https://www.dseinternational.org/research-directions/ Received: 14 January 2008; Accepted: 21 January 2008; Published online: 2 July 2008<\|endoftext\|>	3.78125
692	Subject Area Lessons ## #1676. GOfigure - patterning exercise Mathematics, level: Elementary Posted Tue Apr 18 17:43:59 PDT 2000 by Rick Newell ([email protected]). GOfigure email numbers game Materials Required: none Activity Time: 10 minutes Concepts Taught: Basics - addition, subtraction, multiplication, division GOfigure, a classroom activity Here is the challenge. Take these eight digits 1, 2, 2, 9, 3, 6, 8, and 5 Now find a pattern by using as many as you can. Rules: the pattern must have one rule that is used at least twice. for example the pattern 1, 2, 3 the pattern 1, 3, 9 There are over 300 patterns and 54 of them use all eight digits, put the above numbers on the board and see how many the class can find. Watch the CRUNCHING begin as they look for the eight digit patterns. You may want to do a lesson patterning before they start searching. Following is an outline of patterns used in this exercise. What is a pattern? The aim of the game is to create numeric patterns from the digits students have available to them. Ask the question Is "2,4" a pattern? Asking this question beforehand could serve as an effective introduction to the game, especially for younger students. If everyone agrees that two numbers are not enough to show a pattern, they're probably all ready to begin. If, however, someone says that "2 , 4" is a pattern, ask that person what number comes next and why. The answer often given is 6, because you're adding 2 each time. But, depending on their math experience and abilities, other students may be quick to point out, for example, that "8" or "16" are two other possibilities, and will be able to explain why. e.g. 2, 4, 6 (add 2 to the previous number each time) 2, 4, 8 (multiply the previous number by 2 each time) 2, 4, 16 (square the previous number each time) Each pattern is formed by using a different rule, but it's not until the third number in the sequence that the rule being used repeatedly becomes clear. Only then can the pattern be identified. Discussions around a few examples will help students see why they need to use at least three digits to form a recognizable pattern and how the rule is applied at least twice. A note about graphic or visual patterns: Some students, especially younger ones, may want to use numbers that create a visual pattern such as 1,1,9,9,3,3. Since no one rule is being followed consistently to create this pattern, it is not a mathematical pattern and therefore does not qualify for this exercise. In the example shown first, you multiply the first number by 1, then add 8 to the second; then multiply the third by 1 and subtract 6 from the fourth then multiply times 1 again (1, +8, 1, -6, *1). Who knows what number might come next?<\|endoftext\|>	4.65625
6,251	We developed a sampling plan to help beekeepers monitor Varroa destructor mite infestations in honey bee colonies. From a sample of 280 adult bees, collected using our novel sampling device, beekeepers can estimate the total mite density in a colony (mites on adults and those reproducing on pupae). Standard sampling will help beekeepers make educated treatment decisions. Field trials of bees bred for both Hygienic Behavior and Varroa Sensitive Hygiene had significantly lower mite levels compared to an unselected line of bees demonstrating that the hygienic trait can reduce mite loads and thus the frequency of pesticide application. The North Central region of the US, particularly MN, ND and SD are the top honey producing states based on yield per colony, together producing over 30% of the total honey production for the nation. The majority of the commercial beekeepers in the Upper Midwest are “migratory,” meaning that they transport their colonies every winter either to southern states where they produce bulk bees and queens for sale, or to CA and other states where the bees pollinate almond orchards and other fruit and vegetable crops. The beekeepers transport their colonies back to the Upper Midwest for the summer to produce large honey crops. The honey bee, Apis mellifera, population has been declining since introduction of the parasitic mite Varroa destructor in 1987. Varroa is a devastating pest. The mites weaken individual honey bees, increasing the bee’s susceptibility to the effects of disease and poor nutrition. Infected colonies die in 1-2 years. To prevent potentially large colony losses, many beekeepers have resorted to using pesticides (organophosphates and pyrethroids) within their colonies. Many treat all their colonies with pesticides once or twice a year, irrespective of mite level. The treatments add a great operating expense for the beekeepers, increase the risk of contamination of hive-products (Frazier et al., 2008), pose health risks to bees (Currie, 1999; Rinderer et al, 1999; Haarman et al., 2002; Collins et al., 2004), and the mites have developed resistance to the synthetic pesticides. To minimize these negative effects, beekeepers have expressed their desire to decrease the number of treatments. Many beekeepers have heard of the integrated pest management practice of treating only when economically necessary, but this requires sampling to determine the current mite levels and having a treatment threshold. Prior to this research, beekeepers lacked a standard, accurate and efficient way to sample mites in a honey bee colony. Our goal is to encourage beekeepers to sample their bee colonies for mite pest populations in a standardized way to help them make wise treatment decisions and reduce pesticide use. Such a sampling plan for a mite pest in an insect colony has never been attempted before. Once a beekeeper samples to find the mite infestation, that beekeeper will need to make a treatment decision. Treatment thresholds of 10-12% colony mite infestation have been developed (Martin 1998, 1999; Delaplane and Hood, 1997, 1999). However, the thresholds were created based on mite levels in relatively small-scale beekeeping operations (hobby or side-line) in which the colonies are maintained in one location year round. Migratory beekeepers, those that move their colonies across the U.S. for honey production and to fulfill pollination contracts, would also benefit greatly from a treatment threshold. These commercial beekeepers will likely treat at a lower threshold because their livelihoods depend on it. Transporting bees en masse leads to increased horizontal transmission of diseases and pests, and nutritional stress, which may decrease the colonies tolerance for the mite. Bees with natural mite resistance due to hygienic behavior may have a different treatment threshold than susceptible colonies. There should be an absolute lowest threshold (e.g. 2-5%) where below that level no beekeeper should need to treat. Between the low threshold and the 10-12% threshold is a gray area where a beekeeper should make a decision based on the nature of their beekeeping operation, and their personal tolerance for colony loss. Monitoring mite loads is a first step in helping beekeepers understand their mite levels and dynamics. With this system and good record keeping, beekeepers can find the treatment level works best for them based on their own operation. The objective of this project was to encourage beekeepers to reduce the use of in-hive chemicals by monitoring mite populations through sampling, helping them to make informed treatment decisions, and demonstrating the benefits of keeping bees with natural resistance to the mite. The reduction of treatments will help beekeepers avoid unnecessary costs, pesticide residues, and help slow the evolution of resistance of the mites to chemicals; thereby fostering more sustainable beekeeping practices, increasing the profitability of beekeeping, improving environmental quality, and promoting pollinators. Objective 1. Develop a simple and standardized sampling plan for commercial beekeepers to help them determine the economic treatment level for Varroa destructor mites. Objective 2. Develop published guidelines for migratory beekeepers on making educated treatment decisions for the mite based on the sampling plan. Objective 3. Compare mite levels between our line of bees bred for both Hygienic Behavior (HYG) and Suppression of Mite Reproduction (SMR) with an unselected, commercial line of bees, to demonstrate that the use of resistant bees can reduce mite loads and thus, the frequency of pesticide application. Objective 1. There are two populations of mites: those on adult bees and those on developing pupae which are in cells under wax cappings. The mites hidden in pupal cells are the reproductive population. The adult bee mite infestation is easier to estimate than the pupae infestation, but a comprehensive sampling plan needs to include both the mites on adult bees and the mites on pupae. The sampling plan was developed to efficiently sample mites on adult bees and, using the adult bee infestation, make an inference about the colony infestation; i.e., the mites on adults and on pupae. Mites on adult bees were sampled from colonies in 31 commercial apiaries with 24-84 colonies per apiary for a total of 594 colonies. The apiaries were owned by five different beekeepers in four states, Minnesota, North Dakota, California, and Texas. All the beekeepers were migratory and the operation sizes ranged from 1,000 to 20,000 colonies. The colonies were sampled in the years 2005, 2006, and 2007. All the commercial apiary sampling was done during March (CA and TX), May-June (MN and ND), and August-September (MN and ND). These times were chosen to coincide with when beekeepers would sample and potentially treat their colonies. Hand drawn maps were created to detail the apiary layout including the colony and pallet placement (commercial beekeepers usually maintain 4 colonies per pallet to facilitate transportation) and the colony numbers (each colony was arbitrarily assigned a number). A sample of 35 adult bees ± 15 (mean ± s.d.) was collected from each comb in 70% ethanol. Each comb was classified as having at least one of the following contents: honey, pollen, open worker brood (egg through uncapped 5th instar larva), sealed worker brood (developing pupae covered by a wax capping), sealed drone brood, or empty. In the lab, the mites were strained from the bees to determine the number of mites and adult bees per sample (De Jong et al., 1982). The infestation for each comb within each colony was recorded and used to determine if the mites tend to congregate within a colony and within an apiary. In addition, 75 colonies were sampled more intensively to examine the relationship between the adult bee infestation (mites on adult bees) and colony-level infestation (mites on adult bees and worker pupae). From each colony, the following populations were estimated: the total number of adult bees and worker pupae, and the total number of mites on adult bees and on worker pupae. The adult bee population was estimated either using a calibrated image (MAFF, 1998) or by shaking the bees into a screened box and weighing them. The number of worker pupae was estimated using a wire grid to count the number of squares of sealed brood and using a conversion factor of the number of pupal cells found in one square. The mite population on adult bees was estimated by finding the infestation on ≈900 adult bees in 70% ethanol or by combining the infestations from the 35-bee unit samples from a colony. The adult bee infestations were multiplied the mite by the number of adult bees. To estimate the total number of mites on the pupae, a minimum of 200 individual sealed pupal cells were opened and examined individually for the presence or absence of mites. The infestation of the pupae was multiplied by the number of pupae to estimate the total number of pupal cells infested. Objective 2. Based on the sampling plan developed in objective 1, we developed a sampling device to collect and monitor mites on adult bees in a standard manner. The aim is to provide beekeepers a chart enabling them to convert the adult bee infestation into the colony infestation (mites on adult bees and in worker pupae). We will publish guidelines for beekeepers on making educated treatment decisions for the mite. Objective 3. M. Spivak and G. Reuter have been breeding honey bees for hygienic behavior since 1994. Hygienic bees are able to detect, uncap and remove diseased young bees, preventing the spread of diseases like chalkbrood and American foulbrood. Hygienic bees are also sensitive to V. destructor infested cells, opening the cell and removing the pupae, which interrupts the mite’s reproductive cycle. Another line of bees was bred by researchers at the USDA-ARS Bee Research Lab in Baton Rouge, LA. They bred for bees that reduced the mite loads over a period of time, and called this trait, “Suppression of Mite Reproduction” or SMR (Harbo and Hoopingarner, 1997; Harbo and Harris 1999a; b). During the course of this field study, we investigated the mechanism of SMR to determine how the bees were able to suppress mite reproduction. Originally, it was thought that mites in the colonies expressing this trait appeared to produce fewer offspring overall, fewer offspring that reached maturity, or failed to initiate egg-laying. However, our research, which was later confirmed by the USDA team, showed that the mites do not necessarily have a lower reproductive success. Instead, SMR bees tended to express hygienic behavior on the cells containing mites that are reproductively successful (Ibrahim and Spivak, 2006; Harbo and Harris, 2005). After this finding, the name SMR was changed to Varroa Sensitive Hygiene, or VSH. We compared the mite levels of colonies selectively bred for both hygienic behavior and Varroa Sensitive Hygiene with colonies bred solely for hygienic behavior and unselected control colonies. Colonies were evaluated for strength, brood viability, removal of freeze-killed brood, honey production, mite loads on adult bees and within worker brood, and mite reproductive success on worker brood for two years in MN and ND. Objective 1. The sampling plan developed for beekeepers to help them estimate mite levels on adult bees was developed by applying sampling statistics normally used in plant and pest agricultural systems. We first examined possible sources of variation in mite infestation at the level of apiary, pallet (grouping of 4 colonies), individual colony, box, comb, and comb content (combs with brood verses combs without brood). A nested analysis of variance showed that the colony explained the most variation in mite prevalence, followed by pallet. The analysis also showed apiary can be a large source of variation, meaning that the mite levels were different among apiaries within a single beekeeper’s outfit. This finding indicates beekeepers should sample and make a treatment decision based on each apiary instead of assuming all the apiaries have a similar mite level. To examine within colony variance, the mite infestation on combs with brood were compared to combs without brood. A logistic regression showed combs with brood tended to be 1.6 times more likely (95% confidence interval =1.44-1.74) to be infested compared to combs without brood (chi-squared test=90.73, d.f=1, P <0.0001). The sampling approach, Resampling for Validation of Sampling Plans (Naranjo and Hutchison, 1997), was used to determine the optimal number of adult bees to sample from each colony and the number of colonies to sample within an apiary. Results showed that the best method is to sample 280 bees per colony and eight colonies per apiary to achieve a precision level of 0.25 for apiaries with 24-84 colonies. For the infestation of a single colony, beekeepers should sample 280 adult bees to obtain a precision of 0.25 if there are ≥ 4 mites per 100 bees. If there are ≤ 4 mites per 100 adult bees, the precision decreases and the reliability of the sampling will be ± 1 mite. The next step was to factor in the number of mites parasitizing the worker brood (developing pupae) to estimate total mite loads within a colony. We compared the adult bee infestation to the colony infestation (adult bees and worker pupae) using a linear regression. The regression showed that multiplying the adult bee infestation by a factor of 1.3 estimates the colony infestation with an R²=0.75. In application, this analysis will allow beekeepers to sample mites on adult bees and using a conversion factor to estimate the total colony density of mites. If a beekeeper finds 10 mites on 280 adult bees, they can divide 10 by 280 and multiply by 100 to find 3.6 mites per 100 adult bees, then multiply by 1.3 to account for the mites in the brood to obtain a total infestation of 4.6 mites per 100 adults and pupae. To make this conversion more accessible, it will be provided in a simple chart that allows beekeepers to count the number of mites in a 280 adult bee sample or from eight 280 adult bee samples (2240 bees) and readily find the colony or apiary infestation. We will recommend beekeepers estimate the apiary infestation by sampling 280 adult bees from a brood comb from each of eight colonies or sample 280 adult bees to estimate colony infestation. Ideally beekeepers would sample eight randomly chosen colonies to estimate apiary infestation, but to simplify to recommendation and to assure multiple pallets will be sampled we will recommend beekeepers sample every 5th colony starting at one end of the apiary until they sample a total of eight colonies. The same statistics were applied to develop colony and apiary level sampling plans for researchers. These plans were developed to achieve a higher level of precision, since researchers are more interested in data gathering instead of making a treatment decision. To estimate the apiary-level mite prevalence for a precision of 0.10 researchers should sample 280 adult bees from each colony, where the number of colonies to sample depends on the apiary size. To find the number of colonies to sample, researchers can apply the formula y=0.403x+10.41, where y is the number of colonies to sample and x is the number of colonies in the apiary. To estimate the mite level on adult bees in a single colony, researchers should sample 17.2 (18 rounded up to an integer) 35-bee sample units to achieve a precision of 0.10 when there are ≥ 5 mites per 100 adult bees. If there are ≤ 5 mites per 100 bees, the reliability will be ± 0.5 mites. To convert the adult bee infestation to the colony infestation, researchers should estimate the adult bee and pupae populations and include the relative number of pupae compared to all bees (adults and pupae) for an R²=0.88. Objective 2. It is not feasible to expect beekeepers to count out 280 adult bees in a sample; however consistently sampling the correct number of adult bees is important in estimating the mite infestation. To aid in sampling the correct number of bees we have developed a sampling device that measures out the bees by volume. The creation of a sampling device will allow beekeepers to sample 280 adult bees from a colony with both ease and accuracy in a standardized manner. The body of the device is a jar with a wire mesh cap that can hold up to four 280 bee samples at a time. Once the bees are in the device, the mites will be dislodged from the bees using powdered sugar (Macedo et al., 2002). Bees are coated with powdered sugar, let set for a minute, and then the jar is inverted and shaken so the loose powdered sugar and any mites fall through the wire mesh into a white dish. The dislodged mites can be counted, and the bees released into one of the sampled colonies or outside the entrance unharmed. The simplicity of the device will increase its use and the use of the sampling plan. Once beekeepers tally the number of mites, they can use the simple chart to estimate the colony infestation. The development of the sampling plan allows for beekeepers to monitor the mite levels in their colonies. Sampling and recording mite levels within apiaries over time can result in beekeepers determining the mite level they need to treat to make it until the next treatment window and the levels they can forgo a treatment. If beekeepers nationwide monitor mite levels in their colonies, a much clearer picture of treatment thresholds and mite dynamics will emerge and can be investigated in future studies. Objective 3. Our line of bees bred for both Hygienic Behavior (HYG) and Varroa Sensitive Hygiene (VSH) had significantly lower mite levels compared to an unselected line of bees demonstrating that this line can reduce mite loads and thus the frequency of pesticide application (Ibrhaim et al., 2007). By autumn, the HYG/VSH colonies had significantly fewer mites on adult bees and in worker brood compared to the control colonies and the HYG colonies had intermediate mite populations. There were no differences among the lines in mite reproductive success. Collins, A.M., Pettis, J.S., Wilbanks, R., and M.F. Feldlaufer. 2004. Performance of honey bee (Apis mellifera) queens reared in beeswax cells impregnated with coumaphos. Journal of Apicultural Research, 43: 128-134. Currie, R.W. 1999. Fluvalinate queen tabs for use against Varroa jacobsoni Oud.: efficacy and impact on honey bee, Apis mellifera L., queen and colony performance. American Bee Journal, 139: 871-876. De Jong, D., De Andrea Roma, D., and L.S. Gonςalves. 1982. A comparative analysis of shaking solutions for the detection of Varroa jacobsoni on adult honeybees. Apidologie, 13: 297-306. Delaplane, K. and W.M. Hood. 1997. Effects of delayed acaricide treatment in honey bee colonies parasitized by Varroa jacobsoni and a late-season treatment threshold for the southern USA. Journal of Apicultural Research, 36: 125-132. Delaplane, K.S. and W.M. Hood. 1999. Economic threshold for Varroa jacobsoni Oud. in the Southeastern USA. Apidologie, 30: 383-395. Fraizer et al., 2008 M, Mullin C, Frazier J, and S. Ashcraft. 2008. What have pesticides got to do with it? American Bee Journal, 148: 521-523. Haarmann, T., Spivak, M., Weaver, D., Weaver, B., and T. Glenn. 2002. Effect of fluvalinate and coumaphos on queen honey bees (Hymenoptera: Apidae) in two commercial queen rearing operations. Journal of Economic Entomology, 95: 28-35. Harbo J.R., Harris J.W. 1999a. Selecting honey bees for resistance to Varroa jacobsoni, Apidologie 30, 183-196. Harbo, J.R. and J.W. Harris. 1999b. Heritability in Honey Bees (Hymenoptera: Apidae) of Charateristics Associated with Resistance to Varroa jacobsoni (Megsostigmata: Varroidae). Journal of Economic Entomology, 92: 261-265. Harbo J.R., Hoopingarner R. (1997) Honey bees (Hymenoptera: Apidae) in the United States that express resistance to Varroa jacobsoni (Mesostigmata: Varroidae), J. Econ. Entomol. 90, 893-898. Ibrahim A, Reuter GS, Spivak M. 2007. Field trial of honey bee colonies bred for mechanisms of resistance against Varroa destructor. Apidologie 38: 67-76. Ibrahim A, Spivak M. 2006.The relationship between hygienic behavior and suppression of mite reproduction as honey bee mechanisms of resistance to Varroa destructor Apidologie. 37: 31-40. Macedo, P.A., Wu, J., and M.D. Ellis. 2002. Using inert dusts to detect and assess varroa infestations in honey bee colonies. Journal of Apicultural Research, 40: 3-7. MAFF. 1998. Varroa jacobsoni: monitoring and forcasting mite populations within honey bee colonies in Britian. MAFF Publications PB 3611 pp.12. Martin, S.J. 1998. A population model of the ectoparasitic mite Varroa jacobsoni in honey bee (Apis mellifera) colonies. Ecological Modelling, 109: 267-281. Martin, S.J. 1999. Population modelling and the production of a monitoring tool for Varroa jacobsoni an ectoparasitic mite of honey bees. Aspects of Applied Biology, 53: 105-112. Naranjo, S. E., and W.D. Hutchison. 1997. Validation of arthropod sampling plans using a resampling approach: software and analysis. American Entomology, 43: 48-57. Rinderer, T.E., de Guzman, L.I., Lancaster, V.A., Delatte, G.T., and J.A. Stelzer. 1999. Varroa in the mating yard: The effect of Varroa jacobsoni and Apistan on drone honey bees. American Bee Journal, 139: 134-139. The immediate goal of this research was to develop a sampling plan to help beekeepers estimate the infestation of V. destructor per apiary or per colony based on well-documented sampling statistics. The long-term goal was to help beekeepers use the sampling plan to make informed treatment decisions so they can reduce pesticide application to avoid unnecessary costs, pesticide residues, and evolution of resistance of the mites to chemicals. It can also help beekeepers find colonies with the desirable traits hygienic behavior through monitoring colonies to examine which tend to have fewer mites. These colonies can be bred from those colonies to increase the frequency of colonies with natural defenses. The sampling device will allow beekeepers to determine mite levels and make a treatment decision in a single trip to an apiary. Because this device is standardized, beekeepers can compare their colony or apiary infestation levels with other beekeepers in a meaningful way. The most important outcome of this project is providing beekeepers with a way to acquire critical information about mite populations in their colonies. Beekeepers will finally be able to reliably monitor mite populations and dynamics within their own colonies. Monitoring mite populations through sampling must be economically advantageous for beekeepers to implement this practice into their operations. The advantages can be communicated by showing that sampling is less expensive (both time and cost) than blanket treating, even if sampling reveals that a treatment is needed. For a hobby beekeeper, testing each colony will probably take 8-10 minutes. This includes the time it takes to open the colony, find the first brood frame, check for the queen, take the sample, and perform the powdered sugar roll. For a single person in a commercial operation testing four colonies at a time, sampling bees with the sampling device should take between three and five minutes per colony, depending on the level of experience. This includes the time it takes to perform two powdered sugar rolls (one for each set of four colonies). Sampling a single apiary should not take over 30 minutes. The time can be reduced by using two sampling devices or having two people sample. Beekeepers can integrate sampling into normal management regimen, such as while giving their colonies nutritional supplements. Since beekeepers are already in the colonies, taking eight samples would add little extra time. The use of this sampling plan gives beekeepers an immediate estimate of the mite population, permitting them to make a treatment decision while they are already, reducing travel time. Monitoring mite levels can be more cost effective than treating without sampling. According to the current prices for the miticide treatments formic acid, thymol, tau-fluvalinate and coumaphos, the cost of treating an apiary with 30 colonies is $78 ($2.60 per colony, two fluvalinate strips per colony). These prices do not include any extra equipment, time and labor for the application, or the superfluous expense if the treatment is not warranted. The treatments for 100 apiaries (3,000 colonies assuming 30 colonies per apiary) would have an approximate cost of $7,800. In contrast, monitoring an operation of 100 apiaries would take approximately 50 hr (assuming 30 minutes per apiary). If labor costs $30 per hr (an employee paid $15 per hr, plus $15 in benefits), then the cost to monitor is approximately $1500 or about 19% of what it would cost to treat with fluvalinate. If 19 out of 100 apiaries are deemed to be below a treatment threshold, then sampling is cost effective. Even when apiaries are tested and found to have high enough mite levels to warrant a treatment, sampling would still be beneficial. Beekeepers can learn about the growth cycle of the mite in their colonies, leading them to determine the best times of year for them to sample and potentially treat M. Spivak, K. Lee, and G. Reuter have presented the sampling plan and device to beekeepers at meetings and in on-on-one conversations and we have received enthusiastic interest from the majority of beekeepers. Beekeepers have expressed a desire to reduce treatments for years. They understand the need to sample, but until now they have lacked a method to estimate apiary infestation. We anticipate a quick acceptance by many beekeepers since they are eager to bring this mite under control. To increase the exposure of beekeepers to the plan, we will publish the sampling plan and guidelines on the University of Minnesota webpage and in the American Bee Journal, we will continue giving presentations at local and national beekeeper meetings, and we will have the sampling device produced along with instructions. We are seeking to have the sampling device manufactured by a beekeeping supply company, which would make the device available to beekeepers across the country and world. Educational & Outreach Activities M. Spivak presented these findings to over 25 different professional and public meetings of beekeepers, scientists and the general public in 11 states across the US including 12 talks to groups within MN. She also presented in Peru, Chile, Argentina, and Nordic-Baltic countries (including Norway, Denmark, Sweden, Finland, Estonia, Latvia, Lithuania). K. Lee presented her findings to the American Bee Research Conference (Sacramento, CA, January, 2008), the International Union for the Study of Social Insects in Puerto Rico (September 2008), the MN Hobby Beekeeping Association (April 2008 and April 2007), Entomological Society of America (December 2007), and the Minnesota Honey Producers (December 2006). There are two papers by A. Ibrahim, G. Reuter and M. Spivak resulting from objective 3, and K. Lee will be submitting two for publication on Objectives 1 and 2. In addition, we will be writing at least two extension papers on the results of Objectives 1 and 2. Ibrahim A, Reuter GS, Spivak M. 2007. Field trial of honey bee colonies bred for mechanism of resistance against Varroa destructor. Apidologie 38: 67-76. Ibrahim A, Spivak M. 2006.The relationship between hygienic behavior and suppression of mite reproduction as honey bee mechanisms of resistance to Varroa destructor Apidologie. 37: 31-40. Areas needing additional study We will follow up with beekeepers that implement the sampling plan to collect feedback and address any issues. We will provide access to the sampling device and encourage beekeepers to monitor their colonies and keep records of the mite levels and the success of the apiary, which can be complied to better determine economic thresholds for commercial beekeepers.<\|endoftext\|>	3.734375
243	# How do you solve 3g - 8 = g? May 26, 2018 $g = 4$ #### Explanation: $3 g - 8 = g$ First, you need to get rid of 3g. You can't get of $g$ by itself on the right side because the equation won't be balanced. So, subtract $3 g$ from both sides: $- 8 = - 2 g$ Now, divide by $- 2$ to isolate for $g$: $- \frac{8}{-} 2 = g$ $g = 4$ because two negatives make a positive. May 26, 2018 $g = 4$ #### Explanation: As per the question, we have $3 g - 8 = g$ $\therefore 3 g - g - 8 + 8 = g - g + 8$ ... [Subtracting $g$ and Adding $8$ in both sides] $\therefore 2 g = 8$ $\therefore g = \frac{8}{2}$ $\therefore g = 4$<\|endoftext\|>	4.65625
866	Protected by the bones of the skull, your eyes lie inside hollow openings called orbits. Each eye is held in place by six eye muscles and ligaments, and is cushioned by fat. The eye muscles work together so that your eyes move to form a single image. Symptoms of double vision (diplopia) may appear when these muscles are not working together. The white outer layer of the eyeball is called the sclera. The conjunctiva is the mucous membrane covering the sclera and lining the eyelids. Inflammation or infection of the conjunctiva or sclera may cause redness, tearing, or discharge of the eye. The clear window covering at the front of the eye is the cornea, which lies in front of the colored part of the eye, the iris. The inside of the eye is divided into two chambers. The front chamber is filled with a watery liquid called aqueous humor. The larger chamber in back is filled with a soft jelly called vitreous humor. An overproduction of aqueous humor can cause a build-up of intraocular pressure which contributes to glaucoma. Degenerative changes in the vitreous humor can cause symptoms of floaters. Between the two chambers is a clear disk called the lens. The lens is held in place by rings of muscles that help the eye focus and see at varying distances. Muscles pull on the lens to make it flatter to view distant objects; they relax, allowing it to take a more curved shape for looking at nearby objects. This change in the curvature of the lens is called accommodation. Cataracts result from a natural aging process which causes the clear lens to become cloudy over time and lose its focusing ability. Patients with cataracts may have difficulty seeing at distance, reading newsprint or experience symptoms of glare. Your eye works somewhat like a camera. Light rays pass through the cornea and enter the eye through the pupil.The pupil can adjust in size to limit the amount of light coming in, much like the diaphragm of a camera. Behind the pupil, the lens bends the light rays. After passing through the fluid in the eyeball, the light rays are focused on the retina. The retina is a layer of light-sensitive nerve tissue lining the back inside of the eye. Made up of millions of cells, the retina reacts to light. There are two types of cells in the retina—rods and cones. Rods detect the amount of light —from very dim to very bright light. Cones detect the color of the light—red, green, or blue. Those who cannot see a full range of colors are “color-blind”. The center of the retina is the macula which allows us to read and see fine detail. As people age, the retinal cells in the macula can slowly break down leading to macular degeneration. This causes blurry and distorted central vision, but does not lead to complete blindness as the peripheral retina and vision are still intact. The retina reacts when light hits it, sending a message along the optic (eye) nerve to the brain. Because the light rays are bent by the lens, the image formed on the retina is upside down. As the signals are conveyed by the optic nerve, the brain analyzes the information turning the image upright and producing the single image that we see. Caring for Your Eyes Even if you are not having difficulty with your vision, periodic eye exams can identify developing problems. The sooner problems are detected, the sooner proper treatment may begin. If you wear glasses or contact lenses, visit the eye physician regularly to monitor your vision and adjust your prescription as needed. If you play sports, wearing safety glasses reduces the risk of an eye injury. People with a family history of eye disease such as glaucoma, macular degeneration or corneal dystrophy should be monitored at regular intervals for the possible development of these conditions. A dilated eye exam allows the physician to monitor the health of not only the eye, but the body as well. Patients with systemic disease such as high blood pressure, diabetes, or heart disease can benefit from annual dilated eye exams.<\|endoftext\|>	4
626	The majority of carbon emanating from fossil fuel combustion and wildfires is quickly released into the air as carbon dioxide. Scientists at the University of Zurich have recently revealed that the leftover residue, so-called black carbon, can age for millennia on land and in rivers heading for to the ocean, and thus represents a huge long-term reservoir of organic carbon. The research provides a key missing piece to the puzzle of understanding the global carbon cycle. Rivers transport black carbon from land to sea. (Credit - Gabriela Santilli, ETH Zürich) Due to its common occurrence and propensity to remain in the environment, black carbon may be one of the keys in predicting and alleviating global climate change. In wildfires, usually, one-third of the burned organic carbon is retained as black carbon deposits rather than released as greenhouse gases. To begin with, black carbon remains stored in lakes and in the soil, and is then eroded from river banks and washed into the ocean. However, black carbon is not taken into consideration in global carbon budget warming simulations, since its role in the global carbon cycle is not properly understood due to a lack of knowledge about stocks, fluxes, and residence times in the environment. First Worldwide Assessment of Black Carbon River Transport Our study is the first to address the flux of black carbon in sediments by rivers on a global scale. We found that a surprisingly large amount of black carbon is exported by rivers. Alysha Coppola, Postdoctoral Researcher, Department of Geography at the University of Zurich (UZH). The research includes some of the largest rivers around the world, such as the Amazon, Brahmaputra, Congo, and major Arctic rivers. It is the first global river assessment of the radiocarbon age values and volume of black carbon transported as particles. The scientists learned that the more total river sediment is washed by rivers to the coast, the more black carbon travels with it and is finally buried in ocean sediments, forming a crucial long-term sink for atmospheric carbon dioxide. Black Carbon can Age in Intermediate Reservoirs To gain an insight into the processes taking place in the world's rivers, the UZH scientists partnered with colleagues from ETH Zurich, and the US-based Global Rivers Observatory at the Woods Hole Oceanographic Institution and the Woods Hole Research Center. They discovered that the black carbon pathway from land to ocean is largely fashioned by erosion in river drainage basins. Astonishingly, they learned that some black carbon can be stored for thousands of centuries before being exported to the ocean via rivers. This overview is new since it was formerly always assumed that following a fire, the remaining black carbon was rapidly eroded by water and wind. However, the team learned that black carbon does not always originate from latest wildfires, but could be up to 17,000 years old, predominantly in the Arctic. This explains the mystery as to why black carbon is continuously present in river waters, regardless of wildfire history. We found that black carbon can age in intermediate reservoirs that act as holding pools before being exported to the ocean.<\|endoftext\|>	4.28125
601	Last updated on April 5th, 2019 46. Sun spots have a very strong magnet field, which prevents the convection of energy, and thus accounts for their lower temperatures. 47. In fact, the sun is about 400 times larger than the moon. 48. A typical sunspot consists of a dark spot in the middle called the umbra, and a lighter region known as the penumbra. 49. The solar maximum is a period of time during the solar cycle when the number of sunspots is at its highest. 50. Sunspot cycles are repeated every 11 years, and coincide with the occurrence of solar flares. 51. When the solar cycle is at a minimum, flares are rare because active regions are far between. 52. It travels at 20 kilometers per second relative to other stars, and 220 kilometers per second around the Milky Way. 53. Different parts of the sun rotate at different speeds, the fastest being at its equator. 54. The sun’s rotation period at the equator is about 27 days while that at the poles is about 36 days. 55. It completes a revolution around the entire galaxy once every 250 million years. Facts about its composition 56. The sun is a ball of gas and has no solid surface. 57. Its composition is 91% Hydrogen, 7.8% Helium, and 1% other gases. 58. Helium is the second most abundant element both in the sun and in the universe, but very hard to find on earth. 59. It contains different layers with varying temperatures; the corona, photosphere, chromosphere, and the core. 60. The chromosphere contains spikes of gas called spicules. 61. It is visible as a flash of color at the starting and ending of total solar eclipses. 62. The photosphere is the opaque layer of gas that makes the sun appear to be solid. 63. It’s also responsible for emitting light, and is cooler than the outer most layer, the corona. 64. The transition region is a very narrow (60 miles / 100 km) layer between the chromosphere and the corona where the temperature rises abruptly from about 8000 to about 500,000 K. 65. The corona releases a stream of charged particles referred to as solar wind. Interesting facts about evolution and life cycle of the Sun 66. The sun is currently in its yellow dwarf stage. 67. It has enough nuclear fuel to stay as it is for 5 billion more years. 68. When its energy (hydrogen) burns out, it will expand into the red giant and consume nearby planets, possibly even earth. 69. Its outer layers will then collapse, and it will become the white dwarf. 70. In the end, it will become a dim and cool celestial body referred to as the black dwarf.<\|endoftext\|>	3.84375
1,062	The Mighty Distace Function – Part 2 So in our previous step, we went over some possible approaches to define a distance function between a point and a set. At the end we saw that calculating distance between the point and the average (mean) point of the set seems to be the best possible solution. I am afraid that it is not always that easy. For the reason, consider the below figure. Both points a and a’ have equal distances from the mean point of the set B shown by the filled circle. Here, we have the usual set B, but instead we have two test points a and a’. Both test points have equal distances from the mean point of set so we have d(a,B) = d(a’,B), but it seems that point a is closer to the set than the other point. In other words a is more similar to B, or a has a higher probability of being a member of B. Therefore, a should have a smaller distance to B compared to a’, but it is not the case, why is that? The reason can be explained with a measurement called the variance, which represents the spread of the data along a given axis. Dataset B has a higher variance along the X axis than the Y axis. In the example given, the data has larger variance over the X-axis than the Y-axis. To solve our problem, we can divide the distance along each axis by the variance of the set along that axis: $d(a,B) = \sqrt{\frac{d_X(a,\mu(B))^2}{Var_X(B)} + \frac{d_Y(a,\mu(B))^2}{Var_Y(B)}}$, where $d_X(a,\mu(B))$ is the distance between a and $\mu(B)$ along the X-axis, and $d_Y(a,\mu(B))$ is the distance along the Y-axis. Division by the variance, moves our points into a new space in which the set has equal variances along both X and Y axes. By normalizing the variance along both axes, it becomes clear that point a is closer to the set than a’. As can be seen, in this new space we have $d(a,\mu(B)) < d(a',\mu(B))$. We can imagine this new space in two ways, either by squeezing the points along the X-axis, or by stretching them along the Y-axis, until the points in the set have equal variances along both axes. Now, lets move on to our next problem, given below: Point a’ has smaller distance with $\mu(B)$ and accordingly to set B, while it seems that point a is more “similar” to set B and therefore should have smaller distance. Here, a’ is closer to the mean of B compared to a, but it seems that a better follows the pattern of B and therefore should have a smaller distance. The problem can not be related to the variance, as B has equal variance along X and Y axes. So what is the problem? This problem can be seen similar to the one we saw before, with only a subtle difference. If we rotate the X and Y axes, we observe the same phenomena as the previous problem. As shown below if we rotate the dimensions, we can use the same equation in the new dimension. We need to rotate the axes and then calculate the variance. $d(a,B) = \sqrt{\frac{d_{X'}(a,\mu(B))^2}{Var_{X'}(B)} + \frac{d_{Y'}(a,\mu(B))^2}{Var_{Y'}(B)}}$, where $d_{X'}(a,\mu(B))$ is the distance between a and $\mu(B)$ along the new X-axis, and $d_{Y'}(a,\mu(B))$ is the distance along the new Y-axis. For the next problem, consider the example given below: The distance between to test points a and a’ and the set B is calculated. Although the point a’ has less distance with $\mu(B)$ (and therefore B), it seems that point a is more similar to the dataset and therefore should have less distance compared to a’. In this Figure also, point a’ is closer to the mean, while point a better follows the pattern of the set B, and therefore a should have a smaller distance while it is not the case (using the formulas discussed so far). Here, the variance along the two axes is equal, and no matter how we rotate the axes the problem remains unchanged. To solve this problem we change the axes in a different way. This time, the value of the points along the new X axis equals $(X-\mu_X(B))^2$ and the new Y values equal $(Y-\mu_Y(B))^2$ 1. This time we change the axes in a different way. The new space is the same as the one presented before, and we can define the distance function in the new space by rotating the axes and calculating the variance in the rotated space. It’s enough for this post. I just wanted to remind you that we are not done yet, so stay tuned for the next post on the mighty distance function.<\|endoftext\|>	4.5
237	# What is 1/2 to the negative fourth power? ## ${\left(\frac{1}{2}\right)}^{-} 4$ Apr 20, 2018 ${2}^{4} = 16$ #### Explanation: Recall that ${\left(\frac{a}{b}\right)}^{x} = {a}^{x} / {b}^{x}$, so ${\left(\frac{1}{2}\right)}^{-} 4 = {1}^{-} \frac{4}{2} ^ - 4$ ${1}^{-} 4 = 1$, as $1$ raised to any power is $1.$ So, we get $\frac{1}{2} ^ - 4.$ Recall that you can get rid of a negative exponent in the denominator by moving it to the numerator, making it a positive exponent, and vice versa. IE, $\frac{1}{a} ^ - x = {a}^{x}$ So, $\frac{1}{2} ^ - 4 = {2}^{4} = 16$<\|endoftext\|>	4.40625
2,268	# Bitwise XOR of rational numbers ## Introduction Every rational number between 0 and 1 can be represented as an eventually periodic sequence of bits. For example, the binary representation of 11/40 is 0.010 0011 0011 0011 ... where the 0011 part repeats indefinitely. One way of finding this representation is the following. Start with r = 11/40, then repeatedly double it and take the fractional part, recording when it goes above 1. When the value of r repeats, you know you have entered a loop. 1. r = 11/40 2. 2r = 11/20 < 1 -> next bit is 0, r = 11/20 3. 2r = 11/10 >= 1 -> next bit is 1, r = 2r - 1 = 1/10 4. 2r = 1/5 < 1 -> next bit is 0, r = 1/5 5. 2r = 2/5 < 1 -> next bit is 0, r = 2/5 6. 2r = 4/5 < 1 -> next bit is 0, r = 4/5 7. 2r = 8/5 >= 1 -> next bit is 1, r = 2r - 1 = 3/5 8. 2r = 6/5 >= 1 -> next bit is 1, r = 2r - 1 = 1/5, same as in 4. The loop 5. -> 6. -> 7. -> 8. now repeats. To get from the binary string back to 11/40, you can use the formula (int(prefix) + int(suffix)/(2^len(suffix) - 1)) / 2^len(prefix) where prefix is the initial part 010, suffix is the repeating part 0011, and int converts a binary string to integer. Given two such representations, we can perform the bitwise XOR operation on them. The resulting sequence will also be periodic, so it represents a rational number. For some rational numbers, there are two binary representations. 1/4 = 0.010000000... = 0.001111111... The choice between them can affect the result of the bitwise XOR. In these cases, we use the former representation, which has infinitely many 0s. Your inputs are two rational numbers in the half-open interval [0,1). Your output shall be the result of the bitwise XOR operation applied to the inputs, expressed as a rational number. Note that the output can be 1, even though neither of the inputs are. The exact formats of input and output are flexible, but each rational number should be represented by two integers, the numerator and denominator (with the exception of 0 and 1, which can be represented as 0 and 1 if desired). You can assume that the inputs are expressed in lowest terms. The output must be expressed in lowest terms. A built-in rational number type is an acceptable format, as long as it satisfies these restrictions. You can ignore any bounds on integers imposed by your language, but your algorithm should theoretically work for all rational numbers. The lowest byte count wins. Standard rules apply. ## Example Consider the inputs 11/40 and 3/7. We write their representations one above the other, delimiting the repeating parts by pipes \|. Then we extract repeating parts of equal lengths, and apply bitwise XOR to them and the parts before them. 11/40 = 0. 0 1 0\|0 0 1 1\|0 0 1 1\|0 0 1 1\|0 0 1 1\|0 0 1 1\|0 0 1 1\|0 0 1 ... 3/7 = 0.\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|0 1 1\|... -> 0. 0 0 1\|0 1 0 1 1 1 1 0 1 0 0 0\|0 1 0 1 1 1 1 0 1 0 0 0\|0 1 0 ... The resulting rational number is 89/520. ## Test cases 0 0 -> 0 1/2 1/2 -> 0 1/2 1/4 -> 3/4 1/3 2/3 -> 1 1/2 3/4 -> 1/4 5/8 1/3 -> 23/24 1/3 1/5 -> 2/5 15/16 3/19 -> 257/304 15/16 257/304 -> 3/19 3/7 11/40 -> 89/520 5/32 17/24 -> 59/96 16/29 16/39 -> 621001733121535520/696556744961512799 • What's the maximum period we need to support? – Neil Nov 14, 2017 at 9:54 • @Neil What makes you think that such a maximum exists? – orlp Nov 14, 2017 at 10:06 • Note: Some numbers have two binary expansions, namely those numbers where the eventual period has length one. It is implicit in the problem definition above that we must choose the representation that ends in 000... in this cases (which is also what we get if we use the algorithm with r). For example, in the case 5/8, 1/3 we get 23/24 because we choose the expansion 0.101000... for 5/8. If we choose instead 0.10011111... as 5/8, the result after XOR becomes 19/24, so this is wrong. Related to Wikipedia: 0.999... Nov 14, 2017 at 15:14 • @JeppeStigNielsen Damn... This means that unlike regular XOR (a ^ b) ^ b == a does not hold. E.g. (19/24 ^ 1/3) ^ 1/3 != 19/24. That made me lose quite a bit of excitement about this :( – orlp Nov 14, 2017 at 16:01 • Nov 14, 2017 at 20:11 # Python 3, 193 164 bytes def x(a,b,z=0): l=[] while(a,b)not in l:l+=[(a,b)];z=2z\|(a<.5)^(b<.5);a=a2%1;b=b2%1 p=l.index((a,b));P=len(l)-p return((z>>P)+z%2Pa0/~-2(P or 1))/2*p Takes input as Python 3's fractions.Fraction type, and outputs it as well. Fun fact (you can easily show this using generating functions), if you change (a<.5)^(b<.5) to ((a>=.5)and(b>=.5)) above you get the binary AND between two rational numbers. Call this nd(a, b). Then we have a + b - 2nd(a, b) = x(a, b)! • Indeed, my mistake. Apologies! (note that a link to tio included in the answer would be great) Nov 14, 2017 at 13:53 # JavaScript, 141 bytes (p,q,r,s)=>(h=(v,u)=>v%u?h(u,v%u):[a/u,b/u])(b=(g=x=>x%q\|\|x%s?g((x\|x/2)+x%2):x)(1),a=(o=b/(b-(b&~-b)),x=pb/q,y=rb/s,(x%o^y%o)+(x/o^y/o)o)) Wont work for last test case (integer overflow). Input 4 numbers for p/q xor r/s, output an array with two numbers. For testcase 0, 0, you should input 0, 1, 0, 1. How: (All numbers described here is in binary form.) 1. find the smallest number b, which b = 10 ^ p - 10 ^ q (p, q are integers, p > q); and b = 0 (mod q); and b = 0 (mod s); 2. Let x = p b / q, y = r * b / q; Convert p / q, r / s to x / b and y / b; 3. Let o = 10 ^ (p - q) - 1; split x, y to [x % o, x / o], [y % o, y / o]; get xor for each part [x % o xor y % o, x / o xor y / o], and join back to (x % o xor y % o) + (x / o xor y / o) * o; Donate it as a; 4. If a = 0, the answer is 0 (or 0 / 1); Otherwise let u = gcd(a, b); the answer is (a/u) and (b/u). f= (p,q,r,s)=>(h=(v,u)=>u%v?h(u%v,v):[a/v,b/v])(b=(g=x=>x%q\|\|x%s?g((x\|x/2)+x%2):x)(1),a=(o=b/(b-(b&~-b)),x=pb/q,y=rb/s,(x%o^y%o)+(x/o^y/o)*o)) <div oninput="[v5.value,v6.value]=['?','?'];[v5.value,v6.value]=f(+v1.value,+v2.value,+v3.value,+v4.value)"> <input id=v1 />/<input id=v2 /> xor <input id=v3 />/<input id=v4 /> = <output id=v5></output> / <output id=v6></output> <style>input{width:40px}</style> </div><\|endoftext\|>	4.5625
733	Okinawa prefecture is made up of dozens of islands that lie at the southern end of the Japanese archipelago. The region was once known as the Ryukyu islands, an independent kingdom trading across Asia. Islanders spoke Ryukyu languages, some linked to southern Japanese dialects. In 1609 the Satsuma clan, based on the southern Japanese island of Kyushu, invaded and made the Ryukyu kingdom a tributary. Two and a half centuries later the kingdom was annexed to the Japanese nation. In the early part of the 20th Century, most residents were fishermen and farmers. The capital, Naha, was on the main island of Okinawa, a long, narrow strip of land covered in mountainous forest in the north. But the outbreak of World War II was to have a dramatic effect. As the war ended Allied forces needed a secure base from which to launch an attack on the Japanese mainland. They landed on the main island of Okinawa in April 1945, sparking fierce fighting. By the time US forces controlled the island three months later, 12,000 US soldiers had been killed and 38,000 injured. More than 100,000 Japanese troops were killed. Okinawan civilians caught up in the fighting paid a very severe price. According to the Okinawa Prefectural Museum, as many as 100,000 died. Some starved to death, others were urged - some say forced - by Japanese troops to commit suicide as US victory neared. After the Japanese surrender, Okinawa was placed under US administration for almost three decades. America took over Japanese bases and began stationing troops in the region. The US then signed a security pact with Japan, undertaking to defend it in return for land for its military forces. The Communist take-over in China and eruption of conflicts in Korea and later Vietnam served to emphasise Okinawa's strategic importance to the US. Planes based in Okinawa flew missions to both Korea and Vietnam. Naval forces were also based there, the region acting as a key hub for the conflicts. In 1972 Okinawa reverted to Japanese control, but the bases remained. Today of the 47,000 US troops stationed in Japan, more than half are based in Okinawa. Regional concern over China means that Okinawa remains a key US strategic asset. Since World War II the main island had changed significantly and anti-base sentiment has been rising. The US bases occupy about 19% of the main island, according to the prefectural government, and population centres in the south are cramped. Some Japanese land owners object to the US use of their land, while those living near the bases complain about aircraft noise. Many jobs are linked to the US bases and tourism has boomed, but otherwise Okinawa remains one of the poorest parts of Japan. The 1995 rape of a 12-year-old Japanese girl by three US servicemen triggered anti-base protests across the island, the largest in more than three decades. Subsequent incidents have meant that the Okinawa issue has remained in the spotlight, and in 2006 Japan and the US agreed a deal on reducing the US troop presence. Under the deal, 8,000 US marines will be relocated to Guam, with the Japanese government to pay 60% of the cost. The controversial US airfield at Futenma will be closed and replaced by two new runways in the north of the island. But some residents say that is not good enough. Residents of the area that would host the new base are opposed to the idea - and their stance has so far prevented the plan moving ahead.<\|endoftext\|>	3.75
431	In the second half of the 4th century BC, Armenia was divided by the Achaemenids into Greater and Lesser Armenia. From the beginning of the 4th century BC, the weakened power of the Achaemenids embarked upon the path of decay. In 334 BC, Macedonian King Alexander the Great crossed the straits of the Black Sea, invaded the possessions of the Persian kings, and within one year seized the western possessions of the Achaemenid state. In 331 BC, in the Battle of Gaugamela, Alexander the Great defeated the Persian army led by the king of Persia Darius III. Soon, Darius III was killed by his own entourage. In the battle of Gaugamela, Armenian troops from Greater and Lesser Armenia fought on the Persian side. A squad from Greater Armenia was led by satrap Yervand. Having conquered the Achaemenid state and made a campaign to the East towards Central Asia and India, Alexander the Great created a gigantic state, choosing Babylon as its capital. As a result of the conquests of Alexander the Great in Western Asia, the Hellenistic period began, the most important feature of which was the widespread and fruitful interaction of the Hellenic (Greek) civilization with the local Eastern civilizations. The troops of Alexander the Great did not invade Armenia. A participant of the Battle of Gaugamela, Yervand, reigning in Armenia, restored the statehood whose traditions had been laid in the 6th century BC. At this time, Lesser Armenia also sought independence. The Armenian ruling dynasty of Yervandunis, which was related to the Achaemenids, has accumulated considerable experience of state governance of Armenia. This significantly strengthened the position of the Armenian Kingdom in the system of the Hellenistic states that formed after the collapse of Alexander the Great’s empire. After the sudden death of Alexander of Macedon in 323 BC, his empire was divided between his generals. Source: History of the Armenian people in questions and answers (Russian-Armenian (Slavic) University)<\|endoftext\|>	3.96875
240	# How do you factor ab + 4 + a + 4b? Mar 16, 2018 Group it together $a b + a + 4 + 4 b$ As you must have noticed by now..... it is the only way $a b + a = a \left(b + 1\right)$ $4 + 4 b = 4 \left(b + 1\right)$ Put them together $a \left(b + 1\right) + 4 \left(b + 1\right)$ $\left(a + 4\right) \left(b + 1\right)$ Mar 16, 2018 $\left(a + 4\right) \left(b + 1\right)$ #### Explanation: $a b + 4 + a + 4 b$ $\implies a b + a + 4 b + 4$ $\implies a \left(b + 1\right) + 4 \left(b + 1\right)$ $\implies \left(a + 4\right) \left(b + 1\right)$<\|endoftext\|>	4.625
736	# How to Easily Teach Comparing and Ordering Numbers in 5 Days Are you an upper elementary teacher looking for engaging ways to teach big kids about comparing and ordering numbers? This blog post outlines easy to implement strategies for teaching grade 4 and grade 5 students ways to compare large numbers. ## Day 1: Introduce the math concept with a PowerPoint Presentation Teaching Tip: I LOVE having students use personal whiteboards during introductory lessons. They are SO MUCH more willing to try out new things and make mistakes if they know that they can simply swipe that error away. ## Days 2, 3 and 4: Math Workshop ### Activity 1: Whole Group Warm Up (10-20 minutes) This fun game will help your students to practice ordering numbers. I like to use this as a warmup. Every student needs a card that has a number on it. Explain to students that you are going to set a stopwatch to time how long it takes them to put themselves in order from LEAST to GREATEST based on the card that they received. Encourage students to explain their reasoning when trying to find their spot in the line. Once students are in the correct order, mix up the cards and play again. This round students are trying to beat their previous time. ### Activity 2: Station Rotations (40- 50 minutes) – I like to give each group between 20-25 minutes at each station) ## Day 5: Review and Assessment Day 5 is for summative assessment. I like to give a short and sweet 2 page check in to ensure that all of the kiddos fully understand the concept of comparing and ordering numbers. ## What centers can I use for teaching Comparing and Ordering Numbers? ### 1) Teacher Time I suggest starting each teacher time rotation with a quick formative assessment. This will allow you to see where the kids are at with the concept. Here is a possible assessment question: Order the following numbers from GREATEST to LEAST. 5 672, 5 627, 5 726 Differentiation: You could make this question easier by asking students to order 726, 672 and 627. You could make this question more difficult by turning the numbers into decimals. Students could compare 5.72, 5.67 and 6.62. The next thing that I do is go over the anchor chart with students. This helps them to remember what was taught during the whole group lesson. Next we start working through the different problems in their flip books together. I introduce a type of problem with an example (on the top of the flip book page). Then I give students around 3-5 minutes to try out some of the problems. After a few minutes we move onto another problem type. It is quite common for students not to finish all of their practice problems but that is okay because they can take it home to practice, finish it up during morning work, etc. As they work, I help individual students and ensure that they get the one on one time that they need. ## Where can I find all of these lessons and activities? If you don’t have time to create all of these games and flip books yourself, I have put together a comparing and ordering numbers mini unit. It will take all of the guess work out of creating a meaningful, rigorous and fun place value unit for your students. Click on the image below to check it out. ## I’m new to Math Workshop. How does it work? Check out this awesome blog post all about launching math workshop in your upper elementary classroom.<\|endoftext\|>	4.53125
711	## Create A Beautiful Blog Easily. Blogdot is simple and easy to use blog theme. It is designed and developed primarily to create professional blogging websites. The set containing all the solutions of an equation is called the solution set for that equation. If an equation has no solutions, we write ∅ for the solution set. However, for the vast majority of the equations and inequalities that we will be looking at will have simple enough solution sets that it's just easier to write down . Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. A solution set is the set of all variables that makes the equation true. ## solution set graph In mathematics, a solution set is the set of values that satisfy a given set of equations or inequalities. For example, for a set { f i } {\displaystyle \{f_{i}\}} \{f_{i}\} . A solution set is the set of values which satisfy the given inequality. It is exclusive, as only the values, and all the values must satisfy the. A linear system with a unique solution has a solution set with one element. A linear We will write them vertically, in one-column wide matrices. { (2 − 1 0 0) +. Learning to solve and graph the solution set of an equation is a skill that will serve you well through the rest of your math education. The same three steps work. Understand the difference between the solution set and the column span. Recipes: parametric vector form, write the solution set of a homogeneous system as a. To find the solution set of an equation with a given domain, you first need to plug each value in Create ordered pairs from these values and write them as a set. ## solution set notation In this tutorial, you'll learn the definition of a solution set and see an example. Take a Create ordered pairs from these values and write them as a set. That set. Interval notation is a common way to express the solution set to an inequality, and it's Writing the set for this figure in interval notation can be confusing. x can . You can think of the solution set as having four free variables (let's say y,z,w,v) and one bound variable (x). The solution is a four-dimensional. Or you can also write the system in the following form and reduce it into r: . got from (a) to any one particular solution to the nonhomogeneous linear system. Guidance on how to write a clear solution for mathematical problems, We set that equal to our other volume expression and get h_a= rS/[A]. Solve the inequality. Graph the solution set, and write the solution set in set- builder notation and interval notation. (See Examples 1–3). − + a ≤ a + Learn the representation, properties, formulas and types of sets with examples. For example: Write the following sets in set builder form: A={2, 4, 6, 8}. Solution. To solve an equation in which an absolute value equals a positive constant, solve it TWICE once using 15, in this case, and the second time. sammy s. asked • 09/07/ Write the solution set using interval notation and graph the solution set on a number line - please provide steps. Solve the compound. Get an answer for 'Write the solution set of /x-3/>5 in set builder notation' and find homework help for other Math questions at eNotes.<\|endoftext\|>	4.5
684	Monday, January 26, 2015 Valentine Math - Simple Steps into Multi-Step Problem Solving Looking at the heart shaped tangram type puzzles, floating around this time of year, I spotted an easy opportunity to practice multi-step problem solving skills with my younger children (ages 8-11). One of the first hurtles students hit in high school geometry is the switch over from simple equations to multi-step problem solving. This transition can be made easier by introducing younger children (long before they reach high school geometry) to the concept in simple ways through puzzles, or games, using the math they already know, or can easily grasp. So, back to the tangram - if you ask a child to find the area of a heart shape, like the one below, they will most likely be stumped. It is not one of the usual shapes found in math books. There is no simple formula for the area of heart. So, we come to the first step of multi-step problem solving, determining what we already know. Depending on the age of the children, they will probably know the formal for finding the area of a square. length x width = area They might also know how to find the area of a triangle. 1/2 base x height = area And, most will be able to put the two together to find the area of a right trapezoid (even if they don't know what it's called, or realize they have just taken a baby step into the kind of problem solving we're talking about). Once they've mastered the trapezoid, finding the area of a parallelogram should be a snap. height x base = area Some children will even already know the formula for the area of a circle. If they don't, they can learn it pretty easily, even if they don't fully understand its meaning. It might be a good time to pull out Cindy Neuschwander's Sir Cumference books. πr² = area So, that's what we know. Be what we need to know is the area of a heart. ? = area How can we use what we know, to find out what we need to know? That's where the tangram comes into play. By matching the colors of the shapes within the tangram to the shapes the children already know how to solve for area, the problem becomes simple. And hopefully, at this point, children will be able to see on their own, that for this particular heart, the problem can be solved with fewer steps, as well. We will do each of the steps above, with paper cut-out of the shapes, too. Just to add an element of familiar, hands on learning to the process. Michelle said... Absolutely love this! Thank you for sharing! Natalie PlanetSmartyPants said... Oh how I love it! Brilliant! I have to show it to my 8 year old! Angelic Scalliwags said... Great stuff! Sunita said... Awesome! Thank you. I will do this with my 4th grader (who hasn't done area of circles, yet, but he can just estimate the curved parts!)! Ticia said... We are doing that next week! Brilliant, I now have a great math activity.<\|endoftext\|>	4.84375
683	### Home > CCG > Chapter 7 > Lesson 7.2.2 > Problem7-69 7-69. Jester started to prove that the triangles at right are congruent. He was only told that point $E$ is the midpoint of segments $\overline{AC}$ and $\overline{BD}$. Copy and complete his flowchart below. Be sure that a reason is provided for every statement. $\enclose{circle}{\; \\ \qquad \text{E is a midpoint} \qquad \\}$ $\huge \swarrow$Bubble 1 has an arrow pointing down to bubble 3. $\huge \searrow$ Given Bubble 1 has an arrow pointing down to bubble 4. $\enclose{circle}{\; \\ \qquad \angle AEB \cong \angle CED \qquad \\}$Angle A, E, B, is congruent to angle C, E, DVertical angles are congruent. $\enclose{circle}{\; \\ \qquad \ \ \ \qquad \qquad \\}$Fill in bubble 3. $\enclose{circle}{\; \\ \qquad \overline{AE} \cong \overline{CE} \qquad \\}$A, E is congruent to C, E.Definition of midpoint. $\huge \searrow$Bubble 2 has an arrow pointing down to bubble 5. $\huge \downarrow$Bubble 3 has an arrow pointing down to bubble 5. $\huge \swarrow$Bubble 4 has an arrow pointing down to bubble 5. $\enclose{circle}{\; \\ \qquad \ \ \ \qquad \qquad \\}$Fill in bubble 5. $\enclose{circle}{\; \\ \qquad \text{E is a midpoint} \qquad \\}$ $\huge \swarrow$Bubble 1 has an arrow pointing down to bubble 3. $\huge \searrow$ Given Bubble 1 has an arrow pointing down to bubble 4. $\enclose{circle}{\; \\ \qquad \angle AEB \cong \angle CED \qquad \\}$Vertical angles are congruent. $\enclose{circle}{\; \\ \qquad \bf{\color{red}{\overline{BE}\cong \overline{DE}}} \qquad \\}$$\bf{\color{red}{\text{Definition of midpoint.}}}$ $\enclose{circle}{\; \\ \qquad \overline{AE} \cong \overline{CE} \qquad \\}$Definition of midpoint. $\huge \searrow$Bubble 2 has an arrow pointing down to bubble 5. $\huge \downarrow$Bubble 3 has an arrow pointing down to bubble 5. $\huge \swarrow$Bubble 4 has an arrow pointing down to bubble 5. $\enclose{circle}{\; \\ \qquad \bf{\color{red}{\triangle AEB \cong \triangle CED}} \qquad \\}$$\quad \quad \quad \quad \quad \quad \quad \bf{\color{red}{SAS \cong}}$<\|endoftext\|>	4.46875
227	Swine smallpox is a vesicular condition caused by a virus and mainly transmitted through lice. Alternative names: poxvirus smallpox This disease is caused by a poxvirus that survives outside the pig during long periods of time and is resistant to environmental changes. - Vesicles characterized by small round reddened areas having a diameter of 10 - 20 mm starting as a vesicle containing a yellowish-colored fluid in the center. - After two or three days the vesicle ruptures and a scab appears that will gradually turn black. - Injuries can be seen anywhere in the body but are most common along flanks, abdomen and occasionally the ears. - It can produce a secondary dermatitis. - Not very common in piglets although it may be congenital. Causes / Contributing Factors - It can be spread by lice or scabies. - Skin abrasions. - Fights and mixing of pigs. - Clinical picture with lab tests confirmation. - There is no treatment. - Eliminate lice.<\|endoftext\|>	3.75
2,080	An emulsion is a blend of two or more immiscible liquids or liquid and gas, with droplets of one phase distributed in the other phase. Examples of food emulsions are mayonnaise, milk, and some salad dressings. Emulsions contain both a dispersed and a continuous phase with a boundary between them called the interface. The boundary is where the emulsifier sits to create stability. Types of Emulsions - Oil in water, - Water in oil - Air in Water - Air in Oil Without emulsifiers, these are unstable systems in which the dispersed phase droplets tend to agglomerate or coalescence and separate out. The best example of this is oil and water. When oil and water are mixed together, the oil will quickly agglomerate and float to the surface. The addition of an emulsifier will allow the oil and water to mix and not separate. Emulsifiers contain both hydrophilic (water-loving) and lipophilic (Fat or Air Loving) parts. The hydrophilic end is easy to hydrate (water soluble) and the lipophilic end is very difficult to hydrate (oil soluble). The emulsifier will arrange itself with the water hydrophilic end pointed towards the water and the lipophilic end pointed at the fat. Emulsifiers arrange themselves at the interface between oil and water and reduce the surface or interfacial tension, or to allow them to mix, thereby making the emulsion more stable. In food emulsifiers, the hydrophilic or water loving part typically is glycerol, sorbitol, sucrose, propylene glycol or polyglycerol. The lipophilic or fat/air loving part is formed by fatty acids derived from fats and oils Emulsion stability refers to the ability of an emulsion to resist change in its properties over time. There are four types of instability in emulsions: Flocculation occurs when there is an attractive force between the droplets, so they form flocs, like bunches of grapes. The droplets rise to the top of the emulsion under the influence of buoyancy Coalescence occurs when droplets bump into each other and combine to form a larger droplet, so the average droplet size increases over time. - Ostwald ripening. Small crystals or sol particles dissolve, and redeposit onto larger crystals or sol particles. The stability of an emulsion depends on: - Droplet size - a smaller droplet size facilitates emulsion stability. The droplet size can be influenced by homogenization. - Smaller droplets are more stable. - Viscosity of continuous phase: - a higher viscosity facilitates emulsion stability. The viscosity of the water phase can be influenced by the addition of thickening agents such as gums and starches. - Specific density of the two phases: - if the difference in density between the two phases is small,the emulsion will be more stable. - Quality of the interfacial film: - the film can consist of emulsifiers and/or proteins. Emulsifiers are mixtures During the esterification process the fatty acids are distributed at random, so a mixture of mono- and diglyceride (esters) are formed. Distillation will reduce but not eliminate the minor components. Commercial emulsifiers are characterized by the major component, but minor amounts of related molecules will also be present. Fatty acid composition The type of fatty acid (the type of fat or oil used) influences the characteristics of the emulsifier. The fatty acid composition influences the melting point of fats and oils, and in the same way influences the melting point of emulsifiers. Fatty acids with a longer chain length and which are more saturated will result in higher melting points. Monoglycerides are also characterized by the fatty acid composition, saturated or unsaturated fatty acids. More saturated emulsifiers have a lower iodine value. Emulsifier Crystal form. Fats and oils can form different crystals which affect the texture and physical properties of the fat. The three major crystalline forms are referred to as alpha-, beta prime- and beta-crystals. The alpha crystal has the lowest melting point and forms fine and flexible agglomerates. Some emulsifiers are referred to as alpha tending emulsifiers; these are most stable in the alpha crystalline form. The alpha crystalline formation is very effective for whipping properties. HLB: Hydrophilic Lipophilic Balance Emulsifiers can be characterized by the Hydrophilic-Lipophilic Balance. The balance is measured on molecular weight and is an indication of the solubility of the emulsifier. The HLB scale varies between 0 and 20. Emulsifiers with a low HLB value are more soluble in oil and creates water-in-oil emulsions. Emulsifiers with a high HLB value are more soluble in water and creates oil-in-water emulsions. The HLB value of an emulsifier can be used as an indication of its possible use. An emulsifiers HLB can vary within one group or type. For example sucrose esters, and lecithins can be made with very high or very low HLB values Typical HLB values of common food emulsifiers Glycerol Monostearate 3.8 Sorbitan Monooleate 4.3 Sorbitan Monostearate 4.7 Sorbitan Monopalmitate 6.7 sodium Stearoyl Lactylate 10-12 Polysorbate 60 14.9 Polysorbate 80 15.0 Sucrose Esters 1-16 Emulsifiers concentrate to the surface between the oil and water phase, two liquid phases. Emulsifiers will also concentrate on any surface of two immiscible phases such as: - the interface of gas and liquid as air bubbles in a whipped cream or cake batter. - the interface of solids and liquids as ice crystals in ice cream or sugar crystals in chocolate. Emulsifiers influence the consistency, the viscosity and the texture of many multi-phase food systems. Interaction with starch Emulsifiers with fully saturated long fatty acid chain form complexes with amylose starch. The fatty acid chain penetrates the amylose helix and prevents starch retrogradation or the recrystallization of the starch. Retrogradation is the mechanism responsible for staling of bread Typically a monoglyceride, SSL and Datem are used for starch complexing in yeast raised baked items. The long saturated chain of the fatty acid will complex in the helix of the straight chain starch. Another attribute of the use of emulsifiers in starch-based products is the reduction of stickiness in reconstituted products such as pasta and instant mashed potatoes, and tortillas. Interaction with protein Emulsifiers with an ionic structure can interact with proteins, particularly wheat gluten. This interaction strengthens the gluten network in yeast raised dough resulting in increased volume, tolerance to freezing and improved crumb structure. Typically SSL and Datem are used to increase gluten strength. This type of protein interaction is the disulfide bond in the gluten matrix. Ethoxylated monoglycerides also reinforce the gluten matrix, but they do with hydrogen bonding. Ionic emulsifiers such as SSL can interact with proteins in sauces and gravies to improve mouthfeel. Interaction with fat Emulsifiers are fat-like substances and they influence fat by promoting or inhibiting crystallization, influence the crystal shape of the fat, and improve the dispersion of fat crystals inside the food product. Fats can exist in more than one crystalline form. The number of forms will vary by fat. Most fats can exist in alpha, beta, and beta prime. Cocoa Butter can exist in many more crystal forms. Certain crystals are preferred in certain products and emulsifiers can be used to promote and retain the desired crystal structure. For example, chocolate bloom is a type of undesirable crystal change that occurs that will give the chocolate a dull appearance or mottling. Lactic acid esters, sorbitan esters, and polysorbates can delay this crystal change. Alpha Crystal State - Melts at a lower temperature than beta or beta prime. Typically this form does not give the best texture. It is also a very short lived crystal. Beta Prime - This is typically the desired crystalline state. It provides a nongrainy and creamy texture. Beta- This is commonly found in confectionery to provide shine and gloss. It also helps promote flaky layers in biscuits and danish. Emulsifiers can also act as: lubricant and release agent, lubrication of extruded products but also lubrication of food processing equipment. Lecithin is commonly used as a flow and release agent. Monoglycerides are commonly used to aid in extrusion of pasta and cereals Release or anti-sticking agent, for de-molding purposes during food processing or anti-sticking to confectionary packaging. Lecithin is commonly used in pan release sprays. Optimal emulsifier functionality is determined by the correct food processing conditions. Common emulsions are unstable and do not form spontaneously. Energy input in the form of mixing, shaking, stirring or homogenizing is needed to form the emulsion. Emulsifiers also need to be heated above their melting temperature to become functional. When it is not possible to provide enough heat or shear in the process a pre-hydrated or melted emulsifier is needed. Emulsifiers can be melted into shortening to create an emulsified shortening or they can be melted/pre-hydrated in water. Pre-hydrated an emulsifier before use can increase the functionality significantly.<\|endoftext\|>	3.796875
879	About Diabetes Mellitus Diabetes mellitus is a chronic disease that affects the body's ability to use glucose, a simple sugar used as the basic form of energy in most tissues, particularly nervous and muscular. The three most common types of diabetes are type 1, type 2 and gestational, and the general risk factors for diabetes depend upon the type. Diabetes presents more complications the longer a patient lives with the disease. Careful management of the disease may slow the progression and lower the possibility of complications until later in life. Many patients, especially those which do not intensively manage their blood glucose levels, will experience a number of complications which may include eye disease (retinopathy), kidney disease (nephropathy), nerve disease (neuropathy) and diseases of the cardiovascular system. Type 1 Diabetes Type 1 diabetes mellitus (T1DM) is an autoimmune disease where the body's immune system attacks and destroys parts of itself. In the case of T1DM, the immune system attacks the β-cells of the pancreas, the cells responsible for the development and secretion of insulin. Insulin allows glucose to enter cells and lowers the levels of circulating blood glucose. Type 1 diabetics require regular injections of exogenous insulin to maintain appropriate blood sugar levels, along with proper management which includes diet and exercise. However, since T1DM tends to develop during childhood and adolescence, patients may begin experiencing complications as early as middle age. Risk factors are not entirely exact, but seem to follow certain patterns. The single greatest risk factor is that of genetics. Certain gene types indicate an increased risk of disease development. Also, if another family member has had T1DM, risk increases. Non-Hispanic whites are at a higher risk of T1DM than other races. Oddly, geography is another risk factor; likelihood of disease increases the further from the equator a patient lives. Other risk factors include exposure to certain viruses and low vitamin D levels, however early intake of cow's milk may increase chances of diabetes as well. Type 2 Diabetes Type 2 diabetes mellitus (T2DM) occurs when chronically high levels of circulating blood glucose (hyperglycemia) causes chronically high levels of circulating insulin (hyperinsulinemia). Over time this causes the cells to no longer recognize insulin in a process called desensitization. As this happens, the pancreas makes less and less insulin, called insufficiency. The body develops a condition called insulin resistance. Complications of T2DM are numerous, and unfortunately, nearly 75 percent of patients who develop the disease will die of a cardiovascular complication. Of the risk factors associated with T2DM, several are entirely dependent upon the individual patient's lifestyle. Patients with excessive fat tissue are at the greatest risk of disease; adipose (fat) tissue increases insulin resistance. Sedentary lifestyles increase risk because muscular activity uses glucose for energy and increases cellular insulin sensitivity. Other risk factors with a degree of control is the development of impaired glucose tolerance, or prediabetes, and gestational diabetes during pregnancy. Patients have no control over some risk factors, such as family history, race and age; older patients and black, Hispanic, American Indian and Asian American patients have increased risk. Gestational diabetes mellitus (GDM) develops in pregnant women as a result of placental hormones which interfere with the actions of insulin. While both the mother's and fetus's blood sugar rise after eating, certain placental hormones impair the insulin reaction in the mother and impaired glucose tolerance occurs. This usually occurs later in the pregnancy and, for most mothers, blood glucose levels return to normal soon after delivery. Some of the risk factors for GDM are beyond the scope of a patient's control, such as family history, race and age; pregnant mothers who are of a minority race or over 25 years of age are at an increased risk. Lifestyle-associated risk factors are being overweight and inactivity. Other risk factors include having GDM during an earlier pregnancy, previous delivery of a child weighing more than nine pounds and experiencing an unexplained stillbirth. American Diabetes Association: Genetics of Diabetes Mayo Clinic: Type 1 Diabetes: Risk Factors Mayo Clinic: Type 2 Diabetes: Risk Factors Mayo Clinic: Gestational Diabetes-Risk Factors<\|endoftext\|>	3.671875
1,144	It has sometimes been nicknamed "doorway to the underworld" and locals tend to stay away from it, but the Batagaika crater, located in the Yana river basin in eastern Siberia, fascinates scientists from around the world. Craters like this one are known by geologists as 'slumps', with the largest (more than 5 hectares in area) referred to as 'megaslumps'. The Batagaika crater is the largest of them all, exceeding 50 hectares in area. It is also 86m deep. Slumps and megaslumps are scattered across Siberia, Alaska and Canada. As the climate continues to warm the thawing of permafrost in these regions accelerates, making them grow bigger and bigger. Russian and German scientists are currently monitoring Batagaika to see how fast it is expanding, but another team, led by Julian Murton from the University of Sussex, has taken a different approach in its study of the site. The researchers are looking at what the megaslump can tell us about the history of ancient climates and what Siberia looked like thousands of years ago. Their findings from a first reconnaissance study were published in February. When and why did the Batagaika crater form? The crater began in the 1960s when locals cut down some trees in the area. Because trees shade the soil from the summer heat, this had the effect of increasing the amount of heat entering the soil. Trees also transpire, which has the effect of cooling the soil, so when their number was reduced, this had an impact on the soil's temperature. The permafrost started thawing from the top down. There was a gully that extended down a slope, it started deepening soon after the trees were cut down and the permafrost started to melt, leading to the creation of the megaslump. Murton told IBTimes UK: "The formation of the megaslump was triggered by human disturbance but it's symptomatic of climate warming – where the climate warms you get more permafrost thawing. This occurred at the end of the last Ice age and previous Ice Ages". What are the consequences of the crater's expansion? Monitoring of the megaslump only started a few years back. In the last two decades, there hadn't really been any kind of surveillance system in place, even though satellite images clearly show that since the 1960s, it has expanded dramatically. However, scientists haven't yet been able to calculate at what rate it is expanding. As it gets wider, the landscape collapses by about 50 or 60 metres and the original forest disappears. New, highly gullied terrains have developed at the bottom of the slump. These terrains are what geologists call "badland". In the south west of the USA you have examples of these very triangular, irregular slopes developed by rapid water erosion. It is thought these badlands will develop extensively at the bottom of the megaslump. So as the megaslump expands, the original fairly smooth forested hills disappear to be replaced by this gullied, irregular topography. What kind of sediments have been exposed by the expanding crater? Murton explains that the goal of his team's work at the crater is to look the region's geological history and to reconstruct it by sampling frozen permafrost. About five layers of sediment have been exposed by the expanding megaslump. They are comprised of ice and a lot of frozen sand. There were also a couple of wood-rich layers that represent former forests. "Our work so far has just been reconnaissance. We identified what the layers are and did a bit of reconnaissance sampling to radiocarbon date organic material, things like roots and wood in the sequences, to give us some indication of their age", Murton explained. What can the crater tell us about past climates? These layers can take them back to 125,000 years ago in their study of past climate – this is the last period in geological time when the climate in the Arctic was substantially warmer than it is now, by a few degrees. "We may even be able to go back to 200,000 years, but there is still a bit of uncertainty in the dating, we need to do more work at the site. One hypothesis is that the lower wood layer represents an ancient forest that developed about 200,000 years ago", Murton added. The scientists also looked some of the sand grains to understand what processes deposited them – finding evidence for past windy climates, with a lot of wind having blown and deposited the sand around. If they get sufficient funding, they want to do some drilling to get a continuous sequence, a core of frozen permafrost extending throughout the deposit. They would then like to do high-resolution analysis of the sediment to develop a clearer picture of what climates looked like over the last 125,000 to 200,000 years. How can this work help us to understand future climate change? The climate is warming particularly rapidly in the Arctic. The scientists want to reconstruct past climates and environments, for example from 125,000 years ago, to understand what the region looked like when it was warmer than it is now – as it will potentially be in the future. "If you know what the ecology was, what the plants and animals were in this part of the Arctic, this can give us some indication of what it may be in the future as climate continues to warm or what some of the vegetation responses we may see in the landscape," Murton concluded.<\|endoftext\|>	4.1875
692	# Triangle missing side example ## Video transkripsjon - [Instructor] The triangle shown below has an area of 75 square units. Find the missing side. So pause the video and see if you can find the length of this missing side. Alright, now let's work through this together. They give us the area, they give us this side right over here, this 11. They give us this length 10, which, if we rotate this triangle you can view it as an altitude. And in fact let me do that. Let me rotate this triangle, because then I think it might jump out at you how we can tackle this. So let me copy and let me paste it. So if I move it here, but I'm gonna rotate it. So if I rotate, that is our rotated triangle and now it might be a little bit clearer what we're talking about, this length x that we want to figure out, this is our base. And they give us our height and they give us our area. And we know how base, height and area relate for a triangle. We know that area is equal to 1/2 times the base times the height, and they tell us that our area is 75 units squared. So this is 75 is equal to 1/2. What is our base? Our base is the variable x. So let's just write that down. 1/2 times x and then what is our height? Well, our height is actually the 10. If x is the length of our base, then the height of our triangle is gonna be 10, we actually don't even need to use this 11. They're putting that there just to distract you. So, this is going to be our height, times 10. So 75 is equal to 1/2 times x times 10, or, let me just rewrite it this way. We can say 75 is equal to 1/2 times 10 is equal to five times x is equal to five, let me do the x in that same color, is equal to five times x. So what is x going to be? There's a couple of ways you could think about it. You could say five times what is equal to 75? And you might be able to figure that out. You might say, OK, five times 10 is 50, and then let's see, I need another 25, so put another five there, so it's really five times 15, or you could do it a little bit more systematically. You can divide both sides by what you're multiplying by x. So if you divide this side by five, five times x divided by five, well, you're just going to have an x left over. But these two things were equal, so you can't just do it to one side, you have to do it to both sides. So you have to divide both sides by five. And what's 75 divided by five? Well that is 15. So you get x is equal to 15. And you can verify that. If x is equal to 15, base times height times 1/2. Well, it's 15 times 10 times 1/2, or 15 times five which is going to be 75 square units.<\|endoftext\|>	4.6875
281	Proofs Involving the Triangle Inequality Theorem — Practice Geometry Questions In geometry, the triangle inequality theorem states that when you add the lengths of any two sides of a triangle, their sum will be greater that the length of the third side. By using the triangle inequality theorem and the exterior angle theorem, you should have no trouble completing the inequality proof in the following practice question. Complete the following proof by adding the missing statement or reason. Prove: ET > TV What is the statement for Reason 2? What is the reason for Statement 3? What is the missing angle in Statement 4? What is the reason for Statement 5? What is the reason for Statement 6? Answers and explanations A bisector divides an angle into two congruent angles. The exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles. are the two nonadjacent interior angles of The exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles of the triangle; therefore, You found that The whole is greater than its parts, which means that the transitive property finds that In a triangle, the longest side is opposite the largest angle. You discovered that In a triangle, the longest side is opposite the largest angle, so ET > TV.<\|endoftext\|>	3.890625
1,091	(@^<>^@) @<(!)>@ #(:<^-^>:)# (^&-^^-&^) (PS: That’s the introduction by the way 🙂 ) So, back to business, our current topic is Measurement & Probability. For measurements, to calculate a triangle’s area, you just times the length with the hight then divide them by two. Like this : length x hight / 2 = Area. Also 1m=100cm, 1cm=10mm; 1000m= 1km. There are 5 main words in probability, they are: Impossible – it will NEVER happen. Unlikely – not likely to happen. Even chance – 50% chance of accuring. Likely – quite likely to occur. Certain – Will occur or happen. Now, question time. Measurements: Q1, Area of this triangle: H- 13 L- 21; Q2, 1243m= ( )mm; Probability: Q1, Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5? Q2, A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? I enjoyed doing the probability unit the most. Because in measurement unit we need to calculate a lot I’m not really good at. We were working on a video for math class in the last couple days. We were learning about areas and measurements. This is video is about my team making and introducing a kind of bottle that contains a special kind of drink. Enjoy. 🙂 The unit question for this unit measurement is “How can we calculate using measurement”. We have a lot of tools of measuring like rulers, clocks, thermometers and watches. I use online stopwatches to calculate time. Factual question: what does the unit “mm” stands for? Conceptual question: how do you calculate the area of a rectangle. Debatable question: is internet a good tool of measuring? There are three different main unit of measuring the length. They are centimeter – cm, meter – m, kilometer – km. (I ordered them from small to big by the way. 🙂 ) For example, using a rule to measure the length of the table. Which has those units on it. If you want to measure a area of a rectangle, you just times the length (blue line) with the width ( red line). If you want to measure a square’s area, it’s really easy. You just times any of the edges of that square with it’s self. If you want to measure a triangle, like this. You just time the length and width and devoid it by 2. Because it’s just a half of this rectangle. Hey guys today I am going to tell you the correct way of reading decimals and the wrong way of reading decimals (which you should not use). 1. Printing Machine 112.5cm in length. As you can see we had measured the printer on the side. We can also say that it is 1.125m. These are the same things but they just change the decimal place and measurement. This is a appropriate way of showing it. Here is an inappropriate way of showing it —-> 0.001125km. This is way to complicated as it is a huge measurement. 77cm in length. This is the correct way of showing the measurement. I you show your measurement like this —> 7.7 decimeter. This is probably not the best because decimeters is a confusing for many people. 3. Earthquake Helmet 26cm in length. This one way of showing the measurement. You can also say 0.26m. Although 26cm is better. 260,000,000 nanometers is also 26 centimeters but this is just you know waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay tooooooooooooo COMPLICATED!!!!!!! So people, use the correct way of showing your measurement otherwise you will be having many comments. Do you know what is fraction? No? Me neither. Just joking. 🙂 Of course I know, and if you ask me which means you are asking the right guy. This is a friction, up is numerator and below is denominator. I’ve got two red marker pen on my right hand and one blue marker pen on my left hand. That’s an example of frictions. Guess what friction is that……………. Yes, YOU ARE RIGHT, it’s 1/2. Now, a more complex example. I have 15 pencils in total and 13 of them is color pencil on the right side of the photo. So the color pencil is 13/15. Now is really important, because I’m going to show a WRONG example. Look, my homework paper is on the left!! And my friend’s paper is on re right!!.<\|endoftext\|>	3.6875
1,073	Occupancy Probability Eight Sided Die (Octahedron) For this problem, we'll calculate the probabilty of getting all 8 numbers after rolling an 8-sided die 20 times. STEP 1 The set of numbers on the die (1 through 8), would be called called "n" and the number of trials or attempts is called "r" which in this case is 20 rolls. Calculating the value of "n" raised to the power of "r": 820 which equals 1,152,921,504,606,850,000 This number represents all the possible results from rolling an 8 sided die In steps 2, 3, 4 and 5, we will determine how many of those 820 rolls, will contain all 8 numbers. STEP 2 We first must calculate each value of "n" raised to the power of "r". Rather than explain, this is much easier to show: 820 = 1,152,921,504,606,850,000 720 = 79,792,266,297,612,000 620 = 3,656,158,440,062,980 520 = 95,367,431,640,625 420 = 1,099,511,627,776 320 = 3,486,784,401 220 = 1,048,576 120 = 1 STEP 3 Next, we calculate how many combinations can be made from "n" objects for each value of "n". This is much easier to show than explain: 8 C 8 = 1 7 C 8 = 8 6 C 8 = 28 5 C 8 = 56 4 C 8 = 70 3 C 8 = 56 2 C 8 = 28 1 C 8 = 8 Basically, this is saying that 8 objects can be chosen from a set of 8 in 1 way 7 objects can be chosen from a set of 8 in 8 ways 6 objects can be chosen from a set of 8 in 28 ways 5 objects can be chosen from a set of 8 in 56 ways 4 objects can be chosen from a set of 8 in 70 ways 3 objects can be chosen from a set of 8 in 56 ways 2 objects can be chosen from a set of 8 in 28 ways 1 object can be chosen from a set of 8 in 8 ways STEP 4 We then calculate the product of the first number of STEP 2 times the first number of STEP 3 and do so throughout all 8 numbers. 1,152,921,504,606,850,000 × 1 = 1,152,921,504,606,850,000 79,792,266,297,612,000 × 8 = 638,338,130,380,896,000 3,656,158,440,062,980 × 28 = 102,372,436,321,763,000 95,367,431,640,625 × 56 = 5,340,576,171,875,000 1,099,511,627,776 × 70 = 76,965,813,944,320 3,486,784,401 × 56 = 195,259,926,456 1,048,576 × 28 = 29,360,128 1 × 8 = 8 STEP 5 Then, alternating from plus to minus, we sum the 8 terms we just calculated. + 1,152,921,504,606,850,000 - 638,338,130,380,896,000 + 102,372,436,321,763,000 - 5,340,576,171,875,000 + 76,965,813,944,320 - 195,259,926,456 + 29,360,128 -8 Total 611,692,004,959,217,000 which equals the total number of ways you can roll an 8 sided die 20 times and have all 8 numbers appear. STEP 6 So, if we take the number 611,692,004,959,217,000 and divide it by 1,152,921,504,606,850,000 (all posiible results of 20 rolls of an 8 sided die) we get the probability of having all 8 numbers appearing after 20 rolls. 611,692,004,959,217,000 ÷ 1,152,921,504,606,850,000 = 0.530558240535038 So, if you had an 8 sided die, you would need to roll it at least 20 times in order to have a better than 50 / 50 chance of rolling all 8 numbers.<\|endoftext\|>	4.90625
830	Prime Produced Interval Sizes When one uses the expression "3/2" to define the size of a perfect fifth, two interval generators are already being used : Prime 3 for the fifth ; and Prime 2 for the octave removed from its full size of "3/1" which would be more accurately and specifically described by using the term "Harmonic 3", also Prime produced, as are all pitch distances (intervals). (Nothing but Primes here, no Fibs) Natural Harmonics, the building blocks of the structure of Timbre and Sound, are typically all the integer multiples of a fundamental tone, called Harmonic 1. However, the basic acoustic ratios in the structure of Pitch limit themselves to the first three Prime Numbers (2, 3, 5). And while it is interesting and useful to consider these basic acoustic ratios with a representation similar to the one used for natural harmonics, it does not imply that the musical structure of Pitch might come directly from the Natural Harmonics themselves, but only that it is mathematically convenient to place them graphically as one would place the harmonics, with the appropriate pitch-distance between them. The Octave - Harmonic 2 - proportion 2/1 between Harmonics 1 and 2, Harmonics 2 and 4, Harmonics 4 and 8 The Perfect Fifth - Harmonic 3 - proportion 3/1 between Harmonics 1 and 3, Harmonics 3 and 9. The Major Third - Harmonic 5 - proportion 5/1 between Harmonics 1 and 5. The Process of Cartesian Division The first octave (Harmonics 1-2) remains undivided, the product of Prime 2. The second octave (Harmonics 2-4) is divided by Prime 3 into two uneven parts, a main part of perfect fifth, 3/2, and a left-over part of perfect fourth 4/3, with a proportion of 9/8 between the two parts, the Pythagorian tone, the essence of diatonic scales (heptatonic, pentatonic, tritonic). In the third octave (Harmonics 4-8), the main part perfect fifth (now between Harmonics 4 and 6) is divided by Prime 5 into two uneven parts, a main part of major third, 5/4, and a left-over part of minor third 6/5, with a proportion of 25/24 between the two parts, the chromatic distance between the two MEDIANs (thirds) in any chord FRAME. In the fourth octave, the main part major third (now between Harmonics 8 and 10) is divided by 9, the square of Prime 3 (typical prime behavior of breaking a sequence), into two uneven parts, a main part of Pythagorian major second, 9/8, and a left-over part of natural major second 10/9, with a proportion of 81/80 between the two parts, the syntonic coma, between Pythagorian (Prime 3) and natural (Prime 5) tuning. There seems to be no need of further primes, including Harmonic 7 which is in a left-over part, and the policy of remaining within the simplest ratios is duly respected. These three fundamental interval sizes (the octave, the perfect fifth, and the major third) will not be used indiscriminately in the Structure of Pitch. The perfect fifth (product of Prime 3) is by far the most fruitful, and what we call the series of 21 fifths, all the way from the Fb to the B#, will be its back-bone with Harmony on one side, by "dividing the fifths", and Scales on the other side, by "multiplying the fifths". The perfect fifth produces all the notes and all the structures of Pitch, both vertical and horizontal.<\|endoftext\|>	3.921875
3,859	# Integrals Involving Rational Functions ## Related Calculator: Integral (Antiderivative) Calculator with Steps Consider integral int (4x-6)/(x^2-3x+2)dx. This integral can be easily evaluated using substitution u=x^2-3x+2. Indeed, du=(2x-3)dx. So, integral becoms 2 int (du)/u=2ln\|u\|+C=2ln\|x^2-3x+2\|+C. This integral is easy, because in numerator there is exact derivative of denominator (or a constant multiply). But in many cases there is no derivative in numerator. For example, how to integrate int 1/(x^2-3x+2)dx? Example 1. Integrate int (dx)/(x^2-3x+2). Note, that (x^2-3x+2)=(x-2)(x-1). So, 1/(x^2-3x+2)=1/((x-2)(x-1)). Now, suppose you can decompose it as A/(x-2)+B/(x-1) . We need to determine constants A and B. So, A/(x-2)+B/(x-1)=(A(x-1)+B(x-2))/((x-2)(x-1)). And this must be equal for any x to 1/((x-2)(x-1)). Denominators are equal, so we require equality of numerators: Thus, A(x-1)+B(x-2)=1. At this point we have one of two ways to proceed. One way will always work, but is often requires more work. The other, while it won’t always work, is often quicker when it does work. First way that always works. We can rewrite equation as (A+B)x+(-A-2B)=0x+1. Equating coefficients near like terms gives: {(A+B=0),(-A-2B=1):} . Adding two equations gives that -B=1 or B=-1 . From first equation A=-B=-(-1)=1. Second way that will not always work. Note that A(x-1)+B(x-2)=1 must hold for any x. Therefore, choosing appropriate x's we got the unknown constants to quickly drop out. If x=1 then x-1=0 and A(1-1)+B(1-2)=1 ,so B=-1. If x=2 then x-2=0 and A(2-1)+B(2-2)=1, so A=1. Therefore 1/((x-2)(x-1))=1/(x-2)-1/(x-1). Finally, int (dx)/(x^2-3x+2)dx=int (1/(x-2)-1/(x-1))dx=ln\|x-2\|-ln\|x-1\|+C. The process described in example 1 is called partial fraction decomposition. Definition. Partial Fraction Decomposition is a process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression. Many integrals are solved quickly with performing partial fraction decomposition. Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. If this doesn't hold, we first need to perform long division. Example 2. Evaluate int (x^3-3x^2+2x+4)/(x^2-3x+2)dx . Here degree of numerator is greater than degree of denominator, so perform long division first: int (x^3-3x^2+2x+4)/(x^2-3x+2)dx=int (x(x^2-3x+2)+4)/(x^2-3x+2)dx=int (x+4 1/(x^2-3x+2))dx= =int (x+4/(x-2)-4/(x-1))dx=1/2x^2+4ln\|x-2\|-4ln\|x-1\|+C . In general if you have in denominator (a_nx^n+a_(n-1)x^(n-1)+...+a_0)^k then corresponding term in partial fraction decomposition is (A_(n-1)x^(n-1)+A_(n-2)x^(n-2)+...+A_0)/(a_nx^n+a_(n-1)x^(n-1)+...+a_0)+(A_(n-1)x^(n-1)+A_(n-2)x^(n-2)+...+A_0)/(a_nx^n+a_(n-1)x^(n-1)+...+a_0)^2+...+ +(A_(n-1)x^(n-1)+A_(n-2)x^(n-2)+...+A_0)/(a_nx^n+a_(n-1)x^(n-1)+...+a_0)^k. In particular, 1. ax+b->A/(ax+b) 2. ax^2+bx+c->(Ax+B)/(ax^2+bx+c) 3. (ax+b)^2->A/(ax+b)+B/(ax+b)^2 4. (ax^2+bx+c)^2->(Ax+B)/(ax^2+bx+c)+(Cx+D)/(ax^2+bx+c)^2 . Often there are quadratic terms that can't be reduced (discriminant is negative). In this case you should complete a square. Example 3. Find int 1/(x^2-6x+10)dx. x^2-6x+10 cannot be factored, so complete a square: x^2-6x+10=x^2-6x+9+1=(x-3)^2+1. int 1/((x-3)^2+1)dx=arctan(x-3)+C . In general you will need following integrals: 1. int 1/(ax+b)dx=1/a ln\|ax+b\|+C 2. int x/(x^2+-a)dx=1/2ln\|x^2+-a\|+C 3. int 1/((x+a)^2+b^2)dx=1/b tan^(-1)((x+a)/b)+C 4. int 1/(ax+b)^2dx=-1/a 1/(ax+b)+C Example 4. Integrate int (2x^2-x+4)/(x^3+4x)dx Since x^3+4x=x(x^2+4) then partial fraction decomposition is (2x^2-x+4)/(x(x^2+4))=A/x+(Bx+C)/(x^2+4). Multiplying by x(x^2+4) we have that (2x^2-x+4)=A(x^2+4)+x(Bx+C)=(A+B)x^2+Cx+4A. Equating coefficients near like terms yields A+B=2 , C=-1 and 4A=4. Thus, A=1 , B=1 , C=-1 . int (2x^2-x+4)/(x^3+4x)dx=int (1/x+(x-1)/(x^2+4))dx=int (1/x+x/(x^2+4)-1/(x^2+4))dx= =ln\|x\|+1/2ln\|x^2+4\|-1/2tan^(-1)(x/2)+C. Example 5. Evaluate int (2x^3-11x^2-2x+2)/(2x^2+x-1)dx . Degree of numerator is greater than degree of denominator, so we must first perform long division. (2x^3-11x^2-2x+2)/(2x^2+x-1)=x-6+(5x-4)/(2x^2+x-2)=x-6+(5x-4)/((2x-1)(x+1)). Partial fraction decomposition is (5x-4)/((2x-1)(x+1))=A/(2x-1)+B/(x+1). Multiplying by (2x-1)(x+1) gives the following: 5x-4=A(x+1)+B(2x-1). If x=-1 then 5(-1)-4=A(-1+1)+B(2(-1)-1) or B=3. If x=1/2 then 51/2-4=A(1/2+1)+B(21/2-1) or A=-1. So, (5x-4)/((2x-1)(x+1))=-1/(2x-1)+3/(x+1). Finally, int (2x^3-11x^2-2x+2)/(2x^2+x-1)dx=int (x-6-1/(2x-1)+3/(x+1))dx= =1/2 x^2-6x-1/2 ln\|2x-1\|+3ln\|x+1\|+C. Example 6. Evaluate int (15x^2-29x+15)/(x(x-3)(4x-5))dx. Partial fraction decomposition is (15x^2-29x+15)/(x(x-3)(4x-5))=A/x+B/(x-3)+C/(4x-5) . Multiplying by x(x-3)(4x-5) yields 15x^2-29x+15=A(x-3)(4x-5)+Bx(4x-5)+Cx(x-3). If x=0 then 150^2-290+15=A(0-3)(40-5)+B0(40-5)+C0(0-3) or A=1. If x=3 then 153^2-293+15=A(3-3)(43-5)+B3(43-5)+C3(3-3) or B=3. If x=5/4 then 15(5/4)^2-29(5/4)+15= =A((5/4)-3)(4(5/4)-5)+B(5/4)(4(5/4)-5)+C(5/4)((5/4)-3) or C=-1. So, (15x^2-29+15)/(x(x-3)(4x-5))=1/x+3/(x-3)-1/(4x-5) . Thus, required integral is int (1/x+3/(x-3)+1/(4x-5))dx=ln\|x\|+ln\|x-3\|+1/4ln\|4x-5\|+C. Example 7. Integrate int x/((x+2)^2(x-1)dx . Partial fraction decomposition is x/((x+2)^2(x-1))=A/(x+2)+B/(x+2)^2+C/(x-1). Multiplying both by denominator gives x=A(x+2)(x-1)+B(x-1)+C(x+2)^2. If x=1 then 1=A(1+2)(1-1)+B(1-1)+C(1+2)^2 or C=1/9. If x=-2 then -2=A(-2+2)(-2-1)+B(-2-1)+C(-2+2)^2 or B=2/3. To find constant A we can't use second method anymore, so we rewrite x=A(x+2)(x-1)+B(x-1)+C(x+2)^2 as x=(A+C)x^2+(A+B+4C)x+(-2A-B+4C) . From this we have that A+C=0. We've already found that C=1/9, so A=-1/9. Therefore, x/((x+2)^2(x-1))=-1/9 1/(x+2)+2/3 1/(x+2)^2+1/9 1/(x-1). Thus, required integral is int (-1/9 1/(x+2)+1/3 1/(x+2)^2+1/9 1/(x-1))dx= =-1/9ln\|x+2\|-2/3 1/(x+2)+1/9ln\|x-1\|+C. Example 8. Find int x^2/(x^2-1)dx. First perform long division: (x^2)/(x^2-1)=1+1/(x^2-1). Partial fraction decomposition is 1/(x^2-1)=1/2 1/(x-1)-1/2 1/(x+1). Therefore, int (x^2)/(x^2-1)dx=int (1+1/2 1/(x-1)- 1/2 1/(x+1))dx=x+1/2 ln\|x-1\|-1/2 ln\|x+1\|+C. Some integral with roots can be expressed as integral of rational function. Example 9. int 1/(x-sqrt(x+2))dx . Let u=sqrt(x+2) then du=1/2 1/(sqrt(x+2))dx or dx=2sqrt(x+2)du=2udu. Also since u=sqrt(x+2) then x=u^2-2. So, int 1/(x-sqrt(x+2))dx=int (2u)/(u^2-u-2)du. Since u^2-u-2=(u-2)(u+1) then partial fraction decomposition is (2u)/((u-2)(u+1))=A/(u-2)+B/(u+1). Multiplying by (u-2)(u+1) yields 2u=A(u+1)+B(u-2). If u=-1 then 2(-1)=A(-1+1)+B(-1-2) or B=2/3. If u=2 then 2*2=A(2+1)+B(2-2) or A=4/3. So, int (2u)/(u^2-u-2)du=int (4/3 1/(u-2)+2/3 1/(u+1))du=4/3ln\|u-2\|+2/3ln\|u+1\|+C= =4/3ln\|sqrt(x+2)-2\|+2/3ln\|sqrt(x+2)+1\|+C . Example 10. Find int (3x+5)/(x^2+10x+34)dx . Factor in denominator is irreducible, so complete the square: x^2+10x+34=x^2+10x+25+9=(x+5)^2+9. Now make substitution u=x+5 then du=dx and x=u-5. Thus, int (3x+5)/(x^2+10x+34)dx=int (3(u-5)+5)/(u^2+9)du=3 int u/(u^2+9)du-10int 1/(u^2+9)du= =3/2ln\|u^2+9\|-10/3arctan(u/3)+C=3/2ln((x+5)^2+9)-10/3arctan((x+5)/3)+C. Now, let's see how to handle integrals where irreducible factor is raised to some power. Example 11. Evaluate int 1/(x^2+4x+13)^3 dx. We first complete the square: x^2+4x+13=x^2+4x+4+9=(x+2)^2+9. Now, make substitution u=x+2 then du=dx. int 1/((x+2)^2+9)^3dx=int 1/(u^2+9)^3 du. We will need trig substitution here. Let u=3tan(t) then u^2+9=9tan^2(t)+9=9sec^2(t) and du=3sec^2(t)dt. So, int 1/(u^2+9)^3du=int (3sec^2(t))/(9sec^2(t))dt=1/243 int 1/(sec^4(t))dt. To find this integral we need to use techniques from Integrals Involving Trig Functions note. 1/243 int 1/(sec^4(t))dt=1/243 int cos^4(t)dt=1/243 int (cos^2(t))^2dt=1/243 int (1/2 (1+cos(2t)))^2dt= =1/972 int (1+2cos(2t)+cos^2(2t))dt=1/972 int (1+2cos(2t)+1/2(1+cos(4t)))dt= =1/972 (3/2 t+sin(2t)+1/8sin(4t))+C. Since u=3tan(t) and u=(x+2) then t=arctan((x+2)/3). So, int 1/(x^2+4x+13)^3dx= =1/972(3/2 arctan((x+2)/3)+sin(2arctan((x+2)/3))+1/8sin(4arctan((x+2)/3)))+C.<\|endoftext\|>	4.5625
758	### Chapter 2 Polynomials Ex 2.4 Question 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case: Solution: (i) Comparing the given polynomial with ax3 + bx2 + cx + d, we get: a = 2, b – 1, c = -5 and d = 2. ∴ p(x) = 2x3 + x2 – 5x + 2 Question 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively. Solution: Let α , β and γ be the zeroes of the required polynomial. Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14. ∴ Cubic polynomial = x3 – (α + β + γ)x2 + (αβ + βγ + γα)x – αβγ = x3 – 2x2 – 1x + 14 Hence, the required cubic polynomial is x3 – 2x2 – 7x + 14. Question 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b. Solution: Question 4. If two zeroes of the polynomial x4 – 6x3 – 26x+ 138x – 35 are 2 ± √3, finnd other zeroes. Solution: Since two zeroes are 2 + √3 and 2 – √3, ∴ [x-(2 + √3)] [x- (2 – √3)] = (x-2- √3)(x-2 + √3) = (x-2)2– (√3)2 x2 – 4x + 1 is a factor of the given polynomial. Now, we divide the given polynomial by x2 – 4x + 1. So, x4 – 6x3 – 26x2 + 138x – 35 = (x2 – 4x + 1) (x2 – 2x – 35) = (x2 – 4x + 1) (x2 – 7x + 5x – 35) = (x2-4x + 1) [x(x- 7) + 5 (x-7)] = (x2 – 4x + 1) (x – 7) (x + 5) Hence, the other zeroes of the given polynomial are 7 and -5. Question 5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a. Solution: We have p(x) = x4 – 6x3 + 16x2 – 25x + 10 Remainder = x + a … (i) Now, we divide the given polynomial 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.<\|endoftext\|>	4.75
1,996	In the News An international team of astronomers has captured the image of a black hole, which was previously considered impossible. Evidence for the existence of black holes – mysterious places in space where nothing, even light, can escape – has been around for some time, and astronomers have long observed the effects of these phenomena on the environment. In the popular notion, it was assumed that taking a picture of a black hole was impossible because an image of something out of which no light could escape would appear completely black. For scientists, the challenge was to capture a picture of the hot, glowing gas falling into a black hole from thousands or even millions of light years away. Both have succeeded in an ambitious team of international astronomers and computer scientists. The team worked for over a decade to achieve the feat, enhancing an existing radio astronomy technique for high resolution imaging, and using it to recognize the silhouette of a black hole ̵ How They Did It Although scientists had theorized, they could portray black holes by capturing their silhouettes against their images, glowing surroundings, the ability to image such a distant object escaped them. A team formed to meet the challenge and create a telescope network known as the Event Horizon Telescope or EHT. They wanted to capture a picture of a black hole by improving a technique that allows the imaging of distant objects known as Very Long Baseline Interferometry or VLBI. Telescopes of all types are used to view distant objects. The larger the diameter or opening of the telescope, the greater the ability to collect more light, and the higher its resolution (or ability to image fine details). To see details in objects that are far away and appear small and weak from the earth, we need to gather as much light as possible with very high resolution, so we need to use a telescope with a large aperture. Therefore, the VLBI technique was essential for capturing the image with the black hole. VLBI creates a series of smaller telescopes that can be synchronized to simultaneously focus on the same object and act as a giant virtual telescope. In some cases, the smaller telescopes also consist of several telescopes. This technique was used to track spacecraft and image distant cosmic radio sources such as quasars. The aperture of a huge virtual telescope such as the Event Horizon Telescope is the same size as the distance between the two most distant telescope stations – the EHT has these two stations at the South Pole and in Spain. to create an opening that almost matches the diameter of the earth. Each telescope in the array focuses on the target, in this case the black hole, and collects data from its location on Earth, displaying part of the overall view of the EHT. The more telescopes in the array are far apart, the better the image resolution. To test VLBI to map a black hole and a set of computer algorithms for sorting and synchronizing data, the Event Horizon Telescope team chose two goals: each of them presents unique challenges. The closest supermassive black hole, Sagittarius A , interested the team because it is located in our galactic backyard – at the center of our Milky Way, 26,000 light years (156 billion miles) away. (A star is the astronomical standard for referring to a black hole.) Although not the only black hole in our galaxy, it is the black hole that appears the largest on Earth. Being in the same galaxy as the Earth, however, meant that the team had to look through the pollution caused by stars and dust to image it, so there was more data to filter in processing the image. However, due to the local interest of the black hole and its relatively large size, the EHT team chose Sagittarius A as one of its two objectives. The second target was the supermassive black hole M87 . One of the largest known supermassive black holes, M87 , is located in the center of the gigantic elliptical galaxy Messier 87 or M87, 53 million light-years (318 miles) away. M87 * is much more massive than Sagittarius A * with 4 million solar masses and 6.5 billion solar masses. A solar mass corresponds to the mass of our sun, about 2×10 30 kilograms. In addition to its size, M87 * interested scientists because, unlike Sagittarius A , it is an active black hole into which matter has collapsed and ejected in the form of particle beams that are accelerated to speeds close to the speed of light. But his removal made it even more difficult than the relatively local Sagittarius A . As described by Katie Bouman, a computer scientist at the EHT, who led the development of one of the algorithms used to sort the telescope data during the processing of the historical image, it is almost like taking a picture of an orange on the lunar surface. By 2017, EHT was a collaboration of eight locations around the world – and more have been added since. Before the team could begin collecting data, it had to find a time when the weather was likely to facilitate telescope observations at each location. In April 2017, the team tried to keep the weather good for M87 , and of the ten days selected for observation, there were a whopping four days to spare at all eight locations! Each telescope used for the EHT had to be synchronized with the M7, others within a fraction of a millimeter with an atomic clock linked to a GPS time standard. Thanks to this precision, the EHT is able to resolve objects around 4000 times better than the Hubble Space Telescope. As each telescope received data from the black target hole, the digitized data and time stamp were recorded on computer disk media. By collecting data for four days around the world, the team was able to process a considerable amount of data. The recorded media was then physically transported to a central location because the amount of data (about 5 petabytes) exceeds the current Internet speeds. At this central location, data from all eight locations was synchronized using the timestamps and merged into a composite image set revealing the unprecedented silhouette of the M87 effect horizon. The team is also working to generate a picture of Sagittarius A * from additional observations from the EHT. Adding more telescopes and considering Earth's rotation may result in more image resolution, and it is expected that future images will have a higher resolution. But we may never have a complete picture, as Katie Bouman explains here (under "Imaging a Black Hole"). To supplement the EHT results, several NASA space probes were at great expense in observing the black hole at different wavelengths of light. As part of this effort, NASA's Chandra X-ray Observatory, the Nuclear Spectroscopic Telescope Array (NuSTAR), and Neil Gehrel's Swift Observatory Space Telescope Missions, which were all designed to detect different types of X-rays, were looking at the black hole M87 at the same time with the EHT in April 2017. NASA's Fermi Gamma-Ray Space Telescope also observed changes in M87 * gamma-ray light during EHT observations. If the EHT observed changes in the structure of the black hole environment, data from these missions and other telescopes could be used to find out what was going on. Although NASA observations did not directly track the historical image, astronomers used data from Chandra and NuSTAR satellites to measure the X-ray brightness of the M87 * jet. The scientists used this information to compare their models of the jet and the disk around the black hole with the EHT observations. Other findings could come as researchers continue to think about this data. Why It Matters Learning mysterious structures in the universe provides insights into physics and allows us to test observation methods and theories, such as Einstein's general theory of relativity. Massive objects deform space-time in their vicinity, and although theory of general relativity has been directly applied to smaller objects such as Earth and Sun, the theory for black holes and other dense-matter regions has not yet been directly proven One of The main outcome of the EHT Black Hole Project is the more direct calculation of the mass of a black hole than ever before. Using the EHT, scientists were able to directly observe and measure the radius of the M87 * horizon or its Schwarzschild radius and calculate the mass of the black hole. This estimate was similar to that derived from a method using the orbiting stars – and thus validated as a method of mass estimation. The size and shape of a black hole, which depends on its mass and spin, can be predicted from general relativity equations. The general theory of relativity predicts that this silhouette is approximately circular, but other theories of gravity predict something different. The image of M87 * shows a circular silhouette and lends credibility to Einstein's general theory of relativity near black holes. The data also provide insight into the formation and behavior of black hole structures, such as For example, the accretion disk that gets matter into the black hole and plasma jets that emanate from its center. Scientists have hypothesized how an accretion disk forms, but so far they have never been able to test their theories with direct observation. The scientists are also curious about the mechanism by which some supermassive black holes emit powerful particle beams at near the speed of light. These and other questions are answered when more data is collected by the EHT and synthesized in computer algorithms. Stay up to date and the next expected picture of a black hole – the own shooter of our Milky Way A *. Capture your students' enthusiasm for black holes by encouraging them to solve them Math problems oriented to standards . Model Black-Hole Interaction with this NGSS-Oriented Lesson: See NASA Space Place Student Resources for more information. TAGS: Black Hole, Teachable Moments , Science, K-12 Education, Teacher, Educator<\|endoftext\|>	4.34375
248	Over the past century unprecedented gains in the prevention, diagnosis and treatment of disease have allowed people to live longer and healthier lives. But our health and wellbeing is being challenged by the emergence of diseases that are resistant to antimicrobial drugs and new infectious diseases that spread quickly in our increasingly connected world. Global responses are necessary to tackle the emergence of new threats, contain outbreaks and coordinate efforts to ensure people have access to healthcare and that current treatments remain effective. One of the UN’s Sustainable Development Goals is to ensure healthy lives and promote wellbeing for all ages. This resource has been developed by the British Council in partnership with the Royal Society. It is centred on how to prevent the spread of infectious diseases and has lots of activity ideas to help students aged 7 to 14 investigate and explore the subject in more depth. Examples of activities include: - When is influenza most common? – research activity. - Coughs & sneezes spread diseases – investigating how diseases can spread. - How effectively do you wash your hands? - Evaluating the effectiveness of health education poster messages. - Finding out more about scientists working in this field. - PDF (free) - Visit website Leave a comment<\|endoftext\|>	4.15625
1,087	Teacher: Jennifer Fox School: Arlington Heights Elementary Grade: 3rd Grade Lesson Plan Type: Focus In Lesson 1 Unit Theme: African American History: Slavery & The Underground Railroad Lesson Plan Topic: Is skin color really that important? We are all different! Lesson 1: Looking at our classroom ancestry STANDARDS SOCIAL STUDIES true History 3.1 Students will describe how significant people, events and developments have shaped their own community and region; compare their community to other communities in the region in other times and places; and use a variety of resources to gather information about the past. Human Systems 3.3.8 Construct maps and graphs that show aspects of human/environmental Interaction in the local community, Indiana and communities within the region. READING READING: Comprehension and Analysis of Nonfiction and Informational Text 3.2 Students read and understand grade-level-appropriate material. The selections in the www.doe.in.gov/standards/readinglist.html illustrate the quality and complexity of the materials to be read by students. At Grade 3, in addition to regular classroom reading, students read a variety of nonfiction, such as biographies, books in many subject areas, children's magazines and periodicals, and reference and technical materials. 3.2.2 Analysis of Grade-Level-Appropriate Nonfiction and Informational Text: Ask questions and support answers by connecting prior knowledge with literal information from the text. Example: When reading informational materials about science topics or social science subjects, compare what is read to background knowledge about the subject. WRITING WRITING: English Language Conventions 3.6 Students write using Standard English conventions appropriate to this grade level. Objectives: Students will use prior knowledge to identify other countries and continents Students will interview family members to learn more about their family history/ancestry Students will share interesting facts about their family ancestry and locate/mark their family history on a world map Materials: Who are Your Ancestors worksheet Classroom World Map Star stickers Time: This lesson will span 2 class periods, each about 30 minutes. Primary Sources: All the Colors of the Earth by Sheila Hamanaka Family Trees examples Scholarly Knowledge: Ancestry is a family’s background, or lineage. Most of us have more than one lineage of ancestry. A family’s lineage can go back many generations. PROCEDURES Engage Student Interest: The lesson will begin when I read the book, All the Colors of the Earth by Sheila Hamanaka. Following the reading of the book, we will discuss the main idea of the story of how the book celebrates the color of children, and their ethnic diversity. We will further discuss what ethnic diversity is and how that may determine our skin color. Essential Question: What do you think your family’s ethnic diversity is? Initial Example: I will explain that our ethnic diversity comes from our “ancestry”. I will briefly explain my ancestry, with my background coming from Germany, Ireland, and Canada. One side of my family is German and Irish, and the other is French Canadian and Irish. I will also share how each of my family traveled from their respective countries to live in the United States in the mid 1800s. I will ask the question, “Could my ancestry determine the color of my skin?” New Example: I will first ask what a family tree is, and how it will help to find our ancestry. I will share a few famous Americans’ family trees. We will discuss the importance of each person to American history, and then share what observations are made from the family tree itself. Apply: The students will be shown more about the ancestry of my family, using the Who are your ancestors? worksheet. I will share my information as I model how to use the worksheet. I will again share my ancestry marking the countries using star stickers on our classroom world map, and I will also include facts about the countries my family have come from. I will also share a bit of history, or stories, of my ancestors as interesting facts. The students will then be assigned to find out about their own ancestry by interviewing family members who may be able to help. Assessment: The students will be asked to share what they discovered about their own ancestry from their interviews. Each student will be able to mark the countries of their ancestors on our classroom map with star stickers. They will also share interesting facts about their ancestry as well. Then, the kids will be asked, “Is it the color of your skin that is most important, or the ancestry and your ethnic diversity that makes you special?” We will share our opinions and generate a classroom discussion which is most important to consider. Finally, I will ask the class what is more important about slaves who had to work the fields so many years ago…the color of their skin, or their ancestry and backgrounds? Did their skin color really tell others that they were not important? Cumulating Activity: As each student shares their family’s ancestry, we will record each of the countries their ancestors derive from. As we create a classroom list of countries and tallies for the number of students who have ancestors in these countries, we will create a classroom graph to show the representation of our ancestry for these areas through a visual aide.<\|endoftext\|>	3.890625
1,866	# Parallelograms Think of the markings in a zebra crossing. These are closed 4-sided shapes with opposite sides that are equal and parallel. Such shapes are known as parallelograms. #### Create learning materials about Parallelograms with our free learning app! • Flashcards, notes, mock-exams and more • Everything you need to ace your exams A four-sided plane figure is known as a quadrilateral. In this article, we are going to look at the special type of quadrilateral known as a parallelogram. ## Definition of a parallelogram A quadrilateral with 2 pairs of opposite parallel sides is called a parallelogram. We know that a quadrilateral has 4 sides. In a parallelogram, these 4 sides consist of 2 pairs of opposite parallel sides. The following diagram illustrates a parallelogram. Fig. 1: Parallelogram illustration. In the above figure: • AB // CD • AC // BD ## Properties of parallelograms In addition to the above, we can identify various properties of parallelograms. We will use the following parallelogram ABDC with diagonals d1=BC and d2=AD to illustrate the properties. Fig. 2: Parallelogram with diagonals, d1 and d2. • In a parallelogram, the opposite side are equal. This means that in the above parallelogram, AB=CD and AC=BD. • In a parallelogram, the opposite angles are equal. This means that in the above parallelogram, ∠CAB=∠CDB and ∠ACD=∠ABD • In a parallelogram, consecutive angles are supplementary. In any parallelogram, you can identify 4 pairs of consecutive angles. These are always supplementary (which means the angles sum up to 180 degrees). In the above parallelogram: ∠CAB + ∠ABD = 180, ∠ABD + ∠BDC = 180, ∠BDC + ∠DCA = 180, ∠DCA + ∠CAB = 180. • If any angle in a parallelogram is a right angle, that means all 4 internal angles are right angles. This is a direct consequence of the above property. If any angle in a parallelogram is a right angle, then the adjacent angle is 180-90=90 (according to the above property). In turn, the next adjacent angle will be a right angle and so on. Therefore, in any parallelogram, if you identify any angle as a right angle, you can directly conclude that all 4 angles are right angles. • The diagonals of a parallelogram bisect each other. In the above parallelogram, the point O is the mid-point of both the diagonals d1 and d2. • Each diagonal of a parallelogram separates the parallelogram into two congruent triangles. In the above parallelogram, the diagonal d1 would divide the parallelogram into two congruent triangles, ΔABC and ΔBCD. Similarly, the diagonal d2 would divide the parallelogram into two congruent triangles, ΔABD and ΔACD. ## Area of parallelograms Consider the following parallelogram: Fig. 3: Parallelogram with base b and height h. The area of a parallelogram is given by the formula: Area = b × h where b = base, h = height Now the value, b, is the length of the side AB, which is considered the base here. Conventionally, one of the longer sides of the parallelogram is taken to be the base. The value h is the height of the parallelogram. It is also sometimes called the altitude. The height is the length of the line drawn from the base to its opposite side. The height is perpendicular to the base. ## Parallelogram: Example problems In this section we explore examples of math problems that you may encounter about parallelograms and their properties. A parallelogram with a base of 8 ft. has an area of 20 ft2. What is the height of the parallelogram? In the following parallelogram, ∠ABD = 47°, ∠CBD = 72°. Find ∠CDA. Fig. 4: Parallelogram illustration. ∠ABC = ∠ABD + ∠CBD = 47 + 72 = 119. Opposite angles of a parallelogram are equal. Hence, ∠CDA = ∠ABC = 119° ## Different types of parallelogram shapes In this section, we will identify 3 special types of parallelograms, each with its own characteristics and properties: 1. Rhombus 2. Rectangle 3. Square ### Rhombus A rhombus is a quadrilateral with all 4 of its sides equal in length (equilateral). It turns out that the opposite pairs of sides of a rhombus are always parallel. This makes every rhombus a parallelogram. Conversely, we can say that an equilateral parallelogram is a rhombus. The diagonals of a rhombus always bisect each other at right angles. Since a rhombus is a special type of parallelogram, a rhombus exhibits all the properties of a parallelogram as well. ### Rectangle A rectangle is a parallelogram with all of its internal angles as right angles. Since all angles in a rectangle are equal, it is equiangular. Since a rectangle is a special type of parallelogram, it exhibits all the properties of a parallelogram as well. ### Square A square is a quadrilateral with all 4 of its sides equal and with all of its angles as right angles. This makes a square a type of parallelogram, a type of rhombus, and a type of rectangle! Thus, a square demonstrates all the properties of parallelograms, rhombuses, and rectangles. ## Parallelograms - Key takeaways • A quadrilateral with 2 pairs of opposite parallel sides is called a parallelogram. • In a parallelogram, the opposite side are equal. • In a parallelogram, the opposite angles are equal. • In a parallelogram, consecutive angles are supplementary. • If any angle in a parallelogram is a right angle, that means all 4 internal angles are right angles. • The diagonals of a parallelogram bisect each other. • Each diagonal of a parallelogram separates the parallelogram into two congruent triangles. • The area of a parallelogram is given by the formula: Area = b × h where b = base, h = height • A rhombus is a parallelogram with 4 equal sides. • A rectangle is a parallelogram with all internal angles as right angles. • A square is a parallelogram with 4 equal sides and all right angles. #### Flashcards in Parallelograms 15 ###### Learn with 15 Parallelograms flashcards in the free StudySmarter app We have 14,000 flashcards about Dynamic Landscapes. What is a parallelogram? A quadrilateral with 2 pairs of opposite parallel sides is called a parallelogram. How to find the area of a parallelogram? The area of a parallelogram is given by the formula: Area = b × h where b=base, h=height. Is a rhombus a parallelogram? A rhombus is a type of parallelogram. Is a trapezoid a parallelogram? A trapezoid can have 1 pair of opposite sides that are not parallel, thus making it not a parallelogram. Is a rectangle a parallelogram? A rectangle is a type of parallelogram. ## Test your knowledge with multiple choice flashcards Which of the following are types of parallelograms? If two lines in a plane are parallel to the same line, then are all the lines parallel to each other? The segments with three parallel lines and two traversals what proportion? StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance. ##### StudySmarter Editorial Team Team Math Teachers • Checked by StudySmarter Editorial Team<\|endoftext\|>	4.84375
240	###### Example4.8.12Undefined Slope What is the slope of a vertical line? Figure 4.8.13 shows three lines passing through the origin, each steeper than the last. In each graph, you can see a slope triangle that uses a “rise” of $$4$$ each time. If we continued making the line steeper and steeper until it was vertical, the slope triangle would still have a “rise” of $$4\text{,}$$ but the “run” would become smaller and smaller, closer to $$0\text{.}$$ And then the slope would be $$m=\frac{4}{\text{very small}}=\text{very large}\text{.}$$ So the slope of a vertical line can be thought of as “infinitely large.” If we actually try to compute the slope using the slope triangle when the run is $$0\text{,}$$ we would have $$\frac{4}{0}\text{,}$$ which is undefined. So we also say that the slope of a vertical line is undefined. Some people say that a vertical line has no slope. in-context<\|endoftext\|>	4.46875
591	# A Math Proportion Quiz Question Question If a hen and a half can lay an egg and a half in a day and a half, how many eggs can six hens lay in seven days? Just write the sentences and you will get the answer!? 1.5 hen lays 1.5 eggs in 1.5 day Now think. Now four ” 1.5 hen ” [ that is 1.54 = 6 hens ] will lay four sets of “1.5 eggs” [ that is 1.54 = 6 eggs ] in just 1.5 day!!! So, 6 hens lay 6 eggs in 1.5 day In four “1.5 days” [ that is 1.54 = 6days], 6 hens would have laid a total of 46 = 24 eggs. That is, 6 hens lay 24 eggs in 6 days For the seventh day, it actually depends on what you assume proportional to what. Its a matter of interpretation and assumption. 3 possiblities only for assumption of proportionality, and each is given below. Note : In each of the below, we will use the fact “6 hens lay 24 eggs in 6 days” because it was derived independent of any other extra assumptions. Interpretation 1.(eggs are laid in proportional to the days discretely) Using 6 hens lay 24 eggs in 6 days. Now, In 6 days, 6 hens can lay 24 eggs. What about one day? In one day, 6 hens can lay 4 eggs by assuming the eggs are layed in proportional to the days… i.e, 4 eggs per day from this statement. Thus in the seventh day, 4 eggs are produced by 6 hens. Hence total eggs = 24 + 4 = 28 eggs. Interpretation 2.(eggs are proportional to the hen discretely) If we assume the eggs are proportional to the hen. then, from 6 hens lay 24 eggs in 6 days. Then, 1 hen lays 4 eggs in 6 days. By further assumption that eggs and hen are proportional continously we get : one hen lays (7/6)4 eggs in (7/6)6 = 7 days; Total eggs = 6* (7/6)*4 = 28 eggs. Interpretation 3 ( hen is proportional to the days discretely ) This makes no sense, since the number of hens is NOT proportional to the days. P/S: If you see carefully, actually, interpretation 1 and 2 are the same because we cannot escape the fact that the eggs are laid out proportional to the hens!<\|endoftext\|>	4.71875
449	# Dev Blog ## Kelly Criterion The Wikipedia article on the Kelly criterion might say more, but here is a simple derivation. Assuming you have a coin with probability $p$ of coming up heads and odds of $b:1$, the Kelly criterion states: $$f^* = \frac{bp-q}{b}$$ Where $f^$ is the fraction of your money pot the Kelly criterion tells you to bet and $q=1-p$. That is: $$f^ = \frac{bp-(1-p)}{b} \ = \frac{p(b+1)-1}{b}$$ Assuming the strategy is to bet a fraction of your bank roll every round, with $W_0$ as the initial bank roll, $n$ time units and $W_n$ as your winnings at time $n$: $$W_n = (1 + br)^{pn} (1 - r)^{(1-p)n} W_0$$ Taking logarithms, setting the derivative with respect to $r$ and solving: $$\begin{array} . & \frac{d}{dr} \ln(W_n) &= \frac{d}{dr} ( pn \ln(1+br) + (1-p)n \ln(1-r) + \ln(W_0) ) \\ \to & 0 &= \frac{pnb}{1+br} - \frac{(1-p)n}{1-r} \\ \to & \frac{(1-p)n}{1-r} &= \frac{pnb}{1+br} \\ \to & \frac{1-p}{1-r} &= \frac{pb}{1+br} \\ \to & (1-p) (1+br) &= pb (1-r) \\ \to & 1 - p + br - pbr &= pb - pbr \\ \to & 1 - p + br &= pb \\ \to & r &= \frac{pb + p - 1}{b} \\ \to & r &= \frac{b(p + 1) - 1}{b} \\ \end{array}$$<\|endoftext\|>	4.4375
393	Color quality is one of the key challenges facing light-emitting diodes (LEDs) as a general light source. This paper reviews the basics regarding light and color and summarizes the most important color issues related to white light LEDs. Unlike incandescent and fluorescent lamps, LEDs are not inherently white light sources. Instead, LEDs emit light in a very narrow range of wavelengths in the visible spectrum, resulting in nearly monochromatic light. This is why LEDs are so efficient for colored light applications such as traffic lights and exit signs. However, to be used as a general light source, white light is needed. The potential of LED technology to produce high-quality white light with unprecedented energy efficiency is the impetus for the intense level of research and development currently being supported by the U.S. Department of Energy. White Light from LEDs White light can be achieved with LEDs in two main ways: 1) phosphor conversion, in which a blue or ultraviolet (UV) chip is coated with phosphor(s) to emit white light; 2) RGB systems, in which light from multiple monochromatic LEDs (red, green, and blue) is mixed, resulting in white light. The phosphor conversion approach is most commonly based on a blue LED. When combined with a yellow phosphor (usually cerium-doped yttrium aluminum garnet or YAG:Ce), the light will appear white to the human eye. A more recently developed approach uses an LED emitting in the near-UV region of the spectrum to excite multi-chromatic phosphors to generate white light. The RGB approach produces white light by mixing the three primary colors red, green, and blue. Color quality of the resulting light can be enhanced by the addition of amber to “fill in” the yellow region of the spectrum. Status, benefits and trade-offs of each approach are explored here after.<\|endoftext\|>	3.9375
207	Language and cultural support Approximately 138 different languages are spoken in Scotland today and growing up with more than one language is becoming increasingly common. Bilingual pupils function in more than one language in their daily lives. There are over 20 languages spoken by bilingual children and young people in our educational establishments. They and their families bring a variety of languages and cultures to our educational environment, which enriches the ethos and learning of all. Bilingualism offers advantages for children’s development: - greater awareness of how language operates - enhanced problem solving skills - experience of more than one culture Children’s home language tends to be the first language that they learn and it is important that they continue to develop skills and concepts in their first language, as this provides the best foundation for learning additional languages and concepts. Many bilingual young people are learning English as an additional language (EAL) and require support to develop this and to access the curriculum. More information on our range of language and cultural support:<\|endoftext\|>	3.671875
1,076	Everyone is the product of generations that have gone before—parents, grandparents, and, before them, lines of ancestors stretching back into remote periods of history. Alex Haley, in his 1976 book, called these ancestral lines Roots. Today many persons, both amateurs and professionals, spend time tracing their family histories. This study is called genealogy, from Greek words meaning “family” and “study.” For many who take an interest in their ancestry, the goal is simply to satisfy curiosity. Some pursue the study out of ethnic loyalty or pride. Others hope to find in their backgrounds some instance of royalty or nobility. Genealogies have also been used to show lines of descent for purposes of inheritance and land ownership. They have long been used by Europe’s royal houses and the nobility to prove the authenticity and fitness of heirs to a throne or estate. Many Americans like to claim descent from passengers who came over on the Mayflower in 1620. European royalty and nobility often claim descent from ancient heroes—or even gods—and in so doing add to the prestige of their families as well as to their rights and privileges. The history of genealogy can be divided into three phases. There are ancient oral, or spoken, traditions; the committing of oral traditions to writing to prove the descent of certain persons or lines; and the broadening, after 1500, of the study to such an extent that it is now possible for the majority of people in Western Europe to trace their ancestries. In the ancient world, especially before writing was invented, oral traditions were necessarily significant. Such traditions usually carried within them the whole history of a people as well as the ancestry of its best-known individuals. Numerous ancient genealogies appear in the Old Testament. The object is to show descent from Adam, Noah, Abraham, and other heroes of ancient Israel. These genealogies first developed as oral traditions and were much later committed to writing. In the New Testament there is a genealogy in the book of Matthew that traces the ancestry of Jesus back to Abraham. Among the Greeks and Romans, heroes and kings were celebrated as having descended from gods. The Greek Achilles, for example, was said to be born of the goddess Thetis. Julius Caesar traced his ancestry back to the Trojan Aeneas and beyond that to the goddess Aphrodite, or Venus. In the modern world one of the most ancient genealogies was that of the emperor of Ethiopia, who traced his ancestry back to the marriage of King Solomon and the queen of Sheba as recorded in the Old Testament. The second phase of genealogical development, committing the ancestries to writing, became significant in Europe after the barbarian tribes were converted to Christianity early in the Middle Ages. The chroniclers, most of whom were monks, recorded the family trees of the various ruling houses and of the nobility. Since that time the keeping of records of all kinds—land transactions, baptismal certificates, marriage certificates, and other documents—has broadened considerably to include much greater numbers of people. Those who trace family trees are engaged in complicated research that requires specific abilities and tools. Searchers must have a knowledge of the country of ancestry and perhaps of neighboring countries as well. They should be able to decipher handwriting that may be very old and in unusual scripts. They must locate the proper records that provide information. Some of these records, which may be in various languages, include birth and death certificates, wedding certificates, land sale documents, wills, tax records, family Bibles, and tombstones. In most countries such records have not been collected in a central location but are found where the ancestors originally lived. For Americans the search is complicated by the need to locate ships’ lists of passengers that tell when emigrants left the Old World for the New. They must also find out the emigrants’ original homes, not just the places from which the ships left. There are also records of arrival in the United States, but in many cases immigrants’ names were changed or spelled wrong by record keepers. Many helpful records are kept at the National Archives in Washington, D.C. To assist the genealogist most Western countries have collections of public records of many kinds. Registration of birth, marriage, and death has been required by law in Great Britain since 1837. Parish records in England date back to 1538. In most other nations they begin later. Census records are very useful. These have been kept in the United States since 1790, in Britain since 1801, and in French Canada since the mid-17th century. Amateur genealogical work has increased greatly since 1945. In the United States there has long been an interest. Among groups that have preserved valuable records are the New England Historic Genealogical Society, the Augustan Society in California, and the Church of Jesus Christ of Latter-day Saints (Mormons) in Utah. In Britain there is a Society of Genealogists, and there are similar associations in other countries. Since the establishment of the State of Israel in 1948, a great effort has been put forth to centralize information about the Jews of continental Europe. This work is under the direction of the Central Archives for the Study of the Jewish Peoples, located in Jerusalem.<\|endoftext\|>	3.65625
854	Extreme winds rule exoplanet's weather Supersonic winds more than six times faster than those on Jupiter are blasting through the atmosphere of a Jupiter-sized planet 60 light years away, say scientists who've analyzed results from NASA's Spitzer Space Telescope. They're part of a team that mapped weather on a planet beyond our solar system, a gas giant planet called HD 189733b. "The exoplanet's wind speeds probably exceed the speed of sound," said Adam Showman of the University of Arizona Lunar and Planetary Laboratory (LPL). "And the speed of sound on these planets is 10 times faster than on Earth, so that's saying something." The speed of sound in HD 189733b's atmosphere is about 3 kilometers per second (6,700 mph). Showman and LPL research associate Curtis Cooper analyzed Spitzer data on planet HD 189733b using the numerical models they've been developing for exoplanet atmospheres. The planet, which is in constellation Vulpecula, is the closest known 'transiting' planet. A transiting planet is seen to cross in front and behind its star when viewed from Earth. The planet is "tidally locked" to its star, so that one side always faces the star and the other side is always dark, just as the moon is tidally locked to the Earth. A team led by Heather Knutson of the Harvard-Smithsonian Center for Astrophysics, Cambridge, Mass., used the Spitzer Space Telescope to measure the infrared light, or heat, as the planet orbited its sun-like star. The result is one of the first-ever temperature maps for an exoplanet. The map shows that dayside and nightside temperatures differ only by about 500 degrees Fahrenheit, ranging from 1,200 degrees F on the nightside to 1,700 degrees F on the dayside. "At these high temperatures, air cools off rapidly when it moves from the dayside to the nightside," Showman said. "That relatively small temperature difference implies that fierce winds redistribute a lot of the heat." "We need to do more detailed modeling to calculate actual wind speeds. At this stage, the numbers are all quite uncertain," he said. "However, we can be certain the speeds are FAST, probably a couple of kilometers per second," or about 4,500 mph. The supersonic exoplanet winds might be as great as 10 kilometers per second, or about 22,000 mph, the UA researchers calculate. "This isn't just the case where you need winds, but winds that are fast enough to move air from one side of the gas giant planet to the other before it has time to cool off," Showman said. "We might have a situation where the winds are moving faster than the rotating planet itself because 'hot Jupiters' like this one rotate slowly," Showman said. The hot Jupiter exoplanet rotates at about 2 kilometers per second (about 4,500 mph) at its equator. Earth and Jupiter's winds are anemic, by comparison. The Earth is spinning at almost 1,000 mph at its equator. A given point at Earth's equator rotates through Earth's 25,000 mile circumference in 24 hours. But its wind speeds are only around 20 to 200 mph. The same is true on Jupiter. Jupiter's equator rotates at almost 27,000 mph, but its wind speeds are only around 70 to 340 mph. Cooper's and Showman's numerical simulations predicted a larger difference between dayside and nightside temperatures, so winds on exoplanet HD 189733b are more complex than their models currently reflect, Showman said. But their simulations match the observations in some other respects. One of their predictions is that the winds distort the temperature pattern, blowing the hottest region downwind from the locations that get maximum starlight. "The exoplanet doesn't emit its greatest energy toward Earth when aimed at Earth," Showman said. "As our models predict, the hottest point is seen a couple of hours before the planet passes behind the star." Source: University of Arizona<\|endoftext\|>	3.796875
314	The First Automotive Assembly Line Was Created By? Answer: Ransom E. Olds When it comes to the history of the automotive industry and automation, Henry Ford—American inventor and founder of the Ford Motor Company—is the guy that gets all the credit for the assembly line. Now don’t get us wrong, Henry Ford does in fact deserve a spot in the history of production automation, but he wasn’t the first person to use an assembly line in a car factory. That distinction belongs to Ransom E. Olds, a pioneer in the American automotive industry and the—as you may have guessed—namesake for the Oldsmobile. Before Henry Ford was using assembly lines in his factory, Ransom E. Olds was; except instead of moving the cars along the line, he moved his workers instead. His early factories used a stationary assembly line wherein dozens of cars were assembled at once, but instead of rolling down the old assembly line as we have now come to understand it, the workers rolled their carts of tools down the line and worked a bit on each car. This method significantly sped up the production process and the output of his factory went from 425 cars per year to 2,500. Henry Ford later revolutionized this concept by moving the car instead of the worker, which allowed for even faster production because both the heavy parts, the workers, and the tools, could be kept in place while only the body of the car was moved along the line. Image Public Domain/Wikimedia.<\|endoftext\|>	3.671875
474	# Conics - Hyperbola #### Everything You Need in One Place Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered. #### Learn and Practice With Ease Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals. #### Instant and Unlimited Help Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now! 0/1 ##### Intros ###### Lessons 1. Hyperbola: horizontal hyperbola vertical hyperbola • Graph looks sort of like two mirrored parabolas, with the two "halves" being called "branches". • "Vertices" are defined similarly to the way of a "vertex" is defined for a parabola. • Just as the focus for a parabola, the two foci for a hyperbola are inside each branch. • The line connecting the two vertices is called the "transverse axis". 0/8 ##### Examples ###### Lessons 1. properties of a hyperbola $-(\frac{x-6}{4})^2 + (\frac{y+5}{3})^2=1$ 1. Identify the type of conic section. 2. State the "center". 3. Set up the guidelines for the conic graph. 4. Find the "vertices". 5. Locate the "foci". 6. Find the "eccentricity". 7. Find the equations of the "asymptotes". 8. Find the lengths of transverse axis and conjugate axis. ###### Topic Notes hyperbola: the difference of the distances from any point on a hyperbola to each focus is constant and equal to the transverse axis $2a$. ellipse: the sum of the distances from any point on an ellipse to each focus is constant and equal to the major axis $2a$. $c= \sqrt{a^2 - b^2}$ $c$: distance from the center to a focus $e= \frac{c}{a}$ $e$: eccentricity; the larger the value of $e$, the straighter the hyperbola<\|endoftext\|>	4.53125
2,436	- About Hunger - How to End Hunger - Our Impact - Get Involved What does a world without hunger look like? It looks like everyone having the nutritious meals they need to flourish. Churches, charities, food banks, and non-profit organizations cannot get there alone. Government programs and policies play an important role too. Federal domestic nutrition programs such as SNAP (formerly known as food stamps), Women, Infants, and Children Program (WIC), and the school lunch program are just a few examples. These programs keep millions of Americans from going hungry. International humanitarian assistance responds to natural and human-caused disasters (the Indian Ocean tsunami, for example) and ongoing worldwide crises such as the situation in Syria. But ending hunger requires more than just giving people a meal today. Addressing the root causes of hunger — primarily poverty — is just as important. As long as people don’t have the resources they need to put food on the table, hunger will continue. Bread also works for policy reforms that ensure economic security and self-sufficiency over the long-term for people in the U.S. and around the world. Countries struggling with extreme poverty do not have the resources to adequately finance their own economic and social development. Development assistance programs are designed to reduce poverty and encourage economic growth in poor countries. They include programs for agriculture, health, education, the environment, and democracy and governance. When disaster strikes, hunger often follows. Emergencies include natural disasters and disease outbreaks, or result from climate or economic conditions that slowly build to a breaking point such as food shortages, droughts, and conflict. These emergencies often have disastrous side effects including refugee crises and gender-based violence. Emergency situations can quickly go from bad to worse if the global community does not respond quickly enough to prevent starvation, poor health, and extreme poverty. Bread advocates for the U.S. government to respond to urgent needs and protect the hungry and malnourished. For decades, the U.S. has lead compassionate responses to emergencies across the world, saving lives and preventing millions of people from falling into hunger and poverty. The federal government provides immediate cash and food assistance, health and sanitation items, and supplies to help communities rebuild. It also funds programs such as education, job training, and counseling services to help refugees adjust and find stability in an unfamiliar environment. The U.S. government and nonprofit organizations that respond to these crises recognize the importance of linking short-term emergency response and long-term development assistance. The world has seen immense progress in recent decades including once weak economies growing stronger, and people moving from hunger and poverty into more stable lives. However, these hard-won gains can deteriorate quickly in humanitarian emergencies, especially if the global community responds slowly or not at all. That’s why Bread advocates for the funding that allows smart, compassionate responses to those at the center of disaster. For more information, see how Bread is advocating to reform U.S. food aid to make it more effective and less expensive, and to strengthen foreign assistance to build stronger communities. The people of developing nations can and should do most of the work in ending hunger themselves, but they need some support and resources. The U.S. government can provide some of it. Assistance from the U.S. government helps people help themselves. It funds tools and training for improved agriculture. It builds roads to get food to market. It supports efforts to empower women to play more active roles in their communities. It helps governments develop plans to better educate, care, and feed their people. Development experts agree that the world has the ability to end extreme hunger by 2030. We have already cut it in half since 1990. With continued and increased funding, U.S. foreign assistance can help cut it to zero. U.S. efforts to end global hunger include: The most direct way to end hunger is through food-assistance programs. These programs weave a vital food safety net for millions of children, seniors, people with disabilities, and struggling families. The nation’s largest anti-hunger program is SNAP (formerly known as food stamps). SNAP gives families and people in need a debit-like card to buy groceries. More than 46 million Americans, or 1 in 7 people, were served by SNAP in 2014. Nearly half were children. One in five children lives at risk of hunger in the U.S. School lunch and breakfast programs provide meals to 21.5 million low-income children so they can focus on learning at school. After school and during the summer months, children can also receive meals through the after-school meals program and the Summer Food Service Program. WIC provides healthy food to low-income pregnant and nursing women and young children. This allows our country's most at-risk infants and toddlers to get the nutrients they need for healthy growth and development. The Commodity Supplemental Food Program, senior congregate, and senior home-delivered nutrition services provide healthy food to older Americans. Nine percent of people over 60 are food- insecure and at risk for poor health. These programs help older Americans afford food and other expenses like medicine and housing. In a low-income budget, food is often the most easily squeezed line item. Rent, child care, utilities – these are fixed expenses. Food is one place where families can cut corners and adjust. Nutrition programs help millions of families, but giving food is not enough. Progress against hunger requires helping families move out of poverty. Tax credits are one way to support families working to get out poverty. The earned income tax credit (EITC) helps families keep more of their income, which they can use for essential expenses. The EITC moves more children above the poverty line than any other government program. In 2013, the EITC lifted 6.2 million people, including 3.2 million children, above the poverty line. The child tax credit (CTC) is worth up to $1,000 for each child under age 17 claimed on a worker’s tax return. Families making as little as $3,000 a year can receive the credit. In 2013, the CTC kept 3.1 million people out of poverty, including 1.7 million children. The United States is an international leader in addressing global hunger. By designing programs that enhance global food security, including both short-term emergency food assistance and longer-term development, hunger has been cut in half since 1990. Most often, the U.S. provides food and agriculture development assistance directly to countries through a specific U.S.-led initiative such as Feed the Future or a legislated program like Food for Peace. Other times, it works through international organizations to deliver, provide, and implement humanitarian aid and related assistance. These multi-government institutions are often formed to work on issues that are important to all those in the organization. For over 70 years, the U.S. has provided resources, technical expertise, and guidance to several multilateral organizations working to combat global food insecurity and malnutrition. Some of these institutions include such as the U.N. Food and Agriculture Organization (FAO) and the U.N. World Food Program (WFP). The Millennium Development Goals are an unprecedented effort to better the lives of people who are hungry and poor around the world. In 2000, at a special session of the United Nations, the member nations committed to upholding human dignity and “making the right to development a reality for everyone.” From this commitment a set of eight goals emerged- the Millennium Development Goals (MDGs). When the MDGs expire in December 2015, the United Nations will launch the Sustainable Development Goals (SDGs). The MDGs build on decades of knowledge and success in development work. Each MDG has targets that can be measured so the global community can track progress and make improvements. MDG 1 aimed to cut the number of people suffering from extreme poverty and hunger in half between 1990 and 2015. The poverty goal was met 5 years early and the world will be close to achieving the hunger goal by December 2015! The Sustainable Development Goals (SDGs) will continue the progress of the MDGs and recommit nations to ending hunger and poverty and improving the lives of vulnerable people wherever they live. Currently there are 17 working goals, including SDGs 1 and 2 which aim to end poverty and hunger by 2030. The MDGs showed us that it’s possible to achieve ambitious goals such as cutting poverty in half. Achieving the SDGs will be possible with strong U.S. support, individual country ownership, and a commitment to leave nobody behind. Hunger among children is a major problem in the U.S. One in 5 children — nearly 16 million in total — live at risk of hunger. While hunger affects people of all ages, it is particularly hard on children. Even short-term hunger during a child’s development can cause lasting damage. Because children are hit especially hard by the effects of hunger and malnutrition, feeding programs aimed at children are particularly important. A healthy start in life — even before a child is born — pays off for years, not only for individual children and families, but for communities and our nation as a whole. We have the tools to end child hunger in our country. Strong child feeding programs provide an immediate and direct way to reduce child hunger and improve health and education outcomes. Programs such as the Special Supplemental Nutrition Program for Women, Infants and Children (WIC), school breakfast and lunch programs, and preschool, summer, and after-school meal programs are vital in providing children the food they need for healthy development. Unfortunately, child feeding programs do not reach every child who needs food. For every 7 low-income child receiving school lunches, only about half also get school breakfasts, and only 1 also gets meals during the summer. Many children lack access to feeding programs or find it difficult to participate. A program may not be offered in a child’s community, or transportation may be limited. Child feeding programs could do far more to reduce hunger simply by giving more children access to them. "We can't 'food-bank' our way out of hunger. The government must do its part." The people of developing nations can do most of the work in ending hunger themselves, but they need some support and resources. The U.S. government can provide some of it. These fact sheets provide a snapshot of hunger and poverty in the United States and in each state plus Washington, D.C. Better nutrition is a necessary component of a country’s capacity to achieve development goals such as economic growth and improved public health. Worldwide, maternal and child malnutrition causes millions of deaths each year. In some countries, it holds entire generations back from reaching their economic potential. Dear Members of Congress, As the president and Congress are preparing their plans for this year, almost 100 church leaders—from all the families of U.S. Christianity—are... This devotional guide invites deepened relationship with and among Pan-African people and elected leaders in the mission to end hunger and poverty. Thank you for inviting me to preach here at Duke University Chapel. And I especially want to thank the Bread for the World members who have come this morning. Bruce Puckett urged... Bread for the World and its partners are asking Congress to provide $250 million for global nutrition in the fiscal year 2020 budget. These fact sheets provide a snapshot of hunger and poverty in the United States and in each state plus Washington, D.C. In 2017, 11.8 percent of households in the U.S.—40 million people—were food-insecure, meaning that they were unsure at some point during the year about how they would provide for their next meal.<\|endoftext\|>	3.71875
584	Processed foods have long been linked to health conditions like obesity, high blood pressure and an increased risk of type 2 diabetes. Research shows that some processed foods may double your chances of developing colon cancer. Still, many Americans keep processed foods as a staple in their daily diets. In the United States, we eat more packaged foods per person than any other country. Processed food is an umbrella term to describe any nourishment that was altered prior to consumption. Thus, anything cooked, canned, packaged or preserved is considered processed. Minimally processed foods include bagged lettuce or pre-cut veggies. Although less processed, frozen fruit and canned tuna aren’t the freshest options. Jarred sauces and salad dressings may be prepared with relatively fresh or organic ingredients, but are still considered processed. Deli meats, granola, crackers and frozen pizza can be moderate-to-heavily processed, depending on the food brand and the ingredients used. Consider the following reasons processed foods aren’t the best food choice for your everyday diet: Vanity weight gain aside, hidden sugars and foods filled with high fructose corn syrup can cause you to become insulin resistant, boost triglyceride levels and increase your risk of heart disease. Many processed foods contain refined fructose, or corn syrups, that metabolize directly into fat.When it comes to your oral health, heavy sugar consumption causes decay, cavities, weakened enamel and tooth loss. Unexpected foods, like spaghetti sauce, can also contain added sugars. Don’t assume just because you aren’t eating dessert that you’re curbing your sugar intake. Even organic and reduced-fat foods contain added sugars, so check your nutrition labels while grocery shopping to remain cognizant of what you put in your body. Canned vegetables and soups are notorious for their high sodium content. Food makers add salt to help preserve and extend shelf life, but the daily recommendation is under 2,300 milligrams per day. Use fresh veggies when possible to get the most nutrients with the least number of salty additives. Try making your own soup with homemade chicken broth, your favorite veggies and a crockpot. Because prepackaged foods need to last enough time to reach your kitchen from the manufacturing center, they are filled with numerous preservatives, chemicals, artificial colors and flavors. Some of these “ingredients” are linked to allergic reactions, Alzheimer’s disease and cancer. Avoid any and all artificial ingredients when possible. On the upside, some processing is beneficial when added nutrients are involved. For instance, some milks and juices pack an extra punch of calcium and/or vitamins, while crackers and cereals might include extra fiber for digestive health. Again, keeping an eye on nutrition labels and ingredients will ultimately decide what you should and shouldn’t consume.<\|endoftext\|>	3.65625
1,190	OFF TOPIC: Announcing the release of my new book: Weight Loss Made a Bit Easier: Realistic and Practical Advice for Healthy Eating and Exercise Home \| My Math and Education Books \| Math Lessons \| Ask a Math Question Site Info \| Contact Info \| Tutoring Info \| LarryZafran.com # Lesson 9: Basic Subtraction Many older students have trouble understanding how basic subtraction works. Sometimes this is because they learned the procedure, but never really learned why the procedure worked. Other times it's because they became used to doing every computation on the calculator. It's very important to have a very clear understanding of how subtraction works. Don't forget to watch the embedded YouTube video clip for this lesson at the bottom of the page. To subtract one number from another, all we really have to do is count backwards starting from the larger number. If we want to subtract 8 - 3, we should start with the bigger number, 8, and then count down three numbers....7...6...5, and that is our answer. For small numbers, it's important that you able to do this quickly and easily. The best way to practice this is with flashcards, either store-bought, or home-made. Keep quizzing yourself until you can subtract two small numbers very easily. Many older students are permitted to use calculators on every exam, but if you have trouble doing very basic math in your head, you'll have a hard time figuring out if you made a mistake with your calculator, which is very easy to do. Subtracting two-digit numbers can be hard to do in your head. There is a procedure for subtracting them, and it's important to understand how the procedure works, as well as knowing how to use it. Look at problem 1 below. To subtract two, two-digit numbers, line them up one on top of the other. Remember to line up the ones places and the tens places right on top of each other. Always start on the right, which is the ones place. In problem 1, we'll subtract 8 - 2 to get 6, which we'll write below the line in the ones column. Then we'll move left to the tens column, and subtract 3 - 1 to get 2, which we'll write in the 10 column, giving us a final answer of 26. Take a look at problem 2. First we need to subtract the ones column, but now there is a problem. We need to do 3 - 5, but since the first number is smaller than the second number, we have to do something special. If we look in the tens column of the first number, we'll see that there is a 6, which means 6 tens. What we can do is "borrow" a ten, and move that ten into the ones column. That is OK because we're not changing the value of the number. We'll change the 6 into a 5 as shown, and then put a little 1 to the left of the 3, making it a 13. We borrowed a ten from the tens column, and increased the value of the ones column by ten. Now we can do 13 - 5 which is 8, and then for the tens column we can do 5 - 1 which is 4, giving us 48. Take a look at problem 3. This one is trickier. First we have to do 3 - 9 in the ones column, but we can't, for the same reason as in problem 2. We'll borrow a ten from tens column, crossing out the 2 and making it a 1. Now we can subtract 13 - 9 in the ones column to get 4. Moving left, we see that we have to subtract 1 - 5 in the tens column. Remember, we borrowed a ten, and so the 2 became a 1. But we can't do 1 - 5, so we'll borrow from the hundreds column, using the same logic. We'll decrease the number in the hundreds column by one, and move that one into the tens column. So the 1 that was in the tens column is now 11, and we can subtract 11 - 5 to get 6. All we have in the hundreds column is a 4, and so we can bring it down below the line, and get a final answer of 464. Look at problem 4. This is a bit different. In the ones column, we have to do 0 - 3, which we can't, so we'll borrow from the tens column to the left. But the problem is that the tens column has a 0 in it, so we can't subtract anything from it. The way to handle that is to immediately borrow from the hundreds column, for the sake of the tens column. We'll borrow a hundred to make the 8 into a 7, and then we'll add that one to the left of the zero in the tens column, making it 10. Now we are able to borrow one from the tens column for the sake of the ones column. We can subtract one from the 10 to get 9 in the tens column, and add it to the ones column, which gives us a 10 in the ones column. The answer then becomes 757. Remember that you can ask a math question if you have additional questions about a topic, or you can contact me if you have any comments or suggestions for this site. Go to Next Lesson<\|endoftext\|>	4.5625
335	Perimeter Word Problems Start Practice ## How to Solve Perimeter Word Problems You want to place a string around a rectangle that measures 12 cm by 8 cm. How long should the string be? We need to figure out the rectangle's perimeter. To do this, we add the length of all the sides. Let's write an equation to solve the problem. 12 + 12 + 8 + 8 = 40 Finally, write a complete sentence for the answer. The string should be 40 cm long. ### Example 2: Grandma May's Garden Grandma May is planning to build a fence around her square garden. The length of one of its sides is 6 ft. What is the perimeter of the fence around Grandma May's garden? We need to find the perimeter of the fence around the garden. A square has 4 equal sides. So the length of each side is 6 ft. ğŸ‘‰ Then, we write an equation. We add all the sides. 6 + 6 + 6 + 6 = 24 We can also multiply. This is much faster! 6 x 4 = 24 The unit of length is ft. We write that in our final answer. ğŸ‘‰ Let's write a complete sentence for the answer. The perimeter of the fence around Grandma May's garden is 24 ft. Now, you're ready to ace the practice! ğŸ˜º Start Practice Complete the practice to earn 1 Create Credit Teachers: Assign to students Duplicate Edit Questions Edit Article Assign Preview Questions Edit Duplicate<\|endoftext\|>	4.625
683	Maple syrup can render bacteria more vulnerable to antibiotics. The syrup, which is produced by concentrating the sap from North American maple trees, is a rich source of phenolic compounds with antioxidant properties. And it is these antioxidant properties that prompted the team, led by Professor Nathalie Tufenkji, investigate the potential of maple syrup. The team began by removing a concentrated extract from the syrup. They tested this extract on several infection-causing strains of bacteria, including E. coli and Proteus mirabilis (a common cause of urinary tract infection). The syrup was mildly effective combating the bacteria on its own. However, once mixed with the antibiotics the maple syrup was particularly effective; seemingly synchronising its assault with the pharmaceutical ingredient. Watch this YouTube video to hear Nathalie explaining her work: The syrup and antibiotic combination attacked the bacteria, making the cell membranes more permeable and porous. This allowed more of the extract to seep through the cell walls. Once inside the bacteria, the maple syrup paralyses components of the bacteria that strive to reject the antibiotic. “Inside the [bacteria] skin there are pumps,” explained Nathalie in the above video. “The job of these is to pump out the antibiotics so they can’t do their job. “What we found is that the maple syrup extract knocks out the function of these pumps so they can’t pump out the antibiotic. The antibiotic is now stuck inside the bacteria and it can do its job of killing off the bacteria.” The maple syrup extract also acted synergistically with antibiotics in destroying resistant communities of bacteria known as biofilms, which are common in difficult-to-treat infections, such as catheter-associated urinary tract infections. Biofilms are a collection or layer of bacteria, for example, plaque on your teeth is a biofilm, (see my blog ‘Blocking bacterial biofilms without antibiotics‘). The team also discovered that the maple syrup extract affects the gene expression of the bacteria, repressing several links with antibiotic resistance and virulence, causing it to be less able to spread. “We would have to do in vivo tests, and eventually clinical trials, before we can say what the effect would be in humans,” Nathalie says. “But the findings suggest a potentially simple and effective approach for reducing antibiotic usage. I could see maple syrup extract being incorporated eventually, for example, into the capsules of antibiotics.” Nathalie and her team’s work has been published in the journal Applied and Environmental Microbiology under the title; ‘Polyphenolic Extract from Maple Syrup Potentiates Antibiotic Susceptibility and Reduces Biofilm Formation of Pathogenic Bacteria‘. The team clearly has more work ahead of them; however, their research will no doubt attract the interest of pharmaceutical companies. Increasing the susceptibility of bacteria to antibiotics could improve the effectiveness of treatments; and in a world where the overuse of antibiotics has become a major public-health concern, this has to be a good thing. Well done and good luck to the team! If you are working to improve the effectiveness of antibiotic please get in touch and tell us about your work.<\|endoftext\|>	3.890625

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